Organic Chemistry – T. W. Graham Solomons, Craig B. Fryhle – 10th edition

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W . G R A H A M S O LO M O N S

CRAIG B. FRYHLE

O R G A N IC CHEM ISTRY

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P e r io d ic T able of the E l e m e n t s 1

18 V IIIA

IA 1

H

2 Atom ic nu m b e r-^ 2

Hydrogen 1.0079

11A

3

4

6

c

Symbol —> Name (IUPAC) Atom ic m ass —>

IUPAC re co m m e n d a tio n s^ Chemical Abstracts Service group notation —>

Carbon 12.011

He

13

14

15

16

17

IMA

IVA

VA

V IA

V IIA

Helium 4.0026

5

6

7

8

9

10

LI

Be

B

c

N

F

Ne

Lithium 6.941

Berylium 9.0122

Boron 10.811

Carbon 12.011

Nitrogen 14.007

Oxygen 15.999

Fluorine 18.998

Neon 20.180

11

12

13

14

15

16

17

18

Na

Mg

Sodium 22,990

Magnesium 24.305

3

4

5

6

7

8

9

10

NIB

IVB

VB

VIB

VI IB

V IIIB

V IIIB

19

20

21

22

23

24

25

26

27

V

Cr

Mn

0

AI

Si

P

S

Cl

Ar

MB

Aluminum 26.982

Silicon 28.086

Phosphorus 30.974

Sulfur 32.065

Chlorine 35.453

Argon 39.948

30

31

32

33

34

35

36

11

12

V IIIB

IB

28

29

K

Ca

Sc

Ti

Fe

Co

Ni

Cu

Zn

Ga

Ge

As

Se

Br

Kr

Potassium 39.098

Calcium 40.078

Scandium 44.956

Titanium 47.867

Vanadium 50.942

Chromium 51.996

Manganese 54.938

Iron 55.845

Cobalt 58.933

Nickel 58.693

Copper 63.546

Zinc 65.409

Gallium 69.723

Germanium 72.64

Arsenic 74.922

Selenium 78.96

Bromine 79.904

Krypton 83.798

37

38

39

40

41

42

43

44

45

46

47

48

49

50

51

52

53

54

Mo

Tc

Rb

Sr

Y

Zr

Nb

Rubidium 85.468

Strontium 87.62

Yttrium 88.906

Zirconium 91.224

Niobium 92.906

55

56

57

72

73

Molybdenum Technetium 95.94 (98)

74

75

Ru

Rh

Pd

Ag

Cd

In

Sn

Sb

Te

I

Xe

Ruthenium 101.07

Rhodium 102.91

Palladium 106.42

Silver 107.87

Cadmium 112.41

Indium 114.82

Tin 118.71

Antimony 121.76

Tellurium 127.60

Iodine 126.90

Xeno 131.29

76

77

78

79

80

81

82

83

84

85

86

Cs

Ba

*La

Hf

Ta

w

Re

Os

Ir

Pt

Au

Hg

TI

Pb

Bi

Po

At

Rn

Caesium 132.91

Barium 137.33

Lanthanum 138.91

Hafnium 178.49

Tantalum 180.95

Tungsten 183.84

Rhenium 186.21

Osmium 190.23

Iridium 192.22

Platinum 195.08

Gold 196.97

Mercury 200.59

Thallium 204.38

Lead 207.2

Bismuth 208.98

Polonium (209)

Astatine (210)

Radon (222)

87

88

89

104

105

106

107

108

109

110

111

112

Fr

Ra

#Ac

Rf

Db

sg

Bh

Hs

Mt

Francium (223)

Radium (226)

Actinium (227)

Rutherfordium

(261)

Dubnium (262)

Seaborgium (266)

Bohrium (264)

Hassium (277)

Meitnerium (268)

58

59

60

61

62

63

69

70

71

Pr

Nd

Pm

‘ Lanthanide Series

Ce Cerium 140.12

90 # Actinide Series

Praseodymium Neodymium Promethium

114

Uun Uuu Uub

Uuq

(281)

(272)

(285)

(289)

64

65

66

67

68

Sm

Eu

Gd

Tb

Dy

Ho

Er

Tm

Yb

Lu

Europium 151.96

Gadolinium 157.25

Terbium 158.93

Dysprosium 162.50

Holmium 164.93

Erbium 167.26

Thulium 168.93

Ytterbium 173.04

Lutetium 174.97

95

96

97

98

99

100

101

102

103

140.91

144.24

(145)

Samarium 150.36

91

92

93

94

Th

Pa

U

Np

Pu

Am

Cm

Bk

Cf

Es

Fm

Md

No

Lr

Thorium 232.04

Protactinium 231.04

Uranium 238.03

Neptunium (237)

Plutonium (244)

Americium (243)

Curium (247)

Berkelium (247)

Californium (251)

Einsteinium (252)

Fermium (257)

Mendelevium

Nobelium (259)

Lawrencium (262)

(258)

Relative S trength o f Selected Acids and Their C onjugate Bases Acid S tro n g est acid

t T O C o 00

■a

W eakest acid

H SbF 6 HI H2SO 4 HBr HCl C 6 H5SO+sH

A p p ro x im ate pK a < -1 2 -1 0

C o n ju g ate Base

-9 -9 -7 - 6 .5

S bF 6Ih s o 4BrClC 6 H5S O 3

(CH 3)2O+H + (CH 3)2C = O H

- 3 .8 - 2 .9

(CH 3)2O (CH 3)2C = O

c h 3o+h2 H3O+ hno3 c f 3c o 2h HF c 6 h 5c o 2h C 6 H5NH3 + c h 3c o 2h h 2c o 3 c h 3c o c h 2c o c h 3 n h 4+ C 6 H5OH h c o 3c h 3 n h 3+ h 2o c h 3c h 2o h (CH 3)3COH c h 3c o c h 3 H C #C H H2 nh3 c h 2= c h 2 c h 3c h 3

- 2 .5 - 1 .7 4 - 1 .4 0.18 3.2 4.21 4.63 4.75 6.35 9.0 9.2 9.9

c h 3o h

10.2 10.6

15.7 16 18 19.2 25 35 38 44 50

W eakest base

h 2o

NO 3 CF 3CO 2 FC 6 H5CO 2 C 6 H5NH 2 c h 3c o 2h c o 3c h 3c o h c o c h :3 NH3 C 6 H5O CO 32c h 3n h 2 OHc h 3c h 2o (CH 3)3C O - c h 2c o c h 3 H C#CHn h 2c h 2= c h c h 3c h 2-

S tro n g est base

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Organic Chemistry

T E NT H

EDITION

Organic Chemistry T.W. GRAHAM SOLOMONS University o f South Florida

CRAIG B. FRYHLE Pacific Lutheran University

JOHN WILEY & SONS, INC.

In m em ory o f m y b e lo ve d son, John A llen Solomons, TWGS To Deanna, in the year o f our 25th anniversary. CBF

ASSOCIATE PUBLISHER Petra Recter PROJECT EDITOR Jennifer Yee MARKETING MANAGER Kristine Ruff SENIOR PRODUCTION EDITOR Elizabeth Swain SENIOR DESIGNER Madelyn Lesure SENIOR MEDIA EDITOR Thomas Kulesa SENIOR ILLUSTRATION EDITOR Sandra Rigby SENIOR PHOTO EDITOR Lisa Gee COVER DESIGNER Carole Anson COVER IMAGE © Don Paulson COVER MOLECULAR ART Norm Christiansen This book was set in 10/12 Times Roman by Preparé and printed and bound by Courier Kendallville. The cover was printed by Courier Kendallville. This book is printed on acid-free paper. Copyright © 2011, 2008, 2004, 2000 John Wiley & Sons, Inc. All rights reserved. No part of this publication may be reproduced, stored in a retrieval system or transmitted in any form or by any means, electronic, mechanical, photocopying, recording, scanning, or otherwise, except as permitted under Sections 107 or 108 of the 1976 United States Copyright Act, without either the prior written permission of the Publisher, or authorization through payment of the appropriate per-copy fee to the Copyright Clearance Center, Inc., 222 Rosewood Drive, Danvers, MA 01923, website www.copyright.com. Requests to the Publisher for permission should be addressed to the Permissions Department, John Wiley & Sons, Inc., 111 River Street, Hoboken, NJ 07030-5774, (201)748-6011, fax (201)748-6008, website http://www.wiley.com/go/permissions. Evaluation copies are provided to qualified academics and professionals for review purposes only, for use in their courses during the next academic year. These copies are licensed and may not be sold or transferred to a third party. Upon completion of the review period, please return the evaluation copy to Wiley. Return instruc­ tions and a free of charge return shipping label are available at www.wiley.com/go/returnlabel. Outside of the United States, please contact your local representative. Library of Congress Cataloging-in-Publication Data Solomons, T. W. Graham. Organic chemistry/T.W. Graham Solomons.—10th ed./Craig B. Fryhle. p. cm. Includes index. ISBN 978-0-470-40141-5 (cloth) Binder-ready version ISBN 978-0-470-55659-7 1. Chemistry, Organic—Textbooks. I. Fryhle, Craig B. II. Title. QD253.2.S65 2011 547—dc22 2009032800

Printed in the United States of America 10 9 8 7 6 5 4 3 2 1

1 T h e B a s ic s Bonding and Molecular Structure 1 2

F a m i l i e s o f C a r b o n C o m p o u n d s Functional G roups, Interm olecular Forces, and Infrared (IR) Spectroscopy 53

3 A n I n t r o d u c t i o n t o O r g a n i c R e a c t i o n s a n d T h e i r M e c h a n i s m s Acids and Bases 98 4

N o m e n c l a t u r e a n d C o n f o r m a t i o n s o f A l k a n e s a n d C y c l o a l k a n e s 137

5 S t e r e o c h e m i s t r y Chiral M olecules 186 6

I o n ic R e a c t i o n s Nucleophilic Substitution and Elimination Reactions of Alkyl Halides 230

7 A l k e n e s a n d A l k y n e s I Properties and Synthesis. Elimination Reactions of Alkyl Halides 285 8 A l k e n e s a n d A l k y n e s II Addition Reactions 331

9

N u c l e a r M a g n e t i c R e s o n a n c e a n d M a s s S p e c t r o m e t r y Tools for Structure D eterm ination 385

1 0 R a d ic a l R e a c t i o n s 459 1 1 A l c o h o l s a n d E t h e r s Synthesis and Reactions 502 1 2 A l c o h o l s F r o m C a r b o n y l C o m p o u n d s O xidation-R eduction and O rganom etallic C om pounds 548 1 3 C o n j u g a t e d U n s a t u r a t e d S y s t e m s 585 1 4 A r o m a t i c C o m p o u n d s 632 1 5 R e a c t i o n s o f A r o m a t i c C o m p o u n d s 676 1 6 A l d e h y d e s a n d K e t o n e s N ucleophilic A ddition to th e Carbonyl G roup 729 1 7 C a r b o x y l i c A c i d s a n d T h e i r D e r i v a t i v e s N ucleophilic Addition-Elim ination at th e Acyl C arbon 7 7 9 1 8 R e a c t i o n s a t t h e a C a r b o n o f C a r b o n y l C o m p o u n d s Enols and Enolates 831 1 9 C o n d e n s a t i o n a n d C o n j u g a t e A d d i t i o n R e a c t i o n s o f C a r b o n y l C o m p o u n d s More Chemistry of Enolates 869 2 0 A m i n e s 911 21

P h e n o l s a n d A ry l H a l i d e s N ucleophilic A rom atic Substitution 964

S p e c i a l T o p ic G C a r b o n - C a r b o n B o n d - F o r m i n g a n d O t h e r R e a c t i o n s o f T r a n s i t i o n M e t a l O r g a n o m e t a l l i c C o m p o u n d s G -1 2 2 C a r b o h y d r a te s 1000 2 3 L ip id s 1050 2 4 A m i n o A c i d s a n d P r o t e i n s 1084 2 5 N u c l e i c A c i d s a n d P r o t e i n S y n t h e s i s 1131 A n s w e r s t o S e l e c t e d P r o b l e m s A -1 G l o s s a r y GL-1 P h o t o C r e d i t s C -1 I n d e x I-1

vii

«

Contents 1

T h e B a s ic s Bonding and Molecular Structure

1

1 2 3 4 5

We Are Stardust 2 A to m ic Structure 2 The Structural Theory o f Organic Chemistry 5 Chemical Bonds: The O cte t Rule 7 H ow to W rite Lewis Structures 9 6 Exceptions to the O cte t Rule 11 7 Formal Charges and H ow to Calculate Them 13 8 Resonance Theory 15 9 Quantum Mechanics and A to m ic Structure 20 10 A to m ic Orbitals and Electron Configuration 21 11 M olecular Orbitals 23 12 The Structure o f Methane and Ethane: sp3 Hybridization

25 C alculated M olecular M odels: Electron Density Surfaces 29 The Structure o f Ethene (Ethylene): sp2 Hybridization 30 The Structure o f Ethyne (Acetylene): sp Hybridization 34 A Summary o f Im portant Concepts that Come from Quantum Mechanics 36 M olecular Geom etry: The Valence Shell Electron Pair Repulsion M odel 38 H ow to Interpret and Write Structural Formulas 41 A pplications o f Basic Principles 46

THE CHEMISTRY O F . . .

1 1 1 1 1 1

F a m i l i e s o f C a r b o n C o m p o u n d s Functional G roups, Interm olecular Forces, and Infrared (IR) S pectroscopy 53

2

2.1 2.2

Hydrocarbons: Representative Alkanes, Alkenes, Alkynes, and Arom atic C om pounds 54 Polar C ovalent Bonds 57

Calculated M olecular M odels: Maps of Electrostatic Potential 59 2.3 Polar and N onpolar Molecules 60 2.4 Functional Groups 62 2.5 A lkyl Halides o r Haloalkanes 64 2.6 Alcohols 65 2.7 Ethers 67 THE CHEMISTRY O F Ethers as G eneral A nesthetics 67 2.8 Amines 68 2.9 Aldehydes and Ketones 69 2.10 Carboxylic Acids, Esters, and Am ides 70 2.11 Nitriles 72 2.12 Summary o f Im portant Families o f Organic C om pounds 72 2.13 Physical Properties and M olecular Structure 73 THE CHEMISTRY O F . . . Fluorocarbons and Teflon 78 2.14 Summary o f A ttractive Electric Forces 82 THE CHEMISTRY O F . . . O rganic Tem plates E ngineered to Mimic Bone Growth 2.15 Infrared Spectroscopy: An Instrumental M ethod fo r D etecting Functional Groups 83 2.16 Interpreting IR Spectra 87 2.17 A pplications o f Basic Principles 92 THE CHEMISTRY O F . . .

Viii

82

Contents

3

A n I n tr o d u c tio n t o O r g a n ic R e a c tio n s a n d T h e ir M e c h a n is m s Acids and Bases 98

3.1 3.2 3.3 3 .4

Reactions and Their Mechanisms 99 Acid-Base Reactions 101 Lewis Acids and Bases 102 Heterolysis o f Bonds to Carbon: Carbocations and Carbanions

THE CHEMISTRY O F . . . H O M O s and LU M O s in R eactions

^

3.5

H o w to Use Curved Arrows in Illustrating Reactions

104

105 106

3.6 3.7 3.8 3.9 3.10

The Strength o f Br0 nsted-Lowry Acids and Bases: Ka and pK a 109 H ow to Predict the O utcom e o f A cid-Base Reactions 113 Relationships between Structure and A c id ity 115 Energy Changes 119 The Relationship between the Equilibrium Constant and the Standard Free-Energy Change, AG° 120 3.11 The A cid ity o f Carboxylic Acids 121 3.12 The Effect o f the Solvent on A c id ity 125 3.13 Organic C om pounds as Bases 126 3.14 A Mechanism fo r an Organic Reaction 127 3.15 Acids and Bases in Nonaqueous Solutions 128 3.16 Acid-Base Reactions and the Synthesis o f Deuterium - and Tritium-Labeled C om pounds 130 3.17 Applications o f Basic Principles 131

4

N o m e n c la tu r e a n d C o n f o r m a tio n s o f A lk a n e s a n d C y c lo a lk a n e s

4.1

Introduction to Alkanes and Cycloalkanes

THE CHEMISTRY O F . . . P etroleum Refining

4.2

4.3 4.4 4.5 4.6 4.7

138

139

Shapes o f Alkanes 140 IUPAC N om enclature o f Alkanes, A lkyl Halides, and Alcohols H ow to Name Cycloalkanes 149 N om enclature o f Alkenes and Cycloalkenes 151 N om enclature o f Alkynes 154 Physical Properties o f Alkanes and Cycloalkanes 154

142

THE CHEMISTRY O F . . . Pherom ones: C o m m u n ica tio n by M eans o f C hem icals

4.8 4 .9 ^

THE CHEMISTRY O F . . . M uscle A c tio n

4 .1 0 4.11

162

The Relative Stabilities o f Cycloalkanes: Ring Strain 162 Conform ations o f Cyclohexane: The Chair and the Boat 163

THE CHEMISTRY O F . . . N anoscale M o to rs and M o le cu la r S w itches

4.12 4.13 4 .1 4 ^

156

Sigma Bonds and Bond Rotation 157 Conform ational Analysis o f Butane 160

166

S ubstituted Cyclohexanes: A xial and Equatorial Hydrogen Groups D isubstituted Cycloalkanes: Cis-Trans Isomerism 171 Bicyclic and Polycyclic Alkanes 175

THE CHEMISTRY O F . . . E lem ental C a rbo n

167

176

4.15 Chemical Reactions o f Alkanes 177 4.16 Synthesis o f Alkanes and Cycloalkanes 177 4.17 H ow to Gain Structural Inform ation from M olecular Formulas and the Index o f Hydrogen Deficiency 178 4.19 A pplications o f Basic Principles 181 PLUaS

See SPECIAL TOPIC A: 13C NMR Spectroscopy - A Practical Introduction in WileyPLUS

137

ix

x

Contents

5

S t e r e o c h e m i s t r y Chiral M olecules

5.1 5.2 5.3 5.4

C hirality and Stereochem istry 186 Isomerism: C onstitutional Isomers and Stereoisomers 188 Enantiomers and Chiral Molecules 190 A Single Chirality C enter Causes a M olecule to Be Chiral 191

^

THE CHEMISTRY O F

5.5 5.6 5.7 5.8 5.9 5.10 5.11

186

. . . Life's Molecular H an d ed n ess

193

M ore abo u t the Biological Im portance o f Chirality 194 H ow to Test fo r Chirality: Planes o f Symmetry 195 N am ing Enantiomers: The R,S-System 196 Properties o f Enantiomers: O ptical A ctivity 201 The Origin o f O ptical A ctivity 205 The Synthesis o f Chiral Molecules 207 Chiral Drugs 209

. . . Selective Binding of Drug Enantiom ers to Left- and Right-H anded Coiled DNA 211 5.12 Molecules with More than One Chirality C enter 211 5.13 Fischer Projection Formulas 215 5.14 Stereoisomerism o f Cyclic C om pounds 217 5.15 Relating Configurations through Reactions in Which N o Bonds to the Chirality C enter Are Broken 219 5.16 Separation o f Enantiomers: Resolution 223 5.17 C om pounds with Chirality Centers O ther than Carbon 224 5.18 Chiral Molecules that Do N o t Possess a Chirality C enter 224 THE CHEMISTRY O F

I o n ic R e a c t i o n s N ucleophilic Substitution and Elimination Reactions of Alkyl Halides 2 30

6

6.1 6.2 6.3 6.4 6.5

Organic Halides 231 N ucleophilic Substitution Reactions 233 Nucleophiles 234 Leaving Groups 237 Kinetics o f a N ucleophilic Substitution Reaction: An Sn 2 Reaction 237 6 .6 A Mechanism fo r the Sn 2 Reaction 238 6.7 Transition State Theory: Free-Energy Diagrams 240 6.8 The Stereochem istry o f Sn2 Reactions 243 6.9 The Reaction o f tert-B utyl C hloride with H ydroxide Ion: An Sn 1 Reaction 246 6.10 A Mechanism fo r the Sn 1 Reaction 247 6.11 Carbocations 248 6.12 The Stereochem istry o f Sn 1 Reactions 251 6.13 Factors A ffecting the Rates o f Sn 1 and Sn 2 Reactions 254 6.14 Organic Synthesis: Functional Group Transformations Using S^2 Reactions 264

. . . Biological M ethylation: A Biological N ucleophilic Substitution Reaction 266 6.15 Elimination Reactions o f A lkyl Halides 268 6.16 The E2 Reaction 269 6.17 The E1 Reaction 271 6.18 H ow to D eterm ine w hether Substitution o r Elimination Is Favored 273 6.19 Overall Summary 276 THE CHEMISTRY O F

7

7.1 7.2 7.3

A l k e n e s a n d A l k y n e s I Properties and Synthesis. Elimination Reactions of A lkyl Halides 2 8 5 Introduction 286 The (E)-(Z) System fo r D esignating Alkene Diastereomers Relative Stabilities o f Alkenes 288

286

Contents

7.4 7.5 7.6 7.7 7.8 7.9 7.10 7.11 7.12 7.13

Cycloalkenes 290 Synthesis o f Alkenes via Elimination Reactions 291 D ehydrohalogenation o f A lkyl Halides 291 Acid-Catalyzed D ehydration o f A lcohols 297 Carbocation S tability and the Occurrence o f M olecular Rearrangements 303 The A cid ity o f Terminal Alkynes 307 Synthesis o f Alkynes by Elimination Reactions 308 Replacement o f the A cetylenic Hydrogen A to m o f Terminal Alkynes 310 Alkylation o f A lkynide Anions: Some General Principles o f Structure and Reactivity Illustrated 312 Hydrogenation o f Alkenes 313

THE CHEMISTRY O F . . . H ydrogenation in th e Food Industry 313 7.14 H ydrogenation: The Function o f the Catalyst 314 7.15 Hydrogenation o f Alkynes 315 7.16 An Introduction to Organic Synthesis 317 THE CHEMISTRY O F . . . From th e Inorganic to th e O rganic 321

A l k e n e s a n d A l k y n e s II A ddition Reactions

8

331

8.1 8.2

A d d itio n Reactions o f Alkenes 332 Electrophilic A d d itio n o f Hydrogen Halides to Alkenes: Mechanism and Markovnikov's Rule 334 8.3 Stereochem istry o f the Ionic A d d itio n to an Alkene 339 8.4 A d d itio n o f Sulfuric A c id to Alkenes 340 8.5 A d d itio n o f W ater to Alkenes: A cid-Catalyzed Hydration 340 8.6 Alcohols from Alkenes through Oxym ercuration-Dem ercuration: Markovnikov A d d itio n 344 8.7 Alcohols from Alkenes through H ydroboration-O xidation: A nti-M arkovnikov Syn Hydration 347 8.8 H ydroboration: Synthesis o f Alkylboranes 347 8.9 O xidation and Hydrolysis o f Alkyboranes 350 8.10 Summary o f Alkene Hydration M ethods 353 8.11 Protonolysis o f Alkyboranes 353 8.12 Electrophilic A d d itio n o f Bromine and Chlorine to Alkenes 354

. . . The Sea: A Treasury of Biologically Active Natural Products 357 8.13 Stereospecific Reactions 358 8.14 Halohydrin Formation 359 8.15 Divalent Carbon Compounds: Carbenes 361 8.16 Oxidations o f Alkenes: Syn 1,2-Dihydroxylation 363 THE CHEMISTRY O F . . . Catalytic Asymmetric Dihydroxylation 365 8.17 O xidative Cleavage o f Alkenes 365 8.18 Electrophilic A d d itio n o f Bromine and Chlorine to Alkynes 368 8.19 A d d itio n o f Hydrogen Halides to Alkynes 369 8.20 O xidative Cleavage o f Alkynes 370 8.21 H ow to Plan a Synthesis: Som e Approaches and Examples 370 THE CHEMISTRY O F

9

9.1 9.2 9.3 9.4 9.5 9.6

N u c l e a r M a g n e t i c R e s o n a n c e a n d M a s s S p e c t r o m e t r y Tools for Structure D eterm ination 385 Introduction 386 Nuclear M agnetic Resonance (NMR) Spectroscopy 386 H ow to Interpret Proton NMR Spectra 392 Nuclear Spin: The Origin o f the Signal 395 D etecting the Signal: Fourier Transform NMR Spectrom eters Shielding and D eshielding o f Protons 399

397

xi

xii

Contents

9.7 The Chemical Shift 400 9.8 Chemical Shift Equivalent and N onequivalent Protons 401 9.9 Signal Splitting: Spin-Spin C oupling 405 9.10 Proton NMR Spectra and Rate Processes 415 9.11 Carbon-13 NMR Spectroscopy 417 9.12 Two-Dimensional (2D) NMR Techniques 422 THE CHEMISTRY O F . . . M agnetic Resonance Imaging in M edicine 425 9.13 An Introduction to Mass Spectrom etry 426 9.14 Formation o f Ions: Electron Im pact Ionization 427 9.15 D epicting the M olecular Ion 427 9.16 Fragmentation 428 9.17 H ow to D eterm ine M olecular Formulas and M olecular W eights Using Mass Spectrom etry 435 9.18 Mass S pectrom eter Instrum ent Designs 440 9.19 GC/MS Analysis 442 9.20 Mass Spectrom etry o f Biomolecules 443 10

R a d ic a l R e a c t i o n s

459

10.1 10.2 10.3 10.4 10.5 10.6 10.7 10.8 10.9

Introduction: How Radicals Form and How They React 460 H om olytic Bond Dissociation Energies (DH°) 461 Reactions o f Alkanes with Halogens 465 Chlorination o f Methane: Mechanism o f Reaction 467 Chlorination o f Methane: Energy Changes 470 Halogenation o f H igher Alkanes 477 The G eom etry o f A lkyl Radicals 480 Reactions that Generate Tetrahedral Chirality Centers 481 Radical A d d itio n to Alkenes: The A nti-M arkovnikov A d d itio n o f Hydrogen Bromide 484 10.10 Radical Polymerization o f Alkenes: Chain-Growth Polymers 486 10.11 O th e r Im portant Radical Reactions 490 THE CHEMISTRY O F

of DNA

. . . Calicheamicin

^ ¡ 1 THE CHEMISTRY O F . . . A ntioxidants THE CHEMISTRY O F PLU°S

A Radical Device for Slicing th e Backbone

492 494

. . . O zone D epletion and Chlorofluorocarbons (CFCs)

495

See SPECIAL TOPIC B: Chain-Grow th Polym ers in WileyPLUS 11

A l c o h o l s a n d E t h e r s Synthesis and Reactions

502

11.1 Structure and Nomenclature 503 11.2 Physical Properties o f Alcohols and Ethers 505 11.3 Im portant Alcohols and Ethers 507 ^ THE CHEMISTRY O F . . . Ethanol as a Biofuel 508 11.4 Synthesis o f Alcohols from Alkenes 509 11.5 Reactions o f A lcohols 511 11.6 Alcohols as Acids 513 11.7 Conversion o f Alcohols into A lkyl Halides 514 11.8 A lkyl Halides from the Reaction o f Alcohols with Hydrogen Halides 514 11.9 A lkyl Halides from the Reaction o f Alcohols with PBr3 o r SOC2 517 11.10 Tosylates, Mesylates, and Triflates: Leaving Group Derivatives o f Alcohols ^ THE CHEMISTRY O F . . . Alkyl P hosphates 521 11.11 Synthesis o f Ethers 522 11.12 Reactions o f Ethers 527 11.13 Epoxides 528 THE CHEMISTRY O F . . . The Sharpless Asymmetric Epoxidation 529

518

Contents

11.14 Reactions o f Epoxides 531 THE CHEMISTRY O F . . . Epoxides, C arcinogens, and Biological Oxidation 533 11.15 A n ti 1,2-Dihydroxylation o f Alkenes via Epoxides 535 THE CHEMISTRY O F . . . E n viro n m e n ta lly F rie n d ly A lke n e O x id a tio n M e th o d s 537 11.16 Crown Ethers 537 THE CHEMISTRY O F . . . Transport A ntibiotics and Crown Ethers 539 11.17 Summary o f Reactions o f Alkenes, Alcohols, and Ethers

12

A l c o h o l s F r o m C a r b o n y l C o m p o u n d s O xidation-R eduction and O rganom etallic C om pounds 5 48

12.1 Structure o f the Carbonyl Group 549 12.2 O xidation-R eduction Reactions in Organic Chemistry 12.3

540

Alcohols by Reduction o f Carbonyl Com pounds

550

552

THE CHEMISTRY O F . . . A lc o h o l D e hyd ro g e n a se - A B iochem ical

Hydride R eagent

554

555 12.4 O xidation o f Alcohols 557 12.5 O rganom etallic C om pounds 561 12.6 Preparation o f O rganolithium and Organomagnesium Com pounds 562 12.7 Reactions o f O rganolithium and Organomagnesium C om pounds 563 12.8 Alcohols from Grignard Reagents 566 12.9 Protecting Groups 575

■ THE CHEMISTRY O F . . . S tere oselective R eductions o f C a rbo nyl G roup s

PLU“S

See the F irst R eview P ro b lem S e t in WileyPLUS 13

C o n ju g a te d U n s a tu r a te d S y s te m s

13.1

Introduction 586 A llylic Substitution and the A lly l Radical

13.2

THE CHEMISTRY O F . . . A lly lic B ro m in a tio n

13.3

The S tability o f the A lly l Radical

585

586 590

590

13.4 The A lly l Cation 594 13.5 Resonance Theory Revisited 595 13.6 Alkadienes and Polyunsaturated Hydrocarbons 13.7 1,3-Butadiene: Electron Delocalization 600 13.8 The S tability o f C onjugated Dienes 602 13.9

U ltraviolet-Visible Spectroscopy

599

604

^ THE CHEMISTRY O F . . . The P h o to ch e m istry o f Vision

609

13.10 Electrophilic A tta ck on C onjugated Dienes: 1,4 A d d itio n

612

13.11 The D ie ls-A ld e r Reaction: A 1,4-Cycloaddition Reaction o f Dienes

616

THE CHEMISTRY O F . . . M o le cu le s w ith th e N o b e l Prize in T h e ir S yn th e tic Lineage

14 14.1 14.2 14.3 14.4 14.5 14.6 14.7 14.8

A ro m a tic C o m p o u n d s

632

The Discovery o f Benzene 633 N om enclature o f Benzene Derivatives 634 Reactions o f Benzene 637 The Kekulé Structure fo r Benzene 638 The Therm odynam ic S tability o f Benzene 639 M odern Theories o f the Structure o f Benzene 640 Hückel's Rule: The 4n + 2 p Electron Rule 643 O ther A rom atic Com pounds 651

THE CHEMISTRY O F . . . N a n o tu b e s

655

14.9 H eterocylic A rom atic Com pounds 655 14.10 A rom atic Com pounds in Biochem istry 657 14.11 Spectroscopy o f A rom atic C om pounds 660

620

xiii

xiv

Contents

THE CHEMISTRY O F . . . Sunscreens (C atching th e Sun's Rays and W h a t

H a pp ens to Them )

15

664

R e a c tio n s o f A ro m a tic C o m p o u n d s

676

15.1 Electrophilic A rom atic Substitution Reactions 677 15.2 A General Mechanism fo r E lectrophilic A rom atic Substitution 678 15.3 Halogenation o f Benzene 680 15.4 N itration o f Benzene 681 15.5 Sulfonation o f Benzene 682 15.6 F riedel-C rafts Alkylation 684 15.7 F riedel-C rafts A cylation 685 15.8 Limitations o f Friedel-C rafts Reactions 687 15.9 Synthetic Applications o f Friedel-C rafts Acylations: The Clemmensen Reduction 690 15.10 Substituents Can A ffe ct Both the Reactivity o f the Ring and the O rientation o f the Incom ing Group 691 15.11 How Substituents A ffe ct Electrophilic Arom atic Substitution: A Closer Look 697 15.12 Reactions o f the Side Chain o f Alkylbenzenes

706

THE CHEMISTRY O F . . . Io d in e In c o rp o ra tio n in T hyroxine B iosynthesis

^

THE CHEMISTRY O F . . . Industrial Styrene Synthesis

15.13 15.14 15.15 15.16 16 16.1 16.2 16.3

709

Alkenylbenzenes 712 Synthetic Applications 714 A llylic and Benzylic Halides in N ucleophilic Substitution Reactions Reduction o f A rom atic C om pounds 719

Introduction 730 Nom enclature o f Aldehydes and Ketones Physical Properties 732

730

THE CHEMISTRY O F . . . A ld e h y d e s and K etones in Perfum es

733

741

THE CHEMISTRY O F . . . A Very Versatile V ita m in , P yrid oxin e (Vitam in B6)

17

717

A l d e h y d e s a n d K e t o n e s N ucleophilic A ddition to th e Carbonyl G roup

16.4 Synthesis o f Aldehydes 733 16.5 Synthesis o f Ketones 738 16.6 N ucleophilic A d d itio n to the C arbon-O xygen D ouble Bond 16.7 The A d d itio n o f Alcohols: Hemiacetals and Acetals 744 16.8 The A d d itio n o f Primary and Secondary Am ines 751 16.9 16.10 16.11 16.12 16.13 16.14

707

753

The A d d itio n o f Hydrogen Cyanide: Cyanohydrins 755 The A d d itio n o f Ylides: The W ittig Reaction 757 Oxydation o f Aldehydes 761 Chemical Analyses fo r Aldehydes and Ketones 761 Spectroscopic Properties o f Aldehydes and Ketones 762 Summary o f A ldehyde and Ketone A d d itio n Reactions 765

C a r b o x y l i c A c i d s a n d T h e i r D e r i v a t i v e s Nucleophilic Addition-Elim ination at th e Acyl C arbon 779

17.1 Introduction 780 17.2 Nom enclature and Physical Properties 780 17.3 Preparation o f Carboxylic A cids 789 17.4 A cyl Substitution: N ucleophilic A d d itio n -E lim in a tio n at the A cyl Carbon 17.5 A cyl Chlorides 794 17.6 Carboxylic A c id Anhydrides 796 17.7 Esters 797 17.8 A m ides 804

792

729

Contents

THE CHEMISTRY O F . . . Penicillins

17.9

17.10 17.11 17.12 17.13 18

18.1 18.2 18.3

811

Derivatives o f Carbonic A d d 812 D ecarboxylation o f Carboxylic A cids 814 Chemical Tests fo r A cyl Com pounds 816 Polyesters and Polyamides: Step-G rowth Polymers 817 Summary o f the Reactions o f Carboxylic Acids and Their Derivatives

818

R e a c t i o n s a t t h e a C a r b o n o f C a r b o n y l C o m p o u n d s Enols and Enolates 831 The A cid ity o f the a Hydrogens o f Carbonyl C om pounds: Enolate Anions Keto and Enol Tautomers 833 Reactions via Enols and Enolates 834

THE CHEMISTRY O F . . .

Chloroform in Drinking W ater

832

839

Lithium Enolates 841 18.5 Enolates o f /3-Dicarbonyl Com pounds 844 18.6 Synthesis o f M ethyl Ketones: The A cetoacetic Ester Snythesis 845 18.7 Synthesis o f S ubstituted A ce tic Acids: The M alonic Ester Synthesis 850 18.8 Further Reactions o f A ctive Hydrogen Com pounds 853 18.9 Synthesis o f Enamines: Stork Enamine Reactions 854 18.10 Summary o f Enolate Chemistry 857 18.4

PLU”S

See SPECIAL TOPIC C: S tep-G row th Polym ers in WileyPLUS 19

C o n d e n s a tio n a n d C o n ju g a te A d d itio n R e a c tio n s o f C a rb o n y l C o m p o u n d s More Chem istry of Enolates 8 6 9

19.1 19.2 19.3 19.4

Introduction 870 The Claisen Condensation: The Synthesis o f 3 -K eto Esters 870 3-D icarbonyl C om pounds by Acylation o f Ketone Enolates 875 A ld o l Reactions: A d d itio n o f Enolates and Enols to A ldehydes and Ketones 876

^ THE CHEMISTRY O F . . . A R e tro -A ld o l R eaction in G lycolysis— D ivid in g Assets to D o u b le th e ATP Y ie ld

19.5 19.6 19.7

878

Crossed A ld o l Condensations 882 Cyclizations via A ld o l Condensations 888 A ddition s to a,/3-Unsaturated Aldehydes and Ketones

889

THE CHEMISTRY O F . . . C alicheam icin y-]1A c tiv a tio n fo r C leavage o f D N A

19.8

The Mannich Reaction

THE CHEMISTRY O F . . .

19.9 PLlTS

894

A Suicide Enzyme S ubstrate

Summary o f Im portant Reactions

895

897

See SPECIAL TOPIC D: Thiols, Sulfur Ylides, and Disulfides in WileyPLUS See SPECIAL TOPIC E: Thiol E sters and Lipid Biosynthesis in WileyPLUS 20 20.1 20.2 20.3

A m in e s

911

N om enclature 912 Physical Properties and Structure o f Am ines Basicity o f Amines: A m ine Salts 915

913

THE CHEMISTRY O F . . . B io lo g ic a lly Im p o rta n t A m in e s

20.4

20.5 20.6

Preparation o f Am ines 924 Reactions o f Am ines 933 Reactions o f Am ines with Nitrous A c id

THE CHEMISTRY O F . . . N -N itro so a m in e s

20.7

20.8

922

935 936

Replacement Reactions o f Arenediazonium Salts 937 C oupling Reactions o f Arenediazonium Salts 941

894

xv

xvi

Contents

20.9

Reactions o f Am ines with Sulfonyl Chlorides

943

THE CHEMISTRY O F . . . C h e m o th e ra p y and Sulfa D rugs

20 .10 Synthesis o f Sulfa Drugs

944

947

20.11 Analysis o f Am ines 947 20.12 Eliminations Involving A m m onium Com pounds 949 20.13 Summary o f Preparations and Reactions o f Am ines 950 F'fi ■y^t>

,, ,

,

See SPECIAL TOPIC F: Alkaloids in WileyPLUS 21

P h e n o l s a n d A ry l H a l i d e s N ucleophilic Arom atic Substitution

21.1 Structure and Nomenclature o f Phenols 965 21.2 Naturally O ccurring Phenols 966 21.3 Physical Properties o f Phenols 966 21.4 Synthesis o f Phenols 967 21.5 Reactions o f Phenols as Acids 969 21.6 O th e r Reactions o f the O — H Group o f Phenols 21.7 Cleavage o f A lkyl A ryl Ethers 973 21.8 Reactions o f the Benzene Ring o f Phenols 973

972

THE CHEMISTRY O F . . . P olyketid e A n tic a n c e r A n tib io tic Biosynthesis

The Claisen Rearrangement 21 .10 Quinones 978

21.9

964

975

977

THE CHEMISTRY O F . . . The B o m b a rd ie r Beetle's N o xiou s Spray

21.11 A ryl Halides and N ucleophilic Arom atic Substitution

979

980

THE CHEMISTRY O F . . . Bacterial D e h a lo g e n a tio n o f a PCB D e rivative

21.12 Spectroscopic Analysis o f Phenols and A ryl Halides

983

988

THE CHEMISTRY O F . . . A ryl Halides: T h e ir Uses and E nvironm ental C oncerns

989

See the S eco n d R eview P ro b lem S e t in WileyPLUS C arbon-C arbon Bond-Form ing and O th er Reactions of Transition M etal O rganom etallic C o m p o u n d s G-1

SPECIAL TOPIC G:

See SPECIAL TOPIC H: Electrocyclic and Cycloaddition Reactions in WileyPLUS 22

C a rb o h y d ra te s

1000

22.1 22.2 22.3 22.4 22.5 22.6 22.7 22.8 22.9 22.10 22.11

Introduction 1001 Monosaccharides 1004 M utarotation 1009 Glycoside Formation 1010 O th e r Reactions o f Monosaccharides 1013 O xidation Reactions o f Monosaccharides 1016 Reduction o f Monosaccharides: A ld ito ls 1022 Reactions o f Monosaccharides with Phenylhydrazine: Osazones Synthesis and D egradation o f Monosaccharides 1023 The D Family o f Aldoses 1025 Fischer's Proof o f the Configuration o f D-(+)-Glucose 1027 22.12 Disaccharides 1029 ^

THE CHEMISTRY O F . . . A rtific ia l Sw eeteners (H ow Sw eet It Is)

22.13 Polysaccharides

1022

1032

1033

22.14 O th e r B iologically Im portant Sugars 1037 22.15 Sugars That Contain N itrogen 1038 22.16 G lycolipids and Glycoproteins o f the Cell Surface: Cell Recognition and the Immune System 1040 22.17 Carbohydrate A n tib io tics 1042 22.18 Summary o f Reactions o f Carbohydrates 1042

Contents

23

L ip id s

23.1

1050

Introduction 1051 Fatty A d d s and Triacylglyœrols

23.2

1052

L THE CHEMISTRY O F . . . O lestra and O th e r Fat S u b stitu te s

1055

THE CHEMISTRY O F . . . S elf-A sse m bled M on olayers— Lipids in M aterials Science and

B io e n g in e e rin g

1060

23.3 Terpenes and Terpenoids 23.4 Steroids 1064 23.5 Prostaglandins 1073

1061

Phospholipids and Cell Membranes

23.6

1074

L THE CHEMISTRY O F . . . STEALTFI® Liposom es fo r D rug D e live ry

23.7

24

W axes

1077

1078

A m in o A c id s a n d P r o te in s

1084

24.1 Introduction 1085 24.2 A m ino A cids 1086 24.3 Synthesis o f a-A m ino Acids 1092 24.4 Polypeptides and Proteins 1094 24.5 Primary Structure o f Polypeptides and Proteins 1097 24.6 Examples o f Polypeptide and Protein Primary Structure 1101 THE CHEMISTRY O F . . . S ickle-C ell A ne m ia 1103 24.7 Polypeptide and Protein Synthesis 1104 24.8 Secondary, Tertiary, and Q uaternary Structure o f Proteins 1110 24.9 Introduction to Enzymes 1115 24.10 Lysozyme: M ode o f Action o f an Enzyme 1116 THE CHEMISTRY O F . . . C a rb o n ic A nhydrase: S h u ttlin g th e Protons 1119 24.11 Serine Proteases 1120 24.12 Flem oglobin: A C onjugated Protein 1122 THE CHEMISTRY O F . . . S om e C a ta lytic A n tib o d ie s 1123 24.13 Purification and Analysis o f Polypeptides and Proteins 1125 24.14 Proteomics 1126 25

N u c le ic A c id s a n d P r o te in S y n th e s is

25.1 25.2 25.3 25.4 25.5 25.6

Introduction 1132 N ucleotides and Nucleosides 1133 Laboratory Synthesis o f Nucleosides and N ucleotides 1137 D eoxyribonucleic Acid: D N A 1139 RNA and Protein Synthesis 1146 D eterm ining the Base Sequence o f DNA: The Chain-Terminating (Dideoxynucleotide) M e th o d 1155 Laboratory Synthesis o f O ligonucleotides 1157 The Polymerase Chain Reaction 1158 Sequencing o f the Human Genome: An Instruction Book fo r the Molecules o f Life 1162

25.7 25.8 25.9

A nsw ers to S elected Problem s G lossary

GL-1

P hoto C redits Index

I-1

C-1

A-1

1131

xvii

xviii

Contents

C h a p te r 3

C h a p te r 11

Reaction of W ater with H ydrogen Chloride: The Use of Curved Arrows 107 Reaction of tert-Butyl Alcohol with C oncentrated A queous HCl 127

Conversion of an Alcohol into a M esylate (an Alkyl M ethanesulfonate) 520 Interm olecular Dehydration of Alcohols to Form an Ether 522 The Williamson Ether Synthesis 523 Ether C leavage by Strong Acids 527 Alkene Epoxidation 529 Acid-Catalyzed Ring O pening of an Epoxide 531 Base-Catalyzed Ring O pening of an Epoxide 531

C h a p te r 6 M echanism for th e S n2 Reaction 239 The Stereochem istry of an S n2 Reaction M echanism for th e S n 1 Reaction 248 The Stereochem istry of an S n 1 Reaction M echanism for th e E2 Reaction 270 M echanism for th e E1 Reaction 272

245 252

C h a p te r 7 E2 Elimination W here There Are Two Axial b H ydrogens 296 E2 Elimination W here th e Only Axial b H ydrogen Is from a Less S table C onform er 296 Acid-Catalyzed Dehydration of S econdary or Tertiary Alcohols: An E1 Reaction 301 Dehydration of a Primary Alcohol: An E2 Reaction 302 Form ation of a R earranged Alkene during Dehydration of a Primary Alcohol 306 D ehydrohalogenation of vic-Dibromides to Form Alkynes 309 The Dissolving Metal Reduction of an Alkyne 316 C h a p te r 8 Addition of a H ydrogen Halide to an Alkene 335 Addition of HBr to 2-M ethylpropene 337 Ionic Addition to an Alkene 339 Acid-Catalyzed Hydration of an Alkene 341 O xym ercuration 345 H ydroboration 349 Oxidation of Trialkylboranes 351 Addition of Bromine to an Alkene 356 Addition of Bromine to cis- and trans-2-Butene 359 Halohydrin Form ation from an Alkene 360 Ozonolysis of an Alkene 368 C h a p te r 10 H ydrogen Atom A bstraction 461 Radical Addition to a p Bond 461 Radical Chlorination of M ethane 468 Radical H alogenation of Ethane 477 The Stereochem istry of Chlorination at C2 of P entane 481 The Stereochem istry of Chlorination at C3 of (S)-2-Chloropentane 482 Anti-Markovnikov A ddition 485 Radical Polymerization of Ethene 487

C h a p te r 12 Reduction of A ldehydes and K etones by Hydride Transfer 554 C hrom ate Oxidations: Form ation of th e Chrom ate Ester 559 The Grignard Reaction 566 C h a p te r 15 Electrophilic Arom atic Bromination Nitration of Benzene 682 Sulfonation of Benzene 683 Friedel-Crafts Alkylation 684 Friedel-Crafts Acylation 687 Benzylic H alogenation 710 Birch Reduction 720

680

C h a p te r 16 Reduction of an Acyl Chloride to an A ldehyde 736 Reduction of an Ester to an A ldehyde 737 Reduction of a Nitrile to an A ldehyde 737 Addition of a Strong N ucleophile to an A ldehyde or K etone 742 Acid-Catalyzed N ucleophilic Addition to an A ldehyde or K etone 742 H em iacetal Form ation 744 Acid-Catalyzed H em iacetal Formation 745 Base-Catalyzed Hem iacetal Formation 746 H ydrate Form ation 746 Acid-Catalyzed Acetal Form ation 748 Imine Formation 751 Enamine Form ation 754 Cyanohydrin Formation 755 The Wittig Reaction 758 C h a p te r 17 Acyl Substitution by N ucleophilic A ddition­ Elimination 792 Synthesis of Acyl C hlorides Using Thionyl Chloride Acid-Catalyzed Esterification 798 Base-Prom oted Hydrolysis of an Ester 801 D CC-Prom oted A m ide Synthesis 807

795

xix

Contents

Acidic Hydrolysis of an A m ide 808 Basic Hydrolysis of an A mide 808 Acidic Hydrolysis of a Nitrile 810 Basic Hydrolysis of a Nitrile 810

The The The The

C onjugate Addition of HCN 891 C onjugate Addition of an Amine 892 Michael Addition 892 Mannich Reaction 895

C h a p te r 18

C h a p te r 20

Base-Catalyzed Enolization 835 Acid-Catalyzed Enolization 835 Base-Prom oted H alogenation of A ldehydes and K etones 837 Acid-Catalyzed H alogenation of A ldehydes and Ketones 837 The Haloform Reaction 839 The Malonic Ester Synthesis of S ubstituted Acetic Acids 850

Alkylation of NH 3 925 Reductive Amination 928 The Hofmann R earrangem ent Diazotization 936

C h a p te r 19

C h a p te r 22

The Claisen C ondensation 871 The Dieckmann C ondensation 873 The A ldol A dd itio n 877 Dehydration of th e Aldol A ddition Product 879 The Acid-Catalyzed A ld o l Reaction 880 A Directed Aldol Synthesis Using a Lithium Enolate The Aldol Cyclization 889

Form ation of a G lycoside 1011 Hydrolysis of a G lycoside 1012 Phenylosazone Form ation 1023

931

C h a p te r 21 The Kolbe Reaction 975 The SNAr M echanism 982 The Benzyne Elimination-Addition M echanism

985

C h a p te r 24 Form ation of an a-Aminonitrile during th e Strecker Synthesis 1093

886

THE CHEMISTRY OF . . . BOXES C h a p te r 1

C h a p te r 6

Calculated M olecular M odels: Electron Density Surfaces 29

Biological M ethylation: A Biological Nucleophilic Substitution Reaction 266

C h a p te r 2

C h a p te r 7

C alculated M olecular M odels: M aps of Electrostatic P otential 59 Ethers as General Anesthetics 67 Fluorocarbons and Teflon 78

H ydrogenation in th e Food Industry 313 From th e Inorganic to th e O rganic 321

Organic Templates Engineered to M im ic Bone Growth

C h a p te r 8 82

C h a p te r 3 HOM Os and LUMOs in Reactions

The Sea: A Treasury of Biologically Active Natural Products 357

Catalytic Asymmetric Dihydroxylation

365

105 C h a p te r 9

C h a p te r 4 Petroleum Refining 139 Pherom ones: Com m unication by M eans of Chemicals Muscle Action 162 N anoscale M otors and M olecular Switches 166 Elemental Carbon 176

M agnetic Resonance Imaging in M edicine 156

C h a p te r 5 Life's Molecular H andedness 193 Selective Binding of Drug Enantiom ers to Left- and Right­ H anded Coiled DNA 211

425

C h a p te r 10 Calicheamicin g - 1: A Radical Device for Slicing the Backbone o f D N A 492 A ntioxidants 494 O zone D epletion and Chlorofluorocarbons (CFCs) C h a p te r 11 Ethanol as a Biofuel 508 Alkyl Phosphates 521

495

xx

Contents

C h a p te r 18

529 Epoxides, C arcinogens, and Biological O xidation 533 Environmentally Friendly Alkene Oxidation M ethods 537 Transport A ntibiotics and Crown Ethers 539 The Sharpless A sym m etric Epoxidation

Chloroform in Drinking W ater

839

C h a p te r 19 A Retro-Aldol Reaction in Glycolysis— Dividing A ssets to D ouble th e ATP Yield 878 Calicheamicin g 1I Activation for C leavage of DNA 894 A Suicide Enzyme Substrate 895

C h a p te r 12 Alcohol D ehydrogenase— A Biochemical Hydride Reagent 554 S tereoselective Reductions of Carbonyl G roups 555 C h a p te r 13

C h a p te r 20 Biologically Im portant Amines N-Nitrosamines 936 Chem otherapy and Sulfa Drugs

Allylic Bromination 590 The Photochem istry of Vision 609 M olecules with th e N obel Prize in Their Synthetic Lineage 620

922 944

C h a p te r 21

C h a p te r 14

Polyketide A nticancer Antibiotic Biosynthesis 975 The Bom bardier Beetle's Noxious Spray 979 Bacterial D ehalogenation of a PCB Derivative 983 Aryl Halides: Their Uses and Environmental C oncerns

655 Sunscreens (Catching th e Sun's Rays and W hat H appens to Them) 664

Artificial S w eeteners (How Sw eet It Is)

C h a p te r 15

C h a p te r 23

N anotubes

Iodine Incorporation in Thyroxine Biosynthesis Industrial Styrene Synthesis 709

707

C h a p te r 16

C h a p te r 22 1032

O lestra and O ther Fat Substitutes 1055 Self-A ssem bled M onolayers— Lipids in Materials Science and Bioengineering 1060 STEALTH® Liposom es for Drug Delivery 1077

A ldehydes and K etones in Perfum es 733 A Very Versatile Vitamin, Pyridoxine (Vitamin B6) 753

C h a p te r 24

C h a p te r 17

Sickle-Cell Anemia 1103 C arbonic A nhydrase: Shuttling th e Protons S om e Catalytic A ntibodies 1123

Penicillins

811

989

1119

Preface " C a p t u r i n g t h e P o w e r f u l a n d E x c it in g S u b je c t o f O r g a n ic C h e m is t r y "

We w ant our students to learn organic chem istry as w ell and as easily as possible. We also w ant students to enjoy this exciting subject and to learn about the relevance o f organic chem istry to their lives. A t the sam e tim e, w e w ant to help students develop the skills of critical thinking, problem solving, and analysis that are so im portant in to d ay ’s w orld, no m atter w hat career paths they choose. T he richness o f organic chem istry lends itself to solutions for our tim e, from the fields o f health care, to energy, sustainability, and the environm ent. G uided by these goals, and by w anting to m ake our book even m ore accessib le to s tu d e n ts than it has ever been before, w e have brought m any changes to this edition.

New To This Edition •

S olved P ro b le m s. We have greatly increased the n um ber o f Solved Problem s. Now over 150 S olved P ro b le m s guide students in their strategies for problem solving. Solved Problem s are u su a lly p a ire d w ith a re la te d R ev iew P ro b le m .

•

R eview P ro b le m s. In-text R eview P ro b le m s, over 10% o f them new, provide students w ith opportunities to check their progress as they study. If they can w ork the review problem , they should m ove on. If not, they should review the preceding presentation.

ILLUSTRATING A MULTISTEP SYNTHESIS Starting w ith brom obenzene and any other needed reagents, outline

a synthesis of the follow ing aldehyde: O H AN SW ER W orking backw ard, w e rem em ber that w e can synthesize the aldehyde from the corresponding alcohol

b y oxidation w ith PCC (Section \2 .4 A ). T he alcohol can be m ade b y treating phenylm agnesium brom ide w ith oxirane. [A dding oxirane to a G rignard reagent is a very useful m ethod for adding a — CH 2 CH 2 OH u n it to an organic group (Section \2.1B ).] Phenylm agnesium brom ide can be m ade in the usual way, b y treating brom obenzene w ith m agnesium in an ether solvent. Retrosynthetic Analysis O

V H

OH

MgBr

Synthesis (1) v (2) H3O+

Mg ELO*

,

PCC

MgBr

Br

OH

Provide retrosynthetic analyses and syntheses for each o f the follow ing alcohols, starting w ith appropriate alkyl or aryl halides. OH

(a)

O

CHCl 2 2

H

R e v ie w P ro b le m 1 2 .

OH (three ways)

(e)

(b)

(three ways)

(c)

(two w ays)

(d)

(f)

(two ways)

OH

(two ways)

(three ways) OH

XXi

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Preface

•

•

E nd-of-chapter problem s have been grouped and labeled b y topic. Students and instructors can m ore easily select problem s for specific purposes.

E n d -o f-C h a p te r P ro b le m s. O ver 15% o f the end-of-chapter problem s are new, and others have been revised. R ELATIVE RATES O F N U C L E O P H IL IC S U B S T IT U T IO N

6 .2 0

W hich alkyl halide w ould you expect to react m ore rapidly b y an S ^2 m echanism ? E xplain your answer. xBr

(a)

or

Cl

or

(d) Cl

B Br ( b ).

(c) 6.21

"C l

X

„01

or

(e)

or

^

V

Cl

'C l

W hich S^-2 reaction o f each pair w ould you expect to take place m ore rapidly in a protic solvent? Cl

(a) (1)

C l-

E tO -

or (2)

Cl

EtOH

HCl S Y N T H E S IS

6.2 3

Show how you m ight use a nucleophilic substitution reaction o f 1-brom opropane to synthesize each o f the following com pounds. (You m ay use any other com pounds that are necessary.) (e)

O

(g)

N +(C H 3)3 Br"

•

T hroughout the book, m ore problem s are cast in a visual form at using structures, eq u a­ tions, and schem es. In addition, w e still provide C h a lle n g e P ro b le m s an d L e a rn in g G ro u p P ro b le m s to serve additional teaching goals.

•

K ey ideas in every section have been rew ritten and em phasized as b u lle t p o in ts to help students focus on the m ost essential topics.

3.2A Br0 nsted-Lowry Acids and Bases Two classes of acid-base reactions are fundamental in organic chemistry: Br0nsted-Lowry and Lewis acid-base reactions. W e start our discussion with Br0nsted-Lowry acid-base reactions. • Br0nsted-Low ry acid-base reactions involve the transfer of protons. • A B r 0 n ste d -L o w ry acid is a substance that can donate (or lose) a proton. • A B r 0 n ste d -L o w ry base is a substance that can accept (or remove) a proton.

xxiii

Preface

•

“H ow to ” S ections give step-by-step instructions to guide students in perform ing im portant tasks, such as using curved arrows, draw ing chair conform ations, planning a G rignard synthesis, determ ining form al charges, w riting L ew is structures, an d using 13C and 1H N M R spectra to determ ine structure.

^

3.5 H o w to Use C u rv e d A rro w s in Illu s tra tin g R e a c tio n s U p to this point w e have not indicated how bonding changes occur in the reactions w e have presented, but this can easily be done using curved-arrow notation. C urved arrows •

show the direction o f electron flow in a reaction mechanism.

• point from the source o f an electron pair to the atom receiving the pair. (Curved arrows can also show the m ovem ent o f single electrons. W e shall discuss reactions o f this type in a later chapter.) •

always show the flow o f electrons from a site o f higher electron density to a site of lower electron density.

• never show the m ovem ent o f atoms. A tom s are assum ed to follow the flow o f the electrons.

•

N ew and updated c h a p te r-o p e n in g v ig n e tte s an d T h e C h e m istry o f . . . boxes bring organic chem istry hom e to everyday life experiences. M o re p h o to s a r e in c lu d e d to help students relate organic chem istry to the w orld around them.

Ionic Reactions N ucleophilic Substitution and Elimination Reactions o f Alkyl Halides

TH E C H E M IS TR Y O F . . . B io lo g ic a l M e th y la tio n : A B io lo g ic a l N u c le o p h ilic S u b s titu tio n R eaction T he ce lls o f liv in g o rg a n is m s s y n th e size m a n y o f th e c o m ­

tra n s fe r ta kes p la ce can be d e m o n s tra te d e x p e rim e n ta lly by

p o u n d s th e y n e e d fro m s m a lle r m o le c u le s . O fte n these b io s y n th e s e s re s e m b le th e syn the ses o rg a n ic ch e m ists ca rry o u t in th e ir la b o ra to rie s . L e t us e x a m in e o n e e x a m ­ p le now. M a n y re a ctio n s ta k in g p la ce in th e cells o f p la n ts a n d a n i­

fe e d in g a p la n t o r a n im a l m e th io n in e c o n ta in in g an is o to p ica lly la b e le d ca rb o n a to m (e.g ., 13C o r 14C ) in its m eth yl g ro u p . Later, o th e r c o m p o u n d s c o n ta in in g th e " la b e le d " m e th yl g r o u p can be is o la te d fro m th e o rg a n is m . S o m e o f th e c o m p o u n d s th a t g e t th e ir m e th yl g ro u p s fro m m e th io ­

m als in v o lv e th e tra n s fe r o f a m e th yl g r o u p fro m an am in e a cid c a lle d m e th io n in e to s o m e o th e r c o m p o u n d . T h a t this

nin e are th e fo llo w in g . T h e is o to p ic a lly la b e le d ca rb o n a tom is sho w n in g re en .

O rg a n ic s ynthe ses, w h e th e r th e y ta k e p la c e in th e gla ssw are o f th e la b o ra to r

-O 2CCHCH2CH2SCH3

ism , o fte n in v o lv e fa irly s im p le processes, such as th e in s ta lla tio n o f a m e th yl e x a m p le , w e m ay w a n t to ins ta ll a m e th y l g r o u p o n th e n itro g e n a to m o f a t

Methionine

an im p o r ta n t c o u n te r p a rt in b io c h e m is try . To d o th is w e o fte n e m p lo y a rea ct

R R— N='

I

H 3C— I

R

R.

R — N — CHo I 3 R

HO H HO

If w e w a n te d to d e s c rib e th is re a c tio n to an o rg a n ic c h e m is t w e w o u ld d e scr t io n r e a c tio n , a k ind o f re a c tio n w e d e s c rib e in d e ta il in th is ch a p te r. O n th e o th e r h a n d , i f w e w a n te d to d e s c rib e th is re a c tio n to a b io c h e m is t f e r r e a c tio n . B io c h e m is ts ha ve d e s c rib e d m a n y s im ila r re a c tio n s th is way, fc tra n s fe rs a m e th y l g r o u p fro m S -a d e n o s y /m e th io n in e (SAM ) to a te rtia ry a m in e to m ake c h o lin e . C h o lin e is in c o rp o ra te d in to th e p h o s p h o lip id s o f o u r c e ll m e m b ra n e s , an d i t is th e h yd ro lysis p r o d u c t o f a c e ty lc h o lin e , an im p o r ta n t n e u ro tra n s m itte r. (C rystals o f a c e ty lc h o lin e are s h o w n in th e p o la riz e d lig h t m ic ro s c o p y im a g e ab ove .) N ow , th e b io lo g ic a l re a c tio n m a y see m m o re c o m p lic a te d , b u t its essence is s im ila r to m a n y n u c le o p h ilic s u b s titu tio n re a c tio n s w e shall s tu d y in th is ch a p te r. F irst w e c o n s id e r alkyl h a lide s, on e o f th e m o s t im p o r ta n t ty p e s o f re a c ta n ts in n u c le o p h ilic s u b s titu tio n rea c tion s .

230

CH - N — CH2CH2OH

I CHS

HO Nicotine

CH CH

Adrenaline

2

Choline

2

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Preface

• B o n d -lin e fo rm u la s replace alm ost all dash and condensed structural form ulas after C hapter O ne w here they are introduced an d explained. B ond-line form ulas are cleaner, sim pler, and faster fo r students to interpret, and they are the form at m ost often used by chem ists to depict organic m olecules. Hydroboration

CH,

B H ,: TH F

CH,

3<

+

enantiom er

A n ti-M a rk o v n ik o v and syn ad d itio n

H

B R

R

R = 2-methylcyclopentyl

Oxidation

CH ,

h 2o

2, h o A M E C H A N IS M FO R THE REA C TIO N

— O H repla ces bo ro n w ith re te n tio n of c o n fig u ra tio n

M e ch a n ism f o r th e S n 1 R e actio n

REACTION CH, I 2 C H ,— C — C h 8

T his edition also offers students m any visually oriented tools to accom m odate diverse learning styles. T hese include S y n th etic C o n n e ctio n s, C o n c e p t M a p s, T h e m a tic M e c h a n ism R eview S u m m arie s, and the detailed M e c h a n ism f o r th e R e a c tio n B oxes already m entioned. W e also offer H e lp fu l H in ts and richly annotated illustrations.

CH, I 8 C H ,— C — O H S I CH,

2 H 2O

I CH,

MECHANISM Step 1

CH, I S C H ,— C — C h S I CH,

h2o

A ided by th e polar solvent, a chlorine departs with the electron p air that bonded it to th e carbon.

C H 3— C+ + 3 \ CH.

¡ C l!

»

T h is s lo w ste p p roduces the 3° carbocation in term e­ diate a nd a chloride ion. A lth o u g h n ot sh o w n here, th e io n s a re s o lv a te d (and s t a b iliz e d ) by w a te r m o l­ ecules.

Step 2

Step 1 AG*(i) is5 much UUT (1) 1 Transition larger than state 1 AG*(2) or A G *^ , hence thi this is 4 s t(i) ,he e s|owes slowest step 3'(i)

Reaction coordinate

CH ----------- v .. C H ,— C + + ^ O— H 3 \ I CH, H

Step 2

fast C H ,— C 8 C

-O — H

I

state 2

.

H

A w a te r m olecule a ctin g as

The product is a tert-

_

j . AG*(2)

1 ■

S u m m a r y o f S ^ 1 v e r s u s S ^ 2 R e a c tio n

m

s --------Reaction coordinate

H elpful H in t

S n 1: T h e F o llo w in g C o n d itio n s F a v o r an S n 1 R eaction :

Step 3

1. A substrate that can form a relatively stable carbocation (such as a substrate w ith a

Sn1 versus Sn2 H— O— H

leaving group at a tertiary position)

I

H

2. A relatively w eak nucleophile

tert-butyl d ro n iu m

3. A polar, protic solvent t h e fe rf-b u ty lo x o n iu m

state 3

« A G °4

V -----

(,:

Reaction coordinate

•

C hapters on c a rb o n y l c h e m is try h av e b e e n re o rg a n iz e d to em phasize m echanistic them es o f nucleophilic addition, acyl substitution, and reactivity at the a-carbon.

•

T he im portant m odern synthetic m ethods o f the G ru b b s , H ec k , S o n o g a sh ira , Stille, a n d S u zu k i transition m etal catalyzed carbon-carbon bond-form ing reactions are p re­ sented in a practical and student-oriented w ay that includes review problem s and m ech ­ anistic context (Special Topic G).

•

T hroughout the book, w e have stre a m lin e d o r re d u c e d c o n te n t to m atch the m odern practice o f organic chem istry, and w e have provided new coverage o f current reactions. W e have m ade our b o o k m o re accessible to students than ever before. W hile m ain tain ­ ing our com m itm ent to an appropriate level an d breadth o f coverage.

Preface

Organization - An Emphasis on the Fundamentals So m uch o f organic chem istry m akes sense and can b e generalized if students m aster and apply a few fundam ental concepts. T herein lays the beauty o f organic chem istry. If stu­ dents learn the essential principles, they w ill see that m em orization is n o t needed to suc­ ceed in organic chem istry. M ost im portant is for students to have a solid understanding o f structure— o f hybridization and geom etry, steric hindrance, electronegativity, polarity, form al charges, and resonance — so that they can m ake intuitive sense o f m echanism s. It is w ith these topics that w e begin in C hapter 1. In C hapter 2 w e introduce the fam ilies o f functional groups - so that students have a platform on w hich to apply these concepts. We also in tro ­ duce interm olecular forces, and infrared (IR) spectroscopy - a key tool for identifying functional groups. T hroughout the book w e include calculated m odels o f m olecular orbitals, electron density surfaces, and m aps o f electrostatic potential. T hese m odels enhance students’ appreciation for the role o f structure in properties and reactivity. We begin our study o f m echanism s in the context o f acid-base chem istry in C hapter 3. A cid-base reactions are fundam ental to organic reactions, and they lend them selves to introducing several im portant topics that students need early in the course: ( 1) curved arrow notation for illustrating m echanism s, ( 2 ) the relationship betw een free-energy changes and equilibrium constants, and (3) the im portance o f inductive and resonance effects and o f solvent effects. In C hapter 3 w e present the first o f m any “M echanism for the R eaction” boxes, using an exam ple that em bodies both B ronsted-L ow ry an d L ew is acid-base principles. All throughout the book, w e use boxes like these to show the details o f key reaction m ech ­ anism s. A ll o f the M echanism for the R eaction boxes are listed in the Table o f C ontents so that students can easily refer to them w hen desired. A central them e o f o ur approach is to em p h asize th e re la tio n sh ip between structure and re a ctivity. T his is w hy w e choose an organ izatio n th at com b in es the m o st u sefu l fe a ­ tures o f a fun ctional group approach w ith on e b ased on reactio n m echanism s. O u r p h i­ losophy is to em phasize m echanism s and fun d am en tal p rinciples, w hile giving students the anchor points o f functional groups to apply th eir m ech an istic k n o w led g e an d in tu ­ ition. T he structural aspects o f ou r approach show students w h a t o rg a n ic c h e m is try is. M echanistic aspects o f our approach show students h o w it w o rk s. A n d w herever an opportunity arises, w e show them w h a t it d o e s in living system s an d the p h y sical w orld around us. In summ ary, our w ork on the 10th edition reflects the com m itm ent w e have as teach­ ers to do the best w e can to help students learn organic chem istry an d to see how they can apply their know ledge to im prove our w orld. T he enduring features o f our b o o k have proven over the years to help students learn organic chem istry. T he changes in our 10th edition m ake organic chem istry even m ore accessible and relevant. Students w ho use the in-text learning aids, w ork the problem s, and take advantage o f the resources and practice available in W ileyPLUS (our online teaching and learning solution) w ill b e assured o f suc­ cess in organic chem istry.

WileyPLUS for Organic Chemistry —A P L U S

P o w e rfu l

T e a ch in g a n d L e a rn in g S o lu tio n

This online teaching and learning environm ent integrates the e n tire d ig ita l te x tb o o k w ith the m ost effective instructor and student resources to fit every learning style. W ith W ileyPLUS (w w w .w ileyplus.com ): •

Students achieve concept m astery in a rich, structured environm ent th at’s available 24/7

•

Instructors personalize and m anage their course m ore effectively w ith assessm ent, assignm ents, grade tracking, and m ore.

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Preface

W ileyPLUS can com plem ent your current textbook or replace the printed text altogether. T he problem types and resources in W ileyPLUS are designed to enable and support problem -solving skill developm ent and conceptual understanding. T h re e u n iq u e re p o s i­ to rie s o f a sse ssm e n t a r e o ffe re d w h ich p ro v id e s b r e a d th , d e p th a n d flexibility:

1. E n d o f c h a p te r ex ercises, m any o f w hich are algorithm ic, feature structure draw ing/ assessm ent functionality using M arvinSketch, an d provide im m ediate answ er feed ­ back. A subset o f these end o f chapter questions are linked to G u id e d O n lin e T u to ria ls w hich are stepped-out problem -solving tutorials that w alk the student through the problem , offering in d iv id u alize d fee d b ack at each step. 2. T est B a n k q u e stio n s con­ sisting o f over 3,000 q u es­ tions. 3. P re b u ilt c o n c e p t m a ste ry a s s ig n m e n ts , o rg an ized b y topic and concept, featuring robust answ er feedback.

W ile y P L U S F o r S t u d e n t s

D ifferent learning styles, different levels o f proficiency, different levels o f preparation— each o f your students is unique. W ileyP LU S o ffe rs a m y ria d o f ric h m u ltim e d ia re so u rc e s fo r stu d e n ts to fa c ilita te le a rn in g . T hese include: •

O ffice H o u r V ideos: T he solved problem s from the b o o k are presented by an organic chem istry professor, using audio and a w hiteboard. It em ulates the experience that a student w ould g et if she or h e w ere to attend office hours an d ask for assistance in w ork­ ing a problem . T he goal is to illustrate good problem solving strategies.

Preface

•

S k illB u ild in g E x ercises: A nim ated exercises, w ith instant feedback, reinforce the key skills required to succeed in organic chemistry.

•

C o re C o n c e p t A n im a tio n s: C oncepts are thoroughly explained using audio and w h ite­ board.

W ile y P L U S F o r I n s t r u c t o r s

W ileyPLU S em pow ers you w ith the tools and resources you need to m ake your teaching even m ore effective: •

You can custom ize your classroom presentation w ith a w ealth o f resources an d func­ tionality from Pow erP oint slides to a database o f rich visuals. You can even ad d your ow n m aterials to your W ileyPLUS course.

•

W ileyPLUS allow s you to hellp students w ho m ight fall behind, by tracking their progress and offering assistance easily, even before they com e to office hours.

•

W ileyPLUS sim plifies and autom ates such tasks as student perform ance assessm ent, creating assignm ents, scoring student work, keeping grades, and m ore.

Supplements S t u d y G u i d e a n d S o l u t i o n s M a n u a l (IS B N 9 7 8 - 0 - 4 7 0 - 4 7 8 3 9 - 4 ) The Study G uide and Solutions M anual for O rganic Chemistry, Tenth E d itio n , authored by R obert Johnson, o f X avier U niversity, Craig Fryhle, G raham Solom ons, w ith contributions from C hristopher Callam , o f T he O hio State University, c o n ta in s ex p la in e d so lu tio n s to all o f th e p ro b le m s in th e tex t. The Study G uide also contains: •

A n introductory essay “Solving the P uzzle— or— Structure is E verything” that serves as a bridge from general to organic chem istry

•

Sum m ary tables o f reactions by m echanistic type and functional group

•

A review quiz for each chapter

•

A set o f hands-on m olecular m odel exercises

•

Solutions to the problem s in the Special Topics sections (m any o f the Special Topics are only available w ithin WileyPLUS)

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Preface

O r g a n i c C h e m i s t r y a s a S e c o n d L a n g u a g e ™ , V o l u m e s I & II B y D a v id K le in ( J o h n s H o p k i n s U n iv e r s it y ) D avid K lein’s series o f course com panions has been an enorm ous success w ith students and instructors (O rganic C hem istry as a Second L anguage, Part I, ISBN: 978-0-470­ 12929-6; O rganic C hem istry as a Second L anguage, P art II, ISBN: 978-0-471-73808-5). Presenting fundam ental principles, problem -solving strategies, an d skill-building exercise in relaxed, student-friendly language, these books have been cited by m any students as integral to their success in organic chemistry. M o l e c u l a r V is io n s ™ M o d e l K its We believe that the tactile experience o f m anipulating physical m odels is key to students’ understanding that organic m olecules have shape an d occupy space. To support our p ed a­ gogy, w e have arranged w ith the D arling Com pany to bundle a special ensem ble o f M olecular V isions™ m odel kits w ith our book (for those w ho choose that option). We use H elpful H int icons and m argin notes to frequently encourage students to use hand-held m odels to investigate the three-dim ensional shape o f m olecules w e are discussing in the book. In s tru c to r R e so u rc e s A ll Instructor R esources are available w ithin W ileyPLUS or they can b e accessed by con­ tacting your local W iley Sales Representative. T est B a n k . A uthored by R obert Rossi, o f G loucester C ounty College, Justin W yatt, o f the C ollege o f C harleston, and M aged H enary, o f G eorgia State University, the Test B ank for this edition has been com pletely revised and updated to include over 3,000 short answer, m ultiple choice, and essay/draw ing questions. It is available in b oth a printed and co m ­ puterized version. P o w e rP o in t L e c tu re slides. A set o f P ow erPoint L ecture Slides have been prepared by Professor W illiam Tam, o f the U niversity o f G uelph an d his w ife, Dr. Phillis Chang. This new set o f Pow erP oint slides includes additional exam ples, illustrations, and presentations that help reinforce an d test students’ grasp o f organic chem istry concepts. A n additional set o f P ow erP oint slides features the illustrations, figures, an d tables from the text. All P ow erP oint slide presentations are custom izable to fit your course. P e rso n a l R e sp o n se S y stem (“ C lic k e r” ) Q u estio n s. A b ank o f questions is available for anyone using personal response system technology in their classroom . T he clicker ques­ tions are also available in a separate set o f P ow erP oint slides. D ig ita l Im a g e L ib ra ry . Im ages from the text are available online in JP E G form at. Instructors m ay use these to custom ize their presentations an d to provide additional visu­ al support for quizzes an d exams.

Acknowledgments We are especially grateful to the follow­ ing people who provided detailed reviews that helped us prepare this new edition o f Organic Chemistry.

Ihsan Erden, San Francisco State University

Angela J. Allen, University of Michigan-Dearborn

Andreas Franz, University o f the Pacific

Karen Aubrecht, State University of New York, Stonybrook

Bill Fowler, State University o f New York, Stonybrook

Sandro Gambarotta, University o f Ottawa

Michael S. Leonard, Washington and Jefferson College Jesse More, Loyola College Ed O’Connell, Fairfield University Cathrine Reck, Indiana UniversityBloomington

Jovica Badjic, Ohio State University

Tiffany Gierasch, University o f Maryland-Baltimore County

Ed Biehl, SMU

David Harpp, McGill University

Kaiguo Chang, University o f Arkansas at Fort Smith

Nina E. Heard, University o f North Carolina-Greensboro

Christopher Callam, Ohio State University

Frederick J. Heldrich, College o f Charleston

Arthur Cammers, University o f Kentucky

James W. Hershberger, Miami University-Oxford

Community College

Sean Hickey, University o f New Orleans

Leyte L. Winfield, Spelman College

Ian Hunt, University o f Calgary

Justin Wyatt, College o f Charleston

Shouquan Huo, East Carolina University

Linfeng Xie, University o f Wisconsin, Oshkosh

D. Scott Davis, Mercer University

Ekaterina N. Kadnikova, University o f Missouri

Peter deLijser, California State University Fullerton

Mohammad R. Karim, Tennessee State University

Clarke W. Earley, Kent State University

Adam I. Keller, Columbus State Community College

Jeremy Cody, Rochester Institute of Technology Arlene R. Courtney, Western Oregon University Shadi Dalili, University o f Toronto

James Ellern, University o f Southern California

We are also grateful to the many people who provided reviews that guided prepara­ tion o f the earlier editions o f our book Chris Abelt, College o f William and Mary; James Ames, University o f Michigan, Flint; Merritt B. Andrus, Brigham Young University; W. Lawrence Armstrong. SUNY College at Oneonta; Steven Bachrach. Trinity University; Winfield M. Baldwin, University o f Georgia; David Ball, California State University, Bill J. Baker. University o f South Florida; Chico; George Bandik, University o f Pittsburgh; Paul A. Barks, North Hennepin State Junior College; Kevin Bartlett. Seattle Pacific University; Ronald Baumgarten, University o f Illinois at Chicago; Harold Bell, Virginia Polytechnic Institute and State University; Kenneth Berlin, Oklahoma State University; Stuart R.

Joel Ressner, West Chester University Harold R. Rogers, California State University-Fullerton Robert Stolow, Tufts University Neal Tonks, College o f Charleston Janelle Torres y Torres, Muscatine

Aleksey Vasiliev, East Tennessee State University Kirk William Voska, Rogers State University

Jennifer Koviach-Cote, Bates College

Regina Zibuck, Wayne State University

Berryhill, California State University, Long Beach; Edward V. Blackburn, University o f Alberta; Brian M. Bocknack, University o f Texas, Austin; Eric Bosch, Southwest Missouri StateUniversity; Newell S. Bowman, The University o f Tennessee; Bruce Branchaud, University o f Oregon; Wayne Brouillette, University o f Alabama; Ed Brusch, Tufts University; Christine Brzezowski, University of Alberta; Edward M. Burgess, Georgia Institute o f Technology; Bruce S. Burnham, Rider University; Robert Carlson, University o f Minnesota; Todd A. Carlson, Grand Valley State University; Lyle W. Castle, Idaho State University; Jeff Charonnat, California State University, Northridge; George Clemans, Bowling Green State University; William D. Closson, State University o f New York

at Albany; Sidney Cohen, Buffalo State College; Randolph Coleman, College o f William & Mary; David Collard, Georgia Institute o f Technology; David M. Collard, Georgia Institute o f Technology; Brian Coppola, University o f Michigan; Phillip Crews, University o f California, Santa Cruz; James Damewood, University o f Delaware; D. Scott Davis, Mercer University; Roman Dembinski, Oakland University; O. C. Dermer, Oklahoma StateUniversity; Phillip DeShong, University o f Maryland; John DiCesare, University o f Tulsa; Trudy Dickneider, University o f Scranton; Marion T. Doig III, College o f Charleston; Paul Dowd, University o f Pittsburgh; Robert C. Duty, Illinois State University; Eric Edstrom, Utah State University; James Ellern, University o f Southern California; Stuart

xxix

xxx

Acknowledgments

Fenton, University o f Minnesota; George Fisher, Barry University; Gideon Fraenkel, The Ohio State University; Jeremiah P. Freeman, University o f Notre Dame; Mark Forman, Saint Joseph’s University; Peter Gaspar, Washington University, St. Louis; Cristina H.Geiger, SUNY Geneseo; M. K. Gleicher, Oregon State University; Brad Glorvigen, University o f St. Thomas; Felix Goodson, West Chester University; Ray A. Goss Jr., Prince George’s Community College; Roy Gratz, Mary Washington College; Wayne Guida, Eckerd College; Frank Guziec, New Mexico State University; Christopher M. Hadad, Ohio State University; Dennis Hall, University o f Alberta; Philip L. Hall, Virginia Polytechnic Institute and State University; Steven A. Hardinger, University of California at Los Angeles; Lee Harris, University o f Arizona; Kenneth Hartman, Geneva College; Bruce A. Hathaway, Southeast Missouri State University; David C. Hawkinson, University o f South Dakota; Michael Hearn, Wellesley College; Rick Heldrich, College o f Charleston; John Helling, University o f Florida; William H. Hersh, Queens College; Paul Higgs, Barry University; Jerry A. Hirsch, Seton Hall University; Carl A. Hoeger, University o f California, San Diego; John Hogg, Texas A & M University; John Holum, Augsburg College; John L. Isidor, Montclair State University; John Jewett, University o f Vermont; A. William Johnson, University o f North Dakota; Robert G. Johnson, Xavier University; Stanley N. Johnson, Orange Coast College; Jeffrey P. Jones, Washington State University, Pullman; John F. Keana, University o f Oregon; John W. Keller, University o f Alaska, Fairbanks; Colleen Kelley, Pima Community College; David H. Kenny, Michigan Technological University; Robert C. Kerber, State University o f New York at Stony Brook; Karl R. Kopecky, The University o f Alberta; Paul J. Kropp, University of North Carolina at Chapel Hill; Michael Kzell, Orange Coast College; Cynthia M. Lamberty, Nicholls State University; John A. Landgrebe, University o f Kansas; Paul Langford, David Lipscomb University; Julie E. Larson, Bemidji State University; Allan K. Lazarus, Trenton State College;

Thomas Lectka, Johns Hopkins University; James Leighton, Columbia University; Philip W. LeQuesne, Northeastern University; Robert Levine, University o f Pittsburgh; Samuel G. Levine, North Carolina State University; James W. Long, University o f Oregon; Eugene Losey, Elmhurst College; Patricia Lutz, Wagner College; Frederick A. Luzzio, University o f Louisville; Javier Macossay, The University o f Texas, Pan American; Ronald M. Magid, University o f Tennessee; Rita Majerle, Hamline University; John Mangravite, West Chester University; Jerry March, Adelphi University; Przemyslaw Maslak, Pennsylvania State University; Janet Maxwell, Angelo State University; Shelli R. McAlpine, San Diego State University; James McKee, University o f the Sciences, Philadelphia; Mark C. McMills, Ohio University; John L. Meisenheimer, Eastern Kentucky University; Gary Miracle, Texas Tech University; Gerado Molina, Universidad de Puerto Rico; Andrew Morehead, University o f Maryland; Andrew T. Morehead Jr., East Carolina University; Renee Muro, Oakland Community College; Jesse M. Nicholson, Howard University; Everett Nienhouse, Ferris State College; John Otto Olson, University o f Alberta; Kenneth R. Overly, Richard Stockton College, N J; Michael J. Panigot, Arkansas State University, Jonesboro; Paul Papadopoulos, University o f New Mexico; Cyril Parkanyi, Florida Atlantic University; Dilip K. Paul, Pittsburg State University, KS; James W. Pavlik, Worcester Polytechnic Institute; Robert Pavlis, Pittsburg State University; John H. Penn, West Virginia University; Christine A. Pruis, Arizona State University; William A. Pryor, Louisiana StateUniversity; Shon Pulley, University of Missouri, Columbia; Eric Remy, Virginia Polytechnic Institute; Joel M. Ressner, West Chester University; Michael Richmond, University o f North Texas; Thomas R. Riggs, University o f Michigan; Frank Robinson, University o f Victoria, British Columbia; Stephen Rodemeyer, California State University, Fresno; Alan Rosan, Drew University; Christine Russell, College o f DuPage; Ralph Salvatore, University o f Massachusetts, Boston;

Vyacheslav V. Samoshin, University o f the Pacific; Tomikazu Sasaki, University o f Washington; Yousry Sayed, University o f North Carolina at Wilmington; Adrian L. Schwan, University o f Guelph; Jonathan Sessler, University o f Texas at Austin; John Sevenair, Xavier University o f Louisiana; Warren Sherman, Chicago State University; Don Slavin, Community College o f Philadelphia; Chase Smith, Ohio Northern University; Doug Smith, University o f Toledo; John Sowa, Seton Hall University; Jean Stanley, Wellesley College; Ronald Starkey, University o f Wisconsin—Green Bay; Richard Steiner, University o f Utah; Robert Stolow, Tufts University; Frank Switzer, Xavier University; Richard Tarkka, George Washington University; James G. Traynham, Louisiana State University; Daniel Trifan, Fairleigh Dickinson University; Jennifer A. Tripp, University o f Scranton; Joseph J. Tufariello, State University o f New York, Buffalo; Kay Turner, Rochester Institute o f Technology; Rik R. Tykwinski, University o f Alberta; James Van Verth, Canisius College; Heidi Vollmer-Snarr, Brigham Young University; George Wahl, North Carolina State University; Rueben Walter, Tarleton State University; Darrell Watson, GMI Engineering and Management Institure; Arthur Watterson, University o f Massachusetts-Lowell; Donald Wedegaertner, University o f the Pacific; Carolyn Kraebel Weinreb, Saint Anselm College; Mark Welker, Wake Forest University; Michael Wells, Campbell University; Desmond M. S. Wheeler, University o f Nebraska; Kraig Wheeler, Delaware State University; James K. Whitesell, The University o f Texas at Austin; David Wiedenfeld, University o f North Texas; John Williams, Temple University; Carlton Willson, University o f Texas at Austin; Joseph Wolinski, Purdue University; Anne M. Wilson, Butler University; Darrell J. Woodman, University o f Washington; Stephen A. Woski, University o f Alabama; Linfeng Xie, University o f Wisconsin, Oshkosh; Viktor V. Zhdankin, University o f Minnesota, Duluth; Regina Zibuck, Wayne State University; Herman E. Zieger, Brooklyn College.

Acknowledgments

M any people have helped w ith this edition, and w e ow e a great deal o f thanks to each one o f them . We w ould especially like to thank R obert G. Johnson (Professor E m eritus, X avier U niversity) for his m eticulous assistance w ith the 10th edition Study G uide and Solutions M anual. B ob also h ad an uncanny ability to spot the m inutest inconsistency or error in the m ain text, and his proofreading has alw ays been valuable. We are thankful to C hristopher C allam (The O hio State U niversity) for m any new problem s contributed to the 10th edition and for his assistance w ith the Solutions M anual. We thank Sean H ickey (U niversity o f N ew O rleans) and Justin W yatt (C ollege o f C harleston) for their review s of the m anuscript and problem s. We thank N eal Tonks (C ollege o f Charleston) for his review o f the problem s. We also thank Jam es E llern (U niversity o f Southern C alifornia) for h elp ­ ful com m ents. We are grateful to A lan Shusterm an (R eed C ollege) and W arren H ehre (W avefunction, Inc.) for assistance in prior editions regarding explanations o f electrostat­ ic potential m aps and other calculated m olecu lar m odels. We w ould also like to thank those scientists w ho allow ed us to use o r adapt figures from their research as illustrations fo r a num ber o f the topics in ou r book. A book o f this scope could n ot be p roduced w ithout the excellent support w e have had from m any people at John W iley and Sons, Inc. Photo E ditor L isa G ee helped obtain p h o ­ tographs that illustrate som e exam ples in our book. Joan K alkut gave valuable assistance follow ing up w ith and tracking dow n sources and attributions. Copy E d ito r C onnie Parks helped to ensure consistency throughout the text and m ad e m any helpful suggestions at a highly detailed level. Jennifer Yee ensured coordination an d cohesion am ong m any aspects o f this project. M adelyn L esure created the captivating new design o f the 10th edition, fur­ ther enhanced by C arole A nson’s creative w ork on the cover. Illustration E d ito r Sandra R igby ensured that the art program m et the high technical standards required for illustra­ tions in a b ook o f this sort. A ssociate P ublisher P etra R ecter help ed steer the project from the outset and provided careful oversight and encouragem ent through all stages o f w ork on this revision. Production E ditor E lizabeth Sw ain oversaw production and printing o f the 10 th edition w ith h er characteristic and am azing skill, efficiency, an d attention to detail. Tom K ulesa and M arc W ezdecki supported developm ent o f W ileyPlus resources fo r the book. K ristine R uff enthusiastically and effectively help ed tell the ‘story’ o f our b o o k to the m any people w e hope w ill consider using it. We are thankful to all o f these people and others behind the scenes at W iley for the skills and dedication that they provided to bring this book to fruition. C B F w ould like to thank his colleagues, students, an d m entors for w hat they have taught h im over the years. M ost o f all, he w ould like to thank h is w ife D eanna fo r the sup­ port and patience she gives to m ake this w ork possible. TW G S w ould like to thank his w ife Judith for h e r support over ten editions o f this book. She jo in s m e in dedicating this edition to the m em ory o f our beloved son, Allen.

T. W. Graham Solomons Craig B. Fryhle

xxxi

About the Authors T. W. Graham Solomons

T. W . G r a h a m S olom ons did his undergraduate w ork a t T he C itadel and received his d oc­ torate in organic chem istry in 1959 from D uke U niversity w here h e w orked w ith C. K. Bradsher. Follow ing this h e w as a S loan Foundation P ostdoctoral Fellow at the U niversity o f R ochester w here h e w orked w ith V. Boekelheide. In 1960 h e becam e a charter m em ber o f the faculty o f the U niversity o f South F lorida an d becam e P rofessor o f C hem istry in 1973. In 1992 he w as m ade P rofessor E m eritus. In 1994 h e w as a visiting professor with the Faculté des Sciences P harm aceutiques et Biologiques, U niversité R ené D escartes (Paris V ). H e is a m em ber o f S igm a X i, Phi L am bda U psilon, an d Sigm a Pi Sigm a. H e has received research grants from the R esearch C orporation an d the A m erican C hem ical Society Petroleum R esearch Fund. F or several years he w as director of an N SF -sponsored U ndergraduate R esearch Participation Program at USF. H is research interests have been in the areas o f heterocyclic chem istry and unusual arom atic com pounds. H e has published papers in the Jo u rn a l o f the A m erican C hem ical Society, the Jo u rn a l o f Organic Chem istry , and the Jo u rn a l o f H eterocyclic Chem istry . H e has received several aw ards for distinguished teaching. H is organic chem istry textbooks have been w idely used fo r 30 years and have been translated into French, Japanese, Chinese, K orean, M alaysian, A rabic, Portuguese, Spanish, Turkish, and Italian. H e and his w ife Judith have a daughter w ho is a building conservator an d a son w ho is a research biochem ist.

Craig Barton Fryhle

C ra ig B a rto n F ry h le is C hair and P rofessor o f C hem istry at Pacific L utheran University. H e earned his B.A. degree from G ettysburg C ollege an d Ph.D . from B row n University. His experiences at these institutions shaped his dedication to m entoring undergraduate stu­ dents in chem istry and the liberal arts, w hich is a passion that burns strongly for him . His research interests have been in areas relating to the shikim ic acid pathway, including m ol­ ecular m odeling an d N M R spectrom etry o f substrates an d analogues, as w ell as structure and reactivity studies o f shikim ate pathw ay enzym es using isotopic labeling and m ass spectrom etry. H e has m entored m any students in undergraduate research, a n um ber o f w hom have later earned their Ph.D . degrees and gone on to academ ic or industrial posi­ tions. H e has participated in w orkshops on fostering undergraduate participation in research, and has been an invited p articipant in efforts by the N ational S cience Foundation to enhance undergraduate research in chem istry. H e has received research and instrum en­ tation grants from the N ational Science Foundation, the M J. M urdock C haritable Trust, and other private foundations. H is w ork in chem ical education, in addition to textbook co­ authorship, involves incorporation o f student-led teaching in the classroom an d technolo­ gy-based strategies in organic chem istry. H e has also developed experim ents for under­ graduate students in organic laboratory and instrum ental analysis courses. H e has been a volunteer w ith the hands-on science program in Seattle public schools, an d C hair o f the Puget S ound S ection o f the A m erican C hem ical Society. H e lives in Seattle w ith his w ife and tw o daughters.

xxxii

C ontrary to w hat you m ay have heard, organic chem isty does not have to be a difficult course. It w ill be a rigorous course, and it w ill offer a challenge. B ut you w ill learn m ore in it than in alm ost any course you w ill take— and w hat you learn w ill have a special relevance to life and the w orld around you. However, because organic chem istry can be approached in a logical and system atic way, you w ill find that w ith the right study habits, m astering organic chem istry can be a deeply satisfying experience. Here, then, are som e suggestions about how to study: 1. K eep u p w ith y o u r w o rk fro m d a y to d a y — n e v e r let y o u rse lf g et b e h in d . O rganic chem istry is a course in w hich one idea alm ost alw ays builds on another that has gone before. It is essential, therefore, that you keep up with, or better yet, be a little ahead o f your instructor. Ideally, you should try to stay one day ahead o f your instructor’s lectures in your ow n class preparations. T he lecture, then, w ill be m uch m ore helpful because you w ill already have som e understanding o f the assigned m aterial. Your tim e in class w ill clarify and expand ideas that are already fam iliar ones. 2. S tu d y m a te r ia l in sm a ll u n its, a n d b e s u re th a t you u n d e r s ta n d ea ch n ew se ctio n b e fo re you go o n to th e n ex t. A gain, because o f the cum ulative nature o f organic chem istry, your studying w ill be m uch m ore effective if you take each new idea as it com es and try to understand it co m ­ pletely before you m ove on to the next concept. 3. W o rk all o f th e in -c h a p te r a n d assig n ed p ro b lem s. O ne w ay to check your progress is to w ork each o f the in ­ chapter problem s w hen you com e to it. T hese problem s have been w ritten ju st for this purpose and are designed to help you decide w hether or not you understand the m aterial that has ju st been explained. You should also carefully study the Solved Problem s. If you understand a S olved P roblem and can w ork the related in-chapter problem , then you should go on; if you cannot, then you should go back and study the preceding m aterial again. W ork all o f the problem s assigned by your instructor from the end o f the chapter, as well. D o all o f y our problem s in a n otebook and bring this book with you w hen you go to see your instructor for extra help. 4. W rite w h en you stu d y . W rite the reactions, m ech a­ nism s, structures, and so on, over and over again. Organic chem istry is best assim ilated through the fingertips by w riting, and not through the eyes by sim ply looking, o r by highlighting m aterial in the text, or by referring to flash cards. T here is a good reason fo r this. O rganic structures,

m echanism s, and reactions are com plex. If you sim ply exam ine them, you m ay think you understand them thor­ oughly, but that w ill b e a m isperception. T he reaction m ech ­ anism m ay m ake sense to you in a certain way, but you need a deeper understanding than this. You need to know the m aterial so thoroughly that you can explain it to som eone else. T his level o f understanding com es to m o st o f us (those o f us w ithout photographic m em ories) through w riting. O nly by w riting the reaction m echanism s do w e pay suffi­ cient attention to their details, such as w hich atom s are con­ n ected to w hich atom s, w hich bonds b reak in a reaction and w hich bonds form, and the three-dim ensional aspects o f the structures. W hen w e w rite reactions and m echanism s, con­ nections are m ade in o u r brains that provide the long-term m em ory needed fo r success in organic chem istry. We v irtu ­ ally guarantee that y o u r grade in the course w ill b e directly proportional to the n um ber o f pages o f p aper that y o u r fill with your ow n w riting in studying during the term. 5. L e a rn b y te a c h in g a n d ex p lain in g . Study w ith your student peers and practice explaining concepts an d m ech a­ nism s to each other. U se the L e a rn in g Group Problem s and o ther exercises your instructor m ay assign as vehicles for teaching and learning interactively w ith your peers. 6 . U se th e a n s w e rs to th e p ro b le m s in th e Study Guide in th e p r o p e r w ay. R efer to the answ ers only in tw o cir­ cum stances: (1) W hen you have finished a problem , use the Study G uide to check your answer. (2) W hen, after m aking a real effort to solve the problem , you find that you are co m ­ pletely stuck, then look at the answ er fo r a clue an d go back to w ork o ut the problem on your own. T he value o f a p ro b ­ lem is in solving it. If you sim ply read the problem and look up the answ er, you w ill deprive y o u rself o f an im portant way to learn.

7. U se m o le c u la r m o d els w h e n you stu d y. B ecause of the three-dim ensional nature o f m ost organic m olecules, m o lecu lar m odels can be an invaluable aid to your under­ standing o f them . W hen you need to see the three-dim en­ sional aspect o f a p articu lar topic, use the M olecular V isions™ m odel set that m ay have been p ackaged w ith your textbook, o r buy a set o f m odels separately. A n appendix to the Study Guide that accom panies this text provides a set o f highly useful m o lecu lar m o d el exercises. 8 . M a k e u se o f th e ric h o n lin e te a c h in g re so u rc e s in W ileyP LU S and do any online exercises that m ay be

assigned by y o u r instructor.

xxxiii

The Basics Bonding and Molecular Structure

Organic chemistry is a part o f our lives at every moment. Organic molecules comprise the tissue o f plants as m ighty as the redwoods, convey signals from one neuron to the next in animals, store the genetic information o f life, and are the food we eat each day. The growth of living things from microbes to elephants rests on organic reactions, and organic reactions provide the energy that drives our muscles and our thought processes. O ur lives depend on organic chemistry in many other ways as well. Every article o f clothing we wear is a product o f organic chemistry, whether the fibers are natural or synthetic. Hardly a minute goes by when we're not using som ething made o f organic molecules, such as a pen, a com puter keyboard, a music player, or a cel­ lular phone. We view display screens made o f organic liquid crystal arrays. Natural organic polymers comprise w ood and the paper we read. Natural and synthetic organic molecules enhance our health. There is not a sin­ gle aspect of our lives that is not in some way dependent on organic chemistry. But what is organic chemistry? • O rganic chem istry is the chemistry o f com pounds th a t contain the elem en t carbon.

Clearly, carbon com pounds are central to life on this planet. Carbon as an element, however, has its origin elsewhere.

1

2

Chapter 1

The Basics— Bonding and Molecular Structure

1.1 W e A re Stardust Som e 14.5 billion years ago the big bang form ed hydrogen an d helium , the lightest ele­ m ents. Further nuclear reactions in stars transm uted these elem ents into heavier ones, includ­ ing carbon, nitrogen, oxygen, sulfur, phosphorus, and m ost others in the periodic table. M assive explosions called supernovae scattered the elem ents in the universe, and over tim e heavy elem ents coalesced to form planets and other celestial bodies. T hrough processes not understood but about w hich there continues to b e m uch research, sim ple m olecules form ed, eventually including organic m olecules that could support life— the nucleic acids that m ake up D N A and RN A , the am ino acids that com prise proteins, carbohydrates such as glucose, and other types o f m olecules. It is from elegant m olecular building blocks like these that the incredible richness o f chem istry and life has evolved. So, in the truest sense w e living creatures are com posed o f stardust, and w ithout supernovae n o t only w ould there b e no organic chem istry, there w ould b e no life.

1 .1 A

Development of the Science of Organic Chemistry

T he science o f organic chem istry began to flow er w ith the dem ise o f a nineteenth century theory called vitalism . A ccording to vitalism , organic com pounds w ere only those that cam e from living organism s, and only living things could synthesize organic com pounds through intervention o f a vital force. Inorganic com pounds w ere considered those com pounds that cam e from nonliving sources. Friedrich W ohler, however, discovered in 1828 that an organic com pound called urea (a constituent o f urine) could b e m ade by evaporating an aqueous solution o f the inorganic com pound am m onium cyanate. W ith this discovery, the synthe­ sis o f an organic com pound, began the evolution o f organic chem istry as a scientific discipline. Q OH O I / O\ _.CHO^»C CH^ \ / /C HO

C\

OH

V ita m in C

Vitamin C is found in various citrus fruits.

NH 4+N CQ -

heat

.a NH

h 2n

Ammonium cy anate

Urea

D espite the dem ise o f vitalism in science, the w ord “organic” is still used today by some people to m ean “com ing from living organism s” as in the term s “organic vitam ins” and “organic fertilizers.” T he com m only used term “organic food” m eans that the food was grow n w ithout the use o f synthetic fertilizers and pesticides. A n “organic vitam in” m eans to these people that the vitam in w as isolated from a natural source an d n o t synthesized by a chem ist. W hile there are sound argum ents to b e m ade against using food contam inated w ith certain pesticides, w hile there m ay b e environm ental benefits to b e obtained from organic farm ing, and w hile “natural” vitam ins m ay contain beneficial substances n o t p re­ sent in synthetic vitam ins, it is im possible to argue that p ure “n atural” vitam in C, for exam ­ ple, is healthier than pure “synthetic” vitam in C, since the tw o substances are identical in all respects. In science today, the study o f com pounds from living organism s is called natural products chem istry.

1.2 A to m ic Structure B efore w e begin our study o f the com pounds o f carbon w e need to review som e basic but fam iliar ideas about the chem ical elem ents and their structure. •

T he c o m p o u n d s w e enco u n ter in chem istry are m ad e up o f e le m e n ts com bined in d ifferent pro p o rtio n s. A n ab rid g ed p erio d ic table o f th e elem ents is given in Table 1.1.

An A b rid g ed Periodic Table o f the Elements

P er io d ic T able of the Ele m e n ts IA

VIIIA

1

2 Atom ic num ber-^

H

6

IIA

3

4

Name ( lU P A C ) ^ Atom ic m ass —>

He

C

Symbol —>

Hydrogen 1.0079

Chemical Abstracts Service group notation —>

Carbon 12.011

IVA

VA

VIA

V IIA

5

6

7

8

9

10

O

Li

Be

B

c

N

F

Ne

Lithium 6.941

Beryllium 9.0122

Boron 10.811

Carbon 12.011

Nitrogen 14.007

Oxygen 15.999

Fluorine 18.998

Neon 20.180

11

12

13

14

15

16

17

18

Na

Mg

AI

Si

P

S

Cl

Ar

Sodium 22,990

Magnesium 24.305

Aluminum 26.982

Silicon 28.086

Phosphorus 30.974

Sulfur 32.065

Chlorine 35.453

Argon 39.948

19

20

31

32

33

34

35

36

21

22

23

24

25

26

27

28

29

30

K

Ca

Sc

Ti

V

Cr

Mn

Fe

Co

Ni

Cu

Zn

Ga

Ge

As

Se

Br

Kr

Potassium 39.098

Calcium 40.078

Scandium 44.956

Titanium 47.867

Vanadium 50.942

Chromium 51.996

Manganese 54.938

Iron 55.845

Cobalt 58.933

Nickel 58.693

Copper 63.546

Zinc 65.409

Gallium 69.723

Germanium 72.64

Arsenic 74.922

Selenium 78.96

Bromine 79.904

Krypton 83.798

37

38

39

40

41

42

43

44

45

46

47

48

49

50

51

52

53

54

Mo

Tc

Rb

Sr

Y

Zr

Nb

Rubidium 85.468

Strontium 87.62

Yttrium 88.906

Zirconium 91.224

Niobium 92.906

55

56

57

72

73

Molybdenum Technetium 95.94 (98 )

74

75

Ru

Rh

Pd

Ag

Cd

In

Sn

Sb

Te

I

Xe

Ruthenium 101.07

Rhodium 102.91

Palladium 106.42

Silver 107.87

Cadmium 112.41

Indium 114.82

Tin 118.71

Antimony 121.76

Tellurium 127.60

Iodine 126.90

Xenon 131.29

76

77

78

79

80

81

82

83

84

85

86

Cs

Ba

Hf

Ta

w

Re

Os

Ir

Pt

Au

Hg

TI

Pb

Bi

Po

At

Rn

Cesium 132.91

Barium 137.33

Lanthanum 138.91

Hafnium 178.49

Tantalum 180.95

Tungsten 183.84

Rhenium 186.21

Osmium 190.23

Iridium 192.22

Platinum 195.08

Gold 196.97

Mercury 200.59

Thallium 204.38

Lead 207.2

Bismuth 208.98

Polonium (209)

Astatine (210)

Radon (222)

88

89

104

105

106

107

108

109

110

111

112

87

La

Fr

Ra

Ac

Rf

Db

sg

Bh

Hs

Mt

Francium (223)

Radium (226)

Actinium (227)

Rutherfordium

Dubnium (262)

Seaborgium (266)

Bohrium (264)

Hassium (277)

Meitnerium (268)

(261)

(Lanthanide series (58-71) and actinide series (90-103) elem ents not shown)

W

IIIA

Helium 4.0026

Uun Uuu Uub (281)

(272)

(285)

114

Uuq (289)

4

Chapter 1

E le c tro n c lo u d

N u c le u s

Figure 1.1 An atom is composed of a tiny nucleus containing protons and neutrons and a large surrounding volume containing electrons. The diameter of a typical atom is about 10,000 times the diameter of its nucleus.

•

The Basics— Bonding and Molecular Structure

E le m e n ts are m ade up o f a to m s. A n atom (Fig. 1.1) consists o f a dense, positively charged nucleus containing p ro to n s and n e u tro n s and a surrounding cloud of electrons.

E ach proton o f the nucleus bears one positive charge; electrons bear one negative charge. N eutrons are electrically neutral; they bear n o charge. Protons an d neutrons have nearly equal m asses (approxim ately 1 atom ic m ass u n it each) and are about 1800 tim es as heavy as electrons. M ost o f the m a ss o f an atom , therefore, com es from the m ass o f the nucleus; the atom ic m ass contributed b y the electrons is negligible. M ost o f the v o lu m e of an atom , how ever, com es from the electrons; the volum e o f an atom occupied b y the electrons is about 1 0 ,0 0 0 tim es larger than that o f the nucleus. T he elem ents com m only found in organic m olecules are carbon, hydrogen, nitrogen, oxy­ gen, phosphorus, and sulfur, as w ell as the halogens: fluorine, chlorine, brom ine, and iodine. E ach elem e n t is distinguished by its a to m ic n u m b e r (Z), a n u m b e r e q u a l to th e n u m b e r of p ro to n s in its nucleus. B ecause an atom is electrically neutral, th e ato m ic n u m ­ b e r also e q u a ls th e n u m b e r of e le c tro n s s u r ro u n d in g th e nu cleu s.

1 .2 A

Isotopes

B efore w e leave the subject o f atom ic structure and the p eriodic table, w e n eed to exam ine one other observation: th e ex isten ce of ato m s of th e sa m e e le m e n t th a t h av e d iffe re n t m asses. For exam ple (Table 1.1), the elem ent carbon has six protons in its nucleus giving it an atom ic n um ber o f 6 . M o st carbon atom s also have six neutrons in their nuclei, and because each proton and each neutron contributes one atom ic m ass unit (1 am u) to the m ass o f the atom , carbon atom s o f this k ind have a m ass num ber o f 12 and are w ritten as 12C. •

A lth o u g h a ll th e n u clei o f all ato m s o f th e sa m e e le m e n t w ill h av e th e sam e n u m b e r o f p ro to n s, som e atom s o f the sam e elem ent m a y h av e d iffe re n t m asses because they have d iffe re n t n u m b e rs o f n e u tro n s. Such atom s are called iso to p es.

For exam ple, about 1% o f the atom s o f elem ental carbon have nuclei containing 7 n eu ­ trons, and thus have a m ass n um ber o f 13. Such atom s are w ritten 13C. A tiny fraction of carbon atom s have 8 neutrons in their nucleus and a m ass n um ber o f 14. U nlike atom s of carbon-12 and carbon-13, atom s o f carbon-14 are radioactive. T he 14C isotope is used in carbon dating. T he three form s o f carbon, 12C, 13C, and 14C, are isotopes o f one another. M ost atom s of the elem ent hydrogen have one proton in their nucleus and have n o n eu ­ tron. They have a m ass num ber o f 1 and are w ritten 1H. A very sm all percentage (0.015%) of the hydrogen atom s that occur naturally, however, have one neutron in their nucleus. These atoms, called deuterium atoms, have a m ass num ber o f 2 and are w ritten 2 H. A n unstable (and radioactive) isotope o f hydrogen, called tritiu m (3 H), has tw o neutrons in its nucleus.

R e v ie w P ro b le m 1.1

T here are tw o stable isotopes o f nitrogen, 14N and 15N. H ow m any protons and neutrons does each isotope have?

1 .2 B

Valence Electrons

We discuss the electron configurations o f atom s in m ore detail in Section 1.10. F or the m om ent w e need only to point out that the electrons that surround the nucleus exist in shells o f increasing energy and at increasing distances from the nucleus. T he m ost im portant shell, called the v alence shell, is the outerm ost shell becau se the electrons o f this shell are the ones that an atom uses in m aking chem ical bonds with other atom s to form com pounds. •

How do w e know how m any electrons an atom has in its valence shell? We look at the periodic table. The num ber o f electrons in the valence shell (called valence

electrons) is equal to the group num ber o f the atom. For example, carbon is in group IVA and carbon has fo u r valence electrons; oxygen is in group V IA and oxygen has six valence electrons. T he halogens o f group V IIA all have seven electrons. H ow m any valence electrons does each o f the follow ing atom s have? (a) N a

(b) Cl

(c) Si

(d) B

(e) N e

R e v ie w P ro b le m 1.2

(f) N

1.3 The Structural Theory o f O rganic Chem istry B etw een 1858 and 1861, A ugust Kekule, A rchibald Scott Couper, and A lexander M. Butlerov, w orking independently, laid the basis for one o f the m o st im portant theories in chem istry: the s tr u c tu r a l th e o ry . Two central prem ises are fundam ental: 1. T he atom s in organic com pounds can form a fixed num ber o f bonds using their o ut­ erm ost shell (valence) electrons. Carbon is tetravalent; that is, carbon atom s have four valence electrons and can form four bonds. O xygen is divalent, an d hydrogen and (usually) the halogens are monovalent: I

— C— I

—O—

Carbon ato m s a re tetravalent.

H—

Oxygen ato m s are divalent.

Cl —

Hydrogen an d halogen ato m s a re m onovalent.

2. A carbon atom can use one or m ore o f its valence electrons to form bonds to other carbon atoms: C a rb o n -c arb o n b o n d s I I

II

— C— C—

Single bond

\

/

C= C

/

\

D ouble bond

— C= C— Triple bond

In his original publication C ouper represented these bonds by lines m uch in the sam e w ay that m ost o f the form ulas in this book are draw n. In his textbook (published in 1861), K ekule gave the science o f organic chem istry its m odern definition: a study o f the com ­ pou nds o f carbon.

1 .3 A

Isomers: The Importance of Structural Formulas

T he structural theory allow ed early organic chem ists to begin to solve a fundam ental p ro b ­ lem that plagued them : the problem o f iso m erism . T hese chem ists frequently found exam ­ ples o f d iffe re n t c o m p o u n d s th a t h av e th e sa m e m o le c u la r f o rm u la . Such com pounds are called is o m e rs . L et us consider an exam ple involving tw o com pounds that have practical uses: acetone, used in nail polish rem over and as a paint solvent, and propylene oxide, used w ith sea­ w eed extracts to m ake food-grade thickeners an d foam stabilizers for b eer (am ong other applications). B oth o f these com pounds have the m o lecular form ula C 3 H6O and therefore the sam e m olecular w eight. Yet acetone and propylene oxide have distinctly different b o il­ ing points and chem ical reactivity that, as a result, len d them selves to distinctly different practical applications. T heir shared m olecular form ula sim ply gives us no basis for u nder­ standing the differences betw een them . We m ust, therefore, m ove to a consideration o f their s tr u c tu r a l fo rm u la s .

H e lp f u l H i n t

Terms and concepts that are fundamentally important to your learning organic chemistry are set in bold blue type. You should learn them as they are introduced. These terms are also defined in the glossary. H e lp f u l H i n t

Build handheld models of these compounds and compare their structures.

ó

Chapter 1

The Basics— Bonding and Molecular Structure

A c e to n e H H

I

O

P ro p y le n e o x id e

H

II

I

-C — C — C-

I

H

H

I

H

H O I / \ H— C — C — C — H H

H

H

Acetone is used in some nail polish removers. Figure 1.2 Ball-and-stick models and chemical formulas show the different structures of acetone and propylene oxide.

O n exam ining the structures o f acetone and propylene oxide several key aspects are clearly different (Fig. 1.2). A cetone contains a double b o n d betw een the oxygen atom and the central carbon atom . Propylene oxide does n o t contain a double bond, b ut has three atom s jo in e d in a ring. T he connectivity o f the atom s is clearly different in acetone and propylene oxide. T heir structures have the sam e m olecular form ula b u t a different consti­ tution. We call such com pounds constitutional isom ers.*

Propylene oxide alginates, made from propylene oxide and seaweed extracts, are used as food thickeners.

•

C o n s titu tio n a l iso m ers are different com pounds that have the sam e m olecular fo r­ m ula but differ in the sequence in w hich their atom s are bonded, that is, in their connectivity.

•

C onstitutional isom ers usually have different physical properties (e.g., m elting point, boiling point, and density) and different chem ical properties (reactivity).

S o lv e d P ro b le m 1.1

T here are tw o constitutional isom ers w ith the form ula C 2 H 6O. W rite structural form ulas for these isom ers. STRATEGY AND ANSWER If w e recall that carbon can form four covalent bonds, oxygen can form two, and hydro­ gen only one, w e can arrive at the follow ing constitutional isom ers.

It should be noted that these tw o isom ers are clearly different in their physical properties. A t room tem perature and 1 atm pressure, dim ethyl ether is a gas. E thanol is a liquid.

1 .3 B

The Tetrahedral Shape of Methane

In 1874, the structural form ulas originated by K ekule, Couper, an d Butlerov w ere expanded into three dim ensions by the independent w ork o f J. H. v an ’t H off and J. A. L e Bel. v an’t H off and L e B el proposed that the four bonds o f the carbon atom in m ethane, for exam ple, are arranged in such a w ay that they w ould point tow ard the corners o f a regular tetrahe*An older term for isomers of this type was structural isomers. The International Union of Pure and Applied Chemistry (IUPAC) now recommends that use of the term “structural” when applied to constitutional isomers be abandoned.

1.4 Chem ical B onds: T he O c te t Rule

7

H Methane

H

Figure 1.3 The tetrahedral structure of methane.

dron, the carbon atom being placed at its center (Fig. 1.3). T he necessity for know ing the arrangem ent o f the atom s in space, taken together w ith an understanding o f the order in w hich they are connected, is central to an understanding o f organic chem istry, and w e shall have m uch m ore to say about this later, in C hapters 4 and 5.

1.4 Chemical Bonds: The O c te t Rule The first explanations o f the nature o f chem ical bonds w ere advanced by G. N. L ew is (of the U niversity o f California, Berkeley) and W. K ossel (of the U niversity o f M unich) in 1916. Two m ajor types o f chem ical bonds w ere proposed: 1. Io n ic (or electrovalent) bonds are form ed by the transfer o f one or m o re electrons from one atom to another to create ions. 2 . C o v a le n t bonds result w hen atom s share electrons.

The central idea in their w ork on bonding is that atom s w ithout the electronic configu­ ration o f a noble gas generally react to produce such a configuration because these config­ urations are know n to b e highly stable. F or all o f the noble gases except helium , this m eans achieving an octet o f electrons in the valence shell. •

T he tendency for an atom to achieve a configuration w here its valence shell co n ­ tains eight electrons is called the o c te t r u le .

The concepts and explanations that arise from the original propositions o f L ew is and K ossel are satisfactory for explanations o f m any o f the problem s w e deal w ith in organic chem istry today. F or this reason w e shall review these tw o types o f bonds in m ore m odern term s. 1 .4 A

Ionic Bonds

A tom s m ay gain or lose electrons and form charged particles called io n s. •

A n io n ic b o n d is an attractive force betw een oppositely charged ions.

O ne source o f such ions is a reaction betw een atom s o f w idely differing electronegativi­ ties (Table 1.2). Electronegativities o f Som e o f the Elements Increasing electronegativity H 2.1

Li 1.C

Be 1.5

B 2.C

C 2.5

N 3.C

O 3.5

F 4.C

Na C.9

Mg 1.2

Al 1.5

Si 1.8

P 2.1

S 2.5

Cl 3.G

K C.a

Br 2 .a

Increasing electronegativity

8

Chapter 1

The Basics— Bonding and Molecular Structure

H e lp f u l H i n t

•

E le c tro n e g a tiv ity is a m easure o f the a b ility o f an atom to a ttra c t electrons.

We will use electronegativity frequently as a tool for understanding the properties and reactivity of organic molecules.

•

E lectronegativity increases as w e go across a horizontal row o f the periodic table from left to rig h t an d it increases as w e go up a vertical colum n (Table 1.2).

A n exam ple o f the form ation o f an ionic bond is the reaction o f lithium and fluorine atoms:

/

/ / " '- , L L \ s

+

N • iF • '

/

\ • • /

+

+ ■+■

iF \

\

/

/

Lithium , a typical m etal, has a very low electronegativity; fluorine, a nonm etal, is the m ost electronegative elem ent o f all. T he loss o f an electron (a negatively charged species) by the lithium atom leaves a lithium cation (Li+); the gain o f an electron by the fluorine atom gives a fluoride anion (F _ ). •

Ions form because atom s can achieve the electronic configuration o f a noble gas by gaining or losing electrons.

T he lithium cation w ith tw o electrons in its valence shell is like an atom o f the noble gas helium , and the fluoride anion w ith eight electrons in its valence shell is like an atom o f the noble gas neon. M oreover, crystalline lithium fluoride form s from the individual lithium and fluoride ions. In this process negative fluoride ions becom e surrounded by positive lithium ions, and positive lithium ions by negative fluoride ions. In this crystalline state, the ions have substantially low er energies than the atom s from w hich they have been form ed. L ithium and fluorine are thus “stabilized” w hen they react to form crystalline lithium fluoride. We represent the form ula for lithium fluoride as LiF, because that is the sim plest form ula for this ionic com pound. Ionic substances, because o f their strong internal electrostatic forces, are usually very high m elting solids, often having m elting points above 1000°C. In polar solvents, such as water, the ions are solvated (see Section 2.13D ), and such solutions usually conduct an elec­ tric current. •

1 .4 B

Ionic com pounds, often called s a lts , form only w hen atom s o f very different electronegativities transfer electrons to becom e ions.

Covalent Bonds and Lewis Structures

W hen tw o or m ore atom s o f the sam e or sim ilar electronegativities react, a com plete trans­ fer o f electrons does n o t occur. In these instances the atom s achieve noble gas configura­ tions by sharing electrons. •

C o v a le n t b o n d s form by sharing o f electrons betw een atom s o f sim ilar electroneg­ ativities to achieve the configuration o f a no b le gas.

•

M o lecu les are com posed o f atom s jo in e d exclusively or predom inantly by covalent bonds.

M olecules m ay be represented by electron-dot form ulas or, m ore conveniently, by bond form ulas w here each p air o f electrons shared by tw o atom s is represented by a line. Som e exam ples are show n here: 1. H ydrogen, being in group IA o f the periodic table, has one valence electron. Two hydrogen atom s share electrons to form a hydrogen m olecule, H2. H2

H- + -H ----- > H =H

usually written

H— H

9

1.5 H ow to W rite Lewis S tru c tu res 2. B ecause chlorine is in group V IIA , its atom s have seven valence electrons. Two ch lo ­ rine atom s can share electrons (one electron from each) to form a m olecule o f Cl2. Cl2

:Cl-

•Cl:

u su ally w ritten

:C |:C |:

:Cl — C |:

3. A nd a carbon atom (group IVA) w ith four valence electrons can share each o f these electrons w ith four hydrogen atom s to form a m olecule o f m ethane, C H 4.

CH4

4 H-

H H :C :H H

usually written

H I H— C — H I H

T hese form ulas are often called L ew is s tr u c tu r e s ; in w riting them w e show only the elec­ trons o f the valence shell. 4. A tom s can share two o r more p a irs o f electrons to form m u ltip le co v a le n t b o n d s. For exam ple, tw o nitrogen atom s possessing five valence electrons each (because nitrogen is in group VA) can share electrons to form a triple b o n d betw een them . N2

:N :: N :

usually written

:N= N :

5. Ions, them selves, m ay contain covalent bonds. C onsider, as an exam ple, the am m o­ nium ion.

NH 4

H H :N"=H H

H usually written

H — N— H I

C onsider the follow ing com pounds and decide w hether the bond in them w ould b e ionic or covalent. (a) LiH (b) KCl (c) F 2 (d) PH 3

R e v ie w P ro b le m 1.3

1.5 H o w to W rite Lewis Structures Several sim ple rules allow us to draw proper L ew is structures: 1. L ew is s tr u c tu re s show th e co n n e ctio n s b e tw e e n a to m s in a m o lecu le o r io n u sin g o n ly th e v alen ce e le c tro n s o f th e a to m s involved. V alence electrons are those o f an atom ’s outerm ost shell. 2. F o r m a in g ro u p elem en ts, th e n u m b e r o f v alen ce e lec tro n s a n e u tra l a to m b rin g s to a L ew is s tr u c tu re is th e sa m e as its g ro u p n u m b e r in th e p e rio d ic tab le. Carbon, for exam ple, is in group IVA and has four valence electrons; the halogens (e.g., flu­ orine) are in group V IIA and each has seven valence electrons; hydrogen is in group IA and has one valence electron. 3. I f th e s tr u c tu r e w e a r e d ra w in g is a n eg a tiv e io n (a n a n io n ), w e a d d o n e e lec tro n fo r ea c h neg ativ e c h a rg e to th e o rig in a l c o u n t o f v alen ce e lec tro n s. I f th e s tr u c ­ tu r e is a p ositive io n (a ca tio n ), w e s u b tr a c t o n e e le c tro n fo r ea ch p o sitiv e c h a rg e. 4. I n d ra w in g L ew is s tr u c tu re s w e tr y to give ea ch a to m th e ele c tro n c o n fig u ra tio n o f a n o b le gas. To do so, w e draw structures w here atom s share electrons to form covalent bonds or transfer electrons to form ions. a. H ydrogen form s one covalent bond by sharing its electron w ith an electron of another atom so that it can have tw o valence electrons, the sam e n um ber as in the noble gas helium .

H e lp f u l H i n t

The ability to write proper Lewis structures is one of the most important tools for learning organic chemistry.

1O

Chapter 1

The Basics— Bonding and Molecular Structure

b. C arbon form s four covalent bonds by sharing its four valence electrons w ith four valence electrons from other atom s, so that it can have eight electrons (the sam e as the electron configuration o f neon, satisfying the octet rule). c. To achieve an octet o f valence electrons, elem ents such as nitrogen, oxygen, and the halogens typically share only som e o f their valence electrons through cova­ lent bonding, leaving others as unshared electron pairs.

T he follow ing problem s illustrate this m ethod.

S o lv e d P ro b le m 1 .2

W rite the L ew is structure o f CH 3 F. STRATEGY AND ANSW ER 1. We find the total num ber o f valence electrons o f all the atoms: 4 + 3(1) + 7 = 14 q q q C

3 H

F

2. We use pairs o f electrons to form bonds betw een all atom s that are b onded to each other. We represent these bonding pairs w ith lines. In our exam ple this requires four pairs o f electrons (8 o f the 14 valence electrons). H I H— C — F I H 3. We then add the rem aining electrons in pairs so as to give each hydrogen 2 electrons (a duet) and every other atom 8 electrons (an octet). In our exam ple, w e assign the rem aining 6 valence electrons to the fluorine atom in three nonbonding pairs. H I H — C — F: I H

S o lv e d P ro b le m 1 .3

W rite the L ew is structure for ethane (C 2 H6). STRATEGY AND ANSW ER 1. We find the total num ber o f valence electrons o f all the atom s. 2(4) + 6(1) = 14 q 2 C

q 6 H

2. We use one pair o f electrons to form a single bond betw een tw o carbon atom s, and six pairs o f electrons to form single bonds from each carbon atom to three hydrogen atoms. H

H

H

.... SC : C .... H

H

: H

H H | | H— C — C— H I I H

H

11

1.6 E x cep tio n s to th e O c te t Rule

S o lv e d P ro b le m 1 .4

W rite a Lew is structure for m ethylam ine (CH 5 N). STRATEGY AND ANSW ER 1. We find the total num ber o f valence electrons for all the atom s. 4

5

5(1) = 14 = 7 pairs

q C

q N

q 5 H

2. We use one electron pair to jo in the carbon and nitrogen. C— N 3. We use three pairs to form single bonds betw een the carbon and three hydrogen atom s. 4. We use tw o pairs to form single bonds betw een the nitrogen atom and tw o hydrogen atom s. 5. This leaves one electron pair, w hich w e use as a lone p air on the nitrogen atom. H H— C — N— H H

H

I f n ecessary, w e use m u ltip le b o n d s to satisfy th e o ctet ru le (i.e., give a to m s th e n o b le g as c o n fig u ra tio n ). The carbonate ion (CO 32~) illustrates this: ’’O ' II

T he organic m olecules ethene (C 2 H4) and ethyne (C 2H2) have a double an d triple bond, respectively: H

H xc - < /

HX

and

H— C = C — H

\

1.6 Exceptions to the O c te t Rule A tom s share electrons, not ju st to obtain the configuration o f an inert gas, b ut because shar­ ing electrons produces increased electron density betw een the positive nuclei. T he resu lt­ ing attractive forces o f nuclei for electrons is the “glue” that holds the atom s together (cf. Section 1.11). •

E lem ents o f the second period o f the periodic table can have a m axim um o f four bonds (i.e., have eight electrons around them ) because these elem ents have only one 2 s and three 2p orbitals available for bonding.

Each orbital can contain tw o electrons, and a total o f eight electrons fills these orbitals (Section 1.10A). T he octet rule, therefore, only applies to these elem ents, and even here, as w e shall see in com pounds o f beryllium and boron, few er than eight electrons are possible. •

E lem ents o f the third period and beyond have d orbitals that can be used for bonding.

12

Chapter 1

The Basics— Bonding and Molecular Structure

These elem ents can accom m odate m ore than eight electrons in their valence shells and there­ fore can form m ore than four covalent bonds. E xam ples are com pounds such as PCl 5 and S F 6. Bonds w ritten as / (dashed w edges) project b ehind the plane o f the paper. Bonds w ritten as / (solid w edges) project in front o f the paper. :Ci:

Cl:

:Ci — Pv "

¿ C"

:F ,F >

S o lv e d P ro b le m 1 .5

W rite a L ew is structure for the sulfate ion (S O 42 ). (Note: T he sulfur atom is b onded to all four oxygen atom s.) ANSW ER 1. We find the total n um ber o f valence electrons including the extra 2 electrons needed to give the ion the double negative charge: 6 + 4(6) + 2 = 32

q S

q q 4 O 2 e~

2. We use four pairs o f electrons to form bonds betw een the sulfur atom and the four oxygen atoms: 0 1 O— S — O I O 3. We add the rem aining 24 electrons as unshared pairs on oxygen atom s and as double bonds betw een the sulfur atom and tw o oxygen atom s. This gives each oxygen 8 electrons and the sulfur atom 12: •‘o ’' II . . :O — S — O: II " .O . ..

R e v ie w P ro b le m 1.4

W rite a L ew is structure for the phosphate ion (P O 43 ).

S om e h ig h ly re a c tiv e m o lecu les o r io n s h av e ato m s w ith few er th a n e ig h t elec tro n s in th e ir o u te r shell. A n exam ple is boron trifluoride (B F3). In a BF 3 m olecule the central boron atom has only six electrons around it: :F : I .B . :F

'F :

Finally, one poin t needs to b e stressed: B e fo re w e c a n w rite so m e L ew is stru c tu re s , we m u st kn o w how the atoms are connected to each other. C onsider nitric acid, for exam ­

ple. E ven though the form ula for nitric acid is often w ritten HNO3, the hydrogen is actu­ ally connected to an oxygen, n o t to the nitrogen. T he structure is H O N O 2 an d n o t HNO3. Thus the correct L ew is structure is

H— O— N

and not

V :

H _ N — O — O: II •.O.

13

1.7 Form al C h a rg e s an d H ow to C alculate Them

This know ledge com es ultim ately from experim ents. If you have forgotten the structures u H in t o f som e o f the com m on inorganic m olecules and ions (such as those listed in Review ^ Problem 1.5), this m ay be a good tim e for a review o f the relevant portions o f your genCheck your progress by doing , , . each Review Problem as you come eral chem istry text. to it in the text.

S o lv e d P ro b le m 1 .6

A ssum e that the atom s are connected in the sam e w ay they are w ritten in the form ula, an d w rite a L ew is structure for the toxic gas hydrogen cyanide (HCN). STRATEGY AND ANSW ER 1. We find the total num ber o f valence electrons on all o f the atom s: 1 + 4 + 5 = 10

q H

q C

q N

2. We use one pair o f electrons to form a single bond betw een the hydrogen atom and the carbon atom (see below), and w e use three pairs to form a triple b o n d betw een the carbon atom and the nitrogen atom . This leaves two electrons. We use these as an unshared p air on the nitrogen atom . N ow each atom has the electronic structure o f a noble gas. T he carbon atom has tw o electrons (like helium ) an d the carbon and nitrogen atom s each have eight electrons (like neon). H — C # N:

W rite a L ew is structure for each o f the follow ing:

R e v ie w P ro b le m 1.5

(a) HF

(c) CH 3 F

(e) H 2 S O 3

(g) H 3 P O 4

(b) F 2

(d) HNO 2

(f) BH4~

(h) H 2C O 3

1.7 Form al Charges and H ow to Calculate Them M any L ew is structures are incom plete until w e decide w hether any o f their atom s have a fo rm a l c h a r g e . C alculating the form al charge on an atom in a L ew is structure is sim ply a bookkeeping m ethod for its valence electrons. •

First, w e exam ine each atom and, using the periodic table, w e determ ine how m any v alen ce e le c tro n s it w ould have if it w ere an atom n ot b onded to any other atom s. T h is is e q u a l to th e g ro u p n u m b e r o f th e a to m in th e p e rio d ic ta b le . For hydrogen this num ber equals 1, for carbon it equals 4, for nitrogen it equals 5, and for oxygen it equals 6.

N ext, w e exam ine the atom in the L ew is structure and w e assign the valence electrons in the follow ing way: •

W e assig n to ea c h a to m h a lf o f th e elec tro n s it is s h a rin g w ith a n o th e r a to m a n d all o f its u n s h a re d (lone) e le c tro n p a irs.

Then w e do the follow ing calculation for the atom: Form al c h a rg e = nu m b er of v a le n c e e le c tro n s - 1/2 n u m b er of s h a re d e le c tro n s nu m b er of u n sh a re d elec tro n s or F = Z - (1 /2 )S - U

w here F is the form al charge, Z is the group num ber o f the elem ent, S equals the num ber o f shared electrons, and U is the num ber o f unshared electrons.

H e lp f u l H i n t

Proper assignment of formal charges is another essential tool for learning organic chemistry.

14

Chapter 1

•

The Basics— Bonding and Molecular Structure

It is im portant to note, too, that th e a rith m e tic su m o f all th e fo rm a l c h a rg e s in a m olecu le o r io n w ill e q u a l th e o v era ll c h a rg e o n th e m o lecu le o r ion.

L et us consider several exam ples show ing how this is done. The A m m onium Ion (NH4+) As w e see below, the am m onium ion has no unshared elec­ tron pairs. We divide all o f the electrons in bonds equally betw een the atom s that share them. Thus, each hydrogen is assigned one electron. We subtract this from one (the num ber o f valence electrons in a hydrogen atom ) to give each hydrogen atom a form al charge o f zero. The nitro­ gen atom is assigned four electrons (one from each bond). We subtract four from five (the num ­ ber o f valence electrons in a nitrogen atom) to give the nitrogen a form al charge o f + 1 . For hydrogen: valence electrons of free atom subtract assigned electrons Formal charge on each hydrogen For nitrogen:

0

valence electrons of free atom subtract assigned electrons

2)8

Formal charge on nitrogen

+1

5

Overall charge on ion = 4(0) + 1 = +1

The N itra te Ion (NO3—) L et us n ex t consider the nitrate ion (NO3—), an ion that has oxygen atom s w ith unshared electron pairs. H ere w e find that the nitrogen atom has a form al charge o f + 1 , that tw o oxygen atom s have form al charges o f —1 , and that one oxygen has a form al charge equal to 0 .

:O=----•• •

,,

— Formal charge = 6 - (1/2)2 - 6 = -1

-----------

■ O' I------------------------P-’ I Formal charge = 5 -

, .

(1/2)8 = +1

-------------------------------Formal charge = 6 - (1/2)4 - 4 = 0 Charge on ion = 2 (-1 ) + 1 + 0 = -1

W a te r and A m m onia T he sum o f the form al charges on each atom m aking up a m olecule m u st be zero. C onsider the follow ing exam ples: Water Formal charge = 6 - (1/2)4 - 4 = 0

H— O — H

Formal charge = 1 - (1/2)2 = 0

or

Charge on m olecule = 0 + 2(0) = 0

Ammonia ■

H— N— H

Formal charge = 5 - (1/2)6 - 2 = 0

H : N : H ---------- Formal charge = 1 -

(1/2)2 = 0

H

H

Charge on m olecule = 0 + 3(0) = 0

R e v ie w P ro b le m 1.6

W rite a L ew is structure for each o f the follow ing negative ions, and assign the form al negative charge to the correct atom: (a) C H 3O —

(c) C N —

(e) H CO 3—

(b) NH2—

(d) H CO 2—

(f) H C2—

15

1.8 Resonance Theory

1 .7 A

A Summary of Formal Charges

W ith this background, it should now be clear that each tim e an oxygen atom o f the type — O : appears in a m olecule or ion, it w ill have a form al charge o f - 1 , and that each tim e an oxygen atom o f the type = O " or — O — appears, it w ill have a form al charge o f 0 . I " .. " Similarly, — N — w ill be + 1, and — N — w ill b e zero. T hese and other com m on strucI I tures are sum m arized in Table 1.3.

A Sum m ary o f Formal Charges Form al C h arg e o f +1

G roup

Form al C h arg e o f 0

Form al C h arg e o f —1

B

IIIA

IVA

VA

"C'

=C —

— N—

= [+ /

— C—

#C

# 1N —

<

— B—

=C

— N—

#C —

— C—

= C

#N :

— N—

=N

\

H e lp fu l H i n t

In later chapters, when you are evaluating how reactions proceed and what products form, you will find it essential to keep track of formal charges.

#C :

I V IA

— Ö—

*O+

— Ö—

=O '

—Ö

I V IIA

—X +

— X = (X = F, Cl, Br, or I)

:X

A ssign the proper form al charge to the colored atom in each o f the follow ing structures: H

•'O'

H

H H I I (e) H— C — N— H

/ (a)

H— C — C H

(b)

H— O — H I H

\

(c) H/ C

H

(g)

R e v ie w P ro b le m 1.7

C H 3— C # N

O H

:O : I (d) H— C — H I H

H

H 9 O9 H I (f) H— C — H

(h) C H 3— N # N

H

1.8 Resonance Theory M any tim es m ore than one equivalent Lew is structure can b e w ritten for a m olecule or ion. C onsider, for exam ple, the carbonate ion (C O 32 - ). We can w rite three d ifferent but equivalent structures, 1-3: •O'*

:O -

:Q -

1

2

3

16

Chapter 1

The Basics— Bonding and Molecular Structure

N otice tw o im portant features o f these structures. First, each atom has the noble gas co n ­ figuration. Second, and this is especially im portant, w e can convert one structure into any other by changing only the p o sitio ns o f the electrons. We do n ot need to change the rela­ tive positions o f the atom ic nuclei. For exam ple, if w e m ove the electron pairs in the m an ­ ner indicated by the c u rv e d a rro w s in structure 1 , w e change structure 1 into structure 2 : H e lp f u l H i n t

Curved arrows (Section 3.5) show movement of electron pairs, not atoms. The tail of the arrow begins at the current position of the electron pair. The head of the arrow points to the location where the electron pair will be in the next structure. Curved-arrow notation is one of the most important tools that you will use to understand organic reactions.

¿O " c ll . 0 . iA "O ',

:Os becomes

0

O'

1

O 2

In a sim ilar w ay w e can change structure 2 into structure 3: sO:_ r> o < v .. h (a)

—, a

(b) H

^ c ^

^ H

H H H H + / N'C < - ' H I

H

(d)

,^ - C ^ N : H

20

Chapter 1

R e v ie w P ro b le m 1.10

The Basics— Bonding and Molecular Structure

W rite the contributing resonance structures and resonance hybrid for each o f the following: O (a) C H ,C H = C H — CH = OH

(f) H2C

(b) CH2= CH — C H — CH = CH 2

CH,

(g) C H ,— S — C H 2+ (h ) C H ,— NO 2

(c) (d) CH 2= C H 9 B r OH (e)

R e v ie w P ro b le m 1.11

F rom each set o f resonance structures that follow, designate the one that w ould contribute m ost to the hybrid and explain your choice: (a) C H 2— N(CH ,) 2 ••.o ' (b) C H ,

C

\

•.O..— H

(c) : NH 2— C # N :

sO : / C H ,9 C O+— H NH 2= C = N

1.9 Q uantum Mechanics and A to m ic Structure A theory o f atom ic and m olecular structure w as advanced independently an d alm ost sim ul­ taneously by three people in 1926: E rw in Schrödinger, W erner H eisenberg, and P aul Dirac. T his theory, called w ave m e ch a n ics by Schrödinger and q u a n tu m m e ch a n ics by H eisenberg, has becom e the basis from w hich w e derive our m odern understanding o f bond­ ing in m olecules. A t the h eart o f quantum m echanics are equations called w ave functions (denoted by the G reek letter psi, C). •

E ach w ave fu n c tio n (C) corresponds to a different energy state for an electron.

•

E ach energy state is a sublevel w here one or tw o electrons can reside.

•

T he en e rg y associated w ith the state o f an electron can b e calculated from the w ave function.

•

T he re la tiv e p ro b a b ility o f finding an electron in a given region o f space can be calculated from the w ave function (Section 1.10).

•

T he solution to a w ave function can b e positive, negative, or zero (Fig. 1.5).

Figure 1.5 A wave moving across a lake is viewed along a slice through the lake. For this wave the wave function, C, is plus (+) in crests and minus (—) in troughs. At the average level of the lake it is zero; these places are called nodes. The magnitude of the crests and troughs is the amplitude (a) of the wave. The distance from the crest of one wave to the crest of the next is the wavelength (l, or lambda).

{

Average level of lake

1.1 0 A tom ic O rb itals an d Electron C onfiguration

•

21

T he p h a s e sign o f a w ave equation indicates w hether the solution is positive or negative w hen calculated for a given poin t in space relative to the nucleus.

W ave functions, w hether they are for sound waves, lake waves, or the energy o f an electron, have the possibility o f constructive interference and destructive interference. •

C o n s tru c tiv e in te rfe re n c e occurs w hen w ave functions w ith the sam e p hase sign interact. T here is a reinforcing effect and the am plitude o f the w ave function increases.

•

D e stru c tiv e in te rfe re n c e occurs w hen w ave functions w ith opposite p hase signs interact. T here is a subtractive effect and the am plitude o f the w ave function goes to zero or changes sign. Constructive interference of waves i

A

i

Destructive interference of waves ! A '

Wave crests and troughs cancel

A = wavelength a = amplitude

E xperim ents have show n that electrons have properties o f w aves and particles, w hich was an idea first p u t forth by L ouis de B roglie in 1923. O ur discussion focuses on the w avelike properties o f electrons, however.

1.10 A to m ic O rbitals and Electron Configuration A physical interpretation related to the electron w ave function w as p ut forth by M ax Born in 1926, as follows. •

T he square o f a w ave function (C2) for a particular x,y,z location expresses the probability o f finding an electron at that location in space.

If the value o f C 2 is large in a unit volum e o f space, the probability o f finding an electron in that volum e is high— w e say that the e le c tro n p ro b a b ility d e n sity is large. Conversely, if C 2 for som e other volum e o f space is sm all, the probability o f finding an electron there

22

Chapter 1

The Basics— Bonding and Molecular Structure

T h e re are th re e 2p o r b ita ls , each w ith a (+ ) and ( - ) lobe, a lig n e d s y m m e tr ic a lly a lo n g th e x, y, a n d z axes.

Figure 1.6 The shapes of some s and p orbitals. Pure, unhybridized p orbitals are almost-touching spheres. The p orbitals in hybridized atoms are lobe-shaped (Section 1.13).

is low.* T his leads to the general definition o f an orbital and, by extension, to the fam iliar shapes o f atom ic orbitals. •

A n o r b ita l is a region o f space w here the probability o f finding an electron is high.

•

A to m ic o rb ita ls are plots o f C 2 in three dim ensions. T hese plots generate the fam iliar s,p , an d d orbital shapes.

The volum es that w e show are those that w ould contain the electron 9 0 -9 5 % o f the tim e. T here is a finite, bu t very small, probability o f finding an electron at greater distance from the nucleus than show n in the plots. The shapes o f 5 and p orbitals are show n in Fig. 1.6. All 5 o rb ita ls are spheres. A 1s orbital is a simple sphere. A 2s orbital is a sphere with an inner nodal surface (C 2 = 0). The inner portion o f the 2s orbital, C2s, has a negative phase sign. The shape o f a p o r b ita l is like that o f alm ost-touching spheres or lobes. T he p hase sign o f a 2p w ave function, C 2p, is positive in one lobe and negative in the other. A nodal plane separates the tw o lobes o f a p orbital, an d the three p orbitals o f a given energy level are arranged in space along the x, y, an d z axes in a C artesian coordinate system. •

T he + and — signs o f w ave functions do n ot im ply positive or negative charge or greater or lesser probability o f finding an electron.

•

C 2 (the probability o f finding an electron) is alw ays positive, because squaring either a positive or negative solution to C leads to a positive value.

Thus, the probability o f finding an electron in either lobe o f a p orbital is the same. We shall see the significance o f the + and — signs later w hen w e see how atom ic orbitals co m ­ bine to form m olecular orbitals. 1 .1 0 A

Electron Configurations

The relative energies o f atom ic orbitals in the first and second principal shells are as follows: •

E lectrons in 1s orbitals have the low est energy because they are closest to the p o si­ tive nucleus.

•

E lectrons in 2s orbitals are next low est in energy.

•

E lectrons o f the three 2p orbitals have equal b ut higher energy than the 2s orbital.

♦Integration of C2 over all space must equal 1; that is, the probability of finding an electron somewhere in all of space is 100%.

•

O rbitals o f equal energy (such as the three 2p orbitals) are called d e g e n e ra te o r b ita ls .

We can use these relative energies to arrive at the electron configuration o f any atom in the first tw o row s o f the periodic table. We need follow only a few sim ple rules. 1. A u fb a u p r in c ip le : O rbitals are filled so that those o f low est energy are filled first. (Aufbau is G erm an for “building up.”) 2. P a u li exclusion p rin c ip le : A m axim um o f two electrons m ay be placed in each orbital but only when the spins o f the electrons are paired. A n electron spins about its own axis. For reasons that w e cannot develop here, an electron is perm itted only one or the other o f just tw o possible spin orientations. We usually show these orientations by arrow s, either 1 or j . T hus tw o spin-paired electrons w ould be designated Jl\ U npaired electrons, w hich are not perm itted in the sam e orbital, are designated 11 (or W). 3. H u n d ’s r u le : W hen w e com e to orbitals o f equal energy (degenerate orbitals) such as the three p orbitals, w e add one electron to each w ith th e ir spins unpaired until each o f the degenerate orbitals contains one electron. (This allows the electrons, w hich repel each other, to be farther apart.) T hen w e begin adding a second electron to each degenerate orbital so that the spins are paired. If w e apply these rules to som e o f the second-row elem ents o f the periodic table, w e get the results show n in Fig. 1.7. 1

2p

1

1

1

—

2.

2P “ H“ ------------I 1 2p2P

11

11

1

2p

11

11

11

2p

* o

c LU

—

2.

— 1. Boron

—

2.

— 1. Carbon

—

— 1. Nitrogen

2.

— 1. Oxygen

—

2.

— 1. Fluorine

—

2.

— 1. Neon

Figure 1.7 The ground state electron configurations of some second-row elements.

1.11 M olecular O rbitals A tom ic orbitals provide a m eans for understanding how atom s form covalent bonds. L et us consider a very sim ple case— form ation o f a b o n d betw een tw o hydrogen atom s to form a hydrogen m olecule (Fig. 1.8). W hen tw o hydrogen atom s are relatively far apart their total energy is sim ply that o f tw o isolated hydrogen atom s (I). Form ation o f a covalent bond reduces the overall energy

IV

I

N u c le a r re p u ls io n

No a ttra c tio n

r = 0 .7 4 À

In te r n u c le a r d is ta n c e ( r)

Figure 1.8 The potential energy of the hydrogen molecule as a function of internuclear distance.

24

Chapter 1

The Basics— Bonding and Molecular Structure

o f the system , however. A s the tw o hydrogen atom s m ove closer together (II), each nucleus increasingly attracts the o th er’s electron. This attraction m ore than com pensates for the repulsive force betw een the tw o n uclei (or the tw o electrons). T he resu lt is a covalent bond (III), such that the internuclear distance is an ideal balance that allow s the tw o electrons to be shared betw een both atom s w hile at the sam e tim e avoiding repulsive interactions betw een their nuclei. T his ideal internuclear distance betw een hydrogen atom s is 0.74 Á, and w e call this the b o n d le n g th in a hydrogen m olecule. If the nuclei are m oved closer together (IV ) the repulsion o f the tw o positively charged nuclei predom inates, and the energy o f the system rises. N otice that each H • has a shaded area around it, indicating that its precise position is uncertain. E lectrons are constantly m oving. •

A ccording to the H e is e n b e rg u n c e rta in ty p rin c ip le , w e cannot sim ultaneously know the position and m om entum o f an electron.

T hese shaded areas in our diagram represent orbitals, and they result from applying the prin­ ciples o f quantum m echanics. Plotting the square o f the w ave function (C 2) gives us a three­ dim ensional region called an orbital w here finding an electron is highly probable. •

A n a to m ic o r b ita l represents the region o f space w here one or tw o electrons o f an isolated atom are likely to b e found.

In the case o f our hydrogen m odel above, the shaded spheres rep resen t the 1s orbital of each hydrogen atom . A s the tw o hydrogen atom s approach each other their 1s orbitals begin to overlap until their atom ic orbitals com bine to form m olecular orbitals. •

A m o le c u la r o r b ita l (M O ) represents the region o f space w here one or tw o elec­ trons o f a m olecule are likely to b e found.

•

A n orbital (atom ic or m olecular) can contain a m axim um o f tw o spin-paired elec­ trons (Pauli exclusion principle).

•

W hen atom ic orb itals co m b in e to form m o lecu lar orbitals, th e n u m b e r o f m o le ­ c u la r o r b ita ls th a t r e s u lt alw a y s e q u a ls th e n u m b e r o f a to m ic o r b ita ls th a t co m b in e.

Thus, in the form ation o f a hydrogen m olecule the tw o C 1s atom ic orbitals com bine to produce tw o m olecular orbitals. Two orbitals resu lt because the m athem atical properties of w ave functions p erm it them to b e com bined by either addition or subtraction. T hat is, they can com bine either in or out o f phase. •

A b o n d in g m o le c u la r o r b ita l (Cmoiec) results w hen tw o orbitals o f the sam e phase overlap (Fig. 1.9).

•

A n a n tib o n d in g m o le c u la r o r b ita l (C*molec) results w hen tw o orbitals o f opposite phase overlap (Fig. 1.10).

T he bonding m olecular orbital o f a hydrogen m olecule in its low est energy (ground) state contains both electrons from the individual hydrogen atom s. T he value o f C (and therefore also C2) is large betw een the nuclei, precisely as expected since the electrons are shared by both nuclei to form the covalent bond.

Figure 1.9 (a) The overlapping of two hydrogen 1s atomic orbitals with the same phase sign (indicated by their identical color) to form a bonding molecular orbital. (b) The analogous overlapping of two waves with the same phase, resulting in constructive interference and enhanced amplitude.

(a )

(b)

^ l S ( atomic orbital)

$ 1 s (atomic orbital)

B o n d in g ^(m o le c u la r orbital)

1.12 The Structure o f M ethane and Ethane: s p3 Hybridization

25

N ode

*

• (a )

1 ^(atomic orbital)

I I 1

A n t i b o n d in g

^(atomic orbital)

(molecular orbital)

C

+

ii

/^ \

t

(b )

1

N ode

Figure 1.10 (a) The overlapping of two hydrogen 1s atomic orbitals with opposite phase signs (indicated by their different colors) to form an antibonding molecular orbital. (b) The analogous overlapping of two waves with the opposite sign, resulting in destructive interference and decreased amplitude. A node exists where complete cancellation by opposite phases makes the value of the combined wave function zero.

T h e a n t i b o n d i n g m o l e c u l a r o r b i t a l c o n t a in s n o e l e c t r o n s i n t h e g r o u n d s t a t e o f a h y d r o ­ g e n m o l e c u l e . F u r t h e r m o r e , t h e v a l u e o f C ( a n d t h e r e f o r e a ls o C 2) g o e s t o z e r o b e t w e e n th e n u c le i, c r e a t in g a n o d e ( C =

0 ) . T h e a n t ib o n d in g o r b i t a l d o e s n o t p r o v id e f o r e le c tr o n

d e n s ity b e tw e e n th e a to m s , a n d th u s i t is n o t in v o lv e d in b o n d in g . W h a t w e h a v e j u s t d e s c r ib e d h a s it s c o u n t e r p a r t i n a m a t h e m a t ic a l t r e a tm e n t c a lle d th e

L C A O (lin e a r c o m b in a tio n o f a to m ic o rb ita ls)

m e th o d . I n th e L C A O

tr e a tm e n t, w a v e

f u n c t io n s f o r th e a t o m ic o r b it a ls a re c o m b in e d i n a lin e a r f a s h io n ( b y a d d it io n o r s u b t r a c ­ t io n ) in o r d e r t o o b t a in n e w w a v e f u n c t io n s f o r th e m o le c u la r o r b ita ls . M o l e c u l a r o r b i t a l s , l i k e a t o m i c o r b i t a l s , c o r r e s p o n d t o p a r t i c u l a r e n e r g y s ta te s f o r a n e le c ­ tr o n . C a lc u la t io n s s h o w t h a t th e r e la t iv e e n e r g y o f a n e le c tr o n i n th e b o n d in g m o le c u la r o r b i t a l o f t h e h y d r o g e n m o l e c u l e i s s u b s t a n t i a l l y le s s t h a n i t s e n e r g y i n a C 1s a t o m i c o r b i t a l . T h e s e c a l c u l a t i o n s a ls o s h o w t h a t t h e e n e r g y o f a n e l e c t r o n i n t h e a n t i b o n d i n g m o l e c u l a r o r b i t a l i s s u b s t a n t i a l l y g r e a t e r t h a n i t s e n e r g y i n a C 1s a t o m i c o r b i t a l . A n e n e r g y d ia g r a m

f o r th e m o le c u la r o r b it a ls o f th e h y d r o g e n m o le c u le is s h o w n in

F ig . 1 .1 1 . N o t ic e t h a t e le c tr o n s a re p la c e d i n m o le c u la r o r b it a ls i n th e s a m e w a y t h a t th e y a r e i n a t o m i c o r b i t a l s . T w o e l e c t r o n s ( w i t h t h e i r s p in s o p p o s e d ) o c c u p y t h e b o n d i n g m o l ­ e c u l a r o r b i t a l , w h e r e t h e i r t o t a l e n e r g y i s le s s t h a n i n t h e s e p a r a t e a t o m i c o r b i t a l s . T h i s is , a s w e h a v e s a id , t h e

lowest electronic state

or

ground state

o f th e h y d r o g e n m o le c u le . A n

e le c tr o n m a y o c c u p y th e a n t ib o n d in g m o le c u la r o r b i t a l i n w h a t is c a lle d a n

excited state

f o r t h e m o l e c u l e . T h i s s ta te f o r m s w h e n t h e m o l e c u l e i n t h e g r o u n d s ta te ( F i g . 1 . 1 1 ) a b s o r b s a p h o to n o f lig h t h a v in g th e p r o p e r e n e r g y ( D E ) .

Figure 1.11 Energy diagram for the hydrogen molecule. Combination of two atomic orbitals, ip1s, gives two molecular orbitals, Cmolec and C*molec. The energy of Cmolec is lower than that of the separate atomic orbitals, and in the lowest electronic energy state of molecular hydrogen the bonding MO contains both electrons.

1.12 The Structure o f M eth an e and Ethane: sp3 H ybridization The s and in

p

S e c tio n

o r b ita ls

do

tetrahedral

o r b i t a l s u s e d i n t h e q u a n t u m m e c h a n i c a l d e s c r i p t i o n o f t h e c a r b o n a t o m , g iv e n

p te tra v a le n t-

1 .1 0 , w e r e b a s e d o n c a lc u la t io n s f o r h y d r o g e n a to m s . T h e s e s im p le s a n d n o t,

w hen

ta k e n

a lo n e , p r o v i d e

a

s a tis fa c to r y

c a r b o n o f m e th a n e ( C H 4, se e R e v ie w P r o b le m

m o d e l fo r

th e

1 .1 2 ). H o w e v e r , a s a tis fa c to r y

m o d e l o f m e th a n e ’ s s tr u c tu r e th a t is b a s e d o n q u a n tu m m e c h a n ic s

can

b e o b t a in e d t h r o u g h

a n a p p r o a c h c a l l e d o r b i t a l h y b r i d i z a t i o n . O r b i t a l h y b r i d i z a t i o n , i n i t s s i m p l e s t t e r m s , is

26

Chapter 1

The Basics— Bonding and Molecular Structure

nothing m ore than a m athem atical approach that involves the com bining o f individual wave functions for s and p orbitals to obtain w ave functions for new orbitals. T he new orbitals have, in varying pro p o rtion s, the properties o f the original orbitals taken separately. These new orbitals are called h y b rid a to m ic o r b ita ls . A ccording to quantum m echanics, the electronic configuration o f a carbon atom in its low est energy state— called the g ro u n d s ta te — is that given here: C u U J ___ 1______ 1s 2 s 2px 2py 2pz Ground state of a carbon atom T he valence electrons o f a carbon atom (those used in bonding) are those o f the o uter level, that is, the 2s and 2p electrons. 1 .1 2 A

The Structure of Methane

H ybrid atom ic orbitals that account for the structure o f m ethane can b e derived from carbon’s second-shell s and p orbitals as follow s (Fig. 1.12): •

Wave functions for the 2s, 2px, 2py, and 2pz orbitals o f ground state carbon are m ixed to form four new and equivalent 2sp3 hybrid orbitals.

•

T he designation sp 3 signifies that the hybrid orbital has one p art s orbital character and three parts p orbital character.

•

T he m athem atical resu lt is that the four 2sp3 orbitals are oriented at angles of 109.5° w ith resp ect to each other. T his is precisely the orientation o f the four hydrogen atom s o f m ethane. E ach H — C — H bond angle is 109.5°.

2pz O rb ita l H y b rid iz a tio n

Figure 1.12 Hybridization of pure atomic orbitals of a carbon atom to produce sp3 hybrid orbitals.

109

27

1.12 The Structure o f M ethane and Ethane: sp3 Hybridization

Figure 1.13 The hypothetical formation of methane from an sp3-hybridized carbon atom and four hydrogen atoms. In orbital hybridization we combine orbitals, not electrons. The electrons can then be placed in the hybrid orbitals as necessary for bond formation, but always in accordance with the Pauli principle of no more than two electrons (with opposite spin) in each orbital. In this illustration we have placed one electron in each of the hybrid carbon orbitals. In addition, we have shown only the bonding molecular orbital of each C — H bond because these are the orbitals that contain the electrons in the lowest energy state of the molecule.

I f , i n o u r i m a g i n a t i o n , w e v i s u a l i z e t h e h y p o t h e t i c a l f o r m a t i o n o f m e t h a n e f r o m a n s p 3h y b r id iz e d c a r b o n a to m a n d f o u r h y d r o g e n a to m s , th e p ro c e s s m ig h t b e l ik e th a t s h o w n in

bonding m o le cu la r o rb ita l f o r tetrahedral struc­

F ig . 1 .1 3 . F o r s i m p l i c it y w e s h o w o n ly th e f o r m a t io n o f th e

e a c h c a r b o n - h y d r o g e n b o n d . W e s e e t h a t a n s p 3- h y b r i d i z e d c a r b o n g iv e s a

ture f o r methane, and one w ith fo u r equivalent C — H

(a)

bonds.

C o n s i d e r a c a r b o n a t o m i n i t s g r o u n d s ta te . W o u l d s u c h a n a t o m o f f e r a s a t i s f a c t o r y m o d e l f o r th e c a r b o n o f m e th a n e ? I f n o t, w h y n o t?

(H in t:

R e v ie w P r o b le m

1 .1 2

C o n s id e r w h e th e r a g r o u n d

s ta te c a r b o n a t o m c o u l d b e t e t r a v a l e n t , a n d c o n s i d e r t h e b o n d a n g le s t h a t w o u l d r e s u l t i f i t w e r e to c o m b in e w it h h y d r o g e n a to m s .)

(b)

C o n s i d e r a c a r b o n a t o m i n t h e e x c i t e d s ta te :

C U J ___ 1___ 1____1 _ 1s 2 s 2px 2py 2pz Excited state of a carbon atom W o u ld s u c h a n a to m o f f e r a s a tis fa c to r y m o d e l f o r th e c a r b o n o f m e th a n e ? I f n o t , w h y n o t?

I n a d d it io n t o a c c o u n tin g p r o p e r ly f o r th e s h a p e o f m e th a n e , th e o r b it a l h y b r id iz a t io n m o d e l a ls o e x p l a i n s t h e v e r y s t r o n g b o n d s t h a t a r e f o r m e d b e t w e e n c a r b o n a n d h y d r o g e n . T o s e e h o w t h i s i s s o , c o n s i d e r t h e s h a p e o f a n i n d i v i d u a l s p 3 o r b i t a l s h o w n i n F i g . 1 .1 4 . B e c a u s e a n s p 3 o r b ita l h a s th e c h a ra c te r o f a la r g e a n d e x te n d s r e la t iv e ly f a r f r o m

p

o r b i t a l , t h e p o s i t i v e l o b e o f a n s p 3 o r b i t a l is

th e c a r b o n n u c le u s .

I t is th e p o s it iv e lo b e o f a n s p 3 o r b i t a l t h a t o v e r la p s w i t h th e p o s it iv e 1 s o r b i t a l o f h y d r o ­ g e n t o f o r m th e b o n d in g m o le c u la r o r b it a l o f a c a r b o n - h y d r o g e n b o n d ( F ig . 1 .1 5 ) . B e c a u s e th e p o s it iv e lo b e o f th e s p 3 o r b i t a l is la r g e a n d is e x te n d e d i n t o s p a c e , th e o v e r la p b e tw e e n i t a n d th e

1s

o r b i t a l o f h y d r o g e n i s a ls o l a r g e , a n d t h e r e s u l t i n g c a r b o n - h y d r o g e n b o n d is

q u ite s tro n g .

Wr i

C4 sp3 O rb ita l

|

^

H+ )

----------- ►

1 s O r b ita l

cK

+

H|

C a rb o n -h y d ro g e n bond (b o n d in g M O)

Figure 1.15 Formation of a C— H bond.

Figure 1.14 The shape of an sp3

orbital.

28

Chapter 1

The Basics— Bonding and Molecular Structure

or

Figure 1.16 A s (sigma) bond.

T h e b o n d f o r m e d f r o m t h e o v e r l a p o f a n sp

sig m a ( s ) b o n d •

A

o r b i t a l a n d a 1 s o r b i t a l is a n e x a m p le o f a

( F ig . 1 .1 6 ).

sig m a ( s ) b o n d

h a s a c ir c u l a r ly s y m m e t r ic a l o r b it a l c ro s s s e c tio n w h e n v ie w e d

a lo n g t h e b o n d b e t w e e n t w o a t o m s . •

A l l p u r e ly

sin g le b o n d s

a re s ig m a b o n d s .

F r o m th is p o in t o n w e s h a ll o f te n s h o w o n ly th e b o n d in g m o le c u la r o r b ita ls b e c a u s e th e y a re th e o n e s t h a t c o n t a in th e e le c tr o n s w h e n th e m o le c u le is i n

i t s l o w e s t e n e r g y s ta te .

C o n s i d e r a t i o n o f a n t i b o n d i n g o r b i t a l s is i m p o r t a n t w h e n a m o l e c u l e a b s o r b s l i g h t a n d i n e x p l a i n i n g c e r t a i n r e a c t io n s . W e s h a l l p o i n t o u t t h e s e in s t a n c e s la t e r . I n F ig . 1 .1 7 w e s h o w a c a lc u la te d s tr u c tu r e f o r m e th a n e w h e r e th e te tr a h e d r a l g e o m e ­ t r y d e r iv e d f r o m

o r b i t a l h y b r i d iz a t io n is c le a r ly a p p a re n t.

Figure 1.17 (a) In this structure of methane, based on quantum mechanical calculations, the inner solid surface represents a region of high electron density. High electron density is found in each bonding region. The outer mesh surface represents approximately the furthest extent of overall electron density for the molecule. (b) This ball-and-stick model of methane is like the kind you might build with a molecular model kit. (c) This structure is how you would draw methane. Ordinary lines are used to show the two bonds that are in the plane of the paper, a solid wedge is used to show the bond that is in front of the paper, and a dashed wedge is used to show the bond that is behind the plane of the paper.

1 .1 2 B

The Sti

T h e b o n d a n g le s a t t h e c a r b o n a t o m s o f e t h a n e , a n d o f a l l a lk a n e s , a r e a ls o t e t r a h e d r a l l i k e th o s e i n m e th a n e . A

s a tis fa c to r y m o d e l f o r e th a n e c a n b e p r o v id e d b y s p 3- h y b r id iz e d c a r ­

b o n a to m s . F ig u r e 1 .1 8 s h o w s h o w w e m i g h t im a g in e th e b o n d in g m o le c u la r o r b it a ls o f a n e th a n e m o le c u le b e in g c o n s t r u c te d f r o m

t w o s p 3- h y b r id iz e d c a r b o n a to m s a n d s ix h y d r o ­

g e n a to m s . T h e c a r b o n - c a r b o n b o n d o f e t h a n e is a s ig m a b o n d w i t h c y l i n d r i c a l s y m m e t r y , f o r m e d b y t w o o v e r l a p p i n g s p 3 o r b i t a l s . ( T h e c a r b o n - h y d r o g e n b o n d s a r e a ls o s i g m a b o n d s . T h e y a re f o r m e d f r o m o v e r la p p in g c a r b o n s p 3 o r b ita ls a n d h y d r o g e n s o r b ita ls .) •

R o t a t io n o f g r o u p s j o in e d b y a s in g le b o n d d o e s n o t u s u a lly r e q u ir e a la r g e a m o u n t o f e n e rg y .

C o n s e q u e n tly , g r o u p s jo in e d b y s in g le b o n d s r o ta te r e la t iv e ly f r e e ly w it h r e s p e c t t o o n e a n o t h e r . ( W e d is c u s s t h i s p o i n t f u r t h e r i n S e c t i o n 4 . 8 . ) I n F i g . 1 . 1 9 w e s h o w a c a lc u l a t e d s tr u c tu r e f o r e th a n e i n w h ic h th e te tr a h e d r a l g e o m e t r y d e r iv e d f r o m is c le a r ly a p p a r e n t.

o r b it a l h y b r id iz a t io n

1.12 The Structure o f M ethane and Ethane: sp3 Hybridization

+

+

sp3 C arbon

29

6

sp3 C arbon

S ig m a b o n d s

Figure 1.18 The hypothetical formation of the bonding molecular orbitals of ethane from two sp3-hybridized carbon atoms and six hydrogen atoms. All of the bonds are sigma bonds. (Antibonding sigma molecular orbitals—called s* orbitals—are formed in each instance as well, but for simplicity these are not shown.)

H

i

(a)

H

*

(b)

(c)

Figure 1.19 (a) In this structure of ethane, based on quantum mechanical calculations, the inner solid surface represents a region of high electron density. High electron density is found in each bonding region. The outer mesh surface represents approximately the furthest extent of overall electron density for the molecule. (b) A ball-and-stick model of ethane, like the kind you might build with a molecular model kit. (c) A structural formula for ethane as you would draw it using lines, wedges, and dashed wedges to show in three dimensions its tetrahedral geometry at each carbon.

THE CHEMISTRY OF . . . C a lc u la te d M o le c u la r M o d e ls : E le c tro n D e n s ity S u rf a c e s W e m ake frequent use in this book of m olecular m odels derived from quantum m echanical calculations. T hese m od­ els will help us visualize th e shap es of m olecules as well as understand their properties and reactivity. A useful ty p e of

m odel is o n e th a t shows a calculated three-dim ensional sur­ face at which a chosen value of electron density is th e sam e all around a m olecule, called an e le c tro n d e n s ity su rfa c e . If we make a plot w here th e value chosen is for low electron density, th e result is a van d er Waals surface, th e surface that represents approxim ately th e overall sh ap e of a m olecule as d eterm in ed by th e furthest ex tent of its electron cloud. On th e other hand, if we make a plot w here th e value of elec­ tron density is relatively high, th e resulting surface is one that approxim ately rep resen ts th e region of covalent bonding in a m olecule. Surfaces of low and high electron density are shown in this box for dimethyl ether. Similar m odels are shown for m eth an e and eth an e in Figs. 1.17 and 1.19.

30

Chapter 1

The Basics— Bonding and Molecular Structure

1.13 The Structure o f Ethene (Ethylene): sp2 H ybridization The carbon atom s o f m any o f the m olecules that w e have considered so far have used their four valence electrons to form four single covalent (sigm a) bonds to four other atom s. We find, however, that m any im portant organic com pounds exist in w hich carbon atom s share m ore than tw o electrons w ith another atom . In m olecules o f these com pounds som e bonds that are form ed are m ultiple covalent bonds. W hen tw o carbon atom s share tw o pairs o f electrons, for exam ple, the result is a carb o n -carb o n double bond: .C : :C . ' •

or

\ /

C=C

/ \

H ydrocarbons w hose m olecules contain a carb o n -carb o n double b o n d are called alk en es. E thene (C2H4) and propene (C3H6) are both alkenes. (Ethene is also called ethylene, and propene is som etim es called propylene.) H \ _

H

/

c

H

H

H3C

'

\

E th en e

H

H x c - < /

/

\

H

P ro p e n e

In ethene the only carbon-carbon bond is a double bond. Propene has one carbon-carbon single bond and one carb o n -carb o n double bond. T he spatial arrangem ent o f the atom s o f alkenes is different from that o f alkanes. The six atom s o f ethene are coplanar, and the arrangem ent o f atom s around each carbon atom is triangular (Fig. 1.20).

Figure 1.20 The structure and bond angles of ethene. The plane of the atoms is perpendicular to the paper. The dashed wedge bonds project behind the plane of the paper, and the solid wedge bonds project in front of the paper.

•

C a rbon-carbo n double bonds are com prised o f sp2-hybridized carbon atom s.

T he m athem atical m ixing o f orbitals that furnish the sp2 o rb ita ls for our m odel can b e visu­ alized in the w ay show n in Fig. 1.21. T he 2s orbital is m athem atically m ixed (or hybridized) w ith tw o o f the 2p orbitals. (The hybridization procedure applies only to the orbitals, not to the electrons.) O ne 2p orbital is left unhybridized. O ne electron is then p laced in each o f the sp2 hybrid orbitals an d one electron rem ains in the 2p orbital. T he three sp2 orbitals that result from hybridization are directed tow ard the corners of a regular triangle (w ith angles o f 120° betw een them ). T he carbon p orbital that is not

Figure 1.21 A process for deriving sp2-hybridized carbon atoms.

1.13 The Structure o f Ethene (Ethylene): sp2 Hybridization

31

hybridized is perpendicular to the plane o f the triangle form ed by the hybrid sp2 orbitals (Fig. 1.22). In our m odel for ethene (Fig. 1.23) w e see the following: •

Two sp2-hybridized carbon atom s form a sigm a ( s ) b o n d betw een them by overlap o f one sp2 orbital from each carbon. T he rem aining carbon sp2 orbitals form s bonds to four hydrogens through overlap w ith the hydrogen 1s orbitals. T hese five s bonds account for 10 o f the 12 valence electrons contributed by the tw o carbons an d fo u r hydrogens, an d com prise the s - b o n d fra m e w o rk o f the m olecule. Figure 1.22 An sp2-hybridized carbon atom.

p O rb ita ls

ct

B onds

ct

B onds

Figure 1.23 A model for the bonding molecular orbitals of ethene formed from two sp2-hybridized carbon atoms and four hydrogen atoms.

•

T he rem aining tw o bonding electrons are each located in an unhybridized p orbital o f each carbon. Sidew ays overlap o f these p orbitals and sharing o f the tw o elec­ trons betw een the carbons leads to a p i ( p ) b o n d . T he overlap o f these orbitals is show n schem atically in Fig. 1.24.

H H

n Bond (a)

(b)

Figure 1.24 (a) A wedge-dashed wedge formula for the sigma bonds in ethene and a schematic depiction of the overlapping of adjacent p orbitals that form the p bond. (b) A calculated structure for ethene. The blue and red colors indicate opposite phase signs in each lobe of the p molecular orbital. A balland-stick model for the s bonds in ethene can be seen through the mesh that indicates the p bond.

T he bond angles that w e w ould p redict on the basis o f sp2-hybridized carbon atom s (120° all around) are quite close to the bond angles that are actually found (Fig. 1.20). We can better visualize how these p orbitals interact w ith each other if w e view a struc­ ture show ing calculated m olecular orbitals for ethene (Fig. 1.24). We see that the parallel p orbitals overlap above and below the plane o f the s fram ew ork. N ote the difference in shape o f the bonding m olecular orbital o f a p b o n d as contrasted to that o f a s bond. A s bond has cylindrical sym m etry about a line connecting the two bonded nuclei. A p bond has a nodal plane passing through the tw o b onded nuclei and betw een the p m olecular orbital lobes. •

W hen tw o p atom ic orbitals com bine to form a p bond, tw o m olecular orbitals form : O ne is a bonding m olecular orbital and the other is an antibonding m olecular orbital.

32

Chapter 1

The Basics— Bonding and Molecular Structure

Figure 1.25 How two isolated carbon p orbitals combine to form two p (pi) molecular orbitals. The bonding MO is of lower energy. The higher energy antibonding MO contains an additional node. (Both orbitals have a node in the plane containing the C and H atoms.)

T h e b o n d in g p a n t ib o n d in g p

m o l e c u l a r o r b i t a l r e s u l t s w h e n p - o r b i t a l l o b e s o f l i k e s ig n s o v e r l a p ; t h e

m o le c u la r o r b i t a l r e s u lts w h e n o p p o s ite s ig n s o v e r la p ( F ig . 1 .2 5 ) .

T h e b o n d in g p

o r b i t a l is th e lo w e r e n e r g y o r b i t a l a n d c o n t a in s b o t h p

e le c tr o n s ( w it h

o p p o s i t e s p in s ) i n t h e g r o u n d s ta te o f t h e m o l e c u l e . T h e r e g i o n o f g r e a t e s t p r o b a b i l i t y o f f i n d i n g t h e e le c t r o n s i n t h e b o n d i n g p o r b i t a l i s a r e g i o n g e n e r a l l y s i t u a t e d a b o v e a n d b e l o w th e p la n e o f th e s - b o n d f r a m e w o r k b e t w e e n th e t w o c a r b o n a to m s . T h e a n t ib o n d in g p * o r b i t a l i s o f h i g h e r e n e r g y , a n d i t i s n o t o c c u p i e d b y e le c t r o n s w h e n t h e m o l e c u l e i s i n t h e g r o u n d s ta te . I t c a n b e c o m e o c c u p i e d , h o w e v e r , i f t h e m o l e c u l e a b s o r b s l i g h t o f t h e r i g h t f r e q u e n c y a n d a n e le c t r o n is p r o m o t e d f r o m th e lo w e r e n e r g y le v e l to th e h ig h e r o n e . T h e a n t i b o n d i n g p * o r b i t a l h a s a n o d a l p la n e b e t w e e n t h e t w o c a r b o n a t o m s . •

T o s u m m a r iz e , a c a r b o n - c a r b o n d o u b l e b o n d c o n s is t s o f o n e s b o n d a n d o n e p b o n d .

T h e s b o n d r e s u lts f r o m t w o s p 2 o r b it a ls o v e r la p p in g e n d to e n d a n d is s y m m e t r ic a l a b o u t a n a x is l i n k i n g th e t w o c a r b o n a to m s . T h e p

b o n d r e s u lts f r o m a s id e w a y s o v e r la p o f t w o

p o r b i t a l s ; i t h a s a n o d a l p la n e l i k e a p o r b i t a l . I n t h e g r o u n d s t a t e t h e e l e c t r o n s o f t h e p b o n d a re lo c a te d b e t w e e n th e t w o c a r b o n a to m s b u t g e n e r a lly a b o v e a n d b e lo w th e p la n e o f th e s - b o n d fr a m e w o r k .

The relative energies of electrons involved in s and p bonds.

E le c tr o n s o f th e p

b o n d h a v e g r e a t e r e n e r g y th a n e le c tr o n s o f th e s b o n d . T h e r e la t iv e

e n e r g ie s o f t h e s a n d p m o l e c u l a r o r b i t a l s ( w i t h t h e e le c t r o n s i n t h e g r o u n d s t a t e ) a r e s h o w n i n th e m a r g in d ia g r a m . ( T h e s * o r b i t a l is th e a n t ib o n d in g s ig m a o r b it a l. )

1 .1 3 A

Restricted Rotation and the Double Bond

The s - p

m o d e l f o r t h e c a r b o n - c a r b o n d o u b l e b o n d a ls o a c c o u n t s f o r a n i m p o r t a n t p r o p e r t y

o f th e d o u b le b o n d : •

T h e r e is a la r g e e n e r g y b a r r ie r t o r o t a t io n a s s o c ia te d w i t h g r o u p s j o i n e d b y a d o u ­ b le b o n d .

M a x im u m o v e r la p b e tw e e n th e p o r b ita ls o f a p b o n d o c c u rs w h e n th e a x e s o f th e p o r b ita ls a re e x a c t ly p a r a lle l. R o t a t in g o n e c a r b o n o f th e d o u b le b o n d 9 0 ° ( F ig . 1 .2 6 ) b r e a k s th e p b o n d , f o r th e n th e a x e s o f th e p o r b it a ls a re p e r p e n d ic u la r a n d th e r e is n o n e t o v e r la p b e tw e e n

1.13 The Structure o f Ethene (Ethylene): sp2 Hybridization

33

Figure 1.26 A stylized depiction of how rotation of a carbon atom of a double bond through an angle of 90° results in breaking of the p bond.

them . E stim ates based on therm ochem ical calculations indicate that the strength o f the p bond is 264 kJ m o l- 1 . This, then, is the barrier to rotation o f the double bond. It is m arkedly higher than the rotational b arrier o f groups jo in e d by carb o n -carb o n single bonds (1 3 -2 6 kJ m o l- 1 ). W hile groups jo in e d by single bonds rotate relatively freely at room tem perature, those jo in e d by double bonds do not.

1 .1 3 B

Cis-Trans Isomerism

R estricted rotation o f groups jo in e d by a double bond causes a new type o f isom erism that w e illustrate w ith the tw o dichloroethenes w ritten as the follow ing structures: Cl \

H \

H C II

/

c Cl C

H

c II c

Cl C

c /s -1 ,2 -D ic h lo ro e th e n e

•

Cl /

H

fra n s -1 ,2 -D ic h lo ro e th e n e

T hese tw o com pounds are isom ers; they are different com pounds that have the sam e m olecular form ula.

We can tell that they are different com pounds by trying to place a m odel o f one com pound on a m odel o f the other so that all parts coincide, that is, to try to su p e rp o se one on the other. We find that it cannot b e done. H ad one been su p e rp o sa b le on the other, all parts of one m odel w ould correspond in three dim ensions exactly w ith the other m odel. (The notion o f superposition is different fro m sim ply superim posing one thing on another. T he latter m eans only to lay one on the other w ithout the necessary condition that all parts coincide.) •

We indicate that they are different isom ers by attaching the prefix cis or trans to their nam es (cis, Latin: on this side; trans, Latin: across).

cis-1,2-D ichloroethene and trans-1,2-dichloroethene are n o t constitutional isom ers because the connectivity o f the atom s is the sam e in each. T he tw o com pounds d iffe r o n ly in the a rra n g em e n t o f th e ir atom s in space. Isom ers o f this k ind are classified form ally as ste re o iso m ers, but often they are called sim ply cis-tran s isom ers. (We shall study stereoiso­ m erism in detail in C hapters 4 and 5.) T he structural requirem ents for c is - tr a n s iso m e rism w ill b ecom e clear if w e consider a few additional exam ples. 1,1-D ichloroethene and 1,1,2-trichloroethene do n o t show this type o f isom erism . Cl

H x C = < /

C l^

XH

1 ,1 -D ic h lo ro e th e n e (n o c is -tra n s is o m e ris m )

Cl

Cl x C = < /

C lX

\

1 ,1 ,2 -T ric h lo ro e th e n e (n o c is -tra n s is o m e ris m )

34

Chapter 1

The Basics— Bonding and Molecular Structure

1 , 2 - D i f l u o r o e t h e n e a n d 1 , 2 - d ic h l o r o - 1 , 2 - d i f l u o r o e t h e n e d o e x i s t a s c i s - t r a n s i s o m e r s . N o t i c e t h a t w e d e s i g n a t e t h e i s o m e r w i t h t w o i d e n t i c a l g r o u p s o n t h e s a m e s id e a s b e i n g c is :

F

F \

C=C

H

F

/

Cl

o

/

\

C=C

H

XH

c /s -1 ,2 -D iflu o ro e th e n e

\ =

H \

/ \

fra n s -1 ,2 -D iflu o ro e th e n e

F

W

Cl

Cl

/

c /s -1 ,2 -D ic h lo ro -1 ,2 -d iflu o ro e th e n e

*

\

F

fra n s -1 ,2 -D ic h lo ro -1 ,2 -d iflu o ro e th e n e

C l e a r l y , t h e n , c is -tra n s isom erism o f this type is n o t possible i f one carbon atom o f the double b o n d bears two id e n tic a l groups.

R e v ie w P ro b le m 1.13

W h i c h o f t h e f o l l o w i n g a lk e n e s c a n e x i s t a s c i s - t r a n s i s o m e r s ? W r i t e t h e i r s t r u c t u r e s . B u i l d h a n d h e ld m o d e ls t o p r o v e t h a t o n e is o m e r is n o t s u p e r p o s a b le o n th e o th e r .

(a) C H 2= CH C H 2C H 3

(c)

(b) C H 3C H = CH C H 3

(d) C H 3CH 2CH = CHCl

C H 2= C (C H 3)2

1.14 The Structure o f Ethyne (Acetylene): sp H ybridization H y d r o c a r b o n s i n w h i c h t w o c a r b o n a t o m s s h a r e t h r e e p a i r s o f e le c t r o n s b e t w e e n t h e m , a n d a r e t h u s b o n d e d b y a t r i p l e b o n d , a r e c a l l e d a l k y n e s . T h e t w o s i m p l e s t a lk y n e s a r e e t h y n e a n d p ro p y n e .

H— C # C — H

CH 3— C # C — H

Ethyne (acetylene) (C2H2)

Propyne (C3H4)

E t h y n e , a c o m p o u n d t h a t is a ls o c a l l e d a c e t y le n e , c o n s i s t s o f a l i n e a r a r r a n g e m e n t o f a to m s . T h e

H— C # C

b o n d a n g le s o f e t h y n e m o l e c u l e s a r e 1 8 0 ° :

H^

S

^

H

180° 180° W e c a n a c c o u n t f o r t h e s t r u c t u r e o f e t h y n e o n t h e b a s is o f o r b i t a l h y b r i d i z a t i o n a s w e d id f o r e th a n e a n d e th e n e . I n o u r m o d e l f o r e th a n e ( S e c tio n 1 .1 2 B ) w e s a w th a t th e c a r b o n o r b it a ls a re s p 3 h y b r id iz e d , a n d in o u r m o d e l f o r e th e n e ( S e c tio n 1 .1 3 ) w e s a w th a t th e y a re s p 2 h y b r id iz e d . I n

o u r m o d e l f o r e t h y n e w e s h a l l s e e t h a t t h e c a r b o n a t o m s a r e sp

h y b r id iz e d . T h e m a t h e m a t i c a l p r o c e s s f o r o b t a i n i n g t h e sp h y b r i d o r b i t a l s o f e t h y n e c a n b e v i s u a l i z e d in th e f o l lo w in g w a y ( F ig . 1 .2 7 ).

Figure 1.27 A process fo r deriving sp-hybridized carbon atoms.

1.14 The Structure of Ethyne (Acetylene): sp Hybridization

•

T h e 2 s o r b it a l a n d o n e 2p o r b it a l o f c a r b o n a re h y b r id iz e d to f o r m

tw o

sp

o r b ita ls .

T h e r e m a in in g t w o 2 p o r b it a ls a re n o t h y b r id iz e d . C a l c u l a t i o n s s h o w t h a t t h e sp h y b r i d o r b i t a l s h a v e t h e i r l a r g e p o s i t i v e lo b e s o r ie n t e d a t a n a n g le o f 1 8 0 ° w i t h r e s p e c t t o e a c h o th e r . T h e t w o 2 p o r b it a ls th a t w e r e n o t h y b r id iz e d

a re e a c h p e r p e n d ic u la r to

th e a x is

th a t passe s

t h r o u g h t h e c e n t e r o f t h e t w o sp o r b i t a l s ( F i g . 1 . 2 8 ) . W e p l a c e o n e e le c t r o n in e a c h o r b ita l. W e e n v is io n th e b o n d in g m o le c u la r o r b ita ls o f e th y n e b e in g f o r m e d in th e f o l lo w in g w a y ( F ig . 1 .2 9 ).

Figure 1.28 An sp-hybridized carbon atom.

Figure 1.29 Formation of the bonding molecular orbitals of ethyne from two sp-hybridized carbon atoms and two hydrogen atoms. (Antibonding orbitals are formed as well, but these have been omitted for simplicity.)

•

T w o c a r b o n a t o m s o v e r l a p sp o r b i t a l s t o f o r m

a s ig m a b o n d b e tw e e n th e m

( th is

i s o n e b o n d o f t h e t r i p l e b o n d ) . T h e r e m a i n i n g t w o sp o r b i t a l s a t e a c h c a r b o n a to m o v e r la p w it h s o r b ita ls f r o m

h y d r o g e n a to m s t o p r o d u c e t w o s ig m a C — H

bonds. •

T h e t w o p o r b i t a l s o n e a c h c a r b o n a t o m a ls o o v e r l a p s id e t o s i d e t o f o r m t w o p b o n d s . T h e s e a re th e o th e r tw o b o n d s o f th e t r ip le b o n d .

•

T h e c a r b o n - c a r b o n t r i p le b o n d c o n s is ts o f t w o p

bonds and one s

bond.

S tr u c tu r e s f o r e th y n e b a s e d o n c a lc u la te d m o le c u la r o r b it a ls a n d e le c tr o n d e n s it y a re s h o w n i n F i g . 1 . 3 0 . C i r c u l a r s y m m e t r y e x is t s a lo n g t h e l e n g t h o f a t r i p l e b o n d ( F i g . 1 . 3 0 f t ) . A s a r e s u lt, th e r e is n o r e s t r ic t io n o f r o t a t io n f o r g r o u p s j o i n e d b y a t r ip le b o n d (a s c o m p a r e d w i t h a lk e n e s ) , a n d i f r o t a t i o n w o u l d o c c u r , n o n e w c o m p o u n d w o u l d f o r m .

(a )

(b )

(c)

Figure 1.30 (a) The structure of ethyne (acetylene) showing the sigma-bond framework and a schematic depiction of the two pairs of p orbitals that overlap to form the two p bonds in ethyne. (b) A structure of ethyne showing calculated p molecular orbitals. Two pairs of p molecular orbital lobes are present, one pair for each p bond. The red and blue lobes in each p bond represent opposite phase signs. The hydrogen atoms of ethyne (white spheres) can be seen at each end of the structure (the carbon atoms are hidden by the molecular orbitals). (c) The mesh surface in this structure represents approximately the furthest extent of overall electron density in ethyne. Note that the overall electron density (but not the p-bonding electrons) extends over both hydrogen atoms.

36

Chapter 1

1 .1 4 A

The Basics— Bonding and Molecular Structure

Bond Lengths of Ethyne, Ethene, and Ethane

T h e c a r b o n - c a r b o n t r ip le b o n d o f e th y n e is s h o r te r th a n th e c a r b o n - c a r b o n d o u b le b o n d o f e th e n e , w h ic h i n t u r n is s h o r t e r th a n th e c a r b o n - c a r b o n s in g le b o n d o f e th a n e . T h e re a s o n i s t h a t b o n d l e n g t h s a r e a f f e c t e d b y t h e h y b r i d i z a t i o n s ta te s o f t h e c a r b o n a t o m s i n v o l v e d . •

T h e g r e a t e r t h e s o r b i t a l c h a r a c t e r i n o n e o r b o t h a t o m s , t h e s h o r t e r is t h e b o n d . T h i s is b e c a u s e s o r b i t a l s a r e s p h e r i c a l a n d h a v e m o r e e l e c t r o n d e n s i t y c l o s e r t o t h e n u c le u s th a n d o p o r b ita ls . T h e g r e a te r th e p o r b it a l c h a r a c te r in o n e o r b o t h a to m s , th e lo n g e r is th e b o n d .

•

T h i s is b e c a u s e p o r b i t a l s a r e l o b e - s h a p e d w i t h e l e c t r o n d e n s i t y e x t e n d i n g a w a y f r o m th e n u c le u s . I n t e r m s o f h y b r i d o r b i t a l s , a n sp h y b r i d o r b i t a l h a s 5 0 % s c h a r a c t e r a n d 5 0 % p c h a r a c t e r . A n s p 2 h y b r id o r b ita l h a s 3 3 % s c h a ra c te r a n d 6 7 % p c h a ra c te r. A n s p 3 h y b r id o r b ita l h a s 2 5 % s c h a r a c te r a n d 7 5 % p c h a ra c te r. T h e o v e r a ll tr e n d , th e r e fo r e , is as fo llo w s : B o n d s i n v o l v i n g sp h y b r i d s a r e s h o r t e r t h a n t h o s e i n v o l v i n g s p 2 h y b r i d s , w h i c h a r e

•

s h o r te r th a n th o s e in v o lv in g s p 3 h y b r id s . T h is tr e n d h o ld s tr u e f o r b o t h C — C a n d C — H bonds. The

bond

le n g th s

and bond

a n g le s

o f e th y n e ,

e th e n e , a n d

e th a n e

a re

s u m m a r iz e d

in

F i g . 1 .3 1 .

180°

H

H 18X

H

« 109 5 ° 109.5°

H*

1.20 Â 1.54 Â

1.34 Â

1.10 Â Figure 1.31 Bond angles and bond lengths of ethyne, ethene, and ethane.

H

1.06 «

1.15 A Sum m ary o f Im p o rtan t Concepts That Com e from Q uantum Mechanics 1.

A n

ato m ic o r b ita l (A O )

g le a to m

c o r r e s p o n d s to a r e g io n o f s p a c e a b o u t th e n u c le u s o f a s in ­

w h e r e th e r e is a h ig h p r o b a b il i t y o f f i n d in g a n e le c tr o n . A t o m i c o r b it a ls

c a l l e d s o r b i t a l s a r e s p h e r i c a l; t h o s e c a l l e d p o r b i t a l s a r e l i k e

t w o a lm o s t- t a n g e n t

s p h e r e s . O r b i t a l s c a n h o l d a m a x i m u m o f t w o e l e c t r o n s w h e n t h e i r s p in s a r e p a i r e d . O r b it a ls a re d e s c r ib e d b y th e s q u a re o f a w a v e fu n c t io n , C 2, a n d e a c h o r b it a l h a s a c h a r a c t e r i s t i c e n e r g y . T h e p h a s e s ig n s a s s o c i a t e d w i t h a n o r b i t a l m a y b e + 2 . W h e n a t o m ic o r b it a ls o v e r la p , th e y c o m b in e t o f o r m

m o le c u la r o r b it a ls

or - . (M O s ).

M o l e c u l a r o r b i t a l s c o r r e s p o n d t o r e g i o n s o f s p a c e e n c o m p a s s in g t w o ( o r m o r e ) n u c l e i w h e r e e le c t r o n s a r e t o b e f o u n d . L i k e a t o m i c o r b i t a l s , m o l e c u l a r o r b i t a l s c a n h o l d u p t o t w o e le c t r o n s i f t h e i r s p in s a r e p a i r e d .

37

1.15 A S um m ary o f Im p o rta n t C o n c ep ts T h at C om e from Q u an tu m M echanics 3. W hen atom ic orbitals w ith the sam e phase sign interact, they com bine to form a b o n d ­ in g m o le c u la r o rb ita l:

The electron probability density o f a bonding m olecular orbital is large in the region o f space betw een the tw o nuclei w here the negative electrons h o ld the positive nuclei together. 4. An a n tib o n d in g m o le cu lar o rb ita l form s when orbitals o f opposite phase sign overlap: N ode

A n antibonding orbital has higher energy than a bonding orbital. T he electron p ro b a­ bility density o f the region betw een the nuclei is sm all and it contains a n o d e — a region w here C = 0. Thus, having electrons in an antibonding orbital does n ot help h old the nuclei together. T he internuclear repulsions tend to m ake them fly apart. 5. T he en e rg y o f e le c tro n s in a bonding m o le cu la r orbital is less than the energy of the electrons in their separate atom ic orbitals. T he energy o f electrons in an an ti­ bonding orbital is greater than that o f electrons in their separate atom ic orbitals. 6. T he n u m b e r o f m o le c u la r o rb ita ls alw ays equals the num ber o f atom ic orbitals from w hich they are form ed. C om bining tw o atom ic orbitals w ill alw ays yield two m olecular orbitals— one bonding and one antibonding. 7. H y b rid a to m ic o rb ita ls are obtained by m ixing (hybridizing) the w ave functions for orbitals o f different types (i.e., s and p orbitals) b ut from the sam e atom. 8. H ybridizing three p orbitals w ith one s orbital yields four sp 3 o rb ita ls. A tom s that are sp3 hybridized direct the axes o f their four sp3 orbitals tow ard the corners o f a tetrahedron. T he carbon o f m ethane is sp3 hybridized an d te tr a h e d r a l. 9. H ybridizing tw o p orbitals w ith one s orbital yields three sp2 o rb ita ls. A tom s that are sp2 hybridized p oint the axes o f their three sp2 orbitals tow ard the corners o f an equilateral triangle. T he carbon atom s o f ethene are sp2 hybridized an d trig o n a l p la n a r. 10. H ybridizing one p orbital w ith one s orbital yields tw o sp o rb ita ls. A tom s that are sp hybridized orient the axes o f their tw o sp orbitals in opposite directions (at an angle o f 180°). The carbon atom s o f ethyne are sp hybridized and ethyne is a lin e a r m olecule. 11. A sig m a ( s ) b o n d (a type o f single bond) is one in w hich the electron density has circular sym m etry w hen view ed along the b o n d axis. In general, the skeletons of organic m olecules are constructed o f atom s linked by sigm a bonds. 12. A p i ( p ) b o n d , p art o f double and triple carb o n -carb o n bonds, is one in w hich the electron densities o f two adjacent parallel p orbitals overlap sideways to form a bond­ ing pi m olecular orbital.

H e lp f u l H i n t

A summary of sp3, sp2, and sp hybrid orbital geometries.

38

Chapter 1

The Basics— Bonding and Molecular Structure

1.16 M olecular G eom etry: The Valence Shell Electron Pair Repulsion M o d e l We can predict the arrangem ent o f atom s in m olecules an d ions on the basis o f a relatively sim ple idea called the v alen ce sh ell e le c tro n p a ir re p u lsio n (V S E P R ) m o d e l. We apply the V S E P R m odel in the follow ing way: 1. We consider m olecules (or ions) in w hich the central atom is covalently b onded to tw o or m ore atom s or groups. 2. We consider all o f the valence electron pairs o f the central atom — both those that are shared in covalent bonds, called b o n d in g p a irs, an d those that are unshared, called n o n b o n d in g p a irs or u n s h a re d p a irs or lo n e p a irs. 3. B ecause electron pairs repel each other, the electron pairs o f the valence shell tend to stay as far apart as possible. T he repulsion betw een nonbonding pairs is generally greater than that betw een bonding pairs. 4. We arrive at the geometry o f the m olecule by considering all o f the electron pairs, bonding and nonbonding, b ut w e describe the shape o f the m olecule or ion by refer­ ring to the positions o f the nuclei (or atom s) and n ot by the positions o f the electron pairs. C onsider the follow ing exam ples. 1 .1 6 A

Methane

T he valence shell o f m ethane contains four pairs o f bonding electrons. O nly a tetrahedral orientation w ill allow four pairs o f electrons to have equal and m axim um p ossible separa­ tion from each other (Fig. 1.32). Any other orientation, for example, a square planar arrange­ m ent, places som e electron pairs closer together than others. Thus, m ethane has a tetrahedral shape. T he bond angles for any atom that has a regular tetrahedral structure are 109.5°. A rep ­ resentation o f these angles in m ethane is show n in Fig. 1.33. Figure 1.32 A tetrahedral shape for methane allows the maximum separation of the four bonding electron pairs.

Figure 1.33 The bond angles of methane are 109.5°.

1 .1 6 B

Ammonia

T he shape o f a m olecule o f am m onia (NH 3) is a trig o n a l p y ra m id . T here are three b o n d ­ ing pairs o f electrons and one nonbonding pair. T he bond angles in a m olecule o f am m o­ nia are 107°, a value very close to the tetrahedral angle (109.5°). We can w rite a general tetrahedral structure for the electron pairs o f am m onia by placing the nonbonding p air at one corner (Fig. 1.34). A tetrahedral arrangem ent o f the electron pairs explains the trig o ­ na l p yra m id a l arrangem ent o f the four atom s. The bond angles are 107° (not 109.5°) because the nonbonding pair occupies m ore space than the bonding pairs.

R e v ie w P ro b le m 1.14

W hat do the bond angles o f am m onia suggest about the hybridization state o f the nitrogen atom o f am m onia?

1.16 Molecular Geometry: The Valence Shell Electron Pair Repulsion Model

39 Figure 1.34 The tetrahedral arrangement of the electron pairs of an ammonia molecule that results when the nonbonding electron pair is considered to occupy one corner. This arrangement of electron pairs explains the trigonal pyramidal shape of the NH3 molecule. Ball-and-stick models do not show unshared electrons.

1 .1 6 C

Water

A m olecule o f w ater has an a n g u la r or b e n t shape. T he H— O — H b o n d angle in a m olecule o f w ater is 104.5°, an angle that is also quite close to the 109.5° b o n d angles of m ethane. We can w rite a general tetrahedral structure for the electron pairs o f a m olecule o f w ater i f we place the two bonding p a irs o f electrons and the two nonbonding electron p a irs at the corners o f the tetrahedron. Such a structure is show n in Fig. 1.35. A tetrahedral arrange­ ment o f the electron pairs accounts for the a n g u la r arrangem ent o f the three atom s. The

bond angle is less than 109.5° because the nonbonding pairs are effectively “larger” than the bonding pairs and, therefore, the structure is n o t perfectly tetrahedral.

Figure 1.35 An approximately tetrahedral arrangement of the electron pairs of a molecule of water that results when the pairs of nonbonding electrons are considered to occupy corners. This arrangement accounts for the angular shape of the H2O molecule.

W hat do the b ond angles o f w ater suggest about the hybridization state o f the oxygen atom o f w ater?

1 .1 6 D

R e v ie w P ro b le m 1.15

Boron Trifluoride

Boron, a group IIIA elem ent, has only three valence electrons. In the com pound boron tri­ fluoride (BF3) these three electrons are shared w ith three fluorine atom s. A s a result, the boron atom in BF3 has only six electrons (three bonding pairs) around it. M axim um sepa­ ration o f three bonding pairs occurs w hen they occupy the corners o f an equilateral trian­ gle. Consequently, in the boron trifluoride m olecule the three fluorine atom s lie in a plane a t the corners o f an equilateral triangle (Fig. 1.36). B oron trifluoride is said to have a tr ig ­ o n a l p la n a r structure. T he bond angles are 120°.

■n /

•

\

• :F : 1 2 0 ' H * ' s1 2 0 ° .F I r S L : .F 120 °

. F F :

Figure 1.36 The triangular (trigonal planar) shape of boron trifluoride maximally separates the three bonding pairs.

40

Chapter 1

R e v ie w P ro b le m 1.16

The Basics— Bonding and Molecular Structure

W hat do the bond angles o f boron trifluoride suggest about the hybridization state o f the boron atom?

Beryllium Hydride

1 .1 6 E

T he central beryllium atom o f BeH 2 has only tw o electron pairs around it; both electron pairs are bonding pairs. T hese tw o pairs are m axim ally separated w hen they are on o p p o ­ site sides o f the central atom , as show n in the follow ing structures. T his arrangem ent of the electron pairs accounts for the lin e a r geometry o f the BeH 2 m olecule and its bond angle o f 180°.

R e v ie w P ro b le m 1.17

W hat do the bond angles o f beryllium hydride suggest about the hybridization state o f the beryllium atom ?

R e v ie w P ro b le m 1.18

U se V S E P R theory to p redict the geom etry o f each o f the follow ing m olecules and ions: (a) BH 4 -

(

(b) B eF 2

(d) H2S

1 .1 6 F

c

)

NH 4 +

(e) BH 3

(g) S iF 4

(f) C F 4

(h) :CCl3_

Carbon Dioxide

T he V S E P R m ethod can also be used to predict the shapes o f m olecules containing m u l­ tiple bonds if w e assum e that a ll o f the electrons o f a m u ltip le bond act as though they were a single u n it and, therefore, are located in the region o f space betw een the tw o atom s jo in ed by a m ultiple bond. This principle can be illustrated w ith the structure o f a m olecule o f carbon dioxide (CO2). T he central carbon atom o f carbon dioxide is bonded to each oxygen atom by a double bond. C arbon dioxide is know n to have a linear shape; the bond angle is 180°.

:o =

c

=

o:

or

:o

o.:

The four electrons of each double bond act as a single unit and are maximally separated from each other.

Such a structure is consistent w ith a m axim um separation o f the tw o groups o f four b o n d ­ ing electrons. (The nonbonding pairs associated w ith the oxygen atom s have no effect on the shape.)

R e v ie w P ro b le m 1.19

Predict the bond angles of (a) F 2C = C F 2

(b) CH 3 C # C C H 3

(c) H C # N

T he shapes o f several sim ple m olecules and ions as p redicted by V S E P R theory are show n in Table 1.4. In this table w e have also included the hybridization state o f the cen ­ tral atom.

1.17 How to Interpret and W rite Structural Formulas

41

Shapes o f M olecules and Ions from VSEPR Theory N u m b er o f E lectron Pairs a t C entral A tom Bonding

N o n b o n d in g

Total

2

Q Q Q

2

3 4 3

sp sp ,^3 sp

3 4 4 4

1 2

2

H ybridization S ta te of C entral A tom

~ sp3 ~ sp3

S hape of M olecule or Iona

E xam ples

Linear Trigonal planar Tetrahedral Trigonal pyramidal A ngular

BeH 2 BF 3 , CH 3+ c h 4, n h 4+ n h 3, c h 3h 2o

aReferring to positions of atoms and excluding nonbonding pairs.

1.17 H o w to In te rp re t and W rite Structural Formulas O r g a n ic c h e m is ts u s e a v a r ie t y o f w a y s to w r it e s tr u c tu r a l f o r m u la s . T h e m o s t c o m m o n ty p e s o f r e p r e s e n t a t i o n s a r e s h o w n i n F i g . 1 .3 7 u s in g p r o p y l a l c o h o l a s a n e x a m p l e . T h e

tu r e

d o t s tr u c ­

s h o w s a l l o f t h e v a le n c e e l e c t r o n s , b u t w r i t i n g i t i s t e d io u s a n d t i m e - c o n s u m i n g . T h e

o t h e r r e p r e s e n ta tio n s a re m o r e c o n v e n ie n t a n d a re , th e r e fo r e , m o r e o f t e n u s e d . S o m e t im e s w e e v e n o m i t u n s h a r e d p a i r s w h e n w e w r i t e f o r m u l a s . H o w e v e r , w h e n w e w r it e c h e m ic a l r e a c t io n s , w e se e t h a t i t is n e c e s s a r y t o in c lu d e th e u n s h a r e d e le c tr o n p a ir s w h e n th e y p a r t ic ip a t e i n th e r e a c t io n . I t is a g o o d id e a , th e r e fo r e , to g e t i n t o th e h a b it o f w r i t i n g th e u n s h a r e d ( n o n b o n d in g ) e le c t r o n p a ir s i n th e s tr u c tu r e s y o u d ra w .

Figure 1.37 Structural formulas for propyl alcohol.

1 .1 7 A

Dash Structural Formulas

I f w e l o o k a t th e m o d e l f o r p r o p y l a lc o h o l g iv e n i n F ig . 1 .3 7 a a n d c o m p a r e i t w i t h th e d o t, d a s h , a n d c o n d e n s e d f o r m u la s in F ig s .

1.3 7 b -d

w e f i n d t h a t t h e c h a i n o f a t o m s is s t r a i g h t

i n th o s e fo r m u la s . I n th e m o d e l, w h ic h c o r r e s p o n d s m o r e a c c u r a t e ly to th e a c tu a l s h a p e o f th e m o le c u le , th e c h a in o f a to m s is n o t a t a ll s t r a ig h t . A l s o o f im p o r t a n c e is th is :

jo in e d by single bonds can rotate re la tiv e ly fre e ly w ith respect to one ano th e r.

A tom s

( W e d is ­

c u s s e d t h is p o i n t b r ie f l y i n S e c tio n 1 .1 2 B .) T h is r e la t iv e ly f r e e r o t a t io n m e a n s t h a t th e c h a in o f a to m s i n p r o p y l a lc o h o l c a n a s s u m e a v a r ie t y o f a r r a n g e m e n ts l ik e th e s e :

H H

H

H

Y

H

H

Y

/

H

or

H

Y

H/

X

Y

V

or

\

Y

C

H

O

H/

H

/ X

'

\ y

C

C

H

C

C H

H/ X

H

E q u iv a le n t d as h fo rm u la s fo r p ro p y l a lcoh ol I t a ls o m e a n s t h a t a l l o f t h e s t r u c t u r a l f o r m u l a s a b o v e a r e e q u iv a le n t a n d a l l r e p r e s e n t p r o p y l a lc o h o l.

D a sh s tr u c tu ra l fo rm u la s

s u c h a s th e s e in d ic a te th e w a y in w h ic h th e a to m s

a r e a t t a c h e d t o e a c h o t h e r a n d a re n o t r e p r e s e n t a t i o n s o f t h e a c t u a l s h a p e s o f t h e m o l e c u l e .

42

Chapter 1

H e lp f u l H i n t

It is important that you be able to recognize when a set of structural formulas has the same connectivity versus when they are constitutional isomers.

The Basics— Bonding and Molecular Structure

(Propyl alcohol does n o t have 90° bond angles. It has tetrahedral bond angles.) D ash struc­ tural form ulas show w hat is called the co n n e ctiv ity o f the atom s. C o nstitutional isomers (Section 1.3A) have different connectivities and, therefore, must have different stru ctu ra l fo rm u la s.

C onsider the com pound called isopropyl alcohol, w hose form ula w e m ight w rite in a variety o f w ays: H H O H I H— C — C — C — H I H H H

H

H H I H— C — C — C — H I H O H

or

or

H O— H I I H— C — C — H I H

H

H— C— H I H

E q u iv a le n t d as h fo rm u la s fo r is o p ro p y l a lcoh ol

Isopropyl alcohol is a constitutional isom er (Section 1.3A) o f propyl alcohol because its atom s are connected in a different order and b oth com pounds have the sam e m olecular fo r­ m ula, C3H8O. In isopropyl alcohol the OH group is attached to the central carbon; in propyl alcohol it is attached to an end carbon. •

R e v ie w P ro b le m 1.20

In problem s you w ill often b e asked to w rite structural form ulas for all the isom ers that have a given m olecular form ula. D o n o t m ake the error o f w riting several equivalent form ulas, like those that w e have just shown, m istaking them for differ­ ent constitutional isom ers.

T here are actually three constitutional isom ers w ith the m olecular form ula C 3H8O. We have seen tw o o f them in propyl alcohol and isopropyl alcohol. W rite a dash form ula for the third isomer.

1 .1 7 B

Condensed Structural Formulas

C o n d e n sed s tr u c tu ra l fo rm u las are som ew hat faster to w rite than dash form ulas and, w hen w e becom e fam iliar w ith them , they w ill im part all the inform ation that is contained in the dash structure. In condensed form ulas all o f the hydrogen atom s that are attached to a par­ ticular carbon are usually w ritten im m ediately after the carbon. In fully condensed form u­ las, all o f the atom s that are attached to the carbon are usually w ritten im m ediately after that carbon, listing hydrogens first. For exam ple,

I I I I H— C— C— C— C— H I I I I H Cl H H H

H

H

H

CH 3CH C H 2CH 3 or

CH 3CH ClCH 2CH 3

Cl C o n d e n s e d fo rm u la s

Dash fo rm u la

T he condensed form ula for isopropyl alcohol can b e w ritten in four different w ays:

H

H H H I I I C C C H I I I H O H H Dash fo rm u la

CH 3CH C H 3 or

CH 3CH (O H )CH 3

OH CH 3CH O H CH 3 or (CH3)2CHOH '------------------------v----------------------- ' C o n d e n s e d fo rm u las

43

1.17 How to Interpret and W rite Structural Formulas

C ftli/û r l D m

W rite a condensed structural form ula for the follow ing com pound.

1 .1 7 C

Bond-Line Formulas

The m ost com m on type o f structural form ula used by organic chem ists, and the fastest to draw, is the b o n d -lin e f o rm u la . T he form ula in Fig. 1.37e is a bond-line form ula for propyl alcohol. T he sooner you m aster the use o f bond-line form ulas, the m ore quickly you will be able to draw m olecules w hen you take notes and w ork problem s. A nd, lacking all o f the sym bols that are explicitly show n in dash and condensed structural form ulas, bond-line for­ m ulas allow you to m ore quickly interpret m olecular connectivity and com pare one m ole­ cular form ula w ith another. The efficiencies o f draw ing bond-line form ulas com e from the fact that no C s are w rit­ ten for carbon atom s, and generally no Hs are show n for hydrogen atom s, unless they are needed to give a three-dim ensional perspective to the m olecule (and in that case w e use solid or dashed w edges for bonds to the out-of-plane atom s, as described in the follow ing section). Instead, in bond-line form ulas ordinary lines represent bonds, and carbon atom s are inferred at each bend in the line and at the ends o f lines. T he nu m b er o f hydrogen atom s bonded to each carbon is also inferred, b y assum ing that as m any hydrogen atom s are p resent as n eed ed to fill the valence shell o f each car­ bon, unless a charge is indicated. W hen an atom o th er than carbon is p resen t, the sym ­ bol for that elem ent is w ritten in the form ula at the ap p ro p riate location, i.e., in p lace o f a ben d or at the term inus o f the line leading to the atom . H ydrogen atom s b onded to atom s other than carbon (e.g., oxygen o r nitrogen) are w ritten explicitly. A nd, as m en tio n ed above, hydrogen atom s are show n w here need ed to help specify three dim ensions using

R e v ie w P ro b le m 1.21

44

Chapter 1

The Basics— Bonding and Molecular Structure

solid or d ashed w edges. C onsider the follow ing exam ples o f m olecules d epicted by bondlin e form ulas. B o n d -lin e fo rm u la s

CH 3 c h 2 \V \ 2 CH CH 3

CH 3CH C lCH 2CH 3

Cl

Ci

CH33 CH2 2 CH 3CH (CH 3)CH2CH 3 = CH CH3 CH

CH 3 / H

H e lp f u l H i n t

(CH3)2NCH2CH3

As you become more familiar with organic molecules, you will find bond-line formulas to be very useful tools for representing structures.

N

2

ch3

N

CH

B ond-line form ulas are easy to draw for m olecules w ith m ultiple bonds and for cyclic m olecules, as well. T he follow ing are som e exam ples.

CH

/Vu — 2= A

h 2c

ch

ch3 \

, ch

y

c

h 2c

—ch2

h 2c

II 2 ^ □ I___ I — cI h

and

\

ch3 /

3

ch2

ch3

c h 2= c h c h 2o h

r

45

1.17 How to Interpret and W rite Structural Formulas

S o lv e d P ro b le m 1 .9

W rite the bond-line form ula for CH3CH CH 2CH2CH2OH I CH , STRATEGY A ND ANSW ER First, for the sake o f practice, w e outline the carbon skeleton, including the OH group, as follows: CH3 c h 2 c h 2 \ V \ V \ 2 CH CH2 OH I CH T hen w e w rite the bond-line form ula as

C

\

C

C I C

/

\

C C

/

\

OH

OH. A s you gain experience you w ill likely skip the inter­

m ediate steps show n above and proceed directly to w riting bond-line form ulas.

W rite each o f the follow ing condensed structural form ulas as a bond-line form ula: (a) (CH 3)2CH C H 2CH 3

R e v ie w P ro b le m 1.22

(f) CH 2= C (C H 2CH 3)2 O

(b) (CH3)2CH C H 2CH 2OH (c) (CH3)2C = CH C H 2CH 3

(g)

(d) C H 3CH 2CH 2C H 2CH 3

ch

3C c h 2c h 2c h 2c h 3

(h) CH 3CHCICH2CH (CH 3)2

(e) C H 3CH 2C H (O H )CH 2CH 3

W hich m olecules in Review Problem 1.22 form sets o f constitutional isom ers?

R e v ie w P ro b le m 1.23

W rite a dash form ula for each o f the follow ing bond-line form ulas:

R e v ie w P ro b le m 1.24

OH Cl (a)

(b)

(c)

O

1 .1 7 D

Three-Dimensional Formulas

N one o f the form ulas that w e have described so far conveys any inform ation about how the atom s o f a m olecule are arranged in space. T here are several types o f representations that do this. T he types o f form ulas that w e shall use m ost o f the tim e are show n in Fig. 1.38. In these representations, bonds that p roject upw ard o ut o f the plane o f the p aper are indi­ cated by a solid w edge ( - - ) , those that lie behind the plane are indicated w ith a dashed w edge ( ... n), and those bonds that lie in the plane o f the page are indicated by a line (— ). For tetrahedral atom s, notice that w e draw the tw o bonds that are in the plane o f the page w ith an angle o f approxim ately 109° betw een them and that proper three-dim ensional p er­ spective then requires the w edge and dashed-w edge bonds to b e draw n near each other on the page (i.e., the atom in front nearly eclipses the atom behind). We can draw trigonal p la­ nar atom s either w ith all bonds in the plane o f the page separated by approxim ately 120° or w ith one o f the three bonds in the plane o f the page, one behind, and one in front (as in Fig. 1.20). A tom s w ith linear bonding geom etry are best draw n w ith all bonds to those atoms in the plane o f the page. Lastly, w hen draw ing three-dim ensional form ulas is it generally best to draw as m any carbon atom s as possible in the plane o f the paper, allow ing substituent groups or hydrogen atom s to be prim arily those for w hich w edge or dashed-w edge bonds

H e lp f u l H i n t

Wedge and dashed-wedge formulas are a tool for unambiguously showing three dimensions.

46

Chapter 1

The Basics— Bonding and Molecular Structure

are used. N ote that w hen using bond-line form ulas w e continue to om it hydrogen atom s unless they are relevant to clarifying the three-dim ensional perspective o f som e other group. F igure 1.38 show s som e exam ples o f three-dim ensional form ulas. H H

C I H ^ CH H

or

HH HV / H

H / C— C H

H I

etc.

or 'B r

h

Br

or

etc. H

H

H

H OH HO

Br

Br

H

Figure 1.38 Some

examples of threedimensional formulas.

Br I

B ro m o m eth an e

Eth an e

JDH

H I -C^-«H Br H

E x a m p le s o f b o n d -lin e fo rm u la s that in c lu d e th re e -d im e n s io n a l re p re s e n ta tio n s

R e v ie w P ro b le m 1.25

An e x a m p le in v o lv in g trig o n al p la n a r g e o m e try

A n e x a m p le in vo lvin g lin e a r g e o m e try

W rite three-dim ensional (w ed g e-d ash ed w ed g e-lin e) representations for each o f the following: (a ) CH 3CI

(b) CH 2CI2

(c) CH 2BrCI

(d) C H 3CH 2CI

1.18 Applications o f Basic Principles T hroughout the early chapters o f this b o o k w e review certain basic principles that under­ lie and explain m uch o f the chem istry w e shall b e studying. C onsider the follow ing p rin ­ ciples and how they apply in this chapter. O p p o site C harges A ttra c t We see this principle operating in our explanations for covalent and ionic bonds (Sections 1.11 an d 1.4A). It is the attraction o f the p o sitive ly charged nuclei for the negatively charged electrons that underlies our explanation for the covalent bond. It is the attraction o f the oppositely charged ions in crystals that explains the ionic bond. Like C harges Repel It is the repulsion o f the electrons in covalent bonds o f the valence shell o f a m olecule that is central to the valence shell electron pair repulsion m odel for explaining m olecular geom etry. A nd, although it is n ot so obvious, this sam e factor u nder­ lies the explanations o f m olecular geom etry that com e from orbital hybridization because these repulsions are taken into account in calculating the orientations o f the hybrid orbitals. N a tu re Tends to w a rd States o f L o w er P o ten tia l E n erg y This principle explains so m uch o f the w orld around us. It explains w hy w ater flows downhill: T he potential energy o f the w ater at the bottom o f the hill is low er than that at the top. (We say that w ater is in a m ore stable state at the bottom .) This principle underlies the aufbau principle (Section 1.10A): In its low est energy state, the electrons o f an atom occupy the low est energy orbitals avail­ able [but H und’s rule still applies, as w ell as the Pauli exclusion principle (Section 1.10A), allow ing only tw o electrons p er orbital]. Sim ilarly in m olecular orbital theory (Section 1.11), electrons fill low er energy bonding m olecular orbitals first because this gives the m olecule low er potential energy (or greater stability). Energy has to be provided to m ove an electron to a higher orbital and provide an excited (less stable) state (Review Problem 1.12). O rb ita l O v erlap Stabilizes M olecules T his principle is part o f our explanation for covalent bonds. W hen orbitals o f the sam e p hase from different n u clei overlap, the elec­ trons in these orbitals can b e shared by both nuclei, resulting in stabilization. T he resu lt is a covalent bond.

47

Problems

| /n This Chapter-} In C hapter 1 you have studied concepts and skills that are absolutely essential to your suc­ cess in organic chem istry. You should now be able to use the periodic table to determ ine the num ber o f valence electrons an atom has in its neutral state or as an ion. You should be able to use the periodic table to com pare the relative electronegativity o f one elem ent w ith another, and determ ine the form al charge o f an atom or ion. E lectronegativity and fo r­ m al charge are key concepts in organic chemistry. You should be able to draw chem ical form ulas that show all o f the valence electrons in a m olecule (Lewis structures), using lines for bonds and dots to show unshared electrons. You should be proficient in representing structures as dash structural formulas, condensed struc­ tural form ulas, and bond-line structural form ulas. In particular, the m ore quickly you becom e skilled at using and interpreting bond-line formulas, the faster you will be able to process struc­ tural inform ation in organic chemistry. You have also learned about resonance structures, the use o f w hich w ill help us in understanding a variety o f concepts in later chapters. Lastly, you have learned to predict the three-dim ensional structure o f m olecules using the valence shell electron pair repulsion (V SEPR ) m odel an d m olecular orbital (M O ) th e­ ory. A n ability to predict three-dim ensional structure is critical to understanding the p ro p ­ erties and reactivity o f m olecules. We encourage you to do all o f the problem s that your instructor has assigned. We also recom m end that you use the sum m ary and review tools in each chapter, such as the con­ cept m ap that follow s. C oncept m aps can help you see the flow o f concepts in a chapter and also help rem ind you o f key points. In fact, w e encourage you to build your ow n con­ cept m aps for review w hen the opportunity arises. W ork especially hard to solidify your know ledge from this and other early chapters in the book. T hese chapters have everything to do w ith helping you learn basic tools you need for success throughout organic chemistry.

K e y T e rm s a n d C o n c e p ts

T he key term s and concepts that are highlighted in b o ld , b lu e te x t w ithin the chapter are defined in the glossary (at the back o f the book) and have hyperlinked definitions in the accom panying W ileyP LU S course (w w w .w ileyplus.com ).

„ W ILEY

_

PLUS

Problems "¡5

PLUS

N ote to Instructors: Many of th e hom ew ork problem s are available for assignm ent via WileyPLUS, an online teaching an d learning solution.

ELECTRON C O N FIG U R A TIO N 1.26

W hich o f the follow ing ions possess the electron configuration o f a noble gas? (a) N a +

(c) F +

(e) C a 2+

(g) O2“

(b) C l“

(d) H“

(f) S 2“

(h) Br+

LEWIS STRUCTURES 1.27

W rite a Lew is structure for each o f the following: (a) S O C l2

1.28

(b) PO C l3

(c) PCl5

(d) HO NO 2 (HNO3)

G ive the form al charge (if one exists) on each atom o f the following: 'O ' II . . (a) C H 3— O— S — O: ..

.O.

:O : I (b) C H 3 s CH 3

'O ' II . . (c) :O — S — O:

'O '

..

.O.

(d) CH 3— S — O : .O.

48

Chapter 1

The Basics— Bonding and Molecular Structure

STRUCTURAL FORMULAS A N D ISOMERISM 1.2 9

W rite a condensed structural form ula for each com pound given here.

1.3 0

W hat is the m olecular form ula for each o f the com pounds given in E xercise 1.29?

1.31

C onsider each pair o f structural form ulas that follow and state w hether the tw o form ulas rep resen t the sam e co m ­ pound, w hether they represent different com pounds that are constitutional isom ers o f each other, or w hether they represent different com pounds that are n o t isom eric. ,Br

Cl

and

(a) C l' H (c) H— C— Cl I Cl

,Br

H I Cl— C— Cl I H

and

3 I 3 (e) CH 3— C— CH 2Cl

(b) Cl

and

(d) F

ClCH 2CH (CH 3)2

and

FCH2CH 2CH 2CH 2F

F

ch

Cl

and

(f) CH 2= C H C H 2CH 3

and

CH O (g)

and

X

and

(h) CH 3CH 2 CH2CH3

(i) CH 3O C H 2CH 3

and

O II C / \ h 2c —

Cl I (m ) H— C — Br ch

H and

Cl— C — Br

1

2

1

H

H CH

(j) C H 2ClCH ClCH 3

and CH 3CH C lCH 2Cl

(k) CH 3CH 2CH C lCH 2Cl

and C H 3C H C H 2Cl C H 2Cl

O Il (l) CH 3C C H 3

O and

II C /V h 2c — c h 2

(n) CH 3— C — H I H

and

H HH (o) H^ C - C ^ * H F

and

H I CH 3— C — CH 3 I H

H ^ ,C

C^

H F

F

H ¿H

\

h

and

C C F

:» h VH

49

P roblem s 1.32

Rew rite each o f the follow ing using bond-line form ulas: O (e) C H 2= C H C H 2C H 2C H = C H C H 3

(a) C H 3C H 2C H 2C C H 3 (b) C H 3C H C H 2C H 2C H C H 2C H 3 Ch3

O

Ch3

(c) (C H 3)3C C H 2C H 2C H 2OH

(f)

O (d)

c h 3c h 2c h c h

HC' II H C.

'C H 2 I 2 CH 'C ' H

2C o h

I ch

3

1.33

W rite structural form ulas o f your choice for all o f the constitutional isom ers w ith the m olecular form ula C 4H8.

1.34

W rite structural form ulas for at least three constitutional isom ers w ith the m olecular form ula CH 3N O2. (In answ er­ ing this question you should assign a form al charge to any atom that bears one.)

RESO N AN C E STRUCTURES 1.35

For the follow ing w rite all possible resonance structures. Be sure to include form al charges w here appropriate.

+

"O" (a)

(b)

(d)

(e)

o:

(f)

.O . •o 1

’’O '

(h) A

1.36

n-

W rite the resonance structure that w ould resu lt from m oving the electrons in the w ay indicated by the curved arrows.

"O ;

h 2n

1.37

Show the curved arrow s that w ould convert A into B. ^ 'N \ A

1.38

JO \

\

Y

r f

o:

B

(a) C yanic acid (H— O — C # N) and isocyanic acid (H — N = C = O) differ in the positions o f their electrons but their structures do n o t represent resonance structures. E xplain. (b) L oss o f a proton from cyanic acid yields the sam e anion as that obtained by loss o f a proton from isocyanic acid. Explain.

50

Chapter 1

The Basics— Bonding and Molecular Structure

1 .3 9

C onsider a chem ical species (either a m olecule or an ion) in w hich a carbon atom form s three single bonds to three hydrogen atom s and in w hich the carbon atom possesses no other valence electrons. (a ) W hat form al charge w ould the carbon atom have? (b) W hat total charge w ould the species have? (c) W hat shape w ould you expect this species to have? (d) W hat w ould you expect the hybridization state o f the carbon atom to be?

1.4 0

C onsider a chem ical species like the one in the previous problem in w hich a carbon atom form s three single bonds to three hydrogen atom s, but in w hich the carbon atom possesses an unshared electron pair. (a) W hat form al charge w ould the carbon atom have? (b) W hat total charge w ould the species have? (c) W hat shape w ould you expect this species to have? (d) W hat w ould you expect the hybridization state o f the carbon atom to be?

1.41

C onsider another chem ical species like the ones in the previous problem s in w hich a carbon atom form s three sin­ gle bonds to three hydrogen atom s but in w hich the carbon atom possesses a single unpaired electron. (a) W hat form al charge w ould the carbon atom have? (b) W hat total charge w ould the species have? (c) G iven that the shape o f this species is trigonal planar, w hat w ould you expect the hybridization state o f the carbon atom to be?

1.42

O zone (O 3) is found in the upper atm osphere w here it absorbs highly energetic ultraviolet (UV) radiation and thereby provides the surface o f E arth w ith a protective screen (cf. Section 10.11E). O ne possible resonance structure for ozone is the following: .O '

*O'

(a) A ssign any necessary form al charges to the atom s in this structure. (b) W rite another equivalent resonance structure for ozone. (c) W hat do these resonance structures predict about the relative lengths o f the tw o o x ygen-ox ygen bonds o f ozone? (d) In the structure above, and the one you have w ritten, assum e an angular shape for the ozone m olecule. Is this shape consistent w ith V S E P R theory? E xplain your answer. 1.43

W rite resonance structures for the azide ion, N3~. E xplain how these resonance structures account for the fact that both bonds o f the azide ion have the sam e length.

1.4 4

W rite structural form ulas o f the type indicated: (a) bond-line form ulas for seven constitutional isom ers w ith the form ula C 4H10O; (b) condensed structural form ulas for tw o constitutional isom ers w ith the form ula C 2H7N; (c) con­ densed structural form ulas for four constitutional isom ers w ith the form ula C3H9N; (d) bond-line form ulas for three constitutional isom ers w ith the form ula C5H12.

1.45

W hat is the relationship betw een the m em bers o f the follow ing pairs? T hat is, are they constitutional isom ers, the sam e, or som ething else (specify)? (a ) %

/

NH,

NH,

NH

( d ) £ NH 2

:O r

•O' CH 3CH 2CH 2CH (CH 3)2

(b)

(e)

O

C 'N H ,

(c)

NH2

NH2

.C l

Cl

"Cl

Cl

(f ) < (

CH

NH

>

Challenge Problems 1.46

In C hapter 15 w e shall learn how the nitronium ion, NO2+, form s w hen concentrated nitric and sulfuric acids are m ixed. (a) W rite a L ew is structure for the nitronium ion. (b) W hat geom etry does V S E P R theory predict for the NO2+ ion? (c) G ive a species that has the sam e num ber o f electrons as NO2+.

51

Learning G roup P roblem s 1.47

G iven the follow ing sets o f atom s, w rite bond-line form ulas for all o f the possible constitutionally isom eric com ­ pounds or ions that could be m ade from them . Show all unshared electron pairs and all form al charges, if any. S et

C a to m s

H ato m s

A B C D E

3 3 3 2 3

6 9 4 7 7

O th e r 2 1 1 1 1

Br atom s N atom and 1 O atom (not on sam e C) O atom N atom and 1 p ro to n extra electron

1.48

O pen com puter m olecular m odels for dim ethyl ether, dim ethylacetylene, an d cis-1,2-dichloro-1,2-difluoroethene from the 3D M olecular M odels section o f the b o o k ’s w ebsite. B y interpreting the com puter m olecular m odel for each one, draw (a) a dash form ula, (b) a bond-line form ula, and (c) a three-dim ensional dashed-w edge form ula. D raw the m odels in w hatever perspective is m ost convenient— generally the perspective in w hich the m ost atom s in the chain o f a m olecule can be in the plane o f the paper.

1.49

Boron is a group IIIA elem ent. O pen the m olecular m odel for boron trifluoride from the 3D M olecular M odels sec­ tion o f the b o o k ’s w ebsite. N ear the boron atom , above an d below the plane o f the atom s in BF3, are tw o relatively large lobes. C onsidering the position o f boron in the periodic table an d the three-dim ensional an d electronic struc­ ture o f BF3, w hat type o f orbital does this lobe represent? Is it a hybridized orbital or not?

1.50

T here are tw o contributing resonance structures for an anion called acetaldehyde enolate, w hose condensed m o le­ cular form ula is CH 2C H O ~. D raw the tw o resonance contributors and the resonance hybrid, then consider the m ap o f electrostatic potential (M EP) show n below for this anion. C om m ent on w hether the M E P is consistent or not w ith predom inance o f the resonance contributor you w ould have predicted to b e represented m ost strongly in the hybrid.

Learning Group Problems C onsider the com pound w ith the follow ing condensed m olecular form ula: CH 3C H O H C H = CH 2 W rite a full dash structural form ula for the com pound.

H e lp f u l H i n t

Your instructor will tell you how to work these problems as a Learning Group.

Show all nonbonding electron pairs on your dash structural form ula. Indicate any form al charges that m ay b e present in the m olecule. L abel the hybridization state at every carbon atom and the oxygen. D raw a three-dim ensional perspective representation for the com pound showing approxim ate bond angles as clearly as possible. U se ordinary lines to indicate bonds in the plane o f the paper, solid w edges for bonds in front o f the paper, and dashed w edges for bonds b ehind the paper. 6.

L abel all the bond angles in your three-dim ensional structure.

7.

D raw a bond-line form ula for the com pound.

8.

D evise tw o structures, each having tw o ip -h y b rid ized carbons and the m olecular form ula C4H6O. C reate one of these structures such that it is linear w ith resp ect to all carbon atom s. R epeat parts 1 -7 above for both structures.

52

Chapter 1

The Basics— Bonding and Molecular Structure

CO NCEPT M AP Organic M olecules , have

ca n be p re d ic te d by

VSEPR Theory

can be p re d ic te d by

T h r e e -d im e n s io n a l

(S e c tio n 1 .1 6 )

Quantum mechanics

sh a p e

(S e c tio n 1 .9 )

re q u ire s c re a tio n o f

Proper Lewis structures

m u s t be

(S e c tio n 1 .5 )

sh ow

Resonance structures (S e c tio n 1 .8 ) s h o w al

sh o w a

Formal charges

V a le n c e e le c tro n s

(S e c tio n 1 .7 ) in c lu d e a are a vera g e d in th e B o n d in g and n o n b o n d in g e le c tro n s

R e s o n a n ce h y b rid (S e c tio n 1 .8 )

re p e l e ach o th e r to a ch ie ve M a x im u m s e p a ra tio n In 3 -D s p a ce

m ay becom e

Alkynes

le a d s to

is p re s e n t

Linear geometry

t w o p o r b ita ls

each tr ip le -

[""n Bond

o f tw o g ro u p s * o f e le c tro n s

T w o sp h y b rid and

are a t

(S e c tio n 1 .1 4 )

i

bonded c a rb o n o f

in

p_°b"als /

H

1 £

Alkenes

o f th re e g ro u p s * o f e le c tro n s le a d s to

is

Trigonal planar geometry

O ve rla p

p re s e n t in

a nd o n e p o r b ita l

each d o u b le ­

I H

c

T h re e sp2 h y b rid

are at

(S e c tio n 1 .1 3 )

n c

bonded c a rb o n o f

.

H

I

sp2 Orbita

' sp2 Orbital pO r b i t a l i

H

1 SPOrbital

\

'

------------ x l ^ s p 2 Orbita

\ / / =\ Alkanes (S e c tio n 1 .1 2 ) H

o f fo u r g ro u p s * o f e le c tro n s le a d s to

s in g le ­ bonded

is p re s e n t

Tetrahedral geometry

in

H//

c a rb o n o f 7

H

H

I * A s in g le b o n d , a d o u b le b o n d , a tr ip le b o n d , and a n o n b o n d in g e le c tro n p a ir e ach re p re s e n t a s in g le ‘ g ro u p ' o f e le c tro n s .

are at each

F o u r sp3 h y b rid o r b ita ls

Families of Carbon Compounds Functional Groups, Intermolecular Forces, and Infrared (IR) Spectroscopy • •

• • 5 •

j j

+ * ^ j

:

*

J V

• 77

Ethanol

j

In this chapter we introduce one o f the great simplifying concepts o f

Benzaldehyde

organic chemistry—the functional group. Functional groups are common

/

and specific arrangements o f atoms that impart predictable reactivity and properties to a molecule. Even though there are millions o f organic compounds, you may be relieved to know that we can readily understand much about whole families of compounds simply by learning about the properties of the common functional groups. For example, all alcohols contain an — OH (hydroxyl) functional group attached to a saturated carbon bear­ ing nothing else but carbon or hydrogen. Alcohols as simple as ethanol in alcoholic beverages and as com plex as ethinyl estradiol (Section 2.1C) in birth control pills have this structural unit in common. All aldehydes have a — C(= O) — (carbonyl) group with one bond to a hydrogen and the other to one or more carbons, such as in benzaldehyde (from almonds). All ketones include a carbonyl group bonded by its carbon to one or more o ther carbons on each side, as in the natural oil m enthone, found in geraniums and spearmint. O H „OH E th a n o l

B e n z a ld e h y d e

53

54

Chapter 2

Families o f Carbon Compounds

Members o f each functional group fam ily share common chemical properties and reactivity, and this fact helps greatly in organizing our knowledge o f organic chemistry. As you progress in this chapter it will serve you well to learn the arrangements of atoms that define the common functional groups. This knowledge will be invalu­ able to your study o f organic chemistry. Toward the end o f this chapter we introduce an instrumental technique called infrared spectroscopy that provides physical evidence for the presence of particular functional groups. You will very likely make use of infrared spectroscopy in your organic laboratory work. Let us begin this chapter with families o f compounds that contain only carbon and hydrogen.

2.1 Hydrocarbons: Representative Alkanes, Alkenes, Alkynes, and A rom atic Compounds H ere w e introduce the class o f com pounds that contains only carbon and hydrogen, and w e shall see how the -ane, -ene, or -yne ending in a nam e tells us w hat kinds of carbon -carb o n bonds are present. •

H y d ro c a rb o n s are com pounds that contain only carbon and hydrogen atom s.

M ethane (CH 4) and ethane (C 2H6) are hydrocarbons, for exam ple. They also belong to a subgroup o f com pounds called alkanes. •

A lk an e s are hydrocarbons that do n o t have m ultiple bonds betw een carbon atom s, and w e can indicate this in the fam ily nam e and in nam es for specific com pounds by the -an e ending.

O ther hydrocarbons m ay contain double or triple bonds betw een their carbon atoms. •

A lk en es contain at least one carb o n -carb o n double bond, and this is indicated in the fam ily nam e and in nam es for specific com pounds by an -en e ending.

•

A lk y n es contain at least one carb o n -carb o n triple bond, and this is indicated in the fam ily nam e and in nam es for specific com pounds by a -yne ending.

•

A rom atic com pounds contain a special type o f ring, the m ost common example of which is a benzene ring. There is no special ending for the general family o f aromatic compounds.

We shall introduce representative exam ples o f each o f these classes o f hydrocarbons in the follow ing sections. G enerally speaking, com pounds such as the alkanes, w hose m olecules contain only sin­ gle bonds, are referred to as s a tu r a te d co m p o u n d s because these com pounds contain the m axim um n um ber o f hydrogen atom s that the carbon com pound can possess. Com pounds w ith m ultiple bonds, such as alkenes, alkynes, and arom atic hydrocarbons, are called u n s a ­ tu r a te d c o m p o u n d s because they possess few er than the m axim um num ber o f hydrogen atom s, and they are capable o f reacting w ith hydrogen under the proper conditions. We shall have m ore to say about this in C hapter 7. 2 .1 A

Alkanes

The prim ary sources of alkanes are natural gas and petroleum . The smaller alkanes (methane through butane) are gases under am bient conditions. M ethane is the principal com ponent of natural gas. H igher m olecular w eight alkanes are obtained largely by refining petroleum. M ethane, the sim plest alkane, was one m ajor com ponent o f the early atm osphere o f this planet. M ethane is still found in E arth’s atmosphere, but no longer in appreciable am ounts. It is, how ­ ever, a m ajor com ponent o f the atmospheres o f Jupiter, Saturn, Uranus, and Neptune. Som e living organism s produce m ethane from carbon dioxide and hydrogen. T hese very prim itive creatures, called methanogens, m ay b e E arth ’s oldest organism s, an d they m ay represent a separate form o f evolutionary developm ent. M ethanogens can survive only in an anaerobic (i.e., oxygen-free) environm ent. They have been found in ocean trenches, in m ud, in sewage, and in co w s’ stom achs.

2.1 Hydrocarbons: Representative Alkanes, Alkenes, Alkynes, and Aromatic Compounds

Alkenes

2 .1 B

E t h e n e a n d p r o p e n e , t h e t w o s i m p l e s t a lk e n e s , a r e a m o n g t h e m o s t i m p o r t a n t i n d u s t r i a l c h e m ­ i c a l s p r o d u c e d i n t h e U n i t e d S ta te s . E a c h y e a r , t h e c h e m i c a l i n d u s t r y p r o d u c e s m o r e t h a n 3 0 b i l l i o n p o u n d s o f e th e n e a n d a b o u t 1 5 b i l l i o n p o u n d s o f p r o p e n e . E t h e n e i s u s e d a s a s t a r t i n g m a t e r i a l f o r t h e s y n t h e s is o f m a n y i n d u s t r i a l c o m p o u n d s , i n c l u d i n g e t h a n o l , e t h y le n e o x id e , e t h a n a l, a n d t h e p o l y m e r p o l y e t h y l e n e ( S e c t i o n 1 0 . 1 0 ) . P r o p e n e is u s e d i n m a k i n g t h e p o l y ­ m e r p o ly p r o p y le n e ( S e c tio n

1 0 . 1 0 a n d S p e c ia l T o p ic B * ) , a n d , i n

a d d itio n to o th e r u ses,

p r o p e n e is t h e s t a r t i n g m a t e r i a l f o r a s y n t h e s is o f a c e t o n e a n d c u m e n e ( S e c t i o n 2 1 . 4 B ) . E t h e n e a ls o o c c u r s i n n a t u r e a s a p l a n t h o r m o n e . I t is p r o d u c e d n a t u r a l l y b y f r u i t s s u c h as t o m a t o e s a n d b a n a n a s a n d is i n v o l v e d i n t h e r i p e n i n g p r o c e s s o f t h e s e f r u i t s . M u c h u s e is n o w m a d e o f e th e n e i n t h e c o m m e r c i a l f r u i t i n d u s t r y t o b r i n g a b o u t t h e r i p e n i n g o f to m a t o e s a n d b a n a n a s p i c k e d g r e e n b e c a u s e t h e g r e e n f r u i t s a r e le s s s u s c e p t ib le t o d a m a g e d u r i n g s h ip p i n g . T h e r e a r e m a n y n a t u r a l l y o c c u r r i n g a lk e n e s . T w o e x a m p l e s a r e t h e f o l l o w i n g :

/3-Pinene (a com ponent of turpentine)

An aphid alarm pherom one Ethene ripens bananas.

S o lv e d P ro b le m 2 .1 P ro p e n e , C H

3C

H =

C H 2 , i s a n a lk e n e . W r i t e t h e s t r u c t u r e o f a c o n s t i t u t i o n a l i s o m e r o f p r o p e n e t h a t i s n o t a n a lk e n e .

( H in t: I t d o e s n o t h a v e a d o u b le b o n d .)

STRATEGY AND ANSW ER

A c o m p o u n d w it h a r in g o f n c a r b o n a to m s w i l l h a v e th e s a m e m o le c u la r f o r m u la as

a n a lk e n e w i t h t h e s a m e n u m b e r o f c a r b o n s . is a c o n s t it u t i o n a l

Cyclopropane has anesthetic properties.

is o m e r o f

C yclopropane C 3 H6

P ropene CaHe

Alkynes

2 .1 C

T h e s i m p l e s t a l k y n e i s e t h y n e ( a ls o c a l l e d a c e t y le n e ) . A l k y n e s o c c u r i n n a t u r e a n d c a n b e s y n t h e s iz e d i n t h e l a b o r a t o r y . T w o e x a m p l e s o f a lk y n e s a m o n g t h o u s a n d s t h a t h a v e a b i o s y n t h e t i c o r i g i n a r e c a p i l l i n , a n a n t if u n g a l a g e n t, a n d d a c t y ly n e , a m a r in e n a t u r a l p r o d u c t th a t is a n i n h i b i t o r o f p e n t o ­ b a r b it a l m e ta b o lis m . E t h i n y l e s t r a d io l is a s y n th e tic a lk y n e w h o s e e s t r o g e n - lik e p r o p e r tie s h a v e f o u n d u s e i n o r a l c o n t r a c e p tiv e s .

/

\

-C = C — C = C -

-C H a

Capillin [17a-ethynyM ,3,5(10)-estratriene-3,17/3 -diol] *Special Topics A-F and H are in WileyPLUS; Special Topic G can be found later in this volume.

56

Chapter 2

2 .1 D

Families o f Carbon Compounds

Benzene: A Representative Aromatic Hydrocarbon

In C hapter 14 w e shall study in detail a group o f unsaturated cyclic hydrocarbons know n as arom atic com pounds. T he com pound know n as b e n z e n e is the prototypical arom atic com pound. B enzene can b e w ritten as a six-m em bered ring w ith alternating single and dou­ ble bonds, called a K ek u lé s tr u c tu re after A ugust K ekulé (Section 1.3), w ho first conceived o f this representation:

H

H "C '

Benzene

II

or

„C ,

H

H H

K e k u lé s t r u c t u r e

B o n d - lin e r e p r e s e n ta tio n

fo r b e n z e n e

o f K e k u lé s t r u c t u r e

Even though the K ekule structure is frequently used for benzene com pounds, there is much evidence that this representation is inadequate and incorrect. For example, if benzene had alter­ nating single and double bonds as the K ekule structure indicates, w e w ould expect the lengths o f the carbon-carbon bonds around the ring to be alternately longer and shorter, as w e typi­ cally find w ith carbon-carbon single and double bonds (Fig. 1.31). In fact, the carb o n -car­ bon bonds o f benzene are all the same length (1.39 A), a value in betw een that o f a carbon-carbon single bond and a carbon-carbon double bond. There are two w ays o f deal­ ing w ith this problem : w ith resonance theory or w ith m olecular orbital theory. If w e use resonance theory, w e visualize benzene as being represented by either o f two equivalent K ekule structures:

T w o c o n t r i b u t i n g K e k u lé s t r u c t u r e s

A r e p r e s e n ta tio n o f th e

fo r b e n ze n e

r e s o n a n c e h y b r id

Based on the principles o f resonance theory (Section 1.8) w e recognize that benzene can­ not be represented adequately by either structure, but that, instead, it should be visualized as a h yb rid o f the two structures. We represent this hybrid by a hexagon w ith a circle in the m id ­ dle. Resonance theory, therefore, solves the problem w e encountered in understanding how all o f the carbon-carbon bonds are the sam e length. A ccording to resonance theory, the bonds are not alternating single and double bonds, they are a resonance hybrid o f the two: Any bond that is a single bond in the first contributor is a double bond in the second, and vice versa. •

A ll o f the carb o n -carb o n bonds in benzene are one and o ne-half bonds, have a bond length in betw een that o f a single b o n d an d a double bond, and have bond angles o f 120°.

In the m olecular orbital explanation, w hich w e shall describe in m uch m ore depth in Chapter 14, w e begin by recognizing that the carbon atom s o f the benzene ring are sp2 hybridized. Therefore, each carbon has a p orbital that has one lobe above the plane o f the ring and one lobe below, as shown here in the schem atic and calculated p orbital representations. H

H

/ C

Hl'09Â\ / H

^ "

120'

______ C1.39 Â C \ ^ H

120C 120°

H

2.2 Polar C o v alen t B onds

H

C a lc u la te d p o r b ita l s h a p e s in b e n z e n e

S c h e m a tic re p re s e n ta tio n o f b e n z e n e p o r b ita ls

C a lc u la te d b e n z e n e m o le c u la r o r b ita l r e s u ltin g fro m fa v o ra b le o v e rla p o f p o r b ita ls a b o ve and b e lo w p la n e o f b e n z e n e rin g

The lobes of each p orbital above and below the ring overlap with the lobes of p orbitals on the atoms to either side of it. This kind of overlap of p orbitals leads to a set of bonding molecular orbitals that encompass all of the carbon atoms of the ring, as shown in the calcu­ lated molecular orbital. Therefore, the six electrons associated with these p orbitals (one elec­ tron from each orbital) are delocalized about all six carbon atoms of the ring. This delocalization of electrons explains how all the carbon-carbon bonds are equivalent and have the same length. In Section 14.7B, when we study nuclear magnetic resonance spectroscopy, we shall present convincing physical evidence for this delocalization of the electrons.

Cyclobutadiene (below) is like benzene in that it has alternating single and double bonds in a ring. However, its bonds are not the same length, the double bonds being shorter than the single bonds; the molecule is rectangular, not square. Explain why it would be incor­ rect to write resonance structures as shown.

R eview P roblem 2.1

2.2 Polar Covalent Bonds In our discussion of chemical bonds in Section 1.4, we examined compounds such as LiF in which the bond is between two atoms with very large electronegativity differences. In instances like these, we said, a complete transfer of electrons occurs, giving the compound an ionic bond:

Lithium fluoride h as an ionic bond. We also described molecules in which electronegativity differences are not large, or in which they are the same, such as the carbon-carbon bond of ethane. Here the electrons are shared equally between the atoms. H H H — C :C— H H H E thane h as a covalent bond. The electro n s are sh a re d equally betw een the carbon atom s. Until now, we have not considered the possibility that the electrons of a covalent bond might be shared unequally.

Lithium fluoride crystal model.

58

Chapter 2

Families o f Carbon Compounds

•

If electronegativity differences exist betw een tw o bonded atom s, and they are not large, the electrons are n o t shared equally and a p o la r co v a le n t b o n d is the result.

•

Rem em ber: O ne definition o f e le c tro n e g a tiv ity is the a b ility o f an atom to attract electrons th a t i t is sharing in a covalent bond .

An exam ple o f such a polar covalent bond is the one in hydrogen chloride. The chlorine atom, w ith its greater electronegativity, pulls the bonding electrons closer to it. This m akes the hydrogen atom som ew hat electron deficient and gives it a p a rtia l positive charge (S + ). The chlorine atom becom es som ew hat electron rich and bears a p a rtia l negative charge (S—): 8+

S—

H : C l: B ecause the hydrogen chloride m olecule has a partially positive end and a partially n eg a­ tive end, it is a d ipo le, and it has a d ip o le m o m e n t. The direction o f polarity o f a polar bond can b e sym bolized by a vector quantity 1--------- >. The crossed end o f the arrow is the positive end and the arrowhead is the negative end: (positive end) 1--------- > (negative end) In HCl, for exam ple, w e w ould indicate the direction o f the dipole m om ent in the follow ­ ing way: H — Cl 1--------- : T he dipole m om ent is a physical property that can be m easured experim entally. It is defined as the product o f the m agnitude o f the charge in electrostatic units (esu) and the distance that separates them in centim eters (cm): Dipole moment = charge (in esu) X distance (in cm) X = e X d

T he charges are typically on the order o f 10—10 esu and the distances are on the order o f 10—8 cm . D ipole m om ents, therefore, are typically on the order o f 10—18 esu cm . For convenience, this unit, 1 X 10—18 esu cm , is defined as one d eb y e and is abbreviated D. (The unit is nam ed after P eter J. W. D ebye, a chem ist born in the N etherlands, w ho taught at Cornell U niversity from 1936 to 1966. D ebye w on the N obel P rize in C hem istry in 1936.) In SI units 1 D = 3.336 X 10—30 coulom b m eter (C ■m). If necessary, the length o f the arrow can b e used to indicate the m agnitude o f the dipole m om ent. D ipole m om ents, as w e shall see in Section 2.3, are very useful quantities in accounting for physical properties o f com pounds. S o lv e d P ro b le m 2 .2

U sing a dipole m om ent arrow as show n above an d the table o f electronegativities (Table 1.2), indicate the direc­ tion o f the dipole m om ent o f lithium hydride (LiH), a covalent com pound. A lso place S + an d S— sym bols near the Li and H as appropriate. STRATEGY A ND ANSW ER In Table 1.2 w e find that lithium (a m etal) has a very low electronegativity o f 1.0. H ydrogen (a nonm etal) has a larger electronegativity o f 2.1. T he hydrogen atom w ill pull the electrons it is shar­ ing w ith lithium in its direction. T he bond betw een lithium and hydrogen w ill b e polar w ith the lithium at the p o s­ itive end and the hydrogen at the negative end. Positive end

N egative end

Li — H U sing S + and S— sym bols w e can show the polarity o f LiF as follows: S+Li — FS—

R e v ie w P ro b le m 2 .2

W rite S + and S — by the appropriate atom s and draw a dipole m om ent vector for any of the follow ing m olecules that are polar: (a) HF

(b) IBr

(c) Br2

(d) F2

59

2.2 Polar Covalent Bonds

P o la r co v a le n t b o n d s strongly influence the physical properties and reactivity o f m ol­ ecules. In m any cases, these polar covalent bonds are part o f fu n c tio n a l g ro u p s , w hich we shall study shortly (Sections 2.5-2.13). Functional groups are defined groups o f atom s in a m olecule that give rise to the function (reactivity or physical properties) o f the m olecule. F unctional groups often contain atom s having different electronegativity values and unshared electron pairs. (Atom s such as oxygen, nitrogen, and sulfur that form covalent bonds and have unshared electron pairs are called h e te ro a to m s .)

2.2A Maps of Electrostatic Potential O ne w ay to visualize the distribution o f charge in a m olecule is w ith a m a p o f e le c tro s ta ­ tic p o te n tia l (M E P ). R egions o f an electron density surface that are m ore negative than others in an M E P are colored red. These regions w ould attract a positively charged species (or repel a negative charge). Regions in the M E P that are less negative (or are positive) are blue. Blue regions are likely to attract electrons from another m olecule. The spectrum of colors from red to blue indicates the trend in charge from m ost negative to least negative (or m ost positive). Figure 2.1 shows a m ap o f electrostatic potential for the low -electron-density surface of hydrogen chloride. We can see clearly that negative charge is concentrated near the chlo­ rine atom and that positive charge is localized near the hydrogen atom , as w e predict based on the difference in their electronegativity values. Furtherm ore, because this M E P is p lo t­ ted at the low -electron-density surface o f the m olecule (the van der W aals surface, Section 2.13B), it also gives an indication o f the m olecu le’s overall shape.

e

Figure 2.1 A calculated map of electrostatic potential for hydrogen chloride showing regions of relatively more negative charge in red and more positive charge in blue. Negative charge is clearly localized near the chlorine, resulting in a strong dipole moment for the molecule.

THE CHEMISTRY OF . . . C a lc u la te d M o le c u la r M o d e ls : M a p s o f E le c tr o s ta tic P o te n tia l

A m ap of electrostatic potential is prep ared by carrying out a quantum m echanical calculation th a t involves moving an imaginary positive point charge at a fixed distance over a given electron density surface of a molecule. As this is done, th e varying potential energy in th e attraction b etw een the electron cloud and th e imaginary positive charge is p lo tted in color-coded fashion. Red in th e MEP indicates strong attraction betw een the electron density surface at th a t loca­ tion and th e probing positive charge— in o th er words, g rea ter negative charge at th a t part of th e surface. Blue regions in th e m ap indicate w eaker attraction b etw een the surface and th e positive charge probe. The overall distrib­ ution of charge is indicated by th e trend from blue (most positive or least negative) to green or yellow (neutral) to red (most negative). M ost often we plot th e MEP at th e van d er

Waals surface of a m olecule since th a t rep resen ts approxi­ m ately th e furthest ex ten t of a m olecule's electron cloud and therefore its overall shape. The m olecular m odel in this box is for dimethyl ether. The MEP shows th e concentration of negative charge w here th e unshared electron pairs are located on th e oxygen atom . It is im portant to n ote th a t when directly com paring the MEP for one m olecule to th a t of another, th e color schem e used to rep resen t th e charge scale in each m odel m ust be th e sam e. W hen we make direct com parisons b etw een mol­ ecules, we will plot their MEPs on th e sam e scale. W e will find th at such com parisons are especially useful becau se they allow us to com pare th e electron distribution in one m olecule to th at in an o th er and predict how one m olecule m ight interact with th e electrons of an o th er molecule.

M o s t p o s itiv e ( L e a s t n e g a tiv e )

"

H e lp f u l H i n t

Electron density surfaces and electrostatic potential maps. M o s t n e g a tiv e

D im e t h y l e t h e r

60

Chapter 2

Families o f Carbon Compounds

2.3 Polar and N o n po lar Molecules In the discussion o f dipole m om ents in the previous section, our attention w as restricted to sim ple diatom ic m olecules. A ny diatom ic m olecule in w hich the tw o atom s are different (and thus have different electronegativities) will, o f necessity, have a dipole m om ent. In general, a m olecule w ith a dipole m om ent is a p o la r m o le cu le. If w e exam ine Table 2.1, however, w e find that a n um ber o f m olecules (e.g., CCI 4 , C O 2) consist o f m ore than two atom s, have p o la r bonds, but have no dipole moment. W ith our know ledge o f the shapes o f m olecules (Sections 1 .12-1.16) w e can understand how this can occur. D ipole M om ents o f Some Sim ple Molecules Form ula h2

Cl 2 HF HCl HBr HI BF 3 CO2

Form ula

m (D)

0 0

ch4

1.83 1.08 0.80 0.42 0 0

C H 2 Cl 2

0 1.87 1.55 1.02 0 1.47 0.24 1.85

m (D)

C H 3Cl CHCl 3 CCl 4 NH3 NF 3 h 2o

C onsider a m olecule o f carbon tetrachloride (CCI4). B ecause the electronegativity of chlorine is greater than that o f carbon, each o f the carb o n -ch lo rin e bonds in CCI 4 is polar. E ach chlorine atom has a partial negative charge, and the carbon atom is considerably positive. B ecause a m olecule o f carbon tetrachloride is tetrahedral (Fig. 2.2), however, the center o f p ositive charge and the center o f negative charge coincide, and the m olecule has no net dipole moment.

Figure 2.2 Charge distribution in carbon tetrachloride. The molecule has no net dipole moment.

This result can b e illustrated in a slightly different way: If w e use arrows ( 1 ------>) to rep­ resent the direction o f polarity o f each bond, we get the arrangem ent o f bond m om ents shown in Fig. 2.3. Since the bond m om ents are vectors o f equal m agnitude arranged tetrahedrally, their effects cancel. Their vector sum is zero. The m olecule has no net dipole moment.

2.3 Polar and Nonpolar Molecules

T h e c h lo r o m e th a n e m o le c u le ( C H

3C l )

h a s a n e t d i p o l e m o m e n t o f 1 .8 7 D . S i n c e c a r b o n

a n d h y d r o g e n h a v e e l e c t r o n e g a t i v i t i e s ( T a b le 1 . 2 ) t h a t a r e n e a r l y t h e s a m e , t h e c o n t r i b u t i o n o f t h r e e C — H b o n d s t o t h e n e t d i p o l e is n e g l i g i b l e . T h e e l e c t r o n e g a t i v i t y d if f e r e n c e b e t w e e n c a r b o n a n d c h lo r in e is la r g e , h o w e v e r , a n d th e h ig h l y p o la r C — C l b o n d a c c o u n ts f o r m o s t o f th e d ip o le m o m e n t o f C ^ C l ( F ig . 2 .4 ) .

H

Figure 2.4 (a) The dipole

H

moment of chloromethane arises mainly from the highly polar carbon-chlorine bond. (b) A map of electrostatic potential illustrates the polarity of chloromethane.

H

p = 1 .8 7 D

(b)

(a )

S o lv e d P ro b le m 2 .3 A lth o u g h

m o le c u le s

o f C O

2

h a v e p o la r b o n d s

(o x y g e n

is

m o re

e le c t r o n e g a t i v e

th a n

c a rb o n ), c a rb o n

d io x id e

( T a b le 2 . 1 ) h a s n o d i p o l e m o m e n t . W h a t c a n y o u c o n c l u d e a b o u t t h e g e o m e t r y o f a c a r b o n d i o x i d e m o l e c u l e ?

STRATEGY AND ANSW ER

For a C O

2

m o le c u le

to

have

a z e ro

d ip o le

m o m e n t, th e b o n d

m o m e n ts o f th e t w o c a r b o n - o x y g e n b o n d s m u s t c a n c e l e a c h o th e r. T h is c a n h a p p e n o n ly i f m o l­ e c u le s o f c a r b o n d i o x i d e a r e l in e a r .

B o r o n tr iflu o r id e

(B F

3)

T e t r a c h lo r o e t h e n e ( C C l 2 = b a s is o f t h e s h a p e o f C C l 2 =

3

c

m=

h a s n o d ip o le m o m e n t (m =

c o n f ir m s th e g e o m e tr y o f B F

:o =

0 D ) . E x p la in h o w th is o b s e r v a tio n

= 0

o

:

D

R e v ie w P ro b le m 2 .3

p r e d ic te d b y V S E P R th e o ry .

C C l2) d o e s n o t h a v e a d ip o le m o m e n t. E x p la in th is fa c t o n th e

R e v ie w P ro b le m 2 .4

C C l2 .

S u l f u r d i o x i d e ( S O 2 ) h a s a d i p o l e m o m e n t (m =

1 .6 3 D ) ; o n t h e o t h e r h a n d , c a r b o n d i o x ­

i d e ( s e e S o lv e d P r o b l e m 2 . 3 ) h a s n o d i p o l e m o m e n t ( m =

0 D ) . W h a t d o th e s e fa c ts i n d i ­

c a te a b o u t th e g e o m e tr y o f s u lf u r d io x id e ?

U n s h a r e d p a i r s o f e le c t r o n s m a k e l a r g e c o n t r i b u t i o n s t o t h e d ip o l e m o m e n t s o f w a t e r a n d a m m o n ia . B e c a u s e a n u n s h a r e d p a ir h a s n o o th e r a to m a tta c h e d to i t to p a r t ia lly n e u t r a liz e i t s n e g a t iv e c h a r g e , a n u n s h a r e d e le c t r o n p a i r c o n t r ib u t e s a l a r g e m o m e n t d i r e c t e d a w a y f r o m t h e c e n t r a l a t o m ( F i g . 2 . 5 ) . ( T h e O — H a n d N — H m o m e n t s a r e a ls o a p p r e c i a b l e . )

Figure 2.5 Bond moments and the resulting dipole moments of water and ammonia.

R e v ie w P ro b le m 2 .5

62

Chapter 2

Families o f Carbon Compounds

R e v ie w P ro b le m 2 .6

U sing a three-dim ensional form ula, show the direction o f the dipole m om ent o f CH 3OH. W rite S + and S — signs n ex t to the appropriate atoms.

R e v ie w P ro b le m 2 .7

T richlorom ethane (CH Cl3, also called chloroform ) has a larger dipole m om ent than C FC l3. U se three-dim ensional structures and bond m om ents to explain this fact.

2 .3 A

Dipole Moments in Alkenes

C is-tra n s isom ers o f alkenes (Section 1.13B) have different physical properties. T hey have different m elting points and boiling points, and often cis-tra n s isom ers differ m arkedly in the m agnitude o f their dipole m om ents. Table 2.2 sum m arizes som e o f the physical p ro p ­ erties o f tw o pairs o f cis-tra n s isom ers. Physical Properties o f Some Cis-Trans Isomers C om p o u n d c/s-1,2-D ichloroethene trans-1,2-D ichloroethene c/s-1,2-D ibrom oethene trans-1,2-D ibrom oethene

M elting P oint (°C)

Boiling P oint (°C)

-8 0 -5 0 -5 3

60 48 112.5 108

-6

D ipole M o m en t (D) 1.90 0

1.35 0

S o lv e d P ro b le m 2 .4

E xplain w hy cis-1,2-dichloroethene (Table 2.2) has a large dipole m om ent w hereas trans-1,2-dichloroethene has a dipole m om ent equal to zero. STRATEGY AND ANSW ER If w e exam ine the n et dipole m om ents (show n in red) for the bond m om ents (black), w e see that in trans-1,2-dichloroethene the bond m om ents cancel each other, w hereas in cis-1,2-dichloroethene they augm ent each other. H Bond moments (black) H are in same general \ / direction. Resultant ,C ^ ~ C dipoIe moment (red) Cl Cl isIarge.

H

Cl C= C

Bond moments cancel each other. Net dipole is zero.

H

Cl

Resultant moment i=>

R e v ie w P ro b le m 2 .8

c is - 1 , 2 - D i c h l o r o e t h e n e

tr a n s - 1 , 2 - D ic h lo r o e th e n e

m = 1 .9 D

m = 0 D

Indicate the direction o f the im portant b o n d m om ents in each o f the follow ing com pounds (neglect C— H bonds). You should also give the direction o f the n et dipole m om ent for the m olecule. If there is no n et dipole m om ent, state that m = 0 D. (a) cis-C H F = C H F

R e v ie w P ro b le m 2 .9

(b) trans-C H F = C H F

(c) CH 2= C F 2

(d) C F 2= C F 2

W rite structural form ulas for all o f the alkenes w ith (a) the form ula C2H2Br2 an d (b) the form ula C 2Br2CI2. In each instance designate com pounds that are cis-tra n s isom ers o f each other. P redict the dipole m om ent o f each one.

2.4 Functional Groups •

F u n c tio n a l g ro u p s are com m on and specific arrangem ents o f atom s that im part predictable reactivity an d properties to a m olecule.

63

2.4 Functional Groups

The functional group o f an alkene, for exam ple, is its c a rb o n -c arb o n double bond. W hen w e study the reactions o f alkenes in g reater detail in C h ap ter 8, w e shall find th at m ost o f the chem ical reactions o f alkenes are the chem ical reactions o f the carb o n -c arb o n d o u ­ ble bond. The functional group of an alkyne is its carbon-carbon triple bond. A lkanes do not have a functional group. Their m olecules have carbon-carbon single bonds and carbon-hydrogen bonds, but these bonds are present in m olecules o f alm ost all organic com pounds, and C — C and C — H bonds are, in general, m uch less reactive than com m on functional groups. We shall introduce other com m on functional groups and their properties in Sections 2.5-2.11. Table 2.3 (Section 2.12) sum m arizes the m ost im portant functional groups. First, however, let us introduce som e com m on alkyl groups, w hich are specific groups o f carbon and hydro­ gen atoms that are not part of functional groups.

2 .4 A

Alkyl Groups and the Symbol R

A lkyl g ro u p s are the groups that w e identify for purposes o f nam ing com pounds. They are groups that w ould be obtained by rem oving a hydrogen atom from an alkane: ALKANE

ALKYL GROUP

CH 3— H

CH 3 -

M e th a n e

M e th y l

CH 3CH 2— H

c h 3c h

2-

ABBREVIATION

BOND-LINE

Me-

H

Et-

\ X

E th y l

E th a n e

C H 3CH 2CH 2— H

c h 3c h 2c h

P ro p a n e

2

Pr-

P ro p y l

C H 3C H 2CH 2CH 2— H

c h 3c h 2c h 2c h

B u ta n e

2—

Bu-

B u ty l

W hile only one alkyl group can be derived from m ethane or ethane (the m e th y l and eth y l groups, respectively), two groups can be derived from propane. Rem oval o f a hydrogen from one of the end carbon atom s gives a group that is called the p ro p y l group; rem oval o f a hydrogen from the m iddle carbon atom gives a group that is called the iso p ro p y l group. T he nam es and structures of these groups are used so frequently in organic chem istry that you should learn them now. See Section 4.3C for nam es and structures o f branched alkyl groups derived from butane and other hydrocarbons. We can sim plify m uch o f our future discussion if, at this point, w e introduce a sym bol that is w idely used in designating general structures o f organic m olecules: the sym bol R . R is used as a g e n e ra l sym bol to represent any a lk y l g ro u p . For exam ple, R m ight be a m ethyl group, an ethyl group, a propyl group, or an isopropyl group: C H 3-

c h 3c h 2— c h 3c h 2c h 2c h 3c h c h 3

Methyl Ethyl Propyl Isopropyl

Thus, the general form ula for an alkane is R — H.

T hese and o th e rs can be d e s ig n a t e d b y R.

MODEL J j

64

Chapter 2

2 .4 B

Families o f Carbon Compounds

Phenyl and Benzyl Groups

W hen a benzene ring is attached to som e other group o f atom s in a m olecule, it is called a p h e n y l g ro u p , and it is represented in several ways:

or

$ —

or

or

Ar —

C 6H 5 —

or

Ph—

or

( if r in g s u b s t it u e n t s a r e p r e s e n t )

W a y s o f r e p r e s e n tin g a p h e n y l g r o u p

The com bination o f a phenyl group and a m e th y le n e g ro u p (— CH 2— ) is called a ben z y l g r o u p :

J

y

CH, or

C 6H 5C H 2 —

or

or

Bn—

W a y s o f r e p r e s e n tin g a b e n z y l g r o u p

J 2.5 A lkyl Halides or Haloalkanes

A lkyl halides are com pounds in w hich a halogen atom (fluorine, chlorine, brom ine, or iodine) replaces a hydrogen atom o f an alkane. For exam ple, C H 3Cl and CH 3C H 2Br are alkyl halides. A lkyl halides are also called h a lo a lk a n e s. T he generic form ula for an alkyl halide is R — X : w here X = fluorine, chlorine, brom ine, or iodine. A lkyl halides are classified as being prim ary (1°), secondary (2°), or tertiary (3°). This cla ssifica tio n is based on the carbon atom to w h ic h the halogen is d ire c tly attached. If the carbon atom that bears the halogen is attached to only one other carbon, the carbon atom is said to be a p r im a r y c a r b o n a to m an d the alkyl halide is classified as a p r im a ry alk y l h alid e. If the carbon that bears the halogen is itself attached to tw o other carbon atom s, then the carbon is a se c o n d a ry c a r b o n an d the alkyl halide is a se c o n d a ry alk y l h alid e. If the carbon that bears the halogen is attached to three other carbon atom s, then the car­ bon is a te r tia r y c a r b o n an d the alkyl halide is a te r tia r y a lk y l h alid e. E xam ples o f p ri­ mary, secondary, and tertiary alkyl halides are the following:

2 - C h lo r o p r o p a n e

H e lp f u l H i n t

Although we use the symbols 1°, 2°, 3°, we do not say first degree, second degree, and third degree; we say primary, secondary, and tertiary. 1° C a rb o n

2 ° C a rb o n

H H H / H — C — C — Cl

or

.C l

A 1 ° a lk y l c h lo r id e

Cl

3 ° C a rb o n

CH 3

h

-C — C — C — H H

H H

H J

H

ch

or

Cl

A 2 ° a lk y l c h lo r id e

3

C — Cl

or

CH 3 A 3 ° a lk y l c h lo r id e

Cl

2.6 Alcohols

S o lv e d P ro b le m 2 .5

W rite the structure of an alkane w ith the form ula C 5 H -12 that has no secondary or tertiary carbon atoms. H in t: The com pound has a quaternary (4°) carbon. STRATEGY AND ANSW ER Follow ing the pattern o f designations for carbon atom s given above, a 4° carbon atom m ust be one that is attached to four other carbon atom s. If w e start w ith this carbon atom , and then add four carbon atom s w ith their attached hydrogens, there is only one possible alkane. The other four carbons are all prim ary carbons; none is secondary or tertiary.

4 ° C a r b o n a to m

CH 3 or

CH3

C— c h 3 ch3

W rite bond-line structural form ulas for (a) tw o constitutionally isom eric prim ary alkyl b ro­ m ides w ith the form ula C 4 H 9 Br, (b) a secondary alkyl brom ide, and (c) a tertiary alkyl brom ide w ith the sam e form ula. B uild handheld m olecular m odels for each structure and exam ine the differences in their connectivity.

R e v ie w P ro b le m 2 .1 0

A lthough w e shall discuss the nam ing o f organic com pounds later w hen w e discuss the indi­ vidual fam ilies in detail, one m ethod o f nam ing alkyl halides is so straightforw ard that it is w orth describing here. We sim ply nam e the alkyl group attached to the halogen and add the w ord flu o rid e , chloride, bromide, or iodide. W rite form ulas for (a) ethyl fluoride and (b) isopropyl chloride.

R e v ie w P ro b le m 2.11

W hat are the nam es for (c)

B r, (d)

F

, and (e) C 6 H 5 I?

2.6 Alcohols M ethyl alcohol (also called m ethanol) has the structural form ula C H 3OH and is the sim ­ p lest m em ber o f a fam ily o f organic com pounds know n as alco h o ls. T he ch aracteristic functional group o f this fam ily is the hydroxyl ( — OH) group attached to an sp 3h ybridized carbon atom . A nother exam ple o f an alcohol is ethyl alcohol, C H 3C H 2OH (also called ethanol).

— C — O — H

E th a n o l

T h is is th e f u n c t io n a l g r o u p o f a n a lc o h o l.

A lcohols m ay be view ed structurally in tw o ways: (1) as hydroxyl derivatives o f alkanes and (2) as alkyl derivatives of water. E thyl alcohol, for exam ple, can be seen as an ethane m olecule in w hich one hydrogen has been replaced by a hydroxyl group or as a w ater m ol­ ecule in w hich one hydrogen has been replaced by an ethyl group: E th y l g r o u p

CH 3 CH 3

' CH 3 C 1V 3 \ 2 109.5o£ o ' ) h

y

H y d ro x y l

H \ 104.5o£ o; h7

g ro u p E th a n e

E th y l a lc o h o l ( e th a n o l)

W a te r

66

Chapter 2

Families o f Carbon Compounds

A s w ith alkyl halides, alcohols are classified into three groups: prim ary (1°), secondary (2°), and tertiary (3°) alcohols. T his cla ssifica tio n is based on the degree o f sub stitu tio n o f the carbon to w h ic h the h y d ro x y l g ro u p is d ire c tly attached. If the carbon has only one other carbon attached to it, the carbon is said to b e a p r im a r y c a r b o n and the alcohol is a p r im a ry alcohol:

-1 ° C a rb o n H H I I* H— C — C — O — H or I I H H

OH

„OH

E th y l a lc o h o l (a

Geraniol is a major component of the oil of roses.

1°

G e r a n io l

a lc o h o l)

(a

1°

B e n z y l a lc o h o l

a lc o h o l)

(a

1°

a lc o h o l)

If the carbon atom that bears the hydroxyl group also has tw o other carbon atom s attached to it, this carbon is called a secondary carbon, and the alcohol is a secondary alcohol: ^ -2 °

H

C a rb o n

H /H

H— C— C— C— H

or

OH

:OH H

O

H

H M e n th o l

I s o p r o p y l a lc o h o l (a

2°

a lc o h o l)

(a

2°

a lc o h o l fo u n d

in p e p p e r m i n t o i l )

If the carbon atom that bears the hydroxyl group h as three other carbons attached to it, this carbon is called a tertiary carbon, and the alcohol is a tertiary alcohol: H H e lp f u l H i n t

Practice with handheld molecular models by building models of as many of the compounds on this page as you can.

=OH

H— C — H 3° C a rb o n

H H

/H

- C — C — C-

H

or

OH

H :O : H

(^orethindr

fe c f- B u ty l a lc o h o l

N o r e th in d r o n e

( a 3 ° a lc o h o l)

( a n o r a l c o n t r a c e p t i v e t h a t c o n t a i n s a 3 ° a lc o h o l g r o u p a s w e ll a s a k e to n e g r o u p a n d c a r b o n ­ c a r b o n d o u b le a n d t r ip le b o n d s )

Norethindrone birth control pills.

R e v ie w P ro b le m 2 .1 2

W rite bond-line structural form ulas for (a) tw o prim ary alcohols, (b) a secondary alcohol, and (c) a tertiary alcohol— all having the m olecular form ula C 4H 10O.

R e v ie w P ro b le m 2 .1 3

O ne w ay o f nam ing alcohols is to nam e the alkyl group that is attached to the — OH and add the w ord alcohol. W rite bond-line form ulas for (a) propyl alcohol and (b) isopropyl alcohol.

67

2.7 Ethers

2 .7 Ethers E thers have the general form ula R — O — R or R — O — R ', w here R ' m ay be an alkyl (or phenyl) group different from R. Ethers can be thought o f as derivatives o f w ater in w hich both hydrogen atom s have been replaced by alkyl groups. The bond angle at the oxygen atom of an ether is only slightly larger than that o f water: R

R'

\ :

o

R

or

/

C H

\

R

110°

3

A 3.. O

V

/

CH3 D im e th y l e th e r

G e n e r a l f o r m u la f o r a n e th e r

HaC — C H

-C — O — C -

(a t y p ic a l e th e r )

2

Y

T h e fu n c tio n a l g r o u p

E th y le n e

T e tra h y d ro fu ra n

o x id e

(T H F )

o f a n e th e r T w o c y c lic e th e r s

R e v ie w P ro b le m 2 .1 4

O ne w ay of nam ing ethers is to nam e the tw o alkyl groups attached to the oxygen atom in alphabetical order and add the w ord ether. If the tw o alkyl groups are the same, w e use the prefix di-, for exam ple, as in dim ethyl ether. W rite bond-line structural form ulas for (a) diethyl ether, (b) ethyl propyl ether, and (c) ethyl isopropyl ether. W hat nam e w ould

you give to

(d)

OMe

O

and

(f) CH 3O C 6H5?

THE CHEMISTRY OF . . .

CN it r o u s o x i d e ( N

(e)

E th e rs a s G e n e ra l A n e s th e tic s 2O ) ,

a s a n a n e s t h e t i c in

a ls o c a ll e d l a u g h in g g a s , w a s f i r s t u s e d 1799, and

i t is s t ill in u s e t o d a y , e v e n

s titu e n ts ,

have

r e p la c e d

d ie th y l

e th e r

as

a n e s th e tic s

of

c h o ic e . O n e r e a s o n : u n lik e d ie t h y l e th e r , w h i c h is h ig h l y f l a m ­

t h o u g h w h e n u s e d a lo n e i t d o e s n o t p r o d u c e d e e p a n e s t h e ­

m a b le , th e

s ia . T h e f i r s t u s e o f a n e t h e r , d ie t h y l e t h e r , t o p r o d u c e d e e p

e t h e r s t h a t a r e c u r r e n t l y u s e d f o r i n h a l a t io n a n e s t h e s ia a r e

h a lo g e n a te d

e th e r s a re n o t. T w o h a lo g e n a te d

a n e s t h e s ia o c c u r r e d in 1 8 4 2 . In t h e y e a r s t h a t h a v e p a s s e d

s e v o f lu r a n e a n d d e s f lu r a n e .

s in c e t h e n , s e v e r a l d i f f e r e n t e t h e r s , u s u a lly w it h h a l o g e n s u b ­

C F3 f 3c D ie t h y l e t h e r

"a

'F

f 3c

D e s f lu r a n e

"a S e v o flu r a n e

•

• 9

•

j

•

f

68

Chapter 2

Families o f Carbon Compounds

2.8 Am ines J u s t a s a l c o h o l s a n d e t h e r s m a y b e c o n s i d e r e d a s o r g a n i c d e r i v a t i v e s o f w a t e r , a m in e s m a y b e c o n s id e r e d as o r g a n ic d e r iv a tiv e s o f a m m o n ia :

H— N— H

R— N— H

NH H2N

H

H

A m m o n ia

A n a m in e

A m p h e ta m in e

P u tr e s c in e

(a d a n g e r o u s s t im u la n t )

( f o u n d in d e c a y i n g m e a t )

A m i n e s a r e c l a s s i f i e d a s p r i m a r y , s e c o n d a r y , o r t e r t i a r y a m in e s . T h is classificatio n is b a s e d o n the n u m b e r o f o rg a n ic groups th a t are attached to the n itro g e n atom:

R— N— H

R— N— H

H

N o t ic e

R— N— R"

R'

R'

A p r i m a r y (1 °)

A s e c o n d a r y (2 °)

A t e r t ia r y (3 °)

a m in e

a m in e

a m in e

th a t th is is q u ite d if f e r e n t f r o m

t h e w a y a lc o h o ls a n d a l k y l h a l i d e s a r e c l a s s if ie d .

I s o p r o p y l a m in e , f o r e x a m p le , is a p r i m a r y a m in e e v e n t h o u g h i t s — N H

2

g r o u p is a tta c h e d to

a s e c o n d a r y c a r b o n a t o m . I t is a p r i m a r y a m in e b e c a u s e o n l y o n e o r g a n i c g r o u p is a tta c h e d to th e n it r o g e n a to m : H

H

H

H — C — C— C — H H

or

:NH„

H

N

:N H ,

H Is o p r o p y la m in e (a

A m p h e t a m in e

( b e lo w ) ,

a

p o w e rfu l

1°

a m in e ) and

d a n g e ro u s

P ip e r id in e (a c y c l ic

2°

a m in e )

s t im u la n t ,

is

a

p r im a r y

a m in e .

D o p a m in e , a n i m p o r t a n t n e u r o t r a n s m it t e r w h o s e d e p le t io n is a s s o c ia te d w i t h P a r k in s o n ’ s d is e a s e , i s a ls o a p r i m a r y a m in e . N i c o t i n e , a t o x i c c o m p o u n d f o u n d i n t o b a c c o t h a t m a k e s s m o k in g a d d ic t iv e , h a s a s e c o n d a r y a m in e g r o u p a n d a t e r t ia r y o n e . HO

n h

2

2 HO A m p h e ta m in e

D o p a m in e

A m in e s a re l ik e a m m o n ia ( S e c tio n 1 .1 6 B ) in h a v in g a t r ig o n a l p y r a m id a l s h a p e . T h e C — N — C b o n d a n g le s o f t r im e t h y la m in e a re 1 0 8 . 7 ° , a v a lu e v e r y c lo s e to th e H — C — H b o n d a n g le s o f m e t h a n e . T h u s , f o r a l l p r a c t i c a l p u r p o s e s , t h e n i t r o g e n a t o m o f a n a m in e c a n b e c o n s i d e r e d t o b e sp3 h y b r i d i z e d w i t h t h e u n s h a r e d e le c t r o n p a i r o c c u p y i n g o n e o r b i t a l (s e e b e lo w ) . T h is m e a n s th a t th e u n s h a r e d p a ir is r e la t iv e ly e x p o s e d , a n d a s w e s h a ll see t h i s i s i m p o r t a n t b e c a u s e i t i s i n v o l v e d i n a l m o s t a l l o f t h e r e a c t io n s o f a m in e s .

2.9 Aldehydes and Ketones

O ne w ay o f nam ing am ines is to nam e in alphabetical order the alkyl groups attached to the nitrogen atom , using the prefixes d i- and tr i- if the groups are the same. A n exam ple is isopropylam ine w hose form ula is show n above. W hat are nam es for (a), (b), (c), and (d)? B uild handheld m olecular m odels for the com pounds in parts (a)-(d).

R e v ie w P ro b le m 2 .1 5

N M e2 N

(b)

'IN I H

'M e (d)

(c)

3

W rite bond-line form ulas for (e) propylam ine, (f) trim ethylam ine, and (g) ethylisopropylm ethylam ine. W hich am ines in Review Problem 2.15 are (a) prim ary am ines, (b) secondary am ines, and (c) tertiary am ines?

R e v ie w P ro b le m 2 .1 6

A m ines are like am m onia in being w eak bases. They do this by using their unshared elec­ tron pair to accept a proton. (a) Show the reaction that w ould take place betw een trim ethyl­ am ine and HCI. (b) W hat hybridization state w ould you expect for the nitrogen atom in the product o f this reaction?

R e v ie w P ro b le m 2 .1 7

2.9 A ldehydes and Ketones A ldehydes and ketones both contain the c a rb o n y l g ro u p — a group in w hich a carbon atom has a double bond to oxygen: •O'II z C^ T h e c a rb o n y l g ro u p

T he carbonyl group o f an aldehyde is bonded to one hydrogen atom and one carbon atom (except for form aldehyde, w hich is the only aldehyde bearing tw o hydrogen atom s). The carbonyl group of a ketone is bonded to tw o carbon atom s. U sing R, w e can designate the general form ulas for aldehydes and ketones as follows:

ALDEHYDES O' || R ^^H

KETONES Ö’

or

or

RC HO

(R = H in f o r m a l d e h y d e )

R C O R'

1 ( w h e r e R ' is a n a lk y l g r o u p t h a t m a y b e th e s a m e o r d if f e r e n t fr o m

Som e specific exam ples o f aldehydes and ketones are the follow ing:

R)

70

Chapter 2

Families o f Carbon Compounds

A ld e h y d e s a n d k e to n e s h a v e a t r ig o n a l p la n a r a r r a n g e m e n t o f g r o u p s a r o u n d th e c a r ­

H e lp f u l H i n t

b o n y l c a r b o n a to m . T h e c a r b o n a to m is s p 2 h y b r id iz e d . I n fo r m a ld e h y d e , f o r e x a m p le , th e

Com puter molecular models can be found in the 3D Models section o f the book's website fo r these and many other compounds we discuss in this book.

b o n d a n g le s a r e a s f o l l o w s :

'O' 121^ c \ 121°

H 118° H

R e v ie w P r o b le m 2 . 1 8

W r it e th e re s o n a n c e s tr u c tu r e f o r c a r v o n e th a t r e s u lts f r o m

m o v in g th e e le c tr o n s a s i n d i ­

c a te d . In c lu d e a ll f o r m a l c h a rg e s .

R e t i n a l ( b e l o w ) i s a n a ld e h y d e m a d e f r o m

v ita m in A

t h a t p la y s a v i t a l r o le i n v is io n . W e

d is c u s s t h i s f u r t h e r i n C h a p t e r 1 3 .

R e v ie w P r o b le m 2 . 1 9

W r i t e b o n d - l i n e f o r m u l a s f o r ( a ) f o u r a ld e h y d e s a n d ( b ) t h r e e k e t o n e s t h a t h a v e t h e f o r ­ m u la c

5 h 10o

.

2.10 Carboxylic Acids, Esters, and A m ides C a r b o x y l i c a c id s , e s te r s , a n d a m id e s a l l c o n t a i n a c a r b o n y l g r o u p t h a t i s b o n d e d t o a n o x y ­ g e n o r n it r o g e n a to m . A s w e s h a ll le a r n in la t e r c h a p te r s , a ll o f th e s e f u n c t io n a l g r o u p s a re i n t e r c o n v e r t i b l e b y a p p r o p r i a t e l y c h o s e n r e a c t io n s .

2 .1 0 A

Carboxylic Acids

C a r b o x y l i c a c id s h a v e a c a r b o n y l g r o u p b o n d e d t o a h y d r o x y l g r o u p , a n d t h e y h a v e t h e

O II ^C

g e n e ra l f o r m u la

R g ro u p (c a rb o n y l +

O . T h e fu n c tio n a l g ro u p ,

O— H

, is c a lle d th e

c a rb o x y l

O— H

h y d r o x y l) :

O

O

11

C.

R

II ^C

.H O

or

O

or

ii C .

R C O 2H

M

r ^ 'O h

or

C O 2H

or

O

A carboxylic acid

The carboxyl group

E x a m p l e s o f c a r b o x y l i c a c id s a r e f o r m i c a c id , a c e t i c a c id , a n d b e n z o i c a c id : O

O or

H

O

or H

ÇH

Formic acid

H

COOH

71

2.10 Carboxylic Acids, Esters, and Amides

O

O’

II a

,H

CH3

CH3C O 2H

or

or

O

OH A c e t ic a c id

O

Ö

II

C

OH

OH

C6H5C O 2H

or

or

B e n z o i c a c id F o r m ic a c id is a n i r r it a t in g l i q u i d p r o d u c e d b y a n ts . ( T h e s tin g o f th e a n t is c a u s e d , i n p a r t, b y f o r m i c a c i d b e i n g i n j e c t e d u n d e r t h e s k i n . F o r m ic i s t h e L a t i n w o r d f o r a n t . ) A c e t i c a c id , t h e s u b s t a n c e r e s p o n s i b l e f o r t h e s o u r ta s t e o f v i n e g a r , is p r o d u c e d w h e n c e r t a i n b a c t e r ia a c t o n t h e e t h y l a l c o h o l o f w i n e a n d c a u s e t h e e t h y l a l c o h o l t o b e o x i d i z e d b y a ir .

W h e n f o r m i c a c i d d o n a t e s t h e p r o t o n f r o m i t s o x y g e n t o a b a s e , a f o r m a t e i o n is t h e r e s u l t .

R e v ie w P ro b le m 2 .2 0

W r i t e a n o t h e r r e s o n a n c e s t r u c t u r e f o r f o r m i c a c id a n d f o r t h e f o r m a t e i o n . W h i c h s p e c ie s , f o r m i c a c id o r t h e f o r m a t e i o n , w o u l d b e m o s t s t a b i l i z e d b y r e s o n a n c e ?

W r i t e b o n d - l i n e f o r m u l a s f o r f o u r c a r b o x y l i c a c id s w i t h t h e f o r m u l a C

2 .1 0 B

R e v ie w P ro b le m 2.21

5 H K 3O 2 .

Esters 2R '

E s te rs h a v e th e g e n e r a l f o r m u la R C O

( o r R C O O R ') , w h e r e a c a r b o n y l g r o u p is b o n d e d

to a n a lk o x y l ( — O R ) g ro u p : O

O

II

C R

or

R

.. O R '

or

RCO

2R '

O — R' G e n e r a l f o r m u la f o r a n e s te r

O

Ö

II

c h 3c o 2c h 2c h

C ch:

3

O C H 2C H 3 E th y l a c e t a te is a n im p o r t a n t s o lv e n t .

O The ester pentyl butanoate has the odor of apricots and pears.

P e n ty l b u ta n o a te h a s th e o d o r o f a p r ic o ts a n d p e a rs .

W r i t e b o n d - l i n e f o r m u l a s f o r t h r e e e s te r s w i t h t h e f o r m u l a C

5 H 10O

R e v ie w P ro b le m 2 .2 2

2.

R e v ie w P ro b le m 2 .2 3

W r i t e a n o t h e r r e s o n a n c e s t r u c t u r e f o r e t h y l a c e t a te . I n c l u d e f o r m a l c h a r g e s .

E s te rs c a n b e m a d e f r o m

a c a r b o x y l i c a c i d a n d a n a l c o h o l t h r o u g h t h e a c i d - c a t a l y z e d lo s s

o f a m o le c u le o f w a te r . F o r e x a m p le :

O

O +

C C H3

HO C H 2C H 3

OH

A c e t ic a c id

a c id -c a ta ly z e d

C C H ::

E th y l a lc o h o l

+

h 2o

O C H 2C H 3

E th y l a c e ta te

Y o u r b o d y m a k e s e s te r s f r o m l o n g - c h a i n c a r b o x y l i c a c id s c a l l e d “ f a t t y a c id s ” b y c o m b i n ­ i n g t h e m w i t h g l y c e r o l . W e d is c u s s t h e i r c h e m i s t r y i n d e t a i l i n C h a p t e r 2 3 .

72

Chapter 2

Families o f Carbon Compounds

2.10C Amides A m ides have the form ulas RCO N H 2, R C O N H R ', or R C O N R 'R " w here a carbonyl group is bonded to a nitrogen atom bearing hydrogen and/or alkyl groups. G eneral form ulas and som e specific exam ples are show n below. O

O

O

H

R'

R

Nylon is a polymer comprised of regularly repeating amide groups.

R'

R

A n u n s u b s t it u t e d a m id e

R

A n N - s u b s t it u t e d a m id e

A n N , N - d i s u b s t i t u t e d a m id e

G e n e r a l f o r m u l a s f o r a m id e s

O

O

O Me

H N

Me

N

1

N I Me

1

H

H

A c e ta m id e

N - M e t h y l a c e t a m id e

N N - D i m e t h y l a c e t a m id e

S p e c i f i c e x a m p l e s o f a m id e s

N - and N ,N - indicate that the substituents are attached to the nitrogen atom.

R e v ie w P ro b le m 2 .2 4

W rite another resonance structure for acetam ide.

2.11 Nitriles A nitrile has the form ula R — C # N: (or R — CN). T he carbon and the nitrogen o f a nitrile are sp hybridized. In IUPAC system atic nom enclature, acyclic nitriles are nam ed by adding the suffix -n itrile to the nam e o f the corresponding hydrocarbon. T he carbon atom o f the — C # N group is assigned num ber 1. T he nam e acetonitrile is an acceptable com m on nam e for CH 3 CN, and acrylonitrile is an acceptable com m on nam e for CH 2 = CHCN: 2

1

2

CH 3— C # N :

CH 3CH 2CH 2— C # N :

E th a n e n it r ile

B u t a n e n it r ile

( a c e to n itr ile )

2

1

CN

3

P r o p e n e n itr ile

5

3 . 4 ^ \ 2

1

N

4 - P e n t e n e n it r ile

( a c r y lo n it r ile )

Cyclic nitriles are nam ed by adding the suffix -carbonitrile to the nam e o f the ring system to which the — CN group is attached. Benzonitrile is an acceptable com mon nam e for C 6 H5CN:

\

/

C=N :

B e n z e n e c a r b o n it r ile

C=N : C y c lo h e x a n e c a r b o n it r ile

( b e n z o n it r ile )

2.12 Sum m ary o f Im p o rtan t Families o f O rganic Compounds A sum m ary o f the im portant fam ilies o f organic com pounds is given in Table 2.3. You should learn to identify these com m on functional groups as they appear in other, m ore com ­ plicated m olecules.

73

2.13 Physical Properties and Molecular Structure

2.13 Physical Properties and Molecular Structure So far, w e have said little about one o f the m ost obvious characteristics o f organic com ­ pounds, that is, th e ir p h ysica l state o r phase. W hether a particular substance is a solid, or a liquid, or a gas w ould certainly be one o f the first observations that w e w ould note in any experim ental work. The tem peratures at w hich transitions occur betw een phases, that is, m elting points (mp) and boiling points (bp), are also am ong the m ore easily m easured p h y sic al p ro p e rtie s. M elting points and boiling points are also useful in identifying and isolating organic com pounds. Suppose, for exam ple, w e have ju st carried out the synthesis o f an organic com pound that is know n to be a liquid at room tem perature and 1 atm pressure. If w e know the boil­ ing point o f our desired product and the boiling points o f by-products and solvents that m ay be present in the reaction m ixture, w e can decide w hether or not sim ple distillation w ill be a feasible m ethod for isolating our product. In another instance our product m ight be a solid. In this case, in order to isolate the substance by crystallization, we need to know its melting point and its solubility in different solvents. T he physical constants o f know n organic substances are easily found in handbooks and other reference books.* Table 2.4 lists the m elting and boiling points o f som e o f the com ­ pounds that w e have discussed in this chapter. O ften in the course of research, however, the product o f a synthesis is a new com pound— one that has never been described before. In these instances, success in isolating the new com pound depends on m aking reasonably accurate estim ates o f its m elting point, boiling point, and solubilities. E stim ations o f these m acroscopic physical properties are based on the m ost likely structure of the substance and on the forces that act betw een m olecules and ions. The tem peratures at w hich phase changes occur are an indication o f the strength of these interm olecular forces.

H e lp f u l H i n t

Understanding how molecular structure influences physical properties is very useful in practical organic chemistry.

Im p ortant Families of O rganic Com pounds Family A lkane

A lkene

Alkyne

A rom atic

H aloalkane

\ = /

CX \

—C ^C —

Arom atic ring

— C—X:

CH2 CHR CHR CR2

ArH

RX

ROH

ROR

CH3CH2Cl

CH3CH2OH

CH3OCH3

i

C— H

Functional group

and C— C bonds

I

Alcohol

E ther

I — C —O H

I I — C—O— C— I I

I

G eneral formula

RH

RCH= RCH= R2C = R 2C

Specific exam ple

CH3CH3

CH2= CH2

H C #C H

IUPAC nam e

Ethane

Ethene

Ethyne

Benzene

C hloroethane

Ethanol

M ethoxym ethane

Com m on

Ethane

Ethylene

A cetylene

B enzene

Ethyl chloride

Ethyl alcohol

Dimethyl eth er

3These names are also accepted by the IUPAC. *Two useful handbooks are Handbook of Chemistry, Lange, N. A., Ed., McGraw-Hill: New York; and CRC Handbook of Chemistry and Physics, CRC: Boca Raton, FL.

74

Chapter 2

Families o f Carbon Compounds

I m p o r t a n t F a m ilie s o f O r g a n ic C o m p o u n d s (c o n t.) Family

O'1

' / - 0 —

C arboxylic Acid

. H

0 = 0;

O' II C

K eto n e

\

Functional 1 / — C— N: group 1 \

A ld eh y d e

1 —0 —

A m ine

/ C N .. OH

E ster

A m ide

■'o 'II / C s .. 1 ^ X I— C — ••

N itrile

■'o 'II

—C #N : N I

O r Cnh2

G eneral formula

rnh2

O

O

O

rCh

RCR'

O

O

r 2n h

R3 N

r Co h

r Co r

*

rCn h r *

RCN

O r C n R'R"

O

Specific exam ple

c h 3n h 2

IUPAC nam e Com m on nam e

O

O

O

O c h 3c # n

c h 3C h

c h 3C c h 3

c h 3C o h

c h 3C o c h 3

c h 3C n h 2

Methanamine

Ethanal

Propanone

Ethanoic acid

Methyl ethanoate

Ethanamide

Ethanenitrile

Methylamine

Acetaldehyde Acetone

Acetic acid

Methyl acetate

Acetamide

Acetonitrile

2 .1 3 A

Ionic Compounds: lon-Ion Forces

T he m e ltin g p o in t o f a substance is the tem perature at w hich an equilibrium exists betw een the w ell-ordered crystalline state an d the m ore random liquid state. If the substance is an ionic com pound, such as sodium acetate (Table 2.4), the io n -io n fo rces that h o ld the ions together in the crystalline state are the strong electrostatic lattice forces that act betw een the positive and negative ions in the orderly crystalline structure. In Fig. 2.6 each sodium ion is surrounded by negatively charged acetate ions, an d each acetate ion is surrounded by positive sodium ions. A large am ount o f therm al energy is required to b reak up the orderly structure o f the crystal into the disorderly open structure o f a liquid. A s a result, the tem perature at w hich sodium acetate m elts is quite high, 324°C. T he b o ilin g p o in ts of ionic com pounds are h igher still, so high that m ost ionic organic com pounds decom pose (are changed by undesirable chem ical reactions) before they boil. Sodium acetate shows this behavior.

Figure 2.6 The melting of sodium acetate.

75

2.13 Physical Properties and Molecular Structure

Physical Properties o f Representative Com pounds C om p o u n d M ethane Ethane Ethene Ethyne C hlorom ethane C hloroethane Ethyl alcohol A cetaldehyde Acetic acid Sodium a c etate Ethylamine Diethyl eth er Ethyl ac etate

S tru c tu re ch4 c h 3c h 3 c h 2"

ch2 HC # CH c h 3ci c h 3c h 2ci c h 3c h 2o h c h 3c h o c h 3c o 2h CH3CO2Na c h 3c h 2n h 2 (c h 3c h 2)2o c h 3c o 2c h 2c h 3

mp (°C)

b p (°C) (1 a tm )a

-1 8 2 .6 -1 7 2 -1 6 9 -8 2 -9 7 - 1 3 8 .7 -1 1 4 -1 2 1 16.6 324 -8 0 -1 1 6 -8 4

-1 6 2 - 8 8 .2 -1 0 2 - 8 4 subl - 2 3 .7 13.1 78.5 20 118 dec 17 34.6 77 An instrum ent used to measure melting point

aIn this table dec = decomposes and subl = sublimes.

From PAVIA/LAMPMAN/KRIEL. 2 .1 3 B

Intermolecular Forces (van der Waals Forces)

The forces that act betw een m olecules are not as strong as those betw een ions, but they account for the fact that even com pletely nonpolar m olecules can exist in liquid and solid states. These interm olecular forces, collectively called v a n d e r W aals forces, are all elec­ trical in nature. We w ill focus our attention on three types: 1. D ipole-dipole forces

In tro d u ctio n to O rganic L a b ora to ry Techniques: A M icroscale A p p ro ach (w ith Periodic Table), 3E. © 1999

Brooks/Cole, a part of Cengage Learning, Inc. Reproduced by permission. www.cengage.com/ permissions.

2. H ydrogen bonds 3. D ispersion forces D ip o le -D ip o le Forces

M ost organic m olecules are not fully ionic but have instead a

perm anent dipole moment resulting from a nonuniform distribution o f the bonding elec­

trons (Section 2.3). A cetone and acetaldehyde are exam ples o f m olecules w ith perm anent dipoles because the carbonyl group that they contain is highly polarized. In these com ­ pounds, the attractive forces betw een m olecules are m uch easier to visualize. In the liquid or solid state, d ip o le -d ip o le attractions cause the m olecules to orient them selves so that the positive end o f one m olecule is directed tow ard the negative end o f another (Fig. 2.7).

Figure 2.7 Electrostatic potential

models for acetone molecules that show how acetone molecules might align according to attractions of their partially positive regions and partially negative regions (dipole-dipole interactions).

H yd ro g en Bonds •

Very strong d ip o le -d ip o le attractions occur b etw een h y d rogen atom s b o nded to sm all, strongly electronegative atom s (O, N, or F) and nonbonding electron pairs on other such electronegative atom s. This type o f in term o lecu lar force is called a h y d ro g e n b o n d .

H ydrogen bonds (bond dissociation energies of about 4 -3 8 kJ m o l_1) are w eaker than ordi­ nary covalent bonds but m uch stronger than the d ip o le-d ip o le interactions that occur above, for exam ple, in acetone.

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H ydrogen bonding explains w hy water, am m onia, and hydrogen fluoride all have far higher boiling points than m ethane (bp —161.6°C), even though all four com pounds have sim ilar m olecular w eights. St

St

H

H

St St

St /

° '

St /

H

S-

H

St

H

S-

S-

St

H~P =

H~P =

H

St

St

H ~ N:

H— N;

S­

S-

S-

H

H

St b p 1 0 0 °C

bp I

9 . 50C

St

b p -3 3 .4 ° C

H y d ro g e n b o n d s a re s h o w n b y th e re d d o ts .

Water molecules associated by attraction of opposite partial charges.

O ne o f the m ost im portant consequences o f hydrogen bonding is that it causes w ater to be a liquid rather than a gas at 25°C. C alculations indicate that in the absence o f hydrogen bonding, w ater w ould have a boiling point near —80°C and w ould n o t exist as a liquid unless the tem perature w ere low er than that tem perature. H ad this been the case, it is highly unlikely that life, as w e know it, could have developed on the p lan et Earth. H ydrogen bonds h old the base pairs o f double-stranded D N A together (see Section 25.4). T hym ine hydrogen bonds w ith adenine. C ytosine hydrogen bonds w ith guanine. H ydrogen bonding accounts for the fact that ethyl alcohol has a m uch h ig h er boiling po in t (+ 7 8 .5 °C ) than dim ethyl ether ( —24.9°C ) even though the tw o com pounds have the H

H

\

H3C (/

N— H

:0;-

\ N | •• --H— H— N

N N

DNA backbone / DNA backbone

DNA backbone

\

DNA backbone

.O - -- H— N

\ H

T h y m in e

A d e n in e

C y t o s in e

G u a n in e

sam e m olecular w eight. M olecules o f ethyl alcohol, because they have a hydrogen atom covalently bonded to an oxygen atom , can form strong hydrogen bonds to each other. T h e re d d o ts re p re s e n t a h y d r o g e n b o n d . S tro n g

s+

c h 3c h 2

s«- s+ .O — H

«-

H

h y d r o g e n b o n d i n g is l im i t e d t o m o le c u le s

■O'c h 2c h 3

h a v in g a h y d r o g e n a to m a tta c h e d to a n O , N , o r F a to m .

M olecules o f dim ethyl ether, because they lack a hydrogen atom attached to a strongly elec­ tronegative atom , cannot form strong hydrogen bonds to each other. In dim ethyl ether the interm olecular forces are w eaker d ip o le-d ip o le interactions. R e v ie w P ro b le m 2 .2 5

The com pounds in each part below have the same (or similar) m olecular weights. W hich com ­ pound in each part w ould you expect to have the higher boiling point? Explain your answers. OH

(a )

OH

(c )

or or

(b) (CH 3)3 N HO

OH .

or

CH3

"N ' H

A factor (in addition to polarity an d hydrogen bonding) that affects the m elting p o in t o f m any organic com pounds is the com pactness an d rigidity o f their individual m olecules. •

M olecules that are sym m etrical generally have abnorm ally high m elting points. te rt -B utyl alcohol, for exam ple, has a m uch higher m elting point than the other

isom eric alcohols show n here:

2.13 Physical Properties and Molecular Structure

OH OH f e r f - B u t y l a lc o h o l (m p 2 5 °C )

OH

OH B u t y l a lc o h o l ( m p —9 0 ° C )

I s o b u t y l a lc o h o l ( m p —1 0 8 ° C )

s e c - B u t y l a lc o h o l (m p - 1 1 4 ° C )

W hich com pound w ould you expect to have the higher m elting point, propane or cyclo­ propane? E xplain your answer. Figure 2.8 Temporary dipoles and induced dipoles in nonpolar molecules resulting from an uneven distribution of electrons at a given instant.

Dispersion Forces If w e consider a substance like m ethane w here the particles are non­ polar m olecules, w e find that the m elting point and boiling point are very low: —182.6°C and —162°C, respectively. Instead o f asking, “W hy does m ethane m elt and boil at low tem ­ peratures?” a m ore appropriate question m ight be “W hy does m ethane, a nonionic, nonpo­ lar substance, becom e a liquid or a solid at all?” The answ er to this question can be given in term s of attractive interm olecular forces called d isp e rsio n fo rces or L ondon forces. A n accurate account of the nature of dispersion forces requires the use of quantum m echan­ ics. We can, however, visualize the origin of these forces in the following way. The average distribution of charge in a nonpolar m olecule (such as m ethane) over a period o f tim e is uni­ form. A t any given instant, however, because electrons move, the electrons and therefore the charge m ay not be uniform ly distributed. Electrons may, in one instant, be slightly accum u­ lated on one part of the molecule, and, as a consequence, a sm all tem porary dipole w ill occur (Fig. 2.8). This tem porary dipole in one m olecule can induce opposite (attractive) dipoles in surrounding molecules. It does this because the negative (or positive) charge in a portion of one m olecule will distort the electron cloud of an adjacent portion o f another m olecule, caus­ ing an opposite charge to develop there. These tem porary dipoles change constantly, but the net result of their existence is to produce attractive forces betw een nonpolar m olecules and thus m ake possible the existence of their liquid and solid states. Two im portant factors determ ine the m agnitude o f dispersion forces. 1. T h e re la tiv e p o la riz a b ility o f e lec tro n s o f th e a to m s involved. By p o la riz a b ility we mean how easily the electrons respond to a changing electric fie ld . The electrons of large atom s such as iodine are loosely held and are easily polarized, w hile the elec­ trons o f sm all atom s such as fluorine are m ore tightly held and are m uch less polarizable. A tom s w ith unshared pairs are m ore easily polarized than atom s w ith only bonding pairs. Table 2.5 gives the relative m agnitudes o f dispersion forces and d ipole-dipole interactions for several sim ple com pounds. N otice w ith HI the disper­ sion forces are far m ore im portant than d ip o le-d ip o le forces, w hereas w ith H2O, d ipole-dipole forces (of the kind w e call hydrogen bonds) are m ore im portant. A ttractive Energies in Simple C ovalent Com pounds A ttrac tiv e E nergies (kJ m o l—1) olecule H2O NH3 HCl HBr HI

D ipole M om ent (D) 1.85 1.47 1.08 0.80 0.42

D ipole­ D ipole

D ispersion

M elting P oint (°C)

Boiling P oint (°C) a t

36a 14a 3a

0

100

15 17

-7 8 -1 1 5

0.8

22

-8 8

0.03

28

-5 1

—33 —85 —67 —35

8.8

a These dipole-dipole attractions are called hydrogen bonds.

R e v ie w P ro b le m 2 .2 6

-------

A m ic r o s c a le d is tilla tio n a p p a r a tu s

From PAVIA/LAMPMAN/KRIEL. In tro d u ctio n to O rganic L a b ora to ry Techniques: A M icroscale A p p ro ach (w ith Periodic Table), 3E. © 1999

Brooks/Cole, a part of Cengage Learning, Inc. Reproduced by permission. www.cengage.com/ permissions.

78

Chapter 2

Families o f Carbon Compounds

C F 4 and C I 4 are both nonpolar molecules. B ut if w e were to consider the interm olecular forces betw een two C I 4 molecules, w hich contain polarizable iodine atoms, w e would find that the dispersion forces are m uch larger than betw een two C F 4 molecules, w hich contains fluorine atom s that are not very polarizable.

2. T h e re la tiv e su rfa c e a r e a o f th e m o lecu les involved. T he larger the surface area, the larger is the overall attraction betw een m olecules caused by dispersion forces. M olecules that are generally longer, flatter, or cylindrical have a greater surface area available for interm olecular interactions than m o re spherical m olecules, and conse­ quently have greater attractive forces betw een them than the tangential interactions betw een bran ch ed m olecules. T his is evident w hen com paring pentane, the unbranched C 5 H 12 hydrocarbon, w ith neopentane, the m o st highly branched C 5 H 12 isom er (in w hich one carbon bears four m ethyl groups). Pentane has a boiling point o f 36.1°C. N eopentane has a boiling po in t o f 9.5°C . T he difference in their boiling points indicate that the attractive forces betw een pentane m olecules are stronger than betw een neopentane m olecules. Dispersion forces are what provides a gecko's grip to smooth surfaces.

For large m olecules, the cum ulative effect o f these sm all and rapidly changing dispersion forces can lead to a large n et attraction.

2.13C Boiling Points The b o ilin g p o in t o f a liquid is the tem perature at w hich the vapor pressure o f the liquid equals the pressure o f the atm osphere above it. F or this reason, the boiling points o f liq ­ uids are pressure dependent, and boiling points are alw ays reported as occurring at a p ar­ ticular pressure, at 1 atm (or at 760 torr), for exam ple. A substance that boils at 150°C at 1 atm pressure w ill b o il at a substantially low er tem perature if the pressure is reduced to, for exam ple, 0.01 torr (a pressure easily obtained w ith a vacuum pum p). T he norm al b o il­ ing point given for a liquid is its boiling po in t at 1 atm. In passing from a liquid to a gaseous state, the individual m olecules (or ions) o f the sub­ stance m u st separate. B ecause o f this, w e can understand w hy ionic organic com pounds often decom pose before they boil. T he therm al energy required to com pletely separate (volatilize) the ions is so great that chem ical reactions (decom positions) occur first.

THE CHEMISTRY OF . . F l u o r o c a r b o n s a n d T e f lo n Fluorocarbons (com pounds containing only carbon and flu­ orine) have extraordinarily low boiling points when com ­ pared to hydrocarbons of th e sam e m olecular weight. The fluorocarbon C 5 F 12 has a slightly lower boiling point than p en tan e (C 5 H 12) even though it has a far higher molecular w eight. The im portant factor in explaining this behavior is th e very low polarizability of fluorine atom s that we m en­ tio n ed earlier, resulting in very small dispersion forces.

The fluorocarbon called Teflon [ C F 2 C F 2] n (see Section 10.10) has self-lubricating p roperties th a t are exploited in making "nonstick" frying pans and lightweight bearings. f

f

f

f

f

f

f

f

f

e tc .

F

F

F

F

F

F

F

T e flo n

F

F

F

f

2.13 Physical Properties and Molecular Structure

N onpolar com pounds, w here the interm olecular forces are very weak, usually boil at low tem peratures even at 1 atm pressure. This is not always true, however, because o f other fac­ tors that we have not yet mentioned: the effects of m olecular w eight and m olecular shape and surface area. H eavier m olecules require greater therm al energy in order to acquire velocities sufficiently great to escape the liquid phase, and because the surface areas o f larger m ole­ cules can be m uch greater, intermolecular dispersion attractions can also be m uch larger. These factors explain why nonpolar ethane (bp —88.2°C) boils higher than m ethane (bp —162°C) at a pressure of 1 atm. It also explains why, at 1 atm, the even heavier and larger nonpolar m olecule decane (C-|0H22) boils at + 174°C . The relationship betw een dispersion forces and surface area helps us understand why neopentane (2,2-dim ethylpropane) has a low er boiling point (9.5°C) than pentane (36.1°C), even though they have the same m olecular weight. The branched structure of neopentane allows less surface interaction betw een neopentane m ole­ cules, hence low er dispersion forces, than does the linear structure o f pentane.

S o lv e d P ro b le m 2 .6

A rrange the follow ing com pounds according to their expected boiling points, w ith the low est boiling point first, and explain your answer. N otice that the com pounds have sim ilar m olecular w eights.

D ie t h y l e t h e r

s e c - B u t y l a lc o h o l

P e n ta n e

STRATEGY AND ANSW ER Pentane < diethyl ether < sec-butyl alcohol In c r e a s in g b o ilin g p o in t

P entane has no polar groups and has only dispersion forces holding its m olecules together. It w ould have the low ­ est boiling point. D iethyl ether has the polar ether group that provides d ip o le-d ip o le forces w hich are greater than dispersion forces, m eaning it w ould have a higher boiling point than pentane. sec-B utyl alcohol has an — OH group that can form strong hydrogen bonds; therefore, it w ould have the highest boiling point.

A rrange the follow ing com pounds in order of increasing boiling point. E xplain your answ er in term s of the interm olecular forces in each com pound.

(a)

2 .1 3 D

(b)

(c)

R e v ie w P ro b le m 2 .2 7

(d)

Solubilities

Interm olecular forces are of prim ary im portance in explaining the solubilities o f substances. D issolution of a solid in a liquid is, in m any respects, like the m elting o f a solid. The orderly crystal structure of the solid is destroyed, and the result is the form ation o f the m ore disor­ derly arrangem ent of the m olecules (or ions) in solution. In the process o f dissolving, too, the m olecules or ions m ust be separated from each other, and energy m ust be supplied for both changes. The energy required to overcom e lattice energies and interm olecular or interionic attractions com es from the form ation of new attractive forces betw een solute and solvent.

H e lp f u l H i n t

Your ability to make qualitative predictions regarding solubility will prove very useful in the organic chemistry laboratory.

80

Chapter 2

Families o f Carbon Compounds

Figure 2.9 The dissolution of an ionic solid in water, showing the hydration of positive and negative ions by the very polar water molecules. The ions become surrounded by water molecules in all three dimensions, not just the two shown here.

C onsider the dissolution o f an ionic substance as an example. H ere both the lattice energy and interionic attractions are large. We find that w ater an d only a few other very polar sol­ vents are capable o f dissolving ionic com pounds. T hese solvents dissolve ionic com pounds by h y d r a tin g or so lv a tin g the ions (Fig. 2.9). W ater m olecules, by virtue o f their great polarity as w ell as their very sm all, com pact shape, can very effectively surround the individual ions as they are freed from the crystal surface. Positive ions are surrounded by w ater m olecules w ith the negative en d o f the w ater dipole pointed tow ard the positive ion; negative ions are solvated in exactly the opposite way. B ecause w ater is highly polar, and b ecause w ater is capable o f form ing strong hy d ro ­ gen bonds, the io n -d ip o le fo rces o f attraction are also large. T he energy supplied by the form ation o f these forces is great enough to overcom e both the lattice energy and interi­ onic attractions o f the crystal. A general rule for solubility is that “like dissolves like” in term s o f com parable polarities. •

P olar and ionic solids are usually soluble in po lar solvents.

•

P olar liquids are usually m iscible.

•

N onpolar solids are usually soluble in nonpolar solvents.

•

N onpolar liquids are usually m iscible.

•

P olar and nonpolar liquids, like oil and water, are usually n o t soluble to large extents.

M ethanol and w ater are m iscible in all proportions; so too are m ixtures o f ethanol and w ater and m ixtures o f both propyl alcohols an d water. In these cases the alkyl groups of the alcohols are relatively small, and the m olecules therefore resem ble w ater m ore than they do an alkane. A nother factor in understanding their solubility is that the m olecules are capa­ ble o f form ing strong hydrogen bonds to each other: c h 3CH2

\s .o

5+ H ¿¡r h

— Hydrogen bond

«+

We often describe m olecules or parts o f m olecules as being hydrophilic or hydropho­ bic. T he alkyl groups o f m ethanol, ethanol, and propanol are hydrophobic. T heir hydroxyl groups are hydrophilic.

2.13 Physical Properties and Molecular Structure

•

H y d ro p h o b ic m eans incom patible w ith w ater (hydro, w ater; phobic, fearing or avoiding).

•

H y d ro p h ilic m eans com patible w ith w ater (p h ilic , loving or seeking).

D ecyl alcohol, w ith a chain o f 10 carbon atom s, is a com pound w hose hydrophobic alkyl group overshadow s its hydrophilic hydroxyl group in term s o f w ater solubility. H y d r o p h o b ic p o r t io n H y d r o p h ilic g ro u p D e c y l a lc o h o l

A n explanation for w hy nonpolar groups such as long alkane chains avoid an aqueous environm ent, that is, for the so-called h y d ro p h o b ic effect, is com plex. The m ost im por­ tant factor seem s to involve an u n fa v o ra b le e n tro p y c h a n g e in the water. Entropy changes (Section 3.10) have to do w ith changes from a relatively ordered state to a m ore disordered one or the reverse. C hanges from order to disorder are favorable, w hereas changes from disorder to order are unfavorable. For a nonpolar hydrocarbon chain to be accom m odated by water, the w ater m olecules have to form a m ore ordered structure around the chain, and for this, the entropy change is unfavorable. We w ill see in Section 23.2C that the presence o f a hydrophobic group and a hydrophilic group are essential com ponents o f soaps and detergents. O "O- Na+ A t y p ic a l s o a p m o le c u le

A t y p ic a l d e t e r g e n t m o le c u le

T he hydrophobic long carbon chains of a soap or detergent em bed them selves in the oily layer that typically surrounds the thing we w ant to w ash away. The hydrophilic ionic groups at the ends o f the chains are then left exposed on the surface and m ake the surface one that w ater m olecules find attractive. Oil and w ater d o n ’t m ix, but now the oily layer looks like som ething ionic and the w ater can take it “right dow n the drain.” 2 .1 3 E

Guidelines for W ater Solubility

O rganic chem ists usually define a com pound as w ater soluble if at least 3 g o f the organic com pound dissolves in 100 m L o f water. We find that for com pounds containing one hydrophilic group— and thus capable o f form ing strong hydrogen bonds— the follow ing approxim ate guidelines hold: C om pounds w ith one to three carbon atom s are w ater solu­ ble, com pounds w ith four or five carbon atom s are borderline, and com pounds w ith six car­ bon atom s or m ore are insoluble. W hen a com pound contains m ore than one hydrophilic group, these guidelines do not apply. P olysaccharides (C hapter 22), proteins (C hapter 24), and nucleic acids (C hapter 25) all contain thousands o f carbon atom s and many are w a te r soluble. They dissolve in w ater because they also contain thousands o f hydrophilic groups. 2 .1 3 F

Intermolecular Forces in Biochemistry

Later, after we have had a chance to examine in detail the properties of the molecules that make up living organisms, we shall see how in te rm o le c u la r forces are extremely im portant in the functioning of cells. H y d ro g en b o n d formation, the hydration of polar groups, and the ten­ dency of nonpolar groups to avoid a polar environm ent all cause com plex protein molecules

Hydrogen bonding (red dotted lines) in the a -helix structure of proteins (Illu stra tio n , Irving Geis. Rights o w n e d b y H o w ard Hughes M edical In stitu te . N o t to be rep ro d u ced w ith o u t perm ission.)

82

Chapter 2

Families o f Carbon Compounds

to fold in precise ways— w ays that allow them to function as biological catalysts o f incredi­ ble efficiency. The sam e factors allow m olecules o f hem oglobin to assum e the shape needed to transport oxygen. They allow proteins and m olecules called lipids to function as cell m em ­ branes. Hydrogen bonding gives certain carbohydrates a globular shape that m akes them highly efficient food reserves in animals. It gives m olecules o f other carbohydrates a rigid linear shape that m akes them perfectly suited to be structural com ponents in plants.

2.14 Sum m ary o f A ttrac tiv e Electric Forces T he attractive forces occurring betw een m olecules and ions that w e have studied so far are sum m arized in Table 2.6.

A ttra ctiv e Electric Forces Electric Force

R elative S tre n g th

Cation -a n io n (in a crystal)

Very strong

C ovalent bonds

Strong (1 4 0 -5 2 3 kJ m o T 1)

E xam ple

Ty p e

Sodium chloride crystal lattice Shared electron pairs

H— H (436 kJ m o r 1) CH3— CH3 (378 kJ m ol-1 ) I— I (151 kJ m ol-1 )

s+

slo n -d ip o le

S+4 'o S - tft+ s­ V+

M oderate

s

s+

N a+ in w ater (see Fig. 2.9)

s-

s-

H ydrogen bonds

M oderate to w eak (4 -3 8 kJ m ol-1 )

R

s+

— Z : - - -H—

\s • O-

O ss + /-\ H R

s+H s+

s-

s+

s-

D ipole-dipole

Weak

CH3Cl

CH3CI

Dispersion

Variable

Transient dipole

Interactions betw een m eth an e m olecules

THE CHEMISTRY OF . . . O r g a n i c T e m p l a t e s E n g i n e e r e d t o M im ic B o n e G r o w t h Interm olecular forces play a myriad of roles in life and in th e world around us. Interm olecular forces hold to g e th e r th e strands of our DNA, provide structure to our cell m em ­ branes, cause th e fee t of gecko lizards to stick to walls and

ceilings, keep w ater from boiling at room te m p eratu re and ordinary pressure, and literally provide th e adhesive forces th a t hold our cells, bones, and tissues to g eth er. As th ese exam ples show, th e world around us provides exquisite

2.15 Infrared Spectroscopy: An Instrumental M ethod for Detecting Functional Groups

instruction in nanotechnology and bioengineering, and sci­ entists throug hout th e ag es have been inspired to create and innovate based on nature. O ne ta rg et of recent research in bioengineering is th e dev elo p m en t of synthetic materials th at mimic nature's tem p late for bo n e growth. A synthetic material with bone-prom oting properties could be used to help repair broken bones, offset osteoporosis, and treat bo n e cancer. Both natural bone growth and th e synthetic system under d ev elo p m en t d e p e n d strongly on interm olecular forces. In living system s, b ones grow by adhesion of specialized cells to a long fibrous natural tem p late called collagen. Certain functional groups along th e collagen prom ote th e binding of bone-grow ing cells, while o ther functional groups facili­ ta te calcium crystallization. Chem ists at N orthw estern University (led by S. I. Stupp) have eng in eered a m olecule th at can be m ade in th e laboratory and th a t mimics this

process. The m olecule shown below spontaneously selfassem bles into a long tubular ag g reg ate, imitating the fibers of collagen. Dispersion forces betw een hydrophobic alkyl tails on th e m olecule cause self-assem bly of th e m olecules into tubules. At th e o th er end of th e m olecule, the researchers included functional groups th at pro m o te cell binding and still o th er functional groups th at en co u rag e cal­ cium crystallization. Lastly, they included functional groups th at allow one m olecule to b e covalently linked to its neigh­ bors after th e self-assem bly process has occurred, thus adding further stabilization to th e initially noncovalent struc­ ture. Designing all of th e se features into th e m olecular struc­ ture has paid off, b ecause the self-assem bled fiber prom otes calcium crystallization along its axis, much like nature's col­ lagen tem plate. This exam ple of m olecular design is just one exciting d ev elo p m en t at th e intersection of nanotechnology and bioengineering. 0

I 1

HO— P— OH O

V ^ O H

_____ Hydrophobic alkyl region

O

O

Flexible linker region

OH

H N Hydrophilic cell adhesion region

(R eprinted w ith perm ission from H artgerink, J. D., Beniash, E., Stupp, S. I., SCIENCE 294: 1684-1688, Figure 1 (2001). C o p yrig h t 2001 AAAS.)

2.15 Infrared Spectroscopy: An Instrum ental M e th o d fo r D etectin g Functional Groups I n f r a r e d (IR ) sp e ctro sc o p y is a sim ple and rapid instrum ental technique that can give evi­ dence for the presence o f various functional groups. If you had a sam ple o f unknow n iden­ tity, am ong the first things you w ould do is obtain an infrared spectrum , along w ith determ ining its solubility in com m on solvents and its m elting and/or boiling point. Infrared spectroscopy, as all form s o f spectroscopy, depends on the interaction o f m o l­ ecules or atom s w ith electrom agnetic radiation. Infrared radiation causes atom s and groups o f atoms of organic com pounds to vibrate w ith increased am plitude about the covalent bonds that connect them . (Infrared radiation is not o f sufficient energy to excite electrons, as is the case w hen som e m olecules interact w ith visible, ultraviolet, or higher energy form s of light.) Since the functional groups o f organic m olecules include specific arrangem ents of bonded atom s, absorption o f IR radiation by an organic m olecule w ill occur at specific fre­ quencies characteristic o f the types of bonds and atom s present in the specific functional groups o f that m olecule. These vibrations are quantized, and as they occur, the com pounds absorb IR energy in particular regions o f the IR portion o f the spectrum .

84

Chapter 2

Families o f Carbon Compounds

Figure 2.10 A diagram of a Fourier transform infrared (FTIR) spectrometer. FTIR spectrometers employ a Michelson interferometer, which splits the radiation beam from the IR source so that it reflects simultaneously from a moving mirror and a fixed mirror, leading to interference. After the beams recombine, they pass through the sample to the detector and are recorded as a plot of time versus signal intensity, called an interferogram. The overlapping wavelengths and the intensities of their respective absorptions are then converted to a spectrum by applying a mathematical operation called a Fourier transform. The FTIR method eliminates the need to scan slowly over a range of wavelengths, as was the case with older types of instruments called dispersive IR spectrometers, and therefore FTIR spectra can be acquired very quickly. The FTIR method also allows greater throughput of IR energy. The combination of these factors gives FTIR spectra strong signals as compared to background noise (i.e., a high signal to noise ratio) because radiation throughput is high and rapid scanning allows multiple spectra to be averaged in a short period of time. The result is enhancement of real signals and cancellation of random noise. (Diagram adapted from the computer program IR Tutor, Columbia University.)

A n infrared spectrom eter (Fig. 2.10) operates by passing a beam o f IR radiation through a sam ple and com paring the radiation transm itted through the sam ple w ith that transm it­ ted in the absence o f the sam ple. A ny frequencies absorbed by the sam ple w ill b e appar­ ent by the difference. T he spectrom eter plots the results as a graph show ing absorbance versus frequency or wavelength. •

T he position o f an absorption b and (peak) in an IR spectrum is specified in units of w a v e n u m b e rs (v ).

W avenum bers are the reciprocal o f w avelength w hen w avelength is expressed in cen ­ tim eters (the u nit is c m ~ 1), and therefore give the n um ber o f w ave cycles per centim eter. The larger the w avenum ber, the h igher is the frequency o f the wave, an d correspondingly the higher is the frequency o f the b o n d absorption. IR absorptions are som etim es, though less com m only, reported in term s o f w av e len g th (A), in w hich case the units are m icrom ­ eters (mm; old nam e m icron, m). W avelength is the distance from crest to crest o f a wave. _

1

v = —

A

(with A in cm)

or

_

10,000

v = ----------A

(with A in ^m )

In their vibrations covalent bonds behave as if they w ere tiny springs connecting the atom s. W hen the atom s vibrate, they can do so only at certain frequencies, as if the bonds w ere “tuned.” B ecause o f this, covalently b onded atom s have only particular vibrational energy levels; that is, the levels are quantized. The excitation o f a m olecule from one vibrational energy level to another occurs only w hen the com pound absorbs IR radiation o f a particular energy, m eaning a particular w ave­ length or frequency. N ote that the energy (E) o f absorption is directly proportional to the fre q u e n c y o f radiation (n) because A E = hn, an d inversely proportional to the w avelength c

hc

(A) because n = — , and therefore A E = -----. A A

2.15 Infrared Spectroscopy: An Instrumental M ethod for Detecting Functional Groups

M o le c u le s c a n v ib r a te i n a v a r ie ty o f w a y s . T w o a to m s jo in e d b y a c o v a le n t b o n d c a n u n d e r g o a s t r e t c h i n g v i b r a t i o n w h e r e t h e a t o m s m o v e b a c k a n d f o r t h a s i f j o i n e d b y a s p r in g . T h r e e a t o m s c a n a ls o u n d e r g o a v a r i e t y o f s t r e t c h i n g a n d b e n d i n g v i b r a t i o n s .

T h e fre q u e n c y o f a g i v e n s t r e t c h i n g v i b r a t i o n in a n I R s p e c tru m c a n b e r e l a t e d t o t w o f a c t o r s . T h e s e a r e th e m asses o f th e b o n d e d a to m s — l i g h t a t o m s v i b r a t e a t h i g h e r f r e q u e n ­ c ie s t h a n h e a v i e r o n e s — a n d th e r e la tiv e s tiffn e s s o f th e b o n d . ( T h e s e f a c t o r s a r e a c c o u n t e d f o r i n H o o k e ’ s la w , a r e la t io n s h ip y o u m a y s tu d y i n in t r o d u c t o r y p h y s ic s . ) T r i p le b o n d s a re s t if f e r ( a n d v ib r a te a t h ig h e r fr e q u e n c ie s ) th a n d o u b le b o n d s , a n d d o u b le b o n d s a re s t if f e r ( a n d v ib r a t e a t h ig h e r fr e q u e n c ie s ) th a n s in g le b o n d s . W e c a n se e s o m e o f th e s e e ffe c ts in T a b le 2 . 7 . N o t i c e t h a t s t r e t c h i n g f r e q u e n c i e s o f g r o u p s i n v o l v i n g h y d r o g e n ( a l i g h t a t o m ) s u c h as C — H , N — H , a n d O — H a ll o c c u r a t r e la t iv e ly h ig h fr e q u e n c ie s :

GROUP

BOND

FREQUENCY RANGE (cm“

Alkyl

C— H

2853-2962

Alcohol

O— H

3590-3650

Amine

N— H

3300-3500

N o t ic e , to o , th a t t r i p le b o n d s v ib r a t e a t h ig h e r fr e q u e n c ie s th a n d o u b le

BOND

FREQUENCY RANGE (cm“ 1)

C# C

2100-2260

C# N

2220-2260

C= C

1620-1680

C= O

1630-1780

•

I n o rd e r f o r a v ib ra tio n to o c c u r w ith the absorption o f IR energy, the dipole m om ent o f the m olecule m ust change as the v ib ra tio n occurs. N o t a ll m o le c u la r v ib r a t io n s r e s u lt in th e a b s o r p tio n o f I R e n e rg y .

T h u s , m e t h a n e d o e s n o t a b s o r b I R e n e r g y f o r s y m m e t r i c s t r e c h in g o f t h e f o u r C — H b o n d s ; a s y m m e t r ic s tr e tc h in g , o n th e o th e r h a n d , d o e s le a d to a n I R a b s o r p tio n . S y m m e t r ic a l v i b r a ­ tio n s o f th e c a r b o n - c a r b o n d o u b le a n d t r ip le b o n d s o f e th e n e a n d e th y n e d o n o t r e s u lt in t h e a b s o r p t i o n o f I R r a d i a t i o n , e it h e r . V i b r a t i o n a l a b s o r p t i o n m a y o c c u r o u t s id e t h e r e g i o n m e a s u r e d b y a p a r t i c u l a r I R s p e c ­ tr o m e te r , a n d v ib r a t io n a l a b s o r p tio n s m a y o c c u r so c lo s e ly to g e th e r th a t p e a k s f a l l o n to p o f peaks.

86

Chapter 2

Families o f Carbon Compounds

C haracteristic Infrared Absorptions o f Groups F req u en cy R ange (cm-1 )

G ro u p A. Alkyl C — H (stretching) Isopropyl, — CH(CH 3)2

and tert-Butyl, — C(CH 3)3 and B. Alkenyl C — H (stretching) C = C (stretching) R— CH = CH 2 R2C = CH 2

(out-of-plane C — H bendings)

and

c/s-RCH= CHR trans-RC H = CHR C. Alkynyl # C — H (stretching) C # C (stretching) D. A rom atic Ar — H (stretching) C —C (stretching) A rom atic substitution type (C— H out-of-plane bendings) M onosubstituted and o-D isubstituted m -D isubstituted and p-D isubstituted E. A lcohols, P h en o ls, an d C arboxylic A cids O — H (stretching) Alcohols, phenols (dilute solutions) Alcohols, phenols (hydrogen bonded) Carboxylic acids (hydrogen bonded) F. E th e rs an d A lcohols C — O — C (stretching) G. A ld eh y d e s, K e to n e s, E th e rs, C arboxylic A cids, a n d A m ides C = O (stretching) A ldehydes K etones Esters Carboxylic acids A m ides H. A m ines N— H I. N itriles C# N

2853-2962 1380-1385 1365-1370 1385-1395 -1365

In ten sity a (m-s) (s) (s) (m) (s)

3010-3095 1620-1680 985-1000 905-920 880-900

(m) (v) (s) (s) (s)

675-730 960-975

(s) (s)

3300 2100-2260

(s) (v)

3030 1450-1600

(v) (m)

690-710 730-770 735-770 680-725 750-810 800-860

(very (very (s) (s) (very (very

3590-3650 3200-3550 2500-3000

(sharp, v) (broad, s) (broad, v)

1020-1275

(s)

1630­ ■1780 1690­ ■1740 1680­ ■1750 1735­ ■1750 1710­ ■1780 1630- ■1690

(s) (s) (s) (s) (s) (s)

3300-3500

(m)

2220-2260

(m)

s) s)

s) s)

A b b re v ia tio n s : s = strong, m = m edium , w = weak, v = variable, ~ = approxim ately.

Other factors bring about even more absorption peaks. Overtones (harmonics) of fun­ damental absorption bands may be seen in IR spectra even though these overtones occur with greatly reduced intensity. Bands called combination bands and difference bands also appear in IR spectra.

2.16 Interpreting IR Spectra

Because IR spectra of even relatively simple compounds contain so many peaks, the pos­ sibility that two different compounds will have the same IR spectrum is exceedingly small. It is because of this that an IR spectrum has been called the “fingerprint” of a molecule. Thus, with organic compounds, if two pure samples give different IR spectra, one can be certain that they are different compounds. If they give the same IR spectrum, then they are very likely to be the same compound.

2.16 Interp retin g IR Spectra IR spectra contain a wealth of information about the structures of compounds. We show some of the information that can be gathered from the spectra of octane and methylbenzene (com­ monly called toluene) in Figs. 2.11 and 2.12. In this section we shall learn how to recognize the presence of characteristic IR absorption peaks that result from vibrations of alkyl and func­ tional groups. The data given in Table 2.7 will provide us with key information to use when correlating actual spectra with IR absorption frequencies that are typical for various groups.

W a v e n u m b e r (c m 1)

Figure 2.11 The IR spectrum o f octane. (N otice th a t, in IR spectra, th e peaks are usually measured in % transm ittance. Thus, th e peak at 2900 cm - 1 has 1 0 % transm ittance, th a t is, an absorbance, A, o f 0.90.)

Figure 2.12 The IR spectrum o f W a v e n u m b e r (c m 1)

2.16A Infrared Spectra of Hydrocarbons •

A ll hydrocarbons give absorption peaks in the 2800-3300-cm-1 region that are associated with carbon-hydrogen stretching vibrations. We can use these peaks in interpreting IR spectra because the exact location of the peak depends on the strength (and stiffness) of the C — H bond, which in turn depends on

m ethylbenzene (toluene).

88

Chapter 2

Families o f Carbon Compounds

the hybridization state of the carbon that bears the hydrogen. The C — H bonds involving ip-hybridized carbon are strongest and those involving sp3-hybridized carbon are weakest. The order of bond strength is 2

sp > sp > sp

3

This, too, is the order of the bond stiffness.

• The carbon-hydrogen stretching peaks of hydrogen atoms attached to sphybridized carbon atoms occur at highest frequencies, about 3300 cm-1 . The carbon-hydrogen bond of a terminal alkyne ( # C — H) gives an absorption in the 3300-cm-1 region. We can see the absorption of the acetylenic (alkynyl) C — H bond of 1-heptyne at 3320 cm - 1 in Fig. 2.13.

Figure 2.13 The IR spectrum of 1

-heptyne.

W a v e n u m b e r (c m 1)

•

The carbon-hydrogen stretching peaks of hydrogen atoms attached to sp2hybridized carbon atoms occur in the 3000-3100-cm-1 region.

Thus, alkenyl C — H bonds and the C — H groups of aromatic rings give absorption peaks in this region. We can see the alkenyl C — H absorption peak at 3080 cm - 1 in the spectrum of 1-octene (Fig. 2.14), and we can see the C — H absorption of the aromatic hydrogen atoms at 3090 cm - 1 in the spectrum of methylbenzene (Fig. 2.12). •

The carbon-hydrogen stretching bands of hydrogen atoms attached to sp3-hybridized carbon atoms occur at lowest frequencies, in the 2800-3000-cm-1 region.

We can see methyl and methylene absorption peaks in the spectra of octane (Fig. 2.11), methylbenzene (Fig. 2.12), 1-heptyne (Fig. 2.13), and 1-octene (Fig. 2.14).

Figure 2.14 The IR spectrum of 1

-octene.

W a v e n u m b e r (c m 1)

2.16 Interpreting IR Spectra

H ydrocarbons also give absorption peaks in their IR spectra that result from carb o n -c ar­ bon bond stretchings. C arbon-carbon single bonds norm ally give rise to very w eak peaks that are usually of little use in assigning structures. M ore useful peaks arise from carbon-carbon m ultiple bonds, however. •

C a rb o n -c arb o n double bonds give absorption peaks in the 1 6 2 0 -1 6 8 0 -c m -1 region, and c a rb o n -c arb o n triple bonds give absorption peaks b etw een 2 1 0 0 and 2260 c m - 1 .

These absorptions are not usually strong ones, and they are absent if the double or triple bond is sym m etrically substituted. (No dipole m om ent change w ill be associated w ith the vibration.) The stretchings o f the carbon-carbo n bonds o f benzene rings usually give a set o f characteristic sharp peaks in the 1 4 5 0-1600 -cm -1 region. •

A bsorptions arising from ca rb o n -h y d ro g en bending vibrations o f alkenes occur in the 6 0 0 -1 0 0 0 -c m -1 region. W ith the aid o f a spectroscopy handbook, the exact location o f these peaks can often be used as evidence for the s u b s titu ­ tio n p a tte rn o f the double b o n d a n d its c o n fig u ra tio n .

2.16B IR Spectra of Some Functional Groups Containing Heteroatoms Infrared spectroscopy gives us an invaluable m ethod for recognizing quickly and sim ply the presence o f certain functional groups in a m olecule. C arbonyl Functional Groups

^ ____ H e lp f u l H i n t IR spectroscopy is an exceedingly useful tool for detecting functional groups.

O ne im portant functional group that gives a prom inent

absorption peak in IR spectra is the c a rb o n y l g ro u p , ^ C = O . This group is present in aldehydes, ketones, esters, carboxylic acids, am ides, and others. •

The carbon-oxygen double-bond stretching frequency o f carbonyl groups gives a strong p eak betw een 1630 and 1780 c m - 1 . The exact location of the absorption depends on w hether it arises from an aldehyde, ketone, ester, and so forth.

O

O

O

O C

.C v R A ld e h y d e 1 6 9 0 -1 7 4 0 c m 1

R

'R

K e to n e 1 6 8 0 -1 7 5 0 c m 1

R

O

OR E s te r

1 7 3 5 -1 7 5 0 c m -

R

C OH

C a r b o x y li c a c id 1 7 1 0 - 1 7 8 0 c m -1

R A m id e 1 6 3 0 - 1 6 9 0 c m -1

S o lv e d P ro b le m 2 .7

A com pound with the molecular formula C 4 H 4O 2 has a strong sharp absorbance near 3300 cm -1 , absorbances in the 2800-3000-cm -1 region, and a sharp absorbance peak near 2200 cm - 1 . It also has a strong broad absorbance in the 2500-3600-cm - 1 region and a strong peak in the 1710-1780-cm - 1 region. Propose a possible structure for the compound. STRATEGY AND ANSW ER The sharp peak near 3300 c m -1 is likely to arise from the stretching o f a hydrogen attached to the ^ -h y b rid iz e d carbon of a triple bond. The sharp peak near 2200 c m - 1 , w here the triple bond o f an alkyne stretches, is consistent w ith this. The peaks in the 2 8 0 0 -3 0 0 0 -cm -1 region suggest stretchings o f the C — H bonds of alkyl groups, either C H 2 or C H 3 groups. The strong, broad absorbance in the 2 5 0 0 -3 6 0 0 -cm -1 region sug­ gests a hydroxyl group arising from a carboxylic acid. The strong peak around 11 ----oh 1710-1780 c m -1 is consistent w ith this since it could arise from the carbonyl group o f a carboxylic acid. Putting all this together w ith the m olecular form ula suggests the o com pound is as shown at the right.

90

Chapter 2

R e v ie w P r o b le m 2 . 2 8

Families o f Carbon Compounds

U s e a r g u m e n ts b a s e d o n re s o n a n c e a n d e le c t r o n e g a t iv it y e ffe c ts to e x p la in th e tr e n d i n c a r ­ b o n y l IR

s tr e tc h in g fr e q u e n c ie s f r o m

h i g h e r f r e q u e n c y f o r e s te r s a n d c a r b o x y l i c a c id s t o

(H in t:

U s e th e ra n g e o f c a r b o n y l s tr e tc h in g fr e q u e n c ie s f o r

l o w e r f r e q u e n c i e s f o r a m id e s .

a ld e h y d e s a n d k e t o n e s a s t h e “ b a s e ” f r e q u e n c y r a n g e o f a n u n s u b s t i t u t e d c a r b o n y l g r o u p a n d c o n s i d e r t h e i n f l u e n c e o f e le c t r o n e g a t i v e a t o m s o n t h e c a r b o n y l g r o u p a n d / o r a t o m s th a t a lt e r th e re s o n a n c e h y b r i d o f th e c a r b o n y l. ) W h a t d o e s t h is s u g g e s t a b o u t th e w a y th e n i t r o g e n a t o m i n f l u e n c e s t h e d i s t r i b u t i o n o f e le c t r o n s i n a n a m i d e c a r b o n y l g r o u p ?

Alcohols and Phenols r e c o g n iz e in I R

The

h y d ro x y l g ro u p s

o f a l c o h o l s a n d p h e n o ls a r e a ls o e a s y t o

s p e c t r a b y t h e i r O — H s t r e t c h i n g a b s o r p t i o n s . T h e s e b o n d s a ls o g i v e u s

d i r e c t e v id e n c e f o r h y d r o g e n b o n d i n g ( S e c t i o n 2 . 1 3 B ) . •

T h e I R a b s o r p t i o n o f a n a l c o h o l o r p h e n o l O — H g r o u p is i n t h e 3 2 0 0 - 3 5 5 0 - c m ^ 1 ra n g e , a n d m o s t o f te n i t is b ro a d .

T h e t y p i c a l b r o a d n e s s o f t h e p e a k is d u e t o a s s o c i a t i o n o f t h e m o l e c u l e s t h r o u g h h y d r o g e n b o n d in g ( S e c tio n 2 .1 3 B ) , w h ic h c a u s e s a w id e r d is t r ib u t io n o f s tr e tc h in g fr e q u e n c ie s f o r th e O — H b o n d . I f a n a lc o h o l o r p h e n o l is p r e s e n t a s a v e r y d il u t e s o lu t io n i n a s o lv e n t th a t c a n n o t c o n t r i b u t e t o h y d r o g e n b o n d i n g ( e . g . , C C I 4) , O — H a b s o r p t i o n o c c u r s a s a v e r y s h a r p p e a k i n th e 3 5 9 0 - 3 6 5 0 - c m ^ 1 r e g io n . I n v e r y d ilu t e s o lu t io n i n s u c h a s o lv e n t o r i n th e g a s p h a s e , f o r m a t i o n o f i n t e r m o l e c u l a r h y d r o g e n b o n d s d o e s n o t t a k e p la c e b e c a u s e m o l e c u l e s o f th e a n a ly te a re t o o w id e ly s e p a ra te d . A s h a rp p e a k i n th e 3 5 9 0 - 3 6 5 0 - c m ~ 1 r e g io n , th e r e ­ fo r e , is a t t r ib u t e d to “ f r e e ” ( u n a s s o c ia te d ) h y d r o x y l g r o u p s . I n c r e a s in g th e c o n c e n tr a tio n o f th e a lc o h o l o r p h e n o l c a u s e s th e s h a rp p e a k to b e r e p la c e d b y

a b ro a d b a n d in

th e

3 2 0 0 - 3 5 5 0 - c m ~ 1 r e g io n . H y d r o x y l a b s o r p tio n s in I R s p e c tra o f c y c lo h e x y lc a r b in o l ( c y c lo h e x y l m e t h a n o l ) r u n i n d i l u t e a n d c o n c e n t r a t e d s o lu t i o n s ( F i g . 2 . 1 5 ) e x e m p l i f y t h e s e e f f e c t s .

(cm 1)

0 .0

4000 II 1

111

3000 III

4000

3000

\

I

'Free'

0 .2

0.4

Figure 2.15 (a) The IR spectrum o f an alcohol (cyclohexylcarbinol) in a d ilu te solution shows th e sharp abso rp tio n o f a " fre e " (non-hydrogen-bonded) hydroxyl g ro u p at 3600 cm - 1 . (b) The IR spectrum o f th e same alcohol as a concentrated solution shows a broad hydroxyl g ro u p abso rp tio n at 3300 cm - 1 due to hydrogen b onding. (R eprinted w ith perm ission o f John W ile y & Sons, Inc., from Silverstein, R., a nd W ebster, F X., S p e ctrom e tric Id e n tifica tio n o f O rga n ic Com pounds, Sixth E dition, p. 89. C o p y rig h t 1998.)

C arboxylic Acids

0 .6

A Interm olecularly. hydrogen bonded-

2.5

3

2.5

3

Cum) (b)

(a) The

c a rb o x y lic a c id g ro u p

c a n a ls o b e d e t e c t e d b y I R s p e c t r o s c o p y .

I f b o t h c a r b o n y l a n d h y d r o x y l s tr e tc h in g a b s o r p tio n s a re p r e s e n t in a n I R

s p e c tr u m , th e re

i s g o o d e v id e n c e f o r a c a r b o x y l i c a c i d f u n c t i o n a l g r o u p ( a l t h o u g h i t i s p o s s i b l e t h a t i s o ­ la t e d c a r b o n y l a n d h y d r o x y l g r o u p s c o u ld b e p r e s e n t i n th e m o le c u le ) . •

T h e h y d r o x y l a b s o r p tio n o f a c a r b o x y l ic a c id is o f t e n v e r y b r o a d , e x te n d in g f r o m 3600 cm

-1

to 2 5 0 0 c m - 1 .

F i g u r e 2 . 1 6 s h o w s t h e I R s p e c t r u m o f p r o p a n o i c a c id .

2.16 Interpreting IR Spectra

W avenum ber (cm 1)

Figure 2.16 The IR spectrum o f propanoic acid. A m ines

IR spectroscopy also gives evidence for N — H bonds (see Figure 2.17).

•

Primary (1°) and secondary (2°) amines give absorptions of moderate strength in the 3300-3500-cm 1 region.

•

Primary amines exhibit two peaks in this region due to symmetric and asymmetric stretching of the two N — H bonds.

•

Secondary amines exhibit a single peak.

•

Tertiary amines show no N — H absorption because they have no such bond.

•

A basic pH is evidence for any class of amines.

1 0 0

90­ 80 -| ^

)

arom atic combination band

__ -a lip h a tic arom atic C — H C— H (stretch) (stretch)

70­

e

N— H (wag)

£ 60ta 'g 50 s prim ary N— H § 4 0 ­ (asym. and sym. stretch)

C— N (stretch)

30 -| 2 0

­

­ 4000

C— H (out-of-plane | bend)

C — C (ring stretch), N— H (bend)

1 0

0

3600

3200

2800

2400

2000

1800

1600

1400

1200

1000

800

650

W avenumber (cm-1)

Figure 20.17 A n n o ta te d IR spectrum o f 4-m ethylaniline.

RNH2 (1° Amine) Two peaks in 3300-3500-cm -1 region

v

*

Sym m etric stretching

v

R2NH (2° Amine) One peak in 3300-3500-cm-1 region

Asym m etric stretching

Hydrogen bonding causes N — H stretching peaks of 1° and 2° amines to broaden. The NH groups of am ides give similar absorption peaks and include a carbonyl absorption as well.

91

92

Chapter 2

Families o f Carbon Compounds

2.17 Applications o f Basic Principles We now review how certain basic principles apply to phenom ena that w e have studied in this chaper. Polar Bonds A re Caused by Electro n eg ativity D ifferences We saw in Section 2.2 that w hen atom s w ith different electronegativities are covalently bonded, the m ore elec­ tronegative atom w ill be negatively charged and the less electronegative atom w ill b e p o s­ itively charged. T he b o n d w ill be a p o la r bond and it w ill have a dipole moment. D ipole moments are important in explaining physical properties o f molecules (as w e shall review below), and in explaining infrared spectra. For a vibration to occur with the absorption of IR energy, the dipole m om ent o f the m olecule m ust change during the course o f the vibration. O p p o site Charges A ttra c t This principle underlies a m ap o f electrostatic potential (M EP) (Section 2.2A). M EPs are generated on the basis o f quantum m echanical calculations that involve m oving an im aginary positive charge over the electron density surface o f a m ol­ ecule. If there is a strong attraction betw een the positive charge and the electron density sur­ face, that region is colored red because it is m ost negative. Regions that are less negative are shaded green to yellow. Regions that are the least negative (or m ost positive) are colored blue. This sam e principle is central to understanding physical properties o f organic com pounds (Section 2.13). A ll o f the forces that operate betw een individual m olecules (and thereby affect boiling points, m elting points, and solubilities) are betw een oppositely charged m olecules (ions) or betw een oppositely charged portions o f m olecules. E xam ples are io n -io n forces (Section 2.13A ) that exist betw een oppositely charged ions in crystals o f ionic com pounds, d ip o le-d ip o le forces (Section 2.13B) that exist betw een oppositely charged portions o f polar m olecules and that include the very strong d ip o le-d ip o le forces that w e call hydrogen bonds, and the w eak dispersion or London fo rce s that exist betw een portions o f m olecules that b ear sm all tem porary opposite charges. M olecular Structure D eterm ines Properties ical properties are related to m olecular structure.

We learned in Section 2.13 how phys-

W nThjsChapte^ In Chapter 2 you learned about families o f organic m olecules, som e o f their physical proper­ ties, and how w e can use an instrum ental technique called infrared spectroscopy to study them. You learned that functional groups define the fam ilies to w hich organic com pounds belong. A t this poin t you should b e able to nam e functional groups w hen you see them in structural form ulas, and, w hen given the nam e o f a functional group, draw a general exam ­ ple o f its structure. You also built on your know ledge o f how electronegativity influences charge distribu­ tion in a m olecule an d how, together w ith three-dim ensional structure, charge distribution influences the overall polarity o f a m olecule. B ased on polarity and three-dim ensional struc­ ture, you should be able to predict the k in d and relative strength o f electrostatic forces betw een m olecules. W ith this understanding you w ill b e able to roughly estim ate physical properties such as m elting point, boiling point, an d solubility. Lastly, you learned to use IR spectroscopy as an indicator o f the fam ily to w hich an organic com pound belongs. IR spectroscopy provides signatures (in the form o f spectra) that suggest w hich functional groups are present in a m olecule. If you know the concepts in C hapters 1 an d 2 w ell, you w ill be on your w ay to having the solid foundation you need for success in organic chem istry. K eep up the good w ork (including your diligent hom ew ork habits)!

Key Terms and Concepts The key term s and concepts that are highlighted in b o ld , b lu e te x t w ithin the chapter are defined in the glossary (at the back o f the book) and have hyperlinked definitions in the accom panying W ileyP LU S course (w w w .w ileyplus.com ).

93

Problems

Problems N ote to Instructors: Many of th e hom ew ork problem s are available for assignm ent via WileyPLUS, an online teaching and learning solution.

FU N C T IO N A L GROUPS A N D STRUCTURAL FORMULAS 2 .2 9

C l a s s i f y e a c h o f t h e f o l l o w i n g c o m p o u n d s a s a n a lk a n e , a lk e n e , a l k y n e , a l c o h o l , a ld e h y d e , a m in e , a n d s o f o r t h .

0

(a)

0

OH

(c)

( b ) C H 3— C # C H

H OH

(e)

(f) 7 ^ ---------"

2 .3 0

12s

Sex attractan t of th e com m on housefly

O btained from oil of cloves

I d e n t if y a ll o f th e fu n c tio n a l g ro u p s in e a c h o f th e f o llo w in g c o m p o u n d s :

O Vitamin D 3

(b )

OMe

HO.

O

n h

A spartam e

2

A m phetam ine

(c)

O

O

(f) Demerol

^

^

^

^

^

h

(g)

O

O \ — O

^

A cockroach repellent found in cu cu m b ers A sy n th etic cockroach repellent

C H 2 .3 1

3

T h e r e a re f o u r a lk y l b r o m id e s w it h th e f o r m u la C

4 H 9B r .

W r it e t h e ir s tr u c tu r a l f o r m u la s a n d c la s s if y e a c h as to

w h e th e r i t is a p r im a r y , s e c o n d a ry , o r t e r t ia r y a lk y l b r o m id e . 2 .3 2

T h e re a re s e v e n is o m e r ic c o m p o u n d s w it h th e f o r m u la C

4 H 10O .

W r it e th e ir s tru c tu re s a n d c la s s ify e a c h c o m p o u n d

a c c o r d in g to its fu n c t io n a l g r o u p . 2 .3 3

C la s s if y th e f o l lo w in g a lc o h o ls as p r im a r y , s e c o n d a r y , o r t e r t ia r y :

1 (a)

J \^ 0 H

1 (b)

°h (c)

OH

^

r~ \y 0 H

/

(d)

(e) /

V

0H

94 2 .3 4

Chapter 2

Families o f Carbon Compounds

C l a s s i f y t h e f o l l o w i n g a m in e s a s p r i m a r y , s e c o n d a r y , o r t e r t i a r y :

H

N

I .N v

(a)

(b)

(c)

2

HN

(e)

2.35

N

(f)

W r it e s tr u c tu r a l fo r m u la s f o r e a c h o f th e f o llo w in g :

4 H 10O .

(a )

T h r e e e th e rs w it h th e f o r m u la C

(b )

T h r e e p r im a r y a lc o h o ls w i t h th e f o r m u la C

(c )

A

s e c o n d a r y a lc o h o l w it h th e f o r m u la C

(d )

A

t e r t ia r y a lc o h o l w it h th e f o r m u la C

3 H 6O

4 H 8O .

3 H 6O .

4 H 8O .

(e )

T w o e s te rs w it h th e f o r m u la C

(f)

F o u r p r im a r y a lk y l h a lid e s w it h th e fo r m u la C

5 H 11B r .

(g )

T h re e

fo r m u la

C

5H

s e c o n d a ry

a lk y l

h a lid e s

2.

w it h

th e

(h)

A te r tia r y a lk y l h a lid e w it h th e fo r m u la C

(i)

T h r e e a ld e h y d e s w i t h t h e f o r m u l a C

(j)

T h r e e k e to n e s w it h th e f o r m u la C

(k) (l)

T w o p r i m a r y a m in e s w i t h t h e f o r m u l a C A

s e c o n d a r y a m in e w i t h th e f o r m u la C

(m )

A

t e r t ia r y a m in e w i t h th e f o r m u la C

(n)

T w o a m id e s w i t h t h e f o r m u l a C

5 H 11 B r .

5 H 10O .

5 H 10O . 3H 9N .

3H 9N .

3H 9N .

2H 5N O .

n B r.

PHYSICAL PROPERTIES 2.36

( a ) I n d ic a t e th e h y d r o p h o b ic a n d h y d r o p h ilic p a r ts o f v it a m in A b e s o lu b le i n w a te r . ( b ) D o th e s a m e f o r v i t a m in B

3

a n d c o m m e n t o n w h e t h e r y o u w o u ld e x p e c t i t to

( a ls o c a l l e d n i a c i n ) .

O OH

Vitamin A 2.37

H y d r o g e n f l u o r i d e h a s a d i p o l e m o m e n t o f 1 .8 3 D ; i t s b o i l i n g p o i n t is 1 9 . 3 4 ° C . E t h y l f l u o r i d e ( C H

3C

H

2F )

h as an

a l m o s t i d e n t i c a l d i p o l e m o m e n t a n d h a s a l a r g e r m o l e c u l a r w e i g h t , y e t i t s b o i l i n g p o i n t is - 3 7 . 7 ° C . E x p l a i n .

2.38

W h y d o e s o n e e x p e c t t h e c i s i s o m e r o f a n a lk e n e t o h a v e a h i g h e r b o i l i n g p o i n t t h a n t h e t r a n s i s o m e r ?

2.39

C e t y le t h y ld im e t h y la m m o n iu m b r o m id e is th e c o m m o n n a m e f o r B r­ ' , a c o m p o u n d w i t h a n t is e p t ic p r o p e r tie s . P r e d ic t

'N +

/

\

i t s s o l u b i l i t y b e h a v i o r i n w a t e r a n d i n d i e t h y l e th e r .

2.40

W h i c h o f t h e f o l l o w i n g s o lv e n t s s h o u l d b e c a p a b l e o f d i s s o l v i n g i o n i c c o m p o u n d s ? (a )

2.41

L iq u id S O

2

(b )

L iq u id N H

3

(c )

B enzene

(d )

C C l4

W r i t e a t h r e e - d i m e n s i o n a l f o r m u l a f o r e a c h o f t h e f o l l o w i n g m o l e c u l e s u s in g t h e w e d g e - d a s h e d w e d g e - l i n e f o r ­ m a lis m . I f th e m o le c u le h a s a n e t d ip o le m o m e n t, in d ic a te its d ir e c t io n w it h a n a r r o w ,

1-------- >.

I f th e m o le c u le h a s

n o n e t d i p o l e m o m e n t , y o u s h o u l d s o s ta te . ( Y o u m a y i g n o r e t h e s m a l l p o l a r i t y o f C — H b o n d s i n w o r k i n g t h i s a n d s im ila r p r o b le m s .)

2.42

(a )

C H 3F

(b )

CH

2F 2

(c )

C HF

(d )

CF

3

4

(e )

C H 2F C l

(g )

BeF

(f)

B C l3

(h )

CH

2

3O

C H

3

( i)

C H 3O H

( j)

C H 2O

C o n s id e r e a c h o f th e f o l lo w in g m o le c u le s i n tu r n : ( a ) d im e t h y l e th e r, ( C H tr im e t h y lb o r o n , ( C H

3) 3B ;

a n d ( d ) d im e t h y lb e r y lliu m , ( C H

3) 2 B e .

3) 2 O ;

( b ) t r im e t h y la m in e , ( C H

3) 3 N ;

(c )

D e s c r ib e th e h y b r id iz a t io n s ta te o f th e c e n t r a l a to m

( i . e . , O , N , B , o r B e ) o f e a c h m o l e c u l e , t e l l w h a t b o n d a n g le s y o u w o u l d e x p e c t a t t h e c e n t r a l a t o m , a n d s ta te w h e t h e r th e m o le c u le w o u ld h a v e a d ip o le m o m e n t.

95

Problems

2 .4 3

A lkenes can interact w ith m etal ions such as A g+ . W hat is the nature o f this interaction?

2 .4 4

A nalyze the statem ent: For a m olecule to be polar, the presence o f polar bonds is necessary, but it is not a suffi­ cient requirem ent.

2 .4 5

Identify all o f the functional groups in Crixivan, an im portant drug in the treatm ent o f AIDS.

C r i x i v a n ( a n H IV p r o t e a s e i n h i b i t o r ) 2 .4 6

W hich com pound in each o f the follow ing pairs w ould have the higher boiling point? E xplain your answers. F

"OH

(a)

O

or

or

(f)

O

"OH

(b)

HO

or

(g)

OH

or

O H

-O .

O

O (c)

„OH

or

(h) H exane, C H 3 (CH 2) 4CH 3, or nonane, CH 3 (CH 2) 7CH 3 (d )

I

O

or

[ > —

OH

O (e)

or

N — CH

or

(i)

3

IR SPECTROSCOPY 2 .4 7

Predict the key IR absorption bands w hose presence w ould allow each com pound in pairs (a), (c), (d), (e), (g), and (i) from Problem 2.46 to be distinguished from each other.

2 .4 8

The infrared spectrum o f 1-hexyne exhibits a sharp absorption p eak near 2100 c m -1 due to C # C stretching. However, 3-hexyne shows no absorption in that region. Explain.

2 .4 9

The IR spectrum of propanoic acid (Fig. 2.17) indicates that the absorption for the O — H stretch o f the carboxylic acid functional group is due to a hydrogen-bonded form. D raw the structure o f tw o propanoic acid m olecules show ­ ing how they could dim erize via hydrogen bonding.

2 .5 0

In infrared spectra, the carbonyl group is usually indicated by a single strong and sharp absorption. However, in the case o f carboxylic acid anhydrides, R — C — O — C — R, tw o peaks are observed even though the two

O

O

carbonyl groups are chem ically equivalent. E xplain this fact, considering w hat you know about the IR absorption o f prim ary am ines. M ULTICONCEPT PROBLEMS 2 .5 1

W rite structural form ulas for four com pounds w ith the form ula C 3 H6O and classify each according to its functional group. Predict IR absorption frequencies for the functional groups you have drawn.

2 .5 2

There are four am ides w ith the form ula C 3 H 7 NO. (a) W rite their structures. (b) O ne o f these am ides has a m elt­ ing and a boiling point that are substantially low er than those o f the other three. W hich am ide is this? E xplain your answer. (c) Explain how these am ides could be differentiated on the basis o f their IR spectra.

96

Chapter 2

Families o f Carbon Compounds

2 .53

Write structures for all compounds with molecular formula C4H60 that would not be expected to exhibit infrared absorption in the 3200-3550-cm-1 and 1620-1780-cm-1 regions.

2 .5 4

Cyclic compounds of the general type shown here are called lactones. What functional group does a lactone contain? =0

Challenge Problems 2 .55

Two constitutional isomers having molecular formula C4H60 are both symmetrical in structure. In their infrared spectra, neither isomer when in dilute solution in CCl4 (used because it is nonpolar) has absorption in the 3600cm-1 region. Isomer A has absorption bands at approximately 3080, 1620, and 700 cm-1 . Isomer B has bands in the 2900-cm-1 region and at 1780 cm-1 . Propose a structure for A and two possible structures for B.

2 .5 6

When two substituents are on the same side o f a ring skeleton, they are said to be cis, and when on opposite sides, trans (analogous to use o f those terms with 1,2-disubstituted alkene isomers). Consider stereoisomeric forms of 1,2-cyclopentanediol (compounds having a five-membered ring and hydroxyl groups on two adjacent carbons that are cis in one isomer and trans in the other). A t high dilution in CCl4, both isomers have an infrared absorption band at approximately 3626 cm-1 but only one isomer has a band at 3572 cm-1 . (a) Assume for now that the cyclopentane ring is coplanar (the interesting actuality w ill be studied later) and then draw and label the two iso­ mers using the wedge-dashed wedge method of depicting the OH groups. (b) Designate which isomer w ill have the 3572-cm-1 band and explain its origin.

2 .5 7

Compound C is asymmetric, has molecular formula C5H10O, and contains two methyl groups and a 3° func­ tional group. It has a broad infrared absorption band in the 3200-3550-cm-1 region and no absorption in the 1620-1680-cm-1 region. (a) Propose a structure for C.

2 .5 8

Examine the diagram showing an a-helical protein structure in section 2.13E. Between what specific atoms and of what functional groups are the hydrogen bonds formed that give the molecule its helical structure?

Learning Group Problems Consider the molecular formula C4H8O2. 1.

Write structures for at least 15 different compounds that all have the molecular formula C4H8O2 and contain functional groups presented in this chapter.

2.

Provide at least one example each of a structure written using the dash format, the condensed format, the bondline format, and the fu ll three-dimensional format. Use your choice of format for the remaining structures.

3.

Identify four different functional groups from among your structures. Circle and name them on the representative structures.

4.

Predict approximate frequencies for IR absorptions that could be used to distinguish the four compounds repre­ senting these functional groups.

5.

I f any o f the 15 structures you drew have atoms where the formal charge is other than zero, indicate the formal charge on the appropriate atom(s) and the overall charge for the molecule.

6.

Identify which types o f intermolecular forces would be possible in pure samples o f all 15 compounds.

7.

Pick five formulas you have drawn that represent a diversity of structures, and predict their order with respect to trend in increasing boiling point.

8.

Explain your order o f predicted boiling points on the basis o f intermolecular forces and polarity.

Concept Map

97

An Introduction to Organic Reactions and Their Mechanisms Acids and Bases

"•

*

•

*

j *

•

* >

i

*

*

*

" Nylon

To the uninitiated, a chemical reaction must seem like an act of magic. A chemist puts one or two reagents into a flask, waits for a time, and then takes from the flask one or more completely different compounds. It is, until we understand the details of the reaction, like a magician who puts apples and oranges in a hat, shakes it, and then pulls out rabbits and parakeets. We see a real-life example of this sort o f "m agic" in the photo above, where a stu­ dent is shown pulling a strand o f solid nylon from a flask that contains two immiscible solutions. This synthesis of nylon is not magic but it is indeed wonderful and amazing, and reactions like it have transformed our world. One o f our goals in this course will be, in fact, to try to understand how this chemical magic takes place. We will want to be able to explain how the products o f the reaction are formed. This explanation will take the form of a reaction mechanism— a description o f the events th a t take place on a molecular level as reactants become products. If, as is often the case, the reaction takes place in more than one step, we will want to know what chem­ ical species, called inte rm e d iate s, intervene between each step along the way. By postulating a mechanism, we may take some of the magic out o f the reaction, but we will put rationality in its place. Any mechanism we propose must be consistent with what we know about the reaction and with what we know about the reactivity o f organic compounds generally. In later chapters we shall see how we can glean evidence for or against a given mechanism from studies of reaction rates, from isolating intermediates, and from spectroscopy. We cannot actually see the molecular events because molecules are too small, but from solid

98

3.1 Reactions and Their Mechanisms

99

evidence and from good chemical intuition, we can propose reasonable mecha­ nisms. If at some later time a valid experiment gives results that contradict our pro­ posed mechanism, then we change it, because in the final analysis our mechanism must be consistent with all our experimental observations. One of the most important things about approaching organic chemistry mech­ anistically is this: It helps us organize what otherwise m ight be an overwhelmingly complex body of knowledge into a form that makes it understandable. There are millions of organic compounds now known, and there are millions of reactions that these compounds undergo. If we had to learn them all by rote memorization, then we would soon give up. But, we don't have to do this. In the same way that func­ tional groups help us organize compounds in a comprehensible way, mechanisms help us organize reactions. Fortunately, too, there is a relatively small number of basic mechanisms.

3.1 Reactions and Their Mechanisms Virtually all organic reactions fall into one of four categories: substitutions, additions, elim inations, or rearrangements. S u b stitu tio n s are the characteristic reactions of saturated compounds such as alkanes and alkyl halides and of aromatic compounds (even though they are unsaturated). In a sub­ stitution, one group replaces another. For example, chloromethane reacts with sodium hydroxide to produce methyl alcohol and sodium chloride: HoO

H 3C — C l

N a +O H -

H 3C — O H

N a + C |-

A s u b s titu tio n reactio n

In this reaction a hydroxide ion from sodium hydroxide replaces the chlorine of methyl chloride. We shall study this reaction in detail in Chapter 6 . A d d itio n s are characteristic of compounds with multiple bonds. Ethene, for example, reacts with bromine by an addition. In an addition a ll p a rts o f the adding reagent appear in the pro duct; two molecules become one: H

H

H

\

/ +

C = C

/

B r— Br

CCl4

'

H — C — C — H

\

H

H

1

I Br

H

I Br

An a d d itio n re a c tio n

E lim in a tio n s are the opposite of additions. In an e lim in a tio n one molecule loses the elements o f another sm all molecule. Elimination reactions give us a method for preparing compounds with double and triple bonds. In Chapter 7, for example, we shall study an important elimination called dehydrohalogenation, a reaction that is used to prepare alkenes. In dehydrohalogenation, as the word suggests, the elements of a hydrogen halide are eliminated. An alkyl halide becomes an alkene: H H

H

H

C— H Br

KOH ( - HBr)

H

\

/ C = C

/ H

A n e lim in a tio n reactio n

\ H

100

Chapter 3

An Introduction to Organic Reactions and Their Mechanisms

In a r e a rr a n g e m e n t a m olecule undergoes a reorganization o f its constituent p a rts. For exam ple, heating the follow ing alkene w ith a strong acid causes the form ation o f another isom eric alkene: H \ ,, ^ H3C\

/

H / C=C

\

P \ / H 3C

H3C \

acid cat. -----------»

/ H 3C

H

CH3 / C=C

\ CH3

CH3 A re a rra n g e m e n t

In this rearrangem ent n o t only have the positions o f the double bond an d a hydrogen atom changed, but a m ethyl group has m oved from one carbon to another. In the follow ing sections w e shall begin to learn som e o f the principles that explain how these kinds o f reactions take place.

3.1A Homolysis and Heterolysis of Covalent Bonds R eactions o f organic com pounds always involve the m aking and breaking o f covalent bonds. A covalent bond m ay b reak in tw o fundam entally different ways. •

W hen a bond breaks such that one fragm ent takes aw ay both electrons o f the bond, leaving the other fragm ent w ith an em pty orbital, this k ind o f cleavage is called hetero ly sis (Greek: hetero, different, + lysis, loosening or cleavage). H eterolysis produces charged fragm ents or ions and is term ed an io n ic re a c tio n . T he broken bond is said to have cleaved h e te ro ly tic a lly : B -------* A + +

A

:B _

H e tero ly tic b on d c le av a g e

'------ »------ ' Ions

H e lp f u l H i n t ____ ^ Notice in these illustrations that we have used curved arrows to show the movement o f electrons. We will have more to say about this convention in Section 3.5, but fo r the moment notice that we use a double-barbed curved arrow to show the m ovement o f a pair of electrons and a single-barbed curved arrow to show the movement o f a single electron.

•

W hen a bond breaks so that each fragm ent takes aw ay one o f the electrons o f the bond, this process is called h o m o ly sis (Greek: homo, the sam e, + lysis). H om olysis produces fragm ents w ith unpaired electrons called r a d ic a ls . r* A

" B ------ * A

+

•B

,----- y----- ,

Kj

H o m o lytic bond c le av a g e

R a d ic a ls

We shall postpone further discussions o f reactions involving radicals and hom olytic bond cleavage until w e reach C hapter 10. A t this p o in t w e focus our attention on reactions involv­ ing ions and heterolytic bond cleavage. H eterolysis o f a bond norm ally requires that the bond be polarized:

O

« + A : B S~ --------> A +

+

:B -

Polarization o f a bond usually results from differing electronegativities (Section 2.2) o f the atom s joined by the bond. T he greater the difference in electronegativity, the greater is the polarization. In the instance ju s t given, atom B is m ore electronegative than A. E ven w ith a highly polarized bond, heterolysis rarely occurs w ithout assistance. The reason: H eterolysis requires separation of oppositely charged ions . B ecause oppositely charged ions attract each other, their separation requires considerable energy. O ften, heterolysis is assisted by a m olecule w ith an unshared p air that can form a bond to one o f the atom s: Y

+ " ^ A — B s - --------> Y — A

+

:B-

Form ation o f the new bond furnishes som e o f the energy required for the heterolysis.

101

3.2 Acid-Base Reactions

3.2 Acid-B ase Reactions We begin our study o f chemical reactions by examining some o f the basic principles of acid-base chemistry. There are several reasons for doing this: •

Many o f the reactions that occur in organic chemistry are either acid-base reac­ tions themselves or they involve an acid-base reaction at some stage.

•

A cid-base reactions are simple fundamental reactions that will enable you to see how chemists use curved arrows to represent mechanisms o f reactions and how they depict the processes of bond breaking and bond making that occur as molecules react.

A cid-base reactions also allow us to examine important ideas about the relationship between the structures of m olecules and their reactivity and to see how certain thermodynamic para­ meters can be used to predict how much of the product w ill be formed when a reaction reaches equilibrium. A cid-base reactions also provide an illustration o f the important role solvents play in chemical reactions. They even give us a brief introduction to organic syn­ thesis. Finally, acid-base chemistry is something that you w ill find familiar because o f your studies in general chemistry. We begin, therefore, with a brief review.

3.2A Bronsted-Lowry Acids and Bases Two classes of acid-base reactions are fundamental in organic chemistry: Br0 nsted-Lowry and Lewis acid-base reactions. We start our discussion with Br0 nsted-Lowry acid-base reactions. •

Br0 nsted-Lowry acid-base reactions involve the transfer o f protons.

•

A B r 0 n s te d -L o w ry a c id is a substance that can donate (or lose) a proton.

•

A B r 0 n s te d -L o w ry b a se is a substance that can accept (or remove) a proton.

Let us consider some examples. Hydrogen chloride (HCl), in its pure form, is a gas. When HCl gas is bubbled into water, the follow ing reaction occurs. H— O

+

H — Cl:

H— O — H

H

+

:Ch -

H

B ase (p ro to n a c c e p to r)

A cid (p ro to n d o n o r)

C o n ju g a te a cid o f H2O

C o n ju g a te b ase of HCl

The color o f hydrangea flow ers depends in pa rt on th e relative a cid ity o f th e ir soil.

In this reaction hydrogen chloride donates a proton; therefore it acts as a Br 0 nsted-Lowry acid. Water accepts a proton from hydrogen chloride; thus water serves as a Br0 nsted-Lowry base. The products are a hydronium ion (H 3 O +) and chloride ion (Cl- ). Just as w e classified the reactants as either an acid or a base, w e also classify the prod­ ucts in a specific way. •

The m olecule or ion that forms when an acid loses its proton is called the c o n ju ­ g a te b ase o f that acid. In the above example, chloride ion is the conjugate base.

•

The m olecule or ion that forms when a base accepts a proton is called the con ju ­ gate a cid . Hydronium ion is the conjugate acid o f water.

Hydrogen chloride is considered a strong acid because transfer o f its proton in water proceeds essentially to completion. Other strong acids that com pletely transfer a proton when dissolved in water are hydrogen iodide, hydrogen bromide, and sulfuric acid. HI

+

H2 O

---- :

H3 O +

+

I-

HBr

+

H2 O

---- :

H3 O +

+

Br-

H2 SO 4

+

H2 O

---- :

H3 O +

+

h so

4- +

H2 O

2—

H3 O +

+

SO

h so

4-

42 -

H e lp f u l H i n t The extent to which an acid

transfers protons to a base, such as water, is a measure of its strength as an acid. Acid strength is therefore a measure of the percentage of ionization and n o tof concentration.

102

Chapter 3

An Introduction to Organic Reactions and Their Mechanisms

Sulfuric acid is called a diprotic acid because it can transfer two protons. Transfer of the first proton occurs completely, while the second is transferred only to the extent of about 1 0 % (hence the equilibrium arrows in the equation for the second proton transfer).

3.2B Acids and Bases in Water •

Hydronium ion is the strongest acid that can exist in water to any significant extent: Any stronger acid will simply transfer its proton to a water molecule to form hydronium ions.

•

Hydroxide ion is the strongest base that can exist in water to any significant extent: Any base stronger than hydroxide will remove a proton from water to form hydroxide ions.

When an ionic compound dissolves in water the ions are solvated. With sodium hydrox­ ide, for example, the positive sodium ions are stabilized by interaction with unshared elec­ tron pairs of water molecules, and the hydroxide ions are stabilized by hydrogen bonding of their unshared electron pairs with the partially positive hydrogens of water molecules. H

H2 Q: N a+ H 2Q :

H — Q :

H

:QH,

H 2Q :

:0 — H

:Q — H

:Q — H

:Q H ,

HQ

H

H H

S o lv a te d s o d iu m ion

-Q :

S o lv a te d h y d ro x id e ion

When an aqueous solution of sodium hydroxide is mixed with an aqueous solution of hydrogen chloride (hydrochloric acid), the reaction that occurs is between hydronium and hydroxide ions. The sodium and chloride ions are called s p e c ta to r io n s because they play no part in the acid-base reaction: Total Ionic Reaction H — Q — H

I

:C l

v _

+

N a+

- :Q — H

~Y--------

-------->

N a+

2 H — Q ;

"

I

H

+ “V

: C l :-

1

H

------------S p e c ta to r ion s

—

N et Reaction H — Q — H

-:Q — H

2 H — Q:

I

I

H

H

What we have just said about hydrochloric acid and aqueous sodium hydroxide is true when solutions of all aqueous strong acids and bases are mixed. The net ionic reaction is simply H 3O +

+

O H -

------- :

2 H 2O

3.3 Lewis Acids and Bases In 1923 G. N. Lewis proposed a theory that significantly broadened the understanding of acids and bases. As we go along we shall find that an understanding of L ew is a c id -b a s e th e o ry is exceedingly helpful to understanding a variety of organic reactions. Lewis pro­ posed the following definitions for acids and bases. •

Acids are electron pair acceptors.

•

Bases are electron pair donors.

103

3.3 Lewis Acids and Bases

In Lew is acid -b ase theory, proton donors are not the only acids; m any other species are acids as well. A lum inum chloride, for exam ple, reacts w ith am m onia in the sam e w ay that a proton donor does. U sing curved arrows to show the donation o f the electron pair of am m onia (the Lew is base), w e have the follow ing exam ples: * s+ 5- Cl— H

+

L ew is acid (ele c tro n pair a c c ep to r)

=NH 3

-

\

3

C l— A l - N H3 3

5-Cl L ew is acid (ele c tro n pair a c c e p to r)

H e lp f u l H i n t Verify for yourself that you can calculate the formal charges in these structures.

Cl

s-C l Cl - A l

C l- + H — NH 3

L ew is b ase (ele c tro n p air d o n o r)

Cl L ew is b ase (ele c tro n pair d o n o r)

In the reaction w ith hydrogen chloride above, notice that the electron pair acceptor (the proton) m ust also lose an electron pair as the new bond is form ed w ith nitrogen. This is necessary because the hydrogen atom had a full valence shell o f electrons at the start. On the other hand, because the valence shell o f the alum inum atom in alum inum chloride was not full at the beginning (it had only a sextet o f valence electrons), it can accept an elec­ tron pair w ithout breaking any bonds. The alum inum atom actually achieves an octet by accepting the pair from nitrogen, although it gains a form al negative charge. W hen it accepts the electron pair, alum inum chloride is, in the Lew is definition, acting as an acid. Bases are m uch the same in the Lewis theory and in the Br 0 nsted-Low ry theory, because in the Br 0 nsted-Low ry theory a base m ust donate a pair of electrons in order to accept a proton. •

The Lewis theory, by virtue of its broader definition of acids, allows acid-base theory to include all of the Br 0 nsted-Low ry reactions and, as we shall see, a great many others. M ost of the reactions we shall study in organic chemistry involve Lewis acid-base interactions, and a sound understanding of Lewis acid-base chemistry will help greatly.

A ny electron-deficient atom can act as a Lew is acid. M any com pounds containing group IIIA elem ents such as boron and alum inum are Lew is acids because group IIIA atom s have only a sextet o f electrons in their outer shell. M any other com pounds that have atom s w ith vacant orbitals also act as Lew is acids. Zinc and iron(III) halides (ferric halides) are fre­ quently used as Lew is acids in organic reactions.

3.3A Opposite Charges Attract •

In Lew is acid -b ase theory, as in m any organic reactions, the attraction o f oppo­ sitely charged species is fundam ental to reactivity.

As one further exam ple, w e consider boron trifluoride, an even m ore pow erful Lew is acid than alum inum chloride, and its reaction w ith am m onia. The calculated structure for boron

Carbonic anhydrase A zinc ion acts as a Lewis acid in th e mechanism o f th e enzyme carbonic anhydrase (C hapter 24).

104

Chapter 3

An Introduction to Organic Reactions and Their Mechanisms

Figure 3.1 E lectrostatic p o te n tia l maps fo r BF 3 and N H 3 and th e p ro d u ct th a t results from reaction b etw een th e m . A ttra c tio n b etw een th e stron g ly p ositive region o f BF 3 and th e negative region of N H 3 causes them to react. The e le ctro sta tic p o te n tia l m ap fo r th e p ro d u ct shows th a t the flu o rin e atoms d raw in th e electron density o f the fo rm a l negative charge, and th e nitro g e n atom , w ith its hydrogens, carries th e fo rm a l p o sitive charge.

H e lp f u l H i n t The need fo r a firm understanding o f structure, formal charges, and electronegativity can hardly be emphasized enough as you build a foundation o f knowledge for learning organic chemistry.

R eview P roblem 3.1

*

C f>

—

BF 3

F3B — NH3

NH 3

trifluoride in Fig. 3.1 shows ele c tro s ta tic p o te n tia l at its van der Waals surface (like that in Section 2.2A for HCl). It is obvious from this figure (and you should be able to predict this) that BF 3 has substantial positive charge centered on the boron atom and negative charge located on the three fluorines. (The convention in these structures is that blue represents relatively positive areas and red represents relatively negative areas.) On the other hand, the surface electrostatic potential for ammonia shows (as you would expect) that substan­ tial negative charge is localized in the region o f ammonia’s nonbonding electron pair. Thus, the electrostatic properties o f these two m olecules are perfectly suited for a Lewis acid-base reaction. When the expected reaction occurs between them, the nonbonding electron pair o f ammonia attacks the boron atom o f boron trifluoride, filling boron’s valence shell. The boron now carries a formal negative charge and the nitrogen carries a formal positive charge. This separation o f charge is borne out in the electrostatic potential map for the product shown in Fig. 3.1. N otice that substantial negative charge resides in the BF 3 part o f the m ol­ ecule, and substantial positive charge is localized near the nitrogen. Although calculated electrostatic potential maps like these illustrate charge distribution and molecular shape well, it is important that you are able to draw the same conclusions based on what you would have predicted about the structures o f BF3 and NH 3 and their reaction product using orbital hybridization (Sections 1.12-1.14), VSEPR m odels (Section 1.16), con­ sideration o f formal charges (Section 1.7), and electronegativity (Sections 1.4A and 2.2).

Write equations showing the Lew is acid-base reaction that takes place when: (a) Methanol (CH 3 OH) reacts with BF3. (b) Chloromethane (CH 3 Cl) reacts with AlCl3. (c) D im ethyl ether (CH 3 OCH3) reacts with BF3.

R eview P roblem 3 .2

Which o f the follow ing are potential Lew is acids and which are potential Lewis bases? CH 3 (a) CH 3 CH 2 — N— CH 3

/ (b) H3 C — C

(e) (CH 3) 3 B

(c) (C6H5)3P:

XCH3

CH3 (d) =Br=-

3

(f) H =~

3.4 Heterolysis o f Bonds to Carbon: Carbocations and Carbanions Heterolysis o f a bond to a carbon atom can lead to either o f two ions: either to an ion with a positive charge on the carbon atom, called a carbocation , or to an ion with a negatively charged carbon atom, called a ca rb an ion : heterolysis

/ C -

:Z -

C arbocation

— C — Z s+

heterolysis — C =

C arbanion

Z

3.4 Heterolysis of Bonds to Carbon: Carbocations and Carbanions

105

THE CHEMISTRY OF . . . H O M O s a n d L U M O s in R e a c t i o n s T h e c a lc u l a t e d l o w e s t u n o c c u p i e d m o l e c u l a r o r b i t a l ( L U M O ) fo r B F

3

is s h o w n b y s o li d

r e d a n d b lu e l o b e s . M o s t o f t h e

v o lu m e r e p r e s e n te d b y th e L U M O c o r r e s p o n d s t o th e e m p t y

p

o r b i t a l in t h e

s p 2- h y b r i d i z e d

s ta te

of BF

3

( lo c a te d

involving th e HOMO of one m olecule with th e LUMO of an o th er is, from a m olecular orbital perspective, th e way reactions occur.

p e r­

p e n d i c u l a r t o t h e p la n e o f t h e a t o m s ) . T h is o r b i t a l is w h e r e e le c t r o n d e n s i t y f i l ls ( b o n d i n g o c c u r s ) w h e n B F

3.

by N H

3

is a t t a c k e d

T h e v a n d e r W a a ls s u r f a c e e le c t r o n d e n s i t y o f B F

3

is i n d i c a t e d b y t h e m e s h . A s t h e s t r u c t u r e s h o w s , t h e L U M O e x t e n d s b e y o n d t h e e le c t r o n d e n s i t y s u r f a c e , a n d h e n c e i t is e a s ily a c c e s s i b l e f o r r e a c t io n . The

h ig h e s t

o c c u p ie d

m o le c u la r

o r b ita l

(H O M O )

of

a m m o n i a , w h e r e t h e n o n b o n d i n g p a i r r e s id e s , is s h o w n b y r e d a n d b lu e l o b e s in its s t r u c t u r e . W h e n t h e r e a c t io n o c c u r s , th e

e le c t r o n

d e n s ity

fr o m

th e

HO M O

of

a m m o n ia

is

The LUMO of BF3 (left) and the HOMO of NH3 (right).

t r a n s f e r r e d t o t h e L U M O o f b o r o n t r i f l u o r i d e . T h is i n t e r a c t io n

•

C arbocations are electron deficient. They have only six electrons in their valence shell, and because o f this, carbocations are Lew is acids.

In this w ay they are like BF 3 and AICI3 . M ost carbocations are also short-lived and highly reactive. They occur as interm ediates in som e organic reactions. C arbocations react rapidly w ith Lew is bases— w ith m olecules or ions that can donate the electron pair that they need to achieve a stable octet of electrons (i.e., the electronic configuration o f a noble gas):

\

— C+

+ ^

C a rb o c a tio n (a L ew is a c id )

-------->

X: B "

I

C — B

A n io n (a L ew is base)

A — C +

:O — H

\

C — O — H

H

C a rb o c a tio n (a L ew is a c id )

•

H

W a te r (a L ew is b as e )

C a rb a n io n s are electron rich. They are anions and have an unshared electron pair. Carbanions, therefore, are L ew is b ases a n d r e a c t a c c o rd in g ly (Section 3.3).

3.4A Electrophiles and Nucleophiles Because carbocations are electron-se eking reagents chem ists call them elec tro p h iles

(m eaning electron-loving). •

E le c tro p h ile s a r e re a g e n ts th a t seek e lec tro n s so as to ach iev e a sta b le shell o f elec tro n s lik e th a t o f a n o b le gas.

•

A ll L ew is ac id s a r e electro p h iles. By accepting an electron pair from a Lew is base, a carbocation fills its valence shell. — C +

\

C a rb o c a tio n L ew is acid and e le ctro p h ile

+

:B

Lew is base

---------->

C — B

1

106

Chapter 3

•

An Introduction to Organic Reactions and Their Mechanisms

C a rb o n ato m s th a t a r e e le c tro n p o o r b ec au s e o f b o n d p o la rity , b u t a r e n o t c a rb o c a tio n s, c a n also b e elec tro p h iles. T hey can react w ith the electron-rich centers o f L ew is bases in reactions such as the following:

Lewis base

L ew is acid (ele c tro p h ile)

C a rb a n io n s a r e L ew is b ases. C arbanions seek a proton or som e other positive center to w hich they can donate their electron p air and thereby neutralize their negative charge. W hen a L ew is b ase seeks a p ositive center other than a proton, especially that o f a carbon atom, chem ists call it a n u cleo p h ile (m eaning nucleus loving; the nucleo- p art of the nam e com es from nucleus , the positive center o f an atom ). •

A n u cleo p h ile is a L ew is b a s e th a t seek s a p o sitiv e c e n te r su c h as a positively c h a rg e d c a rb o n ato m .

Since electrophiles are also Lew is acids (electron p air acceptors) and nucleophiles are Lewis bases (electron pair donors), w hy do chem ists have tw o term s for them ? T he answ er is that Lew is acid and Lew is base are term s that are used generally, but w hen one or the other reacts to form a bond to a carbon atom , w e usually call it an electrophile or a nucleophile. N uT

N u c le o p h ile

o

:

N u

E le c tro p h ile

:N u

E le c tro p h ile

^

C — O:

C — Nu

N u c le o p h ile

3.5 H o w to Use Curved Arrow s in Illustrating Reactions U p to this p oint w e have n o t indicated how bonding changes occur in the reactions w e have presented, but this can easily be done using curved-arrow notation. C u rv e d a rro w s •

show the direction o f electron flow in a reaction m echanism .

•

point from the source o f an electron p air to the atom receiving the pair. (Curved arrow s can also show the m ovem ent o f single electrons. We shall discuss reactions o f this type in a later chapter.)

•

alw ays show the flow o f electrons from a site o f higher electron density to a site of low er electron density.

•

n e v e r show the m ovem ent o f atom s. A tom s are assum ed to follow the flow o f the electrons.

The reaction o f hydrogen chloride w ith w ater provides a sim ple exam ple o f how to use curved arrow notation. H ere w e invoke the first o f m any “A M echanism for the R eaction” boxes, in w hich w e show every key step in a m echanism using color-coded form ulas accom ­ panied by explanatory captions.

3.5 How to Use Curved Arrows in Illustrating Reactions

R e a c tio n o f W a te r w ith H y d ro g e n C h lo rid e : T h e U se o f C u rv e d A rro w s R EA C TIO N

H2O

M E C H A N IS M

H — O

+

+

HCl ----- :

H3O + +

Cl-

H — O

H

8+ ^ ^ . . 8— H — C l:

I H

Curved arrows point from electrons to the atom receiving the electrons.

-c y

H

A w a te r m o le c u le uses o n e o f th e n o n b o n d in g e le ctro n p airs to form a b ond to a p ro to n of H Cl. T h e bond b etw e e n th e h yd ro g e n and c h lo rin e b reaks, and th e e le ctro n p air g o e s to th e c h lo rin e ato m .

H e lp f u l H i n t

+

T h is lea d s to the fo rm a tio n o f a h yd ro n iu m ion and a c h lo rid e ion.

T h e c u rv e d a r r o w b eg in s w ith a c o v a le n t b o n d o r u n s h a r e d e le c tro n p a i r (a site of h ig h e r e le c tro n d e n s ity ) a n d p o in ts to w a r d a site of e le c tro n d eficien cy . We see here that as the w ater m olecule collides w ith a h y drogen chloride m olecule, it uses one o f its unshared electron pairs (show n in blue) to form a b ond to the p roton o f HCl. This bond form s because the negatively charged electrons o f the oxygen atom are attracted to the positively charged proton. As the bond betw een the oxygen and the p roton form s, the h y d ro g en -ch lo rin e bond o f HCl breaks, and the chlorine o f HCl departs w ith the elec­ tron pair that form erly bonded it to the proton. (If this did not happen, the p roton w ould end up form ing tw o covalent bonds, w hich, o f course, a p roton cannot do.) We, th e re­ fore, use a curved arrow to show the bond cleavage as w ell. By p o inting from the bond to the chlorine, the arrow indicates that the bon d breaks and the electron p air leaves w ith the chloride ion. The follow ing acid -b ase reactions give other exam ples o f the use o f the curved-arrow notation: H — O v H

+ ~ ""^ O — H

--------»

H — O

I H

H — O — H

H

A c id

B ase

O II

O jT | /

CH3

+

I

+

:O — H

+ CH3

O — H

H — O — H

O :—

H

A cid

H

B ase

• 'O '

O '

.C C H

3 A c id

+ ^ = O — H O :

H

-------- >

C C H

B ase

3

+ O :_

H — O — H

107

108

Chapter 3

An Introduction to Organic Reactions and Their Mechanisms

S o lv e d P ro b le m 3 .2

Add curved arrows to the follow ing reactions to indicate the flow o f electrons for all o f the bond-forming and bondbreaking steps. (a)

H •OH

H +

H— Cl:

+

=Cl:

(b) 'O'

N

H |+ 'NT

'O' O:

■O, (c) 'O' +

(d)

'O' II .. H— O — S — O— H " II " .O.

■V H O

..

+

'O'

fY

'O' II .. :O — S — O — H " II " .O.

_ ..

:O : H

+

H :C = N:

----- >

I|

]

1

STRATEGY AND ANSW ER Recall the rules for use o f curved arrows presented at the beginning o f Section 3.5. Curved arrows point from the source o f an electron pair to the atom receiving the pair, and always point from a site o f higher electron density to a site o f lower electron density. We must also not exceed two electrons for a hydro­ gen atom, or an octet o f electrons for any elements in the second row o f the periodic table. We m ust also account for the formal charges on atoms and write equations w hose charges are balanced. In (a), the hydrogen atom o f HCl is partially positive (electrophilic) due to the electronegativity o f the chlorine atom. The alcohol oxygen is a source o f electrons (a Lew is base) that can be given to this partially positive pro­ ton. The proton must lose a pair o f electrons as it gains a pair, however, and thus the chloride ion accepts a pair o f electrons from the bond it had with the hydrogen atom as the hydrogen becom es bonded to the alcohol oxygen. (a)

H H =C|:

In (b), the carboxylic acid hydrogen is partially positive and therefore electrophilic, and the amine provides an unshared pair o f electrons that forms a bond with the carboxylic acid hydrogen, causing departure o f a carboxylate anion. (b) 'O'

\ je-H

H ^ ^ ,

■O' —

^ O

r

'

N

3.6 The Strength of Br0 nsted-Lowry Acids and Bases: Ka and pKa

Use the curved-arrow notation to w rite the reaction that w ould take place betw een dim ethylam ine (CH3)2NH and boron trifluoride. Identify the Lew is acid and Lew is base and assign appropriate form al charges.

109

R eview P roblem 3.3

3.6 The Strength o f B ronsted-Low ry Acids and Bases: Ka and pK a In contrast to the strong acids, such as HCl and H 2S O 4 , acetic acid is a m uch w eaker acid. W hen acetic acid dissolves in water, the follow ing reaction does not proceed to com pletion: O

O +

CH3

H2O

OH

+ 0H 3

H 3O+

o-

Experim ents show that in a 0.1M solution o f acetic acid at 25°C only about 1% o f the acetic acid m olecules ionize by transferring their protons to water. Therefore, acetic acid is a w eak acid. As w e shall see next, a c id s tre n g th is characterized in term s o f acidity constant (Ka) or pK a values.

3.6A The Acidity Constant, Ka B ecause the reaction that occurs in an aqueous solution o f acetic acid is an equilibrium , we can describe it w ith an expression for the e q u ilib riu m c o n s ta n t ( K eq) : = [H3p +] [CH 3C O 2-]

eq

[CH 3C O 2H][H2O]

For dilute aqueous solutions, the concentration o f w ater is essentially constant (~ 5 5 .5 M ), so w e can rew rite the expression for the equilibrium constant in term s o f a new constant ( K a) called the a c id ity c o n s ta n t: r i, ™ [H3O +] [CH 3C O 2- ] 20 i = [ 3 [c H3c o 2h ] 2 1

k = [ H

A t 25°C, the acidity constant for acetic acid is 1.76 X 10- 5 . We can w rite sim ilar expressions for any w eak acid dissolved in water. U sing a gene­ ralized hypothetical acid (HA), the reaction in w ater is HA

+

H2O

H3 O

+

A-

110

Chapter 3

An Introduction to Organic Reactions and Their Mechanisms

and the expression fo r the acidity constant is K = [H3O+][A-]

a

[HA]

B ecause the concentrations o f the products o f the reaction are w ritten in the num erator and the concentration o f the undissociated acid in the denom inator, a la rg e v alu e o f K a m ean s th e a c id is a s tro n g a c id a n d a sm a ll v alu e o f K a m e a n s th e a c id is a w ea k acid . If the K a is greater than 10, the acid w ill be, for all practical purposes, com pletely dissociated in w ater at concentrations less than 0.01M.

R eview P roblem 3.4

Form ic acid (HCO 2 H) has Ka = 1.77 X 10~ 4. (a) W hat are the m olar concentrations o f the hydronium ion and form ate ion (H CO 2~) in a 0 .1 M aqueous solution o f form ic acid? (b) W hat percentage o f the form ic acid is ionized?

3.6B Acidity and pKa C hem ists usually express the acidity constant, Ka, as its negative logarithm , pK a: pKa = r !og Ka T his is analogous to expressing the hydronium ion concentration as pH: pH = -lo g [H 3O+] For acetic acid the pKa is 4.75: pKa = —log(1.76 X 10—5) = - ( - 4 . 7 5 ) = 4.75 H e lp f u l H i n t ____ ^ Ka and pKa are indicators of acid strength.

N otice that there is an inverse relationship betw een the m agnitude o f the pKa and the strength o f the acid. •

T h e la rg e r th e v alu e o f th e pK a, th e w e a k e r is th e acid.

For exam ple, acetic acid w ith pK a = 4.75 is a w eaker acid than trifluoroacetic acid w ith pK a = 0 (Ka = 1). H ydrochloric acid w ith pK a = —7 (Ka = 107) is a far stronger acid than trifluoroacetic acid. (It is understood that a positive pKa is larger than a negative pK a.) CH 3CO2H < CF 3CO2H < HCl pK a

= 4.75

pKa = 0

W eak acid

pKa = —7 Very stro n g acid

Increasing acid strength Table 3.1 lists pK a values for a selection o f acids relative to w ater as the base. T he val­ ues in the m iddle p K a ran g e o f the table are the m ost accurate because they can be m ea­ sured in aqueous solution. Special m ethods m u st b e used to estim ate the pK a values for the very strong acids at the top o f the table and for the very w eak acids at the bottom .* The p K a values fo r these very strong and w eak acids are therefore approxim ate. A ll o f the acids that w e shall consider in this book w ill have strengths in betw een that o f ethane (an extrem ely w eak acid) and that o f H SbF 6 (an acid that is so strong that it is called a “superacid”). A s you exam ine Table 3.1, take care n o t to lose sight o f the vast range of acidities that it represents (a factor o f 10 62).

* A c id s th a t are stro n g er than a h y d ro n iu m io n and bases th a t are stro n g er than a h y d ro x id e io n re act c o m p le te ly w ith w a te r (a phe n o m en o n c a lle d th e le v e lin g e ffe c t; see S ections 3 .2 B and 3 .1 5 ). T h e re fo re , i t is n o t possible to m easure a c id ity constants fo r these acids in w ater. O th e r solvents and sp ecia l te ch n iq u es are used, b u t w e do n o t have th e space to describe those m ethods here.

111

3.6 The Strength of Br0 nsted-Lowry Acids and Bases: Ka and pKa

Relative Strength of Selected Acids and Their Conjugate Bases Acid H SbF 6 HI H2S O 4 HBr HCl C6H5SO+3H

S trongest acid

Î c < D 3_ W TJ Ô (0 c W (0 p

< -12 -10

C6H5SO 3 (CH 3)2O (C H ^ C " O

(CH 3)2O+H + (CH 3)2C = o h

- 3 .8 - 2 .9

c h 3o+h 2 H3O + hno3 c f 3c o 2h HF c 6h 5c o 2h C 6H5 NH3 + c h 3c o 2h h 2c o 3 c h 3c o c h 2c o c h 3 n h 4+ C 6H5OH h c o 3c h 3n h 3+ h 2o c h 3c h 2o h (CH 3)3COH c h 3c o c h 3 H C #C H

- 2 .5 - 1 .7 4 - 1 .4 0.18 3.2 4.21 4.63 4.75 6.35 9.0 9.2 9.9

CH 3OH h 2o NO3CF 3CO 2 FC 6H5CO 2 C 6H5NH2 c h 3c o 2h c o 3CH 3COHCOCH 3 NH3 C 6H5O CO 32c h 3n h 2 OHc h 3c h 2o (CH 3)3C O - c h 2c o c h 3 H C # C­ Hn h 2c h 2= c h c h 3c h 2-

10.2 10.6

15.7 16 18 19.2 25 35 38 44 50

(a) A n acid (HA) has Ka = 10 7. W hat is its pK a? w hat is its pK a? (c) W hich is the stronger acid?

W eakest base

S bF 6 Ih s o 4BrCl-

-9 -9 -7 - 6 .5

H2 NH3 c h 2= c h 2 CH 3CH 3

W eakest acid

C o n ju g ate Base

A p p ro x im ate p K a

(b) A nother acid (HB) has K a = 5;

o (D Q> W 3 CD

CD 3

CQ

\ S tro n g est base

R e v ie w P ro b le m 3 .5

Water, itself, is a very w eak acid and undergoes self-ionization even in the absence of acids and bases: H — O

+

H

H — O H

^

H — O ±

H

+

O — H

H

In pure water at 25°C, the concentrations of hydronium and hydroxide ions are equal to 10- 7M. Since the concentration of water in pure w ater is 55.5M, we can calculate the Ka for water. Ka = a

[H3 O+][OH- ] — [H2 O]

( 1 0 - 7 )( 1 0 -7) Ka = -------- -------- - = 1.8 X 10 - 1 6 a 55.5

pKa = 15.7 F a

Show calculations proving that the p K a of the hydronium ion (H 3O + ) is —1.74 as given in Table 3.1.

R e v ie w P ro b le m 3 .6

112

Chapter 3

An Introduction to Organic Reactions and Their Mechanisms

3.6C Predicting the Strength of Bases In our discussion so far w e have dealt only w ith the strengths o f acids. A rising as a natural corollary to this is a principle that allow s us to estim ate the s tre n g th s o f b a s e s . Simply stated, the principle is this: •

T h e s tro n g e r th e ac id , th e w e a k e r w ill b e its c o n ju g a te b a s e .

We can, therefore, re la te th e s tre n g th o f a b a se to th e p K a o f its c o n ju g a te acid. •

T h e la rg e r th e pK a o f th e c o n ju g a te ac id , th e s tro n g e r is th e base.

C onsider the follow ing as exam ples: Increasing base strength Cl-

CH 3C O 2-

OH-

V e ry w e a k b ase p K a o f c o n ju g a te ac id (H C l) = - 7

W e a k b ase pKa o f c o n ju g a te a c id (C H 3 C O 2 H) = 4.75

S tro n g b ase p K a o f c o n ju g a te a c id (H 2 O ) = 15.7

We see that the hydroxide ion is the strongest in this series o f three bases because its conjugate acid, water, is the w eakest acid. (We know that w ater is the w eakest acid because it has the largest p ^ a.) A m ines are like am m onia in that they are w eak bases. D issolving am m onia in w ater brings about the follow ing equilibrium : H N H

3

o -

H — O — H

H — N±

H

=O — H

H

B ase

Acid

C o n ju g a te acid pKa = 9.2

C o n ju g a te b ase

D issolving m ethylam ine in w ater causes the establishm ent o f a sim ilar equilibrium . H c h

3n

h

2

H — O — H

C H

3—

N±

H

=O — H

H

B ase

Acid

C o n ju g a te acid pK , =

C o n ju g a te b ase

1 0 .6

Again w e can relate the basicity o f these substances to the strength o f their conjugate acids. The conjugate acid o f am m onia is the am m onium ion, NH4+. The pK a o f the am m onium ion is 9.2. The conjugate acid o f m ethylam ine is the CH 3 NH3+ ion. This ion, called the methylaminium ion, has pKa = 10.6. Since the conjugate acid o f m ethylam ine is a w eaker acid than the conju­ gate acid o f ammonia, w e can conclude that m ethylam ine is a stronger base than ammonia. r

S o lv e d P ro b le m 3 .3

1

U sing the p K a values in Trable 3.1 decide w hich is the stronger base, C H 3OH or H 2O. S T R A T E G Y A N D A N S W ER W ea aci 3 "

F rom Table 3.1, w e find the pKa values o f the conjugate acids o f w ater an d m ethanol. H

O+" H

H3 C - O + - H

H

H

pKa = - 1 .7 4

p K a = - 2 .5

■ S 3 " '

113

3.7 How to Predict the Outcome of Acid-Base Reactions

B e c a u s e w a t e r is t h e c o n j u g a t e b a s e o f t h e w e a k e r a c id , i t i s t h e s t r o n g e r b a s e .

S tro n g e r b as e

H3 C \ o .

H \O |

W eaker b ase

|

H

H

U s i n g t h e p K a v a lu e s o f a n a l o g o u s c o m p o u n d s i n T a b le 3 .1 p r e d i c t w h i c h w o u l d b e t h e

R e v ie w P r o b le m 3 . 7

s tro n g e r b a se .

,a ) ^

(C )

^

c

H _ /N \

r

or

or



• •

(d )

(C H 3 )3 C ° r

or

:

T

or

O

H O v

T

"

°

7

O

+ T h e p K a o f th e a n ilin iu m io n ( C a n ilin e ( C

6H 5N H 2 )

6 H 5N

H 3 ) i s 4 . 6 3 . O n t h e b a s is o f t h i s f a c t , d e c i d e w h e t h e r

R e v ie w P r o b le m 3 . 8

is a s tr o n g e r o r w e a k e r b a s e th a n m e t h y la m in e .

3.7 H o w to Predict the O utcom e o f A cid-B ase Reactions T a b le 3 .1 g iv e s t h e a p p r o x i m a t e p K a v a lu e s f o r a r a n g e o f r e p r e s e n t a t i v e c o m p o u n d s . W h i l e y o u p r o b a b l y w i l l n o t b e e x p e c t e d t o m e m o r i z e a l l o f t h e p K a v a lu e s i n T a b le 3 . 1 , i t i s a g o o d id e a to b e g in t o le a r n th e g e n e r a l o r d e r o f a c id it y a n d b a s ic it y f o r s o m e o f th e c o m ­ m o n a c id s a n d b a s e s . T h e e x a m p l e s g i v e n i n T a b le 3 .1 a r e r e p r e s e n t a t i v e o f t h e i r c la s s o r f u n c t i o n a l g r o u p . F o r e x a m p l e , a c e t ic a c i d h a s a p K a =

4 . 7 5 , a n d c a r b o x y l i c a c id s g e n e r ­

a l l y h a v e p K a v a lu e s n e a r t h i s v a lu e ( i n t h e r a n g e p K a =

3 - 5 ) . E t h y l a lc o h o l is g iv e n as a n

e x a m p l e o f a n a l c o h o l , a n d a l c o h o l s g e n e r a l l y h a v e p K a v a lu e s n e a r t h a t o f e t h y l a l c o h o l ( i n t h e p K a r a n g e 1 5 - 1 8 ) , a n d s o o n . ( T h e r e a r e e x c e p t io n s , o f c o u r s e , a n d w e s h a l l l e a r n w h a t th e s e e x c e p t io n s a r e a s w e g o o n . ) B y l e a r n i n g t h e r e l a t i v e s c a le o f a c i d i t y o f c o m m o n a c id s n o w , y o u w i l l b e a b le t o p r e d i c t w h e t h e r o r n o t a n a c id - b a s e r e a c t io n w i l l o c c u r as w r it t e n . •

T h e g e n e r a l p r in c ip le to a p p ly is th is : A c i d - b a s e r e a c t io n s a lw a y s f a v o r t h e f o r ­ m a t io n o f th e w e a k e r a c id a n d t h e w e a k e r b a s e .

T h e r e a s o n f o r t h i s i s t h a t t h e o u t c o m e o f a n a c i d - b a s e r e a c t i o n is d e t e r m i n e d b y t h e p o s i ­ t i o n o f a n e q u i l i b r i u m . A c i d - b a s e r e a c t io n s a r e s a id , t h e r e f o r e , t o b e u n d e r e q u i l i b r i u m c o n t r o l , a n d r e a c t io n s u n d e r e q u i l i b r i u m

c o n t r o l a lw a y s f a v o r t h e f o r m a t i o n o f t h e m o s t

s t a b le ( l o w e s t p o t e n t i a l e n e r g y ) s p e c ie s . T h e w e a k e r a c i d a n d w e a k e r b a s e a r e m o r e s t a b le ( lo w e r i n p o t e n t ia l e n e r g y ) th a n th e s tr o n g e r a c id a n d s tr o n g e r b a s e . U s i n g t h i s p r i n c i p l e , w e c a n p r e d i c t t h a t a c a r b o x y l i c a c id ( R C O

2H )

w i l l re a c t w it h a q u e ­

o u s N a O H i n t h e f o l l o w i n g w a y b e c a u s e t h e r e a c t io n w i l l le a d t o t h e f o r m a t i o n o f t h e w e a k e r a c id ( H

2O )

a n d w e a k e r b a s e ( R C O 2 —) :

•O

*‘O ‘*

O — H

S tro n g e r acid pKa = 3 - 5

N a + " :O — H

S tro n g e r b ase

-------->

R

O .." N a +

W e a k e r b ase

+

H — O — H

W e a k e r acid pKa = 15.7

B e c a u s e t h e r e is a l a r g e d if f e r e n c e i n t h e v a lu e o f t h e p K a o f t h e t w o a c id s , t h e p o s i t i o n o f e q u i l i b r i u m w i l l g r e a t l y f a v o r t h e f o r m a t i o n o f t h e p r o d u c t s . I n in s t a n c e s l i k e th e s e w e c o m ­ m o n l y s h o w th e r e a c t io n w i t h a o n e - w a y a r r o w e v e n t h o u g h th e r e a c t io n is a n e q u ilib r iu m .

H e lp f u l H i n t Formation of the weaker acid and base is an im portant general principle for predicting the outcome of acid-base reactions.

114

Chapter 3

An Introduction to Organic Reactions and Their Mechanisms

S o lv e d P ro b le m 3 .4 C o n s i d e r t h e m i x i n g o f a n a q u e o u s s o l u t i o n o f p h e n o l, C

6H 5 O H

( s e e T a b le 3 . 1 ) , a n d N a O H . W h a t a c i d - b a s e r e a c t io n ,

i f a n y , w o u l d t a k e p la c e ? STRATEG Y

C o n s id e r th e r e la t iv e a c id it ie s o f th e r e a c t a n t ( p h e n o l) a n d o f th e a c id t h a t m i g h t b e f o r m e d ( w a t e r ) b y

a p r o to n tr a n s fe r to th e b a s e ( th e h y d r o x id e io n ) . ANSW ER

T h e f o l lo w in g r e a c t io n w o u l d ta k e p la c e b e c a u s e i t w o u l d le a d t o th e f o r m a t io n o f a w e a k e r a c id ( w a t e r )

f r o m t h e s t r o n g e r a c i d ( p h e n o l ) . I t w o u l d a ls o l e a d t o t h e f o r m a t i o n o f a w e a k e r b a s e , C

6 H 5O N a ,

f r o m th e s tr o n g e r

base, N a O H . C 6H 5 — O — H

+

N a +^ : O — H

S tro n g e r a c id

--------»

S tro n g e r b a s e

C 6H

5—

O r N a+

+

H — O — H

W eaker base

W e a k e r a c id

pKa = 9 .9

R eview P roblem 3 .9

pKa = 15.7

P r e d i c t t h e o u t c o m e o f t h e f o l l o w i n g r e a c t io n . / --------=

+

NH

2

3.7A Water Solubility as the Result of Salt Formation A l t h o u g h a c e t ic a c i d a n d o t h e r c a r b o x y l i c a c id s c o n t a i n i n g f e w e r t h a n f i v e c a r b o n a t o m s a r e s o lu b l e i n w a t e r , m a n y o t h e r c a r b o x y l i c a c id s o f h i g h e r m o l e c u l a r w e i g h t a r e n o t a p p r e c i a b l y s o lu b l e i n w a t e r . B e c a u s e o f t h e i r a c id i t y , h o w e v e r , w a te r-in s o lu b le c a rb o x y lic a c id s d is s o lv e

in a q u e o u s s o d iu m h y d ro x id e ; t h e y d o s o b y r e a c t in g t o f o r m w a t e r - s o l u b l e s o d i u m s a lts :

O

O' C '0 — H

Na+ - = 0— H

In s o lu b le in w a te r

' 0 .- N a +

H —

0

— H

S o lu b le in w a te r (d u e to its p o la rity as a salt)

W e c a n a ls o p r e d i c t t h a t a n a m i n e w i l l r e a c t w i t h

a q u e o u s h y d r o c h lo r ic a c id in

th e

f o llo w in g w a y : H R— N H ,

H — 0 ^ H

S tro n g e r base P seudoephedrine is an amine th a t is sold as its hydrochloride salt.

C l-

R — N

H

Cl -

=0 — H

I

I

I

H

H

H

W e a k e r acid

S tro n g e r acid PK, = -1 .7 4

W e a k e r b ase

PK, = 9 -1 0

W h i l e m e t h y l a m i n e a n d m o s t a m in e s o f l o w m o l e c u l a r w e i g h t a r e v e r y s o l u b l e i n w a t e r , a m in e s w i t h h i g h e r m o l e c u l a r w e i g h t s , s u c h a s a n i l i n e ( C

6 H 5 N H 2) ,

h a v e l im it e d w a t e r s o l­

u b i l i t y . H o w e v e r , th e s e w a te r- in s o lu b le a m in e s d is s o lv e r e a d ily in h y d r o c h lo r ic a c id b e c a u s e t h e a c i d - b a s e r e a c t io n s c o n v e r t t h e m i n t o s o l u b l e s a lt s : H c

6h

5-

N H ,

“

h

— 0 ^ H

C l-

C 6H

5—

N

^ H

C l-

0

H

I H

H

W a te r in s o lu b le

R eview P roblem 3 .1 0

M o s t c a r b o x y lic

a c id s

H

W a te r-s o lu b le salt

d is s o l v e

in

aqueous

s o lu t i o n s

o f s o d iu m

b ic a r b o n a t e

( N a H C O 3)

b e c a u s e , a s c a r b o x y l a t e s a lt s , t h e y a r e m o r e p o l a r . W r i t e c u r v e d a r r o w s s h o w i n g t h e r e a c t io n

115

3.8 Relationships between Structure and Acidity

between a generic carboxylic acid and sodium bicarbonate to form a carboxylate salt and H2 CO 3 . (Note that H2 CO 3 is unstable and decomposes to carbon dioxide and water. You do not need to show that process.)

3.8 Relationships b etw een Structure and A cidity The strength o f a Br0 nsted-Lowry acid depends on the extent to which a proton can be separated from it and transferred to a base. Removing the proton involves breaking a bond to the proton, and it involves making the conjugate base more electrically negative. When w e compare compounds in a single column o f the periodic table, the strength o f the bond to the proton is the dominating effect. •

Bond strength to the proton decreases as w e m ove down the column, increasing its acidity.

This phenomenon is mainly due to decreasing effectiveness o f orbital overlap between the hydrogen 1s orbital and the orbitals of successively larger elements in the column. The less effective the orbital overlap, the weaker is the bond, and the stronger is the acid. The acidi­ ties of the hydrogen halides furnish an example:

pKa

G ro u p V IIA

3 .2

H— F

—7

H — Cl

—9

H — Br

—10

A c i d i t y i n c r

H— I

Comparing the hydrogen halides with each other, H — F is the weakest acid and H — I is the strongest. This follow s from the fact that the H — F bond is by far the strongest and the H — I bond is the weakest. Because HI, HBr, and HCl are strong acids, their conjugate bases (I- , Br- , Cl- ) are all weak bases. HF, however, which is less acidic than the other hydrogen halides by 10-13 orders o f magnitude (compare their pKa values), has a conjugate base that is corre­ spondingly more basic than the other halide anions. The fluoride anion is still not nearly as basic as other species w e com m only think o f as bases, such as the hydroxide anion, however. A comparison o f the pKa values for HF (3.2) and H2O (15.7) illustrates this point. The same trend o f acidities and basicities holds true in other columns o f the periodic table. Consider, for example, the column headed by oxygen:

pKa 1 5 .7

H 2O

G ro u p V IA 7 .0

H 2S

3 .9

H 2S e

A c i d i t y i n c r e a s e s

H e lp f u l H i n t Proton acidity increases as we descend a column in the periodic table due to decreasing bond strength to the proton.

116

Chapter 3

An Introduction to Organic Reactions and Their Mechanisms

H ere the strongest bond is the O — H bond and H2O is the w eakest acid; the w eakest bond is the S e — H bond an d H2S e is the strongest acid. H e lp f u l H i n t Proton acidity increases from left to right in a given row o f the periodic table due to increasing stability o f the conjugate base.

•

A cidity increases from left to rig h t w hen w e com pare com pounds in a given row o f the periodic table.

B ond strengths vary som ew hat, but the predom inant factor becom es the electronegativity o f the atom bonded to the hydrogen. T he electronegativity o f the atom in question affects acidity in tw o related w ays. It affects the polarity o f the b o n d to the proton and it affects the relative stability o f the anion (conjugate base) that form s w hen the proton is lost. We can see an exam ple o f this effect w hen w e com pare the acidities o f the com pounds CH 4, NH3, H 2O, and HF. T hese com pounds are all hydrides o f first-row elem ents, and elec­ tronegativity increases across a row o f the periodic table from left to rig h t (see Table 1.2): Electronegativity increases C

N

O

F

B ecause fluorine is the m ost electronegative, the b o n d in H — F is m ost polarized, and the proton in H — F is the m ost positive. Therefore, H — F loses a proton m ost readily an d is the m ost acidic in this series: Acidity increases 5-

5+

5-

5+

5-

5+

5-

5+

H3C — H

H2N — H

HO— H

F— H

pKa = 48

pKa = 3 8

pKa = 1 5 .7

pKa = 3.2

E lectrostatic potential m aps for these com pounds directly illustrate this trend based on electronegativity and increasing polarization o f the bonds to hydrogen (Fig. 3.2). A lm ost no positive charge (indicated by extent o f color trending tow ard blue) is evident at the hydro­ gens o f m ethane. Very little positive charge is present at the hydrogens o f am m onia. This is consistent w ith the w eak electronegativity o f b oth carbon and nitrogen and hence w ith the behavior o f m ethane and am m onia as exceedingly w eak acids ( p ^ a values o f 48 and 38, respectively). W ater show s significant positive charge at its hydrogens (pA"a m ore than 2 0 units low er than am m onia), and hydrogen fluoride clearly has the highest am ount of positive charge at its hydrogen (pA"a o f 3.2), resulting in strong acidity. B ecause H— F is the strongest acid in this series, its conjugate base, the fluoride ion (F - ), w ill be the w eakest base. Fluorine is the m ost electronegative atom and it accom ­ m odates the negative charge m ost readily: Basicity increases CH 3-

H 2N-

ho-

f-

T he m ethanide ion (CH 3 - ) is the least stable anion o f the four, because carbon being the least electronegative elem ent is least able to accept the negative charge. T he m ethanide ion, therefore, is the strongest b ase in this series. [The m ethanide ion, a c a rb a n io n , and the am ide ion (NH2- ) are exceedingly strong bases because they are the conjugate bases o f extrem ely w eak acids. We shall discuss som e uses o f these pow erful bases in Section 3.15.]

Figure 3.2 The effe ct o f increasing e le c tro n e g a tiv ity am ong elem ents from le ft to rig h t in th e firs t ro w o f th e p e rio d ic ta b le is evident in these maps of e le ctro sta tic p o te n tia l fo r m ethane, amm onia, w ater, and hydrogen flu o rid e .

117

3.8 Relationships between Structure and Acidity

Trends in acidity w ithin the periodic table are sum m arized in Fig. 3.3. C E J3 O o

Acidity increases within a given row (electronegativity effect)

Hydride PKa

C

N

O

F

(H3 C—H) 48

(H2 N—H) 38

(HO— H) 15.7

(F— H) 3.2

S

Cl

(HS— H) 7.0

(Cl H) -7

Se

Br

(HSe— H) 3.9

(Br— H) -9 I

CÄ

> o ■i

+

:N H 2-

liquid, NH3

S tro n g e r b ase (from N a N H 2)

S tro n g e r acid pKa = 25

H— C = C

+

W eaker b ase

:N H 3 W eaker acid pKa = 38

M ost te rm in a l alk y n es (alkynes w ith a proton attached to a triply bonded carbon) have p ^ a values of about 25; therefore, all react w ith sodium am ide in liquid am m onia in the sam e w ay that ethyne does. The general reaction is

A

R— C = C — H

*

:N H 2-

liquid, nh 3

S tro n g e r b ase

S tro n g e r acid pKa ^ 25

R— C = C

: NH 3

+

W eaker acid pKa = 38

W eaker b ase

Alcohols are often used as solvents for organic reactions because, being somewhat less polar than water, they dissolve less polar organic compounds. Using alcohols as solvents also offers the advantage of using R O —ions (called alkoxide ions) as bases. Alkoxide ions are somewhat stronger bases than hydroxide ions because alcohols are weaker acids than water. For example, we can create a solution of sodium ethoxide (CH 3CH 2O N a) in ethyl alcohol by adding sodium hydride (NaH) to ethyl alcohol. We use a large excess of ethyl alcohol because we want it to be the solvent. Being a very strong base, the hydride ion reacts readily with ethyl alcohol: c h 3c h

20

— H

+

S tro n geracid

: H-

ethyl alcohol

S tro n g e r b ase (fro m NaH )

pKa = 1B

CH 3CH 2O*

H2

W eaker b ase

W eaker acid pKa = 35

T he tert-butoxide ion, (CH3)3C O , in te rt-b u tyl alcohol, (CH 3) 3CO H , is a stronger base than the ethoxide ion in ethyl alcohol, and it can be prepared in a sim ilar way: (C H ^ C O ^ H -

+

S tro n g e r acid

tert-butyl

'= H -

alcohol

S tro n g e r b ase (from N aH )

pKa = i a

(C H ^ C O : — + W eaker b ase

H2 W eaker acid pKa = 35

A lthough the carbon-lithium bond of an alkyllithium (RLi) has covalent character, it is polarized so as to m ake the carbon negative: 8+

R ----- -Li A lkyllithium jug ate bases E thyllithium carbanion. It H

reagents react as though they contain alkanide (R :—) ions and, being the con­ of alkanes, alkanide ions are the strongest bases that w e shall encounter. (CH 3CH 2 Li), for exam ple, acts as though it contains an ethanide (CH 3CH 2:—) reacts w ith ethyne in the follow ing way:

- C = C — H~

S tro n g e r acid pKa = 25

+

- : CH 2CH 3 S tro n g e r b ase (from C H 3 C H 2 Li)

hexane

H — C = C :— W eaker b ase

+

C H 3CH 3 W eaker acid pKa = 50

A lkyllithium s can be easily prepared by allow ing an alkyl brom ide to react w ith lithium m etal in an ether solvent (such as diethyl ether). See Section 12.6.

H e lp f u l H i n t

We shall use this reaction as part of our introduction to organic synthesis in Chapter 7.

130

Review Problem 3.16

Chapter 3

An Introduction to Organic Reactions and Their Mechanisms

W rite equations for the a c id -b ase reaction that w ould occur w hen each o f the follow ing com pounds or solutions are m ixed. In each case label the stronger acid and stronger base, and the w eaker acid and w eaker base, by using the appropriate p K a values (Table 3.1). (If no appreciable a c id -b ase reaction w ould occur, you should indicate this.) (a) NaH is added to CH 3 OH.

(d) NH4Cl is added to sodium am ide in

(b) N aN H 2 is added to CH 3 CH 2 OH.

liquid a m m ^ ra .

(c) G aseous NH 3 is added to ethyllithium in hexane.

(e) (CH 3)3 C O N a is added to H 2° (f) NaOH is added to (CH 3 )3 COH.

3.16 A cid-B ase Reactions and the Synthesis o f Deuterium and Tritium -Labeled Compounds C hem ists often use com pounds in w hich deuterium or tritium atom s have replaced one or m ore hydrogen atom s o f the com pound as a m ethod o f “labeling” or identifying particular hydrogen atom s. D euterium ( 2 H) and tritium ( 3 H) are isotopes o f hydrogen w ith m asses of 2 and 3 atom ic m ass units (am u), respectively. O ne w ay to introduce a deuterium or tritium atom into a specific location in a m olecule is through the ac id -b ase reaction that takes p lace w hen a very strong b ase is treated w ith D2O or T2O (w ater that has deuterium or tritium in place o f its hydrogens). For exam ple, treating a solution containing (CH3)2CHLi (isopropyllithium ) w ith D2O results in the for­ m ation o f propane labeled w ith deuterium at the central atom: CH3

CH 3

CH 3CH :-L b

D2O

Is o p ro p y l­ lith ium ( s tr o n g e r base)

hexane

( s tr o n g e r a c id )

C H 3CH — D

OD-

2 -D e u te rio p ro p a n e (w e a k e r a c id )

(w e a k e r base)

S o lv e d P ro b le m 3 .6 A ssum ing you have available propyne, a solution o f sodium am ide in liquid am m onia, an d T 2 O, show how you w ould prepare the tritium -labeled com pound CH 3 C = CT. ANSW ER F irst add propyne to sodium am ide in liquid am m onia. T he follow ing a c id -b ase reaction w ill take place: CH3C = CH S tro n g e r a c id

+

------------- :

NH2 -

liq. ammonia

S tro n g e r b ase

CH3C = C : -

+

W eaker bas e

NH 3 W eaker acid

T hen adding T2O (a m uch stronger acid than NH3) to the solution w ill produce CH 3C =

Review Problem 3.17

CH 3C = C :

T2O

------------- :

S tro n g e r b ase

S tro n g e r a c id

liq. ammonia

CH3C # C T W eaker acid

+

CT:

OTW eaker b ase

C om plete the follow ing a c id -b ase reactions: (a) H C = CH + N a H ------- >

(d) CH 3CH2OH + N a H ------- >

(b) T he solution obtained in (a) + D2O ----- :

(e) T he solution obtained in (d) + T2O ----- :

(c) CH 3CH2Li + D2O ------- >

(f) CH 3CH 2CH2Li + D2O ------- >

hexane

hexane

hexane

hexane

3.17 Applications of Basic Principles

131

3.17 Applications o f Basic Principles A g a in

w e r e v ie w

how

c e r ta in b a s ic p r in c ip le s

a p p ly

to

to p ic s

w e have

s t u d ie d i n

th is

c h a p te r.

E lec tro n e g ativity D ifferences Polarize Bonds

In

S e c tio n

3 .1 A

w e le a r n e d t h a t

h e te ro ly s is o f a c o v a l e n t b o n d is a id e d w h e n t h e b o n d is p o l a r i z e d b y a d if f e r e n c e i n e le c t r o n e g a t i v i t y o f t h e b o n d e d a to m s . W e s a w h o w t h i s p r i n c i p l e a p p lie s t o t h e h e t e r o l y s i s o f b o n d s t o c a r b o n i n S e c t io n 3 . 4 a n d i n e x p l a i n i n g t h e s t r e n g t h o f a c id s i n S e c tio n s 3 .8 a n d 3 . 1 1 B .

Polarized Bonds Underlie Inductive Effects

I n S e c tio n 3 .1 1 B w e s a w h o w p o la r iz e d

b o n d s e x p l a i n e f f e c t s t h a t w e c a l l in d u c tiv e e ffe cts a n d h o w

th e s e e ffe c ts a re p a r t o f th e

e x p l a n a t i o n f o r w h y c a r b o x y l i c a c id s a r e m o r e a c i d i c t h a n c o r r e s p o n d i n g a lc o h o ls .

O p p o site Charges A ttra c t

T h is

p r in c ip le

is

fu n d a m e n ta l to

u n d e r s ta n d in g

L e w is

a c id - b a s e th e o r y a s w e s a w i n S e c t i o n 3 . 3 A . P o s i t i v e l y c h a r g e d c e n t e r s i n m o l e c u l e s t h a t a r e e le c t r o n p a i r a c c e p t o r s a r e a t t r a c t e d t o n e g a t i v e l y c h a r g e d c e n t e r s i n e le c t r o n p a i r d o n o r s . I n S e c tio n 3 .4 w e s a w th is p r in c ip le a g a in i n th e r e a c t io n o f c a r b o c a tio n s ( p o s it iv e ly c h a r g e d L e w i s a c id s ) w i t h a n io n s ( w h i c h a r e n e g a t i v e ly c h a r g e d b y d e f i n i t i o n ) a n d o t h e r L e w i s b a s e s .

N atu re Prefers States o f Low er Potential Energy

I n S e c tio n 3 . 9 A w e s a w h o w

e n e rg y

c h a n g e s — c a lle d

e n th a lp y ch a n g e s — t h a t t a k e p la c e

w h e n c o v a le n t b o n d s fo r m , a n d in

S e c tio n 3 .1 0 w e

s a w th e r o le e n t h a lp y c h a n g e s p la y

th is

in

p r in c ip le

e x p la in in g

e x p la in s

how

th e

la r g e

or how

s m a ll th e

e q u ilib r iu m

c o n s ta n t f o r

l o w e r t h e p o t e n t i a l e n e r g y o f t h e p r o d u c t s , t h e l a r g e r is t h e e q u i l i b r i u m

a r e a c t io n

is .

The

c o n s ta n t, a n d th e

m o r e f a v o r e d is t h e f o r m a t i o n o f t h e p r o d u c t s w h e n e q u i l i b r i u m is r e a c h e d . T h i s s e c t i o n a ls o i n t r o d u c e d a r e la t e d p r i n c i p l e : N a t u r e p r e f e r s d i s o r d e r t o o r d e r — o r , t o p u t i t a n o t h e r w a y ,

a p o s itiv e e n tro p y ch a n g e f o r a r e a c t i o n f a v o r s t h e f o r m a t i o n o f t h e p r o d u c t s a t e q u i l i b r i u m .

Resonance Effects Can Stabilize M olecules and Ions

W h e n a m o le c u le o r io n c a n

b e r e p r e s e n te d b y t w o o r m o r e e q u iv a le n t r e s o n a n c e s tr u c tu r e s , th e n th e m o le c u le o r io n w i l l b e s t a b iliz e d ( w i l l h a v e it s p o t e n t ia l e n e r g y lo w e r e d ) b y d e lo c a liz a t io n o f c h a r g e . I n S e c t i o n 3 . 1 1 A w e s a w h o w t h i s e f f e c t h e l p s e x p l a i n t h e g r e a t e r a c i d i t y o f c a r b o x y l i c a c id s w h e n c o m p a r e d t o c o r r e s p o n d i n g a l c o h o ls .

t In This ChaPte^ I n C h a p t e r 3 y o u s t u d ie d a c i d - b a s e c h e m i s t r y , o n e o f t h e m o s t i m p o r t a n t t o p i c s n e e d e d t o l e a r n o r g a n i c c h e m i s t r y . I f y o u m a s t e r a c i d - b a s e c h e m i s t r y y o u w i l l b e a b le t o u n d e r s t a n d m o s t o f t h e r e a c t io n s t h a t y o u s t u d y i n o r g a n i c c h e m i s t r y , a n d b y u n d e r s t a n d i n g h o w r e a c ­ t i o n s w o r k , y o u w i l l b e a b l e t o l e a r n a n d r e m e m b e r t h e m m o r e e a s ily .

(

Y o u h a v e r e v ie w e d t h e B r 0 n s t e d - L o w r y d e f i n i t i o n o f a c id s a n d b a s e s a n d t h e m e a n in g s o f p H a n d pK,d. Y o u h a v e le a r n e d t o i d e n t i f y t h e m o s t a c id i c h y d r o g e n a t o m s i n a m o le c u le b a s e d o n a c o m p a r i s o n o f pK,d v a lu e s . Y o u w i l l s e e i n m a n y c a s e s t h a t B r 0 n s t e d - L o w r y a c id - b a s e r e a c t io n s e it h e r i n i t i a t e o r c o m p l e t e a n o r g a n i c r e a c t io n , o r p r e p a r e a n o r g a n i c m o le c u le f o r f u r t h e r r e a c t io n . T h e L e w i s d e f i n i t i o n o f a c id s a n d b a s e s m a y h a v e b e e n n e w t o y o u . H o w e v e r , y o u w i l l s e e o v e r a n d o v e r a g a in t h a t L e w i s a c id - b a s e r e a c t io n s w h i c h i n v o l v e e it h e r t h e d o n a ­ t i o n o f a n e le c t r o n p a i r t o f o r m

a n e w c o v a l e n t b o n d o r t h e d e p a r t u r e o f a n e le c t r o n p a i r t o

b r e a k a c o v a l e n t b o n d a r e c e n t r a l s te p s i n m a n y o r g a n i c r e a c t io n s . T h e v a s t m a j o r i t y o f o r g a n i c r e a c t io n s y o u w i l l s t u d y a r e e it h e r B r 0 n s t e d - L o w r y o r L e w i s a c id - b a s e r e a c t io n s . Y o u r k n o w le d g e o f o r g a n ic s tr u c tu r e a n d p o la r it y f r o m C h a p te rs 1 a n d 2 h a s b e e n c r u c ia l t o y o u r u n d e r s t a n d i n g o f a c id - b a s e r e a c t io n s . Y o u h a v e s e e n t h a t s t a b i l i z a t i o n o f c h a r g e b y d e l o c a l i z a t i o n is k e y t o d e t e r m i n i n g h o w r e a d i l y a n a c id w i l l g iv e u p a p r o t o n , o r h o w r e a d ­ i l y a b a s e w i l l a c c e p t a p r o t o n . I n a d d i t i o n , y o u h a v e l e a r n e d t h e e s s e n t ia l s k i l l o f d r a w i n g c u r v e d a r r o w s t o a c c u r a t e l y s h o w t h e m o v e m e n t o f e le c t r o n s i n t h e s e p r o c e s s e s . W i t h t h e s e c o n c e p t s a n d s k i l l s y o u w i l l b e p r e p a r e d t o u n d e r s t a n d h o w o r g a n i c r e a c t io n s o c c u r o n a s te p b y - s t e p b a s is — s o m e t h in g o r g a n i c c h e m i s t s c a l l “ a m e c h a n i s m f o r t h e r e a c t io n . ”

â

132

Chapter 3

An Introduction to Organic Reactions and Their Mechanisms

So, continue to w ork hard to m aster acid-base chemistry and other fundamentals. Your tool­ box is quickly filling w ith the tools you need for overall success in organic chemistry!

Key Terms and Concepts

PLUS

T he key term s and concepts that are highlighted in b o ld , b lu e te x t w ithin the chapter are defined in the glossary (at the back o f the book) and have hyperlinked definitions in the accom panying W ileyP LU S course (w w w .w ileyplus.com ).

Problems T 'i'i» N ote to Instructors: Many of th e hom ework problem s are available for assignm ent via WileyPLUS, an online teaching and learning solution. B R 0N S T E D -L O W R Y AC ID S A N D BASES 3 .18

W hat is the conjugate base o f each o f the follow ing acids? (a) NH 3

(b) H2O

(c) H 2

(d) H C = CH

(e) CH 3OH

3 .1 9

L ist the bases you gave as answ ers to E xercise 3.18 in order o f decreasing basicity.

3 .2 0

W hat is the conjugate acid o f each o f the follow ing bases? (a) H SO 4-

3.21

(b) H2O

(c) CH 3 NH 2

(d) NH2-

(f) H 3O +

(e) CH 3CH 2-

(f) CH 3 C O 2~

L ist the acids you gave as answ ers to E xercise 3.20 in order o f decreasing acidity.

LEWIS AC ID S A N D BASES 3.2 2

D esignate the L ew is acid and L ew is base in each o f the follow ing reactions: Cl (a) CH 3 CH2— Cl + AlCl 3

CH 3 CH2— Cl— Al— Cl Cl F

CH

CH3 3

(b) CH 3— OH + BF 3

C H 3— O — B — F

(c) CH 3— 3 C+ 2 + H2O

C H 3— C — O H 2+

CH

H F

CH

C U RVED -A R R O W NO TA TIO N 3 .2 3

Rew rite each o f the follow ing reactions using curved arrow s and show all nonbonding electron pairs: (a) C H 3OH

+

(b) C H 3 NH 2 +

HI s HCl s

CH 3O H 2+

+

C H 3 NH 3 + +

I-

(c) H \

Cl-

/ H 3 .2 4

H C=C

'

H

/

'

H +

HF

----- >

\

Follow the curved arrow s and w rite the products. ‘O (a)

(b)

BF 3

JO :

BF 3

( c > A O (d)

CH 3 CH 2 CH 2 C H ^ L i

/ H

H C+ -C -H I H

F

Problems

3.25

133

Write an equation, using the curved-arrow notation, for the acid-base reaction that w ill take place when each of the follow ing are mixed. If no appreciable acid-base reaction takes place, because the equilibrium is unfavorable, you should so indicate. (a) Aqueous NaOH and CH 3 CH 2 CO2H

(d) CH 3 CH2Li in hexane and ethyne

(b) Aqueous NaOH and C 6 H5 S O 3 H

(e) CH 3 CH2Li in hexane and ethyl alcohol

(c) CH 3 CH2ONa in ethyl alcohol and ethyne

A C ID -B A S E STRENGTH A N D EQUILIBRIA 3 .2 6

When methyl alcohol is treated with NaH, the product is CH 3 O - Na+ (and H2) and not Na+ - CH2OH (and H2). Explain why this is so.

3 .2 7

What reaction w ill take place if ethyl alcohol is added to a solution o f HC= C:- Na+ in liquid ammonia?

3 .2 8

(a) The Ka of formic acid (HCO 2 H) is 1.77 X 10- 4 . What is the pKa?

3 .2 9

A cid HA has pKa = 20; acid HB has pKa = 10.

(b) What is the Ka of an acid whose pKa = 13?

(a) Which is the stronger acid? (b) W ill an acid-base reaction with an equilibrium lying to the right take place if N a+A - is added to HB? Explain your answer. 3 .3 0

Starting with appropriate unlabeled organic compounds, show syntheses o f each o f the following: (a) C 6 H5- C = C — T

(b) CH 3 - C H — O — D

(c) CH 3 CH 2 CH2OD

CH 3 3 .3 1

(a) Arrange the follow ing compounds in order o f decreasing acidity and explain your answer: CH 3 CH 2 NH2, CH 3 CH 2 OH, and CH 3 CH 2 CH3. (b) Arrange the conjugate bases o f the acids given in part (a) in order o f increas­ ing basicity and explain your answer.

3 .3 2

Arrange the follow ing compounds in order o f decreasing acidity: (a) CH 3 CH= CH2, CH 3 CH 2 CH3, CH 3 C = CH

(b) CH 3 CH 2 CH 2 OH, CH 3 CH 2 CO 2 H, CH 3 CHClCO2H

(c) CH 3 CH 2 OH, CH 3 CH 2 OH2+, CH 3 OCH 3 3 .3 3

Arrange the follow ing in order of increasing basicity: (a) CH 3 NH2, CH 3 NH3+, CH 3 NH-

(b) CH 3 O - , CH 3 NH- , CH 3 CH2-

(c) CH 3 C H= CH- , CH 3 CH 2 CH2- , CH 3 C = C -

GENERAL PROBLEMS 3 .3 4

Whereas H3 PO 4 is a triprotic acid, H3 PO 3 is a diprotic acid. Draw structures for these two acids that account for this difference in behavior.

3 .3 5

Supply the curved arrows necessary for the follow ing reactions: O (a)

X . H

O .. Ö— H

+

:Ö — H '*

/C .

Ö

0

:

:Ö : +

(b)

:Ö— H

Ö— CH,

H

+

h'

=Ö= (c) H— C — Ö — H ■Ö— CH,

H— C— Ö— H I " : 0 — CH,

Ö /C ^ .. IT Ö— H

+

=Ö— CH,

H

Ö: I H

134

Chapter 3

(d) H — O

+

An Introduction to Organic Reactions and Their Mechanisms

CH3— h

H — O — CH,

+

CH3 (e) H— O

+

;l: CH3

H— CH2— C— C^: H2C

CH3

+

:C ':"

H— O— H

CH 3.36

Glycine is an amino acid that can be obtained from m ost proteins. In solution, glycine exists in equilibrium between two forms: , H2 NCH2 CO2H e f H3 NCH2 C O 2 (a) Consult Table 3.1 and state which form is favored at equilibrium. (b) A handbook gives the melting point o f glycine as 262°C (with decom position). W hich o f the structures given above best represents glycine?

3 .37

M alonic acid, HO2 CCH 2 CO 2 H, is a diprotic acid. The p K a for the loss o f the first proton is 2.83; the pKa for the loss o f the second proton is 5.69. (a) Explain why m alonic acid is a stronger acid than acetic acid (pKa = 4.75). (b) Explain why the anion, —O 2 CCH 2 CO 2 H, is so much less acidic than malonic acid itself.

3 .38

The free-energy change, AG°, for the ionization o f acid HA is 21 kJ m o F 1; for acid HB it is —21 kJ m o l — 1 Which is the stronger acid?

3 .39

At 25°C the enthalpy change, AH°, for the ionization o f trichloroacetic acid is N 6.3 kJ m ol—1 and the entropy change, AS°, is N 0.0084 kJ m ol — 1 K —1. What is the pKa o f trichloroacetic acid?

3 .40

The compound at right has (for obvious reasons) been given the trivial name sq uaric acid. Squaric acid is a diprotic acid, with both protons being m ore acidic than acetic acid. In the dianion obtained after the loss o f both protons, all o f the carbon-carbon bonds are the same length as w ell as all o f the carbon-oxygen bonds. Provide a resonance explanation for these observations.

O \

OH

O OH S quaric acid

Challenge Problems 3.41

CH 3 CH2SH + CH 3 O~ A

c h 2— c h n o—

C

—» A (contains sulfur) + B C (which has the partial structure A— CH 2 CH 2 O)

2 2

h 2o

E (which is inorganic)

(a) Given the above sequence o f reactions, draw structures for A through E. (b) Rewrite the reaction sequence, showing all nonbonding electron pairs and using curved arrows to show electron pair movements. 3.42

First, com plete and balance each o f the equations below. Then, choosing among ethanol, hexane, and liquid ammo­ nia, state which (there may be more than one) m ight be suitable solvents for each o f these reactions. Disregard the practical limitations that com e from consideration o f “like dissolves like” and base your answers only on relative acidities. (a) C H 3 (CH 2 )bOD

n

CH 3 (CH 2)aLI ----- -

(b) N aN H 2 N CH 3 C # CH

(c) HCl

----- -

N

CH3 CH2 C # N N NaBr

135

Learning Group Problems

Suggest an explanation for this effect o f D M F on the basis o f Lew is ac id -b ase considerations. (H in t: A lthough w ater or an alcohol solvates both cations and anions, D M F is only effective in solvating cations.) 3 .4 4

As noted in Table 3.1, the pK a of acetone, C H 3C O C H 3 , is 19.2. (a) D raw the bond-line form ula o f acetone and o f any other contributing resonance form. (b) Predict and draw the structure of the conjugate base o f acetone and o f any other contributing resonance form. (c) W rite an equation for a reaction that could be used to synthesize CH 3C O C H 2 D.

3.4 5

F orm am ide (H C O N H 2) has a p K a o f approxim ately 25. Predict, based on the m ap of electrostatic potential for form am ide show n here, w hich hydrogen atom (s) has this pK a value. Support your conclusion w ith argum ents having to do w ith the electronic structure o f form am ide.

Learning Group Problems Suppose you carried out the follow ing synthesis o f 3-m ethylbutyl ethanoate (isoam yl acetate): O

O

H2SO4 (trace) OH E th an o ic acid (e x c e s s )

+

HO" 3 -M e th y l-1 b utano l

"O'

^

\

+

H2O

3 -M e th y lb u ty l e th a n o a te

As the chem ical equation shows, 3-m ethyl-1-butanol (also called isoam yl alcohol or isopentyl alcohol) was m ixed w ith an excess o f acetic acid (ethanoic acid by its system atic nam e) and a trace o f sulfuric acid (w hich serves as a catalyst). This reaction is an equilibrium reaction, so it is expected that not all o f the starting m aterials w ill be consum ed. The equilibrium should lie quite far to the right due to the excess o f acetic acid used, but not com pletely. A fter an appropriate length o f tim e, isolatio n o f the desired p ro d u ct from the reactio n m ix tu re w as begun by adding a volum e o f 5% aqueous sodium bicarbonate (N aH C O 3 has an effective pK a o f 7) roughly equal to the volum e o f the reaction m ixture. B ubbling occurred and a m ixture consisting o f tw o layers resulted— a basic aqueous layer and an organic layer. The layers w ere separated and the aqueous lay er w as rem oved. T he addition o f aqueous sodium bicarb o n ate to the layer o f organic m aterials and separation o f the layers w ere rep eated tw ice. E ach tim e the pred o m in an tly aqueous layers w ere rem oved, they w ere com bined in the sam e co llectio n flask. T he organic lay er th at rem ain ed after the three b ic ar­ bonate extractions w as dried and then subjected to d istillatio n in order to obtain a pure sam ple o f 3-m ethylbutyl ethanoate (isoam yl acetate). 1.

L ist all the chem ical species likely to be present at the end o f the reaction but before adding aqueous N aH C O 3 . N ote that the H2S O 4 w as not consum ed (since it is a catalyst), and is thus still available to donate a proton to atom s that can be protonated.

2.

U se a table of pK a values, such as Table 3.1, to estim ate pK a values for any potentially acidic hydrogens in each o f the species you listed in part 1 (or for the conjugate acid).

3.

W rite chem ical equations for all the acid -b ase reactions you w ould predict to occur (based on the pK a values you used) w hen the species you listed above encounter the aqueous sodium bicarbonate solution. (H int: C onsider w hether each species m ight be an acid that could react w ith N aH C O 3.)

4.

(a) Explain, on the basis o f polarities and solubility, w hy separate layers form ed w hen aqueous sodium bicarbon­ ate w as added to the reaction m ixture. (H in t: M ost sodium salts o f organic acids are soluble in water, as are neutral oxygen-containing organic com pounds o f four carbons or less.) (b) L ist the chem ical species likely to be present after the reaction w ith N aH C O 3 in (i) the organic layer and (ii) the aqueous layer. (c) W hy was the aqueous sodium bicarbonate extraction step repeated three tim es?

136

Chapter 3

An Introduction to Organic Reactions and Their Mechanisms

CONCEPT MAP

■vjl ymJr!gm P3 Nomenclature and Conformations of Alkanes

When your muscles contract to do work, like the person shown exercising above, it is largely because many carbon-carbon sigma (single) bonds are undergoing rotation (conformational changes) in a muscle protein called myosin. But when a diam ond-tipped blade cuts, as shown above, the carbon-carbon single bonds comprising the diam ond resist all the forces brought to bear on them, such that the material yields to the diamond. This remarkable contrast in properties, from the flexibility o f muscles to the rigidity o f diam ond, depends on many things, but central to them is w hether or not rotation is possible around carbon-carbon bonds. In this chapter we shall consider such bond rotations. We learned in Chapter 2 that our study o f organic chemistry can be organized around functional groups. Now we consider the hydrocarbon framework to which functional groups are attached— the framework that consists o f only carbon and hydrogen atoms. From the standpoint of an architect, hydrocarbon frameworks present a dream o f limitless possibilities, which is part o f what makes organic chemistry such a fascinating discipline. Buckminsterfullerene, named after the visionary architect Buckminster Fuller, is just one example o f a hydrocarbon with intriguing molecular architecture.

B uckm insterfullerene

137

13S

Chapter 4

Nomenclature and Conformations o f Alkanes and Cycloalkanes

Even though there are vast possibilities for the structures o f organic molecules, fortunately there is a well-defined system for naming carbon molecules. We study the essentials o f this system here in Chapter 4, and then build on it as we study the chemistry o f functional groups in later chapters. When chemists talk about structure in organic chemistry, they mean not only the connectivity o f the atoms, but also the shapes that molecules can adopt due to rotations o f groups joined by single bonds. The examination o f these proper­ ties is called conformational analysis, which we also discuss in this chapter as we consider the carbon framework o f organic molecules. We also consider the properties and reactivity o f hydrocarbons. Under am bi­ ent conditions, hydrocarbons containing only carbon-carbon single bonds are rel­ atively inert. Polyethylene, for example, is a hydrocarbon that is used for household Polyethylene is a hydrocarbon m acrom olecule th a t is in e rt to m ost chemicals w e use in dayto -d a y life.

containers, tubing, and many other items where lack o f reactivity is important. Hydrocarbons are com bustible, however, and o f course we make use o f this prop­ erty every tim e we burn hydrocarbon fuels such as natural gas, gasoline, or diesel. The release o f greenhouse gases by combustion o f hydrocarbons is a concern regarding climate change, o f course.

4.1 Introduction to Alkanes and Cycloalkanes We n oted earlier that the fam ily o f organic com pounds called hydrocarbons can b e divided into several groups on the basis o f the type o f bond that exists betw een the individual car­ bon atom s. T hose hydrocarbons in w hich all o f the carb o n -carb o n bonds are single bonds are called alk an e s, those hydrocarbons that contain a carbon-carbon double bond are called alk en es, and those w ith a carb o n -carb o n triple b o n d are called alk y n es. C y c lo a lk an e s are alkanes in w hich all or som e o f the carbon atom s are arranged in a ring. A lkanes have the general form ula C nH2n+2; cycloalkanes containing a single ring have tw o few er hydrogen atom s an d thus have the general form ula C nH2n. A lkanes and cycloalkanes are so sim ilar that m any o f their properties can be considered side by side. Som e differences rem ain, however, and certain structural features arise from the rings o f cycloalkanes that are m o re conveniently studied separately. We shall point out the chem ical and physical sim ilarities o f alkanes and cycloalkanes as w e go along.

4.1A Sources of Alkanes: Petroleum

Petroleum is a fin ite resource whose origin is under debate. A t th e La Brea Tar Pits in Los Angeles, many p re h isto ric animals perished in a natural vat containing hydrocarbons.

The prim ary source o f alkanes is petroleum . P etroleum is a com plex m ixture o f organic com pounds, m ost o f w hich are alkanes an d arom atic hydrocarbons (cf. C hapter 14). It also contains sm all am ounts o f oxygen-, nitrogen-, and sulfur-containing com pounds. Som e o f the m olecules in petroleum are clearly o f biological origin. T he natural origin o f petroleum is still under debate, however. M any scientists believe petroleum originated w ith decay o f prim ordial biological m atter. R ecent theories suggest, however, that organic m olecules m ay have been included as E arth form ed by accretion o f interstellar m aterials. A nalysis o f asteroids and com ets has show n that they contain a significant am ount and vari­ ety o f organic com pounds. M ethane and other hydrocarbons are found in the atm ospheres o f Jupiter, Saturn, and U ranus. S aturn’s m oon Titan has a solid form o f m eth an e-w ater ice at its surface and an atm osphere rich in m ethane. E arth ’s petroleum m ay therefore have orig­ inated sim ilarly to the w ay m eth an e becam e part o f these other bodies in our solar system . The discovery o f m icrobial life in high-tem perature ocean vents and the grow ing evidence for a deep, hot biosphere w ithin E arth suggest that com pounds in petroleum o f biological origin m ay sim ply b e contam inants introduced by prim itive life into a nonbiologically form ed petroleum reserve that w as present from E arth ’s beginning.

139

4.1 Introduction to Alkanes and Cycloalkanes

THE CHEMISTRY OF P e t r o l e u m R e f in in g The first ste p in refining petroleum is distillation; th e object here is to sep arate th e petroleum into fractions based on the volatility of its com ponents. C om plete separation into frac­ tions containing individual com pounds is economically impractical and virtually im possible technically. More than 500 different com pounds are contained in th e petroleum distillates boiling below 200°C, and many have alm ost the sam e boiling points. Thus th e fractions taken contain mix­ tures of alkanes of similar boiling points (see th e table below). Mixtures of alkanes, fortunately, are perfectly suit­ able for uses as fuels, solvents, and lubricants, th e primary uses of petroleum . The d em an d for gasoline is much g re a te r than th a t su p ­ plied by th e gasoline fraction of petroleum . Im portant pro cesses in th e petroleum industry, therefo re, are co n ­ cern ed with converting hydrocarbons from o th e r fractions into gasoline. W hen a mixture of alkanes from th e gas oil fraction (C 12 and higher) is h ea te d at very high te m p e ra ­ tures (~500°C) in th e p rese n ce of a variety of catalysts, th e m olecules break ap a rt and rearrange to smaller, m ore highly b ranched hydrocarbons containing 5 -1 0 carbon atom s. This process is called catalytic cracking. C racking can also b e d o n e in th e ab sen c e of a catalyst— called th e r­ m al cracking — bu t in this process th e p rodu cts te n d to

CH3 CH 3

C

A petroleum refinery. The tall to w e rs are fractio n in g colum ns used to separate com ponents o f crude oil according to th e ir boiling points.

have u n b ran ch ed chains, and alkanes with unbran ch ed chains have a very low "o c ta n e rating." The highly branched com pound 2,2,4-trim ethylpentane (called isooctane in th e petroleum industry) burns very sm oothly (without knocking) in internal com bustion engines and is used as one of th e standards by which th e octane rating of gasolines is established.

CH3 CH 2

CH

CH 3

or

or

CH3 2,2,4-Trim ethylpentane (“isooctane”)

According to this scale, 2,2,4-trim ethylpentane has an octane rating of 100. H eptane, CH 3 (CH 2)5CH 3 , a com ­ pound th at produces much knocking when it is burned in an internal com bustion engine, is given an o ctane rating of 0 . Mixtures of 2,2,4-trim ethylpentane and h ep tan e are used as

standards for o ctane ratings betw een 0 and 100. A gasoline, for exam ple, th at has th e sam e characteristics in an engine as a mixture of 87% 2,2,4-trim ethylpentane and 13% h e p ­ tan e would b e rated as 87-octane gasoline.

Typical Fractions O b tain e d by Distillation o f Petroleum

C 1- C 4 C5—C6 C6- Cy C5- C 10 2 1 1 8

Below 20 20-60 60-100 40-200 175-325 250-400 Nonvolatile liquids Nonvolatile solids

N u m b er o f C arbon A tom s p e r M olecule

1C

Boiling R ange of Fraction (°C)

C 12 and higher C 20 and higher C 20 and higher

A d a p te d w ith permission o f John W iley & Sons, Inc., from Holum , J. R., p. 213. C o p yrigh t 1995.

Use Natural gas, b o ttled gas, petrochem icals Petroleum ether, solvents Ligroin, solvents Gasoline (straight-run gasoline) K erosene and je t fuel Gas oil, fuel oil, and diesel oil Refined mineral oil, lubricating oil, and g rease Paraffin wax, asphalt, and tar General, Organic, and Biological Chemistry, Ninth Edition,

140

Chapter 4

Nomenclature and Conformations o f Alkanes and Cycloalkanes

4.2 Shapes o f Alkanes

Figure 4.1 Ball-and-stick m odels fo r th re e simple alkanes.

A general tetrahedral orientation o f groups— and thus sp3 hybridization— is the rule for the carbon atom s o f all alkanes an d cycloalkanes. We can represent the shapes o f alkanes as show n in Fig. 4.1. B utane and pentane are exam ples o f alkanes that are som etim es called “straight-chain” alkanes. O ne glance at three-dim ensional m odels, however, shows that because o f their tetra­ hedral carbon atom s the chains are zigzagged an d n o t at all straight. Indeed, the structures that w e have depicted in Fig. 4.1 are the straightest p ossible arrangem ents o f the chains because rotations about the carb o n -carb o n single bonds produce arrangem ents that are even less straight. A better description is u n b ra n c h e d . T his m eans that each carbon atom w ithin the chain is bonded to no m o re than tw o other carbon atom s and that unbranched alkanes contain only prim ary and secondary carbon atom s. Prim ary, secondary, and tertiary carbon atom s w ere defined in Section 2.5. Isobutane, isopentane, an d neopentane (Fig. 4.2) are exam ples o f branched-chain alkanes. In neopentane the central carbon atom is b onded to four carbon atom s. B utane and isobutane have the sam e m olecular form ula: C 4 H10. T he tw o com pounds have their atom s connected in a different order an d are, therefore, c o n s titu tio n a l iso m ers (Section 1.3). Pentane, isopentane, and neopentane are also constitutional isom ers. They, too, have the sam e m olecular form ula (C 5 H12) b ut have different structures.

Propane

Butane

ch 3ch 2ch3

ch3ch2ch 2ch

or

Figure 4.2 Ball-and-stick m odels fo r three branched-chain alkanes. In each o f th e com pounds one carbon atom is attached to more than tw o o th er carbon atom s.

H e lp f u l H i n t You should build your own molecular models o f the compounds in Figs. 4.1 and 4.2. View them from different perspectives and experiment with how their shapes change when you tw ist various bonds. Make drawings of your structures.

Pentane ch3ch2ch 2ch 2ch3

3

or

or

Isob utane

Is o p en tan e

CH3 CHCH3

c h 3c h c h 2c h 3

I

CH 3

or

CH 3

or

N eo p e n ta n e

ch3 I 3 c h 3c c h 3 I ch3 or

141

4.2 Shapes of Alkanes

Physical Constants of the Hexane Isomers M olecular Form ula

C o n d e n se d S tructural Form ula

C6H14

CH 3CH 2CH 2CH 2CH 2CH 3

C6H14

CH 3CHCH 2CH 2CH 3

Bond-Line Form ula

mp (°C)

b p (°C)a (1 atm )

D ensityb (g mL-1 )

Index of R efractionc (nD 20°C)

-9 5

ó8.7

G.ó5942G

1.3748

-1 5 3 .7

óG.S

G.ó5S22G

1.3714

-1 1 8

óS.S

G.óó4S2G

1.37ó5

-1 2 8 .8

58

G.óó1ó2G

1.S75G

49.7

G.ó4922G

1.3ó88

ch3 C 6H 14

c h 3c h 2c h c h 2c h CH 3

C 6H 14

CH 3CH—CHCH CH3 CH3

C 6H 14

CH3

-9 8

c h 3— c — c h 2c h 3 I 2 CH3 aUnless otherw ise indicated, all b oiling points given in this b o o k are at 1 atm or 760 torr. bThe superscript indicates the te m p e ra tu re at which the density was measured. cThe index o f refraction is a measure o f the a b ility o f the alkane to bend (refract) lig h t rays. The values rep o rte d are fo r lig h t o f the D line o f the sodium spectrum (nD).

W rite condensed and bond-line structural form ulas for all o f the constitutional isom ers w ith the m olecular form ula C 7 H16. (There is a total o f nine constitutional isom ers.) C onstitutional isom ers, as stated earlier, have different physical properties. The differ­ ences m ay not always be large, but constitutional isom ers are alw ays found to have differ­ ent m elting points, boiling points, densities, indexes o f refraction, and so forth. Table 4.1 gives som e o f the physical properties of the CeH 14 isom ers. As Table 4.2 shows, the num ber o f constitutional isom ers that is possible increases dram atically as the num ber o f carbon atom s in the alkane increases. The large num bers in Table 4.2 are based on calculations that m ust be done w ith a com puter. Sim ilar calculations, w hich take into account stereoisom ers (C hapter 5) as w ell as constitutional isom ers, indicate that an alkane w ith the form ula C-|67 H33e would, in th e­ ory, have m ore possible isom ers than there are particles in the observed universe!

N u m b er o f A lkane Isomers M olecular Form ula

P ossible N u m b er of C o n stitu tio n al Isom ers

C4H10 C 5H 12 C6H14 C 7H 16 C8H18 C 9H20 C 10H22 C 15H32 C20H42 C30H62 C40H82

3 5 9 18 35 75 4,347 366,319 4,111,846,763 62,481,801,147,341

2

R eview P roblem 4.1

142

Chapter 4

Nomenclature and Conformations o f Alkanes and Cycloalkanes

4.3 IUPAC N om enclature o f Alkanes, A lkyl Halides, and Alcohols

The Chemical A b stra cts Service assigns a CAS Registry N um ber to every com pound. CAS num bers make it easy to find in fo rm a tio n about a com pound in th e chemical literature. The CAS num bers fo r ingredients in a can o f latex p aint are shown here.

Prior to the developm ent near the end o f the nineteenth century o f a form al system for nam ing organic com pounds, m any organic com pounds had already been discovered or syn­ thesized. E arly chem ists nam ed these com pounds, often on the basis o f the source o f the com pound. A cetic acid (system atically called ethanoic acid) is an exam ple; it w as obtained by distilling vinegar, and it g ot its nam e from the L atin w ord for vinegar, acetum. Form ic acid (system atically called m ethanoic acid) had been obtained by the distillation o f the b o d ­ ies o f ants, so it got the nam e from the L atin w ord for ants, fo rm ica e . M any o f these older nam es for com pounds, called com m on or trivial nam es, are still in w ide use today. Today, chem ists use a system atic nom enclature developed and updated by the International U nion o f P ure and A pplied C hem istry (IUPAC). U nderlying the IUPAC sys­ tem is a fundam ental principle: ea ch d iffe re n t c o m p o u n d sh o u ld h av e a d iffe re n t a n d u n a m b ig u o u s nam e.* The IU P A C sy stem for nam ing alkanes is n o t difficult to learn, and the principles involved are used in nam ing com pounds in other fam ilies as well. For these reasons w e begin our study o f the IUPAC system w ith the rules for nam ing alkanes and then study the rules for alkyl halides and alcohols. The nam es for several o f the unbranched alkanes are listed in Table 4.3. T he ending for all o f the nam es o f alkanes is -ane. T he stems o f the nam es o f m ost o f the alkanes (above C4) are o f G reek and Latin origin. Learning the stems is like learning to count in organic chem ­ istry. Thus, one, two, three, four, and five becom e m eth-, eth-, prop-, but-, and pent-.

The Unbranched Alkanes

N am e M ethane Ethane Propane Butane P entane Hexane H eptane O ctane N onane D ecane

N u m b er o f C arbon A tom s 1 2

3 4 5 6

7 8

9 10

S tru c tu re ch4 CH 3CH 3 CH 3CH 2CH 3 CH3(CH 2)2CH3 CH3(CH 2)3CH3 CH3(CH 2)4CH3 CH 3 (CH 2)sCH 3 CH 3 (CH 2)eCH 3 CH 3 (CH 2)7CH 3 CH 3 (CH 2)SCH 3

N am e U ndecane D odecane Tridecane T etradecane P en tad ecan e H exadecane H ep tad ecan e O cta d ec an e N o n ad ecan e Eicosane

N u m b er o f C arbon A to m s 11 12

13 14 15 16 17 18 19 20

S tru c tu re CH3(CH2)gCH3 CH 3(CH 2)ioCH 3 CH3(CH2)iiCH3 CH3(CH2)i2CH3 CH3(CH2)i3CH3 CH3(CH2)i4CH3 CH3(CH2)i5CH3 CH3(CH2)ieCH3 CH3(CH2)i7CH3 CH 3(CH 2)iSCH 3

4.3A Nomenclature of Unbranched Alkyl Groups If w e rem ove one h y d ro g en ato m fro m an alkane, w e o b tain w h at is ca lled an alk y l g ro u p . T hese alkyl g roups have n am es th a t en d in -yl. W hen th e alk an e is u n b r a n c h e d , an d th e h y d ro g en ato m th a t is rem o v ed is a te r m in a l h y d ro g en atom , th e n am es are straightforw ard:

*T h e co m p le te IU P A C ru le s fo r n o m e n c la tu re can be fo u n d th ro u g h lin k s at th e IU P A C w eb site (w w w .iu p a c .o rg ).

4.3 lUPAC Nomenclature of Alkanes, Alkyl Halides, and Alcohols

C H 3— H

CH 3CH 2— H

C H 3C H 2CH 2— H

C H 3C H 2CH 2C H 2— H

Methane

E th an e

P ro p an e

B u tan e

C H 3—

CH 3CH 2—

c h 3c h 2c h 2—

c h 3c h 2c h 2c h 2—

M ethyl

Ethyl

Propyl

B utyl

M e-

Et-

Pr-

Bu-

4.3B Nomenclature of Branched-Chain Alkanes B ranched-chain alkanes are nam ed according to the follow ing rules: 1. L o c a te th e lo n g e st c o n tin u o u s c h a in o f c a rb o n a to m s; th is c h a in d e te rm in e s th e p a r e n t n a m e fo r th e a lk a n e . We designate the follow ing com pound, for exam ple, as a hexane because the longest continuous chain contains six carbon atoms: |------ > C H 3 CH 2 CH 2 CH 2 (CHCH3 Longest chain

or

CH 3

T he longest continuous chain m ay not alw ays be obvious from the w ay the form ula is w ritten. N otice, for exam ple, that the follow ing alkane is designated as a heptane because the longest chain contains seven carbon atoms:

2. N u m b e r th e lo n g e st c h a in b eg in n in g w ith th e e n d o f th e c h a in n e a r e r th e s u b stitu e n t. A pplying this rule, w e num ber the tw o alkanes that w e illustrated previously in the follow ing way:

th e s u b s titu e n t g ro u p . The parent nam e is placed last, and the substituent group, preceded by the num ber designating its location on the chain, is placed first. N um bers are separated from w ords by a hyphen. O ur tw o exam ples are 2-m ethylhexane and 3-m ethylheptane, respectively:

143

144

Chapter 4

Nomenclature and Conformations o f Alkanes and Cycloalkanes

4. W h e n tw o o r m o re su b stitu e n ts a r e p re se n t, give ea c h s u b s titu e n t a n u m b e r c o rre sp o n d in g to its lo c a tio n o n th e lo n g e st ch a in . For exam ple, w e designate the follow ing com pound as 4-ethyl-2-m ethylhexane:

4 - E th yl-2 -m e th y lh ex a n e

T he substituent groups should b e listed alp h ab e tica lly (i.e., ethyl before m ethyl).* In deciding on alphabetical order, disregard m ultiplying prefixes such as “d i” and “tri.” 5. W h en tw o su b stitu en ts a re p rese n t o n th e sam e ca rb o n atom , use th a t n u m b e r twice:

3 - E th yl-3 -m e th y lh ex a n e

6 . W h e n tw o o r m o re su b stitu e n ts a r e id e n tic a l, in d ic a te th is b y th e u se o f th e p r e ­

fixes di-, tri-, te tra -, and so on. T hen m ake certain that each an d every substituent has a num ber. C om m as are used to separate num bers from each other:

XX 2 ,3 -D im e th y lb u ta n e

2 ,3 ,4 -T rim e th y lp e n ta n e

2 ,2 ,4 ,4 -T e tra m e th y lp en tan e

A pplication o f these six rules allow s us to nam e m ost o f the alkanes that w e shall encounter. Two other rules, however, m ay b e required occasionally: 7. W h e n tw o c h a in s o f e q u a l le n g th c o m p e te fo r se lectio n as th e p a r e n t ch a in , choose th e c h a in w ith th e g r e a te r n u m b e r o f su b stitu e n ts:

1

2 ,3 ,5 -T rim e th y l-4 -p ro p y lh ep tan e (fo u r s u b s titu e n ts )

8 . W h e n b ra n c h in g first o cc u rs a t a n e q u a l d ista n c e fro m e ith e r en d o f th e longest

chain, choose th e n a m e th a t gives th e low er n u m b e r a t th e first p o in t o f difference: 6

5

4

2

3

1

2 ,3 ,5 -T rim e th y lh exan e

(n o t 2 ,4 ,5 -trim e th y lh e x a n e )

S o lv e d P ro b le m 4 .1

Provide an IUPAC name for the following alkane.

*S o m e h andbooks also lis t th e groups in o rd e r o f inc re a s in g size o r c o m p le x ity (i.e ., m e th y l b e fo re e th y l). A n a lp h a b e tica l lis tin g , how e ve r, is n o w b y fa r th e m o st w id e ly used system .

145

4.3 IUPAC Nomenclature of Alkanes, Alkyl Halides, and Alcohols

We find the longest chain (shown in blue) to be seven carbons; therefore the parent name is heptane. There are two methyl substituents (shown in red). We number the chain so as to give the first methyl group the lower number.

S T R A T E G Y A N D S O L U T IO N

The correct name, therefore, is 3,4-dimethylheptane. Numbering the chain from the other end to give 4,5-dimethylheptane would have been incorrect.

Two methyl groups

(b) I

I

2

R eview P roblem 4 .2

Which structure does not represent 2-methylpentane? (a)

Longest chain

(c)

(d )

Write the structure and give the IUPAC name for an alkane with formula CeH 1 4 that has only primary and secondary carbon atoms.

R eview P roblem 4.3

Draw bond-line formulas for all of the isomers of CgH-ig that have (a) methyl substituents, and (b) ethyl substituents.

R eview P roblem 4.4

4.3C Nomenclature of Branched Alkyl Groups In Section 4.3A you learned the names for the unbranched alkyl groups such as methyl, ethyl, propyl, and butyl, groups derived by removing a terminal hydrogen from an alkane. For alkanes with more than two carbon atoms, more than one derived group is possible. Two groups can be derived from propane, for example; the p ro p y l g r o u p is derived by removal of a terminal hydrogen, and the 1 -m eth y leth y l or iso p ro p y l g ro u p is derived by removal of a hydrogen from the central carbon: Three-Carbon Groups CH 3 CH 2 CH 3 P ro p a n e

c h 3 c h 2 c h 2—

c h 3c h c h 3

P ropyl

Is o p ro p y l

Pr-

i-Pr-

1-Methylethyl is the systematic name for this group; isopropyl is a common name. Systematic nomenclature for alkyl groups is similar to that for branched-chain alkanes, with the provision that num bering always begins at the p o in t where the group is attached to the m ain chain. There are four C 4 groups.

146

Chapter 4

Nomenclature and Conformations o f Alkanes and Cycloalkanes

Four-Carbon Groups CH 3 CH 2 CH 2 CH 3 B u ta n e

CH3

c h 3 c h 2 c h 2 c h 2—

c h 3 c h c h 2—

c h 3 c h 2 c;h ch 3

(C H 3 )3 C -

B u ty l

Is o b u ty l

s e c -B u ty l

fe r f- B u ty l ( o r f-B u )

The follow ing examples show how the names o f these groups are employed:

4 - ( 1 - M e th y le th

)h e p ta n e o r 4 -

o p ro p y l h e p ta n e

4 -(1 ,1 - D im e th y le th y l) o c ta n e o r 4 - fe r t- b u ty lo c ta n e

The common names isopropyl, isobutyl, se c -butyl, and te rt-b u ty l are approved by the IUPAC for the unsubstituted groups, and they are still very frequently used. You should learn these groups so w ell that you can recognize them any w ay that they are written. In deciding on alphabetical order for these groups you should disregard structure-defining prefixes that are written in italics and separated from the name by a hyphen. Thus tert-butyl precedes ethyl, but ethyl precedes isobutyl.* There is one five-carbon group with an IUPAC approved common name that you should also know: the 2 , 2 -dimethylpropyl group, com m only called the neopentyl group:

CH

J J

CH3— C — CH2— ch

3

2 ,2 - D im e th y lp r o p y l o r n e o p e n ty l g r o u p

R e v ie w P ro b le m 4 .5

(a) In addition to the 2,2-dimethylpropyl (or neopentyl) group just given, there are seven other five-carbon groups. Draw bond-line formulas for their structures and give each struc­ ture its systematic name. (b) Draw bond-line formulas and provide IUPAC names for all o f the isomers o f C 7 H16.

* T h e a b b re via tio n s i,

5, and t are

som etim es used fo r is o -, sec-, and tert-, re spe ctive ly.

4.3 IUPAC Nomenclature of Alkanes, Alkyl Halides, and Alcohols

147

4.3D Classification of Hydrogen Atoms T h e h yd ro g e n atom s o f an a lkan e are classified on the basis o f the carbon a to m to w h ic h they are attached. A h yd ro g e n a to m attached to a p rim a ry carbon ato m is a p rim a ry ( 1 ° ) h yd ro g e n ato m , and so fo rth . T h e fo llo w in g c o m po un d, 2 -m e th y lb u ta n e , has p rim a ry , secondary ( 2 ° ) , and te rtia ry ( 3 ° ) h yd ro g e n atoms: 1° H y d ro g e n a to m s -

C H3— C H — C H2— C H3

3° H y d ro g e n atom -'

H y d ro g e n a to m s

O n the o th er hand, 2 ,2 -d im e th y lp ro p a n e , a com p o u n d that is o ften c a lle d n e o p e n ta n e , has o n ly p rim a ry h yd ro g en atoms:

4.3E Nomenclature of Alkyl Halides A lk a n e s b earin g h alo g en substituents are n am ed in the IU P A C substitutive system as haloalkanes:

•

C H 3 C H 2 Cl

CH3CH2CH2 F

C H 3 C H B rC H 3

C hloroethane

1 -F lu o ro p ro p an e

2 -B rom opropane

W h e n the parent chain has both a h alo and an a lk y l substituent attached to it, n um ber the chain fro m the end nearer the first substituent, regardless o f w h e th e r it is h alo or a lk y l. I f tw o substituents are at equal distance fro m the end o f the chain, then n u m ­ b er the chain fro m the end nearer the substituent that has alphabetical precedence:

Cl

Cl

2 -C h lo ro -3 -m e th y lp e n ta n e

1

2 -C h lo ro -4 -m e th y lp e n ta n e

C o m m o n nam es fo r m a n y sim p le h alo alkan es are s till w id e ly used, how ever. In this c o m ­ m o n n o m en c la tu re system , c a lle d f u n c tio n a l class n o m e n c la tu re , h alo alkan es are nam ed as a lk y l h alides. (T h e fo llo w in g nam es are also accepted b y the IU P A C .)

Ethyl c h lo rid e

Is o p ro p y l b ro m id e

ferf-B u tyl b ro m id e

Is o b u ty l c h lo rid e

N e o p e n ty l b ro m id e

D r a w b o n d -lin e fo rm u las and g iv e IU P A C substitutive nam es fo r a ll o f the isom ers o f (a ) C 4 H 9C l and (b ) C g H ^ B r .

R e v ie w P ro b le m 4 .6

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4.3F Nomenclature of Alcohols In w h a t is ca lled IU P A C s u b s titu tiv e n o m e n c la tu re , a n am e m a y have as m a n y as fo u r fea ­ tures: lo c a n ts , p re fix e s , p a r e n t c o m p o u n d , and suffixes. C o nsider the fo llo w in g com pound as an illustration w ith o u t, fo r the m o m e n t, b eing concerned as to h o w the n am e arises:

Locant Prefix Locant Parent Suffix T h e locant 4 - tells th a t the substituent m e t h y l g ro up , n a m e d as a prefix, is attached to the p arent compound a t C 4 . T h e p are n t c o m p o u n d contains six carb on atom s and n o m u l­ tip le bonds, hen ce the p are n t n a m e h e x a n e , a n d it is an a lc o h o l; th e re fo re it has the suffix -o l. T h e lo c a n t 1 - tells th a t C1 bears the h y d ro x y l g roup. I n g e n e ra l, n u m b e r in g o f th e c h a in a lw a y s b e g in s a t th e e n d n e a r e r th e g r o u p n a m e d as a s u ffix . T h e lo c a n t fo r a s u ffix (w h e th e r it is fo r an a lc o h o l o r a no ther fu n c tio n a l g ro u p ) m a y be p la c e d b e fo re the p a re n t n a m e as in the abo ve e x a m p le or, accordin g to a 1 9 9 3 IU P A C re v is io n o f the rules, im m e d ia te ly b e fo re the suffix. B o th m etho d s are IU P A C approved. T h e re fo re , the abo ve c o m p o u n d c o u ld also b e n a m e d 4 -m e th y lh e x a n -1 -o l. T h e fo llo w in g procedure should be fo llo w e d in g ivin g alcohols IU P A C substitutive nam es: 1. S elect the longest con tin uo us carbon chain to which the hydroxyl is directly attached. C h an g e the n a m e o f the a lk an e corresp on din g to this chain b y d ro p p in g the fin a l -e and add ing the s u ffix - o l . 2 . N u m b e r the longest continuous carbon chain so as to give the carbon ato m bearing the h y d ro x y l group the lo w e r num ber. In d ic a te the position o f the h y d ro x y l group b y using this n um ber as a locant; ind icate the positions o f other substituents (as prefixes) b y using the num bers corresponding to th eir positions along the carbon chain as locants. T h e fo llo w in g exam p les show h o w these ru les are a p p lie d : 3

1

1

3

5

3

1

OH 1 -P ro p a n o l

2 -B u ta n o l

4 - M e th y l- 1 - p e n ta n o l o r 4 - m e t h y lp e n t a n - 1 - o l

( n o t 2 - m e t h y l- 5 - p e n ta n o l) 3

5

3 3 - C h lo r o - 1 - p r o p a n o l

4 ,4 - D im e th y l- 2 - p e n ta n o l

o r 3 - c h lo r o p r o p a n - 1 - o l

o r 4 ,4 - d im e th y lp e n ta n - 2 - o l

G iv e an IU P A C n am e fo r the fo llo w in g com po un d.

HO

S T R A T E G Y A N D A N S W E R W e fin d th a t the lon gest carbon chain (in re d a t rig h t) has

HO

fiv e carbons and it bears a h y d ro x y l gro up on the firs t carb on . S o w e n a m e this p a rt o f the m o le c u le as a 1 -p e n ta n o l. T h e re is a p h e n y l g ro u p on carb o n -1 a n d a m e th y l group on c a rb o n -3 , so the f u ll n a m e is 3 -m e th y l-1 -p h e n y l-1 -p e n ta n o l.

1

1

2 3

4 I

5

149

4.4 How to Name Cycloalkanes

Draw bond-line formulas and give IUPAC substitutive names for all of the isomeric alcohols with the formulas (a ) C 4 H-10 O and (b ) C 5 H-12 O.

R eview P roblem 4 .7

Simple alcohols are often called by common functional class names that are also approved by the IUPAC. We have seen several examples already (Section 2.6). In addition to methyl alcohol, ethyl alcohol, and isopropyl alcohol, there are several others, including the following:

OH

P r o p y l a lc o h o l

B u ty l a lc o h o l

s e c - B u t y l a lc o h o l

OH

OH

OH f e r f - B u t y l a lc o h o l

I s o b u t y l a lc o h o l

N e o p e n t y l a lc o h o l

Alcohols containing two hydroxyl groups are commonly called glycols. In the IUPAC substitutive system they are named as diols:

OH

HO S u b s titu tiv e

OH

HO

OH

1 ,2 - E th a n e d io l

1 ,2 - P r o p a n e d io l

o r e t h a n e - 1 ,2 - d i o l

o r p r o p a n e - 1,2- d io l

E th y le n e g ly c o l

P ro p y le n e g ly c o l

Com m on

OH

1 ,3 - P r o p a n e d io l o r p r o p a n e - 1 ,3 - d io l

T rim e th y le n e g ly c o l

4.4 H o w to N am e Cycloalkanes 4.4A Monocyclic Compounds Cycloalkanes with only one ring are named by attaching the prefix cyclo- to the names of the alkanes possessing the same number of carbon atoms. For example, H2C --------CH 2

y

v----------- 7

- V

H2C ------CH 2

-

M

H2

j

\h \ c

^ H 2

Cyclopropane

Naming substituted cycloalkanes is straightforward: We name them as alkylcycloalkanes, halocycloalkanes, alkylcycloalkanols, and so on. If only one substituent is present, it is not necessary to designate its position. When two substituents are present, we number the ring beginning with the substituent first in the alphabet and number in the direction that gives the next substituent the lower number possible. When three or more substituents are pre­ sent, we begin at the substituent that leads to the lowest set of locants:

$

Is o p r o p y lc y c lo h e x a n e

1 - E t h y l- 3 - m e th y lc y c lo h e x a n e

4 - C h lo r o - 2 - e t h y l- 1 - m e t h y lc y c lo h e x a n e

( n o t 1 - e t h y l- 5 - m e t h y lc y c lo h e x a n e )

( n o t 1 - c h lo r o - 3 - e t h y l- 4 - m e t h y lc y c lo h e x a n e )

150

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Nomenclature and Conformations o f Alkanes and Cycloalkanes

OH Cl

C h lo ro c y c lo p e n ta n e

2 -M e th y lc y c lo h e x a n o l

When a single ring system is attached to a single chain with a greater number of carbon atoms, or when more than one ring system is attached to a single chain, then it is appropriate to name the compounds as cycloalkylalkanes. For example,

1 -C y c lo b u ty lp e n ta n e

R eview P roblem 4 .8

1 ,3 -D ic y c lo h e x y lp ro p a n e

Give names for the follow ing substituted alkanes:

(a)

(b)

Cl

OH Cl (e)

(c)

OH

4.4B Bicyclic Compounds We name compounds containing two fused or bridged rings as bicycloalkanes, and w e use the name o f the alkane corresponding to the total number o f carbon atoms in the rings as the parent name. The following compound, for example, contains seven carbon atoms and is, therefore, a bicycloheptane. The carbon atoms common to both rings are called bridgeheads, and each bond, or each chain o f atoms, connecting the bridgehead atoms is called a bridge: One-carbon bridge

Bridgehead

H e lp f u l H i n t Explore the structures of these bicyclic compounds by building handheld molecular models.

Two-carbon— bridge

H,C

CH,

h, c

CH,

C H

—Two-carbon bridge

Bridgehead A bicycloheptane

Then w e interpose an expression in brackets within the name that denotes the number o f carbon atoms in each bridge (in order o f decreasing length). Fused rings have zero carbons in the bridge. For example,

151

4.5 Nomenclature of Alkenes and Cycloalkenes

I f substituents are present, w e n u m b e r the b rid g e d rin g system b eg in n in g at one b rid g e ­ head, p ro ce e d in g first along the longest b rid g e to the o th er b ridg ehead , then along the n ext lon gest b rid g e b ac k to the first b ridg ehead . T h e shortest b rid g e is n u m b e re d last: B rid g ed

F used

8 -M e th y lb ic y c lo [3 .2 .1 ]o c ta n e

8 -M e th y lb ic y c lo [4 .3 .0 ]n o n a n e

1r

S o lv e d P r o b le m 4 . 3

^

W r ite a structural fo rm u la fo r 7 ,7 -d ic h lo ro b ic y c lo [2 .2 .1 ]h e p ta n e .

STRATEGY AND ANSWER F irs t w e w rite a b ic y c lo [2 .2 .1 ]h e p ta n e rin g and n u m b e r

Cl

Cl

it. T h e n w e add the substituents (tw o c h lo rin e atom s) to the p ro p e r carbon.

5

3

R eview P roblem 4 .9

G iv e nam es fo r each o f the fo llo w in g b ic y c lic alkanes: Cl Cl

(b)

(a)

(d)

(c)

(e)

CH3

( f) W r ite the structure o f a b ic y c lic com p o u n d that is a c o n s titu tio n a l is o m e r o f b ic y c lo [2 .2 .0 ]h e x a n e and g iv e its nam e.

4.5 N om enclature o f Alkenes and Cycloalkenes M a n y o ld e r nam es fo r alkenes are s till in c o m m o n use. P rop ene is o fte n c a lle d p ro p y le n e , and 2 -m e th y lp ro p e n e fre q u e n tly bears the n am e isobutylene:

CH3 c h

2=

c h

2

c h

3c

h

=

c h

2

CH3

> IUPAC:

Common:

<

^ ch

* •

Ethene

Propene

2-M ethylpropene

Ethylene

Propylene

Isobutylene

2

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Nomenclature and Conformations o f Alkanes and Cycloalkanes

T h e IU P A C rules fo r n am ing alkenes are s im ila r in m a n y respects to those fo r n am ing alkanes:

1. D eterm ine the parent nam e by selecting the longest chain that contains the double bond and change the ending o f the nam e o f the alkane o f identical length from -ane to -ene. Thus, if the longest chain contains five carbon atoms, the parent name for the alkene is pentene; if it contains six carbon atoms, the parent name is hexene, and so on. 2. N um ber the ch ain so as to in clu d e both carbon atom s o f the d ouble bond, and begin n um berin g at the end o f the chain nearer the d ouble bond. D esignate the location o f the d ouble bond by u sing the num ber o f the first atom o f the double bond as a prefix. T h e locant for the alk en e suffix m ay precede the p arent nam e or b e placed im m ediately before the suffix. We w ill show examples o f both styles: 1 2 3 4 c h 2= c h c h 2c h

c h 3 c h = c h c h 2c h 2c h

3

1 -B u te n e (n ot 3 -b u ten e)

3

2 -H ex e n e (n o t 4 -h e x e n e )

3. Indicate the locations o f the su bstituent groups by the n um bers o f the carbon atom s to w hich they are attached: 4

2 -M e th y l-2 -b u te n e o r 2 -m e th y lb u t- 2 -en e

2 ,5 -D im e th y l-2 -h e x e n e o r 2 ,5 -d im e th y lh e x -2 -e n e

CH, C H 3C H = C H C H 2C— c h 1 32 3 4 2|5 6

3 3

C H 3C H = C H C H 2Cl

CH3 5 ,5 -D im e th y l-2 -h e x e n e o r 5 ,5 -d im e th y lh e x -2 -e n e

1 -C h lo ro -2 -b u te n e o r 1 -c h lo ro b u t- 2 -en e

N um ber substituted cycloalkenes in the w ay that gives the carbon atom s o f the double bond the 1 and 2 positions and that also gives the substituent groups the low er num bers at the first point o f difference. With substituted cycloalkenes it is not necessary to specify the position o f the double bond since it w ill always begin with C1 and C2. The two examples shown here illustrate the application o f these rules:

5

2 4

1 -M e th y lc y c lo p e n te n e (n o t 2 -m e th y lc y c lo p e n te n e )

3 ,5 -D im e th y lc y c lo h e x e n e (n o t 4 ,6 -d im e th y lc y c lo h e x e n e )

5. N a m e com pounds c o n tain in g a d o u b le b o n d and an a lc o h o l gro up as alk en o ls (o r c y c lo a lk e n o ls ) a n d g iv e the a lc o h o l carbon the lo w e r n um ber: OH OH 4 5

2 3

1

4 -M e th y l-3 -p e n te n -2 -o l o r 4 -m e th y lp e n t-3 -e n -2 -o l 6

2 -M e th y l-2 -c y c lo h e x e n -1 -o l or 2 -m e th y lc y c lo h e x - 2 -e n - 1 -ol

. T w o fre q u e n tly enco un tered a lk e n y l groups are the v in y l g r o u p and the a lly l g r o u p :

T h e v in y l g ro u p

T h e allyl gro up

153

4.5 Nomenclature of Alkenes and Cycloalkenes

Using substitutive nomenclature, the vinyl and allyl groups are called ethenyl and prop2-en-1-yl, respectively. The follow ing examples illustrate how these names are employed: OH

Cl

Br

B rom oethene or vinyl brom ide (com m on) 7.

E thenylcyclopropane or vinylcyclopropane

3-C hloropropene or allyl chloride (com m on)

3-(Prop-2-en-1-yl)cyclohexan-1-ol or 3-allylcyclohexanol

If two identical or substantial groups are on the same side o f the double bond, the com ­ pound can be designated cis; if they are on opposite sides it can be designated trans: Cl Cl

Cl

Cl

c/s-1,2-D ichloroethene

frans-1,2-D ichloroethene

(In Section 7.2 we shall see another method for designating the geometry of the double bond.)

S o lv e d P ro b lem 4 .4 Give an IUPAC name for the follow ing molecule. OH

We number the ring as shown below starting with the hydroxyl group so as to give the double bond the lower possible number. We include in the name the substituent (an ethenyl group) and the dou­ ble bond (-ene-), and the hydroxyl group (-o l) with numbers for their respective positions. Hence the IUPAC name is 3-ethenyl-2-cyclopenten-1-ol. STRATEGY A N D A N S W E R

5

OH t

2

Ethenyl group

Give IUPAC names for the follow ing alkenes: (a)

(b )

R eview P roblem 4 .1 0

(c)

> =

Write bond-line formulas for the following: (a )

cis-3-Octene

(f)

1-Brom o-2-m ethyl-1-(prop-2-en-1-yl)cyclopentane

(b )

trans-2-Hexene

(g )

3,4-Dim ethylcyclopentene

(c )

2,4-Dim ethyl-2-pentene

(h )

Vinylcyclopentane

(d )

trans-1-Chlorobut-2-ene

( i)

(e )

4,5-Dibrom o-1-pentene

CD

1,2-Dichlorocyclohexene trans-1,4-Dichloro-2-pentene

R eview P roblem 4.11

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Nomenclature and Conformations o f Alkanes and Cycloalkanes

4.6 N om enclature o f Alkynes Alkynes are named in much the same way as alkenes. Unbranched alkynes, for example, are named by replacing the -ane of the name of the corresponding alkane with the ending -yne. The chain is numbered to give the carbon atoms of the triple bond the lower possible num­ bers. The lower number of the two carbon atoms of the triple bond is used to designate the location of the triple bond. The IUPAC names of three unbranched alkynes are shown here: H— C = C — H E th yn e or a c e ty le n e *

2 -P e n ty n e

l-P e n te n -4 -y n e t o r p e n t-1 -e n -4 -y n e

The locations of substituent groups of branched alkynes and substituted alkynes are also indicated with numbers. An — OH group has priority over the triple bond when number­ ing the chain of an alkynol:

Cl

Cl 3

2

4

3 -C h lo ro p ro p y n e

3

-OH

2

2

1 -C h lo ro -2 -b u ty n e o r 1 -c h lo ro b u t- 2 -yn e

3 -B u ty n -1 -o l o r b u t-3 -y n -1 -o l

OH 6

5 -M e th y l-1 -h e x y n e o r 5 -m e th y lh e x -1 -y n e

R eview P roblem 4 .1 2

4 ,4 -D im e th y l-1 -p e n ty n e o r 4 ,4 -d im e th y lp e n t-1 -y n e

2 -M e th y l-4 -p e n ty n -2 -o l or 2 -m e th y lp e n t-4 -y n -2 -o l

Give the structures and IUPAC names for all the alkynes with the formula C6H10. Monosubstituted acetylenes or 1-alkynes are called terminal alkynes, and the hydro­ gen attached to the carbon of the triple bond is called the acetylenic hydrogen atom: A c e ty le n ic hyd ro gen

R— C = C — H A term in a l alk yn e

When named as a substituent, the HC # C— group is called the ethynyl group. The anion obtained when the acetylenic hydrogen is removed is known as an a lk y n id e io n or an acetylide ion. As we shall see in Section 7.11, these ions are useful in synthesis: R— C = C : or An a lk y n id e ion (an a c e ty lid e io n )

CH3C = C : or T h e p ro p y n id e ion

4.7 Physical Properties o f Alkanes and Cycloalkanes If we examine the unbranched alkanes in Table 4.3, we notice that each alkane differs from the preceding alkane by one — CH2— group. Butane, for example, is CH3(CH2)2CH3 and pentane is CH3(CH2)3CH3. A series of compounds like this, where each member differs

* T h e nam e a cetylene is re ta in e d b y th e IU P A C syste m fo r th e c o m p o u n d H C = C H and is used fre q u e n tly . fW h e r e th e re is a choice , th e d o u b le b o n d is g iv e n th e lo w e r num ber.

4.7 Physical Properties of Alkanes and Cycloalkanes

F ig u re 4 .3 (a) B o ilin g p o in ts o f u n b r a n c h e d a lk a n e s (in re d ) a n d c y c lo a lk a n e s (in w h ite ) . (b) M e lt in g p o in ts o f u n b r a n c h e d a lk a n e s .

fro m the n ex t m e m b e r b y a constant u n it, is c a lle d a h o m o lo g o u s series. M e m b e rs o f a h om olog ou s series are c a lle d h o m o lo g u e s . A t ro o m tem p e ra tu re ( 2 5 ° C ) and 1 a tm pressure the first fo u r m em bers o f the h o m o lo ­ gous series o f unbranch ed alkanes are gases (F ig . 4 .3 ), the C 5 — C

17

unbranch ed alkanes

(pen tan e to h ep tad ecan e) are liq u id s , and the unbranch ed alkanes w ith 18 and m o re ca r­ bon atom s are solids.

Boiling Points

T h e b o ilin g points o f the unbranch ed alkanes show a re g u la r increase

w ith increasin g m o le c u la r w e ig h t (F ig . 4 .3 a ) in the h o m olog ou s series o f straigh t-chain alkanes. B ran c h in g o f the alk an e chain , how ever, low e rs the b o ilin g p o in t. A s exam ples, con sider the C e H 1 4 isom ers in T ab le 4 .1 . H e x a n e b oils at 6 8 .7 ° C and 2 -m e th y lp e n ta n e and 3 -m e th y lp e n ta n e , each h av in g one branch, b o il lo w e r, at 6 0 .3 and 6 3 .3 ° C , respectively. 2 ,3 -D im e th y lb u ta n e and 2 ,2 -d im e th y lb u ta n e , each w ith tw o branches, b o il lo w e r s till, at 5 8 and 4 9 .7 ° C , resp ectively. P art o f the e x p la n atio n fo r these effects lies in the dispersion forces that w e studied in Section 2 .1 3 B . W ith unbranch ed alkanes, as m o le c u la r w e ig h t increases, so too do m o le c u la r size and, even m o re im p o rta n tly , m o le c u la r surface area. W it h increasin g sur­ face area, the dispersion forces b etw een m olecules increase; therefore, m o re energy (a h ig her tem p e ra tu re ) is re q u ire d to separate m o lecules fro m one another and p roduce b o ilin g . C h a in b ranch ing , on the o th er hand, m akes a m o le c u le m o re com pact, re d u c in g its surface area and w ith it the strength o f the dispersion forces o perating b etw e e n it and adjacent m o le ­ cules; this has the e ffe c t o f lo w e rin g the b o ilin g p oint.

M eltin g Points

T h e u nbranched alkanes do n ot show the same sm ooth increase in m e lt­

in g p oints w ith increasin g m o le c u la r w e ig h t (b lu e lin e in F ig . 4 .3 b ) that they show in th e ir b o ilin g points. T h e re is an a lte rn a tio n as one progresses fro m an unbranch ed a lkan e w ith an even n u m b e r o f carbon atom s to the n ex t one w ith an odd n u m b e r o f carbon atom s. F o r e x a m p le , pro pane (m p — 1 8 8 ° C ) m elts lo w e r than ethane (m p — 1 8 3 °C ) and also lo w e r than m ethan e (m p — 1 8 2 °C ). B u ta n e (m p — 1 3 8 ° C ) m elts 5 0 ° C h ig h e r than pro pane and o n ly 8

° C lo w e r than p entane (m p — 1 3 0 °C ). If , how ever, the even- and o d d -n u m b e red alkanes

are p lo tte d on separate curves (w h ite and re d lines in F ig . 4 .3 b ), there is a sm ooth increase in m e ltin g p o in t w ith increasin g m o le c u la r w e ig h t. X - R a y d iffra c tio n studies, w h ic h p ro v id e in fo rm a tio n about m o le c u la r structure, have re v e a le d the reason fo r this apparent a no m aly. A lk a n e chains w ith an even n u m b e r o f car­ b on atom s p a c k m o re c lo sely in the c ry s ta llin e state. A s a result, attractive forces b etw een in d iv id u a l chains are g reater and m e ltin g points are higher. T h e e ffe c t o f chain b ra n c h in g on the m e ltin g p oints o f alkanes is m o re d iffic u lt to p re ­ d ict. G e n e ra lly , how ever, b ra n c h in g that produces h ig h ly s y m m e tric a l structures results in

155

156

Chapter 4

Nomenclature and Conformations o f Alkanes and Cycloalkanes

abnormally high m elting points. The compound 2,2,3,3-tetramethylbutane, for example, melts at 100.7°C. Its boiling point is only six degrees higher, 106.3°C: C H CH 3 c h 3— C----- C— c h ch

3

ch

3

3

2 ,2 ,3 ,3 -T e tra m e th y lb u ta n e

Cycloalkanes also have much higher m elting points than their open-chain counterparts (Table 4.4). Physical Constants o f Cycloalkanes N um ber o f C a rb o n A to m s

3 4 5 6 7 8

Nam e

C yclopropane Cyclobutane C yclopentane Cyclohexane C ycloheptane C yclooctane

mp

D e n s ity , d 20(g m L -1)

b p (°C) ( 1 a tm )

(°C)

-3 3 13 49 81 118.5 149

-1 2 6 .6 -9 0 -9 4 6.5 -1 2 13.5

—

— 0.751 0.779 0.811 0.834

R e fra c tiv e In d e x (nD0)

— 1.4260 1.4064 1.4266 1.4449 —

D en s ity A s a class, the alkanes and cycloalkanes are the least dense o f all groups o f organic compounds. A ll alkanes and cycloalkanes have densities considerably less than 1.00 g m L _ 1 (the density o f water at 4°C). A s a result, petroleum (a mixture o f hydrocar­ bons rich in alkanes) floats on water. S o lu b ility Alkanes and cycloalkanes are alm ost totally insoluble in water because o f their very low polarity and their inability to form hydrogen bonds. Liquid alkanes and cycloalkanes are soluble in one another, and they generally dissolve in solvents o f low polarity. Good solvents for them are benzene, carbon tetrachloride, chloroform, and other hydrocarbons.

THE CHEMISTRY OF . . . P h e r o m o n e s : C o m m u n ic a tio n b y M e a n s o f C h e m ic a ls

Many animals communicate with other members of their species using a language based not on sounds or even visual signals but on the odors of chemicals called p h e ro m o n e s that these animals release. For insects, this appears to be the chief method of communication. Although pheromones are secreted by insects in extremely small amounts, they can cause profound and varied biological effects. Some insects use pheromones in courtship as sex attractants. Others use pheromones as warning substances, and still others secrete chemicals called "aggregation compounds" to cause mem­ bers of their species to congregate. Often these pheromones are relatively simple compounds, and some are hydrocarbons. For example, a species of cockroach uses undecane as an aggregation pheromone: CH3(CH2)gCH3 Undecane (cockroach aggregation pheromone)

(CH3)2CH(CH2)14CH3 2-M ethylheptadecane (sex attractant o f fem ale tiger moth)

When a female tiger moth wants to mate, she secretes 2-methylheptadecane, a perfume that the male tiger moth apparently finds irresistible. The sex attractant of the common housefly (Musca domestica) is a 23-carbon alkene with a cis double bond between atoms 9 and 10 called muscalure: CH3(CH2)7

,(CH2)i2CH3

xc = c h'

Xh

M uscalure (sex a ttra c ta n t o f co m m o n ho usefly)

Many insect sex attractants have been synthesized and are used to lure insects into traps as a means of insect con-

4.8 Sigma Bonds and Bond Rotation

trol, a much more environmentally sensitive method than the use of insecticides. Research suggests there are roles for pheromones in the lives of humans as well. For example, studies have shown that the phenomenon of menstrual synchronization among women who live or work with each other is likely caused by pheromones. Olfactory sensitivity to musk, which includes steroids such as androsterone, large cyclic ketones, and lac­ tones (cyclic esters), also varies cyclically in women, differs between the sexes, and may influence our behavior. Some of these compounds are used in perfumes, including civetone, a natural product isolated from glands of the civet cat, and pentalide, a synthetic musk.

O

HO* H Androsterone O

Pentalide

Civetone

4.8 Sigma Bonds and Bond Rotation Two groups bonded by only a single bond can undergo rotation about that bond with respect to each other.

^

•

The temporary molecular shapes that result from such a rotation are called con­ form ations o f the molecule.

•

Each possible structure is called a c o n fo r m e r .

•

An analysis o f the energy changes that occur as a m olecule undergoes rotations about single bonds is called a c o n fo r m a tio n a l a n a ly sis.

4.8A Newman Projections and How to Draw Them H e lp f u l H i n t

When w e do conformational analysis, w e w ill find that certain types o f structural formu1 t --.-.i las are especially convenient to use. One or these types is called a N ew m an projection form ula and another type is a saw horse form u la. Sawhorse formulas are much like dash-w edge three-dimensional formulas w e have used so far. In conformational analyses, w e w ill make substantial use o f Newman projections.

N ew m a n p ro je c tio n fo rm u la

, Learn to draw Newman projections and sawhorse formulas. Build handheld molecular models and compare them w'th your draw) between these hydrogens is 0 °. H H

F ig u re 4 .5

(a) T h e e c lip s e d

c o n f o r m a t io n o f e th a n e .

(b) T h e N e w m a n p r o je c tio n fo r m u la f o r t h e e c lip s e d c o n f o r m a t io n .

4.8B How to Do a Conformational Analysis N ow let us consider a conformational analysis o f ethane. Clearly, infinitesimally small changes in the dihedral angle between C — H bonds at each end o f ethane could lead to an infinite number o f conformations, including, o f course, the staggered and eclipsed confor­ mations. These different conformations are not all o f equal stability, however, and it is known that the staggered conformation o f ethane is the m ost stable conformation (i.e., it is the conformation o f lowest potential energy). The fundamental reason for this has recently com e to light. Quantum mechanical calculations by L. Goodman and V. T. Pophristic (Rutgers University) have shown that the greater stability o f the staggered conformation in ethane over the eclipsed conformation is mainly due to favorable overlap between sigm a ( s ) bond­ ing orbitals from the C — H bonds at one carbon and unfilled antibonding sigma ( s * ) orbitals at the adjacent carbon. In ethane’s staggered conformation, electrons from a given bond­ ing C — H orbital on one carbon can be shared with an unfilled s * orbital at the adjacent carbon. This phenomenon o f electron delocalization (via orbital overlap) from a filled bond­ ing orbital to an adjacent unfilled orbital is called h y p e r c o n ju g a tio n (and w e shall see in

159

4.8 Sigma Bonds and Bond Rotation

later chapters that it is a general stabilizing effect). F igure 4 .6 a shows the favorable over­ lap of s and s * in ethane by color coding of the orbital phases. If w e now consider the eclipsed conform ation o f ethane (Fig. 4.6£), w here the C — H bonds at each carbon are directly opposed to each other, w e see that the bonding s C — H orbital at one carbon does not overlap to as great an extent w ith the adjacent antibonding orbital as in the staggered conform ation. The possibility for hyperconjugation is dim inished, and therefore the potential energy o f this conform ation is higher. T he s - s * interactions in ethane are present in m ore com plicated m olecules as well. H owever, w here atom s and groups larger than hydrogen are involved in a conform ational analysis, such as our exam ple in Section 4.9, it is likely that repulsion o f the electron clouds involved in the bonding o f those groups increases in im portance as the cause o f the stag­ gered conform ation being m ost stable. T he energy difference betw een the conform ations o f ethane can be represented graphi­ cally in a p o te n tia l e n e rg y d ia g ra m , as show n in Figure 4.7. In ethane the energy differ­ ence betw een the staggered and eclipsed conform ations is about 12 kJ m o l- 1 . This sm all barrier to rotation is called the to rsio n a l b a r r ie r o f the single bond. B ecause o f this bar­ rier, som e m olecules w ill w ag back and forth w ith their atom s in staggered or nearly stag­ gered conform ations, w hile others w ith slightly m ore energy w ill rotate through an eclipsed conform ation to another staggered conform ation. A t any given m om ent, unless the tem per­ ature is extrem ely low ( —250°C), m ost ethane m olecules w ill have enough energy to undergo bond rotation from one conform ation to another.

F ig u re 4 .6

T h e b o n d in g C — H

o r b it a l is t h e o r b it a l w h e r e o n e lo b e o f a s in g le p h a s e ( r e p r e s e n te d b y o n e c o lo r ) e n v e lo p s t h e C — H a to m s . T h e a d ja c e n t u n f ille d a n t ib o n d in g

H H

o r b it a l is t h e o r b it a l w h e r e t h e p ha se c h a n g e s b e tw e e n th e c a r b o n a n d its h y d r o g e n a to m .

H

T h e s ta g g e r e d c o n f o r m a t io n

H

o f e th a n e (a) h as g r e a t e r o v e r la p

Eclipsed

b e t w e e n th e b o n d in g C — H o r b it a l a n d th e a d ja c e n t a n t ib o n d in g o r b it a l th a n in th e e c lip s e d c o n f o r m a t io n (b). T h e o r b it a l o v e r la p s h o w n in (a) le a d s t o t h e lo w e r p o t e n t ia l e n e r g y o f th e s ta g g e r e d c o n f o r m a t io n o f e th a n e .

H

H H

H

H

H H H F ig u re 4 .7

P o te n tia l e n e r g y

c h a n g e s th a t a c c o m p a n y r o t a t io n o f g r o u p s a b o u t th e c a r b o n - c a r b o n b o n d o f e th a n e .

H H

H Staggered

H Staggered Rotation

W hat does all this m ean about ethane? We can answ er this question in tw o different ways. If w e consider a single m olecule o f ethane, w e can say, for exam ple, that it w ill spend m ost o f its tim e in the low est energy, staggered conform ation, or in a conform ation very close to being staggered. M any tim es every second, however, it w ill acquire enough energy through collisions w ith other m olecules to surm ount the torsional barrier and it w ill rotate through an eclipsed conform ation. If w e speak in term s o f a large num ber o f ethane m ol­ ecules (a m ore realistic situation), w e can say that at any given m om ent m ost o f the m olecules w ill be in staggered or nearly staggered conform ations. If w e consider m ore highly substituted ethanes such as G C H 2 C H 2 G (w here G is a group or atom other than hydrogen), the barriers to ro tatio n are som ew hat larger, but they are still far too sm all to allow isolation o f the different staggered conform ations. T he fac­ tors involved in this rotational barrier are tog eth er called t o r s io n a l s t r a in and include the orbital considerations discussed above as w ell as repulsive interactions called s te ric

e

T h e id e a th a t c e rta in c o n fo rm a tio n s o f m o le c u le s are fa v o re d o rig in a te s fr o m th e w o r k o f J.H . v a n 't H o ff.

H e w as a lso w in n e r o f th e fir s t N o b e l Prize in C h e m is try (1901)

fo r his w o r k in c h e m ic a l k in e tic s .

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Chapter 4

Nomenclature and Conformations o f Alkanes and Cycloalkanes

h in d r a n c e betw een electron clouds o f bonded groups. In the next section w e consider a conformational analysis o f butane, where groups larger than hydrogen are involved in the analysis. G

G

G

H

C o n fo rm e rs lik e th e s e c a n n o t be is o la te d e x c e p t at e x tre m e ly lo w te m p e ra tu re s .

4.9 Conform ational Analysis o f Butane If w e consider rotation about the C 2 — C3 bond o f butane, w e find that there are six impor­ tant conformers, shown as I -V I below: CH

CH

HCH CH H CH

3

CH ,

H e lp f u l H i n t

You should build a molecular model of butane and examine its various conformations as we discuss their relative potential energies.

I A n anti c o n fo rm a tio n

H H

II A n e c lip s e d c o n fo rm a tio n

III A g au c h e c o n fo rm a tio n

V A gauche c o n fo rm a tio n

VI An e c lip s e d c o n fo rm a tio n

H H

IV A n e c lip s e d c o n fo rm a tio n

The a n t i c o n fo r m a tio n ( I ) does not have torsional strain from steric hindrance because the groups are staggered and the methyl groups are far apart. The anti conformation is the m ost stable. The methyl groups in the g a u c h e c o n fo rm a tio n s I I I and V are close enough to each other that the dispersion forces between them are repulsive; the electron clouds of the two groups are so close that they repel each other. This repulsion causes the gauche con­ formations to have approximately 3.8 kJ m o F 1 more energy than the anti conformation. The eclipsed conformations (II, IV, and V I) represent energy m axima in the potential energy diagram (Fig. 4.8). Eclipsed conformations II and V I have repulsive dispersion forces arising from the eclipsed methyl groups and hydrogen atoms. Eclipsed conforma­ tion IV has the greatest energy o f all because o f the added large repulsive dispersion forces between the eclipsed methyl groups as compared to II and VI. Although the barriers to rotation in a butane m olecule are larger than those o f an ethane m olecule (Section 4.8), they are still far too small to permit isolation o f the gauche and anti conformations at normal temperatures. Only at extremely low temperatures would the m olecules have insufficient energies to surmount these barriers.

4.9 Conformational Analysis of Butane

ch3 Anti I 0°

h Gauche III 60°

120°

h Gauche V 180°

240°

ch3 Anti I 300°

360°

Rotation

Figure 4.8 Energy changes th a t arise fro m ro ta tio n about th e C 2 — C3 bond o f butane.

We saw earlier (Section 2.16C) that dispersion forces can be attractive. Here, how­ ever, w e find that they can also be repulsive, leading to steric hindrance. Whether dispersion interactions lead to attraction or to repulsion depends on the distance that separates the two groups. A s two nonpolar groups are brought closer and closer together, the first effect is one in which a momentarily unsymmetrical distribution o f electrons in one group induces an opposite polarity in the other. The opposite charges induced in those portions o f the two groups that are in closest proximity lead to attraction between them. This attraction increases to a maximum as the internuclear distance of the two groups decreases. The internuclear distance at which the attractive force is at a maximum is equal to the sum o f what are called the van der Waals radii o f the two groups. The van der Waals radius o f a group is, in effect, a measure o f its size. If the two groups are brought still closer— closer than the sum o f their van der Waals radii— the interaction between them becom es repulsive. Their electron clouds begin to penetrate each other, and strong electron-electron interactions begin to occur.

4.9A Stereoisomers and Conformational Stereoisomers Gauche conformers I I I and V o f butane are examples o f stereoisomers. •

S te re o is o m e rs have the same molecular formula and connectivity but different arrangements of atoms in three-dimensional space.

•

C o n f o r m a t io n a l s te re o is o m e rs

are related to one another by bond rotations.

Conformational analysis is but one o f the ways in which w e w ill consider the three­ dimensional shapes and stereochemistry of m olecules. We shall see that there are other types o f stereoisomers that cannot be interconverted simply by rotations about single bonds. Among these are cis-trans cycloalkane isomers (Section 4.13) and others that w e shall con­ sider in Chapter 5. Sketch a curve similar to that in Fig. 4.8 showing in general terms the energy changes that arise from rotation about the C 2 — C 3 bond o f 2-methylbutane. You need not concern yourself with the actual numerical values o f the energy changes, but you should label all maxima and minima with the appropriate conformations.

Review Problem 4.13

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Chapter 4

Nomenclature and Conformations o f Alkanes and Cycloalkanes

THE CHEMISTRY OF M u s c le A c ti o n

Muscle proteins are essentially very long linear molecules (folded into a compact shape) whose atoms are connected by single bonds in a chainlike fashion. Relatively free rota­ tion is possible about atoms joined by single bonds, as we have seen. In muscles, the cumulative effect of rotations about many single bonds is to move the tail of each myosin molecule 60 A along the adjacent protein (called actin) in a step called the "power stroke." This process occurs over and over again as part of a ratcheting mechanism between many myosin and actin molecules for each muscle movement.

Power stroke in muscle

4.10 The Relative Stabilities o f Cycloalkanes: Ring Strain C y c lo a lk a n e s do n o t a ll h ave the sam e re la tiv e stability. E x p e rim e n ts h ave show n that cyclo h exan e is the m o s t stable c y c lo a lk a n e and that, in com pariso n , cy c lo p ro p a n e and c y c lo b u ­ tane are m u c h less stable. T h is d iffe re n c e in re la tiv e s ta b ility is d ue to r in g s t r a in , w h ic h com prises a n g le s tr a in a n d to r s io n a l s tr a in . •

A n g le s tr a in is the re s u lt o f d e v ia tio n fro m id e a l b o n d angles caused b y in h eren t structural constraints (such as rin g size).

H H

•

T o r s io n a l s tr a in is the re s u lt o f d isp ersion forces th a t c ann ot b e re lie v e d d ue to re stricted c o n fo rm a tio n a l m o b ility .

4.10A Cyclopropane H '"”7 H

H H

sp3-h y b rid iz e d a to m is 1 0 9 .5 °. In cy c lo p ro p a n e (a m o le c u le w ith the shape o f a re g u la r tria n g le ), the in te rn a l angles m u s t b e 6 0 ° a nd th e re fo re th e y m u s t d ep art fro m this id eal

(a)

H

T h e carb on atom s o f alkanes are sp3 h y b rid iz e d . T h e n o rm a l tetrah e d ra l b o n d ang le o f an

v a lu e b y a v e ry la rg e a m o u n t— b y 4 9 .5 °:

1.510 Â

H

HH if

H

C 60°

h A X , H 1.089 Â H /

u .C ­

— C, u

H

(b)

H

A n g le strain exists in a cyclopropane rin g because the sp 3 orbitals o f the carbon atoms cannot o verlap as e ffe c tiv e ly (F ig . 4 .9 a ) as they do in alkanes (w h e re p erfe c t end -on overlap

CH2

is possible). T h e c a rb o n -c a rb o n bonds o f cyclopropane are often described as b eing “bent.” O rb ita l o verlap is less effective. (T h e orbitals used fo r these bonds are n o t p u rely sp3; they contain m o re p character.) T h e c a rb o n -c a rb o n bonds o f cyclopropane are w eaker, and as a

(c)

resu lt the m o le c u le has greater p o te n tial energy. W h ile angle strain accounts fo r m ost o f the rin g strain in cyclopropane, it does n o t account fo r it a ll. Because the rin g is (o f necessity) planar, the C — H bonds o f the rin g are a ll eclipsed (Fig s. 4 .9 b ,c ), and the m o le c u le has torsional strain fro m repulsive dispersion forces as w e ll.

F ig u re 4 .9

(a) O r b it a l o v e r la p in t h e c a r b o n - c a r b o n b o n d s o f c y c lo p r o p a n e c a n n o t o c c u r

p e r f e c t ly e n d - o n . T h is le a d s t o w e a k e r " b e n t " b o n d s a n d t o a n g le s tr a in . (b) B o n d d is ta n c e s a n d a n g le s in c y c lo p r o p a n e . (c) A N e w m a n p r o je c t io n fo r m u la as v ie w e d a lo n g o n e c a r b o n - c a r b o n b o n d s h o w s t h e e c lip s e d h y d r o g e n s . ( V ie w in g a lo n g e it h e r o f th e o t h e r t w o b o n d s w o u ld s h o w t h e s a m e p ic t u r e . ) (d) B a ll- a n d - s tic k m o d e l o f c y c lo p r o p a n e .

163

4.11 Conformations of Cyclohexane: The Chair and the Boat

F ig u re 4 .1 0

(a) T h e " f o l d e d " o r

" b e n t " c o n f o r m a t io n o f H

H ^ Z ------- -- ------- ^ H

c y c lo b u ta n e . (b) T h e " b e n t " o r

H

A t

/ H H

" e n v e lo p e " f o r m o f c y c lo p e n ta n e . In th is s tr u c tu r e t h e f r o n t c a r b o n a to m is b e n t u p w a r d . In a c tu a lity , t h e m o le c u le is f le x ib le a n d s h ifts

(b)

c o n f o r m a t io n s c o n s ta n tly .

4.10B Cyclobutane Cyclobutane also has considerable angle strain. The internal angles are 8 8 °— a departure o f more than 21° from the normal tetrahedral bond angle. The cyclobutane ring is not pla­ nar but is slightly “folded” (Fig. 4.10a). If the cyclobutane ring were planar, the angle strain would be somewhat less (the internal angles would be 90° instead o f 8 8 °), but torsional strain would be considerably larger because all eight C — H bonds would be eclipsed. By folding or bending slightly the cyclobutane ring relieves more o f its torsional strain than it gains in the slight increase in its angle strain.

4.10C Cyclopentane The internal angles o f a regular pentagon are 108°, a value very close to the normal tetra­ hedral bond angles of 109.5°. Therefore, if cyclopentane m olecules were planar, they would have very little angle strain. Planarity, however, would introduce considerable torsional strain because all 10 C — H bonds would be eclipsed. Consequently, like cyclobutane, cyclopentane assumes a slightly bent conformation in which one or two o f the atoms o f the ring are out of the plane of the others (Fig. 4.10^). This relieves some o f the torsional strain. Slight twisting o f carbon-carbon bonds can occur with little change in energy and causes the out-of-plane atoms to m ove into plane and causes others to m ove out. Therefore, the m olecule is flexible and shifts rapidly from one conformation to another. With little tor­ sional strain and angle strain, cyclopentane is almost as stable as cyclohexane.

H e lp f u l H i n t A n u n d e rs ta n d in g o f th is and s u b s e q u e n t d is c u s s io n s o f c o n fo rm a tio n a l a nalysis can b e a id e d im m e a s u ra b ly th r o u g h th e use o f m o le c u la r m o d e ls . W e s u g g e s t y o u " fo llo w a lo n g " w ith m o d e ls as y o u re a d S e c tio n s 4 .1 1 -4 .1 3 .

4.11 Conform ations o f Cyclohexane: The Chair and the Boat Cyclohexane is more stable than the other cycloalkanes w e have discussed, and it has several conformations that are important for us to consider. •

The m ost stable conformation of cyclohexane is the chair con form ation .

•

There is no angle or torsional strain in the chair form o f cyclohexane.

In a chair conformation (Fig. 4.11), all of the carbon-carbon bond angles are 109.5°, and are thereby free of angle strain. The chair conformation is free o f torsional strain,

F ig u re 4.11

R e p r e s e n ta tio n s o f t h e c h a ir c o n f o r m a t io n o f c y c lo h e x a n e : (a) t u b e f o r m a t ; (b) b a ll-

a n d - s t ic k f o r m a t ; (c) lin e d r a w in g ; (d) s p a c e - fillin g m o d e l o f c y c lo h e x a n e . N o t ic e t h a t t h e r e a re t w o t y p e s o f h y d r o g e n s u b s t it u e n ts — th o s e t h a t p r o je c t o b v io u s ly u p o r d o w n (s h o w n in re d ) a n d th o s e t h a t lie a r o u n d t h e p e r im e t e r o f t h e r in g in m o r e s u b t le u p o r d o w n o r ie n t a t io n s (s h o w n in b la c k o r g r a y ). W e s h a ll d is c u s s th is f u r t h e r in S e c tio n 4 .1 2 .

164

Chapter 4

Nomenclature and Conformations o f Alkanes and Cycloalkanes

H

H

H

H

H

H

H

H

(b) F ig u re 4 .1 2

(a) A N e w m a n p r o je c tio n o f t h e c h a ir c o n f o r m a t io n o f c y c lo h e x a n e . ( C o m p a r is o n s

w it h a n a c tu a l m o le c u la r m o d e l w ill m a k e th is fo r m u la t io n c le a r e r a n d w ill s h o w t h a t s im ila r s t a g g e r e d a r r a n g e m e n ts a re s e e n w h e n o t h e r c a r b o n - c a r b o n b o n d s a re c h o s e n f o r s ig h tin g .)

(b) Illu s t r a t io n o f la r g e s e p a r a tio n b e t w e e n h y d r o g e n a to m s a t o p p o s it e c o r n e r s o f t h e r in g ( d e s ig n a t e d C1 a n d C 4 ) w h e n t h e r in g is in t h e c h a ir c o n f o r m a t io n .

as w ell. When view ed along any carbon-carbon bond (viewing the structure from an end, Fig. 4.12), the bonds are seen to be perfectly staggered. Moreover, the hydrogen atoms at opposite corners o f the cyclohexane ring are m aximally separated.

F ig u re 4 .1 3

•

B y partial rotations about the carbon-carbon single bonds o f the ring, the chair conformation can assume another shape called the b o a t c o n fo r m a tio n (Fig. 4.13).

•

The boat conformation has no angle strain, but it does have torsional strain.

(a) T h e b o a t c o n f o r m a t io n o f c y c lo h e x a n e is

fo r m e d b y " f l i p p i n g " o n e e n d o f t h e c h a ir f o r m u p ( o r d o w n ) . T h is f l i p r e q u ir e s o n ly r o ta tio n s a b o u t c a r b o n - c a r b o n s in g le b o n d s . (b) B a ll- a n d - s tic k m o d e l o f t h e b o a t c o n f o r m a t io n .

(c) A s p a c e - fillin g m o d e l o f t h e b o a t c o n f o r m a t io n .

H e lp f u l H i n t You w ill b e s t a p p re c ia te th e d iffe re n c e s b e tw e e n th e c h a ir and b o a t fo rm s o f c y c lo h e x a n e b y b u ild in g a n d m a n ip u la tin g

When a m odel o f the boat conformation is viewed down carbon-carbon bond axes along either side (Fig. 4.14a), the C — H bonds at those carbon atoms are found to be eclipsed, causing torsional strain. Additionally, two o f the hydrogen atoms on C1 and C4 are close enough to each other to cause van der Waals repulsion (Fig. 4.14b). This latter effect has

m o le c u la r m o d e ls o f each. HH

HH H

H

H

(a)

H

(b)

F ig u re 4 .1 4 (a) Illu s t r a t io n o f t h e e c lip s e d c o n f o r m a t io n o f t h e b o a t c o n f o r m a t io n o f c y c lo h e x a n e . (b) F la g p o le in t e r a c t io n o f t h e C1 a n d C 4 h y d r o g e n a to m s o f t h e b o a t c o n f o r m a t io n . T h e C 1 - C 4 f la g p o le in t e r a c t io n is a ls o r e a d ily a p p a r e n t in F ig . 4 .1 3 c .

4.11 Conformations of Cyclohexane: The Chair and the Boat

been called the “flagpole” interaction of the boat conformation. Torsional strain and flag­ pole interactions cause the boat conformation to have considerably higher energy than the chair conformation. Although it is more stable, the chair conformation is much more rigid than the boat con­ formation. The boat conformation is quite flexible. By flexing to a new form— the twist conformation (Fig. 4.15)— the boat conformation can relieve some o f its torsional strain and, at the same time, reduce the flagpole interactions.

F ig u re 4 .1 5

(a) T u b e m o d e l a n d (b) lin e

d r a w in g o f t h e t w i s t c o n f o r m a t io n o f (b )

•

c y c lo h e x a n e .

The twist boat conformation has a lower energy than the pure boat conformation, but is not as stable as the chair conformation.

The stability gained by flexing is insufficient, however, to cause the tw ist conformation o f cyclohexane to be more stable than the chair conformation. The chair conformation is esti­ mated to be lower in energy than the twist conformation by approximately 23 kJ m ol_1. The energy barriers between the chair, boat, and twist conformations o f cyclohexane are low enough (Fig. 4.16) to make separation o f the conformers im possible at room temper­ ature. At room temperature the thermal energies o f the m olecules are great enough to cause approximately 1 m illion interconversions to occur each second. •

Because o f the greater stability o f the chair, more than 99% o f the m olecules are estim ated to be in a chair conformation a t any given moment.

chair F ig u re 4 .1 6

boat

boat

chair

T h e r e la tiv e e n e r g ie s o f t h e v a r io u s c o n f o r m a t io n s o f c y c lo h e x a n e . T h e p o s itio n s o f

m a x im u m e n e r g y a re c o n f o r m a t io n s c a lle d h a lf-c h a ir c o n f o r m a t io n s , in w h ic h t h e c a r b o n a to m s o f o n e e n d o f t h e r in g h a v e b e c o m e c o p la n a r.

165

166

Chapter 4

Nomenclature and Conformations o f Alkanes and Cycloalkanes

THE CHEMISTRY OF . . . N a n o s c a le M o to r s a n d M o le c u la r S w itc h e s

Molecular rings that interlock with one another and com­ pounds that are linear molecules threaded through rings are proving to have fascinating potential for the creation of mol­ ecular switches and motors. Molecules consisting of interlock­ ing rings, like a chain, are called catenanes. The first catenanes were synthesized in the 1960s and have come to include examples such as olympiadane, as mentioned in Section 4.11a. Further research by Stoddart (UCLA) and col­ laborators on interlocking molecules has led to examples such as the catenane molecular switch shown here in (i). In an appli­ cation that could be useful in design of binary logic circuits, one ring of this molecule can be made to circumrotate in con­ trolled fashion about the other, such that it switches between two defined states. As a demonstration of its potential for application in electronics fabrication, a monolayer of these molecules has been "tiled" on a surface (ii) and shown to have characteristics like a conventional magnetic memory bit. Molecules where a linear molecule is threaded through a ring are called rotaxanes. One captivating example of a

rotaxane system is the one shown here in (iii), under devel­ opment by V. Balzani (University of Bologna) and collabora­ tors. By conversion of light energy to mechanical energy at the molecular level, this rotaxane behaves like a "fourstroke" shuttle engine. In step (a) light excitation of an elec­ tron in the P group leads to transfer of the electron to the initially + 2 At group, at which point At is reduced to the + 1 state. Ring R, which was attracted to At when it was in the +2 state, now slides over to a 2 in step (b), which remains +2. Back transfer of the electron from A-i to P+ in step (c) restores the +2 state of A-i, causing ring R to return to its original location in step (d). Modifications envisioned for this system include attaching binding sites to R such that some other molecular species could be transported from one loca­ tion to another as R slides along the linear molecule, or link­ ing R by a springlike tether to one end of the "piston rod" such that additional potential and mechanical energy can be incorporated in the system.

R

(ii) (F ig u re s r e p r in t e d w it h p e r m is s io n f r o m P e a se e t a l., A cco u n ts o f Chem ical Research, Vol. 34, n o .

6,

p p . 4 3 3 - 4 4 4 , 2 0 0 1 . C o p y r ig h t 2 0 0 1 A m e r ic a n

C h e m ic a l S o c ie ty ; a n d r e p r in t e d w it h p e rm is s io n fr o m B a lla rd in i e t a l., A cco u n ts o f Chemical

Research, Vol. 34, n o . 6 , p p . 4 4 5 - 4 5 5 , 2 0 0 1 . C o p y r ig h t 2 0 0 1 A m e r ic a n C h e m ic a l S o c ie ty .) ( iii )

(C)

167

4.12 Substituted Cyclohexanes: Axial and Equatorial Hydrogen Groups

4.11A Conformations of Higher Cycloalkanes Cycloheptane, cyclooctane, and cyclononane and other higher cycloalkanes also exist in nonplanar conformations. The small instabilities o f these higher cycloalkanes appear to be caused primarily by torsional strain and repulsive dispersion forces between hydrogen atoms across rings, called transannular strain. The nonplanar conformations o f these rings, how­ ever, are essentially free o f angle strain. X-Ray crystallographic studies of cyclodecane reveal that the m ost stable conformation has carbon-carbon-carbon bond angles o f 117°. This indicates som e angle strain. The wide bond angles apparently allow the m olecules to expand and thereby m inimize unfavorable repulsions between hydrogen atoms across the ring. There is very little free space in the center o f a cycloalkane unless the ring is quite large. Calculations indicate that cyclooctadecane, for example, is the smallest ring through which a — CH 2 CH 2 CH 2 — chain can be threaded. M olecules have been synthesized, however, that have large rings threaded on chains and that have large rings that are interlocked like links in a chain. These latter m olecules are called catenanes:

(CH2)n

Derek H. R. Barton (1918-1998) and Odd Hassel (1897-1981) shared the Nobel Prize in 1969 "fo r developing and applying the principles of conformation in chemistry." Their work led to fundamental understanding of not only the conformations of cyclohexane rings but also the structures of steroids (Section 23.4) and other compounds containing cyclohexane rings.

e

(CH2)„ A c a te n a n e

(n > 1 8 )

In 1994 J. F. Stoddart and co-workers, then at the University o f Birmingham (England), achieved a remarkable synthesis o f a catenane containing a linear array o f five interlocked rings. Because the rings are interlocked in the same way as those o f the olym pic symbol, they named the compound olym piadane.

4.12 S ubstituted Cyclohexanes: A xial and Equatorial Hydrogen Groups The six-membered ring is the m ost common ring found among nature’s organic m olecules. For this reason, w e shall give it special attention. We have already seen that the chair con­ formation of cyclohexane is the most stable one and that it is the predominant conforma­ tion of the m olecules in a sample o f cyclohexane. The chair conformation of a cyclohexane ring has two distinct orientations for the bonds that project from the ring. These positions are called axial and equatorial, as shown for cyclo­ hexane in Fig. 4.17. H

H

H

H H

H

•

The a x ia l b o n d s of cyclohexane are those that are perpendicular to the average plane o f the ring. There are three axial bonds on each face o f the cyclohexane ring, and their orientation (up or down) alternates from one carbon to the next.

•

The e q u a to r ia l b o n d s o f cyclohexane are those that extend from the perimeter of the ring. The equatorial bonds alternate from slightly up to slightly down in their orientation from one carbon to the next.

•

When a cyclohexane ring undergoes a chair-chair conformational change (a ring flip), all o f the bonds that were axial becom e equatorial, and all bonds that were equatorial becom e axial.

Figure 4.17 The chair co n form a tion o f cyclohexane. Axial hydrogen atom s are shown in red, equatorial hydrogens are shown in black.

168

Chapter 4

Nomenclature and Conformations o f Alkanes and Cycloalkanes

^ — Axial

Equatorial

T* •

3

J \2 • A

Equatorial

A xia l------- 1

4.12A How to Draw Chair Conformational Structures A set o f guidelines w ill help you draw chair conformational structures that are clear and that have unambiguous axial and equatorial bonds. •

N otice in Fig. 4 .18 a that sets o f parallel lines define opposite sides o f the chair. Notice, too, that equatorial bonds are parallel to ring bonds that are one bond away from them in either direction. When you draw chair conformational structures, try to make the corresponding bonds parallel in your drawings.

•

When a chair formula is drawn as shown in Fig. 4.18, the axial bonds are all either up or down, in a vertical orientation (Fig. 4.18b). When a vertex o f bonds in the ring points up, the axial bond at that position is also up, and the equatorial bond at the same carbon is angled slightly down. When a vertex o f ring bonds is down, the axial bond at that position is also down, and the equatorial bond is angled slightly upward.

(a) F ig u re 4 .1 8

Axial bond up Vertex of ring up

(a) S e ts o f p a ra lle l

lin e s t h a t c o n s t it u t e t h e r in g a n d e q u a to r ia l

C— H bonds

o f th e

c h a ir c o n f o r m a t io n . (b) T h e a xia l

Vertex of ring down Axial bond down

b o n d s a re a ll v e r tic a l. W h e n th e v e r t e x o f t h e r in g p o in ts u p , th e a x ia l b o n d is u p a n d v ic e v e rs a .

(b) Now, try to draw som e chair conformational structures for yourself that include the axial and equatorial bonds. Then, compare your drawings with those here and with actual mod­ els. You w ill see that with a little practice your chair conformational structures can be perfect.

4.12B A Conformational Analysis of Methylcyclohexane N ow let us consider m ethylcyclohexane. M ethylcyclohexane has two possible chair con­ formations (Fig. 4.19), and these are interconvertible through the bond rotations that con­ stitute a ring flip. In one conformation (Fig. 4 .19 a) the methyl group (with yellow

4.12 Substituted Cyclohexanes: Axial and Equatorial Hydrogen Groups

F ig u re 4 .1 9

(a) T h e

c o n f o r m a t io n s o f

(less stable)

(more stable by 7.6 kJ mol 1) (a)

H 3l H 3 H5

h

H

IS T *

H

H H

■ H H | H

v ^

m e th y l g r o u p a x ia l ( 1) a n d e q u a to r ia l (2). (b) 1 ,3 - D ia x ia l i n t e r a c t io n s b e t w e e n t h e t w o

H

CH3 H X . rH

m e th y lc y c lo h e x a n e w it h t h e

H H ^ \\\ \ T H * 1H H

a x ia l h y d r o g e n a to m s a n d th e

H H .— ■—r - H

a x ia l m e th y l g r o u p in t h e a x ia l c o n f o r m a t io n o f

CH,

H H

(b)

m e th y lc y c lo h e x a n e a re s h o w n w it h d a s h e d a r r o w s . L ess c r o w d in g o c c u r s in t h e e q u a to r ia l c o n f o r m a t io n .

(c) S p a c e - fillin g m o le c u la r m o d e ls f o r t h e a x ia l- m e t h y l a n d e q u a t o r ia l- m e t h y l c o n f o r m e r s o f m e th y lc y c lo h e x a n e . In t h e a x ia l- m e t h y l c o n f o r m e r th e m e th y l g r o u p (s h o w n w it h y e llo w h y d r o g e n a to m s ) is c r o w d e d b y th e 1 ,3 - d ia x ia l h y d r o g e n a to m s (re d ), as c o m p a r e d t o t h e e q u a t o r ia l- m e t h y l c o n fo r m e r , w h ic h h as n o 1 ,3 - d ia x ia l

Axial-m ethyl conformation

Equatorial-m ethyl conformation

h y d r o g e n s ) o c c u p i e s a n a x ia l p o s i t i o n , a n d i n t h e o t h e r t h e m e t h y l g r o u p o c c u p i e s a n e q u a ­

t o r ia l p o s i t i o n . •

T h e m o s t s t a b le c o n f o r m a t i o n f o r a m o n o s u b s t i t u t e d c y c l o h e x a n e r i n g ( a c y c l o h e x a n e r i n g w h e r e o n e c a r b o n a to m b e a rs a g r o u p o th e r th a n h y d r o g e n ) is th e c o n f o r m a t io n w h e r e th e s u b s t it u e n t is e q u a to r ia l.

S t u d i e s i n d i c a t e t h a t t h e c o n f o r m a t i o n w i t h t h e e q u a t o r i a l m e t h y l g r o u p i s m o r e s t a b le th a n th e c o n f o r m a t io n w it h th e a x ia l m e t h y l g r o u p b y a b o u t 7 .6 k J m o l _ 1 . T h u s , in th e e q u i­ lib r iu m

m i x t u r e , t h e c o n f o r m a t i o n w i t h t h e m e t h y l g r o u p i n t h e e q u a t o r i a l p o s i t i o n is t h e

p r e d o m in a n t o n e , c o n s t it u t in g a b o u t 9 5 %

o f th e e q u ilib r iu m m ix tu r e .

T h e g r e a te r s t a b ilit y o f m e th y lc y c lo h e x a n e w it h a n e q u a to r ia l m e t h y l g r o u p c a n b e u n d e r ­ s to o d t h r o u g h a n in s p e c tio n o f th e t w o fo r m s as th e y a re s h o w n in F ig s . 4 . 1 9 a - c . •

S t u d ie s d o n e w i t h m o d e l s o f t h e t w o c o n f o r m a t io n s s h o w t h a t w h e n t h e m e t h y l g r o u p is a x ia l , i t is s o c lo s e t o t h e t w o a x i a l h y d r o g e n s o n t h e s a m e s id e o f t h e r i n g ( a t t a c h e d to th e C 3 a n d C 5 a to m s ) th a t t h e d is p e r s io n fo r c e s b e t w e e n t h e m a r e r e p u ls iv e .

•

T h i s t y p e o f s t e r i c s t r a i n , b e c a u s e i t a r is e s f r o m g ro u p o n c a rb o n a to m

a n in t e r a c t io n b e t w e e n a n a x ia l

1 a n d a n a x ia l h y d r o g e n o n c a r b o n a to m 3 ( o r 5 ) is c a lle d a

1 ,3 - d ia x ia l in t e r a c t io n . •

S t u d i e s w i t h o t h e r s u b s t it u e n t s s h o w t h a t t h e r e i s g e n e r a l l y le s s r e p u l s i o n w h e n a n y g r o u p l a r g e r t h a n h y d r o g e n is e q u a t o r i a l r a t h e r t h a n a x ia l .

T h e s t r a i n c a u s e d b y a 1 , 3 - d i a x i a l i n t e r a c t i o n i n m e t h y l c y c l o h e x a n e is t h e s a m e a s th e s tr a in c a u s e d b y th e c lo s e p r o x i m i t y o f th e h y d r o g e n a to m s o f m e t h y l g r o u p s i n th e g a u c h e f o r m

in t e r a c t io n s w it h th e m e th y l g ro u p .

(c)

o f b u ta n e ( S e c tio n 4 .9 ) . R e c a ll th a t th e in t e r a c t io n in g a u ch e -

b u t a n e ( c a l l e d , f o r c o n v e n i e n c e , a g a u c h e in te r a c tio n ) c a u s e s g a u c h e - b u t a n e t o b e le s s s t a b le t h a n a n t i - b u t a n e b y 3 . 8 k J m o l _ 1 . T h e f o l l o w i n g N e w m a n p r o j e c t i o n s w i l l h e lp y o u to se e th a t th e t w o s te r ic in t e r a c t io n s a re th e s a m e . I n th e s e c o n d

170

Chapter 4

Nomenclature and Conformations o f Alkanes and Cycloalkanes

projection w e view axial m ethylcyclohexane along the C1 — C2 bond and see that what w e call a 1,3-diaxial interaction is simply a gauche interaction between the hydrogen atoms o f the methyl group and the hydrogen atom at C3:

H 2 _H

j

g a u c h e -B u ta n e (3 .8 kJ m o l _ 1 s te ric s tra in )

A xial m e th y lc y c lo h e x a n e (tw o g a u c h e in te ra c tio n s = 7.6 kJ m ol 1 s te ric s tra in )

E q u a to ria l m e th y lc y c lo h e xa n e (m o re s ta b le by 7.6 kJ m o l-1)

Viewing m ethylcyclohexane along the C1 — C 6 bond (do this with a m odel) shows that it has a second identical gauche interaction between the hydrogen atoms o f the methyl group and the hydrogen atom at C5. The methyl group o f axial m ethylcy­ clohexane, therefore, has two gauche interactions and, consequently, it has 7.6 kJ m o l - 1 o f strain. The methyl group o f equatorial m ethylcyclohexane does not have a gauche interaction because it is anti to C3 and C5.

R eview P roblem 4 .1 4

Show by a calculation (using the formula AG° = - R T ln Keq) that a free-energy difference o f 7.6 kJ m ol - 1 between the axial and equatorial forms o f methylcyclohexane at 25°C (with the equatorial form being more stable) does correlate with an equilibrium mixture in which the concentration o f the equatorial form is approximately 95%.

4.12C 1,3-Diaxial Interactions of a tert-Butyl Group In cyclohexane derivatives with larger alkyl substituents, the strain caused by 1,3-diaxial interactions is even more pronounced. The conformation o f tert-butylcyclohexane with the tert-butyl group equatorial is estimated to be approximately 21 kJ m ol - 1 more stable than the axial form (Fig. 4.20). This large energy difference between the two conformations means that, at room temperature, 99.99% o f the m olecules o f tert-butylcyclohexane have the tert-butyl group in the equatorial position. (The m olecule is not conformationally “locked,” however; it still flips from one chair conformation to the other.) CH 3

F ig u re 4 .2 0

(a) D ia x ia l in t e r a c t io n s w it h t h e la rg e

t e r t - b u t y l g r o u p a x ia l c a u s e t h e c o n f o r m a t io n w it h th e t e r t - b u t y l g r o u p e q u a to r ia l t o b e th e p r e d o m in a n t o n e t o t h e e x t e n t o f 9 9 .9 9 % . (b) S p a c e - fillin g m o le c u la r m o d e ls o f te r t- b u ty lc y c lo h e x a n e in t h e a x ia l (a x) a n d e q u a to r ia l (e q ) c o n f o r m a t io n s , h ig h lig h t in g th e p o s itio n o f t h e 1 ,3 - h y d r o g e n s (r e d ) a n d t h e t e r t - b u t y l g r o u p ( s h o w n w it h y e llo w h y d r o g e n a to m s ).

171

4.13 Disubstituted Cycloalkanes: Cis-Trans Isomerism

Figure 4 .2 0

(continued)

4.13 D isubstituted Cycloalkanes: Cis-Trans Isomerism The presence of two substituents on different carbons of a cycloalkane allows for the possibil­ ity of c is -tra n s is o m e ris m similar to the kind we saw for alkenes in Section 1.13B. These cis-trans isomers are also stereoisom ers because they differ from each other only in the arrange­ ment of their atoms in space. Consider 1,2-dimethylcyclopropane (Fig. 4.21) as an example.

ch3

ch3

ch3

c / 's - 1 , 2 - D im e t h y lc y c lo p r o p a n e

CH 3

H

f r a n s - 1 , 2 - D im e t h y lc y c lo p r o p a n e

Figure 4.21

The cis- and trans1,2-dim ethylcyclopropane

The planarity o f the cyclopropane ring makes the cis-trans isomerism obvious. In the first structure the methyl groups are on the same side o f the ring; therefore, they are cis. In the second structure, they are on opposite sides o f the ring; they are trans. Cis and trans isomers such as these cannot be interconverted without breaking carbon­ carbon bonds. They will have different physical properties (boiling points, melting points, and so on). As a result, they can be separated, placed in separate bottles, and kept indefinitely.

Write structures for the cis and trans isomers o f ( a ) 1,2-dichlorocyclopentane and ( b ) 1,3dibromocyclobutane. (c ) Are cis-trans isomers possible for 1,1-dibromocyclobutane?

4.13A Cis-Trans Isomerism and Conformational Structures of Cyclohexanes Trans 1,4-D isub stitu ted Cyclohexanes If w e consider dimethylcyclohexanes, the structures are somewhat more com plex because the cyclohexane ring is not planar. Beginning with frans-1,4-dimethylcyclohexane, because it is easiest to visualize, w e find

R eview P roblem 4 .1 5

172

Chapter 4

Nomenclature and Conformations o f Alkanes and Cycloalkanes

th e r e a re t w o p o s s ib le c h a ir c o n f o r m a t io n s ( F ig . 4 .2 2 ) . I n o n e c o n f o r m a t io n b o t h m e t h y l g r o u p s a r e a x i a l ; i n t h e o t h e r b o t h a r e e q u a t o r i a l . T h e d i e q u a t o r i a l c o n f o r m a t i o n is , a s w e w o u l d e x p e c t i t t o b e , t h e m o r e s t a b le c o n f o r m a t i o n , a n d i t r e p r e s e n t s t h e s t r u c t u r e o f a t le a s t 9 9 %

o f th e m o le c u le s a t e q u ilib r iu m .

T h a t th e d ia x ia l f o r m o f t r a n s - 1 , 4 - d im e th y lc y c lo h e x a n e is a tr a n s is o m e r is e a s y to see; t h e t w o m e t h y l g r o u p s a r e c l e a r l y o n o p p o s i t e s id e s o f t h e r i n g . T h e t r a n s r e l a t i o n s h i p o f th e m e t h y l g r o u p s in th e d ie q u a t o r ia l f o r m is n o t a s o b v io u s , h o w e v e r.

F ig u re 4 .2 2 The tw o chair conform ations o f trans-1,4dim ethylcyclohexane: tra n s -d ie q u a to ria l and tran s-d ia xia l. The tra n s -d ie q u a to ria l fo rm is m ore stable by 15.2 kJ m o l- 1 .

ring , flip

fr a n s - D ia x ia l

H ow

do w e know

tr a n s - D ie q u a t o r ia l

t w o g r o u p s a re c is o r tra n s ? A

g e n e ra l w a y to r e c o g n iz e a tr a n s ­

d is u b s t i t u t e d c y c l o h e x a n e i s t o n o t i c e t h a t o n e g r o u p i s a t t a c h e d b y t h e u p p e r b o n d ( o f t h e t w o t o i t s c a r b o n ) a n d o n e b y t h e lo w e r b o n d :

U pper bond G r o u p s a r e tra n s

3

if o n e is c o n n e c te d by an u pper bond a n d th e o th e r b y a lo w e r b o n d

bond f r a n s - 1 , 4 - D im e th y lc y c lo h e x a n e I n a c is 1 , 4 - d is u b s titu te d c y c lo h e x a n e b o t h g r o u p s a re a tta c h e d b y a n u p p e r b o n d o r b o t h b y a lo w e r b o n d . F o r e x a m p le ,

U pper bond

G r o u p s a r e cis

C H,

Upper bond

if b o th a re c o n n e c te d b y u p p e r b o n d s o r if b o th a re c o n n e c te d

H3C

b y lo w e r b o n d s

H c is - 1 , 4 - D im e t h y lc y c lo h e x a n e

Cis 1 ,4-D isu b stitu ted Cyclohexanes

r is - 1 ,4 - D im e th y lc y c lo h e x a n e

e x is t s

in

tw o

e q u iv a le n t c h a i r c o n f o r m a t i o n s ( F i g . 4 . 2 3 ) . I n a c i s 1 , 4 - d is u b s t i t u t e d c y c l o h e x a n e , o n e g r o u p is a x ia l a n d th e o th e r is e q u a t o r ia l i n b o t h o f th e p o s s ib le c h a ir c o n f o r m a t io n s .

chair-chair ring '

E q u a t o r i a l- a x i a l

F ig u re 4 .2 3

flip

A x ia l- e q u a to r ia l

Equivalent conform ations o f cis-1,4-dim ethylcyclohexane.

173

4.13 Disubstituted Cycloalkanes: Cis-Trans Isomerism

S o lv e d P ro b lem 4 .5 C o n s id e r e a c h o f th e f o l l o w i n g c o n f o r m a t io n a l s tr u c tu r e s a n d t e l l w h e t h e r e a c h is c is o r tr a n s :

H H

H Cl

Cl

H Cl (a)

(b)

(c)

A N S W E R (a ) E a c h c h l o r i n e i s a t t a c h e d b y t h e u p p e r b o n d a t i t s c a r b o n ; t h e r e f o r e , b o t h c h l o r i n e a t o m s a r e o n t h e s a m e s id e o f t h e m o l e c u l e a n d t h i s i s a c i s is o m e r . T h i s i s a c i s - 1 , 2 - d i c h l o r o c y c l o h e x a n e . ( b ) H e r e b o t h c h l o r i n e a t o m s a r e a t t a c h e d b y a l o w e r b o n d ; t h e r e f o r e , i n t h i s e x a m p l e , t o o , b o t h c h l o r i n e a t o m s a r e o n t h e s a m e s id e o f t h e m o l e c u l e a n d t h i s , t o o , i s a c i s is o m e r . I t i s c i s - 1 , 3 - d i c h l o r o c y c l o h e x a n e . (c ) H e r e o n e c h l o r i n e a t o m i s a t t a c h e d b y a l o w e r b o n d a n d o n e b y a n u p p e r b o n d . T h e t w o c h l o r i n e a t o m s , t h e r e f o r e , a r e o n o p p o s i t e s id e s o f t h e m o l e ­ c u le , a n d t h i s i s a t r a n s i s o m e r . I t i s t r a n s - 1 , 2 - d i c h l o r o c y c l o h e x a n e . V e r i f y th e s e f a c t s b y b u i l d i n g m o d e l s .

T h e t w o c o n f o r m a t i o n s o f c i s 1 , 4 - d i s u b s t i t u t e d c y c l o h e x a n e s a re n o t e q u iv a le n t i f o n e g r o u p is la r g e r th a n th e o th e r. C o n s id e r c is - 1 - t e r t - b u t y l- 4 - m e t h y lc y c lo h e x a n e :

3

( M o r e s t a b le b e c a u s e la r g e

( L e s s s t a b le b e c a u s e la r g e

g r o u p is e q u a t o r ia l)

g r o u p is a x ia l)

c / s - 1 - t e r t - B u t y l- 4 - m e t h y lc y c lo h e x a n e H e r e t h e m o r e s t a b le c o n f o r m a t i o n i s t h e o n e w i t h t h e l a r g e r g r o u p e q u a t o r i a l . T h i s i s a g e n e r a l p r in c ip le : •

W h e n o n e r i n g s u b s t it u e n t g r o u p is la r g e r th a n th e o th e r a n d th e y c a n n o t b o t h b e e q u a t o r i a l , t h e c o n f o r m a t i o n w i t h t h e l a r g e r g r o u p e q u a t o r i a l w i l l b e m o r e s t a b le .

(a ) W r i t e s t r u c t u r a l f o r m u l a s f o r t h e t w o c h a i r c o n f o r m a t i o n s o f c i s - 1 - i s o p r o p y l - 4 - m e t h y l c y c l o h e x a n e . ( b ) A r e t h e s e t w o c o n f o r m a t i o n s e q u i v a le n t ? (c ) I f n o t , w h i c h w o u l d b e m o r e s t a b le ? ( d ) W h i c h w o u l d b e t h e p r e f e r r e d c o n f o r m a t i o n a t e q u i l i b r i u m ?

Trans 1,3-D isu b stituted Cyclohexanes

t r a n s - 1 , 3 - D im e t h y lc y c lo h e x a n e is l ik e th e

c i s 1 ,4 c o m p o u n d i n t h a t e a c h c o n f o r m a t i o n h a s o n e m e t h y l g r o u p i n a n a x i a l p o s i t i o n a n d o n e m e t h y l g r o u p i n a n e q u a t o r ia l p o s it io n . T h e f o l l o w i n g t w o c o n f o r m a t io n s a re o f e q u a l e n e r g y a n d a re e q u a lly p o p u la t e d a t e q u ilib r iu m :

ring 'flip

(eq) H3C

CH 3 (ax) f r a n s - 1 , 3 - D im e t h y lc y c lo h e x a n e E q u a l e n e r g y a n d e q u a lly p o p u la t e d c o n f o r m a t io n s

R eview P roblem 4.16

174

Chapter 4

Nomenclature and Conformations o f Alkanes and Cycloalkanes

The situation is different for trans-l-tert-butyl-3-m ethylcyclohexane (shown below) because the two ring substituents are not the same. Again, w e find that the lower energy conformation is that with the largest group equatorial.

( M o r e s ta b le b e c a u s e la r g e

( L e s s s t a b le b e c a u s e la r g e

g r o u p is e q u a t o r ia l)

g r o u p is a x ia l)

t r a n s - 1 - t e r t - B u t y l- 3 - m e t h y lc y c lo h e x a n e

Cis 1,3-D isu b stitu ted Cyclohexanes cis-1,3-Dim ethylcyclohexane has a conforma­ tion in which both methyl groups are equatorial and one in which both methyl groups are axial. A s w e w ould expect, the conform ation w ith both m ethyl groups equatorial is the m ore stable one. Trans 1 ,2-D isu b stitu ted Cyclohexanes trans-1,2-Dim ethylcyclohexane has a con­ formation in which both methyl groups are equatorial and one in which both methyl groups are axial. A s w e w ould expect, the conform ation w ith both m ethyl groups equatorial is the m ore stable one.

CH3 (ax) ( D i e q u a t o r i a l i s m u c h m o r e s t a b le )

( D ia x i a l i s m u c h l e s s s t a b l e )

tr a n s - 1 ,2 - D im e t h y lc y c lo h e x a n e

Cis 1 ,2-D isu b stitu ted Cyclohexanes cis-1,2-Dim ethylcyclohexane has one methyl group that is axial and one methyl group that is equatorial in each o f its chair conforma­ tions, thus its two conformations are o f equal stability.

(ax)

ring flip ( E q u a t o r ia l- a x ia l)

( A x ia l- e q u a to r ia l)

c /s - 1 , 2 - D im e t h y lc y c lo h e x a n e E q u a l e n e r g y a n d e q u a lly p o p u la t e d c o n f o r m a t io n s

R eview P roblem 4 .1 7

Write a conformational structure for 1-bromo-3-chloro-5-fluorocyclohexane in which all the substituents are equatorial. Then write its structure after a ring flip.

R eview P roblem 4 .1 8

(a) Write the two conformations o f ds-l-teri-butyl-2-m ethylcyclohexane. (b) Which conformer has the low est potential energy?

175

4.14 Bicyclic and Polycyclic Alkanes

4.14 Bicyclic and Polycyclic Alkanes Many of the m olecules that w e encounter in our study o f organic chemistry contain more than one ring (Section 4.4B). One of the most important bicyclic systems is bicyclo [4.4.0]decane, a compound that is usually called by its common name, decalin: 10

2

9

8 7

'6 \ ^ 4 5

D e c a l in ( b i c y c l o [ 4 . 4 . 0 ] d e c a n e ) (c a rb o n a to m s

1

and

6

a r e b r id g e h e a d c a r b o n a to m s )

Decalin shows cis-trans isomerism: H H

H e lp f u l H i n t Chemical Abstracts Service (CAS) determines the number of rings by the formula S —A + 1 = N, where S is the number of single bonds in the ring system, A is the number of atoms in the ring system, and N is the calculated number of rings (see Problem 4.30).

In cis-decalin the two hydrogen atoms attached to the bridgehead atoms lie on the same side of the ring; in trans-decalin they are on opposite sides. We often indicate this by writ­ ing their structures in the follow ing way: H

H

H c /s - D e c a lin

t r a n s - D e c a lin

Simple rotations of groups about carbon-carbon bonds do not interconvert cis- and transdecalins. They are stereoisomers and they have different physical properties. Adamantane is a tricyclic system that contains a three-dimensional array o f cyclohexane rings, all of which are in the chair form.

One goal of research in recent years has been the synthesis o f unusual, and sometimes highly strained, cyclic hydrocarbons. Among those that have been prepared are the com ­ pounds that follow:

r or B ic y c lo [ 1 . 1 . 0 ] b u t a n e

Cubane

B

P r is m a n e

176

Chapter 4

Nomenclature and Conformations o f Alkanes and Cycloalkanes

THE CHEMISTRY OF E le m e n ta l C a rb o n

Carbon as the pure element exists in several forms that are as different from one another as it is possible to imagine. Different forms of a pure element are called allotropes. One allotrope of pure carbon is the very soft and totally black substance called graphite, the main substance at the cen­ ter of pencils and the main component of charcoal and chim­ ney soot. Another allotropic form of carbon is diamond, the colorless brilliant gem that is the hardest of all substances found in nature. Still another allotrope, perhaps the most exotic, is called buckminsterfullerene after the inventor of the geodesic dome, Buckminster Fuller. The different properties of these allotropic forms arise from different structural arrangements of the carbon atoms in each form, and these arrangements result, in part, from different hybridization states of their carbon atoms. The car­ bon atoms of diamond are all sp3 hybridized with tetrahe­ drally oriented bonds. The structure of diamond is what you would get if you extended the structure of adamantane in three dimensions. The great hardness of diamond results from the fact that the entire diamond crystal is one large molecule—a network of interconnecting rings that is held together by millions of strong covalent bonds. In graphite the carbon atoms are sp2 hybridized. Because of the trigonal planar orientation of their covalent bonds, the carbon atoms of graphite are in sheets. The sheets are actu­

ally huge molecules consisting of fused benzene rings (see below). While all of the covalent bonds of each sheet lie in the same plane, the sheets are piled one on another and the p orbitals of their benzene rings keep them apart. Although these p orbitals interact, their interactions are very weak, much weaker than those of covalent bonds, allowing the individual sheets to slide past one another and accounting for graphite's usefulness as a lubricant. Buckminsterfullerene (shown on the next page) is a rep­ resentative of a new class of carbon compounds discovered in 1985 consisting of carbon clusters called fullerenes (see Section 14.8C for the story of their discovery and synthesis). Buckminsterfullerene (also called a "buckyball") is a hollow cluster of 60 carbon atoms, all of which are sp2 hybridized, and which are joined together in a pattern like the seams of a soccer ball. The center of the buckyball is large enough to hold an atom of helium or argon, and such compounds are known. In the buckyball there are 32 interlocking rings: 20 are hexagons and 12 are pentagons, producing a highly symmetrical molecule. A smaller symmetrical molecule, syn­ thesized in 1982 by Leo A. Paquette and co-workers at Ohio State University, is dodecahedrane. One final point: We began this book telling of how all of the carbon atoms of the universe are thought to have been formed in the interiors of stars and to have been dispersed

A p o rtio n o f th e d ia m o n d s tru c tu re

A

9 • 9 •9 • • • •

e • C a rb o n is s h o w n h e re in its d ia m o n d a n d g ra p h ite fo rm s

•

9 •9 • 9 • 9 •9 • • 9 • • • • •

A p o rtio n o f th e s tru c tu re o f g ra p h ite

4.16 Synthesis of Alkanes and Cycloalkanes

177

throughout the universe when some of those stars exploded as supernovae. Consider this evidence. Sediments on our planet, known to be 251 million years old and which were formed at the time of a great extinction caused by the colli­ sion of a comet with Earth, have been found to contain buckyballs with helium atoms in their centers. The isotopic ratio of 3He/4He in them is much larger than the ratio in ordinary helium found on Earth now, indicating that the helium was of extraterrestrial origin. So in these discoveries we have fascinating evidence for the origin of elemental carbon and how some of it got here. Most carbon atoms were produced when Earth was formed billions of years ago. But the carbon atoms of the buckyballs found in this sediment, formed originally in the interior of a star somewhere in the universe, probably made their way here 251 million years ago in a comet or meteorite.

4.15 Chemical Reactions o f Alkanes Alkanes, as a class, are characterized by a general inertness to many chemical reagents. Carbon-carbon and carbon-hydrogen bonds are quite strong; they do not break unless alkanes are heated to very high temperatures. Because carbon and hydrogen atoms have nearly the same electronegativity, the carbon-hydrogen bonds o f alkanes are only slightly polarized. As a consequence, they are generally unaffected by most bases. M olecules of alkanes have no unshared electrons to offer as sites for attack by acids. This low reactivity o f alkanes toward many reagents accounts for the fact that alkanes were originally called p a r a ff in s (parum affinis, Latin: little affinity). The term paraffin, however, was probably not an appropriate one. We all know that alkanes react vigorously with oxygen when an appropriate mixture is ignited. This combus­ tion occurs, for example, in the cylinders o f automobiles, in furnaces, and, more gently, with paraffin candles. When heated, alkanes also react with chlorine and bromine, and they react explosively with fluorine. We shall study these reactions in Chapter 10.

4.16 Synthesis o f Alkanes and Cycloalkanes A chemical synthesis may require, at som e point, the conversion o f a carbon-carbon double or triple bond to a single bond. Synthesis o f the follow ing compound, used as an ingredient in some perfumes, is an example.

(used in s o m e p erfu m e s )

This conversion is easily accomplished by a reaction called h ydrogen ation . There are sev­ eral reaction conditions that can be used to carry out hydrogenation, but among the com ­ mon ways is use o f hydrogen gas and a solid metal catalyst such as platinum, palladium, or nickel. Equations in the follow ing section represent general examples for the hydrogen­ ation of alkenes and alkynes.

4.16A Hydrogenation of Alkenes and Alkynes Alkenes and alkynes react with hydrogen in the presence o f metal catalysts such as nickel, palladium, and platinum to produce alkanes. The general reaction is one in which the atoms

178

Chapter 4

Nomenclature and Conformations o f Alkanes and Cycloalkanes

o f the hydrogen m olecule add to each atom o f the carbon-carbon double or triple bond o f the alkene or alkyne. This converts the alkene or alkyne to an alkane: General Reaction \

/

C C

/ H H

\

Pt, Pd, or Ni solvent, pressure

— C— __ C C

A lk e n e

C

H

2

H H

C I

A lk a n e

Pt H

2 solvent,

H

— C—

H

H

C

H

pressure

A lk y n e

A lk a n e

The reaction is usually carried out by dissolving the alkene or alkyne in a solvent such as ethyl alcohol (C 2 H5 OH), adding the metal catalyst, and then exposing the mixture to hydrogen gas under pressure in a special apparatus. One molar equivalent o f hydrogen is required to reduce an alkene to an alkane. Two molar equivalents are required to reduce an alkyne. (We shall discuss the mechanism o f this reaction in Chapter 7.) Specific Examples

CH c h 3— c = c h

CH Ni H

2

2

EtOH (25°C, 50 atm)

2 -M e th y lp ro p e n e

Pd

C y c lo h e x a n e

2 H

2

R e v ie w P ro b le m 4 .1 9

H

EtOH (25°C, 1 atm)

C y c lo h e x e n e

C y c lo n o n y n - 6 -on e

H

2

Is o b u ta n e

H2

O

c h 3— c — c h

Pd

O

ethyl acetate C y c lo n o n a n o n e

Show the reactions involved for hydrogenation o f all the alkenes and alkynes that would yield 2 -methylbutane.

4.17 H o w to Gain Structural Inform ation from M olecular Formulas and the Index o f H ydrogen Deficiency A chem ist working with an unknown com pound can obtain considerable information about its structure from the com pound’s m olecular formula and its in d ex o f h ydrogen d eficien cy (IH D ). •

The index o f h ydrogen deficiency (IH D )* is defined as the difference in the n u m ­ o f hydrogen atoms between the compound under study and an acyclic alkane having the same number o f carbons.

b e r o f p a ir s

Saturated acyclic hydrocarbons have the general molecular formula C nH2n+2. Each dou­ ble bond or ring reduces the number o f hydrogen atoms by two as compared with the formula for a saturated compound. Thus each ring or double bond provides one unit o f *S o m e o rg a n ic chem ists re fe r to th e in d e x o f h y d ro g e n d e fic ie n c y as th e “ degree o f u n s a tu ra tio n ” o r “ the n u m b e r o f d o u b le -b o n d e quivalencies.”

4.17 How to Gain Structural Information from Molecular Formulas

hydrogen deficiency. For example, 1-hexene and cyclohexane have the same molecular formula (CgH-^) and they are constitutional isomers.

1 -H ex e n e

C y c lo h e x a n e (C 6 H iJ

(CeH 1 2 )

Both 1-hexene and cyclohexane (C6H12) have an index o f hydrogen deficiency equal to 1 (meaning one pair o f hydrogen atoms), because the corresponding acyclic alkane is hexane (C 6 H1 4 ). C6H14 = formula of corresponding alkane (hexane) C6H12 = formula of compound (1-hexene or cyclohexane) H2 = difference = 1 pair o f hydrogen atoms Index o f hydrogen deficiency = 1 Alkynes and alkadienes (alkenes with two double bonds) have the general formula CnH2n_ 2. Alkenynes (hydrocarbons with one double bond and one triple bond) and alkatrienes (alkenes with three double bonds) have the general formula CnH2n_ 4, and so forth.

1 ,3 -B u ta d ie n e IH D = 2

B u t-1 -e n -3 -y n e IH D = 3

1 ,3 ,5 -H e x a trie n e IH D = 3

The index o f hydrogen deficiency is easily determined by comparing the molecular formula of a given compound with the formula for its hydrogenation product. •

Each double bond consumes one molar equivalent o f hydrogen and counts for one unit o f hydrogen deficiency.

•

Each triple bond consumes two molar equivalents o f hydrogen and counts for two units of hydrogen deficiency.

•

Rings are not affected by hydrogenation, but each ring still counts for one unit of hydrogen deficiency.

Hydrogenation, therefore, allows us to distinguish between rings and double or triple bonds. Consider again two compounds with the molecular formula C6H12: 1-hexene and cyclo­ hexane. 1-Hexene reacts with one molar equivalent o f hydrogen to yield hexane; under the same conditions cyclohexane does not react: +

H2

Pt 25°C

no reaction

H2

Or consider another example. Cyclohexene and 1,3-hexadiene have the same m olecu­ lar formula (C6H10). Both compounds react with hydrogen in the presence o f a catalyst, but cyclohexene, because it has a ring and only one double bond, reacts with only one molar equivalent. 1,3-Hexadiene adds two molar equivalents: +

H2

+

2 H0

Pt 25°C

C y c lo h e x e n e

1 ,3 -H e x a d ie n e

Pt t 25°C

179

180

Review Problem 4.20

Chapter 4

Nomenclature and Conformations o f Alkanes and Cycloalkanes

( a ) W h a t is the in d e x o f h yd ro g e n d e fic ie n c y o f 2 -h exene? (b ) O f m e th y lc y c lo p e n ta n e ? (c) D o e s the in d e x o f h yd ro gen d efic ie n c y re veal a nythin g abo ut the lo c a tio n o f the d o u b le bond in the chain? (d ) A b o u t the size o f the rin g? (e ) W h a t is the in d e x o f h yd ro g e n d efic ie n c y o f 2 -h e x y n e ? ( f ) In g en e ra l term s, w h a t structural p o s s ib ilitie s e x ist fo r a c o m p o u n d w ith the m o le c u la r fo rm u la C 1 0 H 16?

R eview P roblem 4.21

Z in g ib e re n e , a fra g ra n t c o m p o u n d is o la te d fro m ginger, has the m o le c u la r fo rm u la C 1 5 H 2 4 a nd is k n o w n n o t to con tain an y trip le bonds. (a ) W h a t is the in d e x o f h yd ro g e n d efic ie n c y o f zin g ib eren e? ( b ) W h e n zin g ib e re n e is subjected to c a ta ly tic h yd ro g e n a tio n using an excess o f h yd ro g e n , 1 m o l o f zin g ib e re n e absorbs 3 m o l o f h yd ro g e n a n d produces a c o m ­ p o u n d w ith the fo rm u la C

1 5

H 30. H o w m a n y d o u b le bonds does a m o le c u le o f zin g ib e re n e

have? (c ) H o w m a n y rings?

4.17A Compounds Containing Halogens, Oxygen, or Nitrogen C a lc u la tin g the in d e x o f h yd ro g e n d e fic ie n c y ( I H D ) fo r com pounds o th e r than h y d ro c a r­ bons is re la tiv e ly easy. F o r c o m p o u n d s c o n ta in in g h a lo g e n a to m s , w e s im p ly c o u n t th e h a lo g e n a to m s as th o u g h th e y w e re h y d ro g e n a to m s . C o n s id e r a c o m p o u n d w ith the fo rm u la C 4 H 6 C l2. To c a lcu la te the I H D , w e change the tw o c h lo rin e atom s to h yd ro g e n atom s, con siderin g the fo rm u la as tho ug h it w e re C 4 H 8. T h is fo rm u la has tw o h y d ro g e n atom s fe w e r than the fo r­ m u la fo r a saturated a lk an e ( C 4 H 10), a n d this tells us th a t the c o m p o u n d has I H D = 1. It c o u ld , the re fo re , h av e e ith e r o ne rin g o r one d o u b le b on d . [W e can te ll w h ic h it has fro m a h y d ro g e n a tio n e x p e rim e n t: I f the c o m p o u n d adds o ne m o la r e q u iv a le n t o f h yd ro g e n ( H 2) on c a ta ly tic h yd ro g e n a tio n a t ro o m tem p eratu re, then it m u s t hav e a d o u b le bond; i f it does n o t add h yd ro g e n , then it m u s t h ave a rin g .] F o r c o m p o u n d s c o n ta in in g o x y g e n , w e s im p ly ig n o r e th e o x y g e n a to m s a n d c a lc u ­ la te th e I H D f r o m th e r e m a in d e r o f th e f o r m u la . C o n s id e r as an e x a m p le a c om po un d w ith the fo rm u la C 4 H 8 O . F o r the purposes o f o u r c a lc u la tio n w e c on sider the c om po un d to b e s im p ly C 4 H 8 and w e ca lcu la te I H D = 1. A g a in , this m eans th a t the c o m p o u n d co n ­ tains e ith e r a rin g o r a d o u b le bon d. S o m e structural p o s s ib ilitie s fo r this c o m p o u n d are show n n ex t. N o tic e th a t the d o u b le b o n d m a y b e p resent as a c a rb o n -o x y g e n d o u b le bond:

O

F o r c o m p o u n d s c o n ta in in g n itr o g e n a to m s w e s u b tra c t o n e h y d ro g e n f o r e a c h n it r o ­ g e n a to m , a n d th e n w e ig n o r e th e n itr o g e n a to m s . F o r e x a m p le , w e tre a t a com p o u n d w ith the fo rm u la C 4 H 9N as tho ug h it w e re C 4 H 8, and again w e g et I H D = 1. S o m e struc­ tu ra l p o s s ib ilitie s are the fo llo w in g :

NH

181

4.18 Applications of Basic Principles

Carbonyl groups also count for a unit of hydrogen deficiency. W hat are the indices o f hydrogen deficiency for the reactant and for the product in the equation show n at the beginning o f Section 4.16 for synthesis o f a perfum e ingredient?

R e v ie w P ro b le m 4 .2 2

4.18 Applications o f Basic Principles In this chapter w e have seen repeated applications o f one basic principle in particular: N atu re Prefers States o f Low er Potential Energy This principle underlies our explanations o f conform ational analysis in Sections 4 .8 -4 .1 3 . The staggered conform ation o f ethane (Section 4.8) is preferred (more populated) in a sample o f ethane because its poten­ tial energy is low est. In the sam e way, the anti conform ation o f butane (Section 4.9) and the chair conform ation o f cyclohexane (Section 4.11) are the preferred conform ations of th ese m o lecu les b ecau se th ese co n fo rm atio n s are o f lo w est p o te n tia l energy. M ethylcyclohexane (Section 4.12) exists m ainly in the chair conform ation w ith its m ethyl group equatorial for the sam e reason. D isubstituted cycloalkanes (Section 4.13) prefer a conform ation w ith both substituents equatorial if this is possible, and, if not, they prefer a conform ation w ith the larger group equatorial. The preferred conform ation in each instance is the one o f low est potential energy. A nother effect that w e encounter in this chapter, and one w e shall see again and again, is how s te ric fa c to r s (spatial factors) can affect the stability and reactivity o f m olecules. U nfavorable spatial interactions betw een groups are central to explaining w hy certain con­ form ations are higher in energy than others. B ut fundam entally this effect is derived itself from another fam iliar principle: lik e c h a rg e s re p e l. Repulsive interactions betw een the elec­ trons of groups that are in close proxim ity cause certain conform ations to have higher poten­ tial energy than others. We call this kind of effect steric hindrance.

One of the reasons we organic chem ists love our discipline is that, besides know ing each m ol­ ecule has a family, we also know that each one has its own architecture, “personality,” and unique name. You have already learned in Chapters 1-3 about m olecular personalities with regard to charge distribution, polarity, and relative acidity or basicity. In this chapter you have now learned how to give unique nam es to sim ple m olecules using the IUPAC system. You also learned m ore about the overall shapes of organic molecules, how their shapes can change through bond rotations, and how we can com pare the relative energies o f those changes using conform ational analysis. You now know that the extent o f flexibility or rigidity in a m olecule has to do w ith the types of bonds present (single, double, triple), and w hether there are rings or bulky groups that inhibit bond rotation. Some organic m olecules are very flexible m em ­ bers of the family, such as the m olecules in our m uscle fibers, w hile others are very rigid, like the carbon lattice of diamond. M ost m olecules, however, have both flexible and rigid aspects to their structures. W ith the know ledge from this chapter, added to other fundam entals you have already learned, you are on your way to developing an understanding of organic chem ­ istry that w e hope will be as strong as diamonds, and that you can flex like a m uscle when you approach a problem . W hen you are finished w ith this chapter’s homework, m aybe you can even take a break by resting your m ind on the chair conform ation of cyclohexane.

Key Terms and Concepts T he key term s and concepts that are highlighted in b o ld , b lu e t e x t w ithin the chapter are defined in the glossary (at the back of the book) and have hyperlinked definitions in the accom panying W ileyP LU S course (w w w .w ileyplus.com ).

PLUS

182

Chapter 4

Nomenclature and Conformations o f Alkanes and Cycloalkanes

Problems Note to Instructors: Many of the homework problems are available for assignment via WileyPLUS, an online teaching and learning solution program.

N O M EN C LA TU R E A N D ISOMERISM 4 .2 3

4 .2 4

W r ite a b o n d -lin e fo rm u la fo r each o f the fo llo w in g com pounds: ( a ) 1 ,4 -D ic h lo ro p e n ta n e

( f) 1 ,1 -D ic h lo ro c y c lo p e n ta n e

(k )

1 ,4 -D ic y c lo p ro p y lh e x a n e

( b ) s e c -B u ty l b ro m id e

(g ) d s -1 ,2 -D im e th y lc y c lo p ro p a n e

( l) N e o p e n ty l alc o h o l

(c ) 4 -Is o p ro p y lh e p ta n e

(h ) ira n s -1 ,2 -D im e th y lc y c lo p ro p a n e

(m ) B ic y c lo [2 .2 .2 ]o c ta n e

( d ) 2 ,2 ,3 -T rim e th y lp e n ta n e

( i) 4 -M e th y l-2 -p e n ta n o l

( n ) B ic y c lo [3 .1 .1 ]h e p ta n e

(e ) 3 -E th y l-2 -m e th y lh e x a n e

( j ) ira n s -4 -Is o b u ty lc y c lo h e x a n o l

(o ) C y c lo p e n ty lc y c lo p e n ta n e

G iv e system atic IU P A C n am es fo r each o f the fo llo w in g

(a)

^

^

(b)

OH

(c)

OH Br

Cl

Br (g)

(f)

4.25 4.26

(h)

T h e n a m e sec-b u ty l a lc o h o l defines a specific structure b u t the n a m e sec-p e n ty l a lc o h o l is am b igu ou s. E x p la in . W r ite the structure a n d g iv e the IU P A C system atic n a m e o f an a lk a n e o r c y c lo a lk a n e w ith the fo rm u la s (a ) C 8 H 1 8 th a t has o n ly p rim a ry h yd ro g e n atom s, (b ) C 6 H 1 2 th a t has o n ly secondary h y d ro g e n atom s, (c ) C 6 H 1 2 th a t has o n ly p rim a ry a n d secondary h yd ro g e n atom s, a n d (d ) C 8 H 1 4 th a t has 12 secondary and 2 te rtia ry h yd ro g e n atom s.

4.27

W r ite the structure(s) o f the sim p lest a lk a n e (s ), i.e ., one(s) w ith the fe w e s t n u m b e r o f carbon atom s, w h e re in each possesses p rim a ry , secondary, tertia ry , a n d q u atern ary carb on atom s. ( A q ua tern a ry carb on is o ne that is b on ded to fo u r o th e r carbon a to m s .) A s s ig n an IU P A C n a m e to each structure.

4.28

Ig n o rin g com pounds w ith d o u b le bonds, w rite structural fo rm u la s a n d g iv e nam es fo r a ll o f the isom ers w ith the fo rm u la C 5 H 10.

4.29

4 .30

W r ite structures fo r the fo llo w in g b ic y c lic alkanes: ( a ) B ic y c lo [1 .1 .0 ]b u ta n e

(c ) 2 -C h lo ro b ic y c lo [3 .2 .0 ]h e p ta n e

( b ) B ic y c lo [2 .1 .0 ]p e n ta n e

( d ) 7 -M e th y lb ic y c lo [2 .2 .1 ]h e p ta n e

U s e the S - A

1 = N m e th o d (H e lp f u l H in t, S ection 4 .1 4 ) to d e te rm in e the n u m b e r o f rin gs in cubane

(S e c tio n 4 .1 4 ).

4.31

A spiro rin g ju n c tio n is one w h e re tw o rin g s th a t share n o bonds o rig in a te fro m a sin g le carbon a to m . A lk a n e s co n ­ ta in in g such a rin g ju n c tio n are c a lle d spiranes. ( a ) F o r the case o f b ic y c lic spiranes o f fo rm u la C 7 H 12, w rite structures fo r a ll p os s ib ilitie s w h e re a ll carbons are in c o rp o ra te d in to rings. ( b ) W r ite structures fo r o th e r b ic y c lic m o le c u le s th a t fit this fo rm u la .

4.32

T e ll w h a t is m e a n t b y an h o m o lo g o u s series and illu s tra te y o u r an s w e r b y w ritin g structures fo r an h om o lo g o u s series o f a lk y l halides.

Problems

183

H YD R O G E N A TIO N 4 .3 3

Four different cycloalkenes w ill all yield m ethylcyclopentane when subjected to catalytic hydrogenation. What are their structures? Show the reactions.

4 .3 4

(a) Three different alkenes yield 2-methylbutane when they are hydrogenated in the presence o f a metal catalyst. Give their structural formulas and write equations for the reactions involved. (b) One o f these alkene isomers has characteristic absorptions at approximately 998 and 914 cm - 1 in its IR spectrum. W hich one is it?

4 .3 5

An alkane with the formula C 6 H1 4 can be prepared by hydrogenation o f either o f two precursor alkenes having the formula C 6 H12. Write the structure o f this alkane, give its IUPAC name, and show the reactions.

C O N F O R M A T IO N S A N D STABILITY 4 .3 6

Rank the follow ing compounds in order o f increasing stability based on relative ring strain.

4 .3 7

Write the structures o f two chair conformations o f 1-tert-butyl-1-methylcyclohexane. Which conformation is more stable? Explain your answer.

4 .3 8

Sketch curves similar to the one given in Fig. 4.8 showing the energy changes that arise from rotation about the C 2 — C3 bond o f (a ) 2,3-dimethylbutane and (b ) 2,2,3,3-tetramethylbutane. You need not concern yourself with actual numerical values of the energy changes, but you should label all maxima and minima with the appropriate conformations.

4 .3 9

Without referring to tables, decide which member o f each o f the follow ing pairs would have the higher boiling point. Explain your answers. (a) Pentane or 2-methylbutane

(c) Propane or 2-chloropropane

(b) Heptane or pentane

(d) Butane or 1-propanol

(e) Butane or CH 3 COCH 3

4 .4 0

One compound w hose molecular formula is C 4 H6 is a bicyclic compound. Another compound with the same for­ mula has an infrared absorption at roughly 2250 cm - 1 (the bicyclic compound does not). Draw structures for each o f these two compounds and explain how the IR absorption allows them to be differentiated.

4 .4 1

Which compound would you expect to be the more stable: ds-1,2-dim ethylcyclopropane or trans-1,2-dimethylcyclopropane? Explain your answer.

4 .4 2

Consider that cyclobutane exhibits a puckered geometry. Judge the relative stabilities o f the 1,2-disubstituted cyclobu­ tanes and of the 1,3-disubstituted cyclobutanes. (You may find it helpful to build handheld molecular models of representative compounds.)

4 .4 3

Write the two chair conformations o f each o f the follow ing and in each part designate which conformation would be the more stable: (a ) ds-1-tert-butyl-3-methylcyclohexane, (b ) trans-1-tert-butyl-3-methylcyclohexane, (c ) trans1-tert-butyl-4-methylcyclohexane, (d ) ds-1-tert-butyl-4-m ethylcyclohexane.

4 .4 4

Provide an explanation for the surprising fact that all-trans-1,2,3,4,5,6-hexaisopropylcyclohexane is a stable m ol­ ecule in which all isopropyl groups are axial. (You may find it helpful to build a handheld molecular model.)

4 .4 5

trans-1,3-Dibromocyclobutane has a measurable dipole moment. Explain how this proves that the cyclobutane ring is not planar. SYNTHESIS

4 .4 6

Specify the m issing compounds and/or reagents in each o f the follow ing syntheses: (a )

trans-5-M ethyl-2-hexene

2

-methylhexane

(b )

(c) Chemical reactions rarely yield products in such initially pure form that no trace can be found o f the starting materials used to make them. What evidence in an IR spectrum o f each o f the crude (unpurified) products from the above reactions would indicate the presence o f one o f the organic reactants used to synthesize each target m olecule? That is, predict one or two key IR absorptions for the reactants that would distinguish it/them from IR absorptions predicted for the product.

184

Chapter 4

Nomenclature and Conformations o f Alkanes and Cycloalkanes

Challenge Problems 4 .4 7

Consider the cis and trans isomers o f 1,3-di-tert-butylcyclohexane (build m olecular m odels). What unusual feature accounts for the fact that one o f these isomers apparently exists in a twist boat conformation rather than a chair conformation?

4 .4 8

U sing the rules found in this chapter, give systematic names for the follow ing or indicate that more rules need to be provided: H-

/C l C

H

Cl

C

(a)

(b) r

/ C\

Br

(c)

(d)

/ C\ Br F

4.49

Open the energy-minimized 3D Molecular Models on the book’s website for trans-1-tert-butyl-3-methylcyclohexane and trans-1,3-di-tert-butylcyclohexane. What conformations o f cyclohexane do the rings in these two compounds resemble m ost closely? How can you account for the difference in ring conformations between them?

4.50

Open the 3D M olecular M odels on the book’s website for cyclopentane and vitamin B12. Compare cyclopentane with the nitrogen-containing five-membered rings in vitamin B12. Is the conformation o f cyclopentane represented in the specified rings o f vitamin B 1 2 ? What factor(s) account for any differences you observe?

4.51

Open the 3D M olecular M odel on the book’s website for buckminsterfullerene. What m olecule has its type o f ring represented 16 times in the surface o f buckminsterfullerene?

Learning Group Problems This is the predominant conformation for D-glucose: H

L W hy is it not surprising that D-glucose is the m ost com m only found sugar in nature? (Hint: Look up structures for sugars such as D-galactose and D-mannose, and compare these with D-glucose.) U sing Newman projections, depict the relative positions o f the substituents on the bridgehead atoms o f cis- and trans-decalin. Which o f these isomers would be expected to be more stable, and why? When 1,2-dimethylcyclohexene (below) is allowed to react with hydrogen in the presence o f a platinum catalyst, the product o f the reaction is a cycloalkane that has a melting point o f - 5 0 ° C and a boiling point o f 130°C (at 760 torr). (a) What is the structure o f the product o f this reaction? (b) Consult an appropriate resource (such as the web or a CRC handbook) and tell which stereoisomer it is. (c) What does this experiment suggest about the mode o f addition o f hydrogen to the double bond? ,C H 3

ch3 1,2-Dimethylcyclohexene When cyclohexene is dissolved in an appropriate solvent and allowed to react with chlorine, the product o f the reac­ tion, C 6 H 1 0 Cl2, has a m elting point o f - 7 ° C and a boiling point (at 16 torr) o f 74°C. (a) W hich stereoisomer is this? (b) What does this experiment suggest about the m ode o f addition o f chlorine to the double bond?

185

CONCEPT MAP Organic nomenclature is guided by The IUPAC system (Sections 4 .3 -4 .6 ) specifies that Each compound shall have an unambiguous name

>

consisting of a

Parent name and one or more

Prefixes

Locants

specify

Suffixes

specify

Substituent or functional group

specify

Numbered positions of groups

Functional groups

CONCEPT MAP Conformers (Section 4.8) are Molecules that differ only by rotation about sigma ( a) bonds can have Ring strain (Section 4.10)

Different potential energies of conformers can be represented by

among conformers is a function of

is caused by

1 ___

and Conformer potential energy diagrams

Torsional strain (Section 4.8)

are a plot o f

is caused by

Dihedral angle vs. potential energy

\f\p j 0

60

Angle strain (Section 4.10)

and loss of

4

Repulsive dispersion forces

Hyperconjugative stabilization (Section 4.8)

result in Steric hindrance (Section 4.8)

120 240 300 360 420 Degrees of Rotation 9

is caused by

4

can be represented

, Deviation from ideal bond angles

involves Favorable overlap of occupied with unoccupied orbitals

of cyclohexane can be represented by

by Newman projection formulas (Section 4.8)

Chair conform ational structures (Section 4.11)

or

Boat

have

can be used to show

Axial positions Eclipsed conformations

and

Staggered conformations

and twist-boat

with substituted groups (G) can be

4

and Equatorial positions

or

Anti

Gauche

X fX G

G

PLUS See Special Topic A in WileyPLUS

conformational structures

Stereochemistry Chiral Molecules

We are all aware o f the fact that certain everyday objects such as gloves and shoes possess the quality o f "hand­ edness." A right-handed glove only fits a right hand; a left-handed shoe only fits a left foot. Objects that can exist in right-handed and left-handed forms are said to be chiral. In this chapter we shall find that molecules can also be chiral and can exist in right- and left-handed forms. For example, one chiral form o f the molecule shown above is a painkiller (Darvon), and the other, a cough suppressant (Novrad)! It is easy to see why it is im portant to understand chirality in molecules.

5.1 Chirality and Stereochem istry Chirality is a phenomenon that pervades the universe. How can w e know whether a par­ ticular object is ch iral or ach iral (not chiral)? •

The glass and its m irror im age are su pe rp ° sab |e.

We can tell if an object has chirality by examining the object and its mirror image.

Every object has a mirror image. Many objects are achiral. B y this w e mean that the object and its m irror image are identical, that is, the object and its mirror im age are superposable one on the other.* Superposable means that one can, in one’s m ind’s eye, place one object on the other so that all parts o f each coincide. Simple geometrical objects such as a sphere or a cube are achiral. So is an object like a water glass. •

A chiral ob ject is one that cannot b e superposed on its m irror im age.

* T o be superposable is d iffe re n t than to be super/m posable. A n y tw o objects can be supe rim p ose d s im p ly by p u ttin g one o b je c t o n to p o f th e other, w h e th e r o r n o t th e objects are th e same. To superpose tw o o bje cts (as in th e p ro p e rty o f s u p e rp o s itio n ) means, on th e o th e r hand, th a t a ll p a r ts o f e ach o b je c t m u s t c o in c id e .

186

T h e c o n d itio n o f s u p e rp o s a b ility m u s t be m e t fo r tw o th in g s to be id e n tic a l.

187

5.1 Chirality and Stereochemistry

Figure 5.1

T h e m ir r o r im a g e o f a r ig h t

Figure 5.2

L e ft a n d r ig h t h a n d s a re n o t

s u p e r p o s a b le .

h a n d is a le f t h a n d .

E a c h o f o u r h a n d s is c h ir a l. W h e n y o u v ie w y o u r r i g h t h a n d i n a m ir r o r , th e im a g e th a t y o u s e e i n t h e m i r r o r is a le ft h a n d ( F i g . 5 . 1 ) . H o w e v e r , a s w e s e e i n F i g . 5 . 2 , y o u r l e f t h a n d a n d y o u r r i g h t h a n d a r e n o t i d e n t i c a l b e c a u s e th e y a re n o t s u p e rp o s a b le . Y o u r h a n d s a r e c h i r a l . I n f a c t , t h e w o r d c h i r a l c o m e s f r o m t h e G r e e k w o r d c h e ir m e a n i n g h a n d . A n o b j e c t s u c h a s a m u g m a y o r m a y n o t b e c h i r a l . I f i t h a s n o m a r k i n g s o n i t , i t is a c h i r a l . I f t h e m u g h a s a l o g o o r i m a g e o n o n e s id e , i t i s c h i r a l .

T h is m u g is c h ir a l.

5.1A The Biological Significance of Chirality T h e h u m a n b o d y is s t r u c t u r a lly c h ir a l, w i t h th e h e a r t l y i n g t o th e l e f t o f c e n t e r a n d th e l iv e r t o t h e r i g h t . H e l i c a l s e a s h e lls a r e c h i r a l a n d m o s t a r e s p i r a l , s u c h a s a r i g h t - h a n d e d s c r e w . M a n y p la n t s s h o w c h i r a l i t y i n t h e w a y t h e y w i n d a r o u n d s u p p o r t i n g s t r u c t u r e s . H o n e y s u c k l e w in d s

as a le f t- h a n d e d h e lix ; b in d w e e d w in d s in

m o le c u le . T h e d o u b le h e lic a l f o r m

a r ig h t- h a n d e d w a y . D N A

is a c h ir a l

o f D N A tu r n s in a r ig h t- h a n d e d w a y .

C h i r a l i t y i n m o le c u le s , h o w e v e r , in v o l v e s m o r e t h a n t h e f a c t t h a t s o m e m o le c u le s a d o p t l e f t o r r i g h t - h a n d e d c o n f o r m a t io n s . A s w e s h a ll s e e i n t h i s c h a p t e r , i t is th e n a t u r e o f g r o u p s b o n d e d a t s p e c i f i c a to m s t h a t c a n b e s t o w c h i r a l i t y o n a m o le c u le . I n d e e d , a l l b u t o n e o f t h e 2 0 a m in o B in d w e e d ( t o p p h o t o ) a c id s t h a t m a k e u p n a t u r a l l y o c c u r r i n g p r o t e in s a r e c h ir a l , a n d a l l o f th e s e a r e c l a s s i f i e d a s b e in g

(Convolvulus sepium ) w in d s in a

le f t - h a n d e d . T h e m o le c u le s o f n a t u r a l s u g a r s a r e a lm o s t a l l c l a s s i f i e d a s b e i n g r i g h t - h a n d e d . I n

r ig h t- h a n d e d fa s h io n , lik e t h e

f a c t , m o s t o f t h e m o le c u le s o f l i f e a re c h ir a l , a n d m o s t a r e f o u n d i n o n l y o n e m i r r o r i m a g e f o r m . *

r ig h t- h a n d e d h e lix o f D N A .

* F o r in te re s tin g re ad in g, see H e g stru m , R. A .; K o n d e p u d i, D . K . Th e H andedness o f the U n ive rse. Sci. Am. 1990, 262(1), 9 8 -1 0 5 , and H o rg a n , J. T he S in is te r C osm os. Sci. Am. 1997, 2 7 6 (5 ), 1 8 -1 9 .

188

Chapter 5

Stereochemistry

Chirality has tremendous importance in our daily lives. M ost pharmaceuticals are chiral. Usually only one mirror-image form o f a drug provides the desired effect. The other mirrorimage form is often inactive or, at best, less active. In som e cases the other mirror-image form o f a drug actually has severe side effects or toxicity (see Section 5.5 regarding thalidomide). Our senses o f taste and smell also depend on chirality. As w e shall see, one mirror-image form o f a chiral m olecule may have a certain odor or taste while its mirror image smells and tastes completely different. The food w e eat is largely made o f molecules o f one mirror-image form. If w e were to eat food that was somehow made o f m olecules with the unnatural mirrorimage form, w e would likely starve because the enzymes in our bodies are chiral and preferentially react with the natural mirror-image form o f their substrates. Let us now consider what causes som e m olecules to be chiral. To begin, w e w ill return to aspects o f isomerism.

5.2 Isomerism: Constitutional Isomers and Stereoisom ers 5.2A Constitutional Isomers Is o m e rs are different compounds that have the same molecular formula. In our study thus far, much o f our attention has been directed toward isomers w e have called constitutional isomers. •

C o n s t itu tio n a l is o m e rs have the same molecular formula but different connectiv­ ity, meaning that their atoms are connected in a different order. Examples o f con­ stitutional isomers are the following:

MOLECULAR FORMULA

CONSTITUTIONAL ISOMERS

C 4 H 10

an d B u tan e

C 3 H 7 Cl

2 -M e th y lp ro p a n e

"Cl

an d Cl

1 -C h lo ro p ro p a n e

-OH

C 4H 10O

2 -C h lo ro p ro p a n e

an d

1 -B u ta n o l

\ ^

O

^

Diethyl e th e r

5.2B Stereoisomers Stereoisomers are not constitutional isomers. •

Stereoisom ers have their atoms connected in the same sequence (the same consti­ tution), but they differ in the arrangement o f their atoms in space. The considera­ tion o f such spatial aspects o f m olecular structure is called stereochem istry.

We have already seen examples o f som e types o f stereoisomers. The cis and trans forms o f alkenes are stereoisomers (Section 1.13B), as are the cis and trans forms o f substituted cyclic m olecules (Section 4.13).

5.2C Enantiomers and Diastereomers Stereoisomers can be subdivided into two general categories: those that are enantiom ers o f each other, and those that are diasterom ers o f each other. •

E n antiom ers are stereoisom ers w hose m olecules are nonsu perp osable m irror im ages o f each other.

5.2 Isomerism: Constitutional Isomers and Stereoisomers

A ll other stereoisomers are diastereomers. •

D iastereom ers are stereoisom ers w hose m olecules are not m irror im ages o f each other.

The alkene isomers cis- and trans--1,2-dichloroethene shown here are stereoisomers that are diastereom ers. C k

Cl

.H

C k

.H

'H

Cl

cis -1 ,2 -D ic h lo ro e th e n e

trans -1 ,2 -D ic h lo ro e th e n e

(C 2 H 2 C l2)

(C 2 H 2 C l 2 )

By examining the structural formulas for cis- and trans-1,2-dichloroethene, w e see that they have the same molecular formula (C 2 H2 CI2 ) and the same connectivity (both compounds have two central carbon atoms joined by a double bond, and both compounds have one chlo­ rine and one hydrogen atom attached to each carbon atom). But, their atoms have a differ­ ent arrangement in space that is not interconvertible from one to another (due to the large barrier to rotation o f the carbon-carbon double bond), making them stereoisomers. Furthermore, they are stereoisomers that are not mirror images o f each other; therefore they are diastereomers and not enantiomers. Cis and trans isomers o f cycloalkanes furnish us with another example o f stereoisomers that are diastereomers. Consider the follow ing two compounds:

H

H

c is -1 ,2 -D im e th y lc y c lo p e n ta n e (C 7 H 1 4 )

H

Me

trans -1 ,2 -D im e th y lc y c lo p e n ta n e (C 7 H 1 4 )

These two compounds have the same molecular formula (C 7 H14), the same sequence of connections for their atoms, but different arrangements o f their atoms in space. In one com ­ pound both methyl groups are bonded to the same face o f the ring, w hile in the other com ­ pound the two methyl groups are bonded to opposite faces o f the ring. Furthermore, the positions o f the methyl groups cannot be interconverted by conformational changes. Therefore, these compounds are stereoisomers, and because they are stereoisomers that are not mirror im ages o f each other, they can be further classified as diastereomers. In Section 5.12 w e shall study other m olecules that can exist as diastereomers but are not cis and trans isomers o f each other. First, however, w e need to consider enantiomers further.

SUBDIVISION OF ISOMERS

189

190

Chapter 5

Stereochemistry

5.3 Enantiomers and Chiral Molecules E n a n t io m e r s a lw a y s h a v e th e p o s s ib ility o f e x is tin g in p a ir s . W e m a y n o t a lw a y s fin d th a t n atu re (o r a re a c tio n ) has p ro d u ce d a p a ir o f enan tio m ers, how ever. In fact, in nature w e o fte n fin d o n ly o ne e n a n tio m e r o f the tw o that are p ossible. W e shall fin d o u t la te r w h y this is o fte n the case. T y p ic a lly , w h e n w e c a rry o u t a c h e m ic a l re a c tio n , w e fin d th a t the re a c tio n produces a p a ir o f enan tio m ers. A g a in , w e w il l e x p la in la te r w h y this occurs. W h a t structural fea tu re m u s t be p resent fo r tw o m o le c u le s to e x ist as enantiom ers? •

E n a n t io m e r s o c c u r o n ly w it h c o m p o u n d s w h o s e m o le c u le s a r e c h ir a l.

H o w do w e re c o g n ize a c h ira l m o le c u le ? •

A c h ir a l m o le c u le is o n e t h a t is n o t s u p e rp o s a b le o n its m i r r o r im a g e .

W h a t is the re la tio n s h ip b e tw e e n a c h ira l m o le c u le and its m ir ro r im a g e ? •

T h e r e la tio n s h ip is o n e t h a t is e n a n tio m e r ic . A c h ira l m o le c u le and its m irro r im a g e are said to be enan tio m ers o f each other.

R eview P roblem 5.1

C la s s ify each o f the fo llo w in g o bjects as to w h e th e r it is c h ira l o r achiral: ( a ) A sc re w d riv e r

( d ) A tennis shoe

( g ) A car

( b ) A b as e b all b at

( e ) A n ear

(h ) A ham m er

( c ) A g o lf club

( f ) A w oo dscrew

T h e c h ira lity o f m o le c u le s can b e d em on strated w ith re la tiv e ly s im p le com pounds. C o nsider, fo r e x a m p le , 2-b u tan ol:

OH 2 -B u ta n o l U n t il n ow , w e h ave p resented the fo rm u la fo r

2

-b u ta n o l as tho ug h it represented o n ly one

c o m p o u n d and w e h ave n o t m e n tio n e d th a t m o le c u le s o f are, there are a c tu a lly tw o d iffe re n t

______ H e lp f u l H i n t Working with models is a helpful study technique whenever three-dimensional aspects of chemistry are involved.

R eview P roblem 5 .2

2

2

-b u ta n o l are chira l. Because they

-bu tan ols and these tw o

2

-bu tan ols are enan tio m ers.

W e can understand this i f w e e x a m in e the d ra w in g s and m o d els in F ig . 5.3. I f m o d e l I is h e ld b e fo re a m irro r, m o d e l I I is seen in the m ir ro r a n d v ic e versa. M o d e ls I and I I are n o t superposable on each o th e r; therefore, they represent d iffe re n t, b ut iso m eric,

B ecau se m odels I a n d I I are nonsu perposable m irror im ages o f each other, the m olecu les th a t they rep resen t are enantiom ers. m o le c u le s .

C o n s tru ct h an d h e ld m o d els o f the 2 -b u ta n o ls represented in F ig . 5 .3 and d em on strate fo r y o u rs e lf th a t they are n o t m u tu a lly superposable. (a ) M a k e s im ila r m o d els o f 2 -b ro m o pro pane. A r e th e y superposable? ( b ) Is a m o le c u le o f 2 -b ro m o p ro p a n e chiral? (c ) W o u ld y o u expect to fin d e n a n tio m e ric fo rm s o f

I

2

-brom op ro pan e?

II

(a) Figure 5.3 (a) Three-dimensional drawings of the 2-butanol enantiomers I and II. (b) Models of the 2-butanol enantiom ers. (c) An unsuccessful attem pt to su perp ose m odels of I and II.

191

5.4 A Single Chirality Center Causes a Molecule to Be Chiral

5.4 A Single Chirality C enter Causes a M olecule to Be Chiral W hat structural feature can w e use to know w hen to expect the possible existence o f a pair o f enantiom ers? •

O ne w ay (but not the only way) is to recognize that a p a i r o f e n a n tio m e rs is a single te tr a h e d r a l a to m w ith fo u r d iffe re n t g ro u p s a tta c h e d to it. a lw a y s p o s s ib le f o r m o le c u le s th a t c o n ta in

Traditionally such atom s have been called a s y m m e tric a to m s , or s te re o g e n ic a to m s , or s te re ­ o c e n te rs . In 1996, however, the IUPAC recom m ended that such atom s be called c h ira lity c e n te rs , and this is the usage that w e shall follow in this text.* It is also im portant to state that chirality is a property o f the m olecule as a w hole, and that a chirality center is a struc­ tural feature that can cause a m olecule to be chiral. Chirality centers are often designated w ith an asterisk (*). In 2-butanol the chirality cen­ ter is C2 (Fig. 5.4). The four different groups that are attached to C2 are a hydroxyl group, a hydrogen atom , a m ethyl group, and an ethyl group. A n ability to find chirality centers in structural form ulas w ill help us in recognizing m ol­ ecules that are chiral, and that can exist as enantiom ers. T h e p re se n c e o f a single c h ira lity c e n te r in a m olecule g u a ra n te e s th a t th e m o lecu le is c h ira l a n d th a t e n a n tio m e ric fo rm s a re possible. H o w e v e r, as w e s h a ll see in S e c tio n 5 .1 2 , th e re a re m o le c u le s w ith m o re th a n o n e c h ir a lit y c e n te r th a t a re n o t c h ir a l, a n d th e re a re m o le c u le s th a t d o n o t c o n ta in a c h ir a l it y c e n te r th a t a re c h ir a l.

Figure 5.5 dem onstrates that enantiom eric com pounds can exist w henever a m olecule contains a single chirality center.

llill III

Mirror (a)

(hydrogen) H (methyl) CH3—C— C h 2 C h 3 (ethyl) OH (hydroxyl) Figure 5.4 The tetrahedral carbon atom of 2-butanol that bears four different groups. [By convention, chirality centers are often designated with an asterisk (*).]

IV

Mirror

A ** £ (rotated)

(b)

(c)

*T h e 1996 IU P A C re com m e n d ed usage can be fo u n d at h ttp ://w w w .c h e m .q m w .a c .u k /iu p a c /s te re o .

Figure 5.5 A demonstration of chirality of a generalized m olecule containing one chirality center. (a) The four different groups around the carbon atom in III and IV are arbitrary. (b) III is rotated and placed in front of a mirror. III and IV are found to be related as an object and its mirror im age. (c) III and IV are not superposable; therefore, the m olecules that they represent are chiral and are enantiomers.

192

Chapter 5

H e lp f u l H i n t ____ j In te rc h a n g in g tw o g ro u p s o f a m o d e l o r th re e -d im e n s io n a l fo rm u la is a u se fu l te s t fo r d e te rm in in g w h e th e r s tru c tu re s o f t w o c h ira l m o le c u le s a re th e sa m e o r d iffe r e n t.

Stereochemistry

An important property o f enantiomers with a single chirality center, such as 2-butanol, is that interchanging any two groups a t the chirality center converts one enantiomer into the other. In Fig. 5.3b it is easy to see that interchanging the methyl and ethyl groups con­ verts one enantiomer into the other. You should now convince yourself that interchanging any other two groups has the same result. •

Any atom at which an interchange o f groups produces a stereoisomer is called a stereogenic center. (If the atom is a carbon atom it is usually called a stereogenic carbon.)

When w e discuss interchanging groups like this, w e must take care to notice that what w e are describing is something we do to a m olecular m odel or something we do on paper. An interchange o f groups in a real m olecule, if it can be done, requires breaking covalent bonds, and this is something that requires a large input o f energy. This means that enan­ tiomers such as the 2 -butanol enantiomers do n o t in te rc o n v e rt spontaneously. The chirality center o f 2-butanol is one example o f a stereogenic center, but there are stereogenic centers that are not chirality centers. The carbon atoms o f cis-1,2-dichloroethene and o f trans-1,2-dichloroethene (Section 5.2c) are stereogenic centers because an inter­ change o f groups at either carbon atom produces the other stereoisomer. The carbon atoms o f cis- 1 ,2 -dichloroethene and trans- 1 , 2 -dichloroethene are not chirality centers, however, because they do not have four different groups attached to them.

R eview P roblem 5 .3

Demonstrate the validity o f what w e have represented in Fig. 5.5 by constructing models. Demonstrate for yourself that I I I and I V are related as an object and its mirror im age and that they are not superposable (i.e., that I I I and I V are chiral m olecules and are enan­ tiomers). (a ) Take I V and exchange the positions o f any two groups. What is the new rela­ tionship between the m olecules? ( b ) N ow take either m odel and exchange the positions of any two groups. What is the relationship between the m olecules now?

If all o f the tetrahedral atoms in a m olecule have two or more groups attached that are the sam e, the m olecule does not have a chirality center. The m olecule is superposable on its mirror im age and is ach iral. An example o f a m olecule o f this type is 2-propanol; carbon atoms 1 and 3 bear three identical hydrogen atoms and the central atom bears two identical methyl groups. If w e write three-dimensional formulas for 2-propanol, w e find (Fig. 5.6) that one structure can be superposed on its mirror image.

HO F ig u re 5 .6

(a) 2 - P r o p a n o l (V ) a n d its m ir r o r

H

H

H3 C ^ C H

CH3

H 3C

3

HO H ho V h H3 C X - C H 3 H3 C CH3

OH

im a g e ( V I) . (b) W h e n e it h e r o n e is r o t a t e d , th e t w o s tr u c tu r e s a re s u p e r p o s a b le a n d s o d o n o t

v

Mirror

r e p r e s e n t e n a n tio m e r s . T h e y r e p r e s e n t t w o m o le c u le s o f t h e s a m e c o m p o u n d . 2 - P r o p a n o l

Superposable

d o e s n o t h a v e a c h ir a lit y c e n te r.

v i

v i

(a)

(b) therefore

Not enantiomers

Thus, w e would not predict the existence o f enantiomeric forms o f 2-propanol, and experimentally only one form o f 2 -propanol has ever been found.

S o lv e d P ro b le m 5.1 D oes 2-bromopentane have a chirality center? If so, write three-dimensional structures for each enantiomer. STRATEGY AND ANSWER First w e write a struc­ tural formula for the m olecule and look for a carbon atom that has four different groups attached to it. In this case, carbon 2 has four different groups: a hydro­ gen, a methyl group, a bromine, and a propyl group. Thus, carbon 2 is a ch irality center.

The chirality c center

Remember: There is a hydrogen here. or

(

c h 3c* h c h 2c h 2c h 3

Br Br 2-B rom opentane

5.4 A Single Chirality Center Causes a Molecule to Be Chiral

193

The enantiomers are CH 3 CH 2C H 2 H

CH3

H 3C

Br B r' V^ Mirror images —^J

C H 2CH 2 CH 3 ”H

Some of the molecules listed here have a chirality center; some do not. Write three-dimensional formulas for both enantiomers of those molecules that do have a chirality center. (a )

2-Fluoropropane

(e )

trans-2-Butene

( i)

2-M ethyl-2-pentene

(b )

2-Methylbutane

(f)

2-Bromopentane

(j)

1-Chloro-2-methylbutane

(c )

2-Chlorobutane

(g )

3-Methylpentane

(d )

2-Methyl-1-butanol

(h )

3-Methylhexane

R eview P roblem 5 .4

5.4A Tetrahedral versus Trigonal Stereogenic Centers It is important to clarify the difference between stereogenic centers, in general, and a chiral­ ity center, which is one type of stereogenic center. The chirality center in 2-butanol is a tetra­ hedral stereogenic center. The carbon atoms o f cis- and trans-1,2-dichloroethene are also stereogenic centers, but they are trigonal stereogenic centers. They are not chirality centers. An interchange of groups at the alkene carbons o f either 1,2-dichloroethene isomer produces a stereoisomer (a molecule with the same connectivity but a different arrangement o f atoms in space), but it does not produce a nonsuperposable mirror image. A chirality center, on the other hand, is one that must have the possibility o f nonsuperposable mirror images. •

Chirality centers are tetrahedral stereogenic centers.

•

Cis and trans alkene isomers contain trigonal stereogenic centers.

THE CHEMISTRY OF . . .

C-

L ife 's M o l e c u l a r H a n d e d n e s s

The amino acids that make up our proteins possess "hand­ edness." They are chiral. Although both mirror image forms are possible, such as those shown below for the amino acid alanine, life on Earth involves amino acids whose chirality is "left-handed" (designated l). The reason that most amino acids are of the left-handed form is not known, however. In the absence of an influence that possesses handedness such as a living system, chemical reactions produce an equal mixture of both mirror-image forms. Since almost all theories about the origin of life presume that amino acids and other molecules central to life were present before self-replicating

Alanine enantiom ers

organisms came into being, it was assumed that they were pre­ sent in equal mirror-image forms in the primordial soup. But could the mirror-image forms of these molecules actu­ ally have been present in unequal amounts before life began, leading to some sort of preference as life evolved? A mete­ orite discovered in 1970, known as the Murchison meteorite, fueled speculation about this topic. Analysis of the meteorite showed that amino acids and other complex molecules asso­ ciated with life were present, proving that molecules required for life could arise outside the confines of Earth. Even more interesting, recent experiments have shown that a 7-9% excess of four L-amino acids is present in the Murchison mete­ orite. The origin of this unequal distribution is uncertain, but some scientists speculate that electromagnetic radiation emit­ ted in a corkscrew fashion from the poles of spinning neutron stars could lead to a bias of one mirror-image isomer over another when molecules form in interstellar space.

194

Chapter 5

Stereochemistry

5.5 M o re ab o u t the Biological Im portance o f Chirality The origin o f biological properties relating to chirality is often likened to the specificity of our hands for their respective gloves; the binding specificity for a chiral m olecule (like a hand) at a chiral receptor site (a glove) is only favorable in one way. If either the m olecule or the biological receptor site had the wrong handedness, the natural physiological response (e.g., neural impulse, reaction catalysis) would not occur. A diagram showing how only one amino acid in a pair o f enantiomers can interact in an optimal way with a hypotheti­ cal binding site (e.g., in an enzym e) is shown in Fig. 5.7. Because o f the chirality center o f the amino acid, three-point binding can occur with proper alignment for only one o f the two enantiomers. R

R CO„

H

-O2C

\

HNH, Figure 5.7 Only one of the two amino acid enantiomers shown (the left-hand one) can achieve three-point binding with the hypothetical binding site (e.g., in an enzyme).

+NH

\

/

Chiral molecules can show their handedness in many ways, including the way they affect human beings. One enantiomeric form o f a compound called limonene (Section 23.3) is pri­ marily responsible for the odor o f oranges and the other enantiomer for the odor o f lemons.

(+)-Lim onene (the enantiom er o f lim onene found in oranges)

(-)-L im o n en e (the enantiom er of lim onene found in lem ons)

One enantiomer o f a compound called carvone (Review Problem 5.14) is the essence of caraway, and the other is the essence o f spearmint. The activity o f drugs containing chirality centers can similarly vary between enantiomers, sometimes with serious or even tragic consequences. For several years before 1963 the drug thalidomide was used to alleviate the symptoms o f morning sickness in pregnant women. In 1963 it was discovered that thalidomide was the cause o f horrible birth defects in many children born subsequent to the use o f the drug.

Thalidom ide (Thalomid®)

5.6 How to Test for Chirality: Planes of Symmetry

195

Even later, evidence began to appear indicating that w hereas one o f the thalidom ide enantiom ers (the right-handed m olecule) has the intended effect o f curing m orning sickness, the other enantiom er, w hich was also present in the drug (in an equal am ount), m ay be the cause o f the birth defects. The evidence regarding the effects o f the tw o enantiom ers is com pli­ cated by the fact that, under physiological conditions, the tw o enantiom ers are intercon­ verted. Now, however, thalidom ide is approved under highly strict regulations for treatm ent o f som e form s o f cancer and a serious com plication associated w ith leprosy. Its potential for use against other conditions including A ID S and rheum atoid arthritis is also under inves­ tigation. We shall consider other aspects o f chiral drugs in Section 5.11. W hich atom is the chirality center of (a ) lim onene and ( b ) o f thalidom ide?

R e v ie w P ro b le m 5 .5

W hich atom s in each o f the follow ing m olecules are chirality centers?

R e v ie w P ro b le m 5 .6

5.6 H o w to Test fo r Chirality: Planes o f Sym m etry T he ultim ate w ay to test for m olecular chirality is to construct m odels o f the m olecule and its m irror im age and then determ ine w hether they are superposable. If the tw o m odels are superposable, the m olecule that they represent is achiral. If the m odels are not superposable, then the m olecules that they represent are chiral. We can apply this test w ith actual m odels, as w e have ju st described, or w e can apply it by draw ing three-dim ensional struc­ tures and attem pting to superpose them in our m inds. T here are other aids, however, that w ill assist us in recognizing chiral m olecules. We have m entioned one already: th e p rese n ce o f a single c h ira lity ce n te r. O ther aids are based on the absence of certain sym m etry elem ents in the m olecule. •

A m olecule w ill not be chiral if it possesses a plane o f symmetry.

•

A plane o f sym m etry (also called a m irror plane) is defined as an im aginary plane that bisects a m olecule in such a w ay that the tw o halves o f the m olecule are m ir­ ror im ages o f each other.

T he plane m ay pass through atom s, betw een atom s, or both. For exam ple, 2-chloropropane has a plane o f sym m etry (Fig. 5.8a), w hereas 2-chlorobutane does not (Fig. 5.8£). •

A ll m olecules w ith a plane o f sym m etry in their m ost sym m etric conform ation are achiral.

196

Chapter 5

Stereochemistry

P la n e o f s y m m e tr y

1 Cl

F ig u re 5 .8

(a) 2 - C h lo r o p r o p a n e

h a s a p la n e o f s y m m e tr y a n d is a c h ir a l. (b) 2 - C h lo r o b u ta n e d o e s n o t p o s s e s s a p la n e o f s y m m e tr y a n d is c h ira l.

A ch ira l

C h iral

(a)

()

S o lv e d P ro b le m 5 .2

G lycerol, CH 2O H C H O H C H 2OH, is an im portant constituent in the biological synthe­ sis o f fats, as w e shall see in C hapter 23. (a) D oes glycerol have a plane o f sym m etry? If so, w rite a three-dim ensional structure for glycerol and indicate w here it is. (b) Is glycerol chiral?

P la n e o f s y m m e tr y

I

(a) Yes, glycerol has a plane sym m etry. N otice w e have to choose the proper conform ation and orientation o f the m olecule to see the plane o f sym m etry. (b) N o, it is achiral because it has a conform ation containing a plane of symm etry. STRATEGY A N D A N S W E R

O HOH2C

! H

CH 2OH

R e v ie w P ro b le m 5 .7

W hich o f the objects listed in Review P roblem 5.1 possess a plane o f sym m etry and are, therefore, achiral?

R e v ie w P ro b le m 5 .8

W rite three-dim ensional form ulas and designate a plane o f sym m etry for all o f the achiral m olecules in Review P roblem 5.4. (In order to be able to designate a plane o f sym m etry you m ay need to w rite the m olecule in an appropriate conform ation.

5.7 N am ing Enantiomers: The R,S-System T he tw o enantiom ers o f 2-butanol are the following:

I

II

If w e nam e these tw o enan tio m ers using o nly the IUPAC system o f n o m en clatu re that w e have learned so far, b oth enantiom ers w ill have the sam e nam e: 2 -butanol (or secbutyl alcohol) (Section 4.3F). T his is u n d esirab le b ecau se each com pound m ust have its own distinct name. M oreover, the n am e th at is given a com p o u n d should allow a chem ist

i3r

F

5.7 Naming Enantiomers: The R,S-System

who is familiar with the rules o f nomenclature to write the structure o f the compound from its name alone. Given the name 2-butanol, a chem ist could write either structure I or structure II. Three chemists, R. S. Cahn (England), C. K. Ingold (England), and V. Prelog (Switzerland), devised a system of nomenclature that, when added to the IUPAC system, solves both o f these problem s. This system , called the R ,S -system or the C ahn-Ingold-Prelog system, is part o f the IUPAC rules. According to this system, one enantiomer o f 2-butanol should be designated (R)-2butanol and the other enantiomer should be designated (S)-2-butanol. [(R) and (S) are from the Latin words rectus and sinister, meaning right and left, respectively.] These m olecules are said to have opposite configurations at C 2 .

5.7A How to Assign (R) and (S) Configurations We assign (R) and (S) configurations on the basis o f the follow ing procedure. 1.

Each o f the four groups attached to the chirality center is assigned a p r i o r i t y or p re fe r e n c e a, b, c, or d. Priority is first assigned on the basis o f the a to m ic n u m b e r o f the atom that is directly attached to the chirality center. The group with the lowest atomic number is given the low est priority, d; the group with next higher atomic number is given the next higher priority, c; and so on. (In the case o f isotopes, the isotope of greatest atomic mass has highest priority.)

We can illustrate the application of the rule with the 2-butanol enantiomer, II: (a)

(d)

HO

.

h 3c

H

X '

(b or c)

..

c h 2c h

3

(b or c) II

Oxygen has the highest atomic number of the four atoms attached to the chirality center and is assigned the highest priority, a. Hydrogen has the low est atomic number and is assigned the low est priority, d. A priority cannot be assigned for the methyl group and the ethyl group by this approach because the atom that is directly attached to the chirality center is a carbon atom in both groups. 2.

When a priority cannot be assigned on the basis o f the atomic number o f the atoms that are directly attached to the chirality center, then the next set o f atoms in the unas­ signed groups is examined. This process is continued until a decision can be made. We assign a priority at the first poin t o f difference.*

When w e examine the methyl group o f enantiomer II, w e find that the next set o f atoms consists of three hydrogen atoms (H, H, H). In the ethyl group o f II the next set o f atoms consists of one carbon atom and two hydrogen atoms (C, H, H). Carbon has a higher atomic number than hydrogen, so w e assign the ethyl group the higher priority, b, and the methyl group the lower priority, c, since (C, H, H) > (H, H, H):

*T h e rules fo r a b ranched ch ain re q u ire th a t w e fo llo w the ch ain w ith the h ig h e st p r io r ity atom s.

197

198

Chapter 5

Stereochemistry

3 . W e n o w ro tate the fo rm u la (o r m o d e l) so th a t the gro up w ith lo w e s t p rio rity (d ) is d ire c te d a w a y fro m us: ( a) OH (a) OH

(d)

(c)

H

(b)

Me

(d )

Et

C H 2CH3

Viewer

(b)

CH3 (c)

New m an p ro je c tio n

II

T h e n w e trace a p ath fro m a to b to c . I f , as w e do this, the d ire c tio n o f o u r fin g er (o r p e n c il) is clockwise, the e n a n tio m e r is d esignated (R ). I f the d ire c tio n is coun­

terclockwise , the e n a n tio m e r is desig nated ( S). O n this basis the

2

-b u ta n o l e n a n tio m e r I I is ( R ) - 2 -butanol:

( a) OH

HO

H

V / C\ CH3 CH2CH3

New m an p ro je c tio n

( c) A rrow s are clockwise. (ff)-2 -B u ta n o l

S o lv e d P ro b le m 5 .3 S h o w n h ere is an e n a n tio m e r o f b ro m o c h lo ro flu o ro io d o m e th a n e . Is it (R ) o r (S)?

Br

I

' Cy '" 'C l F

STRATEGY A N D A N S W E R

Br

This rotation results in

F Im a g in e h o ld in g the m o le c u le b y th e B r an d ro tatin g it as s ho w n s o th a t th e lo w e s t p rio rity g ro u p (F, in this c a s e ) lies in th e p la n e o f th e p ap er.

(b) Br

(d) F

' C.

» Low est p rio rity

(b) Br

"»«I (a) Cl

or

L o o k d ow n th e C — F bond

(c) T h e p ath w e tra c e fro m h ig h e s t to lo w e s t is c o u n te rc lo c k w is e , s o th e e n a n tio m e r is (S ).

5.7 Naming Enantiomers: The R,S-System

199

Write the enantiomeric forms o f bromochlorofluoromethane and assign each enantiomer its correct (R) or (S) designation.

R e v ie w P ro b le m 5 .9

Give (R) and (S) designations for each pair o f enantiomers given as answers to Review Problem 5.4.

R e v ie w P ro b le m 5 .1 0

The first three rules o f the C ahn-Ingold-Prelog system allow us to make an (R) or (S) designation for m ost compounds containing single bonds. For compounds containing multiple bonds one other rule is necessary: 4.

Groups containing double or triple bonds are assigned priorities as if both atoms were duplicated or triplicated, that is,

\

C= Y

I —C—Y

as if it were

and

—C# Y

as if it were

(Y) (C) —C—Y

/ (Y) (C)

(Y) (C)

where the symbols in parentheses are duplicate or triplicate representations o f the atoms at the other end o f the multiple bond. Thus, the vinyl group, — CH= CH2, is o f higher priority than the isopropyl group, CH(CH3)2. That is,

— C H= CH 2

is treated as though it were

H H i i — C— C— H

H H I I — C ------ C — H

which has higher priority than

(C) (C)

H H— C — H H

because at the second set of atoms out, the vinyl group (see the follow ing structure) is C , H, H , whereas the isopropyl group along either branch is H , H, H . (At the first set o f atoms both groups are the same: C , C , H .) H

H

— C— C— H

H >

H

— C ------ C — H

(C) (C )

H H— C — H H

C, H, H V in y l g ro u p

>

H, H, H Is o p ro p y l g ro u p

Other rules exist for more complicated structures, but w e shall not study them here.* List the substituents in each of the following sets in order o f priority, from highest to lowest: (a)

(e)

— C l, — O H , — S H , — H

(b) — C H 3 , — C H 2B r, — C H 2C l, — C H 2 O H

— H , — N (C H

(f) — O H , — O

PO

3

(c )

— H, — O H , — C H O , — C H

(d )

— C H ( C H 3 ) 2 , — C ( C H 3) 3 , — H , C H= CH 2

3) 2 ,

*F u rth e r in fo rm a tio n can be fo u n d in the C h e m ic a l A b stra c ts S ervice Index Guide.

— O C H

3H 2 , —

H,

3,

— C H

—CHO

3

R e v ie w P ro b le m 5 .1 1

200

Chapter 5

R eview P roblem 5 .1 2

Stereochemistry

A s s i g n (R ) o r (S ) d e s i g n a t io n s t o e a c h o f t h e f o l l o w i n g c o m p o u n d s :

D -G ly c e ra ld e h y d e -3 -p h o s p h a te (a g ly c o ly s is in te rm e d ia te )

S o lv e d P ro b le m 5 .4 C o n s id e r th e f o l lo w in g p a i r o f s tr u c tu r e s a n d t e l l w h e t h e r th e y r e p r e s e n t e n a n tio m e r s o r t w o m o le c u le s o f th e s a m e c o m p o u n d i n d if f e r e n t o r ie n ta tio n s :

Cl

Cl CH3

h 3 c A ; B|-

h\ Cl

B

A

S T R A T E G Y O n e w a y t o a p p r o a c h t h is k i n d o f p r o b le m is to ta k e o n e s tr u c tu r e a n d , i n y o u r m in d , h o l d i t b y o n e g r o u p . T h e n r o t a t e t h e o t h e r g r o u p s u n t i l a t l e a s t o n e g r o u p i s i n t h e s a m e p la c e a s i t i s i n t h e o t h e r s t r u c t u r e . ( U n t i l y o u c a n d o t h i s e a s i l y i n y o u r m i n d , p r a c t i c e w i t h m o d e l s . ) B y a s e r ie s o f r o t a t i o n s l i k e t h i s y o u w i l l b e a b l e t o c o n v e r t th e s tr u c tu r e y o u a re m a n ip u la t in g i n t o o n e t h a t is e it h e r i d e n t ic a l w i t h o r th e m i r r o r im a g e o f th e o th e r . F o r e x a m p le , ta k e A , h o ld i t b y th e C l a to m a n d th e n r o ta te th e o th e r g r o u p s a b o u t th e C * — C l b o n d u n t il th e h y d r o ­ g e n o c c u p ie s th e s a m e p o s it io n a s i n B . T h e n h o l d i t b y th e H a n d r o t a t e th e o t h e r g r o u p s a b o u t th e C * — H b o n d . T h is w i l l m a k e B id e n tic a l w it h A :

Cl

orHC 3

Cl

I

rotate

I

% CH

H , < C H .? r

Ide n tic a l w ith B

rotate ^ '" " B r

h ' r c H3 h ' \ B r-^

Cl

A A n o t h e r a p p r o a c h i s t o r e c o g n i z e t h a t e x c h a n g i n g t w o g r o u p s a t t h e c h i r a l i t y c e n t e r in v e rts th e c o n fig u r a tio n o f t h a t c a r b o n a t o m a n d c o n v e r t s a s t r u c t u r e w ith o n ly o n e c h ir a lit y c e n te r i n t o i t s e n a n t i o m e r ; a s e c o n d e x c h a n g e r e ­ c r e a te s th e o r ig i n a l m o le c u le . S o w e p r o c e e d t h is w a y , k e e p in g t r a c k o f h o w m a n y e x c h a n g e s a re r e q u ir e d to c o n ­ v e rt A

i n t o B . I n t h is in s t a n c e w e f i n d th a t t w o e x c h a n g e s a re r e q u ir e d , a n d , a g a in , w e c o n c lu d e t h a t A

a n d B a re

th e s a m e :

Cl h

3c

0 ...""Br H

Cl

CH

exchange

3

exchange

CH3 and H

CH3 and Cl

W

^ " " ""'Br 'B

r

Cl B

A

u s e f u l c h e c k i s t o n a m e e a c h c o m p o u n d i n c l u d i n g i t s ( R ,S ) d e s i g n a t i o n . I f t h e n a m e s a r e t h e s a m e , t h e n t h e

s tr u c tu r e s a re th e s a m e . I n t h is in s t a n c e b o t h s tr u c tu r e s a r e ( R ) - 1 - b r o m o - 1 - c h lo r o e t h a n e . A n o t h e r m e t h o d f o r a s s ig n i n g ( R ) a n d ( S ) c o n f i g u r a t i o n s u s in g o n e ’ s h a n d s a s c h i r a l t e m p la t e s h a s b e e n d e s c r i b e d ( H u h e e y , J . E . J. C h e m . E d u c . 1 9 8 6 , 6 3 , 5 9 8 - 6 0 0 ) . G r o u p s a t a c h i r a l i t y c e n t e r a r e c o r r e l a t e d f r o m l o w e s t t o h i g h ­ e s t p r i o r i t y w i t h o n e ’ s w r is t , th u m b , in d e x fin g e r , a n d s e c o n d f in g e r , r e s p e c t iv e ly . W i t h th e r i n g a n d l i t t l e f in g e r c lo s e d a g a in s t th e p a lm a n d v ie w in g o n e ’ s h a n d w i t h th e w r is t a w a y , i f th e c o r r e la t io n b e t w e e n th e c h ir a l it y c e n ­ te r is w it h th e l e f t h a n d , th e c o n f ig u r a t io n is (S ), a n d i f w it h th e r i g h t h a n d , (R ) .

A N S W E R A a n d B a r e t w o m o le c u le s o f th e s a m e c o m p o u n d o r ie n t e d d if f e r e n t ly .

201

5.8 Properties of Enantiomers: Optical Activity

R eview P roblem 5 .1 3

Tell whether the two structures in each pair represent enantiomers or two molecules of the same compound in different orientations.

(a )

Ck

B rF

and

Ck

> Br

™ (b)

H ^ /C H s

F H . f .C ! an d

C!

H

r

H (c)

JOH an d

CH 3

5.8 Properties o f Enantiomers: O ptical A ctivity The molecules of enantiomers are not superposable and, on this basis alone, we have con­ cluded that enantiomers are different compounds. How are they different? Do enantiomers resemble constitutional isomers and diastereomers in having different melting and boiling points? The answer is no. Pure enantiomers have identical melting and boiling points. Do pure enantiomers have different indexes of refraction, different solubilities in common sol­ vents, different infrared spectra, and different rates of reaction with achiral reagents? The answer to each of these questions is also no. Many of these properties (e.g., boiling points, melting points, and solubilities) are depen­ dent on the magnitude of the intermolecular forces operating between the molecules (Section 2.13), and for molecules that are mirror images of each other these forces will be identical. We can see an example of this if we examine Table 5.1, where boiling points of the 2-butanol enantiomers are listed.

P h y sica l P r o p e r t i e s o f 2 -B u ta n o l a n d T a rta ric A c id E n a n tio m e r s C om pound

B o ilin g P o in t (b p ) o r M e ltin g P o in t (m p )

(R)-2-Butanol (S)-2-Butanol

99.5°C (bp) 99.5°C (bp)

(+)-(R/R)-Tartaric acid (-)-(S,S)-Tartaric acid (+ /-)-T a rtaric acid

168-170°C (mp) 168-170°C (mp) 210-212°C (mp)

Mixtures of the enantiomers of a compound have different properties than pure samples of each, however. The data in Table 5.1 illustrate this for tartaric acid. The natural isomer, (+)-tartaric acid, has a melting point of 168-170°C, as does its unnatural enantiomer, ( —)-tartaric acid. An equal mixture tartaric acid enantiomers, ( + / —)-tartaric acid, has a melting point of 210-212°C, however. Enantiomers show different behavior only when they interact with other chiral sub­ stances, including their own enantiomer, as shown by the melting point data above. Enantiomers show different rates of reaction toward other chiral molecules, that is, toward reagents that consist of a single enantiomer or an excess of a single enantiomer. Enantiomers also show different solubilities in solvents that consist of a single enantiomer or an excess of a single enantiomer. One easily observable way in which enantiomers differ is in their behavior tow ard planepo la rized light. When a beam of plane-polarized light passes through an enantiomer, the plane of polarization rotates. Moreover, separate enantiomers rotate the plane of planepolarized light equal amounts but in opposite directions. Because of their effect on planepolarized light, separate enantiomers are said to be optically active compounds. In order to understand this behavior of enantiomers, we need to understand the nature of plane-polarized light. We also need to understand how an instrument called a polarime­ ter operates.

Tartaric acid is fo u nd naturally in grapes and many o th e r plants. Crystals o f ta rta ric acid can be som etim es be fo u nd w ith wine.

202

F ig u re 5 .9

Chapter 5

Stereochemistry

T h e o s c illa tin g

e le c t r ic a n d m a g n e tic fie ld s o f a b e a m o f o r d in a r y lig h t in o n e p la n e . T h e w a v e s d e p ic t e d h e re o c c u r in a ll p o s s ib le p la n e s in o r d in a r y lig h t.

5.8A Plane-Polarized Light

F ig u re 5 .1 0

O s c illa tio n o f th e

e le c t r ic f ie ld o f o r d in a r y lig h t o c c u rs in a ll p o s s ib le p la n e s p e r p e n d ic u la r t o t h e d ir e c tio n o f p r o p a g a t io n .

F ig u re 5.11

Light is an electromagnetic phenomenon. A beam o f light consists o f two mutually per­ pendicular oscillating fields: an oscillating electric field and an oscillating magnetic field (Fig. 5.9). If w e were to view a beam o f ordinary light from one end, and if w e could actually see the planes in which the electrical oscillations were occurring, w e would find that oscilla­ tions o f the electric field were occurring in all possible planes perpendicular to the direc­ tion o f propagation (Fig. 5.10). (The same would be true o f the m agnetic field.) When ordinary light is passed through a polarizer, the polarizer interacts with the electric field so that the electric field o f the light that emerges from the polarizer (and the magnetic field perpendicular to it) is oscillating only in one plane. Such light is called p la n e -p o la r iz e d lig h t (Fig. 5.11a). If the plane-polarized beam encounters a filter with perpendicular polar­ ization, the light is blocked (Fig. 5.11b). This phenomenon can readily be demonstrated with lenses from a pair o f polarizing sunglasses or a sheet o f polarizing film (Fig. 5.11c).

(a) O r d in a r y lig h t p a s s in g t h r o u g h th e

f ir s t p o la r iz in g f i l t e r e m e r g e s w it h an e le c t r ic w a v e o s c illa t in g in o n ly o n e p la n e (a n d a p e r p e n d ic u la r m a g n e tic w a v e p la n e n o t s h o w n ) . W h e n t h e s e c o n d f i l t e r is a lig n e d w it h its p o la r iz in g d ir e c tio n t h e s a m e as t h e f ir s t f ilt e r , as s h o w n , t h e p la n e - p o la r iz e d lig h t c a n p a s s th r o u g h . (b) If t h e s e c o n d f i l t e r is t u r n e d 9 0 ° , t h e p la n e - p o la r iz e d lig h t is b lo c k e d . (c) T w o p o la r iz in g s u n g la s s le n s e s o r ie n t e d p e r p e n d ic u la r t o e a c h o t h e r b lo c k t h e lig h t b e a m .

(c)

5.8 Properties of Enantiomers: Optical Activity

5.8B The Polarimeter The device that is used for measuring the effect of optically active compounds on plane-polar­ ized light is a polarim eter. A sketch of a polarimeter is shown in Fig. 5.12. The principal working parts of a polarimeter are ( 1 ) a light source (usually a sodium lamp), (2 ) a polarizer, (3) a cell for holding the optically active substance (or solution) in the light beam, (4) an ana­ lyzer, and (5) a scale for measuring the angle (in degrees) that the plane of polarized light has been rotated. The analyzer of a polarimeter (Fig. 5.12) is nothing more than another polarizer. If the cell of the polarimeter is empty or if an optically inactive substance is present, the axes of the planepolarized light and the analyzer will be exactly parallel when the instrument reads 0 °, and the observer will detect the maximum amount of light passing through. If, by contrast, the cell con­ tains an optically active substance, a solution of one enantiomer, for example, the plane of polar­ ization of the light will be rotated as it passes through the cell. In order to detect the maximum brightness of light, the observer will have to rotate the axis of the analyzer in either a clockwise or counterclockwise direction. If the analyzer is rotated in a clockwise direction, the rotation, a (measured in degrees), is said to be positive (+). If the rotation is counterclockwise, the rota­ tion is said to be negative ( —). A substance that rotates plane-polarized light in the clockwise direction is also said to be dextrorotatory, and one that rotates plane-polarized light in a coun­ terclockwise direction is said to be levorotatory (Latin: dexter, right, and laevus, left). + Analyzer (can be rotated)

Observer

0 As the arrows indicate, the optically active substance in solution in the cell is causing the plane of the polarized light to rotate.

Degree scale (fixed) The plane of polarization of the emerging light is not the same as that of the entering polarized light.

Polarimeter sample cell

Polarizer (fixed) Light source

(a)

(b)

IIF IIF Polarizer

Figure 5.12

,

!r V = ==ii

^ ?

-Polarizer and analyzer are parallel. -No optically active substance is present. -Polarized light can get through analyzer.

-Polarizer and analyzer are crossed. -No optically active substance is present. -No polarized light can emerge from analyzer.

K -4 0 °

-Substance in cell between polarizer and analyzer is optically active. -Analyzer has been rotated to the left (from observer’s point of view) to permit rotated polarized light through (substance Analyzer Observer is levorotatory).

T h e p r in c ip a l w o r k in g p a r ts o f a p o la r im e t e r a n d t h e m e a s u r e m e n t o f o p t ic a l

r o t a t io n . (R e p r in te d w it h p e r m is s io n o f J o h n W ile y & S o n s , In c . f r o m H o lu m , J. R., O rganic

Chem istry: A B rie f Course, p . 3 1 6 . C o p y r ig h t 1 9 7 5 .)

203

204

Chapter 5

Stereochemistry

5.8C Specific Rotation T h e n u m b e r o f d e g r e e s t h a t t h e p la n e o f p o l a r i z a t i o n i s r o t a t e d a s t h e l i g h t p a s s e s t h r o u g h a s o lu t io n o f a n e n a n tio m e r d e p e n d s o n th e n u m b e r o f c h ir a l m o le c u le s t h a t i t e n c o u n te r s . T h is , o f c o u r s e , d e p e n d s o n th e le n g t h o f th e tu b e a n d th e c o n c e n t r a t io n o f th e e n a n tio m e r . I n o r d e r t o p la c e m e a s u r e d r o t a t i o n s o n a s t a n d a r d b a s is , c h e m i s t s c a lc u l a t e a q u a n t i t y c a ll e d th e s p e c if ic r o t a t i o n , [ a ] , b y th e f o l lo w in g e q u a tio n :

w h e re [a ] =

th e s p e c ific r o ta tio n

a = th e o b s e rv e d r o ta tio n c = th e c o n c e n tr a tio n o f th e s o lu t io n i n g r a m s p e r m i l l i l i t e r o f s o lu tio n ( o r d e n s ity in g m L

_1

f o r n e a t liq u id s )

l = t h e l e n g t h o f t h e c e l l i n d e c i m e t e r s (1 d m =

10

cm )

T h e s p e c i f i c r o t a t i o n a ls o d e p e n d s o n t h e t e m p e r a t u r e a n d t h e w a v e l e n g t h o f l i g h t t h a t is e m p lo y e d . S p e c if ic r o t a t io n s a re r e p o r t e d s o a s t o in c o r p o r a t e th e s e q u a n t it ie s a s w e l l . A s p e c if ic r o t a t io n m i g h t b e g iv e n a s f o llo w s :

[a]D5 = +3.12 T h is m e a n s th a t th e D lin e o f a s o d iu m la m p ( l =

5 8 9 .6 n m ) w a s u s e d f o r th e lig h t, th a t

a t e m p e r a t u r e o f 2 5 ° C w a s m a in t a i n e d , a n d t h a t a s a m p le c o n t a i n i n g 1 . 0 0 g m L

~ 1o

f th e o p t i­

c a lly a c tiv e s u b s ta n c e , i n a 1 - d m tu b e , p r o d u c e d a r o t a t io n o f 3 .1 2 ° i n a c lo c k w is e d ir e c t io n . * T h e s p e c i f i c r o t a t i o n s o f ( R ) - 2 - b u t a n o l a n d ( S) - 2 - b u t a n o l a r e g i v e n h e r e :

(R )-2 -B u ta n o l [a]D5 = - 1 3 . 5 2

(S )-2 -B u ta n o l [a]2 5 = + 1 3 .5 2

T h e d ir e c t io n o f r o t a t io n o f p la n e - p o la r iz e d l i g h t is o f t e n in c o r p o r a t e d i n t o th e n a m e s o f o p t i c a l l y a c t i v e c o m p o u n d s . T h e f o l l o w i n g t w o s e ts o f e n a n t i o m e r s s h o w h o w t h i s i s d o n e :

(R )-(+ )-2 -M e th y l-1 -b u ta n o l [a]2 5 = + 5 .7 5 6

(R )-(-)-1 -C h lo r o -2 -m e th y lb u ta n e M 2 5 = -1 .6 4

(S )-(-)-2 -M e th y l-1 -b u ta n o l [a]2 5 = - 5 .7 5 6

(S )-(+ )-1 -C h lo ro -2 -m e th y lb u ta n e [a]2 >5 = + 1 .6 4

T h e p r e v i o u s c o m p o u n d s a ls o i l l u s t r a t e a n i m p o r t a n t p r i n c i p l e : •

N o o b v i o u s c o r r e l a t i o n e x is t s b e t w e e n t h e ( R ) a n d ( S ) c o n f i g u r a t i o n s o f e n a n t io m e r s a n d th e d ir e c t io n [ ( + ) o r ( — ) ] in w h ic h th e y r o ta te p la n e - p o la r iz e d lig h t .

( R ) - ( + ) - 2 - M e t h y l - 1 - b u t a n o l a n d ( R ) - ( — ) - 1 - c h l o r o - 2 - m e t h y l b u t a n e h a v e t h e s a m e c o n fig u ­

r a tio n ; t h a t is , t h e y h a v e t h e s a m e g e n e r a l a r r a n g e m e n t o f t h e i r a t o m s i n s p a c e . T h e y h a v e , h o w e v e r , a n o p p o s i t e e f f e c t o n t h e d ir e c t i o n o f r o t a t i o n o f t h e p la n e o f p la n e - p o l a r i z e d l i g h t :

S am e c o n fig u ra tio n (R )-(+ )-2 -M e th y l-1 -b u ta n o l

( R )-(-)-1 -C h lo ro -2 -m e th y lb u ta n e

* T h e m a g n itu d e o f ro ta tio n is dependent on th e s o lv e n t used w h e n s o lu tio n s are m easured. T h is is th e reason the s o lv e n t is sp ecifie d w h e n a ro ta tio n is re p o rte d in th e ch e m ic a l lite ra tu re .

ôr I 5.9 The Origin of Optical Activity

4 Cl

205

These same compounds also illustrate a second important principle: •

N o necessary correlation exists between the (R) and (S) designation and the direc­ tion of rotation o f plane-polarized light.

(R)-2-M ethyl-1-butanol is dextrorotatory ( + ), and (R)-1-chloro-2-m ethylbutane is levorotatory ( —). A method based on the measurement of optical rotation at many different wave­ lengths, called optical rotatory dispersion, has been used to correlate configurations o f chiral m olecules. A discussion of the technique o f optical rotatory dispersion, however, is beyond the scope of this text.

Shown below is the configuration o f ( + )-carvone. (+)-C arvone is the principal component o f caraway seed oil and is responsible for its char­ acteristic odor. ( —)-Carvone, its enantiomer, is the main component of spearmint oil and gives it its characteristic odor. The fact that the carvone enantiomers do not smell the same suggests that the receptor sites in the nose for these compounds are chiral, and that only the correct enantiomer binds w ell to its particular site (just as a hand requires a glove o f the correct chirality for a proper fit). Give the correct (R) and (S) des­ ignations for ( + ) - and ( —)-carvone.

R eview P roblem 5 .1 4 O.

h

r

(+ )-C a rv o n e

5.9 The O rigin o f O ptical A ctivity Optical activity is measured by the degree of rotation o f plane-polarized light passing through a chiral medium. The theoretical explanation for the origin o f optical activity requires con­ sideration of circularly-polarized light, however, and its interaction with chiral m olecules. While it is not possible to provide a full theoretical explanation for the origin o f optical activ­ ity here, the following explanation w ill suffice. A beam o f plane-polarized light (Fig. 5.13a)

Figure 5.13

(a) P la n e - p o la r iz e d

lig h t. (b) C ir c u la r ly - p o la r iz e d lig h t. (c, n e xt page) T w o c ir c u la r ly - p o la r iz e d b e a m s c o u n t e r r o t a t in g a t t h e s a m e v e lo c it y (in p h a s e ) a n d t h e ir v e c t o r s u m . T h e n e t r e s u lt is lik e

(a). (d, n e xt page) T w o c ir c u la r ly p o la r iz e d b e a m s c o u n t e r r o t a t in g a t d if f e r e n t v e lo c itie s , s u c h as a f t e r in t e r a c t io n w it h a c h ira l m o le c u le , a n d t h e ir v e c t o r s u m . T h e n e t r e s u lt is lik e (b). P a rts c a n d d: F ro m A D A M S O N . A T E X T B O O K O F P H Y S IC A L C H E M IS T R Y , 3 E . © 1 9 8 6 B r o o k s / C o le , a p a r t o f C e n g a g e L e a r n in g , In c . R e p r o d u c e d b y p e r m is s io n .

w w w .cengage.com /perm issions

206

Chapter 5

Stereochemistry

(c) Two circularly-polarized beams counter-rotating at the same velocity (in phase), and their vector sum. The net result is like (a) on the previous page. Figure 5.13

(continued)

(d) Two circularly-polarized beams

counter-rotating at different velocities, such as after interaction with a chiral molecule, and their vector sum. The net result is like (b) on the previous page.

can be described in term s o f circularly-polarized light. A beam o f circularly-polarized light rotating in one direction is shown in Fig. 5.13ft. The vector sum o f two counterrotating in-phase circularly-polarized beam s is a beam o f plane-polarized light (Fig. 5.13c). The optical activ­ ity o f chiral m olecules results from the fact that the two counterrotating circularly-polarized beam s travel with different velocities through the chiral medium. A s the difference between the two circularly-polarized beam s propagates through the sample, their vector sum describes a plane that is progressively rotated (Fig. 5.13d). W hat w e m easure w hen light em erges from the sam ple is the net rotation o f the plane-polarized light caused by differences in velocity of the circularly-polarized beam com ponents. The origin o f the differing velocities has ultimately to do w ith interactions betw een electrons in the chiral m olecule and light. M olecules that are n o t chiral cause no difference in velocity o f the tw o circularly-polar­ ized beam s; hence there is no rotation o f the plane o f polarized light described by their vector sum. A chiral m olecules, therefore, are n o t optically active.

5.9A Racemic Forms A sample that consists exclusively or predom inantly o f one enantiom er causes a net rotation o f plane-polarized light. Figure 5.14a depicts a plane o f polarized light as it encounters a m ol­ ecule o f (R)-2-butanol, causing the plane o f polarization to rotate slightly in one direction. (For the rem aining purposes o f our discussion w e shall lim it our description o f polarized light to the resultant plane, neglecting consideration o f the circularly-polarized components from which plane-polarized light arises.) Each additional m olecule o f (R)-2-butanol that the beam encoun­ ters w ould cause further rotation in the same direction. If, on the other hand, the m ixture con­ tained m olecules o f (S)-2-butanol, each m olecule o f that enantiom er w ould cause the plane of polarization to rotate in the opposite direction (Fig. 5.14ft). If the (R) and (S) enantiom ers were present in equal amounts, there w ould be no net rotation o f the plane o f polarized light. •

Figure 5.14 (a) A beam of plane-polarized light encounters a molecule of (R)-2-butanol, a chiral molecule. This encounter produces a slight rotation of the plane of polarization. (b) Exact cancellation of this rotation occurs if a molecule of (S)-2-butanol is encountered. (c) Net rotation of the plane of polarization occurs if (R)-2-butanol is present predominantly or exclusively.

A n equim olar m ixture o f two enantiom ers is called a ra c e m ic m ix tu re (or ra c e m a te or rac em ic fo rm ). A ra c e m ic m ix tu re causes n o n e t ro ta tio n o f p la n e-p o lariz ed light.

ch3

-» H-

/

-C„. OH

VI

C2H5 (R)-2-butanol

(a)

Vi Rotation

X There is net (S)-2-butanol Equal and rotation if (if present) opposite rotation (R)-2-butanol by the enantiomer is present predominantly or exclusively. (b) (c)

207

5.10 The Synthesis of Chiral Molecules

In a racemic mixture the effect of each molecule of one enantiomer on the circularly-polarized beam cancels the effect of molecules of the other enantiomer, resulting in no net optical activity. The racemic form of a sample is often designated as being ( ± ) . A racemic mixture of (R )-(—)- 2 -butanol and (S )-(+ )- 2 -butanol might be indicated as (±)-2-butanol

or

(±)-C H 3 CH 2 CHOHCH 3

5.9B Racemic Forms and Enantiomeric Excess A sample o f an optically active substance that consists o f a single enantiomer is said to be enantiom erically pure or to have an enantiom eric excess o f 100%. An enantiomerically pure sample o f (S)-(+)-2-butanol shows a specific rotation o f + 1 3 .5 2 ([a]D5 — + 1 3.52). On the other hand, a sample o f (S)-(+)-2-butanol that contains less than an equimolar amount of (R )-(—)-2-butanol w ill show a specific rotation that is less than + 1 3 .5 2 but greater than zero. Such a sample is said to have an enantiomeric excess less than 100%. The enantiom eric excess (ee), also known as the optical p u rity , is defined as follows: % Enantiomeric excess —

moles of one enantiomer — moles of other enantiomer X , , ^ ^ • total moles of both enantiomers

100

The enantiomeric excess can be calculated from optical rotations: % Enantiomeric excess* =

observed specific rotation specific rotation of the pure enantiomer

X 100

Let us suppose, for example, that a mixture o f the 2-butanol enantiomers showed a spe­ cific rotation o f + 6 .7 6 . We would then say that the enantiomeric excess o f the (S )-(+ )-2 butanol is 50%: Enantiomeric excess =

+6.76 + 13.52

X 100 = 50%

When w e say that the enantiomeric excess o f this mixture is 50%, w e mean that 50% o f the mixture consists o f the ( + ) enantiomer (the excess) and the other 50% consists of the racemic form. Since for the 50% that is racemic the optical rotations cancel one another out, only the 50% of the mixture that consists o f the ( + ) enantiomer contributes to the observed optical rotation. The observed rotation is, therefore, 50% (or one-half) o f what it would have been if the mixture had consisted only o f the ( + ) enantiomer. 1

S o lv e d P ro b lem 5 .5

"1

What is the actual stereoisomeric com position o f the mixture referred to above? ANSWER Of the total mixture, 50% consists of the racemic form, which contains equal numb ers of the two enantiomers. Therefore, half of this 50%, or 25%, is the ( —) enantiomer and 25% is the ( + ) enantiomer. rEhe other 50% of the mix­ ture (the excess) is also the ( + ) enantiomer. Consequently, the mixture is 75% ( + ) enantiome : and 25% ( —) enantiomer.

A sample o f 2-methyl-1-butanol (see Section 5.8C) has a specific rotation, [a]D5, equal to + 1.151. (a) What is the percent enantiomeric excess o f the sample? (b) W hich enantiomer is in excess, the (R) or the (S)?

R e v ie w Pro b le m 5 1 5

5.10 The Synthesis o f Chiral M olecules 5.10A Racemic Forms Reactions carried out with achiral reactants can often lead to chiral products. In the absence of any chiral influence from a catalyst, reagent, or solvent, the outcome of such a reaction is a racemic mixture. In other words, the chiral product is obtained as a 50:50 mixture of enantiomers. * T h is c a lc u la tio n sh ou ld be a p p lie d to a sin g le e n a n tio m e r o r to m ix tu re s o f ena n tio m ers o nly. I t sh ou ld n o t be a p p lie d to m ix tu re s in w h ic h som e o th e r co m p o u n d is present.

208

Chapter 5

Stereochemistry

An example is the synthesis of 2-butanol by the nickel-catalyzed hydrogenation of butanone. In this reaction the hydrogen molecule adds across the carbon-oxygen double bond in much the same way that it adds to a carbon-carbon double bond. C H 3CH2C C H 3 +

H— H

Ni

------->

O B u ta n o n e (ac h ira l m o le c u le s )

* (± )-C H 3 C H 2C HC H 3 OH

H y d ro g e n (ac h ira l m o le c u le s )

(± )-2 -B u ta n o l [ch iral m o le c u le s b ut 5 0 :5 0 m ix tu re (R ) a n d (S )]

Figure 5.15 illustrates the stereochemical aspects of this reaction. Because butanone is achi­ ral, there is no difference in presentation of either face of the molecule to the surface of the metal catalyst. The two faces of the trigonal planar carbonyl group interact with the metal surface with equal probability. Transfer of the hydrogen atoms from the metal to the car­ bonyl group produces a chirality center at carbon 2. Since there has been no chiral influ­ ence in the reaction pathway, the product is obtained as a racemic mixture of the two enantiomers, ( ^ ) - ( - ) - 2 -butanol and (S )-(+ )- 2 -butanol.

Figure 5.15 The reaction of butanone with hydrogen in the presence of a nickel catalyst. The reaction rate by path (a) is equal to that by path (b). (R-15)-(—)-2-Butanol and (S)-(+)2-butanol are produced in equal amounts, as a racemate.

We shall see that when reactions like this are carried out in the presence of a chiral influ­ ence, such as an enzyme or chiral catalyst, the result is usually not a racemic mixture.

5.10B Stereoselective Syntheses S te re o s ele c tive re a c tio n s are reactions that lead to a preferential formation of one stereoiso­ mer over other stereoisomers that could possibly be formed.

•

If a reaction produces preferentially one enantiomer over its mirror image, the reaction is said to be enantioselective.

•

If a reaction leads preferentially to one diastereomer over others that are possible, the reaction is said to be diastereoselective.

For a reaction to be either enantioselective or diastereoselective, a chiral reagent, catalyst, or solvent must assert an influence on the course of the reaction. In nature, where most reactions are stereoselective, the chiral influences come from protein molecules called enzymes. Enzymes are biological catalysts of extraordinary efficiency. Not only do they have the ability to cause reactions to take place much more rapidly than they would oth­ erwise, they also have the ability to assert a dramatic chiral influence on a reaction. Enzymes do this because they, too, are chiral, and they possess an active site where the reactant molecules are momentarily bound while the reaction takes place. The active site is chiral (See Fig. 5.7), and only one enantiomer of a chiral reactant fits it properly and is able to undergo the reaction.

209

5.11 Chiral Drugs

Many enzymes have found use in the organic chemistry laboratory, where organic chemists take advantage of their properties to bring about stereoselective reactions. One of these is an enzyme called lipase. Lipase catalyzes a reaction called hydrolysis, whereby an ester (Section 2.10B) reacts with a molecule of water to produce a carboxylic acid and an alcohol. O

O hydrolysis

HOH R

HO — R'

E ster

OH

R

O R '

W a te r

A lc o h o l

C a rb o x y lic acid

If the starting ester is chiral and present as a mixture of its enantiomers, the lipase enzyme reacts selectively with one enantiomer to release the corresponding chiral carboxylic acid and an alcohol, while the other ester enantiomer remains unchanged or reacts much more slowly. The result is a mixture that consists predominantly of one stereoisomer of the reactant and one stereoisomer of the product, which can usually be separated easily on the basis of their different physical properties. Such a process is called a kinetic resolution, where the rate of a reaction with one enantiomer is different than with the other, leading to a preponderance of one product stereoisomer. We shall say more about the resolution of enantiomers in Section 5.16. The following hydrolysis is an example of a kinetic resolution using lipase: O

O

O

lipase O Et

+

"O E t

H— OH

F

OH

F

Ethyl (± )-2 -flu o ro h e x a n o a te [an e s te r th a t is a ra c e m ate o f (R ) and (S) form s]

H — O Et

F

Ethyl (R )-(+ )-2 -flu o ro h e x a n o a te (> 9 9 % e n a n tio m e ric e x c es s )

(S )-(-)-2 -F lu o ro h e x a n o ic acid (> 6 9 % e n a n tio m e ric e x c es s )

Other enzymes called hydrogenases have been used to effect enantioselective versions of carbonyl reductions like that in Section 5.10A. We shall have more to say about the stereo­ selectivity of enzymes in Chapter 12.

5.11 Chiral Drugs The U.S. Food and Drug Administration and the pharmaceutical industry are very inter­ ested in the production of chiral drugs, that is, drugs that contain a single enantiomer rather than a racemate. The antihypertensive drug methyldopa (Aldomet), for example, owes its effect exclusively to the (S) isomer. In the case of penicillam ine, the (S) isomer is a highly potent therapeutic agent for primary chronic arthritis, while the (R ) isomer has no thera­ peutic action and is highly toxic. The anti-inflammatory agent ibuprofen (Advil, Motrin, Nuprin) is marketed as a racemate even though only the (S) enantiomer is the active agent. The (R) isomer of ibuprofen has no anti-inflammatory action and is slowly converted to the (S) isomer in the body. A formulation of ibuprofen based on solely the (S) isomer, how­ ever, would be more effective than the racemate. A t the beginning of this chapter we showed the formulas for two enantiomeric drugs, Darvon and Novrad. Darvon (also called dextropropoxyphene) is a painkiller. Novrad (levopropoxyphene) is a cough suppressant.

___ CH3

H C H 3— c — C H

c h

3

CH3

CH3

2

c h

—

c

—

O

^

^ C

H

2—

c — C n h

O

h

Ibu profen

0

;H

c h

3—

c — c h — c o 2h

o h

2

s h

n h

2

h o

M eth yld o p a

P e n ic illa m in e

210

Chapter 5

Review Problem 5.16 R eview P roblem 5 .1 7

Stereochemistry

W rite three-dimensional formulas for the (S) isomers of ( a ) methyldopa, ( b ) penicillamine, and (c ) ibuprofen. The antihistamine Allegra (fexofenadine) has the following structural formula. For any chirality centers in fexofenadine, draw a substructure that would have an (R ) configuration.

F e x o fe n a d in e (A lleg ra )

R eview P roblem 5 .1 8

e

Assign the (R,S) configuration at each chirality center in Darvon (dextropropoxyphene).

There are many other examples of drugs like these, including drugs where the enantiomers have distinctly different effects. The preparation of enantiomerically pure drugs, therefore, is one factor that makes stereoselective synthesis (Section 5.10B) and the resolution of racemic drugs (separation into pure enantiomers, Section 5.16) major areas of research today. Underscoring the importance of stereoselective synthesis is the fact that the 2001 Nobel Prize in Chemistry was given to researchers who developed reaction catalysts that are now widely used in industry and academia. W illiam Knowles (Monsanto Company, retired) and Ryoji Noyori (Nagoya University) were awarded half of the prize for their development of reagents used for catalytic stereoselective hydrogenation reactions. The other half of the prize was awarded to Barry Sharpless (Scripps Research Institute) for development of catalytic stere­ oselective oxidation reactions (see Chapter 11). An important example resulting from the work of Noyori and based on earlier work by Knowles is a synthesis of the anti-inflammatory agent n a p ro x e n , involving a stereoselective catalytic hydrogenation reaction: CH

H3 C COOH + H2

(S)-BI NAP-Ru(OCOCH3)2 (0.5 mol %) MeOH

H3CO

H COOH

*

H3CO ( S )-N a p ro x en (an a n ti-in fla m m a to ry a g e n t) (9 2 % y ie ld , 9 7 % ee)

The hydrogenation catalyst in this reaction is an organometallic complex formed from ruthe­ nium and a chiral organic ligand called (S)-BINAP. The reaction itself is truly remarkable because it proceeds with excellent enantiomeric excess (97%) and in very high yield (92%). We w ill have more to say about BINAP ligands and the origin of their chirality in Section 5.18.

211

5.12 Molecules with More than One Chirality Center

THE CHEMISTRY OF . . . Selective Binding of Drug Enantiomers to Left- and Right-Handed Coiled DNA Would you like left- or right-handed DNA with your drug? That's a question that can now be answered due to the recent discovery that each enantiomer of the drug daunorubicin selectively binds DNA coiled with opposite handed­ ness. (+)-Daunorubicin binds selectively to DNA coiled in the typical right-handed conformation (B-DNA). (—)Daunorubicin binds selectively to DNA coiled in the left­ handed conformation (Z-DNA). Furthermore, daunorubicin is capable of inducing conformational changes in DNA from one coiling direction to the other, depending on which coil­ ing form is favored when a given daunorubicin enantiomer binds to the DNA. It has long been known that DNA adopts a number of secondary and tertiary structures, and it is pre­ sumed that some of these conformations are involved in turning on or off transcription of a given section of DNA. The discovery of specific interactions between each daunorubicin enantiomer and the left- and right-handed coil forms of DNA will likely assist in design and discovery of new drugs with anticancer or other activities.

E n a n tio m e r ic fo r m s o f d a u n o r u b ic in b in d w it h D N A a n d c a u s e it t o c o il w it h o p p o s it e h a n d e d n e s s . [G r a p h ic c o u r t e s y J o h n O . T r e n t, B r o w n C a n c e r C e n te r , D e p a r tm e n t o f M e d ic in e , U n iv e r s ity o f L o u is v ille , KY. B a s e d o n w o r k f r o m Q u , X . T r e n t, J .O ., F o k t, I., P r ie b e , W ., a n d C h a ire s , J .B ., A lloste ric, Chiral-Selective D rug

Building to D N A , Proc. Natl. Acad. Sci. U.S.A., 2 0 0 0 ( O c t. 2 4 ): 9 7 (2 2 ), 1 2 0 3 2 - 1 2 0 3 7 .]

5.12 M olecules w ith M o re than O ne Chirality C enter So far we have mainly considered chiral molecules that contain only one chirality center. Many organic molecules, especially those important in biology, contain more than one chirality center. Cholesterol (Section 23.4B), for example, contains eight chirality centers. (Can you locate them?) We can begin, however, with simpler molecules. Let us consider 2,3-dibromopentane, shown here in a two-dimensional bond-line formula. 2,3Dibromopentane has two chirality centers: Br

Br

HO'

2 ,3 -D ib ro m o p e n ta n e

C h o le s te ro l

A useful rule gives the maximum number of stereoisomers: •

In compounds whose stereoisomerism is due to chirality centers, the total num ber o f stereoisomers will not exceed 2n, where n is equal to the num ber o f chirality centers.

For 2,3-dibromopentane we should not expect more than four stereoisomers (2 2 = 4). Our next task is to write three-dimensional bond-line formulas for the possible stereoiso­ mers. When doing so it is helpful to follow certain conventions. First, it is generally best to write as many carbon atoms in the plane of the paper as possible. Second, when needing to compare the stereochemistry at adjacent carbon atoms, we usually draw the molecule in a fashion that shows eclipsing interactions, even though this would not be the most stable con­ formation of the molecule. We do so because, as we shall see later, eclipsed conformations make it easy for us to recognize planes of symmetry when they are present. (We do not mean to imply, however, that eclipsed conformations are the most stable ones— they most certainly are not. It is important to remember that free rotation is possible about single bonds, and that molecules are constantly changing conformations.) Third, if we need to draw the enantiomer

H e lp f u l H i n t C h o le s te ro l, h a v in g e ig h t c h ira lity c e n te rs , h y p o th e tic a lly c o u ld e x is t in 2 8 (256) s te re o is o m e ric fo rm s , y e t b io s y n th e s is via e n zym e s p ro d u c e s o n ly one s te re o is o m e r.

H e lp f u l H i n t U s e fu l c o n v e n tio n s w h e n w r itin g th re e -d im e n s io n a l fo rm u la s

212

Chapter 5

Stereochemistry

of the first stereoisomer, we can easily do so by drawing a mirror image of the first formula, using as our guide an imaginary mirror perpendicular to the page and between the molecules. The following are two three-dimensional bond-line formulas for 2,3-dibromopentane. Notice that in drawing these formulas we have followed the conventions above. H

H

H H

Br

Br

Br

Br

1

2 M irro r

Since structures 1 and 2 are not superposable, they represent different compounds. Since structures 1 and 2 differ only in the arrangement of their atoms in space, they represent stereoisomers. Structures 1 and 2 are also mirror images of each other; thus 1 and 2 rep­ resent a pair of enantiomers. Structures 1 and 2 are not the only ones possible for 2,3-dibromopentane, however. If we interchange the bromine and hydrogen at C2 (invert the configuration), we find that we have 3, which has a different structural formula than either 1 or 2. Furthermore, we can write a formula for a structure (4) that is a nonsuperposable mirror image of 3, and which is also different from 1 and 2 .

H Br^ !

Br

Br H

a

4 M irro r

Structures 3 and 4 correspond to another pair of enantiomers. Structures 1-4 are all differ­ ent, so there are, in total, four stereoisomers of 2,3-dibromopentane. Essentially what we have done above is to write all the possible structures that result by successively interchanging two groups at all chirality centers. At this point you should convince yourself that there are no other stereoisomers by writing other structural formulas. You w ill find that rotation about the single bonds (or of the entire structure) of any other arrangement of the atoms w ill cause the struc­ ture to become superposable with one of the structures that we have written here. Better yet, using different colored balls, make molecular models as you work this out. The compounds represented by structures 1 -4 are all optically active compounds. Any one of them, if placed separately in a polarimeter, would show optical activity. The compounds represented by structures 1 and 2 are enantiomers. The compounds rep­ resented by structures 3 and 4 are also enantiomers. But what is the isomeric relation between the compounds represented by 1 and 3? We can answer this question by observing that 1 and 3 are stereoisomers and that they are not mirror images of each other . They are, therefore, diastereomers . •

Diastereomers have different physical properties— different melting points and boiling points, different solubilities, and so forth. H

H

Br _

y

Br

\ 1

Review Problem 5.19

H Br

Br r ~

2

H

H Br^ =

v _

Br =^H

Br H H

Br / ^ C

3

_

4

(a) If 3 and 4 are enantiomers, what are 1 and 4? (b) What are 2 and 3, and 2 and 4? (c) Would you expect 1 and 3 to have the same melting point? (d) The same boiling point? (e) The same vapor pressure?

213

5.12 Molecules with More than One Chirality Center

5.12A Meso Compounds A structure with two chirality centers does not always have four possible stereoisomers. Sometimes there are only three. As we shall see: •

Some molecules are achiral even though they contain chirality centers.

To understand this, let us write stereochemical formulas for 2,3-dibromobutane. We begin in the same way as we did before. We write formulas for one stereoisomer and for its mirror image:

H

Br

^

B r

H

H

"

K

B r

Structures A and B are nonsuperposable and represent a pair of enantiomers. When we write the new structure C (see below) and its mirror image D, however, the situation is different. The two structures are superposable. This means that C and D do not represent a pair of enantiomers. Formulas C and D represent identical orientations of the same compound: H e lp f u l H i n t H

H

B r

H B r^ f

Br

H f ^ Br

If a molecule has an internal plane of symmetry it is achiral.

D

The molecule represented by structure C (or D ) is not chiral even though it contains two chirality centers. •

A m e s o c o m p o u n d is an achiral molecule that contains chirality centers. Meso compounds are not optically active.

The ultimate test for molecular chirality is to construct a model (or write the structure) of the molecule and then test whether or not the model (or structure) is superposable on its mirror image. If it is, the molecule is achiral: If it is not, the molecule is chiral. We have already carried out this test with structure C and found that it is achiral. We can also demonstrate that C is achiral in another way. Figure 5.16 shows that structure C has an internal plane o f sym m etry (Section 5.6). The following two problems relate to compounds A - D in the preceding paragraphs.

A pure sample of A

(b )

A pure sample of B

(c )

A pure sample of

(d )

An equimolar mixture of A and B

Br

H ^ Br

(b )

H

Br

\ Figure 5.16 The plane of symmetry of meso-2,3dibromobutane. This plane divides the molecule into halves that are mirror images of each other.

C

The following are formulas for three compounds, written in noneclipsed conformations. In each instance tell which compound (A, B, or C above) each formula represents. (a)

H

R eview P roblem 5 .2 0

Which of the following would be optically active? (a )

Br

B r.

...y ^ H Br

Br

(c)

H

Br H

Review Problem 5.21

214

Chapter 5

Stereochemistry

S o lv e d P ro b le m 5 .6 Which of the following is a meso compound? STRATEGY AND ANSWER In each molecule, rotating the groups joined by the C 2— C3 bond by 180° brings the two methyl groups into comparable position. In the case of compound Z , a plane of symmetry results, and there­ fore, Z is a meso compound. No plane of symmetry is possible in X and Y. H3 C .

H f*O H

OH H 3 ^ f^H

OH H 3 ^ £*H HO'"7 H

hT

HO ^ C H ,3

H HO

CH3

CH3

Rot at e t he gr oups about t he C 2 — C3 bond by 180° as shown.

HO H ^ ./C H 3

H H O ^.

H HO

,C H 3

CH 3 Pl ane of s y mme t r y

H HO

H""7 /''C H 3 HO

HO”"7 ' \ c h H CHa

CH3

A meso compound

R eview P roblem 5 .2 2

W rite three-dimensional formulas for all of the stereoisomers of each of the following compounds. Label pairs of enantiomers and label meso compounds. F

Cl (a)

„CI

(b) OH

(c) C l'

OH

F

Cl F

^

OH

5.12B How to Name Compounds with More than One

Chirality Center I f a compound has more than one chirality center, we analyze each center separately and decide whether it is ( R) or ( S). Then, using numbers, we tell which designation refers to which carbon atom. Consider stereoisomer A of 2,3-dibromobutane: H Br B rN lU H l

A

A

,

A

2,3-Dibromobutane

5.13 Fischer Projection Formulas

215

W h e n th is f o r m u la is r o ta te d s o th a t th e g r o u p o f lo w e s t p r i o r i t y a tta c h e d to C 2 is d ir e c te d a w a y f r o m th e v ie w e r , i t r e s e m b le s th e f o l lo w in g :

T h e o r d e r o f p r o g r e s s io n f r o m th e g r o u p o f h ig h e s t p r i o r i t y to th a t o f n e x t h ig h e s t p r i o r i t y (fro m

— B r, to — C H B r C H

3,

to — C H

3)

is c lo c k w is e . S o C 2 h a s th e ( R ) c o n f ig u r a t io n .

W h e n w e r e p e a t t h i s p r o c e d u r e w i t h C 3 , w e f i n d t h a t C 3 a ls o h a s t h e ( R ) c o n f i g u r a t i o n :

C o m p o u n d A , t h e r e fo r e , is ( 2 Æ ,3 Æ ) - 2 ,3 -d ib r o m o b u ta n e .

G i v e n a m e s t h a t i n c l u d e ( R ) a n d ( S ) d e s i g n a t io n s f o r c o m p o u n d s B a n d C i n S e c t i o n 5 . 1 2 A .

R e v ie w P r o b le m 5 . 2 3

G i v e n a m e s t h a t i n c l u d e ( R ) a n d ( S ) d e s i g n a t io n s f o r y o u r a n s w e r s t o R e v i e w P r o b l e m 5 . 2 2 .

R e v ie w P r o b le m 5 . 2 4

C h lo r a m p h e n ic o l ( a t r i g h t ) is a p o t e n t a n t ib io t ic , is o la te d f r o m

R e v ie w P r o b le m 5 . 2 5

Streptom yces venezuelae, ty p h o id

fe v e r. I t

w as

th e

n o th a t is p a r t ic u l a r ly e f f e c t iv e a g a in s t f ir s t n a tu r a lly

o c c u r r in g

2

s u b s ta n c e

s h o w n to c o n t a in a n it r o ( — N O 2) g r o u p a tta c h e d to a n a r o m a tic r in g . B o th c h ir a lit y h a v e th e

(R)

c e n te rs in

c h lo r a m p h e n ic o l a re k n o w n to

c o n f ig u r a t io n . I d e n t i f y th e t w o c h ir a l it y c e n te r s a n d H O — C — H

w r it e a t h r e e - d im e n s io n a l f o r m u la f o r c h lo r a m p h e n ic o l. H — C — N H C O C H C l2 c h

2o

h

Chloramphenicol

5.13 Fischer Projection Formulas S o f a r i n w r i t i n g s t r u c t u r e s f o r c h i r a l m o le c u le s w e h a v e o n l y u s e d f o r m u l a s t h a t s h o w t h r e e d im e n s io n s w i t h s o lid a n d d a s h e d w e d g e s , a n d w e s h a ll la r g e ly c o n t in u e to d o s o u n t i l w e s tu d y c a rb o h y d ra te s in

C h a p te r 2 2 . T h e r e a s o n is th a t f o r m u la s

w it h

s o lid

and dashed

w e d g e s u n a m b i g u o u s l y s h o w t h r e e d im e n s io n s , a n d t h e y c a n b e m a n i p u l a t e d o n p a p e r i n a n y w a y t h a t w e w i s h s o l o n g a s w e d o n o t b r e a k b o n d s . T h e i r u s e , m o r e o v e r , te a c h e s u s to se e m o le c u le s ( i n o u r m i n d ’ s e y e ) i n th r e e d im e n s io n s , a n d t h is a b i l it y w i l l s e rv e u s w e ll.

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Chemists, however, sometimes use formulas called Fischer projections to show three dimensions in chiral molecules such as acyclic carbohydrates. Fischer projection formulas are useful in cases where there are chirality centers at several adjacent carbon atoms, as is often the case in carbohydrates. Use of Fischer projection formulas requires rigid adher­ ence to certain conventions, however. Used carelessly, these projection form ulas can easily lead to incorrect conclusions.

5.13A How to Draw and Use Fischer Projections Let us consider how we would relate a three-dimensional formula for 2,3-dibromobutane using solid and dashed wedges to the corresponding Fischer projection formula. First, it is neces­ sary to note that in Fischer projections the carbon chain is always drawn from top to bottom, rather than side to side as is often the case with bond-line formulas. We consider the mole­ cule in a conformation that has eclipsing interactions between the groups at each carbon. For 2,3-dibromobutane we turn the bond-line formula so that the carbon chain runs up and down and we orient it so that groups attached to the main carbon chain project out of the plane like a bow tie. The carbon-carbon bonds of the chain, therefore, lie either in the plane of the paper or project behind it. Then to draw the Fischer projection we simply “project” all of the bonds onto the paper, replacing all solid and dashed wedges with ordinary lines. Having done this, the vertical line of the formula now represents the carbon chain, each point of intersection between the vertical line and a horizontal line represents a carbon atom in the chain, and we understand the horizontal lines to be bonds that project out toward us. n 13

CH 3 Br

H

Br

Bn H

'C -

H

H

Br

H

Br

H

Br CH

Ch A

A

A F isch er p ro je c tio n fo rm u la

To test the superposability of two structures represented by Fischer projections we are allowed to rotate them in the plane of the paper by 180°, but by no other angle. We must always keep the Fischer projection formulas in the plane of the paper, and we are not allowed to flip them over. If we flip a Fischer projection over, the horizontal bonds pro­ ject behind the plane instead of in front, and every configuration would be misrepresented as the opposite of what was intended. n 13

H

H e lp f u l H i n t

3

Br

3

H

rotate

Br

H

H

Br

Br

180° in plane

H

Br

Br

H

CH 3

CH 3

CH

A

A

B

*

(F lip p in g th e p ro jection fo rm u la o v e r s id e w a y s c re a te s th e p ro jection fo rm u la fo r the enantiom er o f A .)

Sam e structure

3

N o t the same

Build handheld models of A and B and relate them to the Fischer projections shown here.

3

H

Br

Br

H CH

N o t the sam e (F lip p in g th e p ro je c tio n fo rm u la o v e r e n d fo r end c re a te s th e p ro je c tio n fo rm u la fo r th e enantiom er o f A .)

217

5.14 Stereoisomerism of Cyclic Compounds

Because Fischer projections must be used with such care, we introduce them now only so that you can understand Fischer projections when you see them in the context of other courses. Our emphasis for most of this book w ill be on the use of solid and dashed wedges to represent three-dimensional formulas (or chair conformational structures in the case of cyclohexane derivatives), except in Chapter 22 when we w ill use Fischer projections again in our discussion of carbohydrates. If your instructor wishes to utilize Fischer projections further, you w ill be so advised.

(a) Give the (R,S) designations for each chirality center in compound A and for compound B. (b) W rite the Fischer projection formula for a compound C that is the diastereomer of A and B. (c) Would C be optically active?

R eview P roblem 5.2 6

5.14 Stereoisom erism o f Cyclic Compounds Cyclopentane derivatives offer a convenient starting point for a discussion of the stereoiso­ merism of cyclic compounds. For example, 1,2-dimethylcyclopentane has two chirality centers and exists in three stereoisomeric forms 5, 6 , and 7:

Me

H

H

Me

E n a n tio m e rs

Me

Me

M es o co m p o u n d 7

The trans compound exists as a pair of enantiomers 5 and 6 . cis-1,2-Dimethylcyclopentane (7) is a meso compound. It has a plane of symmetry that is perpendicular to the plane of the ring:

P lan e o f s y m m e try 7

(a) Is trans-1,2-dimethylcyclopentane (5) superposable on its mirror image (i.e., on compound 6 )? (b) Is cis-1,2-dimethylcyclopentane (7) superposable on its mirror image? (c) Is cis-1,2-dimethylcyclopentane a chiral molecule? (d) Would cis- 1,2-dimethylcyclopentane show optical activity? (e) What is the stereoisomeric relationship between 5 and 7? (f) Between 6 and 7?

R eview P roblem 5 .2 7

W rite structural formulas for all of the stereoisomers of 1,3-dimethylcyclopentane. Label pairs of enantiomers and meso compounds if they exist.

R eview P roblem 5 .28

5.14A Cyclohexane Derivatives 1.4-D im eth ylcyclo hexan es If we examine a formula of 1,4-dimethylcyclohexane, we find that it does not contain any chirality centers. However, it does have two stereogenic centers. As we learned in Section 4.13, 1,4-dimethylcyclohexane can exist as cis-trans isomers. The cis and trans forms (Fig. 5.17) are diastereomers . Neither compound is chi­ ral and, therefore, neither is optically active. Notice that both the cis and trans forms of 1.4-dimethylcyclohexane have a plane of symmetry.

H e lp f u l H i n t Build handheld molecular models of the 1,4-, 1,3-, and 1,2-dimethylcyclohexane isomers discussed here and examine their stereochemical properties. Experiment with flipping the chairs and also switching between cis and trans isomers.

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Stereochemistry

Plane of symmetry

Me

Me

Me

Me

or

or Me

H H

Figure 5.17 The cis and trans fo rm s o f 1,4dim ethylcyclohexane are diastereom ers o f each other. B oth com pounds are achiral, as th e internal plane o f sym m etry (blue) shows fo r each.

Me

Me

Me H

H

trans-1,4-

cis-1,4Dimethylcyclohexane

Dimethylcyclohexane

1,3-D im eth ylcyclo hexan es 1,3-Dimethylcyclohexane has two chirality centers; we can, therefore, expect as many as four stereoisomers (2 2 — 4). In reality there are only three. cis-1,3-Dimethylcyclohexane has a plane of symmetry (Fig. 5.18) and is achiral.

Figure 5.18 cis-1,3-Dim ethylcyclohexane has a plane o f sym m etry, shown in blue, and is th e re fo re achiral.

irans-1,3-Dimethylcyclohexane does not have a plane of symmetry and exists as a pair of enantiomers (Fig. 5.19). You may want to make models of the irans-1,3-dimethylcyclohexane enantiomers. Having done so, convince yourself that they cannot be superposed as they stand and that they cannot be superposed after one enantiomer has undergone a ring flip. Me

Me

Me

Me

Figure 5.19

trans-1,3-D im ethylcyclohexane does not have a plane o f sym m etry and exists as a pair o f enantiom ers. The tw o structures (a and b) shown here are not superposable as th e y stand, and flip p in g th e ring o f either structu re does n o t make it superposable on th e other. (c) A sim p lifie d representation o f (b).

hM

Me

Me (no plane of sym m etry)

( a)

( c)

(b)

1,2-D im eth ylcyclo hexan es 1,2-Dimethylcyclohexane also has two chirality centers, and again we might expect as many as four stereoisomers. Indeed there are four, but we find that we can isolate only three stereoisomers. irans-1,2-Dimethylcyclohexane (Fig. 5.20) exists as a pair of enantiomers. Its molecules do not have a plane of symmetry. Figure 5.20 trans-1,2-D im ethylcyclohexane has no plane o f sym m etry and exists as a pair o f enantiom ers (a and b). [N o tice th a t w e have w ritte n th e m ost stable conform ations fo r (a) and (b). A ring flip o f e ith e r (a) or (b) w o u ld cause bo th m ethyl groups to becom e axial.]

( a)

(b)

5.15 Relating Configurations through Reactions

(c)

219

Figure 5.21 c/s-1,2Dimethylcyclohexane exists as tw o rapidly interconverting chair conformations (c) and (d).

(d)

cis-1,2-Dimethylcyclohexane, shown in Fig. 5.21, presents a somewhat more complex situation. If we consider the two conformational structures (c) and (d), we find that these two mirror-image structures are not identical. Neither has a plane of symmetry and each is a chiral molecule, but they are interconvertible by a ring flip . Therefore, although the two structures represent enantiomers, they cannot be separated because they rapidly intercon­ vert even at low temperature. They simply represent different conformations o f the same compound. Therefore, structures (c) and (d) are not configurational stereoisomers; they are co n fo rm a tio n a l ste re o iso m ers (see Section 4.9A). This means that at normal temperatures there are only three isolable stereoisomers of 1 ,2 -dimethylcyclohexane. As we shall see later, there are some compounds whose conformational stereoisomers can be isolated in enantiomeric forms. Isomers of this type are called atropisomers (Section 5.18). W rite formulas for all of the isomers of each of the following. Designate pairs of enantiomers and achiral compounds where they exist. (a) 1-Bromo-2-chlorocyclohexane

R eview P roblem 5 .2 9

(b) 1-Bromo-3-chlorocyclohexane

(c) 1-Bromo-4-chlorocyclohexane Give the ( R,S) designation for each compound given as an answer to Review Problem 5.29.

R eview P roblem 5 .3 0

5.15 Relating Configurations through Reactions in Which N o Bonds to the Chirality C enter A re Broken • I f a reaction takes place in a way so that no bonds to the chirality center are broken, the product w ill of necessity have the same general configuration of groups around the chirality center as the reactant. Such a reaction is said to proceed w ith retention of configuration. Consider as an exam­ ple the reaction that takes place when (S )-(—)-2-methyl-1-butanol is heated with concen­ trated hydrochloric acid: Same configuration

OH + H — Cl (S )-(-)-2 -M e th y l-1 -b u ta n o l H D 5 = -5 .7 5 6

heat

Cl + H — OH (S )-(+ )-1 -C h lo ro -2 -m e th y lb u ta n e M 5 = + 1 .6 4

We do not need to know now exactly how this reaction takes place to see that the reaction must involve breaking the CH2— OH bond of the alcohol because the — OH group is replaced by a — Cl. There is no reason to assume that any other bonds are broken. (We shall study how this reaction takes place in Section 11.8A.) Since no bonds to the chirality center are broken, the reaction must take place with retention of configuration, and the product of the reaction must have the same configuration o f groups around the chirality center that the reac­ tant had. By saying that the two compounds have the same configuration, we simply mean that comparable or identical groups in the two compounds occupy the same relative positions in space around the chirality center. (In this instance the — CH2OH group and the — CH2Cl are comparable, and they occupy the same relative position in both compounds; all the other groups are identical and they occupy the same positions.)

220

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N otice that in this example w hile the (R,S) designation does not change [both reactant and product are (S)], the direction o f optical rotation does change [the reactant is ( —) and the product is ( + )]. Neither occurrence is a necessity when a reaction proceeds with reten­ tion o f configuration. In the next section w e shall see examples o f reactions in which con­ figurations are retained and where the direction o f optical rotation does not change. The following reaction is one that proceeds with retention o f configuration but involves a change in the (R,S) designation: H

>O H

Zn, H+ (—ZnB r2)

H

retention of configuration

B r

(ff)-1 -B ro m o -2 -b u ta n o l

>0 H , H

(S )-2-B utano l

In this example the (R,S) designation changes because the — CH2Br group o f the reactant changes to a — CH 3 group in the product ( — CH2Br has a higher priority than — CH 2 CH3, and — CH 3 has a lower priority than — CH 2 CH3).

Solved Problem 5.7 When (R)-1-bromo-2-butanol reacts with KI in acetone the product is 1-iodo-2-butanol. Would the product be (R) or (S)? STRATEGY AND ANSWER N o bonds to the chirality center would be broken, so w e can reason that the product would be the following.

(R )-1 -B ro m o -2 -b u ta n o l

(R )-1 -Io d o -2 -b u ta n o l

The configuration o f the product would still be (R) because replacing the bromine at C1 with an iodine atom does not change the relative priority o f C 1.

5.15A Relative and Absolute Configurations Reactions in which no bonds to the chirality center are broken are useful in relating con­ figurations o f chiral m olecules. That is, they allow us to demonstrate that certain compounds have the same relative configuration. In each o f the examples that w e have just cited, the products o f the reactions have the same relative configurations as the reactants. •

Chirality centers in different m olecules have the same r e la t iv e c o n fig u r a tio n if they share three groups in common and if these groups w it h the central carbon can be superposed in a pyramidal arrangement. Y C B I

T h e c h ira lity ce n ters in I a nd I I have the sa m e relative co n fig ura tio n . T h e ir co m m o n g ro u p s and ce n tra l ca rb o n can be su p e rp o se d .

II

Before 1951 only relative configurations o f chiral m olecules were known. N o one prior to that time had been able to demonstrate with certainty what the actual spatial arrange­ ment o f groups was in any chiral m olecule. To say this another way, no one had been able to determine the absolute configuration o f an optically active compound. •

The a b s o lu te c o n fig u r a tio n o f a chirality center is its (R) or (S) designation, which can only be specified by know ledge o f the actual arrangement o f groups in space at the chirality center.

221

5.15 Relating Configurations through Reactions

Prior to any known absolute configurations, the configurations of chiral molecules were related to each other through reactions o f known stereochemistry . Attempts were also made to relate all configurations to a single compound that had been chosen arbitrarily to be the standard. This standard compound was glyceraldehyde: OH HO

JO

, H

G ly c e ra ld e h y d e

Glyceraldehyde has one chirality center; therefore, glyceraldehyde exists as a pair of enantiomers:

H

OH

and

O

HO

HO

H O

HO

H

H

(R )-G ly c e ra ld e h y d e

(S )-G ly c e ra ld e h y d e

In one system for designating configurations, (R)-glyceraldehyde is called D-glyceraldehyde and (S)-glyceraldehyde is called L-glyceraldehyde. This system of nomenclature is used with a specialized meaning in the nomenclature of car­ bohydrates. (See Section 22.2B.) One glyceraldehyde enantiomer is dextrorotatory (+ ) and the other, of course, is levorotatory ( —). Before 1951 no one was sure, however, which configuration belonged to which enantiomer. Chemists decided arbitrarily to assign the (R) configuration to the ( + )enantiomer. Then, configurations of other molecules were related to one glyceraldehyde enantiomer or the other through reactions of known stereochemistry. For example, the configuration of ( —)-lactic acid can be related to (+)-glyceraldehyde through the following sequence of reactions in which no bond to the chirality center is broken: H

(+ )-G ly c e ra ld e h y d e

HgO

hno2

(oxidation)

h 2o

T h is bond is b ro ken .

OH

T h is b ond is b ro ken .

(-) -G ly c e r ic acid

H N o te th a t no b on ds d ire c tly to th e c h ira lity c e n te r are b ro ken .

Br

h 2n

^O

hno2

O

C

HBr

OH ( + )-'s o s e rin e

H

OH C

OH

OH O

C

Zn, H3O +

H3 C

T h is bond -----is bro ken .

OH

(-)-3 -B r o m o -2 -h y d r o x y p ro p a n o ic acid

OH ( - ) - L a c t ic acid

The stereochemistry of all of these reactions is known. Because none of the bonds to the chirality center (shown in red) has been broken during the sequence, its original con­ figuration is retained. If the assumption is made that (+)-glyceraldehyde is the (R) stereoiso­ mer, and therefore has the following configuration, H ho

/ OH a

jo

H (R )-(+ )-G ly c e ra ld e h y d e

222

Chapter 5

Stereochemistry

then (-)-lactic acid is also an (R) stereoisomer and its configuration is H H 3 C""

( R )- ( -) -L a c tic acid

R eview P roblem 5.31

W rite bond-line three-dimensional formulas for the starting compound, the product, and all of the intermediates in a synthesis similar to the one just given that relates the configu­ ration of ( —)-glyceraldehyde with (+ )-lactic acid. Label each compound with its proper (R) or (S) and ( + ) or ( —) designation.

The configuration of ( —)-glyceraldehyde was also related through reactions of known stereochemistry to (+ )-tartaric acid: H O H H ^o h HO2C

CO2 H

(+ )-T a rta ric acid

In 1951 J. M . Bijvoet, the director of the van’t H off Laboratory of the University of Utrecht in the Netherlands, using a special technique of X-ray diffraction, was able to show conclusively that (+)-tartaric acid had the absolute configuration shown above. This meant that the original arbitrary assignment of the configurations of (+ )- and ( —)-glyceraldehyde was also correct. It also meant that the configurations of all of the compounds that had been related to one glyceraldehyde enantiomer or the other were now known with certainty and were now absolute configurations.

R eview P roblem 5 .3 2

C rtlw o rl

D m

Fischer projection formulas are often used to depict compounds such as glyceraldehyde, lactic acid, and tartaric acid. Draw Fischer projections for both enantiomers of (a) glycer­ aldehyde, (b) tartaric acid, and (c) lactic acid, and specify the (R) or (S) configuration at each chirality center. [Note that in Fischer projection formulas the terminal carbon that is most highly oxidized is placed at the top of the formula (an aldehyde or carboxylic acid group in the specific examples here).]

223

5.16 Separation o f Enantiomers: Resolution

5.16 Separation o f Enantiomers: Resolution So far we have left unanswered an important question about optically active compounds and racemic forms: How are enantiomers separated? Enantiomers have identical solubili­ ties in ordinary solvents, and they have identical boiling points. Consequently, the con­ ventional methods for separating organic compounds, such as crystallization and distillation, fail when applied to a racemic form.

5.16A Pasteur's Method for Separating Enantiomers It was, in fact, Louis Pasteur’s separation of a racemic form of a salt of tartaric acid in 1848 that led to the discovery of the phenomenon called enantiomerism. Pasteur, consequently, is often considered to be the founder of the field of stereochemistry. (+)-Tartaric acid is one of the by-products of wine making (nature usually only syn­ thesizes one enantiomer of a chiral molecule). Pasteur had obtained a sample of racemic tartaric acid from the owner of a chemical plant. In the course of his investigation Pasteur began examining the crystal structure of the sodium ammonium salt of racemic tartaric acid. He noticed that two types of crystals were present. One was identical with crystals of the sodium ammonium salt of (+)-tartaric acid that had been discovered earlier and had been shown to be dextrorotatory. Crystals of the other type were nonsuperposable mirror images of the first kind. The two types of crystals were actually chiral. Using tweezers and a mag­ nifying glass, Pasteur separated the two kinds of crystals, dissolved them in water, and placed the solutions in a polarimeter. The solution of crystals of the first type was dextro­ rotatory, and the crystals themselves proved to be identical with the sodium ammonium salt of (+)-tartaric acid that was already known. The solution of crystals of the second type was levorotatory; it rotated plane-polarized light in the opposite direction and by an equal amount. The crystals of the second type were of the sodium ammonium salt of ( —)-tartaric acid. The chirality of the crystals themselves disappeared, of course, as the crystals dis­ solved into their solutions, but the optical activity remained. Pasteur reasoned, therefore, that the molecules themselves must be chiral. Pasteur’s discovery of enantiomerism and his demonstration that the optical activity of the two forms of tartaric acid was a property of the molecules themselves led, in 1874, to the proposal of the tetrahedral structure of carbon by van’t H off and Le Bel. Unfortunately, few organic compounds give chiral crystals as do the (+ )- and ( —)-tartaric acid salts. Few organic compounds crystallize into separate crystals (containing separate enan­ tiomers) that are visibly chiral like the crystals of the sodium ammonium salt of tartaric acid. Pasteur’s method, therefore, is not generally applicable to the separation of enantiomers.

5.16B Current Methods for Resolution of Enantiomers One of the most useful procedures for separating enantiomers is based on the following: •

When a racemic mixture reacts with a single enantiomer of another compound, a mix­ ture of diastereomers results, and diastereomers, because they have different melting points, boiling points, and solubilities, can be separated by conventional means.

Diastereomeric recrystallization is one such process. We shall see how this is done in Section 20.3F. Another method is re s o lu tio n by enzymes, whereby an enzyme selectively converts one enantiomer in a racemic mixture to another compound, after which the unreacted enantiomer and the new compound are separated. The reaction by lipase in Section 5.10B is an example of this type of resolution. Chromatography using chiral media is also widely used to resolve enantiomers. This approach is applied in high-performance liquid chromatogra­ phy (HPLC) as well as in other forms of chromatography. Diastereomeric interactions between molecules of the racemic mixture and the chiral chromatography medium cause enantiomers of the racemate to move through the chromatography apparatus at different rates. The enantiomers are then collected separately as they elute from the chromatogra­ phy device. (See “The Chemistry o f . . . HPLC Resolution of Enantiomers,” Section 20.3.)

Tartaric acid crystals

2 24

Chapter 5

Stereochemistry

5.17 Com pounds w ith Chirality Centers O th e r than Carbon Any tetrahedral atom with four different groups attached to it is a chirality center. Shown here are general formulas of compounds whose molecules contain chirality centers other than carbon. Silicon and germanium are in the same group of the periodic table as carbon. They form tetrahedral compounds as carbon does. When four different groups are situated around the central atom in silicon, germanium, and nitrogen compounds, the molecules are chiral and the enantiomers can, in principle, be separated. Sulfoxides, like certain exam­ ples of other functional groups where one of the four groups is a nonbonding electron pair, are also chiral. This is not the case for amines, however (Section 20.2B): R4

aR3

S i' R^

^ R3

R^

G e' R2

R^

^ R3 V

R2

Ri

R2

. R1

Xr2

( v

0

5.18 Chiral M olecules That Do N o t Possess a Chirality C enter A molecule is chiral if it is not superposable on its mirror image. The presence of a tetrahe­ dral atom with four different groups is only one type of chirality center, however. While most of the chiral molecules we shall encounter have chirality centers, there are other structural attrib­ utes that can confer chirality on a molecule. For example, there are compounds that have such large rotational barriers between conformers that individual conformational isomers can be sep­ arated and purified, and some of these conformational isomers are stereoisomers. Conformational isomers that are stable, isolable compounds are called atropisomers. The existence of chiral atropisomers has been exploited to great effect in the development of chiral catalysts for stereoselective reactions. An example is BINAP, shown below in its enantiomeric forms:

P (P h )2

(P h )2P

P (P h )2

(P h )2P

(S > B IN A P

(R )-B IN A P

The origin of chirality in BINAP is the restricted rotation about the bond between the two nearly perpendicular naphthalene rings. This torsional barrier leads to two resolvable enan­ tiomeric conformers, (S)- and (R)-BINAP. When each enantiomer is used as a ligand for metals such as ruthenium and rhodium (bound by unshared electron pairs on the phosphorus atoms), chiral organometallic complexes result that are capable of catalyzing stereoselec­ tive hydrogenation and other important industrial reactions. The significance of chiral lig ­ ands is highlighted by the industrial synthesis each year of approximately 3500 tons of (-)-m en th o l using an isomerization reaction involving a rhodium (S)-BINAP catalyst. Allenes are compounds that also exhibit stereoisomerism. Allenes are molecules that contain the following double-bond sequence: \= C = C ,1 / The planes of the ^ bonds of allenes are perpendicular to each other:

225

Problems

H

H H CH «"'C= C= C

«H *Cl

Cl

Cl Mirror

Figure 5.22 Enantiomeric forms of 1,3-dichloroallene. These two m olecules are nonsuperposable mirror im ages of each other and are therefore chiral. They do not possess a tetrahedral atom with four different groups, however.

This geometry of the p bonds causes the groups attached to the end carbon atoms to lie in perpendicular planes, and, because of this, allenes with different substituents on the end carbon atoms are chiral (Fig. 5.22). (Allenes do not show cis-trans isomerism.)

l /n This Chapter) In this chapter you learned that the handedness of life begins at the molecular level. Molecular recognition, signaling, and chemical reactions in living systems often hinge on the handedness of chiral molecules. Molecules that bear four different groups at a tetrahe­ dral carbon atom are chiral if they are nonsuperposable with their mirror image. The atoms bearing four different groups are called chirality centers. M irror planes of symmetry have been very important to our discussion. If we want to draw the enantiomer of a molecule, one way to do so is to draw the molecule as if it were reflected in a mirror. If a mirror plane of symmetry exists w ithin a molecule, then it is achiral (not chiral), even if it contains chirality centers. Using mirror planes to test for sym­ metry is an important technique. In this chapter you learned how to give unique names to chiral molecules using the Cahn-Ingold-Prelog R,S-system. You have also exercised your mind’s eye in visualizing mol­ ecular structures in three dimensions, and you have refined your skill at drawing three-dimen­ sional molecular formulas. You learned that pairs of enantiomers have identical physical properties except for the equal and opposite rotation of plane-polarized light, whereas diastereomers have different physical properties from one another. Interactions between each enantiomer of a chiral molecule and any other chiral material lead to diastereomeric interactions, which lead to different physical properties that can allow the separation of enantiomers. Chemistry happens in three dimensions. Now, with the information from this chapter building on fundamentals you have learned about molecular shape and polarity in earlier chapters, you are ready to embark on your study of the reactions of organic molecules. Practice drawing molecules that show three dimensions at chirality centers, practice nam­ ing molecules, and label their regions of partial positive and negative charge. Paying atten­ tion to these things w ill help you learn about the reactivity of molecules in succeeding chapters. Most important of all, do your homework!

Key Terms and Concepts The key terms and concepts that are highlighted in b o ld , b lu e t e x t within the chapter are defined in the glossary (at the back of the book) and have hyperlinked definitions in the accompanying WileyPLUS course (www.wileyplus.com).

0

PLUS

Problems Note to Instructors: Many of the homework problems are available for assignment via WileyPLUS, an online teaching and learning solution. CHIRALITY A N D STEREO ISO M ERISM 5.33

Which of the following are chiral and, therefore, capable of existing as enantiomers? 1,3-Dichlorobutane

(d )

3-Ethylpentane

(g )

2-Chlorobicyclo[2.1.1]hexane

(b )

1,2-Dibromopropane

(e )

2-Bromobicyclo[1.1.0]butane

(h )

5-Chlorobicyclo[2.1.1]hexane

(c )

1,5-Dichloropentane

( f)

2-Fluorobicyclo[2.2.2]octane

(a )

226

Chapter 5

Stereochemistry

5.34

(a) How many carbon atoms does an alkane (not a cycloalkane) need before it is capable o f existing in enantiomeric forms? (b) Give correct names for two sets o f enantiomers with this minimum number o f carbon atoms.

5.35

Albuterol, shown here, is a com m only prescribed asthma medication. For either enantiomer o f albuterol, draw a three-dimensional formula using dashes and wedges for bonds that are not in the plane o f the paper. Choose a perspective that allows as many carbon atoms as possible to be in the plane o f the paper, and show all unshared electron pairs and hydrogen atoms (except those on the m ethyl groups labeled M e). Specify the (R,S) configuration o f the enantiomer you drew.

i

OH Me HO Me HO

Albuterol

5.36

(a) Write the structure o f 2,2-dichlorobicyclo[2.2.1]heptane. (b) How many chirality centers does it contain? (c) How many stereoisomers are predicted by the 2n rule? (d) Only one pair o f enantiomers is possible for 2,2-dichlorobicyclo[2.2.1]heptane. Explain.

5.37

Shown below are Newman projection formulas for (R,R)-, (S,S)-, and (R,S)-2,3-dichlorobutane. (a) Which is which? (b) Which formula is a m eso compound? c h

3

c h

H

Cl

Cl

H CH

3

c h Cl

H

Cl

Cl

H

H

3

A

3

-C l

CH

CH

B

C

3

5.38

Write appropriate structural formulas for (a) a cyclic m olecule that is a constitutional isomer o f cyclohexane, (b) m ol­ ecules with the formula C 6 H1 2 that contain one ring and that are enantiomers o f each other, (c) m olecules with the formula C 6 H 1 2 that contain one ring and that are diastereomers o f each other, (d) m olecules with the formula C 6 H1 2 that contain no ring and that are enantiomers o f each other, and (e) m olecules with the formula C 6 H 1 2 that contain no ring and that are diastereomers o f each other.

5.39

Consider the follow ing pairs o f structures. D esignate each chirality center as (R) or (S) and identify the relation­ ship between them by describing them as representing enantiomers, diastereomers, constitutional isomers, or two m olecules o f the same compound. U se handheld molecular m odels to check your answers. F

Br

H

H

H

and

(a )

Br

F

CH3

(g )

and

„C ? » C H 3

H 3C

3

Br

H3 C

C /

Br

H

and

(b) F^*B r

Br

F

(h)

“ *

Cl

(i)

Cl

Cl^

f~^2-ci Cl

and

£

^

and

(j)

C lp

Cl H 3C F « "" y H

CH3 •H Br

H

Br

CH3

n 13

Cl

Cl

H

H

-C l

and

(k)

H

and

(f)

7

Cl

n 13î

Cl H 3C

Br H

Cl

Cl

Cl

and

T

CH

CH

227

Problems

Cl

CO

JX ‘3

-Cl

Br

-H

Cl

-H

and

(o)

and

(i) -Br CH

Cl

Cl

Cl

CH

Cl

and

(m )

Cl H

Cl

Br

Cl

Ciy

Cl

\

Br C =C =C

/

Br H

and

/

p C

H C^ B r

H

D iscuss the anticipated stereochemistry o f each o f the follow ing compounds. (a )

5.41

(q)

and

(n )

C\ and

(p )

5.40

Cl

CICH= C = C = CHCl

(b )

CH 2 = C = C = CHCl

(c )

CICH= C = C = CCl2

Tell whether the compounds o f each pair are enantiomers, diastereomers, constitutional isomers, or not isomeric. (a)

CHO H-

CHO

-O H

(b) H

HO

CHO

CHO

H-

-O H

H

-O H

H

CH 2 OH

CHO H

CHO

-OH

HO

-O H

H

-OH CH 2 OH

CH 2 OH

CHO

CH 2 OH

(d) H

H

HO

CH 2 OH

-O H

and H

-OH

and

CH2OH

(c)

H

H

HO

and

HO

H

HO

H

and HO-

H

H

CH2OH

-O H CHO

CH 2 OH

5.42

A compound D with the molecular formula CgH 1 2 is optically inactive but can be resolved into enantiomers. On catalytic hydrogenation, D is converted to E (CgH-^) and E is optically inactive. Propose structures for D and E.

5.43

Compound F has the molecular formula C 5 H8 and is optically active. On catalytic hydrogenation F yields G (C 5 H12) and G is optically inactive. Propose structures for F and G.

5.44

Compound H is optically active and has the molecular formula CgH 1 o. On catalytic hydrogenation H is converted to I (C 6 H12) and I is optically inactive. Propose structures for H and I.

5.45

Aspartame is an artificial sweetener. Give the (R,S) designation for each chirality center o f aspartame.

O A s p a rta m e

5.46

There are four dimethylcyclopropane isomers. (a ) Write three-dimensional formulas for these isomers. (b ) Which o f the isomers are chiral? (c ) If a mixture consisting o f 1 m ol o f each o f these isomers were subjected to simple gas chromatography (an analytical method that can separate compounds according to boiling point), how many fractions would be obtained and which compounds would each fraction contain? (d ) How many o f these fractions would be optically active?

228

Chapter 5

Stereochemistry

5.47

(U se m odels to solve this problem.) (a) Write a conformational structure for the m ost stable conformation o f trans1,2-diethylcyclohexane and write its mirror image. (b) Are these two m olecules superposable? (c) Are they inter­ convertible through a ring “flip”? (d) Repeat the process in part (a) with cis-1,2-diethylcyclohexane. (e) Are these structures superposable? (f) Are they interconvertible?

5.48

(Use models to solve this problem.) (a) Write a conformational structure for the m ost stable conformation o f trans-1,4diethylcyclohexane and for its mirror image. (b) Are these structures superposable? (c) D o they represent enantiomers? (d) D oes trans-1,4-diethylcyclohexane have a stereoisomer, and if so, what is it? (e) Is this stereoisomer chiral?

5.49

(U se m odels to solve this problem.) Write conformational structures for all o f the stereoisomers o f 1,3-diethylcyclohexane. Label pairs o f enantiomers and m eso compounds if they exist.

Challenge Problems 5.50

Tartaric acid [HO 2 CCH(OH)CH(OH)CO 2 H] was an important compound in the history o f stereochemistry. Two nat­ urally occurring forms o f tartaric acid are optically inactive. One optically inactive form has a melting point of 210-212°C , the other a melting point o f 140°C. The inactive tartaric acid with a melting point o f 210-212°C can be separated into two optically active forms o f tartaric acid with the same melting point (168-170°C ). One optically active tartaric acid has [a]D5 = + 1 2 , and the other, [a]D5 = —12. A ll attempts to separate the other inactive tartaric acid (melt­ ing point 140°C) into optically active compounds fail. (a) Write the three-dimensional structure o f the tartaric acid with melting point 140°C. (b) Write structures for the optically active tartaric acids with melting points o f 168-170°C . (c) Can you determine from the formulas which tartaric acid in (b) has a positive rotation and which has a negative rotation? (d) What is the nature o f the form o f tartaric acid with a melting point o f 210-212°C ?

5.51

(a) An aqueous solution o f pure stereoisomer X o f concentration 0.10 g m L—1 had an observed rotation o f —30° in a 1.0-dm tube at 589.6 nm (the sodium D line) and 25°C. What do you calculate its [a]D to be at this temperature? (b) Under identical conditions but with concentration 0.050 g m L —1, a solution o f X had an observed rotation o f + 165°. Rationalize how this could be and recalculate [a ]D for stereoisomer X. (c) If the optical rotation o f a substance studied at only one concentration is zero, can it definitely be concluded to be achiral? Racemic?

5.52

If a sample o f a pure substance that has two or more chirality centers has an observed rotation o f zero, it could be a racemate. Could it possibly be a pure stereoisomer? Could it possibly be a pure enantiomer?

5.53

Unknown Y has a molecular formula o f C 3 H6 O2. It contains one functional group that absorbs infrared radiation in the 32 0 0 -3 5 5 0 -cm —1 region (when studied as a pure liquid; i.e., “neat”), and it has no absorption in the 1620-1780-cm —1 region. N o carbon atom in the structure o f Y has more than one oxygen atom bonded to it, and Y can exist in two (and only two) stereoisomeric forms. What are the structures o f these forms o f Y?

Learning Group Problems Streptomycin is an antibiotic that is especially useful against penicillin-resistant bacteria. The structure o f strepto­ m ycin is shown in Section 22.17. (a) Identify all o f the chirality centers in the structure o f streptomycin. (b) Assign the appropriate (R) or (S) designation for the configuration o f each chirality center in streptomycin. D-Galactitol is one o f the toxic compounds produced by the disease galactosemia. Accumulation o f high levels o f D-galactitol causes the formation o f cataracts. A Fischer pro­ jection for D-galactitol is shown at right:

CH2OH H-

OH

(a) Draw a three-dimensional structure for D-galactitol.

HO­

H

(b) Draw the mirror im age o f D-galactitol and write its Fischer projection formula.

HO-

H

(c) What is the stereochemical relationship between D-galactitol and its mirror image?

H-

OH CH2OH

Cortisone is a natural steroid that can be isolated from the adrenal cortex. It has anti-inflammatory properties and is used to treat a variety o f disorders (e.g., as a topical application for common skin diseases). The structure o f cor­ tisone is shown in Section 23.4D . (a) Identify all o f the chirality centers in cortisone. (b) Assign the appropriate (R) or (S) designation for the configuration o f each chirality center in cortisone.

Ionic Reactions Nucleophilic Substitution and Elimination Reactions of Alkyl Halides

Organic syntheses, whether they take place in the glassware o f the laboratory or in the cells o f a living organ­ ism, often involve fairly simple processes, such as the installation o f a methyl group in just the right place. For example, we may want to install a methyl group on the nitrogen atom o f a tertiary amine, a reaction that has an im portant counterpart in biochemistry. To do this we often em ploy a reaction like the following: R R— N-

R +

P 1-

h 3c — u

R

R— N — CH 3 R

If we wanted to describe this reaction to an organic chemist we would describe it as a nucleop hilic s u b s titu ­ tio n reaction, a kind o f reaction we describe in detail in this chapter. On the other hand, if we wanted to describe this reaction to a biochemist, we m ight call it a m e th yl trans­ fe r reaction. Biochemists have described many similar reactions this way, for example, the reaction below that transfers a methyl group from S -aden o sylm eth ion in e (SAM) to a tertiary amine to make choline. Choline is incorporated into the phospholipids o f our cell membranes, and it is the hydrolysis product o f acetylcholine, an im portant neurotransmitter. (Crystals o f acetylcholine are shown in the polarized light microscopy image above.) Now, the biological reaction may seem more com plicated, but its essence is similar to many nucleophilic substitution reactions we shall study in this chapter. First we consider alkyl halides, one o f the most im portant types o f reactants in nucleophilic substitution reactions.

230

231

6.1 Organic Halides

CH 3

3 HO R

HO"

NH3

'N + \ CH 3 CH 3

OoC. R NH3

C h o lin e

S - A d e n o s y lm e t h io n in e (S A M )

6.1 O rganic Halides The halogen atom of an alkyl halide is attached to an sp3-hybridized carbon. The arrange­ ment of groups around the carbon atom, therefore, is generally tetrahedral. Because halo­ gen atoms are more electronegative than carbon, the carbon-halogen bond of alkyl halides is polarized; the carbon atom bears a partial positive charge, the halogen atom a partial neg­ ative charge: \s+

s-

Halogen atom size increases as we go down the periodic table: fluorine atoms are the small­ est and iodine atoms the largest. Consequently, the carbon-halogen bond length increases and carbon-halogen bond strength decreases as we go down the periodic table (Table 6.1). Maps of electrostatic potential (see Table 6.1) at the van der Waals surface for the four methyl

H H

H C---- F

H C — X Bond length (Ä) C — X Bond strength (kJ m o T 1)

1.39 472

H

H C-Cl

H

H C-Br

H— C-I

H

H

H

1.78 350

1.93 293

2.14 239

232

Chapter 6

Ionic Reactions

halides, with ball-and-stick models inside, illustrate the trend in polarity, C — X bond length, and halogen atom size as one progresses from fluorine to iodine substitution. Fluoromethane is highly polar and has the shortest C — X bond length and the strongest C — X bond. Iodomethane is much less polar and has the longest C — X bond length and the weakest C — X bond. In the laboratory and in industry, alkyl halides are used as solvents for relatively non­ polar compounds, and they are used as the starting materials for the synthesis of many com­ pounds. As we shall learn in this chapter, the halogen atom of an alkyl halide can be easily replaced by other groups, and the presence of a halogen atom on a carbon chain also affords us the possibility of introducing a multiple bond. A lkyl halides are classified as primary (1°), secondary (2°), or tertiary (3°) according to the number of carbon atoms directly bonded to the carbon bearing the halogen (Section 2.5). Compounds in which a halogen atom is bonded to an sp2-hybridized carbon are called vinylic halides or phenyl halides. The compound C H 2 = C H C l has the common name vinyl chloride, and the group C H 2 = C H — is commonly called the vinyl group. Vinylic halide, therefore, is a general term that refers to a compound in which a halogen is attached to a carbon atom that is also forming a double bond to another carbon atom. Phenyl halides are compounds in which a halogen is attached to a benzene ring (Section 2.4B). Phenyl halides belong to a larger group of compounds that we shall study later, called aryl halides. > = <

( 3 X

A v in y lic h a lid e

^

x

x = /

A p h en y l h a lid e or aryl h a lid e

Together with alkyl halides, these compounds comprise a larger group of compounds known simply as organic halides or organohalogen compounds. The chemistry of vinylic and aryl halides is, as we shall also learn later, quite different from that of alkyl halides, and it is on alkyl halides that we shall focus most of our attention in this chapter.

6.1A Physical Properties of Organic Halides Most alkyl and aryl halides have very low solubilities in water, but as we might expect, they are miscible with each other and with other relatively nonpolar solvents. Dichloromethane (CH 2 Cl2, also called methylene chloride), trichloromethane (CHCl3, also called chloroform ), and tetrachloromethane (CCl4, also called carbon tetrachloride) are sometimes used as solvents for nonpolar and moderately polar compounds. Many chloroalkanes, including CH 2 Cl2, CHCl3, and CCl4, have a cumulative toxicity and are carcinogenic, however, and should therefore be used only in fume hoods and with great care. Table 6.2 lists the physical properties of some common organic halides.

Dichloromethane (CH2 Cl2), a common laboratory solvent

r TABLE

6

.2 ^

O rg an ic H alides F lu o r id e

C h lo r id e

D e n s ity 3 G ro u p

Methyl Ethyl Propyl Isopropyl Butyl sec-Butyl Isobutyl tert-Butyl

b p (°C)

(g m L- 1 )

b p (°C)

D e n s ity 3 (g m L- 1 )

- 7 8 .4 -3 7 .7 - 2 .5 - 9 .4 32

0.84-60 0.7220 0.78-3 0.7220 0.7820 — — 0.7512

- 2 3 .8 13.1 46.6 34 78.4 68 69 51

0 .9115 0.8920 0.8620 0.8920 0.8720 0.8720 0.8420

— —

12

0.9220

aDensities were measured at temperature (°C) indicated in superscript. d e c o m p o s e s is abbreviated dec.

B r o m id e

Io d id e D e n s ity 3

b p (°C)

D e n s ity 3 (g m L- 1 )

b p (°C)

(g m L- 1 )

3.6 38.4 70.8 59.4 101 91.2 91 73.3

1.730 1.4620 1.3520 1.3120 1.2720 1.2620 1.2620 1.2220

42.5 72 102 89.4 130 120 119 100 d e c b

2.2820 1.9520 1.7420 1.7020 1.6120 1.6020 1.6020 1.570

233

6.2 Nucleophilic Substitution Reactions

R eview P roblem 6.1

Give IUPAC names for each of the following. ►Br (a)

(c)

O

C

«

bP

' "

-

Classify each of the following organic halides as primary, secondary, tertiary, vinylic, or aryl.

R eview P roblem 6 .2

6.2 Nucleophilic Substitution Reactions Nucleophilic substitution reactions, like the examples mentioned at the beginning of this chapter, are among the most fundamental types of organic reactions. In general, we can depict nucleophilic substitution reactions in the following way: Nu

+

Nucleophile

R —

L G

R— Nu

-

Substrate

:L G -

Product

Leaving group

In this type of reaction a nucleophile (Nu:) replaces a leaving group (LG) in the molecule that undergoes substitution (called the substrate). The nucleophile is always a Lewis base, and it may be negatively charged or neutral. The leaving group is always a species that takes a pair of electrons with it when it departs. Often the substrate is an alkyl halide (R — X=) and the leaving group is a halide anion ( =X= ), as in the examples of nucleophilic substitution below. R

R R — N :'

R — N— C H 3

R

H O :C H 3 Ö :-

+

+

R

+

CH3 —

CH3CH2—

a:

Br:

------ :

CH3 —

OH

H e lp f u l H i n t

:I: -

+

----- : C H 3 C H 2 — Ö CH 3

+

:Br:

In Section 6.14 we shall see examples of biological nucleophilic substitution.

I: Later we shall see examples of leaving groups other than halide anions. Some of these leaving groups depart as neutral species. For the time being, however, our examples w ill involve alkyl halides, which we represent generally as R— X : ■ In nucleophilic substitution reactions the bond between the substrate carbon and the leav­ ing group undergoes heterolytic bond cleavage, and the unshared electron pair of the nucle­ ophile forms a new bond to the carbon atom. N u

The nucleophile donates an electron pair to the substrate.

R —

L G

—

The bond between the carbon and the leaving group breaks, giving both electrons from the bond to the leaving group.

R —

N u

+

The nucleophile uses its electron pair to form a new covalent bond with the substrate carbon.

=L G -

The leaving group gains the pair of electrons that originally bonded it in the substrate.

H e lp f u l H i n t In color-coded reactions of this chapter, we will use red to indicate a nucleophile and blue to indicate a leaving group.

234

Chapter 6

Ionic Reactions

A key question w e shall want to address later in this chapter is this: When does the bond between the leaving group and the carbon break? D oes it break at the same time that the new bond between the nucleophile and carbon forms, as shown below? s-

R — X:

Nu :

s-

R-

Nu-

-X

Nu—

R

:X:-

Or, does the bond to the leaving group break first? :X:

R — X: ----- : R+ + Follow ed by Nu :

R+

Nu— R

We shall find that the answer depends greatly on the structure o f the substrate.

S o lv e d P ro b le m 6.1 (a) A solution containing methoxide ions, CH 3 O~ ions (as NaOCH3), in methanol can be prepared by adding sodium hydride (NaH) to methanol (CH 3 OH). A flammable gas is the other product. Write the acid-base reaction that takes place. (b) Write the nucleophilic substitution that takes place when CH3I is added and the resulting solu­ tion is heated. STRATEGY AND ANSWER (a) We recall from Section 3.15 that sodium hydride consists o f N a + ions and hydride ions ( H r ions), and that the hydride ion is a very strong base. [It is the conjugate base o f H2, a very weak acid (pA"a = 35, see Table 3.1).] The acid-base reaction that takes place is

C H 3 O9

H

Methanol (stronger acid)

Na+ : H

-

H3 C— O

H : H

Na+

Sodium hydride

Sodium methoxide

(stronger base)

(weaker base)

Hydrogen (weaker acid)

(b) The m ethoxide ion reacts with the alkyl halide (CH 3 I) in a nucleophilic substitution:

CH3— O

Na+

C H

n. 3— h

ch 3oh

H3 C — O — CH 3

+

Na+ +

=I :

6.3 Nucleophiles A nucleophile is a reagent that seeks a positive center. • H e lp f u l H i n t You may wish to review Section 3.3A, "Opposite Charges Attract."

Any negative ion or uncharged m olecule with an unshared electron pair is a poten­ tial nucleophile.

When a nucleophile reacts with an alkyl halide, the carbon atom bearing the halogen atom is the positive center that attracts the nucleophile. This carbon carries a partial positive charge because the electronegative halogen pulls the electrons o f the carbon-halogen bond in its direction.

\s+ „ C -^

T h is is th e p o s itiv e c e n te r th a t th e n u c le o p h ile s e e k s .

s-

X :•

T h e e le c tro n e g a tiv e h a lo g e n p o la r iz e s th e C— X b o n d .

L

w

Let us look at two examples, one in which the nucleophile (a hydroxide ion) bears a negative charge, and one in which the nucleophile (w ater) is uncharged. In the first exam­ ple below involving a negative nucleophile, the reaction produces an alcohol directly. This is because the formal negative charge of the nucleophile is neutralized when the nucleophile uses one of its unshared electron pairs to form a covalent bond. General Reaction for Nucleophilic Substitution o f an Alkyl Halide by Hydroxide Ion H — O :N u c le o p h ile

+

----- :

R— X:

H— O — R +

Alkyl h a lid e

A lc o h o l

:X :L eavin g grou p

In the second example, involving a neutral nucleophile (water), the reaction leads to a product that initially bears a formal positive charge. This is because use of an unshared elec­ tron pair from the neutral nucleophile leaves the nucleophilic atom with a formal positive charge after the bond is formed. The initial product in this case is called an alkyloxonium ion. In a subsequent step a proton is removed from the alkyloxonium ion to form the neu­ tral alcohol. General Reaction for Nucleophilic Substitution o f an Alkyl Halide by Water

H — Q:

R— X:

H— Q ——R

Alkyl h a lid e

A lk yloxon iu m ion

H

:X

:-

H

N u c le o p h ile

H e lp f u l H i n t h 2o

■X-

H— Q— R

In a reaction like this the nucleophile is a solvent molecule (as is often the case when neu­ tral nucleophiles are involved). Since solvent molecules are present in great excess, the equilibrium favors transfer of a proton from the alkyloxonium ion to a water molecule. (This type of reaction is an example of solvolysis, which we shall discuss further in Section 6.12B.) The reaction of ammonia (NH 3) with an alkyl halide, as shown below, provides another example where the nucleophile is uncharged. An excess of ammonia favors equilibrium removal of a proton from the alkylaminium ion to form the neutral amine. H H— N

H -

R— X :

->

H— N— R

H N u c le o p h ile

:X =

H Alkyl h a lid e

A lkylam inium ion :NH3 (excess)

H— N— R H An a m in e

NH

:X :

A deprotonation step is always required to complete the reaction when the nucleophile was a neutral atom that bore a proton.

236

Chapter 6

Ionic Reactions

S o l v e d P r o b le m 6 .2 Write the follow ing as net ionic equations and designate the nucleophile, substrate, and leaving group in each case. (a )

,§ : Na+

,

V CH 3 — I =

^

A

+

CH3

Na+ : I :

(b) =: Na+

CH3

CH3 — I;

(c )

2 H3 N :

n h 4+

'NH 2

: Br:

STRATEGY A net ionic equation does not include spectator ions but is still balanced in terms o f charges and the remaining species. Spectator ions are those ions which have not been involved in covalent bonding changes dur­ ing a reaction, and which appear on both sides o f a chem ical equation. In reactions (a) and (b) the sodium ion is a spectator ion, thus the net ionic equation would not include them, and their net ionic equations would have a net negative charge on each side o f the arrow. Equation (c) has no ions present among the reactants, and thus the ions found with the products are not spectator ions— they have resulted from covalent bonding changes. Equation (c) cannot be simplified to a net ionic equation. N ucleop h iles use a pair o f electrons to form a covalent bond that is present in a product m olecule. In all o f the above reactions w e can identify a species that used a pair o f electrons in this way. These are the nucleophiles. L eaving groups depart from one o f the reactant m olecules and take a pair o f electrons with them. In each reaction above w e can identify such a species. Lastly, the reactant to which the nucleophile became bonded and from which the leaving groups departed is the substrate. ANSWER The net ionic equations are as follow s for (a) and (b), and there is no abbreviated equation possible for (c). N ucleophiles, substrates, and leaving groups are labeled accordingly. (a )

,S : 3 N u c le o p h ile

S u b str a te

L eavin g group

(b) CH3

N u c le o p h ile

CH3 — I:

:I:

S u b s tr a te

L ea v in g group

(c) 2 H3 N : N u c le o p h ile

R e v ie w P ro b le m 6 .3

+

'B r:

:B r:

NH4

NH 2

S u b s tr a te

L ea v in g group

Write the following as net ionic equations and designate the nucleophile, substrate, and leav­ ing group in each reaction: (a) CH3I + (b) NaI

+

CH 3 CH2ONa ----- > CH 3 OCH 2 CH 3 CH 3 CH2Br ----- > CH 3 CH2I +

(c) 2 CH3OH

+

+

NaI

NaBr

(CH3)3CCl ----- > (CH 3 )3 COCH 3

+

CH 3 OH2+ +

Cl-

237

6.5 Kinetics o f a Nucleophilic Substitution Reaction: An SN2 Reaction

Br

(d)

Br

(e)

CN

NaCN

NaBr

nh2

2 NH 3

NH4Br

6.4 Leaving Groups To act as the substrate in a nucleophilic substitution reaction, a molecule must have a good leaving group. •

A good le a v in g g r o u p is a substituent that can leave as a relatively stable, weakly basic molecule or ion.

In the examples shown above (Sections 6.2 and 6.3) the leaving group has been a halogen. Halide anions are weak bases (they are the conjugate bases of strong acids, HX), and there­ fore halogens are good leaving groups. Some leaving groups depart as neutral molecules, such as a molecule of water or an alco­ hol. For this to be possible, the leaving group must have a formal positive charge while it is bonded to the substrate. When this group departs with a pair of electrons the leaving group’s formal charge goes to zero. The following is an example where the leaving group departs as a water molecule.

CH3 -

a

c h 3-

o

—

CH 3

h

H

H

O -C H

:O -

3

H

H

H

H e lp f u l H i n t Note that the net charge is the same on each side of a properly written chemical equation.

As we shall see later, the positive charge on a leaving group (like that above) usually results from protonation of the substrate by an acid. However, use of an acid to protonate the substrate and make a positively charged leaving group is feasible only when the nucleophile itself is not strongly basic, and when the nucleophile is present in abundance (such as in solvolysis). Let us now begin to consider the mechanisms of nucleophilic substitution reactions. How does the nucleophile replace the leaving group? Does the reaction take place in one step or is more than one step involved? If more than one step is involved, what kinds of inter­ mediates are formed? Which steps are fast and which are slow? In order to answer these questions, we need to know something about the rates of chemical reactions.

6.5 Kinetics o f a Nucleophilic Substitution Reaction: An SN2 Reaction To understand how the rate of a reaction (kinetics) might be measured, let us consider an actual example: the reaction that takes place between chloromethane and hydroxide ion in aqueous solution: CH 3 - C l 3

+

OH-

— C h 2o

CH3- O H 3

+

C l-

Although chloromethane is not highly soluble in water, it is soluble enough to carry out our kinetic study in an aqueous solution of sodium hydroxide. Because reaction rates are known to be temperature dependent (Section 6.7), we carry out the reaction at a constant temperature.

6.5A How Do W e Measure the Rate of This Reaction? The rate of the reaction can be determined experimentally by measuring the rate at which chloromethane or hydroxide ion disappears from the solution or the rate at which methanol or chloride ion appears in the solution. We can make any of these measurements by withdraw­ ing a small sample from the reaction mixture soon after the reaction begins and analyzing it

238

Chapter 6

Ionic Reactions

for the concentrations of CH3Cl or OH- and CH3OH or Cl- . We are interested in what are called initial rates, because as time passes the concentrations of the reactants change. Since we also know the initial concentrations of reactants (because we measured them when we made up the solution), it w ill be easy to calculate the rate at which the reactants are disappearing from the solution or the products are appearing in the solution. We perform several such experiments keeping the temperature the same but varying the initial concentrations of the reactants. The results that we might get are shown in Table 6.3. R ate Study o f Reaction o f CH3Cl w ith OH~ at 60°C _____________ E x p e r im e n t N um ber

Initial [C H 3 Cl]

Initial [O H - ]

1 2 3 4

0.0010 0.0020 0.0010 0.0020

1.0 1.0 2.0 2.0

Initial R ate (m ol L- 1 s - 1 )

4.9 9.8 9.8 19.6

X X X X

1 0-7 1 0-7 1 0-7 10-7

Notice that the experiments show that the rate depends on the concentration of chloromethane a nd on the concentration of hydroxide ion. When we doubled the concen­ tration of chloromethane in experiment 2, the rate doubled. When we doubled the con­ centration of hydroxide ion in experiment 3, the rate doubled. When we doubled both concentrations in experiment 4, the rate increased by a factor of fo u r. We can express these results as a proportionality, Rate « [CH3 Cl][OH- ] and this proportionality can be expressed as an equation through the introduction of a proportionality constant (k) called the rate constant: Rate = k[CH3 Cl][OH- ] For this reaction at this temperature we find that k = 4.9 X 10 - 4 L m ol- 1 s-1 . (Verify this for yourself by doing the calculation.)

6.5B What Is the Order of This Reaction? This reaction is said to be second order overall.* It is reasonable to conclude, therefore, thatf o r the reaction to take p la ce a hydroxide ion and a chlorom ethane m olecule m ust col­ lide. We also say that the reaction is bim olecular. (By bim olecular we mean that two species are involved in the step whose rate is being measured. In general the number of species involved in a reaction step is called the m olecularity of the reaction.) We call this kind of reaction an Sn2 reaction, meaning substitution, nucleophilic, bimolecular.

6.6 A Mechanism fo r th e SN2 Reaction A schematic representation of orbitals involved in an SN2 reaction— based on ideas pro­ posed by Edward D . Hughes and Sir Christopher Ingold in 1937— is outlined below. — Antibonding orbital

^

C

1

/

G

* -------- *

^ N

u

LG+

I— Bonding orbital * I n general, th e o v e ra ll o rd e r o f a re a c tio n is equal to th e su m o f th e exponents a and b in th e rate e q u a tio n Rate = k [A ]a [ B ]b. I f in som e o th e r re actio n , fo r e xam ple, w e fo u n d th a t Rate = k [ A ]2 [B ], th e n w e w o u ld say th a t th e re a c tio n is second o rd e r w ith respect to [A ] , fir s t o rd e r w ith respect to [B ], and th ir d o rd e r o v e ra ll.

239

6.6 A Mechanism for the Sn2 Reaction

According to this mechanism: •

The nucleophile approaches the carbon bearing the leaving group from the back side, that is, from the side directly opposite the leaving group.

The orbital that contains the electron pair of the nucleophile (its highest occupied molec­ ular orbital, or H O M O ) begins to overlap with an empty orbital (the lowest unoccupied mol­ ecular orbital, or LU M O ) of the carbon atom bearing the leaving group. As the reaction progresses, the bond between the nucleophile and the carbon atom strengthens, and the bond between the carbon atom and the leaving group weakens. •

As the nucleophile forms a bond and the leaving group departs, the substrate carbon atom undergoes inversion*— its tetrahedral bonding configuration is turned inside out.

The formation of the bond between the nucleophile and the carbon atom provides most of the energy necessary to break the bond between the carbon atom and the leaving group. We can represent this mechanism with chloromethane and hydroxide ion as shown in the box “Mechanism for the SN2 Reaction” below. •

The Sn2 reaction proceeds in a single step (without any intermediates) through an unstable arrangement of atoms called the transition state.

A MECHANISM FOR THE REACTION M e c h a n i s m f o r t h e S N2 R e a c t i o n R E A C T IO N HO -

+

CH3Cl ----- :

c h 3o h

+

or

M E C H A N IS M

H

O:

-

hH h4 d+A - s­ C— C l: / " H

HH

s-

H— O ..

■■ s-

C! — C. . l:

H— O

H ..^H C + : C l: H

H T ran sition s t a t e

The n egative hydroxide ion brings a pair of e le ctr o n s to the partially p o sitiv e carbon from the back s id e with r esp e ct to th e leaving group. The chlorine b e g in s to m o v e aw ay with the pair of e le ctr o n s that b on d ed it to the carbon.

In th e transition sta te, a bond betw een oxy g en and carbon is partially form ed and the bond betw een carbon and chlorine is partially broken. The configuration o f the carbon atom b e g in s to invert.

N ow the bond b etw een the o xygen and carbon h as form ed and the chloride ion h as departed. The configuration o f the carbon h a s inverted.

*C o n sid e ra b le e vidence had appeared in th e years p rio r to H ughes and In g o ld ’s 1937 p u b lic a tio n in d ic a tin g th a t in re actio n s lik e th is an in v e rs io n o f c o n fig u ra tio n o f th e ca rbo n b e a rin g th e le a v in g g ro u p takes place. T h e firs t ob se rva tio n o f such an in v e rs io n was m ade b y th e L a tv ia n c h e m is t Paul W a ld e n in 1896, and such in ve rsio n s are c a lle d

Walden inversions in

his hon o r. W e sh a ll stu d y th is aspect o f th e S n 2 re a c tio n fu rth e r in S e ctio n 6.8.

240

Chapter 6

Ionic Reactions

The transition state is a fleeting arrangement o f the atoms in which the nucleophile and the leaving group are both partially bonded to the carbon atom undergoing substitution. Because the transition state involves both the nucleophile (e.g., a hydroxide ion) and the substrate (e.g., a m olecule o f chloromethane), this mechanism accounts for the second-order reaction kinetics that w e observe. •

The Sn2 reaction is said to be a con certed reaction , because bond forming and bond breaking occur in concert (simultaneously) through a single transition state.

The transition state has an extremely brief existence. It lasts only as long as the time required for one molecular vibration, about 10 “ 12 s. The structure and energy o f the tran­ sition state are highly important aspects o f any chemical reaction. We shall, therefore, exam­ ine this subject further in Section 6.7.

6.7 Transition S tate Theory: Free-Energy Diagram s •

A reaction that proceeds with a negative free-energy change (releases energy to its surroundings) is said to be exergon ic; one that proceeds with a positive freeenergy change (absorbs energy from its surroundings) is said to be en d ergon ic.

The reaction between chloromethane and hydroxide ion in aqueous solution is highly exergonic; at 60°C (333 K), AG° = - 1 0 0 k J m o l- 1 . (The reaction is also exothermic, AH° = - 7 5 kJ m o l- 1 .) CH3- C l

+

OH-

----- > CH3-O H

+

Cl-

AG° = - 1 0 0 k J m ol - 1

The equilibrium constant for the reaction is extremely large, as w e show by the following calculation: AG° = - R T ln K eq ln K eq ln K

-AG° RT - ( - 1 0 0 kJ m ol“1)

eq

0.00831 kJ K - 1 m ol - 1 X 333 K

ln Keq = 36.1 c4 Keq = 5.0 X 10 15 c4 An equilibrium constant as large as this means that the reaction goes to completion. Because the free-energy change is negative, w e can say that in energy terms the reac­ tion goes dow nhill. The products o f the reaction are at a lower level o f free energy than the reactants. However, if covalent bonds are broken in a reaction, the reactants must go up an energy hill first, before they can go downhill. This w ill be true even if the reaction is exergonic. We can represent the energy changes in a reaction using a graph called a free-energy d iagram , where w e plot the free energy o f the reacting particles (y-axis) against the reac­ tion coordinate (x-axis). Figure 6.1 is an example for a generalized SN2 reaction. •

The re a c tio n c o o rd in a te indicates the progress o f the reaction, in terms o f the conversion o f reactants to products.

•

The top o f the energy curve corresponds to the t r a n s it io n s ta te for the reaction.

•

The fre e e n e rg y o f a c tiv a tio n (AG*) for the reaction is the difference in energy between the reactants and the transition state.

•

The fre e e n e rg y c h a n g e f o r th e re a c tio n (AG°) is the difference in energy between the reactants and the products.

6.7 Transition State Theory: Free-Energy Diagrams

P roducts

R eaction coordinate

241

Figure 6.1 A free-energy diagram for a hypothetical exergonic Sn2 reaction (i.e., that takes place with a negative AG°, releasing energy to the surroundings).

The top of the energy h ill corresponds to the transition state. The difference in free energy between the reactants and the transition state is the free energy o f activation, AG*. The difference in free energy between the reactants and products is the free-energy change f o r the reaction, AG°. For our example in Fig. 6.1, the free-energy level of the products is lower than that of the reactants. In terms of our analogy, we can say that the reactants in one energy valley must surmount an energy h ill (the transition state) in order to reach the lower energy valley of the products. If a reaction in which covalent bonds are broken proceeds with a positive free-energy change (Fig. 6.2), there w ill still be a free energy of activation. That is, if the products have greater free energy than reactants, the free energy of activation w ill be even higher. (AG* w ill be larger than AG °.) In other words, in the uphill (endergonic) reaction an even larger energy h ill lies between the reactants in one valley and the products in a higher one.

R eactan ts

R eaction coordinate

Figure 6.2 A free-energy diagram for a hypothetical endergonic Sn2 reaction (i.e., that takes place with a positive AG°, absorbing energy from the surroundings).

242

Chapter 6

Ionic Reactions

Figure 6.3 The distribution of energies at two different tem peratures, T|_ow and Tuigh- The number of collisions with energies greater than the free energy of activation is indicated by the corresponding shaded area under each curve.

6.7A Temperature, Reaction Rate, and the Equilibrium Constant Most chemical reactions occur much more rapidly at higher temperatures. The increase in reaction rate for SN2 reactions relates to the fact that at higher temperatures the number of collisions between reactants with sufficient energy to surmount the activation energy (AG*) increases significantly (see Fig. 6.3). •

A 10°C increase in temperature w ill cause the reaction rate to double for many reactions taking place near room temperature.

This dramatic increase in reaction rate results from a large increase in the number of col­ lisions between reactants that together have sufficient energy to surmount the barrier at the higher temperature. The kinetic energies of molecules at a given temperature are not all the same. Figure 6.3 shows the distribution of energies brought to collisions at two temperatures (that do not differ greatly), labeled TLow and THigh. Because of the way energies are distrib­ uted at different temperatures (as indicated by the shapes of the curves), increasing the tem­ perature by only a small amount causes a large increase in the number of collisions with larger energies. In Fig. 6.3 we have designated an arbitrary minimum free energy of activation as being required to bring about a reaction between colliding molecules. There is also an important relationship between the rate of a reaction and the magnitude of the free energy of activation. The relationship between the rate constant (k) and AG* is an exponential one: k = k0e-AGt/RT

In this equation, e = 2.718, the base of natural logarithms, and k0 is the absolute rate constant, which equals the rate at which all transition states proceed to products. At 25°C, k0 = 6.2 X 10 12 s-1 . •

A reaction with a lower free energy of activation (AG*) w ill occur exponentially faster * than a reaction with a higher free energy of activation, as dictated by k = k0e .

Generally speaking, if a reaction has a AG* less than 84 kJ m ol-1 , it w ill take place read­ ily at room temperature or below. If AG* is greater than 84 kJ m ol-1 , heating w ill be required to cause the reaction to occur at a reasonable rate. A free-energy diagram for the reaction of chloromethane with hydroxide ion is shown in Fig. 6.4. A t 60°C, AG* = 103 kJ m ol-1 , which means that at this temperature the reac­ tion reaches completion in a matter of a few hours.

R eview P roblem 6 .4

Draw a hypothetical free-energy diagram for the SN2 reaction of iodide anion with 1-chlorobutane. Label the diagram as in Fig. 6.4, and assume it is exergonic but without specific values for AG* and AG°.

243

6.8 The Stereochemistry o f Sn2 Reactions

Transition state

Ô-HQ

C H 3 - - CI

Free energy of activation

Figure 6.4 A free-energy diagram for the reaction of chloromethane with hydroxide ion at 60°C.

R eaction coordinate

6.8 The Stereochem istry o f SN2 Reactions The stereochemistry of SN2 reactions is directly related to key features of the mechanism that we learned earlier: •

The nucleophile approaches the substrate carbon from the back side with respect to the leaving group. In other words, the bond to the nucleophile that is forming is opposite (at 180°) to the bond to the leaving group that is breaking.

•

Nucleophilic displacement of the leaving group in an SN2 reaction causes inver­ sion of configuration at the substrate carbon.

We depict the inversion process as follows. It is much like the way an umbrella is inverted in a strong wind.

Transition state for an S n2 reaction.

A n inversion of configuration

H S—

^ C — C l' H "V H

HO

I

* s—

C— C l : C HH

/ HO— C

V /7H H

:C l

244

Chapter 6

Ionic Reactions

With a molecule such as chloromethane, however, there is no way to prove that attack by the nucleophile has involved inversion of configuration of the carbon atom because one form of methyl chloride is identical to its inverted form. W ith a molecule containing chirality centers such as cis-1-chloro-3-methylcyclopentane, however, we can observe the results of an inversion of configuration by the change in stereochemistry that occurs. When cis-1-chloro-3-methylcyclopentane reacts with hydroxide ion in an SN2 reaction, the prod­ uct is trans-3-methylcyclopentanol. The hydroxide ion ends up being bonded on the oppo­ site side o f the ring from the chlorine it replaces:

c /s - 1 - C h lo r o - 3 -

fr a n s - 3 - M e th y lc y c lo p e n ta n o l

m e t h y lc y c lo p e n ta n e

Presumably, the transition state for this reaction is like that shown here.

L e a v in g g r o u p d e p a r ts f r o m t h e t o p s id e .

N u c le o p h ile a tta c k s

: OH s-

f r o m t h e b o t t o m s id e .

Give the structure of the product that would be formed when trans-1-bromo-3-methylcyclobutane undergoes an Sn2 reaction with N aI. STRATEGY AND ANSWER First, write the formulas for the reactants and identify the nucleophile, the substrate, and the leaving group. Then, recognizing that the nucleophile w ill attack the back side of the substrate carbon atom that bears the leaving group, causing an inversion of configuration at that carbon, write the structure of the product.

I n v e r s io n o f c o n f ig u r a t

L e a v in g

N u c le o -

g ro u p

p h ile

fr a n s - 1 - B r o m o - 3 -

c /s - 1 - Io d o - 3 -

m e t h y lc y c l o b u t a n e

m e t h y lc y c l o b u t a n e

(s u b s tra te )

(p ro d u c t)

R eview P roblem 6 .5

Using chair conformational structures (Section 4.11), show the nucleophilic substitution reaction that would take place when trans-1-bromo-4-tert-butylcyclohexane reacts with iodide ion. (Show the most stable conformation of the reactant and the product.)

245

6.8 The Stereochemistry o f Sn2 Reactions

• Sn2 reactions always occur with inversion of configuration. We can also observe inversion of configuration when an SN2 reaction occurs at a chirality center in an acyclic molecule. The reaction of (Æ)-(—)-2-bromooctane with sodium hydroxide provides an example. We can determine whether or not inversion of configura­ tion occurs in this reaction because the configurations and optical rotations for both enantiomers of 2 -bromooctane and the expected product, 2 -octanol, are known.

(R )-(-)-2 -B r o m o o c ta n e M B 5 = - 3 4 .2 5

(S )-(+ )-2 -B ro m o o c ta n e H D 5 = + 34.25

(R )-(-)-2 -O c ta n o l H B 5 = - 9.9 0

(S )-(+ )-2 -O c ta n o l H £ 5 = + 9.90

When the reaction is carried out, we find that enantiomerically pure (R)-(-)-2-brom ooctane ([a ] 2 5 = - 3 4 .2 5 ) has been converted to enantiomerically pure (S )-( + )-2-octanol ([a]D5= +9.90).

A MECHANISM FOR THE REACTION T h e S t e r e o c h e m i s t r y o f a n S N2 R e a c t i o n The reaction of (R)-(-)-2-brom ooctane with hydroxide is an SN2 reaction and takes place with com plete inversion of

configuration: ch3 \

HO

3

C— Br: H "V C6H13

aHO -

CH 3

t

s­

.. HO

C-- Br: H C 6 H 13 J

ch3 / 3 C + V%H C6H13

.. Br:-

An inversion of configuration (R )-(-)-2 -B r o m o o c ta n e [ « ] £ 5 = - 34.25° E n a n tio m e ric p u rity = 1 0 0 %

(S )-(+ )-2 -O c ta n o l [ « ] £ 5 = + 9.90° E n a n tio m e ric p u rity = 1 0 0 %

Sn2 reactions that involve breaking a bond to a chirality center can be used to relate configurations of molecules because the stereochemistry of the reaction is known. (a )

Illustrate how this is true by assigning configurations to the 2-chlorobutane enantiomers based on the following data. [The configuration of ( —)-2-butanol is given in Section 5.8C.] (+)-2-Chlorobutane [a]D 5 = + 3 6 .0 0 E n a n tio m e ric a lly p u re

OH—

Q >

o n2

(—)-2-Butanol [a ]D = - 1 3 .5 2 E n a n tio m e ric a lly pure

An S n 2 reaction has one transition state.

R eview P roblem 6.6

246

Chapter 6

Ionic Reactions

(b) When optically pure (+)-2-chlorobutane is allowed to react with potassium iodide in

acetone in an SN2 reaction, the 2-iodobutane that is produced has a minus rotation. What is the configuration of (-)-2-iodobutane? O f (+)-2-iodobutane?

6.9 The Reaction o f tert-B u tyl Chloride w ith H ydroxide Ion: An SN1 Reaction Let us now consider another mechanism for nucleophilic substitution: the SN1 reaction. When tert-butyl chloride reacts with sodium hydroxide in a mixture of water and acetone, the kinetic results are quite different than for the reaction of chloromethane with hydroxide. The rate of formation of tert -butyl alcohol is dependent on the concentration of tert-butyl chloride, but it is independent o f the concentration o f hydroxide ion. Doubling the tert-butyl chloride concen­ tration doubles the rate of the substitution reaction, but changing the hydroxide ion concentra­ tion (within limits) has no appreciable effect. tert-Butyl chloride reacts by substitution at virtually the same rate in pure water (where the hydroxide ion is 10~ 7 M) as it does in 0.05M aqueous sodium hydroxide (where the hydroxide ion concentration is 500,000 times larger). (We shall see in Section 6.10 that the important nucleophile in this reaction is a molecule of water.) Thus, the rate equation for this substitution reaction is first order with respect to tertbutyl chloride and first order overall: CH3

ch

I

h 2o

c H 3— C — ch

Cl

+

OH

3

I

acetone*

C H 3— C

OH

9

ch

3

+

Cl

3

R a te « [(C H 3 ) 3 C C l] R a te = k [(C H 3 ) 3 C C l]

We can conclude, therefore, that hydroxide ions do not participate in the transition state of the step that controls the rate of the reaction and that only molecules of tert-butyl chlo­ ride are involved. This reaction is said to be unimolecular (first order) in the rate-determining step. We call this type of reaction an Sn1 reaction (substitution, nucleophilic, unimolecular). (In Section 6.15 we shall see that elimination reactions can compete with SN1 reac­ tions, leading to the formation of alkenes, but in the case of tert-butyl chloride in the absence of base and at room temperature, SNl is the dominant process.) How can we explain an SN1 reaction in terms of a mechanism? To do so, we shall need to consider the possibility that the mechanism involves more than one step. But what kind of kinetic results should we expect from a multistep reaction? Let us consider this point further.

6.9A Multistep Reactions and the Rate-Determining Step •

If a reaction takes place in a series of steps, and if one step is intrinsically slower than all the others, then the rate of the overall reaction w ill be essentially the same as the rate of this slow step. This slow step, consequently, is called the ratelim iting step or the rate-determ ining step.

Consider a multistep reaction such as the following: Reactant

k

i

— >

intermediate 1

Step 1

k-2

—

> intermediate 2

Step 2

k-3

—>

> product

Step 3

When we say that the first step in this example is intrinsically slow, we mean that the rate constant for step 1 is very much smaller than the rate constant for step 2 or for step 3. That is, k1< < k2 or k3. When we say that steps 2 and 3 are fast, we mean that because their rate constants are larger, they could (in theory) take place rapidly if the concentrations of the two intermediates ever became high. In actuality, the concentrations of the intermediates are always very small because of the slowness of step 1 .

6.10 A Mechanism for the Sn 1 Reaction

247

Reactant Slow (rate determ ining)

Intermediate 1 Fast

Intermediate 2 Fast

Product

Figure 6.5 A modified hourglass that serves as an analogy for a multistep reaction. The overall rate is limited by the rate of the slow step.

As an analogy, imagine an hourglass modified in the way shown in Fig. 6.5. The open­ ing between the top chamber and the one just below is considerably smaller than the other two. The overall rate at which sand falls from the top to the bottom of the hourglass is lim ­ ited by the rate at which sand passes through the small orifice. This step, in the passage of sand, is analogous to the rate-determining step of the multistep reaction.

6.10 A Mechanism fo r th e SN1 Reaction The mechanism for the reaction of teri-butyl chloride with water (Section 6.9) can be described in three steps. See the box “Mechanism for the SN1 Reaction” below, with a schematic free-energy diagram highlighted for each step. Two distinct intermediates are formed. The first step is the slow step— it is the rate-determining step. In it a molecule of ieri-butyl chloride ionizes and becomes a ieri-butyl cation and a chloride ion. In the tran­ sition state for this step the carbon-chlorine bond of ieri-butyl chloride is largely broken and ions are beginning to develop: CH 3 CH3— C dt— -C l8CH 3 The solvent (water) stabilizes these developing ions by solvation. Carbocation forma­ tion, in general, takes place slowly because it is usually a highly endothermic process and is uphill in terms of free energy. The first step requires heterolytic cleavage of the carbon-chlorine bond. Because no other bonds are formed in this step, it should be highly endothermic and it should have a high free energy of activation, as we see in the free-energy diagram. T h at departure of the halide takes place at all is largely because of the ionizing ability of the solvent, w ater. Experiments indicate that in the gas phase (i.e., in the absence of a solvent), the free energy of activation is about 630 kJ m ol~1! In aqueous solution, however, the free energy of activation is much lower— about 84 kJ m ol_1. W ater molecules surround and stabi­ lize the cation and anion that are produced (cf. Section 2.13D). In the second step the intermediate tert -butyl cation reacts rapidly with water to pro­ duce a ieri-butyloxonium ion, (CH 3 )3 COH2+, which in the third step, rapidly transfers a proton to a molecule of water producing tert -butyl alcohol.

248

Chapter 6

Ionic Reactions

A MECHANISM FOR THE REACTION Mechanism for the SN1 Reaction R EA C TIO N CH3 I 3 .. CH 3— C— Cl

+

CH3 I 3 .. 2 H2Q ----- > CH 3— C— QH

CH3

+ H3 Q + +

+

.. Xh

CH3

M E C H A N IS M Step 1

CH3 I 3 CH3 — C — C l: 3 I CH

slow

' h, q

ch3 / 3 CH 3— C+

Step 1

:C l;

\

CH

A id ed b y th e polar s o lv e n t, a c h lo r in e d e p a r ts w ith th e e le c tr o n pair th at b o n d e d it to th e ca rb o n .

T h is s l o w s t e p p r o d u c e s th e 3° c a r b o c a tio n in term e­ d ia te an d a c h lo r id e ion. A lth o u g h n o t s h o w n h e r e , th e io n s a r e s o lv a t e d (an d s ta b iliz e d ) by w a te r m o l­ e c u le s .

Step 2

AG*

fast

H

CH3— C CH 3

H

A w ater m o le c u le a c tin g a s a L ew is b a s e d o n a t e s an e le c tr o n pair to th e c a r b o ­ c a tio n (a L ew is a cid ). T h is g i v e s th e c a tio n ic c a rb o n e ig h t e le c tr o n s .

hence this is the slowest step

R eaction coord in ate

CH3

CH

AGi(1) is much

larger than Transition sta te 1 AG*(2) or AG*(3),

Step 2

1 3 "+ CH3— C----- O — H 3 I CH3 H T h e p r o d u ct is a tertb u ty lo x o n iu m ion (or p r o to n a te d tert-butyl a lc o h o l).

Step 3

Step 3

CH3 CH3— C ch3

CH3 •Q “ I H

1*0 h

A w a ter m o le c u le a c tin g a s a B r0 n ste d b a s e a c c e p t s a p r o to n fr o m t h e f e r f - b u t y lo x o n iu m ion.

fast

CH3— C CH3

Q: I H

H

-Q ­

Transition sta te 3

I

H

T h e p r o d u c ts a r e tert-butyl a l c o h o l a n d a h y d r o n iu m ion.

(3) R eaction coord in ate

6.11 Carbocations Beginning in the 1920s much evidence began to accumulate implicating simple alkyl cations as intermediates in a variety of ionic reactions. However, because alkyl cations are highly unstable and highly reactive, they were, in all instances studied before 1962, very short-lived,

249

6.11 Carbocations

t r a n s ie n t s p e c ie s t h a t c o u l d n o t b e o b s e r v e d d i r e c t l y . * H o w e v e r , i n

1 9 6 2 G e o r g e A . O la h

( U n i v e r s i t y o f S o u t h e r n C a l i f o r n i a ) a n d c o - w o r k e r s p u b l i s h e d t h e f i r s t o f a s e r ie s o f p a p e r s d e s c r i b in g e x p e r i m e n t s i n w h i c h a l k y l c a t i o n s w e r e p r e p a r e d i n a n e n v i r o n m e n t i n w h i c h t h e y

©

Olah was awarded the 1994 Nobel Prize in Chemistry.

w e r e r e a s o n a b l y s t a b le a n d i n w h i c h t h e y c o u l d b e o b s e r v e d b y a n u m b e r o f s p e c t r o s c o p ic t e c h n iq u e s .

6.11A The Structure of Carbocations •

C a r b o c a t i o n s a r e t r i g o n a l p la n a r .

J u s t a s t h e t r i g o n a l p la n a r s t r u c t u r e o f B F o f sp2 h y b r i d i z a t i o n , s o , t o o ( F i g .

6 .6 ) ,

3

( S e c t i o n 1 . 1 6 D ) c a n b e a c c o u n t e d f o r o n t h e b a s is

c a n t h e t r i g o n a l p la n a r s t r u c t u r e o f c a r b o c a t i o n s .

Figure 6.6 (a) A stylized orbital structure of the methyl cation. The bonds are sigma (ct) bonds formed by overlap of the carbon atom's three sp 2 orbitals with the 1s orbitals of the hydrogen atom s. The p orbital is vacant. (b) A dashed line-w edge representation of the tert-butyl cation. The bonds betw een carbon atoms are formed by overlap of sp 3 orbitals of the methyl groups with sp2 orbitals of the central carbon atom.

a bond (a)

•

(b)

T h e c e n t r a l c a r b o n a to m i n a c a r b o c a tio n is e le c t r o n d e f ic ie n t ; i t h a s o n l y s ix e le c ­ t r o n s i n i t s v a le n c e s h e l l .

In o u r m o d e l ( F ig .

6 .6 )

th e s e s ix e le c tr o n s a re u s e d to f o r m

th r e e s ig m a c o v a le n t b o n d s to

h y d r o g e n a to m s o r a lk y l g r o u p s . •

T h e p o r b i t a l o f a c a r b o c a t i o n c o n t a in s n o e le c t r o n s , b u t i t c a n a c c e p t a n e le c t r o n p a i r w h e n th e c a r b o c a tio n u n d e r g o e s f u r t h e r r e a c t io n .

N o t a ll ty p e s o f c a r b o c a tio n s h a v e th e s a m e r e la t iv e s t a b ilit y a s w e s h a ll le a r n in th e n e x t s e c tio n .

K

6.11B The Relative Stabilities of Carbocations T h e r e la t iv e s t a b ilit ie s o f c a r b o c a tio n s a re r e la te d to th e n u m b e r o f a l k y l g r o u p s a tta c h e d to th e p o s it iv e ly c h a r g e d t r iv a le n t c a r b o n . •

T e r t i a r y c a r b o c a t i o n s a r e t h e m o s t s t a b le , a n d t h e m e t h y l c a r b o c a t i o n i s t h e l e a s t s t a b le .

•

T h e o v e r a ll o r d e r o f s t a b ilit y is as fo llo w s :

I

I

R

R

I

R

R

3° > (m o st sta b le)

W

W

H

2°

I

H

>

H

H

1°

IT

>

H

Methyl (lea st stab le)

T h i s o r d e r o f c a r b o c a t i o n s t a b i l i t y c a n b e e x p l a i n e d o n t h e b a s is o f h y p e r c o n j u g a t i o n . •

H y p e r c o n ju g a tio n i n v o l v e s e l e c t r o n d e l o c a l i z a t i o n ( v i a p a r t i a l o r b i t a l o v e r l a p ) f r o m a f i l le d b o n d in g o r b it a l to a n a d ja c e n t u n f ille d o r b it a l ( S e c tio n 4 .8 ) .

* A s w e sh a ll le a rn later, ca rbo ca tio ns b e a rin g a ro m a tic groups can be m u c h m ore stable; one o f these had been stu d ie d as e a rly as 1901.

H e lp fu l H i n t

An understanding of carbocation structure and relative stability is important for learning a variety of reaction processes.

250

Chapter 6

Ionic Reactions

Figure 6.7 How a methyl group helps stabilize the positive charge o f a carbocation. Electron density from one of the carbon-hydrogen sigma bonds of the methyl group flows into the vacant p orbital of the carbocation because the orbitals can partly overlap. Shifting electron density in this way makes the sp 2-hybridized carbon of the carbocation som ewhat less positive, and the hydrogens of the methyl group assum e som e of the positive charge. Delocalization (dispersal) of the charge in this way leads to greater stability. This interaction of a bond orbital with a p orbital is called hyperconjugation.

In the case of a carbocation, the unfilled orbital is the vacant p orbital of the carbocation, and the filled orbitals are C — H or C — C sigma bonds at the carbons adjacent to the p orbital of the carbocation. Sharing of electron density from adjacent C — H or C — C sigma bonds with the carbocation p orbital delocalizes the positive charge. •

Any time a charge can be dispersed or delocalized by hyperconjugation, inductive effects, or resonance, a system w ill be stabilized.

Figure 6.7 shows a stylized representation of hyperconjugation between a sigma bonding orbital and an adjacent carbocation p orbital. Tertiary carbocations have three carbons with C — H bonds (or, depending on the spe­ cific example, C — C bonds instead of C — H) adjacent to the carbocation that can overlap partially with the vacant p orbital. Secondary carbocations have only two adjacent carbons with C — H or C — C bonds to overlap with the carbocation; hence, the possibility for hyper­ conjugation is less and the secondary carbocation is less stable. Primary carbocations have only one adjacent carbon from which to derive hyperconjugative stabilization, and so they are even less stable. A methyl carbocation has no possibility for hyperconjugation, and it is the least stable of all in this series. The following are specific examples: «+

«+

CH3 +

is m o re

3

ns+ C

s+C H 3 3

CH3

s+

fert-Butyl c a tio n (3°) (m o s t sta b le )

s ta b le than

CH3 + 3 C s+

«+ch 3

H

is m o re

h

Iso p ro p y l ca tio n ( 2 °)

s ta b le th a n

C 8+

s+CH

H I

is m o re

H

C H 3

s ta b le than

Ethyl c a tio n ( 1 °)

H

H

M ethyl ca tio n (le a s t sta b le )

In summary: •

The relative stability of carbocations is 3° > 2° > 1° > methyl.

This trend is also readily seen in electrostatic potential maps for these carbocations (Fig. 6 .8 ).

Figure 6.8 Maps of electrostatic potential for (a) tert-butyl (3°), (b) isopropyl (2°), (c) ethyl (1°), and (d) methyl carbocations show the trend from greater to lesser delocalization (stabilization) of the positive charge in th ese structures. Less blue color indicates greater delocalization of the positive charge. (The structures are m apped on the sam e scale of electrostatic potential to allow direct comparison.)

251

6.12 The Stereochemistry o f Sn 1 Reactions

C ftli/û r l D m

A A

R eview P roblem 6 .7

Rank the following carbocations in order of increasing stability:

6.12 The Stereochem istry o f SN1 Reactions Because the carbocation formed in the first step of an SN1 reaction has a trigonal planar structure (Section 6.11A), when it reacts with a nucleophile, it may do so from either the front side or the back side (see below). W ith the tert-butyl cation this makes no difference; since the tert-butyl group is not a chirality center, the same product is formed by either mode of attack. (Convince yourself of this result by examining models.) CH3 JL

C H O 2

U V /7CH 3 CH 3 3

CH3 b a c k -s id e a tta c k a tta c k

,, ^

H O

2

_ ,!+

C

/ '~ \ *—

A H3C € H 3

CH3 ^ ,,

¡O H

, 1,

fro n t-s id e

2

---------- ;— * a tta c k a tta c k

, C^

u

h 3c 3

oh 2

ch3

2

Sam e product

With some cations, however, stereoisomeric products arise from the two reaction possi­ bilities . We shall study this point next.

6.12A Reactions That Involve Racemization A reaction that transforms an optically active compound into a racemic form is said to pro­ ceed with racem ization. If the original compound loses all of its optical activity in the course of the reaction, chemists describe the reaction as having taken place with complete racemization. If the original compound loses only part of its optical activity, as would be the case if an enantiomer were only partially converted to a racemic form, then chemists describe this as proceeding with partial racemization. •

Racemization takes place whenever the reaction causes chiral molecules to be con­ verted to an achiral intermediate.

Examples of this type of reaction are SN1 reactions in which the leaving group departs from a chirality center. These reactions almost always result in extensive and sometimes complete racemization. For example, heating optically active (S)-3-bromo-3-methylhexane

252

Chapter 6

Ionic Reactions

with aqueous acetone results in the formation of 3-methyl-3-hexanol as a mixture of 50% (R) and 50% (S). CH 3 CH 2 CH 2

CH2CH2CH3

CH3CH2CH2

C— B r H 3 C"" i CH3CH2

h 2o

C— O H H3 C"" i CH3CH2

a c e to n e

/ HO

H Br

i'"C H 3 CH2CH3

50%

50%

(S )-3 -B ro m o -3 -

(S ) - 3 - M e t h y l-

( ff ) - 3 - M e th y l-

m e th y lh e x a n e

3 -h e x a n o l

3 -h e x a n o l

( o p t ic a lly a c tiv e )

( o p t ic a lly in a c tiv e , a r a c e m ic fo r m )

The reason: The SN1 reaction proceeds through the formation of an intermediate car­ bocation and the carbocation, because of its trigonal planar configuration, is achiral. It reacts with water at equal rates from either side to form the enantiomers of 3-methyl-3-hexanol in equal amounts.

A MECHANISM FOR THE REACTION The Stereochemistry of an SN1 Reaction R EA C TIO N ch 2c h 2ch 3

ch2ch2ch3 H3 C""y h 3 c h 2c

h

2o

B r:

a c e to n e (co so lv e n t)

ch 2ch 2c h 3

H3 C'""^ OH

h 3 c h 2c

HO

L 2 2 Y '" c h 3 ch2ch3

3

H Br

M E C H A N IS M

Step 1

CH 2 CH 2 CH 3 H3CX H3 c « " y h 3 c h 2C

CH 2 CH 2 CH 3 B.r:

B r:

h3 c h 2 c ^

Departure of the leaving group (assisted by hydrogen bonding with solvent) leads to the carbocation.

Step 2

H3 C""'^ h 3 c h 2c

c h 2 c h 2c h 3

H

A racemic mixture of protonated alcohols results.

-C H 2 CH 2 CH 3 > 0 : H

h 3 c h 2c

CH 2 CH 2 CH 3

A t t a c k a t e it h e r fa c e :

The carbocation is an achiral intermediate. Because both faces of the carbocation are the same, the nucleophile can bond with either face to form a mixture of stereoisomers.

(mechanism continues on the next page)

253

6.12 The Stereochemistry o f Sn1 Reactions

Step 3

H

H

.

/ :°:

°o H3 o''"’y

50%

\

H3 o » y

H

O H 2O H2O H3

H 3 OH2O 'OH 2 OH 2 OH3

H

H3 CH 2 C H H 3C

OH 2 OH 2 OH 3

H 3 O H 2O

50% H

A d d itio n a l s o lv e n t m o le c u le s (w a te r)

T h e p ro d u ct is a ra c e m ic m ixtu re.

d e p ro to n a te th e a lk y lo x o n iu m ion.

The Sn1 reaction of (S)-3-bromo-3-methylhexane proceeds with racemization because the intermediate carbocation is achiral and attack by the nucleophile can occur from either side.

Keeping in mind that carbocations have a trigonal planar structure, (a ) write a structure for the carbocation intermediate and (b ) write structures for the alcohol (or alcohols) that you would expect from the following reaction:

R eview P roblem 6.8

6.12B Solvolysis The SN1 reaction of an alkyl halide with water is an example of solvolysis. A solvolysis reaction is a nucleophilic substitution in which the nucleophile is a molecule o f the solvent (solvent + lysis: cleavage by the solvent). Since the solvent in this instance is water, we could also call the reaction a hydrolysis. If the reaction had taken place in methanol, we would call it a methanolysis. Examples of Solvolysis (CH 3 )3 C — B r

+

H 2O

(CH 3 )3 C — C l

+

C H 3O H

----- :

(CH 3 )3 C — O H

+ H Br

----- : (CH 3 )3 C — O C H 3 +

H Cl

Solved Problem 6.5 What product(s) would you expect from the following solvolysis? Br c h 3o h

254

Chapter 6

Ionic Reactions

STRATEGY AND ANSWER We observe that this cyclohexyl bromide is tertiary, and therefore in methanol it should lose a bromide ion to form a tertiary carbocation. Because the carbocation is trigonal planar at the positive carbon, it can react with a solvent molecule (methanol) to form two products.

Br

CH 3 OH 2

hoch3

from

(b)

OCH 3

c h 3o h

2

O CH 3

R eview P roblem 6 .9

What product(s) would you expect from the methanolysis of the iodocyclohexane deriva­ tive given as the reactant in Review Problem 6 .8 ?

6.13 Factors A ffecting the Rates o f SN1 and SN2 Reactions Now that we have an understanding of the mechanisms of SN2 and SN1 reactions, our next task is to explain why chloromethane reacts by an SN2 mechanism and tert-butyl chloride by an SN1 mechanism. We would also like to be able to predict which pathway— SN1 or Sn 2— would be followed by the reaction of any alkyl halide with any nucleophile under varying conditions. The answer to this kind of question is to be found in the relative rates o f the reactions that occur. If a given alkyl halide and nucleophile react rapidly by an SN2 mechanism but slowly by an SN1 mechanism under a given set of conditions, then an SN2 pathway w ill be followed by most of the molecules. On the other hand, another alkyl halide and another nucleophile may react very slowly (or not at all) by an SN2 pathway. If they react rapidly by an SN1 mechanism, then the reactants w ill follow an SN1 pathway. •

A number of factors affect the relative rates of SN1 and SN2 reactions. The most important factors are 1

.

the structure of the substrate,

2

.

the concentration and reactivity of the nucleophile (for bimolecular reactions only),

3.

the effect of the solvent, and

4.

the nature of the leaving group.

6.13A The Effect of the Structure of the Substrate S n 2 R eactions Sn2 reactions:

Simple alkyl halides show the following general order of reactivity in

Methyl > primary > secondary > > (tertiary— unreactive)

255

6.13 Factors Affecting the Rates o f Sn1 and Sn2 Reactions

M e t h y l h a lid e s r e a c t m o s t r a p id ly a n d t e r t ia r y h a lid e s r e a c t s o s lo w ly as to b e u n r e a c tiv e b y t h e S N 2 m e c h a n i s m . T a b le 6 . 4 g iv e s t h e r e l a t i v e r a t e s o f t y p i c a l S N 2 r e a c t io n s .

R elative Rates o f Reactions o f A lkyl H alides in SN2 Reactions Sub Csotitu m peonutn d

Methyl 1° 2° N eopentyl 3°

A p p r o x im a te R e la tiv e R ate

CH3X c h 3 c h 2x (CH3)2CHX (CH3 )3 CCH2X (CH3 )3 CX

30 1 0.03 0.00001 ~0

N e o p e n t y l h a lid e s , e v e n th o u g h th e y a re p r im a r y h a lid e s , a re v e r y u n r e a c tiv e :

CH3 C -X

C H /" i " CH3

C H

H

A n e o p e n ty l h a lid e T h e im p o r t a n t f a c t o r b e h in d t h is o r d e r o f r e a c t iv i t y is a s te r ic e ff e c t , a n d i n t h is c a s e , s t e r i c h in d r a n c e . •

A

steric effect

is a n e f f e c t o n th e r e la t iv e r a te s c a u s e d b y th e s p a c e - f illin g p r o p e r ­

t i e s o f t h o s e p a r t s o f a m o l e c u l e a t t a c h e d a t o r n e a r t h e r e a c t i n g s ite . •

Steric hind ran ce

is w h e n t h e s p a t i a l a r r a n g e m e n t o f a t o m s o r g r o u p s a t o r n e a r a

r e a c t i n g s i t e o f a m o l e c u l e h i n d e r s o r r e t a r d s a r e a c t io n . F o r p a r t ic le s ( m o le c u le s a n d io n s ) t o r e a c t, t h e ir r e a c t iv e c e n te r s m u s t b e a b le to c o m e w i t h i n b o n d i n g d is t a n c e o f e a c h o t h e r . A l t h o u g h m o s t m o l e c u l e s a r e r e a s o n a b l y f l e x i b l e , v e r y l a r g e a n d b u l k y g r o u p s c a n o f t e n h i n d e r t h e f o r m a t i o n o f t h e r e q u i r e d t r a n s i t i o n s ta te . I n s o m e c a s e s t h e y c a n p r e v e n t i t s f o r m a t i o n a lt o g e t h e r . A n S N 2 r e a c t i o n r e q u i r e s a n a p p r o a c h b y t h e n u c l e o p h i l e t o a d is t a n c e w i t h i n t h e b o n d ­ i n g r a n g e o f t h e c a r b o n a t o m b e a r i n g t h e l e a v i n g g r o u p . B e c a u s e o f t h i s , b u l k y s u b s t it u e n t s o n o r n e a r t h a t c a r b o n a to m h a v e a d r a m a t ic in h i b it in g e f f e c t ( F ig . 6 .9 ) . T h e y c a u s e th e f r e e e n e r g y o f t h e r e q u i r e d t r a n s i t i o n s t a t e t o b e i n c r e a s e d a n d , c o n s e q u e n t ly , t h e y in c r e a s e

H e lp f u l H i n t

th e f r e e e n e r g y o f a c t iv a t io n f o r th e r e a c t io n . O f th e s im p le a l k y l h a lid e s , m e t h y l h a lid e s

You can best appreciate the steric effects in these structures by building models.

r e a c t m o s t r a p i d l y i n S N 2 r e a c t io n s b e c a u s e o n l y t h r e e s m a l l h y d r o g e n a t o m s i n t e r f e r e w i t h th e a p p r o a c h in g n u c le o p h ile . N e o p e n t y l a n d t e r t ia r y h a lid e s a re th e le a s t r e a c t iv e b e c a u s e b u l k y g r o u p s p r e s e n t a s t r o n g h in d r a n c e t o t h e a p p r o a c h i n g n u c l e o p h i l e . ( T e r t i a r y s u b s t r a te s , f o r a ll p r a c t ic a l p u rp o s e s , d o n o t r e a c t b y a n S N 2 m e c h a n is m .)

\ / H -C

Nu

^ C- X

Nu

_ C- X

Nu

H I ^H

^ C- X H— C H

Methyl

1°

(30)

(1)

2

'h °

(0.03) Relative rate

Neopentyl (0 .0 0 0 0 1 )

■0 )

Figure 6.9 Steric effects and relative rates in the Sn2 reaction.

256

Chapter 6

Ionic Reactions

S o lv e d P ro b le m 6 .6 Rank the following alkyl bromides in order of decreasing reactivity (from fastest to slowest) as a substrate in an Sn2 reaction.

A

B

C

D

STRATEGY AND ANSWER We examine the carbon bearing the leaving group in each instance to assess the steric hindrance to an Sn2 reaction at that carbon. In C it is 3°; therefore, three groups would hinder the approach of a nucleophile, so this alkyl bromide would react most slowly. In D the carbon bearing the leaving group is 2° (two groups hinder the approach of the nucleophile), while in both A and B it is 1° (one group hinders the nucleophile’s approach). Therefore, D would react faster than C, but slower than either A or B. But, what about A and B? They are both 1° alkyl bromides, but B has a methyl group on the carbon adjacent to the one bearing the bromine, which would provide hindrance to the approaching nucleophile that would not be present in A . The order of reactivity, therefore, is A > B > D > > C.

_______H e lp f u l H i n t The primary factor that determines the reactivity of organic substrates in an S^1 reaction is the relative stability of the carbocation that is formed.

SN1 Reactions •

Except for those reactions that take place in strong acids, which we shall study later, th e o n ly o r g a n ic c o m p o u n d s t h a t u n d e r g o re a c tio n b y a n S ^ 1 p a th a t a re a s o n a b le r a t e a r e th ose th at a re capable o f fo rm in g relatively stable carbocations.

O f the simple alkyl halides that we have studied so far, this means (for all practical pur­ poses) that only tertiary halides react by an Sn1 mechanism. (Later we shall see that certain organic halides, called allylic halides and benzylic halides, can also react by an Sn1 mecha­ nism because they can form relatively stable carbocations; see Sections 13.4 and 15.15.) Tertiary carbocations are stabilized because sigma bonds at three adjacent carbons contribute electron density to the carbocation p orbital by hyperconjugation (Section 6.11B). Secondary and primary carbocations have less stabilization by hyperconjugation. A methyl carbocation has no stabilization. Formation of a relatively stable carbocation is important in an Sn1 reac­ tion because it means that the free energy of activation for the slow step of the reaction (e.g., R— L ----- > R+ + L_) w ill be low enough for the reaction to take place at a reasonable rate.

R eview P roblem 6 .1 0

Which o f the following alkyl halides is most likely to undergo substitution by an Sn 1 mechanism?

'B r Br

(c)

The H am m o n d -L effler Postulate If you review the free-energy diagrams that accom­ pany the mechanism for the SN1 reaction of tert-butyl chloride and water (Section 6.10), you w ill see that step 1 , the ionization of the leaving group to form the carbocation, is uphill in terms o f free energy (AG° for this step is positive). It is also uphill in terms of enthalpy (AH° is also positive), and, therefore, this step is endothermic. According to the H a m m o n d - L e f f le r p o s tu la te , th e tra n s itio n -s ta te s tr u c tu r e f o r a step t h a t is u p h ill in e n e rg y s h o u ld sho w

257

6.13 Factors Affecting the Rates o f Sn 1 and Sn 2 Reactions

a strong resem blance to the structure o f the product o f that step. Since the product of this step (actually an intermediate in the overall reaction) is a carbocation, any factor that sta­ bilizes the carbocation— such as dispersal o f the positive charge by electron-releasing groups— should also stabilize the transition state in which the positive charge is developing. Ionization o f the Leaving Group CH3

CH 3 CH,— C— C l

H 2O

CH

CH 3 R ea cta n t

-C«+

CH -C l5

h2o

CH,— C -1

\

CH

C l-

CH3

T ran sition s ta te

P ro d u c t o f s te p

R e s e m b le s p r o d u c t o f s te p b e c a u s e A G ° is p o s itiv e

S ta b iliz e d b y th re e e le c tro n - re le a s in g g ro u p s

A methyl, primary, or secondary alkyl halide would have to ionize to form a methyl, primary, or secondary carbocation to react by an SN1 mechanism. These carbocations, how ­ ever, are much higher in energy than a tertiary carbocation, and the transition states lead­ ing to these carbocations are even higher in energy. The activation energy for an SN1 reaction o f a sim ple methyl, primary, or secondary halide, consequently, is so large (therefore the reaction is so slow) that, for all practical purposes, an SN1 reaction with a methyl, primary, or secondary halide does not com pete with the corresponding SN2 reaction. The Hammond-Leffler postulate is quite general and can be better understood through consideration o f Fig. 6.10. One w ay that the postulate can be stated is to say that the struc­ ture o f a transition state resem bles the stable species that is nearest it in free energy. For example, in a highly endergonic step (blue curve) the transition state lies close to the prod­ ucts in free energy, and w e assume, therefore, that it resem bles the products o f that step in structure. Conversely, in a highly exergonic step (red curve) the transition state lies close to the reactants in free energy, and w e assume it resem bles the reactants in structure as well. The great value o f the Hammond-Leffler postulate is that it gives us an intuitive way o f visualizing those important, but fleeting, species that w e call transition states. We shall make use o f it in many future discussions.

t

Hi9h|y exergonic step

Transition state

Transition state

Reactants

Products

Figure 6.10

T h e t r a n s it io n s ta te f o r a h ig h ly

e x e r g o n ic s te p ( r e d c u rv e ) lie s c lo s e t o a n d re s e m b le s t h e r e a c ta n ts . T h e t r a n s it io n s ta te f o r an e n d e r g o n ic s te p ( b lu e c u rv e ) lie s c lo s e

Highly Reactants endergonic — step

R eaction coordinate

t o a n d r e s e m b le s t h e p r o d u c ts o f a r e a c tio n . ( R e p r in te d w it h p e r m is s io n o f T h e M c G r a w - H ill C o m p a n ie s f r o m P ry o r, Free Radicals, p . 1 5 6 , C o p y r ig h t 1 9 6 6 .)

The relative rates o f ethanolysis o f four primary alkyl halides are as follows: CH 3 CH 2 Br, 1.0; CH 3 CH 2 CH 2 Br, 0.28; (CH 3 )2 CHCH 2 Br, 0.030; (CH 3 )3 CCH 2 Br, 0.00000042. (a) Is each of these reactions likely to be SN1 or SN2? (b) Provide an explanation for the relative reactivities that are observed.

R e v ie w P ro b le m 6. 11

258

Chapter 6

Ionic Reactions

6.13B The Effect of the Concentration and Strength of the Nucleophile Since the nucleophile does not participate in the rate-determining step of an SN1 reaction, the rates of SN1 reactions are unaffected by either the concentration or the identity of the nucleophile. The rates of SN2 reactions, however, depend on both the concentration and the identity of the attacking nucleophile. We saw in Section 6.5 how increasing the con­ centration of the nucleophile increases the rate of an SN2 reaction. We can now examine how the rate of an SN2 reaction depends on the identity of the nucleophile. •

The relative strength of a nucleophile (its nucleophilicity) is measured in terms of the relative rate of its SN2 reaction with a given substrate.

A good nucleophile is one that reacts rapidly in an SN2 reaction with a given substrate. A poor nucleophile is one that reacts slowly in an SN2 reaction with the same substrate under comparable reaction conditions. (As mentioned above, we cannot compare nucleophilicities with regard to SN1 reactions because the nucleophile does not participate in the rate-determining step of an SN1 reaction.) Methoxide anion, for example, is a good nucleophile for a substitution reaction with iodomethane. It reacts rapidly by an SN2 mechanism to form dimethyl ether: C H 3O -

+ C H 3I —

CH3O CH3

+ I-

Methanol, on the other hand, is a poor nucleophile for reaction with iodomethane. Under comparable conditions it reacts very slowly. It is not a sufficiently powerful Lewis base (i.e., nucleophile) to cause displacement of the iodide leaving group at a significant rate: c h 3o h

+

c h 3i

very slow>

c h 3o c h 3 + I ­

H

The relative strengths of nucleophiles can be correlated with three structural features: 1. A negatively charged nucleophile is always a more reactive nucleophile than its con­ jugate acid. Thus HO- is a better nucleophile than H2O and RO- is better than ROH. 2. In a group of nucleophiles in which the nucleophilic atom is the same, nucleophilicities parallel basicities. Oxygen compounds, for example, show the follow­ ing order of reactivity: RO-

> H O - > > R C O 2 - > R O H > H 2O

This is also their order of basicity. An alkoxide ion (R O - ) is a slightly stronger base than a hydroxide ion (H O - ), a hydroxide ion is a much stronger base than a carboxylate ion (RCO2- ), and so on. 3. W hen the nucleophilic atoms are different, nucleophilicities may not parallel basicities. For example, in protic solvents HS- , C N - , and I- are all weaker bases than HO - , yet they are stronger nucleophiles than HO- . HS- > CN - > I- > HON u cle o p h ilic ity versus B asicity W hile nucleophilicity and basicity are related, they are not measured in the same way. Basicity, as expressed by p ^ a, is measured by the posi­ tion o f an equilibrium involving an electron pair donor (base), a proton, the conjugate acid, and the conjugate base. Nucleophilicity is measured by relative rates o f reaction, by how rapidly an electron pair donor reacts at an atom (usually carbon) bearing a leaving group. For example, the hydroxide ion (O H - ) is a stronger base than a cyanide ion (C N - ); at equi­ librium it has the greater affinity for a proton (the p ^ a of H2O is ~ 1 6 , while the p ^ a of HCN is —10). Nevertheless, cyanide ion is a stronger nucleophile; it reacts more rapidly with a carbon bearing a leaving group than does hydroxide ion.

R eview P roblem 6 .1 2

Rank the following in terms of decreasing nucleophilicity: CH 3 CO2-

CH3OH

CH 3 O -

CH 3 CO2H

CN-

6.13 Factors Affecting the Rates o f Sn1 and Sn2 Reactions

259

6.13C Solvent Effects on SN2 Reactions: Polar Protic and Aprotic Solvents A molecule of a solvent such as water or an alcohol— called a p r o tic s o lv e n t (Section 3.12)— has a hydrogen atom attached to a strongly electronegative element (oxygen). Molecules of protic solvents can, therefore, form hydrogen bonds to nucleophiles in the following way.

O-

H

1 =Q : \

H “

H -. ; X ’ v :O — H ‘ H H / H 'O '

H

/ •H — O: •H >O :

M o le c u le s o f th e protic s o lv e n t, w ater, s o lv a te a h a lid e ion b y form in g h y d r o g e n b o n d s to it.

a

*

'*

* .>

/ H

t m

• Hydrogen bonding encumbers a nucleophile and hinders its reactivity in a substitu­ tion reaction. For a strongly solvated nucleophile to react, it must shed some of its solvent molecules so that it can approach the carbon of the substrate that bears the leaving group. This is one type of important s o lv e n t e ffe c t in nucleophilic reactions. •

Hydrogen bonds to a small nucleophilic atom are stronger than those to larger nucleophilic atoms among elements in the same group (column) o f the periodic table.

For example, fluoride anion is more strongly solvated than the other halides because it is the smallest halide anion and its charge is the most concentrated. Hence, in a protic sol­ vent fluoride is not as effective a nucleophile as the other halide anions. Iodide is the largest halide anion and it is the most weakly solvated in a protic solvent; hence, it is the strongest nucleophile among the halide anions. •

In a protic solvent, the general trend in nucleophilicity among the halide anions is as follows: I- > B r - > C l - > F -

H alide n u c le o p h ilic ity in p rotic s o lv e n t s

The same effect holds true when we compare sulfur nucleophiles with oxygen nucleophiles. Sulfur atoms are larger than oxygen atoms and hence they are not solvated as strongly in a protic solvent. Thus, thiols ( R — S H ) are stronger nucleophiles than alcohols, and R S anions are better nucleophiles than R O - anions. The greater reactivity of nucleophiles with large nucleophilic atoms is not entirely related to solvation. Larger atoms have greater p o l a r i z a b ili t y (their electron clouds are more easily distorted); therefore, a larger nucleophilic atom can donate a greater degree of electron density to the substrate than a smaller nucleophile whose electrons are more tightly held. The relative nucleophilicities of some common nucleophiles in protic solvents are as follows: S H - > C N - > I - > O H - > N 3- > B r - > C H 3 C O 2- > C l - > F - > H 2O

R e la tiv e n u c le o p h ilic ity in p rotic s o lv e n t s

Review Problem 6.13

Rank the following in terms of decreasing nucleophilicity: C H 3 C O 2-

ch

3

o -

c h 3s -

c h 3s h

c h 3o h

260

Chapter 6

Ionic Reactions

P olar A p ro tic Solvents •

do not have a hydrogen atom bonded to an electronegative atom, and therefore do not hinder nucleophiles through hydrogen bonding.

A p r o t ic s o lv e n ts

A number o f p o la r a p r o t ic s o lv e n ts have com e into wide use by chem ists because they are especially useful in SN2 reactions. Several exam ples are the following: •Q-

H

•Ü-

•Q'

/C H ,

•Q '

C s

*N CH

ch,

CH

CH

/C H , N

(C H ^ N -P -^ C H ^ =N ( C H , ) 2

CH

DMF (W ,W -dim ethylform am ide)

DMSO (d im eth yl su lfo x id e )

DMA (d im e th y la c e ta m id e )

HMPA (h e x a m e th y lp h o s p h o r a m id e )

A ll o f these solvents (DMF, DM SO, D M A, and HMPA) dissolve ionic compounds, and they solvate cations very w ell. They do so in the same way that protic solvents solvate cations: by orienting their negative ends around the cation and by donating unshared elec­ tron pairs to vacant orbitals o f the cation:

However, because they cannot form hydrogen bonds and because their positive centers are w ell shielded by steric effects from any interaction with anions, aprotic solvents do not solvate anions to an y appreciable extent. In these solvents anions are unencumbered by a layer o f solvent m olecules and they are therefore poorly stabilized by solvation. These “naked” anions are highly reactive both as bases and nucleophiles. In DM SO, for exam ­ ple, the relative order o f reactivity o f halide ions is opposite to that in protic solvents, and it follow s the same trend as their relative basicity: F - > C l - > B r - > I-

H alide n u c le o p h ilic ity in a p ro tic s o lv e n t s

H e lp f u l H i n t Polar aprotic solvents increase S n2 rates.

•

The rates o f SN2 reactions generally are vastly increased when they are carried out in p o la r aprotic solvents. The increase in rate can be as large as a m illionfold.

R eview P roblem 6 .1 4

Classify the follow ing solvents as being protic or aprotic: formic acid, HCO 2 H; acetone, CH 3 COCH3; acetonitrile, CH 3 C # N ; formamide, HCONH2; sulfur dioxide, S O 2; ammo­ nia, NH3; trimethylamine, N(CH3)3; ethylene glycol, HOCH 2 CH 2 OH.

R eview P roblem 6 .1 5

Would you expect the reaction o f propyl bromide with sodium cyanide (NaCN), that is, CH 3 CH 2 CH2Br + N a C N ----- > CH 3 CH 2 CH2CN + NaBr to occur faster in DM F or in ethanol? Explain your answer.

261

6.13 Factors Affecting the Rates o f Sn1 and Sn2 Reactions

Which would you expect to be the stronger nucleophile in a polar aprotic solvent? ( a ) CH 3 CO 2 - or CH 3 O~; (b ) H2O or H2 S; (c ) (CH3)3P or (CH3)3N

R eview P roblem 6.16

6.13D Solvent Effects on Sn1 Reactions: The Ionizing Ability of the Solvent •

Use of a p o la r p r o tic s o lv e n t w ill greatly increase the rate of carbocation forma­ tion of an alkyl halide in any SN1 reaction because of its ability to solvate cations and anions so effectively.

H e lp f u l H i n t Polar protic solvents favor S n1 reactions.

Solvation stabilizes the transition state leading to the intermediate carbocation and halide ion more than it does the reactants; thus the free energy of activation is lower. The transi­ tion state for this endothermic step is one in which separated charges are developing, and thus it resembles the ions that are ultimately produced: CH3 CH — C — Cl CH 3 R e a cta n t

CH3 H 2O

CH

-C

Í C l8

CH3 h 2o

CH T ran sition s ta te S e p a ra te d c h a rg e s a re d e v e lo p in g .

3

CH 3— C 4

C l-

CH3 C arb o ca tio n in te rm e d ia te

A rough indication of a solvent’s polarity is a quantity called the dielectric constant. The dielectric constant is a measure of the solvent’s ability to insulate opposite charges (or separate ions) from each other. Electrostatic attractions and repulsions between ions are smaller in solvents with higher dielectric constants. Table 6.5 gives the dielectric constants of some common solvents. D ielectric C onstants o f C om m on Solvents

Increasing solvent polarity

S o lv e n t

F orm u la

D ie le c tr ic C o n s ta n t

W ater Formic acid Dimethyl sulfoxide (DMSO) N,N -Dim ethylform am ide (DMF) Acetonitrile M ethanol H exam ethylphosphoram ide (HMPA) Ethanol A cetone A cetic acid

H2 O HCO2 H CH3 SOCH 3 HCON(CH3 ) 2 c h 3c # n CH3 OH [(CH3)2N]3P=O CH3 CH2 OH CH3 COCH3 CH3 CO 2 H

SG 59 49 37 36 SS SG 24 21 6

Water is the most effective solvent for promoting ionization, but most organic compounds do not dissolve appreciably in water. They usually dissolve, however, in alcohols, and quite often mixed solvents are used. Methanol-water and ethanol-water are common mixed sol­ vents for nucleophilic substitution reactions.

When tert-butyl bromide undergoes solvolysis in a mixture of methanol and water, the rate of solvolysis (measured by the rate at which bromide ions form in the mixture) increases when the percentage of water in the mixture is increased. (a ) Explain this occurrence. (b ) Provide an explanation for the observation that the rate of the SN2 reaction of ethyl chlo­ ride with potassium iodide in methanol and water decreases when the percentage of water in the mixture is increased.

Review Problem 6.17

262

Chapter 6

Ionic Reactions

6.13E T h e N a tu re o f th e Leaving G ro u p •

H e lp f u l H i n t Good leaving groups are weak bases.

Leaving groups depart with the electron pair that was used to bond them to the substrate.

The best leaving groups are those that become either a relatively stable anion or a neutral molecule when they depart. First, let us consider leaving groups that become anions when they separate from the substrate. Because weak bases stabilize a negative charge effectively, leaving groups that become weak bases are good leaving groups. •

In general, the best leaving groups are those that can be classified as weak bases after they depart.

The reason that stabilization of the negative charge is important can be understood by considering the structure of the transition states. In either an SN1 or SN2 reaction the leav­ ing group begins to acquire a negative charge as the transition state is reached: SN1 Reaction (Rate-Limiting Step) s- t

^ .« + ^ C -

X

■V

+ X-

/ T ra n s itio n sta te

Sn2 Reaction v

Nu

s-

A

Nu-

" -» C — X

s­

C

X

Nu— C

X-

\

/ T ra n s itio n sta te

Stabilization of this developing negative charge at the leaving group stabilizes the transi­ tion state (lowers its free energy); this lowers the free energy of activation and thereby increases the rate of the reaction. •

Among the halogens, an iodide ion is the best leaving group and a fluoride ion is the poorest: I - > B r - > C l- > > F -

The order is the opposite of the basicity: F - > > C l- > B r

> r

Other weak bases that are good leaving groups, which we shall study later, are alkanesulfonate ions, alkyl sulfate ions, and the p -toluenesulfonate ion: O O — S— R

O

O

O — S— O — R

-O — S -

\ O A n a lk a n e s u lfo n a te ion

O

/

C H

0

An alk yl s u lfa te ion

p -T o lu e n e s u lfo n a te ion

These anions are all the conjugate bases of very strong acids. The trifluoromethanesulfonate ion (C F 3 SO3- , commonly called the triflate ion) is one of the best leaving groups known to chemists. It is the conjugate base of CF 3 SO 3 H, an exceedingly strong acid (pA"a ------5 to - 6 ): O -O — S— CF3 O T rifla te ion (a “s u p e r” le a vin g g ro u p )

263

6.13 Factors Affecting the Rates o f Sn1 and Sn2 Reactions

•

Strongly basic ions rarely act as leaving groups.

The hydroxide ion, for example, is a strong base and thus reactions like the follow ing do not take place:

f\

Nu *-

R— OH

R — Nu + O H T h is re a c tio n d o e s not ta k e p la c e b e c a u s e the le a v in g g ro u p is a s tro n g ly b a s ic h y d ro x id e ion.

However, when an alcohol is dissolved in a strong acid, it can undergo substitution by a nucleophile. Because the acid protonates the — OH group o f the alcohol, the leaving group no longer needs to be a hydroxide ion; it is now a m olecule o f water— a much weaker base than a hydroxide ion and a good leaving group:

Nu-

R

A

I

A l

OH H

R — Nu

H2 O

T h is re a c tio n ta k e s p lace b e c a u s e th e le a vin g g ro u p is a w e a k base.

List the follow ing compounds in order o f decreasing reactivity toward CH3O in an SN2 reaction carried out in CH 3 OH: CH 3 F, CH 3 Cl, CH 3 Br, CH 3 I, CH 3 O SO 2 CF3, 1 4 CH 3 OH.

•

R eview P roblem ó.1S

Very powerful bases such as hydride ions (H:- ) and alkanide ions (R:- ) virtually never act as leaving groups.

Therefore, reactions such as the follow ing are not feasible: Nu

P 1H , ---------» CH,CH2— Nu CH3 CH2—

Nu :-

H

T h e s e are n ot leavin g g ro up s.

CH,— C H ,---------> CH,— Nu + CH,

Rem em ber: The best leaving groups are weak bases after they depart.

S o lv e d P ro b le m 6 .7 Explain why the follow ing reaction is not feasible as a synthesis o f butyl iodide. Na

!

„OH

h 2o

\

/I

Na+ ÜH

STRATEGY AND ANSWER The strongly basic OH- ion (hydroxide ion) virtually never acts as a leaving group, something this reaction would require. This reaction would be feasible under acidic conditions, in which case the leaving group would be a water m olecule.

S u m m a ry o f Sn 1 v e rsu s SN2 R ea ctio n s S n 1: The Follow ing C onditions Favor an S n 1 Reaction: 1. A substrate that can form a relatively stable carbocation (such as a substrate with a leaving group at a tertiary position) 2. A relatively weak nucleophile 3. A polar, protic solvent

H e lp f u l H i n t S n1 versus Sn 2

264

Chapter 6

Ionic Reactions

The SN1 mechanism is, therefore, important in solvolysis reactions of tertiary alkyl halides, especially when the solvent is highly polar. In a solvolysis reaction the nucleophile is weak because it is a neutral molecule (of the polar protic solvent) rather than an anion.

Sn 2: The Follow ing C onditions Favor an SN2 Reaction: 1.

A substrate with a relatively unhindered leaving group (such as a methyl, primary, or secondary alkyl halide). The order of reactivity is R C H 3— X

>

R — C H 2— X

M ethyl

>

1

>

R— CH— X

>

°

Tertiary halides do not react by an SN2 mechanism. 2.

A strong nucleophile (usually negatively charged)

3.

High concentration of the nucleophile

4.

A polar, aprotic solvent

The trend in reaction rate among halogens as the leaving group is the same in SN1 and SN2 reactions: R -

I > R -

Br > R -

Cl

SN1 or SN2

Because alkyl fluorides react so slowly, they are seldom used in nucleophilic substitution reactions. These factors are summarized in Table 6 .6 .

Factors Favoring SN1 versus SN2 Reactions F a cto r

Sn 2

Sn1

S ubstrate

3° (requires form ation of a relatively stab le carbocation)

Methyl > 1° > 2° (requires unhindered substrate)

Nucleophile

W eak Lewis base, neutral m olecule, nucleophile may b e th e solvent (solvolysis)

Strong Lewis base, rate increased by high concentration of nucleophile

Solvent

Polar protic (e.g., alcohols, water)

Polar aprotic (e.g., DMF, DMSO)

Leaving group

I > B r > C l > F for both SN1 and SN2 (the w eaker th e b ase after th e gro u p d ep arts, th e b e tte r th e leaving group)

6.14 O rganic Synthesis: Functional Group Transformations Using Sn2 Reactions SN2 reactions are highly useful in organic synthesis because they enable us to convert one functional group into another— a process that is called a functional group transform a­ tion or a functional group interconversion. W ith the SN2 reactions shown in Fig. 6.11, methyl, primary, or secondary alkyl halides can be transformed into alcohols, ethers, thi­ ols, thioethers, nitriles, esters, and so on. (Note: The use of the prefix thio- in a name means that a sulfur atom has replaced an oxygen atom in the compound.) A lkyl chlorides and bromides are also easily converted to alkyl iodides by nucleophilic substitution reactions. R — Cl or R — Br

R — I ( + C l- or

B r-)

265

6.14 Organic Synthesis: Functional Group Transformations Using Sn2 Reactions

OHA lc o h o l

E th e r

T h io l

T h io e t h e r

N it r ile

E s te r

Q u a t e r n a r y a m m o n iu m h a lid e

A lk y l a z id e

Figure 6.11 Functional group interconversions of methyl, primary, and secondary alkyl halides using Sn2 reactions.

One other aspect of the SN2 reaction that is of great importance is stereochemistry (Section 6 .8 ). SN2 reactions always occur with inversion of configuration at the atom that bears the leaving group. This means that when we use SN2 reactions in syntheses we can be sure of the configuration of our product if we know the configuration of our reactant. For example, suppose we need a sample of the following nitrile with the (S) configuration: CH3 / 3 :N = C — C , ^ H

o h 2c h 3 (S )-2 -M e th y lb u ta n e n itrile

If we have available (R)-2-bromobutane, we can carry out the following synthesis: ch3 ,______ \

^

=N = C =- + ,, ,C— B r

ch3 S 2

N . ,> !^ ^ = cC

/

C.,, ^H

+ B r-

c h 2c h 3

c h 2c h 3

(S )-2 -M e th y lb u ta n e n itrile

(ff)-2 -B ro m o b u ta n e

Starting with (5)-2-bromobutane, outline syntheses of each of the following compounds: (a) (R)-CH 3 CHCH 2 CH 3 OCH 2 CH 3 (b) (R)-CH 3 CHCH 2 CH 3 occh3 O

(c) (R)-CH 3 CHCH 2 CH 3 SH (d) (R)-CH 3 CHCH 2 CH 3 sch3

R eview P roblem 6 .1 9

266

Chapter 6

Ionic Reactions

THE CHEMISTRY OF . . . B io lo g ic a l M e t h y l a t i o n : A B io lo g ic a l N u c l e o p h i l i c S u b s t i t u t i o n R e a c t i o n The cells of living organisms synthesize many of the com ­ pounds they need from smaller m olecules. Often these biosyntheses resem ble the syntheses organic chemists carry out in their laboratories. Let us exam ine one exam ­ ple now. Many reactions taking place in the cells of plants and ani­ mals involve the transfer of a methyl group from an amino acid called methionine to som e other com pound. That this

transfer takes place can be demonstrated experimentally by feeding a plant or animal methionine containing an isotopically labeled carbon atom (e.g., 13C or 1 4 C) in its methyl group. Later, other com pounds containing the "labeled" methyl group can be isolated from the organism. Som e of the com pounds that get their methyl groups from methio­ nine are the following. The isotopically labeled carbon atom is shown in green.

- o 2c c h c h 2c h 2s c h 3 Methionine

Choline is important in the transmission of nerve impulses, adrenaline causes blood pressure to increase, and nicotine is the compound contained in tobacco that makes smoking tobacco addictive. (In large d o ses nicotine is poi­ sonous.)

The transfer of the methyl group from methionine to these other com pounds d o es not take place directly. The actual methylating agent is not methionine; it is S-adenosylmethionine,* a compound that results when methionine reacts with adenosine triphosphate (ATP):

Triphosphate group O

O-

O-

Leaving O— P— O— P— O— P— OH group The sulfur atom Cc acts as a nucleophile. O O O - o 2c c h c h 2c h 2s c h 3 + c h 2 o

Adenine

NH'3

—

CH„

- o 2c c h c h 2c h 2— S ^ c h 2 JO.

Adenine

NH,

OOO­ I I I -O — P— O— P—O— P— OH

Il

O

O

I

O

I

Triphosphate ion

Methionine OH

OH ATP

OH OH S-Adenosylmethionine

* T h e p re fix S is a lo c a n t m ean in g “ o n th e s u lfu r a to m ” and s h o u ld n o t be c o nfu sed w ith th e (S) used to defin e a bsolute c o n fig u ra tio n . A n o th e r e xam p le o f th is k in d o f lo c a n t is N , m e an in g “ o n th e n itro g e n atom .”

267

6.14 Organic Synthesis: Functional Group Transformations Using Sn2 Reactions

This reaction is a nucleophilic substitution reaction. The nucleophilic atom is the sulfur atom of methionine. The leav­ ing group is the weakly basic triphosphate group of ATP. The product, S-adenosylmethionine, contains a methyl-sulfonium

S-Adenosylmethionine then acts as the substrate for other nucleophilic substitution reactions. In the biosynthesis of choline, for example, it transfers its methyl group to a nucle­ ophilic nitrogen atom of 2-(N,N-dimethylamino)ethanol:

group, CH3—S— .

CH,

+

CH■,— CH*2^* 2 CH2OH 3 1 N— 1 '2

3

-O^ 22CCHCH CH2 O lwl '22CH^S+— ^' '2

Adenine

NH,

CH,

2-(W,W-Dimethylamino)ethanol OH

OH

CH, CH,— N+— C H C H O H

+

I CH,

O^22CCHCH ^yi IUI 12^C1 H'2^S— ~ CH y '2 O

Adenine

NH

Choline OH

T hese reactions appear com plicated only b ecau se the structures of the nucleophiles and substrates are complex. Yet conceptually they are sim ple, and they illustrate many of the principles w e have encountered thus far in Chapter 6 . In them w e se e how nature m akes use of the high nucleophilicity of sulfur atom s. We also se e how a weakly basic group (e.g., the triphosphate group of ATP) func­

OH

tions as a leaving group. In the reaction of 2-(N,Ndimethylamino) ethanol w e se e that the more basic (CH 3 ) 2 N — group acts as the nucleophile rather than the less basic — OH group. And when a nucleophile attacks S-adenosylm ethionine, w e se e that the attack takes place at the less hindered CH 3 — group rather than at on e of the more hindered — CH2— groups.

S tu d y P roblem (a) What is the leaving group when 2-(N,N-dimethylamino)ethanol reacts with S-adenosylmethionine? (b) What would the leaving group have to be if methionine itself were to react with 2-(N,N-dimethylamino)ethanol? (c) Of what special significance is this difference?

6.14A The Unreactivity of Vinylic and Phenyl Halides As we learned in Section 6.1, compounds that have a halogen atom attached to one carbon atom of a double bond are called vinylic halides; those that have a halogen atom attached to a benzene ring are called aryl or phenyl halides:

A v in y lic h a lid e

•

A p h en y l h alid e

Vinylic and phenyl halides are generally unreactive in SN1 or SN2 reactions.

They are unreactive in SN1 reactions because vinylic and phenyl cations are relatively unsta­ ble and do not form readily. They are unreactive in SN2 reactions because the carbon-halo­ gen bond of a vinylic or phenyl halide is stronger than that of an alkyl halide (we shall see why later), and the electrons of the double bond or benzene ring repel the approach of a nucleophile from the back side.

268

Chapter 6

Ionic Reactions

6.15 Elimination Reactions o f Alkyl Halides Elimination reactions o f alkyl halides are important reactions that compete with substitution reac­ tions. In an elim ination reaction the fragments o f some molecule (YZ) are removed (eliminated) from adjacent atoms o f the reactant. This elimination leads to the creation o f a multiple bond: Y

6.15A Dehydrohalogenation A widely used method for synthesizing alkenes is the elimination o f HX from adjacent atoms o f an alkyl halide. Heating the alkyl halide w ith a strong base causes the reaction to take place. The follow ing are two examples: CH3CHCH3 ----- — ---------Î CH2= C H — CH3 3I 3 c 2h 5o h , 5 5 °C

NaBr

c 2h 5o h

(79% )

Br CH,

|H 3

CH3— C— Br

C 2H 5O N a

c

C 2H 5O H , 55°C

CH^

CH

NaBr

c 2h 5o h

% CH 2

(91%)

Reactions like these are not lim ited to the elim ination o f hydrogen bromide. Chloroalkanes also undergo the elim ination o f hydrogen chloride, iodoalkanes undergo the elim ination o f hydrogen iodide, and, in a ll cases, alkenes are produced. When the elements o f a hydrogen halide are eliminated from a haloalkane in this way, the reaction is often called dehydrohalogenation: H -C — C -

:B-

\ - c

y

/

\

H

B

:X :

-

:X :

A b ase D ehydrohalogenation

In these eliminations, as in SN1 and SN2 reactions, there is a leaving group and an attack­ ing Lewis base that possesses an electron pair. Chemists often call the carbon atom that bears the leaving group (e.g., the halogen atom in the previous reaction) the alpha (a ) carbon atom and any carbon atom adjacent to it a beta (B) carbon atom. A hydrogen atom attached to the b carbon atom is called a B hydro­ gen atom. Since the hydrogen atom that is eliminated in dehydrohalogenation is from the b carbon atom, these reactions are often called B elim inations. They are also often referred to as 1,2 elim inations. The b hydrogen ' and b carbon H

I/3

1a

C— C— LG

The a carbon and leaving group

We shall have more to say about dehydrohalogenation in Chapter 7, but we can exam­ ine several important aspects here.

269

6 .16 The E2 Reaction

6.15B Bases Used in Dehydrohalogenation V a r io u s

s tro n g

bases h a ve

been

used fo r

d e h y d r o h a lo g e n a t io n s .

P o ta s s iu m

h y d r o x id e

d i s s o l v e d i n e t h a n o l ( K O H / E t O H ) i s a r e a g e n t s o m e t im e s u s e d , b u t t h e c o n j u g a t e b a s e s o f a lc o h o ls , s u c h a s s o d iu m e t h o x id e ( E t O N a ) , o f t e n o f f e r d is t in c t a d v a n ta g e s . T h e c o n ju g a t e b a s e o f a n a lc o h o l ( a n a lk o x id e ) c a n b e p r e p a r e d b y t r e a tin g a n a lc o h o l w i t h a n a lk a l i m e t a l. F o r e x a m p le :

2 R— O H -

2 Na

2 R — O :-

A lcoh ol

N a-

H,

So diu m a lk o x id e

T h is r e a c t io n is a n o x i d a t i o n - r e d u c t i o n r e a c t io n . M e t a l l i c s o d iu m r e a c ts w i t h h y d r o g e n a to m s t h a t a re b o n d e d to o x y g e n a to m s to g e n e r a te h y d r o g e n g a s , s o d iu m c a tio n s , a n d th e a lk o x id e a n io n . T h e r e a c t io n w i t h w a t e r is v ig o r o u s a n d a t tim e s e x p lo s iv e .

2 HO H

2 Na

HO : -

2

N a-

H,

So diu m h yd ro xide S o d i u m a l k o x i d e s c a n a ls o b e p r e p a r e d b y a l l o w i n g a n a l c o h o l t o r e a c t w i t h s o d i u m h y d r i d e ( N a H ) . T h e h y d r id e io n ( H : - ) is a v e r y s tr o n g b a s e . ( T h e p K a o f H

R— O— H +

Na+: H-

R— Q: -

2

is 3 5 .)

H9 H

N a+

S o d i u m ( a n d p o t a s s i u m ) a l k o x i d e s a r e u s u a l l y p r e p a r e d b y u s in g a n e x c e s s o f t h e a lc o h o l , a n d th e e x c e s s a lc o h o l b e c o m e s th e s o lv e n t f o r th e r e a c t io n . S o d iu m e t h o x id e is f r e q u e n t ly p r e p a r e d i n t h is w a y u s in g e x c e s s e th a n o l. 2

CH3CH2Q H

2 Na -----> 2 CH3CH2Q :- Na+

H e lp f u l H i n t

H2

E tO N a /E tO H is a c o m m o n

S o diu m e th o xid e d is s o lv e d in e x c e s s e th a n o l

Ethanol (ex c e s s )

a b b re v ia tio n f o r s o d iu m e th o x id e d is s o lv e d in e th a n o l.

P o ta s s iu m t e r t - b u t o x id e ( t - B u O K ) is a n o t h e r h i g h l y e f f e c t iv e d e h y d r o h a lo g e n a t in g r e a g e n t. I t c a n b e m a d e b y t h e r e a c t i o n b e l o w , o r p u r c h a s e d a s a s o li d .

CH

CH3 2 CH3C— OH

2 K

2

CH3C — O : -

K+

+

H e lp f u l H i n t

H,

t-B u O K /t-B u O H re p re s e n ts

CH3

CH

ferf-B u tan ol (ex c e s s )

p o ta s s iu m te r t- b u to x id e d is s o lv e d

3

in te r t- b u ta n o l.

P o tas s iu m ferf-b u to x id e

6.15C Mechanisms of Dehydrohalogenations E l i m i n a t i o n r e a c t io n s o c c u r b y a v a r i e t y o f m e c h a n i s m s . W i t h a l k y l h a l i d e s , t w o m e c h a ­ n is m s a re e s p e c ia lly i m p o r t a n t b e c a u s e th e y a re c lo s e ly r e la t e d to th e S N 2 a n d S N 1 r e a c ­ t i o n s t h a t w e h a v e j u s t s t u d ie d . O n e m e c h a n i s m , c a l l e d t h e E 2 r e a c t i o n , i s b i m o l e c u l a r i n t h e r a t e - d e t e r m i n i n g s t e p ; t h e o t h e r m e c h a n i s m is t h e E 1 r e a c t i o n , w h i c h i s u n i m o l e c u l a r i n t h e r a t e - d e t e r m i n i n g s te p .

6.16 The E2 Reaction W h e n i s o p r o p y l b r o m id e is h e a te d w i t h s o d iu m e t h o x id e i n e th a n o l t o f o r m

p r o p e n e , th e

r e a c t io n r a te d e p e n d s o n th e c o n c e n t r a t io n o f is o p r o p y l b r o m id e a n d th e c o n c e n t r a t io n o f e t h o x id e io n . T h e r a te e q u a tio n is f i r s t o r d e r i n e a c h r e a c t a n t a n d s e c o n d o r d e r o v e r a ll:

R a te = & [C H

3C

H B rC H

3] [ C 2 H 5O

- ]

270

Chapter 6

Ionic Reactions

• From the reaction order we infer that the transition state for the rate-determining step must involve both the alkyl halide and the alkoxide ion: The reaction must be bimolecular. Considerable experimental evidence indicates that the reaction takes place in the follow ­ ing way:

A MECHANISM FOR THE REACTION M e c h a n is m f o r t h e E 2 R e a c tio n R E A C T IO N C

2

H

5

CH3CHBrCH 3

O-

c h 2= c h c h 3

c 2 h 5o h

B r

M E C H A N IS M

s­

C2H5 C 2 H 5— O

CH 3

H

H

.c b - C

S j b a

H H

O.

QBr:

CH3 >H 'C ^ C

H H

H V C=C / \ H H

CH3 3

T ran sition s ta te T h e b a s ic e th o x id e ion b e g in s to r e m o v e a p roton from th e b c a r b o n u s in g its e le c tr o n pair to form a b o n d to it. At th e s a m e tim e, th e e le c tr o n pair of th e b C— H b o n d b e g in s to m o v e in to b e c o m e th e p b o n d o f a d o u b le b o n d , a n d th e b r o m in e b e g in s to d e p a r t w ith th e e le c tr o n s th at b o n d e d it to th e a c a rb o n .

Partial b o n d s in th e tra n sitio n s t a t e e x te n d from th e o x y g e n a to m th at is r e m o v in g th e b h y d r o g e n , th r o u g h th e c a rb o n s k e le t o n o f th e d e v e lo p in g d o u b le b o n d , to th e d e p a r tin g le a v in g g r o u p . T h e flo w o f e le c tr o n d e n s ity is from th e b a s e tow ard th e le a v in g g r o u p a s an e le c tr o n pair fills th e p b o n d in g orbital o f th e a lk e n e .

C 2 H 5 - OH + =Br = At c o m p le tio n o f th e r ea c tio n , th e d o u b le b o n d is fu lly fo r m e d a n d th e a lk e n e h a s a trigon al planar g e o m e tr y at e a c h c a rb o n a tom . T h e o th er p r o d u c ts a re a m o le c u le o f e th a n o l a n d a b r o m id e ion .

An E2 r ea c tio n h a s o n e tra n sitio n s ta te

When we study the E2 reaction further in Section 7.6D, we shall find that the orientations o f the hydrogen atom being removed and the leaving group are not arbitrary and that an orientation where they are a ll in the same plane, like that shown above and in the example that follows, is required.

6.17 The E1 Reaction

271

B: H

-*■

+ BH + LG:

LG N ew m an p rojection

Anti-coplanar transition state of alkyl halide

Alkene

N otice that the geometry required here is sim ilar to that o f the SN2 reaction. In the SN2 reaction (Section 6 .6) the nucleophile must push out the leaving group from the opposite side. In the E2 reaction the electron p a ir o f the C — H bond pushes the leav­ ing group away from the opposite side as the base removes the hydrogen. (We shall also fin d in Section 7.7C that a syn-coplanar E2 transition state is possible, though not as favorable.)

6.17 The E1 Reaction Elim ination reactions may follow a different pathway from that given in Section 6.16. Treating tert-butyl chloride w ith 80% aqueous ethanol at 25°C, for example, gives substi­ tution products in 83% yield and an elim ination product (2-methylpropene) in 17% yield: CH Sn1 ,

CH

CH3— C — O H + CH3— C — O C H 2C H 3 CH3

CH, CH3— C — Cl

CH3

fert-B utyl a lc o h o l

80°/c C2H5OH

v

20°/c H2O

(83%)

25°C

CH3

fert-B utyl eth yl eth er

CH

te rt-B utyl ch lo rid ee

El

c h 2= c .

CH3 2 -M eth y lp ro p en e (17%)

• The in itia l step for both reactions is the formation o f a tert-butyl cation as a com­ mon intermediate. This is also the rate-determining step fo r both reactions; thus both reactions are unimolecular: ch3

CH 3 CH3— C 9 C l: CH

s lo w

CH3

C+

CH 3 (s o lv a te d )

(s o lv a te d )

Whether substitution or elim ination takes place depends on the next step (the fast step).

272

Chapter 6

Ionic Reactions

• I f a solvent molecule reacts as a nucleophile at the positive carbon atom o f the tert-butyl cation, the product is tert-butyl alcohol or tert-butyl ethyl ether and the reaction is SN1: CH 3 . '3

CH, fa s t

CH3— C +u Sol— OH 3 \ CH 3 (S o l = H —

or

•

CH3— C^ Oj 3 I Ch h

H

CH3

Sol

I

: C H — C — O — Sol + H— O 9 Sol

S n1 rea ctio n

CH

H— O — Sol

CH3CH2— )

If, however, a solvent molecule acts as a base and removes one of the b hydrogen atoms as a proton, the product is 2-methylpropene and the reaction is E1. CH / Sol— O ^ H— CH,— C

CH fa s t

\

Sol— O ^ H + CH2= C

CH

H

2

I

H

/

\

E1 rea ctio n

CH 3

2-M eth y lp ro p en e

E1 reactions almost always accompany SN1 reactions.

A MECHANISM FOR THE REACTION M e c h a n i s m f o r t h e E1 R e a c t i o n

R E A C T IO N (C H

3

h 2o

)3 C Cl

CH 2 = C (C H 3 ) 2

H

3

O +

C h

M E C H A N IS M Step 1

Step 1

ch3

I 3 CH 3— C— CM

CH3 A id ed by th e p olar s o lv e n t, a c h lo r in e d e p a r ts w ith th e e le c tr o n pair th at b o n d e d it to th e c a r b o n .

s lo w h 2o

CH / c h 3— C+ :C l;

3

\

CH 3

T h is s lo w s t e p p r o d u c e s th e rela tiv ely s ta b le 3° c a r b o c a tio n an d a c h lo r id e ion. T h e io n s are s o lv a te d (and sta b iliz e d ) by su rr o u n d in g w ater m o le c u le s .

(m echanism continues on the next page)

273

6.18 How to Determ ine W hether Substitution or Elimination is Favored

Step 2

Step 2

Transition state2 +

H— O

H K

CH3 3

H— C— C+ b

H

/

H

H v C =C

-

“ VCH3 u

H

Amolecule of water removes one of the hydrogensfrom theb carbon of the carbocation. These hydrogens areacidic due tothe adjacent positive charge. At thesame time an electron pair moves in toforma double bond between the a and b carbon atoms.

CH3 3

_! AGi(2)

CH3

This step produces the alkene and a hydroniumion.

Reactioncoordinate

6.18 H o w to D eterm in e W h eth er Substitution or Elimination Is Favored A ll nucleophiles are potential bases and a ll bases are potential nucleophiles. This is because the reactive part o f both nucleophiles and bases is an unshared electron pair. It should not be surprising, then, that nucleophilic substitution reactions and elim ination reactions often compete w ith each other. We shall now summarize factors that influence which type o f reac­ tion is favored, and provide some examples. H e lp f u l H i n t

6.18A Sn2 versus E2 Sn2 and E2 reactions are both favored by a high concentration o f a strong nucleophile or base. When the nucleophile (base) attacks a b hydrogen atom, elim ination occurs. When the nucleophile attacks the carbon atom bearing the leaving group, substitution results: (a) elimination E2 ^ H - r - C—

V Nu— H

x-

C

/

\

N ur

\ ( b ^^ ^ C— X (b) substitution Sn2

H — C—

X-

N u— C —

The follow ing examples illustrate the effects o f several parameters on substitution and elimination: relative steric hindrance in the substrate (class o f alkyl halide), temperature, size o f the base/nucleophile (EtONa versus i-BuOK), and the effects o f basicity and polarizability. In these examples we also illustrate a very common way o f w riting organic reac­ tions, where reagents are written over the reaction arrow, solvents and temperatures are written under the arrow, and only the substrate and major organic products are written to the le ft and right o f the reaction arrow. We also employ typical shorthand notations o f organic chemists, such as exclusive use o f bond-line formulas and use o f commonly accepted abbreviations for some reagents and solvents.

This section draws together the various factors that influence the com petition between substitution and elimination.

274

Chapter 6

Ionic Reactions

P rim a ry S u b s tra te When the substrate is a primary halide and the base is strong and unhindered, like ethoxide ion, substitution is highly favored because the base can easily approach the carbon bearing the leaving group:

EtONa

Primary

Sn2 Major (90%)

E2 Minor (10% )

W ith secondary halides, however, a strong base favors elim ina­ tion because steric hindrance in the substrate makes substitution more difficult:

S e c o n d a ry S u b s tra te

Secondary

E2 Major (79%)

S n2 Minor (21%)

W ith tertiary halides, steric hindrance in the substrate is severe and an Sn2 reaction cannot take place. Elim ination is highly favored, especially when the reac­ tion is carried out at higher temperatures. Any substitution that occurs must take place through an Sn 1 mechanism:

T e rtia ry S u b s tra te

Without Heating

Tertiary

E2 Major (91%)

Sn1 Minor (9%)

With Heating EtONa

Tertiary

E2 + E1 Only (100%)

Increasing the reaction temperature favors elim ination (E1 and E2) over substitution. Elim ination reactions have greater free energies o f activation than substitution reactions because more bonding changes occur during elim ination. When higher tempera­ ture is used, the proportion o f molecules able to surmount the energy o f activation barrier for elim ination increases more than the proportion o f molecules able to undergo substitu­ tion, although the rate o f both substitution and elim ination w ill be increased. Furthermore, elim ination reactions are entropically favored over substitution because the products o f an elim ination reaction are greater in number than the reactants. Additionally, because tem­ perature is the coefficient o f the entropy term in the Gibbs free-energy equation AG° = AH° - TAS°, an increase in temperature further enhances the entropy effect. T e m p e ra tu re

S iz e o f t h e B a s e /N u c le o p h ile Increasing the reaction temperature is one way o f favor­ ably influencing an elim ination reaction o f an alkyl halide. Another way is to use a stro n g s te ric a lly h in d e re d b a s e such as the tert-butoxide ion. The bulky methyl groups o f the

6.18 How to Determ ine W hether Substitution or Elimination Is Favored

tert-butoxide ion inhibit its reaction by substitution, allowing elim ination reactions to take precedence. We can see an example o f this effect in the follow ing two reactions. The rel­ atively unhindered methoxide ion reacts w ith octadecyl bromide prim arily by substitution, whereas the bulky tert-butoxide ion gives m ainly elimination. Unhindered (Small) Base/Nucleophile ,B r

C H O Na

o c h

3

CH3OH, 65°C

E2 (1%)

S n2 (99% )

Hindered Base/Nucleophile -B r

f-BuOK

O B u -f

t-BuOK, 40°C E2 (85%)

S n2 (15%)

B a s ic ity a n d P o la r iz a b ility Another factor that affects the relative rates o f E2 and SN2 reactions is the relative basicity and polarizability o f the base/nucleophile. Use o f a strong, slightly polarizable base such as hydroxide ion, amide ion (NH2_), or alkoxide ion (espe­ cially a hindered one) tends to increase the likelihood o f elim ination (E2). Use o f a weakly basic ion such as a chloride ion (CU) or an acetate ion (CH3CO2~) or a weakly basic and highly polarizable one such as Br~, U , or RS~ increases the likelihood of substitution (SN2). Acetate ion, for example, reacts w ith isopropyl bromide almost exclusively by the SN2 path:

O

O

O-

Br

S n2

O

(~ 100 %)

The more strongly basic ethoxide ion (Section 6.15B) reacts w ith the same compound m ainly by an E2 mechanism.

6.18B Tertiary Halides: SN1 versus E1 Because E1 and SN1 reactions proceed through the formation o f a common intermediate, the two types respond in sim ilar ways to factors affecting reactivities. E1 reactions are favored w ith substrates that can form stable carbocations (i.e., tertiary halides); they are also favored by the use o f poor nucleophiles (weak bases) and they are generally favored by the use o f polar solvents. It is usually d iffic u lt to influence the relative partition between SN1 and E1 products because the free energy o f activation for either reaction proceeding from the carbocation (loss o f a proton or combination w ith a molecule o f the solvent) is very small. In most unimolecular reactions the SN1 reaction is favored over the E1 reaction, especially at lower temperatures. In general, however, substitution reactions of tertiary halides do not find wide use as synthetic methods. Such halides undergo eliminations much too easily. Increasing the temperature o f the reaction favors reaction by the E1 mechanism at the expense o f the SN1 mechanism. •

I f a n e lim in a tio n p r o d u c t is d e s ire d fro m a te r tia r y s u b s tra te , it is ad v isa b le to use a stro n g b a se so as to e n c o u ra g e a n E 2 m e c h a n ism o v e r th e c o m p e tin g E1 a n d Sn1 m e ch a n ism s.

275

2 76

Chapter 6

Ionic Reactions

6.19 O verall Sum m ary T h e m o s t i m p o r t a n t r e a c t i o n p a t h w a y s f o r t h e s u b s t i t u t i o n a n d e l i m i n a t i o n r e a c t io n s o f s i m p l e a l k y l h a l i d e s a r e s u m m a r i z e d i n T a b le 6 . 7 .

H e lp f u l H i n t

O verall Sum m ary o f SN1, SN2, E1, and E2 Reactions

Overall summary

H 1 R— C—X 1 H

c h 3x

Methyl

Gives S n2 reactions

1

°

R 1 R— C—X 1 H

R 1 R— C—X 1 R

°

3°

2

Bimolecular (SN2/E2) Reactions Only

Sn 1/E1 or E2

Gives mainly S n2 ex cep t with a hindered strong b ase [e.g., (CH 3) 3C O ] and th en gives mainly E2.

N o SN2 reaction. In solvolysis gives S n 1/E1, and at lower te m p e ra tu re s SN1 is favored. W hen a strong b ase (e.g., RO~) is used, E2 predom inates.

L e t u s e x a m in e

Gives mainly SN2 with w eak b ases (e.g., I_ , CN ~, R C O 2 ) and mainly E2 with strong b ases (e.g., RO~).

s e v e r a l s a m p l e e x e r c is e s t h a t w i l l i l l u s t r a t e h o w

th e in f o r m a tio n

T a b le 6 . 7 c a n b e u s e d .

S o lv e d P ro b le m 6 .8 G i v e t h e p r o d u c t ( o r p r o d u c t s ) t h a t y o u w o u l d e x p e c t t o b e f o r m e d i n e a c h o f t h e f o l l o w i n g r e a c t io n s . I n e a c h c a s e g iv e th e m e c h a n is m ( S N 1, S N 2 , E 1 , o r E 2 ) b y w h ic h th e p r o d u c t is f o r m e d a n d p r e d ic t th e r e la tiv e a m o u n t o f e a c h ( i.e ., w o u ld th e p r o d u c t b e th e o n ly p r o d u c t, th e m a jo r p r o d u c t, o r a m in o r p r o d u c t? ) .

Br (a )

(b)

HO-

CHO 'B r

CH3O , 50°C

'B r

t-BuO t-BuOH, 50°C

(c )

CH3OH, 50°C

CH3OH, 25°C

HS CH3OH, 50°C

STRATEGY AND ANSWER ( a ) T h e s u b s tr a te is a 1 ° h a lid e . T h e b a s e / n u c le o p h ile is C H

3O

~ , a s tr o n g b a s e ( b u t n o t a h in d e r e d o n e ) a n d a g o o d

n u c l e o p h i l e . A c c o r d i n g t o T a b le 6 . 7 , w e s h o u l d e x p e c t a n S N 2 r e a c t i o n m a i n l y , a n d t h e m a j o r p r o d u c t s h o u l d b e ^ " ''- '" '^ V ''O C H 3 . A m in o r p r o d u c t m ig h t b e \

(b)

b y a n E 2 p a th w a y .

A g a i n t h e s u b s t r a t e i s a 1 ° h a l i d e , b u t t h e b a s e / n u c l e o p h il e , r - B u O th e r e fo r e , th e m a jo r p r o d u c t to b e b y a n S N2 p a th w a y .

, is a s tr o n g h in d e r e d b a s e . W e s h o u ld e x p e c t,

b y a n E 2 p a th w a y a n d a m in o r p ro d u c t to b e

/O -f-B u

in

277

Problems

(c) The reactant is (S)-2-bromobutane, a 2° halide and one in which the leaving group is attached to a chirality cen­

ter. The base/nucleophile is HS- , a strong nucleophile but a weak base. We should expect m ainly an Sn2 reac­ tion, causing an inversion o f configuration at the chirality center and producing the (R) stereoisomer: SH H

(d) The base/nucleophile is OH , a strong base and a strong nucleophile. The substrate is a 3° halide; therefore,

we should not expect an Sn2 reaction. The m ajor product should be

via an E2

reaction. A t this higher temperature and in the presence o f a strong base, we should not expect an appreciable amount o f the Sn 1 solvolysis, product, OCH3 .

(e) This is solvolysis; the only base/nucleophile is the solvent, CH3OH, which is a weak base (therefore, no E2

reaction) and a poor nucleophile. The substrate is tertiary (therefore, no SN2 reaction). A t this lower tempera­ ture we should expect m ainly an SN1 pathway leading to OCH3 . A m inor product, by an E1 pathway, would be

Key Terms and Concepts The key terms and concepts that are highlighted in b o ld , b lu e te x t w ithin the chapter are defined in the glossary (at the back o f the book) and have hyperlinked definitions in the accompanying WileyPLUS course (www.wileyplus.com).

PLUS

Problems Note to Instructors: Many of the homework problems are available for assignment via WileyPLUS, an online teaching and learning solution. RELATIVE RATES O F N U CLEO PHILIC SUBSTITUTION 6.20

W hich alkyl halide would you expect to react more rapidly by an Sn2 mechanism? Explain your answer. ,Br

(a)

or

or

(d)

Cl

Cl

Br Br "Cl

(b)

(c) 6.21

Cl

or

or

(e)

or

Cl

W hich Sn2 reaction o f each pair would you expect to take place more rapidly in a protic solvent? (a) (1)

Cl

EtO-

Cl

EtOH

C l-

or (2)

O

HCl

Cl

278

Chapter 6

„Cl

(b) (1)

Ionic Reactions

+ EtO -

+ C l-

+ E tS - ----- *

+ C l-

Br + (C6H5)3N ----- *

+ N(C6H5)3 + Br-

or ,C!

(2)

(c) (1) or

' Br +' (C6H5)3P ----- *

(2)

Br

(d) (1)

P(C6H5)3 + B r■ OMe

(1.0M) + M eO- (1.0M )

+ B r"

or Br

(2)

6 .2 2

OMe

(1.0M ) + M eO - (2.0M )

+ B r"

W hich SN1 reaction o f each pair would you expect to take place more rapidly? Explain your answer. (a) (1)

+ H2O

+ HC!

Cl

OH

or (2)

+ H2O

+ HBr

Br

OH + HC!

(b) (1) ^ k C| + H 2O

OH

or (2)

+ MeOH

+ HC!

Cl (c) (1) ^

OMe

\ C| (1.0M) + E tO - (1.0M) —

+ Cl-

h

OEt

or (2)

(d) (1)

Cl

(2.0M) + EtO - (1.0M) ----- *

+ Cl-

EtOH

c i (1m

+ b ^ o .^

OEt

itO-

+ ClOEt

or (2)

(e) (1)

Cl

(1.0M) + EtO - (2.0M) ----- *

+ H2O

OEt

+ HC!

Cl

or

+ Cl-

EtOH

OH

Cl

(2)

OH + H2O

+ HC!

SYNTHESIS 6.23

Show how you m ight use a nucleophilic substitution reaction o f 1-bromopropane to synthesize each o f the following compounds. (You may use any other compounds that are necessary.) (a) ^ ^ O

H

(e)

(b ) 1-Iodopropane (c) (d) C H 3CH 2CH 2— S — CH 3

(g)

O

(h)

O

(f)

, / N + (CH 3)3 Br-

N3

(i)

N

SH

279

Problems 6.24

W ith methyl, ethyl, or cyclopentyl halides as your organic starting materials and using any needed solvents or inor­ ganic reagents, outline syntheses o f each o f the follow ing. More than one step may be necessary and you need not repeat steps carried out in earlier parts o f this problem.

6.25

(e)

m H C

(a)

(i) ' ' H 3 O C H 3

(b)

(f) ^ ^ S H

(c) CH 3 OH

(g) C H 3CN

(d) "'/ ~'X'OH

(h) ""/ X ''CN

(j)

"~x'-OMe

(k)

Listed below are several hypothetical nucleophilic substitution reactions. None is synthetically useful because the product indicated is not formed at an appreciable rate. In each case provide an explanation for the failure o f the reaction to take place as indicated. (a)

+ OH- - X *

(b)

+ OH- - * - »

(c) I

I + OH

"Br

6.26

c h 3s h

X

"OH

+

CH,

OH

*

+

H-

OH

+ CN-

CN

+

Br-

(e) NH 3 + CH 3OCH 3 ----- » CH 3NH 2 +

CH3OH

(f) NH 3 + CH 3OH2+ ----- » CH 3NH3+ +

H2O

Your task is to prepare styrene by one o f the follow ing reactions. Which reaction would you choose to give the bet­ ter yield o f styrene? Explain your answer. Br ,Br KOH

KOH

or

EtOH, A

(2 )

EtOH, A

S tyrene 6.27

Styrene

Your task is to prepare isopropyl m ethyl ether by one o f the follow ing reactions. Which reaction would give the better yield? Explain your answer. I (1)

ONa

och3 or

+ CH3ONa

(2)

OCH 3 + CH 3 I -------->

Isopropyl methyl ether 6.28

Isopropyl methyl ether

Starting with an appropriate alkyl halide and using any other needed reagents, outline syntheses o f each o f the follow ­ ing. When alternative possibilities exist for a synthesis, you should be careful to choose the one that gives the better yield. (a) B utyl sec-butyl ether

(f)

O

(b)

(k)

(c) M ethyl neopentyl ether (d) M ethyl phenyl ether CN

H CN

(l) ira«s-1-Iodo-4-methylcyclohexane (g) (S)-2-Pentanol (h) (R)-2-Iodo-4-methylpentane

(e)

(j) ds-4-Isopropylcyclohexanol

(i)

280

Chapter 6

Ionic Reactions

GENERAL SN1, SN2, A N D ELIM IN A TIO N 6.29

Which product (or products) would you expect to obtain from each o f the follow ing reactions? In each part give the mechanism (SN1, SN2, E1, or E2) by which each product is formed and predict the relative amount o f each product (i.e., would the product be the only product, the major product, a m inor product, etc.?). -Br

Cl

EtO" EtOH, 50°C

(a )

-Br

(f )

t -BuOK

(b )

t -BuOH, 50°C>

t-Bu

MeOH, 25 °C

MeOMeOH, 50°C CH3CO2 3-Chloropentane CH3CO2H, 50°C HO(Æ)-2-bromobutane 25°C 25°C (5)-3-Bromo-3-methylhexane MeOH I(5)-2-Bromooctane MeOH, 50°C

(g) 3-Chloropentane (c)

MeOMeOH, 50°C

Br

(d )

(h)

t -BuOK t-BuOH, 50°C>

Br

(i) (j)

Cl (e)

6.30

I-

t-Bu

(k)

acetone, 50°C

W rite conformational structures for the substitution products o f the follow ing deuterium-labeled compounds: Cl

Cl I-

I-

MeOH

MeOH

D Cl (b )

H

D

(d)

MeOH

CH3

H

H

MeOH, H2O

D

6.31

Although ethyl bromide and isobutyl bromide are both primary halides, ethyl bromide undergoes SN2 reactions more than 10 times faster than isobutyl bromide does. When each compound is treated w ith a strong base/nucle­ ophile (EtO- ), isobutyl bromide gives a greater yield o f elim ination products than substitution products, whereas w ith ethyl bromide this behavior is reversed. What factor accounts for these results?

6.32

Consider the reaction o f I- w ith CH3CH2Cl. (a) Would you expect the reaction to be SN1 or SN2? The rate constant for the reaction at 60°C is

5 X 10-5 L m ol-1 s-1 . (b) What is the reaction rate if [I- ] = 0.1 m ol L -1 and [CH3CH2Cl] = 0.1 m ol L -1 ? (c) I f [ H = 0.1 m ol L -1 and [CH3CH2Cl] = 0.2 m ol L -1 ? (d) I f [ H = 0.2 mol L -1 and [CH3CH2Cl] = 0.1 m ol L -1 ? (e) I f [I- ] = 0.2 m ol L -1 and [CH3CH2Cl] = 0.2 m ol L -1 ? 6.33

W hich reagent in each pair listed here would be the more reactive nucleophile in a polar aprotic solvent? (a) CH3NH-

6.34

(d) (C6Hs)3N

or

(b) CH3O-

or

or

CH3CO2- (- OAc)

CH3NH2

(e) H2O

or

H3O

^ H s fe P

(c) CH3SH

or

CH3OH

(f) NH3 (i)

or

NH4 +

(g) H2S

W rite mechanisms that account for the products o f the follow ing reactions: (a) HO

-Br

ohh 2o

s— V

7

O

7

(b) H2N '

or

HS-

(h) CH3CO2- (- OAc)

/B r

OHh 2o

N H

or

OH-

281

P roblem s

6.35

Draw a three-dimensional representation for the transition state structure in the SN2 reaction o f N=C: anion) w ith bromoethane, showing a ll nonbonding electron pairs and fu ll or partial charges.

(cyanide

6.36

Many SN2 reactions o f alkyl chlorides and alkyl bromides are catalyzed by the addition o f sodium or potassium iodide. For example, the hydrolysis o f methyl bromide takes place much faster in the presence o f sodium iodide. Explain.

6.37

Explain the follow ing observations: When teri-butyl bromide is treated w ith sodium methoxide in a m ixture of methanol and water, the rate o f formation o f teri-butyl alcohol and teri-butyl methyl ether does not change appre­ ciably as the concentration o f sodium methoxide is increased. However, increasing the concentration o f sodium methoxide causes a marked increase in the rate at which teri-butyl bromide disappears from the mixture.

6.38

(a) Consider the general problem o f converting a tertiary alkyl halide to an alkene, for example, the conversion of

teri-butyl chloride to 2-methylpropene. What experimental conditions would you choose to ensure that elim i­ nation is favored over substitution? (b) Consider the opposite problem, that o f carrying out a substitution reaction on a tertiary alkyl halide. Use as

your example the conversion o f teri-butyl chloride to teri-butyl ethyl ether. What experimental conditions would you employ to ensure the highest possible yield o f the ether? 6.39

1-Bromobicyclo[2.2.1]heptane is extremely unreactive in either SN2 or SN1 reactions. Provide explanations for this behavior.

6.40

When ethyl bromide reacts w ith potassium cyanide in methanol, the major product is CH3CH2CN. Some CH3CH2NC is formed as well, however. W rite Lewis structures for the cyanide ion and for both products and pro­ vide a mechanistic explanation o f the course o f the reaction.

6.41

Give structures for the products o f each o f the follow ing reactions:

N aC l

-S N a N aS"

/■ (c)^ Br'

' B r (1

C

m o l) -O H

C

E t2O N aN H 2 (-N H 3) liq. NH 3

+

2 N aBr

N aH (—H 2)

(d) C l' (e)

4H 8S 2

*

C 3 H3 Na

4H

8 C lO N a

C H 3I --------:

C

4H

E t2O , heat

+

C 4H 8O

N aC l

NaI

6.42

When teri-butyl bromide undergoes SN1 hydrolysis, adding a “ common ion” (e.g., NaBr) to the aqueous solution has no effect on the rate. On the other hand, when (C6H5)2CHBr undergoes SN1 hydrolysis, adding NaBr retards the reaction. Given that the (C6H5)2CH+ cation is known to be much more stable than the (CH3)3C+ cation (and we shall see why in Section 15.12A), provide an explanation for the different behavior o f the two compounds.

6.43

When the alkyl bromides (listed here) were subjected to hydrolysis in a m ixture o f ethanol and water (80% EtOH/20% H2O) at 55°C, the rates o f the reaction showed the follow ing order: (CH3)3CBr > CH3Br > CH3CH2Br > (CH3)2CHBr Provide an explanation for this order o f reactivity.

6.44

The reaction o f 1° alkyl halides w ith nitrite salts produces both RNO2 and RONO. Account for this behavior.

6.45

What would be the effect o f increasing solvent polarity on the rate o f each o f the follow ing nucleophilic substitu­ tion reactions? (a) Nu:

6.46

+

R— L ----- > R — Nu+

+

:L~

(b) R— L+ ----- > R+

+

:L

Competition experiments are those in which two reactants at the same concentration (or one reactant w ith two reactive sites) compete for a reagent. Predict the major product resulting from each o f the following competition experiments: Cl

(a)

H 2O Cl

DMF

(b) Cl

Cl

a c e to n e

282 6.47

Chapter 6

Ionic Reactions

In contrast to SN2 reactions, SN1 reactions show relatively little nucleophile selectivity. That is, when more than one nucleophile is present in the reaction medium, SN1 reactions show only a slight tendency to discriminate between weak nucleophiles and strong nucleophiles, whereas SN2 reactions show a marked tendency to discriminate. (a) Provide an explanation for this behavior. (b) Show how your answer accounts for the follow ing:

Major p r o d u ct

Major p ro d u ct

Challenge Problems 6.48

The reaction o f chloroethane w ith water in the gas phase to produce ethanol and hydrogen chloride has AH° = + 26.6 kJ m o F 1 and AS° = +4.81 J K ^ m o l^ a t 25°C. (a) Which o f these terms, if either, favors the reaction going to completion? (b) Calculate AG° fo r the reaction. What can you now say about whether the reaction w ill proceed to completion? (c) Calculate the equilibrium constant fo r the reaction. (d) In aqueous solution the equilibrium constant is very much larger than the one you just calculated. How can you

account for this fact? 6.49

When (S)-2-bromopropanoic acid [(S)-CH3CHBrCO2H] reacts with concentrated sodium hydroxide, the product formed (after acidification) is (Æ)-2-hydroxypropanoic acid [(Æ)-CH3CHOHCO2H, commonly known as (Æ)-lactic acid]. This is, o f course, the normal stereochemical result for an SN2 reaction. However, when the same reaction is carried out with a low concentration o f hydroxide ion in the presence o f Ag2O (where Ag+ acts as a Lewis acid), it takes place with overall retention o f configuration to produce (S)-2-hydroxypropanoic acid. The mechanism o f this reaction involves a phenomenon called neighboring-group participation. W rite a detailed mechanism for this reaction that accounts for the net retention o f configuration when Ag+ and a low concentration o f hydroxide are used.

6.50

The phenomenon o f configuration inversion in a chemical reaction was discovered in 1896 by Paul Walden (Section 6.6). Walden’s proof o f configuration inversion was based on the follow ing cycle: O

I

I

O

Cl

(- ) - C h lo r o s u c c in ic a cid

HO OH O

OH

(-)- M a lic a cid

(+ )-M a lic a c id

HO OH O

Cl

(+ )-C h lo r o s u c c in ic a cid T h e W a ld e n C y c le

L earning G roup P roblem s

283

(a) Basing your answer on the preceding problem, which reactions o f the Walden cycle are likely to take place w ith overall inversion o f configuration and which are like ly to occur w ith overall retention o f configuration? (b) M alic acid w ith a negative optical rotation is now known to have the (S) configuration. What are the configu­ rations o f the other compounds in the Walden cycle? (c) Walden also found that when (+ )-m alic acid is treated w ith thionyl chloride (rather than PCl5), the product of the reaction is (+)-chlorosuccinic acid. How can you explain this result? (d) Assuming that the reaction o f (-)-m a lic acid and thionyl chloride has the same stereochemistry, outline a Walden cycle based on the use o f thionyl chloride instead o f PCl5. 6.51

(R)-(3-Chloro-2-methylpropyl) methyl ether (A) on reaction w ith azide ion (N3~) in aqueous ethanol gives (S)-(3-azido-2-methylpropyl) methyl ether (B). Compound A has the structure ClCH2CH(CH3)CH2OCH3. (a) Draw wedge-dashed wedge-line formulas o f both A and B. (b) Is there a change o f configuration during this reaction?

6.52

Predict the structure o f the product o f this reaction:

HSV

'C l

N a O H in a q u e o u s EtO H

C6H1oS

The product has no infrared absorption in the 1620-1680-cm^1 region. 6.53

ris-4-Bromocyclohexanol

f-BuOH

> racemic C6H10O (compound C)

Compound C has infrared absorption in the 1620-1680-cm^1 and in the 3590-3650-cm ^1regions. Draw and label the (R) and (S) enantiomers o f product C. 6.54

1-Bromo[2.2.1]bicycloheptane is unreactive toward both SN2 and SN1 reactions. Open the computer molecular model at the book’s website titled “ 1-Bromo[2.2.1]bicycloheptane” and examine the structure. What barriers are there to substitution o f 1-bromo[2.2.1]bicycloheptane by both SN2 and SN1 reaction mechanisms?

6.55

Open the computer molecular model titled “ 1-Bromo[2.2.1]bicycloheptane LU M O ” at the book’s website for the lowest unoccupied molecular orbital (LUM O ) o f this compound. Where is the lobe o f the LUM O w ith which the HOMO o f a nucleophile would interact in an SN2 reaction?

6.56

In the previous problem and the associated molecular model at the book’s website, you considered the role of HOMOs and LUMOs in an SN2 reaction. (a) What is the LU M O in an SN1 reaction and in what reactant and species is it found? (b) Open the molecular model at the book’s website titled “ Isopropyl M ethyl Ether Carbocation LUM O.” Identify the lobe o f the LU M O in this carbocation model w ith which a nucleophile would interact. (c) Open the model titled “ Isopropyl M ethyl Ether Carbocation HOMO.” W hy is there a large orbital lobe between the oxygen and the carbon o f the carbocation?

Learning Group Problems Consider the solvolysis reaction o f (1S,2R)-1-bromo-1,2-dimethylcyclohexane in 80% H20/20% CH3CH2OH at room temperature. (a) W rite the structure o f a ll chemically reasonable products from this reaction and predict which would be the major one. (b) W rite a detailed mechanism for formation o f the major product. (c) W rite the structure o f a ll transition states involved in formation o f the major product.

284 2

.

Chapter 6

Ionic Reactions

C o n s i d e r t h e f o l l o w i n g s e q u e n c e o f r e a c t io n s , t a k e n f r o m t h e e a r l y s te p s i n a s y n t h e s is o f v - f l u o r o o l e i c a c id , a t o x i c n a t u r a l c o m p o u n d f r o m a n A f r i c a n s h r u b . ( v - F l u o r o o l e i c a c id , a ls o c a l l e d “ r a t s b a n e , ” h a s b e e n u s e d t o k i l l r a t s a n d a ls o a s a n a r r o w p o i s o n i n t r i b a l w a r f a r e . T w o m o r e s te p s b e y o n d t h o s e b e l o w a r e r e q u i r e d t o c o m p l e t e i t s s y n t h e s is . ) ( i)

1 - B r o m o - 8 - flu o r o o c ta n e +

( ii)

C om pound A

+

N aN H

( iii)

C om pound B

+

I — ( C H 2 ) 7 — C l -------- : c o m p o u n d C ( C

( iv )

C om pound C

+

N a C N -------- : c o m p o u n d D ( C

2

s o d i u m a c e t y l i d e ( t h e s o d i u m s a lt o f e t h y n e ) -------- : c o m p o u n d A

-------- : c o m p o u n d B ( C

10H 16 F N

a)

17H 30 C

lF )

18H 30 N F )

(a )

E lu c id a te th e s tru c tu re s o f c o m p o u n d s A , B , C , a n d D a b o v e .

(b)

W r i t e t h e m e c h a n i s m f o r e a c h o f t h e r e a c t io n s a b o v e .

(c)

W r i t e t h e s t r u c t u r e o f t h e t r a n s i t i o n s t a t e f o r e a c h r e a c t io n .

Summary and R eview Tools Mechanism Review: Substitution versus Elim ination SN1 and E1

S n2

Primary substrate Back side attack of Nu: with respect to LG Strong/polarizable unhindered nucleophile

Tertiary substrate Carbocation intermediate Weak nucleophile/base (e.g., solvent)

Bimolecular in rate-determining step Concerted bond forming/bond breaking Inversion of stereochemistry Favored by polar aprotic solvent

Unimolecular in rate-determining step Racemization if SN1 Removal of ß-hydrogen if E1 Protic solvent assists ionization of LG Low temperature (SN1) / high temperature (E1)

V

Nu/B ■■■■■

LGs-

H/ V Sn2 and E2

Secondary or primary substrate Strong unhindered base/nucleophile leads to SN2 Strong hindered base/nucleophile leads to E2 Low temperature (SN2) / high temperature (E2)

E2

Tertiary or secondary substrate Concerted anti-coplanar transition state Bimolecular in rate-determining step Strong hindered base High temperature

lgs

Nu/Bs

H

(C

10H 17 F )

Alkenes and Alkynes I Properties and Synthesis. Elimination Reactions of Alkyl Halides /■

\

Three very dissim ilar substances— vitam in A from sources including dark green leafy vegetables, cholesterol from anim als, and rubber from certain trees— have som ething in com m on. Their m olecules all have carbon-carbon double bonds— the alkene functional group.

( Z ) \ ----- f ( Z ) Natural rubber

G utta percha, another natural latex from the sap o f some trees, is sim ilar to natural rubber, ye t also d iffe r­ ent in an im portant way.

'( E ) G utta p e r ch a

285

286

Chapter 7

Alkenes and Alkynes I

Natural rubber and gutta percha d iffer in the directions taken by the main chain at each double bond. As we learn in this chapter, according to the (E)-(Z system, the double bonds o f rubber are all designated (Z) and those o f gutta percha are all (E). This seem ingly slight difference in stereochem istry, however, renders gutta per­ cha useless fo r many applications o f rubber. For exam ple, gutta percha is inelastic. A ll four o f these substances give characteristic reactions o f alkenes th a t we w ill study in C hapter 8 . M oreover, when nature puts these m olecules together, the double bonds are m ade by a reaction th a t we have ju st studied and w ill study fur­ ther in this chapter, the elim ination reaction.

7.1 Introduction Alkenes are hydrocarbons whose molecules contain the carbon-carbon double bond. An old name fo r this fam ily o f compounds that is s till often used is the name olefins. Ethene (ethylene), the simplest olefin (alkene), was called olefiant gas (Latin: oleum, o il + facere, to make) because gaseous ethene (C2H4) reacts w ith chlorine to form C2H4Cl2, a liquid (oil). Hydrocarbons whose molecules contain the carbon-carbon triple bond are called alkynes. The common name for this fam ily is acety len es, after the simplest member, HC# CH:

7.1A Physical Properties of Alkenes and Alkynes Alkenes and alkynes have physical properties sim ilar to those o f corresponding alkanes. Alkenes and alkynes up to four carbons (except 2-butyne) are gases at room temperature. Being relatively nonpolar themselves, alkenes and alkynes dissolve in nonpolar solvents or in solvents o f low polarity. Alkenes and alkynes are only very slightly soluble in water (with alkynes being slightly more soluble than alkenes). The densities o f alkenes and alkynes are lower than that o f water.

7.2 The (E)-(Z) System fo r Designating A lkene Diastereom ers In Section 4.5 we learned to use the terms cis and tr a n s to designate the stereochemistry o f alkene diastereomers. These terms are unambiguous, however, only when applied to di­ substituted alkenes. I f the alkene is trisubstituted or tetrasubstituted, the terms cis and trans are either ambiguous or do not apply at all. Consider the follow ing alkene as an example: Br

H

\ /

C=C

/ \

Cl

F

A

It is impossible to decide whether A is cis or trans since no two groups are the same. A system that works in a ll cases is based on the priorities o f groups in the Cahn-Ingold-Prelog convention (Section 5.7). This system, called the (E)-(Z) system, applies to alkene diastereomers o f a ll types. In the (E)-(Z) system, we examine the two

H

Ft

7.2 The (E)-(Z) System for Designating Alkene Diastereomers

287

g r o u p s a tta c h e d to o n e c a r b o n a to m o f th e d o u b le b o n d a n d d e c id e w h ic h h a s h ig h e r p r i ­ o r it y . T h e n w e r e p e a t th a t o p e r a tio n a t th e o th e r c a r b o n a to m :

J r

H ig h e r p rio rity

F

Cl —

C

C

H ig her p rio rity

Cl

>

F

Br

>

H

C

Br

H ig h e r' p rio rity

H

Br

H ig h e rp riority

(Z )-2 -B ro m o -1 -c h lo ro -1 flu o ro e th e n e

H

(E )-2 -B ro m o -1 -c h lo ro -1 flu o ro e th e n e

W e ta k e th e g r o u p o f h ig h e r p r i o r i t y o n o n e c a r b o n a to m a n d c o m p a r e i t w it h th e g r o u p o f h ig h e r p r i o r i t y o n th e o t h e r c a r b o n a to m . I f th e t w o g r o u p s o f h ig h e r p r i o r i t y a re o n th e

zu sa m ­

s a m e s i d e o f t h e d o u b l e b o n d , t h e a lk e n e i s d e s i g n a t e d ( Z ) ( f r o m t h e G e r m a n w o r d

men,

m e a n i n g t o g e t h e r ) . I f t h e t w o g r o u p s o f h i g h e r p r i o r i t y a r e o n o p p o s i t e s id e s o f t h e

d o u b l e b o n d , t h e a lk e n e is d e s i g n a t e d ( E ) ( f r o m t h e G e r m a n w o r d

entgegen,

m e a n in g o p p o ­

s ite ) . T h e f o l lo w in g e x a m p le illu s t r a t e s th is :

h 3c 3

ch3 3

CH3

C = C

/

H

/ C = C

3

\

H

3 \ >

H

CH3

H

(E )-2 -B u te n e o r (£ )-b u t-2 -e n e (fra n s - 2 -b u te n e )

(Z )-2 -B u te n e o r (Z )-b u t-2 -e n e (c /s - 2 -b u te n e )

S o l v e d P r o b l e m 7 .1 T h e t w o s t e r e o is o m e r s o f 1 - b r o m o - 1 , 2 - d i c h l o r o e t h e n e c a n n o t b e d e s i g n a t e d a s c i s a n d t r a n s i n t h e n o r m a l w a y b e c a u s e t h e d o u b l e b o n d i s t r i s u b s t i t u t e d . T h e y c a n , h o w e v e r , b e g i v e n ( E ) a n d ( Z ) d e s i g n a t io n s . W r i t e a s t r u c t u r a l f o r m u la f o r e a c h is o m e r a n d g iv e e a c h th e p r o p e r d e s ig n a t io n .

STRATEGY AND ANSWER

W e w r it e th e s tru c tu re s ( b e lo w ) , th e n n o te th a t c h lo r in e h a s a h ig h e r p r io r it y th a n h y d r o ­

g e n , a n d b r o m i n e h a s a h i g h e r p r i o r i t y t h a n c h l o r i n e . T h e g r o u p w i t h h i g h e r p r i o r i t y o n C 1 is b r o m i n e a n d t h e g r o u p w it h h ig h e r p r io r i t y a t C 2 is c h lo r in e . I n th e f i r s t s tr u c tu r e th e h ig h e r p r io r i t y c h lo r in e a n d b r o m in e a to m s a re o n o p p o s i t e s id e s o f t h e d o u b l e b o n d , a n d t h e r e f o r e t h i s i s o m e r i s ( E ) . I n t h e s e c o n d s t r u c t u r e t h o s e c h l o r i n e a n d b r o m i n e a t o m s a r e o n t h e s a m e s id e , s o t h e l a t t e r i s o m e r i s ( Z ) .

Cl

Cl x C

- <

Cl > B r>

/

H

Cl

H

x c - c '

Cl

H

X Br

(E )-1 -B ro m o -1 ,2 -d ic h lo ro e th e n e

(R )-(S )

d e s ig n a t io n a s w e l l ] g iv e

I U P A C n a m e s f o r e a c h o f th e f o llo w in g : H

(a )

C = C

H 33 C C N

\

/ Cl

C H 2C H 2C H

3

(c )

/

Cl

/

\

C H 2C H

3

3

CH

(d)

C = C I

CH

Cl

Br

(b )

C H 2C H ( C H 3 ) 2

/C = C ^ H

C = C

/ I

(f)

\ C H

\ i

(Z )-1 -B ro m o -1 ,2 -d ic h lo ro e th e n e

U s in g th e ( E ) - ( Z ) d e s ig n a t io n [a n d i n p a r ts ( e ) a n d ( f ) th e

Br

Br

2

C

H

3

H h 3c cis-2-butene > 1-butene.

A H ° = -1 2 7 kJ mol1 - 1

-120 kJ mol' 115 kJ mol

H

Ft

7.3 Relative Stabilities of Alkenes

289

is more stable than the cis isomer, since less energy is released when the trans isomer is converted to butane. Furthermore, it shows that the terminal alkene, 1-butene, is less sta­ ble than either o f the disubstituted alkenes, since its reaction is the most exothermic. O f course, alkenes that do not yield the same hydrogenation products cannot be compared on the basis o f their respective heats o f hydrogenation. In such cases it is necessary to com­ pare other thermochemical data, such as heats o f combustion, although we w ill not go into analyses o f that type here. 7 .3 B

Overall Relative Stabilities of Alkenes

Studies o f numerous alkenes reveal a pattern o f stabilities that is related to the number o f alkyl groups attached to the carbon atoms o f the double bond. • The greater the number o f attached alkyl groups (i.e., the more highly substituted the carbon atoms o f the double bond), the greater is the alkene’s stability. This order o f stabilities can be given in general terms as follow s:* Relative Stabilities o f Alkenes

Rx R

R R

Tetrasubstituted

>

V c R R

R

Trisubstituted

>

Rx

R

H > R

Rx R

H > R

Disubstituted

Rx R

R

R >

Rx R

R

H

M onosubstituted

>

Hx R

R

H

Unsubstituted

S olved P roblem 7.2 Consider the two alkenes 2-methyl-1-pentene and 2-methyl-2-pentene and decide which would be most stable. STRATEGY AND ANSWER First w rite the structures o f the two alkenes, then decide how many substituents the

double bond o f each has.

2 -M eth y l-1 -p e n te n e (d isu b s titu te d , l e s s s ta b le )

2 -M eth y l-2 -p e n te n e (trisu b stitu ted , m o r e sta b le )

2-Methyl-2-pentene has three substituents on its double bond, whereas 2-methyl-1-pentene has only one, and there­ fore it is the more stable.

Rank the follow ing cycloalkenes in order o f increasing stability.

* T h is o rd e r o f s ta b ilitie s m a y seem c o n tra d ic to ry w he n co m pa re d w ith th e e x p la n a tio n g iv e n fo r th e re la tiv e s ta b ilitie s o f cis and trans isom e rs. A lth o u g h a d e ta ile d e x p la n a tio n o f th e tre n d g iv e n here is b e y o n d o u r scope, th e re la tiv e s ta b ilitie s o f su bstitu te d alkenes can be ra tio n a liz e d . P art o f th e e x p la n a tio n can be g iv e n in term s o f th e e le ctro n -re le a sin g e ffe c t o f a lk y l groups (S e ctio n 6 .1 1 B ), an e ffe c t th a t satisfies th e e le c tro n -w ith d ra w in g p rop e rtie s o f th e sp2-h y b rid iz e d ca rbo n atom s o f th e d o u b le bond.

Review Problem 7.2

290

Chapter 7

Review Problem 7.3

Alkenes and Alkynes I

Heats o f hydrogenation o f three alkenes are as follows: 2-methyl-1-butene (-1 1 9 kJ m ol-1 ) 3-methyl-1-butene (-1 2 7 kJ m ol-1 ) 2-methyl-2-butene (-1 1 3 kJ m ol-1 ) (a) W rite the structure o f each alkene and classify it as to whether its doubly bonded atoms are monosubstituted, disubstituted, trisubstituted, or tetrasubstituted. (b) W rite the struc­ ture o f the product formed when each alkene is hydrogenated. (c) Can heats o f hydro­ genation be used to relate the relative stabilities o f these three alkenes? (d) I f so, what is the predicted order o f stability? I f not, why not? (e) What other alkene isomers are possi­ ble for these alkenes? W rite their structures. (f) What are the relative stabilities among just these isomers?

R e v ie w P ro b le m 7 .4

Predict the more stable alkene o f each pair: (a) 2-methyl-2-pentene or 2,3-dimethyl2-butene; (b) ds-3-hexene or trans-3-hexene; (c) 1-hexene or ds-3-hexene; (d) trans2-hexene or 2-methyl-2-pentene.

R e v ie w P ro b le m 7 .5

Reconsider the pairs o f alkenes given in Review Problem 7.4. Explain how IR spectroscopy can be used to differentiate between the members o f each pair.

7.4 Cycloalkenes The rings o f cycloalkenes containing five carbon atoms or fewer exist only in the cis form (Fig. 7.3). The introduction o f a trans double bond into rings this small would, if it were possible, introduce greater strain than the bonds o f the ring atoms could accommodate.

F ig u re 7 3



R—

R

H

C — OH

>

R — C — OH

H

3° A lcoh ol

H

2° A lcoh ol

1° A lcoh ol

This behavior, as we shall see in Section 7.7B, is related to the relative stabilities o f carbocations. 2. S om e p r im a ry a n d se c o n d a ry alco h o ls also u n d e rg o re a rra n g e m e n ts o f th e ir c a r ­ b o n sk e leto n s d u r in g d e h y d ra tio n . Such a rearrangement occurs in the dehydration

o f 3,3-dimethyl-2-butanol: CH

CH3

H3C

C-

-C H — C H

C H

3

80°C

C =C

/

,c — c h c h

3

\ CH

H3C

OH

2 ,3 -D im e th y l-2 -b u te n e

3 ,3 -D im e th y l-2 -b u ta n o l

CH

CH3

85% H3PO4

(8 0 % )

H2C 2 ,3 -D im e th y l-1 -b u te n e (2 0 %)

Notice that the carbon skeleton o f the reactant is C C — C — C — C C

C\ w h i l e t h a t o f t h e p r o d u c t s is

/ C C — C

C

C

We shall see in Section 7.8 that this reaction involves the m igration o f a methyl group from one carbon to the next so as to form a more stable carbocation. (Rearrangements to carbocations o f approximately equal energy may also be pos­ sible w ith some substrates.)

ii^V

H

7.7 Acid-Catalyzed Dehydration o f Alcohols

7.7A Mechanism for Dehydration of Secondary and Tertiary Alcohols: An E1 Reaction Explanations for these observations can be based on a stepwise mechanism originally pro­ posed by F. Whitmore (o f Pennsylvania State University). T h e m e c h a n ism is a n E 1 re a c tio n in w h ich th e s u b s tra te is a p ro to n a te d alcohol.

Consider the dehydration o f teri-butyl alcohol as an example:

Step 1

CH

-*

CH3— C— O — H

secondary > primary > methyl:

R >

3°

+'

/

o

R

>

\

H —

H /‘ + R— C \ R

>

R / R— C 4

>

\ H >

1°

(m ost stable)

>

H / H— C 4 \ H Methyl (least stable)

In the dehydration o f secondary and tertiary alcohols the slow est step is formation of the carbocation as shown in step 2 o f the “A M echanism for the Reaction” box in this sec­ tion. The first and third steps involve sim ple acid-base proton transfers, which occur very rapidly. The second step involves loss o f the protonated hydroxyl as a leaving group, a highly endergonic process (Section 6.7), and hence it is the rate-determining step. Because step 2 is the rate-determining step, it is this step that determines the overall reac­ tivity o f alcohols toward dehydration. With that in mind, w e can now understand why ter­ tiary alcohols are the most easily dehydrated. The formation o f a tertiary carbocation is easiest because the free energy o f activation for step 2 o f a reaction leading to a tertiary carboca­ tion is lowest (see Fig. 7.7). Secondary alcohols are not so easily dehydrated because the free energy o f activation for their dehydration is higher— a secondary carbocation is less sta­ ble. The free energy o f activation for dehydration o f primary alcohols via a carbocation is so high that they undergo dehydration by another mechanism (Section 7.7C).

Forming the transition state for the 3° carbocation has the lowest energy of activation

H —+1 5+ R— C — O H2 2 H

R 5+

R— C+ + H2O

5+

R— C — O H2

\

H

R R 5+

H

/

R— C+ + H2O

5+

\ H

R— C — O H2 R

2

R R

C+ + H2O 2

R R I

R

R C —OH I R

R I

H

R

C — OH

I

I

H 3

I C — OH H

2 A G (3°) < A G (2°) R — CH2— CH2— R

+

h ea t

Although the process is exothermic, there is usually a high free energy o f activation for uncatalyzed alkene hydrogenation, and therefore, the uncatalyzed reaction does not take place at room temperature. However, hydrogenation w ill take place readily at room tem­ perature in the presence o f a catalyst because the catalyst provides a new pathway for the reaction that involves lower free energy o f activation (Fig. 7.9).

No catalyst (hypothetical)

Catalyst present (usually multistep)

Figure 7.9 Free-energy diagram fo r the h yd rogenation o f an alkene in th e presence o f a catalyst and th e h yp o th e tica l reaction in the absence o f a catalyst. The free energy o f activation fo r th e uncatalyzed reaction (AG*(i)) is very much larger than th e largest fre e energy o f activation fo r th e catalyzed reaction (AG^(2)) (the uncatalyzed h yd rogenation reaction does not occur).

Reaction coordinate

Heterogeneous hydrogenation catalysts typically involve finely divided platinum, pal­ ladium, nickel, or rhodium deposited on the surface o f powdered carbon (charcoal). Hydrogen gas introduced into the atmosphere o f the reaction vessel adsorbs to the metal by a chemical reaction where unpaired electrons on the surface o f the metal p air w ith the electrons o f hydrogen (Fig. 7.10a) and bind the hydrogen to the surface. The collision of an alkene w ith the surface bearing adsorbed hydrogen causes adsorption o f the alkene as

A

^

4fe “

^

v

4fl}

Surface of metal catalyst

Figure 7.10 The mechanism fo r th e hydro g e n a tio n o f an alkene as catalyzed by fin e ly divided platinum m etal: (a) hydrogen a dsorption; (b) adso rp tio n o f th e alkene; (c, d) stepw ise tran sfe r of bo th hydrogen atom s to th e same face o f th e alkene (syn addition).

ii^V

H

7.15 Hydrogenation o f Alkynes

Ft

315

w ell (Fig. 7.10b). A stepwise transfer o f hydrogen atoms takes place, and this produces an alkane before the organic molecule leaves the catalyst surface (Figs. 7.10c,d). As a consequence, both hydrogen atoms usually add from the same side o f the molecule. This mode o f addition is called a sy n addition (Section 7.14A):

^C=C^

Pt v

c — Cv

/ H

\

H

H— H C atalytic h y d r o g e n a tio n is a s y n ad d ition .

7.14A Syn and Anti Additions An addition that places the parts o f the adding reagent on the same side (or face) o f the reactant is called sy n a d d itio n . We have just seen that the platinum-catalyzed addition of hydrogen (X = Y = H) is a syn addition: X— Y

;C = C <

V

/ X

C—

6* \ Y

A sy n a d d ition

The opposite o f a syn addition is an a n ti a d d itio n . An anti addition places the parts of the adding reagent on opposite faces o f the reactant. Y X

'C = C -

Y

^ C — C /..„ / V"

An anti ad d ition

X

In Chapter 8 we shall study a number o f important syn and anti additions.

7.15 H ydrogenation o f Alkynes Depending on the conditions and the catalyst employed, one or two molar equivalents of hydrogen w ill add to a carbon-carbon triple bond. When a platinum catalyst is used, the alkyne generally reacts w ith two molar equivalents o f hydrogen to give an alkane: c h 3c

Pt H

Pt H

# CCH3 — H2-> [CH3CH = CHCH3] — H^

c h 3c h 2c h 2c h 3

However, hydrogenation o f an alkyne to an alkene can be accomplished through the use o f special catalysts or reagents. Moreover, these special methods allow the preparation of either (E)- or (Z)-alkenes from disubstituted alkynes.

7.15A Syn Addition of Hydrogen: Synthesis of c/s-Alkenes A heterogeneous catalyst that permits hydrogenation o f an alkyne to an alkene is the nickel boride compound called P-2 catalyst. The P-2 catalyst can be prepared by the reduction of nickel acetate w ith sodium borohydride: O ... i , I N ^occhJ 2

NaBH. EtOH >

„ „ Ni2B P-2

• Hydrogenation o f alkynes in the presence o f P-2 catalyst causes syn addition of hydrogen. The alkene formed from an internal alkyne has the (Z) or cis configuration. The hydrogenation o f 3-hexyne illustrates this method. The reaction takes place on the sur­ face o f the catalyst (Section 7.14), accounting for the syn addition:

31Ó

Chapter 7

Alkenes and Alkynes I

/

H e lp f u l H i n t

H2/Ni2B (P-2)'

^

Syn addition o f hydrogen to an alkyne.

(syn addition) 3 -H e x y n e

(Z )-3 -H ex e n e (c /s -3 -h e x e n e ) (97% )

Other specially conditioned catalysts can be used to prepare ds-alkenes from disubsti­ tuted alkynes. M etallic palladium deposited on calcium carbonate can be used in this way after it has been conditioned w ith lead acetate and quinoline (an amine, see Section 20.1B). This special catalyst is known as Lindlar’s catalyst: H2, Pd/CaCO3 (Lindlar's catalyst)

R— C = C — R

R

R

\

/ C = C

quinoline

/

(syn addition)

\

H

H

7.15B Anti Addition of Hydrogen: Synthesis of trans-Alkenes o f hydrogen to the triple bond o f alkynes occurs when they are treated with lithium or sodium metal in ammonia or ethylamine at low temperatures.

•

A n t i a d d itio n

This reaction, called a d i s s o l v i n g m e t a l r e d u c t i o n , takes place in solution and produces an (£)- or trans-alkene. The mechanism involves radicals, which are molecules that have unpaired electrons (see Chapter 10). H e lp f u l H i n t

-------

(1) Li, EtNH2, —78°C (2) NHCl

Anti addition o f hydrogen to an alkyne.

4 -O cty n e

(anti addition)

(E )-4-O ctene (tra n s-4 -o c ten e) (52%)

A MECHANISM FOR THE REACTION T h e D is s o lv in g M e ta l R e d u c tio n o f a n A lk y n e R

R

P* Li

C = C



+C— C— / I

\

+

•*=X =

T h e p ele c tro n s of th e a lk e n e fo rm a bond w ith a p roton fro m H X to fo rm a c a rb o c a tio n an d a h a lid e ion.

:X :

Step 2

C— C—

fast

H I I — C— C— I I =X:

T h e h alid e ion re a c ts w ith th e c a rb o c a tio n by d o n a tin g an e le ctro n p a ir; th e re s u lt is an alkyl h a lid e .

The important step— because it is the rate-d eterm in in g step— is step 1. In step 1 the alkene donates a pair o f electrons to the proton o f the hydrogen halide and forms a carbo­ cation. This step (Fig. 8.1) is highly endergonic and has a high free energy o f activation. Consequently, it takes place slowly. In step 2 the highly reactive carbocation stabilizes itself by combining with a halide ion. This exergonic step has a very low free energy o f activa­ tion and takes place very rapidly.

X s-

Carbocation

Step~2 Reaction coordinate Free-energy diagram fo r th e a d d itio n o f HX to an alkene. The free energy of activation fo r ste p 1 is much larger than th a t fo r step 2.

Figure 8.1

x

335

336

Chapter 8 Alkenes and Alkynes II

8.2A Theoretical Explanation of Markovnikov's Rule If the alkene that undergoes addition o f a hydrogen halide is an unsymmetrical alkene such as propene, then step 1 could conceivably lead to two different carbocations: H CH 3 C H = C H 2

CH 3 CH— C H2 1° C arbocation (less stable)

CH 3 C H = C H 2 + H 9 X

CH3CH

-CH

H

X-

2° C arbocation (m o re stable)

These two carbocations are not o f equal stability, however. The secondary carbocation is more stable, and it is the greater stability o f the secondary carbocation that accounts for the correct prediction o f the overall addition by Markovnikov’s rule. In the addition o f HBr to propene, for example, the reaction takes the follow ing course: * -

CH 3 CH 2 CH 2

Br-

HBr slow

CH 3 CH= =CH0

► CH 3 CH 2 CH2Br 1 -B ro m o p ro p a n e (little fo rm e d )

1

+ BrCH 3 CHCH 3 -fa.rs^ CH 3 CHCH, Br 2 -B ro m o p ro p a n e (m a in p ro d u ct)

2

h

Step 1

Step 2

+

H

The ch ief product o f the reaction is 2-bromopropane because the more stable secondary carbocation is formed preferentially in the first step. •

The more stable carbocation predominates because it is formed faster.

We can understand why this is true if w e examine the free-energy diagrams in Fig. 8.2. Br

I

s­

H s+ 1

/

/

/

Brs-

HS+ s+ ch3ch =-Ch2 /

This transition state resembles a 2° carbocation.

s+

CH3CH = CH 2

This transition state resembles a 1° carbocation.

+ CH3CH2CH2 V». + Br-

•»*

CH3CH = CH

AG* (2°)

Figure 8.2 Free-energy diagram s fo r th e a d d itio n of HBr to propene. SG^(2°) is less than SG*(1°).

CH3CH2CH2Br

CH3CHBrCH3 Reaction coordinate

S+

rT*

8.2 Electrophilic Addition of Hydrogen Halides to Alkenes •

The reaction leading to the secondary carbocation (and ultimately to 2-bromopropane) has the lower free energy of activation. This is reasonable because its transition state resembles the more stable carbocation.

•

The reaction leading to the primary carbocation (and ultimately to 1-bromopropane) has a higher free energy of activation because its transition state resem­ bles a less stable primary carbocation. This second reaction is much slower and does not compete appreciably with the first reaction.

The reaction of HBr with 2-methylpropene produces only 2-bromo-2-methylpropane and for the same reason. Here, in the first step (i.e., the attachment of the proton) the choice is even more pronounced— between a tertiary carbocation and a primary carbo­ cation. Thus, 1-bromo-2-methylpropane is not obtained as a product of the reaction because its formation would require the formation of a primary carbocation. Such a reac­ tion would have a much higher free energy of activation than that leading to a tertiary carbocation. Because carbocations are formed in the addition of HX to an alkene, rearrangements invariably occur when the carbocation initially formed can rearrange to a more stable one (see Section 7.8 and Review Problem 8.3).

A MECHANISM FOR THE REACTION Addition of HBr to 2-Methylpropene This reaction takes place: CH 3 V C=CH2 / CH, H — Br =

ch3 I 3 . C H __C __CH M a jo r CH3 c cH 3 p ro d u c t

ch3 V C— CH2— H / CH 3 :B r;

: B r: 2 -B ro m o -2 - m e th y lp r o p a n e

3° C a rb o c a tio n ( m o re s ta b le c a r b o c a tio n )

This reaction d o e s not occur to any appreciable extent: CH 3

CH C=CH2 -----> / CH H— B r

CH 3 — C — C h 2 H

CH3 ----->

CH3— C h — CH 2 — Br

=Br =

1° C a rb o c a tio n

1 -B ro m o -2 -m e th y lp ro p a n e

( le s s s ta b le c a r b o c a tio n )

8.2B Modern Statement of Markovnikov's Rule With this understanding of the mechanism for the ionic addition of hydrogen halides to alkenes, we can now give the following modern statement of M arkovnikov’s rule. •

In the ionic addition of an unsymmetrical reagent to a double bond, the positive portion of the adding reagent attaches itself to a carbon atom of the double bond so as to yield the more stable carbocation as an intermediate.

S'

Little

fo rm e d

337

338

Chapter 8 Alkenes and Alkynes II B ecause addition o f the electrophile occurs first (before the addition o f the nucleophilic portion o f the adding reagent), it determines the overall orientation o f the addition. N otice that this formulation o f Markovnikov’s rule allow s us to predict the outcom e of the addition o f a reagent such as ICI. Because o f the greater electronegativity o f chlorine, the positive portion o f this m olecule is iodine. The addition o f ICI to 2-methylpropene takes place in the follow ing way and produces 2 -chloro- 1 -iodo- 2 -methylpropane: CH3 V C=CH „ / CH

: I — C l:

CH3 V C — CH,— I: CH

:C l =

2 -M e th y lp ro p e n e

CH 3 CH 3— C — CH,— I :Cl: 2 -C h lo ro -1 -io d o 2 -m e th y lp ro p a n e

R e v ie w P ro b le m 8.1

Give the structure and name o f the product that would be obtained from the ionic addition o f IBr to propene.

R e v ie w P ro b le m 8 .2

Outline mechanisms for the follow ing addition reactions:

R e v ie w P ro b le m 8 .3

Provide mechanistic explanations for the follow ing observations:

8.2C Regioselective Reactions Chemists describe reactions like the Markovnikov additions o f hydrogen halides to alkenes as being regioselective. Regio com es from the Latin word regionem meaning direction. •

When a reaction that can potentially yield two or more constitutional isomers actu­ ally produces only one (or a predominance o f one), the reaction is said to be regioselective.

The addition o f HX to an unsymmetrical alkene such as propene could conceivably yield two constitutional isomers, for example. A s w e have seen, however, the reaction yields only one, and therefore it is regioselective.

8.2D An Exception to Markovnikov's Rule In Section 10.9 w e shall study an exception to Markovnikov’s rule. This exception con­ cerns the addition o f HBr to alkenes when the addition is carried o u t in the p resen ce o f p eroxides (i.e., compounds with the general formula ROOR).

S+

8.3 Stereochemistry of the Ionic Addition to an Alkene •

x

S'

rT*

339

When alkenes are treated with HBr in the presence of peroxides, an antiM arkovnikov ad dition occurs in the sense that the hydrogen atom becomes attached to the carbon atom with the fewer hydrogen atoms.

With propene, for example, the addition takes place as follows: CH 3 C H = CH 2 + HBr

ROOR > CH 3 CH 2 CH2Br

In Section 10.9 we shall find that this addition occurs by a radical mechanism, and not by the ionic mechanism given at the beginning of Section 8.2. •

This anti-Markovnikov addition occurs on ly when HBr is u sed in the p resen ce o f p eroxides and does not occur significantly with HF, HCl, and HI even when perox­ ides are present.

8.3 Stereochem istry o f the Ionic A d d itio n to an A lkene Consider the following addition of HX to 1-butene and notice that the reaction leads to the formation of a product, 2 -halobutane, which contains a chirality center: c h 3 c h 2c h = c h

2

+

h x — > c h 3c h 2C h c h 3 X

The product, therefore, can exist as a pair of enantiomers. The question now arises as to how these enantiomers are formed. Is one enantiomer formed in greater amount than the other? The answer is no; the carbocation that is formed in the first step of the addition (see the following scheme) is trigonal planar and is a ch ira l (a model will show that it has a plane of symmetry). When the halide ion reacts with this achiral carbocation in the second step, reaction is equ ally likely a t eith er fa c e . The reactions leading to the two enantiomers occur at the same rate, and the enantiomers, therefore, are produced in equal amounts as a racem ic fo rm .

VjL

THE STEREOCHEMISTRY OF THE REACTION I o n ic A d d i t i o n t o a n A l k e n e

(a)

(a)

■X:'"H CH 3

C2 H 5 — C H = C H 2

C2 H5— C

(S )-2 -H a lo b u ta n e (50% )

X=

H^ X : (b) A c h ira l, trig o n al p la n a r c a rb o c a tio n

H »CH3 C2 H5- ^ c :X : (ff)-2 -H a lo b u ta n e (50% )

1 -B u te n e d o n a te s a p air o f e le ctro n s to th e p ro to n of HX to fo rm an a c h iral c a rb o c a tio n .

T h e c a rb o c a tio n re a c ts w ith th e h alid e ion a t equ al rates by path (a) o r (b) to fo rm th e e n a n tio m e rs as a racem ate.

340

Chapter S Alkenes and Alkynes II

8.4 Addition o f Sulfuric Acid to Alkenes •

When alkenes are treated with cold concentrated sulfuric acid, they dissolve because they react by electrophilic addition to form alkyl hydrogen sulfates.

The m echanism is similar to that for the addition o f HX. In the first step o f this reaction the alkene donates a pair o f electrons to a proton from sulfuric acid to form a carbocation; in the second step the carbocation reacts with a hydrogen sulfate ion to form an alkyl hydro­ gen sulfate:

/

II

C=C

H^ O

-S ­

HO3SO H I I —c—c—

O

O

\

: O'

O— H

-S ­

O— H

II

O

O

S u lfu r ic a c id

A lk e n e

C a rb o c a tio n

H y d ro g e n s u lfa te io n

A lk y l h y d ro g e n s u lfa te

S o lu b le in s u lf u r ic a c id

The addition o f sulfuric acid is also regioselective, and it follows Markovnikov’s rule. Propene, for example, reacts to yield isopropyl hydrogen sulfate rather than propyl hydrogen sulfate: H

H

H C=CH ,

c h 3— c — c h

C— CH2— H

CH

CH h

3

: O S O 3H

— O s o 3h

:O S O 3 H Is o p ro p y l h y d ro g e n s u lfa te

20 C a rb o c a tio n (m o re s ta b le c a rb o c a tio n )

8.4A Alcohols from Alkyl Hydrogen Sulfates Alkyl hydrogen sulfates can be easily hydrolyzed to alcohols by h eating them with water. The overall result o f the addition o f sulfuric acid to an alkene follow ed by hydrolysis is the Markovnikov addition o f H — and — OH: c h 3c h = c h 3

h, s o . 2 2

2

cold

ho

c h 3c h c h 3 | o s o

R e v ie w P ro b le m 8 .4

3 3

2 ^ heat

c h 3c h c h 3

3h

3

3

+

H2 S O 4 2

4

o h

In one industrial synthesis o f ethanol, ethene is first dissolved in 95% sulfuric acid. In a second step water is added and the mixture is heated. Outline the reactions involved.

8.5 A d d itio n o f W ater to Alkenes: A cid-C atalyzed Hydration The acid-catalyzed addition o f water to the double bond o f an alkene (h ydration o f an alkene) is a method for the preparation o f low-molecular-weight alcohols. This reaction has its greatest utility in large-scale industrial processes. The acids m ost com m only used to cat­ alyze the hydration o f alkenes are dilute aqueous solutions o f sulfuric acid and phosphoric acid. These reactions, too, are usually regioselective, and the addition o f water to the dou­ ble bond follow s Markovnikov’s rule. In general, the reaction takes the form that follows: C =C /

+ \

H O+ HOH

I

I

— C— C—

I I H

OH

S'

8.5 Addition of Water to Alkenes: Acid-Catalyzed Hydration

X

rT*

An example is the hydration o f 2-methylpropene: h3o

HOH

H

25 °C

2 -M e th y lp ro p e n e ( is o b u ty le n e )

HO 2 -M e th y l-2 -p ro p a n o l ( fe r f- b u ty l a lc o h o l)

Because the reactions follow Markovnikov’s rule, acid-catalyzed hydrations o f alkenes do not yield primary alcohols except in the special case o f the hydration o f ethene: H PO H— PoO4> c h 3 c h 2o h 300°C 3 2

CH2 = CH2 + HOH 2 2

8.5A Mechanism The m echanism for the hydration o f an alkene is simply the reverse o f the mechanism for the dehydration o f an alcohol. We can illustrate this by giving the mechanism for the h ydra­ tion o f 2 -methylpropene and by comparing it with the mechanism for the d ehydration of 2-methyl-2-propanol given in Section 7.7A.

A MECHANISM FOR THE REACTION A c id -C a ta ly z e d H y d ra tio n o f a n A lk e n e H

CH2 Step l

i

o

CH 2

H— O— H

CH3— C+

CH3

CH3

H

slow

Pal

H I =O — H

CH

T h e a lk e n e d o n a te s an e le c tro n p a ir to a p ro to n to fo rm th e m o re s ta b le 30 c a rb o c a tio n .

H I =O — H

CH3 c

Step 2

fast

CH3 H I 3 CH 3 — C------O — H

CH

CH

T h e c a rb o c a tio n re a c ts w ith a m o le c u le o f w a te r to fo r m a p r o to n a te d a lc o h o l.

CH3 Step S

H

CH3— C------O — H 3 I CH

H I :O — H

fast

CH3 I 3 CH 3— C— O — H

H I H— O— H

CH

A tr a n s fe r o f a p r o to n to a m o le c u le o f w a te r le a d s to th e p r o d u c t.

The rate-determining step in the hydration mechanism is step 1: the formation o f the car­ bocation. It is this step, too, that accounts for the Markovnikov addition o f water to the dou­ ble bond. The reaction produces 2-methyl-2-propanol because step 1 leads to the formation o f the more stable tertiary (3°) cation rather than the much less stable primary (1°) cation: CH / " C\ CH CH 3 CH3

H * r^l H— O— H

CH verv ‘ slow

CH 3— c CH

H :O — H

F o r a ll p ra c tic a l p u rp o s e s th is re a c tio n d o e s n o t ta k e p la c e b e c a u s e it p r o d u c e s a 1 ° c a rb o c a tio n .

341

342

Chapter 8 Alkenes and Alkynes II The reactions whereby alkenes are hydrated o r alcohols are dehydrated are reactions in which the ultimate product is governed by the position o f an equilibrium. Therefore, in the dehydration o f an alcohol it is best to use a concentrated acid so that the concentration o f water is low. (The water can be removed as it is formed, and it helps to use a high tem­ perature.) In the hydration o f an alkene it is best to use dilute acid so that the concentra­ tion o f water is high. (It also usually helps to use a lower temperature.)

C r tlw o r l D m

R e v ie w P ro b le m 8 .5

(a) Write a m echanism for the follow ing reaction. h 2s o 4 h 2° (b) What general conditions would you use to ensure a good yield o f the product? (c) What general conditions would you use to carry out the reverse reaction, i.e., the dehy­ dration o f cyclohexanol to produce cyclohexene? (d) What product would you expect to obtain from the acid-catalyzed hydration o f 1-methylcyclohexene? Explain your answer.

8.5B Rearrangements •

One complication associated with alkene hydrations is the occurrence of rearrangem ents.

Because the reaction involves the formation o f a carbocation in the first step, the carboca­ tion formed initially invariably rearranges to a more stable one (or possibly to an isoener-

S'

S.5 Addition o f W ater to Alkenes: Acid-Catalyzed Hydration

X

rT*

343

getic one) if such a rearrangement is possible. An illustration is the formation o f 2,3dimethyl-2-butanol as the major product when 3,3-dimethyl-1-butene is hydrated: H2SO4 H2O 3 ,3 -D im e th y l-1 -b u te n e

2 ,3 -D im e th y l-2 -b u ta n o l (m a jo r p ro d u ct)

Outline all steps in a mechanism showing how 2,3-dimethyl-2-butanol is formed in the acidcatalyzed hydration o f 3,3-dimethyl-1-butene.

R e v ie w P ro b le m 8 .6

The follow ing order o f reactivity is observed when the follow ing alkenes are subjected to acid-catalyzed hydration:

R e v ie w P ro b le m 8 .7

(CH 3 )2 C = C H , > CH 3 CH= CH 2 > CH 2 = CH 2 Explain this order o f reactivity. Write a mechanism for the follow ing reaction.

R e v ie w P ro b le m 8 .8

S o l v e d P r o b le m 8 .2 Write a mechanism that w ill explain the course o f the follow ing reaction cat. H2 SO4

och

3

CH, OH STRATEGY AND ANSWER A s w e have learned, in a strongly acidic medium such as methanol containing cat­ alytic sulfuric acid, an alkene can accept a proton to becom e a carbocation. In the reaction above, the 2° carboca­ tion formed initially can rearrange as shown below to becom e a 3° carbocation, which can then react with the solvent (methanol) to form an ether. H H — O — CH ,

+

H\ — 0 s o 3h

H — O+— CH ,

+ -0 S 0 3H

+ +

H —O — CH,

^O t- CH3

CH3OH

+

H— O — CH3

+

CH30H2

och3

344

Chapter 8 Alkenes and Alkynes II

8.6 Alcohols from Alkenes through O xym ercuration-D em ercuration: M arkovnikov A d d ition A useful laboratory procedure for synthesizing alcohols from alkenes that avoids rearrange­ ment is a two-step method called o x y m ercu ration -d em ercu ration . •

Alkenes react with mercuric acetate in a mixture o f tetrahydrofuran (THF) and water to produce (hydroxyalkyl)mercury compounds. These (hydroxyalkyl)mercury compounds can be reduced to alcohols with sodium borohydride.

Step 1: O xym ercuration

\

/

C=C /

o

)

( ? H g \O C C H 3 /

h 2o

2

—c—c—

THF

\

HO

o

CH3COH

Hg — OCCH 3

Step 2 : D e m e rcu ra tio n

o — C — C— HO

Mercury compounds are extremely hazardous. Before you carry out a reaction involving mercury or its compounds, you should familiarize yourself with current procedures for its use and containment. There are no satisfactory methods for disposal of mercury.

O

O H - + NaBH4

— C— C— HO

Hg— OCCH 3

Hg

CH 3 C O-

H

•

In the first step, oxym ercuration, water and mercuric acetate add to the double bond.

•

In the second step, dem ercuration, sodium borohydride reduces the acetoxymercury group and replaces it with hydrogen. (The acetate group is often abbreviated — OAc.)

Both steps can be carried out in the same vessel, and both reactions take place very rapidly at room temperature or below. The first step— oxymercuration— usually goes to com pletion within a period o f 20 s to 10 min. The second step— demercuration— normally requires less than an hour. The overall reaction gives alcohols in very high yields, usually greater than 90%.

8.6A Regioselectivity of Oxymercuration-Demercuration Oxymercuration-demercuration is also highly regioselective. •

In oxymercuration-demercuration, the net orientation o f the addition o f the ele­ ments o f water, H — and — OH, is in accordan ce with M ark o vn ik o v’s rule. The H — becom es attached to the carbon atom o f the double bond with the greater number o f hydrogen atoms. H \

H H

H /

(1) Hg(OAc)2/THF-H2O '

\

(2) NaBH4, OH-

C=C / R

R— C — C — H

H

HO

H

HO— H The follow ing are specific examples: Hg(OAc)2 thf- h2o (15 s) 1-Pentene

OH HgOAc

NaBH4 OH( 1 h)

OH

2-Pentanol (93%)

S+

rT*

8.6 Alcohols from Alkenes through Oxymercuration-Demercuration: Markovnikov ... OH

NaBH4 OH- ' ( 6 min)

Hg 1 -M e th y lc y c lo p e n ta n o l

1 -M e th y lc y c lo p e n te n e

8.6B Rearrangements Seldom Occur in Oxymercuration-Demercuration •

H e lp fu l H in t

Rearrangements o f the carbon skeleton seldom occur in oxym ercurationdemercuration.

Oxymercuration-demercuration is not prone to hydride or alkanide rearrangements.

The oxymercuration-demercuration o f 3,3-dimethyl-1-butene is a striking example illus­ trating this feature. It is in direct contrast to the hydration o f 3,3-dimethyl-1-butene w e stud­ ied previously (Section 8.5B). CHo

CH

H3 C — C — CH= CH 2

(1) Hg(OAc)2/THF-H2O (2) NaBH4, OH-

CH 9 C H

H 3 C — CCH 3

CH3

OH

3 ,3 -D im e th y l-2 -b u ta n o l (94 % )

3 ,3 -D im e th y l-1 -b u te n e

Analysis o f the mixture o f products by gas chromatography failed to reveal the presence o f any 2,3-dimethyl-2-butanol. The acid-catalyzed hydration o f 3,3-dimethyl-1-butene, by contrast, gives 2,3-dimethyl-2-butanol as the major product.

8.6C Mechanism of Oxymercuration A m echanism that accounts for the orientation o f addition in the oxymercuration stage, and one that also explains the general lack o f accompanying rearrangements, is shown below. •

Central to this mechanism is an electrophilic attack by the mercury species, HgOAc, at the less substituted carbon o f the double bond (i.e., at the carbon atom that bears the greater number o f hydrogen atoms), and the formation o f a bridged intermediate.

We illustrate the mechanism using 3,3-dim ethyl-1-butene as the example:

A MECHANISM FOR THE REACTION O x y m e rc u ra tio n Hg(OAc ) 2 e F

Step 1

figOAc

OAc-

M e rc u ric a c e ta te d is s o c ia te s to fo rm a H g O A c c a tio n a n d an a c e ta te a n io n .

Step 2

345

OH

HgOAc

Hg(OAc) 2 THF-H2O ( 2 0 s)

S'

CH 3 I 3 CH 3— C— C H = C H 2 CH 3 3 ,3 -D im e th y l-1 -b u te n e

CH, + +

HgOAc

CH 3 — C— CH— CH 2 3 I I 2 CH 3 ' HgOAc M e rc u ry -b rid g e d c a rb o c a tio n

The alkene donates a pair o f electrons to the electrophilic H gO A c+ cation to form a m ercury-bridged carbocation. In this carbocation, the positive charge is shared between the 2 ° (m ore substituted) carbon atom and the m ercury atom . The charge on the carbon atom is large enough to account for the M arkovnikov orientation of the addition, but not large enough for a rearrangem ent to occur.

346

Chapter 8 Alkenes and Alkynes II

H ' :0 — H

CH, Step 3

H3C +:0 — H

I

H3 C— C— CH— CH2

H3 C— C— CH— CH2

^H g O A c

CH3

H3C

HgOAc

A w a te r m o le c u le a tta c k s th e c a rb o n o f th e b rid g e d m e r c u r in iu m io n t h a t is b e tte r a b le to b e a r th e p a rtia l p o s itiv e c h a rg e . H

H

+ 1*9

Step 4

H3C +: O — H 3I H3 C— C— CH— CH2

:0 — H

I

H

H3C : 0 : 3| H3 C— C— CH— CH2

H

O— H

I H g0A c

H3C

HgOAc

H3C

H

( H y d ro x y a lk y l) m e r c u ry com pound A n a c id - b a s e re a c tio n tr a n s fe r s a p ro to n to a n o th e r w a te r m o le c u le ( o r to an a c e ta te io n ). T h is s te p p ro d u c e s th e ( h y d r o x y a lk y l) m e r c u r y c o m p o u n d .

Calculations indicate that mercury-bridged carbocations (termed mercurinium ions) such as those formed in this reaction retain much o f the positive charge on the mercury moiety. Only a small portion o f the positive charge resides on the more substituted carbon atom. The charge is large enough to account for the observed Markovnikov addition, but it is too small to allow the usual rapid carbon skeleton rearrangements that take place with more fully developed carbocations. Although attack by water on the bridged mercurinium ion leads to anti addition o f the hydroxyl and mercury groups, the reaction that replaces mercury with hydrogen is not stereocontrolled (it likely involves radicals; see Chapter 10). This step scrambles the over­ all stereochemistry.

R e v ie w P ro b le m 8 .9

•

The net result o f oxymercuration-demercuration is a mixture o f syn and anti addi­ tion o f — H and — OH to the alkene.

•

A s already noted, oxymercuration-demercuration takes place with Markovnikov regiochemistry.

Write the structure o f the appropriate alkene and specify the reagents needed for synthesis o f the follow ing alcohols by oxymercuration-demercuration: (a)

\

OH

(b)

OH

(c) /

oh

When an alkene is treated with mercuric trifluoroacetate, Hg(O 2 CCF3)2, in THF contain­ ing an alcohol, ROH, the product is an (alkoxyalkyl)mercury compound. Treating this prod­ uct with NaBH 4 /OH~ results in the formation o f an ether. •

When a solvent m olecule acts as the nucleophile in the oxymercuration step the over­ all process is called solvom ercuration-dem ercuration: R0

\= C y

/

\

Hg(02CCF3)2/THF-R0H solvomercuration

R0

— C — C— H g 0 2 CCF 3

A lk e n e

( A lk o x y a lk y l) m e r c u r ic tr iflu o r o a c e ta te

NaBH4, OH demercuration

— C — CH E th e r

8.8 Hydroboration: Synthesis of Alkylboranes

347

x

(a) Outline a likely mechanism for the solvomercuration step o f the ether synthesis just shown.

R e v ie w P r o b le m 8 .1 0

(b) Show how you would use solvomercuration-demercuration to prepare ieri-butyl methyl ether. (c) Why would one use Hg(O 2 CCF 3 ) 2 instead o f Hg(OAc)2?

8.7 Alcohols from Alkenes through H ydro b o ratio n -O xid atio n : A n ti-M arkovnikov Syn Hydration •

A nti-M arkovnikov hydration o f a double bond can be achieved through the use of diborane (B 2 H6) or a solution o f borane in tetrahydrofuran (BH 3 :THF).

The addition o f water is indirect in this process, and two reactions are involved. The first is the addition o f a boron atom and hydrogen atom to the double bond, called hydroboration ; the second is oxidation and hydrolysis o f the alkylborane intermediate to an alcohol and boric acid. The anti-Markovnikov regiochemistry o f the addition is illustrated by the hydroboration-oxidation o f propene: BHjMHF hydroboration

B

P ro p e n e

•

Hj.O./OHoxidation

OH

3

T rip ro p y lb o ra n e

1-P ro p a n o l

Hydroboration-oxidation takes place with syn stereochemistry, as w ell as antiMarkovnikov regiochemistry.

This can be seen in the follow ing example with 1-methylcyclopentene: CH3

CH 3

(1) BH3 :THF (2) H20 2, HO-

H

OH

In the following sections w e shall consider details o f the m echanism that lead to the antiMarkovnikov regiochemistry and syn stereochemistry o f hydroboration-oxidation.

8.8 Hydroboration: Synthesis o f Alkylboranes Hydroboration o f an alkene is the starting point for a number o f useful synthetic proce­ dures, including the anti-Markovnikov syn hydration procedure w e have just mentioned. Hydroboration was discovered by Herbert C. Brown (Purdue University), and it can be rep­ resented in its simplest terms as follows: ' /

c

-

c'

H — B/

hydroboration

— C — C—

\

\

H / B\ A lk e n e

B o ro n h yd rid e

A lk y lb o ra n e

Hydroboration can be accom plished with diborane (B 2 H6), which is a gaseous dimer of borane (BH3), or more conveniently with a reagent prepared by dissolving diborane in THF. When diborane is introduced to THF, it reacts to form a Lewis acid-base com plex o f borane (the Lew is acid) and THF. The com plex is represented as BH3 :THF. H B2 H6

2

:Q:

2 H— B — Q H

D ib o ra n e

THF (te tra h y d ro fu ra n )

B H 3:T H F

e

Brown's discovery of hydroboration led to his

being named a co-winner of the 1979 Nobel Prize in Chemistry.

348

Chapter 8 Alkenes and Alkynes II Solutions containing the BH3:THF com plex can be obtained commercially. Hydroboration reactions are usually carried out in ethers: either in diethyl ether, (CH 3 CH 2 )2 O, or in some higher molecular w eight ether such as “diglym e” [(CH 3 OCH 2 CH 2 )2 O, ¿¿ethylene glycol dimethyl ether]. Great care m ust be used in handling diborane and alkylboranes because they ignite spontaneously in air (with a green flame). The solution o f BH3:THF must be used in an inert atmosphere (e.g., argon or nitrogen) and with care.

8.8A Mechanism of Hydroboration When a terminal alkene such as propene is treated with a solution containing BH3 :THF, the boron hydride adds successively to the double bonds o f three m olecules o f the alkene to form a trialkylborane: M o re s u b s titu te d c a rb o n

Less s u b s titu te d carb o n

H T rip ro p y lb o ra n e

•

In each addition step the boron atom becom es a ttach ed to the less su b stitu ted car­ bon atom o f the dou ble bond, and a hydrogen atom is transferred from the boron atom to the other carbon atom o f the double bond.

•

Hydroboration is regioselective and it is anti-M arkovnikov (the hydrogen atom becom es attached to the carbon atom with fewer hydrogen atoms).

Other exam ples that illustrate the tendency for the boron atom to becom e attached to the less substituted carbon atom are shown here. The percentages designate where the boron atom becom es attached.

T h e s e p e rc e n ta g e s , in d ic a tin g w h e re b oron b e c o m e s a tta c h e d in reaction s u sin g th e s e s ta rtin g m a te ria ls , illu s tra te th e te n d e n c y fo r boron to bond a t th e less s u b s titu te d c a rb o n o f th e d o u b le b o n d .

This observed attachment o f boron to the less substituted carbon atom o f the double bond seem s to result in part from steric factors— the bulky boron-containing group can approach the less substituted carbon atom more easily. In the mechanism proposed for hydroboration, addition o f BH3 to the double bond begins with a donation o f p electrons from the double bond to the vacant p orbital o f BH3 (see the m echanism on the follow ing page). In the next step this com plex becom es the addition product by passing through a four-atom transition state in which the boron atom is partially bonded to the less substituted carbon atom o f the double bond and one hydrogen atom is partially bonded to the other carbon atom. A s this transition state is approached, electrons shift in the direction o f the boron atom and away from the more substituted carbon atom o f the double bond. This makes the more substituted carbon atom develop a partial posi­ tive charge, and because it bears an electron-releasing alkyl group, it is better able to accom m odate this p o sitive charge. Thus, electronic factors also favor addition o f boron at the least substituted carbon. •

Overall, both electronic and steric fa cto rs account for the anti-Markovnikov orien­ tation o f the addition.

S+ $-

8.8 Hydroboration: Synthesis of Alkylboranes

x

rT*

349

A MECHANISM FOR THE REACTION H y d ro b o ra tio n

H

H 3 C(.... .

H 3 C ".> = = < .>"H H H

+

H

H— B<

H — B-

-

H

H

H H — B'

H

H3 C /// H

H

1

" H

H”

H

H

n Complex

j.

I

s

B$ - 'H H

Four-atom concerted tra nsition state

H3 C > H ¿r H

^

„\'H - H B -.(H V H H

Syn addition of H and B

Addition takes place through the in itia l form ation of a n complex, w hich changes into a cyclic four-atom tra nsition state w ith the boron adding to the less hindered carbon atom. The dashed bonds in the tra nsition state represent bonds that are p a rtia lly formed or pa rtia lly broken. The transition state results in syn addition of the hydrogen and boron group, leading to an alkylborane. The other B-H bonds of the alkylborane can undergo sim ila r additions, leading fin a lly to a tria lkylb ora ne. An orbital view of hydroboration

f

t

H 3 C/,,,

H

H

H

H3C H

W ...i" f

ond

H3 C . 3 yo H fc i

H ¿ -- H

ovH H

5

f

H Syn addition of hydrogen and boron

n Complex (donation of alkene n electron density to the vacant boron p orbital)

Borane vacant p orbital

Four-atom concerted tra nsition state (results in syn addition)

8.8B Stereochemistry of Hydroboration •

The transition state for hydroboration requires that the boron atom and the hydro­ gen atom add to the same face o f the double bond: Stereochemistry o f Hydroboration /(/,. a\\\ syn addition

H

H H — B

H We can see the results o f a syn addition in our example involving the hydroboration o f 1-methylcyclopentene. Formation o f the enantiomer, which is equally likely, results when the boron hydride adds to the top face o f the 1 -methylcyclopentene ring:

CH,

CH3

syn addition ^ anti-Markovnikov

enantiom er

H H

^ H

350

Review Problem 8.11

Chapter 8 Alkenes and Alkynes II

S p ecify the alkene needed for synthesis o f each o f the follow in g alkylboranes by hydroboration: (a)

(c) B

(d) Show the stereochemistry involved in the hydroboration o f 1-methylcyclohexene.

R e v ie w P ro b le m 8 .1 2

Treating a hindered alkene such as 2-methyl-2-butene with BH3:THF leads to the forma­ tion o f a dialkylborane instead o f a trialkylborane. When 2 m ol o f 2-methyl-2-butene is added to 1 m ol o f BH3, the product formed is bis(3-methyl-2-butyl)borane, nicknamed “disiamylborane.” Write its structure. Bis(3-methyl-2-butyl)borane is a useful reagent in certain syntheses that require a sterically hindered borane. (The name “disiam yl” com es from “disecondary-iso-am yl,” a com pletely unsystematic and unacceptable name. The name “amyl” is an old common name for a five-carbon alkyl group.)

8.9 O xidation and Hydrolysis o f Alkylboranes The alkylboranes produced in the hydroboration step are usually not isolated. They are o xi­ dized and hydrolyzed to alcohols in the same reaction vessel by the addition o f hydrogen peroxide in an aqueous base: ^ , H2O2 aq. NaOH, 25°C ... . R3B ■ 2 2 ------------- 1----------- » 3R — OH + B(ONa ) 3 3 (oxdidation and hydrolysis) •

T h e o x id a tio n and h y d ro ly sis step s tak e p la ce w ith reten tio n o f co n fig u ra ­ tion at the carbon initially bearing boron and ultim ately bearing the hydroxyl group.

We shall see how this occurs by considering the m echanism s o f oxidation and hydrolysis. Alkylborane oxidation begins with addition o f a hydroperoxide anion (HOO- ) to the trivalent boron atom. An unstable intermediate is formed that has a formal negative charge on the boron. Migration o f an alkyl group with a pair o f electrons from the boron to the adjacent oxygen leads to neutralization o f the charge on boron and displacement o f a hydrox­ ide anion. The alkyl migration takes place with retention o f configuration at the migrating carbon. Repetition o f the hydroperoxide anion addition and migration steps occurs twice more until all o f the alkyl groups have becom e attached to oxygen atoms, resulting in a tri­ alkyl borate ester, B(OR)3. The borate ester then undergoes basic hydrolysis to produce three m olecules o f the alcohol and an inorganic borate anion.

S+

S'

rT*

8.9 Oxidation and Hydrolysis of Alkylboranes

351

A MECHANISM FOR THE REACTION O x id a tio n o f T ria lk y lb o ra n e s R -I •• ÇM.. R— B— 0 — 0. . — H

R R— B

\

=0 — 0 — H

\

R— 0 ¡0 — H

B— 0 — R

/

R Unstable intermediate TrialkylHydroperoxide borane ion The boron atom accepts an electron pair from the hydroperoxide ion to form an unstable intermediate.

R epeat

B— 0 — R

sequence tw ice

R— 0 Borate ester

An alkyl group migrates from boron to the adjacent oxygen atom as a hydroxide ion departs. The configuration at the migrating carbon remains unchanged.

Hydrolysis of the Borate Ester R R" \

O/ R B— O:

R— 0 R— 0 :

R- O/ B o ;Trialkyl borate | ester h

0: B*. .. 0:

R— 0 :

R— 0 = ..

B— 0

R— 0

+ :0 H^—

R

B— 0 : R— 0 :

H

Hydroxide anion attacks the boron atom of the borate ester.

An alkoxide anion departs from the borate anion, reducing the formal change on boron to zero.

Proton transfer completes the formation of one alcohol molecule. The sequence repeats until all three alkoxy groups are released as alcohols and inorganic borate remains.

8.9A Regiochemistry and Stereochemistry of Alkylborane Oxidation and Hydrolysis •

Hydroboration-oxidation reactions are regioselective; the net result o f hydroboration-oxidation is anti-M arkovnikov addition o f water to an alkene.

•

A s a consequence, hydroboration-oxidation gives us a method for the preparation o f alcohols that cannot normally be obtained through the acid-catalyzed hydration o f alkenes or by oxymercuration-demercuration.

For example, the acid-catalyzed hydration (or oxymercuration-demercuration) o f 1-hexene yields 2-hexanol, the Markovnikov addition product. 0H H30 +, H20 1-Hexene

2-Hexanol

In contrast, hydroboration-oxidation o f 1-hexene yields 1-hexanol, the anti-Markovnikov product. ( 1) BH3 : THF^ (2) H 2 0 2 , H 0 1-Hexene

^

^

^

1-Hexanol (90%)

(1 ) BH 3 : TH F, (2) H 2 0 2, H 0 2-M ethyl-2-butene

OH

0H 3-Methyl-2-butanol

+

:0 — R H Alcohol

352

Chapter 8 Alkenes and Alkynes II •

Figure 8.3 The h y d ro b o ra tio n -o x id a tio n of 1-m ethylcyclopentene. The first reaction is a syn a d d itio n of borane. In this illustration w e have shown th e boron and hydrogen entering from th e b o tto m side of 1-m ethylcyclopentene. The reaction also takes place from th e to p side at an equal rate to produce th e enantiom er. In th e second reaction the boron atom is replaced by a hydroxyl g ro u p w ith re te n tio n of co n fig ura tio n . The p ro d u ct is trans-2-m ethylcyclopentanol, and th e overall result is th e syn a d d itio n o f — H and — OH.

R e v ie w P ro b le m 8 .1 3

Hydroboration-oxidation reactions are stereospecific; the net addition of — H and — OH is syn, and if chirality centers are formed, their configuration depends on the stereochemistry of the starting alkene.

Because the oxidation step in the hydroboration-oxidation synthesis of alcohols takes place with retention of configuration, the hydroxyl group replaces the boron atom where it stands in the alkylboron compound. The net result of the two steps (hydroboration and oxidation) is the syn addition of — H and — OH. We can review the anti-Markovnikov and syn aspects of hydroboration-oxidation by considering the hydration of 1 -methylcyclopentene, as shown in Fig. 8.3. BH,:THF Hydroboration: Anti-Markovnikov and syn addition

CH H

3

enantiomer

CH3

, HOOxidation: Boron group is replaced with retention of configuration h2o

enantiomer OH

Specify the appropriate alkene and reagents for synthesis of each of the following alcohols by hydroboration-oxidation. (a)

(b)

„OH

(c) OH

(d)

(e)

OH

C r tli/ o r l D m

a -3

>ch3

OH

(f)

OH

OH

8.11 Protonolysis of Alkylboranes

x

353

8.1 0 Summary o f Alkene Hydration Methods The three methods w e have studied for alcohol synthesis by addition reactions to alkenes have different regiochemical and stereochemical characteristics. 1. Acid-catalyzed hydration o f alkenes takes place with Markovnikov regiochemistry but may lead to a mixture o f constitutional isomers if the carbocation intermediate in the reaction undergoes rearrangement to a more stable carbocation. 2. Oxymercuration-demercuration occurs with Markovnikov regiochemistry and results in hydration o f alkenes without complication from carbocation rearrangement. It is often the preferred choice over acid-catalyzed hydration for Markovnikov addition. The overall stereochemistry o f addition in acid-catalyzed hydration and oxymercuration-demercuration is not controlled— they both result in a mixture o f cis and trans addition products. 3. H ydroboration-oxidation results in anti-M arkovnikov and syn hydration o f an alkene. The com plem entary regiochem ical and stereochem ical aspects o f these m ethods provide useful alternatives when w e desire to synthesize a sp ecific alcohol by hydration o f an alkene. We summarize them here in Table 8.1.

Sum m ary o f M eth ods fo r C onverting an A lken e to an Alcohol R eaction

R egiochem istry

S te re o c h e m istry 3

O ccu rren ce o f R e a rra n g e m e n ts

A cid-catalyzed hydration O xym ercuration-dem ercuration H ydroboration-oxidation

Markovnikov addition Markovnikov addition A nt i- M a rkovn ikov addition

N ot controlled N ot controlled Stereospecific: syn addition of H— and — OH

F requent Seldom Seldom

aA ll o f these m ethods produce racem ic m ixtures in the absence o f a chiral influence.

8.11 Protonolysis o f Alkylboranes Heating an alkylborane with acetic acid causes cleavage o f the carbon-boron bond and replacement with hydrogen: R— B \ A lk y lb o r a n e

heat

R— H

+ c h 3 c o 2— b 3 2 \

A lk a n e

•

Protonolysis o f an alkylborane takes place with retention o f configuration; hydro­ gen replaces boron w here it stands in the alkylborane.

•

The overall stereochemistry o f hydroboration-protonolysis, therefore, is syn (like that o f the oxidation o f alkylboranes).

Hydroboration follow ed by protonolysis o f the resulting alkylborane can be used as an alter­ native method for hydrogenation o f alkenes, although catalytic hydrogenation (Section 7.13) is the more common procedure. Reaction o f alkylboranes with deuterated or tritiated acetic acid also provides a very useful way to introduce these isotopes into a compound in a spe­ cific way.

354

R e v ie w P ro b le m 8 .1 4

Chapter 8 Alkenes and Alkynes II

Starting with any needed alkene (or cycloalkene) and assuming you have deuterioacetic acid (CH 3 CO 2 D) available, outline syntheses o f the follow ing deuterium-labeled compounds. CH 3 (a) (CH 3 )2 CHCH 2 CH 2 D

(b) (C H ^ C H C H D C H

(c)

(+ enantiomer) D

(d) Assuming you also have available BD3:THF and CH 3 CO 2 T, can you suggest a synthesis o f the following? D C H 3

(+ enantiomer)

H

8.12 Electrophilic A d d itio n o f Bromine and Chlorine to Alkenes A lkenes react rapidly with bromine and chlorine in nonnucleophilic solvents to form v ic­ inal dihalides. An example is the addition o f chlorine to ethene. H H

Cl

H \ = / C

H

Cl,

H

H— C — C — H

Cy C\

H

E th en e

Cl

1 ,2 -D ic h lo ro e th a n e

This addition is a useful industrial process because 1,2-dichloroethane can be used as a sol­ vent and can be used to make vinyl chloride, the starting material for poly(vinyl chloride). Cl

H

H— C — C — H H

Cl

H E2 base (-HCl)

1 ,2 -D ic h lo ro e th a n e

\ /

H H

H __ / \

H

polymerization ^ (see Section 10.10)

-C — C -

Cl

H

V in yl c h lo rid e

Cl

Other exam ples o f the addition o f halogens to a double bond are the following: CI CL -9°C Cl fra n s -2 -B u te n e

C y c lo h e x e n e

n

P o ly (v in yl c h lo rid e )

m e s o -1 ,2 -D ic h lo ro b u ta n e

fra n s -1 ,2 -D ib ro m o c y c lo h e x a n e (ra c e m ic )

S+

8.12 Electrophilic Addition of Bromine and Chlorine to Alkenes These two examples show an aspect o f these additions that w e shall address later when w e examine a m echanism for the reaction: T he addition o f halogens is an anti addition to the d ouble bond. When bromine is used for this reaction, it can serve as a test for the presence o f car­ bon-carbon multiple bonds. If w e add bromine to an alkene (or alkyne, see Section 8.18), the red-brown color o f the bromine disappears almost instantly as long as the alkene (or alkyne) is present in excess: Br

Br i_>i 2

room temperature in the dark, CCl,

— C— C— Br

An alk en e (c o lo rle s s )

W c-D ibrom ide (a c o lo rle s s com pound)

R apid d e c o lo riza tio n o f Br 2/C C l 4 is a p o s itiv e te s t for a lk e n e s an d a lkyn es.

This behavior contrasts markedly with that o f alkanes. Alkanes do not react appreciably with bromine or chlorine at room temperature and in the absence o f light. When alkanes do react under those conditions, however, it is by substitution rather than addition and by a m echanism involving radicals that w e shall discuss in Chapter 10: R— H A lk a n e (c o lo rle s s )

H

room temperature in the dark, CCl/

B r2

no appreciable reaction

B ro m in e (re d -b ro w n )

8.12A Mechanism of Halogen Addition A possible mechanism for the addition o f a bromine or chlorine to an alkene is one that involves the formation o f a carbocation. Br •*.. r * . . I + ¡Br— Bn — > ----- C — C

\

Br Br . . + =Br: — > ------ C — C-

Although this mechanism is similar to ones w e have studied earlier, such as the addition o f H — X to an alkene, it does not explain an important fact. A s w e have just seen (in Section 8 . 1 2 ) the addition o f bromine or chlorine to an alkene is an anti ad d ition . The addition o f bromine to cyclopentene, for example, produces irans-1,2-dibromocyclopentane, not d s - 1 ,2 -dibromocyclopentane. Br +

¡Br - Bn

addition

Br

^ --------->

not

Br

± -------- b

Br trans -1 ,2 -D ib ro m o c y c lo p e n ta n e (as a ra c e m ic m ix tu re )

c /s -1 ,2 -D ib ro m o c y c lo h e x a n e (a m e s o c o m p o u n d )

A m echanism that explains anti addition is one in which a bromine m olecule transfers a bromine atom to the alkene to form a cyclic b rom onium ion and a bromide ion, as shown in step 1 o f “A M echanism for the Reaction” that follow s. The cyclic bromonium ion causes net anti addition, as follows. In step 2, a bromide ion attacks the back side o f either carbon 1 or carbon 2 o f the bromo­ nium ion (an SN2 process) to open the ring and produce the irans-1,2-dibromide. Attack occurs from the side op posite the brom ine o f the b rom onium ion because attack from this direction is unhindered. Attack at the other carbon o f the cyclic bromonium ion pro­ duces the enantiomer.

S'

rT*

355

356

Chapter 8 Alkenes and Alkynes II

A MECHANISM FOR THE REACTION A d d itio n o f B r o m in e t o a n A lk e n e Step 2

Step 1

:B r:

/ C— C.

0 /,.

^ Cr s+

s-

n

C^

:B r:

C— C

:B r:

../

\ / Br

Br

+

enantiomer

V'""

:B r

l~2

: B r:

B ro m o n iu m io n

B ro m o n iu m io n

B ro m id e io n

A s a b ro m in e m o le c u le a p p ro a c h e s an a lke n e , th e e le c tro n d e n s ity o f th e a lk e n e p b o n d re p e ls e le c tro n d e n s ity in th e c lo s e r b ro m in e , p o la riz in g th e b ro m in e m o le c u le a n d m a k in g th e c lo s e r b ro m in e a to m e le c tro p h ilic . T h e a lk e n e d o n a te s a p a ir o f e le c tro n s to th e c lo s e r b ro m in e , c a u s in g d is p la c e m e n t o f th e d is ta n t b ro m in e a to m . A s th is o c c u rs , th e n e w ly b o n d e d b ro m in e a to m , d u e to its s iz e a n d p o la riz a b ility , d o n a te s an e le c tro n p a ir to th e c a rb o n th a t w o u ld o th e rw is e be a c a rb o c a tio n , th e re b y s ta b iliz in g th e p o s itiv e c h a rg e b y d e lo c a liz a tio n . T he re s u lt is a b rid g e d b ro m o n iu m io n in te rm e d ia te .

B ro m id e io n

W c-D ib ro m id e

A b ro m id e a n io n a tta c k s a t th e b a c k s id e o f o n e c a rb o n (o r th e o th e r) o f th e b ro m o n iu m io n in an S N2 re a c tio n , c a u s in g th e r in g to o p e n a n d re s u ltin g in th e fo rm a tio n o f a W c-d ib ro m id e .

This process is shown for the addition o f bromine to cyclopentene below.

S - :B>: *

^

S+ =Br=N

+ 'Br' + C y c lic b ro m o n iu m io n

:Br = B ro m id e io n

P la n e o f s y m m e tr y

. .

I ; Br'

Br

Br

=Br=

C y c lic b ro m o n iu m io n

Br

Br

fr a n s - 1 ,2 - D ib r o m o c y c lo p e n ta n e (a s a ra c e m ic m ix tu r e )

Attack at either carbon o f the cyclopentene bromonium ion is equally likely because the cyclic bromonium ion is symmetric. It has a vertical plane o f symmetry passing through the bromine atom and halfway between carbons 1 and 2. The trans-dibromide, therefore, is formed as a racemic mixture. The mechanisms for addition o f Cl2 and I2 to alkenes are similar to that for Br2, involv­ ing formation and ring opening o f their respective h alonium io n s. A s with bridged mercurinium ions, the bromonium ion does not necessarily have sym ­ m etrical charge distribution at its two carbon atoms. If one carbon o f the bromonium ion

8.12 Electrophilic Addition of Bromine and Chlorine to Alkenes

357

x

THE CHEMISTRY OF . . . T h e S e a : A T r e a s u r y o f B io lo g ic a lly A c t i v e N a t u r a l P r o d u c t s

Dactylyne, a halogenated m arine natural product.

The world's oceans are a vast storehouse of dissolved halide ions. The concentration of halides in the ocean is approxi­ mately 0.5 M in chloride, 1 mM in bromide, and 1 mM in iodide ions. Perhaps it is not surprising, then, that marine organisms have incorporated halogen atoms into the struc­ tures of many of their metabolites. Among these are such

CU

Br J

CL

intriguing polyhalogenated com pounds as halomon, dacty­ lyne, tetrachloromertensene, (3E)-laureatin, and (3R)- and (3S)-cyclocymopol. Just the sheer number of halogen atoms in these m etabolites is cause for wonder. For the organisms that make them, som e of these m olecules are part of defense mechanisms that serve to promote the species' sur­ vival by deterring predators or inhibiting the growth of com ­ peting organisms. For humans, the vast resource of marine natural products shows ever-greater potential as a source of new therapeutic agents. Halomon, for example, is in preclinical evaluation as a cytotoxic agent against certain tumor cell types, dactylyne is an inhibitor of pentobarbital m etab­ olism, and the cyclocymopol enantiomers show agonistic or antagonistic effects on the human progesterone receptor, depending on which enantiomer is used.

Br

Cl Br

Br OH Halomon

The biosynthesis of certain halogenated marine natural products is intriguing. Som e of their halogens appear to have been introduced as electrophiles rather than as Lewis bases or nucleophiles, which is their character when they are solutes in seawater. But how do marine organisms transform nucleophilic halide anions into electrophilic species for incor­ poration into their metabolites? It happens that many marine organisms have enzymes called haloperoxidases that convert nucleophilic iodide, bromide, or chloride anions into elec­ trophilic species that react like l+ , Br+ , or Cl+. In the biosyn­ thetic schem es proposed for som e halogenated natural products, positive halogen intermediates are attacked by

electrons from the p bond of an alkene or alkyne in what is called an addition reaction. The final Learning Group Problem for this chapter asks you to propose a schem e for biosynthesis of the marine nat­ ural product kumepaloxane by electrophilic halogen addi­ tion. Kumepaloxane is a fish antifeedant synthesized by the Guam bubble snail Haminoea cymbalum, presumably as a d efen se mechanism for the snail. In later chapters we shall see other exam ples of truly remarkable marine natural prod­ ucts, such as brevetoxin B, associated with deadly "red tides," and eleutherobin, a promising anticancer agent.

is more highly substituted than the other, and therefore able to stabilize positive charge better, it m ay bear a greater fraction o f positive charge than the other carbon (i.e., the positively charged bromine draws electron density from the two carbon atoms o f the ring,

358

Chapter 8 Alkenes and Alkynes II but not equally if they are o f different substitution). Consequently, the more positively charged carbon m ay be attacked by the reaction nucleophile more often than the other carbon. However, in reactions with symmetrical reagents (e.g., Br2, Cl2, and I2) there is no observed difference. (We shall discuss this point further in Section 8.14, where we w ill study a reaction where w e can discern regioselectivity o f attack on a halonium ion by the nucleophile.)

8.13 Stereospecific Reactions The anti addition o f a halogen to an alkene provides us with an example o f what is called a stereospecific reaction . •

A reaction is stereospecific when a particular stereoisomeric form o f the starting mater­ ial reacts by a mechanism that gives a specific stereoisomeric form o f the product.

Consider the reactions o f cis- and trans-2-butene with bromine shown below. When trans-2-butene adds bromine, the product is the m eso compound, (2R,3S)-2,3-dibromobutane. When cis-2-butene adds bromine, the product is a racemic mixture o f (2R,3R)-2,3dibromobutane and (2S,3S)-2,3-dibromobutane: R e a c tio n 1

H

CH3

CCL ch

h

3

Br

C H3

h^

c

Br

Br,

\""H

^

H' j ^ b v CH 3 CH 3 C

Br

C h3

Br E

(2R ,3 S )-2 ,3 -D ib ro m o b u ta n e

fra n s -2 -B u te n e

(a m e s o co m p o u n d )

R e a c tio n 2

H

/

H Br,

CCI4 CH 3

W

c/s -2 -B u te n e

Br H \ ___ ^ C H 3 H"7 ch3 CH

\

H Br C H 3 ^ ___ / Br +

Br

(2 R ,3 R )

/ Br

\ H ch 3

(2 S ,3 S )

(a p air o f e n a n tio m e rs)

The reactants cis-2-butene and trans-2-butene are stereoisomers; they are diastereomers. The product o f reaction 1, (2R,3S)-2,3-dibromobutane, is a m eso compound, and it is a stereoisomer o f both o f the products o f reaction 2 (the enantiomeric 2,3-dibromobutanes). Thus, by definition, both reactions are stereospecific. One stereoisomeric form o f the reactant (e.g., trans-2 -butene) gives one product (the meso compound), whereas the other stereoisomeric form o f the reactant (cis- 2 -butene) gives a stereoisomerically different product (the enantiomers). We can better understand the results o f these two reactions if w e exam ine their m echa­ nisms. The first mechanism in the follow ing box shows how cis-2-butene adds bromine to yield intermediate bromonium ions that are achiral. (The bromonium ion has a plane o f symmetry.) These bromonium ions can then react with bromide ions by either path (a) or path (b). Reaction by path (a) yields one 2,3-dibromobutane enantiomer; reaction by path (b) yields the other enantiomer. The reaction occurs at the same rate by either path; there­ fore, the two enantiomers are produced in equal amounts (as a racemic form). The second m echanism in the box shows how trans-2-butene reacts at the bottom face to yield an intermediate bromonium ion that is chiral. (Reaction at the other face would produce the enantiomeric bromonium ion.) Reaction o f this chiral bromonium ion (or its enantiomer) with a bromide ion either by path (a) or by path (b) yields the same achiral product, meso-2,3-dibromobutane.

8.14 Halohydrin Formation

X

rT *

359

THE STEREOCHEMISTRY OF THE REACTION Addition of Bromine to cis- and trans-2-Butene c/s-2-Butene reacts with bromine to yield the enantiomeric 2,3-dibromobutanes by the following mechanism: Br

:Br:#H ' O=O " OH OH

(â) (b)

H

OH

OH 3

OH X O— O ^ H 'V \ OH Br

(a)

^C(a)

(b)

Br

: Br i«+

(2 R ,3 R )-2 ,3 -D ib ro m o b u ta n e (c h ira l)

O

OH

(b)

B ro m o n iu m ion (a c h ira l)

Ç B r :«-

H

Br

3^ OO— O^ ^„H Br

^OH 3

( 2 S ,3 S )-2 ,3 -D ib ro m o b u ta n e (ch ira l)

c /s -2 -B u te n e re acts w ith b ro m in e to y ie ld an a c h ira l b ro m o n iu m ion a n d a b ro m id e ion. [R e a c tio n a t the o th e r fa c e o f th e a lk e n e (to p ) w o u ld y ie ld th e s a m e b ro m o n iu m ion.]

T h e b ro m o n iu m ion re a c ts w ith th e b ro m id e io n s a t e q u a l ra te s b y p ath s (a) a n d (b ) to y ie ld th e tw o e n a n tio m e rs in equ al a m o u n ts (i.e., as th e ra c e m ic fo rm ).

trans-2-Butene reacts with bromine to yield meso-2,3-dibromobutane. Br (a)

; Br :H^

.«»OHß O=O ' OH^ O ^H :B r-

d: B r

( a (b)

H OH O

O

OH3 H

Br 4

4

(b)

B ro m o n iu m ion (c h ira l)

OH 3

_ O ^ *H HX ^ O\ Br OH3 ( R , S )-2 ,3 -D ib ro m o b u ta n e (m e s o )

(b)

(a)

3

H Br OH 3< : O O^ O— ^ „ O H 3 Br H ( R , S )-2 ,3 -D ib ro m o b u ta n e (m e s o )

trans -2 -B u te n e re a c ts w ith b ro m in e to y ie ld c h ira l b ro m o n iu m ion s and b ro m id e ion s. [R e a c tio n a t th e o th e r fa c e (to p ) w o u ld y ie ld th e e n a n tio m e r o f th e b ro m o n iu m ion as s h o w n h ere.]

W h en th e b ro m o n iu m ion s re a c t by e ith e r path (a) o r path (b), th e y y ie ld th e sam e a c h iral m e s o c o m p o u n d . (R e ac tio n o f th e e n a n tio m e r o f th e in te rm e d ia te b ro m o n iu m ion w o u ld p ro d u c e th e s a m e re s u lt.)

8.14 Halohydrin Form ation •

When the halogenation o f an alkene is carried out in aqueous solution, rather than in a non-nucleophilic solvent, the major product is a halohydrin (also called a halo alcohol) instead o f a vic-dihalide.

M olecules o f water react with the halonium ion intermediate as the predominant nucleophile because they are in high concentration (as the solvent). The result is formation o f a halo-

360

Chapter 8 Alkenes and Alkynes II hydrin as the major product. If the halogen is bromine, it is called a bromohydrin, and if chlorine, a chlorohydrin. OH \ =

C

/

+

X 22

+

H 22O

------->

X

— C— C— X

X = Cl o r Br

+

— C— C—

+

HX

X

H a lo h y d rin (m a jo r)

w 'c-Dihalide (m in o r)

Halohydrin formation can be described by the following mechanism.

A MECHANISM FOR THE REACTION H a lo h y d rin F o r m a tio n f ro m a n A lk e n e ///,.

Step 1

.\\\\

v c— C ^ \ /

^ C = ,C ^

\X-s+ ■■X*B-

=X =

X H alo n iu m ion

H a lid e ion

T h is step is th e s a m e as fo r h alo g en a d d itio n to an a lk e n e (s e e S e c tio n 8.12A ).

H

/

l+ /

■O^t H

Steps 2 and 3

■O

C C \ A) X

H

H a lo n iu m ion

./

X.

V" P ro to n a te d h alo h y d rin

H ere, h o w e v e r, a w a te r m o le c u le a c ts as th e n u c le o p h ile a n d a tta c k s a c a rb o n o f the ring, c a u s in g th e fo rm a tio n o f a p ro to n a te d h alo h y d rin .

H

H

I

I

:O :

H

/ kC— C

C— C..

H

I

:O

.

/ X.

H— O — H

I H

H a lo h y d rin

T h e p ro to n a te d h alo h y d rin lo s e s a p ro to n (it is tra n s fe rre d to a m o le c u le o f w a te r). T h is step p ro d u c e s th e h alo h y d rin and h y d ro n iu m ion.

The first step is the same as that for halogen addition. In the second step, however, the two mechanisms differ. In halohydrin formation, water acts as the nucleophile and attacks one carbon atom of the halonium ion. The three-membered ring opens, and a protonated halohydrin is produced. Loss of a proton then leads to the formation of the halohydrin itself.

ReviewProblem8.15

Write a mechanism to explain the following reaction.

(as a ra c e m ic m ix tu re )

• If the alkene is unsymmetrical, the halogen ends up on the carbon atom with the greater number of hydrogen atoms.

8.15 Divalent Carbon Compounds: Carbenes

x

361

Bonding in the intermediate bromonium ion is unsymmetrical. The more highly substituted carbon atom bears the greater positive charge because it resembles the more stable carbo­ cation. Consequently, water attacks this carbon atom preferentially. The greater positive charge on the tertiary carbon permits a pathway with a lower free energy o f activation even though attack at the primary carbon atom is less hindered:

(73 % )

When ethene gas is passed into an aqueous solution containing bromine and sodium chloride, the products o f the reaction are the following: Br\

Bl\

Br

Br\

OH

R e v ie w P ro b le m 8 .1 6

Cl

Write mechanisms showing how each product is formed.

8.15 D ivalent Carbon Compounds: Carbenes There is a group o f compounds in which carbon forms only two bonds. These neutral diva­ lent carbon compounds are called carb en es. M ost carbenes are highly unstable compounds that are capable o f only fleeting existence. Soon after carbenes are formed, they usually react with another m olecule. The reactions o f carbenes are especially interesting because, in many instances, the reactions show a remarkable degree o f stereospecificity. The reac­ tions o f carbenes are also o f great synthetic use in the preparation o f compounds that have three-membered rings, for example, bicyclo[4.1.0]heptane, shown at right.

8.15A Structure and Reactions of Methylene The simplest carbene is the compound called methylene (:CH2). M ethylene can be prepared by the decom position o f diazomethane (CH 2 N2), a very poisonous yellow gas. This decom position can be accom plished by heating diazomethane (thermolysis) or by irradi­ ating it with light o f a wavelength that it can absorb (photolysis): hpot :CH^ N ^ n

*C

D iazom ethane

H 2

+

:N= N :

M ethylene

Nitrogen

The structure o f diazomethane is actually a resonance hybrid o f three structures: :CH2— S # N : ;---- : CH 2 = S = N: ;— : :CH2— N = S : I

II

III

We have chosen resonance structure I to illustrate the decom position o f diazomethane because with I it is readily apparent that heterolytic cleavage o f the carbon-nitrogen bond results in the formation o f methylene and molecular nitrogen. M ethylene reacts with alkenes by adding to the double bond to form cyclopropanes:

\ ^ C

+

V . ___ ^

y '•CH 2 2

----- »

C/ \ C

A lkene

M ethylene

H/ X H C yclopropane

Bicyclo[4.1.0]heptane.

362

Chapter 8 Alkenes and Alkynes II

8.15B Reactions of Other Carbenes: Dihalocarbenes Dihalocarbenes are also frequently em ployed in the synthesis o f cyclopropane derivatives from alkenes. M ost reactions o f dihalocarbenes are stereospecific:

\=

T h e a d d itio n of :C X 2 is s te re o s p e c ific . If th e R g ro u p s o f th e a lk e n e a re tra n s , they w ill b e tra n s in th e p ro du ct. (If th e R g ro u p s w e re in itia lly cis, th e y w o u ld b e c is in th e p ro d u ct.)

:CCl„

A

R

C l'

Cl

+ e n a n tio m e r

Dichlorocarbene can be synthesized by the a elim ination o f the elem ents o f hydrogen chloride from chloroform. [The hydrogen o f chloroform is m ildly acidic (pAa ~ 24) due to the inductive effect o f the chlorine atoms.] This reaction resembles the b-elimination reac­ tions by which alkenes are synthesized from alkyl halides (Section 6.15): R— 0 : - K +

H :CCl3

R— O : H

:C C U

K+

slow

:CCl2 + :Cl C

D ic h lo ro c a rb en e

Compounds with a b hydrogen react by b elimination preferentially. Compounds with no b hydrogen but with an a hydrogen (such as chloroform) react by a elimination. A variety o f cyclopropane derivatives have been prepared by generating dichlorocarbene in the presence o f alkenes. Cyclohexene, for example, reacts with dichlorocarbene gener­ ated by treating chloroform with potassium tert-butoxide to give a bicyclic product: H .Cl

KOC(CH3) 3 CHCL

Cl H

7 ,7 -D ic h lo ro b ic y c lo [4 .1 .0 ]h e p ta n e (59 % )

8.15C Carbenoids: The Simmons-Smith Cyclopropane Synthesis A useful cyclopropane synthesis was developed by H. E. Simmons and R. D. Smith o f the DuPont Company. In this synthesis diiodomethane and a zinc-copper couple are stirred together with an alkene. The diiodomethane and zinc react to produce a carbene-like species called a carb en oid : c h 22 Ii2

Zn(Cu)

► ICH2ZnI A c a rb e n o id

The carbenoid then brings about the stereospecific addition o f a CH 2 group directly to the double bond. R e v ie w P ro b le m 8 .1 7

What products would you expect from each o f the follow ing reactions? (a)

,

t-BuOK CHCl3

(b)

(c) t-BuOK CHBr3

CH2 I2/Zn(Cu) _ diethyl ether

R e v ie w P ro b le m 8 .1 8

Starting with cyclohexene and using any other needed reagents, outline a synthesis o f 7,7dibromobicyclo[4.1.0]heptane.

R e v ie w P ro b le m 8 .1 9

Treating cyclohexene with 1,1-diiodoethane and a zinc-copper couple leads to two isomeric products. What are their structures?

8.16 Oxidation of Alkenes: Syn 1,2-Dihydroxylation

x

363

8 .1 6 Oxidation o f Alkenes: Syn 1,2-Dihydroxylation Alkenes undergo a number o f reactions in which the carbon-carbon double bond is oxidized. •

1,2-D ihydroxylation is an important oxidative addition reaction o f alkenes.

Osmium tetroxide is w idely used to synthesize 1,2-diols (the products o f 1,2-dihydroxylation, som etim es also called glycols). Potassium permanganate can also be used, although because it is a stronger oxidizing agent it is prone to cleave the diol through further o xi­ dation (Section 8.17). (1) OsO4, pyridine (2) NaHSO3/H 2O

^

OH OH

1 ,2 -P r o p a n ed io l (p r o p y le n e g ly c o l)

P r o p en e

c h 2= c h

2

OHcold

KM nO „

H2O,

HO

E th en e

OH

1 ,2 -E th a n e d io l (e th y le n e g ly c o l)

8.16A Mechanism for Syn Dihydroxylation of Alkenes •

The mechanism for the formation o f a 1,2-diol by osmium tetroxide involves a cyclic intermediate that results in syn addition o f the oxygen atoms (see below).

After formation o f the cyclic intermediate with osmium, cleavage at the oxygen-m etal bonds takes place without altering the stereochemistry o f the two new C — O bonds. xc=cx / \

pyridine

•

O

O

ox

X

O

H2O

— C------ C— I I OH

X

OH

+ O

^O ss

rO /

NaHSO,

— C— C— / \

✓

° \

O

Os

An o s m a t e e s te r

OV

The syn stereochemistry o f this dihydroxylation can readily be observed by the reaction o f cyclopentene with osmium tetroxide. The product is cis-1,2-cyclopentanediol.

+

O s°

4

NaHSO, h2o

25°C, pyridine O

O

\

OH

O sx

/O /

OH

/ VO

c is -1 ,2 -C y clo p e n ta n ed io l (a m e s o c o m p o u n d )

Osmium tetroxide is highly toxic, volatile, and very expensive. For these reasons, methods have been developed that permit O sO 4 to be used catalytically in conjunction with a co-oxi­ dant.* A very small molar percentage o f O sO 4 is placed in the reaction mixture to do the dihydroxylation step, while a stoichiometric amount o f co-oxidant reoxidizes the O sO 4 as it

*S ee N e lso n , D . W ., et al., J. Am. Chem. Soc. 1997, 119, 1 8 4 0 -1 8 5 8 ; and C orey, E. J., et al., J. Am. Chem. Soc. 1996, 118, 3 1 9 -3 2 9 .

364

Chapter S

Alkenes and Alkynes II

is used in each cycle, allowing oxidation o f the alkene to continue until all has been converted to the diol. N-Methylmorpholine N-oxide (NMO) is one o f the m ost commonly used co-oxi­ dants with catalytic O sO 4. The method was discovered at Upjohn Corporation in the context o f reactions for synthesis o f a prostaglandin* (Section 23.5): C atalytic O sO 4 1,2-D ihydroxylation Ph

O

OsO4 (0.2%), NMO 25o0

Ph

O

H HO H3

0

,o -

.

OH

> 9 5 % Y ie ld (u s e d in s y n th e s is o f a p ro s ta g la n d in )

"O" NMO ( s to ic h io m e tr ic c o - o x id a n t f o r c a ta ly tic d ih y d r o x y la tio n )

R e v ie w P ro b le m 8 .2 0

Specify the alkene and reagents needed to synthesize each o f the follow ing diols.

( ra c e m ic )

(ra c e m ic )

S o l v e d P r o b le m 8 .4 Explain the follow ing facts: Treating (Z)-2-butene with OsO 4 in pyridine and then NaHSO 3 in water gives a diol that is optically inactive and cannot be resolved. Treating (£)-2-butene with the same reagents gives a diol that is optically inactive but can be resolved into enantiomers. STRATEGY A ND ANSWER Recall that the reaction in either instance results in syn hydroxylation o f the double bond o f each compound. Syn hydroxylation o f (£)-2-butene gives a pair o f enantiomers, w hile syn hydroxylation o f (Z)-2-butene gives a single product that is a m eso compound. H H30

X

CH-. 3

syn-hydroxylation (from top and bottom)

h

HO O

")

H3 0

Ì" " ' 0 Hc H

H30

0H3

H

OH

■ f H OH

HO

E n a n tio m e rs ( s e p a ra b le a n d w h e n s e p a ra te d , o p t ic a lly a c tiv e )

( E )-2 -B u te n e

Symmetry plane i H

H syn-hydroxylation (from top and bottom)

H3 0 ^

ÜH3

(Z )-2 -B u te n e

HO \

H ""y H30

OH :

/ V "H 0H3

+

H

H H3 0 ^ \ .

:

Í + 0H 3

/

:

\

HO

Id e n tic a l (a m e s o c o m p o u n d )

*V a n Rheenan, V., K e lle y , R. C., and C ha, D . Y., Tetrahedron Lett. 1976, 25, 1973.

OH

S+

S'

rT*

8.17 Oxidative Cleavage of Alkenes

365

THE CHEMISTRY OF . . . C a t a l y t i c A s y m m e t r i c D ih y d r o x y l a t i o n M ethods for catalytic asym m etric syn dihydroxylation have been d eveloped that significantly extend the synthetic util­ ity of dihydroxylation. K. B. Sharpless (The Scripps Research Institute) and co-workers discovered that addi­ tion of a chiral amine to the oxidizing mixture leads to enantioselective catalytic syn dihydroxylation. Asymmetric dihydroxylation has b ecom e an important and widely used tool in the synthesis of com plex organic m olecules. In recognition of this and other advances in asymmetric oxi­ dation procedures d ev elo p ed by his group (Section 11.13), Sharpless was awarded half of the 2001 Nobel

Prize in Chemistry. (The other half of the 2001 prize was awarded to W. Knowles and R. Noyori for their d ev elo p ­ ment of catalytic asymmetric reduction reactions; see Section 7 .1 4A.) The following reaction, involved in an enantioselective synthesis of the side chain of the anti­ cancer drug paclitaxel (Taxol), serves to illustrate Sharpless's catalytic asym metric dihydroxylation. The exam ple utilizes a catalytic amount of K2 O sO 2 (OH)4, an O sO 4 equivalent, a chiral amine ligand to induce enantioselectivity, and NMO as the stoichiom etric co-oxidant. The product is obtained in 99% enantiom eric ex cess (ee):

A s ym m etric C atalytic O SO 4 1,2 -D ih y d ro x ylatio n * :■ \ fl^ M

Sharpless shared the 2001 Nobel Prize in Chemistry fo r his development o f asymmetric Ph oxidation methods.

M u ltip le s te p s

O

OH OMe

K2OsO2(OH)4, (0.2%), NMO chiral amine ligand (0.5%) (see below)

O

Ph^ X " " " "'OMe HO 9 9 % ee (7 2 % y ie ld )

O

x

P h ^ NH Ph

O _

'OH

OH P a c litax e l s id e chain

O

H3COCO A chiral a m in e lig a n d u se d in c a ta ly tic a s y m m e tric d ih y d ro x y la tio n A d a p te d w ith perm ission from Sharpless e t al., The Jo urn a l o f O rg a n ic Chem istry, Vol. 59, p. 5104, 1994. C o p y rig h t 1994 A m erican C hem ical Society.

8.17 O xidative Cleavage o f Alkenes Alkenes can be oxidatively cleaved using potassium permanganate or ozone (as w ell as by other reagents). Potassium permanganate (KMnO4) is used when strong oxidation is needed. Ozone (O3) is used when m ild oxidation is desired. [Alkynes and aromatic rings are also oxidized by KMnO4 and O 3 (Sections 8.20 and 15.13D).]

8.17A •

Cleavage with Hot Basic Potassium Permanganate

Treatment with hot basic potassium permanganate oxidatively cleaves the double bond o f an alkene.

Cleavage is believed to occur via a cyclic intermediate similar to the one formed with osmium tetroxide (Section 8.16A ) and intermediate formation o f a 1,2-diol. Alkenes with monosubstituted carbon atoms are oxidatively cleaved to salts o f carboxylic acids.

36 6

Chapter 8

Alkenes and Alkynes II

Disubstituted alkene carbons are oxidatively cleaved to ketones. Unsubstituted alkene carbons are oxidized to carbon dioxide. The following examples illustrate the results o f potassium per­ manganate cleavage o f alkenes with different substitution patterns. In the case where the prod­ uct is a carboxylate salt, an acidification step is required to obtain the carboxylic acid. O c h 3c h

CHCH,

O H3O+

KMnO4, OH H2O heat

Z C CH3

CH/ C ^ O A c e ta te ion

(cis o r tra n s )

CH,

(1) KMnO4, OHheat (2 ) H3O+

OH

A c e tic acid

O + O = C=O

H2O

One o f the uses o f potassium permanganate, other than for desired oxidative cleavage, is as a chem ical test for the presence o f unsaturation in an unknown compound. Solutions o f potassium permanganate are purple. If an alkene is present (or an alkyne, Section 8.20), the purple color is discharged and a brown precipitate o f manganese dioxide (MnO2) forms as the oxidation takes place. The oxidative cleavage o f alkenes has also been used to establish the location o f the dou­ ble bond in an alkene chain or ring. The reasoning process requires us to think backward much as w e do with retrosynthetic analysis. Here w e are required to work backward from the products to the reactant that might have led to those products. We can see how this might be done with the follow ing example.

Solved Problem 8.5 An unknown alkene with the formula C 8 H1 6 was found, on oxidation with hot basic permanganate, to yield a threecarbon carboxylic acid (propanoic acid) and a five-carbon carboxylic acid (pentanoic acid). What was the struc­ ture o f this alkene?

C8H16

O

(1) KMnO4, H2i OH', heat (2 ) H3O+

O OH

HO

P ro p a n o ic acid

P e n ta n o ic acid

STRATEGY AND ANSWER The carbonyl groups in the products are the key to seeing where the oxidative cleav­ age occurred. Therefore, oxidative cleavage must have occurred as follow s, and the unknown alkene must have been cis- or trans-3-octene, which is consistent with the molecular formula given. O

C le a v a g e o c c u rs h ere

4, H2O, OH' , heat (2 ) H3O+

O OH

HO

U n kn o w n a lk e n e (e ith e r cis- o r trans-3-octene)

8.17B •

Cleavage with Ozone

The m ost useful method for cleaving alkenes is to use ozone (O3).

O zonolysis consists o f bubbling ozone into a very cold ( —78°C ) solution o f the alkene in CH2 Cl2, followed by treatment o f the solution with dimethyl sulfide (or zinc and acetic acid). The overall result is as follows: R

R" (1) O3, CH2Cl2, —78°C ’ J' 2 2 --------- > (2 ) Me2S

1

C=^C / ' \ R'

H

R \ ^ C=O / R'

R" +

^ / O =C \ H

S+

S'

rT*

8.17 Oxidative Cleavage of Alkenes

367

The reaction is useful as a synthetic tool, as w ell as a method for determining the location o f a double bond in an alkene by reasoning backward from the structures o f the products. •

The overall process (above) results in alkene cleavage at the double bond, with each carbon o f the double bond becom ing doubly bonded to an oxygen atom.

The follow ing examples illustrate the results for each type o f alkene carbon. O

O

(1) O3, CH2Cl2, —78°C (2) Me2S * 2 -M e th y l-2 -b u te n e

H A c e to n e

A c e ta ld e h y d e

O (1) O3, CH2Cl2, —78°C (2) Me2S

O H Is o b u ty ra ld e h y d e

3 -M e th y l-1 -b u te n e

H

F o rm a ld e h y d e

S o lv e d P ro b le m 8 .6 Give the structure o f an unknown alkene with the formula C 7 H1 2 that undergoes ozonolysis to yield, after acidifi­ cation, only the follow in g product:

C 7 H 12

O

(1) O3, CH2Cl2, -78°C (2) Me2S

H O

STRATEGY AND ANSWER Since there is only a single product containing the same number o f carbon atoms as the reactant, the only reasonable explanation is that the reactant has a double bond contained in a ring. Ozonolysis o f the double bond opens the ring: (1) O3, CH2Cl2, -78°C (2) Me2S

O H O

U n kn o w n a lk e n e ( 1 -m e th y lc y c lo h e x e n e )

R e v ie w P ro b le m 8 .21

Predict the products o f the follow ing ozonolysis reactions. (a)

(1 ) O3 (2) Me2S

(b)

(1 ) O3 (2) Me2S

(c) (1 ) O3 (2) Me2S

The mechanism o f ozone addition to alkenes begins with formation o f unstable com ­ pounds called initial ozonides (sometimes called m olozonides). The process occurs vigor­ ously and leads to spontaneous (and som etim es noisy) rearrangement to compounds known as ozonides. The rearrangement is believed to occur with dissociation o f the initial ozonide into reactive fragments that recombine to yield the ozonide. Ozonides are very unstable com ­ pounds, and low-molecular-weight ozonides often explode violently.

368

Chapter 8 Alkenes and Alkynes II

A MECHANISM FOR THE REACTION O z o n o ly s is o f a n A lk e n e

C=C 7' (^

II

:O

>rf\?O O ■

■O “ ^ -

'C

O-

T h e in itia l o z o n id e fr a g m e n ts .

° \ C C . / \ / \ :O ----O:

/

■O—Oj

:O

In itia l o z o n id e

O z o n e a d d s to th e a lk e n e to fo r m an in itia l o z o n id e .

Me2S

\ /

O z o n id e

= O + ^=c/

Me2SO

\

A ld e h y d e s a n d /o r k e to n e s

T h e fr a g m e n ts r e c o m b in e to fo r m th e o z o n id e .

R e v ie w P ro b le m 8 .2 2

O :+

D im e th y l s u lfo x id e

Write the structures o f the alkenes that would yield the following carbonyl compounds when treated with ozone and then with dimethyl sulfide. (a)

O

and

(b)

O

O

m ol is produced from m ol o f alkene)

(2 1

H

H

and

(c)

O

O H

H

8.18 Electrophilic A d d itio n o f Bromine and Chlorine to Alkynes •

Alkynes show the same kind o f addition reactions with chlorine and bromine that alkenes do.

•

With alkynes the ad dition m ay occur once or twice, depending on the number of molar equivalents o f halogen w e employ: Br

Br

-C = C -

Br2 CCL

Br2 CCL >

\= C X

b /

X

Cl

Cl2 CCL ■

Cl2 CCl4

c/

N

D ic h lo r o a lk e n e

Br

T e tr a b ro m o a lk a n e

Cl

—C=C—

— C— C — Br

D ib ro m o a lk e n e

Br

Cl

— C— C— Cl

Cl

T e tr a c h lo ro a lk a n e

S+

8.19 Addition of Hydrogen Halides to Alkynes

x

S'

rT*

369

It is usually possible to prepare a dihaloalkene by simply adding one molar equivalent of the halogen: Br

OH

(1 mo!)' c c i4, q°c 5

OH

Br 2

Br

•

Addition o f one molar equivalent o f chlorine or bromine to an alkyne generally results in anti addition and yields a trans-dihaloalkene.

Addition o f bromine to acetylenedicarboxylic acid, for example, gives the trans isomer in 70% yield:

HO2 C— C = C — CO2H 2

2

(1

HO2C

Br, 2 mo!)

Br

\ = < /

2

CO2H

Br A c e ty le n e d ic a rb o x y lic acid

(7 0 % )

A lkenes are more reactive than alkynes toward addition o f electrophilic reagents (i.e., Br2, Cl2, or HCl). Yet when alkynes are treated with one molar equivalent o f these same elec­ trophilic reagents, it is easy to stop the addition at the “alkene stage.” This appears to be a paradox and yet it is not. Explain.

R e v ie w P ro b le m 8 .2 3

8.19 A d d itio n o f Hydrogen Halides to Alkynes •

Alkynes react with one molar equivalent o f hydrogen chloride or hydrogen bromide to form haloalkenes, and with two molar equivalents to form geminal dihalides.

•

Both additions are regioselective and follow Markovnikov’s rule: H

HX

-C = C -

\ /

-

HX

cx

\

X

H a lo a lk e n e

H

X

C

C

H

X

g e m -D ih a lid e

The hydrogen atom o f the hydrogen halide becom es attached to the carbon atom that has the greater number o f hydrogen atoms. 1-Hexyne, for example, reacts slow ly with one molar equivalent o f hydrogen bromide to yield 2 -bromo- 1 -hexene and with two molar equivalents to yield 2 , 2 -dibromohexane: HBr

HBr Br

Br 2 -B ro m o -1 -h e x e n e

Br

2 ,2 -D ib ro m o h e x a n e

The addition o f HBr to an alkyne can be facilitated by using acetyl bromide (CH 3 COBr) and alumina instead o f aqueous HBr. A cetyl bromide acts as an HBr precursor by reacting with the alumina to generate HBr. For example, 1-heptyne can be converted to 2-bromo1 -heptene in good yield using this method: Br

‘‘HBr’’ CH3COBr/alumina CH2Cl2

(82%)

370

Chapter 8 Alkenes and Alkynes II Anti-Markovnikov addition o f hydrogen bromide to alkynes occurs when peroxides are present in the reaction mixture. These reactions take place through a free-radical m echa­ nism (Section 10.9):

8.20 O xidative Cleavage o f Alkynes Treating alkynes with ozone follow ed by acetic acid, or with follow ed by acid, leads to cleavage at the carbon-carbon carboxylic acids: ( 1) O3 RCO2H + (2) HOAc or R — C # C — R' 91) nH°O +°H: RCO 2 H

R e v ie w P ro b le m 8 .2 4

basic potassium permanganate triple bond. The products are R 'C O2H

+ R 'C O 2 H

A, B, and C are alkynes. Elucidate their structures and that o f D using the follow ing reac­ tion roadmap. H2, Pt

A

B

(C.HJ

(C.H J IR: 3300 cm-1

(1 ) O3 (2) HOAc O OH

C (C8H12)

H2, Pt

D (C.H1J

(1) O3 (2) HOAc O HO OH O

8.21 H o w to Plan a Synthesis: Some Approaches and Examples In planning a synthesis w e often have to consider four interrelated aspects: 1

. construction o f the carbon skeleton,

2

. functional group interconversions,

3. control o f regiochemistry, and 4. control o f stereochemistry.

S+

8.21 How to Plan a Synthesis: Some Approaches and Examples You have had som e experience with certain aspects o f synthetic strategies in earlier sections. •

In Section 7.16B you learned about retrosynthetic analysis and how this kind of thinking could be applied to the construction o f carbon skeletons o f alkanes and cycloalkanes.

•

In Section 6.14 you learned the meaning o f a functional group interconversion and how nucleophilic substitution reactions could be used for this purpose.

In other sections, perhaps without realizing it, you have begun adding to your basic store o f methods for construction o f carbon skeletons and for making functional group inter­ conversions. This is the time to begin keeping a card file for all the reactions that you have learned, noting especially their applications to synthesis. This file w ill becom e your Tool K it for O rganic Synthesis. N ow is also the time to look at som e new exam ples and to see how w e integrate all four aspects o f synthesis into our planning.

8.21A

Retrosynthetic Analysis

Consider a problem in which w e are asked to outline a synthesis o f 2-bromobutane from com ­ pounds o f two carbon atoms or fewer. This synthesis, as w e shall see, involves construction o f the carbon skeleton, functional group interconversion, and control o f regiochemistry. H o w

to

S y n th e s iz e 2 -B r o m o b u ta n e

Br We begin by thinking backward. The final target, 2-bromobutane, can be made in one step from 1-butene by addition o f hydrogen bromide. The regiochemistry o f this functional group interconversion m ust be Markovnikov addition: R e tr o s y n th e tic A n a ly s is

H— Br

Q Br

Markovnikov addition

S y n th e s is

HBr

no peroxides Br

Remember: The open arrow is a symbol used to show a retrosynthetic process that relates the target m olecule to its precursors: Target molecule

precursors

Continuing to work backward one hypothetical reaction at a time, w e realize that a syn­ thetic precursor o f 1-butene is 1-butyne. Addition o f 1 m ol o f hydrogen to 1-butyne would lead to 1-butene. With 1-butyne as our new target, and bearing in mind that w e are told that w e have to construct the carbon skeleton from compounds with two carbons or fewer, w e realize that 1 -butyne can be formed in one step from ethyl bromide and acetylene by an alkynide anion alkylation. •

The key to retrosynthetic analysis is to think o f how to synthesize each target m olecule in one reaction from an immediate precursor, considering first the ulti­ mate target m olecule and working backward.

x

S'

rT*

371

372

Chapter 8 Alkenes and Alkynes II R e tr o s y n th e tic A n a ly s is

H2

Q

Q

Na+

— H

'Br

Q

+

Na+ —== — H

H ^ = — H

+

NaNH 2

S y n th e s is

X-N H . 2 HI--------^H ^ = — hydrogen by Cl

C h lo ro e th a n e

C IC H 2 C H 2CI 2

2

1 ,2 -D ic h lo ro e th a n e

H owever, if w e com pare the set o f hydrogens o f the C H 2 group o f chloroethane with those o f its C H 3 set w e find that the hydrogens o f the C H 3 and C H 2 groups are h e te ro to p ic w ith respect to each other. R eplacing either o f the tw o hydrogens o f the C H 2 set by ch lo ­ rine yields 1,1-dichloroethane, w hereas replacing any one o f the set o f three C H 3 h y d ro ­ gens yields a different com pound, 1,2-dichloroethane. replacement of any CH3

C lC H 2 C H 2 C l

hydrogen by Cl

1 ,2 -D ich lo ro e th a n e

C H 3 C H 2C l C h lo ro e th a n e

replacem ent of either CH2 C H 3 C H C l2

hydrogen by Cl

1 ,1 -D ich lo ro e th a n e

Chloroethane, therefore, has tw o sets o f hydrogens that are heterotopic w ith respect to each other, the C H 3 hydrogens and the C H 2 hydrogens. T he hydrogens o f these tw o sets are not chem ical shift equivalent, and chloroethane gives tw o 1H N M R signals. C onsider 2-m ethylpropene as a further example:

(a) H \

C H 3 (b) / C = C

/ (a) H

H \

3

'----------------------- ^ C H 3 (b)

/ H

Cl \

2

C = C

Replace H with Cl

\

C H 2C l /

CH3 / C =C

+

\

/ CH3

3 -C h lo ro -2 -m e th y lp ro p e n e

H

3

\ CH3

1 -C h lo ro -2 -m e th y lp ro p e n e

T he six m ethyl hydrogens (b) are one set o f hom otopic hydrogens; replacing any one of them w ith chlorine, for exam ple, leads to the sam e com pound, 3-chloro-2-m ethylpropene. T he tw o vinyl hydrogens (a) are another set o f hom otopic hydrogens; replacing either o f these leads to 1-chloro-2-m ethylpropene. 2-M ethylpropene, therefore, gives tw o 1H N M R signals.

9.8 Chemical Shift Equivalent and Nonequivalent Protons

U sing the m ethod of Section 9.8A, determ ine the num ber o f expected signals for the fol­ low ing com pounds. (a)

CH3

403

Review Problem 9.3

(b )

H3 C

H ow m any signals w ould each com pound give in its 1H N M R spectrum ? (c)

(a) CH 3 OCH 3

(e) (Z)-2-B utene OCH3

(b )

(d )

2,3-D im ethyl-2-butene

( f)

(£)-2-B utene

t o 13C N M R S p e c tr o s c o p y As a preview o f w hat is to com e later in this chapter w hen w e study 13C N M R spectroscopy, let us look briefly at the carbon atom s o f ethane to see w hether w e can use a sim ilar m ethod to decide w hether they are h om o­ topic or heterotopic, and w hether ethane w ould give one or tw o 13C signals. H ere w e can m ake our im aginary replacem ents using a silicon atom.

A p p lic a tio n

pu CH3CH3 3

replacement of either carbon atom by^ ¡r ~ ^ an Si atom 3 3

3

o :lj ^ lj S iH 3 C H 3

E th an e

O nly one product is possible; therefore, the carbons o f ethane are h o m o to p ic, and e th a n e w o u ld give o nly o ne sig n al in its 13C sp e c tru m . O n the other hand, if w e consider chloroethane, replacem ent o f a carbon atom by a sil­ icon atom gives tw o possibilities: replacement of the CH 3

SiH 3 CH2Cl

carbon by an Si atom

CH3CH2Cl replacement of the CH 2

CH3SiH2Cl

carbon by an Si atom

We do not get the sam e com pounds from each replacem ent. We can conclude, therefore, that the tw o carbon atom s o f chloroethane are h etero to p ic . T hey are not chem ical shift equivalent, and each carbon atom o f chloroethane w ould give a 13C signal at a different chem ical shift. C h lo ro e th a n e gives tw o 1 3 C N M R signals.

9.8B Enantiotopic and Diastereotopic Hydrogen Atoms If replacem ent o f each o f tw o hydrogen atom s by the sam e group yields com pounds that are enantiom ers, the tw o hydrogen atom s are said to be e n a n tio to p ic . •

E nantiotopic hydrogen atom s have the sam e chem ical shift and give only one 1H N M R signal:* H

^ H

H C ^ B r

replacement obfyesaocmH group Q

H

Q

H C ^ B r

Q ^ C

^

^ H ^ B

r

^ E n a n tio m e rs

* E n a n tio to p ic h ydro g e n atom s m a y n o t have the same ch em ica l s h ift i f the co m po u n d is dissolved in a c h ira l solvent. H ow ever, m ost 1H N M R spectra are d eterm ined u sing a c hira l solvents, and in th is situ atio n ena n tio to p ic p roto ns have the same ch e m ica l sh ift.

ReviewProblem9.4

404

Chapter 9

The

tw o

Nuclear Magnetic Resonance and Mass Spectrometry

h y d ro g e n

a to m s o f th e

— C H 2 B r g r o u p o f b r o m o e t h a n e a r e e n a n t io t o p i c .

B r o m o e t h a n e , t h e n , g iv e s t w o s ig n a ls i n it s 1 H N M R s p e c t r u m . T h e t h r e e e q u i v a l e n t p r o ­ to n s o f t h e — C H 3 g r o u p g i v e o n e s ig n a l ; t h e t w o e n a n t io t o p i c p r o t o n s o f t h e — C H 2 B r g r o u p g i v e t h e o t h e r s ig n a l . [ T h e 1H N M R s p e c t r u m o f b r o m o e t h a n e , a s w e s h a ll s e e , a c t u ­ a l l y c o n s is t s o f s e v e n p e a k s ( t h r e e i n o n e s ig n a l , f o u r i n t h e o t h e r ) . T h i s is a r e s u l t o f s ig ­ n a l s p l i t t i n g , w h i c h is e x p l a i n e d i n S e c t i o n 9 . 9 . ] I f r e p l a c e m e n t o f e a c h o f t w o h y d r o g e n a t o m s b y a g r o u p , Q , g iv e s c o m p o u n d s t h a t a r e d ia s te re o m e rs , th e t w o h y d r o g e n s a re s a id to b e d i a s t e r e o t o p ic . •

E x c e p t f o r a c c i d e n t a l c o in c i d e n c e , d i a s t e r e o t o p ic p r o t o n s d o n o t h a v e t h e s a m e c h e m i c a l s h i f t a n d g i v e r is e t o d i f f e r e n t 1H N M R

T h e t w o m e t h y l e n e h y d r o g e n s l a b e l e d aH a n d

bH a t

s ig n a ls . C 3 in 2 -b u ta n o l a re d ia s te r e o to p ic .

W e c a n i l l u s t r a t e t h is b y i m a g i n i n g r e p l a c e m e n t o f a H o r bH w i t h s o m e i m a g i n a r y g r o u p Q . T h e r e s u l t is a p a i r o f d i a s t e r e o m e r s . A s d i a s t e r e o m e r s , t h e y h a v e d i f f e r e n t p h y s i c a l p r o p ­ e r t ie s , in c l u d i n g c h e m i c a l s h if t s , e s p e c i a l l y f o r t h o s e p r o t o n s n e a r t h e c h i r a l i t y c e n te r .

H JDH

b aH 2-B u ta n o l (o n e e n a n tio m e r)

D iaste re o m ers

T h e d i a s t e r e o t o p ic n a t u r e o f a H a n d bH a t C 3 i n 2 - b u t a n o l c a n a ls o b e a p p r e c ia t e d b y v i e w i n g N e w m a n p r o j e c t io n s . I n t h e c o n f o r m a t io n s s h o w n b e l o w ( F i g . 9 . 1 6 ) , a s is t h e c a s e f o r e v e r y p o s s ib le c o n f o r m a t i o n o f 2 - b u t a n o l , a H a n d bH e x p e r i e n c e d i f f e r e n t e n v ir o n m e n t s b e c a u s e o f th e a s y m m e t r y f r o m t h e c h i r a l i t y c e n t e r a t C 2 . T h a t is , t h e “ m o l e c u l a r la n d s c a p e ” o f 2 - b u t a n o l a p p e a r s d i f f e r e n t t o e a c h o f t h e s e d i a s t e r e o t o p ic h y d r o g e n s . a H a n d bH e x p e ­ r ie n c e d i f f e r e n t m a g n e t ic e n v ir o n m e n t s , a n d a r e t h e r e f o r e n o t c h e m i c a l s h if t e q u i v a le n t . T h i s is t r u e i n g e n e r a l: d i a s t e r e o t o p i c h y d r o g e n s a r e n o t c h e m i c a l s h i f t e q u i v a l e n t .

b

aH

bH

120° rotation of the front carbon

120°

a CH 3

Figure 9.16

aH and bH (on C3, th e fro n t carbon in th e Newm an pro je ctio n ) experience diffe re n t environm ents in these th re e conform ations, as well as in every other possible conformation of 2-butanol, because o f th e chirality center at C2 (the back carbon in th e Newm an projection). In o th er w ords, th e m olecular landscape as view ed from one d ia ste re o to p ic hydrogen w ill always appear d iffe re n t from th a t view ed by th e other. Hence, aH and bH experience d iffe re n t m agnetic environm ents and th e re fo re should have d iffe re n t chemical shifts (though th e difference may be small). They are n o t chemical sh ift equivalent.

A l k e n e h y d r o g e n s c a n a ls o b e d i a s t e r e o t o p ic . T h e t w o p r o t o n s o f t h e =

C H 2 g ro u p o f

c h lo r o e t h e n e a r e d i a s t e r e o t o p ic :

H

Cl

V C

/

y

=C

\ H

Cl Q replacement rS ; ^ C— C by Q H H

Cl

H/

H

X Q

D ia s te re o m ers C h l o r o e t h e n e , t h e n , s h o u ld g i v e s ig n a ls f r o m t h r e e n o n e q u i v a l e n t p r o t o n s : o n e f o r t h e p r o ­ to n o f t h e C I C H = g ro u p .

g r o u p , a n d o n e f o r e a c h o f t h e d i a s t e r e o t o p ic p r o t o n s o f t h e =

CH 2

9.9 Signal Splitting: Spin-Spin Coupling

(a) Show that replacing each o f the tw o m ethylene protons by Q in the other enantiom er o f 2 -butanol also leads to a pair o f diastereom ers.

405

Review Problem 9.5

(b) How m any chem ically different kinds o f protons are there in 2-butanol? (c) How m any 1H N M R signals w ould you expect to find in the spectrum o f 2-butanol? H ow m any 1H N M R signals w ould you expect from each o f the follow ing com pounds? (a)

(i)

ReviewProblem9.6

(n) O

(b)

OH

(j) (o)

(c)

O

(d) (e)

Br Br

9.9 Signal Splitting: Spin-Spin Coupling Signal splitting arises from a phenom enon know n as spin-spin coupling. S pin-spin coupling effects are transferred prim arily through bonding electrons and lead to s p in - s p in sp littin g . •

V icinal co u p lin g is coupling betw een hydrogen atom s on adjacent carbons (vicinal hydrogens), w here separation betw een the hydrogens is by three s bonds.

T he m ost com m on occurrence of coupling is vicinal coupling. H ydrogens bonded to the sam e carbon (gem inal hydrogens) can also couple, but only if they are diastereotopic. Longrange coupling can be observed over m ore than three bond lengths in very rigid m olecules such as bicyclic com pounds, and in system s w here p bonds are involved. We shall lim it our discussion to vicinal coupling, however.

9.9A Vicinal Coupling •

V icinal coupling betw een heterotopic protons generally follow s the n + 1 rule (Section 9.2C). Exceptions to the n + 1 rule can occur w hen diastereotopic hydro­ gens or conform ationally restricted system s are involved.

We have already seen an exam ple o f vicinal coupling and how the n + 1 rule applies in our discussion o f the spectrum of 1,1,2-trichloroethane (Fig. 9.4). To review, the signal from the tw o equivalent protons o f the — CH2Cl group o f 1,1,2-trichloroethane is split into a doublet by the proton of the CHCl2 — group. The signal from the proton o f the CHCl2— group is split into a triplet by the tw o protons o f the — CH2Cl group. Before we explain the origin of signal splitting, however, let us also consider two exam ­ ples w here signal splitting w ould not be observed. Part o f understanding signal splitting is recognizing w hen you w ould not observe it. Consider ethane and m ethoxyacetonitrile. All of the hydrogen atoms in ethane are equivalent, and therefore they have the same chem ical shift and do not split each other. The 1H N M R spectrum o f ethane consists o f one signal that is a

406

Chapter 9

Nuclear Magnetic Resonance and Mass Spectrometry

5h (ppm)

Figure 9.17 The 300-M Hz 1H NM R spectrum o f m eth o xya ce to nitrile . The signal o f the e n a n tio to p ic p rotons (b) is not split.

aH (split by one bH proton)

singlet. The spectrum o f m ethoxyacetonitrile is show n in Fig. 9.17. W hile there are tw o sig­ nals in the spectrum o f m ethoxyacetonitrile, no coupling is observed and therefore both sig­ nals are singlets because (1) the hydrogens labeled (a) and (b) are m ore than three single bonds apart, and (2) the hydrogens labeled (a) are hom otopic and those labeled (b) are enantiotopic. •

S ig n al s p littin g is n o t o b se rv e d fo r p ro to n s th a t a r e h o m o to p ic (ch e m ica l sh ift eq u iv ale n t) o r e n a n tio to p ic.

L et us now explain how signal splitting arises from coupled sets o f protons that are not hom otopic.

9.9B Splitting Tree Diagrams and the Origin of Signal Splitting Applied m agnetic field, B 0

Figure 9.18 S p littin g tree diagram fo r a d o u b le t. The signal from th e observed hydrogen (aH) is sp lit in to tw o peaks o f 1 : 1 in te n sity by the a d d itive and su b tra ctive effects o f th e m agnetic fie ld from one adjacent hydrogen (bH) on B0 (the applied field). Jab, the spacing betw een th e peaks (measured in hertz), is called the coupling constant. bH aH I I — C— C—

Signal splitting is caused by the m agnetic effect o f protons that are nearby and nonequiv­ alent to those protons producing a given signal. N earby protons have m agnetic m om ents that can either add to or subtract from the m agnetic field around the proton being observed. T his effect splits the energy levels o f the protons w hose signal is being observed into a sig­ nal w ith m ultiple peaks. We can illustrate the origin o f signal splitting using sp littin g tre e d ia g ra m s and by show ­ ing the possible com binations o f m agnetic m om ent alignm ents for the adjacent protons. In F igures 9.18, 9.19, an d 9.20 w e apply this sort o f analysis to splitting that w ould cause the patterns o f a doublet, triplet, and quartet. S p lit t in g A n a ly s is f o r a D o u b le t Figure 9.18 shows a splitting tree diagram for a dou­ blet. The signal from the observed hydrogen (aH) is split into tw o peaks o f 1 : 1 in te n sity by the additive and subtractive effects o f the m agnetic field from a single adjacent hydrogen (bH) on the applied m agnetic field, B 0. The two possible m agnetic orientations for the adjacent hydrogen (bH) that align either against or w ith the applied m agnetic field are shown under­ neath the splitting tree using arrows. Jab, the spacing betw een the peaks (m easured in hertz), is called the coupling constant. (We shall have m ore to say about coupling constants later.)

F igure 9.19 shows a splitting tree diagram for a triplet. T he signal from the observed hydrogen (aH) is split into three peaks o f 1 : 2 : 1 in te n sity by the m agnetic effects o f tw o adjacent equivalent hydrogens (bH). T he upper level in the diagram represents splitting from one o f the adjacent bH hydrogens, leading initially to two S p lit t in g A n a ly s is f o r a T r ip le t

9.9 Signal Splitting: Spin-Spin Coupling

40 7

Figure 9.19 S p littin g tre e diagram fo r a trip le t. The signal fro m th e observed hydrogen (aH) is sp lit in to th re e peaks o f 1 : 2 : 1 in te n sity by tw o adjacent equivalent hydrogens (bH). The up p er level o f sp littin g in th e diagram represents sp littin g fro m one o f the adjacent bH hydrogens, producing a d o u b le t shown as tw o legs. The second bH hydrogen splits each o f these legs again, as shown at th e next level. The center legs at this level overlap however, because Jab is th e same* fo r th e coupling o f b o th bH hydrogens w ith aH. This analysis accounts fo r th e observed 1 : 2 : 1 ratio o f intensities in a spectrum (sim ulated in blue). In any sp littin g tree diagram , th e lo w e rm o st level m ost closely represents w hat we observe in th e actual spectrum . The possible m agnetic orien ta tio n s o f th e tw o bH hydrogens are shown under th e tre e diagram w ith arrows indicating th a t b o th o f th e adjacent hydrogens may be aligned w ith th e applied fie ld , or one may be aligned w ith and th e o th e r against (in tw o equal energy com binations, hence tw ice th e intensity), or b o th may be aligned against th e applied field.

II — C— C— bH aH

I I I

bH aH bH — C— C — C —

bH * I n th is exam p le , J ab is the same fo r b o th bH h ydrogens because w e assume th e m to be h o m o to p ic o r e n a n tio to p ic (c h e m ic a l s h ift e q u iv a le n t). I f th e y w ere d ia s te re o to p ic o r o th e rw is e c h e m ic a l s h ift n o n e q u iv a le n t, each m a y have had a d iffe re n t c o u p lin g constant w ith aH , and the s p littin g pattern w o u ld n o t have been a p ure tr ip le t (o r n o t even a tr ip le t at a ll). F o r exam p le , i f the tw o c o u p lin g constants had been s ig n ific a n tly d iffe re n t, the p atte rn w o u ld have been a d o u b le t o f d ou b lets instead o f a trip le t. D ia s te re o to p ic g e m in a l hydro g e ns th a t c o up le w ith a v ic in a l h y d ro g e n ty p ic a lly prod u ce a d o u b le t o f dou b lets, because g e m in a l c o u p lin g constants are o fte n la rg e r than v ic in a l c o u p lin g constants.

legs that appear like the diagram for a doublet. Each o f these legs is split by the second bH hydrogen, as shown at the next level. The center legs at this level overlap, however, because Jab is the same for coupling o f both of the bH hydrogens with aH. This overlap o f the two center legs reflects the observed 1 : 2 : 1 ratio o f intensities in a spectrum, as shown in the simulated triplet in Fig. 9.19. (Note that in any splitting tree diagram, the lowermost level schematically represents the peaks w e observe in the actual spectrum.) The possible magnetic orientations o f the two bH hydrogens that cause the triplet are shown under the splitting diagram with arrows. The arrows indicate that both o f the adja­ cent hydrogens may be aligned with the applied field, or one may be aligned with and the other against (in two equal energy combinations, causing a doubling o f intensity), or both may be aligned against the applied field. Diagraming the possible combinations for the nuclear magnetic moments is another way (in addition to the splitting tree diagram) to show the origin of the 1 : 2 : 1 peak intensities that w e observe in a triplet. S p lit t in g A n a ly s is f o r a Q u a r t e t Figure 9.20 shows the splitting tree diagram for a quar­ tet. The signal from the observed hydrogen (aH) is split into three peaks of 1 : 3 : 3 : 1 in te n ­ sity by the magnetic effects of three equivalent hydrogens (bH). The same method o f analysis used for the triplet pattern applies here, wherein each successively lower level in the tree dia­ gram represents splitting by another one of the coupled hydrogens. Again, because in this case Jab is the same for the splitting of aH by all three o f the adjacent bH hydrogens, the internal legs of the diagram overlap, and the intensities are additive each time this occurs. The result is a pattern of 1 : 3 : 3 : 1 intensities, as we would observe for a quartet in an actual spectrum. The possible magnetic orientations shown under the tree diagram for the three bH hydro­ gens indicate that all three o f the adjacent hydrogens may be aligned with the applied field, or two may be aligned with the field and one against (in three equal energy combinations), or two against and one with the field (again, in three equal energy combinations), or all

408

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Nuclear Magnetic Resonance and Mass Spectrometry

Figure 9.20 S p littin g tre e diagram fo r a q u a rte t. The signal from th e observed hydrogen (aH) is sp lit in to fo u r peaks o f 1 : 3 : 3 : 1 in te n sity by th re e equivalent hydrogens (bH). The same m eth o d o f analysis w e used fo r th e trip le t p a tte rn applies here, w herein each successively low er level in th e tre e diagram represents sp littin g by another one o f th e coupled hydrogens. Again, because in this case Jab is th e same fo r sp littin g o f th e aH th e signal by all th re e bH hydrogens, internal legs o f th e diagram overlap, and intensities a m plify each tim e this occurs. The result is a p a ttern o f 1 : 3 : 3 : 1 intensities, as w e w o u ld observe fo r a q u a rte t in an actual spectrum . The possible m agnetic orientations shown under th e tre e diagram fo r th e th re e bH hydrogens indicate th a t all th re e o f th e adjacent hydrogens may be aligned w ith th e applied fie ld , or tw o may be aligned w ith th e fie ld and one against (in three equal energy com binations), or tw o against and one w ith th e fie ld (in three equal energy com binations), or all th re e may be aligned against th e applied field.

b

bH

aH

1

1

bH — C — C — bH

three may be aligned against the applied field. This analysis shows how a quartet results from three hydrogens that split the signal of another. Let us conclude this section with two last examples. The spectrum of 1,1,2,3,3pentachloropropane (Fig. 9.21) is similar to that of 1,1,2-trichloroethane in that it also consists of a 1:2:1 triplet and a 1:1 doublet. The two hydrogen atoms bH of 1,1,2,3,3-pentachloropropane are equivalent even though they are on separate carbon atoms.

ReviewProblem9.7

The relative positions of the doublet and triplet of 1,1,2-trichloroethane (Fig. 9.4) and 1,1,2,3,3-pentachloropropane (Fig. 9.21) are reversed. Explain this. Finally, returning to the spectrum of bromoethane that we used as the opening example in this chapter, the signal from two equivalent protons of the — CH2Br group (Fig. 9.1) appears as a 1:3:3:1 quartet because of the type of signal splitting shown in Fig. 9.20. The three equivalent protons of the CH 3 — group are split into a 1:2:1 triplet by the two pro­ tons of the — CH2Br group. The kind of analysis that we have just given can be extended to compounds with even larger numbers of equivalent protons on adjacent atoms. These analyses also show that if there are n equivalent p ro to n s on a d ja cen t atom s, these w ill sp lit a sig n a l into n + 1 p ea ks.

9.9 Signal Splitting: Spin-Spin Coupling

8

7

6

5

4

3

2

1

40 9

0

Sh (ppm)

Figure 9.21 The 300-M Hz 1H NM R spectrum o f 1,1,2,3,3-pentachloropropane. Expansions o f th e signals are shown in th e o ffse t plots.

(We may not always see all of these peaks in actual spectra, however, because some of them may be very small.)

Sketch the 1H NM R spectrum you would expect for the follow­ ing compound, showing the splitting patterns and relative position of each signal.

Cl

ReviewProblem9.8

Cl

Cl

Cl Cl

Cl

Propose a structure for each of the compounds whose spectra are shown in Fig. 9.22, and account for the splitting pattern of each signal.

8

7

6

5

4 Sh (PPm)

3

2

1

Figure 9.22 The 300-M Hz 1H NM R spectra fo r com pounds A , B, and C in Review Problem 9.9. Expansions are shown in th e o ffse t plots. (continues on the next page)

ReviewProblem9.9

410

Chapter 9

8

7

Nuclear Magnetic Resonance and Mass Spectrometry

6

5

4

3

2

1

0

3

2

1

0

Sh (ppm)

8

7

6

5

4 Sh (PPm)

Figure 9.22

(continued).

9.9C Coupling Constants— Recognizing Splitting Patterns

----------H e lp f u l H i n t Reciprocity of coupling constants.

Signals from coupled protons share a com m on coupling constant value. Coupling constants are determ ined by m easuring the separation in hertz betw een each peak o f a signal. A typ­ ical vicinal coupling constant is 6 - 8 hertz. We show ed how coupling constants are m ea­ sured in Figs. 9.1 8 -9 .2 0 , w here J ab denotes the cou p lin g constant betw een coupled hydrogens aH and bH. Coupling constants are also used w hen draw ing splitting tree d ia­ gram s, as show n in Figs. 9.1 8 -9 .2 0 . If w e w ere to m easure the separation o f peaks in both the quartet an d the triplet in the N M R spectrum o f brom oethane (Fig. 9.1), w e w ould find that they have the sam e coupling constant. T his phenom enon is called th e re c ip ro c ity o f co u p lin g c o n stan ts. A sim ulation o f the reciprocity o f coupling constants for brom oethane is represented in Fig. 9.23. W hile it is easy to assign the splitting patterns in brom oethane w ithout the analysis o f coupling constants, i.e., using solely the n + 1 ru le (as is also the case for the spectra show n in Fig. 9.22), the reciprocity o f coupling constants can b e very helpful w hen assign­ ing sets o f coupled protons in the spectra o f m ore com plicated m olecules.

411

9.9 Signal Splitting: Spin-Spin Coupling

aH bH b bH

X -C -C

1 I I jjuUL J [}1 aH bH

| Jah 1, Jah J Jah I

[ Jab J

Signal for aH

Signal for bH

Figure 9.23 A th e ore tica l s p littin g p a ttern fo r an ethyl g roup. For an actual exam ple, see th e spectrum o f brom oethane (Fig. 9.1).

Other techniques in FTNMR spectroscopy also facilitate the analysis of coupling rela­ tionships. One such technique is 1 H - 1H correlation spectroscopy, also known as 1 H - 1H COSY (Section 9.12A).

9.9D The Dependence of Coupling Constants on Dihedral Angle The magnitude of a coupling constant can be indicative of the dihedral angle between cou­ pled protons. This fact has been used to explore molecular geometry and perform conforma­ tional analysis by NMR spectroscopy. The dependence of the coupling constant on dihedral angles was explored by Martin Karplus (Harvard University), and has become well known as the Karplus correlation. A diagram showing the Karplus correlation is given in Fig. 9.24.

Figure 9.24 The Karplus co rrelation defines a

0

20

40

60

80

100 120 140 160 180 f

relationship betw een dihedral angle ( f ) and coupling constant fo r vicinal protons. (R eprinted w ith perm ission o f John W iley & Sons, Inc. from Silverstein, R., and W ebster, F. X., Spectrometric Identification of Organic Compounds, Sixth Edition, p. 186. C o p yrig h t 1998.)

The influence of dihedral angles on coupling constants is often evident in the 1H NMR spectra of substituted cyclohexanes. The coupling constant between vicinal axial protons (Jax ax) is typically 8-10 Hz, which is larger than the coupling constant between vicinal axial and equatorial protons (Jax,eq), which is typically 2-3 Hz. Measuring coupling constants in the NMR spectrum of a substituted cyclohexane can therefore provide information about low energy conformations available to the compound. What is the dihedral angle and expected coupling constant between the labeled protons in each of the following molecules? (a)

(b )

Hh

Hh

Hc Ha

Review Problem 9.10

412

ReviewProblem9.11 ReviewProblem9.12

Chapter 9

Nuclear Magnetic Resonance and Mass Spectrometry

D raw the m ost stable chair conform ation o f 1-brom o-2-chlorocyclohexane, if the coupling constant betw een hydrogens on C1 and C 2 w as found to b e 7.8 H z (J 12 = 7.8 Hz). E xplain how you could distinguish betw een the follow ing tw o com pounds using N M R cou­ pling constants. (These com pounds are derived from glucose, by a reaction w e shall study in C hapters 16 and 22.)

OH

O

HO

OCH 3

HO OH

9.9E Complicating Features Proton N M R spectra have other features, however, that are n o t at all helpful w hen w e try to determ ine the structure o f a com pound. F or exam ple: 1.

Signals m ay overlap. T his happens w hen the chem ical shifts o f the signals are very nearly the sam e. In the 60-M H z spectrum o f ethyl chloroacetate (Fig. 9.25, top) we see that the singlet o f the — CH 2Cl group falls directly on top o f one o f the outer­ m ost peaks o f the ethyl quartet. U sing N M R spectrom eters w ith higher m agnetic field strength (corresponding to 1H resonance frequencies o f 300, 500, or 600 M H z) often allow s separation o f signals that w ould overlap at low er m agnetic field strengths (for an exam ple, see Fig. 9.25, n ex t page).

2.

S pin -sp in couplings betw een the protons o f nonadjacent atom s m ay occur. T his longrange coupling happens frequently in com pounds w hen p -b o n d e d atom s intervene betw een the atom s bearing the coupled protons, and in som e cyclic m olecules that are rigid.

3.

T he splitting patterns o f arom atic groups can be difficult to ana­ lyze. A m onosubstituted benzene ring (a phenyl group) has three different kinds o f protons:

The chem ical shifts o f these protons m ay b e so sim ilar that the phenyl group gives a signal that resem bles a singlet. O r the chem ical shifts m ay b e different and, because o f long-range couplings, the phenyl group signal m ay appear as a very com plicated m ultiplet.

9.9F Analysis of Complex Interactions In all o f the 1H N M R spectra that w e have considered so far, w e have restricted our atten­ tion to signal splittings arising from interactions o f only tw o sets o f equivalent protons on adjacent atom s. W hat kinds o f patterns should w e expect from com pounds in w hich m ore than tw o sets o f equivalent protons are interacting? We cannot answ er this question co m ­ pletely because o f lim itations o f space, b ut w e can give an exam ple that illustrates the kind o f analysis that is involved. L et us consider a 1-substituted propane: (a) (b) (c) c h 3 — c h 2 — CH2 — Z H ere, there are three sets o f equivalent protons. We have no problem in deciding w hat kind o f signal splitting to expect from the protons o f the CH 3 — group or the — CH 2Z group.

413

9.9 Signal Splitting: Spin-Spin Coupling

Hz 500

400

300

200

100

0

(b) O II (c) ClC H2— C — O CH2CH3 (b) (a)

(a)

(c)

TMS

JI 8.0

7.0

6.0

5.0

1------- T

4.0 Sh (ppm)

3.0

2.0

1.0

0

(a)

(b)

(a)

(c) TMS

1.4

1.2

I ................... I ..................... I ..................... I ..................... I ..................... I ..................... I ..................... I ..................... I

8

7

6

5

4

3

2

1

0

dH (PPm)

Figure 9.25 (Top) The 60-M Hz 1H NM R spectrum o f ethyl chloroacetate. N o te th e overlapping signals at d 4. (Bottom) The 300-M Hz 1H NM R spectrum o f ethyl chloroacetate, show ing resolution at higher m agnetic fie ld stren g th o f th e signals th a t o ve rlapped at 60 MHz. Expansions o f the signals are shown in th e o ffse t plots.

T he m ethyl group is sp in -sp in coupled only to the tw o protons o f the central — CH 2— group. T herefore, the m ethyl group should appear as a triplet. The protons o f the — C H 2Z group are sim ilarly coupled only to the tw o protons o f the central — CH 2 — group. Thus, the protons of the — CH 2Z group should also appear as a triplet. B ut w hat about the protons o f the central — CH 2 — group (b)? They are sp in -sp in cou­ pled w ith the three protons at (a) and w ith tw o protons at (c). The protons at (a) and (c), moreover, are not equivalent. If the coupling constants J ab and Jbc have quite different val­ ues, then the protons at (b) could be split into a quartet by the three protons at (a) and each line of the quartet could be split into a triplet by the tw o protons at (c), resulting in 12 peaks (Fig. 9.26).

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Chapter 9

Nuclear Magnetic Resonance and Mass Spectrometry

(b) (a) (b) (c) CH3CH2CH2Z

/ /

/

/In I I

ab

ab

ab

N

/ N

/In I

/

N N

/

N

I

N

Figure 9.26 The sp littin g p a tte rn th a t w o u ld occur fo r the (b) p rotons o f CH 3 CH 2 CH 2 Z if Jab w ere much larger than Jbc. Here Jab =

3

J.

Jb c ^ ^ Jb c ^

Jbc-

Jb c^

Jb c ^ ^ Jb c ^

A

^ Jb c ^

Jb c ^

^ Jbc~^

I t is u n lik e ly , h o w e v e r, th a t w e w o u ld o b s e rv e as m a n y as 12 p e a ks in an a c tu a l s p e c tru m b e ca u se th e c o u p lin g c o n s ta n ts a re su ch th a t p e a ks u s u a lly f a ll o n to p o f p e a ks. T h e 1H N M R s p e c tru m o f 1 -n itro p ro p a n e ( F ig . 9 .2 7 ) is ty p ic a l o f 1 -s u b s titu te d p ro p a n e c o m p o u n d s , in th a t th e c e n tra l — C H 2 — g ro u p “ sees” fiv e a p p ro x im a te ly e q u iv a le n t a d ja c e n t p ro to n s . H e n c e , b y th e

n + 1 r u le , w e see th a t th e (b ) p ro to n s a re s p lit in to s ix m a jo r pe a ks.

(a) (a)

(b)

(c) (a)

(c)

TMS

(b)

1.2

1.0

1 .............................. ............... ............... ............... ............... ............... ............... . 8

7

6

5

4

3

2

1

0

5h (ppm)

Figure 9.27 The 300-M Hz 1H NM R spectrum o f 1-nitropropane. Expansions o f th e signals are shown in th e o ffset plots.

ReviewProblem9.13

C a r r y o u t an a n a ly s is lik e th a t s h o w n in F ig . 9 .2 6 a n d s h o w h o w m a n y p e a k s th e s ig n a l f r o m (b ) w o u ld b e s p lit in t o i f

Jab = 2Jbc a n d i f J ab = Jbc. [Hint: I n b o th cases p e a k s w i l l

f a l l o n to p o f p e a ks so th a t th e to ta l n u m b e r o f p e a k s in th e s ig n a l is fe w e r th a n 1 2 .]

first-order spectra. I n fir s t- o r d e r s p e ctra , th e d is ta n c e in h e rtz ( A n ) th a t se p a ra te s th e c o u p le d T h e p re s e n ta tio n w e h a v e g iv e n h e re a p p lie s o n ly to w h a t a re c a lle d

s ig n a ls is v e r y m u c h la r g e r th a n th e c o u p lin g c o n s ta n t, J. T h a t is , A n > > J . In

second-order spectra ( w h ic h w e h a v e n o t d is c u s s e d ), A n a p p ro a c h e s J in m a g n i­ tu d e a n d th e s itu a tio n b e c o m e s m u c h m o r e c o m p le x . T h e n u m b e r o f p e a ks in c re a s e s a n d th e in te n s itie s a re n o t th o s e th a t m ig h t b e e x p e c te d f r o m fir s t- o r d e r c o n s id e ra tio n s .

9.10 Proton NMR Spectra and Rate Processes

415

9.10 Proton N M R Spectra and Rate Processes J. D. Roberts (Emeritus Professor, California Institute of Technology), a pioneer in the appli­ cation of N M R spectroscopy to problems of organic chemistry, has compared the N M R spectrometer to a camera with a relatively slow shutter speed. Just as a camera with a slow shutter speed blurs photographs of objects that are moving rapidly, the N M R spectrometer blurs its picture of molecular processes that are occurring rapidly. What are some of the rapid processes that occur in organic molecules? Two processes that we shall mention are chemical exchange of hydrogen atoms bonded to heteroatoms (such as oxygen and nitrogen), and conformational changes. C h e m ic a l E x c h a n g e C a u s e s S p in D e c o u p lin g An example of a rapidly occurring process can be seen in 1H N M R spectra of ethanol. The 1H N M R spectrum of ordinary ethanol shows the hydroxyl proton as a singlet and the protons of the — CH 2 — group as a quartet (Fig. 9.28). In ordinary ethanol we observe no signal splitting arising fro m cou­ plin g betw een the hydroxyl proton and the protons o f the — CH 2 — group even though they are on adjacent atoms.

8

7

6

5

4

3

2

1

0

dH (PPm)

Figure 9.28 The 300-M Hz 1H NMR spectrum o f o rdinary ethanol. There is no signal sp littin g by th e hydroxyl p ro to n due to rapid chemical exchange. Expansions o f th e signals are shown in the o ffse t plots.

If we were to examine a 1H N M R spectrum of very pure ethanol, however, we would find that the signal from the hydroxyl proton was split into a triplet and that the signal from the protons of the — CH 2 — group was split into a multiplet of eight peaks. Clearly, in very pure ethanol the spin of the proton of the hydroxyl group is coupled with the spins of the protons of the — CH 2 — groups. Whether coupling occurs between the hydroxyl protons and the methylene protons depends on the length of time the proton spends on a particular ethanol molecule. •

Protons attached to electronegative atoms with lone pairs such as oxygen (or nitro­ gen) can undergo rapid chemical exchange. That is, they can be transferred rapidly from one molecule to another and are therefore called exchangeable protons.

The chemical exchange in very pure ethanol is slow and, as a consequence, we see the sig­ nal splitting of and by the hydroxyl proton in the spectrum. In ordinary ethanol, acidic and basic impurities catalyze the chemical exchange; the exchange occurs so rapidly that the hydroxyl proton gives an unsplit signal and that of the methylene protons is split only by coupling with the protons of the methyl group.

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•

R apid exchange causes sp in d e c o u p lin g .

•

Spin decoupling is found in the 1H N M R spectra o f alcohols, am ines, and carboxylic acids. T he signals o f OH an d NH protons are norm ally unsplit and broad.

•

Protons that undergo rap id chem ical exchange (i.e., those attached to oxygen or nitrogen) can b e easily detected by placing the com pound in D 2O. T he protons are rapidly replaced by deuterons, and the proton signal disappears from the spectrum .

C o n fo r m a tio n a l C h a n g e s A t tem peratures near room tem perature, groups connected by carbon -carb o n single bonds rotate very rapidly (unless rotation is prevented by som e structural constraint, e.g., a rig id ring system ). B ecause o f this, w hen w e determ ine spec­ tra o f com pounds w ith single bonds that allow rotation, the spectra that w e obtain often reflect the individual hydrogen atom s in their average environm ent— that is, in an envi­ ronm ent that is an average o f all the environm ents that the protons have as a result o f con­ form ational changes. To see an exam ple o f this effect, let us consider the spectrum o f brom oethane again. The m ost stable conformation is the one in w hich the groups are perfectly staggered. In this stag­ gered conformation one hydrogen o f the m ethyl group (in red in the following structure) is in a different environm ent from that o f the other two m ethyl hydrogen atoms. If the N M R spec­ trom eter w ere to detect this specific conform ation o f brom oethane, it w ould show the protons o f the m ethyl group at a different chem ical shift. We know, however, that in the spectrum of brom oethane (Fig. 9.1), the three protons o f the m ethyl group give a single signal (a signal that is split into a triplet by spin-spin coupling with the two protons o f the adjacent carbon).

H

\ ,.C --C

H

H

H £H \

H

H

Br

H

Br

The m ethyl protons o f brom oethane give a single signal because at room tem perature the groups connected by the carbon-carbon single bond rotate approxim ately 1 m illion tim es each second. T he “shutter speed” o f the N M R spectrom eter is too slow to “photograph” this rapid rotation; instead, it photographs the m ethyl hydrogen atom s in their average envi­ ronm ents, and in this sense, it gives us a blurred picture o f the m ethyl group. R otations about single bonds slow dow n as the tem perature o f the com pound is low ­ ered. Som etim es, this slowing o f rotations allow s us to “see” the different conform ations o f a m olecule w hen w e determ ine the spectrum at a sufficiently low tem perature. A n exam ple o f this phenom enon, an d one that also show s the usefulness o f deuterium labeling, can be seen in the low -tem perature 1H N M R spectra o f cyclohexane an d o f undecadeuteriocyclohexane. (These experim ents originated w ith F. A. L. A net, Em eritus Professor, U niversity o f California, L os A ngeles, another pioneer in the applications of N M R spectroscopy to organic chem istry, especially to conform ational analysis.) H

D

d

V

D D

^ A d D D D D

U n d e c a d e u te rio c y c lo h e x a n e

A t room tem perature, ordinary cyclohexane gives one signal because interconversion o f chair form s occurs very rapidly. A t low tem peratures, however, ordinary cyclohexane gives a very com plex 1H N M R spectrum . A t low tem peratures interconversions are slow; the chem ical shifts o f the axial and equatorial protons are resolved, and com plex sp in -sp in cou­ plings occur.

9.11 Carbon-13 NMR Spectroscopy

4 17

A t —100°C, however, undecadeuteriocyclohexane gives only tw o signals o f equal inten­ sity. These signals correspond to the axial and equatorial hydrogen atom s o f the follow ing tw o chair conform ations. Interconversions betw een these conform ations occur at this low tem perature, but they happen slow ly enough for the N M R spectrom eter to detect the in d i­ vidual conform ations. [The nucleus o f a deuterium atom (a deuteron) has a m uch sm aller m agnetic m om ent than a proton, and signals from deuteron absorption do not occur in 1H N M R spectra.] H

D ^

D,n

^

^

D10

^

H

D

H ow m any signals w ould you expect to obtain in the 1H N M R spectrum o f undecadeuteriocyclohexane at room tem perature?

Re v ie w P ro b le m 9 .1 4

9.11 C arbon-13 N M R Spectroscopy 9.11A Interpretation of 13C NMR Spectra We begin our study o f 13C N M R sp e ctro sc o p y w ith a b rief exam ination o f som e special features o f spectra arising from carbon-13 nuclei. A lthough 13C accounts for only 1.1% of naturally occurring carbon, the fact that 13C can produce an N M R signal has profound im portance for the analysis o f organic com pounds. In som e im portant w ays 13C spectra are usually less com plex and easier to interpret than 1H N M R spectra. The m ajor isotope o f car­ bon, on the other hand, carbon-12 (12C), w ith a natural abundance o f about 99% , has no net m agnetic spin and therefore cannot produce N M R signals.

9.11B One Peak for Each Magnetically Distinct Carbon Atom T he interpretation o f 13C N M R spectra is greatly sim plified by the follow ing facts: •

E ach distinct carbon produces a single peak in a 13C N M R spectrum .

•

Splitting of signals into multiple peaks is not observed in routine 13C N M R spectra.

Recall that in 1H N M R spectra, hydrogen nuclei that are near each other (within a few bonds) couple w ith each other and cause the signal for each hydrogen to be split into a m ultiplet of peaks. Coupling is not observed for adjacent carbons because only one carbon atom of every 100 carbon atoms is a 13C nucleus (1.1% natural abundance). Therefore, the probability of there being two 13C atoms adjacent to each other in a m olecule is only about 1 in 10,000 (1.1% X 1.1%), essentially eliminating the possibility of two neighboring carbon atoms splitting each other’s signal into a m ultiplet of peaks. The low natural abundance of 13C nuclei and its inher­ ently low sensitivity also have another effect: Carbon-13 N M R spectra can be obtained only on pulse FTN M R spectrometers, where signal averaging is possible. W hereas carbon-carbon signal splitting does not occur in 13C N M R spectra, hydrogen atom s attached to carbon can split 13C N M R signals into m ultiple peaks. However, it is useful to sim plify the appearance o f 13C N M R spectra by initially elim inating signal split­ ting for 1H - 13C coupling. This can be done by choosing instrum ental param eters that decou­ ple the proto n -carb o n interactions, and such a spectrum is said to be b r o a d b a n d (BB) p ro to n d e c o u p le d . •

In a broadband proton-decoupled 13C N M R spectrum , each carbon atom in a dis­ tinct environm ent gives a signal consisting o f only one peak.

M ost 13C N M R spectra are obtained in the simplified broadband decoupled m ode first and then in modes that provide information from the 1H - 13C couplings (Sections 9.11D and 9.11E).

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9.11C 13C Chemical Shifts As we found with 1H spectra, the chemical shift of a given nucleus depends on the relative electron density around that atom. •

Decreased electron density around an atom d esh ield s the atom from the magnetic field and causes its signal to occur further d o w n field (higher ppm, to the left) in the N M R spectrum.

•

Relatively higher electron density around an atom sh ield s the atom from the mag­ netic field and causes the signal to occur u p field (lower ppm, to the right) in the N M R spectrum.

For example, carbon atoms that are attached only to other carbon and hydrogen atoms are relatively shielded from the magnetic field by the density of electrons around them, and, as a consequence, carbon atoms of this type produce peaks that are upfield in 13C N M R spectra. On the other hand, carbon atoms bearing electronegative groups are deshielded from the magnetic field by the electron-withdrawing effects of these groups and, therefore, produce peaks that are downfield in the N M R spectrum. Electronegative groups such as halogens, hydroxyl groups, and other electron-withdrawing functional groups deshield the carbons to which they are attached, causing their 13C N M R peaks to occur further downfield than those of unsubstituted carbon atoms. Reference tables of approximate chem­ ical shift ranges for carbons bearing different substituents are available. Figure 9.29 and Table 9.2 are examples. [The reference standard assigned as zero ppm in 13C N M R spec­ tra is also tetramethylsilane (TM S), Si(CH3)4.]

A p p ro x im a te C arb o n -13 Chemical Shifts T yp e o f C a rb o n A t o m

C h e m ic a l S h ift (8 , p p m )

1° Alkyl, RC H3 2° Alkyl, RC H2R 3° Alkyl, RC HR2

0 -4 0 10-50 15-50

Alkyl halide or am ine, — C— X I X = Cl, Br, or N — I

10-65

Alcohol or ether, — C — O—

50-90

Alkyne, — C^

60-90

\ A lkene,

C=

Aryl, Nitrile, —

c

100-170

C—

100-170

^ N

120-130

O II I A m ide, — C — N—

150-180

O II Carboxylic acid or ester, — C — O—

160-185

O A ldehyde or ketone, — C —

182-215

41 9

9.11 Carbon-13 NMR Spectroscopy

13C NMR Approximate Chemical Shift Ranges C-Ci, Br C N C-OR O il ^ C-N

1

1

O il 0

O II C-R,H

O 1

O il C OR

C

C- OH

C= N CH2

G C \ / C c

CH3

-C = C

— i----- 1----- 1----- 1----- 1----- 1----- 1----- 1----- 1----- 1----- 1----- 1----- 1----- 1----- 1----- 1----- 1----- 1----- 1----- 1----- 1----- 1----- r 220

200

180

160

140

120

100

80

60

40

20

0

Sc (ppm)

Figure 9.29 A p p ro xim a te 13C chemical shifts.

As a first example of the interpretation of trum o f 1-chloro-2-propanol (Fig. 9.30a): (a)

C NM R spectra, let us consider the

(b)

C spec

(c)

Cl— c h 2— c h — c h 3 2 I 3 OH 1 -C h lo ro -2 -p ro p a n o l

Figure 9.30

dC (PPm)

(a) The broadband p ro to n d ecoupled 13C NMR spectrum o f 1-chloro-2propanol. (b) These three spectra show th e DEPT 13C NM R data fro m 1chloro-2-propanol (see Section 9.11E). (This w ill be th e only full display o f a DEPT spectrum in th e te x t. O th e r 13C NMR figures will show th e full broadband p ro to n d ecoupled spectrum but w ith in fo rm a tio n fro m the DEPT 13C NMR spectra indicated near each peak as C, CH, CH2, o r CH3.)

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1 -C h lo ro -2 -p ro p a n o l c o n ta in s th re e c a rb o n a to m s in d is tin c t e n v iro n m e n ts , a n d th e re fo re p r o ­ du ce s th re e p e a ks in its b ro a d b a n d d e c o u p le d 13C N M R s p e c tru m : a p p ro x im a te ly a t 8 2 0 ,

8

5 1 , a n d 8 6 7 . F ig u r e 9 .3 0 a ls o s h o w s a c lo s e g ro u p in g o f th re e p e a ks a t 8 7 7 . T h e se peaks c o m e fr o m th e d e u te r io c h lo ro fo r m ( C D C l3) u se d as a s o lv e n t f o r th e sa m p le . M a n y 13C N M R sp e ctra c o n ta in these p e a ks. A lth o u g h n o t o f c o n c e rn to us h e re , th e s ig n a l f o r th e s in g le c a r­ b o n o f d e u te r io c h lo ro fo r m is s p lit in t o th re e p e a ks b y a n e ffe c t o f th e a tta c h e d d e u te riu m . •

T h e C D C l3 s o lv e n t p e a k s a t 8 7 7 s h o u ld b e d is re g a rd e d w h e n in te r p r e tin g 13C sp e ctra .

A s w e ca n see, th e c h e m ic a l s h ifts o f th e th re e p e a ks f r o m 1 - c h lo r o -2 - p ro p a n o l a re w e ll se p a ra te d f r o m o n e a n o th e r. T h is s e p a ra tio n re s u lts f r o m d iffe re n c e s in s h ie ld in g b y c ir c u ­ la tin g e le c tro n s in th e lo c a l e n v iro n m e n t o f e a ch c a rb o n . R e m e m b e r: T h e lo w e r th e e le c ­ tr o n d e n s ity in th e v ic in it y o f a g iv e n c a rb o n , th e le ss th a t c a rb o n w i l l b e s h ie ld e d , a n d th e m o r e d o w n fie ld w i l l b e th e s ig n a l f o r th a t c a rb o n . T h e o x y g e n o f th e h y d r o x y l g ro u p is th e m o s t e le c tro n e g a tiv e a to m ; i t w ith d r a w s e le c tro n s m o s t e ffe c tiv e ly . T h e re fo re , th e c a rb o n b e a rin g th e — O H g ro u p is th e m o s t

deshielded c a rb o n , a n d so th is c a rb o n g iv e s th e s ig ­

n a l th a t is th e fu rth e s t d o w n fie ld , a t 8 6 7 . C h lo r in e is less e le c tro n e g a tiv e th a n o x y g e n , ca u s ­ in g th e p e a k f o r th e c a rb o n to w h ic h i t is a tta c h e d to o c c u r m o re u p fie ld , a t 8 5 1 . T h e m e th y l g ro u p c a rb o n has n o e le c tro n e g a tiv e g ro u p s d ir e c tly a tta c h e d to it , so i t o c c u rs th e fu rth e s t u p fie ld , a t 8 2 0 . U s in g ta b le s o f ty p ic a l c h e m ic a l s h ift v a lu e s (s u c h as F ig . 9 .2 9 a n d T a b le 9 .2 ), o n e ca n u s u a lly a s s ig n 13C N M R s ig n a ls to e a ch c a rb o n in a m o le c u le , o n th e b a s is o f th e g ro u p s a tta c h e d to e a ch c a rb o n .

9.11D Off-Resonance Decoupled Spectra A t tim e s , m o r e in f o r m a t io n th a n a p r e d ic te d c h e m ic a l s h if t is n e e d e d to a s s ig n a n N M R p e a k to a s p e c ific c a rb o n a to m o f a m o le c u le . F o rtu n a te ly , N M R s p e c tro m e te rs c a n d if f e r ­ e n tia te a m o n g c a rb o n a to m s o n th e b a s is o f th e n u m b e r o f h y d ro g e n a to m s th a t a re a tta c h e d to e a ch c a rb o n . S e v e ra l m e th o d s to a c c o m p lis h th is a re a v a ila b le . O n e m e th o d n o t w id e ly u se d a n y m o re is c a lle d o ff- r e s o n a n c e d e c o u p lin g . In an o ff-re s o n a n c e d e c o u p le d 13C N M R s p e c tru m , e a ch c a rb o n s ig n a l is s p lit in t o a m u ltip le t o f p e a k s , d e p e n d in g o n h o w m a n y h y d ro g e n s a re a tta c h e d to th a t c a rb o n . A n

n + 1 r u le a p p lie s , w h e re n is th e n u m b e r o f

h y d ro g e n s o n th e c a rb o n in q u e s tio n . T h u s , a c a rb o n w it h n o h y d ro g e n s p ro d u c e s a s in g le t (n = 0 ), a c a rb o n w it h o n e h y d ro g e n p ro d u c e s a d o u b le t ( tw o p e a k s ), a c a rb o n w it h tw o h y d ro g e n s p ro d u c e s a tr ip le t (th re e p e a k s ), a n d a m e t h y l g ro u p c a rb o n p ro d u c e s a q u a rte t ( f o u r p e a k s ). In te rp r e ta tio n o f o ff-re s o n a n c e d e c o u p le d 13C s p e ctra , h o w e v e r, is o fte n c o m ­ p lic a te d b y o v e rla p p in g p e a ks f r o m th e m u ltip le ts .

9.11E DEPT 13C Spectra D E P T 13C N M R s p e c tra a re v e ry s im p le to in te rp re t. •

DEPT

13C N M R s p e c t r a in d ic a te h o w m a n y h y d ro g e n a to m s a re b o n d e d to each

c a rb o n , w h ile a ls o p r o v id in g th e c h e m ic a l s h ift in f o r m a t io n c o n ta in e d in a b ro a d ­ b a n d p r o to n - d e c o u p le d 13C N M R s p e c tru m . T h e c a rb o n s ig n a ls in a D E P T sp e c ­ tr u m a re c la s s ifie d as C H 3 , C H 2 , C H , o r C a c c o rd in g ly . A D E P T ( d is to r tio n le s s e n h a n c e m e n t b y p o la r iz a tio n tr a n s fe r ) s p e c tru m is a c tu a lly p r o ­ d u c e d u s in g d a ta f r o m se v e ra l 13C s p e c tra o f th e sa m e s a m p le ( F ig . 9 .3 0 b ), w it h th e n e t s p e c tru m r e s u lt p r o v id in g th e in f o r m a t io n a b o u t th e h y d ro g e n s u b s titu tio n a t e a ch c a rb o n ( F ig . 9 .3 0 a ). In th is te x t w e s h o w th e 13C p e a ks la b e le d a c c o rd in g to th e in f o r m a t io n g a in e d f r o m th e D E P T sp e c tra f o r th e c o m p o u n d u n d e r c o n s id e ra tio n , r a th e r th a n re p r o d u c in g th e e n tire f a m ily o f s p e c tra th a t le a d to th e f in a l re s u lt. A s a f u r th e r e x a m p le o f in te r p r e tin g 13C N M R s p e ctra , le t us lo o k a t th e s p e c tru m o f m e t h y l m e th a c r y la te ( F ig . 9 .3 1 ). ( T h is c o m p o u n d is th e m o n o m e r ic s ta rtin g m a te r ia l f o r th e c o m m e r c ia l p o ly m e r s L u c it e a n d P le x ig la s , see C h a p te r 1 0 .) T h e fiv e c a rb o n s o f m e th y l m e th a c r y la te re p re s e n t c a rb o n ty p e s f r o m s e v e ra l c h e m ic a l s h if t re g io n s o f 13C sp e ctra .

421

9.11 Carbon-13 NMR Spectroscopy

220

200

180

160

140

120

100

80

60

40

20

0

d c (PPm)

Figure 9.31

The broadband p ro to n -de co u p le d 13C NM R spectrum o f m ethyl m ethacrylate. In form ation fro m th e DEPT 13C NMR spectra is given above th e peaks.

Furthermore, because there is no symmetry in the structure of methyl methacrylate, all of its carbon atoms are chemically unique and so produce five distinct carbon NMR signals. Making use of our table of approximate 13C chemical shifts (Fig. 9.29 and Table 9.2), we can read­ ily deduce that the peak at d 167.3 is due to the ester carbonyl carbon, the peak at d 51.5 is for the methyl carbon attached to the ester oxygen, the peak at d 18.3 is for the methyl attached to C2, and the peaks at d 136.9 and d 124.7 are for the alkene carbons. Additionally, employ­ ing the information from the DEPT 13C spectra, we can unambiguously assign signals to the alkene carbons. The DEPT spectra tell us definitively that the peak at d 124.7 has two attached hydrogens, and so it is due to C3, the terminal alkene carbon of methyl methacrylate. The alkene carbon with no attached hydrogens is then, of course, C2.

ReviewProblem9.15

Compounds A, B, and C are isomers with the formula C5H11Br. Their broadband protondecoupled 13C NMR spectra are given in Fig. 9.32. Information from the DEPT 13C NMR spectra is given near each peak. Give structures for A, B, and C.

220

200

180

160

140

120

100

80

60

40

20

dC (PPm)

Figure 9.32 The broadband p ro to n -de co u p le d 13C NM R spectra o f com pounds A , B, and C, Review Problem 9.15. Inform ation fro m th e DEPT 13C NM R spectra is given above th e peaks.

(continues on the next page)

0

422

Chapter 9

Nuclear Magnetic Resonance and Mass Spectrometry

Sc (ppm)

5c (ppm)

Figure 9.32

(continued).

9.12 Two-Dimensional (2D) N M R Techniques M any N M R techniques are now available that greatly simplify the interpretation o f N M R spec­ tra. Chem ists can now readily glean inform ation about spin -sp in coupling and the exact con­ nectivity o f atom s in m olecules through techniques called m u ltid im e n sio n a l N M R spectroscopy. These techniques require an N M R spectrom eter o f the pulse (Fourier trans­ form) type. The m ost com m on m ultidim ensional techniques utilize tw o -d im en sio n al N M R (2D N M R ) and go by acronym s such as COSY, H ETCO R, and a variety o f others. [Even three-dimensional techniques (and beyond) are possible, although com putational requirem ents can lim it their feasibility.] T he tw o-dim ensional sense o f 2D N M R spectra does not refer to the w ay they appear on paper but instead reflects the fact that the data are accum ulated using two radio frequency pulses w ith a varying tim e delay betw een them. Sophisticated applica­ tion o f other instrum ental param eters is involved as well. D iscussion o f these param eters and the physics behind m ultidim ensional N M R is beyond the scope o f this text. The result, how ­ ever, is an N M R spectrum w ith the usual one-dim ensional spectrum along the horizontal and vertical axes and a set o f correlation peaks that appear in the x - y field o f the graph.

423

9.12 Two-Dimensional (2D) NMR Techniques

W hen 2D N M R is applied to 1H N M R it is called 1 H - 1 H c o r r e la t io n s p e c tr o s c o p y (or C O S Y for short). CO SY spectra are exceptionally useful for deducing p ro to n -p ro to n cou­ pling relationships. T w o-dim ensional N M R spectra can also be obtained that indicate cou­ pling betw een hydrogens and the carbons to w hich they are attached. In this case it is called h e te ro n u c le a r c o rre la tio n sp e ctro sc o p y (H E T C O R , or C - H H E T C O R ). W hen am bi­ guities are present in the one-dim ensional 1H and 13C N M R spectra, a H E T C O R spectrum can be very useful for assigning precisely w hich hydrogens and carbons are producing their respective peaks.

9.12A COSY Cross-Peak Correlations

Sh (ppm)

Figure 9.33 shows the C O SY spectrum for 1-chloro-2-propanol. In a C O SY spectrum the ordinary one-dim ensional 1H spectrum is show n along both the horizontal and the vertical axes. M eanw hile, the x - y field of a CO SY spectrum is sim ilar to a topographic m ap and can be thought of as looking dow n on the contour lines o f a m ap o f a m ountain range. A long the diagonal o f the CO SY spectrum is a view that corresponds to looking dow n on the ordi­ nary one-dim ensional spectrum of 1-chloro-2 -propanol as though each peak w ere a m oun­ tain. The one-dim ensional counterpart of a given p eak on the diagonal lies directly below that peak on each axis. The peaks on the diagonal provide no new inform ation relative to that obtained from the one-dim ensional spectrum along each axis.

dH (PPm)

Figure 9.33 COSY spectrum o f 1-chloro-2-propanol.

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The im portant and new inform ation from the C O SY spectrum , however, com es from the correlation peaks (“m ountains”) that appear o ff the diagonal (called “cross peaks”). If one starts at a given cross peak an d im agines tw o perpendicular lines (i.e., parallel to each spec­ trum axis) leading back to the diagonal, the peaks intersected on the diagonal by these lines are coupled to each other. H ence, the peaks on the one-dim ensional spectrum directly below the coupled diagonal peaks are coupled to each other. T he cross peaks above the diagonal are m irror reflections o f those below the diagonal; thus the inform ation is redundant and only cross peaks on one side o f the diagonal need b e interpreted. T he x - y field cross-peak correlations are the resu lt o f instrum ental param eters used to obtain the C O SY spectrum . L e t’s trace the coupling relationships in 1-chloro-2-propanol m ade evident in its CO SY spectrum (Fig. 9.33). (Even though coupling relationships from the ordinary one-dim en­ sional spectrum for 1-chloro-2 -propanol are fairly readily interpreted, this com pound m akes a good beginning exam ple for interpretation o f C O SY spectra.) First, one chooses a start­ ing point in the C O S Y spectrum from w hich to begin tracing the coupling relationships. A peak w hose assignm ent is relatively apparent in the one-dim ensional spectrum is a good point o f reference. F or this com pound, the doublet from the m ethyl group at 1.2 p pm is quite obvious and readily assigned. If w e find the peak on the diagonal that corresponds to the m ethyl doublet (labeled cH in Fig. 9.33 an d directly above the one-dim ensional m ethyl doublet on both axes), an im aginary line can b e draw n parallel to the vertical axis that inter­ sects a correlation peak (labeled bH -cH) in the x - y field o ff the diagonal. F rom here a p er­ pendicular im aginary line can b e draw n back to its intersection w ith the diagonal peaks. A t its intersection w e see that this diagonal peak is directly above the one-dim ensional spec­ trum peak at d 3.9. Thus, the m ethyl hydrogens at d 1.2 are coupled to the hydrogen w hose signal appears at d 3.9. It is now clear that the peaks at d 3.9 are due to the hydrogen on the alcohol carbon in 1-chloro-2-propanol (bH on C2). Returning to the peak on the diagonal above d 3.9, w e can trace a line back parallel to the horizontal axis that intersects a pair o f cross peaks betw een d 3.4 and d 3.5. M oving back up to the diagonal from each o f these cross peaks (aH '- bH and aH "-bH) indicates that the hydro­ gen w hose signal appears at d 3.9 is coupled to the hydrogens w hose signals appear at d 3.4 and d 3.5. T he hydrogens at d 3.4 and d 3.5 are therefore the tw o hydrogens on the carbon that bears the chlorine (aH' and aH"). O ne can even see that aH' and aH" couple w ith each other by the cross peak they have in com m on betw een them right next to their diagonal peaks. (aH' and aH" are diastereotopic. See Section 9.8B.) Thus, from the CO SY spectrum w e can quickly see w hich hydrogens are coupled to each other. Furtherm ore, from the reference start­ ing point, w e can “w alk around” a m olecule, tracing the neighboring coupling relationships along the m olecule’s carbon skeleton as w e go through the CO SY spectrum.

9.12B HETCOR Cross-Peak Correlations In a H E TC O R spectrum a 13C spectrum is presented along one axis an d a 1H spectrum is shown along the other. Cross peaks relating the two types o f spectra to each other are found in the x -y field. Specifically, the cross peaks in a H E TC O R spectrum indicate w hich hydro­ gens are attached to w hich carbons in a m olecule, or vice versa. These cross-peak correla­ tions are the result o f instrum ental param eters used to obtain the H ETC O R spectrum. There is no diagonal spectrum in the x -y field like that found in the CO SY spectrum. If im aginary lines are draw n from a given cross peak in the x -y field to each respective axis, the cross peak indicates that the hydrogen giving rise to the 1H N M R signal on one axis is coupled (and attached) to the carbon that gives rise to the corresponding 13C N M R signal on the other axis. Therefore, it is readily apparent w hich hydrogens are attached to w hich carbons. L et us take a look at the H ETC O R spectrum for 1-chloro-2-propanol (Fig. 9.34). Having interpreted the CO SY spectrum already, w e know precisely w hich hydrogens o f 1-chloro-2propanol produce each signal in the 1H spectrum. If an im aginary line is taken from the methyl doublet o f the proton spectrum at 1.2 ppm (vertical axis) out to the correlation peak in the x -y field and then dropped down to the 13C spectrum axis (horizontal axis), it is apparent that the 13C peak at 20 ppm is produced by the m ethyl carbon o f 1-chloro-2-propanol (C3). Having assigned the 1H N M R peak at 3.9 ppm to the hydrogen on the alcohol carbon o f the m olecule (C2), tracing out to the correlation peak and down to the 13C spectrum indicates that the 13C

9.12 Two-Dimensional (2D) NMR Techniques

425

dc (ppm)

Figure 9.34 1H - 13C HETCOR NM R spectrum o f 1-chloro-2-propanol. The 1H NM R spectrum is shown in blue and th e 13C NMR spectrum is shown in green. C orrelations o f th e 1H - 13C cross peaks w ith th e one-dim ensional spectra are in dicated by red lines.

N M R signal at 67 ppm arises from the alcohol carbon (C2). Finally, from the 1H N M R peaks at 3.4-3.5 ppm for the two hydrogens on the carbon bearing the chlorine, our interpretation leads us out to the cross peak and down to the 13C peak at 51 ppm (C 1 ). Thus, by a combination of COSY and HETCOR spectra, all 13C and 1H peaks can be unambiguously assigned to their respective carbon and hydrogen atoms in 1 -chloro-2 propanol. (In this simple example using 1-chloro-2-propanol, we could have arrived at complete assignment of these spectra without COSY and HETCOR data. For many com­ pounds, however, the assignments are quite difficult to make without the aid of these 2D N M R techniques.)

THE CHEMISTRY OF . . . M a g n e t i c R e s o n a n c e I m a g i n g in M e d i c i n e An important application of 1H NMR spectroscopy in m ed­ icine today is a technique called m agnetic resonance im ag­ ing, or MRI. One great advantage of MRI is that, unlike X-rays, it d oes not use dangerous ionizing radiation, and it d oes not require the injection of potentially harmful chem ­ icals in order to produce contrasts in the image. In MRI, a portion of the patient's body is placed in a powerful m ag­ netic field and irradiated with RF energy. A typical MRI image is shown at the right. The instru­ ments used in producing im ages like this one use the pulse method (Section 9.5) to excite the protons in the tissue under observation and use a Fourier transformation to trans­ late the information into an image. The brightness of various regions of the image is related to two things. ^ ^ ^

An jmage obtained by magnetic resonance imaging. , . ^ . (continues on the next page)

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The first factor is the number of protons in the tissue at that particular place. The second factor arises from what are called the relaxation tim es of the protons. When protons are excited to a higher energy state by the pulse of RF energy, they absorb energy. They must lose this energy to return to the lower energy spin state before they can be excited again by a second pulse. The process by which the nuclei lose this energy is called relaxation, and the time it takes to occur is the relaxation time. There are two basic m odes of relaxation available to pro­ tons. In one, called spin-lattice relaxation, the extra energy is transferred to neighboring m olecules in the surroundings (or lattice). The time required for this to happen is called T1 and is characteristic of the time required for the spin sys­ tem to return to thermal equilibrium with its surroundings. In solids, T1 can be hours long. For protons in pure liquid water, T is only a few seconds. In the other type of relax­ ation, called spin-spin relaxation, the extra energy is dissi­ pated by being transferred to nuclei of nearby atoms. The

time required for this is called T2- In liquids the magnitude of T2 is approximately equal to T-|. In solids, however, the T1 is very much longer. Various techniques based on the time between pulses of RF radiation have been d eveloped to utilize the differences in relaxation times in order to produce contrasts between different regions in soft tissues. The soft tissue contrast is inherently higher than that produced with X-ray techniques. Magnetic resonance imaging is being used to great effect in locating tumors, lesions, and edem as. Improvements in this technique are occurring rapidly, and the method is not restricted to observation of proton signals. One important area of medical research is based on the observation of signals from 3 1 P. Compounds that contain phosphorus as phosphate esters (Section 11.10) such as adenosine triphosphate (ATP) and adenosine diphosphate (ADP), are involved in most metabolic processes. By using techniques based on NMR, researchers now have a nonin­ vasive way to follow cellular metabolism.

9.13 An Introduction to Mass Spectrom etry M a ss sp e c tro m e try (M S) involves form ation o f ions in a m ass spectrom eter follow ed by separation and detection o f the ions according to m ass and charge. A m ass spectrum is a graph that on the x-axis represents the form ula w eights o f the detected ions, and on the yaxis represents the abundance o f each detected ion. T he x-axis is labeled m/z, w here m = m ass and z = charge. In exam ples w e shall consider, z equals + 1 , and hence the x-axis effec­ tively represents the form ula w eight o f each detected ion. T he y-axis expresses relative ion abundance, usually as a percentage o f the tallest peak or directly as the num ber o f detected ions. The tallest peak is called the b a se p e a k . A s a typical exam ple, the m ass spectrum of propane is show n in Fig. 9.35.

m/z, represents the formula weight of the m/z is the mass (m) to charge (z) ratio. Because z is typically +1, m/z represents the formula weight of each ion.

A The x-axis, in units of 100

detected ions. C

8 80

B The y-axis represents the relative abundance of each detected ion.

ID

C The most abundant ion (tallest peak) is called the base peak. The

C CO

60

base peak is usually an easily formed fragment of the original compound. In this case it is an ethyl fragment (C2H5+,

cd

> 12 CD QC

40

D One of the higher value D

m/z 29).

m/z peaks may or may not represent the

m o le c u la r io n (the ion with the formula weight of the original compound). W hen present, the molecular ion

2 0

(m/z 44 in the case of

propane) is usually not the base peak, because ions from the

B E

0

10

A

20

30

40

m/z

original molecule tend to fragment, resulting in the other 50

m/z peaks

in the spectrum. E Small peaks having

m/z values 1 or 2 higher than the formula weight

of the compound are due to 13C and other isotopes (Section 9.17).

Figure 9.35 A mass spectrum o f propane. (NIST Mass Spec Data Center, S. E. Stein, director, "M ass S pectra" in NIST C h e m istry W e b B o o k, NIST S ta n d a rd R eference D atabase N u m b e r 69, Eds. P. J. Linstrom and W. G. M allard, June 2005, National In stitu te o f Standards and Technology, G aithersburg, M D, 20899 (h ttp ://w e b b o o k .n is t.g o v ).)

4 27

9.15 Depicting the Molecular Ion

9 .1 4 Formation o f Ions: Electron Impact Ionization The ions in mass spectrometry m ay be form ed in a variety of ways. One m ethod for convert­ ing m olecules to ions (ionization) in a mass spectrometer is to place a sample under high vac­ uum and bom bard it with a beam o f high-energy electrons (—70 eV, or —6.7 X 103 kJ m o l_1). This m ethod is called electron im p a c t (EI) ionization mass spectrometry. The im pact o f the electron beam dislodges a valence electron from the gas-phase molecules, leaving them with a + 1 charge and an unshared electron. This species is called the m o le cu lar ion (M+). We can represent this process as follows: Note the + , representing an unshared electron and the net +1 charge. M

t

M o le c u le

e-

----- >

H ig h -e n e rg y e le c tro n

Mt

+

2 e-

M o le c u la r ion

The m olecular ion is a ra d ic a l c a tio n because it contains both an unshared electron and a positive charge. U sing propane as an exam ple, w e can w rite the follow ing equation to rep ­ resent form ation o f its m olecular ion by electron im pact ionization: A radical cation CH 3CH 2CH 3 t e - :

[CH 3CH 2CH3]t

t

2e-

9.15 Depicting the M olecular Ion N otice that w e have w ritten the above form ula for the propane radical cation in brackets. This is because w e do not know precisely from w here the electron w as lost in propane. We only know that one valence electron in propane was dislodged by the electron im pact process. However, depicting the m olecular ion w ith a localized charge and odd electron is som etim es useful (as we shall discuss in Section 9.16 w hen considering fragm entation reac­ tions). O ne possible form ula representing the m olecular ion from propane w ith a localized charge and an odd electron is the following: CH 3CH 2+ c h 3 In many cases, the choice of ju st where to localize the odd electron and charge is arbitrary, however. This is especially true if there are only carbon-carbon and carbon-hydrogen single bonds, as in propane. W hen possible, though, we write the structure showing the m olecular ion that would result from the removal of one of the m ost loosely held valence electrons of the original molecule. Just which valence electrons are m ost loosely held can usually be estimated from ionization potentials (Table 9.3). [The ionization potential of a m olecule is the am ount of energy (in electron volts) required to remove a valence electron from the molecule.] As w e m ight expect, ionization potentials indicate that the nonbonding electrons o f nitro­ gen, oxygen, and halogens and the p electrons o f alkenes and arom atic m olecules are held m ore loosely than the electrons o f carbon-carbon and carbon-hydrogen s bonds. Therefore, the convention o f localizing the odd electron and charge is especially applicable w hen the m olecule contains oxygen, nitrogen, or a p bond. The follow ing are exam ples o f these cases. Radical cations from ionization of nonbonding or p electrons. *+ CH 3— OH

*+ CH 3 — N— CH 3

•+ CH 2 — CH C H 2C H 3

CH 3 M eth an o l

T rim e th y la m in e

1 -B u te n e

Ionization Potentials o f Selected Molecules

Com pound

CH3(CH2)3NH2 CgHg (benzene) C2H4 c h 3o h C 2 H6 ch4

Io n iz a tio n P o te n tia l (eV )

8.7 9.2 10.5 10.8 11.5 12.7

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9 .1 6 Fragmentation Molecular ions formed by E I mass spectrometry are highly energetic species, and in the case of complex molecules, a great many things can happen to them. A molecular ion can break apart in a variety of ways, the fragments that are produced can undergo further frag­ mentation, and so on. We cannot go into all of the processes that are possible, but we can examine a few of the more important ones. As we begin, let us keep three important principles in mind: 1. The reactions that take place in a mass spectrometer are unimolecular, that is, they do not involve collisions between molecules or ions. This is true because the pres­ sure is kept so low ( 1 0 6 torr) that reactions involving bimolecular collisions do not occur. 2. We use single-barbed arrows to depict mechanisms involving single electron move­ ments (see Section 3.1A). 3. The relative ion abundances, as indicated by peak intensities, are very important. We shall see that the appearance of certain prominent peaks in the spectrum gives us key information about the structures of the fragments produced and about their original locations in the molecule.

9.16A Fragmentation by Cleavage at a Single Bond One important type of fragmentation is the simple cleavage of a single bond. With a radi­ cal cation this cleavage can take place in at least two ways; each way produces a cation and a radical . Only the cations are detected in a positive ion mass spectrometer. (The rad­ icals, because they are not charged, are not detected.) With the molecular ion obtained from propane by loss of one carbon-carbon s bonding electron, for example, two possible modes of cleavage are [c h 3 c h 2 c h 3]+

CH 3 CH2+

+

-CH 3

CH 3 CH 2

+

+CH3

These two modes of cleavage do not take place at equal rates, however. Although the rel­ ative abundance of cations produced by such a cleavage is influenced by the stability of both the carbocation and the radical, the carbocation’s stability is more important.* In the spectrum of propane shown earlier (Fig. 9.35), the peak at m/z 29 (CH 3 C H +) is the most intense peak; the peak at m/z 15 (CH+) has a relative abundance of only 5.6%. This reflects the greater stability of CH 3 CH+ as compared to C H +. When drawing mechanism arrows to show cleavage reactions it is convenient to choose a localized representation of the radical cation, as we have done above for propane. (When showing only an equation for the cleavage and not a mechanism, however, we would use the convention of brackets around the formula with the odd electron and charge shown out­ side.) Fragmentation equations for propane would be written in the following way (note the use of single-barbed arrows): _______H e lp f u l H i n t Recall that we use single-barbed arrows to show the movement of single electrons, as in the case of these homolytic bond cleavages and other processes involving radicals (see Section 3.1A).

CH 3 CH 2+CH 3 c h 3c h 2c h 3

-e

or c h 3 c h 2| c h 3

CH 3 CH2+ +

• CH 3

m /z 29

CH 3 CH2- +

+CH 3 m /z 1 5

* T h is can be d em onstrated th ro u g h th e rm o c h e m ic a l c a lc u la tio n s th a t w e ca nn o t go in to here. T h e interested stu d e nt is re fe rre d to M c L a ffe rty , F. W ., Interpretation o f Mass Spectra, 2 n d ed.; B e n ja m in : R eading, M A , 1973; pp. 41, 2 1 0 -2 1 1 .

42 9

9.16 Fragmentation

S o lv e d P ro b lem 9 .6 The m ass spectrum of CH 3F is given in Fig. 9.36. (a) D raw a likely structure for the m olecular ion (m/z 34). (b ) A ssign structural form ulas to the tw o other high abundance peaks (m /z 33 and m /z 15) in the spectrum . (c) P ropose an explanation for the low abundance o f the peak at m /z 19.

100

m/z Relative Ion Abundance C OD 80 c CÖ ■cO -Q

<

60

40 CD oc 20

II

1 10

15

20

25

30

14

17.2

15

100.0

19

2.0

31

10.4

32

9.4

33

89.5

34

95.4

35

1.1

Figure 9.36 Mass spectrum fo r Solved Problem 9.6.

35

m/z

STRATEGY AND ANSWER (a) N onbonding electrons have low er ionization energies than bonding electrons, so w e can expect that the m ole­ cular ion for CH 3F w as form ed by loss o f an electron from the fluorine atom. electron impact ionization

e

(b )

CH 3 - F=

(EI)

:

CH 3 — F

+

2

e

The ion w ith m /z 33 differs from the m olecular ion by one atom ic m ass unit, thus a hydrogen atom m ust have been lost. C leavage w ith loss o f a hydrogen atom could occur as follow s, leaving both the carbon and fluorine w ith full valence electron shells, but as a cationic species overall. H

H

H— C — F:

H H

H

■'c— F+

+ •. C = F,

H

H

The ion w ith m /z 15 m ust be a m ethyl carbocation form ed by loss o f a fluorine atom , as show n below. The fleet­ ing existence of a m ethyl carbocation is possible in electron im pact ionization (EI) m ass spectrom etry (M S) because electrons w ith high kinetic energy bom bard the species undergoing analysis, allow ing higher energy pathw ays to be follow ed than occur w ith reactions that take place in solution. H h

H +

— c — F+ h

A

:F:

,

H (c) The m /z 19 peak in this spectrum w ould have to be a fluorine cation. The presence o f only 6 valence electrons in an F+ ion and the strong electronegativity o f fluorine w ould create a very high energy barrier to form ation o f F+ and hence, cause it to be form ed in very low abundance relative to other ionization and cleavage p ath ­ w ays for C H 3F+.

430

Chapter 9

Nuclear Magnetic Resonance and Mass Spectrometry

M+ - 29 57

100

-

-

0

o 80 (b ase)

43

-

-

60 -

M

40

-

57

29

_TO

86

-

20 -

Figure 9.37

1

0 10

Mass spectrum

-

5

S

+

20

30

.1 40

r

. .1 1 50 60

i

70

80

M+ + 1 “

90

100

m/z

o f hexane.

9.16B Fragmentation of Longer Chain and Branched Alkanes The mass spectrum of hexane shown in Fig. 9.37 illustrates the kind of fragmentation a longer chain alkane can undergo. Here we see a reasonably abundant molecular ion at m/z 8 6 accompanied by a small M+ + 1 peak. There is also a smaller peak at m/z 71 (M+ - 15) corresponding to the loss of -CH3, and the base peak is at m/z 57 (M+ - 29) cor­ responding to the loss of -CH2 CH3. The other prominent peaks are at m/z 43 (M+ - 43) and m/z 29 (M+ - 57), corresponding to the loss of -CH 2 CH 2 CH 3 and -CH 2 CH 2 CH 2 CH3, respectively. The important fragmentations are just the ones we would expect: CH 3 CH 2 CH 2 CH 2 CH 2 4

•CH,

m /z 71

CH 3 CH 2 CH 2 CH 2 4 [CH 3 CH 2 CH 2 CH 2 CH 2 CH3]t

•CH 2 CH 3

m /z 5 7

—

CH 3 CH 2 CH 2 4

•c h 2 c h 2 c h 3

m /z 4 3

•c h 2 c h 2 c h 2 c h 3

CH 3 CH 2 4 m /z 2 9

Chain branching increases the likelihood of cleavage at a branch point because a more stable carbocation can result. When we compare the mass spectrum of 2-methylbutane (Fig. 9.38) with the spectrum of hexane, we see a much more intense peak at M+ - 15.

M+ - 29 43

100 -

-

! 80

j J

-

-

! 60

M+ - 5 57

M+ - 43 29

-

-

Ì4 . -

JS CD M+ - 57 CL 20 15

Figure 9.38 Mass spectrum of 2 -m ethylbutane.

0

10

1 20

1 1 30

1 40

50

.1 l 60

m/z

M+ 72 .1

70

-

80

90

100

431

9.16 Fragmentation

L o s s o f a m e th y l r a d ic a l f r o m

th e m o le c u la r io n o f 2 - m e t h y lb u t a n e c a n g iv e a s e c o n d a r y

c a r b o c a tio n : C H

3

c h 3c h c h 2c h 3

c h 3 c h c h 2 c h 3+ -c h 3

m /z 7 2 M t

m /z 5 7 M t -

15

w h e r e a s w i t h h e x a n e lo s s o f a m e t h y l r a d i c a l c a n y i e l d o n l y a p r i m a r y c a r b o c a t i o n . W i t h n e o p e n ta n e ( F ig . 9 .3 9 ) , th is e f f e c t is e v e n m o r e d r a m a tic . L o s s o f a m e t h y l r a d i ­ c a l b y t h e m o l e c u l a r i o n p r o d u c e s a te r t ia r y c a r b o c a t i o n , a n d t h i s r e a c t i o n ta k e s p la c e s o r e a d ily th a t v i r t u a l l y n o n e o f th e m o le c u la r io n s s u r v iv e lo n g e n o u g h to b e d e te c te d :

ch3

ch3 /

C H 3— C — C H C H

3

3

+

-c h 3

ch3

3 m /z 5 7

m /z 7 2 M t

M t -

15

M+ - 15 57

100 o

c h 3— C +

-

-

-

-

80

cs ■o 60

-

M+ 29

40

4 3

- 31 M+ 441

-

20 M+ - 57 15 .1,

1 10

M+ 72

20

30

■1 ■■ 40

-

I

■ ■ il 1 50 60

70

80

90

100

m/z

Figure 9.39

ReviewProblem9.16

I n c o n tr a s t to 2 - m e t h y lb u ta n e a n d n e o p e n ta n e , th e m a s s s p e c tr u m o f 3 - m e t h y lp e n t a n e ( n o t g iv e n ) h a s a p e a k o f v e r y l o w r e la t iv e a b u n d a n c e a t

— 1 5 . I t h a s a p e a k o f v e r y h ig h

r e la tiv e a b u n d a n c e a t M ^ — 2 9 , h o w e v e r . E x p la in .

9.16C Fragmentation to Form Resonance-Stabilized Cations C a r b o c a tio n s s t a b iliz e d b y r e s o n a n c e a re u s u a lly p r o m in e n t i n m a s s s p e c tr a . S e v e ra l w a y s th a t r e s o n a n c e - s t a b iliz e d c a r b o c a tio n s c a n b e p r o d u c e d a re o u t lin e d i n th e f o l lo w in g lis t . T h e s e e x a m p l e s b e g i n b y i l l u s t r a t i n g t h e l i k e l y s ite s f o r i n i t i a l i o n i z a t i o n ( p a n d n o n b o n d i n g e le c tr o n s ) , as w e ll.

1. A l k e n e s i o n i z e a n d f r e q u e n t l y u n d e r g o f r a g m e n t a t i o n s t h a t y i e l d r e s o n a n c e - s t a b i li z e d a l l y l i c c a tio n s :

C H 2= C H — C H 2— R 2 2

~e~

- ^ 0 ! ^

Mass spectrum of

neopentane.

C H 2^ C ^ C H ) R 2 2\ u

fragmentation> n

+ C H 2- C H = C H 2 2 a 2 m /z 41

+ •R

c h 2= c h - -C+H2

432

Chapter 9

Nuclear Magnetic Resonance and Mass Spectrometry

2. C a rbon-carbo n bonds n ex t to an atom w ith an unshared electron p air usually break readily because the resulting carbocation is resonance stabilized:

R— Z— CH 2— CH 3

ionization

-e

R

-C H 2:C H 3 2U 3

fragmentation

R — Z = CH„

•CH ,

R — Z — CH 0

w here Z = N, O, or S; R m ay also b e H. 3. C arbon-carbon bonds next to the carbonyl group o f an aldehyde or ketone break read ­ ily because resonance-stabilized ions called a c y liu m io n s are produced:

e

/ R'

R '— C = O < R'

R

/ I

R '— C = o : A c y liu m ion

or

R \

p+

C = O:

fragmentation

R — C = O¡

R'-

R' R— C = O. A c y liu m ion

4.

A lkyl-substituted benzenes ionize by loss o f a p electron an d undergo loss o f a hydrogen atom or m eth y l group to yield the relatively stable tropylium ion (see Section 14.7C). T his fragm entation gives a prom inent peak (som etim es the base peak) at m /z 91:

5. M onosubstituted benzenes w ith other than alkyl groups also ionize by loss o f a p electron and then lose their substituent to yield a phenyl cation w ith m /z 77:

433

9.16 Fragmentation

R e v ie w P r o b le m 9 . 1 7

P r o p o s e s tr u c tu r e s a n d fr a g m e n t a t io n m e c h a n is m s c o r r e s p o n d in g t o io n s w i t h m /z 5 7 a n d 4 1 in th e m a s s s p e c tr u m o f 4 - m e t h y l- l- h e x e n e .

m /z 57

m /z 41

S o l v e d P r o b le m

9 .7

E x p l a i n t h e f o l l o w i n g o b s e r v a t i o n s t h a t c a n b e m a d e a b o u t t h e m a s s s p e c t r a o f a lc o h o ls :

(a ) T h e m o l e c u l a r i o n p e a k o f a p r i m a r y o r s e c o n d a r y a l c o h o l i s v e r y s m a l l ; w i t h a t e r t i a r y a l c o h o l i t i s u s u a l l y u n d e te c ta b le .

(b ) P r i m a r y a l c o h o l s s h o w a p r o m i n e n t p e a k a t m/z 3 1 . (c ) S e c o n d a r y a l c o h o l s u s u a l l y g i v e p r o m i n e n t p e a k s a t m/z 4 5 , 5 9 , 7 3 , a n d s o o n . (d ) T e r t i a r y a l c o h o l s h a v e p r o m i n e n t p e a k s a t m/z 5 9 , 7 3 , 8 7 , a n d s o o n .

STRATEGY AND ANSWER (a ) A l c o h o l s u n d e r g o r a p i d c le a v a g e o f a c a r b o n - c a r b o n b o n d n e x t t o o x y g e n b e c a u s e t h i s l e a d s t o a r e s o n a n c e s t a b iliz e d c a tio n .

1 ° a lc o h o l

R : / C H2 ^ O ..H

2°

R — CH —

cR^ a lc o h o l

/-'•+

^ R^ 3 ° a lc o h o l

^

O .. H

—R .

OH

C H

[R— CH= CH2] + + H2O Mt

M t - 18

9.17 How to Determ ine Molecular Formulas and Molecular Weights

2.

435

C ycloalkenes can undergo a retro-D iels-A ld er reaction (Section 13.11) that produces an alkene and an alkadienyl radical cation:

w hich can also be w ritten as

3.

C arbonyl com pounds w ith a hydrogen on their g carbon undergo a fragm entation called the M cLafferty rearrangem ent.

w here Y = R, H, O R, OH, and so on. In addition to these reactions, w e frequently find peaks in m ass spectra that result from the elim ination o f other sm all stable neutral m olecules, for exam ple, H2, NH3, CO, HCN, H2S, alcohols, and alkenes.

9.17 H o w to D eterm in e M olecular Formulas and M olecular W eights Using Mass Spectrom etry 9.17A Isotopic Peaks and the Molecular Ion M ost o f the com m on elem ents found in organic com pounds have naturally occurring h eav­ ier isotopes. Table 9.4 lists som e elem ents and their isotopes, w ith the natural abundance o f each isotope given as the num ber of isotopic atom s per 100 atom s o f the m ost abundant

Principal Stable Isotopes o f Com m on Elem ents 3

E le m e n t

12C 1H 14n 16O 19F 28Si 31p 32S 3SCl 79Br 127I

1GG 1GG 1GG 1GG 1GG 1GG 1GG 1GG 1GG 1GG 1GG

N a tu r a l A b u n d a n c e o f O t h e r Is o to p e s (B ased on 1 0 0 A to m s o f M o s t C o m m o n Is o to p e )

13C 2H 1SN 17O

1.11 G.G1Ó G.3B G.G4

18O

G.2G

29Si

5.1G

3GSi

3.35

S 33

C arbon Hydrogen N itrogen O xygen Fluorine Silicon Phosphorus Sulfur Chlorine Bromine Iodine

M ost Com m on Is o to p e

G.7B 32.5 9B.G

34S

4.4G

37Cl 81Br

aReprinted w ith perm ission o f John W iley & Sons, Inc. from Silverstein, R. and W ebster, F. X., Spectrometric Identification of Organic Compounds, Sixth Edition, p. 7. C o p yrig h t 1998.

436

Chapter 9

Nuclear Magnetic Resonance and Mass Spectrometry

isotope. For three o f the elem ents— carbon, hydrogen, an d nitrogen— the principal heavier isotope is one m ass u n it greater than the m o st com m on isotope. •

T he presence o f isotopes o f carbon, hydrogen, and nitrogen in a com pound gives rise to a sm all M+ + 1 peak.

For four o f the elem ents— oxygen, sulfur, chlorine, an d brom ine— the principal heavier isotope is tw o m ass units greater than the m o st com m on isotope. 100

•

T he presence o f oxygen, sulfur, chlorine, or brom ine in a com pound gives rise to an M1 2 peak.

ce sC

M•

1 Elements:

C, H, N

M1

2 Elements:

O, S, Br, Cl

eC I 4C

e 2C R C 1C

15

2C

m/z m/z Relative Ion Abundance 12

2

1S

s.e

.e

14

17.1

15

s5.e

1B

1CC.C

17

•

T he M1

1 peak can be used to determ ine the n um ber o f carbons in a m olecule.

•

T he M1

2 peak can indicate w hether brom ine or chlorine is present.

•

T he isotopic peaks, in general, give us one m ethod for determ ining m olecular form ulas.

To understand how w e can determ ine the num ber o f carbons, let us begin by noticing that the isotope abundances in Table 9.4 are based on 100 atom s o f the norm al isotope. Now let us suppose, as an exam ple, that w e have 100 m olecules o f m ethane (C H 4). O n the aver­ age there w ill b e 1.11 m olecules that contain a 13C atom an d 4 X 0.016 m olecules that con­ tain a 2H atom . A ltogether, then, these heavier isotopes should contribute an M+ + 1 peak w hose intensity is about 1.17% o f the intensity o f the peak for the m olecular ion: 1.11 + 4(0.016) = 1.17%

1.15

Figure 9.42

Mass spectrum fo r

methane.

T his correlates w ell w ith the observed intensity o f the M+ trum o f m ethane given in Fig. 9.42.

1 peak in the actual spec-

9.17B How to Determine the Molecular Formula For m olecules w ith a m odest num ber o f atom s w e can determ ine m olecular form ulas in the follow ing way. If the M+ peak is n o t the base peak, the first thing w e do w ith the m ass spec­ trum o f an unknow n com pound is to recalculate the intensities o f the M+ + 1 and M+ + 2 peaks to express them as percentages o f the intensity o f the M+ peak. C onsider, for exam ple, the m ass spectrum o f an unknow n com pound given in Fig. 9.43. T he M+ peak at m/z 72 is n ot the base peak. Therefore, w e need to recalculate the inten­ sities o f the peaks in our spectrum at m/z 72, 73, and 74 as percentages o f the peak at m/z 72. We do this by dividing each intensity by the intensity o f the M+ peak, w hich is 73% , and m ultiplying by 100. T hese results are show n here an d in the second colum n o f Fig. 9.43. m /z 72 73 74 H e lp fu l H in t How to determine a molecular formula using MS.

In te n s ity (% o f

M+ )

73.0/73 X 100 = 100 3.3/73 X 100 = 4.5 0.2/73 X 100 = 0.3

T hen w e use the follow ing guides to determ ine the m olecular formula: 1. Is M+ o d d o r even? A cc o rd in g to th e n itro g e n r u le , if it is even, th e n th e co m ­ p o u n d m u s t c o n ta in a n ev en n u m b e r o f n itro g e n a to m s (zero is a n even n u m ­ b e r). F or our unknow n, M+ is even. T he com pound m u st have an even n um ber of nitrogen atom s.

9.17 How to Determ ine Molecular Formulas and Molecular Weights

43 7

Intensity (as percent of M + )

Intensity

m/z (as percent of base peak) m/z 27

59.0

72

M+

28

15.0

73

M+ + 1

4.5

29

54.0

74

M+ + 2

0.3

39

23.0

41

60.0

100.0

Recalculated to base on M+V____

42

12.0

43

79.0

44

100.0 (base)

72

73.C

J

V

1 1 1 1 1 1

2. T h e

73

3.3

74

C.2

N um ber

of

C

1

y

)

I

r e la tiv e a b u n d a n c e o f th e

a to m s .

1

M +l

a to m s =

M-

+

Figure 9.43

Mass spectrum o f an unknown

com pound.

1 p e a k in d ic a te s t h e n u m b e r o f c a r b o n

r e la tiv e

abundance

of

(M + +

1 )/1 .1 .

For

our

u n k n o w n (F ig . 9 .4 3 ),

N um ber o f C atom s =

4 .S =

4

1 .1

(This form ula w orks because 13C is the m ost im portant contributor to the peak and the approxim ate natural abundance o f 13C is 1.1%.)

+ 1

3. T h e re la tiv e a b u n d a n c e o f th e M^ + 2 p e a k in d ic a te s th e p re se n c e (o r ab sen ce) o f S (4.4% ), Cl (33% ), o r Br (98% ) (see Table 9.4). For our unknown, M^ + 2 = 0.3%; thus, w e can assum e that S, Cl, and Br are absent. 4. T he m olecular fo rm u la can now be established by d eterm in in g th e n u m b e r of h y d ro ­ gen atom s a n d ad d in g th e a p p ro p ria te n u m b e r o f oxygen atom s, if necessary. For our unknow n the M^ peak at m/z 72 gives us the m olecular w eight. It also tells us (since it is even) that nitrogen is absent because a com pound w ith four carbons (as estab­ lished above) and tw o nitrogens (to get an even m olecular w eight) w ould have a m olecu­ lar w eight (76) greater than that of our com pound. F o r a molecule composed o f

Cand Honly, H = 72 - (4 X 12) = 24

but C 4H24 is im possible. F o r a molecule composed o f

C,H,

and one

O,

H = 72 - (4 X 12) - 16 = 8 and thus our unknow n has the m olecular form ula C 4H8O. D eterm ine the m olecular form ula for a com pound that gives the following mass spectral data: I n t e n s ity

m iz

(a s % o f b a s e p e a k )

Só M +

1G.GG G.5ó G.G4

87 ss

ReviewProblem9.19

438

Chapter 9

Nuclear Magnetic Resonance and Mass Spectrometry

S o lv e d P ro b le m 9 .8

(a) W hat approxim ate intensities w ould you expect for the M+ and M+ + 2 peaks o f CH 3Cl? (b) F or the M+ and M+ + 2 peaks o f CH 3Br? (c) A n organic com pound gives an M+ peak at m/z 122 and a peak o f nearly equal intensity at m/z 124. W hat is a likely m olecular form ula for the com pound? STRATEGY AND ANSWER (a) T he M+ + 2 peak due to CH 3— 37Cl (at m/z 52) should be alm ost one-third (32.5% ) as large as the M+ peak at m/z 50 because o f the relative natural abundances o f 35Cl and 37Cl. (b) T he peaks due to CH 3— 79Br and CH 3— 81Br (at m/z 94 an d m/z 96, respectively) should b e o f nearly equal intensity due to the relative natural abundances o f 79Br an d 81 Br. (c) T hat the M+ and M+ + 2 peaks are o f nearly equal intensity tells us that the com pound contains brom ine. C3H7Br is therefore a likely m olecular form ula. C3 =

36

C3 =

36

H7 = 79Br =

7

H7 =

7

81Br = _81

m/z = 122

R e v ie w P ro b le m 9 .2 0

m/z = 124

U se the m ass spectral data below to calculate the M+ + 1 and M+ + 2 ratios w ith respect to the m olecular ion (M+). T hen determ ine the m olecular form ula for the com pound by con­ sulting Table 9.5.

m /z

Intensity (as % of base peak)

14 15 18 28 29 42

8.0 38.6 16.3 39.7 23.4 46.6

Relative Intensities o f M+ + 1 and M+ fo r Masses 72 and 73

m /z

43 44 73 74 75

Intensity (as % of base peak) 10.7 100.0 86.1 3.2 0.2

(base) M+ M+ + 1 M+ + 2

2 Peaks fo r Various C om binations o f C, H, N, and O

Percentage of M+ Intensity

Percentage of M+ Intensity

M+

Formula

M+ + 1

M+ + 2

M+

Formula

M+ + 1

M+ + 2

72

CH 2 N3O CH 4 N4 C 2 H2NO 2 C 2 H4N2O

2.30 2.67 2.65 3.03 3.40 3.38 3.76 4.13 4.49 4.86 5.60

0.22 0.03 0.42 0.23 0.04 0.44 0.25 0.07 0.28 0.09 0.13

73

CHN 2O 2 CH3N3O CH5N4 C2HO3 C2H3NO2 C2H5N2O C2H7N3 C3H5O2 C3H7NO C3H9N2 C4H9O c 4h 11n

1.94 2.31 2.69 2.30 2.67 3.04 3.42 3.40 3.77 4.15 4.51 4.88 6.50

0.41 0.22 0.03 0.62 0.42 0.23 0.04 0.44 0.25 0.07 0.28 0.10 0.18

C2H6N3 C 3 H4O 2 C 3 H6NO C 3 H8N2 C4H8O C 4 H 10N C 5 H 12

C 6H

9.17 How to Determ ine Molecular Formulas and Molecular Weights

What are the expected ratios of the (a)

Br

+ 2, and

(b)

Br

+ 4 peaks for the following compounds?

43 9

Review Problem 9.21

Cl

Cl

(a) Determine the molecular formula o f the compound w hose mass spectrum is given in the follow ing tabulation:

(b)The 1H NM R spectrum o f this compound consists only o f a large doublet and a small septet. What is the structure o f the compound?

As the number o f atoms in a m olecule increases, molecular weight calculations like this becom e more and more com plex and time-consuming. Fortunately, however, these calcu­ lations can be done readily with computers, and tables are now available that give relative values for the M^ + 1 and M^ + 2 peaks for all combinations o f common elements with molecular formulas up to mass 500. Part o f the data obtained from one o f these tables is given in Table 9.5. U se Table 9.5 to check the results o f our example (Fig. 9.43).

9.17C High-Resolution Mass Spectrometry A ll o f the spectra that w e have described so far have been determined on what are called “low-resolution” mass spectrometers. These spectrometers, as w e noted earlier, measure m/z values to the nearest whole-number mass unit. Many laboratories are equipped with this type of mass spectrometer. Some laboratories, however, are equipped with the more expensive “high-resolution” mass spectrometers. These spectrometers can measure m/z values to three or four decimal places and thus provide an extremely accurate method for determining molecular weights. And because molecular weights can be measured so accurately, these spectrometers also allow us to determine molecular formulas. The determination o f a molecular formula by an accurate measurement o f a molecular w eight is possible because the actual m asses o f atomic particles (nuclides) are not inte­ gers (see Table 9.6). Consider, as exam ples, the three m olecules O2, N 2 H4 , and CH 3 OH. Exact Masses o f Nuclides Iso to p e

M ass

Iso to p e

M ass

1H 2H 12C 13C 14n 15n 16o 17O 18O

1.00783 2.01410 12.00000 (std) 13.00336 14.0031 15.0001 15.9949 16.9991 17.9992

19f 32s 33S 34s 35ci 37Cl 79Br 81Br 127I

18.9984 31.9721 32.9715 33.9679 34.9689 36.9659 78.9183 80.9163 126.9045

ReviewProblem9.22

440

Chapter 9

Nuclear Magnetic Resonance and Mass Spectrometry

The actual atomic masses of the molecules are all different (though nominally they all have atomic mass of 32): O2

= 2(15.9949) = 31.9898

N 2 H 4 = 2(14.0031) + 4(1.00783) = 32.0375 CH4O = 12.00000 + 4(1.00783) + 15.9949 = 32.0262

High-resolution mass spectrometers are available that are capable of measuring mass with an accuracy of 1 part in 40,000 or better. Thus, such a spectrometer can easily distinguish among these three molecules and, in effect, tell us the molecular formula. The ability of high-resolution instruments to measure exact masses has been put to great use in the analysis of biomolecules such as proteins and nucleic acids. For example, one method that has been used to determine the amino acid sequence in oligopeptides is to mea­ sure the exact mass of fragments derived from an original oligopeptide, where the mixture of fragments includes oligopeptides differing in length by one amino acid residue. The exact mass difference between each fragment uniquely indicates the amino acid residue that occu­ pies that position in the intact oligopeptide (see Section 24.5E). Another application of exact mass determinations is the identification of peptides in mixtures by comparison of mass spectral data with a database of exact masses for known peptides. This technique has become increasingly important in the field of proteomics (Section 24.14).

9.18 Mass S p ectro m eter Instrum ent Designs There are two principal components of mass spectrometers: the ionization chamber, where ionization of the sample occurs, and the mass analyzer, where ion sorting and detection occur. Mass spectrometer instruments vary in design with regard to both of these components. Thus far we have mentioned only one ionization technique, electron impact (EI). In Section 9.18A we discuss EI ionization in more detail, as well as discuss two other important ionization meth­ ods: electrospray ionization (ESI) and matrix-assisted laser desorption ionization (M ALDI). A variety of mass analyzer designs are in use as well, including magnetic focusing, quadrupole, ion trap, and time-of-flight (TOF). In Section 9.18B we explain the classical method for ion sorting (magnetic focusing), and briefly mention the other methods. The student is referred to textbooks of spectroscopy and instrumental analysis for further information.

9.18A Ionization Techniques: Electron Impact, Electrospray, and MALDI ______ H e lp f u l H i n t A classic and highly useful reference on MS, NMR, and IR methods is Silverstein, R. M.; and Webster, F. X. Spectrometric

Identification of Organic Compounds, 6th ed.; Wiley: New York, 1998.

E le c tr o n Im p a c t Io n iz a tio n Electron impact ionization can be described as a “bruteforce” method because it involves striking an organic molecule with 70 eV electrons, a tech­ nique akin to firing a howitzer at a house made of matchsticks. It is no wonder that significant fragmentation takes place. Figure 9.44 shows a schematic diagram of a mass spectrometer that employs electron impact ionization. Ionization occurs in the ionizing chamber as gas-phase analyte molecules are struck by the electron beam. Positive ions formed from the analyte are accelerated and focused into the mass analyzer by passage through slits in negatively charged plates. Electron impact ionization requires molecules of the analyte to be sufficiently volatile that they can be transferred to the gas phase in the high vacuum conditions of the ionizing chamber. This requirement for volatility essentially limits E I mass spectrometry to mole­ cules that have formula weights of less than 1000 daltons (atomic mass units) and to mol­ ecules that are not very polar. While EI mass spectrometry is suitable for most of the types of organic molecules we shall study, it is not generally suitable for biomolecules that have high molecular weights, high polarity, or both. Fortunately, very effective methods have been developed for ionization of biomolecules and other molecules not suited to EI mass spec­ trometry. Among these techniques are electrospray ionization (ESI), ion spray, matrixassisted laser desorption ionization (M A LD I), and fast atom bombardment (FAB).

9.18 Mass Spectrom eter Instrument Designs

Figure 9.44

M a s s s p e c t r o m e t e r . S c h e m a tic d ia g r a m o f C E G m o d e l 2 1 -1 0 3 . T h e m a g n e tic f ie ld

t h a t b r in g s io n s o f v a r y in g m a s s -to -c h a rg e ra tio s (m/z) in t o r e g is te r is p e r p e n d ic u la r t o t h e p a g e . ( R e p r in te d w it h p e r m is s io n o f J o h n W ile y & S o n s , In c . f r o m H o lu m , J. R., O rganic Chem istry: A B rie f

Course, C o p y r ig h t 1 9 7 5 .)

E le c tr o s p r a y

Io n iz a tio n — A

T e c h n iq u e

E s p e c ia lly

U s e fu l

fo r

B io m o le c u le s

E lectrospray ionization w orks especially w ell for m ass spectrom etry o f proteins, carbohy­ drates, and nucleic acids. ESI m ass spectrom etry has been used to study protein m olecular w eights and sequence, enzym e-substrate com plexes, antibody-antigen binding, drug-receptor interactions, and D N A oligonucleotide sequence, as w ell as sim ply for sm all m olecules that cannot be ionized by electron im pact. In e le c tro s p ra y io n iz atio n (E S I) a solution o f the analyte is sprayed into the vacuum cham ber o f the m ass spectrom eter from the tip o f a high-voltage needle. The extrem e elec­ trical potential im parts charge to the m ixture, w hich on evaporation o f the solvent in the m ass spectrom eter affords charged species of the analyte. The analyte m ay actually acquire m ultiple charges through E SI (i.e., z can have a range o f values in the m /z ratio), and hence a fam ily o f m /z peaks typically results for each analyte. This distribution can be converted by the m ass spectrom eter softw are to the form ula w eight o f the original analyte. A nother advantage o f ESI M S is that a high-perform ance liquid chrom atograph (HPLC) can be used to introduce the sam ple to the m ass spectrometer. L inking chrom atographic separation tech­ niques w ith m olecular spectroscopy, as in tandem H PL C and ESI M S analysis, affords a pow erful analytical com bination. We shall see another exam ple below w hen w e consider GC/M S (gas chrom atography w ith m ass spectrometry). We shall also have m ore to say about E SI M S w hen w e study proteins in C hapter 24. M A L D I— A

T e c h n iq u e

U s e fu l fo r

B o th

B io m o le c u le s

and

S y n th e tic

P o ly m e r s

M a trix -a ssisted la se r d eso rp tio n -io n iz a tio n (M A L D I) works very w ell for synthetic poly­ m ers, such as polybutadiene and polystyrene, as w ell as for other classes o f m olecules that do not ionize w ell by electrospray ionization. M A L D I is also useful for biom olecules, and hence can com plem ent ESI m ethods. In M A LD I mass spectrometry the analyte is mixed with low-molecular-weight organic m ol­ ecules that are known for their ability to absorb and transfer energy from the laser in the mass

441

442

Chapter 9

Nuclear Magnetic Resonance and Mass Spectrometry

spectrometer. After evaporation of the solvent, this mixture is called the sample matrix. The matrix is placed in the mass spectrometer under high vacuum and pulsed with laser radiation. Molecules of the matrix absorb radiation from the laser and transfer energy to the analyte. Many of the analyte molecules acquire a +1 charge through this process and are transferred to the vapor phase, after which the ions are drawn into the mass analyzer for separation and detection.

9.18B Mass Analysis: Ion Sorting and Detection Once ions of the sample have been formed by one of the methods above, they are sepa­ rated and detected by the mass analyzer component of the spectrometer. Several common ways exist to accomplish mass analysis. We shall first describe the classic method of mag­ netic focusing, and then briefly mention several other important approaches. M a g n e t ic F o c u s in g Classic mass spectrometers accelerate ions formed in the ioniza­ tion chamber into a curved tube (see Fig. 9.44). This curved tube passes through a variable magnetic field that exerts an influence on the moving ions. Depending on the magnetic field strength at a given moment, ions with a particular m /z will follow a curved path that exactly matches the curvature of the tube. These ions are said to be “in register.” Because they are in register, these ions pass through another slit and impinge on an ion collector where the inten­ sity of the ion beam is measured electronically. The intensity of the beam is simply a mea­ sure of the relative abundance of the ions with a particular m/z. Some mass spectrometers are so sensitive that they can detect the arrival of a single ion. The actual sorting of ions takes place in the magnetic field, and this sorting takes place because laws of physics govern the paths followed by charged particles when they move through magnetic fields. Generally speaking, a magnetic field such as this w ill cause ions moving through it to move in a path that represents part of a circle. The radius of curva­ ture of this circular path is related to the m /z of the ions, to the strength of the magnetic field (B0, in tesla), and to the accelerating voltage. If we keep the accelerating voltage con­ stant and progressively increase the magnetic field, ions whose m /z values are progressively larger w ill travel in a circular path that exactly matches that of the curved tube. Hence, by steadily increasing B 0 , ions with progressively increasing m /z w ill be brought into register and so w ill be detected at the ion collector. Since, as we said earlier, the charge on nearly all of the ions is unity, this means that ions o f progressively increasing m ass arrive a t the

collector and are detected.

A variety of other methods are used for io n s o rtin g in mass spectrometers, including quadrupole mass filtering, ion trapping, and time-of-flight mass analyzers. In a quadrupole mass analyzer, ions are filtered by varying the electrical signal in four parallel charged rods. At any given instant, only ions of certain mass-to-charge ratio are able to travel through the quadrupole region to the detector. Other ions collide with the rods and are neutralized. By varying the electrical state of the four rods in pairs, a range of masses can be scanned. Ion trap mass analyzers involve a ring electrode charged with a varying radio frequency voltage. Ions enter the cavity enclosed by the ring, and those of appropriate mass take up a stable orbit. As the voltage state of the ring varies, so does the mass for which a stable orbit is possible. Varying the ring voltage allows a range of masses to be scanned by progressively trapping and releas­ ing them to the detector. In time-of-flight mass analyzers, ions are accelerated into a tube that is free of electrical fields. The ions drift toward the detector, and the time it takes to traverse the tube is correlated with their respective masses. Q u a d r u p o l e , Io n T r a p , a n d T im e - o f - F lig h t ( T O F ) M a s s A n a ly z e r s

9.19 G C /M S Analysis Gas chromatography is often coupled with mass spectrometry in a technique called GC/MS analysis. The gas chromatograph separates components of a mixture, while the mass spec­ trometer then gives structural information about each one (Fig. 9.45). GC/MS can also pro­ vide quantitative data when standards of known concentration are used with the unknown.

443

Key Terms and Concepts

Electron beam

ionization occurs here.

Figure 9.45 Schem atic o f a typical capillary gas chrom atograph/m ass sp e ctro m e te r (GC/MS).

In GC analysis, a m inute am ount of a m ixture to be analyzed, typically 0.001 m L (1.0 mL) or less o f a dilute solution containing the sam ple, is injected by syringe into a heated port o f the gas chrom atograph. The sam ple is vaporized in the injector port and sw ept by a flow of inert gas into a capillary colum n. The capillary colum n is a thin tube usually 10-30 m eters long and 0 .1 -0 .5 m m in diameter. It is contained in a cham ber (the “oven”) w hose tem perature can be varied according to the volatility o f the sam ples being analyzed. The inside o f the capillary colum n is typically coated w ith a “stationary phase” o f low polarity (essentially a high-boiling and very viscous liquid that is often a nonpolar silicon-based polym er). A s m olecules o f the m ixture are sw ept by the inert gas through the colum n, they travel at different rates according to their boiling points and the degree o f affinity for the stationary phase. M aterials w ith higher boiling points or stronger affinity for the stationary phase take longer to pass through the colum n. Low -boiling and nonpolar m aterials pass through very quickly. The length of tim e each com ponent takes to travel through the col­ um n is called the retention tim e. R etention tim es typically range from 1 to about 30 m in­ utes, depending on the sam ple and the specific colum n used. A s each com ponent of the m ixture exits the GC colum n it travels into a m ass spec­ trom eter. H ere, m olecules o f the sam ple are bom barded by electrons; ions and fragm ents o f the m olecule are form ed, and a m ass spectrum results sim ilar to those w e have studied earlier in this chapter. The im portant thing, however, is that m ass spectra are obtained for each com ponent of the original m ixture that is separated. This ability o f G C/M S to sepa­ rate m ixtures and give inform ation about the structure o f each com ponent m akes it a vir­ tually indispensable tool in analytical, forensic, and organic synthesis laboratories.

9.20 Mass S pectrom etry o f Biomolecules A dvances in m ass spectrom etry have m ade it a tool o f exceptional pow er for analysis of large biom olecules. Electrospray ionization, M A LD I, and other “soft ionization” techniques for nonvolatile com pounds and m acrom olecules m ake possible analyses o f proteins, nucleic acids, and other biologically relevant com pounds w ith m olecular w eights up to and in excess o f 100,000 daltons. Electrospray ionization w ith quadrupole m ass analysis is now routine for biom olecule analysis as is analysis using M A L D I-T O F instrum ents. E xtrem ely high resolution can be achieved using Fourier tran sfo rm -io n cyclotron resonance (FT ICR, or FTM S). We shall discuss E SI and M A L D I applications o f m ass spectrom etry to protein sequencing and analysis in Sections 24.5E, 24.13B, and 24.14.

Key Terms and Concepts The key term s and concepts that are highlighted in b o ld , b lu e te x t w ithin the chapter are defined in the glossary (at the back of the book) and have hyperlinked definitions in the accom panying W ileyP LU S course (w w w .w ileyplus.com ).

PLUi

444

Chapter 9

Nuclear Magnetic Resonance and Mass Spectrometry

Problems Note to Instructors: Many of the homework problems are available for assignment via WileyPLUS, an online teaching and learning solution.

N M R SPECTROSCOPY The following are some abbreviations used to report spectroscopic data: 1H N M R : s = singlet, d = doublet, t = triplet, q = quartet, bs = broad singlet, m = multiplet IR absorptions: s = strong, m = moderate, br = broad 9.23

How many 1H NMR signals (not peaks) would you predict for each of the following compounds? (Consider all protons that would be chemical shift nonequivalent.) (a)

(b)

(d)

(e)

Br

(h)

OH

OH

9.24

How many 13C NMR signals would you predict for the compounds shown in Problem 9.23?

9.25

Propose a structure for an alcohol with molecular formula C 5 H12O that has the 1H NMR spectrum given in Fig. 9.46. Assign the chemical shifts and splitting patterns to specific aspects of the structure you propose.

C5H8O

6H

1H

............................................................................ 1

2H

3H

W aJU Ia J . ..................................................................................... 1

.....................................................................................

ill

2

1 dH (PPm)

Figure 9.46 The 1H NM R spectrum (simulated) o f alcohol C5H8O, Problem 9.25

0

445

Problems

9.26

Propose structures for the com pounds G and H w hose 1H N M R spectra are show n in Figs. 9.47 and 9.48.

TMS

ì I .....................I ....................... I ....................... I ....................... I ....................... I ....................... I ,

8 7

6

5

4

3

21

dH (PPm)

Figure 9.47 The 300-M Hz 1H NMR spectrum o f com pound G, Problem 9.26. Expansions o f the signals are shown in th e o ffse t plots.

dH (PPm)

Figure 9.48 The 300-M Hz 1H NMR spectrum o f com pound H, Problem 9.26. Expansions o f the signals are shown in th e o ffse t plots. 9.27

A ssum e that in a certain 1H N M R spectrum you find tw o peaks o f roughly equal intensity. You are not certain w hether these tw o peaks are singlets arising from uncoupled protons at different chem ical shifts or are tw o peaks o f a doublet that arises from protons coupling w ith a single adjacent proton. W hat sim ple experim ent w ould you perform to distinguish betw een these tw o possibilities?

9.28

Propose structures for com pounds O and P that are consistent w ith the follow ing inform ation. H2(2 Equiv.) C6H

8

pt

C H * CcH i2

O

13C N M R

for C om pound O

d (p p m )

DEPT

26.0 124.5

CH2 CH

446

Chapter 9

Nuclear Magnetic Resonance and Mass Spectrometry

9.29

C om pound Q has the m olecular form ula C7H8. T he broad-band proton decoupled 13C spectrum o f Q has signals at 5 50 (CH), 85 (CH 2), and 144 (CH). O n catalytic hydrogenation Q is converted to R (C7H12). Propose structures for Q and R.

9.30

E xplain in detail how you w ould distinguish betw een the follow ing sets o f com pounds using the indicated m ethod o f spectroscopy. (b) 13C and 1H N M R

(a) 1H N M R O

O O

O

(c) 13C N M R CH3 3

9.31

C om pound S (C 8H 16) reacts w ith one m ole o f brom ine to form a com pound w ith m olecular form ula C8H16Br2. T he broadband proton-decoupled 13C spectrum o f S is given in Fig. 9.49. P ropose a structure for S.

220

200

180

160

140

120

100

80

60

40

20

0

5C (ppm)

Figure 9.49 The broadband p ro to n -d e co u p le d 13C NM R spectrum o f com pound S, Problem 9.31. In form ation from th e DEPT 13C NM R spectra is given above each peak.

MASS SPECTROMETRY 9.32

A com pound w ith m olecular form ula C4H8O has a strong IR absorption at 1730 cm - 1 . Its m ass spectrum is tabu­ lated in Fig. 9.43, and includes key peaks at m /z 44 (the b ase peak) an d m /z 29. Propose a structure for the co m ­ pound and w rite fragm entation equations show ing how peaks having these m /z values arise.

9.33

In the m ass spectrum o f 2,6-dim ethyl-4-heptanol there are prom inent peaks at m /z 87, 111, and 126. Propose reasonable structures for these fragm ent ions.

9.34

In the m ass spectrum o f 4-m ethyl-2-pentanone a M cL afferty rearrangem ent and tw o other m ajor fragm entation pathw ays occur. Propose reasonable structures for these fragm ent ions and specify the m /z value for each.

44 7

Problems

9.35

9.36

W hat are the m asses and structures o f the ions produced in the follow ing cleavage pathw ays? (a )

a-cleavage o f 2-m ethyl-3-hexanone (two pathw ays)

(b )

dehydration of cyclopentanol

(c )

M cLafferty rearrangem ent of 4-m ethyl-2-octanone (two pathw ays)

P redict the m asses and relative intensities o f the peaks in the m olecular ion region for the follow ing com pound. Br

Cl 9.37

Ethyl brom ide and m ethoxybenzene (show n below ) have the sam e nom inal m olecular w eights, displaying a sig­ nificant peak at m /z 108. R egarding their m olecular ions, w hat other features w ould allow the tw o com pounds to be distinguished on the basis o f their m ass spectra? .O C H 3 .Br

9.38

The hom ologous series of prim ary am ines, CH 3(CH2)nNH2, from CH 3NH2 to C H 3(CH2)13NH2 all have their base (largest) peak at m /z 30. W hat ion does this peak represent, and how is it form ed? INTEGRATED STRUCTURE ELUCIDATION

9.39

Propose a structure that is consistent w ith each set o f (a)

(b )

(c)

(d)

(e)

(f)

C4 H1 0 O

C 3 H7Br

C 4 H8O

C 7 H8O

C 4 H9Cl

C 15H 14O

1H N M R

data.

IR

data is provided for som e com pounds.

d (p p m )

S p littin g

In t e g r a t io n

1.28

s

9H

1.35

s

1H

d (p p m )

S p littin g

In t e g r a t io n

1.71

d

6H

4.32

Septet

1H

d (p p m )

S p littin g

In t e g r a t io n

1.05

t

3H

2.13

s

3H

2.47

q

2H

d (p p m )

S p littin g

In t e g r a t io n

2.43

s

1H

4.58

s

2H

7.28

m

5H

d (p p m )

S p littin g

In te g r a tio n

1.04

d

6H

1.95

m

1H

3.35

d

2H

d (p p m )

S p littin g

In te g r a tio n

2.20

s

3H

5.08

s

1H

7.25

m

10H

IR 1720 cm 1 (strong)

IR 3 2 0 0 -3 5 5 0 c m -1 (broad)

IR 1720 cm 1 (strong)

448

Chapter 9

(g)

(h)

(i)

(j)

(k)

(l)

(m )

C4^BrO 2

C8H10

° 4 H8°3

C 3 H 7 NO 2

C 4 H-I0 O 2

C5 H1 0 O

C 8 H 9 Br

S

S

S

S

S

S

S

Nuclear Magnetic Resonance and Mass Spectrometry

S p littin g

In te g ra tio n

1.08

t

3H

2500-3500 cm

2.07

1715 cm- 1 (strong)

1

(broad)

m

2H

4.23

t

1H

10.97

s

1H

(p p m )

S p littin g

In te g ra tio n

1.25

t

3H

2 .6 8

q m

2H

7.23 (p p m )

S p littin g

In te g ra tio n

1.27

t

3H

2500-3550 cm

3.66

q

2H

1715 cm- 1 (strong)

4.13

s

2H

10.95

s

1H

(p p m )

S p littin g

In te g ra tio n

1.55

d

4.67

Septet

1H

(p p m )

S p littin g

In te g ra tio n

3.25

s

6

3.45

s

4H

(p p m )

S p littin g

In te g ra tio n

1 .1 0

d

6

2 .1 0

s

3H

2.50

Septet

1H

(p p m )

S p littin g

In te g ra tio n

d

3H

q m

1H

2 .0

5.15 7.35

5H

6

IR 1

(broad)

H

H

H

IR 1720 cm

1

(strong)

5H

Propose structures for compounds E and F. Compound E (C 8 H6) reacts with 2 molar equivalents of bromine to form F (C 8 H 6 Br4). E has the IR spectrum shown in Fig. 9.50. What are the structures of E and F?

Transmittance

(%)

9.40

IR

(p p m )

W avenum ber (cm 1)

Figure 9.50 The IR spectrum of com pound E, Problem 9.40. (Spectrum courtesy o f Sadtler Research Laboratories, Inc., Philadelphia. © BioRad Laboratories, Inc., Inform ation Division, Sadtler Softw are & Databases. A ll rights reserved. Permission fo r th e publication herein o f Sadtler Spectra has been g ra n te d by BioRad Laboratories, Inc., Inform atics Division.)

4 49

Problems

9.41

R egarding com pound J , C 2 HxCly, use the 1H N M R and IR data below to propose a stereochem ical form ula that is consistent w ith the data. Splitting

1H N M R

6.3 IR

9.42

3125 1625 1280 820 695

Integration

s

cm" cm" cm" cm" cm"

W hen dissolved in CDCI 3 , a com pound (K ) w ith the m olecular form ula C 4 H 8O 2 gives a 1H N M R spectrum that consists o f a doublet at d 1.35, a singlet at d 2.15, a broad singlet at d 3.75 (1H), and a quartet at d 4.25 (1H). W hen dissolved in D 2O, the com pound gives a sim ilar 1H N M R spectrum , w ith the exception that the signal at d 3.75 has disappeared. The IR spectrum of the com pound shows a strong absorption p eak near 1720 c m - 1 . (a ) Propose a structure for com pound K. (b )

E xplain w hy the N M R signal at d 3.75 disappears w hen D 2O is used as the solvent.

9.43

Com pound T (CsHgO) has a strong IR absorption band at 1745 c m - 1 . The broad-band proton decoupled 13C spec­ trum o f T shows three signals: at d 220 (C), 23 (CH 2), and 38 (CH 2). Propose a structure for T.

9.44

D educe the structure o f the com pound that gives the follow ing 1H, 13C, and IR spectra (Figs. 9.5 1 -9 .5 3 ). A ssign all aspects of the 1H, and 13C spectra to the structure you propose. U se letters to correlate protons w ith signals in the 1H N M R spectrum , and num bers to correlate carbons w ith signals in the 13C spectrum . The m ass spectrum of this com pound shows the m olecular ion at m/z 96.

6

5

4

3 dH (PPm)

Figure 9.51

The 1H NMR spectrum (simulated) fo r Problem 9.44.

2

1

0

450

Chapter 9

Nuclear Magnetic Resonance and Mass Spectrometry

. . i ............. i ...............i ...............i ...............i ...............i ...............i ...............i ...............i ...............i ...............i ............... 220

200

180

160

140

120

100

80

60

40

dC (PPm)

Figure 9.52 A sim ulated broadband p ro to n -de co u p le d 13C NM R spectrum fo r Problem 9.44. In form ation fro m th e DEPT 13C spectra is given above each peak.

Transmittance (%)

100

Wavenumber (cm 1) Figure 9.53 The IR spectrum fo r Problem 9.44. Spectra a d apted fro m Sigm a-Aldrich Co. © Sigm a-Aldrich Co.

20

0

451

Problems 9.45

D educe the structure o f the com pound that gives the follow ing 1H, 13C, and IR spectra (Figs. 9.5 4 -9 .5 6 ). A ssign all aspects o f the 1H and 13C spectra to the structure you propose. U se letters to correlate protons w ith the signals in the 1H N M R spectrum , and num bers to correlate carbons w ith the signals in the 13C spectrum . The m ass spec­ trum o f this com pound shows the m olecular ion at m/z 148.

6H

1H

2H

2H 1H

, , , , 1 .................................... 1 .........................................1 .........................................1 .........................................1 .........................................

10

8

4

6

2

0

dH (PPm)

Figure 9.54 The 300-M Hz 1H NM R spectrum (simulated) fo r Problem 9.45.

CH

CH

C

CH

CHo

CH

C

, , , , .................................... ................................... ...................................................................... ...................................................................... ................................... ................................... ................................... ................................

200

180

160

140

120

100

80

60

40

dc (PPm)

Figure 9.55 A sim ulated broadband p ro ton-decoupled 13C NMR spectrum fo r Problem 9.45. Inform ation from th e DEPT 13C spectra is given above each peak.

20

0

452

Chapter 9

Nuclear Magnetic Resonance and Mass Spectrometry

100

0 4000

3000

2000

1500

1000

500

W avenum ber (cm-1)

Figure 9.56 The IR spectrum for Problem 9.45. SDBSWeb : http://riodb01.ibase.aist.go.jp/sdbs/ (National Institute of Advanced Industrial Science and Technology, Septem ber 24, 2009).

9.46

D educe the structure o f the com pound that gives the follow ing 1H, 13C, and IR spectra (Figs. 9.5 7 -9 .5 9 ). A ssign all aspects o f the 1H and 13C spectra to the structure you propose. U se letters to correlate protons w ith signals in the 1H N M R spectrum , and num bers to correlate carbons w ith signals in the 13C spectrum . The m ass spectrum of this com pound shows the m olecular ion at m/z 204.

dH (PPm)

Figure 9.57

The 300-MHz 1H NMR spectrum (simulated) for Problem 9.46.

453

Problems

200

180

160

140

120

100

80

60

40

20

0

dc (PPm)

Figure 9.58

A sim ulated broadband p ro to n -de co u p le d 13C NM R spectrum fo r Problem 9.46.

100 90 80

Transmittance

(%)

70 60 50 40 30 20 10H 0 4000

3000

2000

1500 W avenum ber (cm 1)

Figure 9.59 The IR spectrum fo r Problem 9.46. SDBSWeb: http://riodb01.ibase.aist.go.jp/sdbs/ (N ational In stitu te o f Advanced Industrial Science and Technology, S eptem ber 24, 2009).

1000

500

45 4 9.47

Chapter 9

Nuclear Magnetic Resonance and Mass Spectrometry

D e d u c e the structure o f the c om po un d ( C 5 H- 1 0 O 3 ) that gives the fo llo w in g

1

H,

13

C , and I R spectra (F ig s. 9 .6 0 - 9 .6 2 ),

A s s ig n a ll aspects o f the 1H and 13C spectra to the structure y o u propose. U s e letters to correlate protons w ith sig­ nals in the 1H N M R spectrum , and n um bers to c o rrelate carbons w ith signals in the 13C spectrum .

3H

3H

2H

2H

1

2H

1

1

7

1 6

1

1

1

1

5

1

4

1

1

3

1

1

2

1 1

1 0

dH (PPm)

Figure 9.60 The 300-M Hz 1H NM R spectrum (simulated) fo r Problem 9.47.

180

160

140

120

100

80

60

40

dc (PPm)

Figure 9.61

A sim ulated broadband p ro to n -de co u p le d 13C NMR spectrum fo r Problem 9.47.

20

0

Challenge Problems

455

Figure 9.62 The IR spectrum for Problem 9.47. SDBSWeb: http://riodb01.ibase.aist.go.jp/sdbs/ (National Institute of Advanced Industrial Science and Technology, Septem ber 24, 2009).

Challenge Problems 9.48

T he 1H N M R exam ination o f a solution o f 1,3-dim ethylcyclopentadiene in concentrated sulfuric acid shows three peaks w ith relative areas o f 6:4:1. W hat is the explanation for the appearance o f the spectrum ?

9.49

A cetic acid has a m ass spectrum show ing a m olecular ion peak at m /z 60. O ther unbranched m onocarboxylic acids w ith four or m ore carbon atom s also have a peak, frequently prom inent, at m /z 60. Show how this can occur.

9.50

T he 1H N M R peak for the hydroxyl proton o f alcohols can be found anyw here from d 0.5 to d 5.4. E xplain this variability.

9.51

The 1H N M R study of D M F (^N -dim ethylform am ide) results in different spectra according to the tem perature of the sample. A t room temperature, two signals are observed for the protons of the two m ethyl groups. On the other hand, at elevated temperatures (> 130°C ) a singlet is observed that integrates for six hydrogens. Explain these differences.

9.52

T he m ass spectra o f m any benzene derivatives show a peak at m /z 51. W hat could account for this fragm ent?

9.53

C onsider the follow ing inform ation.

Jab = 5.3 Hz Jac = 8.2 Hz Jbc = 10.7 Hz (a) How m any total 1H N M R signals w ould you expect for the above m olecule? (b) Ha appears as a doublet o f doublets (dd) at 1.32 ppm in the 1H N M R spectrum . D raw a labeled splitting tree diagram for Ha using the coupling constant values given above.

456

Chapter 9

Nuclear Magnetic Resonance and Mass Spectrometry

Learning Group Problems 1.

G iven the follow ing inform ation, elucidate the structures o f com pounds A an d B. Both com pounds are soluble in dilute aqueous HCl, and both have the sam e m olecular form ula. T he m ass spectrum o f A has M+ 149 (intensity 37.1% o f base peak) and M+ + 1 150 (intensity 4.2% o f b ase peak). O ther spectroscopic data for A and B are given below. Justify the structures you propose by assigning specific aspects o f the data to the structures. M ake sketches o f the N M R spectra. (a) T he IR spectrum for com pound A show s tw o bands in the 3 3 0 0 -3 5 0 0 -c m ^ 1 region. T he broadband protondecoupled 13C N M R spectrum displayed the follow ing signals (inform ation from the D E PT 13C spectra is given in parentheses w ith the 13C chem ical shifts): 13C NMR: 5 140 (C), 127 (C), 125 (CH), 118 (CH), 24 (CH2), 13 (CH3) (b) T he IR spectrum for com pound B show s no b an d s in the 3 3 0 0 -3 5 0 0 -c m -1 region. T h e b ro ad b an d protond ecoup led 13C N M R spectrum display ed th e follow ing signals (inform ation from the D E P T 13C sp ectra is given in p arentheses w ith the 13C chem ical shifts): 13C NMR: 5 147 (C), 129 (CH), 115 (CH), 111 (CH), 44 (CH2), 13 (CH3)

2.

Two com pounds w ith the m olecular form ula C 5H 10O have the follow ing 1H and 13C N M R data. Both com pounds have a strong IR absorption band in the 17 1 0 -1 7 4 0 -cm -1 region. E lucidate the structure o f these tw o com pounds and interpret the spectra. M ake a sketch o f each N M R spectrum . (a) 1H N M R: 5 2.55 (septet, 1H), 2.10 (singlet, 3H), 1.05 (doublet, 6H) 13C N M R: 5 212.6, 41.5, 27.2, 17.8 (b) 1H N M R: 5 2.38 (triplet, 2H), 2.10 (singlet, 3H), 1.57 (sextet, 2H), 0.88 (triplet, 3H) 13C N M R: 5 209.0, 45.5, 29.5, 17.0, 13.2

CONCEPT MAP

O o 3 n

CD

"O Q) "D

4^ cn si

458

CONCEPT MAP

Absence of signal splitting and peak integration information in routine 13C NMR spectra

Chapt er 9 Nuclear M agnetic

200

180

160

140

120

100

80

60

40

20

0

Sc (PPm)

Resonance and Mass S p e c tr o m e tr y

1H NMR Approximate Chemical Shift Ranges

i 0 =0 1

-C 5C -H

a™

30,2°,1°

0 II

11

10

9

8

7

/

X,0,N -CH \

a H

-C-H

12

-s -

220

6

5

dH(ppm)

4

3

2

1

0

Radical Reactions

Unpaired electrons lead to many burning questions about radical types of reactivity. In fact, species with unpaired electrons are called radicals, and they are involved in the chemistry o f burning, aging, disease, as well as in reactions related to destruction o f the ozone layer and the synthesis o f products that enhance our everyday lives. For example, polyethylene, which can have a molecular weight from the thousands to the millions, and practical uses ranging from plastic films and wraps to water bottles, bulletproof vests, and hip and knee replace­ ments, is made by a reaction involving radicals. Oxygen that we breathe and nitric oxide that serves as a chem­ ical signaling agent for some fundamental biological processes are both molecules with unpaired electrons. Highly colored natural compounds like those found in blueberries and carrots react with radicals and may pro­ tect us from undesirable biological radical reactions. Large portions of the economy hinge on radicals, as well, from reactions used to make polymers like polyethylene, to the target action of pharmaceuticals like Cialis, Levitra, and Viagra, which act on a nitric oxide biological signaling pathway. Reactions with radicals also play a role in organic synthesis. In this chapter we study the properties and reactivity of species with unpaired electrons, and we shall find that they are radically important to chemistry and life.

459

460

Chapter 10

Radical Reactions

10.1 Introduction: H o w Radicals Form and H o w They React So far alm ost all o f the reactions w hose m echanism s w e have studied have been i o n i c r e a c ­ Ionic reactions are those in w hich covalent bonds b reak h e t e r o l y t i c a l l y and in w hich ions are involved as reactants, interm ediates, or products. A nother broad category o f reactions has m echanism s that involve h o m o ly sis o f cova­ lent bonds w ith the production o f interm ediates possessing unpaired electrons called r a d ­ icals (or fre e rad ic als): t io n s .

Each atomtakes one electron from the covalent bond that joined them. r*

A :B

homolysis

K>

H e lp f u l H i n t A single-barbed curved arrow shows movement of one electron.

A - + -B R adicals

T his sim ple exam ple illustrates the w ay w e use s i n g l e - b a r b e d curved arrow s to show the m ovem ent o f a s i n g l e e l e c t r o n (not o f an electron p air as w e have done earlier). In this instance, each group, A and B, com es aw ay w ith one o f the electrons o f the covalent bond that jo in e d them.

10.1A Production of Radicals •

E nergy in the form o f h eat or light m ust b e supplied to cause hom olysis o f cova­ lent bonds (Section 10.2).

For exam ple, com pounds w ith an o x y g en -o x y g en single bond, called p e ro x id e s, undergo hom olysis readily w hen heated, because the o x y g en -o x y g en b o n d is w eak. T he products are tw o radicals, called alkoxyl radicals:

Homolysis of a dialkyl peroxide. Dialkyl peroxide

Alkoxyl radicals

H alogen m olecules (X2) also contain a relatively w eak bond. A s w e shall soon see, halo ­ gens undergo hom olysis readily w hen h eated or w hen irradiated w ith light o f a w avelength that can be absorbed by the halogen molecule:

. X fX . ' „

homolysis heat or light (hn)

2 •X . X

Homolysis of a halogen molecule.

T he products o f this hom olysis are halogen atom s, an d because halogen atom s contain an unpaired electron, they are radicals.

10.1B Reactions of Radicals •

A lm ost all sm all radicals are short-lived, highly reactive species.

W hen radicals collide w ith other m olecules, they tend to react in a w ay that leads to p air­ ing o f their unpaired electron. O ne w ay they can do this is by abstracting an atom from another m olecule. F or exam ple, a halogen atom m ay abstract a hydrogen atom from an alkane. This h y d r o g e n a b s t r a c t i o n gives the halogen atom an electron (from the hy d ro ­ gen atom ) to pair w ith its unpaired electron. N otice, however, that the other pro d u ct o f this abstraction is another radical interm ediate, in this case, an alkyl radical, R-, w hich goes on to react further, as w e shall see in this chapter.

10.2 Homolytic Bond Dissociation Energies (DH°)

461

A MECHANISM FOR THE REACTION H y d ro g e n A to m A b s tr a c tio n G ene ra l R e a ctio n

:X •

V=R

:X : H

Ci

Alkane Reactive radical interm ediate

+

R-

Alkyl radical interm ediate (reacts further)

S p e c ific E xa m p le

:C !'

:C h H +

h ch3 l* 3

C hlorine M ethane atom (a radical)

CH3-

Methyl radical interm ediate (reacts further)

T his behavior is characteristic o f r a d ic a l re a c tio n s . C onsider another exam ple, one that show s another w ay in w hich radicals can react: They can com bine w ith a com pound con­ taining a m ultiple bond to produce a new radical, w hich goes on to react further. (W e shall study reactions o f this type in Section 10.10.)

A MECHANISM FOR THE REACTION R a d ic a l A d d i t i o n t o a p B o n d

R R

I

■/

|

\

— C— C /

Reactive alkyl radical interm ediate

\ Alkene

Further reaction (Section 10.10)

New radical interm ediate

10.2 H om olytic Bond Dissociation Energies (DH°) W hen atom s com bine to form molecules, energy is released as covalent bonds form. The m ol­ ecules o f the products have low er enthalpy than the separate atoms. W hen hydrogen atoms com bine to form hydrogen molecules, for example, the reaction is exothermic; it evolves 436 kJ o f heat for every m ole o f hydrogen that is produced. Similarly, when chlorine atom s com ­ bine to form chlorine molecules, the reaction evolves 243 kJ m o l-1 o f chlorine produced:

H- + H-

----- >

H— H

A H ° = - 436 kJ moM

Cl- + Cl-

----- »

Cl— Cl

A H ° = - 243 kJ moM

}

Bond form ation is an exotherm ic p ro cess.

R eactions in w hich only bond breaking occurs are alw ays endotherm ic. T he energy required to break the covalent bonds o f hydrogen or chlorine hom olytically is exactly equal

462

Chapter 10

Radical Reactions

to that evolved w hen the separate atom s com bine to form m olecules. In the b o n d cleavage reaction, however, AH° is positive:

H— H C l— Cl

----- >

H- + H-

AH ° = + 436 kJ mol-

Cl-

AH °

Ch

Bond breaking is an endotherm ie p ro cess.

243 kJ mol

•

E nergy m ust be supplied to break covalent bonds.

•

T he energies required to b reak covalent bonds hom olytically are called h o m o ly tic b o n d d isso cia tio n e n e rg ie s, and they are usually abbreviated by the sym bol D H ° .

The hom olytic bond dissociation energies of hydrogen and chlorine, for exam ple, can be w ritten in the follow ing w ay:

H— H

Cl— Cl

(DH° = 436 kJ m ol“ 1)

(DH° = 243 kJ m ol“ 1)

T he hom olytic bond dissociation energies of a variety of covalent bonds have been deter­ m ined experim entally or calculated from related data. Som e o f these D H ° values are listed in Table 10.1. ^

Single-Bond H om olytic Dissociation Energies (DH°) at 2 5 °C a A=B

Bond

■A- + B-

Bond kJ

B ro k e n

( s h o w n in r e d )

m o l“ 1

( s h o w n in r e d )

H— H D— D F— F Cl — Cl Br — Br I— I H— F H — Cl H — Br H— I CH 3 — H CH 3 — F CH 3 — Cl CH 3 — Br CH 3 — I CH 3 — OH CH 3 — o c h 3 CH 3CH2— H CH 3CH2— F CH 3CH 2 — Cl CH 3CH 2 — Br CH 3CH 2 — I CH 3CH 2 — OH

436 443 159 243 193 151 570 432 366 298 440 461 352 293 240 387 348 421 444 353 295 233 393

CH 3CH 2 — OCH 3 CH 3CH 2CH 2 — H CH 3CH 2CH 2 — F CH 3CH 2CH2— Cl CH 3CH 2CH 2 — Br CH 3CH 2CH 2 — I CH 3CH 2CH 2 — OH CH 3CH 2CH 2 — OCH 3 (CH3)2CH — H (CH3)2CH — F (CH3)2CH — Cl (CH3)2CH — Br (CH 3)2CH — I (CH3)2CH — OH (CH 3)2CH — o c h 3 (CH 3)2CHCH 2 — H (CH3)3C— H (CH3)3C — Cl (CH 3)3C — Br (CH 3)3C — I (CH 3)3C — OH (CH3)3C— OCH3 C 6H5CH 2 — H

Bond kJ m o l“ 1

352 423 444 354 294 239 395 355 413 439 355 298 222

402 359 422 400 349 292 227 400 348 375

B ro k e n ( s h o w n in r e d )

CH2= CHCH 2 — H CH2= CH — H C 6H5 — H H C # C— H H C — H C

B ro k e n

CH 3CH 2 — CH 3 CH 3CH 2CH 2 — CH 3 CH 3CH 2 — CH 2CH 3 (CH3)2CH — CH 3 (CH3)3C — CH 3 HO — H HOO — H HO — OH (CH3)3CO — OC(CH3)3 O O Il II Q H 5C O - O C C 6H 5 CH 3CH 2O — OCH 3 CH 3CH2O — H

kJ“ m o l“

1

369 465 474 547 378 371 374 343 371 363 499 356 214 157

139 184 431

O CH3C — H

364

aData compiled from the National Institute of Standards (NIST) Standard Reference Database Number 69, July 2001 Release, accessed via NIST Chemistry WebBook (http://webbook.nist.gov/chemistry/). Copyright 2000. From CRC Handbook of Chemistry and Physics, Updated 3rd Electronic Edition; Lide, David R., ed. Reproduced by permission of Routledge/Taylor & Francis Group, LLC. DH° values were obtained directly or calculated from heat of formation (Hf) data using the equation DH°[A— B] = Hf [A-] + Hf [B-] —Hf [A—B].

10.2A How to Use Homolytic Bond Dissociation Energies to Calculate Heats of Reaction B ond dissociation energies have, as w e shall see, a variety o f uses. T hey can b e used, for exam ple, to calculate the enthalpy change (AH°) for a reaction. To m ake such a calcula-

463

10.2 Homolytic Bond Dissociation Energies (DH°)

tion (see follow ing reaction), w e must remember that for bond breaking AH ° is positive and for bond form ation AH° is negative. AH ° = -(n e t DHproducts) + (net DH°eactants) Negative sig n b e c a u s e energy is released in bond form ation

or

AH°

- 2 DH°nproducts

E DH°reactants

is the mathematical \ \sym bol fo r summation/

L et us consider, for exam ple, the reaction o f hydrogen and chlorine to produce 2 m ol of hydrogen chloride. F rom Table 10.1 w e get the follow ing values o f DH°:

H— H

Cl — Cl

:

(DH° = 436 kJ m ol“ 1) (DH° = 243 kJ m ol“ 1) + 679 kJ is required to cleave 1 mol of H2 bonds and 1 mol of Cl2 bonds.

2 H — Cl (DH° = 432 kJ m ol“ 1) x 2 - 8 6 4 kJ is evolved in form ation of the bonds in 2 mol of HCl.

O verall, the reaction o f 1 m ol o f H2 and 1 m ol o f Cl2 to form 2 m ol o f HCl is exotherm ic: Two m oles of product form ed

AH° = - 2 (432 kJ mol-1 )

+ (436 kJ mol-1 + 243 kJ mol-1 )

Bond forming (exotherm ic; negative sign)

Bond breaking (endotherm ie; positive sign)

864 kJ mol 1 h 679 kJ mol-1 185 kJ m oP1 Overall AH° for 2 mol HCl p ro d u ced from H2 + Cl2 For the purpose o f our calculation, w e have assumed a particular pathway, which amounts to

-2 H

H— H and

Cl— Cl

then

■2 Cl•2 H— Cl

2 H- + 2 Cl--

This is n o t the w ay the reaction actually occurs. N evertheless, the h eat o f reaction, A H °, is a therm odynam ic quantity that is dependent only on the initial and final states o f the react­ ing m olecules. H ere, AH° is independent o f the path follow ed (H ess’s law), and, for this reason, our calculation is valid.

R e v ie w P ro b le m 10.1

C alculate the heat o f reaction, AH°, for the follow ing reactions:

(a) H2 + F2 ----- : 2 HF

(f) CH3CH3

Cl2

(b) CH4

F2 — >CH3F + HF

(g)

Cl2

(c) CH4

Cl2

: CH3Cl + HCl

(d) CH4

Br2

: CH3Br + HBr

(e) CH4

I2 —

CH3I + HI

■CH3CH2Cl

HCl HCl

Cl (h)

Cl2

Cl

HCl

10.2B How to Use Homolytic Bond Dissociation Energies to Determine the Relative Stabilities of Radicals H om olytic bond dissociation energies also provide us w ith a conve­ n ient w ay to estim ate the relative stabilities o f radicals. If w e exam ine the data given in Table 10.1, w e find the follow ing values o f D H ° for the prim ary and secondary C — H bonds o f propane:

H H (DH ° = 423 kJ mol-1)

(DH ° = 413 kJ mol-1)

464

Chapter 10

Radical Reactions

This m eans that for the reaction in w hich the designated C — H bonds are b roken hom olytically, the values o f AH° are those given here.

H

H-

AH°

+423 kJ mol-

H-

AH °

+413 kJ mol-

Propyl radical (a 1 radical)

Isopropyl radical (a 2 radical)

H

T hese reactions resem ble each other in tw o respects: They both begin w ith the sam e alkane (propane), and they b oth produce an alkyl radical and a hydrogen atom. They differ, ho w ­ ever, in the am ount o f energy required and in the type o f carbon radical produced. These tw o differences are related to each other. •

A lkyl radicals are classified as being 1°, 2°, or 3° b ased on the carbon atom that has the unpaired electron, the sam e w ay that w e classify carbocations based on the carbon atom w ith the positive charge.

M ore energy m ust be supplied to produce a prim ary alkyl radical (the propyl radical) from propane than is required to produce a secondary carbon radical (the isopropyl rad i­ cal) from the sam e com pound. This m ust m ean that the prim ary radical has absorbed m ore energy and thus has greater potential energ y. B ecause the relative stability o f a chem ical species is inversely related to its potential energy, the secondary radical m ust be the m ore stable radical (Fig. 10.1a). In fact, the secondary isopropyl radical is m ore stable than the prim ary propyl radical by 10 kJ m ol~ *. We can use the data in Table 10.1 to m ake a sim ilar com parison o f the tert-butyl rad i­ cal (a 3° radical) and the isobutyl radical (a 1° radical) relative to isobutane: H-

AH ° = +400 kJ mol-

H-

AH ° = +422 kJ mol-

ferf-Butyl radical (a 3 radical) H e lp f u l H i n t

H

Knowing the relative stability o f radicals is im portant for predicting reactions.

H

Isobutyl radical (a 1 radical) H ere w e find (Fig. 10.1b) that the difference in stability o f the tw o radicals is even larger. T he tertiary radical is m ore stable than the prim ary radical by 22 kJ m o l_1. The kind o f pattern that w e find in these examples is found w ith alkyl radicals generally. •

O verall, the relative stabilities o f radicals are 3° > 2° > 1° > m ethyl. T ertiary (3°)

>

S eco n d ary (2°)

C C— C Xc •

>

Prim ary (1°)

C >

c— c

-

>

Methyl

H >

c— c

-

H >

H— C -

-H

\

T he order o f stability o f alkyl radicals is the sam e as for carbocations (Section 6.11B).

A lthough alkyl radicals are uncharged, the carbon that bears the odd electron is electron deficient. Therefore, alkyl groups attached to this carbon provide a stabilizing effect through hyperconjugation, and the m ore alkyl groups b onded to it, the m ore stable the radical is. Thus, the reasons for the relative stabilities o f radicals an d carbocations are similar.

465

10.3 Reactions o f Alkanes with Halogens

Figure 10.1 (a) Comparison of the potential energies of the propyl radical (+H-) and the isopropyl radical ( + H-) relative to propane. The isopropyl radical (a 2° radical) is more stable than the 1° radical by 10 kJ m ol—1. (b) Comparison of the potential energies of the tert-butyl radical ( + H-) and the isobutyl radical (+H-) relative to isobutane. The 3° radical is more stable than the 1° radical by 22 kJ m ol—1.

S o lv e d P r o b le m 1 0 .1

Classify each of the following radicals as being 1°, 2°, or 3°, and rank them in order o f decreasing stability.

A

B

C

STRATEGY AND ANSWER We examine the carbon bearing the unpaired electron in each radical to classify the radical as to its type. B is a tertiary radical (the carbon bearing the unpaired electron is tertiary) and is, therefore, most stable. C is a primary radical and is least stable. A, being a secondary radical, falls in between. The order of stability is B > A > C.

L i s t t h e f o l l o w i n g r a d ic a ls i n o r d e r o f d e c r e a s in g s t a b i l i t y :

R e v ie w P ro b le m 1 0 .2

CH,

10.3 Reactions o f Alkanes w ith Halogens • Alkanes react with molecular halogens to produce alkyl halides by a substitution reaction called radical halogenation. A g e n e r a l r e a c t io n s h o w in g f o r m a t i o n o f a m o n o h a lo a lk a n e b y r a d ic a l h a lo g e n a t io n is s h o w n b e lo w . I t is c a l le d r a d ic a l h a lo g e n a t io n b e c a u s e , a s w e s h a ll s e e , th e m e c h a n is m in v o lv e s s p e c ie s w i t h u n p a ir e d e le c tr o n s c a l le d r a d ic a ls . T h i s r e a c t io n is n o t a n u c l e o p h i lic s u b s titu tio n r e a c t io n .

R— H + X2 ----- > R— X + HX

466

Chapter 10

Radical Reactions

I n th e s e r e a c t io n s a h a lo g e n a t o m r e p la c e s o n e o r m o r e o f t h e h y d r o g e n a t o m s o f t h e a lk a n e a n d t h e c o r r e s p o n d in g h y d r o g e n h a l i d e is f o r m e d a s a b y - p r o d u c t . O n l y f l u o r i n e , c h lo r i n e , a n d b r o m i n e r e a c t t h is w a y w i t h a lk a n e s . I o d i n e is e s s e n t i a l ly u n r e a c t i v e , f o r a r e a s o n t h a t w e s h a ll e x p l a i n la t e r .

10.3A Multiple Halogen Substitution O n e c o m p l i c a t i n g f a c t o r o f a lk a n e h a lo g e n a t io n s is t h a t m u l t i p l e s u b s t it u t io n s a lm o s t a lw a y s o c c u r . T h e f o l l o w i n g e x a m p l e il lu s t r a t e s t h is p h e n o m e n o n . I f w e m i x a n e q u i m o l a r r a t i o o f m e t h a n e a n d c h l o r i n e ( b o t h s u b s ta n c e s a r e g a s e s a t r o o m t e m p e r a t u r e ) a n d t h e n e i t h e r h e a t t h e m i x t u r e o r ir r a d i a t e i t w i t h l i g h t o f t h e a p p r o p r i a t e w a v e l e n g t h , a r e a c t io n b e g in s to o c c u r v ig o r o u s ly a n d u lt im a t e ly p ro d u c e s th e f o llo w in g m ix t u r e o f p ro d u c ts .

H H— C— H

Cl,

heat

H

Cl

Cl

Cl

H — C — Cl

H — C — Cl

H — C — Cl

Cl— C — Cl

light

H Methane

Chlorine

H

H

Chloromethane

Dichloromethane

Cl

H

Cl

Cl

Trichloromethane

Tetrachloromethane

Hydrogen chloride

(T h e s u m o f the n u m b e r o f m o le s o f e a ch c h lo rin a te d m e th a n e p ro d u c e d eq u als the n u m b e r o f m o le s o f m e th a n e th a t re a cte d .)

T o u n d e r s t a n d t h e f o r m a t i o n o f t h is m i x t u r e , w e n e e d t o c o n s id e r h o w t h e c o n c e n t r a t i o n o f r e a c ta n ts a n d p ro d u c ts c h a n g e s as th e r e a c t io n p r o c e e d s . A t th e o u ts e t, th e o n ly c o m ­ p o u n d s th a t a re p re s e n t in th e m ix t u r e a re c h lo r in e a n d m e th a n e , a n d th e o n ly r e a c tio n th a t c a n t a k e p l a c e is o n e t h a t p r o d u c e s c h lo r o m e t h a n e a n d h y d r o g e n c h l o r i d e :

H

H

H— C— H

+

C l2

H

----- »

H— C— C

+

l

H— C l

H

A s th e r e a c tio n p ro g re s s e s , h o w e v e r , th e c o n c e n tr a tio n o f c h lo r o m e th a n e in th e m ix t u r e in c r e a s e s , a n d a s e c o n d s u b s t it u t io n r e a c t i o n b e g i n s t o o c c u r . C h l o r o m e t h a n e r e a c t s w i t h c h lo r in e to p r o d u c e d ic h lo r o m e th a n e :

H

Cl

H— C—

+

C l

C l2

H The

----- »

H— C—

+

C l

H—

Cl

H

d ic h lo r o m e th a n e

p ro d u c e d

can

th e n

re a c t

to

fo rm

tr ic h lo r o m e th a n e ,

and

t r i c h l o r o m e t h a n e , a s i t a c c u m u la t e s i n t h e m i x t u r e , c a n r e a c t w i t h c h l o r i n e t o p r o d u c e t e t r a c h lo ro m e th a n e . E a c h

H — Cl

r

S o lv e d P ro b le m 1 0 .2

t i m e a s u b s t it u t io n o f

2

fo r

—H

ta k e s p la c e , a m o le c u le o f

1

I f t h e g o a l o f a s y n t h e s is is t o p r e p a r e c h l o r o m e t h a n e ( C H of CH

— Cl

is p r o d u c e d .

3

C l ) , it s f o r m a t i o n c a n b e m a x i m i z e d a n d t h e f o r m a t i o n

C l 2 , C H C l 3 , a n d C; C l 4 m i n i m i z e d b y u s in g a la r g e e x c e s s o f m e t h a n e i n t h e r e a c t io n m i x t u r e . E x p l a i n w h y

th is is p o s s ib le .

ANSWER

T h e u s e o f a l a rg e e x c e s s o f m e t h a n e m a x i m i z e s t h e p r o b a b i l i t y t h a t c h l o r i n e w i l l a t t a c k m e t h a n e m o l -

e c u le s b e c a u s e t h e c o n c e n t r a t i o n o f m e t h a n e i n t h e m i x t u r e w i l l a l w a y s b e r e l a t i v e l y la r g e . I t a ls o m i n i m i z e s t h e p r o b a b i l i t y t h a t c h lo r i n e w i l l a t t a c k m o le c u le s o f C H

3

C l, C H

2

C l2 , a n d C H C l 3 , b e c a u s e t h e ir c o n c e n tr a tio n s w i l l

a l w a y s b e r e l a t i v e l y s m a l l . A f t e r t h e r e a c t io n is o v e r , t h e u n r e a c t e d e x c e s s m e t h a n e c a n b e r e c o v e r e d a n d r e c y c le d .

467

10.4 Chlorination of Methane: Mechanism of Reaction

10.3B Lack of Chlorine Selectivity C h lo rin a tio n o f m o st higher alkanes gives a m ixture o f isom eric m onochloro products as w ell as m ore highly halogenated com pounds. •

C hlorine is relatively unselective; it does n o t discrim inate greatly am ong the different types o f hydrogen atom s (prim ary, secondary, and tertiary) in an alkane.

A n exam ple is the light-prom oted chlorination o f isobutane: Cl,

Polychlorinated products

Cl

light

Cl Isobutane

•

•

Isobutyl chloride (48%)

HCl

(23%)

fert-Butyl chloride (29%)

B ecause alkane chlorinations usually yield a com plex m ixture o f products, they are n o t useful as synthetic m ethods w hen the goal is preparation o f a specific alkyl chloride.

H e lp f u l H i n t

Q i b m ^ n is Lin^bc^re.

A n exception is the halogenation o f an alkane (or cycloalkane) w hose hydrogen atom s are all equivalent. [Equivalent hydrogen atom s are defined as those w hich on replacem ent by som e other group (e.g., chlorine) yield the sam e com pound.]

N eopentane, for exam ple, can form only one m onohalogenation product, and the use o f a large excess o f neopentane m inim izes polychlorination:

Cl

heat

N eopentane (excess) •

Cl

or light

HCl

N eopentyl chloride

B rom ine is generally less reactive tow ard alkanes than chlorine, an d brom ine is m ore selective in the site o f attack w hen it does react.

We shall exam ine the selectivity o f brom ination further in Section 10.6A.

10.4 Chlorination o f M eth an e: Mechanism o f Reaction T he reaction o f m ethane w ith chlorine (in the gas phase) provides a good exam ple for study­ ing the m echanism o f radical halogenation.

CH4 + Cl2 ----- : CH3Cl + HCl

(+ CH2Cl2, CHCl3, and CCl4)

Several experim ental observations help in understanding the m echanism o f this reaction: 1. T h e re a c tio n is p ro m o te d b y h e a t o r lig h t. A t room tem perature m ethane and ch lo ­ rine do n o t react at a perceptible rate as long as the m ixture is k ep t aw ay from light. M ethane and chlorine do react, however, at room tem perature if the gaseous reaction m ixture is irradiated w ith U V light at a w avelength absorbed by Cl2, an d they react in the dark if the gaseous m ixture is heated to tem peratures greater than 100°C. 2. T h e lig h t-p ro m o te d re a c tio n is h ig h ly efficient. A relatively sm all num ber o f light photons perm its the form ation o f relatively large am ounts o f chlorinated product. A m echanism that is consistent w ith these observations has several steps, show n below. The first step involves the dissociation o f a chlorine m olecule, by h eat or light, into two chlorine atom s. T he second step involves hydrogen abstraction by a chlorine atom.

468

Chapter 10

Vf!,

Radical Reactions

A MECHANISM FOR THE REACTION R a d ic a l C h l o r i n a t i o n o f M e t h a n e

R E A C T IO N

CH4 + CL

heat or light

CH3Cl + HCl

M E C H A N IS M C hain In itia tio n

Step 1: Halogen dissociation

heat ..K^ ■■ or light Under the influence of h eat or light a m olecule of chlorine d isso cia tes; each atom tak es o n e of th e bonding electrons.

: Cl

•C h

This s te p p ro d u ces two highly reactive chlorine atom s.

H C hain P ro p a g a tio n

Step 2: H ydrogen abstraction

=Cl

'o f C — H 'I H A chlorine atom a b s tra c ts a hydrogen atom from a m ethane m olecule.

\

:C l- H

•C— H

/

H This ste p p ro d u ces a m olecule of hydrogen chloride and a methyl radical.

H Step 3: Halogen abstraction

H— C

\

H

H — C Cl

Cl • Cl = V7

I

H

A methyl radical a b s tra c ts a chlorine atom from a chlorine m olecule.

+

:Cl ■

"

H e lp f u l H i n t

Remember. These conventions are

This ste p p ro d u ce s a m olecule of methyl chloride and a chlorine atom . The chlorine atom can now c a u s e a repetition of ste p 2.

used in illustrating reaction mecha­ nisms in this text. 1. Curved arrows or always show the direction of movement of electrons. 2. Single-barbed arrows show the attack (or movement) of an unpaired electron. 3. Double-barbed arrows show the attack (or movement) of an electron pair.

H C hain T erm ination

H— C •

\

H

Ch

I

..

I

■■

H — C -C l: H

Coupling of any two radicals d ep lete s th e su p p ly of reactive interm ediates and term in ates th e chain. Several pairings are p o ssib le for radical coupling term ination s te p s (see text).

In step 3 the highly reactive methyl radical reacts with a chlorine molecule by abstracting a chlorine atom. This results in the formation of a molecule of chloromethane (one of the ultimate products of the reaction) and a chlorine atom. The latter product is particularly sig­ nificant, for the chlorine atom formed in step 3 can attack another methane molecule and cause

10.4 Chlorination of Methane: Mechanism of Reaction a repetition of step 2. Then, step 3 is repeated, and so forth, for hundreds or thousands of times. (With each repetition of step 3 a molecule of chloromethane is produced.) This type of sequential, stepwise mechanism, in which each step generates the reactive intermediate that causes the next cycle of the reaction to occur, is called a chain reaction. Step 1 is called the chain-initiating step. In the chain-initiating step radicals are cre­ ated. Steps 2 and 3 are called chain-propagating steps. In chain-propagating steps one radical generates another. C hain In itia tio n : c re a tio n o f ra d ic a ls

heat

Step 1 Cl2 or light

2 Cl ■

C hain P ro p a g a tio n : re a c tio n a n d re g e n e ra tio n o f ra d ic a ls

Step 2

CH 4 + Cl- ----- : c h 3. + H— Cl

Step 3

CH3- + Cl2 ----- : CH3Cl + Cl-

The chain nature of the reaction accounts for the observation that the light-promoted reac­ tion is highly efficient. The presence of a relatively few atoms of chlorine at any given moment is all that is needed to cause the formation of many thousands of molecules of chloromethane. What causes the chain reaction to terminate? Why does one photon of light not promote the chlorination of all of the methane molecules present? We know that this does not happen because we find that, at low temperatures, continuous irradiation is required or the reaction slows and stops. The answer to these questions is the existence of ch a in-term inating step s: steps that happen infrequently but occur often enough to use up one or both o f the reactive intermediates . The continuous replacement of intermediates used up by chain-terminating steps requires continuous irradiation. Plausible chain-terminating steps are as follows. C hain T e rm in a tio n : c o n s u m p tio n o f r a d ic a ls (e .g ., b y c o u p lin g )

H

H H — C-'

H — C -C l;

• C l:

H H

H H H

H

H_C • C __H

H— C ^ + C— H \ / H H : Cl

(E thane by-product)

H H : Cl Cl:

■C l :

Our radical mechanism also explains how the reaction of methane with chlorine pro­ duces the more highly halogenated products, CH 2 Cl2, CHCl3, and CCl4 (as well as addi­ tional HCl). As the reaction progresses, chloromethane (CH 3 Cl) accumulates in the mixture and its hydrogen atoms, too, are susceptible to abstraction by chlorine. Thus chloromethyl radicals are produced that lead to dichloromethane (CH 2 Cl2). S ide R e a c tio n s : m u ltih a lo g e n a te d b y -p r o d u c t fo rm a tio n

S tep 2

Cl

-N,

H

Cl P/I C­ I H

Cl H

H

/A C l. Cl

H-

H

•C — H / H Cl

Cl

Step 3

Cl

—»

H— C " C l

+

H (D ichlorom ethane)

Cl-

469

470

Chapter 10

Radical Reactions

Then step 2 is repeated, then step 3 is repeated, and so on. Each repetition of step 2 yields a molecule o f HCl, and each repetition of step 3 yields a molecule of CH2Cl2. r

S o lv e d P ro b le m 1 0 .3

1

When methane is chlorin iated, among the products found are traces o f chloroethane. How is it formed? O f what significance is its formati on? STRATEGY AND ANSW ER A small amount o f ethane is formed by the combination of two methyl radicals:

2 CH3- ----- > CH3 : CH3 The ethane byproduct formed by coupling then reacts with chlorine in a radical halogenation reaction (see Section 10.6) to form chloroethai e. The significance of th is observation is that it is evidence for the proposal that the combination o f two methyl radicals is one o f the cha in-terminating steps in the chlorination o f methane.

R e v ie w P ro b le m 10 .3

Suggest a method for separating and isolating the CH3Cl, CH2Cl2, CHCl3, and CCl4 that may be formed as a mixture when methane is chlorinated. (You may want to consult a hand­ book.) What analytical method could be used to separate this mixture and give structural information about each component?

R e v ie w P ro b le m 10 .4

How would the molecular ion peaks in the respective mass spectra of CH3Cl, CH2Cl2, CHCl3, and CCl4 differ on the basis of the number of chlorines (remember that chlorine 37 has isotopes 35Cl and 37Cl found in a 3 : 1 ratio)?

R e v ie w P ro b le m 10 .5

I f the goal is to synthesize CCl4 in maximum yield, this can be accomplished by using a large excess of chlorine. Explain.

10.5 Chlorination o f M eth an e: Energy Changes We saw in Section 10.2A that we can calculate the overall heat o f reaction from bond dissociation energies. We can also calculate the heat o f reaction for each individual step of a mechanism: C hain In itia tio n

Step 1 Cl— Cl ----- > 2 Cl-

AH° = +243 kJ mol-1

(DH° = 243) C hain P ro p a g a tio n

CH3— H + Cl- ----- > CH3- + H— Cl

Step 2

(DH° = 440)

Step 3

AH° = + 8 kJ mol-1

(DH° = 432)

CH3- + Cl— Cl ----- > CH3— Cl + Cl(DH° = 243)

AH° = -1 0 9 kJ m oP1

(DH° = 352)

C hain T erm ination

m I O

+ Cl-

CH3 — Cl

- :

AH° = -3 5 2 kJ mol 1

(DH° = 352) I o

CH3 — CH3

:

AH° = -3 7 8 kJ mol-1

3

m I O

+

A °O

1 l

-

I

Cl- + Cl-

C l

(DH° = 378)

-2 4 3 kJ mol-1

( DH ° = 243)

In the chain-initiating step only one bond is broken— the bond between two chlorine atoms— and no bonds are formed. The heat o f reaction for this step is simply the bond dis­ sociation energy for a chlorine molecule, and it is highly endothermic.

471

10.5 Chlorination of Methane: Energy Changes In the chain-terminating steps bonds are formed, but no bonds are broken. As a result, all o f the chain-terminating steps are highly exothermic. Each of the chain-propagating steps, on the other hand, requires the breaking of one bond and the formation of another. The value of AH0 for each of these steps is the difference between the bond dissociation energy o f the bond that is broken and the bond dissociation energy for the bond that is formed. The first chain-propagating step is slightly endothermic (AH° = + 8 kJ m o F 1), but the second is exothermic by a large amount (AH° = -1 0 9 kJ m o F 1).

Assuming the same mechanism applies, calculate AH° for the chain-initiating, chain-prop­ agating, and chain-terminating steps involved in the fluorination of methane.

R e v ie w P ro b le m 1 0.6

The addition o f the chain-propagating steps, cancelling species that appear on both sides o f the arrows, yields the overall equation for the chlorination o f methane: Cl- + CH3 — H ----- > CH3- + H — Cl

AH° = + 8 kJ mol-1

C H r + Cl — C l----- : CH3— Cl +

AH° = —109 kJ mol-1

------- H d p f u l H m t:

--------------------------------------------------------------------------------------------------------------------------------------- "—

CH3 — H + Cl— C l----- > CH3 — Cl + H — Cl

AH° = —101 kJ mol-1

Calculating overall AH° fo r a chain reaction.

and the addition o f the values o f AH° for the individual chain-propagating steps yields the overall value of AH° for the reaction.

Show how you can use the chain-propagating steps (see Review Problem 10.6) to calculate the overall value of AH° for the fluorination of methane.

R e v ie w P ro b le m 1 0 .7

10.5A The Overall Free-Energy Change For many reactions the entropy change is so small that the term T AS° in the expression AG° = AH ° — T AS ° is almost zero, and AG° is approximately equal to AH°. This happens when the reaction is one in which the relative order or disorder of reactants and products is about the same. Recall (Section 3.10) that entropy measures the relative disorder or randomness o f a system. For a chemical system the relative disorder o f the molecules can be related to the number of degrees o f freedom available to the molecules and their constituent atoms. Degrees o f free­ dom are associated with ways in which movement or changes in relative position can occur. Molecules have three sorts o f degrees of freedom: translational degrees of freedom asso­ ciated with movements of the whole molecule through space, rotational degrees o f free­ dom associated with the tumbling motions o f the molecule, and vibrational degrees of freedom associated with the stretching and bending motion o f atoms about the bonds that connect them (Fig. 10.2). I f the atoms o f the products o f a reaction have more degrees of freedom available than they did as reactants, the entropy change (AS°) for the reaction w ill be positive. If, on the other hand, the atoms o f the products are more constrained (have fewer degrees o f freedom) than the reactants, a negative AS° results.

Figure 10.2 Translational, rota tio n a l, and vib ratio n a l degrees o f free d o m fo r a simple d ia to m ic molecule.

472

Chapter 10

Radical Reactions

Consider the reaction of methane with chlorine: CH4

+

Cl2 ----- : CH3CI

+

HCl

Here, 2 mol of the products are formed from the same number of moles o f the reactants. Thus the number of translational degrees of freedom available to products and reactants is the same. Furthermore, CH3CI is a tetrahedral molecule like CH4, and HCl is a diatomic molecule like Cl2. This means that vibrational and rotational degrees of freedom available to products and reactants should also be approximately the same. The actual entropy change for this reaction is quite small, AS° = +2.8 J K -1 mol-1 . Therefore, at room temperature (298 K) the T AS° term is 0.8 kJ m ol-1, and thus the enthalpy change for the reaction and the free-energy change are almost equal: AH° = -101 kJ m ol-1 and AG° = -1 0 2 k J mol“ 1. In situations like this one it is often convenient to make predictions about whether a reac­ tion w ill proceed to completion on the basis o f AH° rather than AG° since AH° values are readily obtained from bond dissociation energies.

10.5B Activation Energies For many reactions that we shall study in which entropy changes are small, it is also often convenient to base our estimates of reaction rates simply on energies o f activation, E act, rather than on free energies o f activation, AG*. Without going into detail, suffice it to say that these two quantities are closely related and that both measure the difference in energy between the reactants and the transition state. • A low energy o f activation means a reaction w ill take place rapidly; a high energy of activation means that a reaction w ill take place slowly. Having seen earlier in this section how to calculate AH ° for each step in the chlorina­ tion o f methane, let us consider the energy o f activation for each step. These values are as follows: Chain Initiation

Step 1

Cl2 ----- > 2 Cl-

Eact

=

+243 kJ mol-1

Chain Propagation

Step 2

Cl-

+

CH4 ----- >

HCl

+

CH3-

Eact

= +

Step 3

CH3-

+

Cl2 ----- > CH3Cl

+

Cl-

Eact

=

16 kJ mol-1

~ 8 kJ mol-1

How does one know what the energy of activation for a reaction w ill be? Could we, for example, have predicted from bond dissociation energies that the energy of activation for the reaction Cl- + CH4 ----- > HCl + CH3- would be precisely 16 kJ m ol-1 ? The answer is no. The energy o f activation must be determined from other experimental data. It cannot be directly measured— it is calculated. Certain principles have been established, however, that enable one to arrive at estimates of energies of activation: 1. Any reaction in which bonds are broken w ill have an energy o f activation greater than zero. This w ill be true even if a stronger bond is formed and the reaction is exothermic. The reason: Bond form ation and bond breaking do not occur simultaneously in the transition state. Bond formation lags behind, and its energy is not all available for bond breaking. 2. Activation energies of endothermic reactions that involve both bond formation and bond rupture w ill be greater than the heat o f reaction, AH °. Two examples illus­ trate this principle, namely, the first chain-propagating step in the chlorination of methane and the corresponding step in the bromination of methane: Cl-

+

CH3— H (D H ° = 4 4 0 )

Br- + CH3— H (D H ° = 440)

CH3-

AH° = + 8 kJ mol-1 Eact = + 16 kJ mol-1

----- > H— Br + CH3-

AH° = +74 kJ mol-1 Eact = +78 kJ mol-1

----- :

H — Cl

+

(D H ° = 4 3 2 )

(D H ° = 366)

473

10.5 Chlorination of Methane: Energy Changes In both of these reactions the energy released in bond formation is less than that required for bond rupture; both reactions are, therefore, endothermie. We can easily see why the energy o f activation for each reaction is greater than the heat of reaction by looking at the potential energy diagrams in Fig. 10.3. In each case the path from reactants to products is from a lower energy plateau to a higher one. In each case the intervening energy h ill is higher still, and since the energy o f activation is the verti­ cal (energy) distance between the plateau of reactants and the top o f this hill, the energy o f activation exceeds the heat o f reaction.

(a)

(b)

Figure 10.3 Potential energy diagrams for (a) the reaction of a chlorine atom with m ethane and (b) the reaction of a bromine atom with methane.

The

e n e rg y

o f a c tiv a tio n

of a

g a s -p h a s e

r e a c tio n

w h e re

bonds

a re

b ro k e n

A H 0.* An example of this type of reaction is the chain-initiating step in the chlorination of methane— the dissocia­ tion o f chlorine molecules into chlorine atoms:

h o m o l y t i c a l l y b u t n o b o n d s a r e f o r m e d is e q u a l t o

Cl— Cl (D H ° = 2 4 3 )

2Cl-

AH° = +243 kJ m oP1 ^Eart a c t = +243 kJ m oP1

The potential energy diagram for this reaction is shown in Fig. 10.4.

Figure 10.4 Potential energy diagram for the dissociation of a chlorine m olecule into chlorine atoms.

4. The energy o f activation fo r a gas-phase reaction in which small radicals com­ bine to form molecules is usually zero. In reactions of this type the problem of nonsimultaneous bond formation and bond rupture does not exist; only one process occurs: that o f bond formation. A ll of the chain-terminating steps in the chlorination * T h is ru le a pp lies o n ly to ra d ic a l re actio n s ta k in g p la ce in th e gas phase. I t does n o t a p p ly to re actio n s ta k in g p la ce in s o lu tio n , e s p e c ia lly w he re io n s are in v o lv e d , because s o lv a tio n energies are also im p o rta n t.

474

Chapter 10

Radical Reactions

of methane fall into this category. An example is the combination of two methyl rad­ icals to form a molecule o f ethane: 2 CH3- ----- > CH3— CH3

AH° = -3 7 8 kJ mol-1 Eact = 0

(D H ° = 3 7 8 )

Figure 10.5 illustrates the potential energy changes that occur in this reaction.

Figure 10.5 Potential energy diagram fo r the com bination o f tw o m ethyl radicals to fo rm a m olecule o f ethane.

R e v ie w P ro b le m 10 .8

When pentane is heated to a very high temperature, radical reactions take place that pro­ duce (among other products) methane, ethane, propane, and butane. This type of change is called therm al cracking. Among the reactions that take place are the following: (1)

CH3-

(2)

CH3CH2-

(3) CH3.

CH3-

+

CH3CH3 ch4

(4) CH3 ' (5) CH3-

+

(6) CH3CH2 -

c h 3c h 2 +

-

c h 3c h 2 ■

(a) For which o f these reactions would you expect Eact to equal zero? (b) To be greater than zero? (c) To equal AH0? R e v ie w P ro b le m 1 0 .9

(a) Consider the chain-propagating steps shown here for the fluorination o f methane and the accompanying data. Sketch a potential energy diagram for each step. Label energy differences quantitatively, and sketch the transition state structures. CH4 + F- ----- >

CH3- + HF

Eact = +5.0 kJ mol 1 AH° = -1 3 0 kJ mol-1

CH3- + F2 ----- : CH3— F + F-

Eact = +1.0 kJ mol 1 AH° = -3 0 2 kJ mol-1

(b) Consider the chain-initiating and chain-terminating steps shown here for the fluorination o f methane. Sketch and label potential energy diagrams for these reactions, in the same way as specified in part (a). F2 ----- > 2FCH3 + F- ----- > CH3 — F

Eact = +159 kJ mol-1 AH° = -461 kJ mol-1

10.5 Chlorination of Methane: Energy Changes (c) Sketch a potential energy diagram for the follow ing reaction. N ote that it is the reverse o f a reaction in p art (a).

CH3- + H — F ----- > CH4 + F-

10.5C Reaction of Methane with Other Halogens The reactivity o f one substance tow ard another is m easured by the rate at w hich the two substances react. A reagent that reacts very rapidly w ith a particular substance is said to be highly reactive tow ard that substance. O ne that reacts slow ly or n o t at all under the same experim ental conditions (e.g., concentration, pressure, and tem perature) is said to have a low relative reactivity or to be unreactive. The reactions o f the halogens (fluorine, chlorine, brom ine, and iodine) w ith m ethane show a w ide spread o f relative reactivities. Fluorine is m ost reactive— so reactive, in fact, that w ithout special precautions m ixtures o f fluorine and m ethane explode. C hlorine is the n ext m ost reactive. However, the chlorination o f m ethane is easily controlled by the judicious control o f heat, light, and concentration. B rom ine is m uch less reactive tow ard m ethane than chlorine, and iodine is so unreactive that for all practical purposes w e can say that no reaction takes place. If the m echanism s for fluorination, brom ination, and iodination o f m ethane are the same as for its chlorination, w e can explain the w ide variation in reactivity o f the halogens by a careful exam ination o f AH° and E act for each step.

FLUORINATION A H ° (k J m o l- 1 )

E a c t ( k J m o l “ 1)

+159

+ 159

-1 3 0 -3 0 2

+ 5.0 Small

C hain In itia tio n

F2 ----- : 2 FC hain P ro p a g a tio n

HF + CH3F- + CH4 ----- : CH3- + F2 ----- : CH3F + F-

Overall AH° = -4 3 2 The chain-initiating step in f l u o r i n a t i o n is highly endotherm ic and thus has a high energy o f activation. If w e did n o t know otherw ise, w e m ig h t carelessly conclude from the energy o f activa­ tion o f the chain-initiating step alone that fluorine w ould b e quite unreactive tow ard methane. (If w e then proceeded to try the reaction, as a resu lt o f this careless assessm ent, the results w ould be literally disastrous.) We know, however, that the chain-initiating step occurs only infrequently relative to the chain-propagating steps. O ne initiating step is able to produce thousands o f fluorination reactions. A s a result, the high activation energy for this step is n o t an im pedim ent to the reaction. C hain-propagating steps, by contrast, cannot afford to have high energies o f activation. If they do, the highly reactive interm ediates are consum ed by chain-term inating steps before the chains progress very far. Both o f the chain-propagating steps in fluorination have very sm all energies o f activation. This allow s a relatively large fraction o f energetically favorable collisions even at room tem perature. M oreover, the overall heat o f reaction, AH°, is very large. This m eans that as the reaction occurs, a large quantity o f h eat is evolved. This heat m ay accum ulate in the m ixture faster than it dissipates to the surroundings, causing the tem ­ perature to rise and w ith it a rapid increase in the frequency o f additional chain-initiating steps that w ould generate additional chains. T hese tw o factors, the low energy o f activation for the chain-propagating steps and the large overall heat o f reaction, account for the high reactivity o f fluorine tow ard m ethane. (Fluorination reactions can be controlled. This is usu­ ally accom plished by diluting both the hydrocarbon and the fluorine w ith an inert gas such as helium before bringing them together. T he reaction is also carried out in a reactor packed w ith copper shot. T he copper, by absorbing the heat produced, m oderates the reaction.)

475

476

Chapter 10

Radical Reactions CHLORINATION A H ° (k J m o i- 1 )

Eact(kJ moi-1 )

C hain In itia tio n

+243

Cl2 ----- : 2 Cl-

A H ° (k J m o i- 1 )

+243 E a c t(k J m o i- 1 )

C hain P ro p a g a tio n

Cl- + c h 4 — CH3- + Cl2 —

+8 109

HCl + CH3CH3Cl + ClOverall AH 0

+ 16 Small

101

T h e h i g h e r e n e r g y o f a c t i v a t i o n o f t h e f ir s t c h a in - p r o p a g a t in g s te p ( t h e h y d r o g e n a b s t r a c ­ t i o n s t e p ) i n t h e c h l o r i n a t i o n o f m e t h a n e ( + 1 6 k J m o l ~ 1) , c o m p a r e d t o t h e l o w e r e n e r g y o f a c t i v a t i o n ( + 5 . 0 k J m o F 1) i n t h e f l u o r i n a t i o n , p a r t l y e x p la in s t h e l o w e r r e a c t i v i t y o f c h l o ­ r i n e . T h e g r e a t e r e n e r g y r e q u i r e d t o b r e a k t h e c h l o r i n e - c h l o r i n e b o n d i n t h e i n i t i a t i n g s te p ( 2 4 3 k J m o l ~ 1 f o r C l2 v e rs u s 1 5 9 k J m o l ~ 1 f o r F 2) h a s s o m e e ffe c t, to o . H o w e v e r , th e m u c h g r e a t e r o v e r a l l h e a t o f r e a c t io n i n f l u o r i n a t i o n p r o b a b l y p l a y s t h e g r e a t e s t r o l e i n a c c o u n t ­ in g f o r t h e m u c h g r e a t e r r e a c t i v i t y o f f l u o r i n e .

BROMINATION A H ° (k J m o i- 1 )

E a c t(k J m o i- 1 )

C hain In itia tio n

Br2 ----- : 2 Br-

193

+ 193

+ 74 100

+78 Small

C hain P ro p a g a tio n

Br- + CH4 — CH3- + Br2 —

HBr CH3Br

CH3Br-

Overall AH°

-2 6

I n c o n t r a s t t o c h lo r i n a t io n , t h e h y d r o g e n a t o m a b s t r a c t io n s te p i n b r o m i n a t i o n h a s a v e r y h i g h e n e r g y o f a c t i v a t i o n ( E act =

7 8 k J m o l - 1 ) . T h is m e a n s th a t o n ly a v e r y tin y fr a c tio n

o f a l l o f t h e c o l l i s i o n s b e t w e e n b r o m i n e a t o m s a n d m e t h a n e m o le c u le s w i l l b e e n e r g e t i ­ c a l l y e f f e c t i v e e v e n a t a t e m p e r a t u r e o f 3 0 0 ° C . B r o m i n e , a s a r e s u l t , is m u c h le s s r e a c t iv e t o w a r d m e t h a n e t h a n c h l o r i n e , e v e n t h o u g h t h e n e t r e a c t i o n is s l i g h t l y e x o t h e r m i c .

IODINATION A H ° (k J m o i- 1 )

E a c t(k J m o i- 1 )

C hain In itia tio n

I2 ----- : 2 I-

+ 151

+ 151

+142 -8 9

+ 140 Small

C hain P ro p a g a tio n

I- + c h 4 — CH3- + I2 —

HI + CH3CH3I + IOverall AH 0

53

T h e t h e r m o d y n a m i c q u a n t it ie s f o r i o d i n a t i o n o f m e t h a n e m a k e i t c le a r t h a t t h e c h a i n - i n i ­ t ia t in g s te p is n o t r e s p o n s ib le f o r t h e o b s e r v e d o r d e r o f r e a c t iv it i e s : F 2 >

C l2 >

B r2 >

I2 . T h e

i o d i n e - i o d i n e b o n d is e v e n w e a k e r th a n t h e f l u o r i n e - f l u o r i n e b o n d . O n t h is b a s is a lo n e , o n e w o u l d p r e d ic t io d in e t o b e t h e m o s t r e a c t iv e o f t h e h a lo g e n s . T h i s c l e a r l y is n o t t h e c a s e . O n c e a g a in , i t is t h e h y d r o g e n a t o m - a b s t r a c t i o n s te p t h a t c o r r e la t e s w i t h t h e e x p e r i m e n t a l l y d e t e r ­ m i n e d o r d e r o f r e a c t iv it i e s . T h e e n e r g y o f a c t i v a t i o n o f t h is s te p i n t h e i o d i n e r e a c t io n ( 1 4 0 k J m o l - 1 ) is s o la r g e t h a t o n l y t w o c o ll is io n s o u t o f e v e r y 1 0 12 h a v e s u f f ic ie n t e n e r g y t o p r o ­ d u c e r e a c t io n s a t 3 0 0 ° C . A s a r e s u lt , io d in a t i o n is n o t a f e a s i b le r e a c t io n e x p e r i m e n t a ll y . B e f o r e w e le a v e t h is t o p ic , o n e f u r t h e r p o i n t n e e d s t o b e m a d e . W e h a v e g i v e n e x p la n a ­ tio n s o f t h e r e l a t i v e r e a c t iv it ie s o f t h e h a lo g e n s t o w a r d m e t h a n e t h a t h a v e b e e n b a s e d o n e n e r g y c o n s id e r a t io n s a lo n e . T h i s h a s b e e n p o s s ib le

thus have sim ilar entropy changes.

only because the reactions are quite sim ilar and

H a d th e r e a c t io n s b e e n o f d i f f e r e n t t y p e s , th is k i n d o f a n a ly ­

sis w o u l d n o t h a v e b e e n p r o p e r a n d m i g h t h a v e g i v e n in c o r r e c t e x p la n a t io n s .

10.6 Halogenation of Higher Alkanes

477

10.6 Halogenation o f Higher Alkanes Higher alkanes react with halogens by the same kind o f chain m echanism as those that we have just seen. Ethane, for example, reacts with chlorine to produce chloroethane (ethyl chloride). The mechanism is as follows:

A MECHANISM FOR THE REACTION R a d ic a l H a l o g e n a t i o n o f E t h a n e C hain In itia tio n Cl 2

2 or heat

Step 1

2 Cl -

C hain P ro p a g a tio n n r \

Step 2

CH3CH2 H + • C l

Step 3

c h 33c h 2 2 +

CH3CH2 + H - C l

3o

CH3CH2. C l

Cl •Cl

+ Cl -

Chain propagation co n tin u es with s te p s 2, 3, 2, 3, an d s o on. C hain T e rm in a tio n C H 3C H 2- ^

CH3CH2. C l

+ A-Cl

c h 3c h 2- ^ + ■ c h 2c h 3 Cl

c h 3c h 2 c h 2c h 3

+ ^- Cl

C l: C l

(a) Consider the hydrogen abstraction step in the chlorination o f ethane.

CH3— CH3 + C l------ : CH3— CH2- + HCl

R e v ie w P ro b le m 1 0 .1 0

Eact = 4.2 kJ mol-1

Calculate AH° for this step using data from Table 10.1, and draw a fully labeled poten­ tial energy diagram, similar to that shown in Fig. 10.3 a . (b) When an equimolar mixture o f methane and ethane is chlorinated, the reaction yields chloroethane and chloromethane in a ratio o f 400 : 1.

CH3— CH3 + CH4 —— :

CH3— CH2Cl + CH3Cl + 2 HCl 400

:

1

Explain the observed ratio o f products. When ethane is chlorinated, 1,1-dichloroethane and 1,2-dichloroethane, as w ell as more highly chlorinated ethanes, are formed in the mixture (see Section 10.3A). Write chain reac­ tion mechanisms accounting for the formation o f 1 , 1 -dichloroethane and 1 ,2 -dichloroethane.

Chlorination o f m ost alkanes w hose m olecules contain more than two carbon atoms gives a mixture o f isomeric monochloro products (as w ell as more highly chlorinated compounds).

R e v ie w P ro b le m 10.11

478

Chapter 10

Radical Reactions

Several exam ples follow. The percentages given are based on the total amount o f monochloro products formed in each reaction. Cl

cl

Cl

light, 25°C

1-C hloropropane (45%)

P ropane

2-C hloropropane (55%)

Cl, Cl

light 25°C

2-M ethylpropane

Cl

1-Chloro-2-methylp ro p an e (63%)

,-C h lo ro -,-m eth y lp ro p an e (37%)

Cl 300°C

Cl Cl

2-M ethylbutane

1 -C hloro-,-m ethylbutane (30%)

,-C h lo ro -,-m eth y lb u tan e (22%)

The ratios o f products that w e obtain from chlorination reactions o f higher alkanes are not identical with what w e would expect if all the hydrogen atoms o f the alkane were equally reactive. We find that there is a correlation between reactivity o f different hydrogen atoms and the type o f hydrogen atom (1°, 2°, or 3°) being replaced. The tertiary hydrogen atoms o f an alkane are most reactive, secondary hydrogen atoms are next most reactive, and pri­ mary hydrogen atoms are the least reactive (see Review Problem 10.12). R e v ie w P ro b le m 1 0 .1 2

(a )

What percentages o f 1-chloropropane and 2-chloropropane would you expect to obtain from the chlorination o f propane if 1 ° and 2 ° hydrogen atoms were equally reactive? Cl

(b )

What percentages o f 1-chloro-2-methylpropane and 2-chloro-2-methylpropane would you expect from the chlorination o f 2-methylpropane if the 1° and 3° hydrogen atoms were equally reactive?

Cl (c )

Compare these calculated answers with the results actually obtained (above in Section 1 0 .6 ) and justify the assertion that the order o f reactivity o f the hydrogen atoms is 3° > 2° > 1°.

479

10.6 Halogenation o f Higher Alkanes

We can account for the relative reactivities o f the primary, secondary, and tertiary hydrogen atoms in a chlorination reaction on the basis o f the hom olytic bond dissocia­ tion energies w e saw earlier (Table 10.1). O f the three types, breaking a tertiary C — H bond requires the least energy, and breaking a primary C — H bond requires the most. Since the step in w hich the C — H bond is broken (i.e., the hydrogen atom-abstraction step) determines the location or orientation o f the chlorination, w e w ould expect the E act for abstracting a tertiary hydrogen atom to be least and the E act for abstracting a primary hydrogen atom to be greatest. Thus tertiary hydrogen atoms should be m ost reactive, sec­ ondary hydrogen atoms should be the next m ost reactive, and primary hydrogen atoms should be the least reactive. The differences in the rates with which primary, secondary, and tertiary hydrogen atoms are replaced by chlorine are not large, however. Chlorine, as a result, does not d is­ criminate among the different types o f hydrogen atoms in a way that m akes chlorination o f higher alkanes a generally useful laboratory synthesis. (Alkane chlorinations do find use in som e industrial processes, especially in those instances where mixtures o f alkyl chlorides can be used.)

S o lv e d P r o b le m 1 0 .4 An alkane with the formula C 5 H1 2 undergoes chlorination to give only one product with the formula C ^ - i Cl. What is the structure o f this alkane? STRATEGY AND ANSWER The hydrogen atoms o f the alkane must all be equivalent, so that replacing any one o f them leads to the same product. The only five-carbon alkane for which this is true is neopentane.

CH3 |

H3C

C

CH3

CH3

Chlorination reactions o f certain alkanes can be used for laboratory preparations. Examples are the preparation o f chlorocyclopropane from cyclopropane and chlorocyclobutane from cyclobutane.

R e v ie w P ro b le m 1 0 .1 3

Cl

^ A

A

(excess)

hv (excess) What structural feature o f these m olecules makes this possible? Each o f the follow ing alkanes reacts with chlorine to give a single m onochloro substitution product. On the basis o f this information, deduce the structure o f each alkane.

(a) C5H10

1 0 .6 A

R e v ie w P ro b le m 1 0 .1 4

(b) C8H18

Selectivity of Bromine H e lp f u l H i n t

•

Bromine is less reactive than chlorine toward alkanes in general but bromine is more selective in the site o f attack.

Bromination is selective.

480

Chapter 10

Radical Reactions

Brom ine show s a m uch greater ability to discrim inate am ong the different types o f hy d ro ­ gen atom s. T he reaction o f 2-m ethylpropane and brom ine, for exam ple, gives alm ost exclu­ sive replacem ent o f the tertiary hydrogen atom:

Br

(>99%)

(trace)

A very different resu lt is obtained w hen 2-m ethylpropane reacts w ith chlorine:

Cl

(37%)

(63%)

Fluorine, being m uch m o re reactive than chlorine, is even less selective than chlorine. B ecause the energy o f activation for the abstraction o f any type o f hydrogen by a fluorine atom is low, there is very little difference in the rate at w hich a 1°, 2°, or 3° hydrogen reacts w ith fluorine. R eactions o f alkanes w ith fluorine give (alm ost) the distribution o f products that w e w ould expect if all o f the hydrogens o f the alkane w ere equally reactive.

r

S o lv e d P ro b le m 1 0 .5

1

E xplain w hy tem perature is an im portant variable to consider w hen using isom er distribution to evaluate the reactivities o f the hydrogens o f an alkane tow ard radical chlorination. STRATEGY AND ANSW ER A t low er tem peratures, isom er distribution accurately reflects the inherent reactivities o f the hydrogens o f the ailkanes. A s the tem perature is raised, m ore chlorine atom s have sufficient energy to surm ount the larger energy orf activation that accom panies hydrogen abstraction at the less substituted carbons. If the tem perature is high enou gh, hydrogens are replaced by chlorine on a purely statistical basis.

10.7 The G eo m etry o f A lkyl Radicals E xperim ental evidence indicates that the geom etric structure o f m ost alkyl radicals is trig ­ onal planar at the carbon having the unpaired electron. This structure can b e accom m odated by an sp2-hybridized central carbon. In an alkyl radical, the p orbital contains the unpaired electron (Fig. 10.6).

Figure 10.6 (a) Drawing of a methyl radical showing the sp 2-hybridized carbon atom at the center, the unpaired electron in the half-filled p orbital, and the three pairs of electrons involved in covalent bonding. The unpaired electron could be shown in either lobe. (b) Calculated structure for the methyl radical showing the highest occupied molecular orbital, where the unpaired electron resides, in red and blue. The region of bonding electron density around the carbons and hydrogens is in gray.

481

10.8 Reactions That Generate Tetrahedral Chirality Centers

10.8 Reactions That G enerate Tetrahedral Chirality Centers •

W hen achiral m olecules react to produce a com pound w ith a single tetrahedral chirality center, the product w ill be a racem ic form .

T his w ill alw ays be true in the absence o f any chiral influence on the reaction such as an enzym e or the use o f a chiral reagent or solvent. L et us exam ine a reaction that illustrates this principle, the radical chlorination o f pentane: Cl

Cl,

Cl

(achiral)

Cl

P entane (achiral)

1-C hloropentane (achiral)

(±)-2-C hloropentane (a racem ic form)

3-C hloropentane (achiral)

The reaction w ill lead to the products show n here, as w ell as m ore highly chlorinated products. (We can use an excess o f p entane to m inim ize m ultiple chlorinations.) N either 1-chloropentane n o r 3-chloropentane contains a chirality center, b ut 2-chloropentane does, and it is obtained as a racem ic fo rm . If w e exam ine the m echanism w e shall see why.

A MECHANISM FOR THE REACTION The Stereochemistry of Chlorination at C2 of Pentane C2

m CH3CH2CH2CH2CH3

I" /C H a Cl- + Cl — C. vX H CH2CH2CH3

CHa

Cl2 (a) H CH2CH2CH3

Cl2

V

T tT

H ^y

,C — Cl + Cl-

CH2CH2CH3

(S )-2-C hloropentane

Trigonal planar

(ff)-2 -C h lo ro p e n ta n e

(5 0 % )

ra d ica l

(5 0 % )

(a c h ira l)

E n a n tio m e rs

A bstraction of a hydrogen atom from C2 p ro d u ce s a trigonal planar radical th at is achiral. This radical then re a c ts with chlorine at either face [by path (a) or path (b)]. B ecau se th e radical is achiral, th e probability of reaction by either path is th e sam e; therefore, th e two en antio m ers are pro d u ced in equal am o u n ts, an d a racem ic form of 2-chloropentane results.

482

Chapter 10

Radical Reactions

10.8A Generation of a Second Chirality Center in a Radical Halogenation L et us now exam ine w hat happens w hen a chiral m olecule (containing one chirality cen ­ ter) reacts so as to yield a pro d u ct w ith a second chirality center. A s an exam ple consider w hat happens w hen (S )-2-chloropentane undergoes chlorination at C 3 (other products are form ed, o f course, by chlorination at other carbon atom s). T he results o f chlorination at C3 are show n in the box below. T he products o f the reactions are (2S ,3S )-2,3-dichloropentane and (2S ,3R )-2,3dichloropentane. T hese tw o com pounds are d ia ste re o m e rs. (They are stereoisom ers but they are n o t m irror im ages o f each other.) T he tw o diastereom ers are n ot p roduced in equal am ounts. B ecause the interm ediate radical itself is chiral, reactions at the tw o faces are not equally likely. T he radical reacts w ith chlorine to a greater extent at one face than the other (although w e cannot easily p redict w hich). T hat is, the presence o f a chirality cen­ ter in the radical (at C 2) influences the reaction that introduces the new chirality center (at C3). B oth o f the 2,3-dichloropentane diastereom ers are chiral and, therefore, each exhibits optical activity. M oreover, because the tw o com pounds are diastereom ers, they have dif­ ferent physical properties (e.g., different m elting points and boiling points) an d are sepa­ rable by conventional m eans (by gas chrom atography or by careful fractional distillation).

A MECHANISM FOR THE REACTION T h e S te r e o c h e m is tr y o f C h lo rin a tio n a t C 3 o f ( S )-2 -C h lo ro p e n ta n e CH3

CH, ch 2

CH3 (S)-2-Chloropentane Clt

CH3 HV C . . C|

CH3 Hv C ^ Cl

Cl-

C C l^ i ^ H Ch2 CH3 (2S,3S)-2,3-Dichloropentane (chiral)

CH3 H^

Cl2

Cl2

(a)

lb "

h2c h

I H3C Trigonal planar radical (chiral)

Cl

ClCl Ch2 CH3 (2S,3S)-2,3-Dichloropentane (chiral)

Diastereomers A bstraction of a hydrogen atom from C3 of (S)-2-chloropentane p ro d u ce s a radical th at is chiral (it co n tain s a chirality ce n te r at C2). This chiral radical can then react with chlorine at o n e face [path (a)] to p ro d u ce (2S, 3S)-2,3-dichloropenta n e an d th e other fac e [path (b)] to yield (2S, 3R) -2,3-dichloropentane. T h ese two co m p o u n d s a re d iastereo m ers, and they are not produced in equal am ounts. Each p ro d u ct is chiral, an d each alo n e w ould be optically active.

10.8 Reactions That Generate Tetrahedral Chirality Centers

C onsider the chlorination o f (S )-2-chloropentane at C 4. (a) W rite structural form ulas for the products, show ing three dim ensions at all ch irality centers. G ive each its p ro p er (R,S) d esignation. (b) W hat is the stereoisom eric relatio n sh ip betw een th ese p ro d u cts? (c) A re both products chiral? (d) A re b oth o p tically active? (e) C o u ld the p ro d u cts b e separated by conventional m eans? (f) W hat other d ichlo ro p en tan es w ould b e o b tain ed b y ch lo ri­ nation o f (S )-2-chloropentane? (g) W hich o f th ese are o p tically active?

h

483

R e v ie w P ro b le m 1 0 .1 5

S o lv e d P r o b le m 1 0 .6 C onsider the brom ination o f butane using sufficient brom ine to cause dibrom ination. A fter the reaction is over, you separate all the dibrom obutane isom ers by gas chrom atography or by fractional distillation. H ow m any fractions w ould you obtain, and w hat com pounds w ould the individual fractions contain? W hich if any o f the fractions w ould be optically active? STRATEGY AND ANSWER T he construction o f handheld m odels w ill help in solving this problem . First, decide how m any constitutional isom ers are possible by replacing tw o hydrogens o f butane w ith tw o brom ine atom s. There are six: 1,1-dibrom obutane, 1,2-dibrom obutane, 2,2-dibrom obutane, 2,3-dibrom obutane, 1,3-dibrom obutane, and 1,4-dibrom obutane. T hen recall that constitutional isom ers have different physical properties (i.e., boiling points, and retention tim es in a gas chrom atograph), so there should b e at least six fractions. In actuality there are seven. See fractions (a )-(g ) below. We soon see w hy there are seven fractions if w e exam ine each constitutional isom er looking for chirality centers and stereoisom ers. Isom ers (a), (c), and (g) have no chirality centers and are, there­ fore, achiral and are optically inactive. 1,2-D ibrom obutane in fraction (b) an d 1,4-dibrom obutane in fraction (f) each have one chirality center and, because there is no chiral influence on the reaction, they w ill be form ed as a 50 : 50 m ixture o f enantiom ers (a racem ate). A racem ate cannot b e separated by distillation or conventional gas chrom atography; therefore, fractions (b) and (f) w ill n o t b e optically active. 2,3-D ibrom obutane has tw o chirality centers and w ill be form ed as a racem ate [fraction (d)] an d as a m eso com pound, fraction (e). Both fractions w ill be optically inactive. T he m eso com pound is a diastereom er o f the enantiom ers in fraction (d) (and has different physical properties from them ); therefore, it is separated from them by distillation or gas chrom atography.

Br

Achiral (g)

484

R e v ie w P ro b le m 1 0 .1 6

Chapter 10

Radical Reactions

Consider the monochlorination of 2-methylbutane.

Cl Assuming that the product mixture was subjected to fractional distillation, which frac­ tions, if any, would show optical activity? ( b ) Could any of these fractions be resolved, theoretically, into enantiomers? ( c ) Could each fraction from the distillation be identi­ fied on the basis of 1H N M R spectroscopy? What specific characteristics in a 1H N M R spec­ trum of each fraction would indicate the identity of the component(s) in that fraction? (a )

10.9 Radical A d d itio n to Alkenes: The A nti-M arkovnikov A d d itio n o f Hydrogen Brom ide Before 1933, the orientation of the addition of hydrogen bromide to alkenes was the sub­ ject of much confusion. A t times addition occurred in accordance with Markovnikov’s rule; at other times it occurred in just the opposite manner. Many instances were reported where, under what seemed to be the same experimental conditions, Markovnikov addi­ tions were obtained in one laboratory and anti-Markovnikov additions in another. At times even the same chemist would obtain different results using the same conditions but on different occasions. The mystery was solved in 1933 by the research of M . S. Kharasch and F. R. Mayo (of the University of Chicago). The explanatory factor turned out to be organic peroxides pre­ sent in the alkenes— peroxides that were formed by the action of atmospheric oxygen on the alkenes (Section 10.11D). •

When alkenes containing peroxides or hydroperoxides react with hydrogen bro­ mide, anti-Markovnikov addition of HBr occurs. r

— O— O— R

A n o r g a n ic p e r o x id e

R— O — O — H A n o r g a n i c h y d r o p e r o x id e

For example, in the presence of peroxides propene yields 1-bromopropane. In the absence of peroxides, or in the presence of compounds that “trap” radicals, normal Markovnikov addition occurs. Br

H B r, ROOR

A n ti- M a r k o v n ik o v a d d itio n

(peroxides present)

Br

HBr (peroxides absent)

•

M a r k o v n ik o v a d d itio n

Hydrogen bromide is the only hydrogen halide that gives anti-Markovnikov addi­ tion when peroxides are present. Hydrogen fluoride, hydrogen chloride, and hydrogen iodide do not give anti-Markovnikov addition even when peroxides are present. The mechanism for a n t i - M a r k o v n i k o v a d d i t i o n o f h y d r o g e n b r o m i d e is a r a d i c a l c h a i n r e a c t i o n initiated by peroxides.

485

10.9 Radical Addition to Alkenes

A MECHANISM FOR THE REACTION A n ti- M a r k o v n ik o v A d d itio n C hain In itia tio n

R— O p O — R

Step 1

2 R— O •

Heat brings ab o u t hom olytic cleav ag e of th e w eak ox y g en -o x y g en bond. ^ Step 2

R— O-

..

H = Br = R— O - H Br(/The alkoxyl radical a b s tra c ts a hydrogen atom from HBr, p ro d u cin g a brom ine radical.

C hain P ro p a g a tio n

:B r

Step 3

+ H2C = C H — CH3

=B r - c h 2— c h — c h 3 2° Radical

A brom ine radical a d d s to th e dou b le bond to p roduce th e m ore sta b le 2° alkyl radical. Step 4

=B r - CH2— CH— CH3 + h T B p

: B r — CH2— CH— CH3 ” H 1-B rom opropane

T he alkyl radical a b s tra c ts a hydrogen atom from HBr. This leads to th e p roduct and reg e n erate s a brom ine radical. Then repetitions of ste p s 3 and 4 lead to a chain reaction.

S t e p 1 is t h e s i m p le h o m o l y t i c c le a v a g e o f t h e p e r o x i d e m o l e c u l e t o p r o d u c e t w o a l k o x y l r a d ic a ls . T h e o x y g e n - o x y g e n b o n d o f p e r o x id e s is w e a k , a n d s u c h r e a c t io n s a r e k n o w n t o o c c u r r e a d ily :

R— O:O — R ----- : 2 R— OPeroxide

AH° = +150 kJ mol-1

Alkoxyl radical

S t e p 2 o f t h e m e c h a n is m , a b s t r a c t io n o f a h y d r o g e n a t o m b y t h e r a d i c a l , is e x o t h e r m i c a n d h a s a lo w e n e rg y o f a c tiv a tio n :

R— O- + H :Br: ----- > R— O : H + : B*r*

AH° = -9 6 kJ mol-1 Eact is low

S t e p 3 o f t h e m e c h a n i s m d e t e r m in e s t h e f i n a l o r i e n t a t i o n o f b r o m i n e i n t h e p r o d u c t . I t

more stable secondary radical is p r o d u c e d at the prim a ry carbon atom is less hindered. H a d t h e b r o m i n e a t t a c k e d

o c c u rs as i t d o e s b e c a u s e a

and because

attack

p ro p e n e a t th e s ec­

o n d a r y c a r b o n a t o m , a le s s s t a b le , p r i m a r y r a d i c a l w o u l d h a v e b e e n t h e r e s u lt ,

Br- + CH2= C H C H 3 ----- > -CH2CHCH3 Br 1° Radical (less stable) a n d a tta c k a t th e s e c o n d a ry c a rb o n a to m w o u ld h a v e b e e n m o r e h in d e re d .

B r:

486

Chapter 10

Radical Reactions

Step 4 of the mechanism is simply the abstraction of a hydrogen atom from hydrogen bromide by the radical produced in step 3. This hydrogen atom abstraction produces a bromine atom (which, o f course, is a radical due to its unpaired electron) that can bring about step 3 again; then step 4 occurs again— a chain reaction.

10.9A Summary of Markovnikov versus Anti-Markovnikov Addition of HBr to Alkenes H e lp f u l H i n t

A tip fo r alkyl halide synthesis.

We can now see the contrast between the two ways that HBr can add to an alkene. In the absence of peroxides, the reagent that attacks the double bond first is a proton. Because a proton is small, steric effects are unimportant. It attaches itself to a carbon atom by an ionic mechanism so as to form the more stable carbocation. The result is Markovnikov addition. Polar, protic solvents favor this process. Io n ic A d d itio n

Addition to form the m ore sta b le carbocation

Markovnikov product

In the presence of peroxides, the reagent that attacks the double bond first is the larger bromine atom. It attaches itself to the less hindered carbon atom by a radical mechanism, so as to form the more stable radical intermediate. The result is anti-Markovnikov addi­ tion. Nonpolar solvents are preferable for reactions involving radicals. R a d ic a l A d d itio n

Addition to form the m ore sta b le alkyl radical

10.10

Anti-Markovnikov pro d u ct

o f Alkenes: Chain-Growth Polymers Polymers are substances that consist of very large molecules called macromolecules that are made up o f many repeating subunits. The molecular subunits that are used to synthe­ size polymers are called monomers, and the reactions by which monomers are joined together are called polymerizations. Many polymerizations can be initiated by radicals. Ethylene (ethene), for example, is the monomer that is used to synthesize the familiar polymer called polyethylene. M onomeric units

m CH2^ C H 2

Ethylene m onom er

polymerization

CH2CH2- fC H 2CH2- ^ CH2CH2 Polyethylene

polym er (m an d n a re large num bers)

Because polymers such as polyethylene are made by addition reactions, they are often called chain-growth polymers or addition polymers. Let us now examine in some detail how polyethylene is made.

487

10.10 Radical Polymerization of Alkenes: Chain-Growth Polymers E t h y l e n e p o l y m e r i z e s b y a r a d i c a l m e c h a n i s m w h e n i t is h e a t e d a t a p r e s s u r e o f 1 0 0 0 a tm w i t h a s m a ll a m o u n t o f a n o r g a n ic p e r o x id e ( c a lle d a d i a c y l p e r o x id e ).

A MECHANISM FOR THE REACTION R a d ic a l P o l y m e r i z a t i o n o f E t h e n e C hain In itia tio n

O

O

O c\

Step 1

R— C— O :O — C— R

>Pi II

2 R :C -rO -

2 CO,

2 R-

V

Diacyl peroxide Step 2

R:CH2— CH2-

c h 2= c h 2

Rv +

The diacyl peroxide d isso c ia te s and re le a se s carbon dioxide g as. Alkyl radicals are produced, w hich in turn initiate chains. C hain P ro p a g a tio n Step 3

R—CH2CH2-

n "CH2" C h 2

R - ^ C ^ C H ^ CH2CH2

Chains p ro p ag a te by adding successive ethylene units, until their grow th is sto p p e d by com bination or disproportionation. C hain T e rm in a tio n combination

Step 4

[ R ^ C ^ C H ^ CH2CH2-fc

2 R -f C ^ C H ^ CH2CH2disproportionation

- R -^C H 2CH2-);C H = C H 2 + R -^C H 2CH2-); CH2CH3

The radical at the en d of th e growing polym er chain can also a b stra c t a hydrogen atom from itself by w hat is called “back biting.” This lead s to chain branching. C hain B ra n c h in g

CH2 H

R— CH2CH

CH2

RCH2C H ^ C H 2CHr )nCH2CH2

> C H 2CH2< ch,

= ch2

RCH2C H ^ C H 2CH2^nCH2CH3 CH, CH,

etc.

T h e p o l y e t h y l e n e p r o d u c e d b y r a d i c a l p o l y m e r i z a t i o n is n o t g e n e r a l l y u s e f u l u n le s s i t h a s a m o le c u la r w e ig h t o f n e a r ly 1 ,0 0 0 ,0 0 0 . V e r y h ig h m o le c u la r w e ig h t p o ly e th y le n e c a n b e o b t a i n e d b y u s in g a l o w c o n c e n t r a t i o n o f t h e in i t i a t o r . T h is i n it ia t e s t h e g r o w t h o f o n l y

H

488

Chapter 10

Radical Reactions

a few chains and ensures that each chain w ill have a large excess o f the m onom er avail­ able. M ore initiator m ay b e added as chains term inate during the polym erization, and, in this way, new chains are begun. Polyethylene has been produced com m ercially since 1943. It is used in m anufacturing flexible bottles, films, sheets, and insulation for electric w ires. Polyethylene produced by radical polym erization has a softening po in t o f about 110°C. Polyethylene can b e produced in a different w ay using (see Special Topic B) cata­ lysts called Z i e g l e r - N a t t a c a t a l y s t s that are organom etallic com plexes o f tran si­ tion m etals. In this process no radicals are produced, no back biting occurs, and, consequently, there is no chain branching. T he polyethylene that is produced is of higher density, has a h igher m elting point, an d has greater strength. A nother fam iliar polym er is polystyrene. T he m onom er used in m aking polystyrene is phenylethene, a com pound com m only know n as styrene.

m polymerization

Styrene

Polystyrene

Table 10.2 lists several other com m on chain-grow th polym ers. F urther inform ation on each is provided in Special Topic B.

2

O th e r C om m on Chain-G row th Polymers

M onom er

P o ly m e r

N am es

Polypropylene

Poly(vinyl chloride), PVC

Cl Polyacrylonitrile, Orlon

CN

Poly(tetrafluoroethene), Teflon

F F F

F

F F

Poly(methyl m ethacrylate), Lucite, Plexiglas, Perspex

CO2Me

CO2Me

10.10 Radical Polymerization of Alkenes: Chain-Growth Polymers

Can you suggest an explanation that accounts for the fact that the radical polymerization o f styrene (C6H5CH— CH2) to produce polystyrene occurs in a head-to-tail fashion, R

c 6h 6

“H ead”

Review Problem 10.17

R

'O ,

C6H5

489

CeHs C6H5

“Tail” Polystyrene

rather than the head-to-head manner shown here?

C6H5 R'

CeH5 R

V

CeHs

CeHs “H ead”

“H ead”

Outline a general method for the synthesis o f each o f the follow ing polymers by radical polymerization. Show the monomers that you would use. (a )

/

^

^

\

R e v ie w P ro b le m 1 0 .1 8

(b )

Cl Cl Cl

och3 och3 och3

Cl Cl Cl

Alkenes also polym erize when they are treated with strong acids. The growing chains in acid-catalyzed polymerizations are cations rather than radicals. The follow ing reactions illustrate the cationic polymerization o f isobutylene: Step 1

H — O:

BF3 ^

H — O — BF3

H

Step 2 p

H— O — BF3

— O

3

H

Step 3

Step 4

Y

etc.

The catalysts used for cationic polymerizations are usually Lew is acids that contain a small amount o f water. The polymerization o f isobutylene illustrates how the catalyst (BF3 and H2O) functions to produce growing cationic chains.

A lkenes such as ethene, vinyl chloride, and acrylonitrile do not undergo cationic polymerization very readily. On the other hand, isobutylene undergoes cationic polymerization rapidly. Provide an explanation for this behavior.

R e v ie w P r o b le m s 1 0 .1 9

490

Chapter 10

Radical Reactions

Alkenes containing electron-withdrawing groups polymerize in the presence o f strong bases. Acrylonitrile, for example, polymerizes when it is treated with sodium amide (NaNH2) in liquid ammonia. The growing chains in this polymerization are anions:

NHa

H2N:

h 2n

CN

H2N

CH2= C H C N ^

CN

H2N'

etc.

CN

CN

Anionic polymerization o f acrylonitrile is less important in commercial production than the radical process illustrated in Special Topic B.

10.11 O th e r Im p o rtan t Radical Reactions Radical mechanisms are important in understanding many other organic reactions. We shall see other examples in later chapters, but let us examine a few important radicals and radi­ cal reactions here: oxygen and superoxide, the combustion of alkanes, DNA cleavage, autoxidation, antioxidants, and some reactions o f chlorofluoromethanes that have threatened the protective layer of ozone in the stratosphere.

10.11A Molecular Oxygen and Superoxide One of the most important radicals (and one that we encounter every moment of our lives) is molecular oxygen. Molecular oxygen in the ground state is a diradical with one unpaired electron on each oxygen. As a radical, oxygen can abstract hydrogen atoms just like other radicals we have seen. This is one way oxygen is involved in combustion reactions (Section 10.11C) and autoxidation (Section 10.11D). In biological systems, oxygen is an electron

491

10.11 Other Important Radical Reactions a c c e p t o r . W h e n m o l e c u l a r o x y g e n a c c e p ts o n e e le c t r o n , i t b e c o m e s a r a d i c a l a n io n c a l l e d s u p e r o x id e ( O 2 ~ ) . S u p e r o x i d e is i n v o l v e d i n b o t h p o s i t iv e a n d n e g a t iv e p h y s i o l o g i c a l r o le s : T h e i m m u n e s y s t e m u s e s s u p e r o x id e i n its d e f e n s e a g a in s t p a t h o g e n s , y e t s u p e r o x id e is a ls o s u s p e c te d o f b e i n g i n v o l v e d i n d e g e n e r a t iv e d is e a s e p r o c e s s e s a s s o c ia t e d w i t h a g in g a n d o x i d a t i v e d a m a g e t o h e a l t h y c e l ls . T h e e n z y m e s u p e r o x id e d is m u t a s e r e g u la t e s t h e l e v e l o f s u p e r o x id e b y c a t a l y z i n g c o n v e r s io n o f s u p e r o x id e t o h y d r o g e n p e r o x i d e a n d m o l e c u l a r o x y ­ g e n . H y d r o g e n p e r o x i d e , h o w e v e r , is a ls o h a r m f u l b e c a u s e i t c a n p r o d u c e h y d r o x y l ( H O - ) r a d ic a ls . T h e e n z y m e c a t a la s e h e lp s t o p r e v e n t r e l e a s e o f h y d r o x y l r a d ic a ls b y c o n v e r t i n g h y d r o g e n p e r o x id e to w a te r a n d o x y g e n :

2 O2 - + 2 H +

superoxide dismutase

H2O2 + O2

2 H2O2

2 H2 O + O 2

10.11B Nitric Oxide N i t r i c o x id e , s y n t h e s iz e d i n th e b o d y f r o m t h e a m i n o a c i d a r g in i n e , s e r v e s a s a c h e m i c a l m e s ­ s e n g e r in a v a r i e t y o f b i o lo g ic a l p ro c e s s e s , in c l u d i n g b l o o d p r e s s u re r e g u la t io n a n d t h e im m u n e r e s p o n s e . It s r o l e i n r e l a x a t i o n o f s m o o t h m u s c l e i n v a s c u la r tis s u e s is s h o w n i n F i g . 1 0 .7 .

GMP

/

T h e 1 9 9 8 N o b e l P rize in P h y s io lo g y o r M e d ic in e w a s a w a rd e d t o R. F.

F u rc h g o tt, L. J. Ig n a rro , a n d F. M u ra d f o r t h e ir d is c o v e ry t h a t N O is an im p o r ta n t s ig n a lin g m o le c u le .

Nitric oxide (NO) activates GC GC catalyzes cGMP formation from GMP

'

Nitric oxide (NO)

GC

Figure 10.7 cGMP

N it r ic o x id e (N O )

a c tiv a te s g u a n y la te c y c la s e (G C ), le a d in g t o p r o d u c t io n o f c y c lic g u a n o s in e m o n o p h o s p h a te

cGMP causes smooth muscle relaxation (increasing blood flow)

(c G M P ). c G M P s ig n a ls p r o c e s s e s t h a t c a u s e s m o o t h m u s c le r e la x a tio n , u lt im a t e ly r e s u ltin g in in c r e a s e d b lo o d f l o w t o c e rta in tis s u e s . P h o s p h o d ie s te r a s e V

Hydrolysis of cGMP allows smooth muscle contraction (decreasing blood flow)

(P D E 5 ) d e g r a d e s c G M P , le a d in g t o s m o o t h m u s c le c o n t r a c tio n a n d a r e d u c t io n o f b lo o d flo w . C ia lis , L e v itr a , a n d V ia g r a ta k e

cGMP

t h e ir e f f e c t b y in h ib it in g P D E 5 ,

PDE5 is inhibited by Cialis®, Levitra®, and Viagra®, prolonging smooth muscle relaxation by maintaining cGMP concentrations

th u s m a in ta in in g c o n c e n tr a tio n s o f c G M P a n d s u s ta in in g s m o o th m u s c le r e la x a tio n a n d tis s u e e n g o r g e m e n t . ( R e p r in te d w it h p e r m is s io n fr o m C h r is tia n s o n ,

A cco u n ts o f Chem ical Research, 38, p . 1 9 7 , F ig u r e 6 b , 2 0 0 5 . C o p y r ig h t 2 0 0 5 b y A m e r ic a n

Cialis

C h e m ic a l S o c ie ty .)

10.11C Combustion of Alkanes W h e n a lk a n e s r e a c t w i t h o x y g e n ( e .g . , i n o i l a n d g a s f u r n a c e s a n d i n i n t e r n a l c o m b u s t io n e n g in e s ) a c o m p l e x s e r ie s o f r e a c t io n s ta k e s p la c e , u l t i m a t e l y c o n v e r t in g t h e a lk a n e t o c a r ­ b o n d i o x i d e a n d w a t e r . A l t h o u g h o u r u n d e r s t a n d in g o f t h e d e t a ile d m e c h a n is m o f c o m b u s ­ tio n is in c o m p le t e , w e d o k n o w th a t t h e im p o r t a n t r e a c t io n s o c c u r b y r a d ic a l c h a in m e c h a n is m s w i t h c h a i n - i n i t i a t i n g a n d c h a in - p r o p a g a t in g s te p s s u c h a s t h e f o l l o w i n g r e a c t io n s :

RH

O2

R- + ■OOH

R-

O2

R— OO-

In itia tin g

}

P r o p a g a tin g

R— OO •

R— H

R— OOH +

R-

492

Chapter 10

Radical Reactions

O n e p r o d u c t o f t h e s e c o n d c h a i n - p r o p a g a t i n g s te p is R — O O H , c a l l e d a n a l k y l h y d r o p e r ­ o x i d e . T h e o x y g e n - o x y g e n b o n d o f a n a l k y l h y d r o p e r o x i d e is q u i t e w e a k , a n d i t c a n b r e a k a n d p r o d u c e r a d ic a ls t h a t c a n i n i t i a t e o t h e r c h a in s :

RO— O H ----- > RO ■+ - OH

THE CHEMISTRY OF C a lic h e a m ic in g 1I: A R a d ic a l D e v i c e f o r S lic in g t h e B a c k b o n e o f D N A

The beautiful architecture of calicheamicin y -1 conceals a lethal reactivity. Calicheamicin y - 1binds to the minor groove of DNA where its unusual enediyne (pronounced en di in) moiety reacts to form a highly effective device for slicing the backbone of DNA. A model of calicheamicin bound to DNA is shown below, along with its structural formula. Calicheamicin y -1 and its analogs are of great clinical inter­ est because they are extraordinarily deadly for cancer cells, having been shown to initiate apoptosis (programmed cell death). Indeed, research on calicheamicin has since led to development of the drug Mylotarg, now used to treat some cases of acute mylogenous leukemia. Mylotarg carries two calicheamicin "warheads" on an antibody that delivers it specifically to the cancerous cells. In nature, bacteria called Micromonospora echinospora synthesize calicheamicin y -1 as part of their normal metabolism, presumably as a chem­ ical defense against other organisms. The laboratory syn­ thesis of this complex molecule by the research group of K. C. Nicolaou (Scripps Research Institute, University of California, San Diego), on the other hand, represents a tour

de force achievement in synthetic organic chemistry. Synthesis of calicheamicin and analogs, as well as investi­ gations by many other researchers, has led to fascinating insights about its mechanism of action and biological prop­ erties. The DNA-slicing property of calicheamicin y -1 arises because it acts as a molecular machine for producing car­ bon radicals. A carbon radical is a highly reactive and unsta­ ble intermediate that has an unpaired electron. Once formed, a carbon radical can become a stable molecule again by removing a proton and one electron (i.e., a hydro­ gen atom) from another molecule. In this way, its unpaired electron becomes part of a bonding electron pair. (Other paths to achieve this are possible, too). The molecule that lost the hydrogen atom, however, becomes a new reactive radical intermediate. When the radical weaponry of each calicheamicin y -1 is activated, it removes a hydrogen atom from the backbone of DNA. This leaves the DNA molecule as an unstable radical intermediate which, in turn, results in double-strand cleavage of the DNA and cell death.

O NHCO2Me Me S OMe

O

H Me

q

OH

C a lic h e a m ic in b o u n d t o D N A . (P D B ID : 2 P IK . K u m a r, R. A .; Ik e m o to , N ., a n d P a te l, D . J., S o lu t io n s tr u c tu r e o f th e c a lic h e a m ic in g 1 - D N A c o m p le x ,

J. M ol. Biol. 1 9 9 7 , 265, 1 8 7 .) [C a lic h e a m ic in g -1s tr u c tu r e fr o m C h e m is tr y a n d B io lo g y , 1 9 9 4 , 1 (1 ). N ic o la o u , K .C ., P its in o s , E .N ., T h e o d o r a k is , A ., S a im o to , H ., a n d W r a s id io , W ., C hem istry and B io lo g y o f the Calicheamicins, p p . 2 5 -3 0 . C o p y r ig h t E ls e v ie r 1 9 9 4 .

MeO

Me OH

MeO Calicheam icin

y.,'

10.11 Other Important Radical Reactions

O

CO2Me

O

,NH

S S

>

MeS

CO2Me _NH

------ ^

(1) nucleophilic attack_____

-------- U

(2) conjugate addition

Sugar

493

Bergman cycloaromatization

HO//,

Sugar

S

Nu

Calicheam icin 7 1I

In Problem 10.29 and in "The Chemistry of . . . Calicheamicin g i1 Activation for Cleavage of DNA" box in Chapter 17, we shall revisit calicheamicin g i1to consider the

reactions that remodel its structure into a machine for producing radicals.

10.11D Autoxidation L inoleic acid is an exam ple o f a polyunsatura ted fa tty acid, the k ind o f polyunsaturated acid that occurs as an ester in p o l y u n s a t u r a t e d f a t s (Section 7.13, “T he C hem istry o f . . . H ydrogenation in the F ood Industry,” and C hapter 23). B y polyunsaturated, w e m ean that the com pound contains tw o or m ore double bonds:

„CO2R' Linoleic acid (as an ester) Polyunsaturated fats occur widely in the fats and oils that are com ponents o f our diets. They are also w idespread in the tissues o f the body where they perform num erous vital functions. The hydrogen atom s o f the — CH 2 — group located betw een the tw o double bonds of linoleic ester (Lin— H) are especially susceptible to abstraction by radicals (w e shall see w hy in C hapter 13). A bstraction o f one o f these hydrogen atom s produces a new radical (Lin-) that can react w ith oxygen in a chain reaction that belongs to a general type o f reac­ tion called a u t o x i d a t i o n (Fig. 10.8). T he resu lt o f autoxidation is the form ation o f a hydroperoxide. A utoxidation is a process that occurs in m any substances; for example, autoxidation is responsible for the developm ent o f the rancidity that occurs w hen fats and oils spoil and for the spontaneous com bustion o f oily rags left open to the air. A utoxidation also occurs in the body, and here it m ay cause irreversible dam age.

494

Chapter 10

Radical Reactions

Chain Initia tion

Step 1 ,CO 2R'

,C O 2R' 4

7

RP Figure 10.8 Autoxidation of a linoleic acid ester. In step 1 the reaction is initiated by the attack of a radical on one of the hydrogen atoms of the — CH2— group between the two double bonds; this hydrogen abstraction produces a radical that is a resonance hybrid. In step 2 this radical reacts with oxygen in the first of two chainpropagating steps to produce an oxygen-containing radical, which in step 3 can abstract a hydrogen from another molecule of the linoleic ester (Lin — H). The result of this second chain-propagating step is the formation of a hydroperoxide and a radical (Lin ) that can bring about a repetition of step 2.

,CO2R' 4

7

Chain Propagation

Step 2

■O9 O - \

•O — O: CO2R'

,CO2R' 4

7

4

7

Another radical

Step 3

L h P ^ - O - Os

H — O— O: CO2R'

CO2R 4

7

Hydrogen abstraction from another m olecule of the linoleic ester

4

Lin-

7

A hydroperoxide

THE CHEMISTRY OF . . . A n tio x id a n ts

Autoxidation is inhibited when compounds called antioxi­ dants are present that can rapidly "trap" peroxyl radicals by reacting with them to give stabilized radicals that do not continue the chain. Vitamin E (a-tocopherol) is capable of acting as a radical trap, and one of the important roles that vitamin E plays in

the body may be in inhibiting radical reactions that could cause cell damage. Vitamin C is also an antioxidant, although recent work indicates that supplements over 500 mg per day may have prooxidant effects. Compounds such as BHT are added to foods to prevent autoxidation. BHT is also known to trap radicals.

Vitamin E is found in vegetable oils. m m

Vitamin C is found in citrus fruits.

10.11 Other Important Radical Reactions

495

THE CHEMISTRY OF . . . Ozone Depletion and Chlorofluorocarbons (CFCs) In the stratosphere at altitudes of about 25 km, very highenergy (very short wavelength) UV light converts diatomic oxygen (O2) into ozone (O3). The reactions that take place may be represented as follows:

Typical freons are trichlorofluoromethane, CFCI3 (called Freon-11), and dichlorodifluoromethane, CF2CI2 (called Freon-12). Chain Initiation

Step 1

O2 + hn -----: O + O

Step 2

O + O2 + M -----: O3 + M + heat

Step 1

CF2CI2 + hn-----: CF2CI- + Cl-

Chain P ro p ag ation

where M is some other particle that can absorb some of the energy released in the second step. The ozone produced in step 2 can also interact with highenergy UV light in the following way: Step 3

O3 + hn-----: O2 + O + heat

The oxygen atom formed in step 3 can cause a repetition of step 2, and so forth. The net result of these steps is to convert highly energetic UV light into heat. This is impor­ tant because the existence of this cycle shields Earth from radiation that is destructive to living organisms. This shield makes life possible on Earth's surface. Even a relatively small increase in high-energy UV radiation at Earth's surface would cause a large increase in the incidence of skin cancers. Production of chlorofluoromethanes (and of chlorofluoroethanes) called chlorofluorocarbons (CFCs) or f r e o n s began in 1930. These compounds have been used as refrigerants, solvents, and propellants in aerosol cans.

By 1974 world freon production was about 2 billion pounds annually. Most freon, even that used in refrigeration, eventually makes its way into the atmosphere where it diffuses unchanged into the stratosphere. In June 1974 F S. Rowland and M. J. Molina published an article indicating, fo r the first tim e, that in the stratosphere freon is able to initiate radical chain reactions that can upset the natural ozone balance. The 1995 Nobel Prize in Chemistry was awarded to P. J. Crutzen, M. J. Molina, and F. S. Rowland fo r their combined work in this area. The reactions that take place are the following. (Freon-12 is used as an example.)

Step 2

Cl- + O3 -----: CIO- + O2

Step 3

CIO- + O -----: O2 + CI-

In the chain-initiating step, UV light causes homolytic cleav­ age of one C— CI bond of the freon. The chlorine atom thus produced is the real villain; it can set off a chain reac­ tion that destroys thousands of molecules of ozone before it diffuses out of the stratosphere or reacts with some other substance. In 1975 a study by the National Academy of Sciences supported the predictions of Rowland and Molina, and since January 1978 the use of freons in aerosol cans in the United States has been banned. In 1985 a hole was discovered in the ozone layer above Antarctica. Studies done since then strongly suggest that chlorine atom destruction of the ozone is a factor in the for­ mation of the hole. This ozone hole has continued to grow in size, and such a hole has also been discovered in the Arctic ozone layer. Should the ozone layer be depleted, more of the sun's damaging rays would penetrate to the sur­ face of Earth. Recognizing the global nature of the problem, the "Montreal Protocol" was initiated in 1987. This treaty required the signing nations to reduce their production and consump­ tion of chlorofluorocarbons. Accordingly, the industrialized nations of the world ceased production of chlorofluorocarbons as of 1996, and over 120 nations have signed the Montreal Protocol. Increased worldwide understanding of stratospheric ozone depletion, in general, has accelerated the phasing out of chlorofluorocarbons.

496

Chapter 10

Radical Reactions

Key Terms and Concepts

PLUS

T he key term s and concepts that are highlighted in b o l d , b l u e t e x t w ithin the chapter are defined in the glossary (at the back o f the book) and have hyperlinked definitions in the accom panying W ileyP LU S course (w w w .w ileyplus.com ).

Problems N ote to Instructors: Many of the homework problems are available for assignment via WileyPLUS, an online teaching and learning solution.

RADICAL M E C H A N IS M S A N D PROPERTIES 10.20

W rite a m echanism for the follow ing radical halogenation reaction.

^ 'B r

Br,

hn

10.21

E xplain the relative distribution o f products below using reaction energy diagram s for the hydrogen abstraction step that leads to each product. (The rate-determ ining step in radical halogenation is the hydrogen abstraction step.) In energy diagram s for the tw o pathw ays, show the relative energies o f the transition states and o f the alkyl ra d i­ cal interm ediate that results in each case.

Br,

Br

hn

Br (92%)

10.22

(8%)

W hich o f the follow ing com pounds can be prepared by radical halogenation w ith little com plication by form ation o f isom eric by-products?

Br Cl

10.23

Br

Br

T he radical reaction o f propane w ith chlorine yields (in addition to m ore highly halogenated com pounds) 1-chloropropane and 2-chloropropane.

Cl Cl2

,

Cl

hn

W rite chain-initiating and chain-propagating steps show ing how each o f the products above is form ed. 10.24

In addition to m ore highly chlorinated products, chlorination o f butane yields a m ixture o f com pounds w ith the form ula C 4H9Cl. Cl,

hn

C4H9Cl (multiple isom ers)

Problems

497

(a) Taking stereochemistry into account, how many different isomers with the formula C4H9Cl would you expect? (b) I f the mixture o f C4H9Cl isomers were subjected to fractional distillation (or gas chromatography), how many fractions (or peaks) would you expect? (c) Which fractions would be optically inactive? (d) Which fractions could theoretically be resolved into enantiomers? (e) Predict features in the 1H and 13C DEPT NM R spectra for each that would differentiate among the isomers separated by distillation or GC. (f) How could fragmentation in their mass spectra be used to differentiate the isomers? 10.25

Chlorination o f (Æ)-2-chlorobutane yields a mixture of dichloro isomers. Cl Cl2

hv

C4H8Cl2 (multiple isom ers)

(a) Taking into account stereochemistry, how many different isomers would you expect? Write their structures. (b) How many fractions would be obtained upon fractional distillation? (c) Which o f these fractions would be optically active? 10.26

Peroxides are often used to initiate radical chain reactions such as in the following radical halogenation. ,Br

Br2

HBr

O— O (Di-fert-butyl peroxide)

(a) Using bond dissociation energies in Table 10.1, explain why peroxides are especially effective as radical initiators. (b) Write a mechanism for the reaction above showing how it could be initiated by di-tert-butyl peroxide. 10.27

List in order of decreasing stability all o f the radicals that can be obtained by abstraction of a hydrogen atom from 2-methylbutane.

10.28

The relative stability of a series of primary, secondary, and tertiary alkyl radicals can be compared using R— CH3 car­ bon-carbon bond dissociation energies instead of R— H bond dissociation energies (the method used in Section 10.2B). Bond dissociation energies (DH°) needed to make such a comparison for various R— CH3 species can be calculated from values for the heat of formation (Hf) of radicals R-, CH3-, and the molecule R— CH3 using the following equa­ tion: DH°[R — R'] = Hf[R-] + H f[CH3-] - Hf[R— CH3]. Using the data below, calculate the R— CH3 bond disso­ ciation energies for the examples given, and from your results compare the relative stabilities o f the respective primary, secondary, and tertiary radicals in this series. Chemical Species

H f (Heat of Formation, k J mol 1)

c h 3c h 2c h 2c h 2— c h 3 c h 3c h 2c h ( c h 3)— c h 3

-146.8

(CH3)3C — CH3

-167.9

c h 3c h 2c h 2c h 2-

-153.7 80.9

c h 3c h 2c h ( c h 3)-

69

(CH3)3CCH3-

48 147

498 10.29

Chapter 10

Radical Reactions

Draw mechanism arrows to show electron movements in the Bergman cycloaromatization reaction that leads to the diradical believed responsible for the DNA-cleaving action o f the antitumor agent calicheamicin (see “ The Chemistry o f . . . Calicheamicin” in Section 10.11C).

DNA diradical

'

10.30

10.31

DNA double- ^ strand cleavage

Find examples o f C— H bond dissociation energies in Table 10.1 that are as closely related as possible to the bonds to Ha, Hb, and Hc in the molecule at right. Use these values to answer the questions below. (a )

What can you conclude about the relative ease o f radical halogenation at Ha?

(b )

Comparing Hb and Hc, which would more readily undergo substitution by radical halogenation?

Write a radical chain mechanism for the following reaction (a reaction called the Hunsdiecker reaction). O

SYNTHESIS 10.32

Starting with the compound or compounds indicated in each part and using any other needed reagents, outline syn­ theses o f each o f the following compounds. (You need not repeat steps carried out in earlier parts of this problem.) (a) CH3CH3

'!

(b) CH3CH3

"O

(e )

CH4

HC

CH OH

(f) CH3CH3 +

HC =

CH

(c) (g) CH3CH3

N3

(d) Br 10.33

Provide the reagents necessary for the following synthetic transformations. More than one step may be required. ((a) )

'

—

OH

Br Br (e )

(b)

(c)

(d)

>

X

/'''A \ / ' ' ' B r

------*•

X

„OCHc

(f)„OH

O

499

Challenge Problems

Challenge Problems 10.34

The following reaction is the first step in the industrial synthesis o f acetone and phenol (C6H5OH). AIBN (2,2 '-azobisisobutyronitrile) initiates radical reactions by breaking down to form two isobutyronitrile radicals and nitrogen gas. Using an isobutyronitrile radical to initiate the reaction, write a mechanism for the following process. .OH O2, AIBN

A IB N

10.35

In the radical chlorination o f 2,2-dimethylhexane, chlorine substitution occurs much more rapidly at C5 than it does at a typical secondary carbon (e.g., C2 in butane). Consider the mechanism o f radical polymerization and then sug­ gest an explanation for the enhanced rate of substitution at C5 in 2,2-dimethylhexane.

10.36

Write a mechanism for the following reaction. H (PhCO2)2, heat

+

O 10.37

CO

Hydrogen peroxide and ferrous sulfate react to produce hydroxyl radical (HO-), as reported in 1894 by English chemist H. J. H. Fenton. When ieri-butyl alcohol is treated with HO- generated this way, it affords a crystalline reaction product X, mp 92°, which has these spectral properties: MS: heaviest mass peak is at m/z 131 IR : 3620, 3350 (broad), 2980, 2940, 1385, 1370 cm-1 1H N M R : sharp singlets at 8 1.22, 1.58, and 2.95 (6:2:1 area ratio) 13C N M R : 8 28 (CH3), 35 (CH2), 68 (C) Draw the structure o f X and write a mechanism for its formation.

10.38

The halogen atom of an alkyl halide can be replaced by the hydrogen atom bonded to tin in tributyltin hydride (Bu3SnH). The process, called dehalogenation, is a radical reaction, and it can be initiated by AIBN (2,2'-azobisisobutyronitrile). A IB N decomposes to form nitrogen gas and two isobutyronitrile radicals, which in i­ tiate the reaction. Write a mechanism for the reaction. I +

Bu3SnH

AIBN

A IB N

10.39

Write a mechanism that accounts for the following reaction. Note that the hydrogen atom bonded to tin in tributyltin hydride is readily transferred in radical mechanisms.

,Br

(PhCO2)2, Bu3SnH

+ ( M a jo r )

10.40

Molecular orbital calculations can be used to model the location o f electron density from unpaired electrons in a radical. Open the molecular models on the book’s website for the methyl, ethyl, and ieri-butyl radicals. The gray wire mesh surfaces in these models represent volumes enclosing electron density from unpaired electrons. What do you notice about the distribution of unpaired electron density in the ethyl radical and ieri-butyl radi­ cal, as compared to the methyl radical? What bearing does this have on the relative stabilities o f the radicals in this series?

500 10.41

Chapter 10

Radical Reactions

I f one were to try to draw the simplest Lewis structure for molecular oxygen, the result might be the following ( 0 = 0 iJ.However, it is known from the properties o f molecular oxygen and experiments that O2 contains two unpaired electrons, and therefore, the Lewis structure above is incorrect. To understand the structure o f O2, it is necessary to employ a molecular orbital representation. To do so, we w ill need to recall (1) the shapes of bonding and antibonding s and p molecular orbitals, (2) that each orbital can contain a maximum of two electrons, (3) that molecular oxygen has 16 electrons in total, and (4) that the two unpaired electrons in oxygen occupy separate degen­ erate (equal-energy) orbitals. Now, open the molecular model on the book’s website for oxygen and examine its molecular orbitals in sequence from the HOMO-7 orbital to the LUMO. [HOMO-7 means the seventh orbital in energy below the highest occupied molecular orbital (HOMO), HOMO-6 means the sixth below the HOMO, and so forth.] Orbitals HOMO-7 through HOMO-4 represent the s1s, s1s*, s2s, and s2s* orbitals, respectively, each containing a pair o f electrons. (a )

What type o f orbital is represented by HOMO-3 and HOMO-2? [H int: What types of orbitals are possible for second-row elements like oxygen, and which orbitals have already been used?]

(b )

What type o f orbital is HOMO-1? [Hint: The s2s and s2s* orbitals are already filled, as are the HOMO-3 and HOMO-2 orbitals identified in part (b). What bonding orbital remains?]

(c )

The orbitals designated HOMO and LUM O in O2 have the same energy (they are degenerate), and each con­ tains one o f the unpaired electrons o f the oxygen molecule. What type o f orbitals are these?

Learning Group Problems 1

.

(a )

Draw structures for all organic products that would result when an excess o f cis-1,3-dimethylcyclohexane reacts with Br2 in the presence o f heat and light. Use three-dimensional formulas to show stereochemistry.

(b )

Draw structures for all organic products that would result when an excess o f cis-1,3-dimethylcyclohexane reacts with Cl2 in the presence o f heat and light. Use three-dimensional formulas to show stereochemistry.

(c )

As an alternative, use cis-1,2-dimethylcyclohexane to answer parts (a) and (b) above.

(a )

Propose a synthesis o f 2-methoxypropene starting with propane and methane as the sole source for carbon atoms. You may use any other reagents necessary. Devise a retrosynthetic analysis first.

(b )

2-Methoxypropene w ill form a polymer when treated with a radical initiator. Write the structure o f this polymer and a mechanism for the polymerization reaction assuming a radical mechanism initiated by a diacyl peroxide.

CONCEPT MAP

Mechanism Review of Radical Reactions Radical H alogénation of A lkanes heat o r light ---------------- » X- + X-

X -X If X = Br, abstraction is If X = Cl, abstraction is not

hydrogen selective. hydrogen selective.

- C —H

Chain initiation

Radical Polym erization

Chain propagation

O / X —X 1 H — X + — C------------> — C - X

\

O

^ ^ R - C —O - O — C - R + X-

heat ------- > R

/ c=c / \

R- + 2 CO 2

\

I

Coupling The s u b s titu tio n p ro d u ct

Chain initiation

I / R-C-CI \ \ /

- C - X

I S o m e p o s sib le chain-term inating s t e p s

/

c=c\

Anti-Markovnikov Addition o f HBr to A lkenes

I RO—O R -^ » RO + H -

I

Chain initiation

B r ----- > ROH + Br-

Addition of the \ / brom ine radical C= C to the alkene / \ occurs so as to form R the more stable carbon radical interm ediate (The alkene reactant shown is meant to / indicate any alkene R where a difference exists in the extent Br-, of alkyl substitution at the initial alkene I carbons.) R - C - C —

\

I

I Br

\

/

I

C-

I

/

\ Chain propagation

Chain propagation

I

I

R - C — C-

I

I

HBr

I

I

R - C - C - -

I

I

I

Br

H

Br

Br-

I Coupling The anti-M arkovnikov 4 a d d itio n p ro d u ct

—c - c - c —c — I

I

I

I I I I

Br

I

n C= C / \

•c—c —

Br

I

R - C - C — C-

RO- + RO-

R

R

Br

S o m e p o s sib le chain-term inating s t e p s

D is p ro p o rtio n a tio n ^ /

I

c - c cI —c-/ I

I

I

I

^C oupling

P o ssib le chain-term inating s te p s

I

\

I V Alcohols and Ethers Synthesis and Reactions

The flavors and scents of nature include many examples of alcohols and ethers. Menthol, found in pepperm int oil, is an alcohol used both for flavoring and for medicinal purposes. Vanillin, isolated from vanilla beans, con­ tains an ether functional group, as does anethole, the licorice flavor associated with fennel. Ethanol, the alco­ hol produced by fermentation, is, of course, another flavor of nature. Borneol, which can be isolated from artemesia, is an alcohol with a fascinating molecular architecture. And eucalyptol, which shares the ending of its name with other alcohols but is actually an ether, comes from eucalyptus leaves (shown in the left photo above) and is used as a flavoring, scent, and medicinal agent. Nature is an abundant source of alcohols and ethers, and we study the chemistry of these important functional groups in this chapter. O

OH

OCH 3

H3 CO. H HO (-)-Menthol (from peppermint)

Vanillin (from vanilla beans)

Anethole (from fennel)

O CH 3 CH2OH

OH Ethanol (from fermentation)

502

Borneol (from artemesia)

Eucalyptol (from eucalyptus)

11.1 Structure and Nomenclature

ß -? -H

503

11.1 Structure and Nomenclature A lcohols have a hydroxyl (— OH) group bonded to a saturated carbon atom . T he alcohol carbon atom m ay be p art o f a sim ple alkyl group, as in som e o f the follow ing exam ples, or it m ay be p art o f a m ore com plex m olecule, such as cholesterol.

c h 3o h

OH

"OH

Methanol (methyl alcohol)

Ethanol (ethyl alcohol), a 1° a lc o h o l

OH

2-Propanol (isopropyl alcohol),

2-Methyl-2-propanol (fecf-butyl alcohol),

a 2 ° a lc o h o l

a 3 ° a lc o h o l

The alcohol carbon atom m ay also be a saturated carbon atom o f an alkenyl or alkynyl group, or the carbon atom m ay be a saturated carbon atom that is attached to a benzene ring:

= 2 -Propenol (or prop-2 -en-1 -ol, or allyl alcohol),

OH

OH

„OH /

2-Propynol (or prop-2 -yn-1 -ol, or propargyl alcohol)

Benzyl alcohol (a b e n z y lic a lc o h o l)

a n a lly lic a lc o h o l

C om pounds that have a hydroxyl group attached d ire ctly to a benzene ring are called p h en o ls. (Phenols are discussed in detail in C hapter 21.)

r s

OH

Phenol

H3 C

j T \

OH

p -Methylphenol, a s u b s t it u te d p h e n o l

Ar— OH General formula for a phenol

E thers differ from alcohols in that the oxygen atom o f an ether is bonded to tw o carbon atom s. The hydrocarbon groups m ay be alkyl, alkenyl, vinyl, alkynyl, or aryl. Several exam ­ ples are show n here:

r s

.O C H 3 /

-"O

^

Diethyl ether

'o c h 3 Allyl methyl ether

ferf-Butyl methyl ether

¿ ^ o ^ Divinyl ether

OCH 3

Methyl phenyl ether

11.1A Nomenclature of Alcohols We studied the IUPAC system o f nom enclature for alcohols in Sections 2.6 and 4.3F. A s a review consider the follow ing problem .

504

Chapter 11

Alcohols and Ethers

Solved Problem 11.1 Give IUPAC substitutive names for the following alcohols: (a)

OH

(b)

(c) OH

CeHs

OH

ANSWER The longest chain to which the hydroxyl group is attached gives us the base name. The ending is -ol. We then number the longest chain from the end that gives the carbon bearing the hydroxyl group the lower num­ ber. Thus, the names, in both of the accepted IUPAC formats, are (a )

5

3

3

(b)

5

3

(c )

5

OH OH 2,4-Dimethyl-1-pentanol (or 2,4-dimethylpentan-1-ol)

C6 H 5

OH

4-Phenyl-2-pentanol (or 4-phenylpentan-2-ol)

4-Penten-2-ol (or pent-4-en-2-ol)

The hydroxyl group has precedence over double bonds and triple bonds in deciding which functional group to name as the suffix [see example (c) above]. In common functional class nomenclature (Section 2.6) alcohols are called a l k y l a l c o ­ h o l s such as methyl alcohol, ethyl alcohol, and so on.

R eview P roblem 11.1

What is wrong with the use of such names as “isopropanol” and “tert-butanol”?

11.1B Nomenclature of Ethers Simple ethers are frequently given common functional class names. One simply lists (in alpha­ betical order) both groups that are attached to the oxygen atom and adds the word ether:

OCeHa Ethyl methyl ether

Diethyl ether

ferf-Butyl phenyl ether

IUPAC substitutive names should be used for complicated ethers, however, and for com­ pounds with more than one ether linkage. In this IUPAC style, ethers are named as alkoxyalkanes, alkoxyalkenes, and alkoxyarenes. The RO — group is an a l k o x y group.

OCH 3

2-Methoxypentane 1-Ethoxy-4-methylbenzene 1,2-Dimethoxyethane (DME) Cyclic ethers can be named in several ways. One simple way is to use replacement nomenclature, in which we relate the cyclic ether to the corresponding hydrocarbon ring system and use the prefix oxa- to indicate that an oxygen atom replaces a CH 2 group. In another system, a cyclic three-membered ether is named oxirane and a four-membered ether is called oxetane. Several simple cyclic ethers also have common names; in the examples below, these common names are given in parentheses. Tetrahydrofuran (THF) and 1,4dioxane are useful solvents:

11.2 Physical Properties o f Alcohols and Ethers

R - ó - h/

h

- O - Ií '

505

O O

O

□ Oxacyclopropane or oxirane (ethylene oxide)

Oxacyclobutane or oxetane

O Oxacyclopentane (tetrahydrofuran or THF)

O.

O' 1,4-Dioxacyclohexane (1,4-dioxane)

of t

Polyethylene oxide (PEO) (a water-soluble polymer made from ethylene oxide)

Polyethylene o xid e is used in some skin creams.

Ethylene oxide is the starting material for polyethylene oxide (PEO, also called poly­ ethylene glycol, PEG). Polyethylene oxide has many practical uses, including covalent attachment to therapeutic proteins such as interferon, a use that has been found to increase the circulatory lifetime of the drug. PEO is also used in some skin creams, and as a laxa­ tive prior to digestive tract procedures.

S o lv e d P r o b le m

1 1 .2

Albuterol (used in some commonly prescribed respiratory medications) and vanillin (from vanilla beans) each contain several functional groups. Name the functional groups in albuterol and vanillin and, if appropriate for a given group, classify them as primary (1°), secondary (2°), or tertiary (3°). OH

O H3 CO.

HO'

H HO Albuterol (an asthma medication)

HO Vanillin (from vanilla beans)

A lb u te ro l is used in some resp ira to ry m edications.

STRATEGY A ND ANSWER Albuterol has the following functional groups: 1° alcohol, 2° alcohol, phenol, and 2 ° amine. Vanillin has aldehyde, ether, and phenol functional groups. See Chapter 2 for a review of how to classify alcohol and amine functional groups as 1°, 2°, or 3°.

Give bond-line formulas and appropriate names for all of the alcohols and ethers with the formulas (a) C 3 H8O and (b) C 4 H 1 0 O.

R eview P roblem 11.2

11.2 Physical Properties o f Alcohols and Ethers The physical properties of a number of alcohols and ethers are given in Tables 11.1 and 11.2. •

Ethers have boiling points that are roughly comparable with those of hydrocarbons of the same molecular weight (M W ).

For example, the boiling point of diethyl ether (M W = 74) is 34.6°C; that of pentane (M W = 72) is 36°C.

506

Chapter 11 •

Alcohols and Ethers

A lcohols have m uch higher boiling points than com parable ethers or hydrocarbons.

T he boiling point o f butyl alcohol (M W = 74) is 117.7°C. We learned the reason for this behavior in Section 2.13C. •

Propylene glycol (1,2-propanediol) is used as an environm entally frie n d ly engine coolant because it is bio d e gra d a ble, has a high b o ilin g p o in t, and is m iscible w ith water.

A lcohol m olecules can associate w ith each other through w hereas those o f ethers an d hydrocarbons cannot.

h y d r o g e n b o n d in g ,

Ethers, however, are able to form hydrogen bonds w ith com pounds such as water. Ethers, therefore, have solubilities in w ater that are sim ilar to those o f alcohols o f the sam e m o le­ cular w eight and that are very different from those o f hydrocarbons. D iethyl ether and 1-butanol, for example, have the sam e solubility in water, approxim ately 8 g per 100 m L at room tem perature. Pentane, by contrast, is virtually insoluble in water. M ethanol, ethanol, both propyl alcohols, and ieri-butyl alcohol are com pletely m iscible w ith w ater (Table 11.1). T he rem aining butyl alcohols have solubilities in w ater betw een 8.3 and 26.0 g per 100 m L. T he solubility o f alcohols in w ater gradually decreases as the hydrocarbon portion o f the m olecule lengthens; long-chain alcohols are m ore “alkane-like” and are, therefore, less like water.

Physical Properties o f Alcohols

N am e

F o r m u la

mp

b p (°C )

W a te r S o lu b ility

(°C )

(1 a t m )

( g / 1 0 0 m L H 2O )

-9 7 -1 1 7 -1 2 6 -8 8 -9 0 -1 0 8 -1 1 4 25 - 7 8 .5 -5 2 -3 4 -1 5

64.7 78.3 97.2 82.3 117.7 108.0 99.5 82.5 138.0 156.5 176 195

M o n o h y d r o x y A lc o h o ls

M ethanol Ethanol Propyl alcohol Isopropyl alcohol Butyl alcohol Isobutyl alcohol sec-Butyl alcohol tert-Butyl alcohol Pentyl alcohol Hexyl alcohol H eptyl alcohol Octyl alcohol Cyclopentanol

c h 3o h c h 3c h 2o h c h 3c h 2c h 2o h

CH3CH(OH)CH3 c h 3c h 2c h 2c h 2o h CH3CH(CH3)CH2OH CH3CH2CH(OH)CH3 (CH3)3COH c h 3(c h 2)3c h 2o h c h 3(c h 2)4c h 2o h c h 3(c h 2)5c h 2o h c h 3(c h 2)6c h 2o h

0 0H

-1 9

œ œ œ

œ 8.3 10.0 26.0 œ 2.4 0.6 0.2 0.05

140

~

Cyclohexanol

O oh

Benzyl alcohol

c 6h 5c h 2o h

24

161.5

3.6

-1 5

205

4

- 1 2 .6 -5 9 -3 0 18

197 187 215 290

œ œ œ œ

D io ls a n d T r io ls

Ethylene glycol P ropylene glycol Trim ethylene glycol Glycerol

c h 2o h c h 2o h c h 3c h o h c h 2o h c h 2o h c h 2c h 2o h c h 2o h c h o h c h 2o h

R -o -k/lt-Ö -tt.'

11.3 Im portant Alcohols and Ethers

507

Physical Properties o f Ethers Name

mp (°C)

Formula

Dimethyl eth er Ethyl methyl eth er Diethyl eth er Dipropyl eth er Diisopropyl eth er Dibutyl eth er 1,2-D im ethoxyethane (DME)

c h 3o c h 3

-1 3 8

c h 3o c h 2c h 3 c h 3c h 2o c h 2c h 3

-1 1 6 -1 2 2 -8 6 - 9 7 .9 - 68

(CH3CH2CH2)20 (CH3)2CH 0C H (CH 3)2 (CH3CH2CH2CH2)20 c h 3o c h 2c h 2o c h 3

/\ Q 0\Zv° O

O xirane Tetrahydrofuran (THF)

1,4-Dioxane

bp (°C) atm)

(1

- 2 4 .9 10.8 34.6 90.5 68 141 83

-1 1 2

12

-1 0 8

65.4

11

101

I

S o lv e d P r o b le m

1 1 .3

1

1 .2 - P r o p a n e d io l ( p r o p y le n e g ly c o l ) a n d 1 ,3 - p r o p a n e d io l ( t r im e t h y le n e g ly c o l) h a v e h ig h e r b o i lin g p o in t s th a n a n y o f t h e b u t y l a l c o h o l s ( s e e T a b le 1 1 . 1 ) , e v e n t h o u g h t h e y a l l h a v e r o u g h l y t h e s a m e m o le c u l a r w e i g h t . P r o p o s e a n e x p la n a t io n .

STRATEGY AND ANSWER

T h e p r e s e n c e o f t w o h y d r o x y l g r o u p s i n e a c h o f th e d io ls a llo w s t h e ir m o le c u le s to

f o r m m o r e h y d r o g e n b o n d s th a n th e b u t y l a lc o h o ls . G r e a te r h y d r o g e n - b o n d f o r m a t io n m e :a n s th a t th e m o le c u le s o f 1 . 2 - p r o p a n e d i o l a n d 1 , 3 - p r o p a n e d i o l a r e m o r e h i g h l y a s s o c i a t e d a n d , c o n s e q u e n t ly , t h e i r b o i l i n g p o i n t s a r e h ig h e r .

11.3 Im p o rtan t Alcohols and Ethers 11.3A Methanol A t o n e tim e , m o s t m e th a n o l w a s p r o d u c e d b y th e d e s t r u c tiv e d is t ill a t io n o f w o o d ( i.e ., h e a t­ in g w o o d t o a h ig h te m p e ra tu r e in th e a b s e n c e o f a ir ) . I t w a s b e c a u s e o f th is m e th o d o f p r e p a r a t io n t h a t m e t h a n o l c a m e t o b e c a lle d “ w o o d a lc o h o l. ” T o d a y , m o s t m e t h a n o l is p r e ­ p a r e d b y th e c a t a ly t ic h y d r o g e n a t io n o f c a r b o n m o n o x id e . T h is r e a c t io n ta k e s p la c e u n d e r h ig h p re s s u re a n d a t a te m p e ra tu r e o f 3 0 0 - 4 0 0 ° C :

CO

+

, 2 H2 2

300-400°C

-------------- :

20 0 -3 0 0 atm Z n O -C r2O3

CH3OH 3

M e t h a n o l is h ig h l y t o x ic . I n g e s t io n o f e v e n s m a ll q u a n titie s o f m e th a n o l c a n c a u s e b l i n d ­ n e s s ; l a r g e q u a n t i t i e s c a u s e d e a t h . M e t h a n o l p o i s o n i n g c a n a ls o o c c u r b y i n h a l a t i o n o f t h e v a p o r s o r b y p r o lo n g e d e x p o s u r e t o th e s k in .

11.3B Ethanol E t h a n o l c a n b e m a d e b y th e f e r m e n t a t io n o f s u g a r s , a n d i t is th e a lc o h o l o f a ll a lc o h o lic b e v e r a g e s . T h e s y n t h e s is o f e t h a n o l i n t h e f o r m

o f w in e b y th e fe r m e n t a tio n o f th e s u g ­

a r s o f f r u i t j u i c e s w a s p r o b a b l y o u r f i r s t a c c o m p l i s h m e n t i n t h e f i e l d o f o r g a n i c s y n t h e s is .

508

Chapter 11

Alcohols and Ethers

Sugars from a wide variety of sources can be used in the preparation of alcoholic bever­ ages. Often, these sugars are from grains, and it is this derivation that accounts for ethanol having the synonym “grain alcohol.” Fermentation is usually carried out by adding yeast to a mixture of sugars and water. Yeast contains enzymes that promote a long series of reactions that ultimately convert a simple sugar (C 6 H 1 2 O6) to ethanol and carbon dioxide: c 6 H1 2 o 6

yeast

2 CH 3 CH2OH

2 CO 2

(~95% yield) Fermentation alone does not produce beverages with an ethanol content greater than 12-15% because the enzymes of the yeast are deactivated at higher concentrations. To pro­ duce beverages of higher alcohol content, the aqueous solution must be distilled. Ethanol is an important industrial chemical. Most ethanol for industrial purposes is pro­ duced by the acid-catalyzed hydration of ethene: Vineyard grapes fo r use in fe rm e n tatio n .

H2O

acid

OH

About 5% of the world’s ethanol supply is produced this way. Ethanol is a hypnotic (sleep producer). It depresses activity in the upper brain even though it gives the illusion of being a stimulant. Ethanol is also toxic, but it is much less toxic than methanol. In rats the lethal dose of ethanol is 13.7 g kg - 1 of body weight.

THE CHEMISTRY OF. . . E t h a n o l a s a B io fu e l

Ethanol is said to be a renewable energy source because it can be made by fermentation of grains and other agricul­ tural sources such as switchgrass and sugarcane. The crops themselves grow, of course, by converting light energy from the sun to chemical energy through photosynthesis. Once obtained, the ethanol can be combined with gasoline in varying proportions and used in internal combustion engines. During the year 2007, the United States led the world in ethanol production with 6.5 billion U.S. gallons, fol­ lowed closely by Brazil with 5 billion gallons. When used as a replacement for gasoline, ethanol has a lower energy content, by about 34% per unit volume. This, and other factors, such as costs in energy required to pro­ duce the agricultural feedstock, especially corn, have cre­ ated doubts about the wisdom of an ethanol-based program as a renewable energy source. Production of ethanol from corn is 5 to 6 times less efficient than producing it from sug-

arcane, and it also diverts production of a food crop into an energy source. World food shortages may be a result.

11.3C Ethylene and Propylene Glycols Ethylene glycol (HOCH 2 CH 2 OH) has a low molecular weight, a high boiling point, and is miscible with water. These properties made ethylene glycol a good automobile antifreeze. Unfortunately, however, ethylene glycol is toxic. Propylene glycol (1,2propanediol) is now widely used as a low-toxicity, environmentally friendly alternative to ethylene glycol.

11.4 Synthesis of Alcohols from Alkenes

R - o-

h

/n -O -ft!

509

11.3D D ie th y l E th e r Diethyl ether is a very low boiling, highly flammable liquid. Care should always be taken when diethyl ether is used in the laboratory, because open flames or sparks from light switches can cause explosive combustion of mixtures of diethyl ether and air. Most ethers react slowly with oxygen by a radical process called a u t o x i d a t i o n (see Section 10.11D) to form hydroperoxides and peroxides: H

Step 1

•O— O

Step 2

•/ -C

OR'

— C— OR'

• O — O— H

\

OO-

OR' O0

\

- C — OR' H

OO-

Step 3a

Hydrogen abstraction adjacent to the ether oxy­ gen occurs readily.

./ —C

— C— OR'

OOH

OR'

— C— OR'

— C — OR'

—C

A hydroperoxide or OO'

Step 3b

OR'

R'O— C— OO— C— OR'

— C— OR'

Hydroperoxides and per­ oxides can be explosive.

A peroxide

These hydroperoxides and peroxides, which often accumulate in ethers that have been stored for months or longer in contact with air (the air in the top of the bottle is enough), are dangerously explosive. They often detonate without warning when ether solutions are distilled to near dryness. Since ethers are used frequently in extractions, one should take care to test for and decompose any peroxides present in the ether before a distillation is carried out. (Consult a laboratory manual for instructions.) Diethyl ether was at one time used as a surgical anesthetic.The most popular modern anesthetic is halothane (CF 3 CHBrCl). Unlike diethyl ether, halothane is not flammable.

11.4 Synthesis o f Alcohols from Alkenes We have already studied the acid-catalyzed h y d r a t i o n o f a lk e n e s , o x y m e r c u r a t i o n - d e m e r c u r a t i o n , and h y d r o b o r a t i o n - o x i d a t i o n as methods for the synthesis of alcohols from alkenes (see Sections 8.5, 8 .6 , and 8.7, respectively). Below, we briefly summarize these methods. Alkenes add water in the presence of an acid catalyst to yield alcohols (Section 8.5). The addition takes place with M a r k o v n i k o v r e g i o s e l e c t i v i t y . The reaction is reversible, and the mechanism for the acid-catalyzed hydration of an alkene is simply the reverse of that for the dehydra­ tion of an alcohol (Section 7.7).

1 . A c id - C a t a ly z e d H y d r a t io n o f A lk e n e s

\ /

C=C

/ \

HA

/ — C— C

A-

+ h 2o h 2o

— C------- C— H H

Alkene

/O k H

A-

— C— C—

+

H :O„ H Alcohol

HA

510

Chapter 11

Alcohols and Ethers

Acid-catalyzed hydration of alkenes has limited synthetic utility, however, because the carbocation intermediate may rearrange if a more stable or isoenergetic carboca­ tion is possible by hydride or alkanide migration. Thus, a mixture of isomeric alco­ hol products may result. 2. Oxymercuration-Demercuration Alkenes react with mercuric acetate in a mixture of water and tetrahydrofuran (THF) to produce (hydroxyalkyl)mercury compounds. These can be reduced to alcohols with sodium borohydride and water (Section 8 .6 ). Oxymercuration Mercury compounds are hazardous. Before you carry out a reaction involving mercury or its compounds, you should familiarize yourself w ith current procedures fo r its use and disposal.

\

/

Hg(OAc)2

C=C /

h 2o

— C— C—

THF

\

HO

AcOH

HgOAc

Demercuration NaBH4, OH

— C — C— HO

______ H e lp f u l H i n t O xymercuration-demercuration and hydroboration-oxidation have complementary regioselectivity.

— C— C—

HgOAc

HO

Hg

AcO-

H

In the oxymercuration step, water and mercuric acetate add to the double bond; in the demercuration step, sodium borohydride reduces the acetoxymercury group and replaces it with hydrogen. The net addition of H — and — OH takes place with Markovnikov regioselectivity and generally takes place without the complica­ tion of rearrangements, as sometimes occurs with acid-catalyzed hydration of alkenes. The overall alkene hydration is not stereoselective because even though the oxymercuration step occurs with anti addition, the demercuration step is not stere­ oselective (radicals are thought to be involved), and hence a mixture of syn and anti products results. 3 Hydroboration-Oxidation An alkene reacts with BH3:THF or diborane to produce an alkylborane. Oxidation and hydrolysis of the alkylborane with hydrogen perox­ ide and base yield an alcohol (Section 8.7). Hydroboration B H 3 THF

3

enantiomer

Anti-Markovnikov and syn addition

2 -methylcyclopentyl Oxidation h 2o 2, h o

— OH replaces boron with retention of configuration

ch3 3

enantiomer

OH

In the first step, boron and hydrogen undergo syn addition to the alkene; in the second step, treatment with hydrogen peroxide and base replaces the boron with — OH with retention of configuration. The net addition of — H and — OH occurs with antiMarkovnikov regioselectivity and syn stereoselectivity. Hydroboration-oxidation, there­ fore, serves as a useful regiochemical complement to oxymercuration-demercuration.

11.5 Reactions o f Alcohols

R - ö - h/

h

- Ö - K .'

S o lv e d P r o b le m

511 1 1 .4

What conditions would you use for each reaction?

(a )

HO

(b)

HO.

STRATEGY AND ANSWER We recognize that synthesis by path (a) would require a Markovnikov addition of water to the alkene. So, we could use either acid-catalyzed hydration or oxymercuration-demercuration. H3O+/H2O or (1) Hg(OAc)2/H2O (2) NaBH4

Markovnikov addition of H— and —OH HO

Synthesis by path (b) requires an anti-Markovnikov addition, so we would choose hydroboration-oxidation. (1) BH3:THF (2) H2O2, OH“

’

anti-Markovnikov addition of H— and —OH

HO.

R e v ie w P ro b le m 1 1 .3

Predict the major products of the following reactions (a )

(1) BH3:THF

(b )

(2) H2O2, NaOH (1) Hg(OAc)2, H2O/THF

(c )

(2) NaBH4

The following reaction does not produce the product shown.

R e v ie w P ro b le m 1 1 .4

cat. H2SO4 h 2o

OH (a) Predict the major product from the conditions shown above, and write a detailed mech­ anism for its formation. (b) What reaction conditions would you use to successfully synthesize the product shown above (3,3-dimethyl-2-butanol).

11.5 Reactions o f Alcohols The reactions of alcohols have mainly to do with the following: •

The oxygen atom of the hydroxyl group is nucleophilic and weakly basic.

•

The hydrogen atom of the hydroxyl group is weakly acidic.

•

The hydroxyl group can be converted to a leaving group so as to allow substitution or elimination reactions.

512

Chapter 11

Alcohols and Ethers

Our understanding of the reactions of alcohols will be aided by an initial examination of the electron distribution in the alcohol functional group and of how this distribution affects its reactivity. The oxygen atom of an alcohol polarizes both the C — O bond and the O — H bond of an alcohol:

5-c-

jo

;

H «+ The C—O and O—H bonds of an alcohol are polarized

An electrostatic potential map for methanol shows partial negative charge at the oxygen and partial positive charge at the hydroxyl proton.

Polarization of the O — H bond makes the hydrogen partially positive and explains why alco­ hols are weak acids (Section 11.6). Polarization of the C — O bond makes the carbon atom partially positive, and if it were not for the fact that OH- is a strong base and, therefore, a very poor leaving group, this carbon would be susceptible to nucleophilic attack. The electron pairs on the oxygen atom make it both basic and nucleophilic . In the pres­ ence of strong acids, alcohols act as bases and accept protons in the following way: H Protonation of an alcohol.

•

l+

C— O — H + H—A I " Alcohol Strong acid

- C — O— H I " Protonated alcohol

A-

Protonation of the alcohol converts a poor leaving group (OH ) into a good one (H 2 O).

Protonation also makes the carbon atom even more positive (because — OH2+ is more electron withdrawing than — OH) and, therefore, even more susceptible to nucleophilic attack. •

Once the alcohol is protonated substitution reactions become possible (SN2 or SN1, depending on the class of alcohol, Section 11.8). H The protonated hydroxyl group is a good leaving group (H2 O).

u Nu =

H

C— O— H

s n2

'

5 Nu— C

=O — H

Protonated alcohol

Because alcohols are nucleophiles, they, too, can react with protonated alcohols. This, as we shall see in Section 11.11A, is an important step in one synthesis of ethers: H R— O =

H R— O ± C— I I H Protonated ether

+

=O— H "

At a high enough temperature and in the absence of a good nucleophile, protonated alco­ hols are capable of undergoing E1 or E2 reactions. This is what happens in alcohol dehy­ drations (Section 7.7). Alcohols also react with PBr3 and SOCl2 to yield alkyl bromides and alkyl chlorides. These reactions, as we shall see in Section 11.9, are initiated by the alcohol using its unshared electron pairs to act as a nucleophile.

11.6 Alcohols as Acids

R - o- h / n - O - ft !

513

11.6 Alcohols as Acids • Alcohols have acidities similar to that of water. Methanol is a slightly stronger acid than water (pA"a = 15.7) but most alcohols are some­ what weaker acids. Values of p K a for several alcohols are listed in Table 11.3. H R— O — H

H

:O — H

H— O— H

R— O =-

Alcohol

Alkoxide ion

[ TABLE 11.3

p K ; Values fo r Some W e a k Acids

Acid c h 3o h h 2o

( If R is bulky, there is le s s sta b iliz a tio n o f the a lko xid e b y solvation, a n d c o n s e q u e n tly the e q u ilib riu m lie s even fu rth e r to w a rd the a lco h o l.)

c h 3c h 2o h

(CH3)3COH

pKa 15.5 15.74 15.9 18.0

• Sterically hindered alcohols such as tert-butyl alcohol are less acidic, and hence their conjugate bases more basic, than unhindered alcohols such as ethanol or methanol. One reason for this difference in acidity has to do with the effect of solvation. With an unhin­ dered alcohol, water molecules can easily surround, solvate, and hence stabilize the alkox­ ide anion that would form by loss of the alcohol proton to a base. As a consequence of this stabilization, formation of the alcohol’s conjugate base is easier, and therefore its acidity is increased. I f the R group of the alcohol is bulky, solvation of the alkoxide anion is hin­ dered. Stabilization of the conjugate base is not as effective, and consequently the hindered alcohol is a weaker acid. Another reason that hindered alcohols are less acidic has to do with the inductive electron-donating effect of alkyl groups. The alkyl groups of a hindered alcohol donate electron density, making formation of an alkoxide anion more difficult than with a less hindered alcohol. •

H e lp f u l H i n t

Remember: Any factor that stabilizes the conjugate base of an acid increases its acidity.

A ll alcohols are much stronger acids than terminal alkynes, and they are very much stronger acids than hydrogen, ammonia, and alkanes (see Table 3.1). Relative Acidity Water and alcohols are the strongest acids in this series.

H2O > ROH > RC# CH > H 2 > NH 3 > RH

Sodium and potassium alkoxides can be prepared by treating alcohols with sodium or potassium metal or with the metal hydride (Section 6.15B). Because most alcohols are weaker acids than water, most alkoxide ions are stronger bases than the hydroxide ion. •

Conjugate bases of compounds with higher p K a values than an alcohol will deprotonate an alcohol. Relative Basicity

R- > NH2- > H - > RC# C - > R O - > H O -

HVdroxide is the wealœst base in this series.

Write equations for the acid-base reactions that would occur (if any) if ethanol were added to solutions of each of the following compounds. In each reaction, label the stronger acid, the stronger base, and so forth (consult Table 3.1). (a) NaNH 2

(b) H ----- =

:~Na+

(c)

O II •"""^ O N a

(d) NaOH

Sodium and potassium alkoxides are often used as bases in organic syntheses (Section 6.15B). We use alkoxides, such as ethoxide and teri-butoxide, when we carry out reactions that require stronger bases than hydroxide ion but do not require exceptionally powerful bases, such as the amide ion or the anion of an alkane. We also use alkoxide ions when (for reasons of solubility) we need to carry out a reaction in an alcohol solvent rather than in water.

R eview P roblem 11.5

5 14

Chapter 11

Alcohols and Ethers

11.7 Conversion o f Alcohols into Alkyl Halides In this and several following sections we will be concerned with reactions that involve substitution of the alcohol hydroxyl group. •

A hydroxyl group is such a poor leaving group (it would depart as hydroxide) that a common theme of these reactions will be conversion of the hydroxyl to a group that can depart as a weak base.

These processes begin by reaction of the alcohol oxygen as a base or nucleophile, after which the modified oxygen group undergoes substitution. First, we shall consider reactions that convert alcohols to alkyl halides. The most commonly used reagents for conversion of alcohols to alkyl halides are the following: •

Hydrogen halides (HCl, HBr, Hl)

•

Phosphorus tribromide (PBr3)

•

Thionyl chloride (SOCl2)

Examples of the use of these reagents are the following. A ll of these reactions result in cleav­ age of the C — O bond of the alcohol. In each case, the hydroxyl group is first converted to a suitable leaving group. We will see how this is accomplished when we study each type of reaction.

+

HCl(concd)

25 °C

- ''''/'O H

+

h 2o

+

H 3 PO3

Cl (94%)

OH

+

HBr

(concd)

Br

reflux

(95%)

3

OH

+

PBr3 3

-------------------->

3

- 1 0 to 0°C, 4 h

Br

3

3

(55-60%)

SO Cl2

pyridine (an organic base)

OH h 3c o

SO Cl

CO

h

3

+ HCl (forms a salt with pyridine)

11.8 Alkyl Halides from the Reaction o f Alcohols w ith Hydrogen Halides When alcohols react with a hydrogen halide, a substitution takes place producing an alkyl halide and water: R— OH

+

HX ----- » R— X

+

H2O

•

The order of reactivity of alcohols is 3° > 2° > 1° < methyl.

•

The order of reactivity of the hydrogen halides is HI > HBr > HCl (HF is gener­ ally unreactive).

The reaction is acid catalyzed. Alcohols react with the strongly acidic hydrogen halides HCl, HBr, and HI, but they do not react with nonacidic NaCl, NaBr, or Nal. Primary and secondary alcohols can be converted to alkyl chlorides and bromides by allowing them to react with a mixture of a sodium halide and sulfuric acid: ROH

+

NaX

H2SO:

RX

+

NaHSO 4 +

H2O

11.8 Alkyl Halides from the Reaction of Alcohols with Hydrogen Halides

11.8A Mechanisms of the Reactions of Alcohols with HX • Secondary, tertiary, allylic, and benzylic alcohols appear to react by a mechanism that involves the formation of a carbocation— a mechanism that we first saw in Section 3.14 and that you should now recognize as an SN1 reaction with the protonated alcohol acting as the substrate. We again illustrate this mechanism with the reaction of tert-butyl alcohol and aqueous hydrochloric acid (H3O +, CP). The first two steps in this SN1 substitution mechanism are the same as in the mecha­ nism for the dehydration o f an alcohol (Section 7.7). The alcohol accepts a proton.

Step 1

H +

Step 2

=0 — H

The protonated hydroxyl group departs as a leaving group to form a carbo­ cation and water.

H

In step 3 the mechanisms for the dehydration of an alcohol and the formation of an alkyl halide differ. In dehydration reactions the carbocation loses a proton in an E1 reaction to form an alkene. In the formation of an alkyl halide, the carbocation reacts with a nucle­ ophile (a halide ion) in an SN1 reaction.

Step 3

A halide anion reacts with the carbocation.

• How can we account for SN1 substitution in this case versus elimination in others? When we dehydrate alcohols, we usually carry out the reaction in concentrated sulfu­ ric acid and at high temperature. The hydrogen sulfate (HSO4~) present after protonation o f the alcohol is a weak nucleophile, and at high temperature the highly reactive carboca­ tion forms a more stable species by losing a proton and becoming an alkene. Furthermore, the alkene is usually volatile and distills from the reaction mixture as it is formed, thus draw­ ing the equilibrium toward alkene formation. The net result is an E1 reaction. In the reverse reaction, that is, the hydration of an alkene (Section 8.5), the carbo­ cation does react with a nucleophile. It reacts with water. Alkene hydrations are carried out in dilute sulfuric acid, where the water concentration is high. In some instances, too, carbocations may react with HSO4~ ions or with sulfuric acid, itself. When they do, they form alkyl hydrogen sulfates (R— OSO2OH). When we convert an alcohol to an alkyl halide, we carry out the reaction in the pres­ ence o f acid and in the presence o f halide ions, and not at elevated temperature. Halide ions are good nucleophiles (they are much stronger nucleophiles than water), and since halide ions are present in high concentration, most o f the carbocations react with an electron pair o f a halide ion to form a more stable species, the alkyl halide product. The overall result is an SN1 reaction. These two reactions, dehydration and the formation o f an alkyl halide, also furnish us another example o f the competition between nucleophilic substitution and elimination (see Section 6.18). Very often, in conversions of alcohols to alkyl halides, we find that the reac­ tion is accompanied by the formation of some alkene (i.e., by elimination). The free ener­ gies of activation for these two reactions o f carbocations are not very different from one another. Thus, not all o f the carbocations become stable products by reacting with nucle­ ophiles; some lose a b proton to form an alkene.

515

516

Chapter 11

Alcohols and Ethers

P r im a r y A lc o h o ls N ot all acid-catalyzed conversions o f alcohols to alkyl halides pro­ ceed through the formation o f carbocations.

•

Primary alcohols and methanol react to form alkyl halides under acidic conditions by an SN2 mechanism.

In these reactions the function o f the acid is to produce a proton ated alcohol. The halide ion then displaces a m olecule o f water (a good leaving group) from carbon; this produces an alkyl halide: H

H

:X

H

H

O — H ----- » : X — C — R H (Protonated 1° alcohol or methanol)

+

:O — H

H (A good leaving group)

A c i d Is R e q u i r e d Although halide ions (particularly iodide and bromide ions) are strong nucleophiles, they are not strong enough to carry out substitution reactions with alcohols themselves.

•

Reactions like the follow ing do not occur because the leaving group would have to be a strongly basic hydroxide ion:

H e lp f u l H i n t The reverse reaction, that is, the reaction of an alkyl halide with hydroxide ion, does occur and is a m ethod for the synthesis of alcohols. We saw this reaction in Chapter 6.

: Br

-■ A — C— OH ----- > :B r— C—

-:QH

We can see now why the reactions o f alcohols with hydrogen halides are acid-promoted. •

A cid protonates the alcohol hydroxyl group, making it a good leaving group.

Because the chloride ion is a weaker nucleophile than bromide or iodide ions, hydro­ gen chloride does not react with primary or secondary alcohols unless zinc chloride or some similar Lewis acid is added to the reaction mixture as w ell. Zinc chloride, a good Lewis acid, forms a com plex with the alcohol through association with an unshared pair o f elec­ trons on the oxygen atom. This enhances the hydroxyl’s leaving group potential sufficiently that chloride can displace it. R— O:

ZnCL

H

H : Cl

+

R— O— ZnCL

R — O — ZnCl 2 U> I 2

: Cl — R + [Zn(OH)Cl2] -

H [Zn(OH)Cl2]•

H3 O+

ZnCL

2 H2O

A s w e might expect, many reactions o f alcohols with hydrogen halides, particu­ larly those in which carbocations are formed, are accom panied by rearrangements.

How do w e know that rearrangements can occur when secondary alcohols are treated with a hydrogen halide? Results like that in Solved Problem 11.5 indicate this to be the case.

OH

Br

11.9 Alkyl Halides from the Reaction of Alcohols with PBr3 or SOCl2

R - o- h/ H - d - f t !

517

R eview P roblem 11.6

Write a detailed mechanism for the following reaction. Br

(a) What factor explains the observation that tertiary alcohols react with HX faster than secondary alcohols? (b) What factor explains the observation that methanol reacts with HX faster than a primary alcohol?

R eview P roblem 11.7

Since rearrangements can occur when some alcohols are treated with hydrogen halides, how can we successfully convert a secondary alcohol to an alkyl halide without rearrange­ ment? The answer to this question comes in the next section, where we discuss the use of reagents such as thionyl chloride (SOCl2) and phosphorus tribromide (PBr3).

11.9 A lkyl Halides from the Reaction o f Alcohols w ith PB r3 or SO Cl2 Primary and secondary alcohols react with phosphorus tribromide to yield alkyl bromides. H e lp f u l H i n t

3 R— OH + PBr33 ----- > 3 R— Br + H33 PO 3 3 (1

Dd a +* . 10 PBr3 : A reagent fo r synthesizing 1°

° or 2 °)

and 2° alkyl bromides.

•

The reaction of an alcohol with PBr3 does not involve the formation of a carboca­ tion and usually occurs without rearrangement of the carbon skeleton (especially if the temperature is kept below 0°C).

•

Phosphorus tribromide is often preferred as a reagent for the transformation of an alcohol to the corresponding alkyl bromide.

The mechanism for the reaction involves attack of the alcohol group on the phosphorus atom, displacing a bromide ion and forming a protonated alkyl dibromophosphite:

Br

Br . ----- > -P QBr

PBr2 R

O

2

+

H Protonated alkyl dibromophosphite

B r'

518

Chapter 11

Alcohols and Ethers

In a second step a bromide ion acts as a nucleophile to displace HOPBr2, a good leaving group due to the electronegative atoms bonded to the phosphorus: PBr, Br

R

CO'

Br

HOPBr

H A good leaving group

H e lp f u l H i n t SOCI2 : A reagent fo r synthesizing 1° and 2° alkyl chlorides.

R— O — H

Cl-----S — Cl

HOPBr2 can react with 2 more moles of alcohol, so the net result is conversion of 3 mol of alcohol to alkyl bromide by 1 mol of phosphorus tribromide. Thionyl chloride (SOCl2) converts primary and secondary alcohols to alkyl chlorides. Pyridine (C 5 H5 N) is often included to promote the reaction. The alcohol substrate attacks thionyl chloride as shown below, releasing a chloride anion and losing its proton to a mol­ ecule of pyridine. The result is an alkylchlorosulfite.

H l+ R— O — ••

Ox

V s Cl S ' "^C l

C sH sN :^ H O c | + II — ^ R— O — S — Cl -CI •• ••

O R— O-

S

-Cl

+

C 5 H 5 NH

The alkylchlorosulfite intermediate then reacts rapidly with another molecule of pyridine, in the same fashion as the original alcohol, to give a pyridinium alkylsulfite intermediate, with release of the second chloride anion. A chloride anion then attacks the substrate car­ bon, displacing the sulfite leaving group, which in turn decomposes to release gaseous SO 2 and pyridine. (In the absence of pyridine the reaction occurs with retention of configura­ tion. See Problem 11.55.) O R— O — S — Cl

O +C5 H5 N: 5 5 > -Cl

R— O — S — NC 5 H 5 U r•• 5 5

Cl

R— Cl

- • ^

° V

' °

H r*+

■O— S — n c 5 h 5 C 5 H5 N

P i 'n k l û m 1 1 A

11.10 Tosylates, M esylates, and Triflates: Leaving Group Derivatives o f Alcohols The hydroxyl group of an alcohol can be converted to a good leaving group by conversion to a sulfonate ester derivative. The most common sulfonate esters used for this purpose are methanesulfonate esters (“mesylates”), p-toluenesulfonate esters (“tosylates”), and trifluoromethanesulfonates (“triflates”).

R -o -k /lt-d -tt.'

11.10 Tosylates, Mesylates, and Triflates: Leaving Group Derivatives of Alcohols O O

y

O

O%

O

'°

or MeSO, ------- 2

CH 3

or Ts-

or Ms—

CH The tosyl group O

O

O

X S^

CH,

/ S\

'ÎV ' " S \

or MeSO3R OR

or MsOR

or CF3SO2_ or T f—

y 'S ^ CF 3

The mesyl group

The trifyl group

O O OR

or TsOR

C F,/

CH 3

^

Methanesulfonyl chloride

OH

^

(-pyr • HCl)

or TfOR

H e lp f u l H i n t A method fo r making an alcohol hydroxyl group into a leaving group.

O Ms

Ethyl methanesulfonate (ethyl mesylate)

Ethanol

Pyridine

SO2Cl

"OH

p-Toluenesulfonyl chloride (TSCl)

OTs

(-pyr • HCl)

Ethyl p -toluenesulfonate (ethyl tosylate)

Ethanol

It is important to note that formation of the sulfonate ester does not affect the stereo­ chemistry of the alcohol carbon, because the C — O bond is not involved in this step. Thus, if the alcohol carbon is a chirality center, no change in configuration occurs on making the sulfonate ester— the reaction proceeds with retention o f configuration. On reaction of the sulfonate ester with a nucleophile, the usual parameters of nucleophilic substitution reac­ tions become involved. S u b s tra te s fo r N u c le o p h ilic S u b s titu tio n Mesylates, tosylates, and triflates, because they are good leaving groups, are frequently used as substrates for nucleophilic substitution reactions. They are good leaving groups because the sulfonate anions they become when they depart are very weak bases. The triflate anion is the weakest base in this series, and is thus the best leaving group among them.

R

OSO 2 R'

An alkyl sulfonate (tosylate, mesylate, etc.)

•

X OR

Pyridine

MsCl

Nu:

or CF 3 SO3R

An alkyl triflate

An alkyl tosylate

The desired sulfonate ester is usually prepared by reaction of the alcohol in pyridine with the appropriate sulfonyl chloride, that is, methanesulfonyl chloride (mesyl chloride) for a mesylate, p-toluenesulfonyl chloride (tosyl chloride) for a tosylate, or trifluoromethanesulfonyl chloride [or trifluoromethanesulfonic anhydride (triflic anhydride)] for a triflate. Pyridine (C 5 H 5 N, pyr) serves as the solvent and to neutralize the HCl formed. Ethanol, for example, reacts with methanesulfonyl chloride to form ethyl methanesulfonate and with ptoluenesulfonyl chloride to form ethyl p-toluenesulfonate:

CH

O

V

3

An alkyl mesylate

519

Nu

R

R S O '3

A sulfonate ion (a very weak basea good leaving group)

To carry out a nucleophilic substitution on an alcohol, we first convert the alcohol to an alkyl sulfonate and then, in a second reaction, allow it to react with a nucleophile.

Pyridine = 'N ' C 5 H5 N

520

Chapter 11

Alcohols and Ethers

• If the mechanism is SN2, as shown in the second reaction of the following exam­ ple, inversion o f configuration takes place at the carbon that originally bore the alcohol hydroxyl group: R

Step 1

H "7\ R OH

R

retention

TsCl

(-pyr • HCl)

H

S n2

TsO“ Nu

R'

The fact that the C — O bond of the alcohol does not break during formation of the sul­ fonate ester is accounted for by the following mechanism. Methanesulfonyl chloride is used in the example.

A MECHANISM FOR THE REACTION C o n v e r s i o n o f a n A lc o h o l i n t o a M e s y l a t e (a n A lk yl M e t h a n e s u l f o n a t e ) \Q C 5 H5 N

H— O— R M e^

X Cl

(pyridine)

O

,.O

3

O i- .

R Me

O ^Cl

Me

Cl

O Me

H Methanesulfonyl chloride

O

z S\ - / R O

C 5 H5 NH

H C 5H5N

Alcohol

The alcohol oxygen attacks the sulfur atom of the sulfonyl chloride.

O

The intermediate loses a chloride ion.

Alkyl methanesulfonate

Loss of a proton leads to the product.

Solved Problem 11.7 Supply the missing reagents. O

O STRATEGY AND ANSWER The overall transformation over two steps involves replacing an alcohol hydroxyl group by a cyano group with inversion of configuration. To accomplish this, we need to convert the alcohol hydroxyl to a good leaving group in the first step, which we do by making it a methanesulfonate ester (a mesylate) using methanesulfonyl chloride in pyridine. The second step is an SN2 substitution of the methanesulfonate (mesyl) group, which we do using potassium or sodium cyanide in a polar aprotic solvent such as dimethylformamide (DMF).

ReviewProblem11.8

Show how you would prepare the following compounds from the appropriate sulfonyl chlorides. (a)

SO 3 CH 3

f

H3C " " '^ '^

(b)

I ^ J ^ ^ O S O 2Me

(c)

OSO2Me

R - ö- h / n - O - ft !

11.10 Tosylates, Mesylates, and Triflates: Leaving Group Derivatives o f Alcohols

521

R eview P roblem 11.9

Write structures for products X , Y, A, and B, showing stereochemistry. ,S O 2Cl

Ha

x

-------- *

Y

(R )-2-Butanol SO2Cl

(b) (pyridine)

OH

R eview P roblem 11.10

Suggest an experiment using an isotopically labeled alcohol that would prove that the formation of an alkyl sulfonate does not cause cleavage at the C — O bond of the alcohol.

THE CHEMISTRY OF. . . A lk y l P h o s p h a t e s

Alcohols react with phosphoric acid to yield alkyl phosphates: O

O

O

0

RO H

R O H + H O — P— OH

( - H 2O)

R O — P— O H

I

OH

R O — P— O H

( - H 2O) '

( - H 2O) '

I

OH

P h o s p h o ric a cid

II

ROH

• R O — P— O R

1

OR

A lk y l d ih y d ro g e n p h o s p h a te

Esters of phosphoric acids are important in biochemical reac­ tions. Especially important are triphosphate esters. Although hydrolysis of the ester group or of one of the anhydride link­ ages of an alkyl triphosphate is exothermic, these reactions occur very slowly in aqueous solutions. Near pH 7, these

OR

D ia lk y l h y d ro g e n p h o s p h a te

T ria lk y l p h o s p h a te

triphosphates exist as negatively charged ions and hence are much less susceptible to nucleophilic attack. Alkyl triphos­ phates are, consequently, relatively stable compounds in the aqueous medium of a living cell.

0

0

0

>RO H + HO — P — O — P — O — P — O P h o s p h a te „ e s te r lin ka g e O

R— O—

O

1

OO

O

P— O — P— O — P— O -

h 2o

slow

O

OP h o s p h o ric a n h y d rid e lin k a g e s

RO— P— O -

Il 1

Enzymes, on the other hand, are able to catalyze reactions of these triphosphates in which the energy made available when their anhydride linkages break helps the cell make

O

O

I

I

O-

O-

0

O

II

• R O — P— O — P— O O-

I

O-

HO — P— O — P— O -

I

O0

I

O-

I

O-

HO

P

O­

I O-

other chemical bonds. We shall have more to say about this in Chapter 22 when we discuss the important triphosphate called adenosine triphosphate (or ATP).

522

Chapter 11

Alcohols and Ethers

11.11 Synthesis o f Ethers

11.11A Ethers by Intermolecular Dehydration of Alcohols Alcohols can dehydrate to form alkenes. We studied this in Sections 7.7 and 7.8. Primary alcohols can also dehydrate to form ethers: R— OH

HO— R

HA

R— O — R

Dehydration to an ether usually takes place at a lower temperature than dehydration to the alkene, and dehydration to the ether can be aided by distilling the ether as it is formed. Diethyl ether is made commercially by dehydration of ethanol. Diethyl ether is the pre­ dominant product at 140°C; ethene is the major product at 180°C: H 2 SO 4 180°C

Ethene

OH H 2 SO 4 140°C

"O" Diethyl ether

The formation of the ether occurs by an SN2 mechanism with one molecule of the alco­ hol acting as the nucleophile and another protonated molecule of the alcohol acting as the substrate (see Section 11.5).

A MECHANISM FOR THE REACTION In te rm o le c u la r D e h y d ra tio n o f A lco h o ls to F o rm a n E th e r -H

Step 1

O'

H H — OSO3H

'O ' I H

OSO3H

This is an acid-base reaction in which the alcohol accepts a proton from the sulfuric acid.

Step 2 H

H

Another molecule of the alcohol acts as a nucleophile and attacks the protonated alcohol in an SN2 reaction.

Step 3

Another acid-base reaction converts the protonated ether to an ether by transferring a proton to a molecule of water (or to another molecule of the alcohol).

C o m p lic a tio n s o f In t e r m o le c u la r D e h y d r a t io n The method of synthesizing ethers by intermolecular dehydration has some important limitations.

•

Attempts to synthesize ethers by intermolecular dehydration of secondary alcohols are usually unsuccessful because alkenes form too easily.

11.11 Synthesis of Ethers •

Attempts to make ethers with tertiary alkyl groups lead exclusively to the alkenes.

•

Intermolecular dehydration is not useful for the preparation of unsymmetrical ethers from primary alcohols because the reaction leads to a mixture of products: ROR + ROR' + R O R'

H2SO4

ROH + R 'OH '------------v------------ ' 1° Alcohols

.

+

R - o- h / n - O - ft !

523

H2O

An exception to what we have just said has to do with syntheses of unsymmetrical ethers in which one alkyl group is a tert-butyl group and the other group is primary. For exam­ ple, this synthesis can be accomplished by adding teri-butyl alcohol to a mixture of the pri­ mary alcohol and H2 SO 4 at room temperature.

R eview P roblem 11.11

Give a likely mechanism for this reaction and explain why it is successful.

11.11B The Williamson Synthesis of Ethers

H e lp f u l H i n t

An important route to unsymmetrical ethers is a nucleophilic substitution reaction known as the Williamson synthesis. •

The Williamson ether synthesis consists of an SN2 reaction of a sodium alkoxide with an alkyl halide, alkyl sulfonate, or alkyl sulfate.

Alexander William Williamson was an English chemist who lived between 1824 and 1904. His method is especially useful for synthesis o f unsymmetrical ethers.

A MECHANISM FOR THE REACTION T h e W illia m so n E th e r S y n th e s is

R — O Na+ Sodium (or potassium) alkoxide

R — LG Alkyl halide, alkyl sulfonate, or dialkyl sulfate

R — O — R' + Na+ : LG Ether

The alkoxide ion reacts with the substrate in an SN2 reaction, with the resulting formation of an ether. The substrate must be unhindered and bear a good leaving group. Typical substrates are 1° or 2° alkyl halides, alkyl sulfonates, and dialkyl sulfates, that is, — LG =

— Br=, —h,

—OSO2 R",

or —OSO,OR"

The following reaction is a specific example of the Williamson synthesis. The sodium alkoxide can be prepared by allowing an alcohol to react with NaH: 'O H + Propyl alcohol

NaH ----- »

O Na Sodium propoxide

O

Ethyl propyl ether (70%)

H— H

Nal

H e lp f u l H i n t Conditions that favor a Williamson ether synthesis.

524

Chapter 11

Alcohols and Ethers

The usual limitations of SN2 reactions apply here. Best results are obtained when the alkyl halide, sulfonate, or sulfate is primary (or methyl). If the substrate is tertiary, elimination is the exclusive result. Substitution is also favored over elimination at lower temperatures.

ReviewProblem11.12

(a) Outline two methods for preparing isopropyl methyl ether by a Williamson synthesis. (b) One method gives a much better yield of the ether than the other. Explain which is the

better method and why.

Solved Problem 11.8 The cyclic ether tetrahydrofuran (THF) can be synthesized by treating 4-chloro-1-butanol with aqueous sodium hydroxide (see below). Propose a mechanism for this reaction. ,C! HO"

OH-

H2O

NaC!

O

h 2o

Tetrahydrofuran STRATEGY AND ANSWER Removal of a proton from the hydroxyl group of 4-chloro-1-butanol gives an alkox­

ide ion that can then react with itself in an intramolecular SN2 reaction to form a ring. C! o

C!

:O :

:OH

O:

Even though treatment of the alcohol with hydroxide does not favor a large equilibrium concentration of the alkox­ ide, the alkoxide anions that are present react rapidly by the intramolecular SN2 reaction. As alkoxide anions are consumed by the substitution reaction, their equilibrium concentration is replenished by deprotonation of additional alcohol molecules, and the reaction is drawn to completion.

ReviewProblem11.13

Epoxides can be synthesized by treating halohydrins with aqueous base. Propose a mech­ anism for reactions (a) and (b), and explain why no epoxide formation is observed in (c). (a)

Cl H

|ï s C l, pyr 'O H k 2c o 3

D

'B r

B

(Hint:B and D are

11.11 Synthesis o f Ethers

R - ö- h / n - O - ft !

525

11.11C Synthesis o f E thers by A lk o x y m e rc u ra tio n -D e m e rc u ra tio n Alkoxymercuration-demercuration is another method for synthesizing ethers. • The reaction of an alkene with an alcohol in the presence of a mercury salt such as mercuric acetate or trifluoroacetate leads to an alkoxymercury intermediate, which on reaction with sodium borohydride yields an ether. When the alcohol reactant is also the solvent, the method is called solvomercuration-demercuration. This method directly parallels hydration by oxymercuration-demercuration (Section 8 .6 ): '

'

//

(1) Hg(O2CCF 3 )2, t-BuOH (2) NaBH4, HO-

/

\

_

(98% Yield)

11.11D tert-Butyl Ethers by Alkylation of Alcohols: Protecting Groups Primary alcohols can be converted to teri-butyl ethers by dissolving them in a strong acid such as sulfuric acid and then adding isobutylene to the mixture. (This procedure minimizes dimerization and polymerization of the isobutylene.)

R'

H2 SO 4

"OH

fert-Butyl protecting group

R'

• A teri-butyl ether can be used to “protect” the hydroxyl group of a primary alcohol while another reaction is carried out on some other part of the molecule. • A teri-butyl protecting group can be removed easily by treating the ether with dilute aqueous acid. Suppose, for example, we wanted to prepare 4-pentyn-1-ol from 3-bromo-1-propanol and sodium acetylide. If we allow them to react directly, the strongly basic sodium acetylide w ill react first with the hydroxyl group: HO^ ^ "Br 3-Bromo-1-propanol

=

=- Na+

NaO'

However, if we protect the — OH group first, the synthesis becomes feasible: HO

Br

(1 ) H 2SO4

. _

Na+

f-B uO'

(2)

H,O+/H,O

f-B uO '

HO

+

f-BuOH

4-Pentyn-1-ol

Propose mechanisms for the following reactions. (a) ,OH (b)

ReviewProblem11.15

526

Chapter 11

Alcohols and Ethers

11.11E Silyl Ether Protecting Groups •

A hydroxyl group can also be protected by converting it to a silyl ether group.

One of the most common silyl ether protecting groups is the tert-butyldimethylsilyl ether group [tert-butyl(Me)2 Si— O — R, or TBS— O — R], although triethylsilyl, triisopropylsilyl, tert-butyldiphenylsilyl, and others can be used. The tert-butyldimethylsilyl ether is sta­ ble over a pH range of roughly 4-12. A TBS group can be added by allowing the alcohol to react with tert-butyldimethylsilyl chloride in the presence of an aromatic amine (a base) such as imidazole or pyridine: Me H N ^N =

\ / Sk

R— O— H o r

Me

Me imidazole

f-Bu

Imidazole

DMF ’ ( - H C!)

fecf-Butylchlorodimethylsilane (TBSCI) "N" Pyridine

•

R O

\ / Sk

Me f-Bu

(R—O—TBS)

The TBS group can be removed by treatment with fluoride ion (tetrabutylammonium fluoride or aqueous HF is frequently used). Me Me \ / Sk O f-Bu

Bu 4 N+FTHF

R— O— H

Me Me \ / Sk ^ f-B u F

(R—O—TBS) Converting an alcohol to a silyl ether also makes it much more volatile. This increased volatility makes the alcohol (as a silyl ether) much more amenable to analysis by gas chro­ matography. Trimethylsilyl ethers are often used for this purpose. (The trimethylsilyl ether group is too labile to use as a protecting group in most reactions, however.)

Solved Problem 11.9 Supply the missing reagents and intermediates A -E .

STRATEGY AND ANSWER We start by noticing several things: a TBS (tert-butyldimethylsilyl) protecting group is involved, the carbon chain increases from four carbons in A to seven in the final product, and an alkyne is reduced to a trans alkene. A does not contain any silicon atoms, whereas the product after the reaction under conditions B does. Therefore, A must be an alcohol that is protected as a TBS ether by conditions specified as B. A is therefore 4-bromo-1-butanol, and conditions B are TBSCl (tert-butyldimethylsilyl chloride) with imidazole in DMF. Conditions C involve loss of the bromine and chain extension by three carbons with incorporation of an alkyne. Thus, the reaction conditions for C must involve sodium propynide, which would come from deprotonation of propyne using an appropriate base, such as NaNH 2 or CH 3 MgBr. The conditions leading from E to the final prod­ uct are those for removal of a TBS group, and not those for converting an alkyne to a trans alkene; thus, E must still contain the TBS ether but already contain the trans alkene. Conditions D, therefore, must be (1) Li, Et2 NH, (2) NH 4 Cl, which are those required for converting the alkyne to a trans alkene. E, therefore, must be the TBS ether of 5-heptyn-1-ol (which can also be named 1-tert-butyldimethylsiloxy-5-heptynol).

R - o- h / n - O - ft !

11.12 Reactions of Ethers

11.12 Reactions o f Ethers Dialkyl ethers react with very few reagents other than acids. The only reactive sites that molecules of a dialkyl ether present to another reactive substance are the C— H bonds of the alkyl groups and the — O — group of the ether linkage. Ethers resist attack by nucleophiles (why?) and by bases. This lack of reactivity coupled with the ability of ethers to solvate cations (by donating an electron pair from their oxygen atom) makes ethers especially use­ ful as solvents for many reactions. Ethers are like alkanes in that they undergo halogenation reactions (Chapter 10), but these reactions are of little synthetic importance. They also undergo slow autoxidation to form explosive peroxides (see Section 11.3D). The oxygen of the ether linkage makes ethers basic. Ethers can react with proton donors to form oxonium salts:

11.12A Cleavage of Ethers Heating dialkyl ethers with very strong acids (HI, HBr, and H2 SO4) causes them to undergo reactions in which the carbon-oxygen bond breaks. Diethyl ether, for example, reacts with hot concentrated hydrobromic acid to give two molecular equivalents of ethyl bromide: „ 2 HBr

----- >

2

Br

+

H2O

Cleavage of an ether

The mechanism for this reaction begins with formation of an oxonium cation. Then, an SN2 reaction with a bromide ion acting as the nucleophile produces ethanol and ethyl bromide. Excess HBr reacts with the ethanol produced to form the second molar equivalent of ethyl bromide.

A MECHANISM FOR THE REACTION E th e r C le a v a g e b y S tro n g A cid s

Step 1 H + H

Br

Ethanol

Ethyl bromide In Step 2, the ethanol (just formed) reacts with HBr (present in excess) to form a second molar equivalent of ethyl bromide. Step 2

H ^ B r:

:Br: -

~O +H H

Br

+

H

H

527

528

Chapter 11

Review Problem 11.16

Alcohols and Ethers

When an ether is treated with cold concentrated HI, cleavage occurs as follows: R— O — R +

H I----- > ROH

+

RI

When mixed ethers are used, the alcohol and alkyl iodide that form depend on the nature of the alkyl groups. Use mechanisms to explain the following observations: (a)

OMe

OH HI

Mel

(b) HI

MeOH

OMe

R eview P roblem 11.17

I

Write a detailed mechanism for the following reaction. HBr (excess) 2

O

R eview P roblem 11.18

'Br

Provide a mechanism for the following reaction.

HCl

Cl

och3

11.13 Epoxides Epoxides are cyclic ethers with three-membered rings. In IUPAC nomenclature epoxides are called oxiranes. The simplest epoxide has the common name ethylene oxide: ‘O' 7 \ ^ C -C ^

I

'O' / \ ^

\

An epoxide

IUPAC name: oxirane Common name: ethylene oxide

11.13A Synthesis of Epoxides: Epoxidation Epoxides can be synthesized by the reaction of an alkene with an organic peroxy acid (RCO 3 H— sometimes called simply a peracid), a process that is called epoxidation. M etaChloroperoxybenzoic acid (MCPBA) is one peroxy acid reagent commonly used for epoxidation. The following reaction is an example. O

O

Cl 1-Octene

H

O

O'" O^ H

O

Cl MCPBA

(81%)

mefa-Chlorobenzoic acid

meta-Chlorobenzoic acid is a by-product of the reaction. Often it is not written in the chem­ ical equation, as the following example illustrates. O

O MCPBA

,O

(77%)

R - ö- h / n - Ö - ft !

11.13 Epoxides

529

As the first example illustrates, the peroxy acid transfers an oxygen atom to the alkene. The following mechanism has been proposed.

A MECHANISM FOR THE REACTION A lk e n e E p o x id a tio n R

R =O O

H \\\W Carboxylic acid Epoxide

O

^H Peroxy acid Alkene

transition state The peroxy acid transfers an oxygen atom to the alkene in a cyclic, single-step mechanism. The result is the syn addition of the oxygen to the alkene, with the formation of an epoxide and a carboxylic acid.

THE CHEMISTRY OF. . . T h e S h a rp le ss A sy m m e tric E p o x id a tio n In 1980, K. B. Sharpless (then at the Massachusetts Institute of Technology, presently at The Scripps Research Institute) and co-workers reported a m eth o d that has since bec om e one of the most valuable tools for chiral synthesis. The Sharpless asymmetric epoxidation is a method for converting allylic alco­ hols (Section 11.1) to chiral epoxy alcohols with very high enantioselectivity (i.e., with preference for one enantiomer rather than formation of a racemic mixture). In recognition of this and other work in asymmetric oxidation methods (see Section 8.16A), Sharpless received half of the 2001 Nobel Prize in Chemistry (the other half was awarded to W. S. Knowles and Co r R. Noyori; see Section 7.14). The Sharpless O 2 H asymmetric epoxidation involves treating the H allylic alcohol with tert-butyl hydroperoxide, titanium(IV) tetraisopropoxide [Ti(O— /-Pr)4J, HO I H and a specific stereoisomer of a tartrate ester. Co 2r (The tartrate stereoisom er th at is chosen A (+)-dialkyl d ep e n d s on the specific enantiomer of the epox­ ta rtra te ester ide desired). The following is an example:

The oxygen that is transferred to th e allylic alcohol to form th e epoxide is derived from tert-butyl hydroperoxide. The enantioselectivity of the reaction results from a titanium complex am ong th e reagents that includes th e enantiomerically pure tartrate ester as one of the ligands. The choice of whether to use the (+)- or (—)-tartrate ester for stereochemical control d e p e n d s on which enantiomer of the epoxide is desired. [The (+)- and (—)-tartrates are either diethyl or diisopropyl esters.] The stereochemical prefer­ ences of th e reaction have bee n well studied, such that it is possible to prepare either enantiom er of a chiral epoxide in high enantiomeric excess, simply by choosing th e a p p ro ­ priate (+)- or (—)-tartrate stereoisomer as th e chiral ligand: O OH (S )-M e th y lg ly c id o l

(+)-dialkyl tartrate

OH OH

OH

i-BuOO H, Ti(O— /-Pr)4

(-)-dialkyl tartrate

CH 2CI2, —20°C (+ )-dieth yl tartrate

G e ra n io !

S h a rp le s s a s y m m e tric e p o x id a tio n

n OH

77% y ie ld (95% e n a n tio m e ric excess)

(R )-M e th y ig iy c id o i

(continues on the next page)

530

Chapter 11

Alcohols and Ethers ra l e p o x y a lc o h o l s y n t h o n s p r o d u c e d b y t h e S h a r p le s s a s y m ­

H

m e tr ic e p o x id a t io n O

in

h a s b e e n d e m o n s tra te d o v e r a n d o v e r

e n a n tio s e le c tiv e

s y n th e s e s

of

m any

im p o r ta n t

com ­

p o u n d s . S o m e e x a m p l e s i n c l u d e t h e s y n t h e s is o f t h e p o l y ­ e t h e r a n t i b i o t i c X - 2 0 6 b y E . J . C o r e y ( H a r v a r d ) , t h e J . T. B a k e r (7 R, 8S )-D is p a rlu re

c o m m e r c i a l s y n t h e s is o f t h e g y p s y m o t h p h e r o m o n e (7 R ,8 S ) -

C o m p o u n d s o f th is g e n e r a l s tr u c tu r e a re e x tr e m e ly u s e ­ f u l a n d v e r s a t il e s y n t h o n s b e c a u s e c o m b i n e d in o n e m o l e ­ c u le

a re

an

e p o x id e

fu n c tio n a l

g ro u p

(a

h ig h l y

r e a c t iv e

e l e c t r o p h i l i c s it e ) , a n a lc o h o l f u n c t i o n a l g r o u p (a p o t e n t i a l l y n u c l e o p h i l i c s ite ) , a n d a t le a s t o n e c h i r a l i t y c e n t e r t h a t is p r e ­ s e n t in h ig h e n a n t i o m e r i c p u r i t y . T h e s y n t h e t i c u t i l i t y o f c h i-

d is p a r lu r e , a n d

s y n t h e s is

C a lifo r n ia

D ie g o

San

b y K . C . N ic o la o u

and

S c r ip p s

( U n i v e r s it y o f

R e s e a rc h

In s titu te )

of

z a r a g o z i c a c id A ( w h ic h is a ls o c a ll e d s q u a l e s t a t i n S1 a n d h a s b e e n s h o w n t o l o w e r s e r u m c h o l e s t e r o l le v e ls in t e s t a n im a ls b y i n h i b i t i o n o f s q u a l e n e b io s y n t h e s is ; s e e " T h e C h e m i s t r y o f. . .C h o le s te r o l B io s y n th e s is ," C h a p te r

8 ).

O

11.13B Stereochemistry of Epoxidation •

The reaction of alkenes with peroxy acids is, of necessity, a syn addition, and it is stereospecific. Furthermore, the oxygen atom can add to either face of the alkene.

For example, trans-2-butene yields racemic irans-2,3-dimethyloxirane, because addition of oxygen to each face of the alkene generates an enantiomer. cis-2-Butene, on the other hand, yields only cis-2,3-dimethyloxirane, no matter which face of the alkene accepts the oxy­ gen atom, due to the plane of symmetry in both the reactant and the product. I f additional chirality centers are present in a substrate, then diastereomers would result. O

c/s-2-Butene

c/s-2,3-Dimethyloxirane (a meso compound)

H frans-2-Butene

O

Enantiomeric frans-2,3-dimethyloxiranes

In Special Topic C (Section C.3) we present a method for synthesizing epoxides from alde­ hydes and ketones.

R - o- h / n - Ó - K '

11.14 Reactions of Epoxides

531

1 1 .1 4 Reactions o f Epoxides • The highly strained three-membered ring of epoxides makes them much more reactive toward nucleophilic substitution than other ethers. Acid catalysis assists epoxide ring opening by providing a better leaving group (an alco­ hol) at the carbon atom undergoing nucleophilic attack. This catalysis is especially impor­ tant if the nucleophile is a weak nucleophile such as water or an alcohol. An example is the acid-catalyzed hydrolysis of an epoxide.

A MECHANISM FOR THE REACTION A c id - C a ta ly z e d R in g O p e n i n g o f a n E p o x id e

I

\

H

.C -C . \ / o

-C -C \ / =0 + I H

H

Epoxide Protonated epoxide The acid reacts with the epoxide to produce a protonated epoxide. H '•■0—H

/ C — C-

=0 — H I H

>0 - r HH

}0

\J

C — CH—

H— O'-

H

oO

H— 0 — H I H

H Protonated epoxide

Weak nucleophile

Protonated 1 ,2 -diol

1,2-Diol

The protonated epoxide reacts with the weak nucleophile (water) to form a protonated 1,2-diol, which then transfers a proton to a molecule of water to form the 1 ,2 -diol and a hydronium ion.

Epoxides can also undergo base-catalyzed ring opening. Such reactions do not occur with other ethers, but they are possible with epoxides (because of ring strain), provided that the attacking nucleophile is also a strong base such as an alkoxide ion or hydroxide ion.

A MECHANISM FOR THE REACTION B a s e - C a ta ly z e d R ing O p e n i n g o f a n E p o x id e

R— 0 :

,\

I

„ C -C .

R0 j

R0 :

//

C— C

:O i

R— 0 ^ H

C — C.' ^O H

Strong Epoxide An alkoxide ion nucleophile A strong nucleophile such as an alkoxide ion or a hydroxide ion is able to open the strained epoxide ring in a direct SN2 reaction.

R 0:

532

Chapter 11

•

H e lp f u l H i n t Regioselectivity in the opening of epoxides.

Alcohols and Ethers

If the epoxide is unsymmetrical, in base-catalyzed ring opening, attack by the alkoxide ion occurs primarily a t the less su b stitu ted carbon atom .

For example, methyloxirane reacts with an alkoxide ion mainly at its primary carbon atom: 1° Carbon atom is less hindered. EtO E tO +

EtO

&

+

\

EtOH^ O-

0

E tO

OH

Methyloxirane

1-Ethoxy-2-propanol

This is just what w e should expect: The reaction is, after all, an SN2 reaction, and, as we learned earlier (Section 6.13A ), primary substrates react more rapidly in SN2 reactions because they are less sterically hindered. •

In the acid-catalyzed rin g op en ing o f an unsymmetrical epoxide the nucleophile attacks primarily a t the m ore su bstitu ted carbon atom .

For example, cat. HA

MeOH O

MeO

OH

The reason: Bonding in the protonated epoxide (see the follow ing reaction) is unsymmet­ rical, with the more highly substituted carbon atom bearing a considerable positive charge; the reaction is SN1 like. The nucleophile, therefore, attacks this carbon atom even though it is more highly substituted: This carbon resembles a 3° carbocation

Protonated epoxide The more highly substituted carbon atom bears a greater positive charge because it resem­ bles a more stable tertiary carbocation. [Notice how this reaction (and its explanation) resem­ bles that given for halohydrin formation from unsymmetrical alkenes in Section 8.14 and attack on mercurinium ions.]

ReviewProblem11.19

Propose structures for each o f the following products derived from oxirane (ethylene oxide): (a )

O

HA MeOH

(b)

O

(c) c 3 h8o 2 C3H8O2 3

8

O O / \

¿ Ax L—

2

Methyl Cellosolve

(d)

C4H1U O2 ~ 4' ' 1U~ 2

(e) (e)

O ¿A

—

K|

c 2 h 5io

H2O

2

NH

5

C 2 H7NO

ha EtOH

Ethyl Cellosolve

O O ¿A

MeONa MeOH

^

C H O 3 8 2

11.14 Reactions of Epoxides

O

MeO

OH

When sodium ethoxide reacts with 1-(chloromethyl)oxirane (also called epichlorohydrin), labeled with 14C as shown by the asterisk in I, the major product is II. Provide a mecha­ nistic explanation for this result. Cl

\ / O

533

ReviewProblem11.20

Provide a mechanistic explanation for the following observation. MeONa MeOH

R - ö- h/ h - Ö - K .'

ReviewProblem11.21

OEt

EtONa O II

I 1-(Chloromethyl)oxirane (epichlorohydrin)

THE CHEMISTRY OF. . . E p o x id e s , C a r c in o g e n s , a n d B io lo g ical O x id a t io n

Certain molecules from the environment become carcino­ genic by "activation" through metabolic processes that are normally involved in preparing them for excretion. This is the case with two of the most carcinogenic compounds known: dibenzo[a,/]pyrene, a polycyclic aromatic hydrocarbon, and aflatoxin B-|, a fungal metabolite. During the course of oxida­ tive processing in the liver and intestines, these molecules undergo epoxidation by enzymes called P450 cytochromes. Their epoxide products, as you might expect, are excep-

tionally reactive electrophiles, and it is precisely because of this that they are carcinogenic. The dibenzo[a,/]pyrene and aflatoxin Bi epoxides undergo very facile nucleophilic sub­ stitution reactions with DNA. Nucleophilic sites on DNA react to open the epoxide ring, causing alkylation of the DNA by formation of a covalent bond with the carcinogen. Modification of the DNA in this way causes onset of the dis­ ease state.

enzym atic epoxidation (‘‘activation’’)

DNA d e o x y a d e n o s in e a d d u c t ( c a u s e s c a n c e r)

H O' OH D ib e n z o [a ,/]p y re n e

D ib e n z o [a ,/]p y re n e 1 1 ,1 2 -d io l-1 3 ,1 4 -e p o x id e

The normal pathway toward excretion of foreign mole­ cules like dibenzo[a,f]pyrene and aflatoxin B-|, however, also involves nucleophilic substitution reactions of their epoxides. One pathway involves opening of the epoxide ring by nucle­ ophilic substitution with glutathione. Glutathione is a rela­

tively polar molecule that has a strongly nucleophilic sulfhydryl group. After reaction of the sulfhydryl group with the epoxide, the newly formed covalent derivative, because it is substantially more polar than the original epoxide, is readily excreted through aqueous pathways. (continues on the next page)

534

Chapter 11

Alcohols and Ethers

Aflatoxin B1-g!utathione adduct (can be excreted)

11.14A Polyethers from Epoxides Treating ethylene oxide with sodium methoxide (in the presence of a small amount of methanol) can result in the formation of a polyether:

Poly(ethylene glycol) (a polyether)

This is an example of anionic p olym erization (Section 10.10). The polymer chains con­ tinue to grow until methanol protonates the alkoxide group at the end of the chain. The aver­ age length of the growing chains and, therefore, the average molecular weight of the polymer can be controlled by the amount of methanol present. The physical properties of the poly­ mer depend on its average molecular weight.

R - ö- h / n - O - ft !

11.15 Anti 1,2-Dihydroxylation o f Alkenes via Epoxides

535

Polyethers have high water solubilities because of their ability to form multiple hydrogen bonds to water molecules. Marketed commercially as carbowaxes, these polymers have a vari­ ety of uses, ranging from use in gas chromatography columns to applications in cosmetics.

11.15 A n ti 1,2-Dihydroxylation o f Alkenes via Epoxides Epoxidation (1) of cyclopentene produces 1,2-epoxycyclopentane: RCO3 H

(1 )

O Cyclopentene

1,2-Epoxycyclopentane

Acid-catalyzed hydrolysis (2) of 1,2-epoxycyclopentane yields a trans diol, trans-1,2cyclopentanediol. Water acting as a nucleophile attacks the protonated epoxide from the side opposite the epoxide group. The carbon atom being attacked undergoes an inversion of configuration. We show here only one carbon atom being attacked. Attack at the other carbon atom of this symmetrical system is equally likely and produces the enantiomeric form of trans-1 ,2 -cyclopentanediol:

H e lp f u l H i n t A synthetic method fo r anti 1,2dihydroxylation.

H

Epoxidation followed by acid-catalyzed hydrolysis gives us, therefore, a method for anti 1,2-dihydroxylation of a double bond (as opposed to syn 1,2-dihydroxylation, Section 8.16). The stereochemistry of this technique parallels closely the stereochemistry of the bromination of cyclopentene given earlier (Section 8.13).

ReviewProblem11.22

Outline a mechanism similar to the one just given that shows how the enantiomeric form of trans-1 ,2 -cyclopentanediol is produced.

1

SolvedProblem11.101

In Section 11.13B we showed the epoxidation of ds-2-butene to yield ds-2,3-dimethylo:xirane and epoxidation of trans-2-butene to yield trans-2,3-dimethyloxirane. Now consider acid-catalyzed hydroly sis of these two epoxides and show what product or products would result from each. Are these reactions stereospecific? ANSWER (a) The meso compound, ds-2,3-dimethyloxirane (Fig. 11.1), yields on hydrolysis (2^,3^)-2,3-butane-

diol and (2S,3S)-2,3-butanediol. These products are enantiomers. Since the attack by waiter at either carbon [path (a) or path (b) in Fig. 11.1] occurs at the same rate, the product is obtained in a racemic form. When either of the trans-2,3-dimethyloxirane enantiomers undergoes acid-catalyzed 1ydrolysis, the only prod­ uct that is obtained is the meso compound, (2^,3S)-2,3-butanediol. The hydrolysis of one enantiomer is shown in

536

Chapter 11

Alcohols and Ethers

Fig. 11.2. (You might construct a similar diagram showing the hydrolysis of the other enantiomer to convince your­ self that it, too, yields the same product.) O'

O‘ ' " / “ V 'H H

/ V

H

h

One trans-2,3-dimethyloxirane enantiomer c/s-2,3-Dimethyloxirane

h 2o, h a

h 2o, h a H H (a )Ç>° + (b) (a )C j° + y

(a)

"h

'^ H

»

(a)

(b) A

A

vh

(b)

,

, (a)

(a)

|

l

(b)

H—O

""H H

H "^

O'-

I

H \

-

1

V '" H \

\ H

I

+ f -

H

I

H

4.

H —O :

+ :O — H

O:

X !0

:O—H

H

(b) H

H '" "J

(b)

H

H i - HA

HA

i ~ HA

h HA H—O

:O — H H

H—O

:O — H

H

H

.H

H H

H "y H“ ^

^

H

^ - H

!O — H

H

(2R,3R )-2,3-Butanediol

(2S,3S )-2,3-Butanediol

These molecules are identical; they both represent the meso compound (2R, 3S)-2,3-butanediol.

E nantiom ers

F igu re 11.1

O:

F ig u re 11.2

A c id - c a ta ly z e d h y d r o ly s is o f c is -2 ,3 -

T h e a c id - c a ta ly z e d h y d r o ly s is o f o n e trans-2,3-

d im e th y lo x ir a n e y ie ld s (2 S ,3 S ) -2 ,3 -b u ta n e d io l b y p a th (a) a n d

d im e th y lo x ir a n e e n a n tio m e r p r o d u c e s t h e m e s o c o m p o u n d ,

(2 R ,3 R ) - 2 ,3 - b u t a n e d io l b y p a th (b ). (U s e m o d e ls t o c o n v in c e

( 2 R ,3 S ) - 2 ,3 - b u ta n e d io l, b y p a th (a) o r b y p a th (b ). H y d r o ly s is

y o u r s e lf.)

o f t h e o t h e r e n a n tio m e r ( o r t h e r a c e m ic m o d ific a tio n ) w o u ld y ie ld t h e s a m e p r o d u c t . (Y ou s h o u ld u s e m o d e ls t o c o n v in c e y o u r s e lf t h a t t h e t w o s tr u c tu r e s g iv e n f o r t h e p r o d u c ts d o re p re s e n t th e sa m e c o m p o u n d .)

(b) Since both steps in this method for the conversion of an alkene to a 1,2-diol (glycol) are stereospecific (i.e., both the epoxidation step and the acid-catalyzed hydrolysis), the net result is a stereospecific anti 1 ,2 -dihydroxylation of the double bond (Fig. 11.3).

H//,,.

,.'V»H

(1)

O RCOOH HA, H2O

(2) (anti 1,2-dihydroxylation)

H

OH

HO

* "'« H

H 'y

HO

c/s-2-Butene

t* * OH

Enantiomeric 2,3-butanediols

F ig u re 11.3

The

o v e r a ll r e s u lt o f e p o x id a t io n f o llo w e d b y a c id - c a ta ly z e d h y d r o ly s is is a s te r e o s p e c ific a n ti

1, 2 - d ih y d r o x y la t io n O /in,,, ____ ^ H H

>

" ^

trans-2-Butene

(1)

RCOOH HA, H2O

(2) (anti 1,2-dihydroxylation)

of

t h e d o u b le b o n d . cis - 2 -

HO

H

HO

OH

meso-2,3-Butanediol

OH

B u te n e y ie ld s th e e n a n tio m e r ic 2 ,3 -

sam e as

\W‘11 H

w h ic h is th e

H

’J ' hii

b u t a n e d io ls ; trans-2-

H

b u t e n e y ie ld s t h e m e s o com pound.

R - ö- h / n - Ö - ft !

11.16 Crown Ethers

ReviewProblem11.23

Supply the missing reagents and intermediates A -E .

A

C4H6

B

C

D

537

HO E

C 4 H8O

och3 (racemic)

THE CHEMISTRY OF. . . E n v iro n m e n ta lly F rie n d ly A lk e n e O x id a tio n M e th o d s

The effort to develop synthetic methods that are environ­ mentally friendly is a very active area of chemistry research. The push to devise "green chemistry" procedures includes not only replacing the use of potentially hazardous or toxic reagents with ones that are more friendly to the environment but also developing catalytic procedures that use smaller quantities of potentially harmful reagents when other alter­ natives are not available. The catalytic syn 1,2-dihydroxylation methods that we described in Section 8.16 (including the Sharpless asymmetric dihydroxylation procedure) are environmentally friendly modifications of the original pro­ cedures because they require only a small amount of OSO4 or other heavy metal oxidant. Nature has provided hints for ways to carry out environ­ mentally sound oxidations as well. The enzyme methane monooxygenase (MMO) uses iron to catalyze hydrogen per­ oxide oxidation of small hydrocarbons, yielding alcohols or epoxides, and this example has inspired development of new laboratory methods for alkene oxidation. A 1,2-dihydroxylation procedure developed by L. Que (University of Minnesota) yields a mixture of 1,2-diols and epoxides by action of an iron catalyst and hydrogen peroxide on an alkene. (The ratio of diol to epoxide formed depends on the reaction conditions, and in the case of dihydroxylation, the procedure shows some enantioselectivity.) Another green reaction is the epoxidation method developed by E. Jacobsen (Harvard University). Jacobsen's procedure uses hydrogen peroxide and a similar iron catalyst to epoxidize alkenes (without the complication of diol formation). Que's and Jacobsen's methods are environmentally friendly because their procedures employ catalysts containing a non­ toxic metal, and an inexpensive, relatively safe oxidizing reagent is used that is converted to water in the course of the reaction.

^ " ^ O H

R R = hexy/

H2O2 (20 equiv.) Q ue’s catalyst (0.1 mol %)

OH 81% (60% ee)

CH 3CN, 30 oc

+ R. O 13%

R

H2O2 (1.5 equiv.) Jacobsen’s catalyst (3 mol %) C H 3CO 2H (30 mol %), C H 3CN, 5°C

p "

O 85%

Q u e 's c a ta ly s t

The quest for more methods in green chemistry, with benign reagents and by-products, catalytic cycles, and high yields, will no doubt drive further research by present and future chemists. In coming chapters we shall see more exam­ ples of green chemistry in use or under development.

11.16 Crown Ethers Crown ethers are compounds having structures like that of 18-crown-6, below. 18-Crown-6 is a cyclic oligomer of ethylene glycol. Crown ethers are named as x-crown-y, where x is the total number of atoms in the ring and y is the number of oxygen atoms. A key property of crown ethers is that they are able to bind cations, as shown below for 18-crown-6 and a potassium ion.

538

Chapter 11

Alcohols and Ethers

18-Crown-6

Crown ethers render many salts soluble in nonpolar solvents. For this reason they are called phase transfer catalysts. When a crown ether coordinates with a metal cation it masks the ion with a hydrocarbon-like exterior. 18-Crown-6 coordinates very effectively with potassium ions because the cavity size is correct and because the six oxygen atoms are ideally situated to donate their electron pairs to the central ion in a Lewis acid-base complex. The relationship between a crown ether and the ion it binds is called a host-guest relationship. Salts such as KF, KCN, and potassium acetate can be transferred into aprotic solvents using catalytic amounts of 18-crown-6. Use of a crown ether with a nonpolar solvent can be very favorable for an SN2 reaction because the nucleophile (such as F_ , CN~, or acetate from the compounds just listed) is unencumbered by solvent in an aprotic solvent, while at the same time the cation is prevented by the crown ether from associating with the nucle­ ophile. Dicyclohexano-18-crown-6 is another example of a phase transfer catalyst. It is even more soluble in nonpolar solvents than 18-crown-6 due to its additional hydrocarbon groups. Phase transfer catalysts can also be used for reactions such as oxidations. (There are phase transfer catalysts that are not crown ethers, as well.)

Dicyclohexano-18-crown-6

© ReviewProblem11.24

The development of crown ethers and other molecules “with structure specific interac­ tions of high selectivity” led to awarding of the 1987 Nobel Prize in Chemistry to Charles J. Pedersen (retired from the DuPont Company), Donald J. Cram (University of California, Los Angeles, deceased 2001), and Jean-Marie Lehn (Louis Pasteur University, Strasbourg, France). Their contributions to our understanding of what is now called “molecular recog­ nition” have implications for how enzymes recognize their substrates, how hormones cause their effects, how antibodies recognize antigens, how neurotransmitters propagate their sig­ nals, and many other aspects of biochemistry.

Write structures for (a) 15-crown-5 and (b) 12-crown-4.

11.16 Crown Ethers

R -0-h

539

THE CHEMISTRY OF. . . T ra n s p o rt A n tib io tics a n d C ro w n E th ers

There are several antibiotics called ionophores. Some notable examples are monensin, nonactin, gramicidin, and valinomycin. The structures of monensin and nonactin are shown below. lonophore antibiotics like monensin and nonactin coordinate with metal cations in a manner similar to crown ethers. Their mode of action has to do with disrupting the nat­ ural gradient of ions on each side of the cell membrane.

3

M o n e n s in

3

The cell membrane, in its interior, is like a hydrocarbon because it consists in this region primarily of the hydrocarbon portions of lipids (Chapter 23). Normally, cells must maintain a gradient between the concentrations of sodium and potas­ sium ions inside and outside the cell membrane. Potassium ions are "pumped" in, and sodium ions are pumped out. This gradient is essential to the functions of nerves, transport of nutrients into the cell, and maintenance of proper cell vol­ ume. The biochemical transport of sodium and potassium ions through the cell membrane is slow, and requires an

expenditure of energy by the cell. (The 1997 Nobel Prize in Chemistry was awarded in part for work regarding sodium and potassium cell membrane transport.*) Monensin is called a carrier ionophore because it binds with sodium ions and carries them across the cell membrane. Gramicidin and valinomycin are channel-forming antibiotics because they open pores that extend through the mem­ brane. The ion-trapping ability of monensin results princi­ pally from its many ether functional groups, and as such, it is an example of a polyether antibiotic. Its oxygen atoms bind with sodium ions by Lewis acid-base interactions, forming the octahedral complex shown here in the molec­ ular model. The complex is a hydrophobic "host" for the cation that allows it to be carried as a "guest" of monensin from one side of the cell membrane to the other. The trans­ port process destroys the critical sodium concentration gradient needed for cell function. Nonactin is another ionophore that upsets the concentration gradient by binding strongly to potas­ sium ions, allowing the membrane to be permeable to potassium ions, also . . .. . . 1 The io nophore a n tib io tic destr°ying the essential con- monensin com plexed w ith centration gradient. a sodium cation.

C a r r ie r ( le ft) a n d c h a n n e l- f o r m in g m o d e s o f t r a n s p o r t io n o p h o r e s . ( R e p r in te d w it h p e r m is s io n o f J o h n W ile y & S o n s , In c . f r o m V o e t, D . a n d V o e t, J. G . Biochem istry, S e c o n d E d itio n . © 1 9 9 5 V o e t, D . a n d V o e t, J . G .)

* D is c o v e r y a n d c h a r a c t e r iz a tio n o f t h e a c tu a l m o le c u la r p u m p t h a t e s ta b lis h e s t h e s o d iu m a n d p o ta s s iu m c o n c e n tr a tio n g r a d i­ e n t ( N a + , K + -A T P a s e ) e a r n e d J e n s S k o u (A a rh u s U n iv e r s ity , D e n m a r k ) h a lf o f t h e 1 9 9 7 N o b e l P riz e in C h e m is tr y . T h e o t h e r h a lf w e n t t o P a u l D . B o y e r (U C L A ) a n d J o h n E. W a lk e r ( C a m b r id g e ) f o r e lu c id a t in g t h e e n z y m a t ic m e c h a n is m o f A T P s y n th e s is .

5 40

Chapter 11

Alcohols and Ethers

11 .1 7 Summary o f Reactions o f Alkenes, Alcohols, and Ethers H e lp f u l H i n t

Some tools for synthesis

We have studied reactions in this chapter and in Chapter 8 that can be extremely useful in designing syntheses. M ost o f these reactions involving alcohols and ethers are summarized in Fig. 11.4 on the last page o f the chapter, after the Problems. •

We can use alcohols to make alkyl halides, sulfonate esters, ethers, and alkenes.

•

We can oxidize alkenes to make epoxides, diols, aldehydes, ketones, and carboxylic acids (depending on the specific alkene and conditions).

•

We can use alkenes to make alkanes, alcohols, and alkyl halides.

•

If w e have a terminal alkyne, such as could be made from an appropriate vicinal dihalide, w e can use the alkynide anion derived from it to form carbon-carbon bonds by nucleophilic substitution.

A ll together, w e have a repertoire o f reactions that can be used to directly or indirectly inter­ convert almost all o f the functional groups w e have studied so far. In Section 11.17A we summarize som e reactions o f alkenes.

^

11.17A How Alkenes Can Be Used in Synthesis •

Alkenes are an entry point to virtually all o f the other functional groups that we have studied.

For this reason, and because many o f the reactions afford us som e degree o f control over the regiochemical and/or stereochemical form o f the products, alkenes are versatile inter­ mediates for synthesis. •

We have two methods to hydrate a d ouble bond in a M arkovnikov orientation: (1) oxym ercuration-dem ercuration (Section 8 .6 ), and (2) acid-catalyzed hydration (Section 8.5).

O f these methods oxymercuration-demercuration is the m ost useful in the laboratory because it is easy to carry out and is not accom panied by rearrangements. •

We can h ydrate a d ouble bond in an anti-M arkovnikov orien tation by hydroboration-oxidation (Section 8.7). With hydroboration-oxidation w e can also achieve a syn addition o f the H — and — OH groups.

Remember, too, the boron group o f an organoborane can be replaced by hydrogen, d eu ­ terium , or tritium (Section 8.11), and that hydroboration, itself, involves a syn addition o f H 9 and 9 B 9 . •

We can add HX to a d ouble bond in a M arkovnikov sense (Section 8.2) using HF, HCl, HBr, or HI.

•

We can add H B r in an anti-M arkovnikov orien tation (Section 10.9), by treating an alkene with HBr and a peroxide. (The other hydrogen halides do not undergo anti-Markovnikov addition when peroxides are present.)

•

We can add brom ine or chlorine to a d ouble bond (Section 8.12) and the addi­ tion is an anti addition (Section 8.13).

•

We can also add X — and — OH to a double bond (i.e., synthesize a halohydrin) by carrying out a bromination or chlorination in water (Section 8.14). This addi­ tion, too, is an anti addition.

•

We can carry out a syn 1,2-dihydroxylation o f a d ouble bond using either KMnO4 in cold, dilute, and basic solution or O sO 4 follow ed by NaH SO 3 (Section 8.16). O f these two methods, the latter is preferable because o f the tendency of KMnO4 to overoxidize the alkene and cause cleavage at the double bond.

R - ö- h / n - Ö - ft !

Problems •

541

We can carry out anti 1,2-dihydroxylation of a double bond by converting the alkene to an epoxide and then carrying out an acid-catalyzed hydrolysis (Section 11.15).

Equations for most of these reactions are given in the Synthetic Connections reviews for Chapters 7 and 8 and this chapter.

Key Terms and Concepts The key terms and concepts that are highlighted in bold, blue text within the chapter are defined in the glossary (at the back of the book) and have hyperlinked definitions in the accompanying WileyPLUS course (www.wileyplus.com).

PLUS

Problems PLUS

Note to Instructors: Many of the homework problems are available for assignment via WileyPLUS, an online teaching and learning solution. N O M ENCLATURE

11.25

Give an IUPAC substitutive name for each of the following alcohols: OH (a)

(e)

OH

(b)

(f) OH

11.26

HO

Write structural formulas for each of the following: (a) (Z)-But-2-en-1-ol

(e) 2-Chlorohex-3-yn-1-ol

(i) Diisopropyl ether

(b) (Ä)-Butane-1,2,4-triol

(f) Tetrahydrofuran

(j) 2-Ethoxyethanol

(c) (1Ä,2Ä)-Cyclopentane-1,2-diol

(g) 2-Ethoxypentane

(d) 1-Ethylcyclobutanol

(h) Ethyl phenyl ether

REACTIONS A N D SYNTHESIS 11.27

Provide the alkene needed to synthesize each of the following by oxymercuration-demercuration. OH (a)

^

JL

OH (b)

^

OH

X

(d) /

11.28

OH

Provide the alkene needed to synthesize each of the following by hydroboration-oxidation. _OH (a) /

""OH

(b)

^

'"OH

(c)

(d) OH

11.29

Starting with each of the following, outline a practical synthesis of 1-butanol: (a) 1-Butene

11.30

(b) 1-Chlorobutane

(c) 2-Chlorobutane

(d) 1-Butyne

(c) 1-Butene

(d) 1-Butyne

Show how you might prepare 2-bromobutane from (a) 2-Butanol

(b) 1-Butanol

542 11.31

Chapter 11

Alcohols and Ethers

Starting with 2-methylpropene (isobutylene) and using any other needed reagents, outline a synthesis of each of the following (T = tritium, D = deuterium): OH

(a)

(c)

T

(b)

T

O

(d)

D 11.32

Show how you might carry out the following transformations:

(a)

(d)

OH Br

(c) 11.33

What compounds would you expect to be formed when each of the following ethers is refluxed with excess con­ centrated hydrobromic acid?

(a) ^

(

b

)

^

"O '

V

( c M ___ /

(d)

(THF) 11.34

O (1,4-dioxane)

Considering A - L to represent the major products formed in each of the following reactions, provide a structure for each of A through L. I f more than one product can reasonably be conceived from a given reaction, include those as well. L

K

HBr

O) CH3- ; ­ CI PBr

O) pyr

OH

„

C

CH3ONa — 3------- * D

SO2CI

(TBSCI)

pyr H'2~w SO 4 140°

NaF I

H

KI

F

E

G

11.35

Write structures for the products that would be formed under the conditions in Problem 11.34 if cyclopentanol had been used as the starting material. If more than one product can reasonably be conceived from a given reaction, include those as well.

11.36

Starting with isobutane, show how each of the following could be synthesized. (You need not repeat the synthesis of a compound prepared in an earlier part of this problem.) (a) ieri-Butyl bromide (b) 2-Methylpropene (c) Isobutyl bromide

O (g) Isobutyl (h) methyl ether

(i)

CN

(j) CH 3 S

(d) Isobutyl iodide (e) Isobutyl alcohol (two ways) (f) ieri-Butyl bromide

(k)

O

(m)

(l)

OH

NH 2

HO

(n)

HO

543

Problems

11.37

Outlined below is a synthesis of the gypsy moth sex attractant disparlure (a pheromone). Give the structure of disparlure and intermediates A -D . ,

___

1-bromo-5-methylhexane

HC = C Na

: C (C i 9 H3 6 ) 11.38

___ , ,

: A (CgHi6)

liq. NH3

___ ,

1-bromodecane

—

9

H2 Ni B (Fl-2)

____ ,

NaNH2

....................

. |u : B (CgHi 5 Na)

liq. NH3

M CPBA

D (Ci9H38)

.

____ _____

: Disparlure (C igH3 8O)

Provide the reagents necessary for the following syntheses. More than one step may be required.

(a)

(d)

OH OH Br

O

(b)

(f)

(c)

11.39

(e)

OH

Predict the major product from each of the following reactions. OH

soci2 pyr

(a)

NaNH2

'OH

(c)

(1) TsCl, pyr

(e)

OH

(2) EtSNa

OH HBr

(b)

PBr3

(d) OH

11.40

Nal, H2SO4

(f)

'OH

Predict the products from each of the following reactions. (O

(e )

MeOH, cat. H2SO4

(i )

O

(1) EtONa (2) Mel

O (f)

H2SO4 , H2O

(c)

(1) EtSNa (2) H2 O

O (j)

(g)

O

(d)

11.41

O

MeONa

(h)

Provide the reagents necessary to accomplish the following syntheses.

O (a)

MeO

t

MeO (b)

\

SEt

+

/

*

. SEt

Hl

544

Chapter 11

Alcohols and Ethers

(c)

(d)

11.42

Provide reagents that would accomplish the follwing syntheses. OH ---------- >

(a)

HO^

^

\ ^

OH

,O

C' \ / < J

(b)

G lycerol

11.43

E p ic h lo ro h y d rin

Write structures for compounds A -J showing stereochemistry where appropriate. (1) BH3:THF (2) H2O2, OH-

(a)

A TsCl. pyr

KOH C OH MsCI ,> „E pyr

(b)

HC=CNa > F _

(c) OH | MsC^ pyr

* *

MeONaa J J

What is the stereochemical relationship between H and J?

M ECHAN ISM S 11.44

Write a mechanism that accounts for the following reaction:

11.45

Propose a reasonable mechanism for the following reaction. cat. H2SO4

OH HA

11.46

Propose a reasonable mechanism for the following reaction. OH

11.48

HOH

Br

O

O

11.47

Propose a reasonable mechanism for the following reaction. r\

Br HBr

OH H3PO4 (cat.), EtOH

Vicinal halo alcohols (halohydrins) can be synthesized by treating epoxides with HX. (a) Show how you would use this method to synthesize 2-chlorocyclopentanol from cyclopentene. (b) Would you expect the product to be ds-2-chlorocyclopentanol or irans-2-chlorocyclopentanol; that is, would you expect a net syn addition or a net anti addition of — Cl and — OH? Explain.

545

Challenge Problems

11.49

Base-catalyzed hydrolysis of the 1,2-chlorohydrin 1 is found to give a chiral glycol 2 with retention of configura­ tion. Propose a reasonable mechanism that would account for this transformation. Include all formal charges and arrows showing the movement of electrons.

HO

H 1

11.50

Compounds of the type HO,

X , called a-haloalcohols, are unstable and cannot be isolated. Propose a mecha

R

R'

nistic explanation for why this is so 11.51

While simple alcohols yield alkenes on reaction with dehydrating acids, diols form carbonyl compounds. Rationalize mechanistically the outcome of the following reaction: HO

11.52

OH

O

HA

J

When the bicyclic alkene I, a trans-decalin derivative, reacts with a peroxy acid, I I is the major product. What fac­ tor favors the formation of I I in preference to III? (You may find it helpful to build a handheld molecular model.) _

0

%.

-

04

MCPBA H I 11.53

H I I (major)

H I I I (minor)

Use Newman projection formulas for ethylene glycol (1,2-ethanediol) and butane to explain why the gauche con­ former of ethylene glycol is expected to contribute more to its ensemble of conformers than would the gauche con­ former of butane to its respective set of conformers.

Challenge Problems 11.54

When the 3-bromo-2-butanol with the stereochemical structure A is treated with concentrated HBr, it yields meso-2,3dibromobutane; a similar reaction of the 3-bromo-2-butanol B yields (±)-2,3-dibromobutane. This classic experiment performed in 1939 by S. Winstein and H. J. Lucas was the starting point for a series of investigations of what are called neighboring group effects. Propose mechanisms that will account for the stereochemistry of these reactions. Me

Br\ H"^

■ .h

Me

OH

Br.

.HBL

>

A

Me H";; Me

OH

B

H "7 Me

\

Br

2,3-Dibromobutane (meso)

Br versus

Me

\ ___£ + h

HBr

H

Br Me H";; Me

Br

Br

Me Br

2,3-Dibromobutane (racemic)

'H Me

546

Chapter 11

Alcohols and Ethers

11.55

Reaction of an alcohol with thionyl chloride in the presence of a tertiary amine (e.g., pyridine) affords replacement of the OH group by Cl with inversion o f configuration (Section 11.9). However, if the amine is omitted, the result is usually replacement with retention of configuration. The same chlorosulfite intermediate is involved in both cases. Suggest a mechanism by which this intermediate can give the chloro product without inversion.

11.56

Draw all of the stereoisomers that are possible for 1,2,3-cyclopentanetriol. Label their chirality centers and say which are enantiomers and which are diastereomers.

[Hint: Some of the isomers contain a “pseudoasymmetric center,” one that has two possible configurations, each affording a different stereoisomer, each of which is identical to its mirror image. Such stereoisomers can only be distinguished by the order of attachment of R versus S groups at the pseudoasymmetric center. O f these the R group is given higher priority than the S, and this permits assignment of configuration as r or s, lowercase letters being used to designate the pseudoasymmetry.] 11.57 Dimethyldioxirane (D M D O ), whose structure is shown below, is another reagent commonly used for alkene epoxidation. Write a mechanism for the epoxidation of (Z)-2-butene by D M D O , including a possible transition state struc­ ture. What is the by-product of a D M D O epoxidation? O\ > C H 3 O^^CHg Dimethyldioxirane (DMDO) 11.58

Two configurations can actually be envisioned for the transition state in the D M D O epoxidation of (Z )-2-butene, based on analogy with geometric possibilities fitting within the general outline for the transition state in a peroxycarboxylic acid epoxidation of (Z )-2-butene. Draw these geometries for the D M D O epoxidation of (Z )-2-butene. Then, open the molecular models on the book’s website for these two possible transition state geometries in the D M D O epoxidation of (Z )-2-butene and speculate as to which transition state would be lower in energy.

Learning Group Problems 1

.

Devise two syntheses for meso-2,3-butanediol starting with acetylene (ethyne) and methane. Your two pathways should take different approaches during the course of the reactions for controlling the origin of the stereochemistry required in the product.

2

.

(a) Write as many chemically reasonable syntheses as you can think of for ethyl 2-methylpropyl ether (ethyl isobutyl ether). Be sure that at some point in one or more of your syntheses you utilize the following reagents (not all in the same synthesis, however): PBr3 , SOCl2, p-toluenesulfonyl chloride (tosyl chloride), NaH, ethanol, 2-methyl-1-propanol (isobutyl alcohol), concentrated H 2 SO4, Hg(OAc)2, ethene (ethylene). (b) Evaluate the relative merits of your syntheses on the basis of selectivity and efficiency. [Decide which ones could be argued to be the “best” syntheses and which might be “poorer” syntheses.] Synthesize the compound shown below from methylcyclopentane and 2-methylpropane using those compounds as the source of the carbon atoms and any other reagents necessary. Synthetic tools you might need include Markovnikov or anti-Markovnikov hydration, Markovnikov or anti-Markovnikov hydrobromination, radical halogenation, elimination, and nucleophilic substitution reactions.

Learning Group Problems

547

ß -? -H

Sum m ary and Review Tools Som e Synthetic Connections of Alkenes, Alkynes, Alcohols, Alkyl Halides, and Ethers OH OH C C

(1) OsO4 (2) NaHSO3 (syn) H2, Ni2B (syn) or — C=

C—

H2O, HA or HO(anti)

C

(1) Li, EtNH2 (2) NH4+ (anti)

/

RCO3H

\

(a peroxy acid)

C

/ /

HA, heat

Anti-Markovnikov (1) BH3:THF (2) H2O2, HO

W

Nu:

— C— C—

— C— C—

(under acidic or basic conditions)

Nu

- C — C — OTBS

TBSCl imidazole O

RO- , E2 elim. (best if RX is 2° or 3°, or use bulky RO- )

// — C — C — OH

t\

O xidation/Reduction (C hapter 12)

NaH or Na°, K° I. HX (1 ° or 3°) or if 1° o r 2 II. PBr3 or III. SO Cl2

HO-

HBr, ROOR A (AntiMarkov.

/ \

\ Markovnikov ' ( 1 ) Hg(OAc) 2 , T HF/H2O (2) NaBH 4 or H2O, HA cat.

HX (Markov.)

OH

O

\

(1° RX) or H2O (3° RX)

— C— C

Y Y = H, R, o r OH (also by alkene oxidation, see Chapter 8)

— C — C — O~

RX o r ROTs (if RX o r ROTs is 1°)

w C

C — C — OR'

(best if RH is 3°)

* * * * * * * * *

Figure 11.4

Alkynes to alkenes Alkenes and alcohols Alcohols and alkyl halides Alkenes and alkyl halides Alcohols and ethers Alkenes, epoxides, and 1,2-diols Alkanes to alkyl halides Alcohol silyl protecting group Alcohols to carbonyl com pounds

Some syn th etic connections o f alkynes, alkenes, alcohols, alkyl halides, and ethers.

\

Alcohols from Carbonyl Compounds Oxidation-Reduction and Organometallic Compounds

Some reactions w ith carbonyl com pounds involve reagents th a t w e tran sfe r by syringe to keep them away from m oisture and air.

Ask an organic chemist about their favorite functional group, and many will probably name a group that con­ tains a carbonyl group. Why? Because carbonyl groups are at the heart of many key functional groups such as aldehydes, ketones, carboxylic acids, amides, and others. The carbonyl group is also very versatile. It serves as a nexus for interconversions between a number of functional groups. Add to these factors that reactions of car­ bonyl groups include two fascinating and related mechanistic pathways— nucleophilic addition and nucleophilic addition-elim ination— and you have one blockbuster group in terms of its chemistry. Another important aspect of carbonyl groups is that many natural and synthetic compounds contain them. W e have previously mentioned a few, such as vanillin, androsterone, and others. Carbonyl groups are intrinsic to synthetic materials such as nylon and certain other polymers, as well. And, carbonyl groups are central to the organic chemistry of life, as well, which we shall see later when we discuss carbohydrates and other aspects of biological chemistry. Now, therefore, is a good time to introduce you to some methods for interconverting car­ bonyl compounds with alcohols, and how we can use carbonyl compounds for carbon-carbon bond-forming reactions with organometallic reagents. This will prepare us for delving into other aspects of carbonyl chemistry later in the book. W e begin with an introduction and some review.

548

12.1 Structure of the Carbonyl Group

549

12.1 Structure o f the Carbonyl Group Carbonyl compounds are a broad group o f compounds that includes aldehydes, ketones, carboxylic acids, esters, and amides. •'o '-

-''O o '-'

-''O o '-'

A

R

T h e c a rb o n y l g ro u p

-'•O' o '-

-'•O' o '-

Ï

Ï R

OH

A n a ld e h y d e acid

-'o 'OR'

R

A Ank ester e to n e

,R'

A c aArb n oaxmyid lice

The carbonyl carbon atom is sp2 hybridized; thus it and the three atoms attached to it lie in the same plane. The bond angles between the three attached atoms are what we would expect of a trigonal planar structure; they are approximately 120°:

The carbon-oxygen double bond consists o f two electrons in a s bond and two elec­ trons in a p bond. The p bond is formed by overlap o f the carbon p orbital with a p orbital from the oxygen atom. The electron pair in the p bond occupies both lobes (above and below the plane o f the s bonds).

T h e p b o n d in g m o le c u la r o rb ita l of fo rm a ld e h y d e (H C H O ). T h e e le c tro n p a ir o f th e p b o n d o c c u p ie s b o th lob es.

• The more electronegative oxygen atom strongly attracts the electrons of both the s bond and the p bond, causing the carbonyl group to be highly polarized; the car­ bon atom bears a substantial positive charge and the oxygen atom bears a substan­ tial negative charge. Polarization of the p bond can be represented by the following resonance structures for the carbonyl group: 5-

:O:

•O'

'O' or

R e s o n a n c e s tru c tu re s fo r th e c a rb o n y l gro up

H y b rid

Evidence for the polarity of the carbon-oxygen bond can be found in the rather large dipole moments associated with carbonyl compounds. •O'

“O'

F o rm a ld e h y d e p = 2 .2 7 D

p = 2 .8 8 D

A c e to n e

A n e le c tro s ta tic p o te n tial m ap fo r a c e to n e in d ic a te s th e p o la rity o f the ca rb o n y l group.

550

Chapter 12 Alcohols from Carbonyl Compounds

12.1A Reactions of Carbonyl Compounds with Nucleophiles One of the most important reactions of carbonyl compounds is nucleophilic addition to the carbonyl group. The carbonyl group is susceptible to nucleophilic attack because, as we have just seen, the carbonyl carbon bears a partial positive charge. • When a nucleophile adds to the carbonyl group, it uses an electron pair to form a bond to the carbonyl carbon atom and an electron pair from the carbon-oxygen double bond shifts out to the oxygen:

o

■'oy Nu:

v

------ *

+

^

Nu

As the reaction takes place, the carbon atom undergoes a change from trigonal planar geometry and sp2 hybridization to tetrahedral geometry and hybridization.

sp3

• Two important nucleophiles that add to carbonyl compounds are hydride ions from compounds such as NaBH4 or LiAlH4 (Section 12.3) and carbanions from compounds such as RLi or RMgX (Section 12.7C). Another related set of reactions are reactions in which alcohols and carbonyl compounds are oxidized and reduced (Sections 12.2-12.4). For example, primary alcohols can be oxi­ dized to aldehydes, and aldehydes can be reduced to alcohols: :OH

•‘O' oxidation

,.

H

H

reduction

A p rim a ry alc o h o l

R

H

An a ld e h y d e

Let us begin by examining some general principles that apply to the oxidation and reduc­ tion of organic compounds.

12.2 O xidation-R eduction Reactions in O rganic Chem istry

• Reduction of an organic molecule usually corresponds to increasing its hydrogen content or to decreasing its oxygen content. For example, converting a carboxylic acid to an aldehyde is a reduction because the oxy­ gen content is decreased: f

Oxygen content decreases

C a rb o x y lic acid

'v

A ld eh y d e

Converting an aldehyde to an alcohol is a reduction: Hydrogen content increases

O

OH

551

12.2 Oxidation-Reduction Reactions in Organic Chemistry

Converting an alcohol to an alkane is also a reduction:

f

Oxygen content decreases

OH [H]

R

H

H

RCH

reduction

In these examples we have used the symbol [H] to indicate that a reduction of the organic compound has taken place. We do this when we want to write a general equation without specifying what the reducing agent is. • The opposite of reduction is oxidation. Increasing the oxygen content of an organic molecule or decreasing its hydrogen content is an oxidation. The reverse of each reaction that we have just given is an oxidation of the organic mol­ ecule, and we can summarize these oxidation-reduction reactions as shown below. We use the symbol [O] to indicate in a general way that the organic molecule has been oxidized. OH

RCH3

O

[O] "W

O

[O]

[O] R

H

H

"W

R

H

H e lp f u l H i n t

"W

R

OH

H ig hest o x id a tio n s ta te

L o w est o xid atio n sta te

Note the general interpretation of oxidation-reduction regarding organic compounds.

• Oxidation of an organic compound may be more broadly defined as a reaction that increases its content of any element more electronegative than carbon. For example, replacing hydrogen atoms by chlorine atoms is an oxidation: Ar— CH,

[O] I hT

Ar— CH2Cl

[O] IT

Ar— CH Cl,

[O] I hT

Ar— CCl

O f course, when an organic compound is reduced, something else— the reducing agent— must be oxidized. And when an organic compound is oxidized, something else— the oxidizing agent— is reduced. These oxidizing and reducing agents are often inorganic compounds, and in the next two sections we shall see what some of them are.

12.2A Oxidation States in Organic Chemistry One method for assigning oxidation states in organic compounds is similar to the method we used for assigning formal charges (Section 1.7). We base the assignment on the groups attached to the carbon (or carbons) whose oxidation state undergoes change in the reac­ tion we are considering. Recall that with formal charges we assumed that electrons in cova­ lent bonds are shared equally. When assigning oxidation states to carbon atoms we assign electrons to the more electronegative element (see Section 1.4A and Table 1.2). For example, a bond to hydrogen (or to any atom less electronegative than carbon) makes that carbon neg­ ative by one unit (—1), and a bond to oxygen, nitrogen, or a halogen (F, Cl, and Br) makes the carbon positive by one unit (+ 1). A bond to another carbon does not change its oxidation state. Using this method the carbon atom of methane, for example, is assigned an oxidation state of —4, and that of carbon dioxide, + 4.

H e lp f u l H i n t A method fo r balancing organic oxidation-reduction reactions is described in the Study Guide that accompanies this text.

1 Solved Problem 12.1 1 Using the method just described, assign oxidation states to the carbon atoms of methanol (CH 3 OH), formaldehyde (HCHO), and formic acid (HCO 2 H) and arrange these compounds along with carbon cioxide and methane (see above) in order of increasing oxidation state. (co.ntinues on the next page)

552

Chapter 12

Alcohols from Carbonyl Compounds

STRATEGY AND ANSWER We calculate the oxidation state of each carbon based on the number of bonds it is forming to atoms more (or less) electronegative than carbon.

H H— C— OH H

O

B o n d s to C 3 to H = - 3 1 to O = + 1 Total - 2 = O xid. s ta te o f C

H

M eth an o l

/C \

OH

O

B o n d s to C 1 to H = - 1 3 to O = + 3 T otal == + 2 = O xid. s ta te o f C

B o n d s to C 2 to H = - 2 2 to O = + 2 Total = 0 = O xid. s ta te o f C

H ^ H

F o rm a ld e h y d e

F o rm ic acid

The order overall, based on the oxidation state of carbon in each compound, is CH4 = - 4 < CH3OH = - 2 < HCHO = 0 < HCOOH = +2 < CO 2 = +4 Low est O x id . s ta te o f c a rb o n

H ig h e s t O x id . s ta te o f ca rb o n

ReviewProblem12.1

O

O [O]

JOL

vo h

H E th an ol

OH

A c e ta ld e h y d e

A c e tic acid

(a) Assign oxidation states to each carbon of ethanol, acetaldehyde, and acetic acid. (b) What do these numbers indicate about the oxidation reactions?

ReviewProblem12.2

(a) Although we have described the hydrogenation of an alkene as an addition reaction, organic chemists often refer to it as a “reduction.” Use the method described in Section 12.2A for assigning oxidation states to explain this usage of the word “reduction”. H2 Pt (b) Provide a similar analysis for this reaction: O 2

H

OH

Ni

12.3 Alcohols b y Reduction o f Carbonyl Compounds Primary and secondary alcohols can be synthesized by the reduction of a variety of com­ pounds that contain the carbonyl group. Several general examples are shown here: O

R

OH

[H] J- L+

C a rb o x y lic acid

R

/

OH

1 ° A lc o h o l

O

Unless special precautions are taken, lithium aluminum hydride reductions can be very dangerous. You should consult an appropriate laboratory manual before attempting such a reduction, and the reaction should be carried out on a small scale.

[H] R

R

^"O H

(+ R O H )

OR' E ster

1 ° A lc o h o l

O

O

OH

[H] ; R

H

A ld e h y d e

”

[H] R

"'"OH

1° A lcoh ol

R

R'

K e to n e

R

R'

2° A lc o h o l

12.3 Alcohols by Reduction of Carbonyl Compounds

12.3A Lithium Aluminum Hydride • Lithium aluminum hydride (LiAlH4, abbreviated LAH) reduces carboxylic acids and esters to primary alcohols. An example of lithium aluminum hydride reduction is conversion of 2,2-dimethylpropanoic acid to 2 ,2 -dimethylpropanol (neopentyl alcohol). (1) LiAlH4 in Et2O c o 2h

,OH

üyH O H sor

2 ,2 -D im e th y lp ro p a n o ic a cid

N e o p e n ty l alcoh ol (92 % )

LAH reduction of an ester yields two alcohols, one derived from the carbonyl part of the ester group, and the other from the alkoxyl part of the ester. O |

(1) LAH in Et2O (2 ) H2O/H2SO4

R ^^O R '

R'

"OH

+

R'OH

Carboxylic acids and esters are more difficult to reduce than aldehydes and ketones. LAH, however, is a strong enough reducing agent to accomplish this transformation. Sodium borohydride (NaBH4), which we shall discuss shortly, is commonly used to reduce alde­ hydes and ketones, but it is not strong enough to reduce carboxylic acids and esters. Great care must be taken when using LAH to avoid the presence of water or any other weakly acidic solvent (e.g., alcohols). LA H reacts violently with proton donors to release hydrogen gas. Anhydrous diethyl ether (Et2 O) is a commonly used solvent for LAH reduc­ tions. After all of the LAH has been consumed by the reduction step of the reaction, however, water and acid are added to neutralize the resulting salts and facilitate isolation of the alco­ hol products. The stoichiometry of the LAH reduction of a carboxylic acid is shown below. 3 LiAlH 4

4 RCO2H

Lithium a lu m in u m h yd rid e

Et2O

[(RCH2O)4Al]Li +

h2o/h2so.

4 H2

4 RCH2OH

2 LiAlO,

Al2(SO 4)3

Li2SO 4

12.3B Sodium Borohydride • Aldehydes and ketones are easily reduced by sodium borohydride (NaBH4). Sodium borohydride is usually preferred over LAH for the reduction of aldehydes and ketones. Sodium borohydride can be used safely and effectively in water as well as alco­ hol solvents, whereas special precautions are required when using LAH. O H B utanal

O

B u ta n o n e

NaBH. h2o

NaBH, h2o

'OH 1 -B u ta n o l (85% )

OH

2 -B u ta n o l (87% )

Aldehydes and ketones can be reduced using hydrogen and a metal catalyst, as well, and by sodium metal in an alcohol solvent. The stoichiometry of NaBH4 reduction of an aldehyde (or ketone) is as follows. O 4 RCH + NaBH4 + 3 H2O ----- > 4 RCH2OH + NaH2BO3

553

554

Chapter 12 Alcohols from Carbonyl Compounds The key step in the reduction of a carbonyl compound by either lithium aluminum hydride or sodium borohydride is the transfer of a hydride ion from the metal to the car­ bonyl carbon. In this transfer the hydride ion acts as a . The mechanism for the reduction of a ketone by sodium borohydride is illustrated here.

nucleophile

A MECHANISM FOR THE REACTION R e d u c tio n o f A ld e h y d e s a n d K e to n e s b y H y d rid e T ransfer

s-

H

O-

O

y

i

H — OH

I.

OH

H — B— H R'

I

H H y d rid e tra n s fe r

RH

R'

A lk o x id e ion

RH

R'

A lc o h o l

These steps are repeated until all hydrogen atoms attached to boron have been transferred.

THE CHEMISTRY OF . . . A lco h o l D e h y d r o g e n a s e — A B io ch e m ica l H y d rid e R e a g e n t

When the enzyme alcohol dehydrogenase converts acetaldehyde to ethanol, NADH acts as a reducing agent by transferring a hydride from C4 of the nicotinamide ring to the carbonyl group of acetaldehyde. The nitrogen of the nicotinamide ring facilitates this process by contributing its nonbonding electron pair to the ring, which together with loss of the hydride converts the ring to the energetically more stable ring found in NAD+ (we shall see why it is more stable in Chapter 14). The ethoxide anion resulting from hydride transfer to acetaldehyde is then protonated by the enzyme to form ethanol. Although the carbonyl carbon of acetaldehyde that accepts the hydride is inherently electrophilic because of its electronegative oxygen, the enzyme enhances this prop­ erty by providing a zinc ion as a Lewis acid to coordinate with the carbonyl oxygen. The Lewis acid stabilizes the negative charge that develops on the oxygen in the tran­ sition state. The role of the enzyme's protein scaffold, then, is to hold the zinc ion, coenzyme, and substrate in the three-dimensional array required to lower the energy of the transition state. The reaction is entirely reversible, of course, and when the relative concentration of ethanol is high, alcohol dehydrogenase carries out the oxidation of ethanol by removal of a hydride. This role of alcohol dehy­ drogenase is important in detoxification. In "The Chemistry

of . . . Stereoselective Reductions of Carbonyl Groups" we discuss the stereochemical aspect of alcohol dehydroge­ nase reactions.

R

NAD+ Ethanol

12.3C Overall Summary of LiAlH4 and NaBH4 Reactivity Sodium borohydride is a less powerful reducing agent than lithium aluminum hydride. Lithium aluminum hydride reduces acids, esters, aldehydes, and ketones, but sodium boro­ hydride reduces only aldehydes and ketones:

12.3 Alcohols by Reduction of Carbonyl Compounds

555

R edu ced by L iA lH 4 Reduced by N aB H 4

O

A t

O < (A

O r

<

R "^ ^ R '

O

1R > 2R in te rm s o f C a h n -In g o ld -P re lo g p rio ritie s )

The preference of many NADH-dependent enzymes for either the re or si face of their respective substrates is known. This knowledge has allowed some of these enzymes to become exceptionally useful stereoselective reagents for synthesis. One of the most widely used is yeast alcohol dehydrogenase. Others that have become important are enzymes from thermophilic bacteria (bacteria that grow at elevated temperatures). Use of heat-stable enzymes (called extremozymes) allows reactions to be completed faster due to the rate-enhancing factor of elevated temperature (over 100 °C in some cases), although greater enantioselectivity is achieved at lower temperatures.

Prochirality

A second aspect of the stereochemistry of NADH reactions results from NADH having two hydrogens at C4, either of which could, in principle, be transferred as a hydride in a reduction process. For a given enzymatic reaction, however, only one specific hydride from C4 in NADH is transferred. Just which hydride is transferred depends on the specific enzyme involved, and we designate it by a useful extension of stereochemical nomenclature. The hydrogens at C4 of NADH are said to be prochiral. We designate one pro-R, and the other pro-S, depending on whether the configura­ tion would be R or S when, in our imagination, each is replaced by a group of higher priority than hydrogen. If this exercise produces the R configuration, the hydrogen "replaced" is pro-R, and if it produces the S configuration it is pro-S. In general, a prochiral center is one for which addi­ tion of a group to a trigonal planar atom (as in reduction of a ketone) or replacement of one of two identical groups at a tetrahedral atom leads to a new chirality center. HR

O R— N

Therm oanaerobium brockii

HS HO

H

9 6% e n a n tio m e ric e x c e s s (85% y ie ld )

'C = O

N ic o tin a m id e rin g o f N A D H , s h o w in g th e p ro -R and p ro -S h y d ro g e n s

1 2 .4 O x id a tio n o f A lc o h o ls

£rf|

557

12.4 O xidation o f Alcohols 12.4A Oxidation of Primary Alcohols to Aldehydes: RCH2O H -----> RCHO Primary alcohols can be oxidized to aldehydes and carboxylic acids:

O r^

- oh

*

-iOL

OH

J

R

1° A lc o h o l

O

L

"H

Jo u

A ld e h y d e

„ A .

R

"OH

C a rb o x y lic acid

•

The oxidation o f aldehydes to carboxylic acids in aqueous solutions is easier than oxidation o f primary alcohols to aldehydes.

•

It is, therefore, difficult to stop the oxidation o f a primary alcohol at the aldehyde stage unless specialized reagents are used.

An excellent reagent to use for converting a primary alcohol to an aldehyde is pyridirnum chlorochrom ate (abbreviated PCC), the compound formed when CrO3 is dissolved in hydrochloric acid and then treated with pyridine: CrO3 +

HCl

+

H P y rid in e (C 5 H 5 N )

•

CrO3C h

P y rid in iu m c h lo ro c h ro m a te (P C C )

PCC, when dissolved in m ethylene chloride (CH2Cl2), w ill oxidize a primary alco­ hol to an aldehyde and stop at that stage: O

OH

h

PCC 2 -E th y l-2 -m e th y l-1 b utano l

•

2 -E th y l-2 -m e th y lb u ta n a l

PCC w ill also oxidize a secondary alcohol to a ketone.

OH

PCC

O

CH2Cl2 Pyridinium chlorochromate does not attack double bonds. One reason for the success o f oxidation with pyridinium chlorochromate is that the o xi­ dation can be carried out in a solvent such as CH2Cl2, in which PCC is soluble. Aldehydes themselves are not nearly so easily oxidized as are the aldehyde hydrates, RCH(OH)2, that form (Section 16.7A) when aldehydes are dissolved in water, the usual medium for oxida­ tion by chromium compounds:

O A R

+

H2O e f

H

HO OH V / R H

We explain this further in Section 12.4D.

12.4B Oxidation of Primary Alcohols to Carboxylic Acids: RCH2O H -----> RCO2H •

Primary alcohols can be oxidized to carboxylic acids by potassium permanganate (KMnO4), or chromic acid (H2CrO4).

558

Chapter 12 Alcohols from Carbonyl Compounds (Both KMnO4 and H2 CrO4 can also be used to oxidize a secondary alcohol to a ketone, as we shall see in Section 12.4C.) The reaction with KMnO4 is usually carried out in basic aqueous solution, from which MnO2 precipitates as the oxidation takes place. After the oxidation is complete, filtration allows removal of the MnO2 and acidification of the fil­ trate gives the carboxylic acid: O II

KMnO4, OH~ R

OH

H2O, a

O II

r ^ S x k+

MnO2

r^ S d h

The following is an example with H2 CrO4. O H2CrO4 'OH

^

^

OH

12.4C Oxidation of Secondary Alcohols to Ketones: OH O RCHR'— > RCR' Secondary alcohols can be oxidized to ketones. The reaction usually stops at the ketone stage because further oxidation requires the breaking of a carbon-carbon bond: OH

O

A

^

2° A lc o h o l

RX

R,

K eto n e

Various oxidizing agents based on Cr(VI) have been used to oxidize secondary alcohols to ketones. The most commonly used reagent is chromic acid (H 2 CrO4). Chromic acid is usually prepared by adding Cr(VI) oxide (CrO3) or sodium dichromate (Na2 Cr2 O7) to aque­ ous sulfuric acid, a mixture known as Jones reagent. Oxidations of secondary alcohols are generally carried out by adding Jones reagent to a solution of the alcohol in acetone or acetic acid. This procedure rarely affects double bonds present in the molecule. The balanced equa­ tion is shown here: OH 3

/k R

H e lp f u l H i n t ____ ^ The color change from orange to green that accompanies this change in oxidation state allows chromic acid to be used as a test fo r primary and secondary alcohols (Section 12.4E).

O + 2 H2CrO4 + R'

2

6

H+ ----- > 3

+ 2 Cr3+ + R

4

R'

8

H2O 2

As chromic acid oxidizes the alcohol to the ketone, chromium is reduced from the + 6 oxi­ dation state (H 2 CrO4) to the +3 oxidation state (Cr3+). Chromic acid oxidations of sec­ ondary alcohols generally give ketones in excellent yields if the temperature is controlled. A specific example is the oxidation of cyclooctanol to cyclooctanone: O acetone 35°C C y c lo o c ta n o l

C y c lo o c ta n o n e (9 2 -9 6 % )

PCC w ill also oxidize a secondary alcohol to a ketone.

12.4D Mechanism of Chromate Oxidations The mechanism of chromic acid oxidations of alcohols has been investigated thoroughly. It is interesting because it shows how changes in oxidation states occur in a reaction between an organic and an inorganic compound. The first step is the formation of a chromate ester of the alcohol. Here we show this step using a 2° alcohol.

559

12.4 Oxidation o f Alcohols

A MECHANISM FOR THE REACTION C h r o m a te O x id a tio n s: F o rm a tio n o f th e C h r o m a te E ste r H

H

■o 'Step 1

H- \ H3 C

; ° / C\

H

2° A lcoh ol

0

H“ °

H

H3C

- I? °+

H — ° — C r— 0 °

\ /

Ö II C r— 0 : /

C

* , •H° ^ O ^ - H

H3C

h

\

'■

0

0

h

h

:^

••+ H

6

.

r

H

H

6 .1

H

O n e o xy g e n lo s e s a p roton; a n o th e r

T h e a lc o h o l d o n a te s an e le c tro n p air to th e

° x y g e n a c c e p ts a p ro to n .

c h ro m iu m ato m , as an o xy g e n a c c e p ts a p roton.

'O'' H3C 3

°

C r^ O ’

/

H3C

\

O'-

H

H3C

■O- -O— r

H

/

•

I

H3C

H

••

O' H

c

H

I

C r= 0

° \

0

H

=0 — H

I H

C h ro m a te e s te r A m o le c u le o f w a te r d e p a rts as a le a vin g g ro u p a s a c h ro m iu m -o x y g e n d o u b le bond fo rm s . T h e c h ro m a te e s te r is u n s ta b le a n d is n ot is o la te d . It tra n s fe rs a p ro to n to a b as e (u s u a lly w a te r) a n d s im u lta n e o u s ly e lim in a te s an H C rO 3~ ion.

C h r o m a te O x id a tio n s: T h e O x id a tio n S te p

.

H3C Step 2

°

3 x/V / c H3C

h 3c

CM II

\

C r= 0 1

:° :

0

H3C

/

^

°

K eto n e

H H

+ C ^O :°:

+ H— O— H H

I H

H— O = I

H T h e c h ro m iu m a to m d e p a rts w ith a p air of e le c tro n s th a t fo rm e rly b elo ng ed to th e a lc o h o l; th e a lc o h o l is th e re b y o xid ize d and th e c h ro m iu m red uced .

The overall result of the second step is the reduction of HCrO4~ to HCrO3~, a two-elec­ tron (2 e~) change in the oxidation state of chromium, from Cr(VI) to Cr(IV). At the same time the alcohol undergoes a 2 e~ oxidation to the ketone. The remaining steps of the mechanism are complicated and we need not give them in detail. Suffice it to say that further oxidations (and disproportionations) take place, ultimately converting Cr(IV) compounds to Cr3+ ions. The requirement for the formation of a chromate ester in step 1 of the mechanism helps us understand why 1 ° alcohols are easily oxidized beyond the aldehyde stage in aqueous solutions (and, therefore, why oxidation with PCC in CH 2 Cl2 stops at the aldehyde stage).

560

Chapter 12 Alcohols from Carbonyl Compounds The aldehyde initially formed from the 1° alcohol (produced by a mechanism similar to the one we have just given) reacts with water to form an aldehyde hydrate. The aldehyde hydrate can then react with HCrO4—(and H+) to form a chromate ester, and this can then be oxidized to the carboxylic acid. In the absence of water (i.e., using PCC in CH 2 Cl2), the aldehyde hydrate does not form; therefore, further oxidation does not take place. R

R

\

C =O

h'

V , Ö - H

/ C

O-

R H

V

O

H

HCrO4- , H-

X O— H

A ld e h y d e h yd ra te

O

x cx

0 = 1 H

/

H I O

R H

C

H

I H

O

\

‘ h ^ C ^ O — C r-

R OH

\

O

C

/

OH +

H3O4

HCrO,

O C a rb o x y lic acid

The elimination that takes place in step 2 of the preceding mechanism helps us to under­ stand why 3° alcohols do not generally react in chromate oxidations. Although 3° alcohols have no difficulty in forming chromate esters, the ester that is formed does not bear a hydro­ gen that can be eliminated, and therefore no oxidation takes place. O

R R

\ /

C

/ \

O— H R

O II Cr II O

II R

O — H + H-

3° A lcoh ol

O — C r— O — H

V

■

■

/

R

1

C

H 2O

O R

T h is c h ro m a te ester c a n n o t u n d erg o e lim in a tio n o f H 2 C rO 3 b e c a u s e th e re is no h yd ro g e n b o n d e d to th e a lc o h o l c arb on .

12.4E A Chemical Test for Primary and Secondary Alcohols The relative ease of oxidation of primary and secondary alcohols compared with the diffi­ culty of oxidizing tertiary alcohols forms the basis for a convenient chemical test. Primary and secondary alcohols are rapidly oxidized by a solution of CrO 3 in aqueous sulfuric acid. Chromic oxide (CrO3) dissolves in aqueous sulfuric acid to give a clear orange solution containing Cr2 O72— ions. A positive test is indicated when this clear orange solution becomes opaque and takes on a greenish cast within 2 seconds:

RCH2OH or + CrO 3 /aqueous H2SO4 ----RCHOH

*Cr3+ and oxidation products

R '---------------------------------v---------------------------------'

'-------------------------------v-------------------------------'

C le a r o ra n g e s o lu tio n

G re e n ish o p a q u e so lu tio n

Not only will this test distinguish primary and secondary alcohols from tertiary alcohols, it will distinguish primary and secondary alcohols from most other compounds except alde­ hydes. This color change, associated with the reduction of Cr2 O72—to Cr3+, is also the basis for “Breathalyzer tubes,” used to detect intoxicated motorists. In the Breathalyzer the dichromate salt is coated on granules of silica gel.

12.5 Organometallic Compounds

561

SolvedProblem12.2 Which reagents would you use to accomplish the following transformations? O ’"""OH f

Y

(a) 5

OH

‘'TbT O

kJ

f Y

OH

(c)

il

l"

H

^IdT

STRATEGY AND ANSWER

(a) To oxidize a primary alcohol to a carboxylic acid, use (1) potassium permanganate in aqueous base, followed by (2) H3 O +, or use chromic acid (H 2 CrO4). (b) To reduce a carboxylic acid to a primary alcohol, use LiAlH4. (c) To oxidize a primary alcohol to an aldehyde, use pyridinium chlorochromate (PCC). (d) To reduce an aldehyde to a primary alcohol, use NaBH4 (preferably) or LiAlH4.

Show how each of the following transformations could be accomplished:

ReviewProblem12.4

12.4F Spectroscopic Evidence for Alcohols • Alcohols give rise to broad O— H stretching absorptions from 3200 to 3600 cm- 1 in infrared spectra. • The alcohol hydroxyl hydrogen typically produces a broad 1H NM R signal of vari­ able chemical shift which can be eliminated by exchange with deuterium from D2O (see Table 9.1). • Hydrogen atoms on the carbon of a primary or secondary alcohol produce a signal in the 1H NM R spectrum between S 3.3 and S 4.0 (see Table 9.1) that integrates for 2 and 1 hydrogens, respectively. • The 13C NM R spectrum of an alcohol shows a signal between S 50 and S 90 for the alcohol carbon (see Table 9.2).

12.5 O rganom etallic Compounds

• Compounds that contain carbon-metal bonds are called organometallic compounds. The nature of the carbon-metal bond varies widely, ranging from bonds that are essentially ionic to those that are primarily covalent. Whereas the structure of the organic portion of

562

Chapter 12 Alcohols from Carbonyl Compounds the organometallic compound has some effect on the nature of the carbon-metal bond, the identity of the metal itself is of far greater importance. Carbon-sodium and carbon-potas­ sium bonds are largely ionic in character; carbon-lead, carbon-tin, carbon-thallium, and carbon-mercury bonds are essentially covalent. Carbon-lithium and carbon-magnesium bonds lie between these extremes. |

|s - s+

— C :- M+

—C

P rim a rily io n ic (M = N a + o r K+)

_____ H e lp f u l H i n t A number o f organometallic reagents are very useful for carbon-carbon bond forming reactions (see Section 12.8, and Special Topic G).

M

—C ~ M P rim a rily c o v a le n t (M = Pb, Sn, Hg, o r T l)

(M = M g o r Li)

The reactivity of organometallic compounds increases with the percent ionic character of the carbon-metal bond. Alkylsodium and alkylpotassium compounds are highly reactive and are among the most powerful of bases. They react explosively with water and burst into flame when exposed to air. Organomercury and organolead compounds are much less reactive; they are often volatile and are stable in air. They are all poisonous. They are generally soluble in nonpolar solvents. Tetraethyllead, for example, was once used as an “antiknock” compound in gasoline, but because of the lead pollution it contributed to the environment it has been replaced by other antiknock agents. tert-Butyl methyl ether is another antiknock additive, though there are concerns about its presence in the environment, as well. Organometallic compounds of lithium and magnesium are of great importance in organic synthesis. They are relatively stable in ether solutions, but their carbon-metal bonds have considerable ionic character. Because of this ionic nature, the carbon atom that is bonded to the metal atom of an organolithium or organomagnesium compound is a strong base and powerful nucleophile. We shall soon see reactions that illustrate both of these properties.

12.6 Preparation o f O rganolithium and Organom agnesium Com pounds 12.6A Organolithium Compounds Organolithium compounds are often prepared by the reduction of organic halides with lithium metal. These reductions are usually carried out in ether solvents, and since organolithium compounds are strong bases, care must be taken to exclude moisture. (Why?) The ethers most commonly used as solvents are diethyl ether and tetrahydrofuran. (Tetrahydrofuran is a cyclic ether.)

D ieth yl e th e r (E t2O )

T e tra h y d ro fu ra n (TH F )

• Organolithium compounds are prepared in this general way: R— X

+

2 Li

Et2°

(o r A r — X )

> RLi +

LiX

(o r A rL i)

The order of reactivity of halides is RI > RBr > RCl. (Alkyl and aryl fluorides are seldom used in the preparation of organolithium compounds.) For example, butyl bromide reacts with lithium metal in diethyl ether to give a solution of butyllithium: Br B utyl b ro m id e

+

2 Li

^

Et2°, -1 0 ° C

Li

+

LiBr

B u ty llith iu m (8 0 -9 0 % )

Several alkyl- and aryllithium reagents are commercially available in hexane and other hydrocarbon solvents.

12.7 Reactions of Organolithium and Organomagnesium Compounds

563

12.6B Grignard Reagents Organomagnesium halides were discovered by the French chemist Victor Grignard in 1900. Grignard received the Nobel Prize for his discovery in 1912, and organomagnesium halides are now called G rignard reagents in his honor. Grignard reagents have great use in organic synthesis. • Grignard reagents are prepared by the reaction of an organic halide with magne­ sium metal in an anhydrous ether solvent: Et2O RX RMgX Mg G r ig n a r d ArX

r e a g e n ts

Et2O

Mg

ArMgX _

The order of reactivity of halides with magnesium is also RI > RBr > RCl. Very few organo­ magnesium fluorides have been prepared. Aryl Grignard reagents are more easily prepared from aryl bromides and aryl iodides than from aryl chlorides, which react very sluggishly. Once prepared, a Grignard reagent is usually used directly in a subsequent reaction. The actual structures of Grignard reagents are more complex than the general formula RMgX indicates. Experiments have established that for most Grignard reagents there is an equilibrium between an alkylmagnesium halide and a dialkylmagnesium. 2 RMgX

R2 Mg

A lk y lm a g n e s iu m h a lid e

D ia lk y lm a g n e s iu m

+

MgX2

For convenience in this text, however, we shall write the formula for the Grignard reagent as though it were simply RMgX. A Grignard reagent forms a complex with its ether solvent; the structure of the complex can be represented as follows: R

R

Y R—

Mg— X O

R/ \ Complex formation with molecules of ether is an important factor in the formation and sta­ bility of Grignard reagents. The mechanism by which Grignard reagents form is complicated and has been a mat­ ter of debate. There seems to be general agreement that radicals are involved and that a mechanism similar to the following is likely: ----- > R-

+

R— X

+

:Mg

R-

+

-MgX ----- > RMgX

-MgX

12.7 Reactions o f O rganolithium and Organom agnesium Compounds 12.7A Reactions with Compounds Containing Acidic Hydrogen Atoms • Grignard reagents and organolithium compounds are very strong bases. They react with any compound that has a hydrogen atom attached to an electronegative atom such as oxygen, nitrogen, or sulfur. We can understand how these reactions occur if we represent the Grignard reagent and organolithium compounds in the following ways: a-

s+

R :MgX

a - s+

and

R =Li

564

Chapter 12 Alcohols from Carbonyl Compounds When w e do this, w e can see that the reactions o f Grignard reagents with water and alco­ hols are nothing more than acid-base reactions; they lead to the formation o f the weaker conjugate acid and weaker conjugate base. •

A Grignard reagent behaves as if it contained the anion o f an alkane, as if it con­ tained a carbanion: s-

5+

5+

R — MgX Stronger base

s-

R -H

Stronger acid (pKa 15.7)

HO -

Mg2

X-

Mg2

X-

Weaker Weaker acid base (pKa 40-50)

5+

R — MgX

R— H

H- o — R

Stronger base

TS o lv e d P ro b le m 1 2 .3

••

H- Q - H

Stronger acid (pKa 15-18)

RQ:-

Weaker Weaker acid base (pKa 40-50)

1

Write an equation for the; reaction that would take place when phenyllithium is treated with water. D esignate the stronger acid and stronge r base. STRATEGY AND ANSW ER Recognizing that phenyllithium, like a Grignard reagent, acts as though it contains a carbanion, a very powerf ul base (pKa = 4 0 -5 0 ), w e conclude that the follow ing acid-base reaction would occur.

Ar :^Li Stronger base

ReviewProblem12.5

^ iH o H

----- »

Stronger acid

Am H + Weaker acid

HOr

+

Li+

Weaker base

Predict the products o f the follow ing acid-base reactions. Using p K a values, indicate which side o f each equilibrium reaction is favored, and label the species representing the stronger acid and stronger base in each case.

(b)

MgBr

H2O

MeOH

(d) OH

'Li OH

ReviewProblem12.6

Provide the reagents necessary to achieve the follow ing transformations

(a)

^

^Br

,D

(b) 5

3 D (D = deuterium)

Grignard reagents and organolithium compounds remove protons that are much less acidic than those o f water and alcohols.

12.7 Reactions of Organolithium and Organomagnesium Compounds •

Grignard reagents react with the terminal hydrogen atoms o f 1-alkynes by an acid-base reaction, and this is a useful m ethod for the preparation o f alkynylmagnesium halides and alkynyllithiums.

R'—# —H

R 9 MgX

T e rm in a l a lk y n e (s tro n g e r acid,

G rig nard re a g e n t (s tro n g e r b ase)

.

R'—# :

MgX

R—

A lk y n y lm a g n e s iu m h alid e (w e a k e r base)

H

A lk a n e (w e a k e r acid, pKa 4 0 -5 0 )

pKa ~ 2 5 )

R'—# —H

R'—#

T e rm in a l a lk yn e (s tro n g e r acid)

A lk y l­ lithium (s tro n g e r b ase)

Li

R

A lk y n y llith iu m (w e a k e r base)

—H

A lk a n e (w e a k e r ac id )

The fact that these reactions go to completion is not surprising when w e recall that alkanes have pK a values o f 4 0 -5 0 , whereas those o f terminal alkynes are ~ 2 5 (Table 3.1). N ot only are Grignard reagents strong bases, they are also pow erful nucleophiles. •

Reactions in which Grignard reagents act as nucleophiles are by far the m ost important and w e shall consider these next.

12.7B Reactions of Grignard Reagents with Epoxides (Oxiranes) •

Grignard reagents react as nucleophiles with epoxides (oxiranes), providing conve­ nient synthesis o f alcohols.

The nucleophilic alkyl group o f the Grignard reagent attacks the partially positive carbon o f the epoxide ring. Because it is highly strained, the ring opens, and the reaction leads to the alkoxide salt o f an alcohol. Subsequent acidification produces the alcohol. (Compare this reaction with the base-catalyzed ring opening w e studied in Section 11.14.) The fol­ low ing are exam ples with oxirane.

R— MgX

„O:-M g 2+X-

¿ 'Ö '

R

H3O

O x ira n e

MgBr

O /

•

\

OH R A p rim a ry alc o h o l

____ ä

OMgBr

h 3o +

OH

Et2O

Grignard reagents react primarily at the less-substituted ring carbon atom o f a sub­ stituted epoxide.

MgBr

O

h 3o +

Et>O

OMgBr

OH

12.7C Reactions of Grignard Reagents with Carbonyl Compounds •

The m ost important synthetic reactions o f Grignard reagents and organolithium compounds are those in which they react as nucleophiles and attack an unsaturated carbon— especially the carbon o f a carbonyl group.

565

5ÓÓ

Chapter 12 Alcohols from Carbonyl Compounds W e saw in S ection 1 2 .1 A th a t c a rb o n y l com pounds are h ig h ly susceptible to n u c le o p h ilic attack. G rig n a rd reagents re a c t w ith c a rb o n y l com pounds (ald ehyd es and keto nes) in the fo llo w in g w ay.

A MECHANISM FOR THE REACTION T h e G rig n a rd R e ac tio n R E A C T IO N

OH

(1 ) ether* (2 ) HsO+ X-

M gX

R

M gX2 R

M E C H A N IS M

Step 1 sR

: O : M g 2+ X -

s+ M gX

R

G rig n a rd re a g e n t

C arb on yl com pound

H a lo m a g n e s iu m a lk o x id e

T h e s tro n g ly n u c le o p h ilic G rig n a rd re a g e n t u ses its e le ctro n p air to form a b o n d to th e c a rb o n ato m . O ne e le ctro n p air o f th e c a rb o n y l g ro u p s h ifts o u t to the o xy g e n . T h is re a c tio n is a n u c le o p h ilic a d d itio n to the c a rb o n y l g ro up , a n d it re s u lts in th e fo rm a tio n o f an a lk o x id e ion a s s o c ia te d w ith M g 2+ a n d X~.

Step 2

:O= M g 2+ X -

H

H

d

=O '

+

H— O— H

H— O— H R

M gX2

R

A lcoh ol

H a lo m a g n e s iu m a lk o x id e

In th e s e c o n d ste p , th e a d d itio n o f a q u e o u s H X c a u s e s p ro to n a tio n o f th e a lk o x id e ion; th is leads to th e fo rm a tio n o f th e a lc o h o l an d M g X 2.

*By writing "(1) ether" over the arrow and "(2) H3O+ X— " under the arrow, we mean that in the first laboratory step the Grignard reagent and the carbonyl com­ pound are allowed to react in an ether solvent. Then in a second step, after the reaction of the Grignard reagent and the carbonyl compound is over, we add aqueous acid (e.g., dilute HX) to convert the salt of the alcohol (ROMgX) to the alcohol itself. If the alcohol istertiary, it will be susceptible to acid-catalyzed dehy­ dration. Inthis case, a solution of NH4CI in water is often used because it is acidic enough to convert ROMgXto ROH while not allowing acid-catalyzed reactions of the resulting tertiary alcohol.

12.8 Alcohols from G rignard Reagents G rig n a rd a dd itio ns to c a rb o n y l com pounds are e s p e c ia lly u se fu l because they can be used to p re p a re p rim a ry , secondary, o r te rtia ry alcohols:

1. G rignard R eagents R eact w ith F orm aldehyde to G ive a P rim ary A lcohol *O ‘

M

i L +J

-

:O H

—

F o rm a ld e h y d e

R

M gX

A h

:O H

^

R

A h

1° A lcoh ol

567

12.8 Alcohols from Grignard Reagents 2. G rignard R eagents R eact w ith A ll O ther A ldehydes to G ive Secondary A lcohols

:O MgX

O sR

:O H

s+

H3O

MgX H

R

H R'

R

H ig her ald e h y d e

H R'

2° A lcoh ol

3. G rignard R eagents R eact w ith K etones to G ive Tertiary A lcohols sR—

'.O'

s+

MgX

R 7%

:Q MgX R

R"

\

:OH NH.Cl H2O

R' R"

R

K eton e

R' R"

3° A lcoh ol

4. Esters R eact with Two M olar Equivalents o f a Grignard R eagent to Form Tertiary A lcohols When a Grignard reagent adds to the carbonyl group o f an ester, the initial product is unstable and loses a magnesium alkoxide to form a ketone. Ketones, how­ ever, are more reactive toward Grignard reagents than esters. Therefore, as soon as a m olecule o f the ketone is formed in the mixture, it reacts with a second m olecule of the Grignard reagent. After hydrolysis, the product is a tertiary alcohol with two iden­ tical alkyl groups, groups that correspond to the alkyl portion o f the Grignard reagent:

;o sR

,:O

s+

MgX R'

OR"

X

MgX -R"OMgX ' spontaneously

OR"

R Initial p ro d u ct (u n s ta b le )

Ester

O'

:O

MgX

R'

R

R

R

:OH

NH.Cl H2O

RMgX

R 3° A lcoh ol (w ith tw o o r th ree id e n tic a l R g ro u p s)

S a lt o f an a lcoh ol (n o t is o la te d )

K eto n e

R

R'

R

Specific examples o f these reactions are shown here. Grignard Reagent

Carbonyl Reactant

Final Product

Reaction with Formaldehyde MgBr

O H

P h e n y lm ag n e s iu m b ro m id e

O M gBr H

OH

H3O

Et2O

F o rm a ld e h y d e

B enzyl a lcoh ol (90% )

Reaction with a Higher Aldehyde O

MgBr

^ E th y lm a g n e s iu m b ro m id e

O M gBr

+ H

A c e ta ld e h y d e

H3O

OH

Et2O 2 -B u ta n o l (80% )

568

Chapter 12 Alcohols from Carbonyl Compounds

Reaction with a Ketone MgBr

NH4Cl h2o

Et2O

B u ty lm a g n e s iu m b ro m id e

OH

O M gBr

O

2 -M e th y l-2 -h e x a n o l (92% )

A c e to n e

Reaction with an Ester O

O M gBr

MgBr

OEt

-EtOMgBr

N

OEt

E th ylm a g n e siu m b ro m id e

Ethyl a c e ta te

OMgBr

O

OH NH4Cl H2O

MgBr

Solved Problem 12.4 How would you carry out the following synthesis? O OH O

HO.

STRATEGY AND ANSWER Here we are converting an ester (a cyclic ester) to a tertiary alcohol with two iden­ tical alkyl groups (methyl groups). So, we should use two molar equivalents of the Grignard reagent that contains

the required alkyl groups, in this case, methyl magnesium iodide. »O CH3—MgI

OH NH.CI H2O

HO

ReviewProblem12.7

Provide a mechanism for the following reaction, based on your knowledge of the reaction of esters with Grignard reagents. O

(1) Cl (2 ) NH4CI

MgBr

OH (2

equiv.)

12.8 Alcohols from Grignard Reagents

12.8A How to Plan a Grignard Synthesis We can synthesize almost any alcohol we wish by skillfully using a Grignard synthesis. In planning a Grignard synthesis we must simply choose the correct Grignard reagent and the correct aldehyde, ketone, ester, or epoxide. We do this by examining the alcohol we wish to prepare and by paying special attention to the groups attached to the carbon atom bear­ ing the — OH group. Many times there may be more than one way o f carrying out the syn­ thesis. In these cases our final choice w ill probably be dictated by the availability o f starting compounds. Let us consider an example. Suppose we want to prepare 3-phenyl-3-pentanol. We examine its structure and we see that the groups attached to the carbon atom bearing the — OH are a phenyl group and two ethyl groups: OH

3 -P h e n y l-3 -p e n ta n o l

This means that we can synthesize this compound in several different ways: 1. We can use a ketone with two ethyl groups (3-pentanone) and allow it to react with

phenylmagnesium bromide: Retrosynthetic Analysis

OH O

MgBr

Synthesis

OH O

MgBr

(1) Et2O (2) NH4Cl, H2 O P h e n y lm a g n e s iu m b ro m id e

3 -P e n ta n o n e

3 -P h e n y l-3 -p e n ta n o l

2. We can use a ketone containing an ethyl group and a phenyl group (ethyl phenyl

ketone) and allow it to react with ethylmagnesium bromide: Retrosynthetic Analysis

OH

O

MgBr

O

OH ( 1 ) Et2 O

MgBr

(2 ) NH4 CI, H2 O

E th y lm a g n e s iu m b ro m id e

E thyl p henyl k e to n e

3 -P h e n y l-3 -p e n ta n o l

569

570

Chapter 12

Alcohols from Carbonyl Compounds

3. We can use an ester o f benzoic acid and allow it to react with two molar equivalents o f ethylmagnesium bromide: Retrosynthetic Analysis

OH

O OMe

2

MgBr

O 2

OH OMe

,MgBr

E th y lm a g n e s iu m b ro m id e

(1) Et2O (2 ) NH4CI, h2o

M eth yl b e n zo a te

3 -P h e n y l-3 -p e n ta n o l

A ll o f these methods w ill likely give us our desired compound in high yields.

Solved Problem 12.5 ILLUSTRATING A MULTISTEP SYNTHESIS Using an alcohol o f no more than four carbon atoms as your only organic starting material, outline a synthesis o f A:

O A ANSWER We can construct the carbon skeleton from two four-carbon compounds using a Grignard reaction. Then oxidation o f the alcohol produced w ill yield the desired ketone. Retrosynthetic Analysis R e tro s y n th e tic d is c o n n e c tio n

H MgBr OH

O C

Synthesis

H MgBr O B

( 1 ) Et2 O

H 2 CrO 4

(2 ) H 3 O+

acetone

OH

C

We can synthesize the Grignard reagent (B) and the aldehyde (C) from isobutyl alcohol:

OH

PBr3

OH

Br

PCC CH2Cl2

C

Mg Et2O '

B

A

571

12.8 Alcohols from Grignard Reagents

Solved Problem 12.6 ILLUSTRATING A MULTISTEP SYNTHESIS Starting with bromobenzene and any other needed reagents, outline

a synthesis of the following aldehyde:

ANSWER Working backward, we remember that we can synthesize the aldehyde from the corresponding alcohol by oxidation with PCC (Section 12.4A). The alcohol can be made by treating phenylmagnesium bromide with oxirane. [Adding oxirane to a Grignard reagent is a very useful method for adding a — CH 2 CH2OH unit to an organic group (Section 12.7B).] Phenylmagnesium bromide can be made in the usual way, by treating bromobenzene with magnesium in an ether solvent. Retrosynthetic Analysis

Provide retrosynthetic analyses and syntheses for each of the following alcohols, starting with appropriate alkyl or aryl halides. .OH

ReviewProblem12.8

OH (three ways)

(a)

(two ways)

(e) OH (three ways)

(b)

Provide a retrosynthetic analysis and synthesis for each of the following compounds. Permitted starting materials are phenylmagnesium bromide, oxirane, formaldehyde, and alco­ hols or esters of four carbon atoms or fewer. You may use any inorganic reagents and oxi­ dizing agents such as pyridinium chlorochromate (PCC). OH

(a) [

O

(b)

OH

OH

H

r

1

(c)

\

ReviewProblem12.9

(d)

^

572

Chapter 12 Alcohols from Carbonyl Compounds

12.8B Restrictions on the Use of Grignard Reagents Although the Grignard synthesis is one of the most versatile of all general synthetic procedures, it is not without its limitations. Most of these limitations arise from the very feature of the Grignard reagent that makes it so useful, its extraordinary reactivity as a nucleophile and a base. The G rign ard rea g en t is a very p o w erfu l base; in effect it contains a carbanion. • It is not possible to prepare a Grignard reagent from a compound that contains any hydrogen more acidic than the hydrogen atoms o f an alkane or alkene. We cannot, for example, prepare a Grignard reagent from a compound containing an — OH group, an — NH — group, an — SH group, a — CO2H group, or an — SO3H group. I f we were to attempt to prepare a Grignard reagent from an organic halide containing any of these groups, the formation of the Grignard reagent would simply fail to take place. (Even if a Grignard reagent were to form, it would immediately be neutralized by the acidic group.) • Since Grignard reagents are powerful nucleophiles, we cannot prepare a Grignard reagent from any organic halide that contains a carbonyl, epoxy, nitro, or cyano (— CN) group. I f we were to attempt to carry out this kind of reaction, any Grignard reagent that formed would only react with the unreacted starting material: H e lp f u l H i n t A protecting group can sometimes be used to mask the reactivity of an incompatible group (see Sections 11.11D, 11.11E, and 12.9).

— O H , — N H 2, — N H R , — C O 2 H , — S O 3 H , — S H , — C = C — H O

O

O

O O NO2,

H

OR,

C

N,

NH2

G rig n a rd re a g e n ts c a n n o t b e p repared in th e p re s e n c e of th e s e g ro u p s b e c a u s e th e y will re a c t w ith them .

This m eans th at when we p rep a re G rign ard reagents, we are effectively lim ited to alkyl halides o r to an alogous organic halides con tain in g carb o n -ca rb o n dou ble bonds, in ter­ n a l triple bonds, eth er linkages, a n d — NR 2 groups.

Grignard reactions are so sensitive to acidic compounds that when we prepare a Grignard reagent we must take special care to exclude moisture from our apparatus, and we must use an anhydrous ether as our solvent. As we saw earlier, acetylenic hydrogens are acidic enough to react with Grignard reagents. This is a limitation that we can use, however. • We can make acetylenic Grignard reagents by allowing terminal alkynes to react with alkyl Grignard reagents (cf. Section 12.7A). We can then use these acetylenic Grignard reagents to carry out other syntheses. For example,

g

,) (+ethane C

12.8 Alcohols from Grignard Reagents

•

W h e n w e p la n a G rig n a rd synthesis, w e m u s t also tak e care th a t any aldeh yde, ke to n e , epo x id e , o r ester that w e use as a substrate does n o t also c o n tain an acid ic g ro up (o th e r than w h e n w e d e lib e ra te ly le t it re a c t w ith a te rm in a l a lk y n e ).

I f w e w e re to do this, the G rig n a rd re a g e n t w o u ld s im p ly re a c t as a base w ith the acid ic h yd ro g e n ra th e r than reacting at the ca rb o n y l o r e p o x id e carbon as a n u c le o p h ile . I f w e w ere to treat 4 -h y d ro x y -2 -b u ta n o n e w ith m e th y lm a g n e s iu m b ro m id e , fo r e x a m p le , the reaction that w o u ld tak e p la c e is

O

O CH3MgBr

HO"

c h 4î

BrMgO"

4 -H y d ro x y -2 -b u ta n o n e ra th e r than

O

OMgBr CH3MgBr

HO

-------------»

HO

I f w e w e re p re p a re d to w aste one m o la r e q u iv a le n t o f the G rig n a rd reag ent, w e can treat 4 -h y d ro x y -2 -b u ta n o n e w ith tw o m o la r e qu ivalen ts o f the G rig n a rd re a g e n t and th e re b y get a d d itio n to the c a rb o n y l group:

O HO

OMgBr 2 CH3MgBr (2 CH 4 )

OH

2 NH4 X

BrMgO

HO'

h 2o

T h is tec h n iq u e is som etim es e m p lo y e d in s m a ll-sc a le reactions w h e n the G rig n a rd reagent is inexp e n s iv e a n d the o th e r re a g e n t is expensive.

12.8C The Use of Lithium Reagents O rg a n o lith iu m reagents (R L i) re a c t w ith c a rb o n y l com pounds in the sam e w a y as G rig n a rd reagents a n d thus p ro v id e an a lte rn a tive m e th o d fo r p re p a rin g alcohols.

O'

OLi

OH h3o +

R O rg a n o ­ lithium re a g e n t

A ld e h y d e or k e to n e

R

Lith iu m a lk o x id e

A lcoh ol

O rg a n o lith iu m reagents h av e the advan tag e o f b ein g som ew hat m o re re active than G rig n a rd reagents a lth o u g h they are m o re d iffic u lt to p repare and h an dle.

12.8D The Use of Sodium Alkynides S o d iu m a lk yn id e s also re a c t w ith aldehydes a n d ketones to y ie ld alcoh ols. A n e x a m p le is the fo llo w in g : NaNH,

Na+

nh3

O

ONa

OH NH4

573

574

Chapter 12

Alcohols from Carbonyl Compounds

Solved Problem 12.7 ILLUSTRATING MULTISTEP SYNTHESES F o r the fo llo w in g com pounds, w rite a re tro s y n th e tic schem e and then synthetic reaction s th a t c o u ld b e used to p re p a re each one. U s e h yd ro carb on s, o rg an ic h alid es, alcoh ols, a ld e h y ­ des, ketones, o r esters c o n tain in g six carbon atom s o r fe w e r and any o th er n eed ed reagents. OH OH

HO. (c)

M gBr

Br

OH

Synthesis

O Br

OH

Mg ; Et,O

(b) Retrosynthetic Analysis OH

OMe

Synthesis OH

O Br

M gBr

(1) ^ OMe (2) NH4Cl, H2O*

Mg Et2O (c)

Retrosynthetic Analysis HO

Synthesis HO

NaNH2

Na+

(1)

(2) NH4Cl, H2O

12.9 Protecting Groups

575

12.9 Protecting Groups •

A p ro te c tin g g ro u p can be used in some cases w here a reactant contains a group that is in co m p atib le w ith the reaction conditions necessary fo r a g iven transform ation.

F o r e x a m p le , i f it is necessary to p repare a G rig n a rd reagent fro m an a lk y l h a lid e that already contains an a lc o h o l h y d ro x y l g ro u p , the G rig n a rd re a g e n t can s till b e p re p a re d i f the alco ­ h o l is firs t p ro te c te d b y con version to a fu n c tio n a l gro up th a t is stable in the presence o f a G rig n a rd reag ent, fo r e x a m p le , a te r t-b u ty ld im e th y ls ily l ( T B S ) e th e r (S e c tio n 1 1 .1 1 E ). T h e G rig n a rd re a c tio n can be con ducted, a n d then the o rig in a l a lc o h o l gro up can b e lib e ra te d b y cleavag e o f the s ily l eth er w ith flu o rid e io n (see P ro b le m 1 2 .3 6 ). A n e x a m p le is the f o l­ lo w in g synthesis o f 1 ,4 -p e n ta n e d io l. T h is sam e strategy can be used w h e n an o rg an o lith iu m re a g e n t o r a lk y n id e a n io n m u st b e p re p a re d in the presence o f an in c o m p a tib le g roup. In la te r chapters w e w ill encounter strategies that can b e used to pro tect o th er fu n c tio n a l groups d u rin g vario us reactions (S e c tio n 1 6 .7 C ). TBSCI, imidazole HO'

'B r

Mg

DMF (-H C I)>

TBSO'

Br

Et2O

TBSO

'M g B r O

(1 ) M ex TBS =

Me

H

(2 ) H 3 O"

< /

i-B u

TBSO OH

N

Im id a z o le =

H

N

Bu4 N+FTHF O

DMF =

Me N'

I Me D im e th y l fo rm a m id e (a p o la r a p ro tic s o lv e n t)

HO OH 1 ,4 -P en tan e d io l

Solved Problem 12.8 S h o w h o w the fo llo w in g synthesis c o u ld b e ac c o m p lis h e d using a p ro te c tin g group.

STRATEGY AND ANSWER F irs t p ro te c t the — O H g ro u p b y c on verting it to a te r t-b u ty ld im e th y ls ily l ( T B S ) ether (S e c tio n 1 1 .1 1 E ), then tre a t the p ro d u ct w ith e th y l m a g n e s iu m b ro m id e fo llo w e d b y d ilu te acid. T h e n re m o v e the p ro te c tin g g roup.

576

Chapter 12

Alcohols from Carbonyl Compounds

Key Terms and Concepts

PLUS

The key terms and concepts that are highlighted in bold, blue text within the chapter are defined in the glossary (at the back o f the book) and have hyperlinked definitions in the accompanying WileyPLUS course (www.wileyplus.com).

Problems Note to Instructors: Many of the homework problems are available for assignment via WileyPLUS, an online teaching and learning solution. REAGENTS A N D REACTIONS 12.10

What products would you expect from the reaction of ethylmagnesium bromide (CH3CH2MgBr) with each o f the following reagents? O (e)

A Ph

(a) H2O

O

(b) D2O (f)

O (c)

, then NH4CI, H2O OMe

Ph

À

, then NH4CI, H2O 4 2

A A , then H3O+ Ph i ' H 3

O , then H3O+

(g) ^ — = = — H, then

O (d) ph/ u\ ph , then NH4CI, H2O 12.11

What products would you expect from the reaction o f propyllithium (CH3CH2CH2Li) with each of the following reagents? O O H , then H3O4

(a )

O (b)

, then NH4Cl, H2O

(c) 1-Pentyne, then A A (d) Ethanol O (e)

12.12

, then NH4Cl, H2O

OD

What product (or products) would be formed from the reaction o f 1-bromo-2-methylpropane (isobutyl bromide) under each o f the following conditions? O (a) OHT, H2O (g) (1) Mg, Et2O; (2) ; (3) NH4CI, H2O (b) CN , ethanol OMe (c) t-BuOK, t-BuOH

O (h) (1) Mg, Et2O; ( 2 ) ^ ; (3) H3O 4

(d) MeONa, MeOH

O

O

(e) (1) Li, Et2O; ( 2 ) ^ _ ^ ; (3) NH4CI, H2O (i)

O (f) Mg, Et2O, then CH3CH, then H3O4

( 1) Mg, Et2O; (2) H^ V H ; (3) NH4CI, H2O

(j) Li, Et2O; (2) MeOH (k) Li, Et2O; (2) H— =

— H

577

Problems

12.13

W h ic h o x id iz in g o r red ucin g agent w o u ld y o u use to c a rry o u t the fo llo w in g transform ations? O (a )

OCH3

"O H OH

O O (b )

O

OCH3

o ch

3

OH

O O

O

(c) HO

OH

HO

OH O

(d )

HO"

OH

O

OH

HO O

(e )

12.14

H O

OH

H

P re d ic t the products o f the fo llo w in g reactions. O

(1) EtMgBr (excess)

(a ) E tO

12.15

O

OEt

(1) MeMgBr (excess)

(2) NH 4 Cl, H2O

(2) NH 4Cl, H2 O H

OEt

P re d ic t the o rg an ic p ro d u ct fro m each o f the fo llo w in g re d u c tio n reactions. OH

O

O ^ O ( 1 ) Lia ih 4 NaBH 4

(a )

12.16

NaBH 4

(2 ) aq. H 2 SO 4

(b )

O

O

P re d ic t the o rg an ic p ro d u ct fro m each o f the fo llo w in g o x id a tio n reactions z \V O H (1) KMnO 4 ,OH-, A

(a )

OH

PCC

(b )

(2 ) H 3 O+

c h 2c i2

/ ' N. > /

OH

O

(d )

OH (c)

PCC C HhCHI2

H 2 CrO 4

(e) H

12.17

P re d ic t the o rg an ic p ro d u ct fro m each o f the fo llo w in g o x id a tio n and re d u c tio n reactions. OH (a )

OH

PCC c h 2c i2

NaBH4 H2 CrO 4

(b ) H O OH

578 12.18

Chapter 12

Alcohols from Carbonyl Compounds

P re d ic t the m a jo r o rg an ic p ro d u ct fro m each o f the fo llo w in g reactions

(c)

(b)

(d)

,MgBr

(1 )

(2) NH4 Cl, H2O

12.19

P re d ic t the m a jo r o rg an ic p ro d u ct fro m each o f the fo llo w in g re a c tio n sequences.

(1) MeMgBr (excess)

(a )

O.

'O H

(1 ) p c c

(d) (2 ) / '" '' " (3) H3 O+

(2) NH 4 Cl, H2O

MgBr

O H

(1) Mg

(1) EtMgBr3 (2) H 3 O+ ^ 3 (3) NaH (4) CH 3 Br

(3) H3O

(c)

(1) ^ C ^ ^ x / M g B r

( 1 ) PBr3 (2 ) Mg

OH

(2) NH 4 Cl, H2O (3) PCC

(3) H3 O 1 2 .2 0

(excess)

P re d ic t the p ro d u ct o f the fo llo w in g reaction . O

O

(1) BrMg''

'MgBr (1 equiv.)

(2) H3 O

M EC H A N IS M S 1 2 .2 1

S yn th e s ize each o f the fo llo w in g com pounds fro m

D

h o

DO

H

DO

.D

cyclo h ex a n o n e . U s e D to spe c ify d e u te riu m in any ap p ro p ria te re a g e n t o r solven t w h e re it w o u ld take the p la c e o f hyd ro gen . 1 2 .2 2

W r ite a m e c h a n is m fo r the fo llo w in g re a c tio n . In c lu d e fo rm a l charges a n d c u rv e d arrow s to show the m o v e m e n t o f electrons in a ll steps. O

MgBr O

(1 )

(excess)

(2) NH 4 Cl, H2O

12.23

W r ite a m e c h a n is m fo r the fo llo w in g re a c tio n . Y o u m a y use H ~ to rep resent h y d rid e ions fro m L iA lH 4 in y o u r m ech a n is m . In c lu d e fo rm a l charges and c u rv e d arrow s to show the m o v e m e n t o f electrons in a ll steps.

O

579

P ro b le m s

12.24

A lth o u g h o x ira n e (o x a c y clo p ro p a n e ) and oxetane (o x a c y clo b u tan e ) re a c t w ith G rig n a rd and o rg a n o lith iu m reagents to fo r m alcoh ols, te tra h y d ro fu ra n (o x a c y clo p e n ta n e ) is so u n reactive that it can be used as the solv e n t in w h ic h these o rg a n o m e ta llic com pounds are p repared. E x p la in the d iffe re n c e in re a c tiv ity o f these o xy g e n heterocycles.

12.25

Studies suggest th a t a tta c k b y a G rig n a rd re a g e n t at a c a rb o n y l g ro u p is fa c ilita te d b y in v o lv e m e n t o f a second m o l­ ecule o f the G rig n a rd reag ent that p articipates in an o v e ra ll c y c lic tern ary c o m p le x . T h e second m o le c u le o f G rig n a rd re a g e n t assists as a L e w is acid. Propose a structure fo r the tern a ry c o m p le x and w rite a ll o f the products th a t result fro m it.

SYNTHESIS 12.26

W h a t o rg an ic products A - H w o u ld y o u e x p e c t fro m each o f the fo llo w in g reactions? O

MeLi H

(1) NaH

. (1)

'

A

Et2O

(2) NH 4 Cl, H2O

B

D (2) ^

OMs

(2) Ni2B (P—2), H 2

C

O

O NaBH4 MeOH

12.27

MsCl E

(1) LAH

ONa

pyr

(2) aq. H2 SO 4

H

O u tlin e a ll steps in a synthesis th a t w o u ld tra n s fo rm 2 -p ro p a n o l (is o p ro p y l a lc o h o l) in to each o f the fo llo w in g :

D (e )

(c)

(a )

'Cl OH

(b )

OH

(d )

OH 12.28

S h o w h o w 1 -p e n ta n o l c o u ld be tra n s fo rm e d in to each o f the fo llo w in g com pounds. (Y o u m a y use a n y need ed in o r­ g an ic reagents and y o u n ee d n o t show the synthesis o f a p a rtic u la r c o m p o u n d m o re than once.) (a )

1-B ro m o p e n ta n e

(b )

1-P entene

(c )

2 -P e n ta n o l

O ( i)

2-P e n ta n o n e ,

(j)

P e n ta n o ic acid,

( d ) Pentane (e )

2 -B ro m o p e n ta n e

(f)

1 -H e x a n o l

(g )

1 -H e p ta n o l

( k ) D ip e n ty l eth er (tw o w ays) ( l)

O ( h ) Pentanal,

1 -P e n ty n e

( m ) 2 -B ro m o -1 -p e n te n e

H

(n ) P e n ty llith iu m (o )

4 -M e th y l-4 -n o n a n o l

OH

580 12.29

12.30

Chapter 12

Alcohols from Carbonyl Compounds

Provide the reagents needed to accomplish transformations (a )-(g ). M ore than one step may be necessary.

A s s u m in g th a t y o u h ave a v a ila b le o n ly alcoh ols o r esters c o n tain in g n o m o re than fo u r carb on atom s, show h o w y o u m ig h t synthesize each o f the fo llo w in g com pounds. B e g in b y w ritin g a re tro s y n th e tic analysis fo r each. Yo u m u s t use a G rig n a rd re a g e n t a t one step in the synthesis. I f needed, y o u m a y use o x ira n e and y o u m a y use b ro m o b en ze n e , b u t y o u m u s t show the synthesis o f a n y o th e r re q u ire d o rg an ic com pounds. A s s u m e y o u h ave a v a il­ a b le an y solvents a n d an y in o rg a n ic com pounds, in c lu d in g o x id iz in g and re d u c in g agents, th a t y o u req uire. (a )

O

(d)

O H

(e)

12.31

O

F o r each o f the fo llo w in g alcoh ols, w rite a re tro s y n th e tic analysis and synthesis th a t in v o lv es an app ro priate o rg a n o m e ta llic re a g e n t (e ith e r a G rig n a rd o r a lk y llith iu m reag ent). (c)

OH

581

Challenge Problems 12.32

S yn th e s ize each o f the fo llo w in g com pounds starting fro m p rim a ry o r secondary a lcoh ols c o n tain in g seven c a r­ bons o r less and, i f app ro p ria te , b ro m o benzen e.

(b)

12.33

OH

(c)

T h e a lc o h o l show n b e lo w is used in m a k in g p erfu m e s . W r ite a re tro s y n th e tic analysis and then synthetic reactions th a t c o u ld b e used to p repare this a lc o h o l fro m b ro m o b e n ze n e and 1

12.34

-butene.

S h o w h o w a G rig n a rd re a g e n t m ig h t b e used in the fo llo w in g synthesis:

'O ^ o

OH ,OH

12.35

W r ite a re tro s y n th e tic analysis and then synthetic reactions th a t co u ld b e used to p repare ra c e m ic

OH

M e p a rfy n o l, a m ild h y p n o tic (sle e p -in d u c in g c o m p o u n d ), starting w ith com pounds o f fo u r c a r­ b o n atom s o r few er.

Meparfynol 12.36

W r ite a re tro s y n th e tic analysis a n d synthesis fo r the fo llo w in g tra n s fo rm a tio n .

OH ,Br HO 12.37

HO

S yn th e s ize the fo llo w in g c o m p o u n d using cyc lo p en tan e a n d e th y n e (a c e ty le n e ) as the sole source o f carbon atom s.

Challenge Problems 12.38

E x p la in h o w 1H N M R , 13C N M R , and I R spectroscopy could be used to differentiate am ong the fo llo w in g com pounds.

OH 2-Phenylethanol

OH

OH

1,2-Diphenylethanol

1,1-Diphenylethanol

O O

Benzyl 2-phenylethanoate

582 12.39

Chapter 12

Alcohols from Carbonyl Compounds

W h e n sucrose (c o m m o n ta b le sugar) is tre a te d w ith aqueous acid , it is cle av e d and y ie ld s s im p le r sugars o f these types:

F o r reasons to be studied later, in the use o f this p ro ced ure fo r the id e n tific a tio n o f the sugars in c o rp o ra te d in a sac­ c ha rid e lik e sucrose, the p ro d u ct m ix tu re s are o fte n treated w ith s o d iu m b o ro h y d rid e b e fo re analysis. W h a t li m i ­ t a t io n ^ ) does this p u t on id e n tific a tio n o f the sugar b u ild in g b lo cks o f the starting saccharide?

12.40

A n u n k n o w n X shows a b ro ad abso rp tio n b an d in the in fra re d a t 3 2 0 0 - 3 5 5 0 c m ~

1

b u t n o n e in the 1 6 2 0 - 1 7 8 0 c m ~

1

re g io n . I t contains o n ly C , H , a n d O . A 1 1 6 -m g sam p le w as treated w ith an excess o f m e th y lm a g n e s iu m b ro m id e , p ro d u cin g 4 8 .7 m L o f m e th a n e gas c o lle c te d o ver m e rc u ry a t 2 0 ° C and 7 5 0 m m H g . T h e mass spectrum o f X has its m o le c u la r io n (b a re ly d ete c ta b le) a t 1 1 6 m /z and a fra g m e n t p e a k at 9 8 . W h a t does this in fo rm a tio n te ll you ab o u t the structure o f X ?

Learning Group Problems T h e p ro b le m b e lo w is d ire c te d to w a rd d evising a h y p o th e tic a l p a th w a y fo r the synthesis o f the a c y c lic c e n tra l p o rtio n o f C rix iv a n ( M e r c k and C o m p a n y ’ s H I V protease in h ib ito r). N o te th a t y o u r synthesis m ig h t n o t adeq u a te ly co n tro l the stere­ o ch e m is try d u rin g each step, b ut fo r this p a rtic u la r exercise th a t is n o t expected.

F il l in m issin g com pounds a n d reagents in the fo llo w in g o u tlin e o f a h y p o th e tic a l synthesis o f the a c y c lic c e n tra l p o rtio n o f C rix iv a n . N o te th a t m o re than o ne in te rm e d ia te c o m p o u n d m a y b e in v o lv e d b e tw e e n som e o f the structures show n b elow .

OH

OR

OR

LG

OR

LG = s o m e le a vin g gro up

(R w o u ld b e H initially. T h e n , by re a c tio n s w h ic h you d o n ot n ee d to specify, it w o u ld be c o n v e rte d to an alk yl g ro u p .)

J*. a

S u m m a ry o f R e a c tio n s

0

583

£KI

Summary of Reactions Sum m aries o f reactions discussed in this chapter are shown below . D e ta ile d conditions fo r the reactions that are sum m arized can be fo u n d in the chapter section w h ere each is discussed.

Synthetic Connections of Alcohols and Carbonyl Compounds Substrate

1. Carbonyl Reduction Reactions

Reducing agent NaBH4

* * * *

Aldehydes to prim ary alcohols Ketones to secondary alcohols Esters to alcohols Carboxylic acids to prim ary alcohols

O

Aldehydes

0X

OH

U

R

Ketones

OH

[H] R

\ H H OH

O

x R ^ ^

LiAlH4 (LAH)

OH

[H] R'

R

\ R' H

R

O

Esters

R

x

x

R

H

OH [H] OR'

R

R'— OH

\ H H

OH

O

Carboxylic acids

\ R'

[H] R

OH

\ H H

(Hydrogen atoms in blue are added during the reaction workup by water or aqueous acid.)

2. Alcohol Oxidation Reactions * Prim ary alcohols to aldehydes * Prim ary alcohols to carboxylic acids * Secondary alcohols to ketones Substrate

O xidizing agent [O] PCC

H2CrO4

x

KMnO4

O

Primary alcohols

[O] R

OH

R

OH

Secondary alcohols

I R ■ " " ^ DR''

jO i

OH

Tertiary alcohols

[O] R

\ R' R"

*

O OH

A

R

OH

O

O

O

x R ^^R

X R ^^R

X R ^^R

'

R

R''

R

R''

584

Chapter 12

Alcohols from Carbonyl Compounds

Synthetic Connections of Alcohols and Carbonyl Compounds

3. Carbon-Carbon Bond Forming Reactions * * * * * *

Alkynide anion formation Grignard reagent formation Alkyllithium reagent formation Nucleophilic addition to aldehydes and ketones Nucleophilic addition to esters Nucleophilic ring-opening of epoxides

R

—

H

x

OH

(1) R '( H f " 'R " ( H )

R'(H)

NaNH 2 (or other strong base) OH

y Mg in ether

R

MgX

(1) R *

OR' ->

Nu:

R

(2) H 3 O+ (or NH4+) R— X

R"(H) Nu

\ Nu Nu

R'— OH

2 Li R— Li + LiX OH

Nu (a substituted or unsubstituted oxirane Nu = alkynyl group, or alkyl group from Grignard or alkyllithium reagent

See First Review Problem Set in WileyPLUS

Conjugated Unsaturated Systems

All of the brilliant colors of fall foliage have one thing in common. The colors of fall are caused by molecules that contain conjugated unsaturated systems. •

A conjugated unsaturated system is a molecular unit containing ^ electrons that can be delocalized over three or more contiguous atoms.

The carotenes, xanthophylls, and anthocyanins are some families of natural pigments that contain conjugated unsaturated systems. A few examples are shown here.

C ry p to x a n th in (a x a n th o p h y ll)

OH

OH C y a n id in (an a n th o c y a n in )

585

586

Chapter 13

Conjugated Unsaturated Systems

In this chapter we shall study conjugated systems and find that they have special aspects of reactivity. Radicals, cations, and anions that are formed as part of a conjugated system have greater stability than their nonconju­ gated counterparts, making them especially important reaction intermediates. Conjugated unsaturated systems also absorb energy in the ultraviolet and visible regions of the spectrum, the latter of which accounts for the col­ ors we observe in organic pigments. And lastly, conjugated dienes undergo a very important ring-forming reac­ tion called the Diels-Alder reaction, for which the Nobel Prize was awarded to O tto Diels and Kurt Alder.

13.1 Introduction A t its essence, a conjugated system involves at least one ato m w ith a p o rb ita l adjacent to at least one p bond. T h e adjacent ato m w ith the p o rb ita l can b e p art o f another ^ bond, as in 1,3-b utadien e, o r a rad ical, cationic, o r an io n ic reaction interm ediate. I f an e x a m p le specifi­ c a lly derives fro m a p ro p e n y l group, a c o m m o n n am e fo r this group is a lly l. M o r e gen erally w h e n w e are considering a rad ical, cation, o r anion that is adjacent to one o r m o re p bonds in a m o le c u le larg er than p ro pene o r p ropene itself, the adjacent p ositio n is c a lled a lly lic .

1,3-butadiene (a conjugated diene)

The allyl radical

The a lly lic carbocation

A s w e shall see n ex t, ra d ic a l sub stitution a t an a lly lic p o s itio n is e s p e c ia lly fav o ra b le because the in te rm e d ia te ra d ic a l is p a rt o f a con jug ated system.

13.2 Allylic Substitution and the Allyl Radical W h e n p ro p e n e reacts w ith b ro m in e o r c h lo rin e at lo w tem peratures, the re a c tio n that takes p la c e is the usual a d d itio n o f h a lo g e n to the d o u b le b o n d :

+

x

+

X 2

low temperature CCl4

X | r" " ' \ I X

A t lo w te m p e ra tu re an ad d itio n re a c tio n o cc u rs .

H o w e v e r, w h e n p ro pene reacts w ith c h lo rin e o r b ro m in e at v e ry h ig h tem peratures o r under con dition s in w h ic h the con centration o f the h alo gen is v e ry sm all, the reactio n that occurs is a s u b s titu tio n . These tw o exam p les illu s tra te h o w w e can o fte n change the course o f an o rganic reactio n s im p ly b y changing the conditions. (T h e y also illu s tra te the n eed fo r spec­ ify in g the con dition s o f a reactio n c a re fu lly w h e n w e re p o rt e x p e rim e n ta l results.) +

high temperature

x 2

X

+

^

and low concentration of X 2

P ro p e n e A t h ig h te m p e ra tu re (o r in th e p re s e n c e o f a rad ical in itia to r) a n d lo w c o n c e n tra tio n o f X 2 a s u b s titu tio n re a c tio n o ccu rs.

587

13.2 Allylic Substitution and the Allyl Radical

In this substitution a h alo g e n a to m replaces one o f the h yd ro g e n atom s o f the m e th y l g ro up o f propene. •

T h e hyd ro gen s on the sp 3 carbon a d ja c e n t to the d o u b le b o n d are c a lle d the a lly lic h y d ro g e n a to m s .

•

T h e re a c tio n is an a lly lic s u b s titu tio n :

3

A lly lic h yd ro gen a to m s

Allylic hydrogen atom and allylic substitution are general terms as w e ll. T h e hydrogen atoms o f any saturated carbon ato m adjacent to a double bon d are ca lled a lly lic hydrogen atoms. A n y re a c tio n in w h ic h an a lly lic h y d ro g e n a to m is re p la ce d is c a lled an a lly lic substitution.

A n a lly lic h yd ro g en a to m th a t can u n d erg o a lly lic s u b s titu tio n .

Solved Problem 13.1 Id e n tify a ll o f the p ositio ns b ea rin g a lly lic h yd ro g e n atom s in c ry p to x a n th in (sh o w n e a rlie r in this chap ter).

STRATEGY AND ANSWER A lly lic h yd ro g e n atom s in cry p to x a n th in are fo u n d on the sp3-h y b rid iz e d carbons a d ja ­ c ent to p bonds. T h e s e p ositio ns are la b e le d b y boxes in the fo rm u la b elo w .

13.2A Allylic Chlorination (High Temperature) P ro p e n e undergoes a lly lic c h lo rin a tio n w h e n p ro p e n e and c h lo rin e re a c t in the gas phase at 4 0 0 ° C . T h is m e th o d fo r s ynthesizing a lly l c h lo rid e is c a lle d the S h e ll process.

+

C l2 2

Cl

gas phase

P ro p e n e

+

HCl

3 -C h lo ro p ro p e n e (allyl c h lo rid e )

T h e m e c h a n is m fo r a lly lic sub stitution is the sam e as the chain m e c h a n is m fo r a lk an e h alo gen atio ns th a t w e saw in C h a p te r 10. In the c h a in -in itia tin g step, the c h lo rin e m o le ­ c u le dissociates in to c h lo rin e atom s.

Chain-Initiating Step =C ^ C l

----- * 2 : C l'

In the firs t c h a in -p ro p a g a tin g step the c h lo rin e a to m abstracts o ne o f the a lly lic h y d ro ­ gen atom s.

588

Chapter 13

Conjugated Unsaturated Systems

First Chain-Propagating Step C ÎH

+ Y

Cl :

+

H — Cl

A llyl radical T h e ra d ic a l th a t is p ro d u ce d in this step is c a lle d an a lly l r a d i c a l. •

A ra d ic a l o f the g en e ra l typ e show n h ere is c a lle d an a llylic ra d ic al.

An a lly lic rad ical In the second chain-propagating step the a lly l ra d ic al reacts w ith a m o le c u le o f chlorine.

Second Chain-Propagating Step

C l^ C !

—

»

^

\ ^

C!

+

: C l-

A llyl c h lo rid e T h is step results in the fo rm a tio n o f a m o le c u le o f a lly l c h lo rid e a n d a c h lo rin e a to m . T h e c h lo rin e a to m then b rin g s ab o u t a re p e titio n o f the firs t c h a in -p ro p a g a tin g step. T h e chain re a c tio n con tin ues u n til the usual c h a in -te rm in a tin g steps (see S e c tio n 1 0 .4 ) con sum e the rad icals. T h e reason fo r substitution at the a lly lic h yd ro g e n atom s o f p ro pene w il l b e m o re u n d er­ standable i f w e e x a m in e the b o n d disso ciation energy o f an a lly lic c a rb o n -h y d ro g e n bon d a nd co m p a re it w ith the b o n d disso ciation energies o f o th e r c a rb o n -h y d ro g e n bonds.

H Propene

+

H-

DH ° = 3 6 9 kJ m o L 1

+

H-

DH ° = 4 0 0 kJ m o T 1

+

H-

DH ° = 4 1 3 kJ m o T 1

+

H-

DH ° = 4 2 3 kJ m o L 1

+

H-

DH ° = 4 6 5 kJ m o L 1

Allyl radical

H Isobutane

3° Radical

H Propane

2° Radical H

Propane

Radical

H

Ethene

Vinyl radical

See Table 10.1 fo r a list o f additional bond dissociation energies.

W e see that an a lly lic c a rb o n -h y d ro g e n b o n d o f p ro pene is b ro ke n w ith greater ease than even the te rtia ry c a rb o n -h y d ro g e n b o n d o f isobutane and w ith fa r greater ease than a v in y lic c a rb o n -h y d ro g e n bond:

13.2 Allylic Substitution and the Allyl Radical

X:

HX

Eactis low.

HX

Eactis high.

589

A llyl rad ical :X V in y lic rad ical T h e ease w ith w h ic h an a lly lic c a rb o n -h y d ro g e n b o n d is b ro ke n m eans th a t re la ­ tiv e to p rim a ry , secondary, tertia ry , and v in y lic fre e rad icals an a lly lic ra d ic a l is the

most stable (F ig . 1 3 .1):

Relative stability: allylic o r allyl > 3° > 2° > 1° > vinyl o r vinylic

Vinyl radical 1o Radical 2° Radical 3° Radical

DH o = 465

Allyl radical

DHo = 423 kJ mol- 1

DH o = 400

<

kJ mol- 1

kJ mol- 1

DH o = 413 kJ mol- 1

DHo = 369 kJ mol- 1

H

H

H H

Figure 13.1 The relative stability of the allyl radical compared to 1°, 2°, 3°, and vinyl radicals. (The stabilities of the radicals are relative to the hydrocarbon from which each was formed, and the overall order of stability is allyl > 3° > 2° > 1° > vinyl.)

H

13.2B Allylic Bromination with N-Bromosuccinimide (Low Concentration of Br2) P rop ene undergoes a lly lic b ro m in a tio n w h e n it is treated w ith N -b ro m o s u c c in im id e (N B S ) in C C l 4 in the presence o f p ero x id e s o r light: H

Br O

I

I

N

N

O

CCL N -B ro m o s u c c in im id e (N B S )

O

Br

light or ROOR

O

* 3 -B ro m o p ro p e n e (allyl b ro m id e )

S u c c in im id e

T h e re a c tio n is in itia te d b y the fo rm a tio n o f a s m a ll a m o u n t o f B r (p o s sib ly fo rm e d b y d is­ sociation o f the N — B r b o n d o f the N B S ). T h e m a in p ro p a g a tio n steps fo r this re a c tio n are the sam e as fo r a lly lic c h lo rin a tio n (S e c tio n 1 3 .2 A ): •B r

HBr

Br

B r— Br

•B r

N -B ro m o s u c c in im id e is a so lid that is n e a rly in s o lu b le in C C l 4 w h ic h p ro vid es a co n ­ stant b ut v e ry lo w co n cen tratio n o f b ro m in e in the re a c tio n m ix tu re . I t does this b y re a c t­ ing v e ry ra p id ly w ith the H B r fo rm e d in the sub stitution re a c tio n . E a c h m o le c u le o f H B r is re p la ce d b y one m o le c u le o f Br2 . Br O

H

I

I

N

N

O

O HBr

O Br

590

Chapter 13

Conjugated Unsaturated Systems

U n d e r these con dition s, th a t is, in a nonpolar solvent and w ith a very low concentra­

tion of brom ine , v e ry little b ro m in e adds to the d o u b le b o n d ; it reacts b y sub stitution and rep laces an a lly lic h y d ro g e n a to m instead. T h e fo llo w in g re a c tio n w ith c yclo h exen e is a n o th e r e x a m p le o f a lly lic b ro m in a tio n w ith N B S .

Br

8 2 -8 7 % •

In g en eral, N B S is a g o o d re a g e n t to use fo r a lly lic b ro m in a tio n .

THE CHEMISTRY OF A lly lic B r o m i n a t i o n

Why, we might ask, does a low concentration of bromine favor allylic substitution over addition? To understand this, we must recall the mechanism for addition and notice that in the first

V

step only one atom of the bromine molecule becomes attached to the alkene in a reversible process. The other atom (now a bromide anion) becomes attached in the second step:

Br

C /

C

VC-

Br solvent

C

B r—

C—

\

C

nonpolar

B r,

Br

B r-

With a low concentration of bromine initially, the concentration of the bromonium ion and bromide anion after the first step will also be low. Consequently, the prob­ ability of a bromide anion finding a bromonium ion in its vicinity for the second step is also low, and hence the overall rate of addition is slow and allylic substitution com­ petes successfully. The use of a nonpolar solvent also slows addition. Since there are no polar molecules to solvate (and thus stabilize) the bromide ion formed in the first step, the bromide ion uses a bromine molecule as a substitute:

2

C

X -

B r— Br

Br

C

This means that in a nonpolar solvent the rate equation is second order with respect to bromine, R ate

\ /

C=C

/ \

[B rJ2

and that the low bromine concentration has an even more pronounced effect in slowing the rate of addition. Understanding why a high temperature favors allylic sub­ stitution over addition requires a consideration of the effect of entropy changes on equilibria (Section 3.10). The addi­ tion reaction, because it combines two molecules into one, has a substantial negative entropy change. At low temper­ atures, the TAS° term in AG° = AH° — TAS° is not large enough to offset the favorable AH° term. But as the tem­ perature is increased, the TAS° term becomes more signifi­ cant, AG° becomes more positive, and the equilibrium becomes more unfavorable.

13.3 The S tab ility o f the Allyl Radical A n e x p la n a tio n o f the s ta b ility o f the a lly l ra d ic a l can be approached in tw o w ays: in term s o f m o le c u la r o rb ita l th e o ry and in term s o f resonance th e o ry (S e c tio n 1 .8 ). A s w e shall see soon, b oth approaches g iv e us e q u iv a le n t d escrip tion s o f the a lly l ra d ic a l. T h e m o le c u la r o rb ita l app roach is easier to v is u a liz e , so w e sh all b eg in w ith it. (A s p re p a ra tio n fo r this section, it m a y h elp the re a d e r to re v ie w the m o le c u la r o rb ita l th e o ry g iv e n in Sections 1.11 a nd 1 .1 3 .)

591

13.3 The Stability o f the Allyl Radical

13.3A Molecular Orbital Description of the Allyl Radical A s an a lly lic h y d ro g e n a to m is abstracted fro m p ro p e n e (see the fo llo w in g d ia g ra m ), the sp3-h y b rid iz e d carb on a to m o f the m e th y l g ro u p changes its h y b rid iz a tio n state to sp 2 (see S ection 1 0 .7 ). T h e p o rb ita l o f this n e w sp2-h y b rid iz e d carbon a to m o verlap s w ith the p o rb ita l o f the c en tral carbon ato m . •

In the a lly l ra d ic a l three p o rb ita ls o ve rlap to fo r m a set o f p m o le c u la r o rbitals that encom pass a ll three carb on atom s.

•

T h e n e w p o rb ita l o f the a lly l ra d ic a l is said to b e conjugated w ith those o f the d o u b le bon d, and the a lly l ra d ic a l is said to b e a conjugated unsaturated system. All carbons are s p 2 hybridized

t 8« X

H

H

H

Hydrogen abstraction

•

Transition state

Delocalized allyl radical

T h e u n p aire d e le ctro n o f the a lly l ra d ic a l and the tw o electrons o f the p b o n d are d e lo c a liz e d o v e r a ll three carb on atom s.

D e lo c a liz a tio n o f the u n p aire d e lectro n accounts fo r the g reater s ta b ility o f the a lly l ra d i­ ca l w h e n com p a re d to p rim a ry , secondary, and te rtia ry ra d ic als . A lth o u g h som e d e lo c a l­ iz a tio n occurs in p rim a ry , secondary, a n d te rtia ry ra d ic als , d e lo c a liz a tio n is n ot as e ffe c tiv e because it occurs o n ly throu gh h y p e rc o n ju g a tio n (S e c tio n 6 .1 1 B ) w ith s bonds. T h e d ia g ra m in F ig . 1 3 .2 illustrates h o w the th r e e p o rb ita ls o f the a lly l ra d ic a l co m b in e to fo r m three p m o le c u la r o rb itals. (Remember. T h e n u m b e r o f m o le c u la r o rb ita ls that results a lw a y s equals the n u m b e r o f a to m ic o rb ita ls th a t c o m b in e ; see S ection 1 .1 1 .) T h e b o n d in g p m o le c u la r o rb ita l is o f lo w e s t energy; i t encom passes a ll three carb on atom s and is o cc u p ie d b y tw o s p in -p a ire d electrons. T h is b o n d in g p o rb ita l is the resu lt o f h av in g

p o rb ita ls w ith lobes o f the sam e sign o ve rlap b e tw e e n a d ja c e n t carb on atom s. T h is type o f o ve rlap , as w e re c a ll, increases the p -e le c tr o n d en s ity in the regions b e tw e e n the atom s w h e re it is n eed ed fo r b o n d in g . T h e non bo nd ing p o rb ita l is o ccu p ied b y one u n p aire d e le c ­ tro n, and it has a n o d e at the c e n tra l carb on ato m . T h is n o d e m eans that the u n p aire d e le c ­ tro n is loc a te d in the v ic in ity o f carbon atom s 1 and 3 o nly. T h e a n tib o n d in g p m o le c u la r o rb ita l results w h e n o rb ita l lobes o f o pp osite sign o ve rlap b e tw e e n adjacent carb on atom s. Such o ve rlap m ean s that in the a n tib o n d in g p o rb ita l there is a nod e b e tw e e n each p a ir o f carb on atom s. T h is a n tib o n d in g o rb ita l o f the a lly l ra d ic a l is o f h ig h e s t en erg y and is em p ty in the g ro u n d state o f the ra d ic al. W e can illu s tra te the p ic tu re o f the a lly l ra d ic a l g iv e n b y m o le c u la r o rb ita l th e o ry w ith the fo llo w in g structure. H 12

H 2

H

I2 H

W e in d ic a te w ith dashed lines th a t b oth c a rb o n -c a rb o n bonds are p a rtia l d o u b le bonds. T h is a ccom m odates one o f the things th a t m o le c u la r o rb ita l th e o ry tells us. that there is a p bond

encompassing a ll three atoms. W e also p la c e the s y m b o l 2 b esid e the C1 and C 3 atom s. T h is denotes a second th in g m o le c u la r o rb ita l th e o ry tells us. that electron density from the

unpaired electron is equal in the vicinity of C 1 a n d C 3 . F in a lly , im p lic it in the m o le c u la r

592

Chapter 13

Conjugated Unsaturated Systems

Figure 13.2 Combination of three atomic p orbitals to form three p molecular orbitals in the allyl radical. The bonding p molecular orbital is formed by the combination of the three p orbitals with lobes of the same sign overlapping above and below the plane of the atoms. The nonbonding p molecular orbital has a node at C2. The antibonding p molecular orbital has two nodes: between C1 and C2 and between C2 and C3. The shapes of molecular orbitals for the allyl radical calculated using quantum mechanical principles are shown alongside the schematic orbitals.

A n tib o n d in g orb ita l

i N ode

N od e

-L-LJL -

_L

Three isolated p o rb ita ls (w ith an electron in each)

M

N o n b o n d in g orb ita l

N od e

JL

B o n d in g orb ita l

A to m ic o rb ita ls

S ch e m a tic m o le cu la r o rb ita ls

C alculated m o le cu la r o rb ita ls

Y

Y

Y

o rb ita l p ic tu re o f the a lly l ra d ic a l is this: T h e tw o ends o f the a lly l ra d ic a l are equivalent . T h is aspect o f the m o le c u la r o rb ita l descrip tion is also im p lic it in the fo rm u la ju s t given .

13.3B Resonance Description of the Allyl Radical In S ection 1 3 .2 A w e w ro te the structure o f the a lly l ra d ic a l as A :

A H o w e v e r, w e m ig h t ju s t as w e ll h ave w ritte n the e q u iv a le n t structure, B:

B In w ritin g structure B , w e do n o t m e a n to im p ly th a t w e h av e s im p ly taken structure A and tu rn e d it over. W h a t w e h ave done is m o v e the electrons in the fo llo w in g w ay:

We have n o t m o ve d th e nuclei. R esonance th e o ry (S e c tio n 1 .8 ) tells us th a t w h e n e v e r w e can w rite tw o structures fo r a c h e m ic a l e n tity

th a t d iffe r o n ly in th e p o sitio n s o f th e electrons, the e n tity cann ot b e re p ­

resented b y e ith e r structure a lo n e b u t is a hybrid o f b oth. W e can rep resent the h y b rid in tw o w ays. W e can w rite b o th structures A a n d B a n d con nect th e m w ith a d ou b le-h ead ed arrow , the special a rro w w e use to in d ic a te that th e y are resonance structures:

A

B

g - W

13.3 The Stability o f the Allyl Radical

593

O r w e can w rite a sing le structure, C , th a t b len ds the features o f b o th resonance structures:

W e see, th e n , th a t reso n an ce th e o ry g ives us e x a c tly the sam e p ic tu re o f the a lly l ra d ­ ic a l

th a t w e

o b ta in e d

fro m

m o le c u la r

o rb ita l

th e o ry .

S tru c tu re

C

describ es

the

c a rb o n -c a rb o n b onds o f the a lly l ra d ic a l as p a r tia l d o u b le bonds. T h e re so n an ce struc­ tures A a n d B also te ll us th a t the u n p a ire d e le c tro n is associated o n ly w ith th e C1 and C 3 ato m s. W e in d ic a te this in s tru ctu re C b y p la c in g a 8 b es id e C1 a n d C 3 .'t' B ecause reso nan ce structures A an d B are e q u iv a le n t, the e le c tro n d e n s ity fro m the u n p a ire d e le c ­ tro n is shared e q u a lly b y C1 a n d C 3 . A n o th e r ru le in resonance th e o ry is the fo llo w in g : •

W h e n e v e r e q u iv a le n t resonance structures can be w ritte n fo r a c h e m ic a l species, the c h e m ic a l species is m u c h m o re stable than an y resonance structure (w h e n taken a lo n e ) w o u ld ind icate.

I f w e w e re to e x a m in e e ith e r A o r B alo n e, w e m ig h t d ec id e in c o rre c tly th a t it resem bled a p rim a ry ra d ic a l. T h u s, w e m ig h t estim ate the s ta b ility o f the a lly l ra d ic a l as a p p ro x im a te ly that o f a p rim a ry ra d ic a l. In d o in g so, w e w o u ld g re a tly u nd erestim ate the s ta b ility o f the

equivalent reso­ nance structures, the a lly l ra d ic a l should b e m u c h m o re stable than either, th a t is, m u ch a lly l ra d ic a l. R esonance th e o ry tells us, h ow ever, th a t since A and B are

m o re stable than a p rim a ry ra d ic a l. T h is correlates w ith w h a t exp e rim e n ts h av e show n to b e true: T h e a lly l r a d ic a l is e v e n m o r e s ta b le t h a n a t e r t ia r y r a d ic a l.

S o lv e d P r o b le m 1 3 .2 S u b je c tin g p ro pene la b e le d w ith 13C a t carb on 1 to a lly lic c h lo rin a tio n (see b e lo w ) leads to a 5 0 : 5 0 m ix tu re o f 1-c h lorop ro pen e la b e le d at C1 and at C 3 . W r ite a m e c h a n is m that explains this result. (A n asterisk * n e x t to a carbon a to m ind icates th a t the carbon a to m is

13

C .)

---------- — --------- > high temperature

* ^

\ ^

+

CI

C^ ^ ^ \ *

STRATEGY AND ANSWER W e re c a ll (S e c tio n 1 3 .2 A ) th a t the m e c h a n is m fo r a lly lic c h lo rin a tio n inv o lv es the fo r­ m a tio n o f a reso nan ce-stab ilized ra d ic al created b y h avin g a c h lo rin e ato m abstract an a lly lic h yd ro gen ato m . Because the ra d ic a l fo rm e d in this case is a h y b rid o f tw o structures (w h ic h are e q u iv a le n t excep t fo r the p o s itio n o f the la b e l), it can re a c t w ith C l 2 at e ith e r end to g iv e a 5 0 : 5 0 m ix tu re o f the d iffe re n tly la b e le d products. ^

^

n

cl

*

^

.

_ _

. / - x

* ^

ja ,

, ^

^

*

+

Ck

^

\

*

50%

C o n s id e r the a lly lic b ro m in a tio n o f cyc lo h ex e n e la b e le d at C 3 w ith

13

C . N e g le c tin g

stereoisom ers, w h a t p roducts w o u ld y o u e x p e c t fro m this reaction?

NBS, ROOR heat (* = 13C -lab e le d p o s itio n ) ^A resonance structure such as the one shown below would indicate that an unpaired electron is associated with C2. This structure is not a proper resonance structure because resonance theory dictates that all resonance structures must have the same number of unpaired electrons (see Section 13.5A).

C|

An incorrect resonance structure

50%

R e v ie w P ro b le m 13.1

5 94

Chapter 13

Conjugated Unsaturated Systems

13.4 The Allyl Cation C arb o c a tio n s can be a lly lic as w e ll. •

T h e a lly l (p r o p e n y l) c a tio n

+ ) is even m o re stable than a secondary c a r­

b o c a tio n and is a lm o s t as stable as a te rtia ry carbocation. In g en e ra l term s, the re la tiv e o rd er o f sta b ilitie s o f carbocations is th a t g iv e n here.

Relative Order o f Carbocation Stability

S u b s titu te d a lly lic

>

3°

>

A llyl

>

2°

>

1°

>

Vinyl

T h e m o le c u la r o rb ita l d es c rip tio n o f the a lly l ca tio n is show n in F ig . 1 3 .3.

(-)

* I

Î

(+ ) +

H

w

p

‘ (M- )

h

3 CH 2

*

+ V i

-

A n tib o n d in g orb ita l

(-)

N od e

N od e

(+ )

(-)

w --------------w

h 2c

^ ^

2

N onbonding o rb ita l

ch2

(+ )

(-) N od e

Figure 13.3 The p molecular orbitals of the allyl cation. The allyl cation, like the allyl radical (Fig. 13.2), is a conjugated unsaturated system . The shapes of molecular orbitals for the allyl cation calculated using quantum mechanical principles are shown alongside the schem atic orbitals.

^

'

■

^

V ^

JL

CH2

B onding orb ita l

.-

S c hem atic m o le cu la r o rb ita ls

C alcula ted m o le cu la r o rb ita ls

T h e b o n d in g p m o le c u la r o rb ita l o f the a lly l catio n, lik e th a t o f the a lly l ra d ic al (F ig . 1 3 .2 ), contains tw o s p in -p a ire d electrons. T h e n o n b o n d in g p m o le c u la r o rb ita l o f the a lly l catio n, h o w ever, is em p ty. Resonance theory depicts the a lly l cation as a h yb rid o f structures D and E represented here: 2-------------------------------2

1^ ^ " 3+ •*---D

*1

3 E

B ecause D a n d E are equivalent resonance structures, resonance th e o ry predicts that the a lly l ca tio n sho uld b e u n u su a lly stable. S in ce the p o s itiv e charge is loc a te d on C 3 in D and on C1 in E , resonance th e o ry also tells us that the p o s itiv e charge should b e d e lo c a liz e d

13.5 Resonance Theory Revisited

8-W

595

o ver both carbon atom s. C a rb o n a to m 2 carries non e o f the p ositive charge. T h e h y b rid struc­ ture F inclu des charge and b o n d features o f b oth D and E :

2 5+

5+ F

S o lv e d P r o b le m 1 3 .3 A l l y l b ro m id e (3 -b ro m o -1 -p ro p e n e ) fo rm s a c arb ocatio n re a d ily . F o r e x a m p le , it undergoes SN 1 reactions. E x p la in this observation.

STRATEGY AND ANSWER Io n iz a tio n o f a lly l b ro m id e ,B r

(a t rig h t) produces an a lly l ca tio n th a t is u n u su a lly sta­

H2 O

b le (fa r m o re stable than a s im p le p rim a ry ca rb o c a tio n )

Resonance-stabilized carbocation

because it is resonance sta b ilized .

R e v ie w P ro b le m 1 3 .2

( a ) D r a w resonance structures fo r the c arb o catio n that c o u ld be fo rm e d fro m ( £ ) - 2 -b u te n y l triflu o ro m e th a n e s u lfo n a te .

OTf

( b ) O n e o f the resonance structures fo r this c arb o catio n should b e a m o re im p o rta n t co n ­ trib u to r to the resonance h y b rid than the other. W h ic h resonance structure w o u ld b e the greater con tribu tor? (c ) W h a t products w o u ld y o u e x p e c t i f this c arb o catio n re acted w ith a c h lo rid e ion?

13.5 Resonance Theory Revisited W e hav e a lre a d y used resonance th e o ry in e a rlie r chapters, a n d w e hav e been using it ex te n ­ s iv ely in this chap ter because w e are d escrib ing ra d ic als and ions w ith d e lo c a lize d e le c ­

H e lp f u l H i n t

trons (an d charges) in p bonds. R esonance th e o ry is e s p e c ia lly u se fu l w ith system s lik e

Resonance is an im portant tool we use frequently when discussing structure and reactivity.

this, and w e shall use it a g a in and again in the chapters th a t fo llo w . In S ection 1.8 w e had an in tro d u c tio n to resonance th e o ry and an in itia l presentatio n o f som e ru les fo r w ritin g resonance structures. I t should n o w b e h e lp fu l, in lig h t o f o u r p re v io u s discussions o f r e l­ a tiv e c arb o catio n s ta b ility and ra d ic als , and o u r g ro w in g u nderstanding o f co n ju g a te d sys­ tem s, to re v ie w and expan d on those rules as w e ll as those fo r the w ays in w h ic h w e estim ate the re la tiv e c o n trib u tio n a g iv e n structure w i ll m a k e to the o v e ra ll h y b rid .

13.5A Rules for Writing Resonance Structures 1. R e s o n a n c e s tr u c tu r e s e x is t o n ly o n p a p e r. A lth o u g h they h ave n o re a l existence o f th e ir o w n , re s o n a n c e s tr u c tu r e s are u se fu l because they a llo w us to describe m o le ­ cules, ra d ic als , and ions fo r w h ic h a sing le L e w is structure is inad equ ate. Instead, w e w rite tw o o r m o re L e w is structures, c a llin g th e m resonance structures o r resonance con tribu tors. W e con nect these structures b y d o u b le -h e a d e d arrow s (•—

•), a n d w e

say th a t the h y b rid o f a ll o f th e m represents the re a l m o le c u le , ra d ic a l, o r ion. 2 . I n w r i t i n g re s o n a n c e s tr u c tu r e s , w e a r e o n ly a llo w e d to m o v e e le c tro n s . T h e p o s i­ tions o f the n u c le i o f the atom s m u s t re m a in the same in a ll o f the structures. Structure 3 is n o t a resonance structure fo r the a lly lic catio n , fo r e x a m p le , because in o rd er to fo r m i t w e w o u ld h ave to m o v e a h yd ro g e n a to m and this is n o t p erm itted :

3 "v— These are resonance structures for the allylic cation formed when 1,3-butadiene accepts a proton.

This is not a proper resonance structure for the allylic cation b ecause a hydrogen atom has been moved.

G e n e ra lly speaking, w h e n w e m o v e electrons w e m o v e o n ly those o f p bonds (as in the e x a m p le a b o v e ) and those o f lo n e pairs.

596

Chapter 13

Conjugated Unsaturated Systems

3 . A l l o f th e s tr u c tu r e s m u s t b e p r o p e r L e w is s tr u c tu r e s . W e should n o t w rite struc­ tures in w h ic h carbon has fiv e bonds, fo r exam p le: This shift would give a carbon 10 electrons

H < 0 -H

X

This is not a proper resonance structure for ferf-butanol because carbon has five bonds. Elem ents o f the first major row of the periodic table cannot have more than eight electrons in their valence shell.

4 . A l l re s o n a n c e s tr u c tu r e s m u s t h a v e th e s a m e n u m b e r o f u n p a ir e d e le c tro n s . T h e structure on the rig h t is n o t a p ro p e r resonance structure fo r the a lly l ra d ic a l because it contains three u n p aire d electrons w hereas the a lly l ra d ic a l contains o n ly one:

This is not a proper resonance structure for the allyl radical because it does not contain the sam e num ber of unpaired electrons as CH2= CHCH2-. 5 . A l l a to m s t h a t a r e p a r t o f th e d e lo c a liz e d p - e le c t r o n s y s te m m u s t lie in a p la n e o r b e n e a r ly p la n a r . F o r ex a m p le , 2 ,3 -d i-te rt-b u ty l-1 ,3 -b u ta d ie n e behaves lik e a non­

conjugated d ie n e because the la rg e te rt-b u ty l groups tw is t the structure and p revent the dou ble bonds fro m ly in g in the same plan e. Because they are n o t in the same plane, the p o rb ita ls at

C2 and C3 do n o t o ve rlap and d e lo c a liza tio n (an d th e re fo re reso­

n an ce) is prevented:

t -Bu

t -Bu 2 ,3 -D i-fe rf-b u ty l-1 ,3 -b u ta d ie n e 6

. T h e e n e r g y o f th e a c tu a l m o le c u le is lo w e r t h a n th e e n e r g y t h a t m ig h t b e e s ti­ m a te d f o r a n y c o n t r ib u t in g s tr u c tu r e . T h e a ctu al a lly l catio n , fo r e x a m p le , is m o re stable than e ith e r resonance structure 4 o r 5 taken separately w o u ld ind icate. Structures 4 and 5 re s e m b le p rim a ry carbocations and y e t the a lly l ca tio n is m o re sta­ b le (has lo w e r e n e rg y ) than a secondary c arb ocatio n . C h em is ts o fte n c a ll this k in d o f s ta b iliza tio n resonance stabilization:

—

/" X

4

5

In C h a p te r 14 w e shall fin d th a t ben zene is h ig h ly resonance s ta b ilize d because it is a h y b rid o f the tw o e q u iv a le n t fo rm s th a t fo llo w :

R e s o n a n c e s tru c tu re s fo r b e n ze n e

R e p re s e n ta tio n o f h yb rid

7 . E q u iv a le n t re s o n a n c e s tr u c tu r e s m a k e e q u a l c o n tr ib u tio n s to th e h y b r id , a n d a s y s te m d e s c rib e d b y t h e m h as a la r g e re s o n a n c e s ta b iliz a tio n . Structures 4 a n d 5 above m a k e equ al con tribu tio ns to the a lly lic catio n because they are equ ivalen t. T h e y also m a k e a la rg e s ta b ilizin g con trib u tio n and account fo r a lly lic cations b ein g unusu­ a lly stable. T h e sam e can be said abo ut the contributions m a d e b y the e qu ivalen t struc­ tures A and B (S e c tio n 1 3 .3 B ) fo r the a lly l ra d ic al.

13.5 Resonance Theory Revisited

8

. The

m o r e s ta b le a s tr u c t u r e is (w h e n t a k e n b y

its e lf), th e

g r e a te r is its

c o n t r ib u t io n to th e h y b r id . Structures th a t are n o t e q u iv a le n t do n o t m a k e equal c on tribu tio ns. F o r e x a m p le , the fo llo w in g ca tio n is a h y b rid o f structures Stru ctu re

6

m akes a g re a te r c o n trib u tio n than 7 because structure

6

6

a n d 7.

is a m o re stable

te rtia ry c a rb ocatio n w h ile structure 7 is a p rim a ry cation:

•-O 7

6 That

6

m akes a la rg e r c o n trib u tio n m eans th a t the p a rtia l p o s itiv e charge on carbon

b o f the h y b rid w ill be larg er than the p a rtia l p ositive charge on carbon d. I t also means th a t the b o n d b e tw e e n carbon atom s c a n d d w il l b e m o re lik e a d o u b le b o n d than the b o n d b e tw e e n carb on atom s b a n d c.

13.5B Estimating the Relative Stability of Resonance Structures T h e fo llo w in g ru les w il l h e lp us in m a k in g decisions ab o u t the re la tiv e stab ilitie s o f reso­ nan ce structures. a . T h e m o r e c o v a le n t b o n d s a s tr u c t u r e h as , th e m o r e s ta b le i t is. T h is is e x a ctly w h a t w e w o u ld expect because w e k n o w that fo rm in g a cov a le n t b o n d lo w e rs the en ergy o f atom s. T h is m eans th a t o f the fo llo w in g structures fo r 1 ,3 -b u ta d ie n e ,

8

is b y fa r the m o st stable and m akes b y fa r the largest c o n trib u tio n because it co n ­ tains o ne m o re bon d. ( I t is also m o re stable fo r the reason g iv e n u n d er ru le c.) r *

-

8

10

This structure is the m ost stable because it contains more covalent bonds. b. S tr u c tu re s in w h ic h a ll o f th e a to m s h a v e a c o m p le te v a le n c e s h e ll o f e le c tro n s (i.e ., th e n o b le gas s tr u c tu r e ) a r e e s p e c ia lly s ta b le a n d m a k e la r g e c o n t r ib u ­ tio n s to th e h y b r id . A g a in , this is w h a t w e w o u ld expect fro m w h a t w e k n o w about b o n d in g . T h is m eans, fo r e x a m p le , th a t 12 m akes a la rg e r s ta b ilizin g c o n trib u tio n to the catio n b e lo w than

1 1

because a ll o f the atom s o f

1 2

h av e a c o m p le te valen ce

shell. (N o tic e too th a t 12 has m o re c o v a le n t bonds than 11; see ru le a .)

\

Z-O. + 11

—

Y

^ O

^

12 Here the carbon atom has eight electrons.

Here this carbon atom has only six electrons.

c. C h a r g e s e p a ra tio n decreases s ta b ility . Separating opposite charges requires energy. T herefore, structures in w h ic h opposite charges are separated have greater energy (lo w e r stab ility) than those that have no charge separation. T h is means that o f the fo l­ lo w in g tw o structures fo r v in y l chloride, structure 13 m akes a larger contribution because it does n ot have separated charges. (T h is does n o t m ean that structure 14 does n ot contribute to the hybrid; it ju s t means that the contribution m ade b y 14 is smaller.)

¿fX ÿl13

«---- »

14

597

598

Chapter 13

Conjugated Unsaturated Systems

S o lv e d P roblem 13.4 W r ite resonance structures fo r

and te ll w h ic h structure w o u ld b e the greater co n trib u to r to the resonance

h yb rid .

STRATEGY AND ANSWER W e w rite the structure fo r the m o le c u le a n d then m o v e the e le ctro n p airs as show n b elo w . T h e n w e e x a m in e the tw o structures. T h e second structure contains separated charges (w h ic h w o u ld m a k e it less stable) and the firs t structure contains m o re bonds (w h ic h w o u ld m a k e it m o re stable). B o th factors te ll us th a t the firs t structure is m o re stable. C o nsequ ently, it should be the greater co n trib u to r to the h y b rid .

More bonds

R e v ie w P ro b le m 1 3 .3

R e v ie w P ro b le m 1 3 .4

Separated charges

W r ite the im p o rta n t resonance structures fo r each o f the fo llo w in g :

F ro m each set o f resonance structures th a t fo llo w , designate the o ne th a t w o u ld con tribute m o s t to the h y b rid and e x p la in y o u r choice:

+

O

(f)

: N H 2— C = N :

O

N H 2= C = N ; -

8-W

13.6 Alkadienes and Polyunsaturated Hydrocarbons

599

R e v ie w P ro b le m 1 3 .5

T h e fo llo w in g eno l (an a lk en e -a lc h o o /) and keto (a ketone) form s o f C 2 H 4O d iffe r in the p osi­ tions fo r th e ir electrons, b ut they are n ot resonance structures. E x p la in w h y they are not. ,H

'O '

:O

C2 H4O

'H

Enol form

Keto form

13.6 Alkadienes and Polyunsaturated Hydrocarbons M a n y hydro carb on s are k n o w n that c o n tain m o re than o ne d o u b le o r trip le bon d. A h y d ro ­ carb on th a t contains tw o d o u b le bonds is c a lle d an a lk a d ie n e ; o ne th a t contains three d o u ­ b le bonds is c a lle d an a lk a t r ie n e , and so on. C o llo q u ia lly , these com pounds are often re fe rre d to s im p ly as dienes o r trienes. A h yd ro c a rb o n w ith tw o trip le bonds is c a lle d an a lk a d iy n e , a n d a h yd ro c a rb o n w ith a d o u b le a n d trip le b o n d is c a lle d an a lk e n y n e . T h e fo llo w in g exam p le s o f p o ly u n s a tu ra te d hyd ro carb on s illu s tra te h o w specific c o m ­ pounds are n am e d . R e c a ll fro m IU P A C ru les (S ectio ns 4 .5 a n d 4 .6 ) th a t the n u m e ric a l locants fo r d o u b le and trip le bonds can b e p la c e d a t the b eg in n in g o f the n a m e o r im m e ­ d ia te ly p recedin g the resp ective suffix. W e p ro v id e e x am p les o f b oth styles.

1 2 3 c h 2= c = c h

4

3

2

12

(a lie n e , o r p r o p a - , - d ie n e )

4

5

1 ,3 - B u ta d ie n e

( 3 Z ) - P e n ta - 1 ,3 - d ie n e

( b u ta - 1 ,3 - d ie n e )

( c /s - p e n ta - 1 ,3 - d ie n e )

2

6

3

2

5 4

1

3 1 4 / " \ 2^

1

2

5

1 ,2 - P r o p a d ie n e

2

4

2

P e n t- 1 - e n - 4 - y n e

4

8

( 2 E ,4 E ) - 2 ,4 - H e x a d ie n e

( 2 Z ,4 E ) - H e x a - 2 ,4 - d ie n e

( 2 E ,4 E , 6 E ) - O c ta - 2 ,4 ,6 - t r ie n e

( tr a n s , t r a n s - 2 , 4 - h e x a d ie n e )

( c / s , t r a n s - h e x a - 2 , 4 - d ie n e )

( t r a n s , t r a n s , t r a n s - o c t a - 2 , 4 , 6 - t r ie n e )

2

3

1 ,3 - C y c lo h e x a d ie n e

1 ,4 - C y c lo h e x a d ie n e

T h e m u ltip le bonds o f p o ly u n s a tu ra te d com pounds are classified as b ein g c u m u la te d , c o n ju g a te d , o r is o la te d . •

6

6

T h e d o u b le bonds o f a 1 ,2 -d ie n e (such as 1 ,2 -p ro p a d ien e , also c a lle d a lle n e ) are said to b e c u m u la te d because o ne carb on (th e c e n tra l c arb on ) p articip ates in tw o d o u b le bonds.

H y d ro c a rb o n s w ho se m o le c u le s h ave c u m u la te d d o u b le bonds are c a lle d c u m u le n e s . T h e n a m e a lle n e (S e c tio n 5 .1 8 ) is also used as a class n a m e fo r m o le c u le s w ith tw o c u m u la te d d o u b le bonds:

C u m u la t e d d o u b le b o n d s

H \

q = q

= 0

/

H

u >H

\

^H

/

A lle n e

A n e x a m p le o f a co n ju g a te d d ien e is 1,3-b u ta d ie n e.

c = C = C

> ^

A c u m u la t e d d ie n e

600

Chapter 13

•

In

Conjugated Unsaturated Systems

conjugated p o lyen es the d o u b le and sing le bonds alternate along the chain:

C o n ju g a te d

/

^

C = C

d o u b le b o n d s

/ C = C

1 ,3 - B u ta d ie n e

A c o n ju g a t e d d ie n e

(2 E ,4 E ,6 E )-O c ta -2 ,4 ,6 -tr ie n e is an e x a m p le o f a co n ju g a te d a lk atrie n e . •

I f o ne o r m o re saturated carbon atom s in te rv e n e b e tw e e n the d o u b le bonds o f an a lk a d ie n e , the d o u b le bonds are said to b e is o la te d .

A n e x a m p le o f an is o la te d d ie n e is 1 ,4-p entadiene:

Is o la t e d d o u b le b o n d s

/

\ „ h

) ^

^

( 2'n A n i s o l a t e d d ie n e

1 ,4 - P e n ta d ie n e

(n * 0)

R e v ie w P ro b le m 13 .6

(a) W h ic h o th e r com pounds in S e c tio n 1 3 .6 are c o n ju g a te d dienes? (b) W h ic h o th e r com pounds are is o la te d dienes? (c) W h ic h c o m p o u n d is an is o la te d enyne?

In C h a p te r 5 w e saw th a t a p p ro p ria te ly substituted c u m u la te d dienes (a lle n e s ) g iv e rise to c h ira l m o le c u le s . C u m u la te d dienes h av e h a d som e c o m m e rc ia l im p o rtan c e , and c u m u ­ la te d d o u b le bonds are o cc a s io n a lly fo u n d in n a tu ra lly o cc u rrin g m o le c u le s . In g eneral, c u m u la te d dienes are less stable than is o la te d dienes. T h e d o u b le bonds o f is o la te d dienes beh ave ju s t as th e ir n a m e suggests— as isolated “enes.” T h e y und ergo a ll o f the reactions o f alkenes, and, e x c ep t fo r the fa c t th a t they are c apab le o f reactin g tw ic e , th e ir b e h a v io r is n ot unusual. C o n ju g a te d dienes are fa r m o re interesting because w e fin d th a t th e ir d o u b le bonds in te ra c t w ith each other. T h is in te ra c ­ tio n leads to u nexpected pro perties and reactions. W e shall th e re fo re con sider the c h e m ­ is try o f c o n ju g a te d dienes in d etail.

13.7 1,3-Butadiene: Electron Delocalization 13.7A Bond Lengths of 1,3-Butadiene T h e c a rb o n -c a rb o n b on d lengths o f 1,3-b utadien e h ave been d eterm ined and are show n here:

D o u b le b o n d s a re s h o rte r

1 .3 4 A

( 2

th a n s in g le bonds

I

3

1 .4 7 A

T h e C1 — C 2 b o n d a n d the C 3 — C 4 b o n d are (w ith in e x p e rim e n ta l e rro r) the sam e length as the c a rb o n -c a rb o n d o u b le b o n d o f ethene. T h e c e n tra l b o n d o f 1 ,3 -b u ta d ie n e (1 .4 7 A ) , h ow ever, is co n s id e rab ly shorter than the sing le b o n d o f ethane (1 .5 4 A ) . T h is should n o t be surprising. A l l o f the carbon atom s o f 1,3-b utadien e are sp 2 h y b rid ize d and, as a result, the c e n tra l b o n d o f b u ta d ie n e results fro m o ve rlap p in g sp 2 o rb itals. A n d , as w e k n o w , a sig m a b o n d that is sp 3 - s p 3 is longer. T h e re is, in fact, a steady decrease in b o n d le n g th o f c a rb o n -c a rb o n sing le bonds as the h y b rid iz a tio n state o f the b o n d e d atom s changes fro m sp 3 to sp (T a b le 1 3 .1 ).

13.7 1,3-Butadiene: Electron Delocalization

^

8-Wu u TT

601

C arb o n -C arb o n Single-Bond Lengths and H ybridization State

Compound

Hybridization State

H3 C— CH3 CH2 = CH— CH3 CH2 = CH— CH= CH2 HC # C— CH3 HC # C— CH= CH2 HC # C— C # CH

sp3-s p 3 sp2-s p 3 sp2-s p 2 sp -s p 3 sp -sp 2 sp -sp

Bond Length (A) 1.54 1.50 1.47 1.46 1.43 1.37

13.7B Conformations of 1,3-Butadiene T here are tw o possible planar conform ations o f 1,3-butadiene: the s-cis and the s-trans con­ form ations. 2

//

____ , 3 Vv

s -cis C o n fo rm a tio n o f 1 ,3 -b u ta d ie n e

rotate about C 2 _C3

^

3

s -tra n s C o n fo rm a tio n o f 1 ,3 -b u ta d ie n e

T hese are n ot true cis and trans form s since the s-cis and s-trans conform ations o f 1,3-buta­ diene can b e interconverted through rotation about the single bond (hence the prefix s). The s-trans conform ation is the predom inant one at room tem perature. We shall see that the scis conform ation o f 1,3-butadiene and other 1,3-conjugated alkenes is necessary for the D iels-A ld e r reaction (Section 13.11).

Solved Problem 13.5 Provide an explanation for the fact that m any m ore m olecules are in the s-trans conform ation o f 1,3-butadiene at equilibrium . STRATEGY AND ANSWER T he s-cis conform ation of 1,3-butadiene is less stable, and therefore less populated, than the s-trans conform er because it has steric interference betw een the hydrogen atom s at carbons 1 and 4. Interference o f this kind does not exist in the s-trans conform ation, and therefore, the s-trans conform ation is m ore stable and m ore populated at equilibrium .

steric interference of hydrogen a to m s ^

H

H

rotation about C2—C3 bond s -tra n s C o n fo rm a tio n is m o re s ta b le

s -c is C o n fo rm a tio n is less s ta b le

13.7C Molecular Orbitals of 1,3-Butadiene T he central carbon atom s o f 1,3-butadiene (Fig. 13.4) are close enough for overlap to occur betw een the p orbitals o f C 2 and C 3. This overlap is n o t as great as that betw een the orbitals o f C1 and C 2 (or those o f C 3 and C4). T he C 2 -C 3 orbital overlap, however, gives the cen ­ tral b ond partial double-bond character and allow s the four p electrons o f 1,3-butadiene to be delocalized over all four atom s. F igure 13.5 show s how the four p orbitals o f 1,3-butadiene com bine to form a set of four p m olecular orbitals. •

Two o f the p m olecular orbitals o f 1,3-butadiene are bonding m olecular orbitals. In the ground state these orbitals hold the four p electrons w ith tw o spin-paired electrons in each.

•

T he other tw o p m olecular orbitals are antibonding m olecular orbitals. In the ground state these orbitals are unoccupied.

Figure 13.4 The p orbitals of 1.3-butadiene. (See Fig. 13.5 for the shapes of calculated molecular orbitals for 1.3-butadiene.)

602

Chapter 13

Conjugated Unsaturated Systems

^ ^ 1 (+ )

(-)

(-) (+) T 4*

(+ )

(+}

(-)

A n tib o n d in g m o le cu la r o rb ita ls

(-)

(+)

t g*

(-)

LUMO

S

_L_L±±_

Four isolated p o rb ita ls (w ith an electron in each)

(-)

(+)

JL

2

HOMO

• (-}

4

(+ ) B onding m o le cular o rb ita ls (+ )

Figure 13.5 Formation of the p molecular orbitals of 1,3-butadiene from four isolated p orbitals. The shapes of mole­ cular orbitals for 1,3-butadiene calculated using quantum mechanical principles are shown alongside the schematic orbitals.

JL (- )

p

S c hem atic m o le cu la r o rb ita ls

C alculated m o le cu la r o rb ita ls

A n electro n can b e e x c ite d fro m the h ig hest o ccu p ied m o le c u la r o rb ita l ( H O M O ) to the lo w e s t u n o ccu p ied m o le c u la r o rb ita l ( L U M O ) w h e n 1 ,3 -b u ta d ie n e absorbs lig h t w ith a w a v e le n g th o f 2 1 7 n m . (W e shall study the abso rp tio n o f lig h t b y unsaturated m o le c u le s in S ection 1 3 .9 .) •

T h e d e lo c a liz e d b o n d in g th a t w e h av e ju s t described fo r 1 ,3 -b u ta d ie n e is character­ is tic o f a ll co n ju g a te d polyenes.

13.8 The S tab ility o f C onjugated Dienes •

C o n ju g a te d alkad ie n e s are th e rm o d y n a m ic a lly m o re stable than is o m e ric isolated a lkadienes.

T w o exam p le s o f this e x tra s ta b ility o f con jug ated dienes can b e seen in an analysis o f the heats o f h y d ro g e n a tio n g iv e n in T a b le 13 .2.

13.8 The Stability o f Conjugated Dienes

^

H eats o f H ydrogenation o f Alkenes and Alkadienes

Com pound

H2 (m o l)

AH° (kJ m o l- 1 )

1-Butene 1-Pentene trans-2-Pentene 1,3-B utadiene trans-1,3-P entadiene 1,4-P entadiene 1,5-H exadiene

1 1 1 2 2 2 2

-1 2 7 -1 2 6 -1 1 5 -2 3 9 -2 2 6 -2 5 4 -2 5 3

In itse lf, 1 ,3 -b u ta d ie n e cannot b e com p a re d d ire c tly w ith an is o la te d d ie n e o f the same c hain len g th . H o w e v e r, a co m p ariso n can b e m a d e b etw e e n the h ea t o f h yd ro g e n a tio n o f 1 ,3 -b u ta d ie n e an d that o b ta in e d w h e n tw o m o la r eq u ivalen ts o f 1-b u ten e are hydrogenated:

AH° (kJ m o l-1) 2 H2

2 X (-1 2 7 )

= -2 5 4

1 - B u te n e

+

2 H2

239 D ifference

1 ,3 - B u ta d ie n e

15 kJ m o l-

B ecause 1-b u ten e has a m o n osub stituted d o u b le b o n d lik e those in 1 ,3 -b u ta d ie n e, w e m ig h t e x p e c t h y d ro g e n a tio n o f 1 ,3 -b u ta d ie n e to lib e ra te the sam e a m o u n t o f h ea t (2 5 4 kJ m o l ~ x) as tw o m o la r equ ivalen ts o f 1-b u ten e. W e fin d , h o w ever, th a t 1 ,3 -b u ta d ie n e liberates o n ly 2 3 9 kJ m o F 1, 15 kJ m o F

1

less than expected. W e con clud e, th e re fo re , th a t c o n ju g a tio n

im p a rts som e e x tra s ta b ility to the c o n ju g a te d system (F ig . 1 3 .6 ).

Figure 13.6 Heats of hydrogenation of 2 mol of 1-butene and 1 mol of 1,3-butadiene. A n assessm ent o f the s ta b iliza tio n th a t c o n ju g a tio n p ro vid e s tra n s -1 ,3 -p e n ta d ie n e can b e m a d e b y c o m p a rin g the h ea t o f h yd ro g e n a tio n o f tra n s -1 ,3 -p e n ta d ie n e to the sum o f the heats o f h yd ro g e n a tio n o f 1 -p entene a n d trans-2-pentene. T h is w a y w e are c o m p a rin g d o u ­ b le bonds o f c o m p a ra b le types:

AH 0

126 kJ m ol-

1 - P e n te n e

fr a n s - 2 - P e n te n e

fr a n s - 1 ,3 - P e n ta d ie n e

AH 0 Sum

115 kJ mol241 kJ mol-

AH° = - 2 2 6 kJ mol-1 Difference = 15 kJ mol-1

8-W

603

604

Chapter 13

Conjugated Unsaturated Systems

W e see fro m these calcu latio ns th a t c o n ju g a tio n affo rd s tra n s -1 ,3 -p e n ta d ie n e an ex tra s ta b ility o f 15 kJ m o l_ 1 , a v a lu e th a t is equ iv a le n t, to tw o s ig n ific a n t figures, to the o ne w e o b ta in e d fo r 1 ,3 -b u ta d ie n e (1 5 kJ m o F 1). W h e n calcu la tio n s lik e these are c a rrie d o ut fo r o th e r co n ju g a te d dienes, s im ila r results are o btain ed; conjugated dienes are fo u n d to be more stable than isolated dienes. T h e ques­ tion , then, is this: W h a t is the source o f the ex tra s ta b ility associated w ith con jugated dienes? T h e re are tw o factors that con tribu te. T h e extra s ta b ility o f co n ju g a te d dienes arises in p art fro m the stronger c e n tra l b o n d th a t th e y con tain and, in p art, fro m the a d d itio n a l d e lo c a l­ iz a tio n o f the p electrons th a t occurs in co n ju g a te d dienes.

1

1B ■

Solved Problem 13 6 1

J U l V C U

r i U U I C l l l

1 J . U

J

W h ic h d ien e w o u ld y o u expect to be m o re stable: 1 ,3 -c y c lo h e x a d ie n e o r 1 ,4 -c y c lo h e xa d ie n e ? W h y ? W h a t e x p e r­ im e n t c o u ld y o u c a rry o u t to c o n firm y o u r answer?

STRATEGY AND ANSWER 1 ,3 -C y c lo h e x a d ie n e is con jug ated , a n d on th a t basis w e w o u ld e x p e c t it to b e m o re stable. W e c o u ld d e te rm in e the heats o f h y d ro g e n a tio n o f the tw o com pounds, and since on h y d ro g e n a tio n each c o m p o u n d y ie ld s the sam e p ro du ct, the d ie n e w ith the s m a lle r h eat o f h yd ro g e n a tio n w o u ld b e the m o re stable one.

^ o ^ 0 1 ,3 -C y c lo h e x a d ie n e (a c o n ju g a te d d ie n e )

C y c lo h e x a n e

1 ,4 -C y c lo h e x a d ie n e (an is o la te d d ie n e )

13.9 U ltravio let-V isib le Spectroscopy T h e e x tra s ta b ility o f co n ju g a te d dienes w h e n c o m p a re d to corresponding u ncon jug ated dienes can also be seen in d ata fro m u lt r a v io le t - v is ib le ( U V - V i s ) s p e c tro s c o p y . W h e n e le c tro m a g n e tic ra d ia tio n in the U V and v is ib le reg ion s passes throu gh a c o m p o u n d co n ­ ta in in g m u ltip le bonds, a p o rtio n o f the ra d ia tio n is u su a lly absorbed b y the com p o u n d . Just h o w m u c h ra d ia tio n is absorbed depends on the w a v e le n g th o f the ra d ia tio n a n d the struc­ ture o f the com po un d. •

T h e abso rp tio n o f U V - V i s ra d ia tio n is caused b y tra n s fe r o f e n erg y fro m the ra d ia ­ tio n b e a m to electrons th a t can b e e x c ite d to h ig h e r energy o rbitals.

In S ection 1 3 .9 C w e shall return to discuss spe c ifica lly h o w data fro m U V - V i s spectroscopy d em on strate the a d d itio n a l s ta b ility o f co n ju g a te d dienes. F irs t, in S ection 1 3 .9 A w e b rie fly re v ie w the p ro p erties o f e le ctro m ag n e tic ra d ia tio n , and then in S ection 1 3 .9 B w e lo o k at h o w d ata fro m a U V - V i s spectro ph otom eter are o btain ed.

13.9A The Electromagnetic Spectrum A c c o rd in g to q u a n tu m m ech anics, e le ctro m ag n e tic ra d ia tio n has a d u a l a n d s e e m in g ly co n ­ tra d ic to ry natu re. •

E le c tro m a g n e tic ra d ia tio n can b e d escrib ed as a w a v e o cc u rrin g s im u lta n e o u sly in e le c tric a l and m a g n e tic field s. I t can also b e described as i f it consisted o f particles c a lle d q uanta o r photons.

D iffe r e n t experim ents disclose these tw o d iffe re n t aspects o f electrom ag netic rad iatio n. T h e y are n o t seen to g e th e r in the sam e e x p e rim e n t. •

A w a v e is u su a lly d escrib ed in term s o f its w a v e le n g th (A ) o r its fre q u e n c y (n).

13.9 Ultraviolet-Visible Spectroscopy

A s im p le w a v e is show n in F ig . 1 3 .7 . T h e distance b e tw e e n consecutive crests (o r tro ughs) is the w a v ele n g th . T h e n u m b e r o f f u ll cycles o f the w a v e th a t pass a g iv e n p o in t each second, as the w a v e m o ves throu gh space, is c a lle d the frequency and is m easu red in cycles p e r second (cp s), o r h e r t z ( H z ) .*

One cycle

Figure 13.7 A simple wave and its wavelength, l .

A l l e le c tro m a g n e tic ra d ia tio n travels throu gh a v a c u u m at the sam e v e lo c ity . T h is v e lo c ­ ity (c ), c a lle d the v e lo c ity o f lig h t, is 2 .9 9 7 9 2 4 5 8 a nd fre q u e n c y as c

=

in m eters (m ), m illim e te rs ters (1 n m

=

10-

9

X

10 8 m s -

1

and relates to w a v ele n g th

I n . T h e w avelen gths o f ele ctro m ag n e tic ra d ia tio n are expressed eith er ( 1

mm

=

1 0 - 3

m ), m ic ro m e te rs

(1

mm

=

1 0 - 6

m ), o r n a n o m e ­

m ). [A n o ld e r term fo r m ic ro m e te r is micron (ab b reviated m) and an o ld er

te rm fo r n a n o m e te r is m illim icron.] T h e e n ergy o f a q u a n tu m o f e le ctro m ag n e tic e n erg y is d ire c tly re la te d to its frequency:

E = hn w h e re

h

=

n=

P la n c k ’s constant, 6 .6 3

X

10-

3 4

J s

fre q u e n c y ( H z ) T h e h ig h e r th e fre q u e n c y (n) of ra d ia tio n , th e g re a te r is its energy.

X -R a y s , fo r e x a m p le , are m u c h m o re e n ergetic than rays o f v is ib le lig h t. T h e frequencies o f X -ra y s are on the o rd er o f 1 0 1 9 H z , w h ile those o f v is ib le lig h t are on the o rd er o f 1 0 1 5 H z . S in ce

n

= c / l , the e n erg y o f e le c tro m a g n e tic ra d ia tio n is in v e rse ly p ro p o rtio n a l to its

w avelen gth:

A w h e re

c = v e lo c ity o f lig h t T h e s h o rte r th e w a v e le n g th ( l ) of ra d ia tio n , th e g re a te r is its energy.

X - R a y s h a v e w a v e le n g th s o n the o rd e r o f 0 .1 n m a n d a re v e ry e n e rg e tic, w h e re as v is i­ b le lig h t has w a v ele n g th s b e tw e e n 4 0 0 a n d 7 5 0 n m a n d is, th e re fo re , o f lo w e r e n e rg y than X - r a y s .f I t m a y b e h e lp fu l to p o in t out, too, th a t fo r v is ib le lig h t, w avelen g th s (an d thus fre ­ q uencies) are re la te d to w h a t w e p erc e iv e as colors. T h e lig h t that w e c a ll re d lig h t has a w a v e le n g th o f a p p ro x im a te ly 6 5 0 n m . T h e lig h t w e c a ll v io le t lig h t has a w a v e le n g th o f a p p ro x im a te ly 4 0 0 n m . A l l o f the o th e r colors o f the v is ib le spectrum (th e ra in b o w ) lie in b e tw e e n these w avelen gths.

*The term hertz (after the German physicist H. R. Hertz), abbreviated Hz, is used in place of the older term

cycles per second (cps). Frequency of electromagnetic radiation is also sometimes expressed in wavenumbers, that is, the number of waves per centimeter. f A convenient formula that relates wavelength (in nm) to the energy of electromagnetic radiation is the following:

E (in kJ mol *)

1.20 X 10- 9 kJ mol- 1 wavelength in nanometers

g - W

60 5

606

Chapter 13

Conjugated Unsaturated Systems

Increasing v 10 1 9 Hz Cosmic and Y - rays

X-rays

0.1

1013 Hz

10 1 5 Hz (UV) Vacuum ultraviolet

nm

2 0 0

(UV) Near ultraviolet nm

(NIR) Near infrared

Visible

400 nm

I

700 nm _ | __

Microwave radio

(IR) Infrared 2

,um

50 ,um

Increasing X

Figure 13.8

The electrom agnetic spectrum.

T h e d iffe re n t reg ion s o f the e le c tr o m a g n e tic s p e c tr u m are show n in F ig . 1 3 .8 . N e a rly e v e ry p o rtio n o f the e le ctro m ag n e tic spectru m fro m the re g io n o f X -ra y s to th a t o f m ic ro w a v e s and ra d io w aves has b een used in e lu c id a tin g structures o f atom s a n d m o le ­ cules. A lth o u g h techniques d iffe r a ccordin g to the p o rtio n o f the e le ctro m ag n e tic spectrum in w h ic h w e are w o rk in g , there is a consistency and u n ity o f basic p rin cip les.

13.9B UV-Vis Spectrophotometers •

A U V - V i s s pectro ph otom eter (F ig . 1 3 .9 ) m easures the a m o u n t o f lig h t absorbed b y a sam p le at each w a v e le n g th o f the U V and v is ib le reg ion s o f the e le c tro m a g ­ n e tic spectrum .

UV light source

Figure 13.9 A diagram of a UV-Vis spectrophotom eter. (Courtesy William Reusch, w w w .cem .m su.edu/~reusch. © 1999)

13.9 Ultraviolet-Visible Spectroscopy

U V and v is ib le ra d ia tio n are o f h ig h e r energy (sh o rter w a v e le n g th ) than in fra re d ra d ia tio n (used in I R spectroscopy) and ra d io fre q u e n c y ra d ia tio n (used in N M R ) b u t n o t as e n e r­ g etic as X -ra d ia tio n (F ig . 1 3 .8 ). In a standard U V - V i s spectro ph otom eter (F ig . 1 3 .9 ) a b e a m o f lig h t is split; o ne h a lf o f the b e a m (th e sam ple b e a m ) is d ire c te d throu gh a transparent c e ll c o n tain in g a solu tio n o f the c o m p o u n d b ein g a n a ly ze d , and o ne h a lf (th e re fe re n c e b e a m ) is d ire c te d th ro u g h an id e n tic a l c e ll that does n o t con tain the com p o u n d b u t contains the solvent. Solvents are ch o ­ sen to be transp aren t in the re g io n o f the spectrum b ein g used fo r analysis. T h e instru m en t is d esigned so th a t it can m a k e a c om pariso n o f the intensities o f the tw o b eam s as it scans o ve r the d esired re g io n o f w avelen gths. I f the c o m p o u n d absorbs lig h t a t a p a rtic u la r w a v e ­ leng th , the in te n s ity o f the sam p le b e a m ( I S) w il l b e less than th a t o f the re fe re n c e b eam ( I R). T h e absorbance at a p a rtic u la r w a v e le n g th is d efin e d b y the eq u a tio n A a = lo g ( IR/ I S). •

D a ta fro m a U V - V i s spectro ph otom eter are p resented as an a b s o r p tio n s p e c tr u m , w h ic h is a graph o f w a v e le n g th ( l ) versus sam p le absorbance (A ) a t each w a v e ­ le n g th in the spectral re g io n o f interest.

( In d io d e -a rra y U V - V i s spectrophotom eters the abso rp tio n o f a ll w avelen g th s o f lig h t in the re g io n o f analysis is m easu red s im u lta n e o u sly b y an a rra y o f p ho to dio des. T h e absorp­ tio n o f the solven t is m easu red o ve r a ll w avelen g th s o f interest first, and then the absorp­ tio n o f the sam p le is re c o rd e d o v e r the sam e ra n g e . D a ta fro m the solven t are e le c tro n ic a lly subtracted fro m the d ata fo r the sam ple. T h e d iffe re n c e is then d is p la y e d as the absorption spectrum fo r the sam p le.) A ty p ic a l U V

abso rp tio n spectrum , th a t o f 2 ,5 -d im e th y l-2 ,4 -h e x a d ie n e , is g iv e n in

F ig . 1 3 .1 0 . I t shows a b ro ad abso rp tio n b an d in the re g io n b etw e e n 2 1 0 an d 2 6 0 n m , w ith the m a x im u m abso rp tio n a t 2 4 2 .5 n m . •

T h e w a v e le n g th o f m a x im u m abso rp tio n in a g iv e n spectrum is u su a lly re p o rte d in the c h e m ic a l lite ra tu re as

1

max.

In a d d itio n to re p o rtin g the w a v e le n g th o f m a x im u m abso rp tio n ( l max), chem ists o fte n rep o rt another q u an tity c a lled the m o la r absorptivity, e. (In o ld e r literatu re, the m o la r absorp­ tiv ity , e, is o fte n re fe rre d to as the m o la r e x tin c tio n c o e ffic ie n t.)

Wavelength (nm)

Figure 13.10 The UV absorption spectrum of 2,5-dim ethyl-2,4-hexadiene in methanol at a concentration of 5.95 X 10~5 M in a 1.00-cm cell. ©Bio-Rad Laboratories, Inc. Informatics Division, Sadtler Software & Databases (1960-2006). All Rights Reserved. Permission for the publication herein of Sadtler Spectra has been granted by Bio-Rad Laboratories, Inc., Informatics Division.

607

60 8

Chapter 13

Conjugated Unsaturated Systems

T h e m o la r a b s o r p tiv it y (e , in units o f M

- 1

c m - 1 ) ind icates the in te n s ity o f the

absorbance fo r a sam ple a t a g iv e n w a v ele n g th . I t is a p ro p o rtio n a lity constant that relates absorbance to m o la r c on centration o f the sam p le ( M ) and the path length ( l , in c m ) o f lig h t throu gh the sam ple. T h e eq u a tio n th a t relates absorbance (A ) to co n cen tratio n ( C ) and p ath le n g th (i) v ia m o la r a b s o rp tiv ity ( e ) is c a lle d B e e r ’s la w .

A

= eX C X I

or

e =

A -----------c

B e e r’s law

x i

F o r 2 ,5 -d im e th y l-2 ,4 -h e x a d ie n e d isso lved in m e th a n o l the m o la r a b s o rp tiv ity at the w a v e ­ absorb le n g th o f m a x im u m absorbance (2 4 2 .5 n m ) is 1 3 ,1 0 0 M

1

cm

*. In the c h e m ic a l lite ra tu re

this w o u ld b e re p o rte d as

(e = 13,100)

2,5-D im eth y l-2 ,4 -h ex ad ien e, im !xian°l 2 4 2 .5 nm

13.9C Absorption Maxima for Nonconjugated and Conjugated Dienes A s w e n o te d e a rlie r, w h e n com pounds absorb lig h t in the U V and v is ib le regions, electrons are e x c ite d fro m lo w e r e le c tro n ic e n erg y levels to h ig h e r ones. F o r this reason, v is ib le and U V spectra are o fte n c a lle d e le c tr o n ic s p e c tra . T h e abso rp tio n spectru m o f 2 ,5 -d im e th y l2 ,4 -h e x a d ie n e is a ty p ic a l e le ctro n ic spectrum because the absorption b an d (o r p e a k ) is very b ro ad . M o s t abso rp tio n bands in the v is ib le and U V re g io n are b ro ad because each e le c ­ tro n ic e n ergy le v e l has associated w ith it v ib ra tio n a l a n d ro ta tio n a l levels. T h u s, electron transitions m a y o ccu r fro m an y o f several v ib ra tio n a l and ro ta tio n a l states o f o ne e le ctro n ic le v e l to any o f several v ib ra tio n a l and ro ta tio n a l states o f a h ig h e r le v e l. •

A lk e n e s and n o n co n ju g a te d dienes u su a lly h ave abso rp tio n m a x im a ( 1 max) b e lo w 2 0 0

nm .

E th en e, fo r exam p le, gives an absorption m a x im u m at 171 n m ; 1 ,4-p entadiene gives an absorption m a x im u m at 1 78 n m . These absorptions occur at w avelengths that are o ut o f the range o f operation o f m o st u ltra v io le t-v is ib le spectrometers because they occur w h e re the oxy g e n in a ir also absorbs. S p ecial a ir-fre e techniques m u st be e m p lo y e d in m easuring them .

______ H e lp f u l H i n t UV-Vis spectroscopic evidence for conjugated p-electron systems.

•

C o m p o u n d s c o n tain in g conjugated m u ltip le bonds h av e abso rp tio n m a x im a ( 1 max) at w avelen g th s lo n g e r than

2 0 0

nm .

1 ,3 -B u ta d ie n e , fo r e x a m p le , absorbs a t 2 1 7 n m . T h is lo n g e r w a v e le n g th abso rp tio n b y co n ­ ju g a te d dienes is a d ire c t consequence o f con ju g atio n . W e can understand h o w c o n ju g a tio n o f m u ltip le bonds b ring s ab o u t abso rp tio n o f lig h t

t

at lo n g e r w a v ele n g th s i f w e e x a m in e F ig . 1 3 .11 .

>

1

*

* i

I excitation

o

p4

p* (LUMO)

P 3 * LUMO

Antibonding molecular orbitals

excitation

c LU

p 2 HOMO p (HOMO)

Figure 13.11 The relative energies of the p molecular orbitals of ethene and 1,3-butadiene (Section 13.7C).

X — =

pi

Bonding > molecular orbitals

13.9 Ultraviolet-Visible Spectroscopy

•

8-W

609

W h e n a m o le c u le absorbs lig h t at its lon gest w a v ele n g th , an electro n is exc ite d fro m its h ig h e s t o c c u p ie d m o le c u la r o r b it a l ( H O M O ) to the lo w e s t u n o c c u p ie d m o le c u la r o r b it a l ( L U M O ) .

•

F o r m o s t alkenes and a lkad ienes the H O M O is a b o n d in g p o rb ita l a n d the L U M O is an a n tib o n d in g p * o rb ita l.

T h e w a v e le n g th o f the abso rp tio n m a x im u m is d e te rm in e d b y the d iffe re n c e in energy b e tw e e n these tw o levels. T h e e n erg y gap b e tw e e n the H O M O and L U M O o f ethene is g re a te r than that b e tw e e n the corresponding o rb ita ls o f 1 ,3 -b u ta d ie n e. T h u s, the p -------: p * electro n e x c ita tio n o f ethene req uires abso rp tio n o f lig h t o f g reater energy (sh o rter w a v e ­ le n g th ) than the corresponding p

2

------- : p 3* e x c ita tio n in 1 ,3 -b u ta d ie n e. T h e e n erg y d if ­

fere n c e b e tw e e n the H O M O s and the L U M O s o f the tw o com pounds is re fle c te d in th eir abso rp tio n spectra. E th e n e has its 1 max at 171 n m ; 1 ,3 -b u ta d ie n e has a 1 max a t 2 1 7 n m . T h e n a rro w e r g ap b e tw e e n the H O M O and the L U M O in 1 ,3 -b u ta d ie n e results fro m the c o n ju g a tio n o f the d o u b le bonds. M o le c u la r o rb ita l calcu la tio n s in d ic a te that a m u c h larg er gap sho uld o cc u r in isolated alkad ienes. T h is is b o rn e o ut e x p e rim e n ta lly . Is o la te d a lk a d i­ enes g iv e abso rp tio n spectra s im ila r to those o f alkenes. T h e ir 1 max are at shorter w a v e ­ lengths, u su a lly b e lo w 2 0 0 n m . A s w e m e n tio n e d , 1 ,4 -p e n ta d ie n e has its 1 max at 1 7 8 n m . C o n ju g a te d alkatrie n e s absorb a t lo n g e r w avelen g th s than co n ju g a te d alkad ienes, and this too can b e accounted fo r in m o le c u la r o rb ita l calcu latio n s. T h e e n ergy g ap b etw e e n the H O M O and the L U M O o f an a lk a trie n e is even s m a lle r than th a t o f an alk ad ie n e . •

In g en eral, the g reater the num ber o f conjugated m ultiple bonds in a molecule, the

longer w ill be its

1

max.

THE CHEMISTRY OF . . . T h e P h o t o c h e m i s t r y o f V is io n

The chemical changes that occur when light impinges on the retina of the eye involve several of the phenomena that we have studied. Central to an understanding of the visual process at the molecular level are two phenomena in par­ ticular: the absorption of light by conjugated polyenes and the interconversion of cis-trans isomers. The conjugated polyene, derived from a compound called retinal, is a part of a molecule called rhodopsin. When rhodopsin absorbs a photon of light, the 11-c/sretinal chromophore isomerizes to the all-trans form, causing

the cyclohexene ring of the chromophore to swing into a dif­ ferent orientation. The first photo-product is an intermedi­ ate called bathorhodopsin, which through a series of steps becomes metarhodopsin II, shown below. It is believed that repositioning of the retinal cyclohexene ring, through the 11cis to all-trans isomerization, causes further conformational changes in the protein that ultimately initiate a cascade of enzymatic reactions and transmission of a neural signal to the brain.

(1 1 -c/s)

hv (a photon of light)

610

Chapter 13

Conjugated Unsaturated Systems

Po lyenes w ith e ig h t o r m o re co n ju g a te d d o u b le bonds absorb lig h t in the v is ib le reg ion o f the spectrum . F o r e x a m p le , b -c a ro te n e , a precursor o f v ita m in A and a c o m p o u n d that im p a rts its orang e c o lo r to carrots, has

1 1

co n ju g a te d d o u b le bonds; b -c a ro te n e has an

abso rp tio n m a x im u m a t 4 9 7 n m , w e ll in to the v is ib le re g io n . L ig h t o f 4 9 7 n m has a b lu e g reen colo r; this is the lig h t that is absorbed b y b -c a ro te n e . W e p erc e iv e the c o m p le m e n ­ ta ry c o lo r o f b lu e green, w h ic h is re d orange.

P -C a ro te n e Lycopene, a com pound p artly responsible fo r the red colo r o f tomatoes, also has 11 conju­ gated double bonds. Lycopene has an absorption m a x im u m at 5 0 5 n m w here it absorbs intensely. (A p p ro x im a te ly 0 .0 2 g o f lycopene can be isolated fro m 1 kg o f fresh, rip e tomatoes.)

L yc o p e n e T a b le 13 .3 gives the values o f 1 max fo r a n u m b e r o f unsaturated com pounds.

^

Long-W avelength A bsorption M axim a o f U nsaturated Hydrocarbons

Com pound

S tru c tu re

(M

f max 1 c m “ 1)

171

15,530

trans-3-H exene

184

10,000

Cyclohexene

182

7,600

1-O ctene

177

12,600

1-Octyne

185

2,000

1.3-B utadiene

217

21,000

c/s-1,3-Pentadiene

223

22,600

trans-1,3-P entadiene

223.5

23,000

But-1-en-3-yne

228

7,800

1.4-P entadiene

178

17,000

1.3-C yclopentadiene

239

3,400

1.3-C yclohexadiene

256

i,000

trans-1,3,5-H exatriene

274

5 0 ,0 0 0

Ethene

c h 2= c h 2

max (n m )

8-W

13.9 Ultraviolet-Visible Spectroscopy

611

C o m p o u n d s w ith c a rb o n -o x y g e n d o u b le bonds also absorb lig h t in the U V reg ion . A c e to n e , fo r e x a m p le , has a b ro ad abso rp tio n p ea k a t 2 8 0 n m th a t corresponds to the e x c i­ tatio n o f an e le ctro n fro m o ne o f the unshared p airs (a n o n b o n d in g o r “ n ” ele ctro n ) to the p * o rb ita l o f the c a rb o n -o x y g e n d o u b le bond: A c e to n e

•O '

n

G ro u n d sta te

:O ■ p

A max =

280 nm

E max =

15

n * E x c ited sta te

C o m p o u n d s in w h ic h the c a rb o n -o x y g e n d o u b le b o n d is con jug ated w ith a c a rb o n -c a rb o n d o u b le b o n d h ave abso rp tio n m a x im a corresp on din g to n ------- : p * e xcitatio n s and p ------- » p * excitatio n s. T h e n -------: p * abso rp tio n m a x im a o ccu r a t lo n g e r w a v ele n g th s b u t are m u c h w e a k e r (i.e ., h ave s m a lle r m o la r a b s o rp tiv ity (e ) values):

O

n­

n-

*n » sTT * *n

*m a x =

3 2 4

nm

^m a x =

2 4

*m a x =

2 1 9

^

^m ax =

3 6 0 0

13.9D Analytical Uses of UV-Vis Spectroscopy U V - V i s spectroscopy can b e used in the structure e lu c id a tio n o f o rg an ic m o le c u le s to in d i­ cate w h e th e r c o n ju g a tio n is present in a g iv e n sam ple. A lth o u g h c o n ju g a tio n in a m o le c u le m a y b e in d ic a te d b y d ata fro m IR , N M R , o r mass spectrom etry, U V - V i s analysis can p ro ­ v id e co rro b o ra tin g in fo rm a tio n . A m o re w id e s p re a d use o f U V - V i s spectroscopy, h o w e v e r, has to d o w ith d e te rm in ­ in g the c o n c e n tra tio n o f an u n k n o w n sam p le. A s m e n tio n e d in S e c tio n 1 3 .9 B , the r e la ­ tio n s h ip A = e C l in d ic a te s th a t th e a m o u n t o f a b s o rp tio n b y a s am p le a t a c e rta in w a v e le n g th is d ep e n d e n t on its c o n c e n tra tio n . T h is re la tio n s h ip is u s u a lly lin e a r o v e r a ra n g e o f c o n c e n tra tio n s s u ita b le fo r a n a ly s is . T o d e te rm in e the u n k n o w n c o n c e n tra tio n o f a sam p le, a g ra p h o f abso rb ance versus c o n c e n tra tio n is m a d e fo r a set o f standards o f k n o w n con centration s. T h e w a v e le n g th used fo r analysis is u s u a lly the 1 max o f the sam ­ p le . T h e c o n c e n tra tio n o f the s am p le is o b ta in e d b y m e a s u rin g its abso rb ance a n d d e te r­ m in in g the corresp on din g v a lu e o f co n c e n tra tio n fro m the g rap h o f k n o w n con centrations. Q u a n tita tiv e a n alysis u sing U V - V i s spectro sco py is ro u tin e ly used in b io c h e m ic a l stud­ ies to m e a s u re the rates o f e n z y m a tic re a c tio n s . T h e c o n c e n tra tio n o f a species in v o lv e d in the re a c tio n (as re la te d to its U V - V i s ab s o rb a n c e ) is p lo tte d versus tim e to d e te rm in e the ra te o f re a c tio n . U V - V i s spectro sco py is also used in e n v iro n m e n ta l c h e m is try to d e te rm in e the c o n c e n tra tio n o f v a rio u s m e ta l ions (s o m e tim e s in v o lv in g a b s o rp tio n spec­ tra fo r o rg a n ic c o m p le x e s w ith the m e ta l) and as a d e te c tio n m e th o d in h ig h -p e rfo rm a n c e liq u id c h ro m a to g ra p h y ( H P L C ) .

1

S o lv e d P r o b le m

1 3 .7

T w o is o m e ric com po un ds, A a n d B , h av e the m o le c u la r fo rm u la C 7 H 12. C o m p o u n d A si low s no abso rp tio n in the U V - v i s i b l e reg io n . T h e 13C N M R spectrum o f A shows o n ly three signals. C o m p o u n d B shows a U V - v i s i b l e p ea k in the re g io n o f 1 8 0 n m , its 13C N M R spectru m shows fo u r signals, and its D E P T 13C NJ M R d ata show th a t none o f its carb on atom s is a m e th y l g roup. O n c a ta ly tic h y d ro g e n a tio n w ith excess h yd ro g e n, B is con verted to h ep tane. Propose structures fo r A and B .

(Co ntinues on the next page)

1

612

Chapter 13

Conjugated Unsaturated Systems

STRATEGY AND ANSWER O n the basis o f th e ir m o le c u la r fo rm u la s , b o th co m p o u n d s h av e an in d e x o f h y d ro ­ gen d e fic ie n c y (S e c tio n 4 .1 7 ) e q u a l to 2 . T h e re fo re on this basis a lo n e , each c o u ld c o n ta in tw o d o u b le bonds, one rin g a n d o n e d o u b le b o n d , tw o rin g s , o r a trip le b o n d . C o n s id e r A firs t. T h e fa c t th a t A does n o t abso rb in the U V - v i s i b l e re g io n suggests th a t it does n o t h av e a n y d o u b le bonds; th e re fo re , it m u s t c o n ta in tw o rin g s . A c o m p o u n d w ith tw o rin g s th a t w o u ld g iv e o n ly th re e signals in its 13C s p ectru m is b ic y c lo [2 .2 .1 ]h e p ta n e (because it has o n ly th ree d is tin c t typ es o f c a rb o n a to m s ). N o w c on sider B . T h e fa c t th a t B is c on verted to h ep tan e on c a ta ly tic h y d ro g e n a tio n suggests th a t B is a hep tad ie n e o r a h ep tyn e w ith an unb ranch ed ch ain . U V - v i s i b l e abso rp tio n in the 1 8 0 -n m re g io n suggests th a t B does n o t c o n tain con jug ated p bonds. G iv e n th a t the D E P T 13C d ata fo r B shows the absence o f an y m e th y l groups, and o n ly fo u r 13C signals in to ta l, B m u s t b e 1 ,6 -h e xa d ie n e . c

Kb

a ^ ^ -~ -^ a

a\ ^

a ^ D ^

b

A B icyclo[2.2.1]heptane (three 13C signals)

R e v ie w P ro b le m 1 3 .7

c

c d

b

B 1,6-Hexadiene (four 13C sig n als, n o n e co m es from a methyl group)

T w o com pounds, A and B , h av e the sam e m o le c u la r fo rm u la , C 6 H 8. B o th A a n d B react w ith tw o m o la r equ ivalen ts o f h yd ro g e n in the presence o f p la tin u m to y ie ld cyclo h exan e. C om pound A

shows three signals in its b ro ad b a n d d eco u p led 13C N M R

spectrum .

C o m p o u n d B shows o n ly tw o 13C N M R signals. C o m p o u n d A shows an abso rp tio n m a x ­ im u m a t 2 5 6 n m , w hereas B shows n o abso rp tio n m a x im u m a t w a v ele n g th s lo n g e r than 2 0 0 n m . W h a t are the structures o f A a n d B?

C6H8 A 13C N M R: 3 signals UV:

Xmax ~ 2 56 nm C6H8 B

13C N M R: 2 sig n a ls UV:

R e v ie w P ro b le m 13 .8

Xmax < 200 nm

Propose structures fo r D , E , and F .

C5H6 D

IR: no p e a k n e a r 3 3 0 0 cm

1

UV: Amax ~ 2 30 nm

H2, Pt H2, Pt

C5H6 ---------E IR: ~ 3 300 c m - 1 , sh a rp UV: Amax ~ 2 30 nm

H2, Pt

C5H6 F IR: ~ 3 30 0 cm 1, sh a rp UV: Amax < 2 0 0 nm

13.10 Electrophilic A tta c k on C onjugated Dienes: 1,4 A d d ition N o t o n ly are c o n ju g a te d dienes s o m ew hat m o re stable than n o n co n ju g a te d dienes, th e y also d is p la y special b e h a v io r w h e n they re a c t w ith e le c tro p h ilic reagents. •

C o n ju g a te d dienes und ergo b oth 1 ,2 and 1 ,4 a d d itio n throu gh an a lly lic in te rm e d i­ ate th a t is c o m m o n to both.

13.10 Electrophilic A ttack on Conjugated Dienes: 1,4 Addition

F o r e x a m p le , 1 ,3 -b u ta d ie n e reacts w ith o ne m o la r e q u iv a le n t o f h yd ro g e n c h lo rid e to p ro ­ duce tw o products, 3 -c h lo ro -1 -b u te n e a n d 1-c h lo ro -2 -b u te n e :

1,2 A d d itio n

1 ,4 A d d itio n

Cl

HCl 25°C 1 ,3 -B u ta d ie n e

3 -C h lo ro -1 -b u te n e (78% )

1 -C h lo ro -2 -b u te n e (22% , p rim a rily E)

I f o n ly the firs t p ro d u ct (3 -c h lo ro -1 -b u te n e ) w e re fo rm e d , w e w o u ld n o t b e p a rtic u la rly sur­ prised. W e w o u ld con clud e th a t h yd ro g e n c h lo rid e h ad add ed to o ne d o u b le b o n d o f 1 .3 -b u ta d ie n e in the usual w ay. I t is the second p ro du ct, 1 -c h lo ro -2 -b u te n e , th a t is in itia lly surprising. Its d o u b le b o n d is b etw een the central atom s, and the elem ents o f h yd ro g e n c h lo ­ rid e h ave add ed to the C1 and C 4 atom s. T o understand h o w b oth

1 ,2 - and

1 ,4 -a d d itio n products re s u lt fro m

re a c tio n o f

1 .3 -b u ta d ie n e w ith H C L , con sider the fo llo w in g m ech anism .

Step 1

:C l 9 h

'■C l A n a lly lic cation e q u iv a le n t to

S+

5+

Step 2 (a)

(a)

5+

5+

Cl 1,2 A d d itio n

(b) :C l : (b) 'C l

1 ,4 A d d itio n

In step 1 a p ro to n adds to o ne o f the te rm in a l carbon atom s o f 1 ,3 -b u ta d ie n e to fo rm , as usual, the m o re stable carbocation, in this case a reso nan ce-stab ilized a lly lic cation. A d d itio n to o ne o f the in n e r carb on atom s w o u ld h ave p ro d u ce d a m u c h less stable p rim a ry cation, o ne th a t c o u ld n o t b e s ta b ilize d b y resonance: H Cl _

A d d itio n of th e e le c tro p h ile in this fa s h io n d o e s n ot lead to an a lly lic (re s o n a n c e -s ta b ilize d ) c a rb o c a tio n .

^ -y C i

In step 2 a c h lo rid e io n fo rm s a b o n d to o ne o f the carbon atom s o f the a lly lic catio n th a t bears a p a rtia l p o s itiv e charge. R e a c tio n a t one carb on a to m results in the 1 ,2 -a d d itio n p ro du ct; re a c tio n at the o th e r gives the 1 ,4 -a d d itio n product. N o te th a t the designations 1 ,2 a n d 1 ,4 o n ly c o in c id e n ta lly re la te to the IU P A C n u m b e r­ ing o f carbon atom s in this e x a m p le . •

C h em ists ty p ic a lly use 1 ,2 a n d 1 ,4 to re fe r to m o d es o f a d d itio n to an y con jug ated d ie n e system , regardless o f w h e re the co n ju g a te d d o u b le bonds are in the o v e ra ll m o le c u le .

Thu s, add itio n reactions o f 2 ,4 -h e xa d ie n e w o u ld s till in v o lv e references to 1 ,2 and 1,4 m odes o f a d d itio n .

8-W

613

614

Review Problem 13.9

Chapter 13

Conjugated Unsaturated Systems

Predict the products o f the following reactions.

1 ,3 -B u ta d ie n e shows 1 ,4 -a d d itio n reactions w ith e le c tro p h ilic reagents o th er than h y d ro ­ gen c h lo rid e . T w o exam p les are show n here, the a d d itio n o f h yd ro g e n b ro m id e (in the absence o f p ero x id e s) and the a d d itio n o f b ro m ine:

Br HBr 40°C 2 0

%

80%

Br Br,

Br

Br

Br

-15°C 54%

46%

R eactio ns o f this typ e are q u ite g en e ra l w ith o th er co n ju g a te d dienes. C o n ju g a te d trienes o fte n show 1,6 a d d itio n . A n e x a m p le is the 1 ,6 a d d itio n o f b ro m in e to 1,3,5-c y c lo o cta trie n e :

Br2 CHCL Br

Br

> 68%

13.10A Kinetic Control versus Thermodynamic Control of a Chemical Reaction T h e a d d itio n o f h yd ro g e n b ro m id e to 1 ,3 -b u ta d ie n e a llo w s the illu s tra tio n o f a no ther im p o r­ tan t aspect o f re a c tiv ity — the w a y tem p e ra tu re affects p ro d u ct d is trib u tio n in a re a c tio n that can tak e m u ltip le paths. In general: •

T h e fa v o re d products in a re a c tio n a t low er temperature are those fo rm e d b y the p a th w a y h av in g the sm allest energy o f a c tiv a tio n b arrier. In this case the re a c tio n is said to b e u n d er k in e tic ( o r r a t e ) c o n tro l, an d the p re d o m in a n t p roducts are called the k in e t ic p ro d u c ts .

•

T h e fa v o re d products at higher temperature in a reversible re a c tio n are those that are m o s t stable. In this case the re a c tio n is said to b e u n d er th e r m o d y n a m ic (o r e q u ilib r iu m ) c o n tro l, a n d the p re d o m in a n t p roducts are c a lle d the t h e r m o d y ­ n a m ic ( o r e q u ilib r iu m ) p ro d u c ts .

L e t ’s c on sider specific re a c tio n con d itio n s fo r the io n ic a d d itio n o f h yd ro g e n b ro m id e to 1 ,3 -b u ta d ie n e. C a s e 1. W h e n 1 ,3 -b u ta d ie n e and h yd ro g e n b ro m id e re a c t a t lo w tem p e ra tu re ( — 8 0 ° C ), the m a jo r p ro d u ct is fo rm e d b y 1 ,2 a d d itio n . W e o b ta in 8 0 % o f the 1 ,2 p ro d ­ u ct an d 2 0 % o f the 1 ,4 p ro du ct. C a s e 2 . W h e n 1 ,3 -b u ta d ie n e and h yd ro g e n b ro m id e re a c t at h ig h tem p e ra tu re (4 0 ° C ), the m a jo r p ro d u ct is fo rm e d b y 1 ,4 a d d itio n . W e o b ta in ab o u t 2 0 % o f the 1,2 p ro d u ct and ab o u t 8 0 % o f the 1 ,4 product.

13.10 Electrophilic A ttack on Conjugated Dienes: 1,4 Addition

C a s e 3 . W h e n the p ro d u ct m ix tu re fro m the lo w tem p e ra tu re re a c tio n is w a rm e d to the h ig h e r tem perature, the pro d u ct distrib ution becom es the same as w h e n the reac­ tio n w as c a rrie d o ut a t h ig h tem p eratu re, th a t is, the 1 ,4 p ro d u ct p redo m in ates. W e s u m m a rize these scenarios here: Br

+

H Br

2 0

%

80%

F u rth e rm o re , w h e n a p u re sam ple o f 3 -b ro m o -1 -b u te n e (th e p re d o m in a n t p ro d u ct at lo w tem p e ra tu re ) is subjected to the h ig h tem p e ra tu re re a c tio n con dition s, an e q u ilib riu m m ix ­ ture results in w h ic h the 1 ,4 a d d itio n p ro d u ct p redo m in ates. Br

1 ,2 -A d d itio n p ro d u ct

1 ,4 -A d d itio n p ro d u ct

B ecause this e q u ilib riu m favors the 1 ,4 -a d d itio n product, that product must be more stable. T h e reactions o f h yd ro g e n b ro m id e w ith 1 ,3 -b u ta d ie n e serve as a s trik in g illu s tra tio n o f the w a y th a t the o u tc o m e o f a c h e m ic a l re a c tio n can b e d ete rm in e d , in o ne instance, b y r e l­ a tive rates o f com p e tin g reactions and, in another, b y the re la tiv e stab ilities o f the fin a l p ro d ­ ucts. A t the lo w e r tem p eratu re, the re la tiv e am ounts o f the products o f the a d d itio n are d e te rm in e d b y the re la tiv e rates a t w h ic h the tw o add itio n s occur;

1 , 2

a d d itio n occurs faster

so the 1 ,2 -a d d itio n p ro d u c t is the m a jo r p ro du ct. A t the h ig h e r tem p eratu re, the re la tive am ounts o f the products are d e te rm in e d b y the p o s itio n o f an e q u ilib riu m . T h e 1 ,4 -a d d i­ tio n p ro d u ct is the m o re stable, so it is the m a jo r product. T h is b e h a v io r o f 1 ,3 -b u ta d ie n e and h yd ro g e n b ro m id e can b e m o re fu lly u nderstood i f w e e x a m in e the d ia g ra m show n in F ig . 1 3 .1 2 .

Figure 13.12 A schematic free-energy versus reaction coordinate diagram for the 1 , 2 and 1,4 addition of HBr to 1,3butadiene. An allylic carbocation is common to both pathways. :nergy barrier for attack of ide ion on the allylic cation rm the 1 , 2 -addition product s than that to form the ddition product. The ddition product is kinetically ed. The 1,4-addition uct is more stable, and is the thermodynamically ed product.

1,2 Addition

1,4 Addition H

Reaction coordinate

61 6

Chapter 13

•

Conjugated Unsaturated Systems

T h e step th a t determ ines the o v e ra ll o u tc o m e o f this re a c tio n is the step in w h ic h the h y b rid a lly lic ca tio n com bin es w ith a b ro m id e ion. Br BrT h e re a c tio n o f th e b ro m id e a nio n w ith th e a lly lic ca tio n d e te rm in e s th e re g io s e le c tiv ity o f th e re a c tio n .

1 ,2 P rod uct

HBr Br-

Br 1 ,4 P rod uct W e see in F ig . 1 3 .1 2 th a t the fre e e n erg y o f a c tiv a tio n le a d in g to the 1 ,2 -a d d itio n pro du ct is less than the fre e e n ergy o f a c tiv a tio n lea d in g to the 1 ,4 -a d d itio n p ro du ct, even though the 1 ,4 p ro d u ct is m o re stable. •

A t lo w t e m p e r a tu r e , the fra c tio n o f c o llis io n s capab le o f s urm o un tin g the h ig h e r e n ergy b a rrie r lea d in g to fo rm a tio n o f the 1 ,4 p ro d u ct is s m a lle r than the fra c tio n th a t can cross the b a rrie r le a d in g to the

•

1 , 2

product.

A t lo w tem p eratu re, fo rm a tio n o f the 1 ,2 and 1 ,4 products is essen tia lly irre ­

versible because there is n o t enough e n erg y fo r e ith e r p ro d u ct to cross b a c k o ver the b a rrie r to re fo rm the a lly lic catio n . T h u s, the 1 ,2 p ro d u ct p red o m in ates at lo w e r tem p e ra tu re because it is fo rm e d faster and it is n o t fo rm e d reversib ly. I t is the k in e tic p ro d u c t o f this reaction . •

A t h ig h e r te m p e r a tu r e , c o llis io n s b e tw e e n the in te rm e d ia te ions are s u fficie n tly energetic to a llo w ra p id fo rm a tio n o f both the 1 ,2 and 1 ,4 products. But, there is also s u fficie n t e n erg y fo r b o th products to re v e rt to the a lly lic carbocation.

•

Because the 1 ,2 p ro d u ct has a s m a lle r e n ergy b a rrie r fo r con version b a c k to the a lly lic ca tio n than does the 1 ,4 p ro du ct, m o re o f the 1 ,2 p ro d u ct reverts to the a lly lic catio n than does the 1 ,4 p ro du ct. B u t since b oth the 1 ,4 and the 1 ,2 products re a d ily fo r m fro m the a lly lic ca tio n a t h ig h tem p eratu re, e v e n tu a lly this e q u ilib ­ riu m leads to a p repo nd eran ce o f the 1 ,4 p ro d u ct because it is m o re stable. T h e 1,4 p ro d u ct is the th e rm o d y n a m ic o r e q u ilib r iu m p ro d u c t o f this reaction .

B e fo re w e leave this subject, one fin a l p o in t should be m ad e. T h is e x a m p le c le a rly d em on ­ strates that predictions o f re la tive reaction rates m ad e on the basis o f pro du ct stabilities alone can be w ro ng . T h is is n ot alw ays the case, how ever. F o r m a n y reactions in w h ic h a com m o n interm ediate leads to tw o o r m o re products, the m o s t stable pro du ct is fo rm e d fastest.

R e v ie w P ro b le m 1 3 .1 0

( a ) Suggest a structural e x p la n a tio n fo r the fa c t th a t the 1 ,2 -a d d itio n re a c tio n o f 1 ,3 -b u ta ­ d ie n e and h yd ro g e n b ro m id e occurs fas te r than 1 ,4 add itio n? [H int: C o n s id e r the re la tiv e c on tribu tio ns th a t the tw o fo rm s

and

m a k e to the

resonance h y b rid o f the a lly lic catio n.] ( b ) H o w can y o u accou nt fo r the fa c t th a t the 1 ,4 -a d d itio n p ro d u ct is m o re stable?

13.11 The D ie ls -A ld e r Reaction: A 1,4-Cycloaddition Reaction o f Dienes In 1 9 2 8 tw o G e rm a n chem ists, O tto D ie ls and K u rt A ld e r, d ev e lo p e d a 1 ,4 -c y c lo a d d itio n ■ ;N :

re a c tio n o f dienes that has since com e to b ear th e ir nam es. T h e re a c tio n p ro v e d to be one o f such great v e rs a tility a n d synthetic u tility that D ie ls and A ld e r w e re a w a rd ed the N o b e l P riz e in C h e m is try in 1 9 5 0. A n e x a m p le o f the D ie ls - A ld e r re a c tio n is the re a c tio n th a t takes p la c e w h e n 1 ,3 -b u ta ­ d ie n e and m a le ic a n h y d rid e are h eated tog eth er at 1 0 0 °C . T h e p ro d u ct is o b ta in e d in q u a n ­ tita tiv e yield:

g - W

13.11 The Diels-Alder Reaction: A 1,4-Cycloaddition Reaction o f Dienes

O

O

O

O

benzene, 100°C O

O 1 ,3 -B u ta d ie n e (diene)

•

61 7

A dduct ( 1 0 0 %)

M aleic a n h y d rid e (d ie n o p h ile )

In g en e ra l term s, the D i e ls - A l d e r re a c tio n is one b e tw e e n a co n ju g a te d d ie n e (a 4 p -e le c tr o n system ) and a c o m p o u n d c o n tain in g a d o u b le b o n d (a 2 p -e le c tr o n sys­ te m ) c a lle d a d ie n o p h ile (d ie n e + p hilein, G re e k : to lo v e ). T h e p ro d u ct o f a D ie ls - A ld e r re a c tio n is o fte n c a lle d an a d d u c t .

In the D ie ls - A ld e r re a c tio n , tw o n e w s bonds are fo rm e d at the expense o f tw o p bonds o f the d ie n e and d ie n o p h ile . T h e add uct contains a n e w s ix -m e m b e re d rin g w ith a double bon d. S in c e s bonds are u su a lly stronger than p bonds, fo rm a tio n o f the add uct is u su ally fav o re d energ e tica lly , but most D ie ls -A ld e r reactions are reversible. W e can accou nt fo r a ll o f the b o n d changes in a D ie ls - A ld e r re a c tio n b y using c urved arrow s in the fo llo w in g w ay:

D ie n e D ie n o p h ile

Adduct

T h e sim p le st e x a m p le o f a D ie ls - A ld e r re a c tio n is the o ne th a t takes p la c e b e tw e e n 1 ,3 b u ta d ie n e and ethene. T h is re a c tio n , h ow ever, takes p la c e m u c h m o re s lo w ly than the re a c ­ tio n o f b u ta d ie n e w ith m a le ic a n h y d rid e and also m u s t be c a rrie d o u t u n d er pressure: -

h r!

sealed ' tube, 200°C 2 0

%

A n o th e r e x a m p le is the p re p a ra tio n o f an in te rm e d ia te in the synthesis o f the a n tican cer drug T a x o l (p a c lita x e l) b y K . C . N ic o la o u (S c rip p s R esearch In s titu te and the U n iv e rs ity o f C a lifo rn ia , San D ie g o ):

AcO AcO U sed in a s y n th e s is o f Taxol

130 °C

8 5% Cl

O

Ac = (ac e ty l)

O Bz =

J U P h "^

(b e n zo y l)

Taxol (b lu e a to m s d e riv e d fro m th e D ie ls -A ld e r a d d u c t a b o v e )

<

H e lp fu l H i n t

The Diels-Alder reaction is a very useful synthetic tool fo r preparing cyclohexene rings.

618

Chapter 13

Conjugated Unsaturated Systems

13.11A Factors Favoring the Diels-Alder Reaction A ld e r o rig in a lly stated th a t the D ie ls - A ld e r re a c tio n is fa v o re d b y the presence o f ele ctro n w ith d ra w in g groups in the d ie n o p h ile and b y e le ctro n -re le a s in g groups in the dien e. M a le ic a n h y d rid e , a v e ry p o te n t d ie n o p h ile , has tw o e le c tro n -w ith d ra w in g c a rb o n y l groups on c a r­ b o n atom s adja c e n t to the d o u b le bond. T h e h e lp fu l e ffe c t o f e le ctro n -re le a s in g groups in the d ie n e can also b e dem onstrated; 2 ,3 -d im e th y l-1 ,3 -b u ta d ie n e , fo r exam p le, is n ea rly five tim es as re active in D ie ls -A ld e r reac­ tions as is 1 ,3 -b u ta d ie n e. T h e m e th y l groups in d u c tiv e ly release electro n density, ju s t as a lk y l groups do w h e n s ta b ilizin g a carb o c a tio n (th o u g h n o carbocations are in v o lv e d here ). W h e n 2 ,3 -d im e th y l-1 ,3 -b u ta d ie n e reacts w ith p ro p e n a l (a c ro le in ) at o n ly 3 0 ° C , the adduct is o b ta in e d in q u a n tita tiv e yield: O

O H

'H

T h e m e th y l g ro u p s d o n a te e le c tro n d en s ity .

30°C

2 ,3 -D im e th y l-1 ,3 b u ta d ie n e

P ropenal

1 0 0

%

R esearch (b y C . K . B rad sh er o f D u k e U n iv e r s ity ) has show n that the lo catio n s o f e le c ­ tro n -w ith d ra w in g a n d electro n -re le a s in g groups in the d ie n o p h ile and d ien e can b e reversed w ith o u t re d u c in g the y ie ld s o f the adducts. D ie n e s w ith e le c tro n -w ith d ra w in g groups have b een fo u n d to re a c t re a d ily w ith d ie n o p h ile s c o n tain in g e le ctro n -re le a s in g groups. B esides the use o f dienes and d ie n o p h ile s th a t h ave c o m p le m e n ta ry e lectro n -releasin g a nd e le ctro n -d o n a tin g p ro perties, o th er factors fo u n d to enhance the ra te o f D ie ls - A ld e r reactions in c lu d e h ig h tem p e ra tu re and h ig h pressure. A n o th e r w id e ly used m e th o d is the use o f L e w is a c id catalysts. T h e fo llo w in g re a c tio n is o ne o f m a n y exam p le s w h ere D ie ls - A ld e r adducts fo r m re a d ily at a m b ie n t tem p e ra tu re in the presence o f a L e w is acid c atalyst. ( In S ection 1 3 .1 1 C w e see h o w L e w is acids can b e used w ith c h ira l ligan ds to in d u c e a s y m m e try in the re a c tio n pro du cts.) O

O ,O H

_

^

^

,O H

AlCl, Et2O 25°C OMe

O

OH

OMe

O

OH

80 %

13.11B Stereochemistry of the Diels-Alder Reaction N o w let us consider some stereochemical aspects o f the D ie ls -A ld e r reaction. T h e fo llo w in g fac­ tors are am ong the reasons w h y D ie ls -A ld e r reactions are so extraordinarily useful in synthesis. 1. T h e D ie l s - A ld e r re a c tio n is s te re o s p e c ific : T h e re a c tio n is a s y n a d d itio n , a n d th e c o n fig u r a tio n o f th e d ie n o p h ile is re tain e d in th e p r o d u c t. T w o exam p le s that illu s tra te this aspect o f the re a c tio n are show n here: O

O OMe OMe O

D im ethyl m a le a te (a c /s -d ie n o p h ile )

D im ethyl c y c lo h e x -4 -e n e -c /s 1 , 2 -d ic a rb o x y la te

1S.11 The Diels-Alder Reaction: A 1,4-Cycloaddition Reaction o f Dienes

O

8-W

619

O

D im e th y l fu m a ra te (a fean s-d ieno ph ile)

D im e th y l c y c lo h ex -4 -e n e -fe a n s 1 , 2 -d ic a rb o x y la te

In the firs t e x a m p le , a d ie n o p h ile w ith cis ester groups reacts w ith 1 ,3 -b u ta d ie n e to g iv e an add uct w ith cis ester groups. In the second e x a m p le ju s t the reverse is true. A trans -d ie n o p h ile gives a trans adduct. 2 . T h e d ie n e , o f necessity, re a c ts in th e s-cis r a t h e r t h a n in th e s -tra n s c o n fo rm a tio n :

s -c is C o n fo rm a tio n

s -tra n s C o n fo rm a tio n

R e a c tio n in the s-trans c o n fo rm a tio n w o u ld , i f it o ccu rred, p ro du ce a s ix -m e m b ere d rin g w ith a h ig h ly strained trans d o u b le bon d. T h is course o f the D ie ls - A ld e r re a c ­ tio n has n ev e r been observed.

O O

R

R

Use handheld molecular models to investigate the strained nature of hypothetical trans-cyclohexene. Hi g h ly s tra in e d

C y c lic dienes in w h ic h the d o u b le bonds are h e ld in the s-cis c o n fo rm a tio n are usu­ a lly h ig h ly re a c tive in the D ie ls - A ld e r re a c tio n . C y c lo p e n ta d ie n e , fo r e x a m p le , reacts w ith m a le ic a n h y d rid e a t ro o m tem p e ra tu re to g iv e the fo llo w in g add uct in q u a n tita ­ tiv e y ie ld :

O

O

O

O

25oC

O

O

C y c lo p e n ta d ie n e is so re a c tiv e th a t on standing at ro o m tem p e ra tu re it s lo w ly u n d er­ goes a D ie ls - A ld e r re a c tio n w ith itself:

D ic y c lo p e n ta d ie n e

T h e re a c tio n is re v e rsib le , how ever. W h e n d ic y c lo p e n ta d ie n e is d is tille d , it disso ci­ ates (is “ c racked ” ) in to tw o m o la r e qu ivalen ts o f cyclo p en tad ien e. T h e reactions o f c y c lo p en tad ie n e illu s tra te a th ird stereoch em ical c h aracteristic o f the D ie ls - A ld e r reaction .

H e lp f u l H i n t

620

Chapter 13

Conjugated Unsaturated Systems

3 . T h e D ie l s - A ld e r re a c tio n o c c u rs p r i m a r i l y in a n e n d o r a t h e r t h a n a n e x o fa s h ­ io n w h e n th e re a c tio n is k in e t ic a lly c o n tr o lle d (see P ro b le m 1 3 .4 2 ). E n d o and exo

H e lp f u l H i n t

are term s used to designate the stereoch em istry o f b rid g e d rin gs such as b ic y -

In general, the exo substituent is always on the side anti to the longer bridge of a bicyclic structure (exo, outside; endo, inside). For example,

c lo [2 .2 .1 ]h e p ta n e . T h e p o in t o f re fe re n c e is the lon gest b rid g e . A gro up th a t is anti

exo

to the longest b rid g e (th e tw o -c a rb o n b rid g e ) is said to b e exo; i f it is on the same side, it is endo:

endo exo

exo endo

THE CHEMISTRY OF . . . M o l e c u l e s w i t h t h e N o b e l P r iz e in T h e i r S y n t h e t i c L i n e a g e

Many organic molecules from among the great targets for synthesis have the Diels-Alder reaction in their synthetic lin­ eage. As we have learned, from acyclic precursors the Diels-Alder reaction can form a six-membered ring, with as many as four new chirality centers created in a single stere-

ospecific step. It also produces a double bond that can be used to introduce other functionalities. The great utility of the Diels-Alder reaction earned Otto Diels and Kurt Alder the Nobel Prize in Chemistry in 1950 for developing the reaction that bears their names.

CH2OH C=O L oh

O OMe R e s e rp in e

C o rtis o n e

'CH„ Morphine, the synthesis of which involved the Diels-Alder reaction.

Molecules that have been synthesized using the Diels-Alder reaction (and the chemists who led the work) include morphine (above, and shown as a model), the hyp­ notic sedative used after many surgical procedures (M. Gates); reserpine (above), a clinically used antihypertensive agent (R. B. Woodward); cholesterol, precursor of all steroids in the body, and cortisone (also above), the anti­ inflammatory agent (both by R. B. Woodward); prostaglandins F2„ and E2 (Section 13.11C), members of a family of hormones that mediate blood pressure, smooth

muscle contraction, and inflammation (E. J. Corey); vitamin B12 (Section 7.16A), used in the production of blood and nerve cells (A. Eschenmoser and R. B. Woodward); and Taxol (chemical name paclitaxel, Section 13.11), a potent cancer chemotherapy agent (K. C. Nicolaou). This list alone is a veritable litany of monumental synthetic accomplish­ ments, yet there are many other molecules that have also succumbed to synthesis using the Diels-Alder reaction. It could be said that all of these molecules have a certain sense of "Nobel-ity" in their heritage.

13.11 The Diels-Alder Reaction: A 1,4-Cycloaddition Reaction o f Dienes

13.11C Molecular Orbital Considerations That Favor an Endo Transition State In th e D i e l s - A l d e r re a c tio n o f c y c lo p e n ta d ie n e w ith m a le ic a n h y d rid e th e m a jo r p ro d ­ u c t is th e o n e in w h ic h th e a n h y d rid e g ro u p ,

^ , has a ssum ed the endo

o

o^ ^

N s^0

JV 'r e ­ c o n fig u ra tio n . T h is fav o re d endo stereoch em istry seems to arise fro m fav o ra b le interactions b e tw e e n the p electrons o f the d ev e lo p in g d o u b le b o n d in the d ien e and the p electrons o f unsaturated groups o f the d ie n o p h ile . In F ig . 1 3 .1 3 w e can see th a t w h e n the tw o m o le ­ cules approach each o th er in the endo o rien tatio n , as show n, o rbitals in the L U M O o f m a le ic a n h y d rid e and the H O M O o f cy c lo p en tad ie n e can in te ra c t at the carbons w h e re the n e w s bonds w i l l fo rm (th e in te ra c tio n o f these o rb ita ls is in d ic a te d b y p u rp le in F ig . 1 3 .1 3 ft). W e can also see th a t this sam e o rie n ta tio n o f approach (e n d o ) has o ve rlap b e tw e e n the L U M O lobes a t the c a rb o n y l groups o f m a le ic a n h y d rid e and the H O M O lobes in c yclo p en tad ien e above th e m (th e inte ra c tio n o f these o rb itals is in d ic a te d b y green). T h is so-called secondary o rb ita l in te ra c tio n is also fav o ra b le, and it leads to a p re fe re n c e fo r endo approach o f the d ie n o p h ile , such th a t the unsaturated groups o f the d ie n o p h ile are tu c k e d in and u n d er the d ien e, ra th e r than o u t and a w a y in the exo o rien tatio n .

HOMO of cyclopentadiene

LUMO of maleic anhydride

(a)

Primary orbital interactions (where — bonding will occur)

H Secondary orbital interactions

Transition state

H Diels-Alder adduct

(b)

Figure 13.13 Diels-Alder reaction of cyclopentadiene and maleic anhydride. (a) When the highest occupied molecular orbital (HOMO) of the diene (cyclopentadiene) interacts with the lowest unoccupied molecular orbital (LUMO) of the dienophile (maleic anhydride), favorable secondary orbital interactions occur involving orbitals of the dienophile. (b) This interaction is indicated by the purple plane. Favorable overlap of secondary orbitals (indicated by the green plane) leads to a preference for the endo transition state shown.

T h e tra n s itio n state fo r the endo p ro d u ct is thus o f lo w e r e n ergy because o f the fa v o r­ a b le o rb ita l interaction s d escribed abo ve, and th e re fo re the endo fo rm is the k in e tic (an d m a jo r) p ro d u c t o f this D ie ls - A ld e r re a c tio n . T h e exo fo rm is the th e rm o d y n a m ic p ro d u ct

621

622

Chapter 13

Conjugated Unsaturated Systems

because steric interaction s are fe w e r in the exo add uct than in the end o ad d u ct (F ig . 1 3 .1 4 ). T h u s, the exo a dd uct is m o re stable o v e ra ll, b u t i t is n o t the m a jo r p ro d u ct because it is fo rm e d m o re slow ly. H e re w e s u m m a rize som e k e y points. •

T h e D ie ls - A ld e r re a c tio n is a stereospecific syn a d d i­ tio n . T h e c o n fig u ra tio n o f the d ie n o p h ile is re ta in e d in the product.

•

T h e D ie ls - A ld e r re a c tio n is stereoselective fo r endo a d d itio n w h e n the re a c tio n is u n d er k in e tic con tro l.

E v e n tho ug h the D ie ls - A ld e r re a c tio n results in fo rm a tio n o f p re d o m in a n tly o ne stereoisom eric fo r m

(en d o w ith

re te n tio n o f the o rig in a l d ie n o p h ile c o n fig u ra tio n ), the p ro d u ct is nevertheless fo rm e d as a ra c e m ic m ix tu re . T h e reason fo r this is th a t e ith e r face o f the d ie n e can in te ra c t w ith the d ie n o p h ile . W h e n the d ie n o p h ile bonds w ith one fac e o f the d ien e, the p ro d u ct is fo rm e d as o ne e nan tio m er, a n d w h e n the d ie n o p h ile bonds at the o th e r face o f the

Endo and exo product formation in the Diels-Alder reaction of cyclopentadiene and maleic anhydride. Figure 13.14

d ien e, the p ro d u ct is the o th er e nan tio m er. In the absence o f c h ira l influen ces, b o th faces o f the d ie n e are e q u a lly lik e ly to b e attacked.

Solved Problem 13.8 T h e d im e riz a tio n o f c y c lo p en tad ie n e occurs p rim a r ily th ro u g h an endo tra n s itio n state, as is ty p ic a l fo r D ie ls - A ld e r reactions.

(a) In the reactants, d ra w re d and b lu e shaded lobes fo r the o rb ita ls th a t h ave fav o ra b le secondary in te r­ (b) U s in g b o n d -lin e fo r­

actions in the d ien e and d ie n o p h ile , causing the p re fe re n c e fo r an endo tra n s itio n state. m u la s , d ra w

c u rv e d arrow s to show the flo w

th re e -d im e n s io n a l fo rm u la fo r the p ro du ct.

o f electrons th a t leads to p ro d u ct fo rm a tio n , a n d d ra w

a

(c) T h e re a c tio n produces a ra c e m ic m ix tu re . S h o w h o w the reactants

a lig n in three d im ensio ns to fo r m each enantiom er.

STRATEGY AND ANSWER (a) In the H O M O o f the d ien e, the re d and b lu e lobes u nd erneath the rin g h ave fa v o r­ a b le sam e-phase interaction s w ith the d ie n e ’ s L U M O . T h e in d ic a te d re d and b lu e lobes in this d ia g ra m are n o t the ones in v o lv e d in b o n d fo rm a tio n , h ow ever. T h e y are the ones in v o lv e d in secondary o rb ita l interaction s.

(b) T w o

p electrons o f the d ie n o p h ile an d a ll fo u r o f the p electrons o f the d ie n e in te ra c t throu gh a c y c lic tra n s itio n state to fo rm the D ie ls -A ld e r add uct show n.

(a)

(b )

Diene (showing HOMO) Diene Dienophile Transition state

Dienophile (showing LUMO)

0

13.11 The Diels-Alder Reaction: A 1,4-Cycloaddition Reaction o f Dienes

if

623

(c ) F o rm a tio n o f the enan tio m ers can b e show n in tw o w ays. C o m p a re the la b e le d fo rm u la s b e lo w w ith the f o l­ lo w in g in fo rm a tio n . ( i) E n a n tio m e r A is fo rm e d i f the d ie n o p h ile approaches fro m b e lo w the d ien e and w ith its C H 2 gro up to w a rd us ( i - a ) . E n a n tio m e r B is fo rm e d i f the d ie n o p h ile approaches w ith its C H 2 gro up to w a rd us and fro m abo ve ( i - b ) . ( ii) A lte rn a tiv e ly , e n a n tio m e r A fo rm s i f the cy c lo p en tad ie n e d ie n o p h ile align s abo ve the d ien e b u t w ith its C H 2 g ro u p p ro je c tin g b e h in d the p ag e ( i i - a ) . E n a n tio m e r B fo rm s i f the d ie n o p h ile align s b e lo w the d ie n e b u t w ith its C H 2 g ro u p p ro je c tin g b e h in d the p ag e ( i i- b ) . Enantiom ers

(c)

Corresponding top and bottom form ulas are equivalent ( ||) by flipping on a

Enantiom er A horizontal axis.

i-b

M irror plane

In doing this analysis it is interesting to note its b in a ry nature. C h an g in g one param eter at a tim e (e.g., top o r b ot­ to m approach but keeping the d ien op hile in the same C H 2 orientation, o r changing the C H 2 o rientation but keeping the top o r b otto m approach the same) gives the tw o enantiom ers. O n the other hand, changing both o f these param eters at the same tim e leads to ju s t one o f the enantiom ers. T h e situation is analogous to interchanging eith er one o r tw o groups at a c h ira lity center. O n e interchange produces an enantiom er. T w o interchanges returns the o rig in a l com pound.

R e v ie w P ro b le m 13.11

W h a t products w o u ld y o u expect fro m the fo llo w in g reactions?

O

O

O

(a)

O

(c)

(e)

O

O

O

O

H

"O M e (b) ,O M e

(d)

\\

+

O

O

W h ic h d ien e a n d d ie n o p h ile w o u ld y o u e m p lo y to synthesize the fo llo w in g com pounds? (b )

O

O

Review Problem 13.12

624

Chapter 13

R e v ie w P ro b le m 1 3 .1 3

Conjugated Unsaturated Systems

D ie ls - A ld e r reactions also take p la c e w ith trip le -b o n d e d (a c e ty le n ic ) d ie n o p h ile s . W h ic h d ie n e and w h ic h d ie n o p h ile w o u ld y o u use to p repare the fo llo w in g ?

/ ^ \ ^ C

O

2Me

C O 2Me R e v ie w P ro b le m 1 3 .1 4

1 ,3 -B u ta d ie n e and the d ie n o p h ile show n b e lo w w e re used b y A . E sch en m o ser in his syn­ thesis o f v ita m in B 1 2 w ith R . B . W o o d w a rd . D r a w the structure o f the e n a n tio m e ric D ie ls - A ld e r adducts that w o u ld fo r m in this re a c tio n and the tw o tra n s itio n states that lead to them .

O

c

•

Key Terms and Concepts T h e k e y term s a n d concepts th a t are h ig h lig h te d in b o ld , b lu e t e x t w ith in the chap ter are

PLUS

d efin e d in the glossary (a t the b a c k o f the b o o k ) and h ave h y p e rlin k e d d e fin itio n s in the acco m p a n y in g W iley P L U S course ( w w w .w ile y p lu s .c o m ).

Problems Note to Instructors: Many of the homework problems are available for assignment via Wiley PLUS, an online teaching and learning solution. C O N JU G A TED SYSTEMS 13.15

P ro v id e the reagents n eed ed to synthesize 1 ,3 -b u ta d ie n e starting fro m ( a ) 1 ,4 -D ib ro m o b u ta n e (b )

,OH HO"

(d )

(f)

(e)

(c)

13.16

(g)

Cl

W h a t p ro d u ct w o u ld yo u expect fro m the fo llo w in g reaction? 2 NaOEt EtOH, heat Cl

13.17

Cl

W h a t products w o u ld y o u e x p e c t fro m the re a c tio n o f 1 m o l o f 1 ,3 -b u ta d ie n e and each o f the fo llo w in g reagents? ( I f n o re a c tio n w o u ld occur, y o u should in d ic a te that as w e ll.) (a ) 1 m o l o f C l2

(d ) 2 m o l o f H 2 , Ni

( f ) H o t K M n 0 4 (excess)

( b ) 2 m o l o f C l2

(e ) 1 m o l o f C l 2 in H 20

(g ) H 2 0 , cat. H 2 S 0

(c ) 2 m o l o f B r 2

4

8-W

Problems

13.18

625

P ro v id e the reagents necessary to tra n s fo rm 2 ,3 -d im e th y l-1 ,3 -b u ta d ie n e in to each o f the fo llo w in g com pounds. (a )

|

(b )

O

(c )

|

(d ) Br

O 13.19

'B r

Provide the reagents necessary fo r each o f the fo llo w in g transformations. In some cases several steps m a y be necessary. ( a ) 1 -B u te n e ------- : 1,3 -b u ta d ie n e

Br

Br

( b ) 1 -P e n te n e ------- : 1 ,3 -p e n ta d ie n e

OH ------:

(c )

^

(e )

------- »

Br

C o n ju g a te d dienes re a c t w ith ra d ic als b y b oth 1 ,2 and 1 ,4 a d d itio n . W r ite a d e ta ile d m e c h a n is m to accou nt fo r this fa c t using the p e ro x id e -p ro m o te d a d d itio n o f o ne m o la r e q u iv a le n t o f

13.21

\

(f)

7

(d )

13.20

Bk

HBr to 1 ,3 -b u ta d ie n e as an illu s tra tio n .

U V - V i s , IR , N M R , and mass spectro m etry are spectroscopic tools w e use to o b ta in structural in fo rm a tio n abo ut c om pounds. F o r each p a ir o f com pounds b elo w , describe at least o ne aspect fro m each o f tw o spectroscopic m e th ­ ods ( U V - V i s , I R , N M R , o r mass s p ectro m etry) th a t w o u ld d isting uish o ne c o m p o u n d in a p a ir fro m the other. ( a ) 1 ,3 -B u ta d ie n e and 1 -b u tyn e

Br

( b ) 1 ,3 -B u ta d ie n e and butane

(e ) Bi^ 'B r

(c ) B u ta n e and

and

Br ( d ) 1 ,3 -B u ta d ie n e and

13.22

W h e n 2 -m e th y l-1 ,3 -b u ta d ie n e (isoprene) undergoes a 1 ,4 add itio n o f hyd ro gen chlo rid e, the m a jo r pro du ct that is fo rm e d is 1 -c h lo ro -3 -m e th y l-2 -b u te n e . L ittle o r n o 1 -c h lo ro -2 -m e th y l-2 -b u te n e is fo rm e d . H o w can y o u e x p la in this?

13.23

W h e n 1-pentene reacts w ith N -b ro m o s u c c in im id e ( N B S ) , tw o products w ith the fo rm u la C 5 H 9B r are o btain ed. W h a t are these p roducts a n d h o w are th e y form ed?

13.24

( a ) T h e h yd ro g e n atom s attached to C 3 o f 1 ,4 -p e n ta d ie n e are u n u su a lly susceptible to abstraction b y rad icals. H o w can y o u accou nt fo r this? ( b ) C a n y o u p ro v id e an e x p la n atio n fo r the fa c t th a t the protons attached to C 3 o f 1,4 p en tad ie n e are m o re a c id ic than the m e th y l h yd ro g e n atom s o f propene?

13.25

P ro v id e a m e c h a n is m th a t exp la in s fo rm a tio n o f the fo llo w in g products. In c lu d e a ll in term ed iates, fo rm a l charges, a nd arrow s s ho w ing e lectro n flow . O N - Br (NBS)

O___

Br

hn

Br

13.26

P ro v id e a m e c h a n is m fo r the fo llo w in g re a c tio n . D r a w a re a c tio n energy coo rd in a te d ia g ra m th a t illustrates the k in e tic and th e rm o d y n a m ic p ath w a y s fo r this reaction .

HBr

Br Br

13.27

P re d ic t the products o f the fo llo w in g reactions. (a )

HBr - 15°C '

(b )

HBr 40 °C

(c) h v , heat

62 6 13.28

Chapter 13

Conjugated Unsaturated Systems

P ro v id e a m e c h a n is m that expla in s fo rm a tio n o f the fo llo w in g products.

Cl HCl (concd)

13.29

P ro v id e a m e c h a n is m that expla in s fo rm a tio n o f the fo llo w in g products.

OMe Cl2

CU

Cl-

MeOH

13.30

T re a tin g e ith e r 1 -c h lo ro -3 -m e th y l-2 -b u te n e o r 3 -c h lo ro -3 -m e th y l-1 -b u te n e w ith A g 2O in w a te r gives (in a d d itio n to A g C l) the fo llo w in g m ix tu re o f a lc o h o l products.

Cl

Ag2 O

OH Ag2 O

HO

15%

85%

Cl ( a ) W r ite a m e c h a n is m th a t accounts fo r the fo rm a tio n o f these products. ( b ) W h a t m ig h t e x p la in the re la tiv e p ro p o rtio n s o f the tw o alkenes th a t are form ed?

13.31

D e h y d ro h a lo g e n a tio n o f 1 ,2 -d ih a lid e s (w ith the e lim in a tio n o f tw o m o la r eq u ivalen ts o f H X ) n o rm a lly leads to an a lk y n e ra th e r than to a co n ju g a te d dien e. H o w e v e r, w h e n 1 ,2 -d ib ro m o c y c lo h e x a n e is d eh ydroh alog en ated , 1 ,3 -c y c lo h e x a d ie n e is p ro d u ce d and n o t c y c lo h e x y n e . W h a t fa c to r accounts fo r this?

13.32

T h e h eat o f h yd ro g e n a tio n o f a lle n e is 2 9 8 kJ m o l“ 1, w hereas th a t o f p ro p y n e is 2 9 0 kJ m o l“ 1. (a ) W h ic h c o m ­ p o u n d is m o re stable? ( b ) T re a tin g a lle n e w ith a strong base causes it to is o m e rize to p ro p y n e . E x p la in .

13.33

A lth o u g h b oth 1 -b ro m o b u ta n e and 4 -b ro m o -1 -b u te n e are p rim a ry h alid es, the la tte r undergoes e lim in a tio n m o re ra p id ly . H o w can this b e h a v io r b e explained?

DIELS-ALDER REACTIONS 13.34

C o m p le te the fo llo w in g m o le c u la r o rb ita l d escrip tion fo r the g ro u n d state o f cy c lo p en tad ie n e . Shade the a p p ro p ria te lobes to in d ic a te phase signs in each m o le c u la r o rb ita l accordin g to increasin g e n ergy o f the m o le c u ­ la r o rb itals. L a b e l the H O M O and L U M O o rb itals, and p la c e the a p p ro ­ p ria te n u m b e r o f electrons in each le v e l, using a straigh t sin g le-b arb ed

n

a rro w to re p re s e n t each electron .

Energy

13.35

13.36

W h y does the m o le c u le show n b elo w , a lth o u g h a c o n ju g a te d d ien e, f a il to und ergo a D ie ls - A ld e r reaction?

R a n k the fo llo w in g dienes in o rd er o f increasing re a c tiv ity in a D ie ls - A ld e r re a c tio n (1 re a c tiv e ). B r ie fly e x p la in y o u r ra n k in g .

least re a c tive , 4 = m o st

8-W

Problems

13.37

G iv e the structures o f the products th a t w o u ld b e fo rm e d w h e n 1 ,3 -b u ta d ie n e reacts w ith each o f the fo llo w in g : (a )

O

(b ) I

O

( c)

OMe 13.38

627

.O N

(d ) \

CN

OMe

C y c lo p e n ta d ie n e undergoes a D ie ls - A ld e r re a c tio n w ith ethene a t 1 6 0 - 1 8 0 ° C . W r ite the structure o f the p ro d u ct o f this reaction .

13.39

A c e ty le n ic com pounds m a y b e used as d ie n o p h ile s in the D ie ls - A ld e r re a c tio n (see R e v ie w P ro b le m 1 3 .1 3 ). W r ite structures fo r the adducts that y o u e x p e c t fro m the re a c tio n o f 1 ,3 -b u ta d ie n e w ith (a )

O

(b )

O

\

—

/

M eO

f 3c — =

— CF3

( h e x a f lu o r o - 2 -b u ty n e ) OMe

(d im e th y l a c e t y le n e d ic a r b o x y la t e )

13.40

P re d ic t the products o f the fo llo w in g reactions. (a )

(e )

O

O

OMe 2

A

(b )

A ’

O (f)

O II

A

"C N

(1 ) A (2) NaBH.

H' (c)

W

O

(3) H2O

Y O

A

O

(g ) (d )

O OMe

'j

O O

MeO

O

A

OMe

0

CN

A

OMe

°

A

O

13.41

CN

W h ic h d ien e a n d d ie n o p h ile w o u ld y o u e m p lo y in a synthesis o f each o f the fo llo w in g ? (a )

(b )

O

(c)

O

O

OMe

H

OMe O (e)

CN

(f)

H

O

OMe

H

628 13.42

Chapter 13

Conjugated Unsaturated Systems

W h e n fu ra n an d m a le im id e und ergo a D ie ls - A ld e r re a c tio n at 2 5 ° C , the m a jo r p ro d u c t is the end o add uct G . W h e n the re a c tio n is c a rrie d o u t a t 9 0 ° C , h ow ever, the m a jo r p ro d u ct is the exo is o m e r H . T h e endo ad d u ct isom erizes to the exo ad d u ct w h e n it is h eated to 9 0 ° C . Propose an e x p la n a tio n th a t w il l accou nt fo r these results.

O

O H Furan

13.43

T w o c o n tro ve rs ia l “h ard ” insecticides are a ld rin and d ie ld rin . [T h e E n v iro n m e n ta l P ro te c tio n A g e n c y (E P A ) h alte d the use o f these insecticides because o f p ossible h a rm fu l side effects and because they are n o t b io d e g ra d a b le .] T h e c o m m e rc ia l synthesis o f a ld rin began w ith h e x a c h lo ro c y clo p e n ta d ie n e and n o rb o rn ad ie n e . D ie ld r in w as synthe­ sized fro m a ld rin . S h o w h o w these syntheses m ig h t h av e been c a rried out.

Cl

Cl

C k / Cl „C l

Cl

//

Cl

H e x a c h lo ro c y c lo p e n ta d ie n e

13.44

N o rb o rn a d ie n e

(a) N o rb o rn a d ie n e fo r the a ld rin synthesis (P ro b le m 1 3 .4 3 ) can b e p re p a re d fro m cy c lo p en tad ie n e and acetylene. (b) I t can also b e p re p a re d b y a llo w in g cy c lo p en tad ie n e to re a c t w ith v in y l c h lo rid e

S h o w the re a c tio n in v o lv e d .

a nd treatin g the p ro d u ct w ith a base. O u tlin e this synthesis.

13.45

T w o o th er h ard insecticides (see P ro b le m 1 3 .4 3 ) are chlo rd an and h e p ­ tachlor. S h o w h o w they c o u ld b e sy n th ize d fro m c y c lo p en tad ie n e and h e x a c h lo ro c y clo p e n ta d ie n e .

Cl I C h lo rd an

13.46

H e p ta c h lo r

Is o d rin , an is o m e r o f a ld rin , is o b ta in e d w h e n cy c lo p en tad ie n e reacts w ith the h e x a c h lo ro n o rb o rn a d ie n e , show n h ere. Propose a structure fo r is od rin .

Cl Cl

-Cl ----- *

Cl Cl

Cl

Isodrin

629

Challenge Problems

13.47

P ro v id e the reagents necessary to achieve the fo llo w in g synthetic transform ation s. M o r e than o ne step m a y be req uired. (a )

(d )

O

Br

O (e)

(b )

/ (c)

O

O Br

Challenge Problems 13.48

E x p la in the p ro du ct d istrib u tio n b e lo w based on the p o la rity o f the d ien e and d ie n o p h ile , as p red icted b y contributing resonance structures fo r each.

OCH3 O H

heat

"O C H 3 O

O M ajo r

13.49

M ino r

M ix in g fu ra n (P ro b le m 1 3 .4 2 ) w ith m a le ic a n h y d rid e in d ie th y l eth er y ie ld s a c ry s ta llin e s o lid w ith a m e ltin g p o in t o f 1 2 5 °C . W h e n m e ltin g o f this c o m p o u n d takes p la c e , h o w ever, one can n o tic e th a t the m e lt evolves a gas. I f the m e lt is a llo w e d to re s o lid ify , one finds th a t it n o lo n g e r m e lts at 1 2 5 °C b u t instead it m e lts a t 5 6 ° C . C o n s u lt an a p p ro p ria te c h e m is try h a n d b o o k and p ro v id e an e x p la n a tio n fo r w h a t is tak in g place.

13.50

D r a w the structure o f the p ro d u ct fro m the fo llo w in g re a c tio n (fo rm e d d u rin g a synthesis o f o ne o f the en d ian d ric acids b y K . C . N ic o la o u ):

O S i( i-B u ) P h 2

13.51

toluene, 110°C

D r a w a ll o f the con trib u tin g resonance structures and the resonance h y b rid fo r the c a rb ocatio n th a t w o u ld resu lt fro m io n iz a tio n o f b ro m in e fro m 5 -b ro m o -1 ,3 -p e n ta d ie n e . O p e n the co m p u ter m o le c u la r m o d e l at the b o o k ’s w e b ­ site d ep ic tin g a m a p o f e lectrostatic p o te n tia l fo r the p e n ta d ie n y l carb ocatio n . B ased on the m o d e l, w h ic h is the m o s t im p o rta n t con trib u tin g resonance structure fo r this cation? Is this consistent w ith w h a t y o u w o u ld h av e p re ­ d ic te d based on y o u r k n o w le d g e o f re la tiv e c arb o catio n stabilities? W h y o r w h y not?

630

Chapter 13

Conjugated Unsaturated Systems

Learning Group Problems 1.

E lu c id a te the structures o f com pounds

A throu gh I in the fo llo w in g “ro a d m a p ” p ro b le m . S p e c ify an y m issing

reagents. CH

A

B

(C5H8)

( C 9 H i 0)

Br2, warm (1 molar equiv.)

mCPBA (or RCO3 H)

CH3

c 6h

CH 3,

5

CH

C sH5

NaOEt, heat (+ c o rre s p o n d in g 1 ,4 -d im e th y l-2 -p h e n y l is o m e rs )

E

F CH3ONa (2 molar equiv.)

NBS, ROOR, heat

D

G

(C 9H 12)

HBr (no ROOR)

H KOC(CH3)3, heat

OH OCHa

I (C7H i A )

2.

reagents?

OCH„

(a) W r ite reactions to show h o w y o u c o u ld con vert 2 -m e th y l-2 -b u te n e in to 2 -m e th y l-1 ,3 -b u ta d ie n e . (b) W r ite reactions to show h o w y o u c o u ld c on vert e th y lb e n ze n e in to the fo llo w in g com pound: „C N

C sH5

(c) W r ite structures fo r the vario us D ie ls - A ld e r add uct(s) th a t c o u ld re s u lt on re a c tio n o f 2 -m e th y l-1 ,3 -b u ta d ie n e w ith the c o m p o u n d show n in p a rt (b ).

CONCEPT MAP h

Delocalization of it electrons

result from

can be

A resonance hybrid

represented by

Contributing resonance structures

is a weighted average of

Overlap of adjacent p orbitals

must ' be

involving three or more atoms is a can be characterized by Conjugated Unsaturated System(s) gives evidence for

UV-Vis Spectroscopy

Quantitative analysis

Reagents*

Allylic systems

Conjugated alkenes

can undergo

examples of

can undergo

\' Radical allylic substitution

\

Kinetic (rate) control

The product(s) which is formed the fastest

Absorption of energy in the ultraviolet-visible (UV-Vis) region of the electromagnetic spectrum

Thermodynamic (equilibrium) control

leads to

depending on structure, may also be

involves promotion of an electron from the

LUMO (lowest unoccupied molecular orbital)

HOMO (highest occupied molecular orbital)

'

Addition reactions to conjugated systems may occur under

‘ Reagents = X2 (low conc.), ROOR, A; or hv or NBS (if X = Br), ROOR, A; or hv

involves

Proper Lewis structures

can be used for

include

(Any allylic system)

Exist only on paper

(Pi) electron systems

The product(s) which is most stable

E is the electrophilic part of the adding reagent.

Nu is the nucleophilic part of the adding reagent.

Aromatic Compounds

In ordinary conversation, the word "aromatic" conjures pleasant associations— the odor of freshly prepared cof­ fee, or of a cinnamon bun. Similar associations occurred early in the history of organic chemistry, when pleas­ antly "aromatic" compounds were isolated from natural oils produced by plants. As the structures of these compounds were elucidated, a number of them were found to contain a highly unsaturated six-carbon struc­ tural unit that is also found in benzene. This special ring structure became known as a benzene ring, and the aromatic compounds containing a benzene ring became part of a larger family of compounds now classified as aromatic on the basis of their electronic structure rather than their odor. The following are a few examples of aromatic compounds including benzene itself. In these formulas we foreshadow our discussion of the special properties of the benzene ring by using a circle in a hexagon to depict the six p electrons and six-membered ring of these compounds, whereas heretofore we have shown benzene rings only as indicated in the left-hand formula for benzene below. O

O

och3

H or OH B enzene

B e n z a ld e h y d e

M e th y l s a lic y la t e

( in o i l o f a lm o n d s )

( in o i l o f w in t e r g r e e n )

O

O H

HO'

H3CO OCH3

632

Eugenol

A n e t h o le

C in n a m a ld e h y d e

V a n illin

( in o il o f c lo v e s )

( in o il o f a n is e )

( in o il o f c in n a m o n )

( in o i l o f v a n illa )

14.1 The Discovery o f Benzene

633

As time passed, chemists found or synthesized many compounds with benzene rings that had no odor, such as benzoic acid and acetylsalicylic acid (aspirin). O

O

OH

B e n zo ic a c id

A c e ty ls a lic y lic acid (a s p irin )

14.1 The Discovery o f Benzene In this c hap ter w e shall discuss in d e ta il the structural p rin c ip le s th a t u n d e rlie h o w the term “ a ro m a tic ” is used today. W e w i ll also see h o w the structure o f b en ze n e p ro v e d so elusive. E v e n tho ug h b en ze n e w as disco vered in 1 8 2 5 , it w as n o t u n til the d e v e lo p m e n t o f q u a n ­ tu m m ech anics in the 1 9 20s th a t a rea s o n a b ly c le a r u nderstanding o f its structure em erged. •

A s w e h av e seen above, tw o fo rm u la types are c o m m o n ly used to d ep ic t b en zene rin g s. T h e tra d itio n a l b o n d -lin e rep resentatio n a llo w s easier d ep ic tio n o f m e c h a ­ nism s in v o lv in g the p electrons, as w e shall n ee d to do in u p c o m in g chapters, w hereas the c irc le in the h exag on n o ta tio n b etter suggests the structure and p ro p e r­ ties o f b en ze n e rings.

T h e study o f the class o f com pounds th a t o rg an ic chem ists c a ll a ro m a tic com pounds (S e c tio n 2 .1 D ) began w ith the d is c o v ery in 1 8 2 5 o f a n e w h yd ro c a rb o n b y the E n g lis h c h e m ist M ic h a e l F a ra d a y (R o y a l In s titu tio n ). F a ra d a y c a lle d this n e w h yd ro c a rb o n “b ic a rb u ret o f h y d ro g e n ” ; w e n o w c a ll it b en zene. F a ra d a y is o la te d b en ze n e fro m a com pressed illu m in a tin g gas th a t h a d b een m a d e b y p y ro ly z in g w h a le o il. In 1 8 3 4 the G e rm a n c h e m ist E ilh a rd t M its c h e rlic h (U n iv e rs ity o f B e rlin ) synthesized b en ze n e b y h ea tin g b e n zo ic a c id w ith c a lc iu m o xid e . U s in g v a p o r d en s ity m easurem ents, M its c h e rlic h fu rth e r show ed th a t b en ze n e has the m o le c u la r fo rm u la C 6 H 6:

heat C 6 H 5 C O 2H B e n zo ic acid

C aO

C6H

6

CaCO o

B e n ze n e

T h e m o le c u la r fo rm u la its e lf w as surprising. B e n ze n e has only as many hydrogen atoms

as it has carbon atoms. M o s t com pounds th a t w e re k n o w n then h ad a fa r g re a te r p ro p o r­ tio n o f h yd ro gen atom s, u su ally tw ic e as m any. B enzen e, havin g the fo rm u la o f C 6 H 6, should b e a h ig h ly u nsaturated c o m p o u n d because it has an in d e x o f h yd ro g e n d e fic ie n c y equal to 4 . E v e n tu a lly , chem ists began to re c o g n ize th a t b en zene w as a m e m b e r o f a n e w class o f o rganic com pounds w ith unusual and interesting properties. A s w e s h a ll see in S e c tio n 1 4 .3, b e n z e n e does n o t s h o w th e b e h a v io r e x p e c te d o f a h ig h ly u n s a tu r a te d c o m p o u n d . D u rin g the la tte r p a rt o f the n in e te e n th ce n tu ry the K e k u le -C o u p e r -B u tle r o v th e o ry o f v alen ce was system atically a p p lie d to a ll k n o w n o rganic com pounds. O n e result o f this e ffo rt w as the p la c in g o f o rg an ic com pounds in e ith e r o f tw o b ro ad categories; com pounds w e re c lassified as b ein g e ith e r a lip h a tic o r a r o m a t ic . T o b e classified as a lip h a tic m e a n t then th a t the c h e m ic a l b e h a v io r o f a c o m p o u n d w as “fa tlik e .” (N o w it m eans th a t the c om po un d reacts lik e an a lk an e , an a lk en e , an a lk y n e , o r one o f th e ir d eriv ativ e s .) T o b e classified as a ro m a tic m e a n t then th a t the c o m p o u n d h ad a lo w h yd ro g e n -to -c a rb o n ra tio an d th a t it was “ fra g ra n t.” M o s t o f the e a rly a ro m a tic com pounds w e re o b ta in e d fro m balsam s, resins, or essential oils.

One of the p molecular orbitals of b en zen e, seen through a mesh representation of its electrostatic potential at its van der Waals surface.

634

Chapter 14 Aromatic Compounds K e k u le w a s th e f i r s t to r e c o g n iz e t h a t th e s e e a r ly a r o m a t ic c o m p o u n d s a l l c o n t a in a s ix c a r b o n u n it a n d th a t th e y r e ta in th is s ix - c a r b o n u n it th r o u g h m o s t c h e m ic a l tr a n s fo r m a tio n s a n d d e g r a d a t io n s . B e n z e n e w a s e v e n t u a l l y r e c o g n i z e d a s b e i n g t h e p a r e n t c o m p o u n d o f t h i s n e w s e r ie s .

14.2 N om enclature o f Benzene Derivatives T w o s y s te m s a re u s e d in n a m in g m o n o s u b s titu te d b e n z e n e s . •

I n m a n y s i m p l e c o m p o u n d s , b e n zen e i s t h e p a r e n t n a m e a n d t h e s u b s t i t u e n t i s s i m ­ p ly in d ic a te d b y a p r e f ix .

W e h a v e , f o r e x a m p le ,

•

F

C l

B r

F lu o r o b e n z e n e

C h lo r o b e n z e n e

N O ,

B ro m o b e n z e n e

N itr o b e n z e n e

F o r o th e r s im p le a n d c o m m o n c o m p o u n d s , th e s u b s t it u e n t a n d th e b e n z e n e r in g ta k e n to g e th e r m a y f o r m

a c o m m o n ly a c c e p te d p a re n t n a m e .

M e t h y l b e n z e n e i s u s u a l l y c a l l e d to lu e n e , h y d r o x y b e n z e n e i s a l m o s t a lw a y s c a l l e d p h e n o l, a n d a m in o b e n z e n e i s a l m o s t a lw a y s c a l l e d a n ilin e . T h e s e a n d o t h e r e x a m p l e s a r e i n d i c a t e d h e re : H

H

H

C H ,

T o lu e n e

Phenol

A n ilin e

C H • O '

S O 3 H

B e n z e n e s u lf o n ic a c id •

A c e to p h e n o n e

A n is o le

W h e n t w o s u b s t it u e n t s a r e p r e s e n t , t h e i r r e l a t i v e p o s i t i o n s a r e i n d i c a t e d b y t h e p r e ­ fix e s

ortho-, m eta-,

and

para-

( a b b r e v ia te d

o-, m -,

and

p -)

o r b y th e u s e o f n u m b e rs .

F o r t h e d ib r o m o b e n z e n e s w e h a v e B r

B r

B r

B r

B r

B r

1 ,2 - D ib r o m o b e n z e n e ( o - d ib r o m o b e n z e n e )

1 ,3 - D ib r o m o b e n z e n e ( m - d ib r o m o b e n z e n e )

1 ,4 - D ib r o m o b e n z e n e ( p - d ib r o m o b e n z e n e )

o r th o

m e ta

p a ra

635

14.2 Nomenclature of Benzene Derivatives and for the nitrobenzoic acids O

O OH

' OH O2N

N O ,

2-Nitrobenzoic acid (o-nitrobenzoic acid)

4-Nitrobenzoic acid (p-nitrobenzoic acid)

3-Nitrobenzoic acid (m -nitrobenzoic acid)

The dimethylbenzenes are often called xylenes: CH,

CH,

CH

CH 3

H 3 C

CH 1,3-D im ethylbenzene (m-xylene)

1,2-D im ethylbenzene (o-xylene) •

1,4-D im ethylbenzene (p-xylene)

I f more than two groups are present on the benzene ring, their positions must be indicated by the use of num bers.

As examples, consider the following two compounds: Cl

Br

L1

li

Cl

Br

I2 Cl 1,2,3-Trichlorobenzene

•

14

Br 1,2,4-Tribrom obenzene (n o t 1,3,4-tribrom obenzene)

The benzene ring is numbered so as to give the lowest possible numbers to the

substituents. • When more than two substituents are present and the substituents are different, they are listed in alphabetical order. •

When a substituent is one that together with the benzene ring gives a new base name, that substituent is assumed to be in position 1 and the new parent name is used. O S O 3 H

F 3,5-D initrobenzoic acid •

F

2,4-D ifluorobenzenesulfonic acid

When the C 6 H5 — group is named as a substituent, it is called a phenyl group. The phenyl group is often abbreviated as C 6 H5— , Ph— , or w— .

A hydrocarbon composed of one saturated chain and one benzene ring is usually named as a derivative of the larger structural unit. However, if the chain is unsaturated, the

H e lp f u l H i n t Note the abbreviations fo r com­ mon aromatic groups.

636

Chapter 14

Aromatic Compounds

c o m p o u n d m a y b e n a m e d a s a d e r i v a t i v e o f t h a t c h a i n , r e g a r d l e s s o f r i n g s iz e . T h e f o l ­ l o w i n g a r e e x a m p le s :

c 6h 5 B u t y lb e n z e n e

|

2 - P h e n y lh e p ta n e

( E ) - 2 - P h e n y l- 2 - b u te n e •

B e n z y l i s a n a lt e r n a t i v e n a m e f o r t h e p h e n y l m e t h y l g r o u p . I t i s s o m e t im e s a b b r e v i ­ a te d B n .

C l

S o lv e d P r o b le m

T h e b e n zyl g ro u p

B e n z y l c h lo r id e

( th e p h e n y lm e t h y l

( p h e n y lm e t h y l c h l o r i d e

g ro u p )

o r B n C l)

1 4 .1

P r o v id e a n a m e f o r e a c h o f th e f o l lo w in g c o m p o u n d s . (a )

O

STRATEGY A ND ANSWER

(b )

NO 2

(c )

(d )

I n e a c h c o m p o u n d w e lo o k f ir s t t o s e e i f a c o m m o n ly n a m e d u n i t c o n t a in in g a b e n ­

z e n e r i n g is p r e s e n t. I f n o t , w e c o n s id e r w h e t h e r th e c o m p o u n d c a n b e n a m e d a s a s im p le d e r iv a t iv e o f b e n z e n e , o r i f th e c o m p o u n d in c o r p o r a te s th e b e n z e n e r in g as a p h e n y l o r b e n z y l g r o u p . I n ( a ) w e r e c o g n iz e th e c o m m o n s tr u c tu r a l u n i t o f a c e to p h e n o n e , a n d f in d a t e r t - b u t y l g r o u p i n th e p a r a p o s it io n . T h e n a m e is th u s p - t e r t - b u t y la c e t o p h e n o n e o r 4 - t e r t - b u t y l a c e t o p h e n o n e . C o m p o u n d ( b ) , h a v i n g t h r e e s u b s t it u e n t s o n t h e r i n g , m u s t h a v e i t s s u b ­ s titu e n ts n a m e d i n a lp h a b e t ic a l o r d e r a n d t h e ir p o s it io n s n u m b e r e d . T h e n a m e is 1 ,4 - d im e t h y l- 2 - n it r o b e n z e n e . I n ( c ) th e r e w o u l d a p p e a r t o b e a b e n z y l g r o u p , b u t th e b e n z e n e r i n g c a n b e c o n s id e r e d a s u b s t it u e n t o n th e a l k y l c h a in , s o i t is c a lle d p h e n y l i n t h is c a s e . T h e n a m e is 2 - c h lo r o - 2 - m e t h y l- 1 - p h e n y lp e n t a n e . B e c a u s e ( d ) c o n t a in s a n e th e r f u n c t io n a l g r o u p , w e n a m e i t a c c o r d in g to th e g r o u p s b o n d e d t o th e e th e r o x y g e n . T h e n a m e is b e n z y l e t h y l e th e r , o r e t h y l p h e n y l m e t h y l e th e r .

14.3 Reactions of Benzene

637

14.3 Reactions o f Benzene In the mid-nineteenth century, benzene presented chemists with a real puzzle. They knew from its formula (Section 14.1) that benzene was highly unsaturated, and they expected it to react accordingly. They expected it to react like an alkene by decolorizing bromine in carbon tetrachloride through addition of bromine . They expected that it would change the color of aqueous potassium permanganate by being oxidized, that it would add hydrogen rapidly in the presence of a metal catalyst, and that it would add water in the presence of strong acids. Benzene does none of these . When benzene is treated with bromine in the dark or with aqueous potassium permanganate or with dilute acids, none of the expected reactions occurs. Benzene does add hydrogen in the presence of finely divided nickel, but only at high tem­ peratures and under high pressures: No addition of brom ine No oxidation

No hydration Slow addition a t high tem p eratu re and p ressu re Benzene does react with bromine but only in the presence of a Lewis acid catalyst such as ferric bromide. Most surprisingly, however, it reacts not by addition but by substitution— benzene substitution. S u b s titu tio n

c 6h 6

FeBr„ B r.

C 6 H 5B r

2

S ubstitution is ob serv ed .

H B r

A d d itio n C 6 H 6

B

r 2

------------ »

C 6 H 6 B r 2

C

6 ^ B

r 4

C 6 H 6B r 6

Addition is not ob serv ed .

When benzene reacts with bromine, only one monobromobenzene is formed. That is, only one compound with the formula C 6 H5Br is found among the products. Similarly, when ben­ zene is chlorinated, only one monochlorobenzene results. Two possible explanations can be given for these observations. The first is that only one of the six hydrogen atoms in benzene is reactive toward these reagents. The second is that all six hydrogen atoms in benzene are equivalent, and replacing any one of them with a substituent results in the same product. As we shall see, the second explanation is correct.

Listed below are four compounds that have the molecular formula C 6 H6. Which of these compounds would yield only one monosubstitution product, if, for example, one hydrogen were replaced by bromine?

R e v ie w P ro b le m 1 4 .2

638

Chapter 14 Aromatic Compounds

14.4 The Kekulé Structure for Benzene I n 1 8 6 5 , A u g u s t K e k u le , th e o r ig in a t o r o f th e s tr u c t u r a l t h e o r y ( S e c tio n 1 .3 ) , p r o p o s e d th e f i r s t d e f in it e s tr u c tu r e f o r b e n z e n e , * a s tr u c tu r e th a t is s t i l l u s e d to d a y ( a lt h o u g h a s w e s h a ll s o o n se e, w e g iv e i t a m e a n in g d if f e r e n t f r o m

th e m e a n in g K e k u le g a v e it ) . K e k u le s u g ­

g e s te d th a t th e c a r b o n a to m s o f b e n z e n e a re in a r in g , th a t th e y a re b o n d e d to e a c h o th e r b y a lt e r n a t in g s in g le a n d d o u b le b o n d s , a n d t h a t o n e h y d r o g e n a to m is a tta c h e d to e a c h c a r ­ b o n a to m . T h is s tr u c tu r e s a tis fie d th e r e q u ir e m e n ts o f th e s tr u c tu r a l t h e o r y t h a t c a r b o n a to m s f o r m f o u r b o n d s a n d t h a t a l l t h e h y d r o g e n a t o m s o f b e n z e n e a r e e q u i v a le n t :

H I ^ C\

H\ I

H

C

/H

II

^ C

or

^ H

C The Kekule form ula for b enzene A p r o b le m

s o o n a ro s e w it h th e K e k u le s t r u c t u r e , h o w e v e r . T h e K e k u le s tr u c tu r e p r e ­

d i c t s t h a t t h e r e s h o u l d b e t w o d i f f e r e n t 1 , 2 - d ib r o m o b e n z e n e s , b u t t h e r e a r e n o t . I n o n e o f th e s e h y p o t h e tic a l c o m p o u n d s ( b e lo w ) , th e c a r b o n a to m s th a t b e a r th e b r o m in e s w o u ld b e s e p a ra te d b y a s in g le b o n d , a n d in th e o t h e r th e y w o u ld b e s e p a ra te d b y a d o u b le b o n d .

Br

Br T h ese 1,2-dibrom obenzenes do not exist a s isom ers.

a n d

Br •

Br

O nly one 1,2-dibrom obenzene has ever been fo u n d , however.

T o a c c o m m o d a te th is o b je c t io n , K e k u le p r o p o s e d th a t th e t w o f o r m s o f b e n z e n e ( a n d o f b e n z e n e d e r i v a t i v e s ) a r e i n a s ta te o f e q u i l i b r i u m a n d t h a t t h i s e q u i l i b r i u m i s s o r a p i d l y e s t a b ­ lis h e d

th a t i t p re v e n ts

is o la tio n

o f th e

s e p a r a te

com pounds.

T h u s , th e

tw o

1 , 2 - d ib r o ­

m o b e n z e n e s w o u l d a ls o b e r a p i d l y e q u i l i b r a t e d , a n d t h i s w o u l d e x p l a i n w h y c h e m i s t s h a d n o t b e e n a b le to is o la te th e t w o f o r m s :

Br

Br T here is no su c h equilibrium betw een b en zen e ring bond isom ers.

Br •

Br

W e n o w k n o w t h a t t h i s p r o p o s a l w a s a ls o i n c o r r e c t a n d t h a t

no such equilibrium

exists. N o n e t h e le s s , t h e K e k u l e f o r m u l a t i o n o f b e n z e n e ’ s s t r u c t u r e w a s a n i m p o r t a n t s t e p f o r w a r d a n d , f o r v e r y p r a c t ic a l re a s o n s , i t is s t i l l u s e d to d a y . W e u n d e r s ta n d it s m e a n in g d if f e r e n t ly , h o w e v e r. T h e te n d e n c y o f b e n z e n e to r e a c t b y s u b s t it u t io n r a th e r th a n a d d it io n g a v e r is e to a n o th e r c o n c e p t o f a r o m a t ic it y . F o r a c o m p o u n d to b e c a lle d a r o m a t ic m e a n t, e x p e r im e n t a lly , t h a t i t g a v e s u b s t it u t io n r e a c t io n s r a t h e r th a n a d d it io n r e a c t io n s e v e n th o u g h i t w a s h ig h l y u n s a tu ra te d . B e f o r e 1 9 0 0 , c h e m i s t s a s s u m e d t h a t t h e r i n g o f a lt e r n a t i n g s i n g l e a n d d o u b l e b o n d s w a s th e s tr u c t u r a l fe a tu r e th a t g a v e r is e t o th e a r o m a t ic p r o p e r tie s . S in c e b e n z e n e a n d b e n z e n e d e r iv a tiv e s ( i.e ., c o m p o u n d s w it h s ix - m e m b e r e d r in g s ) w e r e th e o n ly a r o m a tic c o m p o u n d s

* I n 1861 th e A u s tria n ch e m is t Johann Josef L o s c h m id t represented th e benzene rin g w ith a c irc le , b u t he m ade n o a tte m p t to in d ic a te h o w th e ca rbo n atom s w ere a c tu a lly arranged in th e rin g .

14.5 The Thermodynamic Stability of Benzene

639

k n o w n , c h e m is ts n a t u r a lly s o u g h t o th e r e x a m p le s . T h e c o m p o u n d c y c lo o c ta t e t r a e n e s e e m e d to b e a l i k e l y c a n d id a te :

C y c lo o c t a t e t r a e n e In

1 9 1 1 , R ic h a r d W ills t a t t e r s u c c e e d e d in

s y n th e s iz in g c y c lo o c ta t e t r a e n e . W ills t a t t e r

f o u n d , h o w e v e r , th a t i t is n o t a t a l l l i k e b e n z e n e . C y c lo o c ta te tr a e n e r e a c ts w i t h b r o m in e b y a d d i t i o n , i t a d d s h y d r o g e n r e a d i l y , i t i s o x i d i z e d b y s o lu t i o n s o f p o t a s s i u m p e r m a n g a n a t e , a n d t h u s i t i s c l e a r l y n o t a ro m a tic . W h i l e t h e s e f i n d i n g s m u s t h a v e b e e n a k e e n d i s a p ­ p o in t m e n t to W ills t a t t e r , th e y w e r e v e r y s ig n if ic a n t f o r w h a t th e y d id n o t p r o v e . C h e m is ts , a s a r e s u lt , h a d t o l o o k d e e p e r t o d is c o v e r th e o r ig i n o f b e n z e n e ’ s a r o m a t ic it y .

14.5 The Therm odynam ic S tab ility o f Benzene W e h a v e s e e n t h a t b e n z e n e s h o w s u n u s u a l b e h a v i o r b y u n d e r g o i n g s u b s t i t u t i o n r e a c t io n s w h e n , o n t h e b a s is o f i t s K e k u l e s t r u c t u r e , w e s h o u l d e x p e c t i t t o u n d e r g o a d d i t i o n . B e n z e n e i s u n u s u a l i n a n o t h e r s e n s e : I t i s m o re s ta b le th e r m o d y n a m ic a lly t h a n t h e K e k u l e s t r u c t u r e s u g g e s ts . T o s e e h o w , c o n s id e r th e f o l l o w i n g th e r m o c h e m ic a l r e s u lts . C y c lo h e x e n e , a s i x - m e m b e r e d r i n g c o n t a i n i n g o n e d o u b l e b o n d , c a n b e h y d r o g e n a t e d e a s ily t o c y c lo h e x a n e . W h e n th e A H ° f o r t h is r e a c t io n is m e a s u r e d , i t is f o u n d t o b e - 1 2 0 k J m o l - 1 , v e r y m u c h l i k e t h a t o f a n y s i m i l a r l y s u b s t i t u t e d a lk e n e :

AH ° = - 1 2 0 kJ mol-1 C y c lo h e x e n e

C y c lo h e x a n e

W e w o u l d e x p e c t t h a t h y d r o g e n a t i o n o f 1 , 3 - c y c lo h e x a d i e n e w o u l d l i b e r a t e r o u g h l y t w i c e as m u c h h e a t a n d th u s h a v e a A H ° e q u a l to a b o u t - 2 4 0 k J m o l - 1 . W h e n th is e x p e r im e n t is d o n e , th e r e s u lt is A H ° =

- 2 3 2 k J m o l - 1 . T h is r e s u lt is q u it e c lo s e to w h a t w e c a lc u ­

la t e d , a n d th e d if f e r e n c e c a n b e e x p la in e d b y t a k in g i n t o a c c o u n t th e f a c t t h a t c o m p o u n d s c o n t a i n i n g c o n j u g a t e d d o u b l e b o n d s a r e u s u a l l y s o m e w h a t m o r e s t a b le t h a n t h o s e t h a t c o n ­ t a in is o la te d d o u b le b o n d s ( S e c tio n 1 3 .8 ):

2

Calculated AH ° = 2 X ( - 1 2 0 ) = - 2 4 0 kJ m ol-1

H2 , Pt

1 , 3 - C y c lo h e x a d ie n e

C y c lo h e x a n e

Observed AH ° = - 2 3 2 kJ m ol-1

I f w e e x te n d t h is k i n d o f t h i n k i n g , a n d i f b e n z e n e is s im p ly 1 ,3 ,5 - c y c lo h e x a t r ie n e , w e w o u ld p r e d ic t b e n z e n e t o lib e r a t e a p p r o x im a t e ly 3 6 0 k J m o l - 1 [3 X

( - 1 2 0 ) ] w h e n i t is

h y d r o g e n a te d . W h e n th e e x p e r im e n t is a c t u a lly d o n e , th e r e s u lt is s u r p r is in g ly d if f e r e n t . T h e r e a c t io n is e x o t h e r m ic , b u t o n l y b y 2 0 8 k J m o l - 1 :

O O bserved B enzene

C y c lo h e x a n e

Calculated AH ° = 3 X ( - 1 2 0 ) = - 3 6 0 kJ m ol-1

AH° = - 2 0 8 kJ m ol-1 D iffe r e n t

=

1 5 2

kJ m°

1

W h e n th e s e r e s u lts a re r e p r e s e n te d as i n F ig . 1 4 .1 , i t b e c o m e s c le a r t h a t b e n z e n e is m u c h m o r e s t a b le t h a n w e c a l c u l a t e d i t t o b e . I n d e e d , i t i s m o r e s t a b le t h a n t h e h y p o t h e t i c a l 1 , 3 , 5 c y c lo h e x a tr ie n e b y

1 5 2 k J m o l - 1 . T h is d if fe r e n c e b e tw e e n th e a m o u n t o f h e a t a c t u a lly

r e le a s e d a n d t h a t c a l c u l a t e d o n t h e b a s is o f t h e K e k u l e s t r u c t u r e i s n o w n a n c e e n e rg y o f th e c o m p o u n d .

c a lle d th e r e s o ­

640

Chapter 14 Aromatic Compounds

Figure 14.1 Relative stabilities of cyclohexene, 1,3-cyclohexadiene, 1,3, 5-cyclohexatriene (hypothetical), and benzene.

14.6 M o d ern Theories o f the Structure o f Benzene I t w a s n o t u n t i l th e d e v e lo p m e n t o f q u a n tu m m e c h a n ic s i n th e 1 9 2 0 s t h a t th e u n u s u a l b e h a v ­ i o r a n d s t a b il i t y o f b e n z e n e b e g a n to b e u n d e r s to o d . Q u a n t u m m e c h a n ic s , a s w e h a v e s e e n , p r o d u c e d t w o w a y s o f v i e w i n g b o n d s i n m o le c u le s : r e s o n a n c e t h e o r y a n d m o le c u la r o r b it a l th e o r y . W e n o w lo o k a t b o t h o f th e s e as th e y a p p ly to b e n z e n e .

14.6A The Resonance Explanation of the Structure of Benzene A b a s i c p o s t u l a t e o f r e s o n a n c e t h e o r y ( S e c t i o n s 1 .8 a n d 1 3 . 5 ) is t h a t w h e n e v e r t w o o r m o r e L e w i s s t r u c t u r e s c a n b e w r i t t e n f o r a m o l e c u l e t h a t d if f e r o n ly in th e p o s itio n s o f t h e ir e le c ­

tro n s , n o n e o f t h e s t r u c t u r e s w i l l b e i n c o m p l e t e a c c o r d w i t h t h e c o m p o u n d ’ s c h e m i c a l a n d p h y s ic a l p r o p e r tie s . I f w e r e c o g n iz e th is , w e c a n n o w u n d e r s ta n d th e tr u e n a tu r e o f th e t w o K e k u lé s tr u c tu r e s ( I a n d I I ) f o r b e n z e n e . •

K e k u lé s tr u c tu r e s I a n d I I b e lo w d i f f e r o n l y i n th e p o s it io n s o f t h e ir e le c tr o n s ; th e y d o n o t r e p r e s e n t t w o s e p a r a te m o le c u le s i n e q u i l ib r i u m a s K e k u lé h a d p r o p o s e d .

In s te a d , s tr u c tu r e s I a n d I I a re th e c lo s e s t w e c a n g e t to a s tr u c tu r e f o r b e n z e n e w it h i n th e l i m i t a t i o n s o f i t s m o l e c u l a r f o r m u l a , t h e c l a s s i c r u le s o f v a le n c e , a n d t h e f a c t t h a t t h e s i x h y d r o g e n a to m s a re c h e m ic a lly e q u iv a le n t. T h e p r o b le m w i t h th e K e k u lé s tr u c tu r e s is th a t t h e y a r e L e w i s s tr u c tu r e s , a n d L e w is s tr u c tu r e s p o r t r a y e le c tr o n s i n lo c a liz e d d is t r ib u t io n s . ( W i t h b e n z e n e , a s w e s h a l l s e e , t h e e le c t r o n s a r e d e l o c a l i z e d . ) R e s o n a n c e t h e o r y , f o r t u n a t e l y , d o e s n o t s t o p w i t h t e l l i n g u s w h e n t o e x p e c t t h i s k i n d o f t r o u b l e ; i t a ls o g iv e s u s a w a y o u t . •

A c c o r d in g t o r e s o n a n c e th e o r y , w e c o n s id e r K e k u lé s tr u c tu r e s I a n d I I b e lo w as

re s o n a n c e c o n tr ib u to r s t o t h e r e a l s t r u c t u r e o f b e n z e n e , a n d w e r e l a t e t h e m t o e a c h o t h e r w i t h o n e d o u b le - h e a d e d , d o u b le - b a r b e d a r r o w ( n o t t w o s e p a ra te a r r o w s , w h ic h w e r e s e rv e f o r e q u ilib r ia ) . R e s o n a n c e c o n t r ib u t o r s , w e e m p h a s iz e a g a in , a r e n o t i n e q u i l ib r i u m . T h e y a re n o t s tr u c ­ tu r e s o f r e a l m o le c u le s . T h e y a re th e c lo s e s t w e c a n g e t i f w e a r e b o u n d b y s im p le r u le s o f v a le n c e , b u t t h e y a r e v e r y u s e f u l i n h e l p i n g u s v i s u a l i z e t h e a c t u a l m o l e c u l e a s a h y b r i d :

I

II

1 4 .6

641

M o d e r n T h e o r ie s o f t h e S t r u c t u r e o f B e n z e n e

L o o k a t th e s tr u c tu r e s c a r e f u lly . A l l o f th e s in g le b o n d s i n s tr u c tu r e I a re d o u b le b o n d s in s tru c tu re I I . •

A h y b r i d (a v e ra g e ) o f K e k u le s tr u c tu r e s I a n d I I w o u ld h a v e n e ith e r p u r e s in g le b o n d s n o r p u r e d o u b le b o n d s b e tw e e n th e c a rb o n s . T h e b o n d o r d e r w o u ld b e b e tw e e n th a t o f a s in g le a n d a d o u b le b o n d .

E x p e r i m e n t a l e v id e n c e b e a r s t h i s o u t . S p e c t r o s c o p i c m e a s u r e m e n t s s h o w t h a t t h e m o l e c u l e o f b e n z e n e is p la n a r a n d th a t a ll o f it s c a r b o n - c a r b o n b o n d s a r e o f e q u a l le n g th . M o r e o v e r , th e c a r b o n - c a r b o n b o n d le n g th s i n b e n z e n e ( F ig . 1 4 .2 ) a re 1 .3 9 A , a v a lu e i n b e t w e e n th a t f o r a c a r b o n - c a r b o n s i n g l e b o n d b e t w e e n s p 2 - h y b r i d i z e d a t o m s ( 1 . 4 7 A ) ( s e e T a b le 1 3 . 1 ) a n d th a t f o r a c a r b o n - c a r b o n d o u b le b o n d ( 1 .3 4 A ) .

H

\ H----- t C 1.09 A

12' •

\

y H

•

V A 20° £ g i- H

/

h on°

c

------- 1.39 A

\

Figure 14.2 Bond lengths and angles in benzene. (Only the s bonds are shown.)

T h e h y b r i d s tr u c tu r e o f b e n z e n e is r e p r e s e n te d b y in s c r ib in g a c ir c le in s id e th e h e x a g o n a s s h o w n i n f o r m u la I I I b e lo w .

ill T h e r e a re tim e s w h e n a n a c c o u n tin g o f th e p

e le c t r o n p a i r s m u s t b e m a d e , h o w e v e r , a n d

f o r th e s e p u r p o s e s w e u s e e it h e r K e k u l e s t r u c t u r e I o r I I . W e d o t h i s s i m p l y b e c a u s e t h e e le c ­ t r o n p a ir s a n d to t a l p p

e le c t r o n c o u n t i s o b v i o u s i n a K e k u l e s t r u c t u r e , w h e r e a s t h e n u m b e r o f

e le c t r o n p a i r s r e p r e s e n t e d b y a c i r c l e c a n b e a m b i g u o u s . A s w e s h a l l s e e l a t e r i n t h i s c h a p ­

te r , t h e r e a r e s y s te m s h a v i n g d i f f e r e n t r i n g s iz e s a n d d i f f e r e n t n u m b e r s o f d e l o c a li z e d p

e le c ­

t r o n s t h a t c a n a ls o b e r e p r e s e n t e d b y a c i r c l e . I n b e n z e n e , h o w e v e r , t h e c i r c l e i s u n d e r s t o o d t o re p re s e n t s ix p •

e le c t r o n s t h a t a r e d e l o c a li z e d a r o u n d t h e s i x c a r b o n s o f t h e r i n g .

A n a c tu a l m o le c u le o f b e n z e n e ( d e p ic t e d b y th e re s o n a n c e h y b r i d I I I ) is m o r e s ta ­ b le th a n e it h e r c o n t r ib u t in g re s o n a n c e s tr u c tu r e b e c a u s e m o r e th a n o n e e q u iv a le n t re s o n a n c e s tru c tu re c a n b e d ra w n f o r b e n z e n e ( I a n d I I a b o v e ).

T h e d i f f e r e n c e i n e n e r g y b e t w e e n h y p o t h e t i c a l 1 , 3 , 5 - c y c l o h e x a t r i e n e ( w h i c h i f i t e x is t e d w o u l d h a v e h i g h e r e n e r g y ) a n d b e n z e n e i s c a l l e d re s o n a n c e e n e rg y , a n d i t i s a n i n d i c a t i o n o f th e e x tr a s t a b il i t y o f b e n z e n e d u e t o e le c tr o n d e lo c a liz a t io n .

I f b e n z e n e w e r e 1 , 3 , 5 - c y c l o h e x a t r i e n e , t h e c a r b o n - c a r b o n b o n d s w o u l d b e a lt e r n a t e l y l o n g a n d s h o r t as in d ic a te d in th e f o l lo w in g s tr u c tu r e s . H o w e v e r , to c o n s id e r th e s tr u c tu r e s h e r e a s r e s o n a n c e c o n t r ib u t o r s ( o r t o c o n n e c t th e m b y a d o u b le - h e a d e d a r r o w ) v io la t e s a b a s ic p r in c ip le o f re s o n a n c e th e o r y . E x p la in .

14.6B The Molecular Orbital Explanation of the Structure of Benzene T h e f a c t th a t th e b o n d a n g le s o f th e c a r b o n a to m s i n th e b e n z e n e r i n g a re a l l 1 2 0 ° s t r o n g ly s u g g e s t s t h a t t h e c a r b o n a t o m s a r e sp 2 h y b r i d i z e d . I f w e a c c e p t t h i s s u g g e s t i o n a n d c o n ­ s t r u c t a p l a n a r s i x - m e m b e r e d r i n g f r o m sp 2 c a r b o n a t o m s , r e p r e s e n t a t i o n s l i k e t h o s e s h o w n

R e v ie w P r o b le m

1 4 .3

642

Chapter 14 Aromatic Compounds in F ig s . 1 4 .3 a a n d b e m e r g e . I n th e s e m o d e ls , e a c h c a r b o n is s p 2 h y b r id iz e d a n d h a s a p o r b it a l a v a ila b le f o r o v e r la p w it h p

o r b ita ls o f its n e ig h b o r in g

c a r b o n s . I f w e c o n s id e r

fa v o r a b le o v e r la p o f th e s e p o r b it a ls a ll a r o u n d th e r in g , th e r e s u lt is th e m o d e l s h o w n in F ig . 1 4 .3 c .

Figure 14.3 (a) Six sp 2-hybridized carbon atom s joined in a ring (each carbon also bears a hydrogen atom). Each carbon has a p orbital with lobes above and below the plane of the ring. (b) A stylized depiction of the p orbitals in (a). (c) Overlap of the p orbitals around the ring results in a molecular orbital encom passing the top and bottom faces of the ring. (Differences in the mathematical phase of the orbital lobes are not shown in these representations.)

•

(a )

(b)

(c)

A s w e r e c a ll f r o m th e p r in c ip le s o f q u a n tu m m e c h a n ic s ( S e c tio n 1 .1 1 ), th e n u m b e r o f m o le c u la r o r b it a ls in a m o le c u le is th e s a m e a s th e n u m b e r o f a t o m ic o r b ita ls f r o m w h ic h th e y a re d e r iv e d , a n d e a c h o r b it a l c a n a c c o m m o d a te a m a x im u m o f t w o e l e c t r o n s i f t h e i r s p in s a r e o p p o s e d .

I f w e c o n s id e r o n l y th e p a t o m ic o r b ita ls c o n t r ib u te d b y th e c a r b o n a to m s o f b e n z e n e , th e re s h o u ld b e s ix p

m o le c u la r o r b ita ls . T h e s e o r b it a ls a re s h o w n in

F ig . 1 4 .4 .

Figure 14.4 How six p atom ic orbitals (one from each carbon of the benzene ring) com bine to form six p molecular orbitals. Three of the molecular orbitals have energies lower than that of an isolated p orbital; th ese are the bonding molecular orbitals. Three of the molecular orbitals have energies higher than that of an isolated p orbital; th ese are the antibonding molecular orbitals. Orbitals C2 and C3 have the sam e energy and are said to be degenerate; the sam e is true of orbitals C4 and 1^5 -

T h e e le c t r o n ic c o n f ig u r a t io n o f th e g r o u n d s ta te o f b e n z e n e is o b t a in e d b y a d d in g th e s ix p

e le c tr o n s to th e p

m o le c u la r o r b it a ls s h o w n in F ig . 1 4 .4 , s ta r tin g w it h th e o r b it a ls o f

lo w e s t e n e rg y . T h e lo w e s t e n e r g y p

m o le c u la r o r b it a l in b e n z e n e h a s o v e r la p o f p o r b ita ls

w i t h th e s a m e m a t h e m a t ic a l p h a s e s ig n a ll a r o u n d th e t o p a n d b o t t o m fa c e s o f th e r in g . I n t h i s o r b i t a l t h e r e a r e n o n o d a l p la n e s ( c h a n g e s i n o r b i t a l p h a s e s i g n ) p e r p e n d i c u l a r t o t h e a t o m s o f t h e r i n g . T h e o r b i t a l s o f n e x t h i g h e r e n e r g y e a c h h a v e o n e n o d a l p la n e . ( I n g e n ­ e r a l, e a c h s e t o f h ig h e r e n e r g y p

m o le c u la r o r b it a ls h a s a n a d d it io n a l n o d a l p la n e .) E a c h

o f t h e s e o r b i t a l s i s f i l l e d w i t h a p a i r o f e le c t r o n s , a s w e l l . T h e s e o r b i t a l s a r e o f e q u a l e n e r g y

14.7 Huckel's Rule: The

4n +

643

2 p Electron Rule

( d e g e n e r a t e ) b e c a u s e t h e y b o t h h a v e o n e n o d a l p la n e . T o g e t h e r , t h e s e t h r e e o r b i t a l s c o m ­ p r is e th e b o n d in g p

m o le c u la r o r b it a ls o f b e n z e n e . T h e n e x t h ig h e r e n e r g y s e t o f p

c u l a r o r b i t a l s e a c h h a s t w o n o d a l p la n e s , a n d t h e h ig h e s t e n e r g y p

m o le ­

m o le c u la r o r b it a l o f

b e n z e n e h a s t h r e e n o d a l p la n e s . T h e s e t h r e e o r b i t a l s a r e t h e a n t i b o n d i n g p m o l e c u l a r o r b i t a l s o f b e n z e n e , a n d t h e y a r e u n o c c u p i e d i n t h e g r o u n d s ta te . B e n z e n e i s s a id t o h a v e a c l o s e d b o n d in g s h e ll o f d e lo c a liz e d p

e le c t r o n s b e c a u s e a l l o f i t s b o n d i n g o r b i t a l s a r e f i l l e d w i t h

e l e c t r o n s t h a t h a v e t h e i r s p in s p a i r e d , a n d n o e l e c t r o n s a r e f o u n d i n a n t i b o n d i n g o r b i t a l s . T h is c lo s e d b o n d in g s h e ll a c c o u n ts , in p a r t, f o r th e s t a b ilit y o f b e n z e n e . H a v in g c o n s id e r e d th e m o le c u la r o r b it a ls o f b e n z e n e , i t is n o w u s e f u l t o v i e w a n e le c ­ t r o s t a t i c p o t e n t i a l m a p o f t h e v a n d e r W a a ls s u r f a c e f o r b e n z e n e , a ls o c a l c u l a t e d f r o m q u a n ­ t u m m e c h a n ic a l p r in c ip le s ( F ig . 1 4 .5 ) . W e c a n s e e t h a t t h is r e p r e s e n ta tio n is c o n s is te n t w it h o u r u n d e r s ta n d in g th a t th e p

e le c tr o n s o f b e n z e n e a r e n o t lo c a liz e d b u t a re e v e n ly d is t r ib ­

u te d a r o u n d th e t o p fa c e a n d b o t t o m fa c e ( n o t s h o w n ) o f th e c a r b o n r in g in b e n z e n e . I t is in t e r e s t in g t o n o t e th e r e c e n t d is c o v e r y th a t c r y s t a llin e b e n z e n e in v o lv e s p e r p e n ­

r

d ic u la r in t e r a c t io n s b e t w e e n b e n z e n e r in g s , s o t h a t th e r e la t i v e ly p o s it iv e p e r ip h e r y o f o n e m o l e c u l e a s s o c ia t e s w i t h

th e r e la t i v e ly n e g a tiv e fa c e s o f th e b e n z e n e m o le c u le s a lig n e d

a b o v e a n d b e lo w it.

Figure 14.5 Electrostatic potential map of benzene.

14.7 Huckel's Rule: The 4n + 2 p Electron Rule In

1 9 3 1 t h e G e r m a n p h y s i c i s t E r i c h H u c k e l c a r r i e d o u t a s e r ie s o f m a t h e m a t i c a l c a l c u l a ­

t io n s b a s e d o n th e k i n d o f t h e o r y t h a t w e h a v e j u s t d e s c r ib e d . H u c k e l ’ s r u l e is c o n c e r n e d w it h c o m p o u n d s c o n t a in in g o n e p la n a r r i n g i n

w h ic h e a c h a to m

has a

p

o r b i t a l as in

b e n z e n e . H i s c a lc u l a t i o n s s h o w t h a t p l a n a r m o n o c y c l i c r i n g s c o n t a i n i n g 4 n + 2 p e le c t r o n s , w h e r e n = 0 , 1, 2 , 3 , a n d s o o n ( i.e ., r in g s c o n t a in in g 2 ,

6,

10, 14, . . . ,

e tc ., p

e le c t r o n s ) ,

h a v e c lo s e d s h e lls o f d e lo c a liz e d e le c tr o n s l ik e b e n z e n e a n d s h o u ld h a v e s u b s t a n t ia l r e s o ­ n a n c e e n e r g ie s . •

I n o t h e r w o r d s , H u c k e l ’ s r u l e s ta te s t h a t p l a n a r m o n o c y c l i c r i n g s w i t h 2 ,

6,

10,

1 4 , . . . , d e lo c a liz e d e le c t r o n s s h o u l d b e a r o m a t i c .

14.7A How to Diagram the Relative Energies of p Molecular Orbitals in Monocyclic Systems Based on Huckel's Rule T h e r e i s a s i m p l e w a y t o m a k e a d i a g r a m o f t h e r e la t i v e e n e r g ie s o f o r b i t a l s i n m o n o c y c l i c c o n j u g a t e d s y s te m s b a s e d o n H u c k e l ’ s c a lc u l a t i o n s . T o d o s o , w e u s e t h e f o l l o w i n g p r o c e d u r e . 1 . W e s ta r t b y d r a w in g a p o ly g o n c o r r e s p o n d in g to th e n u m b e r o f c a r b o n s i n th e r in g ,

p la c in g a c o r n e r o f th e p o ly g o n a t th e b o tto m . 2 . N e x t w e s u r r o u n d th e p o ly g o n w it h a c ir c le th a t to u c h e s e a c h c o r n e r o f th e p o ly g o n . 3 . A t th e p o in t s w h e r e th e p o ly g o n to u c h e s th e c ir c le , w e d r a w

s h o r t h o r i z o n t a l l in e s

o u t s id e th e c ir c le . T h e h e ig h t o f e a c h l in e r e p r e s e n ts th e r e la t iv e e n e r g y o f e a c h p m o le c u la r o r b it a l. 4 . N e x t w e d r a w a d a s h e d h o r i z o n t a l l i n e a c r o s s a n d h a l f w a y u p t h e c i r c l e . T h e e n e r g ie s o f b o n d in g p

m o le c u la r o r b it a ls a re b e lo w t h is lin e . T h e e n e r g ie s o f a n t ib o n d in g p

m o le c u la r o r b it a ls a re a b o v e , a n d th o s e f o r n o n b o n d in g o r b it a ls a re a t th e le v e l o f th e d a s h e d lin e . 5 . B a s e d o n th e n u m b e r o f p

e le c tr o n s i n th e r in g , w e th e n p la c e e le c tr o n a r r o w s o n th e

l i n e s c o r r e s p o n d i n g t o t h e r e s p e c t iv e o r b i t a l s , b e g i n n i n g a t t h e l o w e s t e n e r g y l e v e l a n d w o r k i n g u p w a r d . I n d o in g s o , w e f i l l d e g e n e r a te o r b it a ls e a c h w i t h o n e e le c tr o n f i r s t , t h e n a d d t o e a c h u n p a i r e d e l e c t r o n a n o t h e r w i t h o p p o s i t e s p in i f i t i s a v a i l a b l e .

644

Chapter 14 Aromatic Compounds A p p ly i n g t h is m e t h o d t o b e n z e n e , f o r e x a m p le ( F ig . 1 4 .6 ) , f u r n is h e s th e s a m e e n e r g y l e v ­ e ls t h a t w e s a w e a r l i e r i n F i g . 1 4 . 4 , e n e r g y l e v e l s t h a t w e r e b a s e d o n q u a n t u m m e c h a n i c a l c a lc u la tio n s .

Figure 14.6 The polygon-and-circle m ethod for deriving the relative energies of the p molecular orbitals of benzene. A horizontal line halfway up the circle divides the bonding orbitals from the antibonding orbitals. If an orbital falls on this line, it is a nonbonding orbital. This m ethod was d evelop ed by C. A. Coulson (of Oxford University).

A n tib o n d in g p o r b ita ls

B o n d in g p o r b ita ls

P o ly g o n in c irc le

E n e rg y le v e ls of MOs

T yp e of p o rb ita l

W e c a n n o w u n d e r s ta n d w h y c y c lo o c ta t e t r a e n e is n o t a r o m a t ic . C y c lo o c ta te tr a e n e h a s e le c t r o n s . E i g h t i s n o t a H u c k e l n u m b e r ; i t i s a 4 n n u m b e r, n o t a 4 n + 2

a t o t a l o f e ig h t p

n u m b e r. U s i n g t h e p o l y g o n - a n d - c i r c l e m e t h o d ( F i g . 1 4 . 7 ) , w e f i n d t h a t c y c l o o c t a t e t r a e n e , i f i t w e r e p la n a r , w o u ld n o t h a v e a c l o s e d s h e l l o f p

e le c tr o n s l i k e b e n z e n e ; i t w o u l d h a v e

a n u n p a ir e d e le c tr o n i n e a c h o f t w o n o n b o n d in g o r b it a ls . M o le c u le s w i t h u n p a ir e d e le c ­ t r o n s ( r a d i c a l s ) a r e n o t u n u s u a l l y s t a b le ; t h e y a r e t y p i c a l l y h i g h l y r e a c t i v e a n d u n s t a b l e . A p la n a r f o r m

o f c y c lo o c ta te tr a e n e , th e r e fo r e , s h o u ld n o t b e a t a ll l i k e b e n z e n e a n d s h o u ld

n o t b e a r o m a tic .

Figure 14.7 The p molecular orbitals that cyclooctatetraene would have if it w ere planar. Notice that, unlike benzene, this molecule is predicted to have tw o nonbonding orbitals, and because it has eight p electrons, it would have an unpaired electron in each of the tw o nonbonding orbitals (Hund's rule, Section 1.10). Such a system would not be exp ected to be aromatic.

A n tib o n d in g p o r b ita ls —

( N o n b o n d in g p o r b ita ls ) B o n d in g p o r b ita ls

B e c a u s e c y c l o o c t a t e t r a e n e d o e s n o t g a i n s t a b i l i t y b y b e c o m i n g p la n a r , i t a s s u m e s t h e tu b s h a p e s h o w n b e lo w . ( I n S e c tio n 1 4 .7 E w e s h a ll s e e th a t c y c lo o c ta te tr a e n e w o u ld a c tu ­ a l l y l o s e s t a b i l i t y b y b e c o m i n g p la n a r . ) T h e b o n d s o f c y c l o o c t a t e t r a e n e a r e k n o w n t o b e a l t e r n a t e l y l o n g a n d s h o r t ; X - r a y s t u d ie s i n d i c a t e t h a t t h e y a r e 1 . 4 8 a n d 1 . 3 4 A , r e s p e c t iv e l y .

14.7B The Annulenes T h e w o r d a n n u l e n e i s i n c o r p o r a t e d i n t o t h e c la s s n a m e f o r m o n o c y c l i c c o m p o u n d s t h a t c a n b e r e p r e s e n t e d b y s t r u c t u r e s h a v i n g a lt e r n a t i n g s i n g l e a n d d o u b l e b o n d s . T h e r i n g s iz e o f a n a n n u le n e is in d ic a t e d b y a n u m b e r i n b r a c k e t s . T h u s , b e n z e n e is [ 6 ]a n n u le n e a n d c y c lo o c ta t e t r a e n e is [ 8 ]a n n u le n e .

•

H u c k e l ’ s r u le p r e d ic t s t h a t a n n u le n e s w i l l b e a r o m a t ic i f t h e ir m o le c u le s h a v e 4n +

2 p

e l e c t r o n s a n d h a v e a p l a n a r c a r b o n s k e le t o n :

O

B enzene

C y c lo o c t a t e t r a e n e

( [ 6 ] a n n u le n e )

( [ 8 ] a n n u le n e )

645

14.7 Huckel's Rule: The 4n + 2 p Electron Rule B e f o r e 1 9 6 0 th e o n l y a n n u le n e s th a t w e r e a v a ila b le to te s t H u c k e l ’ s p r e d ic t io n s w e r e b e n z e n e a n d c y c lo o c ta te tr a e n e . D u r in g th e 1 9 6 0 s , a n d la r g e ly as a r e s u lt o f re s e a rc h b y F. S o n d h e im e r , a n u m b e r o f l a r g e - r i n g a n n u l e n e s w e r e s y n t h e s iz e d , a n d t h e p r e d i c t i o n s o f H u c k e l’ s r u le w e r e v e r if ie d . C o n s i d e r t h e [ 1 4 ] , [ 1 6 ] , [ 1 8 ] , [ 2 0 ] , [ 2 2 ] , a n d [ 2 4 ] a n n u l e n e s a s e x a m p l e s . O f t h e s e , as

H u c k e l’s r u le p re d ic ts , t h e [ 1 4 ] , [ 1 8 ] , a n d [ 2 2 ] a n n u l e n e s ( 4 n + 2 w h e n n = 3 , 4 , 5 , r e s p e c ­ t iv e ly ) h a v e b e e n fo u n d to b e a r o m a tic . T h e [1 6 ]a n n u le n e a n d th e [2 4 ]a n n u le n e a re n o t a r o ­ m a t i c ; t h e y a r e a n tia r o m a tic ( s e e S e c t i o n

1 4 .7 E ). T h e y a re 4 n c o m p o u n d s , n o t 4 n +

2

com pounds:

H e lp f u l H i n t These names are often used for conjugated rings of 10 or more carbon atoms, b u t they are seldom used for benzene and cyclooctatetraene.

[1 4 ] A n n u le n e ( a r o m a t ic )

( a n t ia r o m a t ic )

( a r o m a t ic )

E x a m p l e s o f [ 1 0 ] a n d [ 1 2 ] a n n u l e n e s h a v e a ls o b e e n s y n t h e s iz e d a n d n o n e i s a r o m a t i c . W e w o u l d n o t e x p e c t [ 1 2 ]a n n u le n e s t o b e a r o m a t ic s in c e th e y h a v e 1 2 p

e le c t r o n s a n d d o

n o t o b e y H u c k e l’ s r u le . T h e f o llo w in g [1 0 ]a n n u le n e s w o u ld b e e x p e c te d to b e a r o m a tic o n t h e b a s is o f e l e c t r o n c o u n t , b u t t h e i r r i n g s a r e n o t p la n a r .

4

5

6

[ 1 0 ] A n n u le n e s N o n e i s a r o m a t ic b e c a u s e n o n e is p la n a r .

T h e [1 0 ]a n n u le n e

4 has

t w o t r a n s d o u b l e b o n d s . I t s b o n d a n g le s a r e a p p r o x i m a t e l y 1 2 0 ° ;

th e r e fo r e , i t h a s n o a p p r e c ia b le a n g le s tr a in . T h e c a r b o n a to m s o f it s r in g , h o w e v e r , a re p r e ­ v e n te d f r o m b e c o m in g c o p la n a r b e c a u s e th e tw o h y d r o g e n a to m s in th e c e n te r o f th e r in g i n t e r f e r e w i t h e a c h o t h e r . B e c a u s e t h e r i n g i s n o t p la n a r , t h e p o r b i t a l s o f t h e c a r b o n a t o m s a re n o t p a r a lle l a n d , th e r e fo r e , c a n n o t o v e r la p e f f e c t iv e ly a r o u n d th e r in g t o f o r m th e p m o l­ e c u la r o r b it a ls o f a n a r o m a tic s y s te m . T h e [ 1 0 ] a n n u l e n e w i t h a l l c i s d o u b l e b o n d s ( 5 ) w o u l d , i f i t w e r e p la n a r , h a v e c o n s i d e r ­ a b l e a n g l e s t r a i n b e c a u s e t h e i n t e r n a l b o n d a n g le s w o u l d b e 1 4 4 ° . C o n s e q u e n t l y , a n y s ta ­ b i l i t y t h is is o m e r g a in e d b y b e c o m in g p la n a r i n o r d e r t o b e c o m e a r o m a t ic w o u l d b e m o r e t h a n o f f s e t b y t h e d e s t a b i l i z i n g e f f e c t o f t h e in c r e a s e d a n g l e s t r a i n . A

s im ila r p r o b le m o f a

la r g e a n g l e s t r a i n a s s o c i a t e d w i t h a p l a n a r f o r m p r e v e n t s m o l e c u l e s o f t h e [

10 ] a n n u l e n e

is o ­

m e r w it h o n e tra n s d o u b le b o n d ( 6) f r o m b e in g a r o m a tic . A f t e r m a n y u n s u c c e s s fu l a tte m p ts o v e r m a n y y e a rs , in 1 9 6 5 [4 ]a n n u le n e ( o r c y c lo b u ta d ie n e ) w a s s y n t h e s iz e d b y R . P e t t i t a n d c o - w o r k e r s a t t h e U n i v e r s i t y o f T e x a s , A u s t i n . C y c lo b u t a d ie n e is a 4 n m o le c u le , n o t a 4 n +

2 m o le c u le , a n d , a s w e w o u l d e x p e c t, i t is a

h i g h l y u n s t a b l e c o m p o u n d a n d i t is a n tia r o m a tic ( s e e S e c t i o n 1 4 . 7 E ) :

C y c lo b u t a d ie n e o r [4 ] a n n u le n e ( a n t ia r o m a t ic )

[18]Annulene.

646 ri

lvce ud jSuoi v

Chapter 14

rPi ruoub il ce m iii

1i 4 ^2

Aromatic Compounds

H ■

U s in g th e p o ly g o n - a n d - c ir c le m e t h o d t o o u t lin e th e m o le c u la r o r b it a ls o f c y c lo b u ta d ie n e , e x p la in w h y c y c lo b u ta d ie n e is n o t a r o m a tic .

STRATEGY AND ANSWER

W e in s c r ib e a s q u a re in s id e a c ir c le w it h o n e c o r n e r a t th e b o t to m . — _L

A n tib o n d in g M O _L

N o n b o n d in g M O s

—

B o n d in g M O

W e s e e th a t c y c lo b u ta d ie n e , a c c o r d in g to t h is m o d e l, w o u l d h a v e h a v e a n u n p a ir e d e le c tr o n i n e a c h o f it s t w o n o n ­ b o n d in g m o le c u la r o r b it a ls . W e w o u ld , th e r e fo r e , n o t e x p e c t c y c lo b u ta d ie n e t o b e a r o m a tic .

14.7C NMR Spectroscopy: Evidence for Electron Delocalization in Aromatic Compounds T h e 1H N M R s p e c t r u m o f b e n z e n e c o n s i s t s o f a s i n g l e u n s p l i t s i g n a l a t

8

7 .2 7 . T h a t o n ly

a s in g le u n s p lit s ig n a l is o b s e r v e d is f u r t h e r p r o o f th a t a l l o f th e h y d r o g e n s o f b e n z e n e a re e q u i v a le n t . T h a t t h e s i g n a l o c c u r s a t r e l a t i v e l y h i g h f r e q u e n c y is , a s w e s h a l l s e e , c o m p e l l i n g e v id e n c e f o r t h e a s s e r t io n t h a t t h e p

e le c t r o n s o f b e n z e n e a r e d e l o c a l i z e d .

W e le a r n e d i n S e c tio n 9 .6 t h a t c ir c u la t io n s o f s e le c tr o n s o f C — H b o n d s c a u s e th e p r o ­ t o n s o f a lk a n e s t o b e s h ie ld e d f r o m

th e a p p lie d m a g n e tic f ie ld o f a n N M R

s p e c tro m e te r

a n d , c o n s e q u e n t ly , t h e s e p r o t o n s a b s o r b a t l o w e r f r e q u e n c y . W e s h a l l n o w e x p l a i n t h e h i g h f r e q u e n c y a b s o r p t i o n o f b e n z e n e p r o t o n s o n t h e b a s is o f d e s h ie ld in g c a u s e d b y c ir c u la tio n

o f th e p e le c tro n s o f b e n z e n e , a n d t h i s e x p l a n a t i o n , a s y o u w i l l s e e , r e q u i r e s t h a t t h e p e l e c ­ tr o n s b e d e lo c a liz e d . W h e n b e n z e n e m o le c u le s a re p la c e d i n th e p o w e r f u l m a g n e t ic f i e ld o f th e N M R

spec­

t r o m e t e r , e le c t r o n s c i r c u l a t e i n t h e d i r e c t i o n s h o w n i n F i g . 1 4 . 8 ; b y d o i n g s o , t h e y g e n e r ­ a te a r i n g c u r r e n t . ( I f y o u h a v e s t u d ie d p h y s i c s , y o u w i l l u n d e r s t a n d w h y t h e e le c t r o n s c ir c u la te in th is w a y .) •

T h e c ir c u la tio n o f p

e le c tr o n s i n b e n z e n e c r e a te s a n in d u c e d m a g n e t ic f i e ld th a t, a t

th e p o s it io n o f th e p ro to n s , re in fo rc e s th e a p p lie d m a g n e tic f ie ld . T h i s r e i n f o r c e ­ m e n t c a u s e s t h e p r o t o n s t o b e s t r o n g l y d e s h ie ld e d a n d t o h a v e a r e l a t i v e l y h i g h f r e ­ quency

(8

~ 7 ) a b s o r p tio n .

B y “ d e s h i e ld e d ” w e m e a n t h a t t h e p r o t o n s s e n s e t h e s u m o f t h e t w o f i e l d s , a n d , t h e r e f o r e , th e n e t m a g n e tic f ie ld s tr e n g th is g r e a te r th a n i t w o u ld h a v e b e e n in th e a b s e n c e o f th e in d u c e d f i e ld . T h is s tr o n g d e s h ie ld in g , w h i c h w e a t t r ib u t e t o a r i n g c u r r e n t c r e a te d b y th e

d e lo c a liz e d p e le c t r o n s , e x p l a i n s w h y a r o m a t i c p r o t o n s a b s o r b a t r e l a t i v e l y h i g h f r e q u e n c y . - In d u c e d m a g n e tic fie ld

P ro to n d e s h ie ld e d b y

H— -

in d u c e d fie ld

S f

V / I

A p p lie d m a g n e tic fie ld

Figure 14.8 The induced magnetic field of the p electrons of benzene deshields the benzene protons. Deshielding occurs because at the location of the protons the induced field is in the sam e direction as the applied field.

_

z ­

_

I

I

B0

/

I

I

I '

/ \\

I \

\

"

\

^~=r

/

V H

_

\

i C ir c u la tin g p e le c tr o n s

\

/

( r in g c u r r e n t)

647

14.7 Hückel's Rule: The 4n + 2 p Electron Rule

The deshielding of external aromatic protons that results from the ring current is one of the best pieces of physical evidence that we have for p-electron delocalization in aromatic rings. In fact, relatively high frequency proton absorption is often used as a criterion for assessing aromaticity in newly synthesized conjugated cyclic compounds. Not all aromatic protons have high frequency absorptions, however. The internal pro­ tons of large-ring aromatic compounds that have hydrogens in the center of the ring (in the p-electron cavity) absorb at unusually low frequency because they are highly shielded by the opposing induced magnetic field in the center of the ring (see Fig. 14.8). An example is [18]annulene (Fig. 14.9). The internal protons of [18]annulene absorb far upfield at 8 - 3.0, above the signal for tetramethylsilane (TMS); the external protons, on the other hand, absorb far downfield at 8 9.3. Considering that [18]annulene has 4n + 2 p electrons, this evidence provides strong support for p-electron delocalization as a criterion for aromaticity and for the predictive power of Huckel’s rule.

14.7D Aromatic Ions In addition to the neutral molecules that we have discussed so far, there are a number of monocyclic species that bear either a positive or a negative charge. Some of these ions show unexpected stabilities that suggest that they are aromatic ions. Huckel’s rule is helpful in accounting for the properties of these ions as well. We shall consider two examples: the cyclopentadienyl anion and the cycloheptatrienyl cation. Cyclopentadiene is not aromatic; however, it is unusually acidic for a hydrocarbon. (The pKa for cyclopentadiene is 16 and, by contrast, the pKa for cycloheptatriene is 36.) Because of its acidity, cyclopentadiene can be converted to its anion by treatment with moderately strong bases. The cyclopentadienyl anion, moreover, is unusually stable, and N M R spec­ troscopy shows that all five hydrogen atoms in the cyclopentadienyl anion are equivalent and absorb downfield. strong base

H

H

C yclopentadiene

H

Cyclopentadienyl anion

The orbital structure of cyclopentadiene (Fig. 14.10) shows why cyclopentadiene, itself, is not aromatic. Not only does it not have the proper number of p electrons, but the p elec­ trons cannot be delocalized about the entire ring because of the intervening sp3-hybridized — CH 2 — group with no available p orbital. On the other hand, if the — CH 2 — carbon atom becomes sp2 hybridized after it loses a proton (Fig. 14.10), the two electrons left behind can occupy the new p orbital that is pro­ duced. Moreover, this new p orbital can overlap with the p orbitals on either side of it and

Figure 14.10 Cyclopentadiene is not aromatic because it has only four p electrons and the sp 3hybridized carbon prevents com plete delocalization around the ring. Removal of a proton produces the cyclopentadienyl anion, which is aromatic because it has 6 p electrons and all of its carbon atom s have a p orbital.

H

H

H H

H

H H

H

H H

H

H

H

H H

H

Figure 14.9 [18]Annulene. The internal protons (red) are highly shielded and absorb at 8 - 3 .0 . The external protons (blue) are highly deshielded and absorb at 8 9.3.

648

Chapter 14 Aromatic Compounds give rise to a ring with six delocalized p electrons. Because the electrons are delocalized, all of the hydrogen atoms are equivalent, and this agrees with what N M R spectroscopy tells us. A calculated electrostatic potential map for cyclopentadienyl anion (Fig. 14.11) also shows the symmetrical distribution of negative charge within the ring, and the overall symmetry of the ring structure. Six, the number of p electrons in the cyclopentadienyl anion is, of course, a Huckel num­ ber (4n + 2, where n = 1).

r Figure 14.11 An electrostatic potential map of the cyclopentadienyl anion. The ion is negatively charged overall, of course, but regions with greatest negative potential are shown in red, and regions with least negative potential are in blue. The concentration of negative potential in the center of the top face and bottom face (not shown) indicates that the extra electron of the ion is involved in the aromatic p-electron system .

•

The cyclopentadienyl anion is, therefore, an aromatic anion, and the unusual acid­ ity of cyclopentadiene is a result of the unusual stability of its anion.

Cycloheptatriene (Fig. 14.12) (a compound with the common name tropylidene) has six p electrons. However, the six p electrons of cycloheptatriene cannot be fully delocalized because of the presence of the — CH 2 — group, a group that does not have an available p orbital (Fig. 14.12). When cycloheptatriene is treated with a reagent that can abstract a hydride ion, it is con­ verted to the cycloheptatrienyl (or tropylium) cation. The loss of a hydride ion from cycloheptatriene occurs with unexpected ease, and the cycloheptatrienyl cation is found to be unusually stable. The N M R spectrum of the cycloheptatrienyl cation indicates that all seven hydrogen atoms are equivalent. I f we look closely at Fig. 14.12, we see how we can account for these observations.

2

sp3 - hybridization here

sp - hybridization at all carbons allows de localization.

dis ru pts de localization.

-H:-

©

C ycloheptatriene

C ycloheptatrienyl cation

Figure 14.12 Cycloheptatriene is not aromatic, even though it has six p electrons, because it has an sp 3-hybridized carbon that prevents delocalization around the ring. Removal of a hydride (H: ) produces the cycloheptatrienyl cation, which is aromatic because all of its carbon atom s now have a p orbital, and it still has 6 p electrons.

H

H H C ycloheptatriene

Figure 14.13 An electrostatic potential map of the tropylium cation. The ion is positive overall, of course, but a region of relatively greater negative electrostatic potential can clearly be seen around the top face (and bottom face, though not shown) of the ring where electrons are involved in the p system of the aromatic ring.

C ycloheptatrienyl cation (tropylium cation)

As a hydride ion is removed from the — CH 2 — group of cycloheptatriene, a vacant p orbital is created, and the carbon atom becomes sp 2 hybridized. The cation that results has seven overlapping p orbitals containing six delocalized p electrons. The cycloheptatrienyl cation is, therefore, an aromatic cation, and all of its hydrogen atoms should be equivalent; again, this is exactly what we find experimentally. The calculated electrostatic potential map for cycloheptatrienyl (tropylium) cation (Fig. 14.13) also shows the symmetry of this ion. Electrostatic potential from the p electrons involved in the aromatic system is indicated by the yellow-orange color that is evenly dis­ tributed around the top face (and bottom face, though not shown) of the carbon framework. The entire ion is positive, of course, and the region of greatest positive potential is indi­ cated by blue around the periphery of the ion.

649

14.7 Hückel's Rule: The 4n + 2 p Electron Rule

1

1 V

S o lv e d rPi Ur owb ilCe m I I I 11 4 3

J U Iv C U

■

A p p ly th e p o ly g o n - a n d - c ir c le m e th o d to e x p la in w h y th e c y c lo p e n ta d ie n y l a n io n is a r o m a tic .

STRATEGY AND ANSWER

W e in s c r ib e a p e n ta g o n in s id e a c ir c le w it h o n e c o r n e r a t th e b o t to m a n d f in d th a t th e

e n e r g y le v e ls o f th e m o le c u la r o r b it a ls a re s u c h th a t th r e e m o le c u la r o r b ita ls a re b o n d in g a n d t w o a re a n t ib o n d in g :

Antibonding MOs

I

I 3l

C y c lo p e n t a d ie n y l a n io n h a s s ix p

Bonding MOs

e le c t r o n s , w h i c h is a H u c k e l n u m b e r , a n d t h e y f i l l a l l th e b o n d in g o r b it a ls .

T h e r e a r e n o u n p a ir e d e le c t r o n s a n d n o e le c t r o n s i n a n t ib o n d i n g o r b i t a l s . T h i s is w h a t w e w o u l d e x p e c t o f a n a r o m a tic io n .

A p p ly th e p o ly g o n - a n d - c ir c le m e th o d to c y c lo p e n ta d ie n y l c a tio n a n d e x p la in w h e th e r i t

R e v ie w P ro b le m 1 4 .4

w o u ld b e a r o m a tic o r n o t.

C y c lo p e n t a d ie n y l c a t io n A p p ly th e p o ly g o n - a n d - c ir c le m e th o d to th e c y c lo h e p t a t r ie n y l a n io n a n d c a tio n a n d e x p la in

R e v ie w P ro b le m 1 4 .5

w h e th e r e a c h w o u ld b e a r o m a tic o r n o t.

(b)

(a )

C y c lo h e p t a t r ie n y l

C y c lo h e p t a t r ie n y l

a n io n

c a t io n

1 , 3 , 5 - C y c l o h e p t a t r i e n e i s e v e n le s s a c i d i c t h a n 1 , 3 , 5 - h e p t a t r i e n e . E x p l a i n h o w t h i s e x p e r i ­

R e v ie w P ro b le m 1 4.6

m e n t a l o b s e r v a tio n m i g h t h e lp t o c o n f ir m y o u r a n s w e r to p a r t ( b ) o f th e p r e v io u s p r o b le m . W h e n 1 ,3 ,5 - c y c lo h e p t a t r ie n e r e a c ts w i t h o n e m o la r e q u iv a le n t o f b r o m in e a t 0 ° C , i t u n d e r ­ g o e s 1 ,6 a d d it io n . ( a ) W r i t e th e s tr u c tu r e o f t h is p r o d u c t . ( b ) O n h e a t in g , t h is 1 , 6 - a d d it io n p r o d u c t lo s e s H B r r e a d i ly to f o r m a c o m p o u n d w i t h th e m o le c u la r f o r m u la C

7H 7B r ,

c a lle d

tr o p y liu m b ro m id e . T r o p y l i u m b r o m i d e i s i n s o l u b l e i n n o n p o l a r s o lv e n t s b u t i s s o l u b l e i n w a t e r ; i t h a s a n u n e x p e c t e d l y h i g h m e l t i n g ( m p 2 0 3 ° C ) , a n d w h e n t r e a t e d w i t h s i l v e r n it r a t e , a n a q u e o u s s o l u t i o n o f t r o p y l i u m b r o m i d e g iv e s a p r e c i p i t a t e o f A g B r . W h a t d o t h e s e e x p e r ­ im e n t a l r e s u lts s u g g e s t a b o u t th e b o n d in g i n t r o p y liu m b r o m id e ?

14.7E Aromatic, Antiaromatic, and Nonaromatic Compounds •

A n a r o m a tic c o m p o u n d h a s its p

e le c t r o n s d e lo c a liz e d o v e r t h e e n t i r e r i n g a n d i t is

s ta b iliz e d b y t h e p - e l e c t r o n d e l o c a l i z a t i o n . A s w e h a v e s e e n , a g o o d w a y to d e t e r m in e w h e th e r th e p d e lo c a liz e d is t h r o u g h th e u s e o f N M R o f w h e th e r o r n o t th e p

e le c tr o n s o f a c y c l ic s y s te m a re

s p e c t r o s c o p y . I t p r o v i d e s d i r e c t p h y s i c a l e v id e n c e

e l e c t r o n s a r e d e l o c a li z e d .

R e v ie w P ro b le m 1 4 .7

1J

650

Chapter 14

Aromatic Compounds

B u t w h a t d o w e m e a n b y s a y in g t h a t a c o m p o u n d i s s t a b i l i z e d b y p - e l e c t r o n d e l o c a l ­ i z a t io n ? W e h a v e a n id e a o f w h a t t h is m e a n s f r o m

o u r c o m p a r is o n o f th e h e a t o f h y d r o ­

g e n a tio n o f b e n z e n e a n d t h a t c a lc u la te d f o r th e h y p o t h e t ic a l 1 ,3 ,5 - c y c lo h e x a t r ie n e . W e s a w th a t b e n z e n e — in w h ic h th e p

e le c t r o n s a r e d e l o c a l i z e d — i s m u c h m o r e s t a b le t h a n 1 , 3 , 5 -

c y c lo h e x a tr ie n e (a m o d e l in w h ic h th e p d if fe r e n c e

b e tw e e n

th e m

e le c t r o n s a r e n o t d e l o c a l i z e d ) . W e c a l l t h e e n e r g y

th e re s o n a n c e

e n e rg y

( d e lo c a liz a t io n

e n e r g y ) o r s ta b iliz a t io n

e n e rg y . I n o r d e r to m a k e s im ila r c o m p a r is o n s f o r o th e r a r o m a tic c o m p o u n d s , w e n e e d to c h o o s e p r o p e r m o d e ls . B u t w h a t s h o u ld th e s e m o d e ls b e ? O n e w a y t o e v a lu a t e w h e t h e r a c y c l i c c o m p o u n d i s s t a b i l i z e d b y d e l o c a l i z a t i o n o f p e le c ­ tr o n s th r o u g h its r in g is to c o m p a r e i t w it h a n o p e n - c h a in c o m p o u n d h a v in g th e s a m e n u m ­ ber o f p

e le c t r o n s . T h i s a p p r o a c h i s p a r t i c u l a r l y u s e f u l b e c a u s e i t f u r n i s h e s u s w i t h m o d e l s

n o t o n l y f o r a n n u le n e s b u t f o r a r o m a t ic c a tio n s a n d a n io n s , a s w e l l . ( C o r r e c t io n s n e e d to b e m a d e , o f c o u r s e , w h e n th e c y c l ic s y s te m is s tr a in e d .) T o u s e th is a p p ro a c h w e d o th e f o llo w in g : 1 . W e t a k e a s o u r m o d e l a l i n e a r c h a i n o f s p 2- h y b r i d i z e d a t o m s h a v i n g t h e s a m e n u m ­ ber o f p

e le c t r o n s a s o u r c y c l i c c o m p o u n d .

2 . T h e n w e im a g in e r e m o v in g a h y d r o g e n a to m f r o m e a c h e n d o f th e c h a in a n d jo in in g th e e n d s to f o r m a r in g . •

I f , b a s e d o n s o u n d c a l c u l a t i o n s o r e x p e r i m e n t s , t h e r i n g h a s lo w e r p - e l e c t r o n e n e r g y , t h e n t h e r i n g is a r o m a t i c .

•

I f t h e r i n g a n d t h e c h a i n h a v e t h e sa m e p - e l e c t r o n e n e r g y , t h e n t h e r i n g is n o n a r o m a t ic .

•

I f t h e r i n g h a s g r e a te r p - e l e c t r o n e n e r g y t h a n t h e o p e n c h a i n , t h e n t h e r i n g is a n t ia r o m a t ic .

T h e a c tu a l c a lc u la t io n s a n d e x p e r im e n t s u s e d i n d e t e r m in in g p - e l e c t r o n e n e r g ie s a re b e y o n d o u r s c o p e , b u t w e c a n s tu d y f o u r e x a m p le s th a t illu s t r a t e h o w

th is a p p ro a c h h a s

been used.

C y c lo b u ta d ie n e

F o r c y c lo b u ta d ie n e w e c o n s id e r th e c h a n g e i n p - e l e c t r o n e n e r g y f o r

t h e f o l l o w i n g h y p o th e tic a l t r a n s f o r m a t i o n :

ir-electron ------------------> energy increases

1,3-Butadiene 4 it electrons

C yclobutadiene 4 7t electro n s (antiarom atic)

C a l c u la t i o n s i n d i c a t e a n d e x p e r i m e n t s a p p e a r t o c o n f i r m

th a t th e p - e le c tr o n

e n e rg y o f

c y c l o b u t a d ie n e i s h i g h e r t h a n t h a t o f i t s o p e n - c h a i n c o u n t e r p a r t . T h u s c y c l o b u t a d ie n e is c l a s ­ s if ie d a s a n t ia r o m a t ic .

B enzene

H e r e o u r c o m p a r is o n is b a s e d o n th e f o l lo w in g h y p o t h e t ic a l tr a n s fo r m a tio n :

1,3,5-Hexatriene 6 tt electro n s

Benzene 6 w electron s (aromatic)

C a l c u la t i o n s i n d i c a t e a n d e x p e r i m e n t s c o n f i r m t h a t b e n z e n e h a s a m u c h l o w e r p - e l e c t r o n e n e r g y t h a n 1 , 3 , 5 - h e x a t r i e n e . B e n z e n e i s c l a s s i f i e d a s b e i n g a r o m a t i c o n t h e b a s is o f t h i s c o m p a r is o n as w e ll.

651

14.8 O ther Aromatic Compounds

C y c lo p e n ta d ie n y l A n io n

Here we use a linear anion for our hypothetical transformation: ir-electron > -----------------energy decreases

6

tt electrons

^

+

Cyclopentadienyl anion 6 tt electro n s (aromatic)

Both calculations and experiments confirm that the cyclic anion has a lower p-electron energy than its open-chain counterpart. Therefore the cyclopentadienyl anion is classified as aromatic. C y c lo o c ta te tr a e n e

For cyclooctatetraene we consider the following hypothetical trans­

formation: ir-electron > ------------------

H2

energy increases

8

tt electrons

Hypothetical planar cy clo o ctatetraen e 8 tt electro n s (antiarom atic)

Here calculations and experiments indicate that a planar cyclooctatetraene would have higher p-electron energy than the open-chain octatetraene. Therefore, a planar form of cyclooctatetraene would, if it existed, be antiarom atic. As we saw earlier, cyclooctatetraene is not planar and behaves like a simple cyclic polyene. 1

S o lv e d P ro b le m 1 4 .4 1

Calculations indicate that the p-electron energy decreases for the hypothetical transformaition from the allyl cation to the cyclopropenyl cation below. What does this indicate about the possible aromaticity o f the cyclopropenyl cation + ir-eiectron energy decreases STRATEGY AND ANSWER Because the p-electron energy of the cyclic cation is less th an that of the allyl cation, we can conclude that the cyclopropenyl cation would be aromatic. (See Review Problern 14.9 for more information on this cation.)

The cyclopentadienyl cation is apparently antiarom atic. Explain what this means in terms of the p-electron energies of a cyclic and an open-chain compound.

R eview P roblem 14.8

In 1967 R. Breslow (of Columbia University) and co-workers showed that adding SbCl5 to a solution of 3-chlorocyclopropene in CH 2 Cl2 caused the precipitation of a white solid with the composition C 3 H3 +SbCl6~. N M R spectroscopy of a solution of this salt showed that all of its hydrogen atoms were equivalent. (a) What new aromatic ion had Breslow and co-workers prepared? (b) How many 13C N M R signals would you predict for this ion?

R eview P roblem 14.9

14.8 O th e r A rom atic Compounds 14.8A Benzenoid Aromatic Compounds In addition to those that we have seen so far, there are many other examples of aromatic compounds. Representatives of one broad class of benzenoid aromatic compounds, called polycyclic aromatic hydrocarbons (PAH), are illustrated in Fig. 14.14.

652

Chapter 14

Aromatic Compounds

9 Figure 14.14 Benzenoid aromatic hydrocarbons. Som e polycyclic aromatic hydrocarbons (PAHs), such as dibenzo[a,(]pyrene, are carcinogenic. (See "The Chemistry of . . . Epoxides, Carcinogens, and Biological Oxidation" in Section 11.14.)

8

Benzo[a]pyrene C20H12

Pyrene C16H10

•

D ibenzo[a,/ ]pyrene C24H14

B e n z e n o id p o l y c y c l ic a r o m a t ic h y d r o c a r b o n s c o n s is t o f m o le c u le s h a v in g t w o o r m o r e b e n z e n e r in g s fu s e d to g e th e r.

A

c l o s e l o o k a t o n e e x a m p l e , n a p h t h a le n e , w i l l i l l u s t r a t e w h a t w e m e a n b y t h i s . A c c o r d i n g t o r e s o n a n c e t h e o r y , a m o l e c u l e o f n a p h t h a le n e c a n b e c o n s i d e r e d t o b e a

h y b r i d o f th r e e K e k u le s tr u c tu r e s . O n e o f th e s e K e k u le s tr u c tu r e s , th e m o s t im p o r t a n t o n e , i s s h o w n i n F i g . 1 4 . 1 5 . T h e r e a r e t w o c a r b o n a t o m s i n n a p h t h a le n e ( C 4 a a n d C

8a)

th a t a re

c o m m o n t o b o t h r i n g s . T h e s e t w o a t o m s a r e s a id t o b e a t t h e p o i n t s o f r in g f u s io n . T h e y d ir e c t a ll o f th e ir b o n d s to w a r d o th e r c a r b o n a to m s a n d d o n o t b e a r h y d r o g e n a to m s .

H

H

I li H \^ C ^ 8 a /C ^ /H C C C2 I || |

Figure 14.15 One Kekule structure for naphthalene.

S o lv e d P r o b le m

| H

or

| H

1 4 .5

H o w m a n y 13C N M R s i g n a l s w o u l d y o u e x p e c t f o r a c e n a p h t h y le n e ?

A cenaphthylene STRATEGY A ND ANSWER

A c e n a p h t h y le n e h a s a p la n e o f s y m m e t r y w h i c h m a k e s th e f iv e

c a r b o n a to m s o n th e l e f t ( a - e , a t r i g h t ) e q u iv a le n t to th o s e o n th e r ig h t . C a r b o n a to m s f a n d g a r e u n i q u e . C o n s e q u e n t l y , a c e n a p h t h y le n e s h o u l d g i v e s e v e n 13C N M R

s ig n a l s .

A cenaphthylene

14.8 Other Aromatic Compounds

H o w m a n y 13C N M R

s i g n a l s w o u l d y o u p r e d i c t f o r ( a ) n a p h t h a le n e , ( b ) a n t h r a c e n e , ( c )

653

R e v ie w P ro b le m 1 4 .1 0

p h e n a n th re n e , a n d ( d ) p y re n e ?

M o l e c u l a r o r b i t a l c a lc u l a t i o n s f o r n a p h t h a le n e b e g i n w i t h t h e m o d e l s h o w n i n F i g . 1 4 . 1 6 . T h e p o r b ita ls o v e r la p a r o u n d th e p e r ip h e r y o f b o t h r in g s a n d a c ro s s th e p o in ts o f r in g fu s io n . W h e n m o l e c u l a r o r b i t a l c a l c u l a t i o n s a r e c a r r i e d o u t f o r n a p h t h a le n e u s in g t h e m o d e l s h o w n i n F ig . 1 4 .1 6 , th e r e s u lts o f th e c a lc u la t io n s c o r r e la t e w e l l w i t h o u r e x p e r im e n t a l k n o w l e d g e o f n a p h t h a le n e . T h e c a l c u l a t i o n s i n d i c a t e t h a t d e l o c a l i z a t i o n o f t h e 1 0 p

e le c ­

tr o n s o v e r th e t w o r in g s p r o d u c e s a s tr u c tu r e w it h c o n s id e r a b ly lo w e r e n e r g y th a n th a t c a l­ c u l a t e d f o r a n y i n d i v i d u a l K e k u l e s t r u c t u r e . N a p h t h a le n e , c o n s e q u e n t ly , h a s a s u b s t a n t ia l re s o n a n c e e n e rg y . B a s e d o n w h a t w e k n o w

a b o u t b e n z e n e , m o r e o v e r , n a p h th a le n e ’ s te n ­

d e n c y t o r e a c t b y s u b s t i t u t i o n r a t h e r t h a n a d d i t i o n a n d t o s h o w o t h e r p r o p e r t i e s a s s o c ia t e d w i t h a r o m a t ic c o m p o u n d s is u n d e r s ta n d a b le .

A n th r a c e n e a n d p h e n a n t h r e n e ( F ig . 1 4 .1 4 ) a re is o m e r s . I n a n th r a c e n e th e th r e e r in g s a re fu s e d in a lin e a r w a y , a n d in p h e n a n th re n e th e y a re fu s e d s o as to p r o d u c e a n a n g u la r m o l­ e c u l e . B o t h o f t h e s e m o l e c u l e s a ls o s h o w l a r g e r e s o n a n c e e n e r g ie s a n d c h e m i c a l p r o p e r ­ tie s t y p i c a l o f a r o m a t ic c o m p o u n d s . P y r e n e ( F i g . 1 4 . 1 7 ) i s a ls o a r o m a t i c . P y r e n e i t s e l f h a s b e e n k n o w n f o r a l o n g t i m e ; a p y r e n e d e r iv a t iv e , h o w e v e r , h a s b e e n th e o b je c t o f r e s e a r c h t h a t s h o w s a n o t h e r in t e r e s t in g a p p l i c a t i o n o f H u c k e l ’ s r u le . T o u n d e r s ta n d t h is p a r t ic u la r r e s e a r c h , w e n e e d t o p a y s p e c ia l a t t e n t io n t o th e K e k u le s tr u c tu r e f o r p y r e n e ( F ig . 1 4 .1 7 ). T h e to ta l n u m b e r o f p b le

bonds =

16

p

e le c tr o n s ) .

S ix te e n

is

e le c tr o n s i n p y r e n e is 1 6

(8

dou­

a n o n - H u c k e l n u m b e r, b u t H u c k e l’s r u le

is

i n t e n d e d t o b e a p p l ie d o n ly t o m o n o c y c lic c o m p o u n d s a n d p y r e n e is c le a r ly t e t r a c y c lic . I f w e d is r e g a r d th e in t e r n a l d o u b le b o n d o f p y r e n e , h o w e v e r , a n d l o o k o n ly a t th e p e r ip h ­ e r y , w e s e e th a t th e p e r ip h e r y is a p la n a r r i n g w i t h 1 4 p

e le c t r o n s . T h e p e r i p h e r y is , i n f a c t ,

v e r y m u c h l i k e t h a t o f [ 1 4 ] a n n u l e n e . F o u r t e e n is a H u c k e l n u m b e r (4 n + 2 , w h e r e n =

3 ),

a n d o n e m ig h t th e n p r e d ic t th a t th e p e r ip h e r y o f p y r e n e w o u ld b e a r o m a tic b y it s e lf , in th e a b s e n c e o f th e in t e r n a l d o u b le b o n d .

[ 1 4 ] A n n u le n e

f r a n s - 1 5 , 1 6 - D im e t h y ld ih y d r o p y r e n e

T h i s p r e d i c t i o n w a s c o n f i r m e d w h e n V . B o e k e l h e i d e ( U n i v e r s i t y o f O r e g o n ) s y n t h e s iz e d ir a n s - 1 5 ,1 6 - d im e t h y ld ih y d r o p y r e n e a n d s h o w e d th a t i t is a r o m a tic .

Figure 14.17 One Kekulé structure for pyrene. The internal double bond is enclosed in a dotted circle for emphasis.

654

Chapter 14 Aromatic Compounds

R eview P roblem 14.11

jn addition to a signal downfield, the 1H N M R spectrum of irans-15,16-dimethyldihydropyrene has a signal far upfield at 8 - 4 .2 . Account for the presence of this upfield signal.

14.8B Nonbenzenoid Aromatic Compounds Naphthalene, phenanthrene, and anthracene are examples of benzenoid aromatic com­ pounds. On the other hand, the cyclopentadienyl anion, the cycloheptatrienyl cation, trans15,16-dimethyldihydropyrene, and the aromatic annulenes (except for [6 ]annulene) are classified as nonbenzenoid aromatic compounds. Another example of a nonbenzenoid aromatic hydrocarbon is the compound azulene. Azulene has a resonance energy of 205 kJ mol~ 1 There is substantial separation of charge between the rings in azulene, as is indicated by the electrostatic potential map for azulene shown in Fig. 14.18. Factors related to aromaticity account for this property of azulene (see Review Problem 14.12).

Figure 14.18 A calculated electrostatic potential map for azulene. (Red areas are more negative and blue areas are less negative.)

R eview P roblem 14.12

Azulene

Azulene has an appreciable dipole moment. Write resonance structures for azulene that explain this dipole moment and that help explain its aromaticity.

14.8C Fullerenes

©

The Nobel Prize in Chemistry was awarded in 1996 to Professors Curl, Kroto, and Smalley fo r their discovery of fullerenes.

In 1990 W. Kratschmer (Max Planck Institute, Heidelberg), D. Huffman (University of Arizona), and their co-workers described the first practical synthesis of C 6 o, a molecule shaped like a soccer ball and called buckminsterfullerene. Formed by the resistive heat­ ing of graphite in an inert atmosphere, C 6 0 is a member of an exciting new group of aro­ matic compounds called fullerenes. Fullerenes are cagelike molecules with the geometry of a truncated icosahedron or geodesic dome, named after the architect Buckminster Fuller, renowned for his development of structures with geodesic domes. The structure of C 6 0 and its existence had been established five years earlier, by H. W. Kroto (University of Sussex), R. E. Smalley and R. F. Curl (Rice University), and their co-workers. Kroto, Curl, and Smalley had found both C 6 0 and C 7 0 (Fig. 14.19) as highly stable components

Figure 14.19 The structures of Cgo and C7 0 . Reprinted with permission from Diederic, F., and Whetten, R. L. Accounts of Chemical Research, Vol. 25, pp. 119-126, 1992. Copyright 1992 by American Chemical Society.

14.9 Heterocyclic Aromatic Compounds

655

of a mixture of carbon clusters formed by laser-vaporizing graphite. Since 1990 chemists have synthesized many other higher and lower fullerenes and have begun exploring their interesting chemistry.

THE CHEMISTRY OF . . . N a n o tu b e s

Nanotubes are a relatively new class of carbon-based mate­ rials related to buckminsterfullerenes. A n a n o t u b e is a struc­ ture that looks as though it were formed by rolling a sheet of graphitelike carbon (a flat network of fused benzene rings resembling chicken wire) into the shape of a tube and cap­ ping each end with half of a buckyball. Nanotubes are very tough—about 100 times as strong as steel. Besides their potential as strengtheners for new composite materials, some nanotubes have been shown to act as electrical con­ ductors or semiconductors depending on their precise form. They are also being used as probe tips for analysis of DNA and proteins by atomic force microscopy (AFM). Many other applications have been envisioned for them as well including use as molecular-size test tubes or capsules for drug delivery.

a

n e tw o rk o f benzene rings, h ig h lig h te d in black on this scanning tu n ne ling m icroscopy (STM) im age, com prise th e wall o f a nanotube.

Like a geodesic dome, a fullerene is composed of a network of pentagons and hexagons. To close into a spheroid, a fullerene must have exactly 12 five-membered faces, but the num­ ber of six-membered faces can vary widely. The structure of C 6 0 has 20 hexagonal faces; C 7 0 has 25. Each carbon of a fullerene is sp2 hybridized and forms s bonds to three other carbon atoms. The remaining electron at each carbon is delocalized into a system of mol­ ecular orbitals that gives the whole molecule aromatic character. The chemistry of fullerenes is proving to be even more fascinating than their synthesis. Fullerenes have a high electron affinity and readily accept electrons from alkali metals to produce a new metallic phase— a “buckide” salt. One such salt, K3 C60, is a stable metallic crystal consisting of a face-centered-cubic structure of “buckyballs” with a potassium ion in between; it becomes a superconductor when cooled below 18 K. Fullerenes have even been synthesized that have metal atoms in the interior of the carbon atom cage.

14.9 Heterocyclic A rom atic Compounds Almost all of the cyclic molecules that we have discussed so far have had rings composed solely of carbon atoms. However, in many cyclic compounds an element other than carbon is present in the ring. •

Cyclic compounds that include an element other than carbon are called hetero­ cyclic compounds.

Heterocyclic molecules are quite commonly encountered in nature. For this reason, and because some of these molecules are aromatic, we shall now describe a few examples of heterocyclic aromatic compounds. Heterocyclic compounds containing nitrogen, oxygen, or sulfur are by far the most com­ mon. Four important examples are given here in their Kekule forms. These four compounds are all aromatic : Observe the following: •

Pyridine is electronically related to benzene.

656

Chapter 14 Aromatic Compounds

•

P y r r o le , fu r a n , a n d th io p h e n e a re r e la te d to th e c y c lo p e n t a d ie n y l a n io n .

4

P y r id in e

P y r r o le

F u ra n

T h io p h e n e

T h e n i t r o g e n a t o m s i n m o le c u l e s o f b o t h p y r i d i n e a n d p y r r o l e a r e s p 2 h y b r i d i z e d . I n p y r i ­ d in e ( F ig . 1 4 .2 0 ) th e s p 2- h y b r id iz e d n it r o g e n d o n a te s o n e b o n d in g e le c t r o n t o th e p

sys­

t e m . T h i s e l e c t r o n , t o g e t h e r w i t h o n e f r o m e a c h o f t h e f i v e c a r b o n a t o m s , g iv e s p y r i d i n e a s e x te t o f e le c tr o n s l i k e b e n z e n e . T h e t w o u n s h a r e d e le c tr o n s o f th e n it r o g e n o f p y r id i n e a re i n a n s p 2 o r b i t a l t h a t l i e s i n t h e s a m e p la n e a s t h e a t o m s o f t h e r i n g . T h i s s p 2 o r b i t a l d o e s n o t o v e r l a p w i t h t h e p o r b i t a l s o f t h e r i n g ( i t is , t h e r e f o r e , s a id t o b e o r th o g o n a l t o t h e p o r b i t a l s ) . T h e u n s h a r e d p a i r o n n i t r o g e n is n o t a p a r t o f t h e p

s y s t e m , a n d t h e s e e le c t r o n s

c o n f e r o n p y r id in e th e p r o p e r tie s o f a w e a k b a s e .

Figure 14.20 Pyridine is aromatic and weakly basic. Its nitrogen atom has an unshared electron pair in an sp2 orbital (shown in gray) that is not part of the aromatic system.

I n p y r r o l e ( F ig . 1 4 .2 1 ) th e e le c tr o n s a re a r r a n g e d d if f e r e n t ly . B e c a u s e o n l y f o u r p

e le c ­

t r o n s a r e c o n t r i b u t e d b y t h e c a r b o n a t o m s o f t h e p y r r o l e r i n g , t h e sp 2- h y b r i d i z e d n i t r o g e n m u s t c o n t r i b u t e t w o e l e c t r o n s t o g i v e a n a r o m a t i c s e x t e t . B e c a u s e t h e s e e le c t r o n s a r e a p a r t o f th e a r o m a t ic s e x te t, th e y a re n o t a v a ila b le f o r d o n a t io n t o a p r o t o n . T h u s , i n a q u e o u s s o l u t i o n , p y r r o l e i s n o t a p p r e c i a b l y b a s ic .

Figure 14.21 Pyrrole is aromatic but not basic. It does not have any unshared electron pairs. The electron pair on nitrogen is part of the aromatic system .

S o lv e d P r o b le m

1 4 .6

Non-basic nitr° 9 en

5

I m i d a z o l e ( a t r i g h t ) h a s t w o n i t r o g e n s . N 3 is r e l a t i v e l y b a s i c ( l i k e t h e n i t r o g e n o f p y r i d i n e ) . N 1 i s r e l a t i v e l y n o n b a s ic ( l i k e t h e n i t r o g e n o f p y r r o l e ) . E x p la i n th e d if f e r e n t b a s ic it ie s o f th e s e t w o n it r o g e n s .

3

I

=N — H

.N

^ ;;/

V

2

1

Basic nitrogen

14.10 Aromatic Compounds in Biochemistry

STRATEGY AND ANSWER a p a r t o f th e p

657

W h e n i m id a z o le a c c e p ts a p r o t o n a t N 3 th e e le c t r o n p a i r t h a t a c c e p ts th e p r o t o n is n o t

s y s te m o f s ix e le c tr o n s th a t m a k e s i m id a z o le a r o m a tic . C o n s e q u e n tly , th e c o n ju g a t e b a s e t h a t is

f o r m e d i s s t i l l a r o m a t i c ( i t is a n a r o m a t i c c a t i o n ) a n d r e t a in s i t s r e s o n a n c e e n e r g y o f s t a b i l i z a t i o n .

, N ; H

'

(Aromatic)

(Aromatic)

O n th e o th e r h a n d , i f im id a z o le w e r e to a c c e p t a p r o t o n a t N 1 th e r e s u lt in g io n ( w h ic h is n o t fo r m e d ) w o u ld n o t b e a r o m a tic a n d w o u ld h a v e m u c h g r e a te r p o t e n t ia l e n e r g y ( it s re s o n a n c e s ta b iliz a t io n w o u ld b e lo s t ) . H e n c e , N 1 i s n o t a p p r e c i a b l y b a s ic .

H ~ X = , H

N

+

: X

:

'H

(Aromatic) 6 p electro n s

(Non-aromatic) 4 p electro n s

F u r a n a n d th io p h e n e a re s t r u c t u r a lly q u it e s im ila r t o p y r r o le . T h e o x y g e n a to m i n fu r a n a n d th e s u lf u r a to m in th io p h e n e a re s p 2 h y b r id iz e d . I n b o t h c o m p o u n d s th e p o r b it a l o f th e h e t e r o a to m d o n a te s t w o e le c tr o n s t o th e p

s y s te m . T h e o x y g e n a n d s u lf u r a to m s o f fu r a n

a n d t h i o p h e n e c a r r y a n u n s h a r e d p a i r o f e le c t r o n s i n a n s p 2 o r b i t a l ( F i g . 1 4 . 2 2 ) t h a t i s o r t h o g ­ o n a l to th e p

s y s te m .

Furan

T hiophene

Figure 14.22 Furan and thiophene are aromatic. In each case, the heteroatom provides a pair of electrons to the aromatic system , but each also has an unshared electron pair in an sp2 orbital that is not part of the aromatic system .

14.10 A rom atic Com pounds in Biochemistry C o m p o u n d s w i t h a r o m a t i c r i n g s o c c u p y n u m e r o u s a n d i m p o r t a n t p o s i t i o n s i n r e a c t io n s t h a t o c c u r i n l i v i n g s y s t e m s . I t w o u l d b e i m p o s s i b l e t o d e s c r i b e t h e m a l l i n t h i s c h a p t e r . W e s h a ll, h o w e v e r , p o i n t o u t a f e w e x a m p l e s n o w a n d w e s h a l l s e e o t h e r s la t e r . T w o a m i n o a c id s n e c e s s a r y f o r p r o t e i n s y n t h e s is c o n t a i n t h e b e n z e n e r i n g : O

O

658

Chapter 14 Aromatic Compounds A

t h i r d a r o m a t i c a m i n o a c id , t r y p t o p h a n , c o n t a in s a b e n z e n e r i n g f u s e d t o a p y r r o l e r i n g .

( T h is a r o m a t ic r i n g s y s te m is c a lle d a n in d o le s y s te m , s e e S e c tio n 2 0 . 1 B . ) O

4 Y

\ j r

2

\ ' N

3

5

O

N H 3 + '

7

H

H T r y p to p h a n

In d o le

I t a p p e a rs th a t h u m a n s , b e c a u s e o f th e c o u rs e o f e v o lu tio n , d o n o t h a v e th e b io c h e m i­ c a l a b i l i t y t o s y n t h e s iz e t h e b e n z e n e r i n g . A s a r e s u l t , p h e n y l a l a n i n e a n d t r y p t o p h a n d e r i v ­ a tiv e s

a re

e s s e n t ia l

in

th e

hum an

d ie t.

B ecause

ty r o s in e

can

be

s y n t h e s iz e d

fro m

p h e n y l a l a n i n e i n a r e a c t i o n c a t a l y z e d b y a n e n z y m e k n o w n a s p h e n y la la n in e h y d ro x y la s e , i t i s n o t e s s e n t ia l i n t h e d i e t a s l o n g a s p h e n y l a l a n i n e i s p r e s e n t .

Dairy products, beans, fish, m eat, and poultry are dietary sources of the essential amino acids.

H e t e r o c y c lic

a r o m a tic

com pounds

a re

a ls o

p re s e n t in

m any

b io c h e m ic a l

s y s te m s .

D e r i v a t i v e s o f p u r i n e a n d p y r i m i d i n e a r e e s s e n t ia l p a r t s o f D N A a n d R N A :

7

6 1

-

4

- N

3

5 8

2

4

2

6 ' N g

H

I P y r im id in e

P u r in e

D N A is th e m o le c u le r e s p o n s ib le f o r th e s to ra g e o f g e n e tic in f o r m a t io n , a n d R N A is p r o m i ­ n e n t l y i n v o l v e d i n t h e s y n t h e s is o f e n z y m e s a n d o t h e r p r o t e i n s ( C h a p t e r 2 5 ) .

R e v ie w P ro b le m 1 4 .1 3

(a ) T h e —

SH

g r o u p i s s o m e t im e s c a l l e d t h e m e rc a p to g ro u p .

6 -M

e r c a p t o p u r in e is u s e d in

th e t r e a tm e n t o f a c u te le u k e m ia . W r i t e it s s tr u c tu r e . ( b ) A l lo p u r i n o l , a c o m p o u n d u s e d to tr e a t g o u t , is

6 - h y d r o x y p u r in e .

W r it e its s tru c tu re .

N icotin am ide adenine dinucleotide,

o n e o f th e m o s t im p o r t a n t c o e n z y m e s ( S e c tio n

2 4 . 9 ) i n b io l o g i c a l o x id a t io n s a n d r e d u c t io n s , in c lu d e s b o t h a p y r id i n e d e r iv a t iv e ( n ic o t i ­ n a m id e ) a n d a p u r in e d e r iv a t iv e ( a d e n in e ) i n it s s tr u c tu r e . I t s f o r m u la is s h o w n i n F ig . 1 4 .2 3 as N A D + , th e o x id iz e d f o r m t h a t c o n t a in s th e p y r id i n i u m a r o m a t ic r in g . T h e r e d u c e d f o r m o f th e c o e n z y m e is N A D H , i n w h ic h th e p y r id in e r in g is n o lo n g e r a r o m a tic d u e to p r e s ­ e n c e o f a n a d d i t i o n a l h y d r o g e n a n d t w o e le c t r o n s i n t h e r i n g .

Figure 14.23 Nicotinamide adenine dinucleotide (NAD+).

659

14.10 Aromatic Compounds in Biochemistry

A k e y r o l e o f N A D + i n m e t a b o l i s m is t o s e r v e a s a c o e n z y m e f o r g ly c e r a ld e h y d e - 3 - p h o s p h a t e d e h y d r o g e n a s e ( G A P D H ) i n g l y c o l y s i s , t h e p a t h w a y b y w h i c h g lu c o s e i s b r o k e n d o w n f o r e n e r g y p r o d u c t i o n . I n t h e r e a c t i o n c a t a ly z e d b y G A P D H ( F i g . 1 4 . 2 4 ) , t h e a ld e h y d e g r o u p o f g l y c e r a l d e h y d e - 3 - p h o s p h a t e ( G A P ) is o x i d i z e d t o a c a r b o x y l g r o u p ( i n c o r p o r a t e d a s a p h o s p h o r ic a n h y d r id e ) i n 1 ,3 - b is p h o s p h o g ly c e r a t e ( 1 , 3 - B P G ) . C o n c u r r e n t ly , th e a r o m a tic p y r id in iu m

r in g o f N A D +

is r e d u c e d to its h ig h e r e n e r g y f o r m , N A D H . O n e o f th e w a y s

th e c h e m ic a l e n e r g y s to r e d i n th e n o n a r o m a t ic r i n g o f N A D H

is u s e d is i n th e m it o c h o n ­

d r ia f o r th e p r o d u c t io n o f A T P , w h e r e c y t o c h r o m e e le c tr o n t r a n s p o r t a n d o x id a t iv e p h o s ­ p h o r y l a t i o n t a k e p la c e . T h e r e , r e le a s e o f c h e m i c a l e n e r g y f r o m N A D H b y o x i d a t i o n t o t h e m o r e s t a b le a r o m a t i c f o r m N A D +

( a n d a p r o t o n ) is c o u p le d w it h th e p u m p in g o f p r o to n s

a c r o s s t h e i n n e r m i t o c h o n d r i a l m e m b r a n e . A n e l e c t r o c h e m i c a l g r a d i e n t is c r e a t e d a c r o s s th e m it o c h o n d r ia l m e m b r a n e , w h ic h

d r iv e s

th e

s y n t h e s is o f A T P

by

th e

enzym e

ATP

s y n th a s e .

O

H

O

O —

P

=

O

O­ I O 9 P= O I OH

O

O

glyceraldehyde-3-phosphate dehydrogenase (GAPDH)

O — P— O -

0

O H

O —

P =

o

O

h

O-

1

o

O-

-

G ly c e r a ld e h y d e -

1 ,3 - B is p h o s p h o g ly c e r a t e

3 - p h o s p h a t e (G A P )

( 1 ,3 -B P G )

\ ^ E le c tr o n ^ ^ / transport chain

R

NAD+ ( a r o m a t ic )

R

NADH ( n o t a r o m a t ic )

Figure 14.24 NAD+, as the coenzym e in glyceraldehyde-3-phosphate dehydrogenase (GAPDH), is used to oxidize glyceraldehyde-3-phosphate (GAP) to 1,3-bisphosphoglycerate during the degradation of glucose in glycolysis. O ne of the ways that NADH can be reoxidized to NAD+ is by the electron transport chain in mitochondria, where, under aerobic conditions, rearomatization of NADH helps to drive ATP synthesis.

T h e c h e m ic a l e n e rg y s to re d in N A D H

i s u s e d t o b r i n g a b o u t m a n y o t h e r e s s e n t ia l b i o ­

c h e m i c a l r e a c t io n s a s w e l l . N A D H i s p a r t o f a n e n z y m e c a l l e d l a c t a t e d e h y d r o g e n a s e t h a t r e d u c e s t h e k e t o n e g r o u p o f p y r u v i c a c i d t o t h e a l c o h o l g r o u p o f l a c t i c a c id . H e r e , t h e n o n a r o m a t ic r in g o f N A D H

is c o n v e r t e d t o t h e a r o m a t i c r i n g o f N A D + . T h i s p r o c e s s is

i m p o r t a n t i n m u s c le s o p e r a t i n g u n d e r o x y g e n - d e p l e t e d c o n d i t i o n s ( a n a e r o b i c m e t a b o l i s m ) , w h e r e r e d u c t io n o f p y r u v i c a c id t o l a c t ic a c id b y N A D H

s e rv e s to re g e n e ra te N A D +

th a t

i s n e e d e d t o c o n t i n u e g l y c o l y t i c s y n t h e s is o f A T P :

H

H

O

O

NADH

NAD+ N H „

N H

O

O

H 3 C .

H 3 C . O H

O

P y r u v ic a c id

lactate dehydrogenase (regenerates NAD+ in muscle under anaerobic conditions)

■ >H

O H

O H

L a c t ic a c id

Y e a s ts g r o w i n g u n d e r a n a e r o b i c c o n d i t i o n s ( f e r m e n t a t i o n ) a ls o h a v e a p a t h w a y f o r r e g e n ­ e r a t in g N A D + f r o m N A D H . U n d e r o x y g e n - d e p r i v e d c o n d i t i o n s , y e a s t s c o n v e r t p y r u v i c a c id

660

Chapter 14 Aromatic Compounds t o a c e t a ld e h y d e b y d e c a r b o x y l a t i o n ( C O

2 is

r e le a s e d , ( s e e “ T h e C h e m i s t r y o f . . . T h i a m i n e ”

i n W ile y P L U S ); t h e n N A D H i n a l c o h o l d e h y d r o g e n a s e r e d u c e s a c e t a ld e h y d e t o e t h a n o l . A s i n o x y g e n - s t a r v e d m u s c le s , t h i s p a t h w a y o c c u r s f o r t h e p u r p o s e o f r e g e n e r a t in g N A D + n e e d e d t o c o n t i n u e g l y c o l y t i c A T P s y n t h e s is :

NAD+

NADH O

CO H

2

H 3C

3C

H O H

pyruvate decarboxylase (with thiamine as a coenzyme)

O

P y r u v ic a c id

3C

Y

alcohol dehydrogenase (regenerates NAD+ in yeast under anaerobic conditions)

O

A c e t a ld e h y d e

/ H >

H O H

E th a n o l

A l t h o u g h m a n y a r o m a t i c c o m p o u n d s a r e e s s e n t ia l t o l i f e , o t h e r s a r e h a z a r d o u s . M a n y a re q u it e t o x ic , a n d s e v e ra l b e n z e n o id c o m p o u n d s , in c lu d in g b e n z e n e it s e lf , a re c a r c in o ­ g e n ic . T w o o th e r e x a m p le s a re b e n z o [ a ] p y r e n e a n d 7 - m e t h y lb e n z [ a ] a n t h r a c e n e : 2 1

r

il

ì

11

10^ '''''-

Y

Y

9

The mechanism fo r the carcino­ genic effects o f compounds like benzo[a]pyrene was discussed in "The Chemistry o f . . . Epoxides, Carcinogens, and Biological O xidation" in Section 11.14.

5

7|

C H

H e lp f u l H i n t

u

Y ^ f

Y -J Y Y e

B e n z o [a ]p y re n e

I3

12

a 3

7 - M e t h y lb e n z [ a ] a n t h r a c e n e

T h e h y d r o c a r b o n b e n z o [a ]p y r e n e h a s b e e n fo u n d in c ig a r e tt e s m o k e a n d in th e e x h a u s t fro m

a u t o m o b i l e s . I t i s a ls o f o r m e d i n t h e i n c o m p l e t e c o m b u s t i o n o f a n y f o s s i l f u e l . I t is

f o u n d o n c h a r c o a l - b r o i l e d s te a k s a n d e x u d e s f r o m

a s p h a lt s tre e ts o n a h o t s u m m e r d a y .

B e n z o [ a ] p y r e n e is s o c a r c in o g e n ic th a t o n e c a n in d u c e s k in c a n c e r s i n m ic e w i t h a lm o s t t o t a l c e r t a in t y s i m p l y b y s h a v in g a n a r e a o f th e b o d y o f th e m o u s e a n d a p p ly in g a c o a t in g o f b e n z o [a ]p y r e n e .

14.11 Spectroscopy o f A rom atic Compounds 14.11A 1H NMR Spectra •

T h e r in g h y d r o g e n s o f b e n z e n e d e r iv a tiv e s a b s o r b d o w n f ie ld in th e r e g io n b e tw e e n S 6 .0 a n d S 9 .5 .

I n S e c tio n 1 4 .7 C w e f o u n d t h a t a b s o r p tio n ta k e s p la c e f a r d o w n f ie ld b e c a u s e a r i n g c u r r e n t g e n e r a te d i n th e b e n z e n e r i n g c r e a te s a m a g n e t ic f ie ld , c a lle d “ th e in d u c e d f i e ld , ” w h ic h r e in f o r c e s th e a p p lie d m a g n e t ic f i e ld a t th e p o s it io n o f th e p r o t o n s o f th e r in g . T h is r e in ­ f o r c e m e n t c a u s e s t h e p r o t o n s o f b e n z e n e t o b e h i g h l y d e s h i e ld e d . W e a ls o l e a r n e d i n S e c t i o n 1 4 . 7 C t h a t i n t e r n a l h y d r o g e n s o f l a r g e - r i n g a r o m a t i c c o m ­ p o u n d s s u c h a s [ 1 8 ]a n n u le n e , b e c a u s e o f t h e ir p o s it io n , a re h ig h l y s h ie ld e d b y th is in d u c e d f i e l d . T h e y t h e r e f o r e a b s o r b a t u n u s u a l l y l o w f r e q u e n c y , o f t e n a t n e g a t i v e d e l t a v a lu e s .

14.11B 13C NMR Spectra •

T h e c a r b o n a to m s o f b e n z e n e r in g s g e n e r a lly a b s o r b i n th e S 1 0 0 - 1 7 0 r e g io n o f 13C N M R s p e c t r a .

F i g u r e 1 4 . 2 5 g iv e s t h e b r o a d b a n d p r o t o n - d e c o u p l e d 13C N M R s p e c t r u m o f 4 - N , N - d i e t h y l a m in o b e n z a ld e h y d e a n d p e r m i t s a n e x e r c is e i n m a k i n g 13C a s s ig n m e n t s o f a c o m p o u n d w i t h b o t h a r o m a tic a n d a lip h a t ic c a r b o n a to m s .

14.11 Spectroscopy of Aromatic Compounds

5C (ppm) Figure 14.25 The broadband proton-decoupled 13C NMR spectrum of 4-N,Ndiethylam inobenzaldehyde. DEPT information and carbon assignm ents are shown by each peak.

The DEPT spectra (not given to save space) show that the signal at 8 45 arises from a CH 2 group and the one at 8 13 arises from a CH 3 group. This allows us to assign these two signals immediately to the two carbons of the equivalent ethyl groups. The signals at 8 126 and 8 153 appear in the DEPT spectra as carbon atoms that do not bear hydrogen atoms and are assigned to carbons b and e (see Fig. 14.25). The greater elec­ tronegativity of nitrogen (when compared to carbon) causes the signal from e to be further downfield (at 8 153). The signal at 8 190 appears as a CH group in the DEPT spectra and arises from the carbon of the aldehyde group. Its chemical shift is the most downfield of all the peaks because of the great electronegativity of its oxygen and because the second resonance structure below contributes to the hybrid. Both factors cause the electron den­ sity at this carbon to be very low, and, therefore, this carbon is strongly deshielded. :O:_

'O'

A .H

H

R e s o n a n c e c o n t r i b u t o r s f o r a n a ld e h y d e g r o u p

This leaves the signals at 8 112 and 8 133 and the two sets of carbon atoms of the ben­ zene ring labeled c and d to be accounted for. Both signals are indicated as CH groups in the DEPT spectra. But which signal belongs to which set of carbon atoms? Here we find another interesting application of resonance theory. I f we write resonance structures A - D involving the unshared electron pair of the amino group, we see that contributions made by B and D increase the electron density at the set of carbon atoms labeled d:

(d)

H

^ O

B

H

^ O

C

H

O D

661

662

Chapter 14

Aromatic Compounds

O n th e o th e r h a n d , w r it in g s tru c tu re s E - H

i n v o l v i n g t h e a ld e h y d e g r o u p s h o w s u s t h a t c o n ­

t r i b u t i o n s m a d e b y F a n d H d e c r e a s e t h e e le c t r o n d e n s i t y a t t h e s e t o f c a r b o n a t o m s l a b e l e d c:

(d) (c)

H '

'O , H

( O t h e r re s o n a n c e s tr u c tu r e s a re p o s s ib le b u t a re n o t p e r t in e n t t o th e a r g u m e n t h e re .) In c r e a s in g th e e le c t r o n d e n s it y a t a c a r b o n s h o u ld in c r e a s e it s s h ie ld in g a n d s h o u ld s h if t i t s s i g n a l u p f i e l d . T h e r e f o r e , w e a s s ig n t h e s i g n a l a t

8

1 1 2 to th e s e t o f c a r b o n a to m s la b e le d

d . C o n v e r s e ly , d e c r e a s i n g t h e e l e c t r o n d e n s i t y a t a c a r b o n s h o u l d s h i f t i t s s i g n a l d o w n f i e l d , s o w e a s s ig n t h e s i g n a l a t

8

1 3 3 to th e s e t la b e le d c.

C a r b o n - 1 3 s p e c tr o s c o p y c a n b e e s p e c ia lly u s e f u l i n r e c o g n iz in g a c o m p o u n d w i t h a h ig h d e g r e e o f s y m m e tr y . T h e f o l lo w in g S o lv e d P r o b le m illu s t r a t e s o n e s u c h a p p lic a t io n .

S o lv e d P r o b le m

1 4 .7

T h e b r o a d b a n d p r o t o n - d e c o u p l e d 13C

s p e c tru m

g iv e n i n F ig . 1 4 .2 6 is o f a tr ib r o m o b e n z e n e ( C

t r ib r o m o b e n z e n e is it?

Sc (ppm) Figure 14.26

ANSWER

The broadband proton-decoupled 13C NMR spectrum of a tribrom obenzene.

T h e r e a re th r e e p o s s ib le tr ib r o m o b e n z e n e s :

Br

1 ,2 ,3 - T r ib r o m o b e n z e n e

Br

1 ,2 ,4 - T r ib r o m o b e n z e n e

Br

1 ,3 ,5 - T r ib r o m o b e n z e n e

6H 3B r 3) .

W h ic h

663

14.11 Spectroscopy of Aromatic Compounds

O u r s p e c t r u m ( F ig . 1 4 .2 6 ) c o n s is ts o f o n l y t w o s ig n a ls , in d ic a t in g th a t o n l y t w o d if f e r e n t ty p e s o f c a r b o n a to m s a re p r e s e n t in th e c o m p o u n d . O n ly 1 ,3 ,5 - tr ib r o m o b e n z e n e h a s a d e g re e o f s y m m e tr y s u c h th a t i t w o u ld g iv e o n ly t w o s i g n a l s , a n d , t h e r e f o r e , i t i s t h e c o r r e c t a n s w e r . 1 , 2 , 3 - T r i b r o m o b e n z e n e w o u l d g i v e f o u r 13C s i g n a l s a n d 1 , 2 , 4 tr ib r o m o b e n z e n e w o u ld g iv e s ix .

E x p la in h o w

13C N M R

s p e c tr o s c o p y c o u ld b e u s e d to d is t in g u is h th e

ortho-, m eta-,

and

p a r a - d ib r o m o b e n z e n e is o m e r s o n e f r o m a n o th e r.

14.11C Infrared Spectra of Substituted Benzenes B e n z e n e d e r iv a tiv e s g iv e c h a r a c t e r is t ic C — H s tr e tc h in g p e a k s n e a r 3 0 3 0 c m

-1

( T a b le 2 . 7 ) .

S tr e tc h in g m o t io n s o f th e b e n z e n e r in g c a n g iv e a s m a n y a s f o u r b a n d s i n th e 1 4 5 0 - 1 6 0 0 cm

-1

r e g io n , w it h t w o p e a k s n e a r 1 5 0 0 a n d 1 6 0 0 c m

-1

b e in g s tro n g e r.

A b s o r p t io n p e a k s in th e 6 8 0 - 8 6 0 - c m - 1 r e g io n f r o m

o u t - o f - p la n e C — H b e n d in g c a n

o f t e n ( b u t n o t a lw a y s ) b e u s e d t o c h a r a c t e r iz e th e s u b s t it u t io n p a t te r n s o f b e n z e n e c o m ­ p o u n d s ( T a b le 1 4 . 1 ) . M o n o s u b s t i t u t e d b e n z e n e s g i v e t w o v e r y s t r o n g p e a k s , b e t w e e n 6 9 0 and 710 cm

-1

a n d b e tw e e n 7 3 0 a n d 7 7 0 c m - 1 .

Infrared Absorptions in the 6 8 0 -8 6 0 -c m 11

12

13

1

Regiona 14

15

/xm

as, strong; vs, very strong.

O r t h o - d is u b s t i t u t e d b e n z e n e s s h o w a s tr o n g a b s o r p tio n p e a k b e tw e e n 7 3 5 a n d 7 7 0 cm

-1

t h a t a r is e s f r o m b e n d i n g m o t i o n s o f t h e C — H b o n d s . M e t a - d i s u b s t i t u t e d b e n z e n e s

s h o w tw o p e a k s : o n e s tro n g p e a k b e tw e e n 6 8 0 a n d 7 2 5 c m

- 1

a n d o n e v e r y s tro n g p e a k

b e t w e e n 7 5 0 a n d 8 1 0 c m - 1 . P a r a - d i s u b s t it u t e d b e n z e n e s g iv e a s in g le v e r y s tr o n g a b s o r p ­ tio n b e tw e e n 8 0 0 a n d 8 6 0 c m - 1 .

R e v ie w P r o b le m

1 4 .1 4

664

Chapter 14 Aromatic Compounds

R e v ie w P r o b le m

1 4 .1 5

F o u r b e n z e n o id c o m p o u n d s , a ll w it h th e f o r m u la C

7H 7B r ,

g a v e th e f o llo w in g I R p e a k s in

th e 6 8 0 - 8 6 0 - c m - 1 r e g io n : A , 740 cm B , 800 cm

-1 -1

( s tro n g )

C , 680 cm

( v e r y s tro n g )

D , 693 c m

-1 -1

( s tro n g ) a n d 7 6 0 c m

-1

( v e r y s tr o n g )

( v e r y s tro n g ) a n d 7 6 5 c m

-1

( v e r y s tro n g )

P ro p o s e s tru c tu re s f o r A , B , C , a n d D .

14.11D Ultraviolet-Visible Spectra of Aromatic Compounds T h e c o n j u g a t e d p e le c t r o n s o f a b e n z e n e r i n g g i v e c h a r a c t e r i s t i c u l t r a v i o l e t a b s o r p t i o n s t h a t in d ic a te th e p re s e n c e o f a b e n z e n e r in g i n a n u n k n o w n c o m p o u n d . O n e a b s o r p tio n b a n d o f m o d e ra te

in te n s ity

2 5 0 -2 7 5 -n m

o c c u rs n e a r 2 0 5 n m

and

a n o t h e r , le s s

r a n g e . C o n ju g a t io n o u t s id e th e b e n z e n e r i n g

in t e n s e b a n d

a p p e a rs

in

th e

le a d s t o a b s o r p t i o n s a t o t h e r

w a v e le n g th s .

THE CHEMISTRY OF . . . S u n s c r e e n s ( C a t c h i n g t h e S u n 's R a y s a n d W h a t H a p p e n s t o T h e m )

The use of sunscreens in recent years has increased due to heightened concern over the risk of skin cancer and other conditions caused by exposure to UV radiation. In DNA, for example, UV radiation can cause adjacent thymine bases to form mutagenic dimers. Sunscreens afford protection from UV radia­ tion because they contain aro­ matic molecules that absorb energy in the UV region of the electromagnetic spectrum. Absorption of UV radiation by these molecules promotes p and nonbonding electrons to higher energy levels (Section 13.9C), after which the energy is dissipated by relaxation through molecular vibration. In essence, the UV radiation is converted to A UV-A and UV-B sunscreen heat (IR radiation). p ro d u ct w hose active Sunscreens are classified ingredients are octyl according to the portion of the 4 -m ethoxycinnam ate and UV spectrum where their maxi­ 2 -hydroxy-4-m ethoxymum absorption occurs. Three benzophenone (oxybenzone). regions of the UV spectrum are

typically discussed. The region from 320 to 400 nm is called UV-A, the region from 280 to 320 nm is called UV-B, and the region from 100 to 280 nm is called UV-C. The UV-C region is potentially the most dangerous because it encompasses the shortest UV wavelengths and is therefore of the highest energy. However, ozone and other components in Earth's atmosphere absorb UV-C wavelengths, and thus we are pro­ tected from radiation in this part of the spectrum so long as Earth's atmosphere is not compromised further by ozonedepleting pollutants. Most of the UV-A and some of the UV­ B radiation passes through the atmosphere to reach us, and it is against these regions of the spectrum that sunscreens are formulated. Tanning and sunburn are caused by UV-B radiation. Risk of skin cancer is primarily associated with UV­ B radiation, although some UV-A wavelengths may be important as well. The specific range of protection provided by a sunscreen depends on the structure of its UV-absorbing groups. Most sunscreens have structures derived from the following par­ ent compounds: p-aminobenzoic acid (PABA), cinnamic acid (3-phenylpropenoic acid), benzophenone (diphenyl ketone), and salicylic acid (o-hydroxybenzoic acid). The structures and Amax for a few of the most common sunscreen agents are given below. The common theme among them is an aro­ matic core in conjugation with other functional groups.

O

O

O'

MeO

Me2N Octyl 4-M,M-dimethylaminobenzoate (Padimate O), Amax 310 nm

2-Ethylhexyl 4-methoxycinnamate (Parsol MCX), Amax 310 nm

14.11E Mass Spectra of Aromatic Compounds The major ion in the mass spectrum of an alkyl-substituted benzene is often m/z 91 (C 6 H5 CH2+), resulting from cleavage between the first and second carbons of the alkyl chain attached to the ring. The ion presumably originates as a benzylic cation that rearranges to a tropylium cation (C 7 H7+, Section 14.7D). Another ion frequently seen in mass spectra of monoalkylbenzene compounds is m/z 77, corresponding to C 6 H5+.

Key Terms and Concepts The key terms and concepts that are highlighted in b o l d , b l u e t e x t within the chapter are defined in the glossary (at the back of the book) and have hyperlinked definitions in the accompanying WileyPLUS course (www.wileyplus.com).

PL US

Problems Note to Instructors: Many of the homework problems are available for assignment via Wiley PLUS, an online teaching and learning solution. N O M ENCLATURE 14.16

14.17

Write structural formulas for each of the following: (a) 3-Nitrobenzoic acid

(g )

3-Chloro-1-ethoxybenzene

(m )

tert-Butylbenzene

(b) p-Bromotoluene

(h )

p-Chlorobenzenesulfonic acid

(n )

p -Methylphenol

(c) o-Dibromobenzene

(i)

Methyl p-toluenesulfonate

(o )

p-Bromoacetophenone

(d) m-Dinitrobenzene

(j)

Benzyl bromide

(p )

3-Phenylcyclohexanol

(e) 3,5-Dinitrophenol

(k )

p-Nitroaniline

(q )

2-Methyl-3-phenyl-1-butanol

(f) p-Nitrobenzoic acid

(l)

o-Xylene

(r)

o-Chloroanisole

Write structural formulas and give acceptable names for all representatives of the following: (a) Tribromobenzenes

(c )

Nitroanilines

(b) Dichlorophenols

(d )

Methylbenzenesulfonic acids

(e) Isomers of C 6 H 5 — C 4 Hg

666

Chapter 14

Aromatic Compounds

A R O M A TIC ITY 14.18

W h i c h o f th e f o l lo w in g m o le c u le s w o u l d y o u e x p e c t to b e a r o m a tic ? (a )

_+

(e )

(b)

I M

(f)

(c )

(i)

N

(j)

(g )

n

(k )

(h )

(d)n 14.19

U s e t h e p o l y g o n - a n d - c i r c l e m e t h o d t o d r a w a n o r b i t a l d ia g r a m f o r e a c h o f t h e f o l l o w i n g c o m p o u n d s .

w 14.20

> ■

N

E>,

( b

^

_

W r it e th e s tr u c tu r e o f th e p r o d u c t f o r m e d w h e n e a c h o f th e f o l lo w in g c o m p o u n d s r e a c ts w it h o n e m o la r e q u iv a ­ le n t o f H C l. (a )

N A

(b)

H

H Cl (1 equiv.)

N

CH 3

14.21

14.22

W h i c h o f th e h y d r o g e n a to m s s h o w n b e lo w is m o r e a c id ic ? E x p la i n y o u r a n s w e r.

T h e r in g s b e lo w a re j o i n e d b y a d o u b le b o n d th a t u n d e rg o e s c is - t r a n s is o m e r iz a t io n m u c h m o r e r e a d ily th a n th e b o n d o f a t y p i c a l a lk e n e . P r o v i d e a n e x p l a n a t i o n .

667

Problems 14.23

A l t h o u g h H u c k e l’ s r u le ( S e c tio n 1 4 .7 ) s t r ic t ly a p p lie s o n ly to m o n o c y c lic c o m p o u n d s , i t d o e s a p p e a r t o h a v e a p p li­ c a t io n to c e r ta in b i c y c l i c c o m p o u n d s , i f o n e a s s u m e s u s e o f r e s o n a n c e s tr u c tu r e s i n v o l v in g o n l y th e p e r im e t e r d o u ­ b l e b o n d s , a s s h o w n w i t h o n e r e s o n a n c e c o n t r i b u t o r f o r n a p h t h a le n e b e l o w .

B o t h n a p h t h a le n e ( S e c t i o n 1 4 . 8 A ) a n d a z u l e n e ( S e c t i o n 1 4 . 8 B ) h a v e 1 0 p ( b e lo w ) is a p p a r e n t ly a n t ia r o m a t ic a n d is u n s t a b le e v e n a t -

e l e c t r o n s a n d a r e a r o m a t i c . P e n t a le n e

1 0 0 ° C . H e p ta le n e h a s b e e n m a d e b u t i t a d d s b r o m in e ,

i t r e a c t s w i t h a c id s , a n d i t i s n o t p la n a r . I s H u c k e l ’ s r u l e a p p l i c a b l e t o t h e s e c o m p o u n d s ? I f s o , e x p l a i n t h e i r l a c k o f a r o m a tic ity .

P e n ta le n e

14.24

(a ) In

H e p ta le n e

1 9 6 0 T . K a t z ( C o l u m b i a U n i v e r s i t y ) s h o w e d t h a t c y c l o o c t a t e t r a e n e a d d s t w o e le c t r o n s w h e n t r e a t e d w i t h

p o t a s s i u m m e t a l a n d f o r m s a s t a b le , p l a n a r d i a n i o n , C

8 H 82 -

( a s t h e d i p o t a s s i u m s a lt ) :

2 equiv. K THF

*

2

K +

C 8 H 82 -

U s e th e m o le c u la r o r b i t a l d ia g r a m g iv e n i n F ig . 1 4 .7 a n d e x p la in t h is r e s u lt. ( b ) I n 1 9 6 4 K a t z a ls o s h o w e d t h a t r e m o v i n g t w o p r o t o n s f r o m t h e c o m p o u n d b e l o w ( u s i n g b u t y l l i t h i u m a s t h e b a s e ) l e a d s t o t h e f o r m a t i o n o f a s t a b le d i a n i o n w i t h t h e f o r m u l a C

2 BuLi

8 H 62 -

(a s th e d il i t h iu m

s a lt ) .

2 Li + C8 H62-

P r o p o s e a r e a s o n a b l e s t r u c t u r e f o r t h e p r o d u c t a n d e x p l a i n w h y i t i s s t a b le .

14.25

A l t h o u g h n o n e o f t h e [ 1 0 ] a n n u le n e s g iv e n i n S e c t io n 1 4 . 7 B i s a r o m a t i c , t h e f o l l o w i n g 1 0 p - e l e c t r o n s y s t e m i s a r o m a t ic :

W h a t f a c t o r m a k e s t h is p o s s ib le ?

14.26

C y c l o h e p t a t r i e n o n e ( I ) is v e r y s t a b le . C y c l o p e n t a d i e n o n e ( I I ) b y c o n t r a s t i s q u i t e u n s t a b l e a n d r a p i d ly u n d e r g o e s a D i e l s - A l d e r r e a c t io n w i t h it s e lf .

O O

( a ) P r o p o s e a n e x p la n a t io n f o r th e d if f e r e n t s t a b ilit ie s o f th e s e t w o c o m p o u n d s . ( b ) W r it e th e s tr u c tu r e o f th e D i e l s - A ld e r a d d u c t o f c y c lo p e n ta d ie n o n e .

14.27

5 - C h lo r o - 1 ,3 - c y c lo p e n ta d ie n e ( b e lo w ) u n d e rg o e s S N 1 s o lv o ly s is in th e p re s e n c e o f s ilv e r io n e x tr e m e ly s lo w ly e v e n t h o u g h th e c h lo r in e is d o u b ly a l l y l i c a n d a l l y l i c h a lid e s n o r m a l ly io n iz e r e a d i ly ( S e c tio n 1 5 .1 5 ) . P r o v id e a n e x p la ­ n a t io n f o r th is b e h a v io r .

0

14.28

^ ci

E x p l a i n t h e f o l l o w i n g : ( a ) C y c l o n o n a t e t r a e n y l a n i o n i s p l a n a r ( i n s p it e o f t h e a n g l e s t r a i n i n v o l v e d ) a n d a p p e a r s t o b e a r o m a t i c . ( b ) A l t h o u g h [ 1 6 ] a n n u l e n e is n o t a r o m a t i c , i t a d d s t w o e le c t r o n s r e a d i l y t o f o r m a n a r o m a t i c d i a n i o n .

14.29

F u r a n p o s s e s s e s le s s a r o m a t i c c h a r a c t e r t h a n b e n z e n e a s m e a s u r e d b y t h e i r r e s o n a n c e e n e r g i e s ( 9 6 k J m o l - 1 f o r f u r a n ; 1 5 1 k J m o l - 1 f o r b e n z e n e ) . W h a t r e a c t i o n h a v e w e s t u d i e d e a r l i e r t h a t s h o w s t h a t f u r a n i s le s s a r o m a t i c t h a n b e n z e n e a n d c a n r e a c t i n a w a y c h a r a c t e r i s t i c o f s o m e d ie n e s ?

668

Chapter 14

Aromatic Compounds

SPECTROSCOPY A N D STRUCTURE ELUCIDATION 14.30

F o r e a c h o f th e p a ir s b e lo w , p r e d ic t s p e c ific a s p e c ts in t h e ir 1H N M R

s p e c tra th a t w o u ld a llo w y o u to d is t in g u is h

o n e c o m p o u n d f r o m th e o th e r. (a )

Q

(c ) B r

B r

(b) O

O

14.31

A s s ig n s tr u c tu r e s to e a c h o f th e c o m p o u n d s A , B , a n d C w h o s e 1H N M R s p e c tra a re s h o w n in F ig . 1 4 .2 7 .

14.32

T h e 1H N M R

s p e c t r u m o f c y c lo o c ta t e t r a e n e c o n s is ts o f a s in g le l in e lo c a te d a t

8

5 .7 8 . W h a t d o e s th e lo c a tio n o f

t h is s ig n a l s u g g e s t a b o u t e le c tr o n d e lo c a liz a t io n i n c y c lo o c ta t e t r a e n e ?

14.33

G iv e a s tr u c tu r e f o r c o m p o u n d F th a t is c o n s is te n t w it h th e 1H N M R a n d I R s p e c tra in F ig . 1 4 .2 8 .

14.34

A

9 H 10

c o m p o u n d ( L ) w it h th e m o le c u la r f o r m u la C

r e a c t s w i t h b r o m i n e i n c a r b o n t e t r a c h l o r i d e a n d g iv e s a n I R

a b s o r p tio n s p e c tr u m th a t in c lu d e s th e f o l lo w in g a b s o r p tio n p e a k s : 3 0 3 5 c m _ 2853 c m _

1( w

), 1 64 0 c m _

1( m

), 9 9 0 c m _

1( s ) ,

915 cm _

1( s ) ,

740 cm _

1( s ) ,

1( m

), 3 0 2 0 c m _

695 cm _

1( s ) .

1( m

) , 2 9 2 5 c m _ 1( m ) ,

T h e 1H N M R

s p e c tru m o f

L c o n s is ts o f: D o u b le t

8

M u lt ip le t The U V

3 .1 ( 2 H )

8

4 .8

M u lt ip le t

8

5 .1

M u lt ip le t

8

5 .8

M u lt ip le t

8

7 .1 ( 5 H )

s p e c t r u m s h o w s a m a x i m u m a t 2 5 5 n m . P r o p o s e a s t r u c t u r e f o r c o m p o u n d L a n d m a k e a s s ig n m e n t s f o r

e a c h o f th e I R p e a k s .

14.35

C om pound M

h a s th e m o le c u la r f o r m u la C

9H

1 2 . T h e 1H N M R

s p e c tru m o f M

is g iv e n i n F ig . 1 4 .2 9 a n d th e I R

s p e c tr u m in F ig . 1 4 .3 0 . P ro p o s e a s tr u c tu re f o r M .

14.36

A c o m p o u n d ( N ) w it h th e m o le c u la r f o r m u la C

9H

10O r e a c t s w i t h o s m i u m t e t r o x i d e . T h e 1 H N M R s p e c t r u m o f N

is s h o w n i n F ig . 1 4 .3 1 a n d th e I R s p e c t r u m o f N is s h o w n i n F ig . 1 4 .3 2 . P r o p o s e a s tr u c tu r e f o r N .

14.37

T h e I R a n d 1H N M R s p e c tra f o r c o m p o u n d X

(C

8 H 10)

a re g iv e n i n F ig . 1 4 .3 3 . P r o p o s e a s tr u c tu r e f o r c o m p o u n d

X.

14.38

T h e I R a n d 1H N M R s p e c t r a o f c o m p o u n d Y

14.39

( a ) H o w m a n y s ig n a ls w o u l d y o u e x p e c t to f i n d i n th e 1H N M R s p e c t r u m o f c a ffe in e ?

(C

9 H 12O )

a re g iv e n i n F ig . 1 4 .3 4 . P r o p o s e a s tr u c tu r e f o r Y .

O

C H

3

H 3C .

O

"N

I CH 3

C a f f e in e

(b)

W h a t c h a r a c t e r is tic p e a k s w o u ld y o u e x p e c t to f in d in th e I R s p e c tr u m o f c a ffe in e ?

P r o b le m s

5 h (ppm)

5 h (ppm)

5 h (ppm)

Figure 14.27 The 300-MHz 1H NMR spectra for Problem 14.31. Expansions of the signals are shown in the offset plots.

669

670

Chapter 14 Aromatic Compounds

W avenum ber (cm 1)

5 h (ppm)

Figure 14.28 The 300-MHz 1H NMR and IR spectra of com pound F, Problem 14.33. Expansions of the signals are shown in the offset plots. IR spectra, SDBSWeb: http://riodb01.ibase.aist.go.jp/sdbs/ (National Institute of Advanced Industrial Science and Technology, Septem ber 24, 2009)

5 h (ppm) Figure 14.29 The 300-MHz 1H NMR spectrum of com pound M, Problem 14.35. Expansions of the signals are shown in the offset plots.

671

P r o b le m s

Wavenumber (cm 1) Figure 14.30

The IR spectrum of compound M, Problem 14.35.

9 10O

N, C H

....1......'......1...... 1 6.2 6.0

_ .... ’.......■............ I"" 5.2 5.0 TMS

............... ............... ............... ............... ............... ............... ............... ............... . 8

7

6

5

4 3 5 h (ppm)

2

1

0

Figure 14.31 The 300-MHz 1H NMR spectrum of com pound N, Problem 14.36. Expansions of the signals are shown in the offset plots.

Wavenumber (cm 1) Figure 14.32

The IR spectrum of compound N, Problem 14.36.

67 2

Chapter 14 Aromatic Compounds

Wavenumber (cm 1)

5 h (ppm) Figure 14.33 The IR and 300-MHz 1H NMR spectra of com pound X, Problem 14.37. Expansions of the signals are shown in the offset plots.

Wavenumber (cm 1) Figure 14.34 The IR and 300-MHz 1H NMR spectra (next page) of com pound Y, Problem 14.38. Expansions of the signals are shown in the offset plots.

673

Challenge Problems

5h (ppm)

Figure 14.34

(Continued)

Challenge Problems 14.40

G iv e n th e f o l lo w in g in f o r m a t io n , p r e d ic t th e a p p e a ra n c e o f th e 1H N M R s p e c t r u m a r is ­ in g fr o m

(c ).

th e v i n y l h y d r o g e n a to m s o f p - c h lo r o s ty r e n e . D e s h ie ld in g b y th e in d u c e d

(b)

m a g n e t ic f i e ld o f th e r i n g is g r e a te s t a t p r o t o n c ( d 6 .7 ) a n d is le a s t a t p r o t o n b ( d 5 .3 ) . T h e c h e m ic a l s h if t o f a is a b o u t d 5 .7 . T h e c o u p lin g a p p r o x i m a t e m a g n i t u d e s : J ac

=

1 8 H z , J bc

=

c o n s ta n ts h a v e th e f o llo w in g

1 1 H z , a n d J ab

=

2 H z . (T h e s e c o u ­

p li n g c o n s ta n ts a re t y p i c a l o f th o s e g iv e n b y v i n y l i c s y s te m s : C o u p lin g c o n s ta n ts f o r

Cl

tr a n s h y d r o g e n a to m s a re la r g e r th a n th o s e f o r c is h y d r o g e n a to m s , a n d c o u p lin g c o n ­ s ta n ts f o r g e m in a l v i n y l i c h y d r o g e n a to m s a r e v e r y s m a ll.)

1 4 .4 1

C o n s i d e r t h e s e r e a c t io n s :

Cl

f-BuOK -K C l

HBr A

-f-BuOH

B

T h e in t e r m e d ia t e A is a c o v a le n t ly b o n d e d c o m p o u n d th a t h a s t y p i c a l 1 H N M R s ig n a ls f o r a r o m a t ic r i n g h y d r o ­ g e n s a n d o n l y o n e a d d it io n a l s ig n a l a t d 1 .2 1 , w i t h a n a re a r a t i o o f 5 :3 , r e s p e c t iv e ly . F i n a l p r o d u c t B is i o n i c a n d h a s o n l y a r o m a t i c h y d r o g e n s ig n a l s . W h a t a re th e s tru c tu re s o f A

14.42

and B?

T h e f i n a l p r o d u c t o f t h i s s e q u e n c e , D , is a n o r a n g e , c r y s t a l l i n e s o l i d m e l t i n g a t 1 7 4 ° C a n d h a v i n g m o l e c u l a r w e i g h t 1 8 6 :

Cyclopentadiene + N a ----- > C + H 2 2 C + FeCl2 ----- > D + 2 NaCI I n its

1H

and

13C

N M R

s p e c tra , p r o d u c t D

s h o w s o n ly o n e k in d o f h y d ro g e n a n d o n ly o n e k in d o f c a rb o n ,

r e s p e c t iv e l y . D r a w th e s tr u c tu r e o f C a n d m a k e a s tr u c tu r a l s u g g e s tio n as t o h o w th e h ig h d e g re e o f s y m m e tr y o f D c a n b e e x p la in e d . ( D b e lo n g s to a g r o u p o f c o m p o u n d s n a m e d a f t e r s o m e t h in g y o u m i g h t g e t a t a d e l i f o r lu n c h . )

674 14.43

Chapter 14

C o m p o u n d E h a s th e s p e c t r a l fe a tu r e s g iv e n b e lo w . W h a t is it s s tr u c tu r e ? M S IR

(m /z ): M + 2 0 2 ( c m - 1 ): 3 0 3 0 -3 0 8 0 , 2 1 5 0 (v e ry w e a k ), 1 6 0 0 , 1 4 9 0 , 7 6 0 , a n d 6 9 0

1H N M R U V

14.44

Aromatic Compounds

(S ): n a r r o w m u lt ip le t c e n te re d a t 7 .3 4

(n m ): 2 8 7 (e =

D r a w a ll o f th e p

2 5 ,0 0 0 ), 3 0 5 (e =

3 6 ,0 0 0 ), a n d 3 2 6 (e =

3 3 ,0 0 0 )

m o le c u la r o r b it a ls f o r ( 3 £ ) - 1 , 3 , 5 - h e x a t r ie n e , o r d e r t h e m f r o m

lo w e s t t o h ig h e s t i n e n e rg y , a n d

in d ic a t e th e n u m b e r o f e le c tr o n s t h a t w o u l d b e f o u n d i n e a c h i n th e g r o u n d s ta te f o r th e m o le c u le . A f t e r d o in g so , o p e n th e c o m p u t e r m o le c u la r m o d e l f o r ( 3 £ ) - 1 ,3 , 5 - h e x a t r ie n e a n d d is p la y th e c a lc u la te d m o le c u la r o r b ita ls . H o w w e ll d o e s th e a p p e a ra n c e a n d s e q u e n c e o f th e o r b ita ls y o u d r e w (e .g ., n u m b e r o f n o d e s , o v e r a ll s y m m e tr y o f e a c h , e t c . ) c o m p a r e w i t h t h e o r b i t a l s i n t h e c a lc u l a t e d m o d e l ? A r e t h e s a m e o r b i t a l s p o p u l a t e d w i t h e l e c t r o n s i n y o u r a n a ly s is as in th e c a lc u la te d m o d e l?

Learning Group Problems 1

.

W r it e m e c h a n is m a r r o w s f o r th e f o l lo w in g

s te p i n t h e c h e m i c a l s y n t h e s is b y A . R o b e r t s o n a n d R . R o b i n s o n (J .

C h e m . S oc. 1 9 2 8 , 1 4 5 5 - 1 4 7 2 ) o f c a l l i s t e p h i n c h l o r i d e , a r e d f l o w e r p i g m e n t f r o m t h e p u r p l e - r e d a s te r . E x p l a i n w h y t h is t r a n s f o r m a t io n is a r e a s o n a b le p r o c e s s .

HCl

C a llis t e p h in c h lo r id e

2

.

T h e f o l l o w i n g r e a c t i o n s e q u e n c e w a s u s e d b y E . J . C o r e y (J. A m . C h e m . S oc. 1 9 6 9 , 9 1 , 5 6 7 5 - 5 6 7 7 ) a t t h e b e g i n ­ n i n g o f a s y n t h e s is o f p r o s t a g l a n d i n F 2 a a n d p r o s t a g l a n d i n E 2 . E x p l a i n w h a t i s i n v o l v e d i n t h i s r e a c t i o n a n d w h y i t is a r e a s o n a b le p ro c e s s .

3.

T h e 1H N M R s ig n a ls f o r th e a r o m a t ic h y d r o g e n s o f m e t h y l p - h y d r o x y b e n z o a t e a p p e a r a s t w o d o u b le t s a t a p p r o x ­ im a t e ly 7 .0 5 a n d 8 .0 4 p p m

( d ) . A s s i g n t h e s e t w o d o u b l e t s t o t h e r e s p e c t iv e h y d r o g e n s t h a t p r o d u c e e a c h s i g n a l .

J u s t i f y y o u r a s s ig n m e n t s u s i n g a r g u m e n t s o f r e l a t i v e e l e c t r o n d e n s i t y b a s e d o n c o n t r i b u t i n g r e s o n a n c e s t r u c t u r e s . O

^

r Y '''^ O

C H

3

H O " " ^ 5^

4.

D r a w th e s tr u c tu r e o f a d e n in e , a h e t e r o c y c lic a r o m a t ic c o m p o u n d in c o r p o r a t e d i n th e s tr u c tu r e o f D N A . I d e n t i f y th e n o n b o n d in g e le c tr o n p a ir s t h a t a re n o t p a r t o f th e a r o m a t ic s y s te m i n th e r in g s o f a d e n in e . W h i c h n it r o g e n a to m s i n t h e r i n g s w o u l d y o u e x p e c t t o b e m o r e b a s i c a n d w h i c h s h o u l d b e le s s b a s ic ? D r a w s tr u c tu r e s o f th e n ic o t in a m id e r in g in N A D H a n d N A D + . I n th e t r a n s fo r m a tio n o f N A D H t o N A D + , i n w h a t f o r m m u s t a h y d r o g e n b e tr a n s fe r r e d in o r d e r to p r o d u c e th e a r o m a tic p y r id in iu m

io n in N A D + ?

CONCEPT MAP

O' V I

Ol

Reactions of Aromatic Compounds

Thyroxine (see the model above ) is an aromatic compound and a key hormone that raises metabolic rate. Low levels of thyroxine (hypothyroidism) can lead to obesity, lethargy, and an enlarged thyroid gland (goiter). The thyroid gland makes thyroxine from iodine and tyrosine, which are two essential components of our diet. Most of us obtain iodine from iodized salt, but iodine is also found in products derived from seaweed, like the kelp shown above. An abnormal level of thyroxine is a relatively common malady, however. Fortunately, low levels of thyroxine are easily corrected by hormone supplements. After we study a new class of reaction in this chap­ ter called electrophilic aromatic substitution, we shall return to see how that reaction is related to thyroxine in "The Chemistry of . . . Iodine Incorporation in Thyroxine Biosynthesis."

676

CP"-

A

15.1 Electrophilic Aromatic Substitution Reactions

677

15.1 Electrophilic Aromatic Substitution Reactions S o m e o f t h e m o s t i m p o r t a n t r e a c t io n s o f a r o m a t i c c o m p o u n d s a r e t h o s e i n w h i c h a n e le c t r o p h ile r e p la c e s o n e o f th e h y d r o g e n a to m s o f th e r in g .

H

E

E— A

H— A

( E — A is a n e l e c t r o p h i l i c r e a c t a n t )

T h e s e r e a c t io n s , c a l l e d e l e c t r o p h i l i c a r o m a t i c s u b s t i t u t i o n s ( E A S ) , a l l o w t h e d i r e c t i n t r o ­ d u c t io n o f g r o u p s o n t o a r o m a tic r in g s s u c h a s b e n z e n e , a n d th e y p r o v id e s y n th e tic r o u te s to m a n y im p o r ta n t c o m p o u n d s . F ig u r e

1 5 .1 o u t lin e s f iv e d if f e r e n t ty p e s o f e le c t r o p h ilic

a r o m a t ic s u b s t it u tio n s t h a t w e w i l l s tu d y i n t h is c h a p te r , in c l u d in g c a r b o n - c a r b o n b o n d f o r m i n g r e a c t io n s a n d h a l o g e n a t i o n s .

Figure 15.1 Electrophilic aromatic substitution reactions.

A

n o t e w o r t h y e x a m p le o f e le c t r o p h ilic a r o m a t ic s u b s t it u t io n i n n a tu r e , a s m e n t io n e d

a b o v e , is b io s y n th e s is o f th e t h y r o id h o r m o n e t h y r o x in e , w h e r e io d in e is in c o r p o r a t e d in t o b e n z e n e r in g s th a t a re d e r iv e d f r o m ty r o s in e .

O

HO.

O-

OI

HO

I

T y r o s in e

T h y r o x in e

(a d ie ta r y a m in o a c id )

(a t h y r o id h o r m o n e )

I n t h e n e x t s e c t i o n w e s h a l l l e a r n t h e g e n e r a l m e c h a n i s m f o r t h e w a y a n e le c t r o p h i l e r e a c ts w i t h a b e n z e n e r in g . T h e n i n S e c tio n s 1 5 . 3 - 1 5 . 7 w e s h a ll s e e s p e c if ic e x a m p le s o f e l e c t r o p h i l e s a n d h o w e a c h is f o r m e d i n a r e a c t i o n m i x t u r e .

678

Chapter 15

Reactions o f Aromatic Compounds

15.2 A G eneral Mechanism fo r Electrophilic A rom atic Substitution The p electrons of benzene react with strong electrophiles. In this respect, benzene has something in common with alkenes. When an alkene reacts with an electrophile, as in the addition of HBr (Section 8.2), electrons from the alkene p bond react with the electrophile, leading to a carbocation intermediate.

:Br; Alkene

Electrophile

C arbocation

The carbocation formed from the alkene then reacts with the nucleophilic bromide ion to form the addition product.

'

-Br

■■Bn

Carbocation

H

----- >

Addition product

The similarity of benzene reactivity with that of an alkene ends, however, at the carboca­ tion stage, prior to nucleophilic attack. As we saw in Chapter 14, benzene’s closed shell of six p electrons give it special stability. •

Although benzene is susceptible to electrophilic attack, it undergoes substitution reactions rather than addition reactions.

Substitution reactions allow the aromatic sextet of p electrons in benzene to be regener­ ated after attack by the electrophile. We can see how this happens if we examine a general mechanism for electrophilic aromatic substitution. Experimental evidence indicates that electrophiles attack the p system of benzene to form a n o n a r o m a t ic c y c lo h e x a d ie n y l c a r b o c a tio n known as an arenium ion. In showing this step, it is convenient to use Kekule structures, because these make it much easier to keep track of the p electrons: H e lp f u l H i n t Resonance structures (like those used here for the arenium ion) will be important for our study of electrophilic aromatic substitution.

/ " ^ i s+ f V

e9 a

Step 1

A f

Arenium ion (a delocalized cyclohexadienyl cation) •

In step 1 the electrophile takes two electrons of the six-electron p system to form a s bond to one carbon atom of the benzene ring.

Formation of this bond interrupts the cyclic system of p electrons, because in the formation of the arenium ion the carbon that forms a bond to the electrophile becomes sp3 hybridized and, therefore, no longer has an available p orbital. Now only five carbon atoms of the ring are sp2 hybridized and still have p orbitals. The four p electrons of the arenium ion are delocalized through these five p orbitals. A calculated electrostatic potential map for the arenium ion formed by electrophilic addition of bromine to benzene indicates that positive charge is distributed in the arenium ion ring (Fig. 15.2), just as was shown in the contributing resonance structures.

Figure 15.2 A calculated structure for the arenium ion intermediate formed by electrophilic addition of bromine to benzene (Section 15.3). The electrostatic potential map for the principal location of bonding electrons (indicated by the solid surface) show s that positive charge (blue) resides primarily at the ortho and para carbons relative to the carbon where the electrophile has bonded. This distribution of charge is consistent with the resonance model for an arenium ion. (The van der Waals surface is indicated by the wire mesh.)

A 15.2 A General Mechanism for Electrophilic Aromatic Substitution •

679

In step 2 a proton is removed from the carbon atom of the arenium ion that bears the electrophile, restoring aromaticity to the ring. +

E

Step 2

'H

----- »

I

+

H— A

f;A

The two electrons that bonded the proton to the ring become a part of the p system. The carbon atom that bears the electrophile becomes sp2 hybridized again, and a benzene deriv­ ative with six fully delocalized p electrons is formed. (The proton is removed by any of the bases present, for example, by the anion derived from the electrophile.) Show how loss of a proton can be represented using each of the three resonance structures for the arenium ion and show how each representation leads to the formation of a benzene ring with three alternating double bonds (i.e., six fully delocalized p electrons).

R e v ie w P ro b le m 15.1

Kekule structures are more appropriate for writing mechanisms such as electrophilic aro­ matic substitution because they permit the use of resonance theory, which, as we shall soon see, is invaluable as an aid to our understanding. If, for brevity, however, we wish to show the mechanism using the hybrid formula for benzene we can do it in the following way. We draw the arenium ion as a delocalized cyclohexadienyl cation:

Step 1

H

E— A

+

:A-

+

H— A

Arenium ion E E

H

Step 2 8+

<

H e lp fu l H i n t

In our color scheme for chemical formulas, blue generally indicates groups that are electrophilic or have electron-withdrawing character. Red indicates groups that are or become Lewis bases, or have electron-donating character.

L a -

There is firm experimental evidence that the arenium ion is a true intermediate in elec­ trophilic substitution reactions. It is not a transition state. This means that in a free-energy diagram (Fig. 15.3) the arenium ion lies in an energy valley between two transition states. The free energy of activation for step 1, A G |1), has been shown to be much greater than the free energy of activation for step 2, AG(2). This is consistent with what we would expect.

Reaction coordinate

Figure 15.3 The free-energy diagram for an electrophilic aromatic substitution reaction. The arenium ion is a true intermediate lying betw een transition states 1 and 2. In transition state 1 the bond betw een the electrophile and one carbon atom of the benzene ring is only partially form ed. In transition state 2 the bond betw een the sam e benzene carbon atom and its hydrogen atom is partially broken. The bond betw een the hydrogen atom and the conjugate base is partially formed.

680

Chapter 15

Reactions of Aromatic Compounds

T h e r e a c t i o n l e a d i n g f r o m b e n z e n e a n d a n e le c t r o p h i l e t o t h e a r e n i u m i o n i s h i g h l y e n d o t h e r ­ m ie , b e c a u s e th e a r o m a t ic s t a b il i t y o f th e b e n z e n e r i n g is lo s t . T h e r e a c t io n le a d in g f r o m t h e a r e n i u m i o n t o t h e s u b s t i t u t e d b e n z e n e , b y c o n t r a s t , is h i g h l y e x o t h e r m i c b e c a u s e i t re s to re s a r o m a t ic it y t o th e s y s te m . O f t h e f o l l o w i n g t w o s te p s , s te p 1 ( t h e f o r m a t i o n o f t h e a r e n i u m i o n ) i s u s u a l l y t h e r a t e d e t e r m i n i n g s te p i n e l e c t r o p h i l i c a r o m a t i c s u b s t i t u t i o n b e c a u s e o f i t s h i g h e r f r e e e n e r g y o f a c tiv a t io n :

Step 1

E

■A

—A H

Step 2

W H

E

H— A

ys+

s + ^

E

Slow, rate determ ining

Fast

E

Ha S t e p 2 , t h e r e m o v a l o f a p r o t o n , o c c u r s r a p i d l y r e l a t i v e t o s te p 1 a n d h a s n o e f f e c t o n t h e o v e r a l l r a t e o f r e a c t io n .

15.3 Halogenation o f Benzene B e n z e n e r e a c t s w i t h b r o m i n e a n d c h l o r i n e i n t h e p r e s e n c e o f L e w i s a c id s t o g i v e h a l o g e n a t e d s u b s t it u tio n p r o d u c ts in g o o d y ie ld .

Cl Cl,

FeCl3

HCl

25°C

C hlorobenzene (90%) Br Br,

FeBr3'

HBr

heat

B rom obenzene (75%) T h e L e w i s a c id s t y p i c a l l y u s e d a r e a l u m i n u m c h l o r i d e ( A l C l 3) a n d i r o n c h l o r i d e ( F e C l 3) f o r c h l o r i n a t i o n , a n d i r o n b r o m i d e ( F e B r 3) f o r b r o m i n a t i o n . T h e p u r p o s e o f t h e L e w i s a c id is t o m a k e th e h a lo g e n a s tr o n g e r e le c t r o p h ile . A

m e c h a n is m

b r o m in a t io n is s h o w n h e re .

A MECHANISM FOR THE REACTION E le c tro p h ilic A ro m a tic B ro m in a tio n

Step 1

=B r — Br="

FeBr,

=B r— Br— FeBr3

Bromine co m b in es with FeBr3 to form a com plex.

f o r e le c t r o p h ilic a r o m a tic

CP"-

S *n ¡s -

A

15.4 Nitration of Benzene

Br: +

-

_

: B r— Br— FeBr3

Step 2

681

H Br:

slow

------- ¡

FeBr4

Arenium ion H e lp f u l H i n t An electrostatic potential map fo r this arenium ion is shown in Fig. 15.2.

The benzene ring d o n a te s an electron pair to th e term inal brom ine, form ing th e arenium ion and neutralizing th e formal positive ch a rg e on th e oth er brom ine.

Br: ■ i0

-

: Br— FeBr3

Step 3

H — Br =

FeBr3

A proton is rem oved from th e arenium ion to form b rom obenzene and reg en erate th e cataly st.

T h e m e c h a n is m o f th e c h lo r in a t io n o f b e n z e n e in th e p re s e n c e o f f e r r i c c h lo r id e is a n a l­ o g o u s to th e o n e f o r b r o m in a tio n . F lu o r in e r e a c ts s o r a p id ly w it h b e n z e n e th a t a r o m a tic f lu o r in a t io n r e q u ir e s s p e c ia l c o n ­ d it io n s a n d s p e c ia l ty p e s o f a p p a r a tu s . E v e n th e n , i t is d i f f i c u l t t o l i m i t th e r e a c t io n to m o n o f lu o r in a t io n . F lu o r o b e n z e n e c a n b e m a d e , h o w e v e r , b y a n i n d i r e c t m e t h o d th a t w e s h a ll se e i n S e c tio n 2 0 . 7 D . I o d in e , o n th e o th e r h a n d , is s o u n r e a c tiv e th a t a s p e c ia l t e c h n iq u e h a s to b e u s e d t o e ffe c t d ir e c t io d in a t io n ; th e r e a c t io n h a s t o b e c a r r ie d o u t i n th e p r e s e n c e o f a n o x id i z in g a g e n t s u c h a s n i t r i c a c id :

86% B io c h e m ic a l io d in a t io n , a s i n th e b io s y n th e s is o f t h y r o x in e , o c c u r s w i t h e n z y m a t ic c a t a l­ y s is .

T h y r o x in e

b io s y n th e s is

is

d is c u s s e d

fu r th e r

in

“The

C h e m is tr y

o f

...

I o d in e

In c o r p o r a t io n in T h y r o x in e B io s y n t h e s is ” b o x in S e c tio n 1 5 .1 1 E .

15.4 N itratio n o f Benzene B e n z e n e u n d e r g o e s n it r a t io n o n r e a c t io n w i t h a m ix t u r e o f c o n c e n tr a te d n i t r i c a c id a n d c o n ­ c e n t r a t e d s u l f u r i c a c id .

85% C o n c e n t r a t e d s u l f u r i c a c i d in c r e a s e s t h e r a t e o f t h e r e a c t i o n b y i n c r e a s i n g t h e c o n c e n ­ t r a t i o n o f t h e e l e c t r o p h i l e , t h e n i t r o n i u m i o n ( N O 2 + ) , a s s h o w n i n t h e f i r s t t w o s te p s o f t h e f o l lo w in g m e c h a n is m .

ÒS2

Chapter 15

Reactions of Aromatic Compounds

A MECHANISM FOR THE REACTION N itra tio n o f B e n z e n e

O: H— O

HOgSO— H

Step l

t/

N

(^SO *)

H— O 1

\ .

N

HSO

\ .

.o:

.g :

In this ste p nitric acid a c c e p ts a proton from th e stro n g er acid, sulfuric acid. ■O'

H o i< + / H— O1 N " *• Co

Step 2

h 2o

I

N+

•O.

Nitronium ion Now that it is protonated, nitric acid can d isso cia te to form a nitronium ion.

Step S

N+

slow

.G. Arenium ion The nitronium ion is the electrophile in nitration; it rea cts with b en zen e to form a resonance-stabilized arenium ion.

Step 4 H The arenium ion then lo se s a proton to a Lewis b a s e and b ec o m e s nitrobenzene.

R eview P roblem 15.2

Given that the p ^ a of H2 SO 4 is - 9 and that of H N O 3 is -1 .4 , explain why nitration occurs more rapidly in a mixture of concentrated nitric and sulfuric acids than in concentrated nitric acid alone.

15.5 Sulfonation o f Benzene Benzene reacts with fuming sulfuric acid at room temperature to produce benzenesulfonic acid. Fuming sulfuric acid is sulfuric acid that contains added sulfur trioxide (SO3). Sulfonation also takes place in concentrated sulfuric acid alone, but more slowly. Under either condition, the electrophile appears to be sulfur trioxide.

O / ~

\

„

O'

i

__________ -

•G ^ ^ G ‘

25°C concd H2SO4

Sulfur trioxide

S— G— H

II

■.O,

"

B enzenesulfonic acid (56%)

CP"-

h n ¡s .

A

15.5 Sulfonation of Benzene

683

In concentrated sulfuric acid, sulfur trioxide is produced in an equilibrium in which H 2 SO 4 acts as both an acid and a base (see step 1 of the following mechanism).

A MECHANISM FOR THE REACTION S u lfo n a tio n o f B e n z e n e 2 H 2 SO4

Step 1

SO 3 + h 3 o+ + h s o 4-

This equilibrium p ro d u ces SO3 in co n cen trated H2SO4.

'G

G’

G-

slow

other resonance structures

^ < G .

Step 2

= /

H

SO3 is th e electrophile th at re a c ts with b en zen e to form an arenium ion.

G Step 3

G:

G'' S_ G: S—

hso4 -

Il

G.

H2SO4

"

A proton is rem oved from th e arenium ion to form th e b en zen esu lfo n ate ion.

•'G'' Step 4

L— G = S

G' fast

\

H~ o O -H

11 G " S—

H

H2 O

O H .O. T he b enzenesulfonate ion a c c e p ts a proton to b eco m e benzenesulfonic acid.

A ll of the steps in sulfonation are equilibria, which means that the overall reaction is reversible. The position of equilibrium can be influenced by the conditions we employ. S O 3 H

H 2 S O 4

H2O

•

I f we want to sulfonate the ring (install a sulfonic acid group), we use concentrated sulfuric acid or— better yet— fuming sulfuric acid. Under these conditions the posi­ tion of equilibrium lies appreciably to the right, and we obtain benzenesulfonic acid in good yield.

•

I f we want to desulfonate the ring ( r e m o v e a sulfonic acid group), we employ dilute sulfuric acid and usually pass steam through the mixture. Under these conditions— with a high concentration of water— the equilibrium lies appreciably to the left and desulfonation occurs.

We shall see later that sulfonation and desulfonation reactions are often used in synthetic work. •

We sometimes install a sulfonate group as a protecting group, to temporarily block its position from electrophilic aromatic substitution, or as a directing group, to influence the position of another substitution relative to it (Section 15.10). When it is no longer needed we remove the sulfonate group.

H e lp f u l H i n t Sulfonation-desulfonation is a useful tool in syntheses involving electrophilic aromatic substitution.

684

Chapter 15

Reactions of Aromatic Compounds

15.6 Friedel-Crafts Alkylation C h a r le s F r ie d e l, a F r e n c h c h e m is t, a n d h is A m e r ic a n c o lla b o r a t o r , J a m e s M . C r a f ts , d is ­ c o v e r e d n e w m e t h o d s f o r t h e p r e p a r a t i o n o f a lk y l b e n z e n e s ( A r R ) a n d a c y lb e n z e n e s ( A r C O R ) i n 1 8 7 7 . T h e s e r e a c t io n s a r e n o w c a l l e d t h e F r i e d e l - C r a f t s a l k y l a t i o n a n d a c y l a t i o n r e a c ­ tio n s .

W e

s h a ll

s tu d y

th e

F r ie d e l- C r a fts

a lk y la tio n

r e a c t io n

h e re

and

ta k e

up

th e

F r i e d e l - C r a f t s a c y la t io n r e a c t io n i n S e c tio n 1 5 .7 . •

T h e f o l l o w i n g i s a g e n e r a l e q u a t i o n f o r a F r i e d e l - C r a f t s a l k y l a t i o n r e a c t io n : R

AlCi, HX

R — X

•

T h e m e c h a n is m f o r th e r e a c t io n s ta r ts w i t h th e f o r m a t io n o f a c a r b o c a tio n .

•

T h e c a r b o c a tio n th e n a c ts a s a n e le c t r o p h ile a n d a tta c k s th e b e n z e n e r in g t o f o r m a n a r e n iu m io n .

•

T h e a r e n iu m i o n th e n lo s e s a p r o t o n .

T h is m e c h a n is m is i llu s t r a t e d b e lo w u s in g 2 - c h lo r o p r o p a n e a n d b e n z e n e .

A MECHANISM FOR THE REACTION F r ie d e l- C r a f ts a lk y la tio n

=C|:

: C l:

:C h ..

Step 1

+

+

, C l— A l— Cl :

.. , A l . : Ci "C h

U-

I

••

I_

..

:Cl — Ai — Cl:

I

: Ch : C.i; The com plex d isso cia te s to form a carbocation and A!C!4_

This is a Lewis acid-base reaction (se e Section 3).

:C l:

Ch

Step 2

/

+

I_

..

I

■■

Ch The carbocation, acting a s an electrophile, reacts with benzene to produce an arenium ion.

+

:C l— A l— Ch

HCl

I

+

-A L :C l"

"C h

A proton is removed from the arenium ion to form isopropylbenzene. This step also regenerates the AlCl3 and liberates HCl.

•

W hen R —

X is a p r im a r y h a lid e , a s im p le c a r b o c a tio n p r o b a b ly d o e s n o t f o r m .

In s te a d , th e a lu m in u m c h lo r id e f o r m s a c o m p le x w it h th e a lk y l h a lid e , a n d th is c o m p le x a c ts a s th e e le c t r o p h ile . T h e c o m p l e x is o n e i n w h i c h t h e c a r b o n - h a l o g e n b o n d i s n e a r l y b r o k e n — a n d o n e i n w h i c h th e c a r b o n a to m h a s a c o n s id e r a b le p o s it iv e c h a rg e :

5+ RCH

.. 52

- - - - C l cA l C l 3

E v e n th o u g h t h is c o m p le x is n o t a s im p le c a r b o c a tio n , i t a c ts a s i f i t w e r e a n d i t tr a n s fe r s a p o s itiv e a lk y l g r o u p to th e a r o m a tic r in g . •

T h e s e c o m p l e x e s r e a c t s o m u c h l i k e c a r b o c a t i o n s t h a t t h e y a ls o u n d e r g o t y p i c a l c a r b o c a tio n r e a r r a n g e m e n ts ( S e c tio n 1 5 .8 ).

•

F r ie d e l- C r a f t s a lk y la t io n s a re n o t r e s tr ic te d t o th e u s e o f a lk y l h a lid e s a n d a lu ­ m i n u m c h l o r i d e . O t h e r p a i r s o f r e a g e n t s t h a t f o r m c a r b o c a t i o n s ( o r s p e c ie s l i k e c a r b o c a tio n s ) m a y b e u s e d in F r ie d e l- C r a f t s a lk y la t io n s a s w e ll.

CP"-

A

15.7 Friedel-Crafts Acylation

685

These possibilities include the use of a mixture of an alkene and an acid:

HF_ 0°C

Propene

r ~

Isopropylbenzene (eum ene) (84%)

(

\

r s

HF , 0°C '

C yclohexene

Cyclohexylbenzene (62%)

A mixture of an alcohol and an acid may also be used: .

^

u n _ HO-

\

60°C

Cyclohexanol

h 2o

Cyclohexylbenzene (56%)

There are several important limitations of the Friedel-Crafts reaction. These are discussed in Section 15.8. Outline all steps in a reasonable mechanism for the formation of isopropylbenzene from propene and benzene in liquid HF (just shown). Your mechanism must account for the prod­ uct being isopropylbenzene, not propylbenzene.

R e v ie w P ro b le m 1 5 .3

15.7 F ried el-C rafts Acylation O The R group is called an acyl group, and a reaction whereby an acyl group is intro­ duced into a compound is called an acylation reaction. Two common acyl groups are the acetyl group and the benzoyl group. (The benzoyl group should not be confused with the benzyl group, — CH 2 C 6 H5; see Section 14.2.) O O

O

CH

Acetyl group (ethanoyl group)

Benzoyl group

The F r i e d e l - C r a f t s a c y l a t i o n reaction is often carried out by treating the aromatic com­ pound with an a c y l h a l i d e (often an acyl chloride). Unless the aromatic compound is one that is highly reactive, the reaction requires the addition of at least one equivalent of a Lewis acid (such as AlCl3) as well. The product of the reaction is an aryl ketone: O O

Cl Acetyl chloride

excess benzene, 80°C

A cetophenone (methyl phenyl ketone) (97%)

HCl

686

Chapter 15

Reactions o f Aromatic Compounds

Acyl chlorides, also called acid chlorides, are easily prepared (Section 18.5) by treat­ ing carboxylic acids with thionyl chloride (SOCl2) or phosphorus pentachloride (PCl5): O

O

OH Acetic acid

SO C I2 Thionyl chloride

Cl Acetyl chloride (80-90%)

8 0 oC

SO 2 + HCI

O

O PCI5

OH

Benzoic acid

CI

P h o sp h o ru s pentachloride

POCI3

HCl

Benzoyl chloride (90%)

Friedel-Crafts acylations can also be carried out using carboxylic acid anhydrides. For example, O O excess benzene, 80°C

A cetic anhydride (a carboxylic acid anhydride)

OH A cetophenone (82-85% )

In most Friedel-Crafts acylations the electrophile appears to be an a c y l i u m from an acyl halide in the following way: • S + - OH

O-

Xs_ O'

O -

O

/ O

=O I

¿ 1 1

r C\ G^

(G =

r C+

H, R, O H , o r O R )

E l e c t r o n - w i t h d r a w i n g g r o u p s w i t h a full o r pa rt ial c h a r g e on th e a t o m a t t a c h e d to t h e ring

R esonance Effects The re s o n a n c e e ffe c t of a substituent G refers to the possibility that the presence of G may increase or decrease the resonance stabilization of the inter­ mediate arenium ion. The G substituent may, for example, cause one of the three contrib­ utors to the resonance hybrid for the arenium ion to be better or worse than the case when G is hydrogen. Moreover, when G is an atom bearing one or more nonbonding electron pairs, it may lend extra stability to the arenium ion by providing a resonance con­ tributor in which the positive charge resides on G:

fourth

E

H

H

This electron-donating resonance effect applies with decreasing strength in the following order:

M o s t electron do nating

A

NH„

f iN R

>

OH,

OR

>

X :

Least electron donating

This is also the order of the activating ability of these groups. • Amino groups are highly activating, hydroxyl and alkoxyl groups are somewhat less activating, and halogen substituents are weakly deactivating. When X = F, this order can be related to the electronegativity of the atoms with the non­ bonding pair. The more electronegative the atom is, the less able it is to accept the positive charge (fluorine is the most electronegative, nitrogen the least). When X = Cl, Br, or I, the relatively poor electron-donating ability of the halogens by resonance is understandable on a different basis. These atoms (Cl, Br, and I) are all larger than carbon, and, therefore, the orbitals that contain the nonbonding pairs are further from the nucleus and do not overlap well with the 2p orbital of carbon. (This is a general phenomenon: Resonance effects are not transmitted well between atoms of different rows in the periodic table.)

699

700

C h a p te r 1 5

R e a c t io n s o f A r o m a t i c C o m p o u n d s

15.11C Meta-Directing Groups • A ll meta-directing groups have either a partial positive charge or a full positive charge on the atom directly attached to the ring. As a typical example let us consider the trifluoromethyl group. The trifluoromethyl group, because of the three highly electronegative fluorine atoms, is strongly electron withdraw­ ing. It is a strong deactivating group and a powerful meta director in electrophilic aromatic substitution reactions. We can account for both of these characteristics of the trifluoromethyl group in the following way. The trifluoromethyl group affects the rate of reaction by causing the transition state lead­ ing to the arenium ion to be highly unstable. It does this by withdrawing electrons from the developing carbocation, thus increasing the positive charge on the ring: s+ OF,

^

OF3

E H E Arenium ion

Trifluoromethylbenzene

We can understand how the trifluoromethyl group affects orientation in electrophilic aro­ matic substitution if we examine the resonance structures for the arenium ion that would be formed when an electrophile attacks the ortho, meta, and para positions of trifluo­ romethylbenzene. O rth o A tta c k

M e ta A tta c k

OF

OF

P ara A tta c k

OF

OF

OF

OF

H E

H E

O

e .

H E

Highly unstable contributor • The arenium ion arising from ortho and para attack each has one contributing structure that is highly unstable relative to a ll the others because the positive charge is located on the ring carbon that bears the electron-w ithdraw ing group.

CP"-

A

15.11 How Substituents A ffect Electrophilic Aromatic Substitution

701

• The arenium ion arising from meta attack has no such highly unstable resonance structure. • By the usual reasoning we would also expect the transition state leading to the meta-substituted arenium ion to be the least unstable and, therefore, that meta attack would be favored. This is exactly what we find experimentally. The trifluoromethyl group is a powerful meta director:

CF

C F

HNO,

H2SO4

NO2 T riflu o ro m e th y lb e n z e n e

- 100%

Bear in mind, however, that meta substitution is favored only in the sense that it is the The free energy of activation for substi­ tution at the meta position of trifluoromethylbenzene is less than that for attack at an ortho or para position, but it is still far greater than that for an attack on benzene. Substitution occurs at the meta position of trifluoromethylbenzene faster than substitution takes place at the ortho and para positions, but it occurs much more slowly than it does with benzene. least unfavorable o f three unfavorable pathways.

• The nitro group, the carboxyl group, and other meta-directing groups (see Table 15.2) are all powerful electron-withdrawing groups and act in a similar way.

Solved Problem 15.4 Write contributing resonance structures and the resonance hybrid for the arenium ion formed when benzaldehyde undergoes nitration at the meta position. STRATEGY AND ANSWER O

O H

O H

O H

H

O= N= O kJ

+

15.11D Ortho-Para-Directing Groups Except for the alkyl and phenyl substituents, all of the ortho-para-directing groups in Table 15.2 are of the following general type: H \

■G At least one nonbonding electron pair

/H

N

:O

M

=Cl =

as in A n ilin e

P henol

O

C h lo ro b e n ze n e

This structural feature— an unshared electron pair on the atom adjacent to the ring— deter­ mines the orientation and influences reactivity in electrophilic substitution reactions. The directive effect of groups with an unshared pair is predominantly caused by an elec­ tron-releasing resonance effect. The resonance effect, moreover, operates primarily in the arenium ion and, consequently, in the transition state leading to it.

H

Chapter 15

Reactions of Aromatic Compounds

Except for the halogens, the primary effect of these groups on relative reactivity of the benzene ring is also caused by an electron-releasing resonance effect. And, again, this effect operates primarily in the transition state leading to the arenium ion. In order to understand these resonance effects, let us begin by recalling the effect of the amino group on electrophilic aromatic substitution reactions. The amino group is not only a powerful activating group, it is also a powerful ortho-para director. We saw earlier (Section 15.10D) that aniline reacts with bromine in aqueous solution at room temperature and in the absence of a catalyst to yield a product in which both ortho positions and the para posi­ tion are substituted. The inductive effect of the amino group makes it slightly electron withdrawing. Nitrogen, as we know, is more electronegative than carbon. The difference between the electronega­ tivities of nitrogen and carbon in aniline is not large, however, because the carbon of the benzene ring is sp2 hybridized and so it is somewhat more electronegative than it would be if it were sp3 hybridized. • The resonance effect of the amino group is far more important than its inductive effect in electrophilic aromatic substitution, and this resonance effect makes the amino group electron releasing. We can understand this effect if we write the resonance structures for the arenium ions that would arise from ortho, meta, and para attack on aniline:

OrthoAttack ;NH,

MetaAttack =NH,

=NH

E

ParaAttack =NH,

NH,

: N H,

I

Î

CN H 2

r

+

+\ m

702

/ E

H

-

n VA E

H

E

H

E

H

Relatively stable contributor

Four reasonable resonance structures can be written for the arenium ions resulting from ortho and para attack, whereas only three can be written for the arenium ion that results from meta attack. This, in itself, suggests that the ortho- and para-substituted arenium ions should be more stable. O f greater importance, however, are the relatively stable structures that contribute to the hybrid for the ortho- and para-substituted arenium ions. In these structures, nonbonding pairs of electrons from nitrogen form an additional covalent bond to the carbon of the ring. This extra bond—and the fact that every atom in each of these structures has a complete outer octet of electrons—makes these structures the most stable of all of the contributors. Because

CP"-

A

15.11 How Substituents A ffect Electrophilic Aromatic Substitution

703

—andstabilizing

these structures are unusually stable, they make a large — contribution to the hybrid. This means, of course, that the ortho- and para-substituted arenium ions themselves are considerably more stable than the arenium ion that results from the meta attack. The tran­ sition states leading to the ortho- and para-substituted arenium ions occur at unusually low free energies. As a result, electrophiles react at the ortho and para positions very rapidly. Use resonance theory to explain why the hydroxyl group of phenol is an activating group and an ortho-para director. Illustrate your explanation by showing the arenium ions formed when phenol reacts with a Br+ ion at the ortho, meta, and para positions. Phenol reacts with acetic anhydride in the presence of sodium acetate to produce the ester phenyl acetate: (CH3CO) 2O

O

CH3CO2Na

OH

O Phenyl a c e ta te

Phenol

The CH3COO — group of phenyl acetate, like the — OH group of phenol (Review Problem 15.8), is an ortho-para director. (a )

What structural feature of the CH3COO — group explains this?

(b )

Phenyl acetate, although undergoing reaction at the ortho and para positions, is less reac­ tive toward electrophilic aromatic substitution than phenol. Use resonance theory to explain why this is so.

(c )

Aniline is often so highly reactive toward electrophilic substitution that undesirable reac­ tions take place (see Section 15.14A). One way to avoid these undesirable reactions is to convert aniline to acetanilide (below) by treating aniline with acetyl chloride or acetic anhydride: O

(CH3CO) 2O

NH2 A n ilin e

NH A c e ta n ilid e

What kind of directive effect would you expect the acetamido group (CH 3 CONH — ) to have? (d )

Explain why it is much less activating than the amino group, — NH2.

The directive and reactivity effects of halo substituents may, at first, seem to be contra­ dictory. (in Table 15.2) [Because of this behavior we have color coded halogen substituents green rather than red (electron donating) or blue (electron withdrawing).] A ll other deactivating groups are meta directors. We can readily account for the behavior of halo substituents, how­ ever, if we assume that their electron-withdrawing inductive effect influences and their electron-donating resonance effect governs . Let us apply these assumptions specifically to chlorobenzene. The chlorine atom is highly electronegative. Thus, we would expect a chlorine atom to withdraw electrons from the ben­ zene ring and thereby deactivate it:

Thehalogroupsaretheonlyortho-paradirectors vatinggroups.

orientation

:Ci:

X

(| |l

\

T h e in d u c tiv e e ffe c t o f th e c h lo rin e a to m d e a c tiv a te s th e ring.

thataredeacti­ reactivity

ReviewProblem15.8 ReviewProblem15.9

704

Chapter 15

Reactions of Aromatic Compounds

On the other hand, when electrophilic attack does take place, the chlorine atom stabilizes the arenium ions resulting from ortho and para attack relative to that from meta attack. The chlorine atom does this in the same way as amino groups and hydroxyl groups do These electrons give rise to relatively stable resonance structures contributing to the hybrids for the ortho- and para-substituted arenium ions.

—bydonat­

inganunsharedpairofelectrons. OrthoAttack

Relatively stable contributor

MetaAttack :CI;

:Ch

■Ci-

:Ch

r O E

ParaAttack

cV :Ch I

:Ch I

E+

/\ E H

:C l ;

E

H

E

H

E

H

Relatively stable contributor

What we have said about chlorobenzene is also true of bromobenzene. We can summarize the inductive and resonance effects of halo substituents in the following way. • Through their electron-withdrawing inductive effect, halo groups make the ring more electron deficient than that of benzene. This causes the free energy of activa­ tion for any electrophilic aromatic substitution reaction to be greater than that for benzene, and, therefore, halo groups are deactivating. • Through their electron-donating resonance effect, however, halo substituents cause the free energies of activation leading to ortho and para substitution to be lower than the free energy of activation leading to meta substitution. This makes halo substituents ortho-para directors. You may have noticed an apparent contradiction between the rationale offered for the unusual effects of the halogens and that offered earlier for amino or hydroxyl groups. That is, oxygen is more electronegative than chlorine or bromine (and especially iodine). Yet the hydroxyl group is an activating group, whereas halogens are deactivating groups. An expla­ nation for this can be obtained if we consider the relative stabilizing contributions made to the transition state leading to the arenium ion by resonance structures involving a group — G (— G = — NH2, — O — H, — F :, — C l:, — Br:, — I :) that is directly attached to the benzene ring in which G donates an electron pair. If — G is — OH or — NH2, these resonance structures arise because of the overlap of a 2 p orbital of carbon with that of oxy­ gen or nitrogen. Such overlap is favorable because the atoms are almost the same size. With

CP"-

A

1 5 .1 1

H ow

S u b s t it u e n t s A f f e c t E le c t r o p h ilic A r o m a t ic S u b s titu tio n

705

chlorine, however, donation of an electron pair to the benzene ring requires overlap of a carbon 2p orbital with a chlorine orbital. Such overlap is less effective; the chlorine atom is much larger and its 3p orbital is much further from its nucleus. With bromine and iodine, overlap is even less effective. Justification for this explanation can be found in the obser­ vation that fluorobenzene (G = — R ) is the most reactive halobenzene in spite of the high electronegativity of fluorine and the fact that — R is the most powerful ortho-para director of the halogens. With fluorine, donation of an electron pair arises from overlap of a 2 orbital of fluorine with a 2p orbital of carbon (as with — NH2 and — O H). This overlap is effective

3p

p

because the orbitals of = C

and — R are of the same relative size.

\

Chloroethene adds hydrogen chloride more slowly than ethene, and the product is 1,1dichloroethane. How can you explain this using resonance and inductive effects? Cl\ ^

Cl-

HCl

Cl

15.11E Ortho-Para Direction and Reactivity of Alkylbenzenes Alkyl groups are better electron-releasing groups than hydrogen. Because of this, they can activate a benzene ring toward electrophilic substitution by stabilizing the transition state leading to the arenium ion: R

R

R

E

H

H

E

T ra n s itio n sta te is s ta b ilized .

E

A re n iu m ion is s ta b ilized .

For an alkylbenzene the free energy of activation of the step leading to the arenium ion (just shown) is lower than that for benzene, and alkylbenzenes react faster. Alkyl groups are ortho-para directors. We can also account for this property of alkyl groups on the basis of their ability to release electrons— an effect that is particularly important when the alkyl group is attached directly to a carbon that bears a positive charge. (Recall the ability of alkyl groups to stabilize carbocations that we discussed in Section 6.11 and in Fig. 6 .8 .) If, for example, we write resonance structures for the arenium ions formed when toluene undergoes electrophilic substitution, we get the results shown below:

OrthoAttack CH

CH

3

CH,

CH,

E+

R e la tiv e ly s ta b le c o n trib u to r

MetaAttack CH

c h

3

r O E

3

CH,

CH,

ReviewProblem15.10

7 06

C h a p te r 1 5

R e a c t io n s o f A r o m a t i c C o m p o u n d s

ParaAttack

R e la tive ly sta b le c o n trib u to r

In ortho attack and para attack we find that we can write resonance structures in which the methyl group is directly attached to a positively charged carbon of the ring. These structures are more stable relative to any of the others because in them the stabilizing influence of the methyl group (by electron release) is most effective. These structures, therefore, make a large (stabilizing) contribution to the overall hybrid for ortho- and para-substituted arenium ions. No such relatively stable structure contributes to the hybrid for the meta-substituted arenium ion, and as a result it is less stable than the ortho- or para-substituted arenium ions. Since the ortho- and para-substituted arenium ions are more stable, the transition states leading to them occur at lower energy and ortho and para substitutions take place most rapidly.

ReviewProblem15.11

ReviewProblem15.12

Write resonance structures for the arenium ions formed when ethylbenzene reacts with a Br+ ion (as formed from Br2 /FeBr3) to produce the following ortho and para products.

Provide a mechanism for the following reaction and explain why it occurs faster than nitra­ tion of benzene.

15.11F Summary of Substituent Effects on Orientation and Reactivity With a theoretical understanding now in hand of substituent effects on orientation and reac­ tivity, we refer you back to Table 15.2 for a summary of specific groups and their effects.

15.12 Reactions o f th e Side Chain o f Alkylbenzenes Hydrocarbons that consist of both aliphatic and aromatic groups are also known as arenes. Toluene, ethylbenzene, and isopropylbenzene are alkylbenzenes:

M eth ylb e n ze n e (to lu e n e )

E th ylb e n ze n e

Is o p ro p y lbenzene (cu m e n e )

P he nylethe ne (s ty re n e o r v in y lb e n z e n e )

Phenylethene, usually called styrene, is an example of an alkenylbenzene. The aliphatic portion of these compounds is commonly called the side chain.

S*nisA

1 5 .1 2

R e a c t io n s o f t h e S id e C h a in o f A lk y lb e n z e n e s

707

THE CHEMISTRY OF . . . I o d i n e I n c o r p o r a t i o n in T h y r o x i n e B i o s y n t h e s i s

The biosynthesis of thyroxine involves introduction of iodine atoms into tyrosine units of thyroglobin. This process occurs by a biochemical version of electrophilic aromatic substitu­ tion. An iodoperoxidase enzyme catalyzes the reaction between iodide anions and hydrogen peroxide to generate an electrophilic form of iodine (presumably a species like I— OH). Nucleophilic attack by the aromatic ring of tyrosine on the electrophilic iodine leads to incorporation of iodine at the 3 and 5 positions of the tyrosine rings in thyroglobulin. These are the positions ortho to the phenol hydroxyl group, precisely where we would expect electrophilic aro­ matic substitution to occur in tyrosine. (Substitution para to the hydroxyl cannot occur in tyrosine because that position is blocked, and substitution ortho to the alkyl group is less favored than ortho to the hydroxyl.) Electrophilic iodine is

also involved in the coupling of two tyrosine units necessary to complete biosynthesis of thyroxine. Electrophilic aromatic substitution also plays a role in the 1927 laboratory synthesis of thyroxine by C. Harington and G. Barger. Their synthesis helped prove the structure of this important hormone by comparison of the synthetic mater­ ial with natural thyroxine. Harington and Barger used elec­ trophilic aromatic substitution to introduce the iodine atoms at the ortho positions in the phenol ring of thyroxine. They used a different reaction, however, to introduce the iodine atoms in the other ring of thyroxine (nucleophilic aromatic substitution—a reaction we shall study in Chapter 21.) (Figure b e lo w a dapted w ith perm ission o f John W ile y & Sons, Inc. from Voet, D. and Voet, J. G., Biochem istry, 2nd editio n . © 1995 Voet D. and Voet, J. G.)

i odoperoxi dase

I- + H2O 2

I- + H2O 2

Thyroglobin (Two tyrosine groups are shown. The rem ainder of the thyroglobin protein is indicated by the shaded area.)

O + H 3N — CH — C — O -

I ch

2

protein hydrol ysi s r

I O

r

I OH

Thyroxine The biosynthesis o f thyroxine in the thyroid gland through the iodination, rearrangem ent, and hydrolysis (proteolysis) of thyroglobin Tyr residues. The relatively scarce I- is actively sequestered by the thyroid gland.

708

Chapter 15

Reactions o f Aromatic Compounds

15.12A Benzylic Radicals and Cations Hydrogen abstraction from the methyl group of methylbenzene (toluene) produces a radi­ cal called the benzyl radical: -C- H

•OH,

/ H . .CH, R

B e n zylic ca rb o n

RH

M ethylbe nzen e (to lu e n e )

A b e n z y lic hyd ro g e n

T he benzyl rad ica l

A b e n zylic radical

The name benzyl radical is used as a specific name for the radical produced in this reaction. The general name benzylic radical applies to all radicals that have an unpaired electron on the side-chain carbon atom that is directly attached to the benzene ring. The hydrogen atoms of the carbon atom directly attached to the benzene ring are called benzylic hydrogen atoms. A group bonded at a benzylic position is called a benzylic substituent. Departure of a leaving group (LG) from a benzylic position produces a benzylic cation: -O — LG

LG-

A b e n z y lic ca tio n

Benzylic radicals and benzylic cations are c onjugatedunsaturatedsystemsand bothareunusu­ allystable. They have approximately the same stabilities as allylic radicals and cations. This exceptional stability of benzylic radicals and cations can be explained by resonance theory. In the case of each entity, resonance structures can be written that place either the unpaired elec­ tron (in the case of the radical) or the positive charge (in the case of the cation) on an ortho or para carbon of the ring (see the following structures). Thus resonance delocalizes the unpaired electron or the charge, and this delocalization causes the radical or cation to be highly stabilized.

B e n z y lic ra d ic a ls are s ta b iliz e d by reso na nce.

B e n z y lic c a tio n s are s ta b iliz e d b y reso na nce.

Calculated structures for the benzyl radical and benzyl cation are presented in Fig. 15.6. These structures show the presence at their ortho and para carbons of unpaired electron den­ sity in the radical and positive charge in the cation, consistent with the resonance struc­ tures above. Figure 15.6

T h e g r a y lo b e s in t h e c a lc u la te d s tr u c tu r e f o r t h e b e n z y l

r a d ic a l (left) s h o w t h e lo c a tio n o f d e n s ity f r o m t h e u n p a ir e d e le c tr o n . T h is m o d e l in d ic a te s t h a t t h e u n p a ir e d e le c tr o n re s id e s p r im a r ily a t th e b e n z y lic , o r t h o , a n d p a ra c a r b o n s , w h ic h is c o n s is te n t w it h t h e r e s o n a n c e m o d e l f o r t h e b e n z y lic r a d ic a l d is c u s s e d e a rlie r. T h e c a lc u la te d e le c t r o s t a t ic p o t e n t ia l m a p f o r t h e b o n d in g e le c tr o n s in t h e b e n z y l c a tio n

(right) in d ic a te s t h a t p o s itiv e c h a rg e ( b lu e r e g io n s ) re s id e s p r im a r ily a t th e b e n z y lic , o r t h o , a n d p a ra c a r b o n s , w h ic h is c o n s is te n t w it h t h e r e s o n a n c e m o d e l f o r t h e b e n z y lic c a tio n . T h e v a n d e r W a a ls s u rfa c e o f b o th s tr u c tu r e s is r e p r e s e n te d b y t h e w ir e m e s h .

CP"-

A

15.12 Reactions of the Side Chain of Alkylbenzenes

709

THE CHEMISTRY OF . . . In d u strial S ty r e n e S y n th e sis

Styrene is one of the most important industrial chemicals— more than 11 billion pounds is produced each year. The starting material for a major commercial synthesis of styrene is ethylbenzene, produced by Friedel-Crafts alkylation of benzene:

catalyst 630°C *

Styrene (90-92% yield) Most styrene is polymerized (Special Topic A) to the famil­ iar plastic, polystyrene:

HCl

CH2= C H 2

AlCl

Ethylbenzene

catalyst CeH5 CeH

Ethylbenzene is then dehydrogenated in the presence of a catalyst (zinc oxide or chromium oxide) to produce styrene.

Polystyrene

15.12B Halogenation of the Side Chain: Benzylic Radicals We have already seen that we can substitute bromine and chlorine for hydrogen atoms on the of toluene and other alkylaromatic compounds using electrophilic aromatic sub­ stitution reactions. Chlorine and bromine can also be made to replace hydrogen atoms that are on a carbon, such as the methyl group of toluene.

ring benzylic

• Benzylic halogenation is carried out in the absence o f Lew is acids and under conditions that favor the formation of radicals. When toluene reacts with N-bromosuccinimide (NBS) in the presence of light, for exam­ ple, the major product is benzyl bromide. N-Bromosuccinimide furnishes a low concen­ tration of Br2, and the reaction is analogous to that for allylic bromination that we studied in Section 13.2B. O c h

3

N— Br

l i ght . CCL

N BS

(64% )

Side-chain chlorination of toluene takes place in the gas phase at 400-600°C or in the presence of U V light. When an excess of chlorine is used, multiple chlorinations of the side chain occur: CH,

H2

C H 2C l

C CL

C H C l,

CL

CL

CL

heat or light

heat or light

heat or light

Benzyl c h lo rid e

D ic h lo ro m e th y lb e n ze n e

T ric h lo ro m e th y lb e n ze n e

These halogenations take place through the same radical mechanism we saw for alkanes in Section 10.4. The halogens dissociate to produce halogen atoms and then the halo­ gen atoms initiate chain reactions by abstracting hydrogens of the methyl group. Benzylic halogenations are similar to allylic halogenations (Section 13.2) in that they involve the formation of (Section 15.12A).

unusuallystableradicals

• Benzylic and allylic radicals are even more stable than tertiary radicals.

CeH5

710

C h a p te r 1 5

R e a c t io n s o f A r o m a t i c C o m p o u n d s

A MECHANISM FOR THE REACTION B e n zy lic H a lo g é n a t io n Chain Initiation Step 1

peroxides, x 9 x

2

or light

X

P eroxides, heat, o r lig h t ca u se ha lo gen m o le c u le s to cle a ve in to ra d ica ls.

Chain Propagation Step 2 I

/H

C 6 H5— C — H

+

X

------ >

C 6 H 5— &

+

H

H— X

H

B enzyl rad ica l A h a lo g e n ra d ica l a b s tra c ts a b e n z y lic h yd ro g e n atom , fo rm in g a b e n z y lic ra d ica l and a m o le c u le o f th e h yd ro g e n ha lid e. Step 3

H

,

I

C eH5— C : '

+

I X— X

------->

C 6 H5— C— X

H H

+

X-

H

B enzyl rad ica l

B enzyl ha lid e

T he b e n z y lic rad ica l rea cts w ith a halo gen m o le c u le to fo rm th e b e n z y lic h a lid e p ro d u c t an d a ha lo gen rad ica l th a t p ro pa gate s the chain.

Chain Termination Step 4 C eH sC H f + a n d C 6H 5C H ^ +

-X -C H 2C 6H

5

-------- >

C

-------->

C 6 H 5 C H 2— C H

6

H 5C H -

X 2

C 6H

5

V ariou s ra d ica l c o u p lin g re a c tio n s te rm in a te th e chain.

The greater stability of benzylic radicals accounts for the fact that when ethylbenzene is halogenated, the major product is the 1-halo-1-phenylethane. The benzylic radical is formed much faster than the 1 ° radical: X X

1-H alo -1-p hen yle th an e (m a jo r p ro d u c t)

X X

1 ° R adical (le ss stab le)

1-H alo -2-p hen yle th an e (m in o r p ro d u c t)

S*nisA

15.12 Reactions o f the Side Chain o f Alkylbenzenes

When propylbenzene reacts with chlorine in the presence of U V radiation, the major prod­ uct is 1-chloro-1-phenylpropane. Both 2-chloro-1-phenylpropane and 3-chloro-1-phenylpropane are minor products. Write the structure of the radical leading to each product and account for the fact that 1 -chloro-1 -phenylpropane is the major product.

711

Review Problem 15.13

SolvedProblem15.5 ILLUSTRATING A MULTISTEP SYNTHESIS Show how phenylacetylene (C6 H5 C # CH) could be synthesized from ethylbenzene (phenylethane). Begin by writing a retrosynthetic analysis, and then write reactions needed for the synthesis. ANSWER Working backward, that is, using retrosynthetic analysis, we find that we can easily envision two syn­

theses of phenylacetylene. We can make phenylacetylene by dehydrohalogenation of 1,1-dibromo-1-phenylethane, which could have been prepared by allowing ethylbenzene (phenylethane) to react with 2 mol of NBS. Alternatively, we can prepare phenylacetylene from 1 ,2 -dibromo-1 -phenylethane, which could be prepared from styrene (phenylethene). Styrene can be made from 1-bromo-1-phenylethane, which can be made from ethylbenzene. Brv

Br

Br Br

Following are the synthetic reactions we need for the two retrosynthetic analyses above: Brs

Br

NBS (2 equiv.), light CCI 4

(1) NaNH2, mineral oil, heat (2 ) H3 O+

or Br NBS, light c c i4

:

[

Br KOH, heat

[i

J

[1

1

L

J

^

Br ^

Br2, CCI4

Br Br (1) NaNH2, mineral oil, heat (2) H 3 O+

:

Show how the following compounds could be synthesized from phenylacetylene (C 6 H5 C # CH): ( a ) 1-phenylpropyne, (b ) 1-phenyl-1-butyne, (c ) (Z)-l-phenylpropene, and ( d ) (£)-1-phenylpropene. Begin each synthesis by writing a retrosynthetic analysis.

ReviewProblem15.14

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15.13 Alkenylbenzenes

15.13A Stability of Conjugated Alkenylbenzenes • Alkenylbenzenes that have their side-chain double bond conjugated with the ben­ zene ring are more stable than those that do not: / \

/ C = C

\

\ is more stable than

/ C = C

/

/%

\

A C o n ju g a te d sy s te m

N o n c o n ju g a te d sys te m

Part of the evidence for this comes from acid-catalyzed alcohol dehydrations, which are known to yield the most stable alkene (Section 7.8A). For example, dehydration of an alco­ hol such as the one that follows yields exclusively the conjugated system:

V/

\

^

C = C

H

i

H

OH

Because conjugation always lowers the energy of an unsaturated system by allowing the p electrons to be delocalized, this behavior is just what we would expect.

15.13B Additions to the Double Bond of Alkenylbenzenes In the presence of peroxides, hydrogen bromide adds to the double bond of 1-phenylpropene to give 2 -bromo-1 -phenylpropane as the major product:

1 -P h e n y lp ro p e n e

2 -B ro m o -1 -p h e n y lp ro p a n e

In the absence of peroxides, HBr adds in just the opposite way: Br

1 -P h e n y lp ro p e n e

1 -B ro m o -1 -p h e n y lp ro p a n e

The addition of hydrogen bromide to 1-phenylpropene proceeds through a benzylic rad­ ical in the presence of peroxides and through a benzylic cation in their absence (see Review Problem 15.15 and Section 10.9).

ReviewProblem15.15

Write mechanisms for the reactions whereby HBr adds to 1-phenylpropene (a ) in the presence of peroxides and (b ) in the absence of peroxides. In each case account for the regiochemistry of the addition (i.e., explain why the major product is 2 -bromo-1 -phenylpropane when per­ oxides are present and why it is 1 -bromo-1 -phenylpropane when peroxides are absent).

15.13 Alkenylbenzenes

(a)

W h a t w o u ld y o u e x p e c t to b e th e m a jo r p ro d u c t w h e n 1 -p h e n y lp ro p e n e re a c ts w it h H C l?

(b)

W h a t p ro d u c t w o u ld y o u e x p e c t w h e n i t is s u b je c te d to o x y m e rc u ra tio n -d e m e rc u ra tio n ?

15.13C Oxidation of the Side Chain S tro n g o x id iz in g a g e n ts o x id iz e to lu e n e to b e n z o ic a c id . T h e o x id a tio n ca n b e c a rr ie d o u t b y th e a c tio n o f h o t a lk a lin e p o ta s s iu m p e rm a n g a n a te . T h is m e th o d g iv e s b e n z o ic a c id in a lm o s t q u a n tita tiv e y ie ld : O

CH

OH

(1) KMnO4, OH-, heat (2 ) H3O+ '

B e n zo ic acid ( ~ 1 0 0 %) A n im p o r ta n t c h a ra c te ris tic o f s id e -c h a in o x id a tio n s is th a t o x id a tio n ta ke s p la c e in it ia lly a t th e b e n z y lic c a rb o n . •

A lk y lb e n z e n e s w it h a lk y l g ro u p s lo n g e r th a n m e t h y l a re u lt im a te ly d e g ra d e d to b e n z o ic a cid s:

c h 2r

OH

(1) KMnO4, OH-, heat (2 ) H3O+ : An a lk y lb e n ze n e

B enzo ic acid

S id e -c h a in o x id a tio n s a re s im ila r to b e n z y lic h a lo g e n a tio n s , b e ca u se in th e f ir s t ste p th e o x id iz in g a g e n t a b s tra c ts a b e n z y lic h y d ro g e n . O n c e o x id a tio n is b e g u n a t th e b e n z y lic c a r­ b o n , i t c o n tin u e s a t th a t s ite . U lt im a t e ly , th e o x id iz in g a g e n t o x id iz e s th e b e n z y lic c a rb o n to a c a r b o x y l g ro u p , a n d , in th e p ro c e s s , i t cle a v e s o f f th e r e m a in in g c a rb o n a to m s o f th e s id e c h a in . ( te r t- B u ty lb e n z e n e is re s is ta n t to s id e -c h a in o x id a tio n . W h y ? ) •

S id e -c h a in o x id a tio n is n o t re s tr ic te d to a lk y l g ro u p s . A lk e n y l, a lk y n y l, a n d a c y l g ro u p s a re a ls o o x id iz e d b y h o t a lk a lin e p o ta s s iu m p e rm a n g a n a te .

15.13D Oxidation of the Benzene Ring T h e b e n z e n e r in g c a rb o n w h e re a n a lk y l g ro u p is b o n d e d c a n b e c o n v e rte d to a c a r b o x y l g ro u p b y o z o n o ly s is , f o llo w e d b y tre a tm e n t w it h h y d ro g e n p e ro x id e .

O (1 ) O„ CH3 CO2H

R — C sH

l iFH O

'

R—

COH

i H P ’i & r —

713

R e v ie w P r o b le m 1 5 .1 6

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15 .1 4 Synthetic Applications The substitution reactions of aromatic rings and the reactions of the side chains of alkyland alkenylbenzenes, when taken together, offer us a powerful set of reactions for organic synthesis. By using these reactions skillfully, we shall be able to synthesize a large num­ ber of benzene derivatives.

• Part of the skill in planning a synthesis is deciding in what order to carry out the reactions. Let us suppose, for example, that we want to synthesize o-bromonitrobenzene. We can see very quickly that we should introduce the bromine into the ring first because it is an ortho-para director: Br

Br

Br NO,

Br,

HNO,

FeBr3

H2 SO 4

NO

o-Bromonitrobenzene

p-Bromonitrobenzene

The ortho and para products can be separated by various methods because they have dif­ ferent physical properties. However, had we introduced the nitro group first, we would have obtained m-bromonitrobenzene as the major product. Other examples in which choosing the proper order for the reactions is important are the syntheses of the and para-nitrobenzoic acids. Because the methyl group of toluene is an electron-donating group (shown in red below), we can synthesize the and para-nitrobenzoic acids from toluene by nitrating it, separating the and nitrotoluenes, and then oxidizing the methyl groups to carboxyl groups:

ortho-,meta-,

orthoortho- para-

ch3

C O 2H

3

,

A

.

(1) (2)

C H

KMnO4, OH-, heat H3O+

V N O

NO

2

p-Nitrotoluene (separate ortho from para)

HNO3 H2 SO 4

p-Nitrobenzoic acid C O

C H 33 N O 2

a

(1)

2

KMnO4, OH- , heat

H NO

(2 ) H3 O+

o-Nitrotoluene

o-Nitrobenzoic acid

m

We can synthesize -nitrobenzoic acid by reversing the order of the reactions. We oxidize the methyl group to a carboxylic acid, then use the carboxyl as an electron-withdrawing group (shown in blue) to direct nitration to the meta position.

Benzoic acid

m-Nitrobenzoic acid

CP"-

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A

15.14 Synthetic Applications

715

SolvedProblem15.6 Starting with toluene, outline a synthesis of (a) 1-bromo-2-trichloromethylbenzene, (b) 1-bromo-3-trichloromethylbenzene, and (c) 1-bromo-4-trichloromethylbenzene. ANSWER Compounds (a) and (c) can be obtained by ring bromination of toluene followed by chlorination of the side chain using three molar equivalents of chlorine:

(c) To make compound (b), we reverse the order of the reactions. By converting the side chain to a — CCl3 group first, we create a meta director, which causes the bromine to enter the desired position:

(b)

Suppose you needed to synthesize m-chloroethylbenzene from benzene. Cl

You could begin by chlorinating benzene and then follow with a Friedel-Crafts alkylation using chloroethane and AlCl3, or you could begin with a Friedel-Crafts alkylation followed by chlorination. Neither method will give the desired product, however. (a) Why will neither method give the desired product? (b) There is a three-step method that will work if the steps are done in the right order. What is this method?

15.14A Use of Protecting and Blocking Groups •

Very powerful activating groups such as amino groups and hydroxyl groups cause the benzene ring to be so reactive that undesirable reactions may take place.

Some reagents used for electrophilic substitution reactions, such as nitric acid, are also strong

oxidizing agents. Both electrophiles and oxidizing agents seek electrons. Thus, amino groups and hydroxyl groups not only activate the ring toward electrophilic substitution but also acti­ vate it toward oxidation. Nitration of aniline, for example, results in considerable destruction of the benzene ring because it is oxidized by the nitric acid. Direct nitration of aniline, conse­ quently, is not a satisfactory method for the preparation of o - and p -nitroaniline.

ReviewProblem15.17

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Treating aniline with acetyl chloride, CH3 COCl, or acetic anhydride, (CH3 CO)2 O, con­ verts the amino group of aniline to an amide, (specifically an acetamido group, — NHCOCH3), forming acetanilide. An amide group is only moderately activating, and it does not make the ring highly susceptible to oxidation during nitration (see Review Problem 15.9). Thus, with the amino group of aniline blocked in acetanilide, direct nitration becomes possible: O

O

O

NH„ NO,

CH3COCl base

NO

A n ilin e

A c e ta n ilid e

p -N itro a ce ta n ilid e (90%)

o-N itroa ce ta n ilid e (trace)

(1) H 2 O, H2 SO4, heat (2) OH-

T h is ste p rem o ves O

NH, O

I th e CH3C— g r o u p "

o-

and re p la ce s it w ith an — H. NO,

p -N itro a n ilin e

Nitration of acetanilide gives p-nitroacetanilide in excellent yield with only a trace of the ortho isomer. Acidic hydrolysis of p-nitroacetanilide (Section 18.8F) removes the acetyl group and gives p-nitroaniline, also in good yield. Suppose, however, that we need o-nitroaniline. The synthesis that we just outlined would obviously not be a satisfactory method, for only a trace of -nitroacetanilide is obtained in the nitration reaction. (The acetamido group is purely a para director in many reactions. Bromination of acetanilide, for example, gives p-bromoacetanilide almost exclusively.) We can synthesize -nitroaniline, however, through the reactions that follow:

o

o

O

O

O

NH NO,

NO,

( 1 ) H2O

A c e ta n ilid e

SO

3

H

SO

3

H

H 2 SO 4 , heat (2) OHo -N itro a n ilin e

(56%)

Here we see how a sulfonic acid group can be used as a “blocking group.” We can remove the sulfonic acid group by desulfonation at a later stage. In this example, the reagent used for desulfonation (dilute H2 SO4) also conveniently removes the acetyl group that we employed to “protect” the benzene ring from oxidation by nitric acid.

15.14B Orientation in Disubstituted Benzenes • When two different groups are present on a benzene ring, the more powerful acti­ vating group (Table 15.2) generally determines the outcome of the reaction.

A 15.15 Allylic and Benzylic Halides in Nucleophilic Substitution Reactions

717

p

Let us consider, as an example, the orientation of electrophilic substitution of -methylacetanilide. The amide group is a much stronger activating group than the methyl group. The following example shows that the amide group determines the outcome of the reaction. Substitution occurs primarily at the position ortho to the amide group: O

CH

O

CH

3

O

CH

3

M ajor p ro d u c t

3

M in o r p ro d u c t

• An ortho-para director takes precedence over a meta director in determining the position of substitution because all ortho-para-directing groups are more activating than meta directors. Steric effects are also important in aromatic substitutions. • Substitution does not occur to an appreciable extent between meta substituents if another position is open. A good example of this effect can be seen in the nitration of m-bromochlorobenzene: Cl

Cl

Cl

Cl

Only 1% of the mononitro product has the nitro group between the bromine and chlorine. Predict the major product (or products) that would be obtained when each of the following compounds is nitrated: OH

CN

ReviewProblem15.18

OCHo

(b)

(c)

SO 3H

no2

C F

15.15 Allylic and Benzylic Halides in Nucleophilic Substitution Reactions Allylic and benzylic halides can be classified in the same way that we have classified other organic halides:

v

c h 2x

\

/

\

C=C /

H

: /

^X

\

C=C \

1° A lly lic

/

/

^X

Ar— CH2X

Ar— C— X

1° B e n zylic

2° B e n zylic

R Ar— C— X

C=C \

2° A lly lic

/

\ 3° A lly lic

R

R1 3° B e n zylic

718

Chapter 15

Reactions o f Aromatic Compounds

A ll of these compounds undergo nucleophilic substitution reactions. As with other tertiary halides (Section 6.13A), the steric hindrance associated with having three bulky groups on the carbon bearing the halogen prevents tertiary allylic and tertiary benzylic halides from react­ ing by an SN2 mechanism. They react with nucleophiles only by an SN1 mechanism. Primary and secondary allylic and benzylic halides can react either by an SN2 mechanism or by an SN1 mechanism in ordinary nonacidic solvents. We would expect these halides to react by an SN2 mechanism because they are structurally similar to primary and secondary alkyl halides. (Having only one or two groups attached to the carbon bearing the halogen does not prevent SN2 attack.) But primary and secondary allylic and benzylic halides can also react by an SN1 mechanism because they can form relatively stable allylic carbocations and benzylic carbocations, and in this regard they differ from primary and secondary alkyl halides.* • Overall we can summarize the effect of structure on the reactivity of alkyl, allylic, and benzylic halides in the ways shown in Table 15.3. A Sum m ary o f A lkyl, Allylic, and Benzylic Halides in SN Reactions These halides give mainly S N2 reactions.

These halides may give either S N1 or S N2 reactions. CH29 X 2 Ar—CH2—X 2

Ar—CH—X 1 R R

9

/

H \

C =C

/

\ X

/

c=c

X

T l

-01

R' Rs

O \

1

R" R

C=C

>

f

R'

R 1

R X

R—CH—X

-0­

R—CH2—X

R

CH3—X

These halides give mainly S N1 reactions.

R' / c. / ^X \

SolvedProblem15.7 When either enantiomer of 3-chloro-1-butene [(R) or (5)] is subjected to hydrolysis, the products of the reaction are optically inactive. Explain these results. ANSWER The solvolysis reaction is SN1. The intermediate allylic cation is achiral and therefore reacts with water

to give the enantiomeric 3-buten-2-ols in equal amounts and to give some of the achiral 2-buten-1-ol: Cl

(R) or (S)

OH

Achiral

Racemic +

Achiral (but two diastereomers are possible)

Review Problem 15.19

Account for the following observations with mechanistic explanations. (a)

0

i

EtONa (in high concentration)

EtO

EtOH

At high concentration of ethoxide, the rate depends on both the allylic halide and ethoxide concentrations. *There is som e dispute as to whether 2° alkyl halides react by an S n 1 m echanism to any appreciable extent in ordinary nonacidic solvents such as mixtures of water and alcohol or acetone, but it is clear that reaction by an SN2 m echanism is, for all practical purposes, the more im portant pathway.

CP"-

A

15.16 Reduction of Aromatic Compounds

(b )

OEt

EtONa (in low concentration)

CU

719

EtOv

At low concentration of ethoxide, the rate depends only on the allylic halide concentration. 1-Chloro-3-methyl-2-butene undergoes hydrolysis in a mixture of water and dioxane at a rate that is more than a thousand times that of 1-chloro-2-butene. (a ) What factor accounts for the difference in reactivity? ( b ) What products would you expect to obtain? [Dioxane is a cyclic ether (below) that is miscible with water in all proportions and is a useful cosol­ vent for conducting reactions like these. Dioxane is carcinogenic (i.e., cancer causing), how­ ever, and like most ethers, it tends to form peroxides.]

ReviewProblem15.20

O D io xan e

Primary halides of the type ROCH2X apparently undergo SN1-type reactions, whereas most primary halides do not. Can you propose a resonance explanation for the ability of halides of the type ROCH2X to undergo SN1 reactions? The following chlorides (Ph = phenyl) undergo solvolysis in ethanol at the relative rates given in parentheses. How can you explain these results? Ph Ph^

Cl

Ph^

(0 .0 8 )

Cl

Ph

(1 )

ReviewProblem15.21 ReviewProblem15.22

Ph Cl

(3 0 0 )

Phy Ph

Cl

(3 X 1 0 6)

15.16 Reduction o f A rom atic Compounds Hydrogenation of benzene under pressure using a metal catalyst such as nickel results in the addition of three molar equivalents of hydrogen and the formation of cyclohexane (Section 14.3). The intermediate cyclohexadienes and cyclohexene cannot be isolated because these undergo catalytic hydrogenation faster than benzene does. H /N i slow B e n ze n e

H /N i N

H

[I

fast

C y c lo h e x a d ie n e s

H /N i |l

I

C y c lo h e x e n e

fast C y c lo h e x a n e

15.16A The Birch Reduction Benzene can be reduced to 1,4-cyclohexadiene by treating it with an alkali metal (sodium, lithium, or potassium) in a mixture of liquid ammonia and an alcohol. This reaction is called the Birch reduction, after A. J. Birch, the Australian chemist who developed it. Na NH3, EtOH B e n ze n e

1 ,4 -C y c lo h e x a d ie n e

The Birch reduction is a dissolving metal reduction, and the mechanism for it resembles the mechanism for the reduction of alkynes that we studied in Section 7.15B. A sequence of electron transfers from the alkali metal and proton transfers from the alcohol takes place, lead­ ing to a 1,4-cyclohexadiene. The reason for formation of a 1,4-cyclohexadiene in preference to the more stable conjugated 1,3-cyclohexadiene is not understood.

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A M E C H A N IS M

Ù L

F O R T H E R E A C T IO N

B irch R e d u c t i o n

T he firs t electron transfer produces a delocalized b en zene radical Benzene

B enzene radical anion EtOH

Protonation produces a cyclohexadienyl radical (also a delocalized species).

etc.

H

H H

H

C yclohexadienyl radical Na-

H H

■etc.

H

H

EtOH

H

H

H

H

1,4-C yclohexadiene

C yclohexadienyl anion

T ransfer of another electron leads to the form ation of a delocalized cyclohexadienyl anion, and protonation o f this p ro du ces the 1,4-cyclohexadiene.

Substituent groups on the benzene ring influence the course of the reaction. Birch reduc­ tion of methoxybenzene (anisole) leads to the formation of 1-methoxy-1,4-cyclohexadiene, a compound that can be hydrolyzed by dilute acid to 2-cyclohexenone. This method pro­ vides a useful synthesis of 2 -cyclohexenones:

M e th o x y b e n ze n e (a n is o le )

ReviewProblem15.23

1 -M e th o x y -1 ,4 c y c lo h e x a d ie n e (84% )

2 -C y c lo h e x e n o n e

Birch reduction of toluene leads to a product with the molecular formula C7 H10. On ozonolysis followed by reduction with dimethyl sulfide, the product is transformed into O O and o O . What is the structure of the Birch reduction product?

Key Terms and Concepts W

eyo

PLUS

The key terms and concepts that are highlighted in b o ld , b lu e t e x t within the chapter are defined in the glossary (at the back of the book) and have hyperlinked definitions in the accompanying course (www.wileyplus.com).

WileyPLUS

CP"-

h r * i-

A

Problems

721

Problem s Note to Instructors: Many of the homework problems are available for assignment via Wiley PLUS, an online teaching and learning solution.

M EC H A N IS M S 15.24

Provide a detailed mechanism for each of the following reactions. Include contributing resonance structures and the resonance hybrid for the arenium ion intermediates. Br (a )

h n o 3, h 2s o 4

Br2, FeBr3

(b)

NO,

(c)

\^ B r AlBr3

15.25

Provide a detailed mechanism for the following reaction.

H,SO.

H,O

15.26

One ring of phenyl benzoate undergoes electrophilic aromatic substitution much more readily than the other. (a) Which one is it? (b) Explain your answer.

15.27

Many polycyclic aromatic compounds have been synthesized by a cyclization reaction known as the Bradsher reac­ tion or aromatic cyclodehydration. This method can be illustrated by the following synthesis of 9-methylphenanthrene:

11 I Ii f Jl

11

] 1 /^ O

HBr acetic acid, heat

9-Methylphenanthrene

An arenium ion is an intermediate in this reaction, and the last step involves the dehydration of an alcohol. Propose a plausible mechanism for this example of the Bradsher reaction. 15.28

Write mechanisms that account for the products of the following reactions:

HA (a )

( - H 2O)

OH

(b)

HA

722 15.29

Chapter 15

Reactions o f Aromatic Compounds

The addition of a hydrogen halide (hydrogen bromide or hydrogen chloride) to 1-phenyl-1,3-butadiene produces (only) 1-phenyl-3-halo-1-butene. (a) Write a mechanism that accounts for the formation of this product. (b) Is this 1,4 addition or 1,2 addition to the butadiene system? (c) Is the product of the reaction consistent with the forma­ tion of the most stable intermediate carbocation? (d) Does the reaction appear to be under kinetic control or equi­ librium control? Explain.

REACTIONS A N D SYNTHESIS 15.30

Predict the major product (or products) formed when each of the following reacts with Cl2 and FeCl3: (a) Ethylbenzene

15.31

(e) Nitrobenzene

(b) Anisole (methoxybenzene)

(f) Chlorobenzene

(c) Fluorobenzene

(g) Biphenyl (C 6 H 5 — C 6 H5)

(d) Benzoic acid

(h) Ethyl phenyl ether

Predict the major product (or products) formed when each of the following reacts with a mixture of concentrated HNO 3 and H2 SO4. (a )

O

(c )

4-Chlorobenzoic acid

(d )

3-Chlorobenzoic acid

N H

O (e)

A c e ta n ilid e

O (b)

B e n zo p h e n o n e

O P h enyl a c etate

15.32

What monobromination product (or products) would you expect to obtain when the following compounds undergo ring bromination with Br2 and FeBr3?

(a)

15.33

(b)

(c)

Predict the major products of the following reactions: (a)

(d )

HCl

Product of (c) + HBr

peroxides

HA

(e) Product of (c) + H2O

S ty re n e

(b) 2-Bromo-1-phenylpropane

EtONa

(c)

heat

(f) Product of (c) + H2 (1 molar equivalent)

(g) Product of (f)

Pt 25°C

(1) KMnO4, O H - , heat (H) H3 O+

HA, heat

15.34

Starting with benzene, outline a synthesis of each of the following: (a) Isopropylbenzene

(f) 1-Phenylcyclopentene

(k) p-Chlorobenzenesulfonic acid

(b) tert-Butylbenzene

(g) irans-2-Phenylcyclopentanol

(l)

(c) Propylbenzene

(h) m-Dinitrobenzene

(m) m -Nitrobenzenesulfonic acid

(d) Butylbenzene

(i) m-Bromonitrobenzene

(e) 1-tert-Butyl-4-chlorobenzene

(j) p-Bromonitrobenzene

o-Chloronitrobenzene

CP"-

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Problems 15.35

723

Starting with styrene, outline a synthesis of each of the following: Cl

X N

Br (k ) C 6 H 5 Cl

(a ) C 6 H 5

( f)

C6 H5 " D OH

(b ) C 6 H 5 '

D

(g ) C 6H 5

( l)

OH

C6H5

D (h ) C 6 H 5

„O H (C) C 6 H 5

Br O

(d ) C 6 H ^

( i)

C6 H5

( j)

C6 H5

^O H

■

\

OH

(e ) C 66H5 H

15.36

15.37

15.38

Starting with toluene, outline a synthesis of each of the following: (a )

m-Chlorobenzoic acid

( f)

p-Isopropyltoluene (p-cymene)

(b )

p-Methylacetophenone

(g )

1-Cyclohexyl-4-methylbenzene

(c )

2-Bromo-4-nitrotoluene

(h )

2,4,6-Trinitrotoluene (TNT)

(d )

p-Bromobenzoic acid

( i)

4-Chloro-2-nitrobenzoic acid

(e )

1-Chloro-3-trichloromethylbenzene

(j)

1-Butyl-4-methylbenzene

Starting with aniline, outline a synthesis of each of the following: (a )

p -Bromoaniline

(d )

4-Bromo-2-nitroaniline

(b )

o -Bromoaniline

(e )

2,4,6-Tribromoaniline

(c )

2-Bromo-4-nitroaniline

Both of the following syntheses will fail. Explain what is wrong with each one. NO (1) NBS, CCl4, light

(1) HNO3/H 2SO4 (2) CH3COCl/AlCl3 (a )

(b )

(3) Zn(Hg), HCl

(2) NaOEt, EtOH, heat (3) Br2, FeBr3

Br

15.39

Propose structures for compounds G -I: OH concd H2SOt 60-65°C

G (C6 H6 S2 O8)

concd HNO3 concd H2SOt

H (C6 H5 NS2 O10)

h 3o +, h 2o

heat

I (C6 H5 NO4)

OH 15.40

2,6-Dichlorophenol has been isolated from the females of two species of ticks (Amblyomma americanum and A. maculatum), where it apparently serves as a sex attractant. Each female tick yields about 5 ng of 2,6-dichlorophenol. Assume that you need larger quantities than this and outline a synthesis of 2,6-dichlorophenol from phenol. [Hint: When phenol is sulfonated at 100°C, the product is chiefly p-hydroxybenzenesulfonic acid.]

724 15.41

Chapter 15

Reactions o f Aromatic Compounds

2-Methylnaphthalene can be synthesized from toluene through the following sequence of reactions. Write the struc­ ture of each intermediate.

Toluene

+

------ ------------» A (C11 H12O3)

B (CiiH 14 O2)

C (C11H13CIO)

heat

15.42

D (C11 H12 O)

F ^

C C i^

G (C- Hi2Br)

EtOH, heat

Show how you might synthesize each of the following starting with a-tetralone (Section 15.9):

(

a

)

^

n

(

b

)

^

^

(C)

OH

15.43

E (C11H14O)

•

OH

C 6H 5

Give structures (including stereochemistry where appropriate) for compounds A -G : O ,

n

(a) Benzene +

II

AICI3

CI

--------- : A

PCI5 o°C

,

2 NaNH2

> B ( C 9H 10C I2) — ------— —: v 9 10

2

(Section 7.10)

mineraI oiI,

.

H2, Ni2B (P-2)

,

C ( C 9H S) ---------------------: D ( C 9H 10) v 9 81

K 9

1 0

heat

[Hint: The 1H N M R spectrum of compound C consists of a multiplet at 8 7.20 (5H) and a singlet at 8 2.0 (3H).]

(b) C

I1) U EtNH2 : _ (C H ^ (2) NH4CI (—ection 7.1 5B) E (C9H10j

Br2 (c) D ——, , —: F + enantiomer (major products) CCI4 , 2-5 C Br2 (d) E ——, , > G + enantiomer (major products) CCI4 , 2-5 C

G EN ER A L PROBLEM S

15.44

Show how you might synthesize each of the following compounds starting with either benzyl bromide or allyl bromide: (a) C6 H5^ ^ C N (b) C6 H5

OMe

O

(e)

(c) C6 H5 / ~'""O'' W 6 5

N3

O

(d) C6 H ^ ^ " I 15.45

/

()

/

■

Provide structures for compounds A and B: Benzene Iiq. NH a EtOH A (C6H8) 9 CCiT B (C6H7Br)

15.46

Ring nitration of a dimethylbenzene (a xylene) results in the formation of only one dimethylnitrobenzene. Which dimethylbenzene isomer was the reactant?

15.47

The compound phenylbenzene (C 6 H5 — C 6 H5) is called biphenyl, and the ring carbons are numbered in the fol­ lowing manner: 3 __ 2 2 __ 3 '

5

6

6' 5'

CP"-

h r * i-

A

Challenge Problems

725

Use models to answer the following questions about substituted biphenyls. (a) When certain large groups occupy three or four of the positions (e.g., 2 , 6 , 2 ', and 6 '), the substituted biphenyl may exist in enantiomeric forms. An example of a biphenyl that exists in enantiomeric forms is the compound in which the following substituents are present: 2-NO2, 6 -CO 2 H, 2'-NO 2, 6 '-CO 2 H. What factors account for this? (b) Would you expect a biphenyl with 2-Br, 6 -CO 2 H, 2'-CO 2 H, 6 '-H to exist in enantiomeric forms? (c) The biphenyl with 2-NO2, 6 -NO2, 2'-CO 2 H, 6 '-Br cannot be resolved into enantiomeric forms. Explain.

ortho

15.48

Treating cyclohexene with acetyl chloride and AlCl3 leads to the formation of a product with the molecular for­ mula C8 H 13 ClO. Treating this product with a base leads to the formation of 1-acetylcyclohexene. Propose mech­ anisms for both steps of this sequence of reactions.

15.49

The group can be used as a blocking group in certain syntheses of aromatic compounds. (a) How would you introduce a tert-butyl group? (b) How would you remove it? (c) What advantage might a tert-butyl group have over a — SO3H group as a blocking group?

15.50

When toluene is sulfonated (concentrated H2 SO4) at room temperature, predominantly (about 95% of the total) ortho and para substitution occurs. If elevated temperatures (150-200°C) and longer reaction times are employed, meta (chiefly) and para substitution account for some 95% of the products. Account for these differences in terms of kinetic and thermodynamic pathways. m-Toluenesulfonic acid is the most stable isomer.]

tert-butyl

[Hint:

15.51

A C — D bond is harder to break than a C— H bond, and, consequently, reactions in which C— D bonds are bro­ ken proceed more slowly than reactions in which C — H bonds are broken. What mechanistic information comes from the observation that perdeuterated benzene, C6 D6, is nitrated at the same rate as normal benzene, C6 H6?

15.52

Heating 1,1,1-triphenylmethanol with ethanol containing a trace of a strong acid causes the formation of 1-ethoxy1,1,1-triphenylmethane. Write a plausible mechanism that accounts for the formation of this product.

15.53

(a) Which of the following halides would you expect to be most reactive in an SN2 reaction? (b) In an SN1 reaction? Explain your answers.

Br Br

Br

^

Challenge Problems 15.54

2

Furan undergoes electrophilic aromatic substitution. Use resonance structures for pos­ sible arenium ion intermediates to predict whether furan is likely to undergo bromination more rapidly at C2 or at C3.

O Br2, FeBr3

3

Furan 15.55

Acetanilide was subjected to the following sequence of reactions: (1) concd H2 SO4; (2) HNO3, heat; (3) H2 O, H2 SO4, heat, then OH- . The 13C NM R spectrum of the final product gives six signals. Write the structure of the final product.

15.56

The lignins are macromolecules that are major components of the many types of wood, where they bind cellulose fibers together in these natural composites. The lignins are built up out of a variety of small molecules (most having phenylpropane skeletons). These precursor molecules are covalently connected in varying ways, and this gives the lignins great complexity. To explain the formation of compound B below as one of many products obtained when lignins are ozonized, lignin model compound A was treated as shown. Use the fol­ lowing information to determine the structure of B.

(1) NaBH 4

(2) O 3

(3) H2O

B

To make B volatile enough for GC/MS (gas chromatography-mass spectrometry, Section 9.19), it was first con­ verted to its tris(O-trimethylsilyl) derivative, which had 308 [“Tris” means that three of the indicated com­ plex groups named (e.g., trimethylsilyl groups here) are present. The capital, italicized means these are attached to oxygen atoms of the parent compound, taking the place of hydrogen atoms. Similarly, the prefix “bis” indicates the presence of two complex groups subsequently named, and “tetrakis” (used in the problem below), means four.] The IR spectrum of B had a broad absorption at 3400 cm-1 , and its 1H NM R spectrum showed a single multiplet at 3.6. What is the structure of B?

m/z.

8

O

726 15.57

Chapter 15

Reactions o f Aromatic Compounds

When compound C, which is often used to model a more frequently occurring unit in lignins, was ozonized, prod­ uct D was obtained. In a variety of ways it has been estab­ lished that the stereochemistry of the three-carbon side chain of such lignin units remains largely if not completely unchanged during oxidations like this.

OMe

OH

o3

h 2o

D

C

m/z.

For GC/MS, D was converted to its tetrakis(O-trimethylsilyl) derivative, which had 424 The IR spectrum of D had bands at 3000 cm- 1 (broad, strong) and 1710 cm- 1 (strong). Its 1H NM R spectrum had peaks at 8 3.7 (multiplet, 3H) and 8 4.2 (doublet, 1H) after treatment with D2 O. Its DEPT 13C NM R spectra had peaks at 8 64 (CH2), 8 75 (CH), 8 82 (CH), and 8 177 (C). What is the structure of D, including its stereochemistry?

Learning Group Problems The structure of thyroxine, a thyroid hormone that helps to regulate metabolic rate, was determined in part by com­ parison with a synthetic compound believed to have the same structure as natural thyroxine. The final step in the laboratory synthesis of thyroxine by Harington and Barger, shown below, involves an electrophilic aromatic sub­ stitution. Draw a detailed mechanism for this step and explain why the iodine substitutions occur ortho to the phe­ nolic hydroxyl and not ortho to the oxygen of the aryl ether. [One reason iodine is required in our diet (e.g., in iodized salt), of course, is for the biosynthesis of thyroxine.] I HO

I

I

'2’ HO > HO

O

CO„

CO„

HsN^

T h y ro x in e

Synthesize 2-chloro-4-nitrobenzoic acid from toluene and any other reagents necessary. Begin by writing a retrosynthetic analysis. Deduce the structures of compounds E -L in the roadmap below. HBr

E (CgH^Br)

F (C8 H14B y meso

(no peroxides)

G (C8 HMB y racemate

f-BuOK’ f-BuOH’ heat

Br

Br2

Br

warm

»CO2Et

H (C8H12)

CO2Et

I

(1 ) O 3 (2) Me2S

O

O O CO2Et

EtO2C OO O

J (C12H14O3 ) AlCl3 ch3

K (C19 H22O3)

(1) Zn(Hg), HCl, reflux (2) SOCl2 (3) AlCl3

'

L

Zn(Hg), HCl, reflux

CH

CONCEPT MAP Summary of Mechanisms

Electrophilic Aromatic Substitution involves reaction between An aromatic ring

An electrophile

and

is s '

can be

Activated toward electrophilic substitution

is formed from

Deactivated toward electrophilic substitution

by

X+ N 0 2+ S03 R+ RCO+

by

>• Electron-donating groups include.

\/ Electron-w ithdraw ing groups

lower

raise

the

the

S03, H2S04 R— X, AICI3 RCOCI, AlClg

Halogénation Nitration Sulfonation Friedel-Crafts Alkylation Friedel-Crafts Acylation

include

Energy of activation (Eact)

Decrease reaction rate are yr

o,p-Directing groups

stabilize

t

Transition state structures G

for arenium ion formation (see diagram at right)

Increase reaction rate

X2, FeX 3 HNOg, h 2 s o 4

Arenium ion intermediates G Slower (electron-withdrawing substituent; higher Eact)

m - Directing

destabilize

groups

An arenium ion Faster (electron-donating substituent; lower Eact)

loses a proton to form the Cyclohexadienyl cation Aromatic substitution product

T

O o 3 o (D "Ö fi> "Ö

Reaction progress

=

'V H

+ H— A

'

E

-a^

h

E

VI N>

VI

728

CONCEPT MAP S o m e S y n th e tic C o n n e c tio n s o f B e n z e n e a n d A r y l D e riv a tiv e s

• • • • • •

Nitration Halogénation Sulfonation/desulfonation Friedel-C rafts alkylation Friedel-C rafts acylation Clem mensen reduction

• • • • • •

Side-chain oxidation Ring oxidation Catalytic hydrogenation of ring Birch reduction Benzylic radical halogénation Benzylic substitution/elim ination

O C hapter 15 Reactions of Arom atic Com pounds

*ln the F riedel-C rafts alkylation example shown here, R1is a prim ary alkyl halide. If carbocation rearrangem ents are likely, then Friedel-C rafts acylation followed by Clem mensen reduction should be used to incorporate a prim ary alkyl group.

Aldehydes and Ketones Nucleophilic Addition to the Carbonyl Group

Everyone has at least some first-hand sensory knowledge of aldehydes and ketones. Some aldehydes are respon­ sible for very pleasant tastes and odors, such as vanillin from vanilla beans, benzaldehyde from almonds, and cinnamaldehyde from cinnamon (shown above). On the other hand, acetaldehyde (ethanal) causes the unpleas­ ant "hangover" feeling that can result from consuming alcoholic beverages, and formaldehyde (methanal) is highly toxic and has a very pungent odor, as do many other aldehydes. O

O

H

O

H

O

H

O

H

HO' OCH 3 V an illin

B en zaldeh yde

C in n a m a ld e h yd e

A ce ta ld e h yd e

F o rm a ld e h yd e

Structurally, the difference between the chemical cause of hangovers and the taste of vanilla ice cream is sim­ ply a methyl group versus a substituted phenyl group. But, what a difference this switch makes! Lovers of vanilla ice cream would not be pleased if the vanillin in their dessert were replaced by acetaldehyde! The family of ketones has similar variation in properties. Acetone, for example, is a solvent with a sharp odor, whereas (R)-carvone is a natural oil that has the odor of spearmint. O O

A ce to n e

(R )-C arvone

729

7 30 16.1

C h a p te r 1 6

A ld e h y d e s a n d

K e to n e s

Introduction •

A ld e h y d e s h a v e a c a r b o n y l g ro u p b o n d e d to a c a rb o n a to m o n o n e s id e a n d a h y d ro g e n a to m o n th e o th e r s id e . (F o r m a ld e h y d e is a n e x c e p tio n b e ca u se i t has h y d ro g e n a to m s o n b o th s id e s .)

•

K e to n e s h a v e a c a rb o n y l g ro u p b o n d e d to c a rb o n a to m s o n b o th sides.

A lth o u g h e a rlie r ch a p te rs h a v e g iv e n us so m e in s ig h t in to th e c h e m is tr y o f c a rb o n y l c o m ­ p o u n d s , w e s h a ll n o w c o n s id e r th e ir c h e m is tr y in d e ta il. T h e re a s o n : T h e c h e m is tr y o f th e c a r b o n y l g ro u p is c e n tra l to th e c h e m is tr y o f m o s t o f th e c h a p te rs th a t fo llo w . I n th is c h a p te r w e fo c u s o u r a tte n tio n o n th e p re p a r a tio n o f a ld e h y d e s a n d k e to n e s , th e ir p h y s ic a l p ro p e rtie s , a n d e s p e c ia lly

16.2

nucleophilic addition reactions a t their carbonyl groups.

Nom enclature o f A ldehydes and Ketones •

A lip h a t ic a ld e h y d e s a re n a m e d s u b s titu tiv e ly in th e IU P A C s y s te m b y r e p la c in g th e f in a l -e o f th e n a m e o f th e c o rre s p o n d in g a lk a n e w it h -a l.

S in c e th e a ld e h y d e g ro u p m u s t b e a t an e n d o f th e c a rb o n c h a in , th e re is n o n e e d to in d i ­ c a te its p o s itio n . W h e n o th e r s u b s titu e n ts a re p re s e n t th e c a rb o n y l g ro u p c a rb o n is a s s ig n e d p o s itio n 1. M a n y a ld e h y d e s a ls o h a v e c o m m o n n a m e s ; th e se a re g iv e n b e lo w in p a re n th e ­ ses. T h e s e c o m m o n n a m e s a re d e riv e d f r o m th e c o m m o n n a m e s f o r th e c o rre s p o n d in g c a rb o x y lic a c id s (S e c tio n 1 7 .2 A ), a n d s o m e o f th e m a re re ta in e d b y th e IU P A C as a c c e p ta b le n a m e s. O

.A ,

H

O H

M ethanal (fo rm a ld e h y d e )

O

^

H

H

Ethanal (a ce ta ld e h yd e )

P ropanal (p ro p io n a ld e h y d e )

O O c r

H

h

5-C h lo ro p e n ta n a l

•

P he nyletha nal (p h e n yla ce ta ld e h yd e )

A ld e h y d e s in w h ic h th e — C H O g ro u p is a tta c h e d to a r in g s y s te m a re n a m e d s u b ­ s tit u tiv e ly b y a d d in g th e s u f fix c a r b a ld e h y d e . S e v e ra l e x a m p le s fo llo w : O

O

H

B e n ze n e ca rb a ld e h yd e (b e nza lde hyd e)

O

H

C yclo h e xa n e ca rb a ld e h yd e

H

2 -N a p h th a le n e ca rb a ld e h yd e

0 I1

731

16.2 Nomenclature o f Aldehydes and Ketones

The com m on nam e

benzaldehyde is fa r m o r e fr e q u e n tly u se d th a n b e n z e n e c a rb a ld e h y d e

f o r C 6H 5C H O , a n d i t is th e n a m e w e s h a ll use in th is te x t. •

A li p h a t ic k e to n e s a re n a m e d s u b s titu tiv e ly b y r e p la c in g th e f in a l -e o f th e n a m e o f th e c o rre s p o n d in g a lk a n e w it h -o n e .

T h e c h a in is th e n n u m b e re d in th e w a y th a t g iv e s th e c a r b o n y l c a rb o n a to m th e lo w e r p o s s ib le n u m b e r, a n d th is n u m b e r is u se d to d e s ig n a te its p o s itio n . O

O

B u ta n o n e (eth yl m e th y l k e to n e )

2 -P e n ta n o n e (m e th y l p ro pyl k e to n e )

P e n t-4 -e n -2 -o n e (n o t 1 -p e n te n -4 -o n e ) (allyl m eth yl ke to n e )

C o m m o n fu n c t io n a l g ro u p n a m e s f o r k e to n e s ( in p a re n th e s e s a b o v e ) are o b ta in e d s im ­ p ly b y s e p a ra te ly n a m in g th e tw o g ro u p s a tta c h e d to th e c a r b o n y l g ro u p a n d a d d in g th e w o r d k e to n e as a se p a ra te w o rd . S o m e k e to n e s h a ve c o m m o n n a m e s th a t a re re ta in e d in th e IU P A C s y s te m : O

O

O

A c e to n e (p ro p a n o n e )

A c e to p h e n o n e ( 1 -p h e n y le th a n o n e or m eth yl p h en yl k e to n e )

B e n zo p h e n o n e (d ip h e n y lm e th a n o n e o r d ip h e n y l ketone)

O W h e n i t is n e ce ssa ry to n a m e th e

H g ro u p as a p r e fix , i t is th e m e th a n o y l o r fo r m y l

O g ro u p . T h e

g ro u p is c a lle d th e e th a n o y l o r a c e ty l g r o u p ( o fte n a b b re v ia te d as A c ).

O W hen

R g ro u p s a re n a m e d as s u b s titu e n ts , th e y a re c a lle d a lk a n o y l o r a c y l g ro u p s .

H O 2C

, S O 3H

O

H O

2 -M e th a n o y lb e n z o ic a c id (o -fo rm y lb e n z o ic a c id )

4 -E th a n o y lb e n z e n e s u lfo n ic acid (p -a c e ty lb e n z e n e s u lfo n ic ac id )

1 B

1

■

S o lv e d rPi ro b ilem 1J U U c i f l 16 I U- 1 I

J U I v c U

W r it e b o n d - lin e fo rm u la s f o r th re e is o m e r ic c o m p o u n d s th a t c o n ta in a c a rb o n y l g ro u p a n d h a ve th e m o le c u la r f o r ­ m u la C 4H 8O . T h e n g iv e t h e ir IU P A C n a m e s.

STRATEGY AND ANSWER W r it e th e fo rm u la s a n d th e n n a m e th e c o m p o u n d s . O

B u tan al

O

B u ta n o n e

O

2 -M e th y lp ro p a n a l

732

C h a p te r 1 6

R e v ie w P r o b le m 16.1

A ld e h y d e s a n d

K e to n e s

( a ) G iv e IU P A C s u b s titu tiv e n a m e s f o r th e seven is o m e r ic a ld e h y d e s a n d k e to n e s w it h th e fo r m u la C 5H 10O . ( b ) G iv e s tru c tu re s a n d n a m e s ( c o m m o n o r IU P A C s u b s titu tiv e n a m e s ) f o r a ll th e a ld e h y d e s a n d k e to n e s th a t c o n ta in a b e n z e n e r in g a n d h a v e th e f o r m u la C 8 H 8O .

16.3

Physical Properties T h e c a r b o n y l g ro u p is a p o la r g ro u p ; th e re fo re , a ld e h y d e s a n d k e to n e s h a v e h ig h e r b o ilin g p o in ts th a n h y d ro c a rb o n s o f th e sa m e m o le c u la r w e ig h t . H o w e v e r, s in c e a ld e h y d e s a n d k e to n e s c a n n o t h a v e s tro n g h y d ro g e n b o n d s between th eir molecules, th e y h a v e lo w e r b o i l ­ in g p o in ts th a n th e c o rre s p o n d in g a lc o h o ls . T h e f o llo w in g c o m p o u n d s th a t h a v e s im ila r m o l­ e c u la r w e ig h ts e x e m p lif y th is tr e n d : O

O OH 'H

B utane bp - 0 . 5 C (M W = 58)

A map of electrostatic potential for acetone shows the polarity of the carbonyl C= O bond.

R e v ie w P r o b le m 1 6 .2

Propanal bp 4 9 C (MW = 58)

A ce to n e bp 56.1 °C (MW = 58)

1-P ro pa nol bp 9 7 .2 C (M W = 60)

W h ic h c o m p o u n d in e a ch o f th e f o llo w in g p a irs has th e h ig h e r b o ilin g p o in t? ( A n s w e r th is p r o b le m w ith o u t c o n s u ltin g ta b le s .)

H

"O: I

H

O H3C

,C„

H \ /H or^ 'CH„

H yd ro g e n b o n d in g (sho w n in red) betw een w a te r m o le c u le s and ace tone

( a ) P e n ta n a l o r 1 -p e n ta n o l

(d ) A c e to p h e n o n e o r 2 -p h e n y le th a n o l

( b ) 2 -P e n ta n o n e o r 2 -p e n ta n o l

(e ) B e n z a ld e h y d e o r b e n z y l a lc o h o l

(c ) P e n ta n e o r p e n ta n a l

T h e c a r b o n y l o x y g e n a to m a llo w s m o le c u le s o f a ld e h y d e s a n d k e to n e s to f o r m s tro n g h y d ro g e n b o n d s to m o le c u le s o f w a te r. A s a r e s u lt, lo w - m o le c u la r - w e ig h t a ld e h y d e s a n d k e to n e s s h o w a p p re c ia b le s o lu b ilitie s in w a te r. A c e to n e a n d a c e ta ld e h y d e a re s o lu b le in w a te r in a ll p ro p o r tio n s .

Physical Properties o f A ldehydes and Ketones

Formula

Name

HCHO CH3CHO c h 3 c h 2ch o CH3 (CH2)2CHO CH3 (CH2)3CHO CH3 (CH2)4CHO c 6 h5c h o c 6 h 5 c h 2ch o c h 3c o c h 3 c h 3c o c h 2c h 3 c h 3c o c h 2c h 2c h 3 c h 3c h 2c o c h 2c h 3 C6 H5 COCH3 C6 H5 COC6 H5

Formaldehyde Acetaldehyde Propanal Butanal Pentanal Hexanal Benzaldehyde Phenylacetaldehyde Acetone Butanone 2-Pentanone 3-Pentanone Acetophenone Benzophenone

mp (°C) -9 2 -125 -81 -9 9 -9 1.5 -51 -2 6 33 -9 5 -8 6

bp (°C) -

21

Solubility in Water Very soluble

21

rc

49 76

Very soluble Soluble Slightly soluble Slightly soluble Slightly soluble Slightly soluble

102

131 178 193 56.1 79.6

-7 8 -3 9

102

21

202

48

306

102

rc

Very soluble Soluble Soluble Insoluble Insoluble

T a b le 16.1 lis ts th e p h y s ic a l p ro p e rtie s o f a n u m b e r o f c o m m o n a ld e h y d e s a n d k e to n e s . S o m e a ld e h y d e s o b ta in e d f r o m n a tu r a l s o u rce s h a v e v e ry p le a s a n t fra g ra n c e s . T h e f o l lo w ­ in g a re s o m e in a d d itio n to th o s e w e m e n tio n e d a t th e b e g in n in g o f th is c h a p te r.

0 I1 1 6 .4

733

S y n th e s is o f A ld e h y d e s

CHO CHO O

OH

H

O O

S a lic y la ld e h y d e (fro m m e a do w sw e et)

C itro n e lla l (th e s c e n t o f le m on in c e rta in p la n ts )

P iperonal (m ad e fro m sa fro le ; o d o r o f h e lio tro p e )

THE CHEMISTRY OF . . . A l d e h y d e s a n d K e t o n e s in P e r f u m e s

Many aldehydes and ketones have pleasant fragrances and, because of this, they have found use in perfumes. Originally, the ingredients for perfumes came from natural sources such as essential oils (Section 23.3), but with the development of synthetic organic chemistry in the nineteenth century, many ingredients now used in perfumes result from the creativity of laboratory chemists. Practitioners of the perfumer's art, those who blend per­ fumes, talk of their ingredients in a language derived from music. The cabinet that holds the bottles containing the compounds that the perfumer blends is called an "organ." The ingredients themselves are described as having certain "notes." For example, highly volatile substances are said to display "head notes," those less volatile and usually associ­ ated with flowers are said to have "heart notes," and the least volatile ingredients, usually with woody, balsamic, or musklike aromas, are described as "base notes."* (Z)-Jasmone (with the odor of jasmine) and a-damascone (odor of roses) have "heart notes," as do the ionones (with the odor of violets). All of these ketones can be obtained from natural sources.

O

a -Ion on e

O

b -Io n o n e

Two ketones from exotic natural sources are muscone (from the Himalayan musk deer) and civetone (from the African civet cat).

O

C iveton e

Stereochemistry has a marked influence on odors. For example, the (R)-enantiomer of muscone (depicted above) is described as having a "rich and powerful musk," whereas the (S)-enantiomer is described as being "poor and less strong." The (R)-enantiomer of a-damascone has a rose petal odor with more apple and fruitier notes than the (S)-enantiomer.

O

*F or an in-depth discussion o f the perfum e industry, see Fortineau, A.-D. "C hem istry Perfumes Your Daily Life," J. Chem. Educ., 2004, 81, 45-50.

16.4 16.4A •

Synthesis o f Aldehydes

Aldehydes by Oxidation of 1° Alcohols

T h e o x id a tio n sta te o f a n a ld e h y d e lie s b e tw e e n th a t o f a 1° a lc o h o l a n d a c a r b o x y lic a c id (S e c tio n 1 2 .4 A ).

O

O

[O] R

OH

1° A lc o h o l

< = *

[H]

[O] R

H

A ld e h y d e

< = *

[H]

„

R

^

OH

C a rb o x y lic acid

734

C h a p te r 1 6

A ld e h y d e s a n d

K e to n e s

Aldehydes can be prepared from 1° alcohols by oxidation with pyridinium chlorochrom a te ( C 5H 5N H + C r O 3C r , o r P C C ): O

PCC

R

OH

R

CH2Cl2

1° A lc o h o l

OH

A ld e h yd e

A n e x a m p le o f th e use o f P C C in th e s y n th e s is o f an a ld e h y d e is th e o x id a tio n o f 1 -h e p ta n o l to h e p ta n a l: O

PCC OH

1-H eptanol

H

CH2Cl2

H eptanal (93%)

16.4B Aldehydes by Ozonolysis of Alkenes •

A lk e n e s ca n b e c le a v e d b y o z o n o ly s is o f t h e ir d o u b le b o n d (S e c tio n 8 .1 7 B ). T h e p ro d u c ts a re a ld e h y d e s a n d k e to n e s .

I n C h a p te r 8 w e a ls o s a w h o w th is p ro c e d u re has u t i l i t y in s tru c tu re d e te r m in a tio n . T h e f o l­ lo w in g e x a m p le s illu s tr a te th e s y n th e s is o f a ld e h y d e s b y o z o n o ly s is o f a lke n e s. H O (1) O3, CH2Cl2, -78°C

+

(2) Me2S

H

O (1) O3, CH2Cl2, -78°C

+

(2) Me2S

16.4C Aldehydes by Reduction of Acyl Chlorides, Esters, and Nitriles T h e o r e tic a lly , i t o u g h t to b e p o s s ib le to p re p a re a ld e h y d e s b y r e d u c tio n o f c a r b o x y lic a c id s . I n p ra c tic e , th is is n o t p o s s ib le w it h th e re a g e n t n o r m a lly u se d to re d u c e a c a r b o x y lic a c id , l it h iu m a lu m in u m h y d r id e ( L iA lH 4 o r L A H ) . •

W h e n a n y c a r b o x y lic a c id is tre a te d w it h L A H , i t is re d u c e d a ll th e w a y to th e 1° a lc o h o l.

•

T h is h a p p e n s b e c a u s e L A H is a v e r y p o w e r fu l r e d u c in g a g e n t a n d a ld e h y d e s are v e r y e a s ily re d u c e d .

A n y a ld e h y d e th a t m ig h t b e fo r m e d in th e r e a c tio n m ix t u r e is im m e d ia te ly r e d u c e d b y L A H to th e 1° a lc o h o l. ( I t d o e s n o t h e lp to use a s to ic h io m e tr ic a m o u n t o f L A H , b e ca u se as so o n as th e fir s t fe w m o le c u le s o f a ld e h y d e a re fo r m e d in th e m ix tu r e , th e re w i l l s t ill b e m u c h u n re a c te d L A H p re s e n t a n d i t w i l l re d u c e th e a ld e h y d e .) O

O LiAlH.

R

OH

C a rb o x y lic a cid

LiAlH.

A

.

A ld e h yd e

R

'O H

1° A lc o h o l

T h e s e c re t to success h e re is n o t to use a c a r b o x y lic a c id its e lf, b u t to use a d e riv a tiv e o f a c a r b o x y lic a c id th a t is m o r e e a s ily re d u c e d , a n d an a lu m in u m h y d r id e d e riv a tiv e th a t is le s s r e a c tiv e th a n L A H . •

A c y l c h lo rid e s ( R C O C I) , e ste rs ( R C O 2 R ') , a n d n it r ile s ( R C N ) a re a ll e a s ily p re ­ p a re d f r o m c a r b o x y lic a c id s (C h a p te r 1 7 ), a n d th e y a ll a re m o r e e a s ily re d u c e d .

1 6 .4

S y n th e s is o f A ld e h y d e s

( A c y l c h lo rid e s , esters, a n d n it r ile s a ll a ls o h a v e th e sam e o x id a tio n state as c a r b o x y lic acid s. C o n v in c e y o u r s e lf o f th is b y a p p ly in g th e p r in c ip le s th a t y o u le a rn e d in S e c tio n 1 2 .2 A ). •

T w o d e riv a tiv e s o f a lu m in u m h y d r id e th a t a re less re a c tiv e th a n L A H , in p a rt b e c a u s e th e y a re m u c h m o r e s te r ic a lly h in d e re d , a re l i t h i u m t r i- t e r i- b u t o x y a lu m i n u m h y d r id e a n d d iis o b u t y la lu m in u m h y d r id e ( D I B A L - H ) :

X O L i4

H — A l— O —

x O

H'

L ith iu m tri-fe rt-b u to x y a lu m in u m h y d rid e •

A

D iis o b u ty la lu m in u m h y d rid e (a b b re via te d ABu2A lH o r DIBAL-H )

T h e f o llo w in g sch e m e s u m m a riz e s h o w lit h iu m tr i- t e r t - b u t o x y a lu m in u m h y d r id e a n d D I B A L - H ca n b e u se d to s y n th e s iz e a ld e h y d e s f r o m a c id d e riv a tiv e s : O (1) LiAlH(O-f-Bu)3, —78°C R '

"C l

(2) H2O

A c y l c h lo rid e O

O (1) DIBAL-H, hexane, —78°C R

OR'

(2) H2O

R

E ste r

A ld e h y d e

__. . =N

R

H

(1) DIBAL-H, hexane

(2) H2O

N itrile

W e n o w e x a m in e e a ch o f th e se a ld e h y d e syn th e se s in m o r e d e ta il.

A ld eh yd es from

•

A cyl C hlorides: R C O C l

s

RCHO

A c y l c h lo rid e s c a n b e re d u c e d to a ld e h y d e s b y tr e a tin g th e m w it h L iA lH [O C ( C H 3)3]3, lit h iu m t r i- t e r t- b u to x y a lu m in u m h y d r id e , a t —7 8 °C .

•

C a r b o x y lic a c id s c a n b e c o n v e rte d t o a c y l c h lo r id e s b y u s in g S O C l2 (see S e c tio n 1 5 .7 ). O

R

L

O

O

SOCl, R^

OH

C

l

(1) LiAlH(O-f-Bu)3, Et2O, —78°C |l (2 ) H2O * r ^ "" h

T h e f o llo w in g is a s p e c ific e x a m p le :

O

O

Cl

(1) LiAlH(O-f-Bu)3, Et2O, —78°C (2 ) H2O

OMe

3 -M e th o xy-4 -m e th ylb e n zo yl c h lo rid e

H

OMe

3 -M e th o xy-4 -m e th ylb e n za ld e h yd e

M e c h a n is tic a lly , th e re d u c tio n is b r o u g h t a b o u t b y th e tra n s fe r o f a h y d r id e io n f r o m th e a lu m in u m a to m to th e c a rb o n y l c a rb o n o f th e a c y l c h lo r id e (see S e c tio n 1 2 .3 ). S u b s e q u e n t h y d r o ly s is fre e s th e a ld e h y d e .

735

736

Chapter 16

Aldehydes and Ketones

A MECHANISM FOR THE REACTION R e d u c tio n o f a n A cyl C h lo r id e t o a n A l d e h y d e

O — Li

/ Q:

^ L i A l H [O C (C H 3 )3 ]3

R— C

A lH [O C (C H 3 )3 ]3

R— C

\

/

:Ch

\ =C h Li

Li C ^°:

R— C

\

:O:

A l[O C (C H 3 )3 ]s

I R— C — H

H

=Ch

=C h

T ra n s fe r o f a h y d rid e ion to th e c a rb o n y l c a rb o n b rin g s ab o u t th e re d u c tio n . O - A l[O C ( C H

A c tin g as a L ew is acid, th e alu m in u m a to m a c c e p ts an e le c tro n p air fro m o xyg en.

O:

O - A l[ O C ( C H 3 ) 3 ] s

3 « 3

/

LiCl R— C— H

A l[O C (C H 3 )3 ]s

I

L i+

R— C

Cl

/

h 2o

R— C H

: C| : T h is in te rm e d ia te loses a c h lo rid e ion as an e le ctro n p air fro m th e o xy g e n as s is ts .

H

T h e a d d itio n o f w a te r c a u s e s h y d ro ly s is o f th is a lu m in u m c o m p le x to ta k e place, p ro d u c in g th e a ld e h y d e . (S e v e ra l s te p s a re in v o lv ed .)

S o lv e d P ro b le m 1 6 .2 P r o v id e th e re a g e n ts f o r tr a n s fo r m a tio n s (1 ), (2 ), a n d (3 ). O

O

O

STRATEGY AND ANSWER I n (1 ), w e m u s t o x id iz e m e th y lb e n z e n e to b e n z o ic a c id . T o d o th is w e use h o t p o ta s ­ s iu m p e rm a n g a n a te in a b a s ic s o lu tio n f o llo w e d b y a n a c id ic w o r k u p (see S e c tio n 1 5 .1 3 C ). F o r (2 ), w e m u s t c o n ­ v e rt a c a r b o x y lic a c id to an a c id c h lo rid e . F o r th is tra n s fo rm a tio n w e use t h io n y l c h lo rid e o r p h o s p h o ru s p e n ta c h lo rid e (see S e c tio n 1 5 .7 ). F o r (3 ), w e m u s t re d u c e an a c id c h lo r id e to an a ld e h y d e . F o r th is w e use lit h iu m tr i- ie r i- b u t o x y a lu m in u m h y d r id e (see a b o v e ).

A ld eh yd es fro m

•

Esters a n d

N itriles:

R CO

2

R ' s

RCHO and R C #

N s

RCHO

B o th e ste rs a n d n it r ile s ca n b e re d u c e d to a ld e h y d e s b y D I B A L - H .

C a r e fu lly c o n tro lle d a m o u n ts o f D I B A L - H m u s t b e u se d to a v o id o v e rre d u c tio n , a n d th e ester re d u c tio n m u s t b e c a rrie d o u t a t lo w te m p e ra tu re s. B o th re d u c tio n s re s u lt in th e fo rm a tio n o f a re la tiv e ly sta b le in te rm e d ia te b y th e a d d itio n o f a h y d rid e io n to th e c a rb o n y l c a rb o n o f the e ste r o r to th e c a rb o n o f th e — C #

N g ro u p o f th e n itr ile . H y d r o ly s is o f th e in te rm e d ia te l i b ­

e rates th e a ld e h y d e . S c h e m a tic a lly , th e re a c tio n s ca n b e v ie w e d in th e f o llo w in g w a y.

0 I1 1 6 .4

737

S y n th e s is o f A ld e h y d e s

A MECHANISM FOR THE REACTION R e d u c tio n o f an E ste r to a n A ld e h y d e

Q:

Al( /-Bu)2

'

■O—Al( /-Bu

+Q—Al( /-Bu)2

O=

)2

/

R— C

\

H

r

-

cÌ

:QR'

ì

r-

ì

^OR'

T h e a lu m in u m a to m a c c ep ts an e le ctro n p air fro m th e ca rb o n y l o xy g e n a to m in a Lew is a c id -b a s e reaction .

C- h

/

R— C

h°

H

:OR'

T ra n s fe r o f a h yd rid e ion to th e c a rb o n y l c a rb o n b ring s a b o u t its re d u c tio n .

A d d itio n of w a te r at the e nd o f th e re a c tio n h yd ro ly ze s th e a lu m in u m c o m p le x and p ro d u c e s th e a ld e h y d e .

A MECHANISM FOR THE REACTION R e d u c tio n o f a N itrile to a n A ld e h y d e

R— C# N :

/

R— C #N — Al( /-Bu)2

Al( -Bu)2 H

H

T h e a lu m in u m a to m a c c e p ts an e le c tro n p air fro m th e n itrile in a L ew is a c id -b a s e reaction .

T ra n s fe r of a h y d rid e ion to th e n itrile c a rb o n b rin g s a b o u t its red uction .

/

O:

N— Al( -Bu) 2 /

/

H2O

R— C

R— C

H

H A d d itio n o f w a te r a t the end o f th e re a c tio n h yd ro ly ze s th e a lu m in u m c o m p le x and p ro d u c e s th e a ld e h y d e . (S e v e ra l s te p s a re in vo lved . S e e S e c tio n 16.8 re la tin g to im in es.)

The following specific examples illustrate these syntheses: OAl(/-Bu)2 O

O

HO

(/-B u)2A lH

''O E t

'H

hexane, - 7 8 ° C

H

OEt

N A l(/-B u )2

O

HO

(/-B u ^ A IH hexane

88%

H

H

738

Chapter 16

Aldehydes and Ketones

c « / 1/û

S o l v e d P r o b le m 1 6 .4 S ta rtin g w it h b e n z y l a lc o h o l, o u tlin e a s y n th e s is o f p h e n y le th a n a l.

STRATEGY AND ANSWER C o n v e rt th e b e n z y l a lc o h o l to b e n z y l b r o m id e w it h P B r 3, th e n re p la c e th e b r o m in e b y c y a n id e in a n SN2 re a c tio n . L a s tly , re d u c e th e n it r i le to p h e n y le th a n a l.

R e v ie w P r o b le m 1 6 .3

S h o w h o w y o u w o u ld s y n th e s iz e p ro p a n a l f r o m e a ch o f th e f o llo w in g : (a ) 1 -p ro p a n o l a n d ( b ) p ro p a n o ic a c id ( C H 3C H 2C O 2H ).

16.5

Synthesis o f Ketones 16.5A

Ketones from Alkenes, Arenes, and 2° Alcohols

W e h a v e seen th re e la b o ra to r y m e th o d s f o r th e p re p a r a tio n o f k e to n e s in e a r lie r c h a p te rs :

1. K e to n e s (a n d a ld e h y d e s ) b y o z o n o ly s is o f a lk e n e s (d is c u s s e d in S e c tio n 8 .1 7 B ). R

R"

R

(1 ) O, (2) Me2S R'

R"

O

H

+

O=

R'

H

K e to n e

A ld e h y d e

2. K e to n e s f r o m a re n e s b y F r ie d e l- C r a ft s a c y la tio n s (d is c u s s e d in S e c tio n 1 5 .7 ). F o r e x a m p le : O

a,c,3

R

'R

"Cl A n alkyl aryl k e to n e

HCl

1 6 .5

739

S y n th e s is o f K e to n e s

A lt e r n a tiv e ly , O

O

Cl

AlCi, HCl

A d ia ry l keto ne 3 . K e to n e s f r o m s e c o n d a ry a lc o h o ls b y o x id a tio n (d is c u s s e d in S e c tio n 1 2 .4 ):

OH

H2CrO4 R ^ 'R '

16.5B

Ketones from Nitriles

T re a tin g a n it r i le ( R — C #

N ) w it h e ith e r a G r ig n a r d re a g e n t o r an o r g a n o lith iu m re a g e n t

fo llo w e d b y h y d r o ly s is y ie ld s a k e to n e .

G e n e ra l R e a c tio n s O

N- M g+X R— ^

N

+

il

R ' — M gX

R

_HO+

R'

R'

^ =

N

+

A

R ' — LI

N H 4+

+

M g 2+

N H 4+

+

Li+

O

N - Li+ R

+

h 3o

■

A

R'

T h e m e c h a n is m f o r th e a c id ic h y d r o ly s is ste p is th e re v e rs e o f o n e th a t w e s h a ll s tu d y f o r im in e fo r m a t io n in S e c tio n 1 6 .8 A .

S p e c ific E xa m p les O ,C N

Li

(1) Et2O (2 ) H3O+

O CN

M gB r

(1 ) Et2O (2 ) H3O+ 2 -C ya n o p ro p a n e

2 -M e th yl-1 -p h e n ylp ro p a n o n e (is o p ro p y l p h e n yl ketone)

E v e n th o u g h a n it r i le h a s a t r ip le b o n d , a d d itio n o f th e G r ig n a r d o r lit h iu m re a g e n t ta ke s p la c e o n ly o n c e . T h e re a so n : I f a d d itio n to o k p la c e tw ic e , th is w o u ld p la c e a d o u b le n e g ­ a tiv e c h a rg e o n th e n itro g e n .

N - Li+ R

N 2 - 2 Li+ R'

Li

=N:

(>

R

\

R'

R'

(The d ia n io n do e s n o t fo rm .)

+

X-

740

Chapter 16

Aldehydes and Ketones

Solved Problem 16.5 ILLUSTRATING A MULTISTEP SYNTHESIS With 1-butanol as your only organic starting compound, devise a syn­ thesis of 5-nonanone. Begin by writing a retrosynthetic analysis. ANSWER Retrosynthetic disconnection of 5-nonanone suggests butylmagnesium bromide and pentanenitrile as

immediate precursors. Butylmagnesium bromide can, in turn, be synthesized from 1-bromobutane. Pentanenitrile can also be synthesized from 1-bromobutane, via SN2 reaction of 1-bromobutane with cyanide. To begin the syn­ thesis, 1 -bromobutane can be prepared from 1 -butanol by reaction with phosphorus tribromide. R e tro syn th e tic A n a ly s is

Butylmagnesium bromide

S ynth esis

ReviewProblem16.4

Provide the reagents and indicated intermediates in each of the following syntheses.

0 I1 1 6 .6

16.6 •

N u c le o p h ilic A d d it io n t o t h e

C a rb o n -O x y g e n

D o u b le

Bond

Nucleophilic A d d itio n to the C arb on -O xygen D ouble Bond nucleophilic addition

T h e m o s t c h a ra c te ris tic r e a c tio n o f a ld e h y d e s a n d k e to n e s is to th e c a r b o n - o x y g e n d o u b le b o n d .

G e n e ra l R e ac tio n O

OH N u— H

J L

'H Nu

S p e c ific E xa m p les O

OH EtO — H

H

H OEt

A h em ia ce ta l (s e e S e c tio n 16.7)

O

OH HCN

H CN

A c y a n o h y d rin (s e e S e c tio n 16.9) A ld e h y d e s a n d k e to n e s a re e s p e c ia lly s u s c e p tib le to n u c le o p h ilic a d d itio n b e ca u se o f th e s tr u c tu r a l fe a tu re s th a t w e d is c u s s e d in S e c tio n 12 .1 a n d w h ic h a re s h o w n b e lo w . Nu ,C \..

R ^ C = O r

A ld e h y d e o r k e to n e (R o r R' m a y be H)

•

T h e n u c le o p h ile m ay a tta c k fro m a b o v e o r below .

A X O= R

R' -C

/ ° 7

Nu

T h e t r ig o n a l p la n a r a rra n g e m e n t o f g ro u p s a ro u n d th e c a r b o n y l c a rb o n a to m m ea n s th a t th e c a r b o n y l c a rb o n a to m is r e la t iv e ly o p e n to a tta c k f r o m a b o v e o r b e lo w th e p la n e o f th e c a r b o n y l g ro u p (see a b o v e ).

•

T h e p o s itiv e c h a rg e o n th e c a r b o n y l c a rb o n a to m m e a n s th a t i t is e s p e c ia lly sus­ c e p tib le to a tta c k b y a n u c le o p h ile .

•

741

T h e n e g a tiv e c h a rg e o n th e c a rb o n y l o x y g e n a to m m e a n s th a t n u c le o p h ilic a d d itio n is s u s c e p tib le to a c id c a ta ly s is .

N u c le o p h ilic a d d itio n to th e c a rb o n - o x y g e n d o u b le b o n d o c c u rs , th e re fo re , in e ith e r o f t w o g e n e ra l w a y s . 1. W h e n th e r e a g e n t is a s t r o n g n u c le o p h ile ( N u : 2 ), a d d itio n u s u a lly ta ke s p la c e in th e f o llo w in g w a y (see th e m e c h a n is m b o x o n th e f o llo w in g p a g e ), c o n v e rtin g th e t r ig o n a l p la n a r a ld e h y d e o r k e to n e in to a te tra h e d ra l p ro d u c t. I n th is ty p e o f a d d itio n th e n u c le o p h ile uses its e le c tro n p a ir to f o r m a b o n d to th e c a r b o n y l c a rb o n a to m . A s th is h a p p e n s th e e le c tro n p a ir o f th e c a rb o n - o x y g e n p b o n d s h ifts o u t to th e e le c tro n e g a tiv e c a rb o n y l o x y g e n a to m a n d th e h y b r id iz a tio n s ta te o f b o th th e c a rb o n a n d th e o x y g e n c h a n g e s f r o m

sp2 to sp 3. T h e im p ortan t

aspect o f this step is the ab ility o f the carb on yl oxygen atom to accom m odate the electron p a ir o f the carb on -o xygen double bond.

742

C h a p te r 1 6

A ld e h y d e s a n d

K e to n e s

A MECHANISM FOR THE REACTION A d d itio n o f a S tro n g N u cle o p h ile to an A ld e h y d e o r K e to n e

(a)

Nu

Nu

\

Nu

\

H — Nu

,.C— O: r

(b)

'7

R T rig o n a l p la n a r

T e tra h e d ra l in te rm e d ia te (p lu s e n a n tio m e r)

In th is s te p th e n u c le o p h ile fo rm s a bon d to th e c a rb o n by d o n a tin g an e le c tro n p air to the top o r b o tto m fa c e o f th e c a rb o n y l g ro u p [p a th (a ) o r (b )]. A n e le ctro n p air s h ifts o u t to the o xyg en .

••

,..C— O R '7 " R

H

: Nu-

T e tra h e d ra l p ro d u ct (p lu s e n a n tio m e r)

In th e s e c o n d s te p th e a lk o x id e o xy g e n , b e c a u s e it is s tro n g ly basic, re m o v e s a p ro to n fro m H— Nu o r s o m e o th e r a c id .

In the second step the oxygen atom accepts a proton. This happens because the oxygen atom is now much more basic; it carries a full negative charge as an alkox­ ide anion. 2. When an acid catalyst is present and the nucleophile is weak, reaction of the car­ bonyl oxygen with the acid enhances electrophilicity of the carbonyl group.

A MECHANISM FOR THE REACTION A c id -C a ta ly z e d N u cle o p h ilic A d d itio n to an A ld e h y d e o r K e to n e

Step 1

R*.V

C =O ,

h

-Q

h -

R*. ^ c= oh ^-*+ R

R*. ,C — OH

A:

R

(o r a L ew is acid) In th is s te p an e le c tro n p air o f th e c a rb o n y l o xy g e n a c c e p ts a p ro to n fro m th e acid (o r a s s o c ia te s w ith a L ew is acid), p ro d u c in g an o x o n iu m c a tio n . T h e c a rb o n o f th e o x o n iu m ca tio n is m o re s u s c e p tib le to n u c le o p h ilic a tta c k than th e c a rb o n y l o f th e s ta rtin g keto ne.

Step 2

Nu:

\

Ri

^C ^Ö H R

^+

: Nu— H

••

,,C— O— H

r

H—A

'7

R In th e firs t o f th e s e tw o s tep s, th e o x o n iu m catio n a c c ep ts th e e le c tro n p air of th e n u c le o p h ile . In th e s e c o n d ste p , a b ase re m o v e s a p ro to n fro m th e p o s itiv e ly c h a rg e d ato m , re g e n e ra tin g th e acid.

0 I1 1 6 .6

N u c le o p h ilic A d d it io n t o t h e

C a rb o n -O x y g e n

D o u b le

strongacids

This mechanism operates when carbonyl compounds are treated with in the pres­ ence of In the first step the acid donates a proton to an electron pair of the carbonyl oxygen atom. The resulting protonated carbonyl compound, an oxonium cation, is highly reactive toward nucleophilic attack at the carbonyl carbon atom because the car­ bonyl carbon atom carries more positive charge than it does in the unprotonated compound.

weaknucleophiles.

16.6A Reversibility of Nucleophilic Additions to the Carbon-Oxygen Double Bond • Many nucleophilic additions to carbon-oxygen double bonds are reversible; the overall results of these reactions depend, therefore, on the position of an equilibrium. This contrasts markedly with most electrophilic additions to carbon-carbon double bonds and with nucleophilic substitutions at saturated carbon atoms. The latter reactions are essen­ tially irreversible, and overall results are a function of relative reaction rates.

16.6B

Relative Reactivity: Aldehydes versus Ketones

• In general, aldehydes are more reactive in nucleophilic additions than are ketones. Both steric and electronic factors favor aldehydes. In aldehydes, where one group is a hydrogen atom, the central carbon of the tetrahedral product formed from the aldehyde is less crowded and the product is more stable. Formation of the product, therefore, is favored at equilibrium. With ketones, the two alkyl substituents at the carbonyl carbon cause greater steric crowding in the tetrahedral product and make it less stable. Therefore, a smaller concentration of the product is pre­ sent at equilibrium. Steric Factors

Electronic Factors Because alkyl groups are electron releasing, aldehydes are more reactive on electronic grounds as well. Aldehydes have only one electron-releasing group to partially neutralize, and thereby stabilize, the positive charge at their carbonyl carbon atom. Ketones have two electron-releasing groups and are stabilized more. Greater stabi­ lization of the ketone (the reactant) relative to its product means that the equilibrium con­ stant for the formation of the tetrahedral product from a ketone is smaller and the reaction is less favorable:

Od

Od

A ld e h y d e

K etone

T h e K e to n e c a rb o n y l c a rb o n is less p o s itiv e b e c a u s e it h as tw o e le c tro n -re le a s in g alkyl g ro up s.

On the other hand, electron-withdrawing substituents (e.g., — CF3 or — CCI3 groups) cause the carbonyl carbon to be more positive (and the starting compound to become less stable), causing the addition reaction to be more favorable.

16.6C Addition Products Can Undergo Further Reactions Nucleophilic addition to a carbonyl group may lead to a product that is stable under the reaction conditions that we employ. If this is the case we are then able to isolate products with the following general structure: OH

Nu

743

Bond

<

H e lp fu l H i n t

Any compound containing a positively charged oxygen atom that forms three covalent bonds is an oxonium cation.

744

C h a p te r 1 6

A ld e h y d e s a n d

K e to n e s

I n o th e r re a c tio n s th e p r o d u c t fo r m e d in i t i a l l y m a y b e u n s ta b le a n d m a y s p o n ta n e o u s ly

elim ination reac­ tion, e s p e c ia lly dehydration. E v e n i f th e in it ia l a d d itio n p r o d u c t is s ta b le , w e m a y d e lib e r ­

u n d e rg o s u b s e q u e n t re a c tio n s . O n e c o m m o n s u b s e q u e n t r e a c tio n is an

a te ly b r in g a b o u t a s u b s e q u e n t r e a c tio n b y o u r c h o ic e o f r e a c tio n c o n d itio n s .

ReviewProblem16.5

T h e r e a c tio n o f an a ld e h y d e o r k e to n e w it h a G r ig n a r d re a g e n t (S e c tio n 1 2 .8 ) is a n u c le o p h ilic a d d itio n to th e c a rb o n - o x y g e n d o u b le b o n d . (a ) W h a t is th e n u c le o p h ile ? (b ) T h e m a g n e s iu m p o r t io n o f th e G r ig n a r d re a g e n t p la y s a n im p o r ta n t p a r t in th is re a c tio n . W h a t is its fu n c tio n ? (c ) W h a t p r o d u c t is fo r m e d in it ia lly ? (d ) W h a t p r o d u c t fo r m s w h e n w a te r is a d d e d ?

ReviewProblem16.6

T h e re a c tio n s o f a ld e h y d e s a n d k e to n e s w it h L iA lH 4 a n d N a B H 4 (S e c tio n 1 2 .3 ) a re n u c le o p h ilic a d d itio n s to th e c a rb o n y l g ro u p . W h a t is th e n u c le o p h ile in th e se re a c tio n s ?

16.7

The A d d itio n o f Alcohols: Hem iacetals and Acetals •

A ld e h y d e s a n d k e to n e s re a c t w it h a lc o h o ls to f o r m h e m ia c e ta ls a n d a c e ta ls b y an e q u ilib r iu m re a c tio n .

16.7A •

Hemiacetals

T h e e s s e n tia l s tr u c tu r a l fe a tu re s o f a h e m ia c e ta l are a n — O H a n d an — O R g ro u p a tta c h e d to th e sa m e c a rb o n a to m .

T h e h e m ia c e ta l re s u lts b y n u c le o p h ilic a d d itio n o f a n a lc o h o l o x y g e n to th e c a rb o n y l c a r­ b o n o f a n a ld e h y d e o r k e to n e .

A MECHANISM FOR THE REACTION H e m ia c e ta l F o rm a tio n

H R

R

HO— R' H

/ H

A ld e h y d e (o r ke to n e )

A lcoh ol

In th is s te p th e alc o h o l a tta c k s th e ca rb o n y l c arb on .

•

O — R'

\ C

\

\~ O

O — R'

R

/"

/ H

/" C

\~ O

H

H em ia c e tal (u s u a lly too u n s ta b le to iso la te )

In tw o in te rm o le c u la r s tep s, a pro to n is re m o v e d fro m th e p o s itiv e o xy g e n and a p ro to n is g ain e d at th e n e g a tiv e o xyg en.

M o s t o p e n -c h a in h e m ia c e ta ls a re n o t s u f fic ie n tly s ta b le to a llo w th e ir is o la tio n . C y c lic h e m ia c e ta ls w it h fiv e - o r s ix - m e m b e re d r in g s , h o w e v e r, a re u s u a lly m u c h m o re s ta b le .

1 6 .7

T h e A d d itio n

745

o f A lc o h o ls : H e m ia c e t a ls a n d A c e ta ls

H H e m ia c e tal: O H a n d O R g ro u p s b o n d e d to th e s a m e carb o n

:

H QH A c yclic h em iacetal

■QB u tan al-4-o l

M o s t s im p le su g a rs (C h a p te r 2 2 ) e x is t p r im a r ily in a c y c lic h e m ia c e ta l fo r m . G lu c o s e is a n e x a m p le : OH

H e m ia c e tal: OH a n d O R g ro u p s b o n d e d to th e s a m e ca rb o n Q

HQ

QH

HQ QH

(+ )-G lu c o s e (a c yclic h em ia ce ta l) W h e th e r th e c a r b o n y l re a c ta n t is a n a ld e h y d e o r a k e to n e , th e p ro d u c t w it h a n — O H a n d an — O R g ro u p is c a lle d a h e m ia c e ta l.

H R"

R"

\

HO— R' R

RÍ

O— R'

X o-

Q :

two steps

O — R'

^

Q- h

H e m ia c e ta l

Keton e •

R"

T h e fo r m a tio n o f h e m ia c e ta ls is c a ta ly z e d b y a c id s a n d bases.

A MECHANISM FOR THE REACTION A cid -C a ta ly ze d H e m ia c e ta l F o rm a tio n

R"

.. H — O— R'

>1

x c = Q ^

/« +

»-■

R

=O— R'

I

H

(R" m a y be H) P roto natio n o f the a ld e h y d e o r k e to n e o xy g e n ato m m akes the ca rb o n y l c a rb o n m o re s u s c e p tib le to n uc le o p h ilic attack. [Th e p ro to nated a lcoh ol resu lts fro m reaction of th e alc o h o l (p re s e n t in ex c es s ) w ith th e acid cata ly s t, e.g., g as e o u s (an h yd ro u s ) HCl.]

An alc o h o l m o le c u le ad d s to the carb o n of th e o xo n iu m catio n.

•

R '— O-P I

-

I

-

R"— C— Q R

O — R' H

H

=O— R' I

-

I

-

R"— C— Q R

H

H

0 1

H

T h e tra n s fe r o f a p roton fro m th e p o s itiv e o xy g e n to a n o th e r m o le c u le o f th e a lcoh ol leads to th e h em iacetal.

R'

746

C h a p te r 1 6

A ld e h y d e s a n d

K e to n e s

A MECHANISM FOR THE REACTION B a se -C a ta ly z e d H e m ia c e ta l F o rm a tio n

R '—

R

~ O — R' C =O '

/S+LXR-

O-

R' — H ^ -Q R '

O­

I

••

R"— C— O

R"— C— O

I

••

I

■■

I

-

H

:O — R'

R

R

(R" m a y b e H)

T h e a lk o x id e a nio n a b s tra c ts a p roton fro m an alc o h o l m o le c u le to p ro d u ce the hem ia ce ta l and re g e n e rate s an a lk o x id e anion.

An a lk o x id e a nio n a c tin g as a n u c le o p h ile a tta c k s th e c arb on yl carb o n ato m . An electro n p air shifts o n to th e o xy g e n atom , p ro d u cin g a n ew a lk o x id e anion.

R

A ld e h y d e H ydrates: gem -D iols

D is s o lv in g an a ld e h y d e su ch as a c e ta ld e h y d e in w a te r

causes th e e s ta b lis h m e n t o f an e q u ilib r iu m b e tw e e n th e a ld e h y d e a n d its h y d r a t e . T h is h y d ra te is in a c tu a lity a 1 ,1 -d io l, c a lle d a g e m in a l d io l ( o r s im p ly a g e m - d io l) . ,0 — H

A c e ta ld e h y d e

H2O

/= O H'

'

O—H

H y d ra te (a g em -d io l)

T h e g e m -d io l re su lts fr o m a n u c le o p h ilic a d d itio n o f w a te r to th e c a rb o n y l g ro u p o f th e a ldehyde.

A MECHANISM FOR THE REACTION H y d ra te F o rm a tio n

h

3c 3 \

. C ^O

/ r+ ^ * r-

h 3c 3

"=OH,

/

H3C

oh2 2

H

H

/

In this s te p w a te r a ttacks th e ca rb o n y l c a rb o n atom .

\

: OH

3V

C ..

^ OH

H

,O -

In tw o in te rm o le c u la r s te p s a p roton is lost fro m th e p o s itiv e o xy g e n a to m and a pro to n is g ain ed a t th e n eg a tiv e o xy g e n ato m .

T h e e q u ilib r iu m f o r th e a d d itio n o f w a te r to m o s t k e to n e s is u n fa v o ra b le , w h e re a s s o m e a ld e h y d e s (e .g ., fo rm a ld e h y d e ) e x is t p r im a r ily as th e g e m - d io l in a q u e o u s s o lu tio n . I t is n o t p o s s ib le to is o la te m o s t g e m - d io ls f r o m th e a q u e o u s s o lu tio n s in w h ic h th e y are fo rm e d . E v a p o r a tio n o f th e w a te r, f o r e x a m p le , s im p ly d is p la c e s th e o v e r a ll e q u ilib r iu m to th e r ig h t a n d th e g e m - d io l ( o r h y d ra te ) re v e rts to th e c a r b o n y l c o m p o u n d : HO R

OH

X

O

> H

R

A

+ H

h 2o 2

C o m p o u n d s w it h s tro n g e le c tr o n - w ith d r a w in g g ro u p s a tta c h e d to th e c a rb o n y l g ro u p ca n f o r m s ta b le

gem -d io ls . A n e x a m p le is th e c o m p o u n d c a lle d c h lo r a l h y d ra te : HO C l3C ^

OH H

C h lo ral h yd ra te

0 I1

747

16.7 The Addition o f Alcohols: Hemiacetals and Acetals

Dissolving formaldehyde in water leads to a solution containing primarily the gem-diol CH2 (OH)2. Show the steps in its formation from formaldehyde. When acetone is dissolved in water containing 18O instead of ordinary 16O (i.e., H218O

Review Problem 16.7

ReviewProblem16.8

18O

instead of H216 O), the acetone soon begins to acquire 18O and becomes

||

. The

C H 3C C H 3

formation of this oxygen-labeled acetone is catalyzed by traces of strong acids and by strong bases (e.g., OH- ). Show the steps that explain both the acid-catalyzed reaction and the basecatalyzed reaction.

16.7B Acetals • An acetal has two — OR groups attached to the same carbon atom. If we take an alcohol solution of an aldehyde (or ketone) and pass into it a small amount of gaseous HCl, a hemiacetal forms, and then the hemiacetal reacts with a second molar equivalent of the alcohol to produce an acetal. O

R'OH ■

R"

HO

OR

HCl(g

R 'O

R

R"

R'OH

R

OR' HO H

A hem iacetal (R" m ay be H)

R"

A n acetal (R" m ay be H)

Shown below is the structural formula for sucrose (table sugar). Sucrose has two acetal groupings. Identify these.

ReviewProblem16.9

OH

• The mechanism for acetal formation involves acid-catalyzed formation of the hemiacetal, then an acid-catalyzed elimination of water, followed by a second addition of the alcohol and loss of a proton. • A ll steps in the formation of an acetal from an aldehyde are reversible. If we dissolve an aldehyde in a large excess of an anhydrous alcohol and add a small amount of an anhydrous acid (e.g., gaseous HCl or concentrated H2 SO4), the equilibrium w ill strongly favor the formation of an acetal. After the equilibrium is established, we can isolate the acetal by neutralizing the acid and evaporating the excess alcohol. If we then place the acetal in water and add a catalytic amount of acid, all of the steps reverse. Under these conditions (an excess of water), the equilibrium favors the formation of the aldehyde. The acetal undergoes

hydrolysis:

R 'O

O

OR'

R ^^H A ceta l

H3O H2O

(several steps)

A A ld e h yd e

2 R 'O H

^ ____ H e lp f u l H i n t Equilibrium conditions govern the form ation and hydrolysis of hemiacetals and acetals.

748

Chapter 16

Aldehydes and Ketones

A MECHANISM FOR THE REACTION A cid -C a ta ly ze d A c e ta l F o rm a tio n

H C=O /s+ s-

H — O — R'

I /

H

C =O

+

6-+

R

Proton transfer to the carbonyl oxygen

I H

H

I

R

I I

H

+

H — O — R'

"

I

R

H

Proton removal from the positive oxygen results in formation of a hemiacetal.

O— R'

Ol

H

+. C=O

H— C— OH

H

"

H— C— O

"

H

H— C— O

:O — R'

R

H

H-^O —R'

=O— R'

-

H— C— O

:O — R'

Nucleophilic addition of the first alcohol molecule

I

:O — R' I H

+/ H R '— O O

H

I

"

R

"

r , /

X

R

h 2o

+

2

R'

P ro to n a tio n o f th e hydroxyl g ro u p leads to e lim in atio n of w a te r and fo rm a tio n o f a h ig h ly re a c tiv e o xo n iu m catio n .

H 3

n c = c'

/

R"'

A ld e h y d e or keto n e

R"

R'

P h o s p h o ru s y lid e (o r p h o sp h o ra n e )

\

R"

A lk e n e [(E) a n d (Z ) is o m e rs ]

T rip h e n y lp h o s p h in e o xid e

T he W ittig reaction is applicable to a w ide variety o f com pounds, an d although a m ix ­ ture o f (E ) an d (Z ) isom ers m ay result, the W ittig reaction offers a great advantage over m ost other alkene syntheses in that no am biguity exists as to the location o f the double bond in the product. (This is in contrast to E1 elim inations, w hich m ay yield m ultiple alkene prod­ ucts by rearrangem ent to m ore stable carbocation interm ediates, and both E1 an d E 2 elim ­ ination reactions, w hich m ay produce m ultiple products w hen different b hydrogens are available fo r rem oval.) Phosphorus ylides are easily prepared from triphenylphosphine and prim ary or secondary alkyl halides. T heir preparation involves tw o reactions: G e n e ra l R e ac tio n

R" CH— X

R eaction 1

^ H ^ P -C h

XR"

A n a lk y ltrip h e n y lp h o s p h o n iu m h alid e

T rip h e n y lp h o s p h in e

R"

R R eaction 2

( C

^ P - C

^ H

\

:B

( C ^ P “ C (R"

R"

H :B

A p h o s p h o ru s y lid e

S p e c ific E xa m p le

R eaction 1

(C 6H5)3P:

CH3Br

c kh „

^ H ^ P - C ^ BrM e th y ltrip h e n y lp h o s p h o n iu m b ro m id e (89% )

R eaction 2

(C6H5)3P-CH3 Br-

C6H5Li

(C 6H 5)3^C H 2:

C6H6

T h e firs t re a c tio n is a n u cleo p h ilic su b s titu tio n re a c tio n . T riphenylphosphine is an excellent nucleophile and a w eak base. It reacts readily w ith 1° and 2° alkyl halides by an Sn 2 m echanism to displace a halide ion from the alkyl halide to give an alkyltriphenylphosphonium salt. T h e seco n d re a c tio n is a n a c id - b a s e r e a c tio n . A strong base (usu­ ally an alkyllithium or phenyllithium ) rem oves a proton from the carbon that is attached to phosphorus to give the ylide.

LiBr

758

C h a p te r 1 6

A ld e h y d e s a n d

K e to n e s

Phosphorus ylides can be represented as a hybrid o f the tw o resonance structures shown here. Q uantum m echanical calculations indicate that the contribution m ade by the first struc­ ture is relatively unim portant. R"

R"

/

(C6H5)3P- c f -

( C 6 H 5)3 P = C R"

R"

Studies by E. Vedejs (University o f M ichigan) indicate that the W ittig reaction takes place in tw o steps. In the first step (below), the aldehyde or ketone com bines w ith the ylide in a cycloaddition reaction to form the four-m em bered ring o f an oxaphosphetane. T hen in a second step, the oxaphosphetane decom poses to form the alkene and triphenylphosphine oxide. T he driving force for the reaction is the form ation o f the very strong (DH° = 540 kJ m o l~ 1) p h o sp h o ru s-o x y g en b o n d in triphenylphosphine oxide.

A MECHANISM FOR THE REACTION T h e W ittig R e a c tio n

R" R'

R'

I I

\

/

■O

P(C6H5)3

+

Aldehyde or ketone

R'

R— C- t C— R" I^ C P(C6H5)3

=C — R " •O. .

R"

cx \

R

R"

Alkene (+ diastereom er)

O xap h o sp h etan e

Ylide

R"

-

T riphenylphosphine oxide

S p e c ific E xa m p le

O+

=Ch2

P(C6H5)3

2

O

P(C6H5)3

M ethylenecyclohexane (86% from cy clohexanone and m ethyltriphenylphosphonium bromide)

W hile W ittig syntheses m ay appear to b e com plicated, in practice they are easy to carry out. M ost o f the steps can b e carried o ut in the sam e reaction vessel, and the entire syn­ thesis can be accom plished in a m atter o f hours. The overall resu lt o f a W ittig synthesis is R

X ^ C = O

R'

16.10A

R"

+

R

several > steps H

Rm

R"

^ C= C^ R'

+

d ia s t e r e o m e r

Rm

How to Plan a W ittig Synthesis

Planning a W ittig synthesis begins w ith recognizing in the desired alkene w hat can be the aldehyde or ketone com ponent and w hat can b e the halide com ponent. A ny or all o f the R groups m ay be hydrogen, although yields are generally b etter w hen at least one group is hydrogen. T he halide com ponent m ust be a prim ary, secondary, or m ethyl halide.

0 I1

759

16.10 The Addition o f Ylides: The W ittig Reaction

Solved Problem 16.7 S y n th e s iz e 2 - m e th y l-1 - p h e n y lp r o p -1 - e n e u s in g a W it t ig re a c tio n . B e g in b y w r it in g a r e tr o s y n th e tic a n a ly s is .

STRATEGY AND ANSWER W e e x a m in e th e s tru c tu re o f th e c o m p o u n d , p a y in g a tte n tio n to th e g ro u p s o n e a ch s id e o f th e d o u b le b o n d :

2 -M eth y l-1 -p h en y lp ro p -1 -en e W e see th a t tw o r e tr o s y n th e tic a n a ly s e s a re p o s s ib le .

Retrosynthetic Analysis O

H

(a) P (C e H 5 )3

Ji =P (C e H 5 )3

X

(C e H 5 )3 P

(b) O

X (C e H 5 )3 P =

Synthesis F o llo w in g re tr o s y n th e tic a n a ly s is (a ), w e b e g in b y m a k in g th e y lid e f r o m a 2 -h a lo p ro p a n e a n d th e n a llo w th e y lid e to re a c t w it h b e n z a ld e h y d e :

H

(a)

:P(C6 H5)3

I

RLi (C e H 5 )3 P O

X

'P (C e H 5 )3

P (C e H 5 )3

F o llo w in g r e tr o s y n th e tic a n a ly s is (b ), w e m a k e th e y lid e f r o m a b e n z y l h a lid e a n d a llo w i t to re a c t w it h a c e to n e :

(b)

..X

(C e H 5 )3 P

(CeH5)3P=

X

J

R L i,

(C e H 5 )3 P

O

A

+

(C e H 5 )3 P O

760

Chapter 16 Aldehydes and Ketones

16.10B The Horner-Wadsworth-Emmons Reaction: A Modification of the W ittig Reaction A w idely used variation o f the W ittig reaction is the H o rn e r- W a d s w o r th -E m m o n s m odification. •

The H orner-W ad sw o rth -E m m o n s reaction involves use o f a phosphonate ester instead o f a triphenylphosphonium salt. T he m ajor pro d u ct is usually the (E)alkene isomer.

Som e bases that are typically used to form the phosphonate ester carbanion include sodium hydride, potassium tert-butoxide, and butyllithium . T he follow ing reaction sequence is an example: Step 1 O

O

NaH

H2 p^ O E t

P\ ^ O E t Na+ O E t

OEt

A p h o sp h o n a te ester Step 2 O O

O

H

.P v E tO / ' O _N a + E tO

.- ' ^ O E t Na+ O E t

T he phosphonate ester is prepared by reaction o f a trialkyl phosphite [(R O )3P] w ith an appropriate halide (a process called the A rbuzov reaction). T he follow ing is an exam ple: OEt

X

E ta

/ P

\

OEt

Triethyl p h o sp h ite

R e v ie w P r o b le m 1 6 .1 7

E tX

OEt OEt

In addition to triphenylphosphine, assum e that you have available as starting m aterials any necessary aldehydes, ketones, and organic halides. Show how you m ig h t synthesize each o f the follow ing alkenes using the W ittig reaction:

0 I1

16.12 Chemical Analyses for Aldehydes and Ketones

761

R e v ie w P r o b le m 1 6 .1 8

T rip h e n y lp h o s p h in e c a n b e u s e d to c o n v e rt e p o x id e s to a lk e n e s , f o r e x a m p le , .O

+

(C 6H 5)sP O

P ro p o s e a lik e ly m e c h a n is m f o r th is re a c tio n .

16.11

O xidation o f Aldehydes

A ld e h y d e s a re m u c h m o r e e a s ily o x id iz e d th a n k e to n e s . A ld e h y d e s a re r e a d ily o x id iz e d b y s tro n g o x id iz in g a g e nts su ch as p o ta s s iu m p e rm a n g a n a te , a n d th e y a re a ls o o x id iz e d b y such m i ld o x id iz in g a g e n ts as s ilv e r o x id e :

O

R

or Ag2O, OH"

O

H

R

O

X „ _

O-

R

JL

OH

N o t ic e th a t in th e se o x id a tio n s a ld e h y d e s lo s e th e h y d ro g e n th a t is a tta c h e d to th e c a rb o n y l c a rb o n a to m . B e c a u s e k e to n e s la c k th is h y d ro g e n , th e y a re m o r e re s is ta n t to o x id a tio n . A ld e h y d e s u n d e rg o s lo w o x id a tio n b y o x y g e n in th e a ir, a n d th u s s to re d sa m p le s o f a ld e ­ h y d e s o fte n c o n ta in th e c o rre s p o n d in g c a r b o x y lic a c id as a n im p u r ity .

16.12 16.12A

Chemical Analyses fo r A ldehyde

Derivatives of Aldehydes and Ketones

A ld e h y d e s a n d k e to n e s ca n b e d iffe r e n tia te d f r o m n o n c a r b o n y l c o m p o u n d s th ro u g h th e ir re a c tio n s w it h d e riv a tiv e s o f a m m o n ia (S e c tio n 1 6 .8 B ). 2 ,4 - D in itr o p h e n y lh y d r a z in e a n d h y d r o x y la m in e re a c t w it h a ld e h y d e s a n d k e to n e s to f o r m p re c ip ita te s . O x im e s a re u s u a lly c o lo rle s s , w h e re a s 2 ,4 -d in itr o p h e n y lh y d ra z o n e s a re u s u a lly o ra n g e . T h e m e ltin g p o in ts o f th e se d e riv a tiv e s ca n a ls o b e u s e d in id e n t if y in g s p e c ific a ld e h y d e s a n d k e to n e s .

16.12B Tollens' Test (Silver Mirror Test) T h e ease w it h w h ic h a ld e h y d e s u n d e rg o o x id a tio n d iffe re n tia te s th e m f r o m m o s t k e to n e s . M ix in g a q u e o u s s ilv e r n itra te w it h a q u e o u s a m m o n ia p ro d u c e s a s o lu tio n k n o w n as T o lle n s ’ re a g e n t. T h e re a g e n t c o n ta in s th e d ia m m in o s ilv e r ( I ) io n , A g ( N H 3)2 + . A lt h o u g h th is io n is a v e ry w e a k o x id iz in g a g e n t, i t o x id iz e s a ld e h y d e s to c a rb o x y la te a n io n s . A s i t d o e s th is , s ilv e r is re d u c e d f r o m th e + 1 o x id a tio n state [ o f A g ( N H 3)2 + ] to m e t a llic s ilv e r. I f th e ra te o f r e a c tio n is s lo w a n d th e w a lls o f th e v e s s e l a re c le a n , m e t a llic s ilv e r d e p o s its o n th e w a lls o f th e te s t tu b e as a m ir r o r ; i f n o t, i t d e p o s its as a g r a y - to - b la c k p re c ip ita te . T o lle n s ’ re a g e n t g iv e s a n e g a tiv e r e s u lt w it h a ll k e to n e s e x c e p t a - h y d r o x y k e to n e s : O R

^

O H

A ldehyde

Ag(NH3)2+ i-i i i H2O

R

"O -

^ Silver mirror

762 16.13

Chapter 16 Aldehydes and Ketones

Spectroscopic Properties o f Aldehydes and Ketones 16.13A •

IR Spectra of Aldehydes and Ketones

C a r b o n y l g ro u p s o f a ld e h y d e s a n d k e to n e s g iv e r is e to v e ry s tro n g C = O s tre tc h ­ in g a b s o rp tio n b a n d s in th e 1 6 6 5 - 1 7 8 0 - c m - 1 re g io n .

T h e e x a c t lo c a tio n o f th e c a r b o n y l I R a b s o rp tio n (T a b le 1 6 .3 ) d e p e n d s o n th e s tru c tu re o f th e a ld e h y d e o r k e to n e a n d is o n e o f th e m o s t u s e fu l a n d c h a ra c te ris tic a b s o rp tio n s in th e I R s p e c tru m . •

S a tu ra te d a c y c lic a ld e h y d e s t y p ic a lly a b s o rb n e a r 1 7 3 0 c m - 1 ; s im ila r k e to n e s a b s o rb n e a r 1 7 1 5 c m - 1 .

IR C a r b o n y l S t r e t c h i n g B a n d s o f A l d e h y d e s a n d K e t o n e s

C = O S tre tc h in g F req u en cies C om pound

R ange (cm -1 )

Com pound

R ange (cm -1 )

A r— CHO

1720-1740 1695-1715

RCOR ArCOR

1705-1720 1680-1700

W

1680-1690

' c = c/

1665-1680

/ ^COR Cyclohexanone Cyclopentanone Cyclobutanone

1715 1751 1785

R — CHO

''Cho

•

C o n ju g a tio n o f th e c a rb o n y l g ro u p w it h a d o u b le b o n d o r a b e n z e n e r in g s h ifts th e C = O a b s o rp tio n to lo w e r fre q u e n c ie s b y a b o u t 4 0 c m - 1 .

T h is s h if t to lo w e r fre q u e n c ie s o c c u rs b e ca u se th e c a r b o n y l d o u b le b o n d o f a c o n ju g a te d c o m p o u n d has m o r e s in g le -b o n d c h a ra c te r (see th e re s o n a n c e s tru c tu re s b e lo w ) , a n d s in ­ g le b o n d s a re e a s ie r to s tre tc h th a n d o u b le b o n d s .

Single bond

T h e lo c a tio n o f th e c a r b o n y l a b s o rp tio n o f c y c lic k e to n e s d e p e n d s o n th e siz e o f th e r in g (c o m p a re th e c y c lic c o m p o u n d s in T a b le 1 6 .3 ).

A s the ring grow s smaller, the C = O stretch­

ing peak is shifted to h igher frequencies . V ib r a tio n s o f th e C — H b o n d o f th e C H O g ro u p o f a ld e h y d e s a ls o g iv e tw o w e a k b a n d s in th e 2 7 0 0 - 2 7 7 5 - a n d 2 8 2 0 - 2 9 0 0 - c m - 1 r e g io n s th a t a re e a s ily id e n tifie d . F ig u r e 16.1 s h o w s th e I R s p e c tru m o f p h e n y le th a n a l.

0 I1

16.13 Spectroscopic Properties of Aldehydes and Ketones

Wavenumber (cm 1) Figure 16.1 The infrared spectrum of phenylethanal.

16.13B 13C

•

N M R

NMR Spectra of Aldehydes and Ketones Spectra

T he carbonyl carbon o f an aldehyde or ketone gives characteristic N M R signals in the S 1 80-220 region o f 13C spectra.

Since alm ost no other signals occur in this region, the presence o f a signal in this region (near S 200) strongly suggests the presence o f a carbonyl group. 1H

N M R

•

Spectra

A n aldehyde proton gives a distinct 1H N M R signal dow nfield in the S 9 -1 2 region w here alm ost no other protons absorb; therefore, it is easily identified.

T he aldehyde proton o f an aliphatic aldehyde show s sp in -sp in coupling w ith protons on the adjacent a carbon, and the splitting pattern reveals the degree o f substitution o f the a carbon. F or exam ple, in acetaldehyde (CH 3C H O ) the aldehyde proton signal is split into a quartet by the three m ethyl protons, and the m ethyl proton signal is split into a doublet by the aldehyde proton. The coupling constant is small, how ever (about 3 H z, as com pared w ith typical vicinal splitting o f about 7 Hz). •

Protons on the a carbon are deshielded by the carbonyl group, an d their signals generally appear in the S 2 .0 -2 .3 region.

•

M ethyl ketones show a characteristic (3H) singlet n ear S 2.1.

F igures 16.2 and 16.3 show annotated 1H and 13C spectra o f phenylethanal.

763

764

Chapter 16 Aldehydes and Ketones

8

7

6

5

4 3 5h (ppm)

2

1

0

Figure 16.2 The 300-MHz 1H NMR spectrum of phenylethanal. The small coupling between the aldehyde and methylene protons (2.6 Hz) is shown in the expanded offset plots.

220

200

180

160

140

120 100 Sc (ppm)

80

60

40

20

0

Figure 16.3 The broadband proton-decoupled 13C NMR spectrum of phenylethanal. DEPT 13C NMR information and carbon assignments are shown near each peak.

16.13C

Mass Spectra of Aldehydes and Ketones

T h e m a s s s p e c tra o f k e to n e s u s u a lly s h o w a p e a k c o rre s p o n d in g to th e m o le c u la r io n . A ld e h y d e s t y p ic a lly p ro d u c e a p r o m in e n t M .+ - 1 p e a k in t h e ir m a ss s p e c tra f r o m c le a v ­ a ge o f th e a ld e h y d e h y d ro g e n . K e to n e s u s u a lly u n d e rg o c le a v a g e o n e ith e r sid e o f th e c a r­ b o n y l g ro u p to p ro d u c e a c y liu m io n s , R C #

O : +, w h e re R ca n b e th e a lk y l g ro u p f r o m e ith e r

s id e o f th e k e to n e c a rb o n y l. C le a v a g e v ia th e M c L a f f e r t y re a rra n g e m e n t (S e c tio n 9 .1 6 D ) is a ls o p o s s ib le in m a n y a ld e h y d e s a n d k e to n e s .

16.13D

UV Spectra

T h e c a r b o n y l g ro u p s o f s a tu ra te d a ld e h y d e s a n d k e to n e s g iv e a w e a k a b s o rp tio n b a n d in th e U V r e g io n b e tw e e n 2 7 0 a n d 3 0 0 n m . T h is b a n d is s h ifte d to lo n g e r w a v e le n g th s ( 3 0 0 - 3 5 0 n m ) w h e n th e c a rb o n y l g ro u p is c o n ju g a te d w it h a d o u b le b o n d .

16.14 Summary of Aldehyde and Ketone Addition Reactions

16.14

Sum m ary o f A ld eh yd e and K etone A d d itio n Reactions

T h e n u c le o p h ilic a d d itio n re a c tio n s o f a ld e h y d e s a n d k e to n e s o c c u r r in g a t th e c a rb o n y l c a r­ b o n a to m th a t w e h a v e s tu d ie d so f a r a re s u m m a riz e d b e lo w . I n C h a p te rs 18 a n d 19 w e s h a ll see o th e r e x a m p le s .

N ucleophilic A d d itio n

R eactions o f A ld e h y d e s a n d

K etones

1. A d d itio n o f O rg a n o m e ta llic Com pounds (S ectio n 12.7C) G en era l R ea c tio n s-

s+

R9

M

765

V'O -'

O

M+

R

=OH H3 °+

R

S p e c ific E x a m p le U sin g a G rignard R e a g e n t (S e ctio n 12.7C) O

(/ 1n)

^MMnRr gBr

(2 ) HO+

H

OH

*

67% 2. A d d itio n o f H y d rid e Io n (S ection 12.3) G en era l R e a c tio n ^O '

O

OH

H-

H3O+

H

H

S p e c ific E x a m p le s U sin g M etal H yd rid es (S e ctio n 12.3)

(90%) O

OH

NaBH4 MeOH

H

(100%) 3. A d d itio n o f H yd ro g e n C yanide (S ection 16.9) G en era l R ea c tio n O­ I

OH

HA

N # C:

A: C

C

\

N

S p e c ific E xa m p le

O

OH

(1) Na CN (2) HA

'

N CN

A cetone cyanohydrin (78%)

766

Chapter 16

4.

Aldehydes and Ketones

A d d itio n o f Ylides ( S e c tio n 1 6 .1 0 ) The Wittig Reaction ( C 6H 5 )3 P O

(C6H5)3P^ 5.

(C6H5)3P- O

O

A d d itio n o f A lc o h o ls ( S e c tio n 1 6 .7 ) General Reaction O

HO

ROH

/

OR

....

RO

ROH, HA

O R

H 2O

Hemiacetal

Acetal

Specific Example O EtOH

HO

/ OEt

EtOH, HA

E tO

OEt

H 6

.

A d d itio n o f D erivatives o f A m m o n ia ( S e c tio n 1 6 .8 ) ¡mines O* h 3 o+ H 2O

R — NH

1° Amine

Aldehyde or ketone

Imine [(E) and (Z) isomers]

Enamines R \

O

- / N

R

cat. HA R— N— H

+

H

h

2o

R

Enamine

2° Amine

Key Terms and Concepts

PLUS

T h e k e y te rm s a n d c o n c e p ts th a t a re h ig h lig h t e d in b o ld , b lu e t e x t w it h in th e c h a p te r are d e fin e d in th e g lo s s a ry (a t th e b a c k o f th e b o o k ) a n d h a v e h y p e r lin k e d d e fin itio n s in th e a c c o m p a n y in g

W ileyP LU S c o u rs e ( w w w .w ile y p lu s .c o m ) .

Problems Note to Instructors: Many of th the homework problems are available for assignment via WileyPLUS, an online teaching and learning solution. R E A C TIO N S A N D

16.19

N O M E N C L A T U R E

G iv e a s tr u c tu r a l fo r m u la a n d a n o th e r a c c e p ta b le n a m e f o r e a ch o f th e f o llo w in g c o m p o u n d s : ( a ) F o r m a ld e h y d e

( f ) A c e to p h e n o n e

( k ) E t h y l is o p r o p y l k e to n e

( b ) A c e ta ld e h y d e

(g ) B e n z o p h e n o n e

( l) D iis o p r o p y l k e to n e

(c ) P h e n y la c e ta ld e h y d e

( h ) S a lic y la ld e h y d e

( m ) D ib u t y l k e to n e

( d ) A c e to n e

( i) V a n illin

( n ) D ip r o p y l k e to n e

(e ) E t h y l m e t h y l k e to n e

( j ) D ie t h y l k e to n e

( o ) C in n a m a ld e h y d e

767

P roblem s 16.20

W r it e s tr u c tu r a l fo rm u la s f o r th e p ro d u c ts fo r m e d w h e n p ro p a n a l re a c ts w it h e a ch o f th e f o llo w in g re a g e n ts : ( a ) N a B H 4 in a q u e o u s N a O H

(h) C H 3 C H - P ( C 6 H5)3

( b ) C 6H 5M g B r, th e n H 3O +

(i) A g ( N H 3 ) 2+

(c ) L iA lH 4, th e n H 2O

(j) H y d r o x y la m in e

( d ) A g 2O , O H ~

(k) P h e n y lh y d ra z in e

(e ) ^ H ^ P = C ^

(l) C o ld d ilu te K M n O 4

( f ) H 2 a n d Pt

HA

SH

„.O H

(n) H S "

and HA

(g ) H O "

16.21

^SH

(m ) H S "

H A , th e n R a n e y n ic k e l

G iv e s tr u c tu r a l fo rm u la s f o r th e p ro d u c ts fo r m e d ( i f a n y ) f r o m th e r e a c tio n o f a c e to n e w it h e a ch re a g e n t in E x e rc is e 1 6 .2 0 .

16.22

W h a t p ro d u c ts w o u ld b e o b ta in e d w h e n a c e to p h e n o n e re a c ts u n d e r e a ch o f th e f o llo w in g c o n d itio n s ? (a ) A c e to p h e n o n e + H N O 3

u

H2 SO 4

>

O

ha

( b ) A c e to p h e n o n e + C 6H 5N H N H 2 ---------- >

( c ) A c e to p h e n o n e + A c e to p h e n o n e

: C H 2— P ( C 6H 5) 3

( d ) A c e to Ph e n o n e + N a B H

4

- OHHO-

(e ) A c e to p h e n o n e + C 6H 5M g B r

16.23

(2) NH4Cl

P r e d ic t th e m a jo r o rg a n ic p r o d u c t f r o m e a ch o f th e f o llo w in g re a c tio n s .

(a)

H

HO

„OH, H2 SO 4 (cat.)

H2 SO 4 (cat.)

(c )

O OH

H2 SO 4 (cat.) H2 O, H2 SO 4 (cat.)

OH (d )

O

(g )

(1 ) H2SO4, H2O

(2)

H, H2 SO 4 (cat.)

O

768 16.24

Chapter 16

Aldehydes and Ketones

P r e d ic t th e m a jo r p r o d u c t f r o m e a ch o f th e f o llo w in g re a c tio n s . (a )

(e)

H

CH3 NH2, cat. HA

( 1 ) HS'

,SH

(2) Raney Ni, H2 O O

(b)

O

O (c )

16.25

- H , cat. HA H

O

P r e d ic t th e m a jo r p r o d u c t f r o m e a ch o f th e f o llo w in g re a c tio n s . (a )

O

A

r~ \Q ( 1 ) ^ ^ ^ ^ /- \^ ,,.M g B r (excess)

r

(2 ) H3 O+ (3) H2 CrO4

'

H2 SO 4 (cat.)

16.26

P r o v id e th e re a g e n ts n e e d e d to a c c o m p lis h e a ch o f th e f o llo w in g tr a n s fo rm a tio n s .

(d) O

O

O

H OH

(e)

OH

H

O

OH

(c) O

= \

(f) O

O

S:>

769

P roblem s 16.27

W rite detailed m echanism s for each o f the follow ing reactions. (a )

O

(b )

O

(c)

(d)

HO

16.28

OH

Provide the reagents necessary for the follow ing synthesis. OH

S Y N T H E S IS

16.29

16.30

16.31

(a )

Synthesize phenyl propyl ketone from benzene an d any other n eeded reagents.

(b )

G ive tw o m ethods for transform ing phenyl propyl ketone into butylbenzene.

Show how you w ould convert benzaldehyde into each o f the follow ing. You m ay use any other need ed reagents, and m ore than one step m ay be required. (a )

B enzyl alcohol

(f)

3-M ethyl-1-phenyl-1-butanol

(k )

C6H5CHDOH

(b )

B enzoic acid

(g )

B enzyl brom ide

( l)

C6H5CH(OH)CN

(c )

B enzoyl chloride

(h )

Toluene

(d )

B enzophenone

( i)

C6H5C H (O C H 3)2

(n )

C6H5C H = N N H C 6H5

(e )

1-Phenylethanol

( j)

C 6H5C H 18O

(o )

C6H5C H = CHCH = CH 2

Show how ethyl phenyl ketone (C 6H5C O C H 2C H 3) could b e synthesized from each o f the following: (a )

16.32

( m ) C 6 H 5C H = N O H

B enzene

(b )

B enzonitrile, C 6H5CN

(c )

B enzaldehyde

Show how benzaldehyde could be synthesized from each o f the following: (a )

B enzyl alcohol

(c )

Phenylethyne

(e ) C 6 H 5 C O 2 C H 3

(b )

B enzoic acid

(d )

P henylethene (styrene)

(f) C 6 H 5 C #

N

770 16.33

Chapter 16

Aldehydes and Ketones

G iv e s tru c tu re s f o r c o m p o u n d s A - E . ~ , H 2 CrO 4 _ C y c lo h e x a n o l ___.___ : A (C 6 H 1 0 O ) acetone

(1) CH3Mgl _ . u „+ : B (C 7 H 1 4 O) (2 ) H3O+

HA heat ( 1 ) O3

C (C 7 H 1 2 )

16.34

(2) Me2S

D ( C 7 H 1 2 O 2 ) (1) ’ OH : E ( C 7 H 1 2 O 3 ) ' 7 1 2 2 (2) H3O

W a r m in g p ip e r o n a l (S e c tio n 1 6 .3 ) w it h d ilu te a q u e o u s H C l c o n v e rts i t to a c o m p o u n d w it h th e fo r m u la C 7 H 6 O 3 W h a t is th is c o m p o u n d , a n d w h a t ty p e o f r e a c tio n is in v o lv e d ?

16.35

S ta rtin g w it h b e n z y l b ro m id e , s h o w h o w y o u w o u ld s y n th e s iz e e a ch o f th e f o llo w in g :

(a)

(b)

(c)

O

(d)

H

16.36

C o m p o u n d s A a n d D d o n o t g iv e p o s itiv e T o lle n s ’ te sts; h o w e v e r, c o m p o u n d C do e s. G iv e s tru c tu re s f o r A - D . HOCH 2 CH 2 OH, HA 4 - B r o m o b u ta n a l

. ___ _ ■A ( C 6 H 1 1 O 2 B r)

Mg, Et2O ( 1 ) c h 3c h o

[B ( C 6 H n M g O 2 B r)j

16.37

(2) H3 O+, H2O

• C ( C 6H 12O 2) ' 6 12 2

HA

D (C7H14O2)

D ia n e a c k e ro n e is a v o la t ile n a tu r a l p r o d u c t is o la te d f r o m s e c re to ry g la n d s o f th e a d u lt A f r ic a n d w a r f c r o c o d ile . T h e c o m p o u n d is b e lie v e d to b e a p h e ro m o n e a s s o c ia te d w it h n e s tin g a n d m a tin g . D ia n e a c k e ro n e is n a m e d a fte r D ia n e A c k e r m a n , a n a u th o r in th e fie ld o f n a tu r a l h is to r y a n d c h a m p io n o f th e im p o r ta n c e o f p re s e rv in g b io d iv e r ­ s ity . T h e IU P A C n a m e o f d ia n e a c k e ro n e is 3 ,7 -d ie th y l- 9 - p h e n y ln o n a n - 2 - o n e , a n d i t is fo u n d as b o th th e (3 S ,7 S ) a n d (3 S ,7 R ) s te re o is o m e rs . D r a w s tru c tu re s f o r b o th s te re o is o m e rs o f d ia n e a c k e ro n e .

16.38

O u tlin e d h e re is a s y n th e s is o f g ly c e ra ld e h y d e (S e c tio n 5 .1 5 A ). W h a t are th e in te rm e d ia te s A - C a n d w h a t s te re o is o m e r ic f o r m o f g ly c e ra ld e h y d e w o u ld y o u e x p e c t to o b ta in ? PCC OH

16.39

CH 2 Cl2

c h 3o h , ha ■A ( C 3 H 4 O )

B ( C 5 H 1 0 O 2 ) KMn0>4,’’0 !> > C (C 5 H 1 2 O 4 ) cold, dilute

H2 O

g ly c e r a ld e h y d e

C o n s id e r th e r e d u c tio n o f ( R )-3 - p h e n y l- 2 -p e n ta n o n e b y s o d iu m b o ro h y d r id e . A f t e r th e r e d u c tio n is c o m p le te , th e m ix t u r e is se p a ra te d b y c h ro m a to g r a p h y in to tw o fr a c tio n s . T h e s e fr a c tio n s c o n ta in is o m e r ic c o m p o u n d s , a n d each is o m e r is o p t ic a lly a c tiv e . W h a t a re th e se tw o is o m e rs a n d w h a t is th e s te re o is o m e ric r e la tio n s h ip b e tw e e n th e m ?

16.40

T h e s tru c tu re o f th e sex p h e ro m o n e (a ttra c ta n t) o f th e fe m a le tse tse f l y has b e e n c o n fir m e d b y th e f o llo w in g s y n ­ th e s is . C o m p o u n d C a p p e a rs to b e id e n tic a l to th e n a tu r a l p h e ro m o n e in a ll re s p e c ts ( in c lu d in g th e re s p o n s e o f th e m a le ts e ts e f ly ) . P r o v id e s tru c tu re s f o r A , B , a n d C .

O 2 ( >) Br

Br

2

H2 ’Pt

(C 6 H5 )3 P

(2 ) 2 RL i

A (C 45H 46P 2 )

B (C 37H 72)

C ( C 37H 76)

7

16.41

P r o v id e re a g e n ts th a t w o u ld a c c o m p lis h each o f th e f o llo w in g syntheses. B e g in b y w r it in g a re tro s y n th e tic a n a ly s is .

(a) ,B r

HO OH

(b)

OH

O

OH

771

P roblem s

M E C H A N I S M S A N D S T R U C T U R E E L U C I D A T IO N

16.42

W rite a detailed m echanism for the follow ing reaction. OH

OH

H2SO4 (cat.), H2O

O

O

OH HO

O

16.43

W hen H2N NHNH2 (semicarbazide) reacts with a ketone (or an aldehyde) to form a derivative know n as a semicarbazone, only one nitrogen atom o f semicarbazide acts as a nucleophile and attacks the carbonyl carbon atom o f the O N'

ketone. T he product o f the reaction, consequently, is

I

R'(H) O NH2 rather than R

NHNH

2

.

R '(H ) H

W hat factor accounts for the fact that tw o nitrogen atom s o f sem icarbazide are relatively non-nucleophilic? 16.44

D utch elm disease is caused by a fungus transm itted to elm trees by the elm b ark beetle. T he fem ale beetle, w hen she has located an attractive elm tree, releases several pherom ones, including m ultistriatin, below. T hese pherom ones attract m ale beetles, w hich bring w ith them the deadly fungus.

OMultistriatin T reating m ultistriatin w ith dilute aqueous acid at room tem perature leads to the form ation o f a product, C 10H20O 3, w hich show s a strong infrared peak near 1715 c m - 1 . Propose a structure for this product. 16.45

The follow ing structure is an interm ediate in a synthesis o f prostaglandins F 2a and E 2 by E. J. C orey (Harvard U niversity). A H orner-W adsw orth-E m m ons reaction w as used to form the (£)-alkene. W rite structures for the phosphonate ester and carbonyl reactant that w ere used in this process. (Note: T he carbonyl com ponent o f the reaction included the cyclopentyl group.)

A c = C H 3C

16.46

C om pounds W and X are isom ers; they have the m olecular form ula CgH8O. T he IR spectrum o f each com pound show s a strong absorption b and near 1715 c m - 1 . O xidation o f either co m ­ pound w ith hot, basic potassium perm anganate follow ed by acidification yields phthalic acid. T he 1H N M R spectrum o f W show s a m ultiplet at 8 7.3 an d a singlet at 8 3.4. T he 1H N M R spectrum o f X show s a m ultiplet at 8 7.5, a triplet at 8 3.1, and a triplet at 8 2.5. Propose struc­ tures for W and X.

c o 2h

c o 2h Phthalic acid

772 16.47

Chapter 16

Aldehydes and Ketones

Compounds Y and Z are isomers with the molecular formula C 1 0 H 1 2 O. The IR spectrum of each compound shows a strong absorption band near 1710 cm-1 . The 1H N M R spectra of Y and Z are given in Figs. 16.4 and 16.5. Propose structures for Y and Z.

5h (ppm) Figure 16.4 The 300-MHz 1H NMR spectrum of compound Y, Problem 16.47. Expansions of the signals are shown in the offset plots.

5h (ppm) Figure 16.5 The 300-MHz 1H NMR spectrum of compound Z, Problem 16.47. Expansions of the signals are shown in the offset plots.

16.48

Compound A (C 9 H 1 8 O) forms a phenylhydrazone, but it gives a negative Tollens’ test. The IR spectrum of A has a strong band near 1710 cm-1 . The broadband proton-decoupled 13C N M R spectrum of A is given in Fig. 16.6. Propose a structure for A.

C h allen g e P roblem s

773

5C(ppm) Figure 16.6 The broadband proton-decoupled 13C NMR spectrum of compound A, Problem 16.48. Information from the DEPT 13C NMR spectra is given above the peaks. 16.49

C o m p o u n d B ( C 8 H 1 2 O 2 ) s h o w s a s tro n g c a rb o n y l a b s o rp tio n in its I R s p e c tru m . T h e b ro a d b a n d p ro to n - d e c o u p le d 13C N M R s p e c tru m o f B has o n ly th re e s ig n a ls , a t S 19 ( C H 3), 71 ( C ), a n d 2 1 6 ( C ). P ro p o s e a s tru c tu re f o r B .

Challenge Problems 16.50

( a ) W h a t w o u ld b e th e fre q u e n c ie s o f th e tw o a b s o rp tio n b a n d s e x p e c te d to b e m o s t p r o m in e n t in th e in fr a r e d sp e c ­ tr u m o f 4 -h y d r o x y c y c lo h e p ta n o n e (C )? ( b ) I n r e a lity , th e lo w e r fr e q u e n c y b a n d o f th e se tw o is v e ry w e a k . D r a w th e s tru c tu re o f an is o m e r th a t w o u ld e x is t in e q u ilib r iu m w it h C a n d th a t e x p la in s th is o b s e rv a tio n .

16.51

O n e o f th e im p o r ta n t re a c tio n s o f b e n z y lic a lc o h o ls , e th e rs , a n d esters is th e ease o f c le a v a g e o f th e b e n z y l- o x y g e n b o n d d u r in g h y d ro g e n a tio n . T h is is a n o th e r e x a m p le o f “ h y d ro g e n o ly s is ,” th e c le a v a g e o f a b o n d b y h y d ro g e n . I t is fa c ilita te d b y th e p re s e n c e o f a c id . H y d r o g e n o ly s is ca n a ls o o c c u r w it h s tr a in e d - rin g c o m p o u n d s . O n h y d ro g e n a tio n o f c o m p o u n d D (see b e lo w ) u s in g R a n e y n ic k e l c a ta ly s t in a d ilu t e s o lu tio n o f h y d ro g e n c h lo ­ r id e in d io x a n e a n d w a te r, m o s t p ro d u c ts h a v e a 3 ,4 - d im e th o x y p h e n y l g ro u p a tta c h e d to a s id e c h a in . A m o n g these, an in te re s tin g p r o d u c t is E , w h o s e fo r m a tio n illu s tr a te s n o t o n ly h y d ro g e n o ly s is b u t a ls o th e m ig r a to r y a p titu d e o f p h e n y l g ro u p s . F o r p ro d u c t E , th e se a re k e y s p e c tra l data: MS

(m/z): 1 9 6 .1 0 8 4 ( M - + , a t h ig h r e s o lu tio n ) , 178

I R ( c m - 1 ): 3 4 0 0 (b ro a d ), 3 0 5 0 , 2 8 5 0 ( C H 3 — O s tre tc h ) * H N M R (S, in C D C l3): 1.21 (d , 3 H , (d , 2 H ,

J = 7 H z ) , 3 .8 2 (s,

6

J = 7 H z ) , 2 .2 5 (s, 1 H ), 2 .8 3 ( m , 1 H ), 3 .5 8

H ), 6 .7 0 (s, 3 H ).

W h a t is th e s tru c tu re o f c o m p o u n d E ? OH

OCH3

H3 C O . Raney Ni H3 CO'

OH

D

E

774

Chapter 16 Aldehydes and Ketones

Learning Group Problems A synthesis o f ascorbic acid (vitam in C, 1) starting from D -(+ )-galactose (2) is show n below (Haw orth, W.N., et al., J. Chem. Soc., 1933, 1419-1423). C onsider the follow ing questions about the design and reactions used in this synthesis:

(a) W hy did H aw orth and co-w orkers introduce the acetal functional groups in 3? (b) W rite a m echanism for the form ation o f one o f the acetals. (c) W rite a m echanism for the hydrolysis o f one o f the acetals (4 to 5). A ssum e that w ater w as present in the reaction m ixture. (d) In the reaction from 5 to 6 you can assum e that there w as acid (e.g., HCl) present w ith the sodium am algam . W hat reac­ tion occurred here and from w hat functional group d id that reaction actually proceed? (e) W rite a m echanism for the form ation o f a phen y lh y d razo n e from the ald eh y d e carbonyl o f 7. [Do n o t b e concerned about the p henylhydrazone group at C2. We shall study the form ation o f bishy d razo n es o f this ty p e (called an osazone) in C hapter 22.] (f) W hat reaction w as used to add the carbon atom that ultim ately becam e the lactone carbonyl carbon in ascorbic acid (1)?

Me2CO/H2SO4

KMnO,

HO

OH

H2SO4

Na/Hg HO

HO H

OH

HO

O

^ ^

O

\^ O H COOH

COOH

OH

6

5

(1)NH3 (2)NaOCl OH

HO

PhNHNH2

HO ^ O OH

OH

HO

HO

PhCHO/H2SO4

y

"

OH

HO HO

y -

"^N N H P h

O

PhHNN

7

8

KCN/CaCl2

OH

H

CN

HO rotation about C3— C4 bond

OH

O'

H OH

SUMMARY OF MECHANISMS Acetals, Imines, and Enamines: Common M echanistic Themes in Their Acid-catalyzed Form ation from Aldehydes and Ketones Many steps are nearly the same in acid-catalyzed reactions of aldehydes and ketones with alcohols and amines. Com pare the m echanism s vertically to see the sim ilarities and differences. Note differences in com pletion of the mechanism for each type of product.

I. Hemiacetal and acetal form ation: reaction w ith ilcohols

~:a

H H -A

(b) CICOCI + excess CH 3 NH 2 ---------> (c) Glycine (H3N CH 2 CO2) + C 6 H5 CH2OCOCl

OH2

(d) Product of (c) + H2, Pd -------- > (e) Product of (c) + cold HBr, CH 3 CO2H ------(f) Urea + OH2 , H 2 O, heat

Although alkyl chloroformates (ROCOCI), dialkyl carbonates (ROCOOR), and carba­ mates (ROCONH2, ROCONHR, etc.) are stable, chloroformic acid (HOCOCI), alkyl hydro­ gen carbonates (ROCOOH), and carbamic acid (HOCONH2) are not. These latter compounds decompose spontaneously to liberate carbon dioxide: O

O HCI

/C ^ HO Cl Chloroform ic acid (unstable)

+

CO 2

.a ----- > ROH RO OH An alkyl hydrogen carb o n ate (unstable)

CO

O z -C ^ HO NH 2 A carb am ic acid (unstable)

NH

CO

This instability is a characteristic that these compounds share with their functional parent, carbonic acid: O /C ^ HO OH C arbonic acid (unstable)

H 2O

C O

2

17.10 Decarboxylation o f Carboxylic Acids The reaction whereby a carboxylic acid loses CO 2 is called a decarboxylation: o

R

A

decarboxylation

----------- -------->

_

R— H

+

C O

2

OH

Although the unusual stability of carbon dioxide means that decarboxylation of most acids is exothermic, in practice the reaction is not always easy to carry out because the reaction is very slow. Special groups usually have to be present in the molecule for decarboxyla­ tion to be rapid enough to be synthetically useful.

815

17 .10 Decarboxylation o f Carboxylic Acids

•

Carboxylic acids that have a carbonyl group one carbon removed from the car­ boxylic acid group, called b-keto acids, decarboxylate readily when they are heated to 100-150°C. Some b-keto acids even decarboxylate slowly at room temperature. O

O

II

II

O 100-150°C

R ^ ^ ^ ^ O H A J -k e to acid

CO„ *

R

There are two reasons for this ease of decarboxylation: 1. When the acid itself decarboxylates, it can do so through a six-membered cyclic tran­ sition state: O

Ketone This reaction produces an enol directly and avoids an anionic intermediate. The enol then tautomerizes to a methyl ketone. 2. When the carboxylate anion decarboxylates, it forms a resonance-stabilized anion:

O

O

"

''O''

1

-CO,

R

''O '

^

A cylacetate ion

=O

R esonance-stabilized anion This type of anion, which we shall study further in chapter 19, is much more stable than simply RCH2:~, the anion that would have been produced by decarboxylation in the absence of a b-carbonyl group. 1 B

1

S o lv e d fiW P ro fC b lem 8 JkJI vfcrU flll 11/7 .0

■ Provide structures for A and B.

OH

-

H2 CrO4 2----- ^

A (C 7 H 12O 3 )

B (C 6 H 12 O)

+

CO 2

O STRATEGY AND ANSWER H 2 CrO 4 oxidizes a primary alcohol to a carboxylic acid, which is consistent with the formula provided for A. Because A is a b-ketocarboxylic acid, it decarboxylates on heating to form B.

OH O

O A

O B

816

Chapter 17

Carboxylic Acids and Their Derivatives

b-D icarboxylic acids (1,3-dicarboxylic acids, also called m alonic acids) decarboxylate read ­ ily for reasons sim ilar to b -k eto acids. O

P

O

A b-dicarboxylic acid b-D icarboxylic acids undergo decarboxylation so readily that they do n o t form cyclic anhy­ drides (Section 17.6A). We shall see in Sections 18.6 and 18.7 how decarboxylation o f b -k eto acids and m alonic acids is synthetically useful.

17.10A Decarboxylation of Carboxyl Radicals A lthough the carboxylate ions (R C O 2~) o f sim ple aliphatic acids do n ot decarboxylate read­ ily, carboxyl radicals (R C O 2 • ) do. T hey decarboxylate by losing C O 2 an d producing alkyl radicals: RCO2- ---------: R* + C O 2

R e v ie w P r o b le m 1 7 .1 6

U sing decarboxylation reactions, outline a synthesis o f each o f the follow ing from appro­ priate starting m aterials: (a) 2-H exanone

(c) C yclohexanone

(b) 2-M ethylbutanoic acid

(d) Pentanoic acid

R e v ie w P r o b le m 1 7 .1 7

O

X

D iacyl peroxides, R

O

O —O

X

R , decom pose readily w hen heated.

(a) W hat factor accounts for this instability? (b) T he decom position o f a diacyl peroxide produces C O 2. H ow is it form ed? (c) D iacyl peroxides are often used to initiate radical reactions, for exam ple, the p olym er­ ization o f an alkene: O

n

O

rA o _o A r >

Show the steps involved.

17.11 Chemical Tests fo r Acyl Com pounds C arboxylic acids are w eak acids, and their acidity helps us to detect them. •

A queous solutions o f w ater-soluble carboxylic acids give an acid test w ith blue lit­ m us paper.

•

W ater-insoluble carboxylic acids dissolve in aqueous sodium hydroxide and aq u e­ ous sodium bicarbonate (see Section 17.2C).

•

Sodium bicarbonate helps us distinguish carboxylic acids from m ost phenols. E xcept for the di- an d trinitrophenols, phenols do n o t dissolve in aqueous sodium bicarbonate. W hen carboxylic acids dissolve in aqueous sodium bicarbonate, they also cause the evolution o f carbon dioxide.

R ^ C

1 7 .1 2 P o ly e s te rs a n d P o ly a m id e s : S te p -G r o w th P o ly m e rs

x' o h

817

17. 12 Polyesters and Polyamides: S tep-G row th Polymers We have seen in Section 17.7A that carboxylic acids react with alcohols to form esters. O

O OH

C a r b o x y l ic

HA cat.

HO'""""

H2O O

A lc o h o l

E s te r

a c id

In a similar way carboxylic acid derivatives (L is a leaving group) react with amines (Sect. 17.8) to form amides. O

O H L

C a r b o x y l ic

HL

'N I H

'N ' I H

A m in e

A m id e

a c i d d e r iv a t iv e

In each reaction the two reactants becom e joined and a small m olecule is lost. Such reac­ tions are often called condensation reactions. Similar condensation reactions beginning with dicarboxylic acids and either diols or diamines can be used to form polymers that are either polyesters or polyam ides. These polymers are called step-growth polym ers. [Recall that in Section 10.10 and Special Topic B, w e studied another group o f polymers called chain-growth polym ers (also called a ddi­ tion polym ers), which are formed by radicals undergoing chain-reactions.] •

Polyesters. When a dicarboxylic acid reacts with a diol under the appropriate con­ ditions, the product is a polyester. For example, the reaction o f 1,4-benzenedicarboxylic acid (terephthalic acid) with 1 , 2 -ethanediol leads to the formation o f the familiar polyesters called Dacron, Terelene or Mylar, and system ically called poly(ethylene terephthalate). O O OH

O

HO

H2O

OH

HO O

n

T e r e p h t h a li c

1 , 2 - E t h a n e d io l

P o l y ( e t h y le n e t e r e p h t h a l a t e )

a c id

•

( a p o ly e s t e r )

P olyam ides. When a dicarboxylic acid or acid chloride or anhydride reacts with a diamine under the appropriate conditions, the product is a polyamide. For example, 1 ,6 -hexanedioic acid (adipic acid) can react with 1 ,6 -hexanediamine with heat in an industrial process to form a familiar polyam ide called N ylon. This example of nylon is called nylon 6 , 6 because both components o f the polymer have six carbon atoms. Other nylons can be made in a similar way. O

O 1 ,6 - H e x a n e d i o i c a c i d

1 ,6 -H e x a n e d ia m in e

818

Chapter 17

Carboxylic Acids and Their Derivatives

H2O

n N y l o n 6 ,6 ( A p o ly a m id e )

Special Topic C continues our discussion o f Step-Growth Polymers.

17. 13 Sum m ary o f th e Reactions o f Carboxylic Acids and Their Derivatives The reactions o f carboxylic acids and their derivatives are summarized here. Many (but not all) o f the reactions in this summary are acyl substitution reactions (they are principally the reactions referenced to Sections 17.5 and beyond). A s you use this summary, you w ill find it helpful to also review Section 17.4, which presents the general nucleophilic addition-elim ­ ination m echanism for acyl substitution. It is instructive to relate aspects o f the specific acyl substitution reactions below to this general mechanism. In som e cases proton transfer steps are also involved, such as to make a leaving group more suitable by prior protonation or to transfer a proton to a stronger base at som e point in a reaction, but in all acyl substitution the essential nucleophilic addition-elim ination steps are identifiable.

R e a c t io n s o f C a r b o x y lic A c id s

1. A s acids (discussed in Sections 3.11 and 17.2C): O

X R

^ O

- N a+

H 2O

O

OH

W

n.

O

'U'(_

5

R ' '"O- Na+ +

H2O +

CO2

2. Reduction (discussed in Section 12.3):

O

+

j 1]_U A lH 4

R^ " O H

(2) HsO+

H^

H

^

OH

3. Conversion to acyl chlorides (discussed in Section 17.5): 0

rA

SOC l2 or PCl5

h

9

-----------------'

Cl

4. Conversion to esters (Fischer esterification) or lactones (discussed in Section 17.7A): O

A

R

O

HA

J

R — OH

R

OH

+

h

2o

OR'

5. Conversion to amides (discussed in Section 17.8E): O

O n h

R^

O

H

3

^ " 'O

O

heat

T -

n h

4

A

R

h

N H2

A n a m id e

2o

17.13 Summary o f the Reactions o f Carboxylic Acids and Their Derivatives

6

. Decarboxylation (discussed in Section 17.10): O

O

O

+

R

OH O

HO

heat

R

O

-

2c

c o

CH 3 O

-¡hear heat

OH

^

+

OH

CO, Î

R e a c t io n s o f A c y l C h lo r id e s

1. Conversion (hydrolysis) to acids (discussed in Section 17.5B):

O O Jl + H,O ----» Jl + HCl R ^Cl R OH 2. Conversion to anhydrides (discussed in Section 17.6A):

O RA Xl

O + R A O- —

O O RA OA R' + ci-

3. Conversion to esters (discussed in Section 17.7A):

R

0 1

0 + R. _ oh

Cl

Jl + Cl- + pyr-H* R OR’

4. Conversion to amides (discussed in Section 17.8B):

0 1 + R'NHR" (excess) R ^ Cl

O * I + R'NH2R"ClR^ NR'R"

---->

R' and/or R" may be H. 5. Conversion to ketones (Friedel-Crafts acylation, Section 15.7-15.9):

O O

'R

1

R

6

Xl

. Conversion to aldehydes (discussed in Section 16.4C): O JT

O (1) LiAlH(f-BuO)3 ^

R^ X l

+

(2 ) H3O+

Ï

*■ R^ ^ H

R e a c t io n s o f A c id A n h y d r id e s

1. Conversion (hydrolysis) to acids (discussed in Section 17.6B): O

A

R

O

A

O

O R

+

h 2o

—

>

2

R

2

A

OH

2. Conversion to esters (discussed in Sections 17.6B and 17.7A): O

O

O

O

R O H

R

o

R

R

OR'

R

OH

819

820

Chapter 17

Carboxylic Acids and Their Derivatives

3. Conversion to amides (discussed in Section 17.8C): O H\

N

z

O

R '

x

R ^""O H R"

R' and/or R" may be H.

4. Conversion to aryl ketones (Friedel-Crafts acylation, Sections 15.7-15.9): O^ O

/R

O

0

1

AlCl3 A

A

R ^"O H

+

R e a c tio n s o f E sters 1. Hydrolysis (discussed in Section 17.7B): O

O

HA

J

+

r'

h

R

OR'

O R

R 'O H

J

2o

OH O

J

+

OH­

OR'

h 2o

A

o-

R 'O H

2. Conversion to other esters: transesterification (discussed in Review Problem 17.10): O J R

+

R"OH

O

HA

J

R

OR'

R 'O H OR"

3. Conversion to amides (discussed in Section 17.8D): O

J

R

R" HN OR'

R'OH

\R "

R

R" and/or R" may be H.

R"

4. Reaction with Grignard reagents (discussed in Section 12.8): O

OMgX

J R

OR'

2 R"MgX

R

'

-R" +

OH R 'O MgX

R-

R"

R"

5. Reduction (discussed in Section 12.3): O

R

(1) L iA lH 4

OR'

(2) ^O+

R"

R— CH2OH

R 'O H

821

K e y T erm s a n d C o n c e p ts

R e a c t io n s

o f A m id e s

1. Hydrolysis (discussed in Section 17.8F): O

H

O N

Z

H3O-

R'

h2o

H— N— R'

r

OH R"

R"

O R

N

/ R

OH-

'

O II

___ „ H 2O

H\

O-

R

N

Z

R'

R" R"

R, R ', a n d /o r R" m a y b e H.

2. Conversion to nitriles: dehydration (discussed in Section 17.8G): O J , ^

Ph4Ol° > N Ha

R— C # N

(heaO)

R e a c t io n s o f N it r ile s

1. Hydrolysis to a carboxylic acid or carboxylate anion (Section 17.8H): O

R— C

#N

Hs°

hear

rs^

OH

R— C

#N

HOH2 O , heat 2

O

R

,

k .O -

2. Reduction to an aldehyde with (¿-Bu)2AlH (DIBAL-H, Section 16.4C):

R— C # N

(1) (i-Bu)2AlH (2 ) H2 O

3. Conversion to a ketone by a Grignard or organolithium reagent (Section 16.5B):

R— C # N

(1) R'MgBr or R'Li (2) H3O+ ’

O R

R'

Key Terms and Concepts The key terms and concepts that are highlighted in bold, blue text within the chapter are defined in the glossary (at the back o f the book) and have hyperlinked definitions in the accompanying W ileyPLU S course (www.wileyplus.com ).

PLU<

822

Chapter 17

Carboxylic Acids and Their Derivatives

Problems Note to Instructors: Many of the homework problems are available for assignment via WileyPLUS, an online teaching and learning solution. STRUCTURE A N D

17.18

N O M EN C LA TU R E

Write a structural formula for each of the following compounds: (a) Octanoic acid

17.19

(h) Acetic anhydride

(b) Propanamide

(i) Isobutyl propanoate

(c) NN-Diethylhexanamide

(j) Benzyl acetate

(d) 2-Methyl-4-hexenoic acid

(k) Ethanoyl chloride (acetyl chloride)

(e) Butanedioic acid

(l) 2-Methylpropanenitrile

(f) 1,2-Benzenedioc acid (phthalic acid)

(m ) Ethyl 3-oxobutanoate (ethyl acetoacetate)

(g) 1,4-Benzenedioic acid (terephthalic acid)

(n) Diethyl propanedioate (diethyl malonate)

Give an IUPAC systematic or common name for each of the following compounds: (a)

O

(d) OH

O

O

(g)

O O"

O (e)

(h) CH3CN

O

(b)

O'"'"'"' Cl (f) (c) O nh2 17.20

Amides are weaker bases than corresponding amines. For example, most water-insoluble amines (RNH2) will dis­ solve in dilute aqueous acids (aqueous HCl, H2 SO4, etc.) by forming water-soluble alkylaminium salts (RNH 3 +X~). Corresponding amides (RCONH2) do not dissolve in dilute aqueous acids, however. Propose an explanation for the much lower basicity of amides when compared to amines.

17.21

While amides are much less basic than amines, they are much stronger acids. Amides have pKa values in the range 14-16, whereas for amines, pKa = 33-35. (a) What factor accounts for the much greater acidity of amides? O

O

(b) Imides, that is, compounds with the structure R

R ', are even stronger acids than amides.

For imides, pKa = 9-10, and as a consequence, water-insoluble imides dissolve in aqueous NaOH by forming sol­ uble sodium salts. What extra factor accounts for the greater acidity of imides? F U N C T IO N A L G R O U P T R A N S F O R M A T IO N S

17.22

What major organic product would you expect to obtain when acetyl chloride reacts with each of the following? (a) H 2 O (b) BuLi (excess) (c)

CH 3 and AICI3

(e) |

(d) NH 3 (excess)

(i) (CH3)2NH (excess) (j) EtOH and pyridine

oh

and pyridine

(h) CH 3 NH 2 (excess)

(f)

L iA |H ( t - B u O ) 3

(g) NaOH/H2O

(k) CH 3 C

O 2

Na+

(l) CH 3 CO 2 H and pyridine

R ^ C

P ro b le m s

17.23

x' o h

823

What major organic product would you expect to obtain when acetic anhydride reacts with each of the following? (a) NH 3 (excess)

(c) CH 3 CH 2 CH2OH

(e) CH 3 CH 2 NH 2 (excess)

(b) H2O

(d) C 6 H 6 + AlCl3

(f) (CH 3 CH2)2NH (excess)

17.24

What major organic product would you expect to obtain when succinic anhydride reacts with each of the reagents given in Problem 17.23?

17.25

What products would you expect to obtain when ethyl propanoate reacts with each of the following?

17.26

(a) H 3 O+, H2O

(c) 1-Octanol, HCI

(e) LiAIH4, then H2O

(b) O H 2 , H2O

(d) CH 3 NH 2

(f) Excess C 6 H5 MgBr, then H 2 O, NH4CI

What products would you expect to obtain when propanamide reacts with each of the following? (a) H 3 O+, H2O

17.27

(b) O H2 , H2O

(c) P4 O 1 0 and heat

What products would you expect to obtain when each of the following compounds is heated? (a) 4-Hydroxybutanoic acid

o

(f)

(b) 3-Hydroxybutanoic acid (c) 2-Hydroxybutanoic acid

O-

(d) Glutaric acid

n h 3+

(e)

n h 3+

o 'O G EN ER A L PRO BLEM S

17.28

Write structural formulas for the major organic products from each of the following reactions. (a)

(e)

O

( 1 ) Mg 0

SOCi,

OH (excess), H2SO4(cat.)

(c)

(d)

N

H2O, H2SO4(cat.)

O

O H

17.29

(g)

H2O, H2SO4(cat.)

(h)

H2CrO4

OH

(1) H2CrO4 (2) SOCl2

Indicate reagents that would accomplish each of following transformations. More than one reaction may be nec­ essary in some cases. (a)

O OH

(b) och3

O

O O OH

824

Chapter 17

Carboxylic Acids and Their Derivatives

(c) O

N

Cl O

H O

(f)

EtNH^Cf

17.30

Write structural formulas for the major organic products from each o f the follow ing reactions. (a)

O

(d)

i

O OH

HO

O

(e)

O O^

17.31

\ , Aids

Write structural formulas for the major organic products from each o f the follow ing reactions. (a)

(d)

Br

(1) KCN (2) H2O, H2SO4 (cat.)

N

(1) DIBAL-H (2) H2O

O NH

(e)

H2O, H2SO4 (cat.)

O

! L OCHs (c)

N

HI (1) CH3MgBr (2) H3O+

M E C H A N IS M S

17.32

H2SO4 (cat.)

Write detailed mechanisms for the acidic and basic hydrolysis o f propanamide.

(1) CH3 CH2 MgBr (excess) (2 ) H3 O+

Problems

17.33

R ^ C

825

x' o h

P r o v id e a d e ta ile d m e c h a n is m f o r e a ch o f th e f o llo w in g re a c tio n s . (a )

O

O

H 2O

H2O, H2SO4 (cat.) OH

O

17.34

O n h e a tin g ,

cis - 4 - h y d r o x y c y c lo h e x a n e c a r b o x y lic a c id fo r m s a la c to n e b u t tra n s - 4 - h y d r o x y c y c lo h e x a n e c a r b o x y lic

a c id d o e s n o t. E x p la in .

S Y N T H E S IS

17.35

Show how

p - c h lo r o to lu e n e c o u ld b e c o n v e rte d to e a ch o f th e fo llo w in g :

(a) p - C h lo r o b e n z o ic a c id

(c) p - C h lo r o p h e n y la c e tic a c id

(b)

(d)

OH

^

_.C O 2H

CO 2H Cl Cl

17.36

In d ic a te th e re a g e n ts n e e d e d f o r e a ch o f th e f o llo w in g syn th e se s. M o r e th a n o n e ste p m a y b e n e e de d .

(a)

O HO

HO OH

OH O

(b)

O OH HO. OH O

17.37

17.38

S h o w h o w p e n ta n o ic a c id ca n b e p re p a re d f r o m e a ch o f th e f o llo w in g :

(a) 1 -P e n ta n o l

(c) 5 -D e c e n e

(b) 1 -B ro m o b u ta n e ( tw o w a y s )

(d) P e n ta n a l

T h e a c tiv e in g r e d ie n t o f th e in s e c t r e p e lle n t O f f is M N - d ie th y l- m - t o lu a m id e , m - C H 3C 6H 4C O N ( C H 2C H 3) 2. O u tlin e a s y n th e s is o f th is c o m p o u n d s ta rtin g w it h 3 -m e th y lb e n z o ic a c id ( m - to lu ic a c id ).

17.39

S ta rtin g w it h b e n z e n e a n d s u c c in ic a n h y d r id e a n d u s in g a n y o th e r n e e d e d re a g e n ts , o u tlin e a s y n th e s is o f 1 -p h e n y ln a p h th a le n e .

17.40

S ta rtin g w it h e ith e r

cis- o r trans-H O 2C — C H = C H — C O 2H (i.e ., e ith e r m a le ic o r fu m a r ic a c id ) a n d u s in g a n y o th e r

n e e d e d c o m p o u n d s , o u tlin e syn th e se s o f e a ch o f th e fo llo w in g :

H

(b)

(a) H

C O 2H

C O 2H H

8 26 17.41

Chapter 17

Carboxylic Acids and Their Derivatives

Give stereochemical formulas for compounds A -Q : , s

(a) ^ m

p-toluenesulfonyl , ch lo ride (TsCI) 1— p -d — ne—

N~ ^ - ^ -^ ™

(+ )-C (CsH 1 0 O 2 ) —( ——L—^

...

: b

(C5H9N)

H2SO4

( - ) - D (CsH 1 2 O)

PBr

(b) (R)-(-)-2-B utanol —

CN _

:A —

CN-

:

H SO

E (C 4 H 9 Br) ------- : F (C 5 H 9 N)

( - ) - C (C 5 H 1 0 O 2 ) ((2 ) H—OH:

(+ )-D (C5 H 1 2 O)

(c) A CH3CO2: G (C 6 H i 2 O 2 )

( + )-H (C 4 H 1 0 O) + CH 3 CO 2 -

(d) ( - ) - D - ^

K (C5HnMgBr) " ( 2 ) ^ ^ L (C6 H 1 2 O 2 )

J (C5 H 1 1 BO - E —:

OH HO

(e)

H

M (C 4 H7 NO3) + N (C4H7 NO3)

O

Diastereomers, separated by fractional crystallization

(R )-(+ )-G ly cer aldehyde H SO

(f) M — f———:

(g) N 17.42

HHSoO

[O]

P (C 4 H 8 O 5 ) ^ n O T - meso-tartaric acid

>Q

(C 4 H 8 O 5 )

h [nO(O3 >

(-)-tartaric acid

(Æ)-(+)-Glyceraldehyde can be transformed into (+)-m alic acid by the following synthetic route. Give stereo­ chemical structures for the products of each step. ( £ ) -( + )-Glyceraldehyde

Br2, H2O ox— d at|on> NaCN

(-)-3-bromo-2-hydroxypropanoic acid ------- > C 4 H5 NO 3 17.43

PBr3

(-)-glyceric acid -------> H3O+ heat >

(+)-m alic acid

(Æ)-(+)-Glyceraldehyde can also be transformed into (-)-m a lic acid. This synthesis begins with the conversion of (Æ)-(+)-glyceraldehyde into (-)-tartaric acid, as shown in Problem 17.41, parts (e) and (g). Then (-)-tartaric acid is allowed to react with phosphorus tribromide in order to replace one alcoholic OH group with Br. This step takes place with inversion of configuration at the carbon that undergoes attack. Treating the product of this reaction with dimethyl sulfide produces (-)-m a lic acid. (a) Outline all steps in this synthesis by writing stere­ ochemical structures for each intermediate. (b) The step in which (-)-tartaric acid is treated with phosphorus tri­ bromide produces only one stereoisomer, even though there are two replaceable — OH groups. How is this possible? (c) Suppose that the step in which (-)-tartaric acid is treated with phosphorus tribromide had taken place with “mixed” stereochemistry, that is, with both inversion and retention at the carbon under attack. How many stereoisomers would have been produced? (d) What difference would this have made to the overall out­ come of the synthesis?

R ^ C

P ro b le m s

17.44

x' o h

827

Cantharidin is a powerful vesicant that can be isolated from dried beetles (Cantharis vesicatoria, or “Spanish fly”). Outlined here is the stereospecific synthesis of cantharidin reported by Gilbert Stork (Columbia University). Supply the missing reagents (a)-(n). O O

O

(a)

(c)

(b)

CO2CH3

CO2CH3

'c o 2 c h 3

CO2CH3 O

O (e)

(d)

CH2OMs

CH2OH CH2OH

CH2OMs M s = m e t h a n e s u lf o n y l

O

OH

(g)

(f)

c h 2s c 2 h 5 c h 2s c 2 h 5 O

(h)

OH

O OH OH

h io 4

(see Section 22.6D)

CHO CHO

—

(i)

(see Section 19.4)

(i)

O OCCHo (-C H 3CO2H)

C a n t h a r i d in

SPEC TR O SC O PY

17.45

The IR and 1H NMR spectra of phenacetin (C 1 0 H 1 3 NO2) are given in Fig. 17.7. Phenacetin is an analgesic and antipyretic compound and was the P of A -P -C tablets (aspirin-phenacetin-caffeine). (Because of its toxicity, phenacetin is no longer used medically.) When phenacetin is heated with aqueous sodium hydroxide, it yields phenetidine (C 8 H 1 1 NO) and sodium acetate. Propose structures for phenacetin and phenetidine.

828

Chapter 17

Carboxylic Acids and Their Derivatives

4 5 h (ppm )

Figure 17.7 The 300-MHz 1H NMR and IR spectra of phenacetin. Expansions of the 1H NMR signals are shown in the offset plots.

17.46

W a v e n u m b e r (cm 1)

Given here are the 1H NM R spectra and carbonyl IR absorption peaks o f five acyl compounds. Propose a structure for each. (a) C8H14O4

1H NM R Spectrum 8 1.2 ( 6 H) Triplet 8 2.5 (4H) Singlet 8 4.1 (4H) Quartet

IR Spectrum 1740 cm 2 1

(b) C 11H14O2

1H NM R Spectrum 8 1.0 ( 6 H) Doublet 8 2.1 (1H) Multiplet 8 4.1 (2H) Doublet 8 7.8 (5H) Multiplet

IR Spectrum 1720 cm - 1

(c) C 1 0 H 1 2 O

1H NM R Spectrum 8 1.2 (3H) Triplet 8 3.5 (2H) Singlet 8 4.1 (2H) Quartet 8 7.3 (5H) Multiplet

IR Spectrum 1740 cm - 1

1H NM R Spectrum Singlet 8 6.0 Singlet 8 11.70

IR Spectrum Broad peak 2500 1705 cm 2 1

2

(d) C 2 H2 Cl2 O 2

1

829

C h a lle n g e P ro b le m s

(e) C4H7ClO2

1H NMR Spectrum

17.47

IR Spectrum 1 1745 cm

8 1.3

Triplet Singlet Quartet

4.0 4.2

8 8

Compound X (C 7 H 1 2 O 4 ) is insoluble in aqueous sodium bicarbonate. The IR spectrum of X has a strong absorp­ tion peak near 1740 cm- \ and its broadband proton-decoupled 13C spectrum is given in Fig. 17.8. Propose a struc­ ture for X .

CH ch x

3

2

CH2

, c 7h 12o 4

TM S

C CDCl3

I

220

1

I

1

I

1

I

200

1

I

1

I

1

180

I

160

1

I

1

I

1

I

140

1

I

1

I

1

I

1

I

120 100 5C (ppm)

1

I

1

I

80

1

I

1

I

60

1

I

1

40

I

1

I

1

I

20

1

I

0

Figure 17.8 Broadband proton-decoupled 13C NMR spectrum of compound X, Problem 17.47. Information from the DEPT 13C NMR spectra is given above each peak. 17.48

Compound Y (C 8 H 4 O3) dissolves slowly when warmed with aqueous sodium bicarbonate. The IR spectrum of Y has strong peaks at 1779 and at 1854 cm-1 . The broadband proton-decoupled 13C spectrum of Y exhibits signals at 8 125 (CH), 130 (C), 136 (CH), and 162 (C). Acidification of the bicarbonate solution of Y gave compound Z. The proton-decoupled 13C N M R spectrum of Z showed four signals. When Y was warmed in ethanol, a compound A A was produced. The 13C N M R spectrum of A A displayed 10 signals. Propose structures for Y, Z , and AA.

Challenge Problems 17.49

Ketene, H 2 C = C = O, is an important industrial chemical. Predict the products that would be formed when ketene reacts with (a) ethanol, (b) acetic acid, and (c) ethylamine. [Hint: Markovnikov addition occurs.]

17.50

Two unsymmetrical anhydrides react with ethylamine as follows: O A

O oA O

O CF

O +

nh2

O O

NH2

'N H

O H

O-

O

O CF

O-

NHs+

Explain the factors that might account for the formation of the products in each reaction. 17.51

Starting with 1-naphthol, suggest an alternative synthesis of the insecticide Sevin to the one given in Section 17.9A.

17.52

Suggest a synthesis of ibuprofen (Section 5.11) from benzene, employing chlorom ethylation as one step. Chloromethylation is a special case of the Friedel-Crafts reaction in which a mixture of HCHO and HCl, in the presence of ZnCl2, introduces a — CH2Cl group into an aromatic ring.

830 17.53

Chapter 17

Carboxylic Acids and Their Derivatives

An alternative synthesis of ibuprofen is given below. Supply the structural formulas for compounds A -D : O O Cl .

---------- ---

„

X '''" ^ C l

A --------------- * B —— — > C AlCl 3

17.54

N aC N .

^

HI

____ > D — rrr*

H 2 SO4

red P

.

ibuprofen (racemic)

,

As a method for the synthesis of cinnamaldehyde (3-phenyl-2-propenal), a chemist treated 3-phenyl-2-propen-1ol with K2 Cr2 O 7 in sulfuric acid. The product obtained from the reaction gave a signal at 8 164.5 in its 13C N M R spectrum. Alternatively, when the chemist treated 3-phenyl-2-propen-1-ol with PCC in CH 2 Cl2, the 13C N M R spec­ trum of the product displayed a signal at 8 193.8. (All other signals in the spectra of both compounds appeared at similar chemical shifts.) (a) Which reaction produced cinnamaldehyde? (b) What was the other product?

Learning Group Problems Carboxylic acids and acyl derivatives of the carboxyl functional group are very important in biochemistry. For example, the carboxylic acid functional group is present in the family of lipids called fatty acids. Lipids called glycerides contain the ester functional group, a derivative of carboxylic acids. Furthermore, the entire class of biopolymers called proteins contain repeating amide functional group linkages. Amides are also derivatives of carboxylic acids. Both laboratory and biochemical syntheses of proteins require reactions that involve substitution at activated acyl carbons. This Learning Group Problem focuses on the chemical synthesis of small proteins, called peptides. The essence of pep­ tide or protein synthesis is formation of the amide functional group by reaction of an activated carboxylic acid derivative with an amine. First we shall consider reactions for traditional chemical synthesis of peptides and then we look at reactions used in auto­ mated solid-phase peptide synthesis. The method for solid-phase peptide synthesis was invented by R. B. Merrifield (Rockefeller University), for which he earned the 1984 Nobel Prize in Chemistry. Solid-phase peptide synthesis reactions are so reliable that they have been incorporated into machines called peptide synthesizers (Section 24.7D). T h e C h e m ic a l S y n th e s is o f P e p t id e s

1.

The first step in peptide synthesis is blocking (protection) of the amine functional group of an amino acid (a com­ pound that contains both amine and carboxylic acid functional groups). Such a reaction is shown in Section 24.7C in the reaction between Ala (alanine) and benzyl chloroformate. The functional group formed in the structure labeled O Z-A la is called a carbamate (or urethane). (Z is a benzyloxycarbonyl group, C 6 H 5 C H 2 O C — ). (a) Write a detailed mechanism for formation of Z-Ala from Ala and benzyl chloroformate in the presence of hydroxide. (b) In the reaction of part (a), why does the amino group act as the nucleophile preferentially over the carboxylate anion? (c) Another widely used amino protecting group is the 9-fluorenylmethoxycarbonyl (Fmoc) group. Fmoc is the pro­ tecting group most often used in automated solid-phase peptide synthesis (see part 4 below). Write a detailed mechanism for formation of an Fmoc-protected amino acid under the conditions given in Section 24.7A.

2.

The second step in the reactions of Section 24.7C is the formation of a mixed anhydride. Write a detailed mecha­ nism for the reaction between Z-Ala and ethyl chloroformate (ClCO 2 C 2 H5) in the presence of triethylamine to form the mixed anhydride. What is the purpose of this step?

3.

The third step in the sequence of reactions in Section 24.7C is the one that actually joins the new amino acid (in this case leucine, abbreviated Leu) by another amide functional group. Write a detailed mechanism for this step (from the mixed anhydride of Z-Ala to Z-Ala-Leu). Show how CO 2 and ethanol are formed in the course of this mechanism.

4.

A sequence of reactions commonly used for solid-phase peptide synthesis is shown in Section 24.7D. (a) Write a detailed mechanism for step 1, in which diisopropylcarbodiimide is used to join the carboxyl group of the first amino acid (in Fmoc-protected form) to a hydroxyl group on the polymer solid support. (b) Step 3 of the automated synthesis involves removal of the Fmoc group by reaction with piperidine (a reaction also shown in Section 24.7A). Write a detailed mechanism for this step.

Reactions at the Carbon of Carbonyl Compounds a

Enols and Enolates

G lyceraldehyde-3-phosphate (GAP).

W hen we exercise vigorously, our bodies rely heavily on the metabolic process of glycolysis to derive energy from glucose. Glycolysis splits glucose into two three-carbon molecules. Only one of these three-carbon mol­ ecules (glyceraldehyde-3-phosphate, GAP, shown above) is directly capable of going further in the glycolytic pathway. The other three-carbon molecule (dihydroxyacetone-3-phosphate, DHAP) is not wasted, however. It is converted to a second molecule of GAP, via a type of interm ediate that is key to our studies in this chapter— an enol (so named because the intermediate is an alkene alcohol). W e shall learn about enols and enolates, their conjugate bases, in this chapter. In Chapter 16, we saw how aldehydes and ketones can undergo nucleophilic addition at their carbonyl groups. For example: H

O'

O

H—Nu

R

R'

R

R'

Nucleophilic addition

Nu

In Chapter 17 we saw how substitution could occur at a carbonyl group if a suitable leaving group is present. This type of reaction is called acyl substitution. For example: O

O Nu:

A

g

LG = R

Nu

(Proton transfer steps are involved in some nucleophilic addition and acyl substitution reactions, as detailed in Chapters 16 and 17.)

831

832

R e a c tio n s a t t h e a C a rb o n o f C a rb o n y l C o m p o u n d s

C h a p te r 1 8

In this chapter we shall discuss reactions that derive from the weak acidity of hydrogen atoms on carbon atoms adjacent to a carbonyl group. These hydrogen atoms are called the a hydrogens, and the carbon to which they are attached is called the a carbon. O

H

a H y d ro g e n s

—

a r e w e a k ly a c id ic (p H ; = 1 9 - 2 0 ) .

18.1 The A cid ity o f the a Hydrogens o f Carbonyl Compounds: Enolate Anions When w e say that the a hydrogens o f carbonyl compounds are acidic, we mean that they are unusually acidic f o r hydrogen atom s attached to carbon. •

The p ^ a values for the a hydrogens o f m ost sim ple aldehydes or ketones are o f the order o f 19-20.

This means that they are more acidic than hydrogen atoms o f ethyne, pK a = 25, and are far more acidic than the hydrogens o f ethene (pKa = 44) or o f ethane (pKa = 50). The reasons for the unusual acidity o f the a hydrogens o f carbonyl compounds are straightforward. •

The carbonyl group is strongly electron withdrawing, and when a carbonyl com ­ pound loses an a proton, the anion that is produced, called an en o la te, is stabi­ lized by delocalization. :d

H ^ B

\

SY

/

V

C — C, „„

-

:O'^

.O'-“

V

C ^ C

/



\

'\

/

/

\

C = C

A

+

H— B

B

'----------------------------y----------------------------'

R e s o n a n c e s tru c tu re s fo r t h e d e lo c a liz e d e n o la te

Two resonance structures, A and B, can be written for the enolate. In structure A the negative charge is on carbon, and in structure B the negative charge is on oxygen. Both structures contribute to the hybrid. Although structure A is favored by the strength o f its carbon-oxygen p bond relative to the weaker carbon-carbon p bond o f B, structure B makes a greater contribution to the hybrid because oxygen, being highly electronegative, is better able to accommodate the negative charge. We can depict the enolate hybrid in the follow ing way: O f-

VC — C /

\

H y b rid r e s o n a n c e s tr u c tu r e f o r a n e n o la te

When this resonance-stabilized enolate accepts a proton, it can do so in either o f two ways: It can accept the proton at carbon to form the original carbonyl compound in what is called the keto form or it may accept the proton at oxygen to form an enol (alkene alcohol). •

The enolate is the conjugate base o f both the enol and keto forms.

833

1 8 .2 K e to a n d E nol T a u to m e rs

A p ro to n c a n a d d h e r e /

or

O

vC— C

A p r o to n c a n a d d h e re . HB

/ \ Enolate

O

HO

B:-

/ \ Enol form

\/ —c/

H B:

Keto form

Both o f these reactions are reversible. A calculated electrostatic potential map for the enolate o f acetone is shown below. The map indicates approximately the outermost extent o f electron density (the van der Waals surface) o f the acetone enolate. Red color near the oxygen is consistent with oxygen being better able to stabilize the excess negative charge o f the anion. Yellow at the carbon where the a hydrogen was removed indicates that som e o f the excess negative charge is localized there as w ell. These implications are parallel with the conclusions above about charge dis­ tribution in the hybrid based on delocalization and electronegativity effects.

O il

H

H ' C ^ '

I H

H

H

18.2 Keto and Enol Tautomers The keto and enol forms o f carbonyl compounds are constitutional isomers, but o f a spe­ cial type. Because they are easily interconverted in the presence o f traces o f acids and bases, chem ists use a special term to describe this type o f constitutional isomerism. •

Interconvertible keto and enol forms are called tautom ers, and their interconver­ sion is called tautom erization .

Under m ost circumstances, w e encounter k eto-enol tautomers in a state o f equilibrium. (The surfaces o f ordinary laboratory glassware are able to catalyze the interconversion and establish the equilibrium.) For simple monocarbonyl compounds such as acetone and acetaldehyde, the amount o f the enol form present at equilibrium is very small. In acetone it is much less than 1 %; in acetaldehyde the enol concentration is too small to be detected. The greater stability o f the following keto forms o f monocarbonyl compounds can be related to the greater strength o f the carbon-oxygen p bond compared to the carbon-carbon p bond ( ~ 3 6 4 versus ~ 2 5 0 kJ m o F 1): Keto Form

O Acetal dehyde

"

E nol Form

OH H

(—1 0 0 %)

H

(extremely small)

^ ____ H e lp f u l H i n t Keto-enol tautom ers are not resonance structures. They are constitutional isomers in equilibrium (generally favoring the keto form).

8 34

Chapter 18

Reactions at the

a Carbon

of Carbonyl Compounds

OH

O

H e lp f u l H i n t See "The Chemistry of... TIM (Triose Phosphate Isomerase) Recycles Carbon via an Enol" in WileyPLUS fo r more information relating to this chapter's opener about an im portant energyyielding biochemical process.

Acet one (> 9 9 % )

( 1 . 5 X 1 0 - 4% )

OH

O

Cycl ohexanone

( 1 .2 %)

(9 8 .8 % )

In compounds w hose m olecules have two carbonyl groups separated by one carbon atom (called b-dicarbonyl compounds), the amount o f enol present at equilibrium is far higher. For example, pentane-2,4-dione exists in the enol form to an extent o f 76%: O

•

O

OH

O

P e n ta n e -2 ,4 -d io n e

E n o l fo rm

(2 4 % )

(7 6 % )

The greater stability o f the enol form o f b-dicarbonyl compounds can be attributed to resonance stabilization o f the conjugated double bonds and (in a cyclic form) through hydrogen bonding.

SolvedProblem18.1 Write bond-line structures for the keto and enol forms o f 3-pentanone. ANSWER

R e v ie w P ro b le m 18.1

■O

:OH

K e to

Enol

For all practical purposes, the compound cyclohexa-2,4-dien-1-one exists totally in its enol form. Write the structure o f cyclohexa-2,4-dien-1-one and o f its enol form. What special factor accounts for the stability o f the enol form?

18.3 Reactions via Enols and Enolates 18.3A Racemization When a solution o f (^)-(+)-2-m ethyl-1-phenylbutan-1-one (see the follow ing reaction) in aqueous ethanol is treated with either acids or bases, the solution gradually loses its opti­ cal activity. After a time, isolation o f the ketone shows that it has been com pletely racemized. The ( + ) form o f the ketone has been converted to an equimolar mixture o f its enantiomers through its enol form.

Si

S'

835

1 8 .3 R e a c tio n s v ia Enols a n d E n o la te s

O

O

O

OH- or H3O

(fl)-(+ )-2 -M e th y l-1 p h e n y lb u ta n -1 -o n e

•

(± )-2 -M e th y l-1 p h e n y lb u ta n -1 -o n e (ra c e m ic form )

Enol (ach iral)

Racemization at an a carbon takes place in the presence of acids or bases because the keto form slowly but reversibly changes to its enol and the enol is achiral. When the enol reverts to the keto form, it can produce equal amounts of the two enantiomers.

A base catalyzes the formation of an enol through the intermediate formation of an enolate anion.

A MECHANISM FOR THE REACTION B a s e - C a ta ly z e d E n o liz a tio n 0*

H — OH

\

C=C

/ HO-

/

\

\

/

C=C

:O — H / =O H "

\

H

E n o la te (ach iral)

Enol (ach iral)

An acid can catalyze enolization in the following way.

A MECHANISM FOR THE REACTION A c id - C a ta ly z e d E n o liz a tio n

O

\ / „„.C— C

"V H

\

H — O — H

Ç O+

H

\

,,C-jrC

H

" 7 ^ H

\

C /

>O C \

+ :O — H

I H

H

+

H — O — H

I H

Enol (ac h ira l)

836

C h a p te r 1 8

R e a c tio n s a t t h e a C a rb o n o f C a rb o n y l C o m p o u n d s

In acyclic ketones, the enol or enolate formed can be (E) or (Z). Protonation on one face of the (E) isomer and protonation on the same face o f the (Z) isomer produces enantiomers. R e v ie w P ro b le m 1 8 .2

Would optically active ketones such as the following undergo acid- or base-catalyzed racemization? Explain your answer.

•

Diastereomers that differ in configuration at only one o f several chirality centers are som etim es called ep im ers.

K eto-enol tautomerization can som etim es be used to convert a less stable epimer to a more stable one. This equilibration process is an example o f ep im erization . An example is the epimerization o f cis-decalone to trans-decalone:

c /s -D e c a lo n e

fra n s -D e c a lo n e

S o lv e d P r o b le m 1 8 .2 Treating racemic 2-methyl-1-phenylbutan-1-one with NaOD in the presence o f D2O produces a deuterium-labeled compound as a racemic form. Write a m echanism that explains this result. O CeHs R acem ic

STRATEGY AND ANSWER Either enantiomer o f the ketone can transfer an a proton to the OD ion to form an achiral enolate which can accept a deuteron to form a racemic mixture o f the deuterium-labeled product. 'O' :OD

R ac e m ic

O

O

Cf6H5 iH

C6H5

R ace m ic

R e v ie w P ro b le m 1 8 .3

Write a mechanism using sodium ethoxide in ethanol for the epimerization o f d s-decalone to trans-decalone. Draw chair conformational structures that show why trans-decalone is more stable than ds-decalone. You may find it helpful to also examine handheld m olecu­ lar m odels o f cis- and trans-decalone.

837

1 S .3 R e a c tio n s v ia Enols a n d E n o la te s

18.3B Halogenation at the a Carbon •

Carbonyl compounds bearing an a hydrogen can undergo halogen substitution at the a carbon in the presence of acid or base. H

\

O

X

/ ,..O — O

"V

X

\

acid or base

2

O

\

/ ,..O — O

"V

HX

\

( R a c e m ic )

B a s e -P r o m o te d H a lo g e n a tio n In the presence of bases, halogenation takes place through the slow formation of an enolate anion or an enol followed by a rapid reaction of the enolate or enol with halogen.

A MECHANISM FOR THE REACTION B a s e -P ro m o te d H a lo g e n a tio n o f A ld e h y d e s a n d K e to n e s

Step 1

B

"V

V

O-

\s -

slow

Hn O O .O— O

B

\

H

+

//

:OH / O =O + / \

fast

\

O— O

/

\

Enol

E n o la t e ( r e s o n a n c e h y b r id )

'o ; -

\

Step 2

X

/ ■

O =O

/

— O— O

\

X — X

O:

\

/ .O— O

fast

"V

E n o la t e

X -

\

(R a c e m ic )

( c o n tr ib u tin g r e s o n a n c e s tr u c tu r e s )

As we shall see in Section 18.3C, multiple halogenations can occur. In the presence of acids, halogenation takes place through the slow formation of an enol followed by rapid reaction of the enol with the halogen.

A c id -C a ta ly z e d H a lo g e n a tio n

A MECHANISM FOR THE REACTION A c id -C a ta ly z e d H a lo g e n a tio n o f A ld e h y d e s a n d K e to n e s

H

Step 1

,...O— O

"V

CBH

O O

fast H ¡B ^

\

„

C/O

.

.. O — O

^

"V

H

slow

\

\

O=O

/

/

+

H =B

\ Enol

Step 2

ÇO O =O

*9

X — X

/ X

Step 3

\

X

fast

O+ +

\

H

/

,..O — O

H<

+

\

X

fast +

O+

\

"V

\

/ ,,..O— O

"V

H

X -

X-

O:

\ / ,,..O— O

"V

\

R a c e m ic

HX

B:

838

C h a p te r 1 8

R e a c tio n s a t t h e a C a rb o n o f C a rb o n y l C o m p o u n d s

Part o f the evidence that supports these mechanisms com es from studies o f reaction kinet­ ics. Both base-promoted and acid-catalyzed halogenations o f ketones show initial rates that are independent o f the halogen concentration. The mechanisms that w e have written are in accord with this observation: In both instances the slow step o f the m echanism occurs before the intervention o f the halogen. (The initial rates are also independent o f the nature o f the halogen; see Review Problem 18.5.)

R e v ie w P ro b le m 1 8 .4

Why do w e say that the halogénation o f ketones in a base is “base promoted” rather than “base catalyzed”?

R e v ie w P ro b le m 1 8 .5

Additional evidence for the halogenation mechanisms that w e just presented com es from the follow ing facts: (a) Optically active 2-methyl-l-phenylbutan-1-one undergoes acid-cat­ alyzed racemization at a rate exactly equivalent to the rate at which it undergoes acid-cat­ alyzed halogenation. (b) 2-M ethyl-1-phenylbutan-1-one undergoes acid-catalyzed iodination at the same rate that it undergoes acid-catalyzed bromination. (c) 2-M ethyl-1phenylbutan-1 -one undergoes base-catalyzed hydrogen-deuterium exchange at the same rate that it undergoes base-promoted halogenation. Explain how each o f these observations sup­ ports the mechanisms that w e have presented.

18.3C The Haloform Reaction When methyl ketones react with halogens in the presence o f excess base, multiple halo­ genations always occur at the carbon o f the methyl group. M ultiple halogenations occur because introduction o f the first halogen (owing to its electronegativity) makes the remain­ ing a hydrogens on the methyl carbon more acidic. The resulting CX 3 group bonded to the carbonyl can be a leaving group, however. Thus, when hydroxide is the base, an acyl sub­ stitution reaction follow s, leading to a carboxylate salt and a haloform (CHX3, e.g., chlo­ roform, bromoform, or iodoform). The follow ing is an example. O

O

3 X2, 3 OH-

O CX3 3

O-

HO-

X-

+

chx3

Ahaloform (X= Cl, Br, I) The haloform reaction is one o f the rare instances in which a carbanion acts as a leav­ ing group. This occurs because the trihalomethyl anion is unusually stable; its negative charge is dispersed by the three electronegative halogen atoms (when X = Cl, the conju­ gate acid, CHCl3, has p K a = 13.6). In the last step, a proton transfer takes place between the carboxylic acid and the trihalomethyl anion. The haloform reaction is synthetically useful as a means o f converting methyl ketones to carboxylic acids. When the haloform reaction is used in synthesis, chlorine and bromine are m ost com m only used as the halogen component. Chloroform (CHCl3) and bromoform (CHBr3) are both liquids which are im m iscible with water and are easily separated from the aqueous solution containing the carboxylate anion. When iodine is the halogen com ­ ponent, the bright yellow solid iodoform (CHI3) results. This version is the basis o f the iod­ oform classification test for methyl ketones and methyl secondary alcohols (which are oxidized to methyl ketones first under the reaction conditions): O R

Amethyl ketone

I2, HO(both in excess)

O -

A

CHI3P

,

Iodoform (a yellow precipitate)

839

1 8 .3 R e a c tio n s v ia Enols a n d E n o la te s

A MECHANISM FOR THE REACTION T h e H a lo fo rm R e a c tio n

H a lo g en atio n S tep •'O'

• ' O’

•'O'

H

■'O'X

Repeat steps

3

twice

■O--

A cyl substitution step

C a rb o x y la te anion

A h alo fo rm

THE CHEMISTRY OF . . . C h l o r o f o r m in D r in k in g W a t e r

When water is chlorinated to purify it for public consump­ tion, chloroform is produced from organic impurities in the water via the haloform reaction. (Many of these organic impurities are naturally occurring, such as humic substances.) The presence of chloroform in public water is of concern for water treatment plants and environmental officers, because

chloroform is carcinogenic. Thus, the technology that solves one problem creates another. It is worth recalling, however, that before chlorination of water was introduced, thousands of people died in epidemics of diseases such as cholera and dysentery.

18.3D a-Halo Carboxylic Acids: The Hell-Volhard-Zelinski Reaction Carboxylic acids bearing a hydrogen atoms react with bromine or chlorine in the presence o f phosphorus (or a phosphorus halide) to give a-halo carboxylic acids through a reaction known as the Hell-V olhard-Zelinski (or HVZ) reaction.

Chapter 18 Reactions at the a Carbon of Carbonyl Compounds

840

General Reaction O

O m X 2, p

R OH

R 'OH

(2) H 2o

X

a-Halo acid Specific Example O

O

O (2) H 2O

(1) B r 2, P

OH

OH

Br Br

Br

2-Bromobutanoic acid (77%)

Butanoic acid

If more than one molar equivalent o f bromine or chlorine is used in the reaction, the prod­ ucts obtained are a,a-dihalo acids or a,a,a-trihalo acids. Important steps in the reaction are formation o f an acyl halide and the enol derived from the acyl halide. The acyl halide is key because carboxylic acids do not form enols readily since the carboxylic acid proton is removed before the a hydrogen. A cyl halides lack the carboxylic acid hydrogen.

R

H

O

O P + Br2

OH (pBr3)

R

R

Br

O

(''O' SI

R

Br

Br—Br Enol form

Acyl bromide

O Br

+ HBr

h 2o

R

OH

Br

+ HBr

Br

An alternative method for a-halogenation has been developed by D . N. Harpp (McGill University). A cyl chlorides, formed in situ by the reaction o f the carboxylic acid with SOCl2, are treated with the appropriate N-halosuccinimide and a trace o f HX to produce a-chloro and a -bromo acyl chlorides. O

O HX

R

SOCl 2

Cl

X

a-Iodo acyl chlorides can be obtained by using molecular iodine in a similar reaction. O

O HI

R

2

Cl

SOCl 2

R Cl I

a-H alo acids are important synthetic intermediates because they are capable o f reacting with a variety o f nucleophiles:

Conversion toa-HydroxyAcids O

O

R OH X

a-Halo acid

J1 _OH~ (2) H3O+

OH OH

a-Hydroxy acid

X

-

1 8 .4 L ith iu m E n o la te s

841

Specific Example O

O (1) K 2C O 3, H 2O , 1 00 °C

OH

OH

(2) H3O±

Br

OH 2 - H y d r o x y b u t a n o i c a c id (6 9 % )

Conversion to a-Amino Acids O

O

R

R OH

2

nh3

X

O

+

n h 4x

NH3

a - H a lo a c id

a - A m in o a c id

Specific Example O

O 2 NH

Br

NH4Br

H3N+

3

OH

O

A m i n o a c e t i c a c id ( g l y c in e ) (6 0 -6 4 % )

18.4 Lithium Enolates The position o f the equilibrium by which an enolate forms depends on the strength o f the base used. If the base employed is a weaker base than the enolate, then the equilibrium lies to the left. This is the case, for example, when a ketone is treated with sodium ethoxide in ethanol. O

O

Na+

EtONa W e a k e r a c id

EtOH

W eaker base

S tro n g e r b a s e

S t r o n g e r a c id

(PK = 1 9 )

(P K , = 1 6 )

On the other hand, if a very strong base is employed, the equilibrium lies far to the right. One very useful strong base for converting carbonyl compounds to enolates is lithium diisopropylam ide (LDA) or LiN(i-Pr)2: O Na+

O

+

LiN(/-Pr) 2

•

HN(/-Pr) 2

K e to n e

LD A

E n o la t e

D iis o p r o p y la m in e

(s tr o n g e r a c id )

(s tro n g e r

(w e a k e r

(w e a k e r a c id )

(p K a = 1 9 )

base)

base)

(p K , = 38)

Lithium diisopropylamide (LDA) can be prepared by dissolving diisopropylamine in a solvent such as diethyl ether or THF and treating it with an alkyllithium: Li+ THF

M

B u t y lli t h iu m

D iis o p r o p y la m in e

L i t h iu m d i i s o p r o p y l a m i d e

B u ta n e

(B u L i)

(p K a = 3 8 )

[ L D A o r L i N ( / - P r ) 2]

(p K a = 5 0 )

842

C h a p te r 1 8

R e a c tio n s a t t h e a C a rb o n o f C a rb o n y l C o m p o u n d s

18.4A Regioselective Formation of Enolates An unsymmetrical ketone such as 2-methylcyclohexanone can form two possible enolates, arising by removal o f an a hydrogen from one side or the other o f the carbonyl group. Which enolate predominates in the reaction depends on whether the enolate is formed under con­ ditions that favor an acid-base equilibrium.

O

2-Methylcyclohexanone

•

The therm odyn am ic enolate is that which is m ost stable among the possible enolates. Enolate stability is evaluated in the same way as for alkenes, meaning that the more highly substituted enolate is the more stable one.

•

The therm odynam ic enolate predominates under conditions where a deprotonationprotonation equilibrium allow s interconversion among the possible enolates, such that eventually the more stable enolate exists in higher concentration. This is the case when the pK a o f the conjugate acid o f the base is similar to the pKa of the a hydrogen o f the carbonyl compound. U se o f hydroxide or an alkoxide in a protic solvent favors formation o f the thermodynamic enolate.

•

The k inetic en olate is that which is formed fastest. It is usually formed by removal o f the least sterically hindered a hydrogen.

•

The k inetic en olate predominates under conditions that do not favor equilibrium among the possible enolates. U se o f a very strong and sterically hindered base in an aprotic solvent, such as LD A in tetrahydrofuran (THF) or dimethoxyethane (DME) favors formation o f the kinetic enolate.

C onditions favoring formation o f the therm odynamic and kinetic enolates from 2 -methylcyclohexanone are illustrated below.

F o r m a tio n

o f a T h e r m o d y n a m ic E n o la te

B This enolate is more stable b eca u se the double bond is more highly substituted. It is the predominant enolate at equilibrium. Kinetic (less stable) enolate

(more stable) enolate

F o r m a tio n

o f a K in e tic

E n o la te

This enolate is formed faster b ecau se the hindered strong base rem oves the le ss hindered proton faster. Kinetic enolate

18.4B Direct Alkylation of Ketones via Lithium Enolates H e lp f u l H i n t ______ ^ Alkylation o f lithium enolates is a useful method fo r synthesis.

The formation o f lithium enolates using lithium diisopropylam ide furnishes a useful way o f alkylating ketones in a regioselective way. For exam ple, the lithium enolate formed from 2 -m ethylcyclohexanone can be m ethylated or benzylated at the less hindered a carbon by allow ing it to react with LD A follow ed by m ethyl iodide or benzyl bromide, respectively:

843

1 8 .4 L ith iu m E n o la te s

O

OH,

HC

oh3 i

Li+ O-

O

H3 C

( LiI)

H3C

56%

O HC

(-LiBr) '

O eH s

4 2 -4 5 %

Alkylation reactions like these have an important limitation, however, because the reactions are SN2 reactions, and also because enolates are strong bases. •

Successful alkylations occur only when primary alkyl, primary benzylic, and pri­ mary allylic halides are used. With secondary and tertiary halides, elimination becomes the main course of the reaction.

18.4C Direct Alkylation of Esters Examples of the direct alkylation of esters are shown below. In the second example the ester is a lactone (Section 17.7C): O

i

O

O

"

LDA "OMe thf

''O M e

'O M e

L

M e th y l b u ta n o a te

M e th y l 2 -e th y lb u ta n o a te (9 6 % )

O

O

O

OH,

H O'

'H

B u ty ro la c to n e

LDA THF

O

O H,— I

H

O

H

2 -M e th y lb u ty r o la c to n e (8 8 % )

c « / 1/Û

H e lp f u l H i n t Proper choice o f the alkylating agent is key to successful lithium enolate alkylation.

844

Review Problem 18.6

Chapter 18

Reactions at the

a Carbon of Carbonyl Compounds

(a) Write a reaction involving a lithium enolate for introduction o f the methyl group in the follow ing compound (an intermediate in a synthesis by E. J. Corey o f cafestol, an anti­ inflammatory agent found in coffee beans): O-.

O

(b) Dienolates can be formed from b-keto esters using two equivalents o f LDA. The dienolate can then be alkylated selectively at the more basic o f the two enolate carbons. Write a reaction for synthesis o f the follow ing compound using a dienolate and the appro­ priate alkyl halide: O

O QCH„

(CH 3 )3 Si

18.5 Enolates o f ß-D icarbonyl Compounds Hydrogen atoms that are between two carbonyl groups, as in a b -d icarbon yl com ­ poun d , have p K a values in the range o f 9 -1 1 . Such a-hydrogen atoms are much more acidic than a hydrogens adjacent to only one carbonyl group, which have pKa values o f 18-20. O

O

O

H pKa 9-11 •

H pKa 18-20

A much weaker base than LDA, such as an alkoxide, can be used to form an eno­ late from a b-dicarbonyl compound. O

O

O

O

RO-

ROH

H We can account for the greater acidity o f b-dicarbonyl systems, as compared to single car­ bonyl systems, by delocalization o f the negative charge to two oxygen atoms instead o f one. We can represent this delocalization by drawing contributing resonance structures for a bdicarbonyl enolate and its resonance hybrid: •'O‘z

^

; O ‘-

O

II

II5 C

^

:O

*11 ^

*> C<

:O

P ^

C\

I <

*■

O

II

/ C^ C/ C \

Contributing resonance structures

= /

''CT

-'O''

li

il

C H Q

OH

(2) H3O+

R

R

Monoalkylmalonic ester H OM

O

( O

HO

O

HO

-C O 2

R

R

Monoalkylacetic acid

or after dialkylation, Q

Q

O (1) HO-, H2O

OEt

E tO R

(2) H3O+

^

HO

R'

O

^ R

OH

heat

R'

Dialkylmalonic ester O R'

HO

-C O 2

R

Dialkylacetic acid

Two specific exam ples o f the m alonic ester synthesis are the syntheses o f hexanoic acid and 2 -ethylpentanoic acid that follow. A M alo n ic E s te r S y n th e s is o f H e x a n o ic A c id O

O

E tO

O OEt

(2)

O

O (1) 50% KOH’ reflux_______ ;

(1) NaOEt

E tO

(2) dil. H2SO4’ reflux ( -C O 2) '

"Br

"OH Hexanoic acid (75%)

Diethyl butylmalonate (80-90%) A M a lo n ic E s te r S y n th e s is o f 2 -E th y lp e n ta n o ic A c id O E tO

O

O OEt

(1) NaOEt (2 ^

>

E tO

O

O OEt

(1) f-BuOK

E tO

O OEt

(2)

Diethyl ethylmalonate

Diethyl ethylpropylmalonate

(1) HO- ’ H2O ' (2) H3O+

*

852 c«/

Chapter 18

Reactions at the

a Carbon of Carbonyl Compounds

r/ Dk«K inm 1 0 Ä

R e v ie w P ro b le m 1 8 .1 2

Outline all steps in a m alonic ester synthesis o f each o f the following: (a) pentanoic acid, (b) 2-methylpentanoic acid, and (c) 4-methylpentanoic acid. Two variations o f the malonic ester synthesis make use o f dihaloalkanes. In the first of these, two molar equivalents o f sodiom alonic ester are allowed to react with a dihaloalkane. Two consecutive alkylations occur, giving a tetraester; hydrolysis and decarboxylation of the tetraester yield a dicarboxylic acid. An example is the synthesis o f glutaric acid: O OEt

CHI

+

2

Na+ OEt O

4 E tO H

Glutaric acid (80% from tetraester) In a second variation, one molar equivalent o f sodiomalonic ester is allowed to react with one molar equivalent o f a dihaloalkane. This reaction gives a haloalkylmalonic ester, which, when treated with sodium ethoxide, undergoes an internal alkylation reaction. This method has been used to prepare three-, four-, five-, and six-membered rings. An example is the synthesis o f cyclobutanecarboxylic acid:

Br

EtONa

Si

S'

853

1 8 .8 F u rth e r R e a c tio n s o f A c tiv e H y d ro g e n C o m p o u n d s

hydrolysis and decarboxylation O

HO C y c lo b u ta n e c a r b o x y lic a c id

•

A s w e have seen, the malonic ester synthesis is a useful method for preparing m ono- and dialkylacetic acids:

O O

H e lp f u l H i n t The malonic ester synthesis is a tool fo r synthesizing substituted acetic acids.

R

OH

R

OH R A m o n o a lk y la c e tic a c id

•

A d ia lk y la c e tic a c id

Thus, the malonic ester synthesis provides us with a synthetic equivalent o f an ester enolate o f acetic acid or acetic acid dianion.

O O Etu

O V

OEt OEt

is the synthetic equivalent of

and O

D ie t h y l m a l o n a t e a n io n

O

Direct formation o f such anions is possible (Section 18.4), but it is often more conve­ nient to use diethyl malonate as a synthetic equivalent because its a hydrogens are more easily removed. In Special Topic E w e shall see biosynthetic equivalents o f these anions.

18.8 Further Reactions o f A ctive Hydrogen Compounds Because o f the acidity o f their m ethylene hydrogens m alonic esters, acetoacetic esters, and similar compounds are often called active hydrogen com pounds or active m ethylene com ­ p ou n d s. Generally speaking, active hydrogen compounds have two electron-withdrawing groups attached to the same carbon atom:

Z

Z'

A c tiv e h y d ro g e n c o m p o u n d (Z a n d Z ' a r e e le c tr o n -w ith d r a w in g g r o u p s .)

854

R e a c tio n s a t t h e a C a rb o n o f C a rb o n y l C o m p o u n d s

C h a p te r 1 8

The electron-withdrawing groups can be a variety of substituents, including O

O

O

O — NO„

N R

H

OR

NR 2 O

O

O

O

— S— R

S R

— S— OR

O

or

— S— NR 0

O

O

The range of pK a values for such active methylene compounds is 3-13. Ethyl cyanoacetate, for example, reacts with a base to yield a resonance-stabilized anion: •'O'

:N

‘°3

•O base

'O Et

:N t) "OEt

-H +

E th y l c y a n o a c e ta te

^

:O -

"OEt

'OEt

R e s o n a n c e s t r u c t u r e s f o r e t h y l c y a n o a c e t a t e a n io n

Ethyl cyanoacetate anions also undergo alkylations. They can be dialkylated with iso­ propyl iodide, for example: O

O

O

NC

(1) EtONa/EtOH

OEt

(1) EtONa/EtOH

(2 ) H3 O+

R eview P roblem 18.13

(2)

The antiepileptic drug valproic acid is 2-propylpentanoic acid (administered as the sodium salt). One commercial synthesis of valproic acid begins with ethyl cyanoacetate. The penul­ timate step of this synthesis involves a decarboxylation, and the last step involves hydrol­ ysis of a nitrile. Outline this synthesis.

18.9 Synthesis o f Enamines: S to rk Enamine Reactions Aldehydes and ketones react with secondary amines to form compounds called enam ines. The general reaction for enamine formation can be written as follows: ‘ ' — C' H A ld e h y d e

OH HN — R I R 2 ° A m in e

■- C — C — N

H

R

\

R I N C " R

h 2o

R E n a m in e

o r k e to n e

See Section 16.8C for the mechanism of enamine formation. Since enamine formation requires the loss of a molecule of water, enamine preparations are usually carried out in a way that allows water to be removed as an azeotrope or by a drying agent. This removal of water drives the reversible reaction to completion. Enamine formation is also catalyzed by the presence of a trace of an acid. The secondary amines most commonly used to prepare enamines are cyclic amines such as pyrrolidine, piperidine, and morpholine:

855

1 8 .9 S yn th e s is o f E n a m in e s: S to rk E n a m in e R e a c tio n s

O

N

N '

N

I H

H

H

P y r r o l id i n e

P ip e r id in e

M o r p h o lin e

Cyclohexanone, for example, reacts with pyrrolidine in the follow ing way:

(a n e n a m in e )

Enamines are good nucleophiles. Examination o f the resonance structures that follow show that w e should expect enamines to have both a nucleophilic nitrogen and a nucleophilic car­ bon. A map o f electrostatic potential highlights the nucleophilic region o f an enamine.

C o n tr ib u tio n to th e

C o n tr ib u tio n to th e

h y b r id m a d e b y th is

h y b r id m a d e b y t h i s

s tru c tu r e c o n fe rs

s tru c tu re c o n fe rs

n u c le o p h ilic ity o n

n u c le o p h ilic ity o n

n itro g e n .

c a rb o n a n d d e c re a s e s

A m a p o f e l e c t r o s t a t i c p o t e n t ia l f o r

th e n u c le o p h ilic ity o f

W - ( 1 - c y c lo h e x e n y l ) p y r r o l id i n e s h o w s

n itro g e n .

th e d is tr ib u tio n o f n e g a tiv e c h a r g e a n d t h e n u c l e o p h i l i c r e g io n o f a n e n a m i n e .

The nucleophilicity o f the carbon o f enamines makes them particularly useful reagents in organic synthesis because they can be acylated, alkylated, and used in M ichael additions (see Section 19.7A). Enamines can be used as synthetic equivalents o f aldehyde or ketone enolates because the alkene carbon o f an enamine reacts the same way as does the a car­ bon o f an aldehyde or ketone enolate and, after hydrolysis, the products are the same. Developm ent o f these techniques originated with the work o f Gilbert Stork o f Columbia University, and in his honor they have com e to be known as Stork en am ine reactions. When an enamine reacts with an acyl halide or an acid anhydride, the product is the Cacylated compound. The iminium ion that forms hydrolyzes when water is added, and the overall reaction provides a synthesis o f b -diketones: O N

H e lp f u l H i n t Enamines are the synthetic equivalents o f aldehyde and ketone enolates.

O

O h 2o

N

C l-

Cl

H I m in iu m s a l t

2 -A c e ty lc y c lo h e x a n o n e (a /J -d ik e to n e )

/

\

ClH

856

C h a p te r 1 8

R e a c tio n s a t t h e a C a rb o n o f C a rb o n y l C o m p o u n d s

Although N-acylation may occur in this synthesis, the N-acyl product is unstable and can act as an acylating agent itself: Cl-

N

Enamine

W-Acylated enamine

C-Acylated iminium salt

Enamine

A s a consequence, the yields o f C-acylated products are generally high. Enamines can be alkylated as w ell as acylated. Although alkylation m ay lead to the for­ mation o f a considerable amount o f N-alkylated product, heating the N-alkylated product often converts it to a C-alkyl compound. This rearrangement is particularly favored when the alkyl halide is an allylic halide, benzylic halide, or a-haloacetic ester:

h 2o

O R

N I H Enamine alkylations are SN2 reactions; therefore, when w e choose our alkylating agents, w e are usually restricted to the use o f methyl, primary, allylic, and benzylic halides. a-Halo esters can also be used as the alkylating agents, and this reaction provides a convenient syn­ thesis o f g-keto esters:

Br-

(75% )

Si 1 8 .1 0 S u m m a ry o f E n o la te C h e m is try

S'

Gl £

857

18.10 Sum m ary o f Enolate Chem istry 1. Form ation o f an E n olate (Section 18.1) ''O' R

:O ;

'O ',

: B~

f t - ”"i, L-l

r

n

+ H ;B~

R

H Resonance-stabilized enolate -B = -OH, :OR, or ;N(/-Pr) 2 (Section 18.4) 2. R acem ization (Section 18.3A)

3. H alogenation o f A ldeh ydes a n d K etones (Sections 18.3B and 18.3C) G e n e ra l R e actio n

O

O

R' R

R'

acid

+ X2 or base

R X

H

S p e c ific E xa m p le — H a lo fo rm R e actio n

O

O

3X2

CeHa H

O OH“ H2O

CeHa X

X

OH-

X

H 2O

CeHs

O- + CHX 3

858

C h a p te r 1 8

R e a c tio n s a t t h e

a C a rb o n

o f C a rb o n y l C o m p o u n d s

4. H alo g en a tio n o f C arboxylic A cid s: T h e H V Z R ea ctio n (Section 18.3D) O

O

(1) X2, P

R

R

(2) h2o

OH

OH X

5. D irect A lk yla tio n via L ith iu m E n o la te s (Section 18.4) G e n e ra l R e actio n O

O -L i -

LDA THF, -78°C (formation of the kinetic enolate)

R H (R ')

O

R"— X

R

R H (R ')

H (R ') R

S p e c ific E xam ple O

O - Li +

O

CH3I

LDA THF, -78°C 6

. D irect A lk yla tio n o f E sters (Section 18.4C) O O

LDA THF

R

OEt

R

OEt

7. A cetoacetic E ste r S yn th esis (Section 18.6) O

O

O

OEt

O

O

(1) NaOEt (2) RBr

(1) OH- , heat ^ (2) H3O+ * (3) heat (-CO2)

OEt R

O

O

O

OEt

(1) f-BuOK (2) R'Br

O

O

R

R,

'O E t

R 8

R

R'

(1) OH- , heat (2) H3O+ (3) heat (-CO2)

R

. M alonic E ste r S yn th esis (Section 18.7) O

O

Et

O

OEt

(1) NaOEt (2) RBr

O

O OEt

Et R

O E tO

O V "

O "O E t

(1) f-BuO^

(2) R'Br

O

O OEt

E t1 R

R

R

(1) OH-, heat > ho(2) H3O+ (3) heat (-CO2)

(1) H O -, heat

,

(2) H3O+ (3) heat (-CO2)

R'

R

HOR

9. S to rk E n a m in e R ea ctio n (Section 18.9)

O

R

nr;

^R +

r ;n h

— »

R'. R E n am in e

O

(1) R"^"- Br (2) heat (3) H2O

R

R

R"

859

P ro b le m s

Key Terms and Concepts The key terms and concepts that are highlighted in bold, blue text within the chapter are defined in the glossary (at the back o f the book) and have hyperlinked definitions in the accompanying W ileyPLU S course (w w w.w ileyplus.com )

PLUS

Problems

NotetoInstructors:

Many of the homework problems are available for assignment via teaching and learning solution.

WileyPLUS,an online

ENOLATES, ENOLS, A N D CARBONYL a-CARBON REACTIVITY 18.15

Rank the follow ing in order o f increasing acidity for the indicated hydrogen atoms (bold) (1 = least acidic; 4 = m ost acidic). (a)

O

O

CH2

H3 C .

H3 C .

'O

(b)

O

O

O

O

O

O H

c h3

O

O

H3C

CH3 H

O

O H

H

O"

och3

H

O

H

O

H

H

H

H

H

H

H

\ /

H3

0

XH

3

H3C"X

X ,

18.16

Treating a solution o f ds-1-decalone with base causes an isomerization to take place. When the system reaches equilibrium, the solution is found to contain about 95% trans-1-decalone and about 5% ds-1-decalone. Explain this isomerization.

18.17

Explain the variation in enol content that is observed for solutions o f acetylacetone (pentane-2,4-dione) in the several solvents indicated: S o lv e n t

H2O c h 3cn

CôHm Gas phase

% Enol

15 58 92 92

860 18.18

Chapter 18

Reactions at the a Carbon of Carbonyl Compounds

Provide a structural formula for the product from each o f the follow ing reactions. (a)

O

(d) O

(1) LDA (2) CH3CH2l

Br2 (excess), NaOH

O

(b)

(e)

O

(1) LDA O H

Br2, CH3CO2H O

(c)

NaH

(f)

(2) Cl

(3) H2 O

O

Cl'

(1) I2, NaOH

N

(2) H3O+

18.19

Write a stepwise m echanism for each o f the follow ing reactions. (a)

O

O ,Br +

(b)

O

HBr

O O- Na+

+

H3O MeO (d)

MeOH H

O

O O

O c h 3o h , h 2s o 4

'OCH 3 18.20

Write a stepwise m echanism for each o f the follow ing reactions. (a)

X

CHI3

OEt '

861

Problems (c)

O

O

(d)

O

+

A C E T O A C E T IC

18.21

18.22

ESTER A N D

M A L O N IC

ESTER SYN TH ESES

Outline syntheses o f each o f the follow ing from acetoacetic ester and any other required reagents: (a) ieri-Butyl methyl ketone

(d) 4-Hydroxypentanoic acid

(b) 2-Hexanone

(e) 2-Ethyl-1,3-butanediol

(c) 2,5-Hexanedione

(f) 1-Phenyl-1,3-butanediol

Outline syntheses o f each o f the follow ing from diethyl malonate and any other required reagents: (a) 2-Methylbutanoic acid

(b) 4-M ethyl-1-pentanol

18.23

Provide a structural formula for the product from each o f the follow ing reactions. (a)

O

(b)

O

O

(1) NaOCH3, CH3OH

heat

(2) (3) HCl, H2O, heat

18.24

The synthesis o f cyclobutanecarboxylic acid given in Section 18.7 was first carried out by W illiam Perkin, Jr., in 1883, and it represented one o f the first syntheses o f an organic compound with a ring smaller than six carbon atoms. (There was a general feeling at the time that such compounds would be too unstable to exist.) Earlier in 1883, Perkin reported what he mistakenly believed to be a cyclobutane derivative obtained from the reaction o f acetoacetic ester and 1,3-dibromopropane. The reaction that Perkin had expected to take place was the following: O

O

O Brv OEt

,B r

O

EtONa OEt

The molecular formula for his product agreed with the formulation given in the preceding reaction, and alkaline hydrolysis and acidification gave a nicely crystalline acid (also having the expected molecular formula). The acid, however, was quite stable to heat and resisted decarboxylation. Perkin later found that both the ester and the acid contained six-membered rings (five carbon atoms and one oxygen atom). Recall the charge distribution in the enolate ion obtained from acetoacetic ester and propose structures for Perkin’s ester and acid.

862 18.25

Chapter 18

Reactions at the a Carbon of Carbonyl Compounds

(a) In 1884 Perkin achieved a successful synthesis o f cyclopropanecarboxylic acid from sodiom alonic ester and 1,2-dibromoethane. Outline the reactions involved in this synthesis. (b) In 1885 Perkin synthesized five-membered carbocyclic compounds D and E in the follow ing way: O

O

Et

OEt

+

Br

B r-

2 EtONa Br2

A (C17H 28O8)

Na+ ( 1 ) o h m h 2o B (C 17H 26O 8) (2

) H3 O+

heat *

C (C 9H 10O 8)

D (C 7H 10O 4)

where D and E are diastereomers; D can be resolved into enantiomeric forms w hile E cannot. What are the structures o f A -E ? (c) Ten years later Perkin was able to synthesize 1,4-dibromobutane; he later used this compound and diethyl malonate to prepare cyclopentanecarboxylic acid. Show the reactions involved. 18.26

Synthesize each o f the follow ing compounds from diethyl malonate or ethyl acetoacetate and any other organic and inorganic reagents. (a)

(b)

O

(c)

O

(e)

O

OH

G EN ER A L PRO BLEM S

18.27

Outline a reaction sequence for synthesis o f each o f the follow ing compounds from the indicated starting material and any other organic or inorganic reagents needed. (a)

O

O

O

863

Problems

(c)

O O

O O

(d) OEt [ I

r

X

r O

OEt O

(e) f ^ B k

r

:

/ Br

O i T

V

A O

’

(f)

O

O

o (g)

18.28

O

O

O

Linalool, a fragrant compound that can be isolated from a variety o f plants, is 3,7-dim ethyl-1,6-octadien-3-ol. Linalool is used in making perfumes, and it can be synthesized in the follow ing way:

HBr

----- >

G

F (C 5 H9 Br)

(C 1 1 H18 O 3 )

(1

sodioacetoacetic ' ester

di'nNaOH> H (C 8 H 14 O)

(2) H3O+ (3) heat

CH> I (C 10 H1eO)

(1 ¡ f L

“rk

Lindlar’s catalyst

(2) H3 linalool

Outline the reactions involved. [Hint: Compound F is the more stable isomer capable o f being produced in the first step.] 18.29

Compound J, a compound with two four-membered rings, has been synthesized by the follow ing route. Outline the steps that are involved. O

O Br-

Et

,Br

OEt

~

^ ~

NaOEt

* C10H17BrO4

> O

C H O C 1 0 H 1 eO ^

( 1) L iA lH ^ ^ *

(2) H2O

CH O C eH 1 2 O 2

C13H20O4

hb:

:

C H Br C eH 1 0 B r 2

(2) H O+

:

Et^

C9H12O4

O ^

o b

^

2 NaOEt

* J (C8H12O2)

+

C O 2

864 18.30

Reactions at the a Carbon of Carbonyl Compounds

Chapter 18

The Wittig reaction (Section 16.10) can be used in the synthesis o f aldehydes, for example, O CH3O

\ = P ( C 6Hs)3

+

'

CH3 O'

CH3 O" 60%

H3O+/H2O

H

(a) How would you prepare CH3OCH = P(C 6 H5 )3? (b) Show with a mechanism how the second reaction produces an aldehyde. (c) How would you use this method to prepare

18.31

^CHO from cyclohexanone?

Aldehydes that have no a hydrogen undergo an intermolecular oxidation-reduction called the C annizzaro reac­ tion when they are treated with concentrated base. An example is the follow ing reaction o f benzaldehyde: O

2

ÍI 1

O

l^ ^ H

O H -^ h 2o '

J

OH

I

^O +

(a) When the reaction is carried out in D2 O, the benzyl alcohol that is isolated contains no deuterium bound to car­ bon. It is C 6 H5 CH 2 OD. What does this suggest about the m echanism for the reaction? (b) W hen (CH3)2CHCHO and Ba(OH) 2 /H2O are heated in a sealed tube, the reaction produces only (CH 3 )2 CHCH2OH and [(CH 3 )2 CHCO 2 ]2 Ba. Provide an explanation for the formation o f these products. 18.32

Shown below is a synthesis o f the elm bark beetle pheromone, multistriatin (see Problem 16.44). Give structures for compounds A , B, C, and D. O

O OH

A (C5 H 1 oO)

TsCl base

B (C12H16O3S)

base

RCOH C (C ioH180 )

Lewis acid D (C10H18O2) Multistriatin

SPEC TR O SC O PY

18.33

(a) A compound U (C 9 H1 0 O) gives a negative iodoform test. The IR spectrum o f U shows a strong absorption peak at 1690 cm - 1 . The 1H N M R spectrum o f U gives the following: Triplet Quartet Multiplet

8 8 8

1.2 (3H) 3.0 (2H) 7.7 (5H)

What is the structure o f U?

Si L e a rn in g G ro u p P ro b le m s

S'

Gl £

865

(b) A compound V is an isomer o f U. Compound V gives a positive iodoform test; its IR spectrum shows a strong peak at 1705 cm - 1 . The 1H NM R spectrum o f V gives the following: Singlet Singlet Multiplet

8 8 8

2.0 (3 H ) 3.5 (2 H ) 7.1 (5 H )

What is the structure o f V? 18.34

Compound A has the molecular formula C 6H 12O 3 and shows a strong IR absorption peak at 1710 cm - 1 . When treated with iodine in aqueous sodium hydroxide, A gives a yellow precipitate. When A is treated with Tollens’ reagent (a test for an aldehyde or a group that can be hydrolyzed to an aldehyde), no reaction occurs; how ­ ever, if A is treated first with water containing a drop o f sulfuric acid and then with Tollens’ reagent, a silver mir­ ror (positive Tollens’ test) forms in the test tube. Compound A shows the follow ing 1H NM R spectrum: Singlet Doublet Singlet Triplet

8 8 8 8

2.1 2.6 3.2 (6 H ) 4.7

Write a structure for A.

Challenge Problem 18.35

The follow ing is an example o f a reaction sequence developed by Derin C. D ’A m ico and M ichael E. Jung (UCLA) that results in enantiospecific formation o f two new chirality centers and a carbon carbon bond. The sequence includes a Horner-W adsworth-Emmons reaction (Section 16.10B), a Sharpless asymmetric epoxidation (Section 11.13), and a novel rearrangement that ultimately leads to the product. Propose a m echanism for rearrangement of the epoxy alcohol under the conditions shown to form the aldol product. [Hint: The rearrangement can also be accom ­ plished by preparing a trialkylsilyl ether from the epoxy alcohol in a separate reaction first and then treating the resulting silyl ether with a Lewis acid catalyst (e.g., BF3).] (1) CH3CHCO2CH3

O

OH

PO(OCH3)2, NaH H

(2) DIBAL-H C H f-BuOOH, Ti(O-/-Pr)4, d-(—)-diisopropyl tartrate TBSOTf (i-butyldimethylsilyl triflate), 1.3 equiv. W,W-ethyldiisopropylamine 1.35 equiv., molecular sieves, —42°C

OH O

CH 94% (9 5 % e n a n tio m e ric e x c es s )

Learning Group Problems B-C A RO TEN E, DEHYDROABEITIC ACID b-Carotene is a highly conjugated hydrocarbon with an orange-red color. Its biosynthesis occurs via the isoprene pathway (Special Topic E), and it is found in, among other sources, pumpkins. One o f the chem ical syntheses of b-carotene was accom plished near the turn o f the twentieth century by W. Ipatiew (Ber. 1901, 34, 59 4 -5 9 6 ). The

866

C h a p te r 1 8

R e a c tio n s a t t h e

a C a rb o n

o f C a rb o n y l C o m p o u n d s

first few steps o f this synthesis involve chemistry that should be familiar to you. Write mechanisms for all o f the reactions from compounds 2 to 5, and from 8 to 9 and 10.

^-Carotene

HBr (2 equiv.)

gr O C O 2Et

4 (1) NaOH (2 ) H3O +, heat _ (1) Zn/BrCH2CO2Et C O 2E t ( (2) Ac2O O

5

6

(1) Ca(OH)2 (2) Ca(O2CH)2

CHO O

7

8

H2SO.

2.

Dehydroabietic acid is a natural product isolated from Pinus palustris. It is structurally related to abietic acid, which com es from rosin. The synthesis o f dehydroabietic acid (J. Am. Chem. Soc. 1962, 84, 2 8 4 -2 9 2 ) was accomplished by Gilbert Stork. In the course o f this synthesis, Stork discovered his famous enamine reaction. (a) Write detailed mechanisms for the reactions from 5 to 7 below. (b) Write detailed mechanisms for all o f the reactions from 8 to 9a in Stork’s synthesis o f dehydroabietic acid. N ote that 9a contains a dithioacetal, which forms similarly to acetals you have already studied (Chapter 16).

867

L e a rn in g G ro u p P ro b le m s

O

CH 2 CO2Et 9 (1) (CH2SH)2/HCl (2) KOH

10

11

9a

(1) c h 2n 2 (2) PhMgBr (3) AcOH/Ac2O

(Structures from Flem ing, I.,

Selected Organic Synthesis, p. 76. C o p yrigh t John W ile y & Sons, Lim ited. Reproduced w ith perm ission.)

00 O'

00

S u m m a ry o f M e c h a n is m s E n o la te s : « - S u b s titu tio n

n3 "

G e n e ra l R e a c t io n ^ O

a>

- :A +

73

-:A

+ stereoisom er (if a carbon is, and/or if E contains, a stereogenic center) T y p ic a l b a s e s (_:A ) a n d s o lv e n ts fo r e n o la te fo r m a tio n

S o m e g r o u p s th a t in c re a s e « - h y d r o g e n a c id ity

I.

HO- in H20 o r ROH; o r RCT in ROH; Useful for reactions involving therm odynam ically favored enolates and equilibrium product control

V H Nitrile (cyano group)

(and in general, other groups that can stabilize an a-carbanion)

II. LDA (lithium diisopropylam ide) in THF o r DME; Useful, in general, for forming enolates in aprotic solvents (especially kinetically favored enolates and direct alkylation)

a>

H ----- A D e p r o to n a tio n - p r o to n a tio n

(m ay lead to racemization or epimerization)

(0 8

: X ----- X : H a lo g é n a tio n

■N

JO (0

P o s s ib le e le c t r o p h ile s ( E - A )

FI'----- X :

Substitution of enolate *■a hydrogen by H, X, o r R

na> cr

o 3

Q ■n cr

A lk y la tio n

o 3

3 O C 3 Q_

"G

(A

Condensation and “ Conjugate Addition Reactions of Carbonyl Compounds More Chemistry of Enolates

In this chapter we shall consider two additional reaction types of carbonyl compounds: condensation reactions and conjugate addition reactions. Both of these types of reactions involve enolates or enols. Carbonyl conden­ sation and conjugate addition reactions are very useful in synthesis, and also have important biological signifi­ cance, as we shall see in due course. One biomedical example relates to the cancer-fighting mechanism of 5-fluorouracil (see molecular model), which masquerades as the natural metabolite uracil in a conjugate addition reaction. In doing so, 5-fluorouracil irreversibly halts biosynthesis of a key D N A building block, thus taking its anticancer effect. Many drugs used in medicine take their effect by acting as imposters for natural compounds. W e shall see how 5-fluorouracil works in "The Chemistry of. . . A Suicide Enzyme Substrate" later in this chapter.

869

870

C h a p te r 1 9

C o n d e n s a tio n a n d C o n ju g a te A d d itio n R e a c tio n s o f C a rb o n y l C o m p o u n d s

19.1 Introduction In carbonyl condensation reactions the enolate or enol o f one carbonyl compound reacts with the carbonyl group o f another to join the two reactants. As part o f the process, a new m ole­ cule that is derived from them “condenses” (forms). Often this m olecule is that o f an alcohol or water. The main types o f condensation reactions w e shall study are the C laisen conden­ sation and the aldol condensation. Aldol condensations are preceded mechanistically by aldol additions, which w e shall also study. The name aldol derives from the fact that aldehyde and alcohol functional groups are present in the products o f many aldol reactions. A n E x a m p le o f a C l a i s e n C o n d e n s a t i o n

O

O (1)

2 R'

'O R

O

NaOR

(2 ) H3 O+

ROH

R'.

OR

5

R' A n E x a m p le o f a n A l d o l A d d i t i o n a n d C o n d e n s a t i o n

O

OH B$

2R

O

O

R

R

H(R')

H(R')

H(R') R

R A ld o l add itio n p ro d u ct

A ld o l c o n d e n s a tio n p ro d u ct

C onjugate addition reactions involve a nucleophile, which is often an enolate, adding to the b position o f an a,b-unsaturated carbonyl compound. One o f the m ost common con­ jugate addition reactions is the M ichael addition. A s w e shall see, the aldol condensation provides a way to synthesize a,b-unsaturated carbonyl compounds that w e can then use for subsequent conjugate addition reactions. A n E x a m p le o f C o n j u g a t e A d d i t i o n

O

O (1) Nu $ (or H— Nu)^

R(OR)

(2) HA

Nu

R(OR) H

19.2 The Claisen Condensation: A Synthesis o f ¡3-Keto Esters The Claisen condensation is a carbon-carbon bond-forming reaction that is useful for synthe­ sizing b-keto esters. In Chapter 18 w e saw how b-keto esters are useful in synthesis. In a Claisen condensation, the enolate o f one ester molecule adds to the carbonyl group o f another, result­ ing in an acyl substitution reaction that forms a b-keto ester and an alcohol molecule. The alco­ hol molecule that is formed derives from the alkoxyl group o f the ester. A classic example is the Claisen condensation by which ethyl acetoacetate (acetoacetic ester) can be synthesized. O

S o d io a c e to a c e tic e s te r

(re m o v e d by d is tilla tio n )

Ethyl a c e to a c e ta te (a c e to a c e tic e s te r) (76 % )

Another example is the Claisen condensation o f two m olecules o f ethyl pentanoate, lead­ ing to ethyl 3-oxo-2-propylheptanoate.

871

1 9 .2 T h e C la is en C o n d e n s a tio n : A S y n th e s is o f j8 -K e to Esters

O

2 OEt

NaOEt

EtOH

Ethyl pentanoate (77 % )

If w e look closely at these examples, w e can see that, overall, both reactions involve a condensation in which one ester loses an a hydrogen and the other loses an ethoxide ion: O R

O R

OEt

O OEt

R

_(1) NaOEt (2 ) H3 O+

O OEt + EtOH

'

H

R

(Rmay also be H)

A^-keto ester

We can understand how this happens if w e examine the reaction m echanism in detail. In doing so, w e shall see that the Claisen condensation mechanism is a classic example of acyl substitution (nucleophilic addition-elim ination at a carbonyl group).

¡to

A MECHANISM FOR THE REACTION T h e C la is e n C o n d e n s a t io n

O Step 1

O-

R

:O

R

:OEt

OEt

R

EtOH

"OEt

An alkoxide base removes an a proton fromthe ester, generating a nucleophilic enolate ion. (The alkoxide base used toformthe enolate should have the same alkyl group as the ester, e.g., ethoxide for an ethyl ester; otherwise transesterification may occur.) Although the a protons of an ester are not as acidic as those of aldehydes and ketones, the resulting enolate is stabilized by resonance in a similar way. 'O'-

•O'

:° 7

Step 2

R

R OEt

EtO R

OEt

:OEt

R

Tetrahedral intermediate and elimination The enolate attacks the carbonyl carbon of another ester molecule, forming a tetrahedral intermediate. The tetrahedral intermediate expels an alkoxide ion, resulting in substitution of the alkoxide by the group derived fromthe enolate. The net result is nucleophilic addition-elimination at the ester carbonyl group. The overall equilibriumfor theprocess is unfavorable thus far, however, but it is drawn toward the final product by removal of the acidic a hydrogen fromthe newb-dicarbonyl system.

872

Step 3

C h a p te r 1 9

O

C o n d e n s a tio n a n d C o n ju g a te A d d itio n R e a c tio n s o f C a rb o n y l C o m p o u n d s

O

-’O OEt

+

; OEt

^

o y"^^ O E t

+

EtO H

R

0-Keto ester (pKa~9; stronger acid)

Ethoxide ion (stronger base)

0-Keto ester anion Ethanol (weaker base) (pKa16; weaker acid)

An alkoxide ion removes an a proton fromthe newlyformed condensation product, resulting in a reso­ nance stabilizedb-keto ester ion. This step is highly favorable and draws the overall equilibriumtoward product formation. The alcohol by-product (ethanol in this case) can be distilled fromthe reaction mixture as it forms, thereby further drawing the equilibriumtoward the desired product. O

O

OH

Keto form

O

Enol form

Addition of acidquenches the reaction byneutralizingthe base and protonating the Claisen condensa­ tion product. Theb-keto ester product exists as an equilibriummixture of its ketoandenol tautomers.

•

When planning a reaction with an ester and an alkoxide ion it is important to use an alkoxide that has the same alkyl group as the alkoxyl group o f the ester.

The alkoxyl group o f the ester and the alkoxide must be the same so as to avoid transesterification (which occurs with alkoxides by the same mechanism as base-promoted ester hydrolysis; Section 17.7B). Ethyl esters and methyl esters, as it turns out, are the m ost com ­ mon ester reactants in these types o f syntheses. Therefore, w e use sodium ethoxide when ethyl esters are involved and sodium methoxide when methyl esters are involved. (There are some occasions when w e shall choose to use other bases, but w e shall discuss these later.) •

Esters that have only one a hydrogen do not undergo the usual Claisen condensation.

An example o f an ester that does not react in a normal Claisen condensation, because it has only one a hydrogen, is ethyl 2 -methylpropanoate:

The a carbon has only —^ one hydrogen.

O OEt

This ester does not undergo a Claisen condensation.

Ethyl 2-methylpropanoate •

Inspection o f the mechanism just given w ill make clear why this is so: an ester with only one a hydrogen w ill not have an acidic hydrogen when step 3 is reached, and step 3 provides the favorable equilibrium that ensures the success o f the reaction.

In Section 19.2B w e shall see how esters with only one a hydrogen can be converted to a b-keto ester by a method that uses a strong base. R e v ie w P ro b le m 19.1

(a) Write a mechanism for all steps o f the Claisen condensation that take place when ethyl propanoate reacts with ethoxide ion. (b) What products form when the reaction mixture is acidified?

R e v ie w P ro b le m 1 9 .2

Since the products obtained from Claisen condensations are b-keto esters, subsequent hydrolysis and decarboxylation o f these products give a general method for the synthesis o f ketones. Show how you would em ploy this technique in a synthesis o f 4-heptanone.

873

1 9 .2 T h e C la is en C o n d e n s a tio n : A S y n th e s is o f j8 -K e to Esters

19.2A Intramolecular Claisen Condensations: The Dieckmann Condensation An intramolecular Claisen condensation is called a D ieckm an n con d en sation . For exam ­ ple, when diethyl hexanedioate is heated with sodium ethoxide, subsequent acidification o f the reaction mixture gives ethyl 2 -oxocyclopentanecarboxylate: O

O

O

,OEt ( 1 ) NaOEt^ (2 ) ^ O +

O

Diethyl hexanedioate (diethyl adipate) •

'oEt

'

Ethyl 2-oxocyclopentanecarboxylate (74-81%)

In general, the Dieckmann condensation is useful only for the preparation o f fiveand six-membered rings.

Rings smaller than five are disfavored due to angle strain. Rings larger than seven are entropically less favorable due to the greater number o f conformations available to a longer chain precursor, in which case intermolecular condensation begins to com pete strongly.

A MECHANISM FOR THE REACTION T h e D ie c k m a n n C o n d e n s a tio n EtO

EtO

n,

O

OEt OEt

OEt

Ethoxide ion removes ana hydrogen. EtO ( OtV ' H

The enolate ion attacks the carbonyl groupat the other end of thechain. O

O OEt OEt

An ethoxide ion isexpelled.

OEt

The ethoxide ion removes the acidic hydrogen located between twocarbonyl groups. This favorable equilibriumdrives the reaction.

Addition of aqueous acid rapidly protonates theanion, giving thefinal product.

(a) What product would you expect from a Dieckmann condensation o f diethyl heptanedioate? (b) Can you account for the fact that diethyl pentanedioate (diethyl glutarate) does not undergo a Dieckmann condensation?

R e v ie w P ro b le m 1 9 .3

874

C h a p te r 1 9

C o n d e n s a tio n a n d C o n ju g a te A d d itio n R e a c tio n s o f C a rb o n y l C o m p o u n d s

19.2B Crossed Claisen Condensations •

Crossed Claisen condensations are possible w h en on e ester co m p o n en t has no a h ydrogen s and, therefore, is unable to form an enolate ion and undergo self­ condensation.

Ethyl benzoate, for example, condenses with ethyl acetate to give ethyl benzoylacetate: O

O

(1) NaOEt

OEt

OEt

(2 ) H3 O+

OEt 3

E th y l b e n z o a te (n o a

O

O

E th y l b e n z o y la c e ta te

h y d ro g e n )

(6 0 % )

Ethyl phenylacetate condenses with diethyl carbonate to give diethyl phenylmalonate: O

O

(1) NaOEt E tO

OEt

(2 ) H3 O+

3

OEt

E th y l p h e n y l a c e t a t e

D ie th y l c a r b o n a te (n o a c a r b o n )

c « / i/a

(6 5 % )

19.3 ß-Dicarbonyl Compounds by Acylation of Ketone Enolates

875

Write mechanisms that account for the products that are formed in the crossed Claisen con­ densation just illustrated o f ethyl phenylacetate with diethyl carbonate.

R e v ie w P ro b le m 1 9 .4

What products would you expect to obtain from each o f the follow ing crossed Claisen con­ densations?

R e v ie w P ro b le m 1 9 .5

(a) Ethyl propanoate (b) Ethyl acetate

+ diethyl oxalate ( 1 Na OEt-, (2 ) H3O+

(1) NaOEt + ethyl formate ___ __ 9 + > (2) H3O

A s w e learned earlier in this section, esters that have only one a hydrogen cannot be con­ verted to b-keto esters by sodium ethoxide. However, they can be converted to b-keto esters by reactions that use very strong bases such as lithium diisopropylamide (LDA) (Section 18.4). The strong base converts the ester to its enolate ion in nearly quantitative yield. This allows us to acylate the enolate ion by treating it with an acyl chloride or an ester. An exam ­ ple o f this technique using LD A is shown here: O

O

O

O OEt

LDA .

O

Cl

OEt

OEt

THF '

C l-

Ethyl 2,2-dimethyl-3-oxo3-phenylpropanoate 19.3 ß-D icarbonyl Com pounds b y Acylation o f K etone Enolates Enolate ions derived from ketones also react with esters in nucleophilic substitution reac­ tions that resemble Claisen condensations. In the follow ing first example, although two anions are possible from the reaction o f the ketone with sodium amide, the major product is derived from the primary carbanion. This is because (a) the primary a hydrogens are slightly more acidic than the secondary a hydrogens and (b) in the presence o f the strong base (NaNH2) in an aprotic solvent (Et2 O), the kinetic enolate is formed (see Section 18.4). LD A could be used similarly as the base. O

O NaNH 2

2-Pentanone

Et2O

Na4 O OEt

O

O

4,6-Nonanedione (76%) O

O

O

OEt

EtO NaOC 2 H5

Na4

(1 ) (2 ) H3 O 4

O

O OEt

O

O

67%

876

Chapter 19

Condensation and Conjugate Addition Reactions o f Carbonyl Compounds

Solved Problem 19.2 Keto esters are capable o f undergoing cyclization reactions similar to the Dieckmann condensation. Write a m ech­ anism for the follow ing reaction. O

O

O

(1) NaOEt _

OEt

(2 ) H3 O + '

O ANSWER E tO

E tO ^ C O E tO '

O'

O

O

OEt

+ O

O

O' O

R e v ie w P ro b le m 19.6

O

Show how you might synthesize each o f the follow ing compounds using, as your starting materials, esters, ketones, acyl halides, and so on: O

O

O CH

H (b)

(a)

O

19.4 A ld o l Reactions: A d d itio n o f Enolates and Enols to A ldehydes and Ketones •

Aldol additions and aldol condensations together represent an important class of carbon-carbon bond-forming reaction.

An aldol reaction begins with addition o f an enolate or enol to the carbonyl group o f an alde­ hyde or ketone, leading to a b -hydroxy aldehyde or ketone as the initial product. A simple example is shown below, whereby two molecules o f acetaldehyde (ethanal) react to form 3-hydroxybutanal. 3-Hydroxybutanal is an “aldol” because it contains both an aldehyde and an alcohol functional group. Reactions o f this general type are known as aldol additions. O 2

OH 10

'H

h 2 o , 5 °c

O J[

% NaOH '

3 -H y d ro x y b u ta n a l (50 % )

877

19.4 Aldol Reactions: Addition o f Enolates and Enols to Aldehydes and Ketones

A s w e shall see, the initial aldol addition product often dehydrates to form an a,b-unsaturated aldehyde or ketone. When this is the result, the overall reaction is an ald ol con ­ den sation . First let us consider the m echanism o f an aldol addition.

19.4A Aldol Addition Reactions An aldol addition is an equilibrium reaction when it is conducted in a protic solvent with a base such as hydroxide or an alkoxide. The m echanism for an aldol addition involving an aldehyde is shown below.

A MECHANISM FOR THE REACTION T h e A ld o l A d d itio n

Step 1

Enolate fo rm a tio n

H

HQ:

H

H

H

HQH

Enolate anion In this step the base (a hydroxide ion) removes a proton from the a carbon of one molecule of acetaldehyde to give a resonance-stabilized enolate.

:O: - 'O’

■O-s,

¿O Step 2

'O'

:O :

¿O'

c

A d d itio n

H An alkoxide anion

o f th e e n o la te

The enolate then acts as a nucleophile and attacks the carbonyl carbon of a second molecule of acetaldehyde, producing an alkoxide anion.

:OH Step 3

H

P ro to natio n o f th e alkoxid e

'O ’ 'H

Stronger base

+

The aldol product

:QH Weaker base

The alkoxide anion now removes a proton from a molecule of water to form the aldol.

With ketones, the addition step leading to the aldol is unfavorable due to steric hindrance, and the equilibrium favors the aldol precursors rather than the addition product (Section 19.4B). However, as w e shall see in Section 19.4C, dehydration o f the aldol addition prod­ uct can draw the equilibrium toward completion, whether the reactant is an aldehyde or a ketone. Enolate additions to both aldehydes and ketones are also feasible when a stronger base (such as LDA) is used in an aprotic solvent (Section 19.5B).

19.4B The Retro-Aldol Reaction Because the steps in an aldol addition mechanism are readily reversible, a retro-aldol reac­ tion can occur that converts a b-hydroxy aldehyde or ketone back to the precursors o f an aldol addition. For example, when 4-hydroxy-4-methyl-2-pentanone is heated with hydroxide in water, the final equilibrium mixture consists primarily o f acetone, the retro-aldol product. OH

O

O HO\

'u n

2

H e lp f u l H i n t See "The Chemistry of... A RetroAldol Reaction in Glycolysis: Dividing Assets to Double the ATP Yield" fo r an im portant biochemical application that increases the energy yield from glucose.

878

Chapter 19

Condensation and Conjugate Addition Reactions of Carbonyl Compounds

THE CHEMISTRY OF . . . A R e tr o - A ld o l R e a c t i o n in G ly c o ly s is — D iv id in g A s s e t s t o D o u b l e t h e A T P Y ie ld

Glycolysis is a fundamental pathway for production of ATP in living systems. The pathway begins with glucose and ends with two molecules of pyruvate and a net yield of two ATP molecules. Aldolase, an enzyme in glycolysis, plays a key role by dividing the six-carbon compound fructose-1,6bisphosphate (derived from glucose) into two compounds that each have three carbons, glyceraldehyde-3-phosphate (GAP) and 1,3-dihydroxyacetone phosphate (DHAP). This process is essential because it provides two three-carbon

:N H 2 H O H CH2OPO32-

units for the final stage of glycolysis, wherein the net yield of two ATP molecules per glucose is realized. (Two ATP mol­ ecules are consumed to form fructose-1,6-bisphosphate, and only two are generated per pyruvate. Thus, two pas­ sages through the second stage of glycolysis are necessary to obtain a net yield of two ATP molecules per glucose.) The cleavage reaction catalyzed by aldolase is a net retroaldol reaction. Details of the mechanism are shown here, beginning at the left with fructose-1,6-bisphosphate.

^

(1) Aldolase and F-1,6-bisP react to form the iminium ion linkage shown to the right

O 3POCH

O 3POCH2

Im in iu m io n

HO

H

(2) Retro-aldol C— C bond cleavage occurs, as the mechanism arrows show above, releasing GAP and forming an enamine enzyme intermediate.

F r u c to s e -1 ,6 -b is p h o s p h a te (F -1 ,6 -b is P )

HO

in te rm e d ia te

O CH2OPO32-

G ly c e ra ld e h y d e -3 p h o s p h a te (G A P ) (fir s t p ro d u c t)

(3) The enamine carbon is protonated and an iminium ion forms.

O3POCH2

OH e n z y m e -s u b s tra te

1 ,3 -D ih y d ro x y a c e to n e p h o s p h a te (D H A P ) (s e c o n d p ro d u c t)

(4) Hydrolysis of the iminium ion releases DHAP and regenerates the free enzyme to begin another catalytic cycle. HO

Im in iu m io n e n z y m e D H A P in te rm e d ia te

Two key intermediates in the aldolase mechanism involve functional groups that we have studied (Chapter 16)— an imine (protonated in the form of an iminium cation) and an enamine. In the mechanism of aldolase, an iminium cation acts as a sink for electron density during C— C bond cleavage (step 2), much like a carbonyl group does in a typ­ ical retro-aldol reaction. In this step the iminium cation is converted to an enamine, corresponding to the enolate or enol that is formed when a carbonyl group accepts electron density during C— C bond cleavage in an ordinary retroaldol reaction. The enamine intermediate is then a source of an electron pair used to bond with a proton taken from the tyrosine hydroxyl at the aldolase active site (step 3).

E n a m in e e n z y m e -D H A P in te rm e d ia te

HO

Lastly, the resulting iminium group undergoes hydrolysis (step 4), freeing aldolase for another catalytic cycle and releasing DHAP, the second product of the retro-aldol reac­ tion. Then, by the process described in "The Chemistry of . . . TIM (Triose Phosphate Isomerase and Carbon Recycling via Enol" (Chapter 18), DHAP undergoes iso­ merization to GAP for processing to pyruvate and synthe­ sis of two more ATP molecules. As we have seen with aldolase, imine and enamine functional groups have widespread roles in biological chemistry. Yet the functions of imines and enamines in biol­ ogy are just as we would predict based on their native chem­ ical reactivity.

879

19.4 Aldol Reactions: Addition o f Enolates and Enols to Aldehydes and Ketones

This result is not surprising, because we know that the equilibrium for an aldol addition (the reverse o f the reaction above) is not favorable when the enolate adds to a ketone. But, as mentioned earlier, dehydration o f an aldol addition product can draw the equilibrium forward. We shall discuss the dehydration o f aldols next (Section 19.4C).

S o l v e d P r o b le m 1 9 .3

The carbon-carbon bond cleavage step in a retro-aldol reaction involves, under basic conditions, a leaving group that is an enolate, or under acidic conditions, an enol. Write a mechanism for the retro-aldol reaction of 4-hydroxy4-methyl-2-pentanone under basic conditions (shown above). STRATEGY AND ANSWER Base removes the proton from the b-hydroxyl group, setting the stage for reversal of

the aldol addition. As the alkoxide reverts to the carbonyl group, a carbon-carbon bond breaks with expulsion of the enolate as a leaving group. This liberates one of the original carbonyl molecules. Protonation o f the enolate forms the other.

19.4C Aldol Condensation Reactions: Dehydration of the Aldol Addition Product Dehydration o f an aldol addition product leads to a conjugated a,b-unsaturated carbonyl system. The overall process is called an aldol condensation, and the product can be called an enal (alkene aldehyde) or enone (alkene ketone), depending on the carbonyl group in the product. The stability o f the conjugated enal or enone system means that the dehydra­ tion equilibrium is essentially irreversible. For example, the aldol addition reaction that leads to 3-hydroxybutanal, shown in Section 19.4, dehydrates on heating to form 2-butenal. A mechanism for the dehydration is shown here.

A MECHANISM FOR THE REACTION D e h y d r a tio n o f t h e A ld o l A d d itio n P r o d u c t T h e d o u b le b o n d s o f th e a lk e n e a n d c a rb o n y l g ro u p s a re c o n ju g a te d , s ta b ilizin g th e p ro d u ct.

T h e a h yd ro g e n s a re acidic.

:QH

:QH =Q:

•Qi

•'O’-

o

H

H 2 -B u te n a l (an e n al )

:OH

Even though hydroxide is a leaving group in this reaction, the fact that each dehydrated mol­ ecule forms irreversibly, due to the stability from conjugation, draws the reaction forward.

HOH

:QH

sso

Chapter 19

Condensation and Conjugate Addition Reactions of Carbonyl Compounds

19.4D Acid-Catalyzed Aldol Condensations Aldol reactions can occur under acid catalysis, in which case the reaction generally leads to the a,b-unsaturated product by direct dehydration of the b-hydroxy aldol intermediate. This is one way by which ketones can successfully be utilized in an aldol reaction. The fol­ lowing is an example, in which acetone forms its aldol condensation product, 4-methylpent3-en-2-one, on treatment with hydrogen chloride.

A MECHANISM FOR THE REACTION T h e A c id - C a ta ly z e d A ld o l R e a c tio n

R E A C T IO N

4-Methylpent-3-en-2-one M E C H A N IS M

The mechanismbeginswith the acid-catalyzed formation of the enol.

Then the enol adds to the protonated carbonyl group of another molecule of acetone.

■'O

=O"

H

•'o ' H — Cl :

HOH

Finally, proton transfers and dehydration lead tothe product.

Acid catalysis can promote further reactions after the aldol condensation. An example is given in Review Problem 19.8. Generally, it is more common in synthesis for an aldol reac­ tion to be conducted under basic rather than acidic conditions.

881

19.4 Aldol Reactions: Addition o f Enolates and Enols to Aldehydes and Ketones

S olved P roblem 19.4 One industrial process for the synthesis o f 1-butanol begins with ethanal. Show how this synthesis might be car­ ried out. STRATEGY AND ANSWER Ethanal can be converted to an aldol via an aldol addition. Then, dehydration would produce 2-buten-1-al, which can be hydrogenated to furnish 1-butanol.

O

OH

O

O

10% NaOH

H

0-10°C

heat '

H

E th an al

H /N i '|_|

H2O

A ld ol

OH

high pressure 1 -B u ta n o l

2 -B u te n -1 -a l

The acid-catalyzed aldol condensation o f acetone (just shown) also produces some 2,6dimethylhepta-2,5-dien-4-one. Give a mechanism that explains the formation of this product.

Review Problem 19.7

Heating acetone with sulfuric acid leads to the formation of mesitylene (1,3,5-trimethylbenzene). Propose a mechanism for this reaction.

Review Problem 19.8

19.4E Synthetic Applications of Aldol Reactions As we are beginning to see, aldol additions and aldol condensations are important meth­ ods for carbon-carbon bond formation. They also result in b-hydroxy and a,b-unsaturated carbonyl compounds that are themselves useful for further synthetic transformations. Some representative reactions are shown below. The A ld ol Reaction in Synthesis

H e lp f u l H i n t

R H

2

base

R H

R

O

OH

A ld e h y d e

NaBH.

H

R

O

OH

An aldo l

OH

A 1,3-diol

HA

R

R

_ Hg/Ni

R OH A s a tu ra te d a lcoh ol

high pressure

R H

R O

LiAlH.*

R OH

A n a , ß -u n s a tu ra te d a ld e h y d e

An a lly lic alc o h o l

H2, Pd-C

R H

R O An a ld e h y d e

* L iA lH reduces the carbonyl group of a,/8-unsaturated aldehydes and ketones cleanly. N aB H 4 often reduces the carbon-carbon double bond as well.

The aldol reaction: a tool for synthesis. See also the Synthetic Connections review at the end of the chapter.

882

Review Problem 19.9

Chapter 19

Condensation and Conjugate Addition Reactions of Carbonyl Compounds

(a) Provide a mechanism for the aldol addition o f propanal shown here.

P ro p a n a l

3 -H y d ro x y -2 -m e th y lp e n ta n a l (5 5 -6 0 % )

(b) How can you account for the fact that the product of the aldol addition is 3-hydroxy-

2-methylpentanal and not 4-hydroxyhexanal? (c) What products would be formed i f the reaction mixture were heated?

Review Problem 19.10

Show how each o f the following products could be synthesized from butanal: (a) 2-Ethyl-3-hydroxyhexanal (b) 2-Ethylhex-2-en-1-ol (c) 2-Ethylhexan-1-ol (d) 2-Ethylhexane-1,3-diol (the insect repellent “ 6-12” )

Thus far we have only considered examples o f aldol reactions where the reactant forms a product by dimerization. In the coming sections we shall discuss the use of aldol reactions to more generally prepare b-hydroxy and a,b-unsaturated carbonyl compounds. We shall then study reactions called conjugate addition reactions (Section 19.7), by which we can further build on the a,b-unsaturated carbonyl systems that result from aldol condensations.

19.5 Crossed A ld o l Condensations An aldol reaction that starts with two different carbonyl compounds is called a crossed aldol reaction. Unless specific conditions are involved, a crossed aldol reaction can lead to a m ix­ ture o f products from various pairings of the carbonyl reactants, as the following example illustrates with ethanal and propanal. O

O

OH

O

OH-

H

Ethanal

h 2o

P ropanal

3 -H y d ro x y b u ta n a l (fro m tw o m o le c u le s o f e th a n a l)

OH

3-H y d ro x y -2 m e th y lp e n ta n a l (fro m tw o m o le c u le s o f p ro p a n a l)

O

H H OH

O

3 -H y d ro x y -2 -m e th y lb u ta n a l 3 -H y d ro x y p e n ta n a l (fro m o n e m o le c u le o f e th a n a l a n d o n e m o le c u le o f p ro p a n a l)

We shall therefore consider crossed aldol condensations by two general approaches that allow control over the distribution of products. The first approach hinges on structural factors of the carbonyl reactants and the role that favorable or unfavorable aldol addition equilibria play in determining the product distribution. In this approach relatively weak bases such as hydrox­ ide or an alkoxide are used in a protic solvent such as water or an alcohol. The second

883

19.5 Crossed Aldol Condensations

approach, called a directed aldol reaction, involves use of a strong base such as LD A in an aprotic solvent. With a strong base, one reactant can be converted essentially completely to its enolate, which can then be allowed to react with the other carbonyl reactant. S o l v e d P r o b le m 1 9 .5

Show how each o f the four products shown at the beginning o f this section is formed in the crossed aldol addition between ethanal and propanal. ANSWER In the basic aqueous solution, four organic entities w ill initially be present: molecules of ethanal, mol­

ecules of propanal, enolate anions derived from ethanal, and enolate anions derived from propanal. We have already seen (Section 19.4) how a molecule o f ethanal can react with its enolate to form 3-hydroxybutanal (aldol). We have also seen (Review Problem 19.9) how propanal can react with its enolate anion to form 3-hydroxy-2-methylpentanal. The other two products are formed as follows. 3-Hydroxy-2-methylbutanal results when the enolate of propanal reacts with ethanal. O

O

HO-

H Ethanal

Enolate of propanal

methylbutanal

And finally, 3-hydroxypentanal results when the enolate of ethanal reacts with propanal. O

O

OH

O

HOPropanal

H 3-Hydroxypentanal

H Enolate of ethanal

19.5A Crossed Aldol Condensations Using Weak Bases Crossed aldol reactions are possible with weak bases such as hydroxide or an alkoxide when one carbonyl reactant does not have an a hydrogen. A reactant without a hydrogens can­ not self-condense because it cannot form an enolate. We avoid self-condensation of the other reactant, that which has an a hydrogen, by adding it slowly to a solution of the first reac­ tant and the base. Under these conditions the concentration of the reactant with an a hydro­ gen is always low, and it is present mostly in its enolate form. The main reaction that takes place is between this enolate and the carbonyl compound that has no a hydrogens. The reac­ tions shown in Table 19.1 on the bottom of the next page illustrate results from this approach. The crossed aldol examples shown in Table 19.1 involve aldehydes as both reactants. A ketone can be used as one reactant, however, because ketones do not self-condense appre­ ciably due to steric hindrance in the aldol addition stage. The following are examples of crossed aldol condensations where one reactant is a ketone. Reactions such as these are sometimes called Claisen-Schmidt condensations. Schmidt discovered and Claisen devel­ oped this type o f aldol reaction in the late 1800s. O

O

4-Phenylbut-3-en-2-one (benzalacetone) (70%)

884

Chapter 19 Condensation and Conjugate Addition Reactions of Carbonyl Compounds O

O

O

H

OH20°C

1,3-Diphenylprop-2-en-1-one (benzalacetophenone) (85%) In these reactions, dehydration occurs readily because the double bond that forms is con­ jugated both with the carbonyl group and with the benzene ring. In general, dehydration o f the aldol is especially favorable when it leads to extended conjugation. As a further example, an important step in a commercial synthesis o f vitamin A makes use o f a crossed aldol condensation between geranial and acetone: H e lp f u l H i n t ____ ^

OH

See "The Chemistry of... Antibody-catalyzed Aldol Condensations" in W iley Plus fo r a method that uses the selectivity o f antibodies to catalyze aldol reactions.

Vitamin A O

O

O

H

EtONa EtOH, ' —5°C

Geranial

Pseudoionone (49%)

C r o s s e d A l d o l R e a c t io n s

This Reactant with No This Reactant with a Hydrogen Is an a Hydrogen Is Placed in Base Added Slowly

Product

O

O

H

O OH-

H

Benzaldehyde

,

H

10°C '

Propanal

2-Methyl-3-phenyl-2-propenal (a-methylcinnamaldehyde) (68%) O

O

H

O OH- _ 2 0 °C '

H

Benzaldehyde

Phenylacetaldehyde

Formaldehyde

2-Methylpropanal

3-Hydroxy-2,2dim ethylpropanal (>64% )

885

19.5 Crossed Aldol Condensations Geranial is a naturally occurring aldehyde that can be obtained from lemongrass oil. Its a hydrogen is vinylic and, therefore, not appreciably acidic. Notice, in this reaction, too, dehy­ dration occurs readily because dehydration extends the conjugated system. C a

/ « / r t /•/ D i - a

l/-\ k v i

1 O

à

Outlined below is a synthesis of a compound used in perfumes, called lily aldehyde. Provide all o f the missing structures. PCC propanal p-tert-Butylbenzyl alcohol ———i > C ^n-^O — ———: CH2CI2 OH H pd-C --------> lily aldehyde (C14H20O) C14H18O — —

Review Problem 19.11

When excess formaldehyde in basic solution is treated with ethanal, the following reaction takes place:

Review Problem 19.12

OH

OH

82%

Write a mechanism that accounts for the formation o f the product. When pseudoionone is treated with BF3 in acetic acid, ring closure takes place and a- and b -ionone are produced. This is the next step in the vitamin A synthesis.

(a) Write mechanisms that explain the formation o f a - and b -ionone. (b) b-Ionone is the major product. How can you explain this? (c) Which ionone would you expect to absorb at longer wavelengths in the UV-visible region? Why?

Review Problem 19.13

886

Chapter 19

Condensation and Conjugate Addition Reactions of Carbonyl Compounds

Nitriles with a hydrogens are also weakly acidic (pA"a = 25) and consequently these nitriles undergo condensations of the aldol type. An example is the condensation o f ben­ zaldehyde with phenylacetonitrile: O

fl R e v ie w P r o b le m 1 9 .1 4

/O N

H +

Ii l J

/

EtO-

CN

( l ^ l

EtOH

(a) Write resonance structures for the anion o f acetonitrile that account for its being much more acidic than ethane. (b) Give a step-by-step mechanism for the condensation o f ben­ zaldehyde with acetonitrile.

19.5B Crossed Aldol Condensations Using Strong Bases: Lithium Enolates and Directed Aldol Reactions H e lp f u l H i n t Lithium enolates are useful for crossed aldol syntheses.

One o f the most effective and versatile ways to bring about a crossed aldol reaction is to use a lithium enolate obtained from a ketone as one component and an aldehyde or ketone as the other. An example o f this approach, called a directed aldol reaction, is shown by the following mechanism.

A MECHANISM FOR THE REACTION A D i r e c t e d A l d o l S y n t h e s i s U s in g a L ith iu m E n o l a t e

OH

O

THF, -7 8 °C T h e k e to n e is a d d e d to L iN (/-P r ) 2 (LD A ), a s tro n g b ase, w h ic h re m o v e s an a h yd ro g e n fro m th e k e to n e to p ro d u c e an e n o la te .

T h e a ld e h y d e is a d d e d a n d the e n o la te re a c ts w ith th e a ld e h y d e a t its c a rb o n y l c a rb o n .

An a c id -b a s e reactio n o c c u rs w h e n w a te r is a d d e d a t th e end, p ro to n a tin g th e lith iu m a lkoxide.

Regioselectivity can be achieved when unsymmetrical ketones are used in directed aldol reactions by generating the kinetic enolate using lithium diisopropylamide (LDA). This ensures production of the enolate in which the proton has been removed from the less sub­ stituted a carbon. The following is an example: A n A ld o l R eaction via the K in etic Enolate (Using L D A )

A s in g le c ro s s e d aldol p ro d u ct results.

887

19.5 Crossed Aldol Condensations

I f this aldol reaction had been carried out in the classic way (Section 19.5A) using hydroxide ion as the base, then at least two products would have been formed in signifi­ cant amounts. Both the kinetic and thermodynamic enolates would have been formed from the ketone, and each o f these would have added to the carbonyl carbon of the aldehyde: A n A ld o l R eaction T h a t Produces a M ixture via Both K in etic and T herm odynam ic Enolates (U sing a W eak er Base under Protic Conditions)

A m ix tu re o f c ro s s e d aldol p ro d u c ts results.

S o l v e d P r o b le m 1 9 .7

Outline a directed aldol synthesis o f the following compound. O

OH

STRATEGY AND ANSWER Retrosynthetic Analysis

O

OH

O- Li + 1

O H

Synthesis O

O

O

O- Li + (1) LDA

(2) H (3) H2O

OH

888

Chapter 19

Review Problem 19.15

Condensation and Conjugate Addition Reactions of Carbonyl Compounds

Starting with ketones and aldehydes of your choice, outline a directed aldol synthesis of each o f the following using lithium enolates: O

OH

(a)

(b) ^

O

OH

A

h

O

OH

5

(c)

19.6 Cyclizations via A ld o l Condensations The aldol condensation also offers a convenient way to synthesize molecules with five- and six-membered rings (and sometimes even larger rings). This can be done by an intramol­ ecular aldol condensation using a dialdehyde, a keto aldehyde, or a diketone as the sub­ strate. For example, the following keto aldehyde cyclizes to yield 1-cyclopentenyl methyl ketone:

This reaction almost certainly involves the formation o f at least three different enolates. However, it is the enolate from the ketone side o f the molecule that adds to the aldehyde group leading to the product. The reason the aldehyde group undergoes addition preferentially may arise from the greater reactivity o f aldehydes toward nucleophilic addition generally. The carbonyl car­ bon atom o f a ketone is less positive (and therefore less reactive toward a nucleophile) because it bears two electron-releasing alkyl groups; it is also more sterically hindered: O

O

H e lp f u l H i n t Selectivity in aldol cyclizations is influenced by carbonyl type and ring size.

K '

Ketones are less reactive toward nucleophiles.

R

H

Aldehydes are more reactive toward nucleophiles.

In reactions o f this type, five-membered rings form far more readily than sevenmembered rings, and six-membered rings are more favorable than four- or eight-membered rings, when possible.

19.7 Additions to a,ß-Unsaturated Aldehydes and Ketones

889

A MECHANISM FOR THE REACTION T h e A ld o l C y c liz a tio n

•■Q' :Q

"Q i

H — QH **

,H H

H — QH Q: •A

Qther enoIate anions

This enolate leads to the main product via an intramolecular aldol reaction.

The alkoxide anion removes a protonfromwater.

H— QH + HQ:

Base-promoted dehydration leads toa product with conjugated double bonds.

Assuming that dehydration occurs, write the structures of the two other products that might have resulted from the aldol cyclization just given. (One of these products w ill have a fivemembered ring and the other w ill have a seven-membered ring.)

Review Problem 19.16

What starting compound would you use in an aldol cyclization to prepare each of the following?

Review Problem 19.17

What experimental conditions would favor the cyclization process in an intramolecular aldol reaction over intermolecular condensation?

Review Problem 19.18

19.7 A dditions to a ß -U n s a tu ra te d A ldehydes and Ketones When a,b-unsaturated aldehydes and ketones react with nucleophilic reagents, they may do so in two ways. They may react by a simple addition, that is, one in which the nucle­ ophile adds across the double bond of the carbonyl group; or they may react by a conju­ gate addition. These two processes resemble the 1,2- and the 1,4-addition reactions of conjugated dienes (Section 13.10):

8 90

Chapter 19

Condensation and Conjugate Addition Reactions of Carbonyl Compounds :Q H

Simple addition

•Q.0

%

II ^0 . 0

Nu Nu

:Q H I 0" Nu

0

I

*A "

Q

" 0 v 'o / 0 '' Nu H

Enol form

Conjugate addition

Ketoform

In many instances both modes of addition occur in the same mixture. As an example, let us consider the Grignard reaction shown here:

Simple addition product

Conjugate addition product (in ketoform) 20%

H e lp f u l H i n t ____ ^ Note the influence of nucleophile strength on conjugate versus simple addition.

In this example we see that simple addition is favored, and this is generally the case with strong nucleophiles. Conjugate addition is favored when weaker nucleophiles are employed. I f we examine the resonance structures that contribute to the overall hybrid for an a,bunsaturated aldehyde or ketone (see structures A -C ), we shall be in a better position to understand these reactions: : I ¿II / c % c / c \ «__ > A

I

O

I

:

O I

:

-

I C\

B

C

Although structures B and C involve separated charges, they make a significant contribu­ tion to the hybrid because, in each, the negative charge is carried by electronegative oxy­ gen. Structures B and C also indicate that both the carbonyl carbon and the b carbon should bear a p a rtia l positive charge. They indicate that we should represent the hybrid in the fol­ lowing way: Q80 This structure tells us that we should expect a nucleophilic reagent to attack either the car­ bonyl carbon or the b carbon.

19.7 Additions to a,j3-Unsaturated Aldehydes and Ketones

891

Almost every nucleophilic reagent that adds at the carbonyl carbon of a simple alde­ hyde or ketone is capable of adding at the b carbon of an a,b-unsaturated carbonyl com­ pound. In many instances when weaker nucleophiles are used, conjugate addition is the major reaction path. Consider the following addition o f hydrogen cyanide:

O "il

CN 1

CNEtOH, HOAc

O

/ V

i l 1 H 9 5%

A MECHANISM FOR THE REACTION T h e C o n ju g a te A d d itio n o f H C N

CN

O-

CN

Enolate intermediate Then, theenolate intermediate accepts a proton ineither of twoways: CN

OH Enol form

CN

O

Ketoform

Another example of this type of addition is the following: O

O

892

Chapter 19

ÙL

Condensation and Conjugate Addition Reactions of Carbonyl Compounds

A MECHANISM FOR THE REACTION T h e C o n j u g a te A d d itio n o f a n A m in e

O

K >

O-

3

^M eN H2

O

M eN H 2

MeNH

MeNH H

The nucleophile attacks Intwoseparate steps, a proton the partially positive b is lost fromthe nitrogen atomand a carbon. proton isgained at theoxygen.

Enol form

Ketoform

We shall see examples of biochemically relevant conjugate additions in “ The Chemistry o f . . . Calicheamicin g 1I Activation for Cleavage o f DNA” (see Section 19.7B) and in “ The Chemistry of . . . A Suicide Enzyme Substrate” (Section 19.8).

19.7A Conjugate Additions of Enolates: Michael Additions Conjugate additions of enolates to a,b-unsaturated carbonyl compounds are known gen­ erally as Michael additions (after their discovery, in 1887, by Arthur Michael, o f Tufts University and later o f Harvard University). The following mechanism box provides an example o f a Michael addition.

A MECHANISM FOR THE REACTION T h e M ic h a e l A d d itio n

O

Abase removes an a proton to forman enolate from one carbonyl reactant.

This enolate adds to thebcarbon of the a,jS-unstaturated carbonyl compound, forming a new carbon-carbon bond between them. As this bond isformed, electron density inthe a,jS-unsaturated compound shifts to its carbonyl oxygen, leading toa new enolate.

Protonation of the resulting enolate leads to the final Michael addition product.

19.7 Additions to a,j3-Unsaturated Aldehydes and Ketones

893

Michael additions take place with a variety of other reagents; these include acetylenic esters and a,b-unsaturated nitriles: O

O

O ^

O

O

"OEt

E tnEtO-

.

EtOH 5

OEt

EtO

"OEt

O O

O

O

EtO-

^ C N

EtOH 1

EtO

CN

EtO

OEt EtO

O

What product would you expect to obtain from the base-catalyzed Michael reaction of (a) 1,3-diphenylprop-2-en-1-one (Section 19.5A) and acetophenone and (b) 1,3-diphenylprop2-en-1-one and cyclopentadiene? Show all steps in each mechanism.

Review Problem 19.19

When acrolein (propenal) reacts with hydrazine, the product is a dihydropyrazole:

Review Problem 19.20

O H

+

H2N — NH2

N 'N I

H Acrolein

Hydrazine

A dihydropyrazole

Suggest a mechanism that explains this reaction. Enamines can also be used in Michael additions. An example is the following:

O CN

CN

19.7B The Robinson Annulation A Michael addition followed by a simple aldol condensation may be used to build one ring onto another. This procedure is known as the R obinson annulation (ring-forming) reaction (after the English chemist, Sir Robert Robinson, who won the Nobel Prize in Chemistry in 1947 for his research on naturally occurring compounds): O

O

O

O

aldol condensation

OH-

O 2-Methylcyclohexane-1,3-dione

Methyl vinyl ketone

c h 3o h (conjugate addition)

base ( - H 2 O)

O

(a) Propose step-by-step mechanisms for both transformations o f the Robinson annulation sequence just shown. (b) Would you expect 2-methylcyclohexane-1,3-dione to be more or less acidic than cyclohexanone? Explain your answer.

O 65%

Review Problem 19.21

894

Chapter 19

Condensation and Conjugate Addition Reactions of Carbonyl Compounds

THE CHEMISTRY OF C a lic h e a m ic in g / A c t i v a t i o n f o r C l e a v a g e o f D N A

by the sulfur nucleophile on the alkene carbon is a conju­ gate addition. Now that the bridgehead carbon is tetrahedral, the geometry of the bicyclic structure favors conversion of the enediyne to a 1,4-benzenoid diradical by a reaction called the Bergman cycloaromatization (after R. G. Bergman of the University of California, Berkeley). Once the calicheamicin diradical is formed it can pluck two hydrogen atoms from the DNA backbone, converting the DNA to a reactive diradical and ultimately resulting in DNA cleavage and the death of the cell.

In "The Chemistry of . . . Calicheamicin g i1" in Chapter 10, we described a potent antitumor antibiotic called calicheam­ icin g i1. Now that we have considered conjugate addition reactions, it is time to revisit this fascinating molecule. The molecular machinery of calicheamicin g i1for destroying DNA is unleashed by attack of a nucleophile on the trisulfide link­ age shown in the accompanying scheme. The sulfur anion that initially was a leaving group from the trisulfide immedi­ ately becomes a nucleophile that attacks the bridgehead alkene carbon. This alkene carbon is electrophilic because it is conjugated with the adjoining carbonyl group. Attack O

CO2Me 1 2 .NH

O

CO2Me 1 2 NH

(1) nucleophilic attack

Bergman cyclo­ arom atization

Sugar M eS

Nu

Calicheamicin g1I CO2Me Hydrogen abstraction from DNA backbone

R ep rinte d

fro m

CURRENT

BIOLOGY,

Vol

Sugar

1,

N ic o la o u , "C h e m is try and b io lo g y o f th e calicheam icins," co p yrig h t 1994 w ith perm ission o f

Sugar Oxidative Radicals in O2 ► double strand DNA backbone cleavage of DNA

Elsevier.

19.8 The Mannich Reaction Compounds capable of forming an enol react with imines from formaldehyde and a pri­ mary or secondary amine to yield b-aminoalkyl carbonyl compounds called Mannich bases. The following reaction of acetone, formaldehyde, and diethylamine is an example: O

O

O

A M an n ic h b ase

The Mannich reaction apparently proceeds through a variety of mechanisms depend­ ing on the reactants and the conditions that are employed. The mechanism below appears to operate in neutral or acidic media. Note the aspects in common with imine formation and with reactions o f enols and carbonyl groups.

895

19.8 The Mannich Reaction

A MECHANISM FOR THE REACTION T h e M a n n ic h R e a c tio n

■O' O

H

Step 1

R

H'O ? H H

R- n R" C R

H

HA

H

H

o

R

Iminiumcation

O H

Step 2

Enol

H

The hemiaminal loses a molecule of water toforman iminiumcation.

H

... HA

R N+

- H 2O

R '^ r

R

Reaction of the secondary aminewith the aldehyde forms a hemiaminal.

o

H

R ^=N +

H

R

NR 2

Mannich base

Iminiumcation The enol formof the active hydrogen compound reactswith the iminiumcation toformab-aminocarbonyl compound (a Mannich base).

Outline reasonable mechanisms that account for the products of the following Mannich reactions:

Review Problem 19.22

O

O O

(a)

A

Me2NH

O (b)

A

h

N H OH

O (c)

2

2 Me2NH A

h

THE CHEMISTRY OF . . . A S u ic id e E n z y m e S u b s tr a te

5-Fluorouracil is a chemical imposter for uracil and a potent clinical anticancer drug. This effect arises because 5-fluorouracil irreversibly destroys the ability of thymidylate syn­ thase (an enzyme) to catalyze a key transformation needed

for DNA synthesis. 5-Fluorouracil acts as a mechanism-based inhibitor (or suicide substrate) because it engages thymidylate synthase as though it were the normal substrate but then leads to self-destruction of the enzyme's activity by its

896

Chapter 19

Condensation and Conjugate Addition Reactions of Carbonyl Compounds

own mechanistic pathway. The initial deception is possi­ ble because the fluorine atom in the inhibitor occupies roughly the same amount of space as the hydrogen atom does in the natural substrate. Disruption of the enzyme's mechanism occurs because a fluorine atom cannot be removed by a base in the way that is possible for a hydro­ gen atom to be removed. O H

F

O' 5-Fluorodeoxyuracil monophosphate covalently bound to tetrahydrofolate in thymidylate synthase, blocking the enzyme's catalytic activity.

5-Fluorouracil The mechanism of thymidylate synthase in both its nor­ mal mode and when it is about to be blocked by the inhibitor involves attack of an enolate ion on an iminium cation. This process is closely analogous to the Mannich reaction discussed in Section 19.8. The enolate ion in this attack arises by conjugate addition of a thiol group from thymidylate synthase to the a,y8-unsaturated carbonyl group of the substrate. This process is analogous to the way an enolate intermediate occurs in a Michael addition. The iminium ion that is attacked in this process derives from the coenzyme N5,N10-methylenetetrahydrofolate (N5,N10-methylene-THF). Attack by the enolate in this step forms the bond that covalently links the substrate to the enzyme. It is this bond that cannot be broken when the fluorinated inhibitor is used. The mechanism of inhibition is shown at right.

Conjugate addition of a thiol group from thymidylate synthase to the P carbon of the a, p-unsaturated carbonyl group in the inhibitor leads to an enolate intermediate.

|T |

U

eP

O

N1° 'R ^H 9 B

C

« 5,« 10-Methylene-THF

lb * H N

X CH2 t-H CH, *N

H

¿ II

Iminium calion

E O'

N

®

=

O ^ '" N ^

\

B -H ' S — Enz^

dRibose — ® phosphate

I

HN

F

=B

=S— E n z '

dRibose — ©

F-dUMP

H,C

H

F

h

"^C H ,

H

|~2| Attack of the enolate ion on the iminium cation of N 5, N 10-methylene-THF forms a covalent bond between the inhibitor and the coenzyme (forming the alkylated enzyme).

H

| 3 | The next step in the normal mechanism would be an elimination reaction involving loss of a proton at the carbon a to the substrate’s carbonyl group, releasing the tetrahydrofolate coenzyme as a leaving group. In the case of the fluorinated inhibitor, this step is not possible because a fluorine atom takes the place of the hydrogen atom needed for removal in the elimination. The enzyme cannot undergo the elimination reaction necessary to free it from the tetrahydrofolate coenzyme. These blocked steps are marked by cross-outs. Neither can the subsequent hydride transfer occur from the coenzyme to the substrate, which would complete formation of the methyl group and allow release of the product from the enzyme thiol group. These blocked steps are shown in the shaded area. The enzyme’s activity is destroyed because it is irreversibly bonded to the inhibitor.

CH

Enolale

897

19.9 Summary of Important Reactions

19.9 Sum m ary o f Im p o rtan t Reactions 1. Claisen Condensation (Section 19.2)

O 2R

O OEt

O

R

(1) NaOEt

OEt

(2 ) H3 O+

R 2. Crossed Claisen Condensation (Section 19.2B): O

O

(1) C6 H5 CO 2 Et/NaOEt (2 ) H3 O+ O II

(1) EtOCOEt/NaOEt

R

(2 ) H3 O+

'

'

OEt

O R

O'"

"OEt

OEt O (1) HCO 2 Et/NaOEt

R

(2 ) H3 O+

OEt H

O (1) EtO 2 CCO 2 Et/NaOEt ^ (2 ) H 3 O+

R

OEt

O OEt 3. Aldol Reaction (Section 19.4) General Reaction

OH

O 2

H

OH

O

R

O H

h 2o >

-H 2O

R

H

R

R

Specific Example

O

O H

+

O H2 O

H

H

898

Chapter 19

Condensation and Conjugate Addition Reactions o f Carbonyl Compounds

4. Directed Aldol Reactions via Lithium Enolates (Section 19.5B) General Reaction

O

O- Li +

(1 )

O Jl

O

LDA

R H(R')

R

THF, -78°C (formation of the kinetic enolate)

'H(R')

(2) NH4CI

Specific Example

O O

O- Li +

LDA

(1)

THF, -7 8 °C

OH

H

O

(2) NH4CI

5. Conjugate Addition (Section 19.7) General Example

R-

O

R'

R-

(1) Nu : (or Nu— H)

R

(2) H — A

R' Nu

O R H

Nu: = CN ; an enolate (Michael addition); R"'MgBr Nu—H = 1° or 2° amines; an enamine Specific Example

CN

O

O

CN

"C6 H5

CeHs'

EtOH, HOAc

C6 H r

y

C6 H5

H Specific Example (Michael Addition) O

O , HO

C6H5

C6 H5 "

MeOH

6

. Mannich Reaction (Section 19.8): O

O

O

R' H— N

R-"O

+

H -^ H

+

R' R"

R

P ro b le m s

899

Key Terms and Concepts The key terms and concepts that are highlighted in bold, blue text within the chapter are defined in the glossary (at the back o f the book) and have hyperlinked definitions in the accompanying WileyPLUS course (www.wileyplus.com)

p ^ y 'j g

Problems Note to Instructors: Many of the homework problems are available for assignment via WileyPLUS, an online teaching and learning solution. C L A IS E N C O N D E N S A T IO N

19.23

R E A C T IO N S

Write a structural formula for the product from each of the following reactions. (a)

O NaOEt EtOH *

O (b)

O O NaOEt EtOH ’

O (c)

O

O NaOEt EtOH *

O (d)

O

O

NaOEt

O (e)

EtOH ’

O

O O

O O

NaOEt EtOH >

19.24

Show all steps in the following syntheses. You may use any other needed reagents but you should begin with the compound given. (a)

O

O OEt

(b)

O

O

OEt OEt O (c)

OEt O

OEt

900 19.25

Chapter 19

Condensation and Conjugate Addition Reactions o f Carbonyl Compounds

Provide the starting materials needed to synthesize each compound by acylation o f an enolate. O

(a)

r"

(b

O

O

(c)

O 1 19.26

J

CO2Et

O

O

Write structural formulas for both o f the possible products from the following Dieckmann condensation, and pre­ dict which one would likely predominate. O O

Il

NaOEt a EtOH, heat

O 19.27

When a Dieckmann condensation is attempted with diethyl succinate, the product obtained has the molecular for­ mula C12H16O6. What is the structure of this compound?

19.28

Show how this diketone could be prepared by a condensation reaction: O

O 19.29

In contrast to the reaction with dilute alkali (Section 18.6), when concentrated solutions of NaOH are used, acetoacetic esters undergo cleavage as shown below. O

O

O

O OEt

oh-

R

O-

O-

EtOH

R Provide a mechanistic explanation for this outcome. 19.30

Write a detailed mechanism for the following reaction. O

O OEt

O .OEt

Et

NaOEt

OEt

O Ethyl c ro to n a te

19.31

O

D ieth yl o xa la te

In the presence o f sodium ethoxide the following transformation occurs. Explain. O

O

O OEt

19.32

O

(1) NaOEt

O OEt

(2) HCl

Thymine is one o f the heterocyclic bases found in DNA. Starting with ethyl propanoate and using any other needed reagents, show how you might synthesize thymine. O 3

901

Problems

A L D O L R E A C T IO N S

19.33

Predict the products from each o f the following crossed aldol reactions. (a) -H

2

NaOH H2O

O (b)

H

NaOH

(c)

h 2o

T

O

O O

O H

NaOH

H

Y

NaOH h 2o

O

(e)

O

h 2o

O H

NaOH h 2o

19.34

What four b -hydroxy aldehydes would be formed by a crossed aldol reaction between the following compounds? O H

19.35

H

Show how each o f the following transformations could be accomplished. You may use any other required reagents. (a)

O

O

(e)

CN c h 3c n

CH3O'

o

(f)

H

(c)

O

(g) NO2

H

O

OH

902 19.36

Chapter 19

Condensation and Conjugate Addition Reactions o f Carbonyl Compounds

What starting materials are needed to synthesize each of the following compounds using an aldol reaction? (a)

O

(d)

O

(g) O

OH (e)

19.37

(h)

O

O

What reagents would you use to bring about each step of the following syntheses? (a)

O O

(b)

O

OH H^

O

°

H

H O O

(d)

903

Problems

19.38

The hydrogen atoms o f the g carbon o f crotonaldehyde are appreciably acidic (pKa = 20). O v ' xa A

H

Crotonaldehyde (a) Write resonance structures that w ill explain this fact. (b) Write a mechanism that accounts for the following reaction:

O

O O

H

H

H

base EtOH 87%

19.39

Provide a mechanism for the following reaction. O O

19.40

(1) NaOEt (2) H3O+

O

3

3

H'

OEt

When the aldol reaction of acetaldehyde is carried out in D2O, no deuterium is found in the methyl group o f unre­ acted aldehyde. However, in the aldol reaction o f acetone, deuterium is incorporated in the methyl group of the unreacted acetone. Explain this difference in behavior.

C O N J U G A T E A D D IT IO N

19.41

O

R E A C T IO N S

Write mechanisms that account for the products of the following reactions: (a)

O

O

O O

O

OEt OEt

EtO

(b)

O

O OMe

CH3NH2

MeO

O N

OMe

base

I

CH3

O OEt

EtONa ^ (-EtOH)

O OEt

Et

O OEt

904 19.42

Chapter 19

Condensation and Conjugate Addition Reactions o f Carbonyl Compounds

Condensations in which the active hydrogen compound is a b-keto ester or a b-diketone often yield products that result from one molecule of aldehyde or ketone and two molecules of the active methylene component. For example, O

O

O

x

O

base

R^"" R'

O R

R'

Suggest a reasonable mechanism that accounts for the formation o f these products. 19.43

The following reaction illustrates the Robinson annulation reaction (Section 19.7A). Provide a mechanism. O

19.44

O

What is the structure of the cyclic compound that forms after the Michael addition o f 1 to 2 in the presence of sodium ethoxide? O

O O

OEt

Et

NaOEt

O

G E N ER A L PRO BLEM S

19.45

Synthesize each compound starting from cyclopentanone. (a)

(b)

O

O 19.46

Provide a mechanism for the following reaction. O O NaOH H2O

3

19.47

Predict the products o f the following reactions (a)

(1) NaOEt, EtOH

O

O

O

(2 )

O

(b)

H O

(c)

O

O

(3) H3 O+

x

O

NaOEt, EtOH H3 O+ NaOH CH 3 CH2Br

O (d)

(1) LDA

(1) (2 ) (3) (4)

(1) NaOEt, EtOH

O

O

(2 )

O

(2 ) C I ^ I O ' ' (3) H3Ü+

O

905

P ro b le m s

19.48

Predict the products from the following reactions. (a)

(c)

O

O KOH H2O/EtOH

(1) I2 (excess), NaOH, H2O (2) HO+ O (d)

O KOH H2O/EtOH

2

O 19.49

(1) LDA (1.1 equiv) (2) O

O

H (3) H3O+

O

The mandibular glands of queen bees secrete a fluid that contains a remarkable compound known as “ queen sub­ stance.” When even an exceedingly small amount o f the queen substance is transferred to worker bees, it inhibits the development of their ovaries and prevents the workers from bearing new queens. Queen substance, a monocarboxylic acid with the molecular formula C10H16O3, has been synthesized by the following route:

~ , Cycloheptanone

(1) CH3MgI _ , j U r j+— > A (C8H16O) (2) H3O+

HA, heat , ----------- > B (C8H14)

(1) O. (2) Me2S O

C (C8H14O2)

HO pyridine

O OH

queen substance (C10H16O3)

On catalytic hydrogenation, queen substance yields compound D, which, on treatment with iodine in sodium hydrox­ ide and subsequent acidification, yields a dicarboxylic acid E; that is, ^ , H2 ^ ^ x (1) I2 in aq. NaOH . Queen substance - p - : D (C1oH18O3) -2 ---O ---------- * E (CgH16O4) Provide structures for the queen substance and compounds A -E . 19.50

(+)-Fenchone is a terpenoid that can be isolated from fennel oil. (±)-Fenchone has been synthesized through the following route. Supply the missing intermediates and reagents. CO2Me (1) (b) (2) (c)

+ (a)

CO2Me CO2Me CO2Me CO2H

mixture of (e) and (f) "ihr

CO2Me

CO2Me

(i) O

CO2Me O

CO2Me

CO2Me

OH

HA heat

... (j)

(k),

906 19.51

Chapter 19

Condensation and Conjugate Addition Reactions o f Carbonyl Compounds

Outline a racemic synthesis o f the analgesic Darvon (below) starting with ethyl phenyl ketone. O

CH^

19.52

Show how dimedone can be synthesized from malonic ester and 4-methyl-3-penten-2-one (mesityl oxide) under basic conditions. OH

19.53

Write the mechanistic steps in the cyclization of ethyl phenylacetoacetate (ethyl 3-oxo-4-phenylbutanoate) in con­ centrated sulfuric acid to form naphthoresorcinol (1,3-naphthalenediol).

19.54

When an aldehyde or a ketone is condensed with ethyl a-chloroacetate in the presence of sodium ethoxide, the product is an a,b-epoxy ester called a glycidic ester. The synthesis is called the Darzens condensation. O

O

R'

O

A glycidic ester (a) Outline a reasonable mechanism for the Darzens condensation. (b) Hydrolysis of the epoxy ester leads to an

epoxy acid that, on heating with pyridine, furnishes an aldehyde. What is happening here? R'

(c) Starting with b-ionone (Review Problem 19.13), show how you might synthesize the following aldehyde. (This

aldehyde is an intermediate in an industrial synthesis of vitamin A.) H

19.55

The Perkin condensation is an aldol-type condensation in which an aromatic aldehyde (ArCHO) reacts with a carboxylic acid anhydride, (RCH2CO)2O, to give an a,b-unsaturated acid (ArCH= CRCO2H). The catalyst that is usually employed is the potassium salt o f the carboxylic acid (RCH2CO2K). (a) Outline the Perkin condensation that takes place when benzaldehyde reacts with propanoic anhydride in the presence of potassium propanoate. (b) How would you use a Perkin condensation to prepare p-chlorocinnamic acid, p-C!C6H4CH= CHCO2H?

SPECTROSCOPY

19.56

(a) Infrared spectroscopy provides an easy method for deciding whether the product obtained from the addition of

a Grignard reagent to an a,b-unsaturated ketone is the simple addition product or the conjugate addition prod­ uct. Explain. (What peak or peaks would you look for?)

907

Challenge Problems (b) How might you follow the rate o f the following reaction using U V spectroscopy?

O

O + CHNH 3

2

h 2o

c h 3n h

19.57

Allowing acetone to react with 2 molar equivalents o f benzaldehyde in the presence of KOH in ethanol leads to the formation of compound X. The 13C NMR spectrum of X is given in Fig. 19.1. Propose a structure for compound X.

_l__ l__ i__ l__ I__ l__ i__ l__ I__ l__ i__ l__ I__ l__ i__ l__ I__ l__ i__ l__ I__ l__ i__ l__ I__ l__ i__ l__ I__ l__ i__ l__ I__ l__ i__ l__ I__ l__ i__ l__ I__ l__ i__ l__ L_

220

200

180

160

140

120

100

80

60

40

20

Sc (ppm)

Figure 19.1 The broadband proton-decoupled 13C NMR spectrum of compound X, Problem 19.57. Information from the DEPT 13C NMR spectra is given above the peaks.

Challenge Problems

0

908 19.59

Chapter 19

Condensation and Conjugate Addition Reactions o f Carbonyl Compounds

(a) Deduce the structure of product A, which is highly symmetrical:

O 2

A

ethanol

The following are selected spectral data for A: M S (m/z): 220 (M-+) IR (cm-1 ): 2930, 2860, 1715 1H N M R (d): 1.25 (m), 1.29 (m), 1.76 (m), 1.77 (m), 2.14 (s), and 2.22 (t); (area ratios 2:1:2:1:2:2, respectively) 13C N M R (d): 23 (CH2), 26 (CH2), 27 (CH2), 29 (C), 39 (CH), 41 (CH2), 46 (CH2), 208 (C) (b) Write a mechanism that explains the formation of A. 19.60

Write the structures of the three products involved in this reaction sequence: HO

COOH ^

Rr CHCl,

B

hcho

^

C

(CH3)2NH in EtOH/HOAc

Raney Ni/H2 water

D

OH Spectral data for B: M S (m/z): 314, 312, 310 (relative abundance 1:2:1) 1H N M R (d): only 6.80 (s) after treatment with D2O Data for C: M S (m/z): 371, 369, 367 (relative abundance 1:2:1) 1H N M R (d): 2.48 (s) and 4.99 (s) in area ratio 3:1; broad singlets at 5.5 and 11 disappeared after treatment with D2 O. Data for D: M S (m/z): 369 (M- + — CH3) [when studied as its tris(trimethylsilyl) derivative] 1H N M R (d): 2.16 (s) and 7.18 (s) in area ratio 3:2; broad singlets at 5.4 and 11 disappeared after treatment with D2 O.

Learning Group Problems Lycopodine is a naturally occurring amine. As such, it belongs to the family of natural products called alkaloids. Its synthesis (J. Am. Chem. Soc. 1968, 90, 1647-1648) was accomplished by one of the great synthetic organic chemists of our time, Gilbert Stork (Columbia University). Write a detailed mechanism for all the steps that occur when 2 reacts with ethyl acetoacetate in the presence of ethoxide ion. Note that a necessary part of the mechanism will be a base-catalyzed isomerization (via a conjugated enolate) of the alkene in 2 to form the corresponding a,bunsaturated ester.

3 L yc o p o d in e

909

S u m m a ry o f M ec h a n is m s

2

Steroids are an extremely important class o f natural and pharmaceutical compounds. Synthetic efforts directed toward steroids have been underway for many years and continue to be an area of important research. The synthesis of cholesterol by R. B. Woodward (Harvard University, recipient of the Nobel Prize in Chemistry for 1965) and co­ workers represents a paramount accomplishment in steroid synthesis, and it is rich with examples of carbonyl chem­ istry and other reactions we have studied. Selected reactions from Woodward’s cholesterol synthesis and the questions for this Learning Group Problem are shown in the WileyPlus materials for this chapter. Access those materials online to complete this problem.

.

Sum m ary o f M echanism s Enolate Reactions with Carbonyl Electrophiles

1

R

R' /

■ A r— NO2

A r— NH2

We studied ring nitration in Chapter 15 and saw there that it is applicable to a wide vari­ ety o f aromatic compounds. Reduction of the nitro group can also be carried out in a num­ ber o f ways. The most frequently used methods employ catalytic hydrogenation, or treatment o f the nitro compound with acid and iron. Zinc, tin, or a metal salt such as SnCl2 can also be used. Overall, this is a 6e~ reduction. General Reaction A r - NO2

A r - N *

Specific Example

NO,

NH„ (1) Fe, HCl (2) OH97%

20.4C Preparation of Primary, Secondary, and Tertiary Amines through Reductive Amination Aldehydes and ketones can be converted to amines through catalytic or chemical reduction in the presence o f ammonia or an amine. Primary, secondary, and tertiary amines can be prepared this way: nh 3 [H]

H.

O , A

R H R'

R"NH2

.

Aldehyde or ketone

[H]

[H]

„R"

K H R' R"

R"R"'NH

N

N

1° Amine

2° Amine

„R"

R H R'

3° Amine

This process, called reductive amination o f the aldehyde or ketone (or reductive alkyla­ tion o f the amine), appears to proceed through the following general mechanism (illustrated with a 1° amine).

927

928

Chapter 20 Amines

A MECHANISM FOR THE REACTION R e d u c tiv e A m in a tio n

R"

¿ 0

H 2 N — R"

Aldehyde or ketone

1° Amine

.

( - H 2 O)

N

'

R

steps

R

1

1

R'

NHR"

y

HO

two

^

Hemiaminal

Imine

[H] NHR"

W

R' H

2° Amine

W h e n a m m o n ia o r a p r im a r y a m in e is u se d , th e re a re t w o p o s s ib le p a th w a y s to th e p r o d ­

H e lp f u l H i n t We saw the importance o f imines in "The Chemistry of . . . Pyridoxal Phosphate" (vitamin B^) in WileyPLUS fo r Chapter 16 (Section 16.8).

u c t: v ia a n a m in o a lc o h o l th a t is s im ila r to a h e m ia c e ta l a n d is c a lle d a

hem iam inal o r v ia

an im in e (S e c tio n 1 6 .8 A ). W h e n s e c o n d a ry a m in e s a re u se d , an im in e c a n n o t fo r m , a n d , th e re fo re , th e p a th w a y is th ro u g h th e h e m ia m in a l o r th ro u g h an im in iu m io n : R"

R" N

R

R'

Iminium ion

T h e r e d u c in g a g e n ts e m p lo y e d in c lu d e h y d ro g e n a n d a c a ta ly s t (s u c h as n ic k e l) o r N a B H 3C N o r L iB H 3C N ( s o d iu m o r lit h iu m c y a n o b o r o h y d rid e ) . T h e la tte r tw o re d u c in g a g e n ts a re s im ila r to N a B H 4 a n d a re e s p e c ia lly e ffe c tiv e in re d u c tiv e a m in a tio n s . T h re e spe­ c if ic e x a m p le s o f re d u c tiv e a m in a tio n f o llo w : O

H

n h 3, h 2,

Ni

n h

2

90 atm 40-70°C

Benzylamine (89%)

Benzaldehyde O

r v

H

N

( 1 ) c h 3 c h ,n h 2

f

'N

|f

\

(2) LiBH3CN

y

Benzaldehyde

N-Benzylethanamine (89%)

.

( 1 ) (CH3 )2 NH O

(2) NaBH3CN

c h

\l

3

/

\ CH,

Cyclohexanone

N,N-Dimethylcyclohexanamine (52-54%)

/

2 0.4 Preparation of Amines

929

S olved P roblem 20.2 Outlined below is a synthesis o f the stimulant amphetamine. Provide the intermediates A and B. H 2N / =

\

\

Br

NaCN >

/

(1) CH 3 U m u ~ * (2) H2O

A

(1) NH 3 B B

(2) LiBH3CN A m p h e ta m in e

ANSWER O CN A =

B =

Show how you might prepare each o f the following amines through reductive amination:

Review Problem 20.5

I

H 1 S o lv e d P ro b le m 2 0 .3 1 Reductive amination o f a ketone is almost always a better method for the synthesis of an amine o f the type r' than treatment o f an alkyl halide with ammonia. Explain why this is true. R ^^N H 2 STRATEGY A ND ANSWER Consider the structure o f the required alkyl halide. Reaction o f a secondary halide

with ammonia would inevitably be accompanied by considerable elimination, thereby d ecreasing the yield o f the secondary amine. Multiple N-alkylations may also occur.

20.4D Preparation of Primary, Secondary, or Tertiary Amines through Reduction of Nitriles, Oximes, and Amides Nitriles, oximes, and amides can be reduced to amines. Reduction of a nitrile or an oxime yields a primary amine; reduction of an amide can yield a primary, secondary, or tertiary amine:

R— C = N

[H]

N itrile RCH =

NOH

r c h 2n h

N itrile s can b e p re p a re d from alkyl h alid e s a n d CN~ (S e c tio n 17.3) o r fro m a ld e h y d e s a n d k e to n e s as c y a n o h y d rin s (S e c tio n 16.9).

2

1° A m in e

[H]

O xim e

r c h 2n h

2

O x im e s can be p re p a re d fro m a ld e h y d e s a n d k e to n e s (S e c tio n 1 6 .8B).

1° A m in e

O R — C — N — R' R" A m id e

[H]

R C H 2 N — R' R" 3° A m in e

{

A m id e s ca n b e p re p a re d fro m a c id c h lo rid e s , a c id a n h y d rid e s , a n d e s te rs (S e c tio n 17.8).

930

Chapter 20

Amines

(In the last example, i f R' = H and R" = H, the product is a 1° amine; i f only R' = H, the product is a 2° amine.) A ll of these reductions can be carried out with hydrogen and a catalyst or with LiAlH4. Oximes are also conveniently reduced with sodium in ethanol. Specific examples follow: /

OH

Na EtOH

NH„

50-60% NH,

[T

CN

2 H2, Raney Ni 140°C *

2-Phenylethanenitrile (phenylacetonitrile)

1 2-Phenylethanamine (71%)

CH3 I 3

CH3

Kl

3

LiAlH4 O

(2) H,O

W-Methylacetanilide

W-Ethyl-W-methylaniline

Reduction o f an amide is the last step in a useful procedure for monoalkylation of an amine. The process begins with acylation o f the amine using an acyl chloride or acid anhy­ dride; then the amide is reduced with lithium aluminum hydride. For example, O

O

IJ f

1

"N H ,

x^cl base

ll

1 J

N

(1)

H

(2)

LiAlH4 H,O

Benzylamine

r

S o lv e d P ro b le m 2 0 .4

il

T

1 J

N

I H

Benzylethylamine

1

Show how you might syi thesize 2-propanamine from a three-carbon starting material that is a ketone, aldehyde, nitrile, or amide. STRATEGY AND ANSW ER We begin by recognizing that 2-propanamine has a primary amine group bonded to a

secondary carbon. Neithe r a three-carbon nitrile nor a three-carbon amide can lead to this structural unit from a C3 starting material. An oxi me can lead to the proper structure, but we must start with a three-carbon ketone rather than an aldehyde. Theref re, we choose propanone as our starting material, convert it to an oxime, and then reduce the oxime to an amine.

R e v ie w P r o b le m 2 0 .6

Show how you might utilize the reduction of an amide, oxime, or nitrile to carry out each o f the following transformations: (a)

O

Br

NH2

931

20.4 Preparation of Amines (c)

(d)

O

O

nh

2

OH

20.4E Preparation of Primary Amines through the Hofmann and Curtius Rearrangements H o fm a n n R e a rra n g e m e n t Amides with no substituent on the nitrogen react with solu­ tions o f bromine or chlorine in sodium hydroxide to yield amines through a reaction known as the Hofmann rearrangement or Hofmann degradation:

O II X . R

H2O 4 NaO H

B r,

R — N H

2 N a Br

Na2C O 3

2 H 2O

'N H c

From this equation we can see that the carbonyl carbon atom o f the amide is lost (as CO32~) and that the R group o f the amide becomes attached to the nitrogen of the amine. Primary amines made this way are not contaminated by 2° or 3° amines. The mechanism for this interesting reaction is shown in the following scheme. In the first two steps the amide undergoes a base-promoted bromination, in a manner analogous to the base-promoted halogenation of a ketone that we studied in Section 18.3B. (The electron-with­ drawing acyl group o f the amide makes the amido hydrogens much more acidic than those of an amine.) The N-bromo amide then reacts with hydroxide ion to produce an anion, which spontaneously rearranges with the loss of a bromide ion to produce an isocyanate (Section 17.9A). In the rearrangement the R— group migrates with its electrons from the acyl carbon to the nitrogen atom at the same time the bromide ion departs. The isocyanate that forms in the mixture is quickly hydrolyzed by the aqueous base to a carbamate ion, which undergoes spontaneous decarboxylation resulting in the formation of the amine.

A MECHANISM FOR THE REACTION T h e H o fm a n n R e a r r a n g e m e n t

•o

•'o '-

R

/C ^ .. . N -rH

: OH

/

C\

W

l* J

H

-O ' ■■ N

Br— Br Br Br

/

Cv

R

'N

I

I

H

H

Br

B r-

H 2O

Amide

W-Bromo amide

Base-promoted W-bromination of the amide occurs.

‘O' R

'O '

N

/C O -

Br OH

R

+

O h

N-Bromo amide >■ Base removes a proton from the nitrogen to give a bromo amide anion.

7 2o

Br

( - Br- )

r

—

n

=

c

=

o

:

Isocyanate The R— group migrates to the nitrogen as a bromide ion departs. This produces an isocyanate.

(continued on the next page)

932

Chapter 20

Amines

•-O -

: OH

)p l^ > C%

H OH

R— N ^"C % O:

R — N = fC = O :

Isocyanate

R— N

. O'

R— NH2

, H — OH

Transfer of a proton leads to a carbamate

!..

CO2

OH -

Amine

=O H

HCO

R — N ^ C^ O -

The isocyanate undergoes hydrolysis and decarboxylation to produce the amine.

A n e x a m in a tio n o f th e f ir s t tw o steps o f th is m e c h a n is m s h o w s th a t, in it ia lly , tw o h y d r o ­ g e n a to m s m u s t b e p re s e n t o n th e n itr o g e n o f th e a m id e f o r th e r e a c tio n to o c c u r. C o n s e q u e n tly , th e H o fm a n n re a rra n g e m e n t is lim it e d to a m id e s o f th e ty p e R C O N H 2. S tu d ie s o f th e H o fm a n n re a rra n g e m e n t o f o p t ic a lly a c tiv e a m id e s in w h ic h th e c h ir a l it y c e n te r is d ir e c tly a tta c h e d to th e c a r b o n y l g ro u p h a v e s h o w n th a t th e se re a c ­ tio n s o c c u r w it h

retention o f configuration. T h u s , th e R g ro u p m ig ra te s to n itro g e n but w ith o u t inversion .

w it h its e le c tro n s ,

C u r tiu s

The

R e a rra n g e m e n t

C urtius rearrangem ent is a re a rra n g e m e n t th a t o c c u rs

w it h a c y l a z id e s . I t re s e m b le s th e H o fm a n n re a rra n g e m e n t in th a t a n R — g ro u p m ig ra te s f r o m th e a c y l c a rb o n to th e n itr o g e n a to m as th e le a v in g g ro u p d e p a rts . I n th is in s ta n c e th e le a v in g g ro u p is N 2 (th e b e s t o f a ll p o s s ib le le a v in g g ro u p s s in c e i t is h ig h ly s ta b le , is v i r ­ t u a lly n o n b a s ic , a n d b e in g a gas, re m o v e s it s e lf f r o m th e m e d iu m ) . A c y l a z id e s a re e a s ily p re p a re d b y a llo w in g a c y l c h lo rid e s to re a c t w it h s o d iu m a zid e . H e a tin g th e a c y l a z id e b rin g s a b o u t th e re a rra n g e m e n t; a fte rw a rd , a d d in g w a te r causes h y d r o ly s is a n d d e c a rb o x y la tio n o f th e is o c y a n a te : •'O'-

•‘ o -

h 2o

/ C R

\ . . C |:

T -N n C r ( NaCl)

r \ ^

Acyl chloride

n

n = n

-rN r (-N2)

Acyl azide

R-

N=

c

=

o

: '

Isocyanate

R — N H2

Amine

S o l v e d P r o b le m 2 0 .5 T h e re a c tio n s e q u e n ce b e lo w s h o w s h o w a m e th y l g ro u p o n a b e n z e n e r in g c a n b e re p la c e d b y an a m in o g ro u p . S u p p ly th e m is s in g re a g e n ts a n d in te rm e d ia te s . O

STRATEGY AND ANSWER A n a c id c h lo r id e re s u lts f r o m tre a tm e n t o f A w it h B . T h e re fo re , A is lik e ly to b e a c a r b o x y lic a c id , a c o n c lu s io n th a t is c o n s is te n t w it h th e o x id iz in g c o n d itio n s th a t le d to fo r m a tio n o f A f r o m m e th y lb e n z e n e ( to lu e n e ). B m u s t b e a re a g e n t th a t ca n le a d to a n a c id c h lo rid e . T h io n y l c h lo r id e o r P C l5 w o u ld s u ffic e . O v e ra ll, C , D , a n d E in v o lv e in tr o d u c tio n o f th e n itr o g e n a to m a n d lo s s o f th e c a r b o n y l c a rb o n . T h is se q u e n ce is c o n s is te n t w it h p re p a r a tio n o f a n a m id e f o llo w e d b y a H o fm a n n re a rra n g e m e n t.

(continued on the next page)

CO2

933

20.5 Reactions of Amines

Using a different method for each part, but taking care in each case to select a g o o d method, show how each o f the following transformations might be accomplished: (a)

(d)

C H 3O

nh

CH3

O 2N

o

(e)

(b)

R e v ie w P r o b le m 2 0 .7

2n

nh

CH3

2

2

C H 3O

(c) CH3 N - CH3

Cl

20.5 Reactions o f Am ines We have encountered a number o f important reactions o f amines in earlier sections. In Section 20.3 we saw reactions in which primary, secondary, and tertiary amines act as bases. In Section 20.4 we saw their reactions as n u cleo p h iles in a lkyla tio n reactions, and in Chapter 17 as n u cleo p h iles in a cylatio n reactions. In Chapter 15 we saw that an amino group on an aromatic ring acts as a powerful a ctiva tin g group and as an o rth o -p a ra director.

The feature of amines that underlies all o f these reactions and that forms a basis for our understanding o f most of the chemistry o f amines is the ability o f nitrogen to share an electron pair: A c id -B a s e R e ac tio n s

\ + N— H

=A -

An amine acting as a base A lk y la tio n N

\ + R — C H 2^ B r

— N — C H 2R

An amine acting as a nucleophile in an alkylation reaction

Br-

934

Chapter 20 Amines A c y la tio n

O V

+

l

II

O)

C

— N— C— R

"Cl

/

I H

R

XN

/

\Q Cl

An amine acting as a nucleophile in an acylation reaction In the p re c e d in g exam p les the a m in e acts as a n u c le o p h ile b y d on ating its electro n p a ir to an e le c tro p h ilic reag ent. In the fo llo w in g e x a m p le , resonance con tribu tio ns in v o lv in g the n itro g e n e le ctro n p a ir m a k e

carbon atom s n u c le o p h ilic :

E le c tro p h ilic A ro m a tic S u b s titu tio n H^

..

,H E

HA

a

9

e

E

The amino group acting as an activating group and as an ortho-para director in electrophilic aromatic substitution

R e v ie w P r o b le m 2 0 .8

R e v ie w the c h e m is try o f am in es g iv e n in e a rlie r sections and p ro v id e a sp ecific e x a m p le f o r each o f the p re v io u s ly illu s tra te d reactions.

20.5A Oxidation of Amines P rim a ry and secondary a lip h a tic am ines are subject to o x id a tio n , alth ou gh in m o s t instances u se fu l p roducts are n o t o btain ed. C o m p lic a te d side reactions o fte n o ccu r, causing the fo r­ m a tio n o f c o m p le x m ixtu res. T e rtia ry am ines can b e o x id iz e d c le a n ly to te rtia ry a m in e o xid es. T h is tra n s fo rm a tio n can b e b ro u g h t ab o u t b y using h yd ro g e n p e ro x id e o r a p e ro x y ac id : O r 3n

h 2 ° 2 or r C o o h

>

R 3N -

O -

A tertiary amine oxide T e rtia ry a m in e o xid es und ergo a u se fu l e lim in a tio n re a c tio n to b e discussed in S e c tio n 2 0 .1 2 B . A ry la m in e s are v e ry e a s ily o x id iz e d b y a v a rie ty o f reagents, in c lu d in g the o x y g e n in air. O x id a tio n is n o t co n fin e d to the a m in o gro up b u t also occurs in the rin g . (T h e am in o gro up throu gh its e le ctro n -d o n a tin g a b ility m a k e s the rin g electro n ric h a n d hen ce espe­ c ia lly susceptible to o x id a tio n .) T h e o x id a tio n o f o th e r fu n c tio n a l groups on an a ro m a tic rin g cannot u su a lly b e a c c o m p lis h e d w h e n an a m in o gro up is present on the rin g , because o x id a tio n o f the rin g takes p la c e first.

20.6 Reactions of Amines with Nitrous Acid

V x

935

2 0 .6 Reactions o f Amines with Nitrous Acid Nitrous acid (HONO) is a weak, unstable acid. It is always prepared in situ, usually by treat­ ing sodium nitrite (NaNO2) with an aqueous solution of a strong acid: HCl(aq) + NaNO2(aq) ----- : H2SO4 + 2 NaNO2(aq) ----- :

HONO(aq) + NaCl(aq) 2 HONO(aq) + Na2SO4(aq)

Nitrous acid reacts with all classes o f amines. The products that we obtain from these reactions depend on whether the amine is primary, secondary, or tertiary and whether the amine is aliphatic or aromatic.

20.6A Reactions of Primary Aliphatic Amines with Nitrous Acid Primary aliphatic amines react with nitrous acid through a reaction called diazotization to yield highly unstable aliphatic diazonium salts. Even at low temperatures, aliphatic dia­ zonium salts decompose spontaneously by losing nitrogen to form carbocations. The car­ bocations go on to produce mixtures of alkenes, alcohols, and alkyl halides by removal of a proton, reaction with H2O, and reaction with X - : G e n e ra l R e ac tio n

R— NH2 + NaNO2 + 2 HX 1 ° A lip h a tic a m in e

(^ ° N O '^ H2 O

[

N # N : X- ]

+ NaX + 2 H2O

A lip h a tic d ia zo n iu m salt (h ig h ly u n s tab le ) -N 2 (j.e .,:N # N ;)

X-

A lk e n e s , a lc o h o ls , alkyl h alid e s

Diazotizations of primary aliphatic amines are of little synthetic importance because they yield such a complex mixture o f products. Diazotizations o f primary aliphatic amines are used in some analytical procedures, however, because the evolution of nitrogen is quanti­ tative. They can also be used to generate and thus study the behavior of carbocations in water, acetic acid, and other solvents.

20.6B Reactions of Primary Arylamines with Nitrous Acid The most important reaction of amines with nitrous acid, by far, is the reaction o f primary arylamines. We shall see why in Section 20.7. Primary arylamines react with nitrous acid to give arenediazonium salts. Even though arenediazonium salts are unstable, they are still far more stable than aliphatic diazonium salts; they do not decompose at an appreciable rate in solution when the temperature o f the reaction mixture is kept below 5°C: Ar— NH2

+

P rim a ry a ry la m in e

NaNO2 +

2 H X ----- > Ar — N # N: X +

NaX +

2 H2O

A re n e d ia zo n iu m sa lt (s ta b le if kept b e lo w 5°C )

Diazotization o f a primary amine takes place through a series of steps. In the presence of strong acid, nitrous acid dissociates to produce +NO ions. These ions then react with the nitrogen of the amine to form an unstable N-nitrosoaminium ion as an intermediate. This intermediate then loses a proton to form an N-nitrosoamine, which, in turn, tautomerizes to a diazohydroxide in a reaction that is similar to keto-enol tautomerization. Then, in the presence o f acid, the diazohydroxide loses water to form the diazonium ion.

H e lp f u l H i n t Primary arylamines can be con­ verted to aryl halides, nitriles, and phenols via aryl diazonium ions (Section 20.7).

936

Chapter 20 Amines

A MECHANISM FOR THE REACTION D ia z o tiz a tio n

HONO

A

H 3 O+

h 2o H ^ :O H

H Ar — N • +

+N = O

C l+ A r— N+

h 2o

n o

2 H 2O

2

■■ N = O

H3 O

Ar — N

N =O ;

I H

H

1° Arylamine (or alkylamine)

H

t

N = O.

+ HA

H— —A A H

H

W-Nitrosoamine

W-Nitrosoaminium ion HA

Ar - N

+ ha Ar — N

N = O

N— OH

Diazohydroxide

.r\ A r— N

N — O H,

HA

A- ^ •A r— N

N + H 2O

Diazonium ion

Diazotization reactions of primary arylamines are of considerable synthetic importance because the diazonium group, — N # N: can be replaced by a variety of other functional groups. We shall examine these reactions in Section 20.7.

THE CHEMISTRY OF N -N itro s o a m in e s

N-Nitrosoamines are very powerful carcinogens which sci­ entists fear may be present in many foods, especially in cooked meats that have been cured with sodium nitrite. Sodium nitrite is added to many meats (e.g., bacon, ham, frankfurters, sausages, and corned beef) to inhibit the growth of Clostridium botulinum (the bacterium that pro­ duces botulinus toxin) and to keep red meats from turning brown. (Food poisoning by botulinus toxin is often fatal.) In the presence of acid or under the influence of heat, sodium nitrite reacts with amines always present in the meat to pro­ duce N-nitrosoamines. Cooked bacon, for example, has been shown to contain N-nitrosodimethylamine and Nnitrosopyrrolidine. There is also concern that nitrites from food may produce nitrosoamines when they react with amines in the presence of the acid found in the stomach. In 1976, the FDA reduced the permissible amount of nitrite allowed in cured meats from 200 parts per million (ppm) to 50-125 ppm. Nitrites (and nitrates that can be converted to nitrites by bacteria) also occur naturally in many foods.

Cigarette smoke is known to contain N-nitrosodimethylamine. Someone smoking a pack of cigarettes a day inhales about 0.8 mg of N-nitrosodimethylamine, and even more has been shown to be present in the sidestream smoke.

A processed fo o d preserved w ith sodium n itrite .

0

20.7 Replacement Reactions of Arenediazonium Salts

937

20.6C Reactions of Secondary Amines with Nitrous Acid S e c o n d a ry a m in e s — b o th a r y l a n d a lk y l— re a c t w it h n itro u s a c id to y ie ld N - n itro s o a m in e s .

N-Nitrosoamines u s u a lly se p a ra te f r o m th e r e a c tio n m ix t u r e as o ily y e llo w liq u id s : S p e c ific E xa m p les (C H ^ N H

-

HCl

N aN O ,

(HONO)

>

H2 O

(C H ^ N

N = O

W-Nitrosodimethylamine (a yellow oil)

Dimethylamine

N O

H N

HCl

\

N aN O ,

(HONO) H2O

'

\

'C H 3

W-Methylaniline

N

/

\ CH

W-Nitroso-W-methylaniline (87-93%) (a yellow oil)

20.6D Reactions of Tertiary Amines with Nitrous Acid W h e n a t e r tia r y a lip h a tic a m in e is m ix e d w it h n itro u s a c id , a n e q u ilib r iu m is e s ta b lis h e d a m o n g th e t e r tia r y a m in e , its s a lt, a n d an N - n itr o s o a m m o n iu m c o m p o u n d : 2 R3 N *

+

NaNO ,

HX

Tertiary aliphatic

R3 N h X

R 3N — N =

O X~

N-Nitrosoammonium compound

Amine salt

A lth o u g h A - n itr o s o a m m o n iu m c o m p o u n d s a re s ta b le a t lo w te m p e ra tu re s , a t h ig h e r te m ­ p e ra tu re s a n d in a q u e o u s a c id th e y d e c o m p o s e to p ro d u c e a ld e h y d e s o r k e to n e s . T h e s e re a c ­ tio n s a re o f l i t t le s y n th e tic im p o rta n c e , h o w e v e r. T e r tia r y a ry la m in e s re a c t w it h n itro u s a c id to f o r m

C - n itro s o a ro m a tic c o m p o u n d s .

N it r o s a t io n ta ke s p la c e a lm o s t e x c lu s iv e ly a t th e p a ra p o s itio n i f i t is o p e n a n d , i f n o t, at th e o r th o p o s itio n . T h e r e a c tio n (see R e v ie w P r o b le m 2 0 .9 ) is a n o th e r e x a m p le o f e le c t r o p h ilic a ro m a tic s u b s titu tio n .

S p ec ific E xa m p le CH3

CH3

\ 3-

\ 3/ N c h 3

\

/

HCl

NaN O ,

N

N = O

CH3

p-Nitroso-W,W-dimethylaniline (80-90%)

P a r a -n itr o s a tio n o f

A ,A - d im e t h y la n ilin e ( C - n itr o s a tio n ) is b e lie v e d to ta k e p la c e th ro u g h

an e le c t r o p h ilic a tta c k b y N O io n s .

R e v ie w P r o b le m 2 0 .9

(a) S h o w h o w N O io n s m ig h t b e fo r m e d in a n a q u e o u s

(b) W r it e a m e c h a n is m f o r p - n it r o s a t io n o f A ,A - d im e th y la n i(c) T e r tia r y a ro m a tic a m in e s a n d p h e n o ls u n d e rg o C - n itro s a tio n re a c tio n s , w h e re a s

s o lu tio n o f N a N O 2 a n d H C l. lin e .

m o s t o th e r b e n z e n e d e riv a tiv e s d o n o t. H o w ca n y o u a c c o u n t f o r th is d iffe re n c e ?

20.7 Replacem ent Reactions o f Arenediazonium Salts •

A re n e d ia z o n iu m salts are h ig h ly u s e fu l in te rm e d ia te s in th e syn th e s is o f a ro m a tic c o m p o u n d s , be ca use th e d ia z o n iu m g ro u p ca n b e re p la c e d b y a n y o n e o f a n u m b e r o f o th e r a to m s o r g ro u p s , in c lu d in g — F, — C l, — Br, — I, — C N , — O H , a n d — H.

D ia z o n iu m s a lts a re a lm o s t a lw a y s p re p a re d b y d ia z o tiz in g p r im a r y a ro m a tic a m in e s . P r im a r y a ry la m in e s ca n b e s y n th e s iz e d th ro u g h r e d u c tio n o f n it r o c o m p o u n d s th a t a re re a d ­ i l y a v a ila b le th ro u g h d ir e c t n itr a t io n re a c tio n s .

938

Chapter 20 Amines

20.7A Syntheses Using D ia zo n iu m Salts M o s t a re n e d ia z o n iu m sa lts a re u n s ta b le a t te m p e ra tu re s a b o v e 5 - 1 0 ° C , a n d m a n y e x p lo d e w h e n d ry . F o rtu n a te ly , h o w e v e r, m o s t o f th e re p la c e m e n t re a c tio n s o f d ia z o n iu m s a lts d o n o t r e q u ir e th e ir is o la tio n . W e s im p ly a d d a n o th e r re a g e n t ( C u C l, C u B r, K I, e tc .) to th e m i x ­ tu re , g e n tly w a r m th e s o lu tio n , a n d th e re p la c e m e n t (a c c o m p a n ie d b y th e e v o lu tio n o f n it r o ­ g e n ) ta ke s p la c e : Cu 2O, Cu2+,

H,O

CuCl

Ar

CuBr

Ar

NH,

0 -5 °C *

r

Ar

2

Arenediazonium salt

Cl

A r— Br

CuCN

HONO

A r— O H

KI

CN

A r— I

( 1 ) HBF 4 (2 ) heat

A r— F

H3 PO,, h,o

Ar

H

O n ly in th e re p la c e m e n t o f th e d ia z o n iu m g ro u p b y — F n e e d w e is o la te a d ia z o n iu m s a lt. W e d o th is b y a d d in g H B F 4 to th e m ix tu r e , c a u s in g th e s p a rin g ly s o lu b le a n d re a s o n ­ a b ly s ta b le a re n e d ia z o n iu m flu o ro b o r a te , A r N 2 + B F 4- , to p re c ip ita te .

20.7B The Sandmeyer Reaction: Replacement of the Diazonium Group by —Cl, —Br, or —CN A r e n e d ia z o n iu m sa lts re a c t w it h c u p ro u s c h lo rid e , c u p ro u s b ro m id e , a n d c u p ro u s c y a n id e to g iv e p ro d u c ts in w h ic h th e d ia z o n iu m g ro u p h a s b e e n re p la c e d b y — C l, — B r, a n d — C N , re s p e c tiv e ly . T h e s e re a c tio n s a re k n o w n g e n e r a lly as

Sandm eyer reactions. S e v e ra l

s p e c ific e x a m p le s f o llo w . T h e m e c h a n is m s o f th e se re p la c e m e n t re a c tio n s a re n o t f u l ly u n d e rs to o d ; th e re a c tio n s a p p e a r to b e r a d ic a l in n a tu re , n o t io n ic . CH,

CH

CH3 N , C l-

NH,

Cl

HCl, NaNO,

CuCl

H2O (0-5°C)

15-60°C

N,

o-Chlorotoluene (74-79% overall)

o-Toluidine

N , B r-

NH,

HBr, NaNO, H2O (0-10°C) Cl

Br

CuBr 10 0

N,

«

Cl

Cl

m-Chloroaniline

m-Bromochlorobenzene (70% overall)

NO2

NO

n o

N , C l-

NH,

HCl, NaNO,

2 CN

CuCN N,

90-10CPC (room temp.)

o-Nitroaniline

o-Nitrobenzonitrile (65% overall)

9 39

2 0 .7 R e p la c e m e n t R e a c tio n s o f A r e n e d ia z o n iu m Salts

20.7C Replacement by —I A r e n e d ia z o n iu m sa lts re a c t w it h p o ta s s iu m io d id e to g iv e p ro d u c ts in w h ic h th e d ia z o n iu m g ro u p has b e e n re p la c e d b y — I. A n e x a m p le is th e s y n th e s is o f p -io d o n itr o b e n z e n e : NO,

NO,

NO,

H2 SO4, NaNO 2

Kl N 2

H,O 0-5°C I

NH2

p -Iodonitrobenzene

p-Nitroaniline

(81% overall)

20.7D Replacement by —F T h e d ia z o n iu m g ro u p ca n b e re p la c e d b y flu o r in e b y tr e a tin g th e d ia z o n iu m s a lt w it h flu o r o b o r ic a c id ( H B F 4). T h e d ia z o n iu m flu o ro b o r a te th a t p re c ip ita te s is is o la te d , d rie d , a n d h e a te d u n t il d e c o m p o s itio n o c c u rs . A n a r y l f lu o r id e is p ro d u c e d : CH

CH

CH

(1) HONO, H + N 2

heat

(,) HBF 4 NH2

n

m -Toluidine

2+

b f

BF3

F

4-

m -Toluenediazonium

m -Fluorotoluene

fluoroborate (79%)

(69%)

20.7E Replacement by —OH T h e d ia z o n iu m g ro u p c a n b e re p la c e d b y a h y d r o x y l g ro u p b y a d d in g c u p ro u s o x id e to a d ilu t e s o lu tio n o f th e d ia z o n iu m s a lt c o n ta in in g a la rg e exce ss o f c u p r ic n itra te :

CH

n

2+

h s o

4-

Cu2O Cu2+, h 2o

p-Toluenediazonium hydrogen sulfate

CH

OH

p -Cresol (93%)

T h is v a r ia tio n o f th e S a n d m e y e r r e a c tio n (d e v e lo p e d b y T . C o h e n , U n iv e r s it y o f P itts b u rg h ) is a m u c h s im p le r a n d s a fe r p ro c e d u re th a n an o ld e r m e th o d f o r p h e n o l p re p a ra tio n , w h ic h r e q u ir e d h e a tin g th e d ia z o n iu m s a lt w it h c o n c e n tra te d a q u e o u s a c id .

I n th e p re c e d in g e x a m p le s o f d ia z o n iu m re a c tio n s , w e h a v e illu s tr a te d syn th e se s b e g in n in g w it h th e c o m p o u n d s ( a ) - ( d ) h e re . S h o w h o w y o u m ig h t p re p a re e a ch o f th e f o llo w in g c o m ­ p o u n d s f r o m b e n ze n e : ( a ) m - C h lo r o a n ilin e

(b )

m - B r o m o a n ilin e

(c )

o - N it r o a n ilin e

( d ) p - N it r o a n ilin e

20.7F Replacement by Hydrogen: Deamination by Diazotization A r e n e d ia z o n iu m sa lts re a c t w it h h y p o p h o s p h o ro u s a c id ( H 3 P O 2) to y ie ld p ro d u c ts in w h ic h th e d ia z o n iu m g ro u p h a s b e e n r e p la c e d b y — H. S in c e w e u s u a lly b e g in a s y n th e s is u s in g d ia z o n iu m sa lts b y n itr a t in g a n a ro m a tic c o m ­ p o u n d , th a t is , r e p la c in g — H b y — N O 2 a n d th e n b y — N H 2, i t m a y se e m s tra n g e th a t w e w o u ld e v e r w a n t to re p la c e a d ia z o n iu m g ro u p b y — H . H o w e v e r, re p la c e m e n t o f th e d ia ­ z o n iu m g ro u p b y — H c a n b e a u s e fu l re a c tio n . W e c a n in tro d u c e a n a m in o g ro u p in t o an a ro m a tic r in g to in flu e n c e th e o r ie n ta tio n o f a s u b s e q u e n t re a c tio n . L a te r w e c a n re m o v e

R e v ie w P ro b le m 2 0 .1 0

940

C h a p te r 2 0

A m in e s

the amino group (i.e., carry out a deam ination) by diazotizing it and treating the diazonium salt with H3 PO2. We can see an example o f the usefulness o f a deamination reaction in the follow ing syn­ thesis o f m-bromotoluene.

a rV V V CH3

CHo

Ia

n h

2

p -Toluidine

)

ch

3

H2 SO4, NaNO2

( 1 ) Br2 (2) OH~, H2O heat

h 2o 0-5°C

Br nh

HN.

2

65% (from p-toluidine)

I O CHo

CH h, po,

N2

h 2o

Br

25 °C

Br

N

m-Bromotoluene (85% from 2-bromo-4methylaniline) We cannot prepare m-bromotoluene by direct bromination o f toluene or by a Friedel-Crafts alkylation o f bromobenzene because both reactions give o- and p-bromotoluene. (Both C H 3 — and B r — are ortho-para directors.) However, if w e begin with p-toluidine (pre­ pared by nitrating toluene, separating the para isomer, and reducing the nitro group), we can carry out the sequence o f reactions shown and obtain m-bromotoluene in good yield. The first step, synthesis o f the N-acetyl derivative o f p-toluidine, is done to reduce the acti­ vating effect o f the amino group. (Otherwise both ortho positions would be brominated.) Later, the acetyl group is removed by hydrolysis.

c « / 1/Û

R e v ie w P ro b le m 2 0 .1 1

( a ) In Section 20.7D w e showed a synthesis o f m-fluorotoluene starting with m-toluidine. How would you prepare m-toluidine from toluene? ( b ) How would you prepare m-chlorotoluene? ( c ) m-Bromotoluene? ( d ) m-Iodotoluene? ( e ) m-Tolunitrile (m-CH 3 C 6 H4 CN)? ( f ) m-Toluic acid?

R e v ie w P ro b le m 2 0 .1 2

Starting with p-nitroaniline [Review Problem 20.10 (d)], show how you might synthesize 1,2,3-tribromobenzene.

/

2 0 .8 C o u p lin g R e a c tio n s o f A r e n e d ia z o n iu m S a lts

941

2 0 .8 Coupling Reactions o f Arenediazonium Salts Arenediazonium ions are weak electrophiles; they react with highly reactive aromatic compounds— with phenols and tertiary arylamines— to yield azo compounds. This electrophilic aromatic substitution is often called a diazo coupling reaction.

General Reaction

Q

N= N

\

Xx

Q = — NR2

Q

// or

Xx

N

H

— OH

HX Q

An azo compound

Specific Exam ples

OH

N

\

/

N 2+ C l x

OH

Benzenediazonium chloride

NaOH H2 O, 0°C

Phenol

p -(Phenylazo)phenol (orange solid) N (C H 3 )2 O II

N

CH3 COx Na+ N (C H 3> 2

Benzenediazonium chloride

0°C h 2o

N, N-Dimethylaniline

N,N-Dimethyl-p -(phenylazo)aniline (yellow solid)

Couplings between arenediazonium cations and phenols take place m ost rapidly in slightly alkaline solution. Under these conditions an appreciable amount o f the phenol is present as a phenoxide ion, ArO~, and phenoxide ions are even more reactive toward elec­ trophilic substitution than are phenols themselves. (Why?) If the solution is too alkaline (pH > 1 0 ), however, the arenediazonium salt itself reacts with hydroxide ion to form a rel­ atively unreactive diazohydroxide or diazotate ion: =O H

=O= OHx HA

Phenol (couples slowly)

Phenoxide ion (couples rapidly)

OHx

Ar— N = N ;

Arenediazonium ion (couples)

HA

OHx

A r — N = N — OH

Diazohydroxide (does not couple)

HA

Ar— N = N — O X

Diazotate ion (does not couple)

942

C h a p te r 2 0

A m in e s

C o u p lin g s b e tw e e n a re n e d ia z o n iu m c a tio n s a n d a m in e s ta k e p la c e m o s t r a p id ly in s lig h t ly a c id ic s o lu tio n s ( p H 5 - 7 ) . U n d e r th e se c o n d itio n s th e c o n c e n tr a tio n o f th e a re n e ­ d ia z o n iu m c a tio n is a t a m a x im u m ; a t th e sa m e tim e a n e x c e s s iv e a m o u n t o f th e a m in e has n o t b e e n c o n v e rte d to a n u n re a c tiv e a m in iu m sa lt:

:N R ,

HNR

HA OH-

Amine (couples)

Aminium salt (does not couple)

I f th e p H o f th e s o lu tio n is lo w e r th a n 5, th e ra te o f a m in e c o u p lin g is lo w . W it h p h e n o ls a n d a n ilin e d e riv a tiv e s , c o u p lin g ta ke s p la c e a lm o s t e x c lu s iv e ly a t th e p a ra p o s itio n i f i t is o p e n . I f i t is n o t, c o u p lin g ta ke s p la c e a t th e o rth o p o s itio n . OH N

2

C l-

NaOH

3

H,O

CH,

4-Methylphenol (p-cresol)

4-Methyl-2-(phenylazo)phenol

A z o c o m p o u n d s a re u s u a lly in te n s e ly c o lo re d b e ca u se th e a z o ( d ia z e n e d iy l) lin k a g e , — N=

N — , b rin g s th e tw o a ro m a tic r in g s in to c o n ju g a tio n . T h is g iv e s an e x te n d e d s y s ­

te m o f d e lo c a liz e d

p e le c tro n s a n d a llo w s a b s o rp tio n o f lig h t in th e v is ib le re g io n . A z o c o m ­

p o u n d s , b e ca u se o f th e ir in te n s e c o lo rs a n d b e ca u se th e y c a n b e s y n th e s iz e d f r o m r e la tiv e ly

d ye s . A zo dyes a lm o s t a lw a y s c o n ta in o n e o r m o r e — S O 3~ N a + g ro u p s to c o n fe r w a te r s o l­

in e x p e n s iv e c o m p o u n d s , a re u s e d e x te n s iv e ly as

u b ilit y o n th e d y e a n d a s s is t in b in d in g th e d y e to th e s u rfa c e s o f p o la r fib e rs ( w o o l, c o t­ to n , o r n y lo n ) . M a n y d ye s a re m a d e b y c o u p lin g re a c tio n s o f n a p h th y la m in e s a n d n a p h th o ls . O ra n g e I I , a d y e in tro d u c e d in 1 8 7 6 , is m a d e f r o m 2 -n a p h th o l:

Orange II

R e v ie w P ro b le m 2 0 .1 3

O u tlin e a s y n th e s is o f o ra n g e I I f r o m 2 -n a p h th o l a n d p -a m in o b e n z e n e s u lfo n ic a c id .

R e v ie w P ro b le m 2 0 .1 4

B u tte r y e llo w is a d y e o n c e u s e d to c o lo r m a rg a rin e .

N(CH3)2

I t has s in c e b e e n s h o w n to b e c a rc in o g e n ic , a n d its use in fo o d is n o lo n g e r p e rm itte d . O u t lin e a s y n ­ th e s is o f b u tte r y e llo w

fro m

benzene and

N ,N -

d im e th y la n ilin e .

Butter yellow

2 0 .9 R e a c tio n s o f A m in e s w ith S u lfo n y l C h lo rid e s

A z o c o m p o u n d s ca n b e re d u c e d to a m in e s b y a v a r ie ty o f re a g e n ts in c lu d in g s ta n n o u s c h lo -

943 R e v ie w P r o b le m 2 0 .1 5

r id e ( S n C l2): SnCI. A r 'N H

2

T h is re d u c tio n ca n b e u s e fu l in s y n th e s is as th e f o llo w in g e x a m p le sh o w s: „ , ... (1) HONO, H 3 O+ NaOH, CH 3 CH2Br 4 - E th o x y a n ilin e — — — , A ( C 1 4 H 1 4 N 2 O 2) ----------------------------- > (2 ) phenol, OH~ SnCI2

two molar equivalents of C

acetic anhydride ( C g H - | iN O ) ------------------------> p h e n a c e tin ( C 1 0 H 1 3 N O 2 )

G iv e a s tru c tu re f o r p h e n a c e tin a n d f o r th e in te rm e d ia te s A , B , a n d C . (P h e n a c e tin , f o r ­ m e r ly u s e d as an a n a lg e s ic , is a ls o th e s u b je c t o f P r o b le m 1 7 .4 5 .)

20.9 Reactions o f Am ines w ith Sulfonyl Chlorides P r im a r y a n d s e c o n d a ry a m in e s re a c t w it h s u lf o n y l c h lo rid e s to f o r m s u lf o n a m id e s : H

O

R— N— H

C l— S — A r

H

HCl

R— N— S— Ar

O

1° A m in e

O

W -Substituted s u lfo n a m id e

S u lfo nyl c h lo rid e

R

O

R— N— H

C l— S — A r

O

R

O

R— N— S— Ar

HCl

O

O

W ,W -Disubstituted s u lfo n a m id e

, ° A m in e

W h e n h e a te d w it h a q u e o u s a c id , s u lfo n a m id e s a re h y d r o ly z e d to a m in e s : R

O

R

O

(1) H3 O+, heat^ R— N— S— A r

R— N— H

(2 ) OH“

O

O— S— Ar O

T h is h y d r o ly s is is m u c h s lo w e r, h o w e v e r, th a n h y d r o ly s is o f c a rb o x a m id e s .

20.9A The Hinsberg Test •

S u lfo n a m id e fo rm a tio n is th e b a sis f o r a c h e m ic a l test, c a lle d th e H in s b e rg test, th a t ca n b e used to d e m o n s tra te w h e th e r an a m in e is p rim a ry , secondary, o r te rtia ry .

A H in s b e rg te s t in v o lv e s tw o steps. F irs t, a m ix t u r e c o n ta in in g a s m a ll a m o u n t o f th e a m in e a n d b e n z e n e s u lfo n y l c h lo r id e is sh a ke n w it h

excess p o ta s s iu m h y d r o x id e . N e x t, a fte r a llo w ­

in g tim e f o r a r e a c tio n to ta k e p la c e , th e m ix t u r e is a c id ifie d . E a c h ty p e o f a m in e — p rim a r y , s e c o n d a ry , o r t e r tia r y — g iv e s a d if fe r e n t se t o f

visible re s u lts a fte r e a ch o f th e se tw o stages

o f th e test. P r im a r y a m in e s re a c t w it h b e n z e n e s u lfo n y l c h lo r id e to f o r m N - s u b s titu te d b e n z e n e s u lfo n a m id e s . T h e s e , in tu rn , u n d e rg o a c id -b a s e re a c tio n s w it h th e exce ss p o ta s s iu m h y d r o x ­ id e to f o r m w a te r-s o lu b le p o ta s s iu m sa lts. (T h e s e re a c tio n s ta k e p la c e b e ca u se th e h y d ro g e n a tta c h e d to n itro g e n is m a d e a c id ic b y th e s tr o n g ly e le c tr o n - w ith d r a w in g — S O 2 — g ro u p .)

944

C h a p te r 2 0

A m in e s

At this stage our test tube contains a clear solution. Acidification o f this solution will, in the next stage, cause the water-insoluble ^-substituted sulfonamide to precipitate: Acidic hydrogen H

O

I

II Cl— S

R— N— H

H

O

oh-

(-H C l)

R- A - S - Y

)

i O

O

1° Amine KOH

H

O

I

II

R— N— S " II

HCl

K+

O

-

II

R — N — S'

"

O

II O

Water insoluble (precipitate)

Water-soluble salt (clear solution)

Secondary amines react with benzenesulfonyl chloride in aqueous potassium hydroxide to form insoluble N,N-disubstituted sulfonamides that precipitate after the first stage. N,NDisubstituted sulfonamides do not dissolve in aqueous potassium hydroxide because they do not have an acidic hydrogen. Acidification o f the mixture obtained from a secondary amine produces no visible result; the nonbasic N,N-disubstituted sulfonamide remains as a precipitate and no new precipitate forms: R'

I R — N— H

R'

O

O

OHCl— S O

(-H C l)

R— N— S O

Water insoluble (precipitate) If the amine is a tertiary amine and if it is water insoluble, no apparent change w ill take place in the mixture as w e shake it with benzenesulfonyl chloride and aqueous KOH. When w e acidify the mixture, the tertiary amine dissolves because it forms a water-soluble salt. R e v ie w P ro b le m 2 0 .1 6

An amine A has the molecular formula C 7 H 9 N . Compound A reacts with benzenesulfonyl chloride in aqueous potassium hydroxide to give a clear solution; acidification o f the solu­ tion gives a precipitate. When A is treated with N a N O 2 and H C l at 0 -5 °C , and then with 2-naphthol, an intensely colored compound is formed. Compound A gives a single strong IR absorption peak at 815 cm - 1 . What is the structure o f A?

R e v ie w P ro b le m 2 0 .1 7

Sulfonamides o f primary amines are often used to synthesize pure secondary amines. Suggest how this synthesis is carried out.

THE CHEMISTRY OF . . . C h e m o t h e r a p y a n d S u lf a D r u g s C h e m o th e ra p y

Chemotherapy is defined as the use of chemical agents selectively to destroy infectious cells without simultaneously destroying the host. Although it may be difficult to believe (in this age of "wonder drugs"), chemotherapy is a relatively

modern phenomenon. Before 1900 only three specific chemical remedies were known: mercury (for syphilis—but often with disastrous results), cinchona bark (for malaria), and ipecacuanha (for dysentery).

0

945

2 0 .9 R e a c tio n s o f A m in e s w ith S u lfo n y l C h lo rid e s

,

\

Paul Ehrlich's w o rk in c h e m o th e ra p y led t o his sharing o n e -h a lf o f th e 1 9 0 8 N o b e l P rize in Physiology o r M e d ic in e w ith Ilya M ech niko v.

Modern chemotherapy began with the work of Paul Ehrlich early in the twentieth century—particularly with his discovery in 1907 of the cura­ tive properties of a dye called trypan red I when used against experimental trypanosomia­ sis and with his discovery in 1909 of salvarsan as a remedy for syphilis. Ehrlich was awarded one-half of the Nobel Prize in Physiology or Medicine in 1908. He invented the term "chemotherapy," and in his research he sought what he called "magic bullets," that is, chemicals that would be toxic to infectious microorganisms but harmless to humans. As a medical student, Ehrlich had been impressed with the ability of certain dyes to stain tissues selectively. Working on the idea that "staining" was a result of a chemical reaction between the tissue and the dye, Ehrlich sought dyes with selective affinities for microorganisms. He hoped that in this way he might find a dye that could be modified so as to ren­ der it specifically lethal to microorganisms. S u lfa D r u g s

Between 1909 and 1935, tens of thousands of chemicals, including many dyes, were tested by Ehrlich and others in

i ■ ^ Gerhard Domagk won f * the 1939 Nobel Prize w in Physiology or

a search for such "magic bul­ lets." Very few compounds, Medicine fo r discovering the however, were found to have antibacterial effects of any promising effect. Then, in prontosil. 1935, an amazing event hap­ pened. The daughter of Gerhard Domagk, a doctor employed by a German dye manufacturer, contracted a streptococcal infection from a pin prick. As his daughter neared death, Domagk decided to give her an oral dose of a dye called prontosil. Prontosil had been developed at Domagk's firm (I. G. Farbenindustrie), and tests with mice had shown that prontosil inhibited the growth of strepto­ cocci. Within a short time the little girl recovered. Domagk's gamble not only saved his daughter's life, but it also initi­ ated a new and spectacularly productive phase in modern chemotherapy. G. Domagk was awarded the Nobel Prize in Physiology or Medicine in 1939 but was unable to accept it until 1947. In 1936, Ernest Fourneau of the Pasteur Institute in Paris demonstrated that prontosil breaks down in the human body to produce sulfanilamide, and that sulfanilamide is the actual active agent against streptococci. Prontosil, therefore, is a prodrug because it is converted to the active compound in vivo. O

O

II S— nh

2

2 2

O H 2N

Sulfanilamide

Fourneau's announcement of this result set in motion a search for other chemicals (related to sulfanilamide) that might have even better chemotherapeutic effects. Literally thousands of chemical variations were played on the sul­ fanilamide theme; the structure of sulfanilamide was varied in almost every imaginable way. The best therapeutic results

were obtained from compounds in which one hydrogen of the — SO2NH2 group was replaced by some other group, usually a heterocyclic ring (shown in blue in the following structures). Among the most successful variations were the following compounds. Sulfanilamide itself is too toxic for general use.

O

nh

2

R=

R =

Sulfapyridine

|= S =

I HN

N

Sulfamethoxazole

O =S =O

I / N\

N. N

Succinylsulfathiazole

N

R=

R = R

S

Sulfadiazene

R= O

Sulfacetamide

Sulfathiazole

946

C h a p te r 2 0

A m in e s

Sulfapyridine was shown to be effective against pneumonia in 1938. (Before that time pneumonia epidemics had brought death to tens of thousands.) Sulfacetamide was first used successfully in treating urinary tract infections in 1941. Succinoylsulfathiazole and the related compound phthalylsulfathiazole were used as chemotherapeutic agents against infec­ tions of the gastrointestinal tract beginning in 1942. (Both compounds are slowly hydrolyzed internally to sulfathiazole.) Sulfathiazole saved the lives of countless wounded soldiers during World War II. In 1940 a discovery by D. D. Woods laid the groundwork for our understanding of how the su lfa d ru g s work. Woods observed that the inhibition of growth of certain microorganisms by sulfanilamide is competitively overcome by p-aminobenzoic acid. Woods noticed the structural similarity between the two compounds (Fig. 20.4) and reasoned that the two compounds compete with each other in some essential metabolic process. E s s e n t ia l N u t r i e n t s a n d A n t i m e t a b o l i t e s

All higher animals and many microorganisms lack the biochemical ability to synthesize certain essential organic compounds. These essential nutrients include vitamins, certain amino acids, unsaturated carboxylic acids, purines, and pyrimidines. The aro­ matic amine p-aminobenzoic acid is an essential nutrient for those bacteria that are sensitive to sulfanilamide therapy. Enzymes within these bacteria use p-aminobenzoic acid to synthesize another essential compound called folic acid: H

H \

N

H

/

H

\

OH

/

N

6.7 Á

6.9 Á

O =S =O h

V

\

RNH

O

t _____ 1 2.3 Á

p-Aminobenzoic acid

2.4 Á

A sulfanilamide

Folic acid

Figure 20.4

The structural sim ila rity o f p-am inobenzoic acid and a sulfanilam ide. (R eprinted w ith perm ission of John W ile y and Sons, Inc. from Korolkovas, Essentials o f Molecular Pharmacology, C o p yrig h t 1970.)

Chemicals that inhibit the growth of microbes are called antimetabolites. The sulfanilamides are antimetabolites for those bacteria that require p-aminobenzoic acid. The sulfanilamides apparently inhibit those enzymatic steps of the bac­ teria that are involved in the synthesis of folic acid. The bacterial enzymes are apparently unable to distinguish between a molecule of a sulfanilamide and a molecule of p-aminobenzoic acid; thus, sulfanilamide inhibits the bacterial enzyme. Because the microorganism is unable to synthesize enough folic acid when sulfanilamide is present, it dies. Humans are unaffected by sulfanilamide therapy because we derive our folic acid from dietary sources (folic acid is a vitamin) and do not synthesize it from p-aminobenzoic acid. The discovery of the mode of action of the sulfanilamides has led to the development of many new and effective antimetabolites. One example is methotrexate, a derivative of folic acid that has been used successfully in treating certain carcinomas as well as rheumatoid arthritis:

O Methotrexate

Methotrexate, by virtue of its resemblance to folic acid, can enter into some of the same reactions as folic acid, but it can­ not serve the same function, particularly in important reactions involved in cell division. Although methotrexate is toxic to all dividing cells, those cells that divide most rapidly—cancer cells—are most vulnerable to its effect.

947

20.11 Analysis of Amines

2 0 .1 0 Synthesis o f Sulfa Drugs Sulfanilamides can be synthesized from aniline through the following sequence o f reactions:

O

NH,

O

O 2 HOSO2Cl

O

H2N— R ( - HCl)

80 °C ( - H 2 O) S O 2C l

Aniline (1 )

A c e ta n ilid e (2)

p -A c e ta m id o b e n z e n e s u lfo n y l c h lo rid e (3) O

NH (1) dilute HCl heat (2) HCO 3s o

2n

h r

s o

2n

h r

A sulfanilamide (5)

(4)

Acetylation o f aniline produces acetanilide ( 2 ) and protects the amino group from the reagent to be used next. Treatment o f 2 with chlorosulfonic acid brings about an electrophilic aromatic substitution reaction and yields p-acetam idobenzenesulfonyl chloride ( 3 ) . Addition o f ammonia or a primary amine gives the diamide, 4 (an amide o f both a carboxylic acid and a sulfonic acid). Finally, refluxing 4 with dilute hydrochloric acid selectively hydrolyzes the carboxamide linkage and produces a sulfanilamide. (Hydrolysis o f carbox­ amides is much more rapid than that o f sulfonamides.)

(a) Starting with aniline and assuming that you have 2-aminothiazole available, show how you would synthesize sulfathiazole. (b) How would you convert sulfathiazole to succinylsulfathiazole?

H2 N \

N \ S-—/

R e v ie w P ro b le m 2 0 .1 8

2-Aminothiazole

20.11 Analysis o f Am ines 20.11A Chemical Analysis Amines are characterized by their basicity and, thus, by their ability to dissolve in dilute aqueous acid (Sections 20.3A , 20.3E). M oist pH paper can be used to test for the presence o f an amine functional group in an unknown compound. If the compound is an amine, the pH paper shows the presence o f a base. The unknown amine can then readily be classified as 1°, 2°, or 3° by IR spectroscopy (see below). Primary, secondary, and tertiary amines can also be distinguished from each other on the basis o f the Hinsberg test (Section 20.9A). Primary aromatic amines are often detected through diazonium salt formation and subse­ quent coupling with 2-naphthol to form a brightly colored azo dye (Section 20.8).

948

C h a p te r 2 0

A m in e s

20.11B S p ectro sco p ic Analysis In fr a r e d

S p e c tra

P r im a r y a n d s e c o n d a ry a m in e s a re c h a ra c te riz e d b y I R a b s o rp tio n

b a n d s in th e 3 3 0 0 - 3 5 5 5 - c m ~ 1 r e g io n th a t a ris e f r o m N — H s tre tc h in g v ib r a tio n s . P r im a ry a m in e s g iv e tw o b a n d s in th is r e g io n (see F ig . 2 0 .5 ) ; s e c o n d a ry a m in e s g e n e r a lly g iv e o n ly on e . T e r tia r y a m in e s , b e ca u se th e y h a v e n o N — H g ro u p , d o n o t a b s o rb in th is re g io n . A b s o r p tio n b a n d s a ris in g f r o m C — N s tre tc h in g v ib r a tio n s o f a lip h a tic a m in e s o c c u r in th e 1 0 2 0 -1 2 2 0 -c m ^

1

r e g io n b u t a re u s u a lly w e a k a n d d if f ic u lt to id e n tify . A r o m a t ic a m in e s

g e n e r a lly g iv e a s tro n g C — N s tre tc h in g b a n d in th e 1 2 5 0 - 1 3 6 0 - c m ^ 1 re g io n . F ig u r e 2 0 .5 s h o w s an a n n o ta te d I R s p e c tru m o f 4 - m e th y la n ilin e .

100 90 arom atic combination band

80 arom atic C— H (stretch)

C? 70 8

60

f

50

|

40

to I-

c—H

(stretch)

N—H (wag)

nh 2

prim ary N— H (asym. and sym. stretch)

C— N (stretch)

30

3

20

C— H (out-of-plane bend)

C— C (ring stretch), N— H (bend)

10 0 4000

~ i----------1--------- 1--------- 1—

~i----- 1----- 1----- 1—

3600

3200

2800

2400

2000

1800

1600

1400

1200

1000

800

650

W avenum ber (cm-1)

Figure 20.5

1H

N M R

r e g io n

8

A nnotated IR spectrum of 4-methylaniline.

S p e c tra

P r im a r y a n d s e c o n d a ry a m in e s s h o w N — H p ro to n s ig n a ls in th e

0 . 5 - 5 . T h e s e s ig n a ls a re u s u a lly b ro a d , a n d th e ir e x a c t p o s itio n d e p e n d s o n th e

n a tu re o f th e s o lv e n t, th e p u r it y o f th e sa m p le , th e c o n c e n tra tio n , a n d th e te m p e ra tu re . B e c a u s e o f p r o to n e xc h a n g e , N — H p ro to n s a re n o t u s u a lly c o u p le d to p ro to n s o n a d ja c e n t c a rb o n s . A s su ch , th e y a re d if f ic u lt to id e n t if y a n d a re b e s t d e te c te d b y p r o to n c o u n tin g o r b y a d d in g a s m a ll a m o u n t o f D 2O to th e s a m p le . E x c h a n g e o f N — D d e u te ro n s f o r th e N — H p ro to n s ta ke s p la c e , a n d th e N — H s ig n a l d is a p p e a rs f r o m th e s p e c tru m . P ro to n s o n th e

a c a rb o n o f an a lip h a tic a m in e a re d e s h ie ld e d b y th e e le c tro n - w ith d ra w in g

e ffe c t o f th e n itr o g e n a n d a b s o rb t y p ic a lly in th e

8

2 .2 - 2 .9 re g io n ; p ro to n s o n th e b c a rb o n

are n o t d e s h ie ld e d as m u c h a n d a b s o rb in th e ra n g e

8

1 .0 -1 .7 .

F ig u r e 2 0 .6 (n e x t p a g e ) s h o w s an a n n o ta te d 1H N M R s p e c tru m o f d iis o p ro p y la m in e .

13C

N M R

S p e c tra

The

a c a rb o n o f a n a lip h a tic a m in e e x p e rie n c e s d e s h ie ld in g b y th e

e le c tro n e g a tiv e n itro g e n , a n d its a b s o rp tio n is s h ifte d d o w n fie ld , t y p ic a lly a p p e a rin g a t 3 0 - 6 0 . T h e s h if t is n o t as g re a t as f o r th e

a c a rb o n o f a n a lc o h o l ( t y p ic a lly

8

8

5 0 -7 5 ), h o w ­

ever, b e ca u se n itr o g e n is le ss e le c tro n e g a tiv e th a n o x y g e n . T h e d o w n fie ld s h if t is e v e n less f o r th e b c a rb o n , a n d so o n d o w n th e c h a in , as th e c h e m ic a l s h ifts o f th e c a rb o n s o f p e n ty la m in e s h o w :

23.0 14.3 1 3,

34.0

29.7

n

H2

42.5

3C N M R c h e m ica l s h ifts (5)

2 0 .1 2 E lim in a tio n s In v o lv in g A m m o n iu m C o m p o u n d s

8

7

6

5

4

3

2

1

949

0

5 h (ppm) Figure 20.6 The 300-MHz 1H NMR spectrum of diisopropylamine. N ote the integral for the broad NH peak at approximately S 0.7. Vertical expansions are not to scale.

M a s s S p e c t r a o f A m in e s The m olecular ion in the m ass spectrum o f an amine has an odd number mass (unless there is an even number o f nitrogen atoms in the m olecule). The peak for the molecular ion is usually strong for aromatic and cyclic aliphatic amines but weak for acyclic aliphatic amines. Cleavage between the a and b carbons o f aliphatic amines is a common m ode o f fragmentation.

20.12 Eliminations Involving Am m onium Compounds 20.12A The Hofmann Elimination A ll o f the eliminations that w e have described so far have involved electrically neutral sub­ strates. However, eliminations are known in which the substrate bears a positive charge. One o f the most important o f these is the E2-type elimination that takes place when a quaternary ammonium hydroxide is heated. The products are an alkene, water, and a tertiary amine: ..

r*

H O H

O'

Cn r .

HOH

heat

A quaternary ammonium hydroxide

An alkene

=N R 3

A tertiary amine

This reaction was discovered in 1851 by August W. von Hofmann and has since com e to bear his name. Quaternary ammonium hydroxides can be prepared from quaternary ammonium halides in aqueous solution through the use o f silver oxide or an ion exchange resin: , N M e 3 XR

A quaternary ammonium halide

Ag 2 Û ) h 2o

>

, NMe3 HOR'

2

AgXp

A quaternary ammonium hydroxide

Silver halide precipitates from the solution and can be removed by filtration. The quater­ nary ammonium hydroxide can then be obtained by evaporation o f the water. Although m ost eliminations involving neutral substrates tend to follow the Zaitsev rule (Section 7.6B), eliminations with charged substrates tend to follow what is called the

950

Chapter 20

Amines

H o f m a n n r u l e and yield mainly the least substituted alkene. We can see an exam ple o f this behavior if w e compare the follow ing reactions:

EtONa EtOH, 25°C

Br

75%

25%

5%

95%

- OH 150°C

NMe3

NaBr

EtOH

NMe3

H 2O

The precise mechanistic reasons for these differences are com plex and are not yet fully understood. One possible explanation is that the transition states o f elimination reactions with charged substrates have considerable carbanionic character. Therefore, these transi­ tion states show little resemblance to the final alkene product and are not stabilized appre­ ciably by a developing double bond: HO.

HO.

+NMe3

Br5

C a rb a n io n -lik e tra n s itio n sta te (g iv e s H o fm ann o rie n ta tio n )

A lk e n e -lik e tra n s itio n sta te (g iv e s Z a its e v o rie n ta tio n )

With a charged substrate, the base attacks the m ost acidic hydrogen instead. A primary hydro­ gen atom is more acidic because its carbon atom bears only one electron-releasing group.

20.12B The Cope Elimination Tertiary amine oxides undergo the elimination o f a dialkylhydroxylamine when they are heated. The reaction is called the Cope elimination, it is a syn elimination and proceeds through a cyclic transition state. R

Me

Me

H

Me

R

150°C

:XO

N Me

A te rtia ry a m in e o xid e

A n a lk e n e

W ,W -D im ethylh yd ro xylam ine

Tertiary amine oxides are easily prepared by treating tertiary amines with hydrogen per­ oxide (Section 20.5A ). The Cope elimination is useful synthetically. Consider the following synthesis o f methylenecyclohexane: 160°C

(C H 3 )2 N O H

\^ C H 3 ch 3

98%

20.13 Sum m ary o f Preparations and Reactions o f Am ines P r e p a r a t i o n o f A m in e s 1.

Gabriel synthesis (discussed in Section 20.4A):

O

O

O

H n h 2n h 2

(1) KOH N— H

O

(2) R— X

N

O

R

ethanol, reflux

R— NH2

H O

2 0 .1 3 S u m m a ry o f P re p a ra tio n s a n d R e a c tio n s o f A m in e s

2.

B y reduction o f alkyl azides (discussed in Section 20.4A): _ R—

_

NaN3

Br

..

_

V

Na/alcohol

_

---------------- »

R— N = N = N

ethanol

R— N H,

or LiAlH.

3. B y amination o f alkyl halides (discussed in Section 20.4A): R — Br

+

NH3

------ »

R N H 3+ B r -

+

R 2 N H 2+ B r -

+

R 3 N H + B r-

I o HRNH,

R 2N H

+

R 3N

+

r

4

n+oh-

(A mixture of products results.) (R = a 1° alkyl group) 4.

By reduction o f nitroarenes (discussed in Section 20.4B): H2, catalyst A r— NO,

5.

A r— NH,

or (1)

Fe/HCl (2) NaOH

By reductive amination (discussed in Section 20.4C): Hv

.H N

nh3

1° Amine

[H]

^~^R' H H^

O

A

,R ’ N

R"NH2

2° Amine

[H] H

Aldehyde or ketone

W

R" N

R"R"'NH

3° Amine

[H] H 6

. B y reduction o f nitriles, oxim es, and amides (discussed in Section 20.4D): H

LiAlH4, Et2O — ----------- 4 2 (2 ) H2O

_ R— C N

(1)

^ >

,

FI

^N ^

^

.

1° A m in e H

OH N

NH

Na/ethanol R '

R'

R

R'

1

° A m in e

O H R

N

( 1 ) LiAlH4, Et2 O_ (2 ) H2O

H R

N

I

I

1° A m in e

H

H O R' R

N

( 1 ) LiAlH4, Et2O _ (2 ) H2O

R' R

2 ° A m in e

N

I

I

H

H O R' R

N

I R"

( 1 ) LiAlH . Et2 O_ (2 ) H2 O

R' R

N'

I R"

3 ° A m in e

R 4 N + B r-

951

952

Chapter 20

Amines

7. Through the Hofmann and Curtius rearrangements (discussed in Section 20.4E): Hofmann Rearrangement O RX

N_ H

r -n h

2

+

C O 3 2-

H

Curtius Rearrangement O

O

R^ C l

— —N (-N a C i)

R

''" N 3

heat

3

r _ N = C =O

- H2^

R— N H2

+

CO2

2

2

R e a c t io n s o f A m in e s 1.

A s bases (discussed in Section 20.3): H R— N — R' + H — A

R— N — R' A-

R"

R"

(R , R ', a n d /o r R" m a y b e alkyl, H, o r Ar) 2.

Diazotization o f 1° arylamines and replacement of, or coupling with, the diazonium group (discussed in Sections 20.7 and 20.8): Cu2O, Cu2+, h2o A r— OH A r — Cl Ar A r— N H 2

Br

A r — CN Ar— I Ar— F

Ar

H Q

Q

N, Q = NR2 or OH Conversion to sulfonamides (discussed in Section 20.9):

■N— H

(2) HCl

R—

N-

O

R'

1 — 1

O A

H (1) ArSO2Cl, O H -

1

H

= < /)=

3.

R

N

R'

H

A rS ° 2 Cl^

R—

O

N- — S — O

953

P ro b le m s

4. Conversion to amides (discussed in Section 17.8): O

0 1 R"^" Cl

H R— N— H

H

O

base

R"

J

N

R

C l-

R"

R— N— H

V

O

O

R

J

H O x R " ^ Cl

R' R— N— H

O

/2

N

R

R". X

OH

H

O

base

R"

J

N

R

C l-

R' 5. Hofmann and Cope eliminations (discussed in Section 20.12): Hofmann Elimination H OH-

h 2o

heat

+

n r

3

NR,

Cope Elimination Me N -M e H

syn elimination heat

N-

M e ^ .. N

Me

I OH

Key Terms and Concepts The key terms and concepts that are highlighted in bold, blue text within the chapter are defined in the glossary (at the back o f the book) and have hyperlinked definitions in the accompanying W ileyPLU S course (www.wileyplus.com )

/WILEY _

PLUS

Problems r W IL E Y

_

PLUS

Note to Instructors: Many of the homework problems are available for assignment via WileyPLUS, an online teaching and learning solution. NO M ENCLATURE

20.19

Write structural formulas for each o f the follow ing compounds: (a) Benzylmethylamine

(k) Dimethylaminium chloride

(b) Triisopropylamine

(l) 2-M ethylim idazole

(c) N-Ethyl-N-methylaniline

(m) 3-Aminopropan-1-ol

(d) m-Toluidine

(n) Tetrapropylammonium chloride

(e) 2-Methylpyrrole

(o) Pyrrolidine

(f) N-Ethylpiperidine

(p) N,N-Dimethyl-p-toluidine

(g) N-Ethylpyridinium bromide

(q) 4-M ethoxyaniline

(h) 3-Pyridinecarboxylic acid

(r) Tetramethylammonium hydroxide

(i) Indole

(s) p-A m inobenzoic acid

(j) Acetanilide

(t) N-M ethylaniline

954 2 0 .2 0

Chapter 20

Amines

Give common or systematic names for each o f the follow ing compounds: ,NH 2

(a )

(f)

( j)

^

^so?nh

2

N N I H

H

(b )

3 (k ) C H

(g)

3

N H

3

C H 3C O 2"

N

(c )

'N

X

(l)

NH2

H O ^ ^ \^ „ .N H

(m )

(h )

n h 3 Cl-

(d)

2

N

N

N \

H (i) (n )

(e )

N N I CH 3

A M I N E S Y N T H E S IS A N D

20.21

W hich is the m ost basic nitrogen in each compound. Explain your choices. (a )

20.22

R E A C T IV IT Y

NH2

(b) H2 N'

(c)

HN

Show how you m ight prepare benzylam ine from each o f the follow ing compounds (a)

^

(c) Benzyl bromide (two ways)

,C N B e n z o n i t r i le

(d) Benzyl tosylate (e) Benzaldehyde

NH 2

(f) Phenylnitromethane (b )

O

(g)

B e n z y la m in e

NH 2

20.23

Benzene

(b )

Bromobenzene

.N H 2 P h e n y la c e ta m id e

Show how you might prepare aniline from each o f the follow ing compounds: (a )

20.24

^

B e n z a m id e

(c )

Benzamide

Show how you might synthesize each o f the follow ing compounds from 1-butanol: (a )

Butylamine (free o f 2° and 3° amines)

(c )

Propylamine

(b )

Pentylamine

(d )

Butylmethylamine

0

955

P ro b le m s

20.25

20.26

Show how you m ight convert aniline into each o f the follow ing compounds. (You need not repeat steps carried out in earlier parts o f this problem.) (a )

Acetanilide

(l)

(b )

N-Phenylphthalimide

(m )

(c )

p-Nitroaniline

(n )

(d )

Sulfanilamide

(e )

N,N-Dimethylaniline

(f)

Fluorobenzene

(g )

Chlorobenzene

(h )

Bromobenzene

(i)

Iodobenzene

(j)

Benzonitrile

(k )

B enzoic acid

Phenol Benzene

(o )

Provide the major organic product from each o f the follow ing reactions. (d )

O

(1) NH 2 OH, cat. HA (2) NaBH3CN (e )

NH

O (1 )

O O

(2) LiAlH 4 (3) H3 O+ (c )

Br

( 1 ) NaN 3 (2) LiAlH 4 (3) H3 O+

20.27

What products would you expect to be formed when each o f the follow ing amines reacts with aqueous sodium nitrite and hydrochloric acid? (a )

Propylamine

(d )

N.N-Dipropylaniline

(b )

Dipropylamine

(e )

p-Propylaniline

(c )

N-Propylaniline

20.28

( a ) What products would you expect to be formed when each o f the amines in the preceding problem reacts with benzenesulfonyl chloride and excess aqueous potassium hydroxide? ( b ) What would you observe in each reaction? ( c ) What would you observe when the resulting solution or mixture is acidified?

20.29

What product would you expect to obtain from each o f the follow ing reactions? (a )

^

NaNO2, HCl

(b )

^

RO2Cl

N

aq. KOH

I H

N

I H

2

956 20.30

Chapter 20

Amines

Give structures for the products of each of the following reactions: (a )

O

(f)

n h

( g ) A n ilin e + p r o p a n o y l c h lo r id e —

O

heat

2

( h ) T e tr a e th y la m m o n iu m h y d r o x id e

O (c )

O

O c h

3

n h

O

H

O 3

O 7

I

2

(b) c h

O

N

Cl ,n h

O

O

(i)

2

p - T o lu id in e + B r 2 (e xce ss)

7

H2O

heat ( d ) P r o d u c t o f (c )

--------- »

(e )

N

I H

20.31

S ta rtin g w it h b e n z e n e o r to lu e n e , o u tlin e a s y n th e s is o f e a ch o f th e f o llo w in g c o m p o u n d s u s in g d ia z o n iu m s a lts as in te rm e d ia te s . ( Y o u n e e d n o t re p e a t syn th e se s c a rr ie d o u t in e a rlie r p a rts o f th is p ro b le m .) (a )

p - F lu o r o to lu e n e

(b )

o - Io d o to lu e n e

( c ) p -C r e s o l (d )

m - D ic h lo ro b e n z e n e O H

(e ) m - C 6 H 4 ( C N ) 2 (f)

m - B r o m o b e n z o n itr ile

( g ) 1 ,3 -D ib r o m o - 5 -n itr o b e n z e n e ( h ) 3 , 5 - D ib r o m o a n ilin e (i)

3 ,4 ,5 - T rib r o m o p h e n o l

(j)

3 , 4 ,5 - T r ib r o m o b e n z o n itr ile C H

3

( k ) 2 ,6 - D ib r o m o b e n z o ic a c id (l)

1 ,3 -D ib r o m o - 2 -io d o b e n z e n e

(m ) C H

Br

CN

20.32

W r it e e q u a tio n s f o r s im p le c h e m ic a l te sts th a t w o u ld d is tin g u is h b e tw e e n ( a ) B e n z y la m in e a n d b e n z a m id e

( f ) C y c lo h e x y la m in e a n d a n ilin e

( b ) A lly la m in e a n d p ro p y la m in e

(g ) T r ie th y la m in e a n d d ie th y la m in e

( c ) p - T o lu id in e a n d N - m e th y la n ilin e

( h ) T r ip r o p y la m in iu m c h lo r id e a n d te tr a p r o p y la m m o n iu m c h lo r id e

( d ) C y c lo h e x y la m in e a n d p ip e r id in e

( i)

( e ) P y r id in e a n d b e n z e n e

20.33

T e tr a p ro p y la m m o n iu m c h lo r id e a n d te tr a p r o p y la m m o n iu m h y d r o x id e

D e s c r ib e w it h e q u a tio n s h o w y o u m ig h t se p a ra te a m ix t u r e o f a n ilin e , p -c r e s o l, b e n z o ic a c id , a n d to lu e n e u s in g o r d in a r y la b o ra to r y re a g e n ts.

957

Problems

M E C H A N IS M S

20.34

Using reactions that w e have studied in this chapter, propose a mechanism that accounts for the following reaction:

H

O

IN

CN H2, Pd

20.35

Provide a detailed m echanism for each o f the following reactions. (a)

O ,NH 2 NH 2

(b )

Br2, NaOH, H2O

O

N / V

20.36

nh2

POCl3 _

il

1

Suggest an experiment to test the proposition that the Hofmann reaction is an intramolecular rearrangement, that is, one in which the migrating R group never fully separates from the amide molecule.

G E N E R A L S Y N T H E S IS

20.37

Show how you m ight synthesize b-am inopropionic acid from succinic anhydride. (b-Aminopropionic acid is used in the synthesis o f pantothenic acid, a precursor o f coenzym e A.) O O ^ H3 N ^

20.38

20.39

^

O

^

"O-

b-Aminopropanoic acid

O

Succinic anhydride

Show how you might synthesize each o f the following from the compounds indicated and any other needed reagents: (a )

Me 3 N+

N+Me3 2Br~ from 1,10-decanediol

(b )

Succinylcholine bromide (see “The Chemistry o f . . . B iologically Important Am ines” in Section 20.3) from succinic acid, 2 -bromoethanol, and trimethylamine

A commercial synthesis o f folic acid consists o f heating the following three compounds with aqueous sodium bicarbonate. Propose reasonable mechanisms for the reactions that lead to folic acid. Hint: The first step involves formation o f an imine between the lower right NH2 group o f the heterocyclic amine and the ketone.

OH

O nh2

N J ,

H2N

-'N

nh2

Br

Br Br

O HCO 3 -, h 2o

Folic acid ( - 1 0 %)

958 20.40

Chapter 20

Amines

Give structures for compounds R -W : N-Methylpiperidine

9

9

:

R ( C 7 H 1 6 NI ) 9 H 0 : S ( C 7 H 1 7 N O )

9 g ° l

T ( C 7 H 1 5 N ) 9 9 : U ( C 5 H 1 8 N I) 9 H g 0 : V (C8H i9NO) 9 20.41

^

W (CsH8) + H 20 + ( C H ^ N

Outline a synthesis o f acetylcholine iodide using dimethylamine, oxirane, iodomethane, and acetyl chloride as starting materials.

q .N(CHs)3 IAcetylcholine iodide

20.42

Ethanolamine, HOCH2CH2NH2, and diethanolamine, (HOCH2CH2)2NH, are used commercially to form emulsi­ fying agents and to absorb acidic gases. Propose syntheses of these two compounds.

20.43

Diethylpropion (shown here) is a compound used in the treatment o f anorexia. Propose a synthesis o f diethylpropion starting with benzene and using any other needed reagents.

20.44

Using as starting materials 2-chloropropanoic acid, aniline, and 2-naphthol, propose a synthesis o f naproanilide, a herbicide used in rice paddies in Asia:

Naproanilide

SPECTROSCOPY

20.45

When compound W (C15H17N) is treated with benzenesulfonyl chloride and aqueous potassium hydroxide, no appar­ ent change occurs. Acidification of this mixture gives a clear solution. The 1H NM R spectrum of W is shown in Fig. 20.7. Propose a structure for W.

offset plots.

5h (ppm)

9 59

P ro b le m s

20.46

Propose structures for compounds X, Y, and Z: X (C 7 H 7 B r) - NaC:

Y (C 8 H 7 N )

T h e 1H N M R s p e c tru m o f X g iv e s tw o s ig n a ls , a m u lt ip le t a t

Z (C 8 H ^ N ) 8

7 .3 ( 5 H ) a n d a s in g le t a t

8 4 .2 5 ( 2 H ) ; th e 6 8 0 -

8 4 0 - c m - 1 r e g io n o f th e I R s p e c tru m o f X s h o w s p e a k s a t 6 9 0 a n d 7 7 0 c m - 1 . T h e 1H N M R s p e c tru m o f Y is s im ­ il a r to th a t o f X : m u lt ip le t a t

8

7 .3 ( 5 H ) , s in g le t a t

8

3 .7 ( 2 H ) . T h e 1H N M R s p e c tru m o f Z is s h o w n in F ig . 2 0 .8 .

Figure 20.8 The 300MHz 1H NMR spectrum of com pound Z, Problem 20.46. Expansion of the signals is shown in the offset plot.

Sh (PPm)

20.47

C om pound A (C cm

220

200

- 1

1 0

H 1 5 N ) is s o lu b le in d ilu te H C l . T h e I R a b s o rp tio n s p e c tru m s h o w s t w o b a n d s in th e 3 3 0 0 - 3 5 0 0 -

re g io n . T h e b ro a d b a n d p r o to n - d e c o u p le d 13C s p e c tru m o f A is g iv e n in F ig . 2 0 .9 . P ro p o s e a s tru c tu re f o r A .

180

160

140

120

100

Sc (ppm)

80

60

40

20

0

Figure 20.9 The broadband protondecoupled 13C NMR spectra of com pounds A, B, and C, Problems 20.47-20.49. Information from the DEPT 13C NMR spectra is given above each peak.

960

Chapter 20

220

200

Amines

180

160

140

120

100

80

60

40

20

0

80

60

40

20

0

5 C (p p m )

220

Figure 20.9

200

180

160

140

(continued)

120

100

sC (p p m )

20.48

Compound B, an isomer o f A (Problem 20.47), is also soluble in dilute HCl. The IR spectrum o f B shows no bands in the 33 0 0 -3 5 0 0 -cm -1 region. The broadband proton-decoupled 13C spectrum o f B is given in Fig. 20.9. Propose a structure for B.

20.49

Compound C (C9H n NO) gives a positive Tollens’ test (can be oxidized to a carboxylic acid) and is soluble in dilute HCl. The IR spectrum o f C shows a strong band near 1695 cm - 1 but shows no bands in the 3 3 0 0 -3 5 0 0 -cm ~ 1 region. The broadband proton-decoupled 13C N M R spectrum o f C is shown in Fig. 20.9. Propose a structure for C.

Challenge Problems 20.50

When phenyl isothiocyanate, C 6 H5N = C = S , is reduced with lithium aluminum hydride, the product formed has these spectral data: M S (m/z): 107, 106 IR (cm -1 ): 3330 (sharp), 3050, 2815, 760, 700 1H N M R (g): 2.7 (s, 3H), 3.5 (broad, 1H),

6 .6

(d, 2H), 6.7 (t, 1H) 7.2 (t, 2H)

13C N M R (g): 30 (CH3), 112 (CH), 117 (CH), 129 (CH), 150 (C) (a) What is the structure o f the product? (b) What is the structure that accounts for the 106 m/z peak and how is it formed? (It is an iminium ion.)

0

961

Challenge Problems

20.51

H I N

When N,N'-diphenylurea (A) is reacted with tosyl chloride in pyridine, it yields product B. The spectral data for B include:

O

M S (m/z): 194 (M+) IR (cm -1 ): 3060, 2130, 1590, 1490, 760, 700 1H 13C

T

H I _N

A

N M R (S): only 6 .9 -7 .4 (m) N M R (S): 122 (CH), 127 (CH), 130 (CH), 149 (C), and 163 (C)

(a) What is the structure o f B? (b) Write a m echanism for the formation o f B. 20.52

Propose a m echanism that can explain the occurrence o f this reaction: O OH

. \ CH3 CH3

CH3 ..

nh2

X Suggest a mechanism for this rearrangement. 21.39

Hexachlorophene was a widely used germicide until it was banned in 1972 after tests showed that it caused brain damage in test animals. Suggest how this compound might be synthesized, starting with benzene.

N I

/C H 3 3

CH3

996 21.40

Chapter 21

Phenols and Aryl Halides

The Fries rearrangement occurs when a phenolic ester is heated with a Friedel-Crafts catalyst such as AlCl3:

(X

R

The reaction may produce both ortho and para acylated phenols, the former generally favored by high tempera­ tures and the latter by low temperatures. (a) Suggest an experiment that might indicate whether the reaction is inter­ or intramolecular. (b) Explain the temperature effect on product formation. 21.41

Compound W was isolated from a marine annelid com m only used in Japan as a fish bait, and it was shown to be the substance that gives this organism its observed toxicity to som e insects that contact it. M S (m/z): 151 (relative abundance 0.09), 149 (M-+, rel. abund. 1.00), 148 IR (cm -1 ): 2960, 2850, 2775 1H N M R (g): 2.3 (s,

6

H), 2.6 (d, 4H), and 3.2 (pentet, 1H)

13C N M R (g): 38 (CH3), 43 (CH2), and 75 (CH) These reactions were used to obtain further information about the structure o f W: NaBH4

C6H5COCl

Raney Ni

W --------- : X - 9 -9 ------ : Y -------------- : Z Compound X had a new infrared band at 2570 cm - 1 and these NM R data: 1H N M R (g): 1.6 (t, 2H), 2.3 (s,

6

H), 2.6 (m, 4H), and 3.2 (pentet, 1H)

13C N M R (g): 28 (CH2), 38 (CH3), and 70 (CH) Compound Y had these data: IR (cm -1 ): 3050, 2960, 2850, 1700, 1610, 1500, 760, 690 1H N M R (g): 2.3 (s,

6

H), 2.9 (d, 4H), 3.0 (pentet, 1H), 7.4 (m, 4H), 7.6 (m, 2H), and 8.0 (m, 4H)

13C N M R (g): 34 (CH2), 39 (CH3), 61 (CH), 128 (CH), 129 (CH), 134 (CH), 135 (C), and 187 (C) Compound Z had M S (m/z): 87 (M-+),

8 6

, 72

IR (cm -1 ): 2960, 2850, 1385, 1370, 1170 1H N M R (g): 1.0 (d,

6

H), 2.3 (s,

6

H), and 3.0 (heptet, 1H)

13C N M R (g): 21 (CH 3 ), 39 (CH 3 ), and 55 (CH) What are the structures o f W and of each of its reaction products X , Y, and Z? 21.42

Phenols generally are not changed on treatment with sodium borohydride follow ed by acidification to destroy the excess, unreacted hydride. For example, the 1,2-, 1,3-, and 1,4-benzenediols and 1,2,3-benzenetriol are unchanged under these conditions. However, 1,3,5-benzenetriol (phloroglucinol) gives a high yield o f a product A that has these properties: M S (m/z): 110 IR (cm -1 ): 3250 (broad), 1613, 1485 1H N M R (g in DMSO): 6.15 (m, 3H), 6.89 (t, 1H), and 9.12 (s, 2H) (a) What is the structure o f A? (b) Suggest a mechanism by which the above reaction occurred. (1,3,5-Benzenetriol is known to have more ten­ dency to exist in a keto tautomeric form than do simpler phenols.)

O'

OH

L e a rn iin g G ro u p P ro b le m s

21.43

997

Open the m olecular m odel file for benzyne and examine the follow ing molecular orbitals: the LUM O (low est unoc­ cupied molecular orbital), the HOMO (highest occupied molecular orbital), the HOMO-1 (next lower energy orbital), the HOM O-2 (next lower in energy), and the HOMO-3 (next lower in energy). (a) W hich orbital best represents the region where electrons o f the additional p bond in benzyne would be found? (b) W hich orbital would accept electrons from a Lew is base on nucleophilic addition to benzyne? (c) W hich orbitals are associated with the six p electrons o f the aromatic system? Recall that each m olecular orbital can hold a maximum o f two electrons.

Learning Group Problems 1

.

Thyroxine is a hormone produced by the thyroid gland that is involved in regulating m etabolic activity. In a pre­ vious Learning Group Problem (Chapter 15) w e considered reactions involved in a chem ical synthesis o f thyrox­ ine. The following is a synthesis o f optically pure thyroxine from the amino acid tyrosine (also see Problem 2, below). This synthesis proved to be useful on an industrial scale. (Schem e adapted from Fleming, I., Selected Organic Syntheses, pp. 3 1 -3 3 . © 1973 John W iley & Sons, Limited. Reproduced with permission.) (a) 1 to 2

What type o f reaction is involved in the conversion o f 1 to 2? Write a detailed m echanism for this trans­ formation. Explain why the nitro groups appear where they do in 2.

(b) 2 to 3

(i) Write a detailed m echanism for step (1) in the conversion o f 2 to 3. (ii) Write a detailed m echanism for step (2) in the conversion o f 2 to 3. (iii) Write a detailed mechanism for step (3) in the conversion o f 2 to 3.

(c) 3 to 4

(i) What type o f reaction mechanism is involved in the conversion o f 3 to 4? (ii) Write a detailed m echanism for the reaction from 3 to 4. What key intermediate is involved?

(d) 5 to

Write a detailed m echanism for conversion o f the m ethoxyl group o f 5 to the phenolic hydroxyl o f 6 .

6

hno3

(1) (CH3CO)2O (2) EtOH, HA (3 ) TsCl, pyridine

Tyrosine (1)

C H

3 Ü

NHCOCH

^ ]^ O

COOEt

-(1)-H2-Pd/CCH3O (2) HONO 3 (3) I2, Nal

HI, CH3CO2H

HO

O

NH

NH2 COOH

H

^

I

COOH

' I

(-)-T h y ro x in e

H

998 2

.

Chapter 21

Phenols and Aryl Halides

Tyrosine is an amino acid with a phenolic side chain. Biosynthesis in plants and microbes o f tyrosine involves enzy­ matic conversion o f chorismate to prephenate, below. Prephenate is then processed further to form tyrosine. These steps are shown here: CO 2

-O2C, O

chorismate mutase

HO Chorismate

h

Prephenate prephenate dehydrogenase

-NAD+

s

NADH + CO 2

CO O aminotransferase a-ketoglutarate

glutamate

OH 4-Hydroxyphenylpyruvate

Tyrosine (a )

There has been substantial research and debate about the enzymatic conversion o f chorismate to prephenate by chorismate mutase. Although the enzymatic mechanism may not be precisely analogous, what laboratory reac­ tion have w e studied in this chapter that resembles the biochem ical conversion o f chorismate to prephenate? Draw arrows to show the movem ent o f electrons involved in such a reaction from chorismate to prephenate.

(b )

When the type o f reaction you proposed above is applied in laboratory syntheses, it is generally the case that the reaction proceeds by a concerted chair conformation transition state. Five o f the atoms o f the chair are car­ bon and one is oxygen. In both the reactant and product, the chair has one bond m issing, but at the point o f the bond reorganization there is roughly concerted flow o f electron density throughout the atoms involved in the chair. For the reactant shown below, draw the structure o f the product and the associated chair conformation transition state for this type o f reaction: c o 2-

CO

(c )

Draw the structure o f the nicotinamide ring o f N A D + and draw m echanism arrows to show the decarboxyla­ tion o f prephenate to 4-hydroxyphenylpyruvate with transfer o f the hydride to N A D + (this is the type o f process involved in the m echanism o f prephenate dehydrogenase).

(d )

Look up the structures o f glutamate (glutamic acid) and a-ketoglutarate and consider the process o f transami­ nation involved in conversion o f 4-hydroxyphenylpyruvate to tyrosine. Identify the source o f the amino group in this transamination (i.e., what is the amino group “donor”?). What functional group is left after the amino group has been transferred from its donor? Propose a mechanism for this transamination. N ote that the m ech­ anism you propose w ill likely involve formation and hydrolysis o f several imine intermediates— reactions sim ­ ilar to others w e studied in Section 16.8.

CO NCEPT MAP Som e Synthetic Connections o f Phenols and Related Arom atic Com pounds

R e a c tio n s o f b e n z e n e re la te d to p h e n o l s y n th e s is :

PLUS See Second Review Problem Set in WileyPLUS

• • • •

Nitration and nitro-group reduction Friedel-C rafts alkylation Sulfonation Chlorination (X = Cl)

S y n th e s is o f p h e n o ls :

R e a c tio n s o f p h e n o ls :

• Laboratory: via diazonium salts • Industrial synthesis of phenols (1) Via cum ene hydroperoxide (2) Dow process (via benzyne)

• Acylation • Alkylation (Williamson ether synthesis) • Electrophilic arom atic substitution

X = F, Cl, Br, I

O th e r re a c tio n s :

• Substitution of diazonium salts • SNA r (Nucleophilic arom atic substitution— requires electronwithdrawing groups on ring) • Substitution via benzyne • Claisen rearrangem ent of allyl aryl ethers

N ur via benzyne (e.g., NaNH2; HO“ , heat)

Benzyne

W hite blood cells continually patrol the circulatory system and interstitial spaces, ready for mobilization at a site of trauma. The frontline scouts for leukocytes are carbohydrate groups on their surface called sialyl Lewisx acids. When injury occurs, cells at the site of trauma display proteins, called selectins, that signal the site of injury and bind sialyl Lewisx acids. Binding between selectins and the sialyl Lewisx acids on the leukocytes causes adhesion of leukocytes at the affected area. Recruitment of leukocytes in this way is an important step in the inflammatory cascade. It is a necessary part of the healing process as well as part of our natural defense against infection. A molecular model of a sialyl Lewisx acid is shown above, and its structural formula is given in Section 22.16. There are some maladies, however, that result from the over-enthusiastic recruitment of leukocytes. Rheumatoid arthritis, strokes, and injuries related to perfusion during surgery and organ transplantation are a few examples. In these conditions, the body perceives that certain cells are under duress, and it reacts accord­ ingly to initiate the inflammatory cascade. Unfortunately, under these circumstances the inflammatory cascade actually causes greater harm than good. A strategy for combating undesirable initiation of the inflammatory cascade is to disrupt the adhesion of leukocytes. This can be done by blocking the selectin binding sites for sialyl Lewisx acids. Chemists have advanced this approach by synthesizing both natural and mimetic sialyl Lewisx acids for studies on the binding process. These compounds have helped identify key functional groups in sialyl Lewisx acids that are required

1000

Patrolling leukocytes bind at th e site of traum a by interactions betw een sialyl Lewis x g lyco p ro te in s on th e ir surface and selectin proteins on th e injured cell. (R eprinted w ith perm ission fro m Sim anek, E.E.; McGarvey, G.J.; Jablonow ski, J.A.; W ong, C.A., Chemical Reviews, 98, p. 835, Figure 1, 1998. C o p yrig h t 1998 Am erican Chemical Society.)

for recognition and binding. Chemists have even designed and synthesized novel compounds that have tighter binding affinities than the natural sialyl Lewisx acids. Among them are polymers with repeating occurrences of the structural motifs essential for binding. These polymeric species presumably occupy multiple sialyl Lewisx acid binding sites at once, thereby binding more tightly than monomeric sialyl Lewisx acid analogs. Efforts like these to prepare finely tuned molecular agents are typical of research in drug discovery and design. In the case of sialyl Lewis* acid analogs, chemists hope to create new therapies for chronic inflammatory diseases by mak­ ing ever-improved agents for blocking undesired leukocyte adhesion.

22.1 Introduction 22.1A Classification of Carbohydrates The group of compounds known as carbohydrates received their general name because of early observations that they often have the formula Cx(H2O)y— that is, they appear to be “hydrates of carbon.” Simple carbohydrates are also known as sugars or saccha­ rides (Latin Greek sugar) and the ending of the names of most sugars is Thus, we have such names as for ordinary table sugar, for the principal sugar in blood, for a sugar in fruits and honey, and for malt sugar.

saccharum, -ose.

sakcharon, sucrose fructose

glucose maltose

• Carbohydrates are usually defined as polyhydroxy aldehydes and ketones or sub­ stances that hydrolyze to yield polyhydroxy aldehydes and ketones. They exist pri­ marily in their hemiacetal or acetal forms (Section 16.7). The simplest carbohydrates, those that cannot be hydrolyzed into simpler carbohydrates, are called monosaccharides. On a molecular basis, carbohydrates that undergo hydroly­ sis to produce only 2 molecules of monosaccharide are called disaccharides; those that yield 3 molecules of monosaccharide are called trisaccharides; and so on. (Carbohydrates that hydrolyze to yield 2-10 molecules of monosaccharide are sometimes called oligosac­ charides.) Carbohydrates that yield a large number of molecules of monosaccharides (> 10) are known as polysaccharides. Maltose and sucrose are examples of disaccharides. On hydrolysis, 1 mol of maltose yields 2 mol of the monosaccharide glucose; sucrose undergoes hydrolysis to yield 1 mol of glucose and 1 mol of the monosaccharide fructose. Starch and cellulose are examples

H elpful H in t You may find it helpful now to review the chemistry of hemiacetals and acetals (Section 16.7).

1002

Chapter 22

Carbohydrates

of polysaccharides; both are glucose polymers. Hydrolysis of either yields a large number of glucose units. The following shows these hydrolyses in a schematic way: -O

/— O -O

'

V

/— O

o

H

- H3° ^

2

mol of m altose

1 A

2

d is a c c h a r id e

A

\

mol of glu cose m o n o s a c c h a r id e

OH + mol of su cr o se

1 A

1

mol of g lu co se +

d is a c c h a r id e

O

1

C

1

} — OH

mol of fructose

M o n o s a c c h a r id e s

O—< 1

OH

^O H

mol of starch or mol of cellu lose

/— O

h 3 o+

— — > n

OH

many m oles of g lu co se

P o ly s a c c h a r id e s

M o n o s a c c h a r id e s

Carbohydrates are the most abundant organic constituents of plants. They not only serve as an important source of chemical energy for living organisms (sugars and starches are important in this respect), but also in plants and in some animals they serve as important constituents of supporting tissues (this is the primary function of the cellulose found in wood, cotton, and flax, for example). We encounter carbohydrates at almost every turn of our daily lives. The paper on which this book is printed is largely cellulose; so, too, is the cotton of our clothes and the wood of our houses. The flour from which we make bread is mainly starch, and starch is also a major constituent of many other foodstuffs, such as potatoes, rice, beans, corn, and peas. Carbohydrates are central to metabolism, and they are important for cell recognition (see the chapter opening vignette and Section 22.16).

22.1B Photosynthesis and Carbohydrate Metabolism

photosynthesis

Carbohydrates are synthesized in green plants by — a process that uses solar energy to reduce, or “fix,” carbon dioxide. Photosynthesis in algae and higher plants occurs in cell organelles called chloroplasts. The overall equation for photosynthesis can be writ­ ten as follows: x CO 2 + y H2O + solar energy s

Cx(H2 O)y + x O 2 C a rb o h y d ra te

Schem atic diagram o f a chloroplast from corn. (Reprinted w ith perm ission o f John W iley & Sons, Inc., from Voet, D. and Voet, J. G., Biochemistry, Second Edition. © 1995 Voet, D. and Voet, J. G.)

Many individual enzyme-catalyzed reactions take place in the general photosynthetic process and not all are fully understood. We know, however, that photosynthesis begins with the absorp­ tion of light by the important green pigment of plants, chlorophyll (Fig. 22.1). The green color of chlorophyll and, therefore, its ability to absorb sunlight in the visible region are due pri­ marily to its extended conjugated system. As photons of sunlight are trapped by chlorophyll, energy becomes available to the plant in a chemical form that can be used to carry out the reac­ tions that reduce carbon dioxide to carbohydrates and oxidize water to oxygen. Carbohydrates act as a major chemical repository for solar energy. Their energy is released when animals or plants metabolize carbohydrates to carbon dioxide and water: Cx(H2 O)y + x O

2

s

x CO 2 + y H2O + energy

The metabolism of carbohydrates also takes place through a series of enzyme-catalyzed reactions in which each energy-yielding step is an oxidation (or the consequence of an oxidation).

1003

22.1 Introduction

F igu re 2 2 .1 C h lo r o p h y ll a. [T h e s t r u c t u r e o f c h lo r o p h y ll a w a s e s ta b lis h e d la r g e ly t h r o u g h t h e w o r k o f H . F is c h e r (M u n ic h ) , R. W ills t a t t e r (M u n ic h ) , a n d J. B. C o n a n t (H a r v a r d ) . A s y n th e s is o f c h lo r o p h y ll a fr o m s im p le o r g a n ic c o m p o u n d s w a s a c h ie v e d b y R. B. W o o d w a r d (H a r v a r d ) in 1 9 6 0 , w h o w o n t h e N o b e l p r iz e in 1 9 6 5 f o r h is o u t s ta n d in g c o n t r ib u t io n s t o s y n th e tic o r g a n ic c h e m is tr y .]

Although some of the energy released in the oxidation of carbohydrates is inevitably con­ verted to heat, much of it is conserved in a new chemical form through reactions that are coupled to the synthesis of adenosine triphosphate (ATP) from adenosine diphosphate (ADP) and inorganic phosphate (Pj) (Fig. 22.2). The phosphoric anhydride bond that forms between the terminal phosphate group of ADP and the phosphate ion becomes another repository of chemical energy. Plants and animals can use the conserved energy of ATP (or very similar substances) to carry out all of their energy-requiring processes: the contraction of a muscle,

Adenine NH2

O

H

ADP

O

/OH HO\CH2— O — P— O — P— OH I I H H OOO

O +

—

P —

O -

I

O -

Diphosphate

Ribose

H O

Hydrogen phosphate ion

chemical energy from oxidation reactions

NH

H

O

H

/OH HO\CH,— O

ATP

h^

o^

O

P— O — P— O — P— O + O

h

O

-

F igu re 2 2 .2

H2O

T h e s y n th e s is o f a d e n o s in e

tr ip h o s p h a t e (A T P ) f r o m a d e n o s in e d ip h o s p h a te (A D P ) a n d h y d r o g e n p h o s p h a te

O-

O-

New phosphoric anhydride bond

io n . T h is r e a c tio n ta k e s p la c e in a ll liv in g o r g a n is m s , a n d a d e n o s in e t r ip h o s p h a t e is th e m a jo r c o m p o u n d in t o w h ic h t h e c h e m ic a l e n e r g y r e le a s e d b y b io lo g ic a l o x id a tio n s is tr a n s fo r m e d .

1004

Chapter 22

Carbohydrates

the synthesis of a macromolecule, and so on. When the energy in ATP is used, a coupled reaction takes place in which ATP is hydrolyzed, ATP + H2O s ADP + Pj + energy

or a new anhydride linkage is created, O

O

,C .

+ A T P ----- >

II

R

OH

II R

.a

O

II

^O —P -O -

+ ADP

1

O-

An acyl phosphate

22.2 Monosaccharides 2 2 .2 A

Classification of Monosaccharides

Monosaccharides are classified according to (1) the number of carbon atoms present in the molecule and (2) whether they contain an aldehyde or keto group. Thus, a monosaccharide containing three carbon atoms is called a one containing four carbon atoms is called a ; one containing five carbon atoms is a and one containing six carbon atoms is a A monosaccharide containing an aldehyde group is called an aldose; one con­ taining a keto group is called a ketose. These two classifications are frequently combined. A C 4 aldose, for example, is called an a C 5 ketose is called a

triose; pentose;

tetrose hexose.

aldotetrose;

ketopentose.

CH2OH

/H

1 C=0

^C 1 H — C — OH

1 C=0 1 H— C — OH

(H— C — OH)n

H — C — OH

H— C — OH

CH2OH

CH2OH

CH2OH

0^. /H C (H — C — OH)n CH2OH

CH2OH

An aldose

Review Problem 22.1

A ketose

An aldotetrose (C4)

A ketopentose (C5)

How many chirality centers are contained in (a) the aldotetrose and (b) the ketopentose just given? (c) How many stereoisomers would you expect from each general structure?

2 2 .2 B d

and

l

Designations of Monosaccharides

The simplest monosaccharides are the compounds glyceraldehyde and dihydroxyacetone (see the following structures). O f these two compounds, only glyceraldehyde contains a chirality center. CHO

0

-O -

1 0

-O -

1

c h 2o h Glyceraldehyde (an aldotriose)

CH2OH

c h 2o h Dihydroxyacetone (a ketotriose)

Glyceraldehyde exists, therefore, in two enantiomeric forms that are known to have the absolute configurations shown here:

22.2 Monosaccharides O

H

H

OH

OV and

CH2OH

1005

H H

HO

CH2OH

(+)-Glyceraldehyde

(-)-Glyceraldehyde

We saw in Section 5.7 that, according to the Cahn-Ingold-Prelog convention, (+)-glyceraldehyde should be designated (R)-(+)-glyceraldehyde and (—)-glyceraldehyde should be designated (S)-(—)-glyceraldehyde. Early in the twentieth century, before the absolute configurations of any organic com­ pounds were known, another system of stereochemical designations was introduced. According to this system (first suggested by M . A. Rosanoff of New York University in 1906), (+)-glyceraldehyde is designated D-(+)-glyceraldehyde and (-)-glyceraldehyde is designated l - ( —)-glyceraldehyde. These two compounds, moreover, serve as configurational standards for all monosaccharides. A monosaccharide (the penultimate carbon) has the same configuration as D-(+)-glyceraldehyde is des­ ignated as a d sugar; one whose highest numbered chirality center has the same configu­ ration as L-glyceraldehyde is designated as an l sugar. By convention, acyclic forms of monosaccharides are drawn vertically with the aldehyde or keto group at or nearest the top. When drawn in this way, d sugars have the — OH on their penultimate carbon on the right:

whosehighestnumberedchirality

center

1

2

CHO *1

H— C— OH H— C— OH •OH 4

C

5

c h 2o h

c h 2o h

2

C=O

3

* H— C— OH

4

H— C— OH

*1 3

i

HO h ig h e s t n u m b e re d c h ira lity c e n te r

-----------------------------------6

A D- a ld o p e n to s e

C5

H

I

c h 2o h

A n L - k e to h e x o s e

The d a n d l n o m e n c l a t u r e designations are like (R) and (S) designations in that they are not necessarily related to the optical rotations of the sugars to which they are applied. Thus, one may encounter other sugars that are d -(+ ) or d -(—) and ones that are l -(+ ) or l -(—). The D-L system of stereochemical designations is thoroughly entrenched in the litera­ ture of carbohydrate chemistry, and even though it has the disadvantage of specifying the configuration of only one chirality center— that of the highest numbered chirality center— we shall employ the d- l system in our designations of carbohydrates.

Write three-dimensional formulas for each aldotetrose and ketopentose isomer in Review Problem 22.1 and designate each as a d or l sugar.

22.2C Structural Formulas for Monosaccharides Later in this chapter we shall see how the great carbohydrate chemist Emil Fischer* was able to establish the stereochemical configuration of the aldohexose D-(+)-glucose, the most abundant monosaccharide. In the meantime, however, we can use D-(+)-glucose as an exam­ ple illustrating the various ways of representing the structures of monosaccharides. *Emil Fischer (1852-1919) was professor of organic chemistry at the University of Berlin. In addition to monumental work in the field of carbohydrate chemistry, where Fischer and co-workers established the configuration of most of the monosaccharides, Fischer also made important contributions to studies of amino acids, proteins, purines, indoles, and stereochemistry generally. As a graduate student, Fischer discovered phenylhydrazine, a reagent that was highly important in his later work with carbohydrates. Fischer was the second recipient (in 1902) of the Nobel Prize in Chemistry.

Review Problem 22.2

1006

Chapter 22

Carbohydrates CHO

CHO H -------

H

OH

H O -------

HO

H

H -------

OH

H

OH

CH2OH

CHO

> "O H

> " H

H

> "O H

H

y

_

OH

H—

C —

OH

HO—

C —

H

H ^ C

—

OH

H ^ C

—

OH

CH2 OH

CH2 OH

Circle-and-line formula

Fischer projection formula

Wedge-linedashed wedge formula

1

+

Haworth formulas

OH Figure 22.3 Formulas 1 -3 are used fo r th e open-chain s tructu re o f D-(+)-glucose. Formulas 4 -7 are used fo r the tw o cyclic hem iacetal fo rm s o f D(+)-glucose.

OH

ho:

+

HOJ HO' OOH

h

a-D-(+)-Glucopyranose 6

HO

O OH

HO HO

b-D-(+)-Glucopyranose 7

Fischer represented the structure of D-(+)-glucose with the cross formulation (1) in Fig. 22.3. This type of formulation is now called a Fischer projection (Section 5.13) and is still useful for carbohydrates. In Fischer projections, by convention, horizontal lines pro­ ject out toward the reader and vertical lines project behind the plane o f the page. When we use Fischer projections, however, we must not (in our mind’s eye) remove them from the plane o f the page in order to test their superposability and we must not rotate them by 90°. In terms of more familiar formulations, the Fischer projection translates into formulas 6 and 7. In IUPAC nomenclature and with the Cahn-Ingold-Prelog system of stereochemi­ cal designations, the open-chain form of D-(+)-glucose is (2R,3S,4R,5R)-2,3,4,5,6-pentahydroxyhexanal.

H elpful H in t _____ Use molecular models to help you learn to interpret Fischer projection formulas.

The meaning of formulas 1, 2, and 3 can be seen best through the use of molecular models: We first construct a chain of six carbon atoms with the — CHO group at the top and a — CH2OH group at the bottom. We then bring the CH2OH group up behind the chain until it almost touches the — CHO group. Holding this model so that the — CHO and — CH2OH groups are directed generally away from us, we then begin placing — H and — OH groups on each of the four remaining carbon atoms. The — OH group of C2 is placed on the right; that of C3 on the left; and those of C4 and C5 on the right.

1007

22.2 Monosaccharides

H\ C^ O I H — 2C — O H H O — 3C — H 4

H H

C — OH

OH

5C 6c

h

2o

h

Glucose (p la n e projectio n fo rm u la ) W h e n a m o d e l of this is m a d e it will coil a s follow s:

H C H 2O H

4

H /

1

OH-

1

,CH =

C

O

2H h

O

\ ?

H ■C

I

3f

OH

H

If th e g ro u p a tta c h e d to C 4 is p ivo ted a s th e a rro w s in dicate,

T h is

w e h a v e th e stru ctu re belo w .

O H g ro u p a d d s

a c ro s s th e ^

C = O

to c lo s e a ring of six a to m s a n d m a k e a cyclic h e m ia c e ta l. C H 2O H I 2 'C H

H

H

CH2O H 5 '

O

'C ' H

H

H

I ch

c ' h

O

\ O

h

C—

4H\ HO

H Z ■C

I

I

H

OH

C H 2O H I 2

O

OH

1V

H

_________ 2 H C

C

H

o

-

OH

/ C

C

/ /

O

h

^

o h

_____ H

I

I

OH

OH

H

a-D-(+)-Glucopyranose

Open-chain form of D-glucose

b-D-(+)-Glucopyranose

(S ta rre d — O H is th e

(T h e proton tra n s fe r s te p o c curs

(S ta rre d — O H is th e

h e m ia c e ta l — O H , w h ic h in a -g lu c o s e is on

b e tw e e n s e p a ra te m o le c u le s .

h e m ia c e ta l — O H ,

It is not in tra m o le c u la r

w h ich in b -g lu c o s e is on

th e

opposite side

o f th e ring

from th e — C H 2O H g ro u p a t C 5.)

o r c o n c e rte d .)

th e

same side

o f th e ring

a s th e — C H 2O H g rou p a t C 5 .’

Figure 22.4 H aw orth form ulas fo r th e cyclic hem iacetal fo rm s o f D-(+)-glucose and th e ir relation to th e open-chain p o lyh yd ro xy aldehyde structure. (R eprinted w ith perm ission o f John W ile y & Sons, Inc., from Holum , J. R., Organic Chemistry: A Brief Course, p. 316. C o p yrig h t 1975.)

Although many of the properties of D-(+)-glucose can be explained in terms of an openchain structure (1, 2, or 3), a considerable body of evidence indicates that the open-chain structure exists, primarily, in equilibrium with two cyclic forms. These can be represented by structures 4 and 5 or 6 and 7. The cyclic forms of D-(+)-glucose are hemiacetals formed by an intramolecular reaction of the — OH group at C5 with the aldehyde group (Fig. 22.4). Cyclization creates a new chirality center at C1, and this chirality center explains how two cyclic forms are possible. These two cyclic forms are that differ only in the configuration of C1 .

diastereomers

1008

Chapter 22

Carbohydrates

• In carbohydrate chemistry diastereomers differing only at the hemiacetal or acetal carbon are called anomers, and the hemiacetal or acetal carbon atom is called the anomeric carbon atom.

H elpful H in t _____ a and b also find common use in steroid nomenclature (Section 23.4A).

Structures 4 and 5 for the glucose anomers are called Haworth formulas* and, although they do not give an accurate picture of the shape of the six-membered ring, they have many practical uses. Figure 22.4 demonstrates how the representation of each chirality center of the open-chain form can be correlated with its representation in the Haworth formula. Each glucose anomer is designated as an a anomer or a b anomer depending on the location of the — OH group of C1. When we draw the cyclic forms of a d sugar in the ori­ entation shown in Figs. 22.3 or 22.4, the anomer has the — OH trans to the — CH2OH group and the b anomer has the — OH cis to the — CH2OH group. Studies of the structures of the cyclic hemiacetal forms of D-(+)-glucose using X-ray analysis have demonstrated that the actual conformations of the rings are the chair forms represented by conformational formulas 6 and 7 in Fig. 22.3. This shape is exactly what we would expect from our studies of the conformations of cyclohexane (Chapter 4), and it is especially interesting to notice that in the b anomer of D-glucose all of the large sub­ stituents, — OH and — CH2OH, are equatorial. In the anomer, the only bulky axial sub­ stituent is the — OH at C1 . It is convenient at times to represent the cyclic structures of a monosaccharide without specifying whether the configuration of the anomeric carbon atom is or b. When we do this, we shall use formulas such as the following:

a

a

a

OH

HO;

O

H O ^ V ^ HO

T h e sym bol w

OH

in d ic a te s a o r p (th re e -d im e n s io n a l v ie w n ot s p e c ified ).

Not all carbohydrates exist in equilibrium with six-membered hemiacetal rings; in sev­ eral instances the ring is five membered. (Even glucose exists, to a small extent, in equi­ librium with five-membered hemiacetal rings.) Because of this variation, a system of nomenclature has been introduced to allow designation of the ring size. • If the monosaccharide ring is six membered, the compound is called a pyranose; if the ring is five membered, the compound is designated as a furanose.** Thus, the full name of compound 4 (or 6) is a-D-(+)-glucopyranose, while that of 5 (or 7) is b-D-(+)-glucopyranose.

*H a w orth form ulas are named after the English chemist W. N. H aw orth (U niversity o f B irm ingham ), who, in 1926, along w ith E. L . H irst, demonstrated that the cyclic fo rm o f glucose acetals consists o f a six-membered ring. H aw orth received the N obel prize fo r his w o rk in carbohydrate chemistry in 1937. For an excellent discussion o f H aw orth form ulas and the ir relation to open-chain form s, see “ The Conversion o f Open Chain Structures o f Monosaccharides in to the Corresponding H aw orth Formulas,” Wheeler, D. M . S., Wheeler, M . M ., and Wheeler, T. S., J. Chem. Educ. 1982, 59, 969-970.

**These names come fro m the names o f the oxygen heterocycles pyran and furan + ose:

A p yran

Furan

Draw the b-pyranose form of (a) in its lowest energy chair conformation, and a Fischer projection for (b). (a )

CHO H-

-O H

H-

-O H

HOH-

-H

(b ) O H

Review Problem 22.3

OH _ --O OH HO

HO OH

-O H C H 2O H

22.3 M u taro tatio n Part of the evidence for the cyclic hemiacetal structure for D-(+)-glucose comes from exper­ iments in which both a and b forms have been isolated. Ordinary D-(+)-glucose has a melt­ ing point of 146°C. However, when D-(+)-glucose is crystallized by evaporating an aqueous solution kept above 98°C, a second form of D-(+)-glucose with a melting point of 150°C can be obtained. When the optical rotations of these two forms are measured, they are found to be significantly different, but when an aqueous solution of either form is allowed to stand, its rotation changes. The specific rotation of one form decreases and the rotation of the other increases, A solution of ordinary D-(+)-glucose (mp 146°C) has an initial specific rotation of +112, but, ultimately, the specific rotation of this solution falls to +52.7. A solution of the second form of D-(+)-glucose (mp 150°C) has an initial specific rotation of +18.7, but, slowly, the specific rotation of this solution rises to +52.7.

untilbothsolutionsshowthesamevalue.

• This change in specific rotation toward an equilibrium value is called mutarotation. The explanation for this mutarotation lies in the existence of an equilibrium between the open-chain form of D-(+)-glucose and the and b forms of the cyclic hemiacetals:

a

« -D-(+)-Glucopyranose (mp 146°C; [«]D5 = +112)

Open-chain form of D-(+)-glucose

£ -D-(+)-Glucopyranose (mp 150°C; [«]D5 = +18.7)

a

X-Ray analysis has confirmed that ordinary D-(+)-glucose has the configuration at the anomeric carbon atom and that the higher melting form has the b configuration. The concentration of open-chain D-(+)-glucose in solution at equilibrium is very small. Solutions of D-(+)-glucose give no observable U V or IR absorption band for a carbonyl group, and solutions of D-(+)-glucose give a negative test with Schiff’s reagent— a special reagent that requires a relatively high concentration of a free aldehyde group (rather than a hemiacetal) in order to give a positive test. Assuming that the concentration of the open-chain form is negligible, one can, by use of the specific rotations in the preceding figures, calculate the percentages of the a and b anomers present at equilibrium. These percentages, 36% anomer and 64% b anomer, are in accord with a greater stability for b-D-(+ )-glucopyranose. This preference is what we might expect on the basis of its having only equatorial groups:

a

1010

Chapter 22

Carbohydrates OH HO; HO

OH O

HO

HO

O OH

HO

HO

(e q u a to ria l)

OH (axial)

a

- D-(+)-G lucopyranose (36% at equilibrium )

ß -D -(+)-G lucopyranose (64% at e quilibrium )

The b anomer of a pyranose is not always the more stable, however. With D-mannose, the equilibrium favors the anomer, and this result is called an

a

OH HO

anomericeffect:

OH

OH O

HO

HO

OH O OH

OH

a -D-Mannopyranose (69% at equilibrium)

ß -D-Mannopyranose (31% at equilibrium)

The anomeric effect is widely believed to be caused by hyperconjugation. An axially ori­ ented orbital associated with nonbonding electrons of the ring oxygen can overlap with a s * orbital of the axial exocyclic C— O hemiacetal bond. This effect is similar to that which causes the lowest energy conformation of ethane to be the anti conformation (Section 4.8). An anomeric effect w ill frequently cause an electronegative substituent, such as a hydroxyl or alkoxyl group, to prefer the axial orientation.

22.4 Glycoside Form ation When a small amount of gaseous hydrogen chloride is passed into a solution of D-(+)-glucose in methanol, a reaction takes place that results in the formation of anomeric methyl

acetals: OH

OH

D-(+)-Glucose

Methyl a-D-glucopyranoside (mp 165°C; [a]D 25 = +158)

OH

Methyl ß-D-glucopyranoside (mp 107°C; [a]D 25 = -33)

• Carbohydrate acetals are generally called g l y c o s id e s (see the following mecha­ nism), and an acetal of glucose is called a . (Acetals of mannose are acetals of fructose are and so on.)

nosides,

glucoside fructosides,

man-

The methyl D-glucosides have been shown to have six-membered rings (Section 22.2C) so they are properly named methyl a-D-glucopyranoside and methyl b-D-glucopyranoside. The mechanism for the formation of the methyl glucosides (starting arbitrarily with D-glucopyranose) is as follows:

b-

1011

22.4 Glycoside Formation

A MECHANISM FOR THE REACTION F o rm a tio n o f a G ly c o s id e

OH

OH

HO

HO

HO

-H—A

OH

HO

CO

- h2o +h2o

OH

^ - D - G lu c o p y r a n o s e

OH (a)

OH

HO

HO

H -H A

^1+

HO

HO

OH

:A

r

OCH 3

(a)

OH

HO

+HA

M e t h y l / 3 - D - g lu c o p y r a n o s id e

HO

+ HOCH,

OH

OH (b)

-HA^ +HA

H

)( b

o f t h e r e s o n a n c e - s t a b i li z e d

H O -T " ^

Ao

o x y g e n o c c u r s o n e it h e r fa c e

OH O

H O - 'T ''^

A t t a c k b y t h e a lc o h o l

c a r b o c a tio n .

I+^ u

oh

H

OH

HO

3

=OCH3

- A

M e th y l a - D - g lu c o p y r a n o s id e

OH O

h 2o , h 3o +

OR

HO HO G ly c o s id e

HO

O OH

HO

ROH

HO A sugar

O

H O -\_

You should review the mechanism for acetal formation given in Section 16.7B and com­ pare it with the steps given here. Notice, again, the important role played by the electron pair of the adjacent oxygen atom in stabilizing the carbocation that forms in the second step. Glycosides are stable in basic solutions because they are acetals. In acidic solutions, how­ ever, glycosides undergo hydrolysis to produce a sugar and an alcohol (again, because they are acetals, Section 16.7B). The alcohol obtained by hydrolysis of a glycoside is known as an aglycone:

HO

och3

OH

A g ly c o n e

( s t a b l e in b a s i c s o l u t i o n s )

For example, when an aqueous solution of methyl b-D-glucopyranoside is made acidic, the glycoside undergoes hydrolysis to produce D-glucose as a mixture of the two pyranose forms (in equilibrium with a small amount of the open-chain form).

1012

Chapter 22

Carbohydrates

S 'f/l

A MECHANISM FOR THE REACTION

///«

H y d ro ly s is o f a G ly c o s id e

OH

OH

,o [ H o HO

H p j+ ( OCH,

" H^O— H+ H

^ o c h3

ch3oh +ch3oh

HO

HO

Methyl /3-D-glucopyranoside

(a). HO

HO /i-D-Glucopyranose

H ,I O— H

OH

occurs on either face of the resonance-stabilized carbocation.

OH HO

(b)

HO

I H(

:OH

a-D-Glucopyranose

Glycosides may be as simple as the methyl glucosides that we have just studied or they may be considerably more complex. Many naturally occurring compounds are glycosides. An example is a compound found in the bark of willow trees:

salicin,

Carbohydrate

Aglycone

As early as the time of the ancient Greeks, preparations made from willow bark were used in relieving pain. Eventually, chemists isolated salicin from other plant materials and were able to show that it was responsible for the analgesic effect of the willow bark prepa­ rations. Salicin can be converted to salicylic acid, which in turn can be converted into the most widely used modern analgesic, (Section 21.8).

aspirin

TSolvedProblem22.1 ) In neutral or basic solutioms, glycosides do not show mutarotation. However, if the solutions are made acidic, glycosides show mutarotatic n. Explain. ANSWER Because glyci:>sides are acetals, they undergo hydrolysis in aqueous acid to form cyclic hemiacetals

that then undergo mutarotation. Acetals are stable to base, and therefore in basic solution they do not show mutarotation.

1013

22.5 Other Reactions of Monosaccharides

(a )

What products would be formed if salicin were treated with dilute aqueous HCl?

(b )

Outline a mechanism for the reactions involved in their formation.

Review Problem 22.4

How would you convert D-glucose to a mixture of ethyl a-D-glucopyranoside and ethyl ßD-glucopyranoside? Show all steps in the mechanism for their formation.

Review Problem 22.5

22.5 O th e r Reactions o f Monosaccharides 22.5A Enolization, Tautomerization, and Isomerization Dissolving monosaccharides in aqueous base causes them to undergo a series of enolizations and keto-enol tautomerizations that lead to isomerizations. For example, if a solution of D-glucose containing calcium hydroxide is allowed to stand for several days, a number of products can be isolated, including D-fructose and D-mannose (Fig. 22.5). This type of reaction is called the L o b r y d e B r u y n - A l b e r d a v a n E k e n s t e i n t r a n s f o r m a t i o n after the two Dutch chemists who discovered it in 1895. When carrying out reactions with monosaccharides, it is usually important to prevent these isomerizations and thereby to preserve the stereochemistry at all of the chirality cen­ ters. One way to do this is to convert the monosaccharide to the methyl glycoside first. We can then safely carry out reactions in basic media because the aldehyde group has been con­ verted to an acetal and acetals are stable in aqueous base. Preparation of the methyl gly­ coside serves to “protect” the monosaccharide from undesired reactions that could occur with the anomeric carbon in its hemiacetal form.

E p im e riz e d a t C -2 H

H

H ^

C ^

^

H

. O -

''C '

O

rela tiv e C

I - C — OH

-O H

C

OH

OHHO-

to D -glucose

HO

H

HO

H

H2O HO

H

^

H

HO

H

H2O

OH-

H-

OH

H

OH

H

OH

H

OH

H-

OH

H

OH

H

OH

H

OH

—

OH D -G lu c o s e

—

OH E n o la t e io n

OH

—

OH

D -M a n n o s e

( o p e n - c h a in fo r m ) h2o

oh

C a rb o n y l grou p

H

is o m e riz e d

C H 2O H

OH 'C '

to C -2 C — OH

C = O

rela tiv e to D -glucose

HO

H

tautomerization

HO

H

H ------------ O H

H

OH

H

H

OH

OH —

OH

D -F r u c to s e

—

OH

E n e d io l

FIGURE 22.5 M onosaccharides undergo isom erizations via enolates and enediols when placed in aqueous base. Here w e show how D-glucose isom erizes to D-mannose and to D-fructose.

1014

Chapter 22

Carbohydrates

22.5B Use o f P ro te c tin g G ro u p s in C a rb o h y d ra te Synthesis Protecting groups are functional groups introduced selectively to block the reactivity of cer­ tain sites in a molecule while desired transformations are carried on elsewhere. After the desired transformations are accomplished, the protecting groups are removed. Laboratory reactions involving carbohydrates often require the use of protecting groups due to the mul­ tiple sites of reactivity present in carbohydrates. As we have just seen, formation of a gly­ coside (an acetal) can be used to prevent undesired reactions that would involve the anomeric carbon in its hemiacetal form. Common protecting groups for the alcohol functional groups in carbohydrates include ethers, esters, and acetals.

22.5C Formation of Ethers • Hydroxyl groups of sugars can be converted to ethers using a base and an alkyl halide by a version of the Williamson ether synthesis (Section 11.11B). Benzyl ethers are commonly used to protect hydroxyl groups in sugars. Benzyl halides are easily introduced because they are highly reactive in SN2 reactions. Sodium or potassium hydride is typically used as the base in an aprotic solvent such as DM F or DMSO. The benzyl groups can later be easily removed by hydrogenolysis using a palladium catalyst. Benzyl Ether Formation OH

OBn

OMe

OMe B n

=

C 6 H 5C H 2

Benzyl Ether Cleavage OBn

OH

OMe

OMe

Methyl ethers can also be prepared. The pentamethyl derivative of glucopyranose, for example, can be synthesized by treating methyl glucoside with excess dimethyl sulfate in aqueous sodium hydroxide. Sodium hydroxide is a competent base in this case because the hydroxyl groups of monosaccharides are more acidic than those of ordinary alcohols due to the many electronegative atoms in the sugar, all of which exert electron-withdrawing inductive effects on nearby hydroxyl groups. In aqueous NaOH the hydroxyl groups are all converted to alkoxide ions, and each of these, in turn, reacts with dimethyl sulfate in an Sn2 reaction to yield a methyl ether. The process is called

exhaustivemethylation:

OH

OH

P e n ta m e th y l d e r iv a t iv e

Although not often used as protecting groups for alcohols in carbohydrates, methyl ethers have been useful in the structure elucidation of sugars. For example, evidence for the pyranose form of glucose can be obtained by exhaustive methylation followed by aqueous hydrolysis of the acetal linkage. Because the C2, C3, C4, and C6 methoxy groups of the pentamethyl derivative are ethers, they are not affected by aqueous hydrolysis. (To cleave them requires heating with concentrated HBr or HI, Section 11.12.) The methoxyl group at C1, however, is part of an acetal linkage, and so it is labile under the conditions of aque­ ous hydrolysis. Hydrolysis of the pentamethyl derivative of glucose gives evidence that the C5 oxygen was the one involved in the cyclic hemiacetal form because in the open-chain form of the product (which is in equilibrium with the cyclic hemiacetal) it is the C5 oxy­ gen that is not methylated:

P e n ta m e th y l d e riv a tiv e

CH 2 OCH 3 ----------------------------------------V-------------------------------------------------------------------- J 2 ,3 ,4 ,6 -T e tra -O -m e th y l-D -g lu c o s e

Silyl ethers, including tert-butyldimethylsilyl (TBS) ethers (Section 11.11E) and phenylsubstituted ethers, are also used as protecting groups in carbohydrate synthesis. Butyldiphenylsilyl (TBDPS) ethers show excellent regioselectivity for primary hydroxyl groups in sugars, such as at C6 in a hexopyranose. (We shall see the use of some related silyl ether groups in Section 22.13D.)

tert-

Regioselective TBDPS Ether Formation OTBDPS

OH O HO

OMe

HO HO

O

TBDPS—Cl, AgNO3 TBDPS = ferf-butyldiphenylsilyl, (CH^C^H^Si—

HO

OCH 3

HO HO

TBDPS Ether Cleavage OTBDPS -O HO: HO

OH OCH 3

HO

Bu4N+F THF

O HO

och

3

HO HO

22.5D Conversion to Esters Treating a monosaccharide with excess acetic anhydride and a weak base (such as pyri­ dine or sodium acetate) converts all of the hydroxyl groups, including the anomeric hydroxyl, to ester groups. If the reaction is carried out at a low temperature (e.g., 0°C), the reaction occurs stereospecifically; the a anomer gives the a-acetate and the b anomer gives the b-acetate. Acetate esters are common protecting groups for carbohydrate hydroxyls.

1016

Chapter 22

Carbohydrates

22.5E Conversion to Cyclic Acetals In Section 16.7B we learned that aldehydes and ketones react with open-chain 1,2-diols to produce cyclic acetals: O

HA O

_OH

HOH

O

HO 1 ,2 - D io l

C y c lic a c e ta l

If the 1,2-diol is attached to a ring, as in a monosaccharide, formation of the cyclic acetals occurs only when the vicinal hydroxyl groups are cis to each other. For example, a-Dgalactopyranose reacts with acetone in the following way: O

OH OH HO

H2SO4 HO

2 H2O

OH

Cyclic acetals are commonly used to protect vicinal cis hydroxyl groups of a sugar while reactions are carried out on other parts of the molecule. When acetals such as these are formed from acetone, they are called acetonides.

22.6 O xidation Reactions o f Monosaccharides A number of oxidizing agents are used to identify functional groups of carbohydrates, in elucidating their structures, and for syntheses. The most important are (1) Benedict’s or Tollens’ reagents, (2) bromine water, (3) nitric acid, and (4) periodic acid. Each of these reagents produces a different and usually specific effect when it is allowed to react with a monosaccharide. We shall now examine what these effects are.

22.6A Benedict's or Tollens' Reagents: Reducing Sugars Benedict’s reagent (an alkaline solution containing a cupric citrate complex ion) and Tollens’ solution [Ag (N H 3)2OH] oxidize and thus give positive tests with The tests are positive even though aldoses and ketoses exist primarily as cyclic hemiacetals. We studied the use of Tollens’ silver mirror test in Section 16.12B. Benedict’s solution and the related Fehling’s solution (which contains a cupric tartrate complex ion) give brickred precipitates of Cu2O when they oxidize an aldose. [In alkaline solution ketoses are con­ verted to aldoses (Section 22.5A), which are then oxidized by the cupric complexes.] Since the solutions of cupric tartrates and citrates are blue, the appearance of a brick-red precip­ itate is a vivid and unmistakable indication of a positive test.

aldosesandketoses.

1017

22.6 Oxidation Reactions of Monosaccharides O

H c h 2o h

(H— C— OH)n

Cu2+ (complex)

c h 2o h

or

C =O (H

C

Cu 2 OJ,

oxidation products

OH)n

CH2OH

A ld o s e

B e n e d ic t’s s o lu tio n (blue )

K etose

(b ric k-re d re d u c tio n p ro d u c t)

• Sugars that give positive tests with Tollens’ or Benedict’s solutions are known as reducing sugars, and all carbohydrates that contain a hemiacetal group give posi­ tive tests. In aqueous solution the hemiacetal form of sugars exists in equilibrium with relatively small, but not insignificant, concentrations of noncyclic aldehydes or a-hydroxy ketones. It is the latter two that undergo the oxidation, perturbing the equilibrium to produce more aldehyde or a-hydroxy ketone, which then undergoes oxidation until one reactant is exhausted. • Carbohydrates that contain only acetal groups do not give positive tests with Benedict’s or Tollens’ solutions, and they are called nonreducing sugars. Acetals do not exist in equilibrium with aldehydes or a-hydroxy ketones in the basic aque­ ous media of the test reagents. Reducing Sugar

Nonreducing Sugar

'A lk y l g ro u p o r a n o th e r s u g a r C— O

O-

H

| V — C/

C— O

O-

R

| V X R'

H e m ia ce ta l (R ' = H o r CH2OH) (g iv e s p o s itiv e T o lle n s ’ o r B e n e d ic t’s te st)

— C/

X R'

A c e ta l (R ' = H o r C H2OH) (do es n o t g iv e a p o s itiv e T o lle n s ’ o r B e n e d ic t’s te st)

How might you distinguish between a-D-glucopyranose (i.e., D-glucose) and methyl a-Dglucopyranoside?

Although Benedict’s and Tollens’ reagents have some use as diagnostic tools [Benedict’s solution can be used in quantitative determinations of reducing sugars (reported as glucose) in blood or urine], neither of these reagents is useful as a preparative reagent in carbohy­ drate oxidations. Oxidations with both reagents take place in alkaline solution, and in alka­ line solutions sugars undergo a complex series of reactions that lead to isomerizations (Section 22.5A).

22.6B Bromine Water: The Synthesis of Aldonic Acids Monosaccharides do not undergo isomerization and fragmentation reactions in mildly acidic solution. Thus, a useful oxidizing reagent for preparative purposes is bromine in water (pH 6.0).

Review Problem 22.6

1018

Chapter 22

Carbohydrates

• Bromine water is a general reagent that selectively oxidizes the — CHO group to a — CO2H group, thus converting an aldose to an aldonic acid: CHO (H

C

c o 2h

Br, h2o

OH)n

(H

C

OH)n

CH2OH

CH2OH

A ld o s e

A ld o n ic acid

Experiments with aldopyranoses have shown that the actual course of the reaction is somewhat more complex than we have indicated. Bromine water specifically oxidizes the b anomer, and the initial product that forms is a . This compound may then hydrolyze to an aldonic acid, and the aldonic acid may undergo a subsequent ring closure to form a

d-aldonolactone

g-aldonolactone: OH

OH O - oh

ho ho

Br2 h2o

+ h 2o

HO

HO

h 2o

HO

h

o

''* O

D -G lu c o n o -dla c to n e

0 -D -G lu c o p y ra n o s e

HO ^

H HO

° -OH H

O H

OH

H

OH

OH D -G luconic acid

D -G lu c o n ic -gla c to n e

22.6C Nitric Acid Oxidation: Aldaric Acids • Dilute nitric acid— a stronger oxidizing agent than bromine water— oxidizes both the — CHO group and the terminal — CH2OH group of an aldose to — CO2H groups, forming dicarboxylic acids are known as aldaric acids: CHO (H -C -O H )„

CO2H (H

C

OH)n

CH2OH

CO2H

A ld o s e

A ld a ric acid

It is not known whether a lactone is an intermediate in the oxidation of an aldose to an aldaric acid; however, aldaric acids form g- and d-lactones readily:

22.6 Oxidation Reactions of Monosaccharides OH

O.

*C " I

hm

H— C— OH I

H — C— OH I

h 2o

H— C— OH I

H_ C

OH

-OH C H— C — OH H

0

C

or

Cv HO

H^ C

H— C— OH

C

C O

/ OH

/ H

C

OH

OH

O

A ld a ric acid (fro m an a ld o h e x o se )

1019

OH O OH

C

C

I

y -L a c to n e s of an a ld a ric acid

The aldaric acid obtained from D-glucose is called D-glucaric acid*: Ox

^

H

V

/ OH

H

-OH

OH -OH

H O HO

OH

HO HO

H

HO

hno

3

HO

H

H

-OH

H

-OH

H

-OH

H

-OH

O

OH

OH D -G lucose

(a ) (b )

(c ) (d )

D -G lucaric acid

Would you expect D-glucaric acid to be optically active? Write the open-chain structure for the aldaric acid (mannaric acid) that would be obtained by nitric acid oxidation of D-mannose. Would you expect mannaric acid to be optically active? What aldaric acid would you expect to obtain from D-erythrose? CHO H--------- OH H --------- OH CH2OH D -E rythrose

Would the aldaric acid in (d) show optical activity? D-Threose, a diastereomer of D-erythrose, yields an optically active aldaric acid when it is subjected to nitric acid oxidation. Write Fischer projection formulas for D-threose and its nitric acid oxidation product. (g) What are the names of the aldaric acids obtained from D-erythrose and D-threose? (e ) (f)

*O lder terms fo r an aldaric acid are a glycaric acid or a saccharic acid.

ReviewProblem22.7

1020

Review Problem 22.8

Chapter 22

Carbohydrates

D-Glucaric acid undergoes lactonization to yield two different g-lactones. What are their structures?

22.6D Periodate Oxidations: Oxidative Cleavage of Polyhydroxy Compounds •

Compounds that have hydroxyl groups on adjacent atoms undergo oxidative cleav­ age when they are treated with aqueous periodic acid (HIO4). The reaction breaks carbon-carbon bonds and produces carbonyl compounds (aldehydes, ketones, or acids).

The stoichiometry o f oxidative cleavage by periodic acid is — C— O H HO

HIO,

H2O

— C— O H Since the reaction usually takes place in quantitative yield, valuable information can often be gained by measuring the number o f molar equivalents o f periodic acid consumed in the reaction as w ell as by identifying the carbonyl products.* Periodate oxidations are thought to take place through a cyclic intermediate:

\

C O

— C— OH

-C -rO < - h 2°> >

y

c ^ id o

/ ^ y \

— C— OH

/

o-

— C— O

IO ,

O

C O / Before w e discuss the use o f periodic acid in carbohydrate chemistry, w e should illus­ trate the course o f the reaction with several sim ple examples. N otice in these periodate o xi­ dations that f o r every C — C bond broken, a C— O bond is fo rm ed a t each carbon. 1. When three or more — CHOH groups are contiguous, the internal ones are obtained as form ic acid. Periodate oxidation o f glycerol, for example, gives two molar equiv­ alents o f formaldehyde and one molar equivalent o f formic acid:

O (formaldehyde)

C

H — C— OH — 4 --------

H — C— OH H

\------

C

H

H

H

O 2 IO,

(formic acid)

HA

OH

OH + O

H Glycerol

(formaldehyde) H

C H

2. Oxidative cleavage also takes place when an — OH group is adjacent to the carbonyl group o f an aldehyde or ketone (but not that o f an acid or an ester). Glyceraldehyde yields two molar equivalents o f formic acid and one molar equivalent o f formalde-

*The reagent lead tetraacetate, Pb(O 2CCH 3) 4 , brings about cleavage reactions sim ilar to those of periodic acid. The two reagents are com plem entary; periodic acid works well in aqueous solutions and lead tetraacetate gives good results in organic solvents but is m ore toxic.

22.6 Oxidation Reactions o f Monosaccharides

1021

hyde, while dihydroxyacetone gives two molar equivalents of formaldehyde and one molar equivalent of carbon dioxide: O / C

^

( f o r m ic a c id )

W

OH

+ O

..........^............ H — C — OH

2

JC^

IQ, H

- - 4 -------H

C

OH

( f o r m ic a c id ) OH

+ O

H G ly c e r a ld e h y d e H

(fo r m a ld e h y d e )

C H O

H H

( fo r m a ld e h y d e )

C H

+

H — C — OH

- - 4 -------O = C

2 IQ

C = O

=

O

( c a r b o n d io x id e )

I ------H

C

OH O

H D ih y d r o x y a c e t o n e H

( fo r m a ld e h y d e )

C H

3. Periodic acid does not cleave compounds in which the hydroxyl groups are separated by an intervening — C H 2 — group, nor those in which a hydroxyl group is adjacent to an ether or acetal function: c h

2o

h

CH

2q

c h

no cleavage

IQ

c h

3

H— C— OH

C H 2O H

c h

no clea v a g e

IQ

2r

What products would you expect to be formed when each of the following compounds is treated with an appropriate amount of periodic acid? How many molar equivalents of H I O 4 would be consumed in each case? (a )

2,3-Butanediol

(b )

1,2,3-Butanetriol

(c )

o c h

O

(d )

ds-1,2-Cyclopentanediol

(g )

HO^ OH

HO'

3

OH

OCHc

HO'

(f)

(e )

Review Problem 22.9

(h) D-Erythrose

O

O

OH

OH

Show how periodic acid could be used to distinguish between an aldohexose and a ketohexose. What products would you obtain from each, and how many molar equivalents of H I O 4 would be consumed?

Review Problem 22.10

1022

Chapter 22

Carbohydrates

2 2 .7 Reduction o f Monosaccharides: Alditols

• Aldoses (and ketoses) can be reduced with sodium borohydride to compounds called alditols: c h 2o h

CHO (H -C -O H )n CH2OH

NaBH, or H2, Pt

(H

C

OH)n

CH2OH A lditol

A ld o s e

Reduction of D-glucose, for example, yields D-glucitol: OH OH HO

H O OH

HO HO

OH

HO

H

H NaBH,

OH

HO

H

H

OH

H

OH

H

OH

H

OH

OH

— OH D-G lucitol (o r D -sorbitol)

Review Problem 22.11

(a) Would you expect D-glucitol to be optically active? (b) Write Fischer projection for­ mulas for all of the D-aldohexoses that would yield

opticallyinactivealditols.

22.8 Reactions o f Monosaccharides w ith Phenylhydrazine: Osazones The aldehyde group of an aldose reacts with such carbonyl reagents as hydroxylamine and phenylhydrazine (Section 16.8B). With hydroxylamine, the product is the expected oxime. With enough phenylhydrazine, however, three molar equivalents of phenylhydrazine are consumed and a second phenylhydrazone group is introduced at C2. The product is called a Phenylosazones crystallize readily (unlike sugars) and are useful deriva­ tives for identifying sugars.

phenylosazone.

H O^

/H

H — C — OH (H— C— OH)n + CH2OH A ld o s e

|

c = n n h c 6h 5

C =N N H C 6H5 3 C6H5NHNH2 ----- > (H— C— OH)„ +

C6H5NH2 +

NH3 +

H2O

c h 2o h P h e n y lo s a zo n e

The mechanism for osazone formation probably depends on a series of reactions in which ^ C = N ^ behaves very much like ^ C = O in giving a nitrogen version of an enol.

1023

22.9 Synthesis and Degradation of Monosaccharides

A MECHANISM FOR THE REACTION P h e n y lo s a z o n e F o rm a tio n

^H 9

H H

a

I

C H = N — NHC6H5

A - A :^ H

tautomerization C — OH

I

C H * - N - r N — C6 H 5 Or 6 5 C - '^ l^ H

I

I

(fo rm e d fro m th e a ld o s e )

^ = CH

a

-

NH

I C O

C H = N N H C 6H5



C = NNHC6H5

+

NH3

h 2o

formation results in a loss of the chirality center at C 2 but does not affect other chirality centers; D-glucose and D-mannose, for example, yield the same phenylosazone:

O sazone

H

H HO

O

c h = n n h c 6h -O H H

H

c = n n h c 6h 5

c 6h5nhnh2

HO

H

O

5

c 6h5nhnh2

HO

H

HO

H

H

-O H

H

-O H

H

-O H

H

-O H

H

-O H

H

-O H

OH

OH D-G lucose

S a m e p h e n y lo s a zo n e

OH D -M an no se

This experiment, first done by Emil Fischer, established that D-glucose and D-mannose have the same configurations about C3, C4, and C5. Diastereomeric aldoses that differ in con­ figuration at only one carbon (such as D-glucose and D-mannose) are called epimers. In gen­ eral, any pair of diastereomers that differ in configuration at only a single tetrahedral chirality center can be called epimers.

Although D-fructose is not an epimer of D-glucose or D-mannose (D-fructose is a ketohexose), all three yield the same phenylosazone. ( a ) Using Fischer projection formulas, write an equation for the reaction of fructose with phenylhydrazine. (b ) What information about the stereochemistry of D-fructose does this experiment yield?

22.9

Review Problem 22.12

Synthesis and D egradation o f M

22.9A Kiliani-Fischer Synthesis In 1885, Heinrich Kiliani (Freiburg, Germany) discovered that an aldose can be converted to the epimeric aldonic acids having one additional carbon through the addition of hydro­ gen cyanide and subsequent hydrolysis of the epimeric cyanohydrins. Fischer later extended this method by showing that aldonolactones obtained from the aldonic acids can be reduced to aldoses. Today, this method for lengthening the carbon chain of an aldose is called the Kiliani-Fischer synthesis.

1024

Chapter 22

Carbohydrates H

O

H

OH — OH

d - G ly c e r a ld e h y d e

I HCN

r

1

CN

CN OH

H-

E p im e r ic

HO

H

c y a n o h y d r in s

OH

H-

(s e p a ra te d )

H

OH

OH

(1) Ba(0H)2 (2) H30+ H0\

^

(1) Ba(0H)2 (2) ^ 0 +

0

H-

-OH

H-

-OH

OH

E p im e r ic a ld o n ic a c id s

H0^

0

HO

H

H

-OH

OH OH

O

0 E p im e r ic y - a ld o n o la c to n e s

Na-Hg, H20 pH 3-5 H\

Figure 22.6

A Kiliani-Fischer synthesis o f d-(—)-erythrose and D(—)-threose from D-glyceraldehyde.

^

0

N a - H g , H 20 pH 3 -5 2

H

H

OH

HO

H

OH

H

— OH D -(-)-E ry th ro s e

0

H OH — OH

D -(-)-T h re o s e

We can illustrate the Kiliani-Fischer synthesis with the synthesis of D-threose and d erythrose (aldotetroses) from D-glyceraldehyde (an aldotriose) in Fig. 22.6. Addition of hydrogen cyanide to glyceraldehyde produces two epimeric cyanohydrins because the reaction creates a new chirality center. The cyanohydrins can be separated eas­ ily (since they are diastereomers), and each can be converted to an aldose through hydrol­ ysis, acidification, lactonization, and reduction with Na-Hg at pH 3-5. One cyanohydrin ultimately yields d -(—)-erythrose and the other yields d -(—)-threose. We can be sure that the aldotetroses that we obtain from this Kiliani-Fischer synthesis are both D sugars because the starting compound is D-glyceraldehyde and its chirality cen­ ter is unaffected by the synthesis. On the basis of the Kiliani-Fischer synthesis, we cannot know just which aldotetrose has both — OH groups on the right and which has the top — OH on the left in the Fischer projection. However, if we oxidize both aldotetroses to aldaric acids, one [d-(—)-erythrose] w ill yield an (meso) product while the other [d-(—)-threose] w ill yield a product that is (see Review Problem 22.7).

opticallyinactive opticallyactive

1025

2 2.1 0 The D Family of Aldoses

Review Problem 22.13

What are the structures of L-(+)-threose and L-(+)-erythrose? ( b ) What aldotriose would you use to prepare them in a Kiliani-Fischer synthesis? (a )

Review Problem 22.14

Outline a Kiliani-Fischer synthesis of epimeric aldopentoses starting with d -(—)-erythrose (use Fischer projections). ( b ) The two epimeric aldopentoses that one obtains are D( —)-arabinose and d -(—)-ribose. Nitric acid oxidation of d -(—)-ribose yields an optically inactive aldaric acid, whereas similar oxidation of d -(—)-arabinose yields an optically active product. On the basis of this information alone, which Fischer projection represents d -(—)arabinose and which represents d -(—)-ribose? (a )

Review Problem 22.15

Subjecting d -(—)-threose to a Kiliani-Fischer synthesis yields two other epimeric aldopen­ toses, D-(+)-xylose and d -(—)-lyxose. D-(+)-Xylose can be oxidized (with nitric acid) to an optically inactive aldaric acid, while similar oxidation of d -(—)-lyxose gives an opti­ cally active product. What are the structures of D-(+)-xylose and d -(—)-lyxose? There are eight aldopentoses. In Review Problems 22.14 and 22.15 you have arrived at the structures of four. What are the names and structures of the four that remain?

Review Problem 22.16

22.9B The Ruff Degradation Just as the Kiliani-Fischer synthesis can be used to lengthen the chain of an aldose by one carbon atom, the Ruff degradation* can be used to shorten the chain by a similar unit. The Ruff degradation involves (1) oxidation of the aldose to an aldonic acid using bromine water and (2) oxidative decarboxylation of the aldonic acid to the next lower aldose using hydro­ gen peroxide and ferric sulfate. d -(—)-Ribose, for example, can be degraded to d -(—)-erythrose: H\ ^

H-

H°

°

-OH

H-

-OH

H-

-OH -O H

D -(-)-R ib o s e

H Br, H,O

°

-OH

H

-OH

H

-OH OH

D -R ibo nic acid

H^

H2O2 Fe2(S°4)3

°

H

-OH

H

-OH

C °

OH D -(-)-E ry th r o s e

lactose,

Review Problem 22.17

The aldohexose D-(+)-galactose can be obtained by hydrolysis of a disaccharide found in milk. When D-(+)-galactose is treated with nitric acid, it yields an optically inac­ tive aldaric acid. When D-(+)-galactose is subjected to Ruff degradation, it yields d -(—)lyxose (see Review Problem 22.15). Using only these data, write the Fischer projection formula for D-(+)-galactose.

22.10 The The Ruff degradation and the Kiliani-Fischer synthesis allow us to place all of the aldoses into families or “family trees” based on their relation to d - or L-glyceraldehyde. Such a tree is constructed in Fig. 22.7 and includes the structures of the D-aldohexoses, 1-8. • Most, but not all, of the naturally occurring aldoses belong to the d family, with d (+)-glucose being by far the most common.

♦Developed by Otto Ruff, 1871-1939, a German chemist.

d

Family o f Aldoses

H\ -O H

-H

HO

H■

/

H\

° -H

H■

-O H

-H

HO

-H

H■

-O H

-O H

-O H

H■

-O H

H■

-O H

H■

-O H

-O H

H■

-O H

H■

-O H

H■

-O H

-O H

-O H

-----OH

D -(+ )-G lucose

D -(+ )-M annose

-O H

D -(+ )-A ltro se

H■

-H

H■ HO

H■

-O H

H■

-O H

HO

-H

-H

HO

-H

-H

HO

-H

HO

-H

H■

H■

-O H

D -(+ )-Talose

7

6

A ld o h e x o se s

-OH

D -(+ )-G alactose

8

O

H

N> Os

-O H

-O H

-OH

o

/ °

°

HO

D -(+ )-ld ose

5

Hx

H\ X

-O H

-O H

-OH

D -(-)-G u lo s e

O

.0

/ °

HO

-H

HO

4

3

2

1

Hx

HO

H■

D -(+ )-A llose

0

-O H

-O H

HO

/

O o

3"

- H «=

¡H O

;H -

-O H

H■

-O H

H■

-O H

H■

-O H

H■

-O H

-O H

-O H

D -(-)-R ib o s e

D -(-)-A ra b in o s e

9

!=>H ■

/

'

-H

HO

H■

¡H O

-H

HO

-H

H■

-O H

¡HO

H■

-O H

cr

o 3" Q_ -? 0r+ ) (D

12

H

-OH

.0 -H

<

A ld o te tro s e s

-O H

-OH

D -(-)-E ry th ro s e

□-( (-Threose

| ___________ lengthen chain shorten chain

A

H\ y °

lengthen chain

A

___ I

shorten chain

-O H

-OH D -(+ )-G ly c e ra ld e h y d e

F igu re 2 2 .7 The D family o f aldohexoses.* (From Copyright © 1956 by International Thompson.)

OrganicChemistry

by Fieser, L. F., and Fieser, M.

*A useful m nem onic for the D-aldohexoses: All altruists gladly m ake gum in gallon tanks. W rite the nam es in a line and above each w rite C H 2 OH. Then, for C 5 w rite OH to the right all the w ay across. For C 4 write OH to the right four tim es, then four to the left; for C 3, w rite OH tw ice to the right, tw ice to the left, and repeat; for C2. alternate OH and H to the right. (From Fieser, L. F., and Fieser, M., Organic Chemistry, Reinhold: New York, 1 956; p 3 5 9 .)

N) N)

9

D -(-)-L y x o s e

11

-O H <

A ld o p e n to se s

-O H

D -(+ )-X ylo se

O

&) ~U

-O H

-OH

10

H\

-O H

A ld o trio s e

22.11 Fischer's Proof of the Configuration of D-(+)-Glucose

1027

D-(+)-Galactose can be obtained from milk sugar (lactose), but L-(-)-galactose occurs in a polysaccharide obtained from the vineyard snail, L-(+)-Arabinose is found widely, but D-(-)-arabinose is scarce, being found only in certain bacteria and sponges. Threose, lyxose, gulose, and allose do not occur naturally, but one or both forms (D or L) of each have been synthesized.

Helixpomatia.

22.11 Fischer's P ro o f o f the Configuration o f D-(+)-Glucose Emil Fischer began his work on the stereochemistry of ( + )-glucose in 1888, only 12 years after van’t Hoff and Le Bel had made their proposal concerning the tetrahedral structure of carbon. Only a small body of data was available to Fischer at the beginning: Only a few monosaccharides were known, including ( + )-glucose, (+)-arabinose, and (+)-mannose. [(+)-Mannose had just been synthesized by Fischer.] The sugars ( + )-glucose and (+)-mannose were known to be aldohexoses; (+)-arabinose was known to be an aldopentose. Since an aldohexose has four chirality centers, 24 (or 16) stereoisomers are possi­ ble Fischer arbitrarily decided to lim it his attention to the eight structures with the d configuration given in Fig. 22.7 (structures 1-8). Fischer realized that he would be unable to differentiate between enantiomeric configurations because methods for determining the absolute configuration of organic compounds had not been developed. It was not until 1951, when Bijvoet (Section 5.15A) determined the absolute configuration of L-(+)-tartaric acid [and, hence, D-(+)-glyceraldehyde], that Fischer’s arbitrary assignment of ( + )-glucose to the family we call the d family was known to be correct. Fischer’s assignment of structure 3 to (+)-glucose was based on the following reasoning:

—oneofwhichis(+)-glucose.

1. Nitric acid oxidation of (+)-glucose gives an optically active aldaric acid. This elim­ inates structures 1 and 7 from consideration because both compounds would yield -aldaric acids.

meso 2. D egradationof(+)-glucosegives(-)-arabinose, andnitricacidoxidationof( —)arabinosegivesanopticallyactivealdaricacid.This means that (—)-arabinose can­ not have configuration 9 or 11 and must have either structure 10 or 12. It also establishes that (+)-glucose cannot have configuration 2, 5, or 6. This leaves struc­ tures 3, 4, and 8 as possibilities for (+)-glucose. 3. Kiliani-Fischer synthesis beginning with ( —)-arabinose gives (+)-glucose and (+ )mannose; nitric acid oxidation of (+ )-mannose gives an optically active aldaric acid. This, together with the fact that (+ )-glucose yields a different but also optically active aldaric acid, establishes 10 as the structure of ( —)-arabinose and eliminates 8 as a possible structure for (+ )-glucose. Had ( —)-arabinose been represented by structure 12, a Kiliani-Fischer synthesis would have given the two aldohexoses, 7 and 8, one of which (7) would yield an optically inactive aldaric acid on nitric acid oxidation. 4. Two structures now remain, 3 and 4; one structure represents (+ )-glucose and one represents (+ )-mannose. Fischer realized that (+ )-glucose and (+ )-mannose were epimeric (at C2), but a decision as to which compound was represented by which structure was most difficult.

interchangingthetwoend groups ofanaldosechain wouldyieldthesamealdohexose:

5. Fischer had already developed a method for effectively (aldehyde and primary alcohol) . And, with brilliant logic, Fischer realized that if (+ )-glucose had structure 4, an interchange of end groups

1027

1028

Chapter 22

Carbohydrates H

OH HO­

H

HO-

H

end-group

H-

-O H

H-

-O H

interchange by chemical reactions

OH

HO

H

HO

H

HO

H

HO

H

H-

-O H

H

-O H

H

-O H

H

-O H

H

4

O

( R e c a ll t h a t it is p e r m is s ib le to tu r n a F is c h e r p r o je c tio n 180° in th e p la n e o f th e p a g e .)

OH

O

4

On the other hand, if (+)-glucose has structure 3, an end-group interchange w ill yield a different aldohexose, 13: OH H

H

-O H

HO

H

H

-O H

H

-O H

end-group interchange by chemical reactions

OH 3

HO

H\ X

O

-O H

HO

H

H

HO

H

H-

-O H

H

H

-O H

HO

H

O

-OH H — OH

13 L-G ulose

This new aldohexose, if it were formed, would be an L sugar and it would be the mirror reflection of D-gulose. Thus its name would be L-gulose. Fischer carried out the end-group interchange starting with (+)-glucose and the prod­ uct was the new aldohexose 13. This outcome proved that (+)-glucose has structure 3. It also established 4 as the structure for (+)-mannose, and it proved the structure of l -(+ )gulose as 13. The procedure Fischer used for interchanging the ends of the (+)-glucose chain began with one of the g-lactones of D-glucaric acid (see Review Problem 22.8) and was carried out as follows: O.

V H

OH 1— I

HO

H

OH O A y -la c to n e o f D-glucaric acid

OH

Li H

OH

O HO Na-Hg

HH-

----

OH

H

H

H

-O H

Li

H

OH

H

HO

O

L -G ulonic acid

OH OH

O A y -a ld o n o la c to n e

Na-Hg pH 3 -5 3

1029

22.12 Disaccharides H

OH H HO (co n tin u e d fro m p re v io u s p ag e)

OH

HO

H

H

HO

H

H

OH

H

OH

H

O

H-

<

See WileyPLUS fo r "The Chemistry of... Stereoselective Synthesis of all the L-Aldohexoses."

OH

HO

H elpful H in t

H OH

O L -(+ )-G u lo s e 13

Notice in this synthesis that the second reduction with N a-H g is carried out at pH 3-5. Under these conditions, reduction of the lactone yields an aldehyde and not a primary alcohol. Fischer actually had to subject both g-lactones of D-glucaric acid (Review Problem 22.8) to the procedure just outlined. What product does the other g-lactone yield?

ReviewProblem22.18 22.12 Disaccharides

22.12A Sucrose Ordinary table sugar is a disaccharide called sucrose. Sucrose, the most widely occurring disaccharide, is found in all photosynthetic plants and is obtained commercially from sug­ arcane or sugar beets. Sucrose has the structure shown in Fig. 22.8.

From D-glucose

From D-fructose

linkage

linkage OH

Figure 22.8 Two representations o f the form ula fo r (+)-sucrose (a-D-glucopyranosyl b-D -fructofuranoside).

The structure of sucrose is based on the following evidence: 1.

Sucrose has the molecular formula C12H22O ''.

2.

Acid-catalyzed hydrolysis of l mol of sucrose yields l mol of D-glucose and l mol of D-fructose. HO

6

O

OH

S — OH Oh

h

F ru cto se (as a ß -fu ra n o s e )

1030

Chapter 22

Carbohydrates

3.

Sucrose is a nonreducing sugar; it gives negative tests with Benedict’s and Tollens’ solutions. Sucrose does not form an osazone and does not undergo mutarotation. These facts mean that neither the glucose nor the fructose portion of sucrose has a hemiacetal group. Thus, the two hexoses must have a glycosidic linkage that involves C1 of glucose and C2 of fructose, for only in this way w ill both carbonyl groups be present as full acetals (i.e., as glycosides).

4.

The stereochemistry of the glycosidic linkages can be inferred from experiments done with enzymes. Sucrose is hydrolyzed by an a-glucosidase obtained from yeast but not by b-glucosidase enzymes. This hydrolysis indicates an a configuration at the glucoside portion. Sucrose is also hydrolyzed by sucrase, an enzyme known to hydrolyze b-fructofuranosides but not a-fructofuranosides. This hydrolysis indicates a b configuration at the fructoside portion.

5.

Methylation of sucrose gives an octamethyl derivative that, on hydrolysis, gives 2,3,4,6-tetra-O-methyl-D-glucose and 1,3,4,6-tetra-O-methyl-D-fructose. The identi­ ties of these two products demonstrate that the glucose portion is a pyranoside and that the fructose portion is a furanoside.

The structure of sucrose has been confirmed by X-ray analysis and by an unambiguous synthesis.

22.12B Maltose When starch (Section 22.13A) is hydrolyzed by the enzyme diastase, one product is a dis­ accharide known as maltose (Fig. 22.9). The structure of maltose was deduced based on the following evidence: 1.

When 1 mol of maltose is subjected to acid-catalyzed hydrolysis, it yields 2 mol of D-(+)-glucose.

2.

Unlike sucrose, maltose is a reducing sugar; it gives positive tests with Fehling’s, Benedict’s, and Tollens’ solutions. Maltose also reacts with phenylhydrazine to form a monophenylosazone (i.e., it incorporates two molecules of phenylhydrazine).

3.

Maltose exists in two anomeric forms: a-(+)-maltose, [a]D25 = +168, and b -(+ )maltose, [a]D25 = +112. The maltose anomers undergo mutarotation to yield an equi­ librium mixture, [a]D25 = +136.

Facts 2 and 3 demonstrate that one of the glucose residues of maltose is present in a hemiacetal form; the other, therefore, must be present as a glucoside. The configuration of this glucosidic linkage can be inferred as a, because maltose is hydrolyzed by a-glucosidase enzymes and not by b-glucosidase enzymes.

OH

OH

4 HO [3 H

|2 OH

O

[3 H

La-Glucosidic linkage OH

HO

|2 OH

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22.12 Disaccharides

Maltonic acid (CH3)2SO4 OH"

2,3,4,6-Tetra-Omethyl-D-glucose (as a pyranose)

HO

2,3,6-Tri-Omethyl-D-glucose (as a pyranose)

F ig u re 2 2 .1 0

(a) O x id a t io n o f

m a lto s e t o m a lt o n ic a c id f o llo w e d b y m e th y la t io n a n d h y d r o ly s is . (b) M e th y la t io n a n d 2,3,4,6-Tetra-O-methylD-glucose (as a pyranose)

2,3,5,6-Tetra-O-methylD-gluconic acid

s u b s e q u e n t h y d r o ly s is o f m a lto s e its e lf.

4.

Maltose reacts with bromine water to form a monocarboxylic acid, maltonic acid (Fig. 22.10a). This fact, too, is consistent with the presence of only one hemiacetal group.

5.

Methylation of maltonic acid followed by hydrolysis gives 2,3,4,6-tetra-O-methylD-glucose and 2,3,5,6-tetra-O-methyl-D-gluconic acid. That the first product has a free — OH at C5 indicates that the nonreducing glucose portion is present as a pyranoside; that the second product, 2,3,5,6-tetra-O-methyl-D-gluconic acid, has a free — OH at C4 indicates that this position was involved in a glycosidic linkage with the nonre­ ducing glucose. Only the size of the reducing glucose ring needs to be determined.

6.

Methylation of maltose itself, followed by hydrolysis (Fig. 22.10ft), gives 2,3,4,6tetra-O-methyl-D-glucose and 2,3,6-tri-O-methyl-D-glucose. The free — OH at C5 in the latter product indicates that it must have been involved in the oxide ring and that the reducing glucose is present as a

pyranose.

22.12C Cellobiose Partial hydrolysis of cellulose (Section 22.13C) gives the disaccharide cellobiose (C 1 2 H2 2 O 11 ) (Fig. 22.11). Cellobiose resembles maltose in every respect except one: the configuration of its glycosidic linkage. Cellobiose, like maltose, is a reducing sugar that, on acid-catalyzed hydrolysis, yields two molar equivalents of D-glucose. Cellobiose also undergoes mutarotation and forms a monophenylosazone. Methylation studies show that C1 of one glucose unit is connected

1032

Chapter 22

Carbohydrates

^-G lycosidic linkage -

5 4

Figure 22.11 Two representations o f th e b anomer o f cellobiose, 4-0-(b-D-glucopyranosyl)-8-Dglucopyranose.

in glycosidic linkage with C4 of the other and that both rings are six membered. Unlike maltose, however, cellobiose is hydrolyzed by enzymes and not by a-glucosidase enzymes: This indicates that the glycosidic linkage in cellobiose is b (Fig. 22.11).

b-glucosidase

THE CHEMISTRY OF . . . A r tif ic ia l S w e e t e n e r s ( H o w S w e e t It Is)

Sucrose (table sugar) and fructose are the most common nat­ ural sweeteners. We all know, however, that they add to our calorie intake and promote tooth decay. For these reasons, many people find artificial sweeteners to be an attractive alter­ native to the natural and calorie-contributing counterparts.

Some p roducts th a t contain the artificial sw eetener aspartam e.

Perhaps the most successful and widely used artificial sweetener is aspartame, the methyl ester of a dipeptide formed from phenylalanine and aspartic acid (Section 24.3D). Aspartame is roughly 100 times as sweet as sucrose. It undergoes slow hydrolysis in solution, however, which lim­ its its shelf life in products such as soft drinks. It also cannot be used for baking because it decomposes with heat. Furthermore, people with a genetic condition known as phenylketonuria cannot use aspartame because their metab­ olism causes a buildup of phenylpyruvic acid derived from aspartame. Accumulation of phenylpyruvic acid is harmful, especially to infants. Alitame, on the other hand, is a com­

pound related to aspartame, but with improved properties. It is more stable than aspartame and roughly 2000 times as sweet as sucrose.

Sucralose is a trichloro derivative of sucrose that is an arti­ ficial sweetener. Like aspartame, it is also approved for use by the U.S. Food and Drug Administration (FDA). Sucralose is 600 times sweeter than sucrose and has many properties desirable in an artificial sweetener. Sucralose looks and tastes like sugar, is stable at the temperatures used for cook­ ing and baking, and it does not cause tooth decay or pro­ vide calories.

1033

22.13 Polysaccharides Cyclamate and saccharin, used as their sodium or calcium salts, were popular sweeteners at one time. A common for­ mulation involved a 10:1 mixture of cyclamate and saccha­ rin that proved sweeter than either compound individually. Tests showed, however, that this mixture produced tumors in animals, and the FDA subsequently banned it. Certain exclusions to the regulations nevertheless allow continued use of saccharin in some products.

SO, SO 3H

N— H

O

Cyclamate

Saccharin

the Sharpless asymmetric epoxidation (Sections 11.13 and 22.11) and other enantioselective synthetic methods. OH

OH O

HO OH

OH

L-Glucose

Much of the research on sweeteners involves probing the structure of sweetness receptor sites. One model proposed for a sweetness receptor incorporates eight binding inter­ actions that involve hydrogen bonding as well as van der Waals forces. Sucronic acid is a synthetic compound designed on the basis of this model. Sucronic acid is reported to be 200,000 times as sweet as sucrose.

Many other compounds have potential as artificial sweet­ eners. For example, L sugars are also sweet, and they pre­ sumably would provide either zero or very few calories because our enzymes have evolved to selectively metabo­ lize their enantiomers instead, the D sugars. Although sources of Lsugars are rare in nature, all eight L-hexoses have been synthesized by S. Masamune and K. B. Sharpless using

.C H

ho 2c

22.12D Lactose Lactose (Fig. 22.12) is a disaccharide present in the milk of humans, cows, and almost all other mammals. Lactose is a reducing sugar that hydrolyzes to yield D-glucose and D-galactose; the glycosidic linkage is

b.

From

Figure 22.12 Two representations o f th e b anom er o f lactose, 4-O-(b-D-galactopyranosyl)-8-Dglucopyranose.

22.13 Polysaccharides

• Polysaccharides, also known as glycans, consist of monosaccharides joined together by glycosidic linkages. Polysaccharides that are polymers of a single monosaccharide are called homopolysac­ charides; those made up of more than one type of monosaccharide are called het­ eropolysaccharides. Homopolysaccharides are also classified on the basis of their monosaccharide units. A homopolysaccharide consisting of glucose monomeric units is called a glucan; one consisting of galactose units is a galactan, and so on. Three important polysaccharides, all of which are glucans, are starch, glycogen, and cellulose.

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Chapter 22

Carbohydrates

• Starch is the principal food reserve of plants, glycogen functions as a carbohydrate reserve for animals, and cellulose serves as structural material in plants. As we examine the structures of these three polysaccharides, we shall be able to see how each is especially suited for its function.

22.13A Starch Starch occurs as microscopic granules in the roots, tubers, and seeds of plants. Corn, pota­ toes, wheat, and rice are important commercial sources of starch. Heating starch with water causes the granules to swell and produce a colloidal suspension from which two major com­ ponents can be isolated. One fraction is called and the other Most starches yield 10-20% amylose and 80-90% amylopectin.

amylose

•

amylopectin. Amylose typically consists of more than 1000 D-glucopyranoside units c onnected inalinkagesbetween C1 of one unit and C4 of the next (Fig. 22.13). «(1 : 4) Glucosidic linkage

Figure 22.13 Partial structu re o f amylose, an unbranched p o lym e r o f D-glucose connected in a(1 s 4) glycosidic linkages.

Thus, in the ring size of its glucose units and in the configuration of the glycosidic link­ ages between them, amylose resembles maltose. Chains of D-glucose units with a-glycosidic linkages such as those of amylose tend to assume a helical arrangement (Fig. 22.14). This arrangement results in a compact shape for the amylose molecule even though its molecular weight is quite large (150,000-600,000).

Figure 22.14 Am ylose. The a(1 s 4) linkages cause it to assume th e shape o f a left-handed helix. (Illustration, Irving Geis. Rights ow ned by H oward Hughes M edical In stitu te . N o t to be reproduced w ith o u t perm ission.)

• Amylopectin has a structure similar to that of amylose [i.e., a(1 s 4) links], except that in amylopectin the chains are branched. Branching takes place between C6 of one glucose unit and C1 of another and occurs at intervals of 20-25 glucose units (Fig. 22.15). Physical measurements indicate that amylopectin has a molecular weight of 1-6 million; thus amylopectin consists of hundreds of interconnecting chains of 20-25 glucose units each.

a (1 : 4) Figure 22.15

Partial structu re o f am ylopectin.

22.13B Glycogen • Glycogen has a structure very much like that of amylopectin; however, in glycogen the chains are much more highly branched. Methylation and hydrolysis of glycogen indicate that there is one end group for every 10-12 glucose units; branches may occur as often as every 6 units. Glycogen has a very high mol­ ecular weight. Studies of glycogens isolated under conditions that minimize the likelihood of hydrolysis indicate molecular weights as high as 100 million. The size and structure of glycogen beautifully suit its function as a reserve carbohydrate for animals. First, its size makes it too large to diffuse across cell membranes; thus, glyco­ gen remains inside the cell, where it is needed as an energy source. Second, because glyco­ gen incorporates tens of thousands of glucose units in a single molecule, it solves an important osmotic problem for the cell. Were so many glucose units present in the cell as individual molecules, the osmotic pressure within the cell would be enormous— so large that the cell membrane would almost certainly break.* Finally, the localization of glucose units within a large, highly branched structure simplifies one of the cell’s logistical prob­ lems: that of having a ready source of glucose when cellular glucose concentrations are low and of being able to store glucose rapidly when cellular glucose concentrations are high. There are enzymes within the cell that catalyze the reactions by which glucose units are detached from (or attached to) glycogen. These enzymes operate at end groups by hydrolyz­ ing (or forming) a(1 s 4) glycosidic linkages. Because glycogen is so highly branched, a very large number of end groups is available at which these enzymes can operate. At the same time the overall concentration of glycogen (in moles per liter) is quite low because of its enormous molecular weight. Amylopectin presumably serves a similar function in plants. The fact that amylopectin is less highly branched than glycogen is, however, not a serious disadvantage. Plants have a much lower metabolic rate than animals— and plants, of course, do not require sudden bursts of energy. Animals store energy as fats (triacylglycerols) as well as glycogen. Fats, because they are more highly reduced, are capable of furnishing much more energy. The metabolism of a typical fatty acid, for example, liberates more than twice as much energy per carbon as glucose or glycogen. Why, then, we might ask, have two different energy repositories evolved? Glucose (from glycogen) is readily available and is highly water soluble.** Glucose, as a result, diffuses rapidly through the aqueous medium of the cell and serves as

* T h e p h e n o m e n o n o f o s m o tic p re s s u re o c c u rs w h e n e v e r t w o s o lu tio n s o f d iff e r e n t c o n c e n tr a tio n s a re se p a ra te d b y a m e m b ra n e t h a t a llo w s p e n e tr a tio n ( b y o s m o s is ) o f th e s o lv e n t b u t n o t o f th e s o lu te . T h e o s m o tic p re s s u re ( p ) o n o n e s id e o f th e m e m b ra n e is r e la te d t o th e n u m b e r o f m o le s o f s o lu te p a r tic le s (n ), th e v o lu m e o f th e s o lu tio n (V ), a n d th e gas c o n s ta n t tim e s th e a b s o lu te te m p e ra tu r e ( R T ) : p V = n R T . * * G lu c o s e is a c t u a lly lib e r a te d as g lu c o s e - 6 - p h o s p h a te ( G 6 P ), w h ic h is a ls o w a te r s o lu b le .

1036

Chapter 22

Carbohydrates

an ideal source of “ready energy.” Long-chain fatty acids, by contrast, are almost insolu­ ble in water, and their concentration inside the cell could never be very high. They would be a poor source of energy if the cell were in an energy pinch. On the other hand, fatty acids (as triacylglycerols), because of their caloric richness, are an excellent energy repos­ itory for long-term energy storage.

22.13C Cellulose When we examine the structure of cellulose, we find another example of a polysaccharide in which nature has arranged monomeric glucose units in a manner that suits its function. • Cellulose contains D-glucopyranoside units linked in (1 s 4) fashion in very long unbranched chains. Unlike starch and glycogen, however, the linkages in cellulose are (Fig. 22.16).

b-glycosidiclinkages

Figure 22.16 A p o rtio n o f a cellulose chain. The glycosidic linkages are b(1 s 4).

The b-glycosidic linkages of cellulose make cellulose chains essentially linear; they do not tend to coil into helical structures as do glucose polymers when linked in an a(1 s 4) manner. The linear arrangement of b-linked glucose units in cellulose presents a uniform distri­ bution of — OH groups on the outside of each chain. When two or more cellulose chains make contact, the hydroxyl groups are ideally situated to “zip” the chains together by form­ ing hydrogen bonds (Fig. 22.17). Zipping many cellulose chains together in this way gives a highly insoluble, rigid, and fibrous polymer that is ideal as cell-wall material for plants. This special property of cellulose chains, we should emphasize, is not just a result of b(1 s 4) glycosidic linkages; it is also a consequence of the precise stereochemistry of d glucose at each chirality center. Were D-galactose or D-allose units linked in a similar fash­ ion, they almost certainly would not give rise to a polymer with properties like cellulose.

Figure 22.17 A pro p o se d structu re fo r cellulose. A fib e r o f cellulose may consist of about 40 parallel strands o f glucose m olecules linked in a b(1 s 4) fashion. Each glucose unit in a chain is tu rn e d over w ith respect to th e preceding glucose unit and is held in this p o sition by hydrogen bonds (dashed lines) betw een th e chains. The glucan chains line up laterally to fo rm sheets, and these sheets stack vertica lly so th a t th e y are staggered by one-half o f a glucose unit. (H ydrogen atom s th a t do n o t p a rticip a te in hydrogen bo n ding have been o m itte d fo r clarity.) (Illustration, Irving Geis. Rights ow ned by H ow ard Hughes M edical Institute. N o t to be rep ro d u ced w ith o u t perm ission.)

22.14 Other Biologically Important Sugars

1037

Thus, we get another glimpse of why D-glucose occupies such a special position in the chem­ istry of plants and animals. Not only is it the most stable aldohexose (because it can exist in a chair conformation that allows all of its bulky groups to occupy equatorial positions), but its special stereochemistry also allows it to form helical structures when a linked as in starches, and rigid linear structures when b linked as in cellulose. There is another interesting and important fact about cellulose: The digestive enzymes of humans cannot attack its b(1 s 4) linkages. Hence, cellulose cannot serve as a food source for humans, as can starch. Cows and termites, however, can use cellulose (of grass and wood) as a food source because symbiotic bacteria in their digestive systems furnish b-glucosidase enzymes. Perhaps we should ask ourselves one other question: Why has D-(+)-glucose been selected for its special role rather than L-(-)-glucose, its mirror image? Here an answer cannot be given with any certainty. The selection of D-(+)-glucose may simply have been a random event early in the course of the evolution of enzyme catalysts. Once this selection was made, however, the chirality of the active sites of the enzymes involved would retain a bias toward D-(+)-glucose and against L-(-)-glucose (because of the improper fit of the latter). Once introduced, this bias would be perpetuated and extended to other catalysts. Finally, when we speak about evolutionary selection of a particular molecule for a given function, we do not mean to imply that evolution operates on a molecular level. Evolution, of course, takes place at the level of organism populations, and molecules are selected only in the sense that their use gives the organism an increased likelihood of surviving and procreating.

22.13D Cellulose Derivatives A number of derivatives of cellulose are used commercially. Most of these are compounds in which two or three of the free hydroxyl groups of each glucose unit have been converted to an ester or an ether. This conversion substantially alters the physical properties of the material, making it more soluble in organic solvents and allowing it to be made into fibers and films. Treating cellulose with acetic anhydride produces the triacetate known as “Arnel” or “acetate,” used widely in the textile industry. Cellulose trinitrate, also called “gun cot­ ton” or nitrocellulose, is used in explosives. is made by treating cellulose (from cotton or wood pulp) with carbon disulfide in a basic solution. This reaction converts cellulose to a soluble xanthate:

Rayon

S C ellulose— OH

C S0

NaOH

cellu lose— O— C— S~ N a 4 C e llu lo s e x a n th a te

The solution of cellulose xanthate is then passed through a small orifice or slit into an acidic solution. This operation regenerates the — OH groups of cellulose, causing it to precipi­ tate as a fiber or a sheet: S C ellu lose— O— C— S~ N a 4

H,O+

c e llu lo se— OH R a y o n o r c e llo p h a n e

The fibers are

rayon;the sheets, after softening with glycerol, are cellophane.

Cellophane on rollers at a manufacturing plant.

22.14 O th e r Biologically Im p o rtan t Sugars Monosaccharide derivatives in which the — CH2OH group at C 6 has been specifically oxi­ dized to a carboxyl group are called uronic acids. Their names are based on the mono­ saccharide from which they are derived. For example, specific oxidation of C 6 of glucose

1038

Chapter 22

Carbohydrates

glucose

to a carboxyl group converts to glucuronic acid. In the same way, specific oxida­ tion of C6 of would yield galacturonic acid:

galactose

*O

H-

-OH

HO­

H

HO-

H

H-

OH CO2H

D -G ala c tu ro n ic acid

Review Problem 22.19

Direct oxidation of an aldose affects the aldehyde group first, converting it to a carboxylic acid (Section 22.6B), and most oxidizing agents that w ill attack 1° alcohol groups w ill also attack 2° alcohol groups. Clearly, then, a laboratory synthesis of a uronic acid from an aldose requires protecting these groups from oxidation. Keeping this in mind, suggest a method for carrying out a specific oxidation that would convert D-galactose to D-galacturonic acid. See Section 22.5E.)

(Hint:

• Monosaccharides in which an — OH group has been replaced by — H are known as deoxy sugars. The most important deoxy sugar, because it occurs in DNA, is deoxyribose. Other deoxy sugars that occur widely in polysaccharides are L-rhamnose and L-fucose:

OH

OH

/3-2-D eoxy-D -ribo se

OH

HO

a -L -R h a m n o s e ( 6 -d e o xy -L -m a n n o s e )

a -L -F u c o s e ( 6 -de o xy -L -g ala c to s e )

22.15 Sugars That Contain N itro g en 22.15A Glycosylamines A sugar in which an amino group replaces the anomeric — OH is called a glycosylamine. Examples are b-D-glucopyranosylamine and adenosine:

OH HO HO

O NH„

HO ^ -D -G lu c o p y ra n o s y la m in e

A d e n o s in e

Adenosine is an example of a glycosylamine that is also called a nucleoside.

1039

22.15 Sugars That Contain Nitrogen • Nucleosides are glycosylamines in which the amino component is a pyrimidine or a purine (Section 20.1B) and in which the sugar component is either D-ribose or 2deoxy-D-ribose (i.e., D-ribose minus the oxygen at the 2 position). Nucleosides are the important components of RNA (ribonucleic acid) and DNA (deoxyri­ bonucleic acid). We shall describe their properties in detail in Section 25.2.

22.15B Amino Sugars • A sugar in which an amino group replaces a nonanomeric — OH group is called an amino sugar. D-Glucosamine is an example of an amino sugar. In many instances the amino group is acetylated as in W-acetyl-D-glucosamine. N-Acetylmuramic acid is an important com­ ponent of bacterial cell walls (Section 24.10). HO,

HO

H HO

OH H_ OH H

O H

y

H

O

H HO \

NH2

/j-D -G lu c o s a m in e

HO

H OH H

/

O

H

OH

H OR H O \

h

NHCOCH3

H

/3-W -A cetyl-D -glucosam ine (NAG)

CH,

OH R= H

H CO2H

NHCOCH3

/3 -W -A cetylm ura m ic acid (NAM )

D-Glucosamine can be obtained by hydrolysis of chitin, a polysaccharide found in the shells of lobsters and crabs and in the external skeletons of insects and spiders. The amino group of D-glucosamine as it occurs in chitin, however, is acetylated; thus, the repeating unit is actually N-acetylglucosamine (Fig. 22.18). The glycosidic linkages in chitin are b(1 s 4). X-Ray analysis indicates that the structure of chitin is similar to that of cellulose. HO O O"'' H H H

H

NHCOCH,

NHCOCH

n

Figure 22.18 A p artial structu re o f chitin. The re p e a tin g units are N -acetylglucosam ines linked b(1 s 4).

D-Glucosamine can also be isolated from heparin, a sulfated polysaccharide that con­ sists predominately of alternating units of D-glucuronate-2-sulfate and N-sulfo-D-glucosamine-6-sulfate (Fig. 22.19). Heparin occurs in intracellular granules of mast cells that line arterial walls, where, when released through injury, it inhibits the clotting of blood. Its purpose seems to be to prevent runaway clot formation. Heparin is widely used in medi­ cine to prevent blood clotting in postsurgical patients.

6 -sulfate

Figure 22.19 A pa rtia l structure o f heparin, a polysaccharide th a t prevents b lo o d clo ttin g .

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Chapter 22

Carbohydrates

22.16 Glycolipids and Glycoproteins o f the Cell Surface: Cell Recognition and the Im m une System Before 1960, it was thought that the biology of carbohydrates was rather uninteresting, that, in addition to being a kind of inert filler in cells, carbohydrates served only as an energy source and, in plants, as structural materials. Research has shown, however, that carbohy­ drates joined through glycosidic linkages to lipids (Chapter 23) and to proteins (Chapter 24), called glycolipids and glycoproteins, respectively, have functions that span the entire spectrum of activities in the cell. Indeed, most proteins are glycoproteins, of which the car­ bohydrate content can vary from less than 1% to greater than 90%. Glycolipids and glycoproteins on the cell surface (Section 23.6A) are now known to be the agents by which cells interact with other cells and with invading bacteria and viruses. The immune system’s role in healing and in autoimmune diseases such as rheumatoid arthritis involves cell recognition through cell surface carbohydrates. Important carbohydrates in this role are sialyl Lewis* acids (see the chapter opening vignette). Tumor cells also have specific carbohydrate markers on their surface as well, a fact that may make it possible to develop vaccines against cancer. (See “The Chemistry of . . . Vaccines Against Cancer” in

WileyPLUS.)

OH R eprinted w ith perm ission of John W ile y & Sons, Inc., from Voet, D., and Voet, J.G. Biochemistry, Second Edition © 1995 Voet, D., and Voet, J.G.

3

A

H elpful H in t _____ ^ See "The Chemistry of... Oligosaccharide Synthesis on a Solid Support-the Glycal Assembly Approach" in WileyPLUS regarding the synthesis of promising carbohydrate anticancer vaccines.

s ia ly l L e w is * a c id

The human blood groups offer another example of how carbohydrates, in the form of glycolipids and glycoproteins, act as biochemical markers. The A, B, and O blood types are determined, respectively, by the A, B, and H determinants on the blood cell surface. (The odd naming of the type O determinant came about for complicated historical reasons.) Type AB blood cells have both A and B determinants. These determinants are the carbo­ hydrate portions of the A, B, and H antigens. Antigens are characteristic chemical substances that cause the production of antibodies when injected into an animal. Each antibody can bind at least two of its corresponding anti­ gen molecules, causing them to become linked. Linking of red blood cells causes them to agglutinate (clump together). In a transfusion this agglutination can lead to a fatal block­ age of the blood vessels. Individuals with type A antigens on their blood cells carry anti-B antibodies in their serum; those with type B antigens on their blood cells carry anti-A antibodies in their serum. Individuals with type AB cells have both A and B antigens but have neither anti-A nor antiB antibodies. Type O individuals have neither A nor B antigens on their blood cells but have both anti-A and anti-B antibodies. The A, B, and H antigens differ only in the monosaccharide units at their nonreducing ends. The type H antigen (Fig. 22.20) is the precursor oligosaccharide of the type A and B antigens. Individuals with blood type A have an enzyme that specifically adds an V-acetylgalactosamine unit to the 3-OH group of the terminal galactose unit of the H antigen. Individuals with blood type B have an enzyme that specifically adds galactose instead. In individuals with type O blood, the enzyme is inactive.

22.16 Cell Recognition and the Immune System

L-Fuc

OH Type Bdeterminant

Type Hdeterminant Figure 22.20 The te rm in a l m onosaccharides o f th e an tig en ic determ inants fo r typ e s A, B, and O b lo o d . The ty p e H d e term in a nt is present in individuals w ith b lo o d ty p e O and is th e precursor of th e ty p e A and B d eterm inants. These oligosaccharide antigens are attached to carrier lipid or p ro te in m olecules th a t are anchored in th e red b lo o d cell m em brane (see Fig. 23.9 fo r a d e p iction o f a cell m em brane). A c = acetyl, Gal = D-galactose, G alN A c = N-acetylgalactosam ine, G lycN A c = N -acetylglucosam ine, Fuc = fucose.

Antigen-antibody interactions like those that determine blood types are the basis of the immune system. These interactions often involve the chemical recognition of a glycolipid or glycoprotein in the antigen by a glycolipid or glycoprotein of the antibody. In “The Chemistry of . . . Antibody-Catalyzed Aldol Condensations” (in Chapter 19), however, we saw a different and emerging dimension of chemistry involving antibodies. We shall explore this topic further in the Chapter 24 opening vignette on designer catalysts and in “The Chemistry of . . . Some Catalytic Antibodies” (Section 24.12).

WileyPLUS,

1041

1042

Chapter 22

Carbohydrates

2 2 .1 7 Carbohydrate Antibiotics One of the important discoveries in carbohydrate chemistry was the isolation (in 1944) of the carbohydrate antibiotic called Streptomycin disrupts bacterial protein syn­ thesis. Its structure is made up of the following three subunits:

streptomycin.

2 S tr e p t id in e

►L - S t r e p t o s e

2 -D e o x y 2 - m e t h y la m in o " a - L - g iu c o p y ra n o s e

A ll three components are unusual: The amino sugar is based on L-glucose; streptose is a branched-chain monosaccharide; and streptidine is not a sugar at all, but a cyclohexane derivative called an amino cyclitol. Other members of this family are antibiotics called kanamycins, neomycins, and gen­ tamicins (not shown). A ll are based on an amino cyclitol linked to one or more amino sug­ ars. The glycosidic linkage is nearly always a. These antibiotics are especially useful against bacteria that are resistant to penicillins.

22.18 Sum m ary o f Reactions o f Carbohydrates The reactions of carbohydrates, with few exceptions, are the reactions of functional groups that we have studied in earlier chapters, especially those of aldehydes, ketones, and alco­ hols. The most central reactions of carbohydrates are those of hemiacetal and acetal for­ mation and hydrolysis. Hemiacetal groups form the pyranose and furanose rings in carbohydrates, and acetal groups form glycoside derivatives and join monosaccharides together to form di-, tri-, oligo-, and polysaccharides. Other reactions of carbohydrates include those of alcohols, carboxylic acids, and their derivatives. Alkylation of carbohydrate hydroxyl groups leads to ethers. Acylation of their hydroxyl groups produces esters. Alkylation and acylation reactions are sometimes used to protect carbohydrate hydroxyl groups from reaction while a transformation occurs else­ where. Hydrolysis reactions are involved in converting ester and lactone derivatives of car­ bohydrates back to their polyhydroxy form. Enolization of aldehydes and ketones leads to epimerization and interconversion of aldoses and ketoses. Addition reactions of aldehydes and ketones are useful, too, such as the addition of ammonia derivatives in osazone for­ mation, and of cyanide in the Kiliani-Fischer synthesis. Hydrolysis of nitriles from the Kiliani-Fischer synthesis leads to carboxylic acids. Oxidation and reduction reactions have their place in carbohydrate chemistry as well. Reduction reactions of aldehydes and ketones, such as borohydride reduction and catalytic hydrogenation, are used to convert aldoses and ketoses to alditols. Oxidation by Tollens’ and Benedict’s reagents is a test for the hemiacetal linkage in a sugar. Bromine water oxi­ dizes the aldehyde group of an aldose to an aldonic acid. Nitric acid oxidizes both the alde­ hyde group and terminal hydroxymethyl group of an aldose to an aldaric acid (a dicarboxylic acid). Lastly, periodate cleavage of carbohydrates yields oxidized fragments that can be use­ ful for structure elucidation.

1043

Problems

Key Terms and Concepts The key terms and concepts that are highlighted in b o l d , b l u e t e x t within the chapter are defined in the glossary (at the back of the book) and have hyperlinked definitions in the accompanying W ileyPLU S course (www.wileyplus.com).

/W IL E Y

_

PLUS

Problems Note to Instructors: Many of the homework problems are available for assignment via Wiley PLUS, an online teaching and learning solution. C A R B O H Y D R ATE STRUCTURE A N D 2 2 .2 0

2 2 .2 1

2 2 .2 2

R E A C T IO N S

Give appropriate structural formulas to illustrate each of the following: (a )

An aldopentose

( g ) A n a ld o n o la c to n e

(b )

A ketohexose

( h ) A p y ra n o s e

(c )

An L-monosaccharide

(i)

A fu ra n o s e

( o ) A p h e n y lo s a z o n e

(d )

A glycoside

(j)

A r e d u c in g su g a r

( p ) A d is a c c h a rid e

(e )

An aldonic acid

( k ) A p y ra n o s id e

(f)

An aldaric acid

(l)

( m ) E p im e rs ( n ) A n o m e rs

( q ) A p o ly s a c c h a rid e ( r ) A n o n re d u c in g su g a r

A fu ra n o s id e

Draw conformational formulas for each of the following: ( c ) methyl 2,3,4,6-tetra-O-methyl-b-D-allopyranoside.

(a )

a-D-allopyranose, ( b ) methyl b-D-allopyranoside, and

Draw structures for furanose and pyranose forms of D-ribose. Show how you could use periodate oxidation to dis­ tinguish between a methyl ribofuranoside and a methyl ribopyranoside.

22.23

One reference book lists D-mannose as being dextrorotatory; another lists it as being levorotatory. Both references are correct. Explain.

22.24

The starting material for a commercial synthesis of vitamin C is L-sorbose (see the following reaction); it can be — OH = O

D-Glucose

H Ni

D-Glucitol

O2

A sucbeotoxbyadcatensr

HO— - H H— - OH HO— - H — OH L-S orbo se

A.suboxydans

The second step of this sequence illustrates the use of a bacterial oxidation; the microorganism accom­ plishes this step in 90% yield. The overall result of the synthesis is the transformation of a D-aldohexose (D-glucose) into an L-ketohexose (L-sorbose). What does this mean about the specificity of the bacterial oxidation? 22.25 22.26

What two aldoses would yield the same phenylosazone as L-sorbose (Problem 22.24)? In addition to fructose (Review Problem 22.12) and sorbose (Problem 22.24), there are two other 2-ketohexoses,

psicoseand tagatose.D-Psicose yields the same phenylosazone as D-allose (or D-altrose); D-tagatose yields the same osazone as D-galactose (or D-talose). What are the structures of D-psicose and D-tagatose?

22.27

A, B, and C are three aldohexoses. Compounds A and B yield the same optically active alditol when they are reduced with hydrogen and a catalyst; A and B yield different phenylosazones when treated with phenylhydrazine; B and C give the same phenylosazone but different alditols. Assuming that all are d sugars, give names and structures for A, B, and C.

1044 22.28

Chapter 22

Carbohydrates

Xylitol is a sweetener that is used in sugarless chewing gum. Starting with an appropriate monosaccharide, outline a possible synthesis of xylitol. OH H —

—

OH

HO —

—

H

H

—

OH OH

X y lito l

22.29

Although monosaccharides undergo complex isomerizations in base (see Section 22.5A), aldonic acids are epimerized specifically at C 2 when they are heated with pyridine. Show how you could make use of this reaction in a syn­ thesis of D-mannose from D-glucose.

22.30

The most stable conformation of most aldopyranoses is one in which the largest group, the — C H 2 O H group, is equatorial. However, D-idopyranose exists primarily in a conformation with an axial — C H 2 O H group. Write for­ mulas for the two chair conformations of a-D-idopyranose (one with the — C H 2 O H group axial and one with the — C H 2 O H group equatorial) and provide an explanation.

S T R U C T U R E E L U C ID A T IO N

anhydrosugar

22.31

Heating D-altrose with dilute acid produces a nonreducing (C6H10O5). Methylation of the anhydro sugar followed by acid hydrolysis yields 2,3,4-tri-O-methyl-D-altrose. The formation of the anhydro sugar takes place through a chair conformation of b-D-altropyranose in which the — CH2OH group is axial. What is the struc­ ture of the anhydro sugar, and how is it formed? ( b ) D-Glucose also forms an anhydro sugar but the conditions required are much more drastic than for the corresponding reaction of D-altrose. Explain.

22.32

Show how the following experimental evidence can be used to deduce the structure of lactose (Section 22.12D):

22.33

(a )

1.

Acid hydrolysis of lactose (C12H22O11) gives equimolar quantities of D-glucose and D-galactose. Lactose under­ goes a similar hydrolysis in the presence of a

2.

Lactose is a reducing sugar and forms a phenylosazone; it also undergoes mutarotation.

3.

Oxidation of lactose with bromine water followed by hydrolysis with dilute acid gives D-galactose and D-gluconic acid.

4.

Bromine water oxidation of lactose followed by methylation and hydrolysis gives 2,3,6-tri-O-methylgluconolactone and 2,3,4,6-tetra-O-methyl-D-galactose.

5.

Methylation and hydrolysis of lactose give 2,3,6-tri-O-methyl-D-glucose and 2,3,4,6-tetra-O-methyl-D-galactose.

b-galactosidase.

melibiosefrom the following data: Melibiose is a reducing sugar that undergoes mutarotation and forms a phenylosazone. Hydrolysis of melibiose with acid or with an a -g alactosidasegives D-galactose and D-glucose. Bromine water oxidation of melibiose gives m elibionicacid. Hydrolysis of melibionic acid gives D-galactose

Deduce the structure of the disaccharide 1. 2. 3.

and D-gluconic acid. Methylation of melibionic acid followed by hydrolysis gives 2,3,4,6-tetra-O-methyl-D-galactose and 2,3,4,5-tetra-O-methyl-D-gluconic acid. 4.

22.34

Methylation and hydrolysis of melibiose give 2,3,4,6-tetra-O-methyl-D-galactose and 2,3,4-tri-O-methylD-glucose.

Trehalose is a disaccharide that can be obtained from yeasts, fungi, sea urchins, algae, and insects. Deduce the struc­ ture of trehalose from the following information: 1.

Acid hydrolysis of trehalose yields only D-glucose.

2.

Trehalose is hydrolyzed by a-glucosidase but not by b-glucosidase enzymes.

3. Trehalose is a nonreducing sugar; it does not mutarotate, form a phenylosazone, or react with bromine water. 4.

Methylation of trehalose followed by hydrolysis yields two molar equivalents of 2,3,4,6-tetra-O-methyl-D-glucose.

Problems 22.35

1045

Outline chemical tests that w ill distinguish between members of each of the following pairs: (a) D-Glucose and D-glucitol

(d) D-Glucose and D-galactose

(b) D-Glucitol and D-glucaric acid

(e) Sucrose and maltose

(c) D-Glucose and D-fructose

(f) Maltose and maltonic acid

(g) Methyl b-D-glucopyranoside and 2,3,4,6-tetra-O-methyl-b-D-glucopyranose (h) Methyl a-D-ribofuranoside (I) and methyl 2-deoxy-a-D-ribofuranoside (II): HO.

HO.

OCH, OH

OH I

22.36

22.37

OCH, OH

H

II

Schardingerdextrins

Bacillusmacerans nonreducing.

A group of oligosaccharides called can be isolated from when the bacil­ lus is grown on a medium rich in amylose. These oligosaccharides are all A typical Schardinger dex­ trin undergoes hydrolysis when treated with an acid or an a-glucosidase to yield six, seven, or eight molecules of D-glucose. Complete methylation of a Schardinger dextrin followed by acid hydrolysis yields only 2,3,6-tri-Omethyl-D-glucose. Propose a general structure for a Schardinger dextrin.

Isomaltoseis a disaccharide that can be obtained by enzymatic hydrolysis of amylopectin. Deduce the structure of isomaltose from the following data:

22.38

1.

Hydrolysis of 1 mol of isomaltose by acid or by an a-glucosidase gives 2 mol of D-glucose.

2.

Isomaltose is a reducing sugar.

3.

Isomaltose is oxidized by bromine water to isomaltonic acid. Methylation of isomaltonic acid and subsequent hydrolysis yields 2,3,4,6-tetra-O-methyl-D-glucose and 2,3,4,5-tetra-O-methyl-D-gluconic acid.

4.

Methylation of isomaltose itself followed by hydrolysis gives 2,3,4,6-tetra-O-methyl-D-glucose and 2,3,4-tri-methyl-D-glucose.

O Stachyoseoccurs in the roots of several species of plants. Deduce the structure of stachyose from the following

data: 1.

Acidic hydrolysis of 1 mol of stachyose yields 2 mol of D-galactose, 1 mol of D-glucose, and 1 mol of D-fructose.

2.

Stachyose is a nonreducing sugar.

3.

Treating stachyose with an a-galactosidase produces a mixture containing D-galactose, sucrose, and a nonre­ ducing trisaccharide called

4.

Acidic hydrolysis of raffinose gives D-glucose, D-fructose, and D-galactose. Treating raffinose with an a-galactosidase yields D-galactose and sucrose. Treating raffinose with invertase (an enzyme that hydrolyzes sucrose) yields fructose and (see Problem 22.33).

raffinose.

melibiose

5.

Methylation of stachyose followed by hydrolysis yields 2,3,4,6-tetra-O-methyl-D-galactose, 2,3,4-tri-O-methylD-galactose, 2,3,4-tri-O-methyl-D-glucose, and 1,3,4,6-tetra-O-methyl-D-fructose.

SPECTROSCOPY

22.39

Arbutin, a compound that can be isolated from the leaves of barberry, cranberry, and pear trees, has the molecular formula C12H16O7. When arbutin is treated with aqueous acid or with a b-glucosidase, the reaction produces dglucose and a compound X with the molecular formula C6H6O2. The 1H NM R spectrum of compound X consists of two singlets, one at 8 6.8 (4H) and one at 7.9 (2H). Methylation of arbutin followed by acidic hydrolysis yields 2,3,4,6-tetra-O-methyl-D-glucose and a compound Y (C7H8O2). Compound Y is soluble in dilute aqueous NaOH but is insoluble in aqueous NaHCO3. The 1H NM R spectrum of Y shows a singlet at 8 3.9 (3H), a singlet at 8 4.8 (1H), and a multiplet (that resembles a singlet) at 8 6.8 (4H). Treating compound Y with aqueous NaOH and (CH3)2SO4 produces compound Z (C8H10O2). The 1H NM R spectrum of Z consists of two singlets, one at 8 3.75 (6H) and one at 8 6.8 (4H). Propose structures for arbutin and for compounds X, Y , and Z.

8

1046 2 2 .4 0

2 2 .4 1

Chapter 22

Carbohydrates

When subjected to a Ruff degradation, a D-aldopentose, A, is converted to an aldotetrose, B. When reduced with sodium borohydride, the aldotetrose B forms an optically active alditol. The 13C NM R spectrum of this alditol dis­ plays only two signals. The alditol obtained by direct reduction of A with sodium borohydride is not optically active. When A is used as the starting material for a Kiliani-Fischer synthesis, two diastereomeric aldohexoses, C and D, are produced. On treatment with sodium borohydride, C leads to an alditol E, and D leads to F. The 13C NMR spectrum of E consists of three signals; that of F consists of six. Propose structures for A-F. Figure 22.21 shows the 13C NM R spectrum for the product of the reaction of D-(+)-mannose with acetone con­ taining a trace of acid. This compound is a mannofuranose with some hydroxyl groups protected as acetone acetals (as acetonides). Use the 13C NM R spectrum to determine how many acetonide groups are present in the compound.

TMS

0

_I__i__I__i__I__i__I__i__I__i__I__i__I__i__I__i__I__i__ I__i__I__i__I__i__I__i__I__i__I__i__I__i__I__i__I__i__I__i__ I__i__I__i__L

220

200

180

160

140

120

100

80

60

40

20

0

5C (ppm)

Figure 22.21

The broadband p ro to n -d e co u p le d

13C NM R

spectrum fo r th e reaction p ro d u ct in

Problem 22.41.

2 2 .4 2

D-(+)-Mannose can be reduced with sodium borohydride to form D-mannitol. When D-mannitol is dissolved in acetone containing a trace amount of acid and the product of this reaction subsequently oxidized with NaIO4, a compound whose 13C NM R spectrum consists of six signals is produced. One of these signals is near 8 200. What is the structure of this compound?

Challenge Problems 22.43

O f the two anomers of methyl 2,3-anhydro-D-ribofuranoside, I, the b form has a strikingly lower boiling point. Suggest an explanation using their structural formulas.

OCH

I

1047

Learning Group Problems 22.44

The following reaction sequence represents an elegant method of synthesis of 2-deoxy-D-ribose, IV , published by D. C. C. Smith in 1955: H^

HHO-

°

-OH H

H-

-OH

H-

-OH

CH3SO2C! C5H5N

anhydrous CuSO.

OH

II

H3O

D -G lucose

TV

H

O

LI

H

H

l-l

n u

LI

nw

H

HOh2o

[III]

OH

(a) What are the structures of I I and III? (b) Propose a mechanism for the conversion of I I I to IV 22.45

D-G lucose

acetic anhydride anhydrous sodium acetate acetic anhydride cat. HA

D-G lucopyranose pentaacetate, anom er V D -G lucopyranose pentaacetate, anom er VI

The 1H NM R data for the two anomers included very comparable peaks in the 8 2.0-5.6 region but differed in that, as their highest 8 peaks, anomer V had a doublet at 8 5.8 (1H, = 12 Hz) while anomer V I had a doublet at 8 6.3 (1H, = 4 Hz).

J

J

(a) Which proton in these anomers would be expected to have these highest 8 values? (b) Why do the signals for these protons appear as doublets? (c) The relationship between the magnitude of the observed coupling constant and the dihedral angle (when mea­ sured using a Newman projection) between C — H bonds on the adjacent carbons of a C— C bond is given by the Karplus equation. It indicates that an axial-axial relationship results in a coupling constant of about 9 Hz (observed range is 8-14 Hz) and an equatorial-axial relationship results in a coupling constant of about 2 Hz (observed range is 1-7 Hz). Which of V and V I is the anomer and which is the b anomer?

a

(d) Draw the most stable conformer for each of V and V I.

Learning Group Problems (a) The members of one class of low-calorie sweeteners are called polyols. The chemical synthesis of one such polyol sweetener involves reduction of a certain disaccharide to a mixture of diastereomeric glycosides. The alcohol (actually polyol) portion of the diastereomeric glycosides derives from one of the sugar moieties in the original disaccharide. Exhaustive methylation of the sweetener (e.g., with dimethyl sulfate in the presence of hydroxide) followed by hydrolysis would be expected to produce 2,3,4,6-tetra-O-methyl-a-D-glucopyranose, 1,2,3,4,5-penta-O-methyl-D-sorbitol, and 1,2,3,4,5-penta-O-methyl-D-mannitol, in the ratio of 2:1:1. On the basis of this information, deduce the structure of the two disaccharide glycosides that make up the diastereomeric mixture in this polyol sweetener.

1048

Chapter 22

Carbohydrates

(b) Knowing that the mixture of two disaccharide glycosides in this sweetener results from reduction of a single dis­ accharide starting material (e.g., reduction by sodium borohydride), what would be the structure of the disaccha­ ride for the reduction step? Explain how reduction of this compound would produce the two glycosides.

reactant

O

(c) Write the lowest energy chair conformational structure for 2,3,4,6-tetra- -methyl-a-D-glucopyranose. Shikimic acid is a key biosynthetic intermediate in plants and microorganisms. In nature, shikimic acid is converted to chorismate, which is then converted to prephenate, ultimately leading to aromatic amino acids and other essen­ tial plant and microbial metabolites (see the Chapter 21 Learning Group problem). In the course of research on biosynthetic pathways involving shikimic acid, H. Floss (University of Washington) required shikimic acid labeled with 13C to trace the destiny of the labeled carbon atoms in later biochemical transformations. To synthesize the labeled shikimic acid, Floss adapted a synthesis of optically active shikimic acid from D-mannose reported earlier by G. W. J. Fleet (Oxford University). This synthesis is a prime example of how natural sugars can be excellent chiral starting materials for the chemical synthesis of optically active target molecules. It is also an excellent exam­ ple of classic reactions in carbohydrate chemistry. The Fleet-Floss synthesis of D -(-)-[1 ,7 -13C]-shikimic acid (1) from D-mannose is shown in Scheme 1.

2.

(a) Comment on the several transformations that occur between D-mannose and 2. What new functional groups are formed? (b) What is accomplished in the steps from 2 to 3, 3 to 4, and 4 to 5? (c) Deduce the structure of compound 9 (a reagent used to convert 5 to 6), knowing that it was a carbanion that displaced the trifluoromethanesulfonate (triflate) group of 5. Note that it was compound 9 that brought with it the required 13C atoms for the final product. (d) Explain the transformation from 7 to 8. Write out the structure of the compound in equilibrium with 7 that would be required for the process from 7 to 8 to occur. What is the name given to the reaction from this intermediate to 8? (e) Label the carbon atoms of D-mannose and 1 by number or letter so as to show which atoms in 1 came from which atoms of D-mannose.

H

— O -P -O O

(from L-serine) R is saturated and R' is unsaturated.

I

3

-

(from 2-am inoethanol) or — O CH2CH2N+(CH3)3 (from choline) R' is an unsaturated fa tty acid.

Phosphatides resemble soaps and detergents in that they are molecules having both polar and nonpolar groups (Fig. 23.8a). Like soaps and detergents, too, phosphatides “ dissolve” in aqueous media by form ing micelles. There is evidence that in biological systems the pre­ ferred micelles consist o f three-dimensional arrays o f “ stacked” bimolecular micelles (Fig. 23.8b) that are better described as lip id bilayers.

1075

1076

C h a p te r 2 3

Lipids

Polar group

Nonpolar group O

CH3CH2CH2CH2CH2CH2CH2CH2CH2CH2CH2CH2CH2CH2CH2CH2CH2— COCH2 O CH3CH2CH2CH2CH2CH2CH2CH2CH=C H C H 2CH2CH2CH2CH2CH2CH2 — COCH O II + CH2OPOCH2CH2N(CH3)3

O(a)

Figure 23.8 (a) Polar and nonpolar sections o f a phosphatide. (b) A phosphatide m icelle or lip id bilayer.

(b) T h e h y d r o p h ilic a n d h y d r o p h o b ic p o rtio n s o f p h o s p h a tid e s m a k e th e m p e r fe c tly s u ite d f o r o n e o f t h e ir m o s t im p o r ta n t b io lo g ic a l fu n c tio n s : T h e y f o r m a p o r tio n o f a s tru c tu ra l u n it th a t cre a te s an in te rfa c e b e tw e e n an o rg a n ic a n d an a q u e o u s e n v iro n m e n t. T h is s tru c ­ tu re ( F ig . 2 3 .9 ) is lo c a te d in c e ll w a lls a n d m e m b ra n e s w h e re p h o s p h a tid e s a re o fte n fo u n d a s s o c ia te d w it h p ro te in s a n d g ly c o lip id s (S e c tio n 2 3 .6 B ).

Figure 23.9 A schem atic diagram o f a plasma m em brane. Integral proteins ( red-orange ), shown fo r cla rity in much g re a te r p ro p o rtio n than th e y are fo u n d in actual b io lo g ica l m em branes, and cholesterol (yellow) are e m b e d d e d in a bilayer com posed o f p h o spholipids (blue spheres with two wiggly tails). The carbohydrate com ponents o f glyco p ro te in s (yellow beaded chains) and g lyco lip id s (green beaded chains) occur only on th e external face o f th e m em brane. (R eprinted w ith perm ission o f John W ile y & Sons, Inc., from Voet, D.; Voet, J. G.; Pratt, C., Fundamentals of Biochemistry, Life at the Molecular Level; © 1999 Voet, D. and Voet, J. G.)

2 3 .6 P h o s p h o lip id s a n d C e ll M e m b r a n e s

Under suitable conditions all of the ester (and ether) linkages o f a phosphatide can be hydrolyzed. What organic compounds would you expect to obtain from the complete hydrol­ ysis o f (see Table 23.5) (a) a lecithin, (b) a cephalin, and (c) a choline-based plasmalogen? [Note: Pay particular attention to the fate o f the a,b-unsaturated ether in part (c).

1077

Review Problem 23.11

THE CHEMISTRY OF . . . S TE A L T H ® L i p o s o m e s f o r D r u g D e l i v e r y

The anticancer drug Doxil (doxorubicin) has been packaged in STEALTH® liposomes that give each dose of the drug extended action in the body. During manufacture of the drug it is ensconced in microscopic bubbles (vesicles) formed by a phospholipid bilayer and then given a special coating that masks it from the immune system. Ordinarily, a foreign particle such as this would be attacked by cells of the immune system and degraded, but a veil of polyethyl­ ene glycol oligomers on the liposome surface masks it from detection. Because of this coating, the STEALTH® liposome

circulates through the body and releases its therapeutic contents over a period of time significantly greater than the life­ time for circulation of the undisguised drug. Coatings like those used for STEALTH® liposomes may also be able to reduce the toxic side effects of some drugs. Furthermore, by attaching specific cell recognition "marker" molecules to the polymer, it may be possible to focus binding of the lipo­ somes specifically to cells of a targeted tissue. One might be tempted to call a targeted liposome a "smart stealth lipo­ some."

Lipid Membrane ^ (Phospholipid+ Cholesterol)

| Doxorubicin A STEALTH® liposom e canning th e d ru g d o xo ru b icin . (© and courtesy o f A LZ A C orporation.)

2 3 .6 B

1

t*p|,«-tiiyIc-ni- Glycnl ...................................

Derivatives of Sphingosine

Another important group of lipids is derived from sphingosine; the derivatives are called sphingolipids. Two sphingolipids, a typical sphingomyelin and a typical cerebroside, are shown in Fig. 23.10. On hydrolysis, sphingomyelins yield sphingosine, choline, phosphoric acid, and a C 2 4 fatty acid called lignoceric acid. In a sphingomyelin this last component is bound to the — N H 2 group o f sphingosine. The sphingolipids do not yield glycerol when they are hydrolyzed. The cerebroside shown in Fig. 23.10 is an example o f a glycolipid. Glycolipids have a polar group that is contributed by a carbohydrate. They do not yield phosphoric acid or choline when they are hydrolyzed. The sphingolipids, together w ith proteins and polysaccharides, make up m yelin, the pro­ tective coating that encloses nerve fibers or axons. The axons o f nerve cells carry electri­ cal nerve impulses. M yelin has a function relative to the axon sim ilar to that of the insulation on an ordinary electric wire (see the chapter opening vignette).

1078

C h a p te r 2 3

Lipids

Figure 23.10

A sphingosine and tw o sphingolipids.

A g a la c to s y l g ro u p

23.7 Waxes Most waxes are esters o f long-chain fatty acids and long-chain alcohols. Waxes are found as protective coatings on the skin, fur, and feathers o f animals and on the leaves and fruits o f plants. Several esters isolated from waxes are the follow ing: O

O O'

C e tyl p a lm ita te (fro m s p e rm a c e ti)

O O

14

n

=

24 o r 26; m = 28 o r 30 (fro m b eesw ax)

HO n

O

= 1 6 -2 8 ; m = 30 o r 32 (fro m c a rn a u b a w a x)

S u m m a ry o f R ea ctio n s o f L ipids The reactions o f lipids represent many reactions that we have studied in previous chapters, especially reactions o f carboxylic acids, alkenes, and alcohols. Ester hydrolysis (e.g., saponi­ fication) liberates fatty acids and glycerol from triacylglycerols. The carboxylic acid group o f a fatty acid can be reduced, converted to an activated acyl derivative such as an acyl chlo­ ride, or converted to an ester or amide. Alkene functional groups in unsaturated fatty acids can be hydrogenated, hydrated, halogenated, hydrohalogenated, converted to a vicinal diol or epoxide, or cleaved by oxidation reactions. Alcohol functional groups in lipids such as terpenes, steroids, and prostaglandins can be alkylated, acylated, oxidized, or used in elim ­ ination reactions. A ll o f these are reactions we have studied previously in the context o f smaller molecules.

«

P ro b le m s

1079

Key Terms and Concepts The key terms and concepts that are highlighted in bold, blue text w ithin the chapter are defined in the glossary (at the back of the book) and have hyperlinked definitions in the accompanying WileyPLUS course (www.wileyplus.com).

PLUS

Problems Note to Instructors: Many of the homework problems are available for assignment via WileyPLUS, an online teaching and learning solution. G E N E R A L R E A C T IO N S

23.12

How would you convert stearic acid, CH3(CH2)16CO2H, into each o f the following? O

O (g) Octadecanal,

H O

(h) Octadecyl stearate,

(c) Stearamide,

(i) 1-Octadecanol,

NH2 O

(d) ^N-Dim ethylstearam ide,

(e) Octadecylamine,

(f) Heptadecylamine, 23.13

16

15

, ^

N

(j)

2-Nonadecanone,

(k) 1-Bromooctadecane,

NH2

nh2

16

JA

^ c o 2h CN

Br (d) c o 2h

OH

16

(l) Nonadecanoic acid,

(c) ,c o 2h

(b )

OH (two ways) O

How would you transform tetradecanal into each o f the following? (a)

O

CO2N H s+

Br

CO2H

1080 23.14

C h a p te r 2 3

Lipids

Using palm itoleic acid as an example and neglecting stereochemistry, illustrate each o f the follow ing reactions of the double bond: (a) A ddition o f bromine

(b) Addition o f hydrogen

(c) Hydroxylation

(d) Addition o f HCl

23.15

When oleic acid is heated to 180-200°C (in the presence o f a small amount o f selenium), an equilibrium is estab­ lished between oleic acid (33%) and an isomeric compound called elaidic acid (67%). Suggest a possible structure for elaidic acid.

23.16

When limonene (Section 23.3) is heated strongly, it yields 2 m ol o f isoprene. What kind o f reaction is involved here?

23.17

Gadoleic acid (C2oH38O2), a fatty acid that can be isolated from cod-liver o il, can be cleaved by hydroxylation and subsequent treatment w ith periodic acid to CH3(CH2)9CHO and OHC(CH2)7CO2H. (a) What two stereoisomeric structures are possible for gadoleic acid? (b) What spectroscopic technique would make possible a decision as to the actual structure o f gadoleic acid? (c) What peaks would you look for?

23.18

a-Phellandrene and b-phellandrene are isomeric compounds that are m inor constituents o f spearmint o il; they have the molecular form ula C10H16. Each compound has a U V absorption maximum in the 230-270-nm range. On cat­ alytic hydrogenation, each compound yields 1-isopropyl-4-methylcyclohexane. On vigorous oxidation w ith potas­ sium permanganate, a-phellandrene yields

R O A D M A P SYNTHESES

23.19

Vaccenic acid, a constitutional isomer o f oleic acid, has been synthesized through the follow ing reaction sequence: ~ - , KM, liq. ,,, , ICH 2 (CH 2 )7 CH2Cl 1 -O c ty n e + N a N H 2 — > : A ( C 8 H 1 3 N a ) -----------------------------> NH 3 B ( C 1 7 H 3 1 C I)

C (C 18H 3 1 N)

KO H’ H2° > D ( C 1 8 H 3 1 O 2 K) ^ E (C 13H 32O 2)

H Pd BaSO>> v a c c e n ic

a c id ( C 1 8 H 3 4 O 2 )

Propose a structure for vaccenic acid and for the intermediates A -E . 23.20

v-Fluorooleic acid can be isolated from a shrub, Dechapetalum toxicarium, that grows in A frica. The compound is highly toxic to warm-blooded animals; it has found use as an arrow poison in tribal warfare, in poisoning enemy water supplies, and by witch doctors “ for terrorizing the native population.” Powdered fru it o f the plant has been used as a rat poison; hence v-fluorooleic acid has the common name “ratsbane.” A synthesis o f v-fluorooleic acid is outlined here. Give structures for compounds F -I: 1 -B ro m o -8 -flu o ro o c ta n e

G

+

( C 1 7 H 3 0 F C I)

s o d iu m a c e ty lid e

H

— >F

(C 18H 3 0 N F )

(1 ] H2 > (2) l(CH2)7Cl

( C 1 0 H 1 7 F)

I K° H+> (2) H3 O+

I

O F

OH w -F lu o ro o le ic acid (4 6 % y ie ld , o ve ra ll)

( C 1 8 H 3 1 O 2 F) ' 18 31 2 '

H 2

Ni2B (P-2)

'

"

P ro b le m s

23.21

1081

Give formulas and names for compounds A and B: O r-

, O

5a-Cholest-2-ene

C 6H 5C O O H - ^ - 5-------

. ,

. , x

HBr

_

> A (an epoxide) ------ > B

(Hint: B is not the most stable stereoisomer.) 23.22

The in itia l steps o f a laboratory synthesis o f several prostaglandins reported by E. J. Corey (Section 7.16B) and co-workers in 1968 are outlined here. Supply each of the missing reagents:

(e)

The in itia l step in another prostaglandin synthesis is shown in the follow ing reaction. What kind o f reaction— and catalyst— is needed here? NO2

CN H

CHO

23.23

och3

O

A useful synthesis o f sesquiterpene ketones, called cyperones, was accomplished through a modification of the fo l­ lowing Robinson annulation procedure (Section 19.7B). O +

N aN H 2 " N R o I-

p y rid in e - E t2O

O

O

D ih yd ro ca rvo n e H A, h e a t

W rite a mechanism that accounts for each step o f this synthesis.

1082

Chapter 23

Lipids

Challenge Problems 23.24

A Hawaiian fish called the pahu or boxfish (Ostracian lentiginosus) secretes a toxin that k ills other fish in its vicin­ ity. The active agent in the secretion was named pahutoxin by P. J. Scheuer, and it was found by D. B. Boylan and Scheuer to contain an unusual combination o f lip id moieties. To prove its structure, they synthesized it by this route: OH

pyridinium BrCH2CO2Et, Zn OH chlorochromate• A

OEt

B Ac2O pyridine

Compound

O

D

SOCl2

(1) HO(2) H3O+

choline chloride pahutoxin ■E ■ pyridine

Selected Infrared Absorption Bands (cm 1)

A B C D E Pahutoxin

1725 3300 (broad), 1735 3300-2500 (broad), 1710 3000-2500 (broad), 1735, 1710 1800, 1735 1735

What are the structures o f A , C, D, and E and o f pahutoxin? 23.25

The reaction illustrated by the equation below is a very general one that can be catalyzed by acid, base, and some enzymes. It therefore needs to be taken into consideration when planning syntheses that involve esters o f polyhy­ droxy substances like glycerol and sugars: HO

OH j.

O^ ^O Y

trace HClO. in CHCl3 -------------4--------- ^ F 10 min., room temp. 90% yield

( C H 2 ) i 4C H 3

Spectral data for F: MS (m/z): (after trim ethylsilylation): 546, 531 IR (cm-1 , in CCl4 solution): 3200 (broad), 1710 1H N M R (d) (after exchange w ith D2O): 4.2 (d), 3.9 (m), 3.7 (d), 2.2 (t), and others in the range 1.7 to 1 13C N M R (d): 172 (C), 74 (CH), 70 (CH2), 67 (CH2), 39 (CH2), and others in the range 32 to 14 (a) What is the structure o f product F? (b) The reaction is intramolecular. W rite a mechanism by which it probably occurs.

Learning Group Problems Olestra is a fat substitute patented by Procter and Gamble that mimics the taste and texture o f triacylglycerols (see “ The Chemistry o f . . . Olestra and Other Fat Substitutes” in Section 23.2B). It is calorie-free because it is neither hydrolyzed by digestive enzymes nor absorbed by the intestines but instead is passed directly through the body unchanged. The FDA has approved olestra for use in a variety o f foods, including potato chips and other snack foods that typically have a high fat content. It can be used in both the dough and the frying process.

L e a rn in g G ro u p P ro b le m s

1083

(a) Olestra consists of a mixture o f sucrose fatty acid esters (unlike triacylglycerols, which are glycerol esters of fatty acids). Each sucrose molecule in olestra is esterified w ith six to eight fatty acids. (One undesirable aspect o f olestra is that it sequesters fat-soluble vitamins needed by the body, due to its high lipophilic character.) Draw the structure o f a specific olestra molecule comprising six different naturally occurring fatty acids esterified to any o f the available positions on sucrose. Use three saturated fatty acids and three unsaturated fatty acids. (b) W rite reaction conditions that could be used to saponify the esters of the olestra molecule you drew and give IUPAC and common names for each o f the fatty acids that would be liberated on saponification. (c) Olestra is made by sequential transesterification processes. The first transesterification involves reaction of methanol under basic conditions w ith natural triacylglycerols from cottonseed or soybean o il (chain lengths o f C8-C 22). The second transesterification involves reaction o f these fatty acid methyl esters w ith sucrose to form olestra. W rite one example reaction, including its mechanism, for each of these transesterification processes used in the synthesis o f olestra. Start w ith any triacylglycerol having fatty acids like those incorporated into olestra. The biosynthesis o f fatty acids is accomplished two carbons at a time by an enzyme complex called fatty acid syn­ thetase. The biochemical reactions involved in fatty acid synthesis are described in Special Topic E (WileyPLUS). Each o f these biochemical reactions has a counterpart in synthetic reactions you have studied. Consider the bio­ chemical reactions involved in adding each CH2CH2 segment during fatty acid biosynthesis (those in Special Topic E that begin w ith acetyl-S-ACP and malonyl-S-ACP, and end w ith butyryl-S-ACP). W rite laboratory syn­ thetic reactions using reagents and conditions you have studied (not biosynthetic reactions) that would accomplish the same sequence of transformations (i.e., the condensation-decarboxylation, ketone reduction, dehydration, and alkene reduction steps). A certain natural terpene produced peaks in its mass spectrum at m/z 204, 111, and 93 (among others). On the basis o f this and the follow ing inform ation, elucidate the structure o f this terpene. Justify each o f your conclusions. (a) Reaction o f the unknown terpene w ith hydrogen in the presence of platinum under pressure results in a com­ pound w ith molecular formula C-15H30. (b) Reaction o f the terpene w ith ozone followed by dimethyl sulfide produces the follow ing mixture of compounds (1 mol of each for each mole of the unknown terpene): O

O

H

O

H O"

'H

(c) A fter w riting the structure o f the unknown terpene, circle each o f the isoprene units in this compound. To what class o f terpenes does this compound belong (based on the number of carbons it contains)? Draw the structure o f a phospholipid (from any o f the subclasses o f phospholipids) that contains one saturated and one unsaturated fatty acid. (a) Draw the structure of all o f the products that would be formed from your phospholipid if it were subjected to complete hydrolysis (choose either acidic or basic conditions). (b) Draw the structure o f the product(s) that would be formed from reaction o f the unsaturated fatty acid moiety o f your phospholipid (assuming it had been released by hydrolysis from the phospholipid first) under each of the follow ing conditions: (i)

Br2 in CCI4

(ii) OsO4, followed by NaHSO3 (iii) HBr (iv) Hot alkaline KMnO4, followed by H3O+ (v) SOCI2, followed by excess CH3NH2

Amino Acids and Proteins

A syn th etic Diels-Alderase catalytic a n tib o d y w ith a bound hapten.

Chemists are capitalizing on the natural adaptability of the immune system to create what we can fittingly call

d e sig n e r catalysts. These catalysts are a n tib o d ies — protein species usually produced by the immune system to capture and remove foreign agents but which, in this case, are elicited in a way that makes them able to cat­ alyze chemical reactions. The creation of the first catalytic antibodies by Richard A. Lerner and Peter G. Schultz (both of Scripps Research Institute) represented an ingenious union of principles relating to enzyme chemistry and the innate capabilities of the immune system. In some respects catalytic antibodies are like enzymes, the protein catalysts we have mentioned many times already and shall study further in this chapter. Unlike enzymes, however, cat­ alytic antibodies can virtually be "m ade to order" for specific reactions by a marriage of chemistry and immunol­ ogy. Examples include catalytic antibodies for Claisen rearrangements, Diels-Alder reactions (such as that shown in the molecular graphic above), ester hydrolyses, and aldol reactions. W e shall consider how catalytic antibodies are produced in "The Chemistry o f . . . Some Catalytic Antibodies" later in this chapter. Designer catalysts are indeed at hand.

1084

2 4 .1

In tro d u c tio n

1085 24.1 Introduction

The three groups o f biological polymers are polysaccharides, proteins, and nucleic acids. We studied polysaccharides in Chapter 22 and saw that they function prim arily as energy reserves, as biochemical labels on cell surfaces, and, in plants, as structural materials. When we study nucleic acids in Chapter 25, we shall find that they serve two major purposes: storage and transmission o f information. O f the three groups o f biopolymers, proteins have the most diverse functions. As enzymes and hormones, proteins catalyze and regulate the reactions that occur in the body; as muscles and tendons they provide the body w ith the means for movement; as skin and hair they give it an outer covering; as hemoglobin m ol­ ecules they transfer all-im portant oxygen to its most remote corners; as antibodies they pro­ vide it w ith a means o f protection against disease; and in combination w ith other substances in bone they provide it w ith structural support. Given such diversity o f functions, we should not be surprised to find that proteins come in all sizes and shapes. By the standard o f most o f the molecules we have studied, even small proteins have very high molecular weights. Lysozyme, an enzyme, is a relatively small protein and yet its molecular weight is 14,600. The molecular weights o f most proteins are much larger. Their shapes cover a range from the globular proteins such as lysozyme and hemoglobin to the helical coils o f a-keratin (hair, nails, and wool) and the pleated sheets o f silk fibroin. And yet, in spite o f such diversity o f size, shape, and function, a ll proteins have com­ mon features that allow us to deduce their structures and understand their properties. Later in this chapter we shall see how this is done. • Proteins are polyamides, and their monomeric units are composed o f about 20 d if­ ferent a-amino acids: NH2

An a -a m in o acid

R is a side chain at the a carbon that determines the identity of the amino acid (Table 24.1).

R1

O

1

'3

O

R.

A p o rtio n o f a p ro tein m o le c u le

Amide (peptide) linkages are shaded. R1 to R5 may be any of the possible side chains. • The exact sequence o f the different a-amino acids along the protein chain is called the p rim a ry structure o f the protein. A protein’s primary structure, as its name suggests, is o f fundamental importance. For the protein to carry out its particular function, the primary structure must be correct. We shall see later that when the primary structure is correct, the protein’s polyamide chain folds in particular ways to give it the shape it needs for its particular task. • Folding o f the polyamide chain gives rise to higher levels o f complexity called the secondary and te rtia ry structures o f the protein. • Q uaternary structure results when a protein contains an aggregate o f more than one polyamide chain. • Hydrolysis o f proteins w ith acid or base yields a m ixture o f amino acids.

1086

C h a p te r 2 4

A m in o A c id s a n d P ro te in s

Although hydrolysis of naturally occurring proteins may yield as many as 22 different amino acids, the amino acids have an important structural feature in common: With the exception of glycine (whose molecules are achiral), almost all naturally occurring amino acids have the l configuration at the carbon.* That is, they have the same relative con­ figuration as L-glyceraldehyde:

a

O

R

O OH

HO

H

NH2

OH L -G lyceraldeh yde [(S )-g ly c e ra ld eh y d e ]

An L -a-am ino acid [u s u ally an (S )-a -a m in o acid]

CHO

CO 2 H

H

H2 N

HO

H CH2OH

R

F is c h e r p ro je c tio n s fo r an L -a -am in o a c id an d L -g lycerald eh yd e

24.2 Am ino Acids 2 4 .2 A

Structures and Names

• The 22 a-amino acids that can be obtained from proteins can be subdivided into three different groups on the basis of the structures of their side chains, R. These are given in Table 24.1.

L - A m i n o A c i d s F o u n d in P r o t e i n s

S tru c tu re

N am e

A b b r e v ia tio n s 2

Glycine

G or Gly

p K ai « -C O 2 H

p K 32 . a -N H 3+

2.3

9.6

p K 33 R g ro u p

p'

N e u tr a l A m in o A c id s

O h 2n

OH

*Some D-amino acids have been obtained fromthe material comprising the cell walls of bacteria and by hydrolysis of certain antibiotics.

6.Q

1

c

2 4 .2 A m in o A cid s

[

table

24 .1 ^

C o n t in u e d

S tru c tu re

P^a-, a -C O 2H

pKa :2 »-NH3

P Ka3 R g ro u p

N am e

A b b r e v ia tio n s 3

Isoleucine6

I or Ile

2.4

9.7

6.1

Phenylalanine6

F or Phe

1.8

9.1

5.5

Tyrosine

Y or Tyr

2.2

9.1

Tryptophan6

W or Trp

2.4

9.4

5.9

Serine

S or Ser

2.2

9.2

5.7

Threonine6

T or Thr

2.6

10.4

6.5

2.0

10.6

6.3

4-Hydroxyproline O or Hyp (cis and trans)

1.9

9.7

6.3

Cysteine

1.7

10.8

Proline

HO....

1087

P or Pro

C or Cys

10.1

8.3

pi

5.7

5.0

(continues on next page)

1088

C h a p te r 2 4

A m in o A c id s a n d P ro te in s

C o n t in u e d

S tru c tu re

nh2

pKa, a -C O 2H

pKa2 + a - N H 3+

P K33 R g ro u p

Pi

N am e

A b b r e v ia tio n s 3

Cystine

Cys-Cys

1.6 2.3

7.9 9.9

5.1

Methionineb

M or Met

2.3

9.2

5.8

Asparagine

N or Asn

2.0

8.

5.4

Glutamine

Q or Gln

2.2

9.1

5.7

Aspartic acid

D or Asp

2.1

9.8

3.9

3.0

Glutamic acid

E or Glu

2.2

9.7

4.3

3.2

Lysinefc

K or Lys

2.2

9.0

10.5°

9.8

Arginine

R or Arg

2.2

9.0

12.5°

10.8

Histidine

H or His

1.8

9.2

6.0°

7.6

O

hoy nh2

O O MeS. "^ Y "^ O h nh2

O h 2n

'O H O

nh2

O

O

h 2n

OH nh2

S id e C h ain s C o n ta in in g an A c id ic (C a rb o x y l) G ro u p

O HO

"OH O

nh2

O

O OH

HO nh2

S id e C h ain s C o n ta in in g a Basic G ro u p

O OH nh2

NH H2N

O OH

N nh2

O OH nh2 aSingle-letter abbreviations are now the m ost com m only used form in current biochem ical literature. bAn essential am ino acid. cpKa is o f p ro to n a te d amine o f R group.

1 2 4 .2 A m in o A cid s

Only 20 o f the 22 a-amino acids in Table 24.1 are actually used by cells when they syn­ thesize proteins. Two amino acids are synthesized after the polyamide chain is intact. Hydroxyproline (present m ainly in collagen) is synthesized by oxidation o f proline, and cystine (present in most proteins) is synthesized from cysteine. The conversion o f cysteine to cystine requires additional comment. The — SH group of cysteine makes cysteine a thiol. One property o f thiols is that they can be converted to disul­ fides by m ild oxidizing agents. This conversion, moreover, can be reversed by m ild reduc­ ing agents: [O ],

2 R— S— H

I hT

R— S— S — R

T hiol

D is u lfid e D is u lfid e lin kag e

O

O [O]

2 HS"

y

OH

HO

IH T

OH

nh2

nh2

C y s te in e

nh2 C y s tin e

We shall see later how the disulfide linkage between cysteine units in a protein chain con­ tributes to the overall structure and shape o f the protein. 2 4 .2 B

Essential Amino Acids

Am ino acids can be synthesized by all living organisms, plants and animals. Many higher animals, however, are deficient in their ability to synthesize a ll o f the amino acids they need for their proteins. Thus, these higher animals require certain amino acids as a part o f their diet. For adult humans there are eight essential amino acids; these are identified in Table 24.1 by a footnote. 2 4 .2 C

Amino Acids as Dipolar Ions

• Amino acids contain both a basic group (— NH2) and an acidic group (— CO2H). • In the dry solid state, amino acids exist as dipo la r ions, a form in which the car­ boxyl group is present as a carboxylate ion, — CO2~, and the amino group is pre­ sent as an aminium ion, — NH3+ (Dipolar ions are also called zw itterions.) • In aqueous solution, an equilibrium exists between the dipolar ion and the anionic and cationic forms o f an amino acid. O

O

R

OH-

R

"O-

OH H3 O

HNH 3 C a tio n ic form (p re d o m in a n t in s tro n g ly a c id ic s o lu tio n s , e.g., at pH 0)

O

+NH3 D ip o la r ion

OH-

R

O-

H3 O

NH 2 A n io n ic form (p re d o m in a n t in s tro n g ly b a s ic s o lu tio n s , e.g., at pH 14)

The predominant form o f the amino acid present in a solution depends on the pH o f the solution and on the nature o f the amino acid. In strongly acidic solutions a ll amino acids are present prim arily as cations; in strongly basic solutions they are present as anions. • The isoelectric p oint (p i) is the pH at which the concentration o f the dipolar ion is at its maximum and the concentrations o f the anions and cations are equal.

c

1089

1090

C h a p te r 2 4

A m in o A c id s a n d P ro te in s

Each amino acid has a particular isoelectric point. These are given in Table 24.1. Proteins have isoelectric points as w ell. As we shall see later (Sections 24.13 and 24.14), this prop­ erty o f proteins is important for their separation and identification. Let us consider first an amino acid with a side chain that contains neither acidic nor basic groups— an amino acid, for example, such as alanine. I f alanine is dissolved in a strongly acidic solution (e.g., pH 0), it is present in mainly a net cationic form. In this state the amino group is protonated (bears a form al +1 charge) and the carboxylic acid group is neutral (has no form al charge). As is typical o f a-amino acids, the pKa for the carboxylic acid hydrogen o f alanine is considerably lower (2.3) than the pKa o f an ordinary carboxylic acid (e.g., propanoic acid, pKa 4.89): O

C a tio n ic fo rm o f a la n in e pKa, = 2.3

P ro p a n o ic acid pKa = 4.89

The reason for this enhanced acidity o f the carboxyl group in an a-amino acid is the inductive effect o f the neighboring aminium cation, which helps to stabilize the carboxylate anion formed when it loses a proton. Loss o f a proton from the carboxyl group in a cationic a-amino acid leaves the molecule electrically neutral (in the form o f a dipolar ion). This equilibrium is shown in the red-shaded portion o f the equation below. The protonated amino group o f an a-amino acid is also acidic, but less so than the carboxylic acid group. The pKa o f the aminium group in alanine is 9.7. The equilibrium for loss o f an aminium proton is shown in the blue-shaded portion o f the equation below. The carboxylic acid proton is always lost before a proton from the aminium group in an a-amino acid. O

O

O

+n h 3

+n h 3

nh2

C a tio n ic fo rm (pKa, = 2.3)

D ip o la r ion (pKa 2 = 9.7)

A n io n ic form

The state o f an a-amino acid at any given pH is governed by a combination o f two equi­ libria, as shown in the above equation for alanine. The isoelectric point (pI) o f an amino acid such as alanine is the average o f pKai and pK%: pi

=2(2.3 + 9.7) = 6.0

(isoelectric point of alanine)

When a base is added to a solution o f the net cationic form o f alanine (in itia lly at pH 0, for example), the first proton removed is the carboxylic acid proton, as we have said. In the case o f alanine, when a pH o f 2.3 is reached, the carboxylic acid proton w ill have been removed from ha lf o f the molecules. This pH represents the pKa o f the alanine carboxylic acid proton, as can be demonstrated using the Henderson-Hasselbalch equation. • The Henderson-Hasselbalch equation shows that fo r an acid (H A) and its conju­ gate base (A _) when [H A ] = [ A ] , then pH = pKa. ,, ,, , [HA] pKa = pH + |og —

H e n d e rs o n -H a s s e lb a lc h e q u a tio n

Therefore, when the acid is half neutralized, [HA] = [A ], log

[A ]

= 0, and thus pH = pKa

1091

2 4 .2 A m in o A cid s

Figure 24.1

Equivalents of OH

A titra tio n curve fo r alanine.

As more base is added to this solution, alanine reaches its isoelectric point (p/), the pH at which a ll o f alanine’s carboxylic acid protons have been removed but not its aminium protons. The molecules are therefore electrically neutral (in their dipolar ion or zwitterionic form ) because the carboxylate group carries a - 1 charge and the aminium group a + 1 charge. The p / for alanine is 6.0. Now, as we continue to add the base, protons from the aminium ions w ill begin to be removed, until at pH 9.7 half o f the aminium groups w ill have lost a proton. This pH rep­ resents the pKa o f the aminium group. Finally, as more base is added, the remaining aminium protons w ill be lost until a ll o f the alanine molecules have lost their aminium protons. A t this point (e.g., pH 14) the molecules carry a net anionic charge from their carboxylate group. The amino groups are now electrically neutral. Figure 24.1 shows a titration curve for these equilibria. The graph represents the change in pH as a function o f the number o f molar equivalents o f base. Because alanine has two protons to lose in its net cationic form, when one molar equivalent o f base has been added, the molecules w ill have each lost one proton and they w ill be electrically neutral (the dipo­ lar ion or zw itterionic form). If an amino acid contains a side chain that has an acidic or basic group, the equilibria become more complex. Consider lysine, for example, an amino acid that has an additional — NH2 group on its e carbon. In strongly acidic solution, lysine is present as a dication because both amino groups are protonated. The first proton to be lost as the pH is raised is a proton o f the carboxyl group (pKa = 2.2), the next is from the a-aminium group (pK% = 9.0), and the last is from the e-aminium group (pKa = 10.5): O

O OH-

H3 N

OH



f a *

3.0-0 A

■

.

* -•• •♦ = #

In D N A this is C 1 ' o f d e o x y rib o s e .

1«------------------------------- 10.8 A ----------

(b )

In D N A this is C 1 ' of d e o x y rib o s e .

Figure 25.9

Base pairing of adenine w ith th ym in e (a) and cytosine w ith guanine (b). The dim ensions o f th e th ym in e -a d en in e and cyto sin e -g u a n in e hydrogenb o n de d pairs are such th a t th e y allo w th e fo rm a tio n o f strong hydrogen bonds and also allow the base pairs to fit inside th e tw o ph o sp h a te -rib o se chains o f the d o u ble helix. (R eprinted from Archives o f Biochemistry and Biophysics, 65, Pauling, I., Corey, R., p. 164-181, 1956. C o p yrigh t 1956, w ith perm ission from Elsevier.) E lectrostatic p o te n tia l maps calculated fo r th e individual bases show th e com plem entary d istrib u tio n o f charges th a t leads to hydrogen bonding.

1144

Chapter 25

Nucleic Acids and Protein Synthesis

-2 0 Â -

M a jo r

M in o r g ro o v e

p /

gro o ve

^R_

ipy \ p ~S ' P

■ R s^

34 Â

a^

„_P Sugar

P h o s p h a te

S

P'

XR

ËJV



DM Tr

I

N N—N

g ro u p c le a v e d

=

w ith e a c h

R— O— P— N(/-Pr)2j

c y c le from

Y

Tetrazole

Phosphoramidite

th e n e w n u c le o tid e

B,

O °3

DM Tr — O — ,

O C H ,

Dimethoxytrityl O B

R — O ~ P ~ O ~ | ,O

B 2

O B

R — O — P— O — , ^O Bl O 2

DM Tr —

.Oxidation .

O— \ O

O

H ~ O ~ |

B3

O

B3

O B

R— O ~ P ~ O ~ |

,O

B

R— O ~ P ~ O ~ |

B 2

3. Detritylation HA

O

,O

O

B 2

Simultaneous cleavage from the solid support and deprotection of phosphate esters th e n l, 2, 3, etc. and bases n h 4 o h , heat

O

O

R — O — P— O — i O

B1

R — O — P— O — i ^O B 1

I O

O

O ^™

c p g

o^™

C P G

Figure 25.19 The step s involved in autom ated synthesis of oligonucleotides using the phosphoramidite coupling m ethod.

S y n th e tic o lig o n u c le o tid e

1160

Chapter 25

Nucleic Acids and Protein Synthesis

help repair and replicate D NA. The PCR makes use o f a particular property o f D N A poly­ merases: their ability to attach additional nucleotides to a short oligonucleotide “primer” when the primer is bound to a complementary strand o f D N A called a template. The nucleotides are attached at the 3' end o f the primer, and the nucleotide that the polymerase attaches w ill be the one that is complementary to the base in the adjacent position on the template strand. If the adjacent template nucleotide is G, the polymerase adds C to the primer; if the adjacent template nucleotide is A, then the polymerase adds T, and so on. Polymerase repeats this process again and again as long as the requisite nucleotides (as triphosphates) are present in the solution, until it reaches the 5' end o f the template. Figure 25.20 shows one PCR cycle. The target D N A , a supply o f nucleotide triphos­ phate monomers, D N A polymerase, and the appropriate oligonucleotide primers (one primer sequence for each 5' to 3' direction o f the target double-stranded D N A) are added to a small reaction vessel. The mixture is briefly heated to approximately 90°C to separate the D N A strands (denaturation); it is cooled to 50-60°C to allow the primer sequences and D N A poly­ merase to bind to each o f the separated strands (annealing); and it is warmed to about 70°C to extend each strand by polymerase-catalyzed condensation o f nucleotide triphosphate monomers complementary to the parent D N A strand. Another cycle o f the PCR begins by heating to separate the new collection o f D N A m olecules into single strands, cooling for the annealing step, and so on.

3 0 - 4 0 c y c le s o f 3 s te p s : S te p 1 : Denaturation of double-stranded

DNA to single strands. 1 minute at approxim ately 90°C

3'

S te p 2 : Annealing of primers to each

single-stranded DNA. Primers are needed with sequences com plem entary to both single strands. 45 seconds at 50-60°C

strands with nucleotide triphosphate monomers from the reaction mixture. 2 minutes at approxim ately 70°C

Figure 25.20 One cycle o f th e PCR. H eating separates th e strands o f D N A o f th e ta rg e t to give tw o single-stranded te m p la te s. Primers, designated to com plem ent th e nucle o tide sequences flanking th e ta rg e ts, anneal to each strand. D N A polym erase, in th e presence o f nucleotide triph o sp h a te s, catalyzes th e synthesis o f tw o pieces o f DNA, each identical to th e original ta rg e t DN A . (Used with permisson from Andy Vierstraete, University o f Ghent.)

1161

25.8 The Polymerase Chain Reaction

4th cycle E x p o n e n tia l a m p lific a tio n

-> 3 5 th cycle

Figure 25.21 Each cycle of th e PCR doubles th e num ber of copies o f th e ta rg e t area. (Used

2 copies

4 copies

8 copies

16 copies

2 35 = 34 billion copies

with permisson from Andy Vierstraete, University of Ghent.)

Each cycle, taking only a few minutes, doubles the amount o f target D N A that existed prior to that step (Fig. 25.21). The result is an exponential increase in the amount o f D N A over time. After n cycles, the D N A w ill have been replicated 2n times— after 10 cycles there is roughly 1000 times as much DNA; after 20 cycles roughly 1 m illion times as much; and so on. Thermal cycling machines can carry out approximately 20 PCR cycles per hour, resulting in billions o f D N A copies over a single afternoon. Each application o f PCR requires primers that are 10 -2 0 nucleotides in length and whose sequences are complementary to short, conveniently located sequences flanking the target D N A sequence. The primer sequence is also chosen so that it is near sites that are cleavable with restriction enzym es. Once a researcher determines what primer sequence is needed, the primers are usually purchased from commercial suppliers who synthesize them on request using solid-phase oligonucleotide synthesis methods like that described in Section 25.7. A s an intriguing adjunct to the PCR story, it turns out that cross-fertilization between dis­ parate research fields greatly assisted development o f current PCR methods. In particular, the discovery o f extremozymes, which are enzym es from organisms that live in hightemperature environments, has been o f great use. D N A polymerases now typically used in PCR are heat-stable forms derived from thermophilic bacteria. Polymerases such as Taq poly­ merase, from the bacterium Thermus aquaticus, found in places such as geyser hot springs, and VentR™ , from bacteria living near deep-sea thermal vents, are used. U se of extremozyme polymerases facilitates PCR by allowing elevated temperatures to be used for the D N A melting step without having to worry about denaturing the polymerase enzym e at the same time. A ll materials can therefore be present in the reaction mixture throughout the entire process. Furthermore, use o f a higher temperature during the chain extension also leads to faster reaction rates. (See “The Chemistry o f . . . Stereoselective Reductions o f Carbonyl Groups,” Section 12.3C, for another example o f the use o f high-temperature enzymes.)

T he rm o p h ilic bacteria, g ro w ing in hot springs like these at Yellow stone N ational Park, produce heatstable enzymes called extrem ozym es th a t have proved useful fo r a va riety o f chemical processes.

1162

Chapter 25

Nucleic Acids and Protein Synthesis

25.9 Sequencing o f the Human G enom e: An Instruction Book fo r the Molecules o f Life

"This structure has novel features, which are of considerable biological importance." James Watson, one o f the scientists who determined the structure o f DNA.

The announcement by scientists from the public Human Genom e Project and Celera G enom ics Company in June 2000 that sequencing o f the approximately 3 billion base pairs in the human genom e was com plete marked achievement o f one o f the m ost important and ambitious scientific endeavors ever undertaken. To accom plish this feat, data were pooled from thousands o f scientists working around the world using tools including PCR (Section 25.8), dideoxynucleotide sequencing reactions (Section 25.6), capillary electrophoresis, laser-induced fluorescence, and supercomputers. What was ultimately produced is a tran­ script o f our chromosomes that could be called an instruction book for the m olecules o f life. But what do the instructions in the genom e say? How can w e best make use o f the m ol­ ecular instructions for life? O f the roughly 35,000 genes in our D N A , the function o f only a small percentage o f genes is understood. Discovering genes that can be used to benefit our human condition and the chem ical means to turn them on or o ff presents som e o f the greatest opportunities and challenges for scientists o f today and the future. Sequencing the genom e was only the beginning o f the story. A s the story unfolds, chem ists w ill continue to add to the molecular archive o f com ­ pounds used to probe our D N A . D N A microchips, with 10,000 or more short diagnostic sequences o f D N A chem ically bonded to their surface in predefined arrays, w ill be used to test D N A samples for thousands o f possible genetic conditions in a single assay. With the map o f our genom e in hand, great libraries o f potential drugs w ill be tested against genetic targets to discover more m olecules that either promote or inhibit expression o f key gene products. Sequencing o f the genom e w ill also accelerate development o f m olecules that interact with proteins, the products o f gene expression. Knowledge o f the genom e sequence w ill expedite identification o f the genes coding for interesting proteins, thus allowing these proteins to be expressed in virtually lim itless quantities. With an ample supply o f target proteins available, the challenges o f solving three-dimensional protein structures and under­ standing their functions w ill also be overcom e more easily. Optimization o f the structures o f small organic m olecules that interact with proteins w ill also occur more rapidly because the protein targets for these m olecules w ill be available faster and in greater quantity. There is no doubt that the pace o f research to develop new and useful organic m olecules for inter­ action with gene and protein targets w ill increase dramatically now that the genom e has been sequenced. The potential to use our chem ical creativity in the fields o f genom ics and proteomics is immense.

K e y T erm s a n d C o n c e p ts

PLUS

The key terms and concepts that are highlighted in bold, blue text within the chapter are defined in the glossary (at the back o f the book) and have hyperlinked definitions in the accompanying W ileyPLU S course (www.wileyplus.com ).

Problems

NotetoInstructors:

Many of the homework problems are available for assignment via WileyPLUS, an online teaching and learning solution.

N U C L E IC A C ID S T R U C T U R E

25.12

Write the structure o f the RNA dinucleotide G -C in which G has a free 5'-hydroxyl group and C has a free 3 '-hydroxyl group.

25.13

Write the structure o f the D N A dinucleotide T -A in which T has a free 5'-hydroxyl group and A has a free 3 '-hydroxyl group.

1163

Problems MECHANISMS 25.14

The example o f a silyl-H ilbert-Johnson nucleosidation reaction in Section 25.3 is presumed to involve an inter­ mediate ribosyl cation that is stabilized by intramolecular interactions involving the C2 benzoyl group. This inter­ mediate blocks attack by the heterocyclic base from the a face o f the ribose ring but allows attack on the b face, as required for formation o f the desired product. Propose a structure for the ribosyl cation intermediate that explains the stereoselective bonding o f the base.

25.15

(a) M itomycin is a clinically used antitumor antibiotic that acts by disrupting D N A synthesis through covalent bondforming reactions with deoxyguanosine in D N A . Maria Tomasz (Hunter College) and others have shown that alky­ lation o f D N A by m itomycin occurs by a com plex series o f m echanistic steps. The process begins with reduction o f the quinone ring in m itomycin to its hydroquinone form, follow ed by elimination o f methanol from the adjacent ring to form an intermediate called leuco-aziridinomitosene. One o f the paths by which leuco-aziridinomitosene alkylates D N A involves protonation and opening o f the three-membered aziridine ring, resulting in an intermediate cation that is resonance stabilized by the hydroquinone group. Attack on the cation by N2 o f a deoxyguanosine residue leads to a monoalkylated D N A product, as shown in the scheme. Write a detailed m echanism to show how the ring opening might occur, including resonance forms for the cation intermediate, followed by nucleophilic attack by DNA. (Intra- or interstrand cross-linking o f D N A can further occur by reaction o f another deoxyguanosine residue to displace the carbamoyl group o f the initial m itosene-D N A monoadduct. A cross-linked adduct is also shown.) (b) 1-Dihydromitosene A is sometimes formed from the cation intermediate in part (a) by loss o f a proton and tautomerization. Propose a detailed mechanism for the formation o f 1-dihydromitosene A from the resonance-stabi­ lized cation o f part (a).

oconh2

O

O H

C H 3O

C H

3

O C O N H

2

O

(a) Write a mechanism for protonation and ring opening.

R e s o n a n c e - s t a b i l i z e d c a t i o n

N H

C H .

i n t e r m e d i a t e

N H

C H .

O

O H

Mitomycin A

Leuco-aziridinomitosene

alkylation by N2 of a deoxyguanosine in DNA

O

N

C H 3O

(b) Write a mechanism for deprotonation and tautomerization.

D N A

\

I

2 ' - D e o x y r i b o s e

C H .

d Na O H

N

H

A monoadduct of mitomycin with DNA

O

O

C

O

H C H .

O

1-Dihydromitosene A

O

I 2 ' - D e o x y r i b o s e ' N H O D N A N H N

N H O H

n h ^ 'n

C H 3O

N

D N A

\

I

2 ' - D e o x y r i b o s e

C H 3^

D N A O H

N H 2

A cross-linked adduct of DNA with mitomycin

H

C H 3O

second alkylation by N2 of another deoxyguanosine in DNA

D N A

N

N

H

1164 25.16

Chapter 25

Nucleic Acids and Protein Synthesis

A s described in Section 25.5B, acid-base catalysis is believed to be the mechanism by which ribosomes catalyze the formation o f peptide bonds in the process o f protein translation. Key to this proposal is assistance by the N3 nitrogen (highlighted in the scheme below ) o f a nearby adenine in the ribosome for the removal o f a proton from the a-am ino group o f the amino acid adding to the growing peptide chain (Fig. 25.14). The ability o f this adenine group to remove the proton is, in turn, apparently facilitated by relay o f charge made possible by other nearby groups in the ribosome. The constellation o f these groups is shown in the scheme. Draw mechanism arrows to show for­ mation o f a resonance contributor wherein the adenine group could carry a formal negative charge, thereby facil­ itating its removal o f the a-am ino proton o f the amino acid. (The true electronic structure o f these groups is not accurately represented by any single resonance contributor, o f course. A hybrid o f the contributing resonance struc­ tures weighted according to stability would best reflect the true structure.) R ibosom e

N G2102 (2061)

NH0

O H G2482 (2447) ,N

H A 2486 (2451)

.H

T h is n itro g en is b elieved to be in v o lv ed in proton tra n s fe rs d u rin g th e p e p tid e b o n d -fo rm in g re a c tio n .

.O‘‘*

N / R ibosom e

H O

N— H

/ O !

- O:

O !

o4

\

OM O O

O

Learning Group Problem Research suggests that expression o f certain genes is controlled by conversion o f som e cytosine bases in the genom e to 5m ethylcytosine by an enzym e called D N A methyltransferase. Cytosine methylation may be a means by which som e genes are turned off as cells differentiate during growth and development. It may also play a role in som e cancer processes and in defending the genom e from foreign D N A such as viral genes. Measuring the level o f methylation in D N A is an important analytical process. One method for measuring cytosine methylation is known as methylation-specific PCR. This technique requires that all unmethylated cytosines in a sample o f D N A be converted to uracil by deamination o f the C4 amino group in the unmethylated cytosines. This is accom plished by treating the D N A with sodium bisulfite (N aH SO 3) to form a bisul­ fite addition product with its unmethylated cytosine residues. The cytosine sulfonates that result are then subjected to hydrol­ ysis conditions that convert the C4 amino group to a carbonyl group, resulting in uracil sulfonate. Finally, treatment with base causes elimination o f the sulfonate group to produce uracil. The modified D N A is then amplified by PCR using primers designed to distinguish D N A with methylated cytosine from cytosine-to-uracil converted bases. Write detailed mechanisms for the reactions used to convert cytosine to uracil by the above sequence o f steps.

Answers to Selected Problems Chapter 1

C h a p te r 4

1.18 (a), (c), (f), (g) are tetrahedral; (e) is trigonal planar; (b) is linear; (d) is angular; (h) is trigonal pyramidal.

4.8

1.23

(a) and (d); (b) and (e); and (c) and (f).

(a) 2-Chlorobicyclo[1.1.0]butane; (c) bicyclo[2.1.1]hexane; (e) 2 -methylbicyclo[2 .2 .2 ]octane.

4.9

1.31 (a), (g), (i), (l), represent different compounds that are not isomeric; (c-e), (h), (j), (m), (n), (o) represent the same compound; (b), (f), (k), (p) represent constitutional isomers. 1.38

(a) (1,1-dimethylethyl)cyclopentane or tert-butyl-cyclopentane; (c) butylcyclohexane; (e) 2 -chlorocyclopentanol.

4.10

(a)

4 .11

(a)

(a) The structures differ in the positions of the nuclei.

trans-3-Heptene; (c) 4-ethyl-2-methyl-1-hexene

1.46 (a) A negative charge; (b) a negative charge; (c) trigonal pyramidal.

C h a p te r 2

2 .11

(c) Propyl bromide; (d) isopropyl fluoride; (e) phenyl iodide.

2.14

(a)

"a

(b)

"a

(e) diisopropyl ether. 2.25

(a)

OH

(b)

XH 3 ' N "

(c)

H HO ^ OH 2.29 (a) ketone; (c) 2° alcohol; (e) 2° alcohol.

(a) 3 alkene, and a 2° alcohol; (c) phenyl and 1° amine; (e) phenyl, ester and 3° amine; (g) alkene and 2 ester groups.

2.30

2.35

(f)

,Br

1-Hexyne, 2-Hexyne, 3-Hexyne, 4-Methyl-1-pentyne, 4-Methyl-2-pentyne 3,3-Dimethyl-1-butyne

4.12

3

(R )-3 -M e th y l1 -p e n ty n e

,Br

H3C

(S )-3 -M e th y l1 -p e n ty n e

(a) 3,3,4-Trimethylhexane; (c) 3,5,7-Trimethylnonane; (e) 2-Bromobicyclo[3.3.1]nonane; (g) Cyclobutylcyclopentane.

4.24

(a) Pentane would boil higher because its chain is unbranched. (c) 2-Chloropropane because it is more polar and has a higher molecular weight. (e) CH3 COCH3 because its molecules are more polar.

4.39

,Br 2.54

Br

Ester

4.43

C h a p te r 3

(a), (c), (d), and (f) are Lewis bases; (b) and (e) are Lewis acids.

3.2

3.4

(a)

ch3

(a) [H3 O+] = [HCO2 ] = .0042 M; (b) Ionization = 4.2%.

Ka,

The pKa of the methylaminium ion is equal to 10.6 (Section 3.6B). Because the pKa of the anilinium ion is equal to 4.6, the anilinium ion is a stronger acid than the methylaminium ion, and aniline (C6 H5 NH2 ) is a weaker base than methylamine (CH3 NH2 ).

3.8

(a) CHCl2 CO2 H would be the stronger acid because the elec­ tron-withdrawing inductive effect of two chlorine atoms would make its hydroxyl proton more positive. (c) CH2 FCO2 H would be the stronger acid because a fluorine atom is more electronegative than a bromine atom and would be more electron withdrawing. 3.14

3.28

CH,

(C H ^ C

(a) pKa = 7; (b) pKa = —0.7; (c) Because the acid with a pKa = 5 has a larger it is the stronger acid.

3.5

ptCH^

More stable conformation because both alkyl groups are equatorial

(b)

C(CH 3 )

(CH 3 )3 C

CH CH

More stable because larger group is equatorial

(a) pKa = 3.752; (b) Ka = 10—13.

A-1

A-2

Answers to Selected Problems

(c)

6.26 The better yield is obtained by using the secondary halide, 1-bromo-1-phenylethane, because the desired reaction is E2. Using the primary halide will result in substantial SN 2 reaction as well, producing the alcohol instead of the desired alkene.

C(CH 3 ) ch3 (CH 3 )3 C CH, More stable conformation because both alkyl groups are equatorial

(d)

CH,

andasthenucleophile,

C(CH3)

(CH 3 )3 C

6.38 (a) You should use a strong base, such as RO , at a higher temperature to bring about an E2 reaction. (b) Here we want an Sn1 reaction. We use ethanol as the solvent and we carry out the reaction at a low temperature so that elimina­ tion is minimized.

ch

3

C h a p te r 7

More stable because larger group is equatorial C h a p te r 5

5.1 (a) achiral; (c) chiral; (e) chiral.

7.4 (a) 2,3-Dimethyl-2-butene would be the more stable because the double bond is tetrasubstituted. (c) -3-Hexene would be more stable because its double bond is disubstituted.

cis

7.7 (a)

Br

5.2 (a) Yes; (c) no. 5.3 (a) They are the same. (b) They are enantiomers. 5.7 The following possess a plane of symmetry and are, therefore, achiral: screwdriver, baseball bat, hammer. 5.11 (a) — Cl > — SH > — OH > — H (c) — OH > — CHO > — CH3 > — H (e) — OCH3 > — N(CH3 ) 2 > — CH3 > — H

(tris u b s titu te d , m o re s ta b le )

(m o n o s u b s titu te d , less s ta b le )

5.13 (a) enantiomers; (c) enantiomers.

M ajo r p ro d u ct

M in o r p ro d u ct

5.19 (a) diastereomers; (c) no; (e) no. 5.21 (a) represents A; (b) represents C; (c) represents B. 5.23 B (2S,3S)-2,3-Dibromobutane; C (2R,3S)-2,3-Dibromobutane. 5.39 (a) same; (c) diastereomers; (e) same; (g) diastereomers; (i) same; (k) diastereomers; (m) diastereomers; (o) diastereomers; (q) same.

7.25 (a) We designate the position of the double bond by using the of the two numbers of the doubly bonded carbon atoms, and the chain is numbered from the end nearer the double bond. The correct name is trans-2-pentene. (c) We use the lower number of the two doubly bonded carbon atoms to designate the position of the double bond. The correct name is 1-methylcyclohexene.

lower

7.26

(a)

C h a p te r 6 6 .6 (a) The reaction is Sn2 and, therefore, occurs with inversion of configuration. Consequently, the configuration of ( + )-2-chlorobutane is opposite [i.e., (5)] to that of ( —)-2-butanol [i.e., (R)]. (b) The configuration of ( —)-2-iodobutane is (R).

6.14 Protic solvents are formic acid, formamide, ammonia, and ethylene glycol. The others are aprotic. 6.16 (a) CH3 O—; (c) (CH3 ^P.

(e)

7.28 (a) (£)-3,5-Dimethyl-2-hexene; (c) 6-methyl-3-heptyne; (e) (3Z,5R)-5-chloro-3-hepten-6-yne. 7.43 Only the deuterium atom can assume the anti coplanar orien­ tation necessary for an E2 reaction to occur.

6.20 (a) 1-Bromopropane would react more rapidly, because, being a primary halide, it is less hindered. (c) 1-Chlorobutane, because the carbon bearing the leaving group is less hindered than in 1-chloro-2-methylpropane. (e) 1-Chlorohexane because it is a primary halide. Phenyl halides are unreactive in SN2 reactions. 6.21 (a) Reaction (1) because ethoxide ion is a stronger nucle­ ophile than ethanol; (c) reaction (2 ) because triphenylphosphine, (CgH^P, is a stronger nucleophile than triphenylamine. (Phosphorus atoms are larger than nitrogen atoms.) 6.22 (a) Reaction (2) because bromide ion is a better leaving group than chloride ion; (c) reaction (2 ) because the concentration of the substrate is twice that of reaction ( 1 ).

H

H Br H

CH,

H3C D

C h a p te r 8

8.1 2-Bromo-1-iodopropane. 8.7 The order reflects the relative ease with which these alkenes accept a proton and form a carbocation. 2 -Methylpropene reacts

A-3

Answers to Selected Problems

fastest because it leads to a 3° cation; ethene reacts slowest because it leads to a 1 ° cation. 8.25 By converting the 3-hexyne to c«-3-hexene using h ^ /^ B (P-2).

C h a p te r 9

9.4 (a) One; (b) two; (c) two; (d) one; (e) two; (f) two. 9.8 A doublet (3H) downfield; a quartet (1H) upfield. 9.9 A, CH3 CHICH3; B, CH3 CHCl2; C, CH2 ClCH2 CH2Cl

\

/

h

9.40

Ni2B (P-2)

Phenylacetylene.

9.26 G,

Br

Then, addition of bromine to c«-3-hexene will yield (3_R,4_R), and (3S,4S)-3,4-dibromohexane as a racemic form. H,

Br anti addition

Br.

Br

H"'7

v '«w H

Br.

Br

Br

\ '" H H

(3R, 4 R )

(3S, 4S)

9.24 Q is bicyclo[2.2.1]hepta-2,5-diene. R is bicyclo[2.2.1]heptane.

Racemic 3,4-dibromohexane

8.26 (a)

(b)

(e)

OH

C h a p te r 1 0

=

10.1 (a) AH° -545 kJ mol-1; (c) AH° = -101 kJ mol-1; (e) AH° = +53 kJ mol-1; (g) AH° = -132kJ mol-1. (i)

Cl

Ci)

O

1 0 .2

>

CHc

1°

OH

10.14 (a) Cyclopentane; (b) 2,2-dimethylpropane. 10.15

HBr (no peroxides)

(c)

M ethyl

(a) .Cl

HF

(d)

F

CI

H H

Cl

(2 S ,4 S )-2 ,4 -D ic h lo ro p en tan e

Cl2 light

H

Cl H

Cl

(2 R ,4 S )-2 ,4 -D ic h lo ro p e n ta n e

(c) No, (2Æ,4S)-2,4-dichloropentane is achiral because it is a meso compound. (It has a plane of symmetry passing through C3.) H

(c)

(e) Yes, by fractional distillation or by gas-liquid chromatogra­ phy. (Diastereomers have different physical properties. Therefore, the two isomers would have different vapor pressures.)

OH CH3

10.16 (a) The only fractions that would contain chiral molecules (as enantiomers) would be those containing 1 -chloro-2 -methylbutane and the two diastereomers of 2-chloro-3-methylbutane. These fractions would not show optical activity, however, because they would contain racemic forms of the enantiomers.

D 8.64

>CHa pt

X D

E

(b) Yes, the fractions containing 1-chloro-2-methylbutane and the two containing the 2-chloro-3-methylbutane diastereomers.

A-4

Answers to Selected Problems

12.9 (a)

10.27

OH

O (3°)

( 2 °)

H

> MgBr

1

(1°)

OH MgBr

O

C h a p t e r 11

11.3 (a)

(b)

'OH

(b)

PC C

OH

'O H

H

c h 2c i 2 5

OH OH

O~

11.10 Use an alcohol containing labeled oxygen. If all of the labeled oxygen appears in the sulfonate ester, then it can be con­ cluded that the alcohol C — O bond does not break during the reaction.

h3o H2O

11.25 (a) 3,3-Dimethyl-1-butanol; (c) 2-methyl-1,4-butanediol; (e) 1 -methyl-2 -cyclopenten-1 -ol. 11.26 , (a)

__

(c)

'OH

OH O

(c)

1

OH

H

+

Me

O 2 C6H5MgBr

MeO

ether

OH

(e)

OMgBr

Cl r\

NH. H2O

(i) /

'O

11.33 (a) CH3Br +

12.10 (a) CH3 CH3; (b) CH3 CH2 D; OH

Br

(c) Br. Br

(c) C6 H5 "

C h a p te r 12

12.3 (a) LiAlH4; (c) NaBH4

(g) CH3CH3 + OH

12.4 (a) NHCrO3Ci-(PCC)/CH2Ci2 (c) H2 CrO4/acetone

(h) CH3CH3 +

MgBr

MgBr

2

A-5

Answers to Selected Problems

14.9 The cyclopropenyl cation.

OH

1 2 .1 1

14.15 A, o-bromotoluene; B, p-bromotoluene; C, m-bromotoluene; D, benzyl bromide.

(a)

14.23 Huckel’s rule should apply to both pentalene and heptalene. Pentalene’s antiaromaticity can be attributed to its having 8 p elec­ trons. Heptalene’s lack of aromaticity can be attributed to its hav­ ing 12 p electrons. Neither 8 nor 12 is a Huckel number.

(b)

14.25 The bridging — CH2 — group causes the 10 p electron ring system (below) to become planar. This allows the ring to become aromatic.

(e) 12.34 r " ’° \ = O . ( 1) 2 CHsMgl, (2) NH

OH OH

14.28 (a) The cycloheptatrienyl anion has 8 p electrons, and does not obey Huckel’s rule; the cyclononatetraenyl anion with 10 p electrons obeys Huckel’s rule. 14.27 A,

B,

C,

C h a p te r 1 3

13.2 (a)

+

+

(c)

Cl

and

(racemic)

14.33 F,

13.6 (b) 1,4-Cyclohexadiene and 1,4-pentadiene are isolated dienes. 13.15 (a) 1,4-Dibromobutane + t-BuOK, and heat; (g) HC# CCH= CH2 + H2, Ni2B (P-2). 13.19 (a) 1-Butene + N-bromosuccinimide, then t-BuOK and heat; (e) cyclopentane + Br2 , then t-BuOK and heat, then Nbromosuccinimide.

hv,

13.42 The endo adduct is less stable than the exo, but is produced at a faster rate at 25°C. At 90°C the Diels-Alder reaction becomes reversible; an equilibrium is established, and the more stable exo adduct predominates.

p

4-Bromobenzoic acid (or -bromobenzoic acid) 2-Benzyl-1.3-cyclohexadiene (2 -chloro-2 -pentyl) benzene Phenyl propyl ether

heat . -H Br '

15.6 If the methyl group had no directive effect on the incoming electrophile, we would expect to obtain the products in purely sta­ tistical amounts. Since there are two ortho hydrogen atoms, two meta hydrogen atoms, and one para hydrogen, we would expect to get 40% ortho (2/5), 40% meta (2/5), and 20% para (1/5). Thus, we would expect that only 60% of the mixture of mononitrotoluenes would have the nitro group in the ortho or para position. And, we would expect to obtain 40% of -nitrotoluene. In actuality, we get 96% of combined and p-nitrotoluene and only 4% m-nitrotoluene. This result shows the ortho-para directive effect of the methyl group.

o-

C h a p te r 1 4

14.1 (a) (b) (c) (d)

C h a p te r 1 5

BrT ro p y liu m b ro m id e

These results suggest that the bonding in tropylium bromide is ionic; that is, it consists of a positive tropylium ion and a negative bromide ion.

m

15.9 (b) Structures such as the following compete with the ben­ zene ring for the oxygen electrons, making them less available to the benzene ring.

(d) Structures such as the following compete with the benzene ring for the nitrogen electrons, making them less available to the benzene ring. ■’O i r>J. N'

N

1

1

H

H

A-6

Answers to Selected Problems

15.33 (a)

Qi

17.6 (a) C g ^C ^B r + Mg in diethyl ether, then CO2 , then H3 O+; (c) CH2 = CHCH2 Br + Mg in diethyl ether, then CO2 , then H3 O+. 17.7 (a), (c), and (e). 17.9 In the carboxyl of benzoic acid. 17.14 (a) (CH3 )3 CCO2 H + SOCI2 , then NH3 , then P4 O10 , heat: A 17.22 (a)

O ^

JL

OH

(b)

+

HCI OH

(I

(c)

(d)

O

O

15.32

C h a p te r 1 6

16.2 (a) 1-Pentanol; (c) pentanal; (e) benzyl alcohol. 16.6 A hydride ion. 16.17 (b) CH3 CH2 Br + (Cgh^^P, then strong base, then C6 H5 COCH3 ; (d) CH3 I + (Cg^^P, then strong base, then cyclopentanone; (f) CH2 = CHCH2 Br + (Cg^^P, then strong base, then C6 H5 CHO. 16.20 (a) CH3 CH2 CH2 OH; (c) CH3 CH2 CH2 OH (h) CH3 CH2 CH= CHCH3; (j) CH3 CH2 CO2 ~NH4+ + AgT (l) CH3 CH2 CH= NNHCONH2; (n) CH3 CH2 CO2~

16.47 Y is 1-phenyl-2-butanone; Z is 4-phenyl-2-butanone.

(j)

O

(k)

O

O

17.46 (a) Diethyl succinate; (c) ethyl phenylacetate; (e) ethyl chloroacetate. 17.47 X is diethyl malonate.

C h a p te r 1 7

17.3 (a) CH2 FCO2 H; (c) CH2 ClCO2 H; (e) CH3 CH2 CHFCO2 H; (C) QF3— (

CH3

x^ Q Ü 2 H

C h a p te r 1 8

18.1 The enol form is phenol. It is especially stable because it is aromatic.

A-7

Answers to Selected Problems

O

(b)

18.2 No.

does not have a hydrogen

attached to its a-carbon atom (which is a chirality center) and thus enol formation involving the chirality center is not possible. With

O

O

H

OEt

19.11

O O

the chirality center is a ß carbon and H

H OH

thus enol formation does not affect it. 18.5 Base is consumed as the reaction takes place. A catalyst, by definition, is not consumed.

O

18.8 (a) Reactivity is the same as with any SN2 reaction. With pri­

mary halides substitution is highly favored, with secondary halides elimination competes with substitution, and with tertiary halides elimination is the exclusive course of the reaction. (b) Acetoacetic ester and 2-methylpropene. (c) Bromobenzene is unreactive toward nucleophilic substitution.

H2

H

Pd-C

O

C 14H 18O

18.10 Working backward

H

O

O O CrHs

(1) O H - , H 2O, heat

Æ

OH O

O

‘ (2) H 3O*

crhs

19.17 O

O

H

O O OEt

(1 )

O

(a) C

'O N a O

O

O

(b)

O

O

(c)

O

OEt Br

(2) CeHs

CeHa 18.17 In a polar solvent, such as water, the keto form is stabi­ lized by solvation. When the interaction with the solvent becomes minimal, the enol form achieves stability by internal hydrogen bonding. 18.25 (b) D is racemic frans-1,2-cyclopentanedicarboxylic acid, E

O Notice that starting compounds are drawn so as to indicate which atoms are involved in the cyclization reaction. 19.19

(a)

O

O oh~3

C eH

CeH eHs

is c«-1 ,2 -cyclopentanedicarboxylic acid, a meso compound. O

O*

C h a p te r 1 9

19.3 (a)

o

CeHs

O

'CeHa

CeHs"

O

OEt

CeHs O~

C e H ^ ^ ^ ^ '^ ^ C e H a

(b) To undergo a Dieckmann condensation, diethyl 1,5-pentanedioatc would have to form a highly strained four-membered ring. 19.5 (a)

OH

HA

o OEt

O

CeHs O

EtO

O

C e H ^ ^ ^ ^ ''^ ''''^ C e H s

A-8

Answers to Selected Problems

(b) H

Chapter 21 H

21.1 The electron-releasing group (i.e., — CH3 ) changes the charge distribution in the molecule so as to make the hydroxyl oxygen less positive, causing the proton to be held more strongly; it also destabi­ lizes the phenoxide anion by intensifying its negative charge. These effects make the substituted phenol less acidic than phenol itself.

:a . ' HA

O +

OH

C6H5 rH

O-

C6H5 rH +

H2O

+

CHc

HA

CH3

:a

Cr H 6H5

C6H5 6H

C6H5 6H

H3 O+

S+

E le c tro n -re le a s in g — C H 3 d e s ta b ilize s the a n io n m o re than th e a cid . pKa is la rg e r than fo r p heno l.

C66H5 H

19.50 (a) CH2 = C(CH3 )CO2 CH3; (b) KMnO4, OH~; H3 O+; (c) CH3 OH, HA; (d) CH3 ONa, then H3 O+

21.4 (a) The para-sulfonated phenol. (b) For ortho sulfonation. 3 2 1 .9

(a)

OCH2CH3

(b)

NH2 ,NO2

c o 2c h 3

CH3O2C

21.10 That o-chlorotoluene leads to the formation of two products (o-cresol and m-cresol), when submitted to the conditions used in the Dow process, suggests that an elimination-addition mechanism takes place.

O

(g) OH , H2 O, then H3 O+; (h) heat ( - CO2 ); (i) CH3 OH, HA; (j)

CO 2 CH 3

c h c o 2c h 3

(k) H2 , Pt; (m) CH3 ONa, then H3 O+; (n) 2 NaNH2 + 2 CH3 I C h a p te r 2 0

20.5 (a) CH3(CH2) 3 CHO + NH3 (c) CH3 (CH2)4CHO + c 6 h 5 nh 2

H2, Ni

CH3 (CH2 )3 CH2 NH2

LiBH3CN

21.11 2-Bromo-1,3-dimethylbenzene, because it has no o-hydrogen atom, cannot undergo an elimination. Its lack of reactivity toward sodium amide in liquid ammonia suggests that those com­ pounds (e.g., bromobenzene) that do react, react by a mechanism that begins with an elimination. 21.14 (a) 4-Fluorophenol because a fluorine substituent is more electron withdrawing than a methyl group. (e) 4-Fluorophenol because fluorine is more electronegative than bromine. 21.16 (a) 4-Chlorophenol will dissolve in aqueous NaOH; 4chloro-1-methylbenzene will not. (c) Phenyl vinyl ether will react with bromine in carbon tetrachloride by addition (thus decolorizing the solution); ethyl phenyl ether will not. (e) 4-Ethylphenol will dissolve in aqueous NaOH; ethyl phenyl ether will not.

CH3 (CH2 )4 CH2 NHC6 H5 20.6 The reaction of a secondary halide with ammonia is almost always accompanied by some elimination. 20.7 (a) Methoxybenzene + HNO3 + H2 SO4 , then Fe + HCI; (b) Methoxybenzene + CH3 COCI + AICI3 , then NH3 + H2 + Ni; (c) toluene + CI2 and light, then (CH3 ^N; (d) p-nitrotoluene + KMnO4 + OH- , then H3 O+, then SOCI2 followed by NH3 , then NaOBr (Br2 in NaOH); (e) toluene + N-bromosuccinimide in CCI4 , then KCN, then LiAIH4 . 20.12 p-Nitroaniline + Br2 + Fe, followed by H2 SO4/NaNO2 followed by CuBr, then H2/Pt, then H2 SO4/NaNO2 followed by H3PO2. 20.45 W is N-benzyl-N-ethylaniline.

C h a p te r 2 2

22.1 (a) Two; (b) two; (c) four. 22.5 Acid catalyzes hydrolysis of the glycosidic (acetal) group. 22.9 (a) 2 CH3 CHO, one molar equivalent HIO4 ; (b) HCHO + HCO2 H + CH3 CHO, two molar equivalents HIO4 ; (c) HCHO + OHCCH(OCH3 )2 , one molar equivalent HIO4 ; (d) HCHO + HCO2 H + CH3 CO2 H, two molar equivalents HIO4 ; (e) 2 CH3 CO2 H + HCO2 H, two molar equivalents HIO4 22.18 D-( + )-Glucose. 22.23 One anomeric form of D-mannose is dextrorotatory ([«Id = + 29.3), the other is levorotatory ([«]d = — 17.0).

A-9

Answers to Selected Problems

22.24 The microorganism selectively oxidizes the — CHOH group of D-glucitol that corresponds to C5 of D-glucose.

24.8 Glutathione is HS

22.27 A is D-altrose; B is D-talose, C is D-galactose 'O" C h a p te r 2 3

O

CO 2-

23.5 Br2 in CCI4 would react with geraniol (discharging the bromine color) but would not react with menthol. 23.12 (a) C2 H5 OH, HA, heat; or SOCI2 , then C2 H5 OH; (d) SOCI2, then (CH3 )2 NH; (g) SOCI2, then LiAIH[OC(CH3 )3]3 23.15 Elaidic acid is

trans-9-octadecenoicacid.

23.19 A is CH3 (CH2 )5 C# CNa B is CH3 (CH2 )5 C# CCH2 (CH2 )7 CH2CI C is CH3 (CH2 )5 C# CCH2 (CH2 )7 CH2CN E is CH3 (CH2 )5 C# CCH2 (CH2 )7 CH2 CO2H

h

/

c=c

24.23 Val-Leu-Lys-Phe-Ala-Glu-Ala C h a p te r 2 5

25.2 (a) The nucleosides have an N-glycosidic linkage that (like an O-glycosidic linkage) is rapidly hydrolyzed by aqueous acid, but one that is stable in aqueous base. 25.4 (a) The isopropylidene group is part of a cyclic acetal. (b) By treating the nucleoside with acetone and a trace of acid.

/ (CH2)9CO2H

25.7 (b) Thymine would pair with adenine, and, therefore, ade­ nine would be introduced into the complementary strand where guanine should occur.

\

25.9

Vaccenic acid is CH3(CH2)n

24.22 Arg-Pro-Pro-Gly-Phe-Ser-Pro-Phe-Arg

H

h

23.20 F is FCH2 (CH2 )6 CH2 C# CH G is FCH2 (CH2 )6 CH2 C# C(CH2)7CI

\

H is FCH2 (CH2 )6 CH2 C# C(CH2)7CN I

is FCH2 (CH2 )7 C# C(CH2 )7 CO2H (in m R N A )

(b) T

! P

K

S C G C A

(c) UGCi GGC UUU

M A UC

24.5 The labeled amino acid no longer has a basic — NH2 group; it is, therefore, insoluble in aqueous acid.

25.10 (a) ACCj CCC AAA AUG

C C U

C h a p te r 2 4

(in D N A )

mRNA Amino acids Anticodons

A Absolute configuration (Section 5.15A): The actual arrangement of groups in a molecule. The absolute configuration of a molecule can be determined by X-ray analysis or by relating the configura­ tion of a molecule, using reactions of known stereochemistry, to another molecule whose absolute configuration is known. Absorption spectrum (Section 13.9B): A plot of the wavelength (l) of a region of the spectrum versus the absorbance (A) at each wavelength. The absorbance at a particular wavelength (Aa) is defined by the equation log(/R//s), where Ir is the intensity of the reference beam and Is is the intensity of the sample beam.

=

Acetal (Section 16.7B): A functional group, consisting of a carbon bonded to alkoxy groups [i.e., RCH(OR') 2 or R2 C(OR')2 ], derived by adding 2 molar equivalents of an alcohol to an aldehyde or ketone. An acetal synthesized from a ketone is sometimes called a ketal. Acetoacetic ester synthesis (Section 18.6): A sequence of reac­ tions involving removal of the a-hydrogen of ethyl 3-oxobutanoate (ethyl acetoacetate, also called “acetoacetic ester”), creating a resonance-stabilized anion which then can serve as a nucleophile in an SN2 reaction. The a-carbon can be substituted twice; the ester functionality can be converted into a carboxylic acid which, after decarboxylation, yields a substituted ketone. Acetonide (Section 22.5E): A cyclic acetal formed from acetone. Acetylene (Sections 1.14, 7.1, and 7.11): A common name for ethyne. Acetylenic hydrogen atom (Sections 3.15, 4.6, and 7.9): A hydro­ gen atom attached to a carbon atom that is bonded to another carbon atom by a triple bond. Achiral molecule (Section 5.3): A molecule that is superposable on its mirror image. Achiral molecules lack handedness and are incapable of existing as a pair of enantiomers. Acid strength (Section 3.6): The strength of an acid is related to its acidity constant, or to its pKa. The larger the value of its Ka or the smaller the value of its pKa, the stronger is the acid.

Ka Acidity constant, K a(Section 3. A): An equilibrium constant 6

related to the strength of an acid. For the reaction, HA + H2O 2=9 h 30 + + A­

K=[H O +][A- ] 3

a

[HA]

Activating group (Sections 15.10, 15.10D, and 15.11A): A group that when present on a benzene ring causes the ring to be more reactive in electrophilic substitution than benzene itself. Activation energy, Eact (See Energy of activation and Section 10.5B) Active hydrogen compounds (Section 18.8): Compounds in which two electron-withdrawing groups are attached to the same carbon atom (a methylene or methane carbon). The electron-with­ drawing groups enhance the acidity of the hydrogens on carbon; these hydrogens are easily removed, creating a resonance-stabi­ lized nucleophilic anion.

Active site (Section 24.9): The location in an enzyme where a sub­ strate binds. Acyl compounds (Section 17.1): A compound containing the group (R— C = O)—, usually derived from a carboxylic acid, such as an ester, acid halide (acyl halide), amide, or carboxylic acid anhydride. Acyl group (Section 15.7): The general name for groups with the structure RCO— or ArCO— .

acidhalide.

Acyl halide (Section 15.7): Also called an A general name for compounds with the structure RCOX or ArCOX. Acyl transfer reactions (Section 17.4): A reaction in which a new acyl compound is formed by a nucleophilic addition-elimination reaction at a carbonyl carbon bearing a leaving group. Acylation (Section 15.7): The introduction of an acyl group into a molecule. Acylium ion (Sections 9.16C and 15.7): The resonance-stabilized cation: r—c

= o:

Addition polymer (Section 10.10 and Special Topic A): A poly­ mer that results from a stepwise addition of monomers to a chain (usually through a chain reaction) with no loss of other atoms or molecules in the process. Also called a chain-growth polymer. Addition reaction (Sections 3.1, 8.1-8.9, 8.12, 8.13, 12.1A, 16.6, and 17.4): A reaction that results in an increase in the number of groups attached to a pair of atoms joined by a double or triple bond. An addition reaction is the opposite of an elimination reaction. Adduct (Section 13.11): The product formed by a Diels-Alder [4+ 2] cycloaddition reaction, so called because two compounds (a and a are added together to form the product.

diene

dienophile)

Aglycone (Section 22.4): The alcohol obtained by hydrolysis of a glycoside. Aldaric acid (Section 22.6 C): An a,v-dicarboxylic acid that results from oxidation of the aldehyde group and the terminal 1 ° alcohol group of an aldose. Alditol (Section 22.7): The alcohol that results from the reduction of the aldehyde or keto group of an aldose or ketose. Aldol (Section 19.4): A common name for 3-hydoxybutanal, which contains both ehyde and an alcoh functional groups. Aldol is formed from the (see below) of ethanal (acetaldehyde) with itself.

ald

aldol reaction

ol

Aldol additions (Section 19.4): See Aldol reaction and aldol condensation. Aldol condensation (Section 19.1, Section 19.4C): An aldol reaction that forms an a,b-unsaturated product by dehydration of the b-hydroxy aldehyde or ketone aldol product. Aldol reactions (Sections 19.4-19.6): Reactions in which the enol or enolate ion of an aldehyde or ketone reacts with the carbonyl group of the same or a different aldehyde or ketone, creating a hydroxy aldehyde or ketone and a new carbon-carbon s-bond.

b-

Gl-1

Gl-2

Glossary

(Section 22.6B): A monocarboxylic acid that results from oxidation of the aldehyde group of an aldose.

A ldonic acid

(Section 14.1): A nonaromatic compound such as an alkane, cycloalkane, alkene, or alkyne.

A lip h atic com pound

(Special Topic E): A naturally occurring basic compound that contains an amino group. Most alkaloids have profound physiological effects. A lk alo id

(Sections 2.2, 4.1-4.3, 4.7, and 4.16): Hydrocarbons having only single (s) bonds between carbon atoms. Acyclic alka­ nes have the general formula CnH2 n+2 . Monocyclic alkanes have the general formula of CnH2n. Alkanes are said to be “saturated” because C—C single bonds cannot react to add hydrogen to the molecule. A lkan es

An alkyl anion, R:_, or alkyl species that reacts as though it were an alkyl anion.

A lkan id e (Section 7.8A):

(Sections 2.2, 4.1, and 4.5): Hydrocarbons having at least one double bond between carbon atoms. Acyclic alkenes have the general formula CnH2n. Monocyclic alkenes have the general for­ mula of CnH2n- 2. Alkenes are said to be “unsaturated” because their C= C double bonds can react to add hydrogen to the mole­ cule, yielding an alkane. A lken es

Alkyl group (See R): (Sections 2.5A and 4.3A) The designation given to a fragment of a molecule hypothetically derived from an alkane by removing a hydrogen atom. Alkyl group names end in “yl.” Example: the methyl group, CH3—, is derived from methane, CH4 . (Sections 7.12, 15.6, and 18.4C): The introduction of an alkyl group into a molecule.

A lkylatio n

(Sections 2.2, 4.1, and 4.6): Hydrocarbons having at least one triple bond between carbon atoms. Acyclic alkynes have the general formula CnH2n_2. Monocyclic alkynes have the general formula of CnH2 n- 4 . Alkynes are said to be “unsaturated” because C # C triple bonds can react to add two molecules of hydrogen to the molecule, yielding an alkane.

A lkyn es

A lly l group

(Section 4.5): The CH2 — CHCH2— group.

(Section 13.4) The carbocation formally related to propene by removal of a proton from its methyl group. The two contributing resonance structures of the delocalized car­ bocation each include a positive charge on a carbon adjacent to the double bond, such that a orbital on each of the three carbons overlaps to delocalize positive charge to each end of the allyl system. A lly l (propenyl cation):

p

(Sections 13.2A and 13.3): The radical formally related to propene by removal of a hydrogen atom from its methyl group. The two contributing resonance structures of the delocal­ ized radical each include an unpaired electron on a carbon adjacent to the double bond, such that a orbital on each of the three carbons overlaps to delocalize the radical to each end of the allyl system. in which the radical carbon is adjacent to a carbon-carbon double bond. A lly l rad ical

p

(Sections 13.4, 13.10, and 15.15): A substruc­ ture involving a three-carbon delocalized carbocation in which the positively charged carbon is adjacent to a carbon-carbon double bond in each of two contributing resonance structures.

(Section 22.2C): In the standard Haworth for­ mula representation for a D-hexopyranose, the a anomer has the hemiacetal hydroxyl or acetal alkoxyl group trans to C6 . Similar usage applies to other carbohydrate forms regarding the stereo­ chemical relationship of the anomeric hydroxyl or alkoxyl group and the configuration at the carbon bearing the ring oxygen that forms the hemiacetal or acetal. A lp h a («) anom er

A lp h a («) carbon

(Section 18.1): A carbon adjacent to a carbonyl

(C = O) group. (Section 24.8A): A secondary structure in proteins where the polypeptide chain is coiled in a right-handed helix.

A lp h a («) helix

(Sections 18.1, 18.5C, 18.5D): A hydrogen atom bonded to an a carbon. These hydrogens are significantly more acidic than the typical alkane hydrogen.

A lp h a («) hydrogens

(Section 20.3D): The product of the reaction of an amine, acting as a Bronsted-Lowry base, with an acid. The amine can be primary, secondary, or tertiary. The positively charged nitro­ gen in an aminium salt is attached to at least one hydrogen atom. (An ammonium salt has no hydrogen atoms bonded directly to the nitrogen.)

A m inium salt

A m ino acid residue

(Section 24.4): An amino acid that is part of

a peptide. (Section 4.10): The increased potential energy of a molecule (usually a cyclic one) caused by deformation of a bond angle away from its lowest energy value. A n gle strain

(Section 14.7B): Monocyclic hydrocarbon that can be represented by a structure having alternating single and double bonds. The ring size of an annulene is represented by a number in brackets, e.g., benzene is [6 ]annulene and cyclooctatetraene is [8 ]annulene. A nnulene

(Section 22.2E): The hemiacetal or acetal carbon in the cyclic form of a carbohydrate. The anomeric carbon can have either the a or b stereochemical configuration (using carbohydrate nomenclature), resulting in diastereomeric forms of the carbohydrate called anomers (a-anomers and b-anomers). Anomers differ in the stereochemistry at the anomeric carbon.

A n o m e ric ca rb o n

only

(Section 22.2C): A term used in carbohydrate chemistry. Anomers are diastereomers that differ only in configuration at the acetal or hemiacetal carbon of a sugar in its cyclic form.

A nom ers

(Sections 7.14A, 7.15B, and 8.13): An addition that places the parts of the adding reagent on opposite faces of the reactant. A n ti addition

(Section 4.9): An anti conformation of butane, for example, has the methyl groups at an angle of 180° to each other:

A n ti conform ation

CH 3

A lly lic carbocation

(Section 13.2): Refers to a substituent on a car­ bon atom adjacent to a carbon-carbon double bond.

A lly lic substituent

CH 3

Anti conformation of butane

Gl-3

Glossary

Antiaromatic compound (Section 14.7E): A cyclic conjugated

system whose p electron energy is greater than that of the corre­ sponding open-chain compound.

momentum of an electron and can have the values of + only.

1/ 2

and — 1/ 2

Atropisomers (Section 5.18): Conformational isomers that are sta­

Antibonding molecular orbital (antibonding MO) (Sections 1.11,

ble, isolable compounds.

1.13, and 1.15): A molecular orbital whose energy is higher than that of the isolated atomic orbitals from which it is constructed. Electrons in an antibonding molecular orbital destabilize the bond between the atoms that the orbital encompasses.

Aufbau principle (Section 1.10A): A principle that guides us in

Anticodon (Section 25.5C): A sequence of three bases on transfer

Autoxidation (Section 10.11D): The reaction of an organic com­

RNA (tRNA) that associates with a codon of messenger RNA (mRNA). anti-Markovnikov addition (Sections 8.2D, 8.7, 8.19, and 10.9):

An addition reaction where the hydrogen atom of a reagent becomes bonded to an alkene or alkyne at the carbon having the fewer hydrogen atoms initially. This orientation is the opposite of that predicted by Markovnikov’s rule.

assigning electrons to orbitals of an atom or molecule in its lowest energy state or ground state. The aufbau principle states that elec­ trons are added so that orbitals of lowest energy are filled first. pound with oxygen to form a hydroperoxide. Axial bond (Section 4.12): The six bonds of a cyclohexane ring

(below) that are perpendicular to the general plane of the ring, and that alternate up and down around the ring.

Aprotic solvent (Section 6.13C): A solvent whose molecules do

not have a hydrogen atom attached to a strongly electronegative element (such as oxygen). For most purposes, this means that an aprotic solvent is one whose molecules lack an — OH group. Arenium ion (Section 15.2): A general name for the cyclohexadi-

enyl carbocations that form as intermediates in electrophilic aro­ matic substitution reactions. Aromatic compound (Sections 2.1, 2.1D, 14.1-14.8, and 14.11):

A cyclic conjugated unsaturated molecule or ion that is stabilized by p electron delocalization. Aromatic compounds are character­ ized by having large resonance energies, by reacting by substitu­ tion rather than addition, and by deshielding of protons exterior to the ring in their 1H NMR spectra caused by the presence of an induced ring current. Aromatic ions (Section 14.7D): Cations and anions that fulfill the

B Base peak (Section 9.13): The most intense peak in a mass spec­

trum. Base strength (Sections 3.6C and 20.3): The strength of a base is

inversely related to the strength of its conjugate acid; the weaker the conjugate acid, the stronger is the base. In other words, if the conjugate acid has a large p a, the base will be strong.

K

Benzene (Section 2.2D): The prototypical aromatic compound

having the formula CgHg. Aromatic compounds are planar, cyclic, and contain 4n + 2 p electrons in contiguous fashion about a ring of electron density in the molecule. Electron delocal­ ization gives aromatic compounds a high degree of stability.

delocalized

criteria for aromaticity (planarity, electron delocalization, and a Huckel number of p-electrons) and thus have additional (aromat­ ic) stability.

Benzenoid aromatic compound (Section 14.8A): An aromatic compound whose molecules have one or more benzene rings.

Aryl amines (Section 20.1A): A compound in which the carbon of

Benzyl group (Section 2.4B): The C6 H5 CH2— group.

an aromatic ring bears the amine nitrogen atom. Aryl amines can be primary, secondary, or tertiary.

Benzylic cation (Section 15.12A): A carbocation where the posi­

Aryl group (Section 6.1): The general name for a group obtained

(on paper) by the removal of a hydrogen from a ring position of an aromatic hydrocarbon. Abbreviated Ar— . Aryl halide (Section 6.1): An organic halide in which the halogen

atom is attached to an aromatic ring, such as a benzene ring. Atactic polymer (Special Topic B.1): A polymer in which the con­

figuration at the stereogenic centers along the chain is random. Atomic orbital (AO) (Sections 1.10, 1.11, and 1.15): A volume of

space about the nucleus of an atom where there is a high probabil­ ity of finding an electron. An atomic orbital can be described math­ ematically by its wave function. Atomic orbitals have characteris­ tic quantum numbers; the is related to the energy of the electron in an atomic orbital and can have the values 1, 2, 3, . . . . The , determines the angular momentum of the electron that results from its motion around the nucleus, and can have the values 0 , 1 , 2 , . . . , (n 1 ). The , determines the orientation in space of the angular momentum and can have values from to — . The , specifies the intrinsic angular

principalquantumnumber, n, azimuthalquantumnumber,l

magneticquantumnumber, m l spinquantumnumber, s

— +l

tive charge is on a carbon bonded to a benzene ring. The positive charge is delocalized into the benzene ring through conjugation, resulting in a relatively stable carbocation. Benzylic radical (Section 15.12): The radical comprised of a methylene (CH2 ) group bonded to a benzene ring, wherein the unpaired electron is over the methylene group and the ring. As a highly system, the benzylic radical has great­ ly enhanced stability.

delocalized conjugated

Benzylic substituent (Sections 15.15): Refers to a substituent on a

carbon atom adjacent to a benzene ring. Benzyne (Section 21.11B): An unstable, highly reactive interme­

diate consisting of a benzene ring with an additional bond resulting from sideways overlap of orbitals on adjacent atoms of the ring.

sp2

Beta (b) anomer (Section 22.2C): In the standard Haworth for­ mula representation for a D-hexopyranose, the b anomer has the hemiacetal hydroxyl or acetal alkoxyl group cis to C6 . Similar usage applies to other carbohydrate forms regarding the stereo­ chemical relationship of the anomeric hydroxyl or alkoxyl group and the configuration at the carbon bearing the ring oxygen that forms the hemiacetal or acetal.

Gl-4

Glossary

B e ta ( b ) - c a r b o n y l c o m p o u n d (Section 18.4C): A compound hav­ ing two carbonyl groups separated by an intervening carbon atom. B ic y c lic c o m p o u n d s (Sections 4.4B and 4.14): Compounds with two fused or bridged rings. B im o le c u la r re a c tio n (Section 6.5): A reaction whose ratedetermining step involves two initially separate species.

(Section 4.11): A conformation of cyclohexane that resembles a boat and that has eclipsed bonds along its two sides:

B o a t c o n fo rm a tio n

It is of higher energy than the chair conformation. B o ilin g p o in t (Sections 2.14A and 2.14D): The temperature at which the vapor pressure of a liquid is equal to the pressure above the surface of the liquid.

(Section 1.12): The angle between two bonds origi­ nating at the same atom. B o n d a n g le

B o n d d iss o c ia tio n e n e rg y e n e rg y

(See

H o m o ly tic b o n d d isso c ia tio n

and Section 10.2)

B o n d le n g th (Sections 1.11 and 1.14A): The equilibrium distance between two bonded atoms or groups.

(Section 8.15C): A carbene-like species. A species such as the reagent formed when diiodomethane reacts with a zinc-copper couple. This reagent, called the Simmons-Smith reagent, reacts with alkenes to add methylene to the double bond in a stereospecific way. C a r b e n o id

C a rb o c a t io n (Sections 3.4, 6.11, and 6.12): A chemical species in which a trivalent carbon atom bears a formal positive charge. C a r b o h y d r a t e (Section 22.1A): A group of naturally occurring compounds that are usually defined as polyhydroxyaldehydes or polyhydroxyketones, or as substances that undergo hydrolysis to yield such compounds. In actuality, the aldehyde and ketone groups of carbohydrates are often present as hemiacetals and acetals. The name comes from the fact that many carbohydrates possess the empirical formula C^^O)^,. C a r b o n - 1 3 N M R s p e c tro s c o p y (Section 9.11): NMR spec­ troscopy applied to carbon. Carbon-13 is NMR active, whereas carbon-12 is not and therefore cannot be studied by NMR. Only 1.1% of all naturally occurring carbon is carbon-13.

(Section 16.1): A functional group consisting of a carbon atom doubly bonded to an oxygen atom. The carbonyl group is found in aldehydes, ketones, esters, anhydrides, amides, acyl halides, and so on. Collectively these compounds are referred to as carbonyl compounds. C a r b o n y l g ro u p

B o n d -lin e f o r m u la (Section 1.17C): A formula that shows the carbon skeleton of a molecule with lines. The number of hydro­ gen atoms necessary to fulfill each carbon’s valence is assumed to be present but not written in. Other atoms (e.g., O, Cl, N) are written in.

C a r b o x y lic a c id d e riv a t iv e s (Section 17.1): Acyl compounds that can be synthesized from a carboxylic acid or another carboxylic acid derivative. Examples include esters, amides, acid halides, anhydrides, etc.

Bonding molecular orbital (bonding MO) (Sections 1.11 and 1.15): The energy of a bonding molecular orbital is lower than the energy of the isolated atomic orbitals from which it arises. When electrons occupy a bonding molecular orbital they help hold together the atoms that the molecular orbital encompasses.

C h a in -g r o w t h p o ly m e r (see A d d it io n p o ly m e r and Special Topic B): Polymers (macromolecules with repeating units) formed by adding subunits (called repeatedly to form a chain.

Broadband (BB) proton decoupling (see Proton decoupling) (Section 9.11B): A method of eliminating carbon-proton coupling by irradiating the sample with a wide-frequency (“broadband”) energy input in the frequencies in which protons absorb energy. This energy input causes the protons to remain in the high energy state, eliminating coupling with carbon nuclei.

chain-initiatingsteps,chain-propagat­ ingsteps, chain-terminatingsteps.

B r o m in a t io n (Sections 8.12, 10.5C, and 10.6 A): A reaction in which one or more bromine atoms are introduced into a molecule.

(Section 8.14): A compound bearing a bromine atom and a hydroxyl group on adjacent (vicinal) carbons.

B r o m o h y d r in

(Section 8.12A): An ion containing a positive bromine atom bonded to two carbon atoms. B r o m o n iu m io n

(Section 3.2A): An acid is a substance that can donate (or lose) a proton; a base is a substance that can accept (or remove) a proton. The c o n ju g a te a c id of a base is the molecule or ion that forms when a base accepts a proton. The c o n ju g a te b ase of an acid is the molecule or ion that forms when an acid loses its proton. B r 0 n s t e d - L o w r y th e o ry o f a c id s a n d b ase s

C

(Sections 3.4 and 12.1A): A chemical species in which a carbon atom bears a formal negative charge. C a r b a n io n

(Section 8.15): An uncharged species in which a carbon atom is divalent. The species :CH2 , called methylene, is a carbene. C a rb e n e

CFC (see Freon): A chlorofluorocarbon.

monomers)

C h a in re a c tio n (Sections 10.4-10.6, 10.10, and 10.11): A reaction that proceeds by a sequential, stepwise mechanism, in which each step generates the reactive intermediate that causes the next step to occur. Chain reactions have and

C h a in -t e rm in a t in g (d id e o x y n u c le o tid e ) m e th o d (Section 25.6): A method of sequencing DNA that involves replicating DNA in a way that generates a family of partial copies, each differing in length by one base pair and containing a nucleotide-specific fluor­ escent tag on the terminal base. The partial copies of the parent DNA are separated by length, usually using capillary electrophore­ sis, and the terminal base on each strand is identified by the cova­ lently attached fluorescent marker. C h a i r c o n fo rm a tio n (Section 4.11): The all-staggered conforma­ tion of cyclohexane that has no angle strain or torsional strain and is, therefore, the lowest energy conformation:

(Sections 9.2A, 9.7, and 9.11C): The position in an NMR spectrum, relative to a reference compound, at which a nucleus absorbs. The reference compound most often used is tetramethylsilane (TMS), and its absorption point is arbitrarily des­ ignated zero. The chemical shift of a given nucleus is proportional to the strength of the magnetic field of the spectrometer. The chem­ ical shift in delta units, S, is determined by dividing the observed C h e m ic a l sh ift, d

Gl-5

Glossary

shift from TMS in hertz multiplied by 106 by the operating fre­ quency of the spectrometer in hertz.

alcohol. Polyesters, polyamides, and polyurethanes are all conden­ sation polymers.

C h ira l au xiliary

(Sections 19.1, 19.2, and 19.4-19.6): A reaction in which molecules become joined through the intermole­ cular elimination of water or an alcohol.

(Section 13.11C): A group, p resentinoneenan­ tiomericformonly, that is appended by a functional group to a reactant so as to provide a chiral influence on the course of the reaction. The chiral auxiliary is removed once the reaction has been completed. (Sections 5.3 and 5.12): A molecule that is not superposable on its mirror image. Chiral molecules have handed­ ness and are capable of existing as a pair of enantiomers. C h ira l m olecule

C h ira lity (Sections 5.1, 5.4, and 5.6): The property of having handedness. C h ira lity center (Sections 5.2, 5.4, and 5.17): An atom bearing groups of such nature that an interchange of any two groups will produce a stereoisomer.

(Section 13.11C): A starting material for a reaction in which a chirality center, is included. The chiral influence of the chirality center in the chiron leads to enantioselective interactions during the synthesis. C h iron

inasingleenantiomericform,

C h lorination (Sections 8.12, 10.3B, 10.4, and 10.5): A reaction in which one or more chlorine atoms are introduced into a molecule.

Condensation reaction

(Sections 5.7, 5.15, and 6 .8 ): The particular arrangement of atoms (or groups) in space that is characteristic of a given stereoisomer. C on figuration

(Section 4.8): A particular temporary orientation of a molecule that results from rotations about its single bonds. Conform ation

(Sections 4.8, 4.9, 4.11, and 4.12): An analysis of the energy changes that a molecule undergoes as its groups undergo rotation (sometimes only partial) about the single bonds that join them. C on form ation al analysis

C on form ation al stereoisom ers (Section 4.9A): Stereoisomers differing in space only due to rotations about single (s) bonds. C o n fo rm a tio n s o f cyclo h exan e (Sections 4.11 and 4.13): Rotations about the carbon-carbon single bonds of cyclohexane can produce different conformations which are interconvertible. The most important are the chair conformation, the boat confor­ mation, and the twist conformation.

(Section 4.8): A particular staggered conformation of

(Section 8.14): A compound bearing a chlorine atom and a hydroxyl group on adjacent (vicinal) carbons.

C on form er

C is-tra n s isom ers (Sections 1.13B, 4.13, and 7.2): Diastereomers that differ in their stereochemistry at adjacent atoms of a double bond or on different atoms of a ring. Cis groups are on the same side of a double bond or ring. Trans groups are on opposite sides of a double bond or ring.

C on jugate acid (Section 3.2A): The molecule or ion that forms when a base accepts a proton.

C h lorohydrin

Claisen condensation (Section 19.1): A reaction in which an enolate anion from one ester attacks the carbonyl function of another ester, forming a new carbon-carbon s-bond. A tetrahedral inter­ mediate is involved that, with expulsion of an alkoxyl group, col­ lapses to a b-ketoester. The two esters are said to into a larger product with loss of an alcohol molecule.

“condense”

(Section 21.9): A [3,3] sigmatropic rearrangement reaction involving an allyl vinyl ether, in which the allyl group of migrates to the other end of the vinyl system, with bond reorganization leading to a g,S-unsaturated carbonyl compound. C laisen rearran gem en t

Codon (Section 25.5C): A sequence of three bases on messenger RNA (mRNA) that contains the genetic information for one amino acid. The codon associates, by hydrogen bonding, with an anti­ codon of a transfer RNA (tRNA) that carries the particular amino acid for protein synthesis on the ribosome. Coenzym e (Section 24.9): A small organic molecule that partici­ pates in the mechanism of an enzyme and which is bound at the active site of the enzyme.

(Section 24.9): A metal ion or organic molecule whose presence is required in order for an enzyme to function. C o fa cto r

(Section 6 .6 ): A reaction where bond forming and bond breaking occur simultaneously (in concert) through a sin­ gle transition state. C on certed reaction

Condensation polymer (see Step-growth polymer, Section 17.12, and Special Topic C): A polymer produced when bifunc­ tional monomers (or potentially bifunctional monomers) react with each other through the intermolecular elimination of water or an

a molecule.

C on jugate addition (Section 19.7): A form of nucleophilic addi­ tion to an a,b-unsaturated carbonyl compound in which the nucle­ ophile adds to the b carbon. Also called Michael addition. C on jugate base (Section 3.6C): The molecule or ion that forms when an acid loses its proton. C on jugated protein (Section 24.12): A protein that contains a non­ protein group (called a prosthetic group) as part of its structure. C on jugated unsaturated system (Section 13.1): Molecules or ions that have an extended p system. A conjugated system has a orbital on an atom adjacent to a multiple bond; the orbital may be that of another multiple bond or that of a radical, carbocation, or carbanion.

p

p

C onnectivity (Sections 1.3 and 1.17A): The sequence, or order, in which the atoms of a molecule are attached to each other. C o n stitu tio n al isom ers (Sections 1.3A, 4.2, and 5.2A): Compounds that have the same molecular formula but that differ in their connectivity (i.e., molecules that have the same molecular for­ mula but have their atoms connected in different ways). C o p lan ar (Section 7.6D): A conformation in which vicinal groups lie in the same plane. C op olym er (Special Topic A): A polymer synthesized by poly­ merizing two monomers.

(Section 9.12) (Correlation Spectroscopy): A two-dimen­ sional NMR method that displays coupling relationships between protons in a molecule. CO SY

(Section 9.2C): In NMR, the splitting of the energy levels of a nucleus under observation by the energy levels of nearby NMRactive nuclei, causing characteristic splitting patterns for the signal of the nucleus being observed. The signal from an NMR-active nucleus will be split into (2nI + 1) peaks, where n = the number of equivalent Coupling

Gl-6

Glossary

neighboring magnetic nuclei and I = the spin quantum number. For hydrogen (I = 1/2) this rule devolves to (n + 1), where n = the num­ ber of equivalent neighboring hydrogen nuclei. C ouplin g constant, / ab (Section 9.9C): The separation in frequen­ cy units (hertz) of the peaks of a multiplet caused by spin-spin coupling between atoms a and b. Covalent bond (Section 1.4B): The type of bond that results when atoms share electrons. C rack in g (Section 4.1A): A process used in the petroleum indus­ try for breaking down the molecules of larger alkanes into smaller ones. Cracking may be accomplished with heat (thermal cracking), or with a catalyst (catalytic cracking).

(Section 19.5): An aldol reaction involv­ ing two different aldehyde or ketone reactants. If both aldol reac­ tants have a hydrogens, four products can result. Crossed aldol reactions are synthetically useful when one reactant has no hydrogens, such that it can serve only as an electrophile that is sub­ ject to attack by the enolate from the other reactant. Crossed-aldol reaction

a

C ro w n ether (Section 11.16): Cyclic polyethers that have the abil­ ity to form complexes with metal ions. Crown ethers are named as -crown where is the total number of atoms in the ring and is the number of oxygen atoms in the ring.

x

-y

x

y

(Sections 1.8, 3.5, and 10.1): Curved arrows show the direction of electron flow in a reaction mechanism. They point from the source of an electron pair to the atom receiving the pair. Double-barbed curved arrows are used to indicate the movement of a pair of electrons; single-barbed curved arrows are used to indi­ cate the movement of a single electron. Curved arrows are never used to show the movement of atoms. C u rved arrow s

(Sections 16.9 and 17.3): A functional group con­ sisting of a carbon atom bonded to a cyano group and to a hydrox­ yl group, i.e., RHC(OH)(CN) or R2 C(OH)(CN), derived by adding HCN to an aldehyde or ketone. C yan oh ydrin

C ycloaddition (Section 13.11): A reaction, like the Diels-Alder reaction, in which two connected groups add to the end of a p sys­ tem to generate a new ring. Also called 1,4-cycloaddition. C ycloalkan es (Sections 4.1, 4.4, 4.7, 4.10-4.12, 4.15, and 4.16): Alkanes in which some or all of the carbon atoms are arranged in a ring. Saturated cycloalkanes have the general formula CnH2 n. D

(Section 22.2B): A method for designat­ ing the configuration of monosaccarides and other compounds in which the reference compound is ( + )- or ( —)-glyceraldehyde. According to this system, ( + )-glyceraldehyde is designated D( + )-glyceraldehyde and ( —)-glyceraldehyde is designated L-( —)glyceraldehyde. Therefore, a monosaccharide whose highest numbered stereogenic center has the same general configuration as D-(+)-glyceraldehyde is designated a D-sugar; one whose highest numbered stereogenic center has the same general configuration as L-( + )-glyceraldehyde is designated an L-sugar. D and L nom enclature

Dash structu ral form ulas (Section 1.17A): Structural formulas in which atom symbols are drawn and a line or “dash” represents each pair of electrons (a covalent bond). These formulas show connec­ tivities between atoms but do not represent the true geometries of the species.

(Sections 15.10, 15.10E, 15.10F, and 15.11A): A group that when present on a benzene ring causes the ring to be less reactive in electrophilic substitution than benzene itself. D e activ atin g group

(Section 7.10): The elimination of two atoms of bromine from a -dibromide, or, more generally, the loss of bromine from a molecule. Debrom ination

vic

(Section 2.2): The unit in which dipole moments are stated. One debye, D, equals 1 X 10—18 esu cm.

D ebye unit

D ecarboxylation (Section 17.10): A reaction whereby a carboxylic acid loses CO2 . D egenerate orbitals (Section 1.10A): Orbitals of equal energy. For example, the three 2 orbitals are degenerate.

p

(Sections 7.7 and 7.8): An elimination that involves the loss of a molecule of water from the substrate. D ehydration reaction

D ehydrohalogenation (Sections 6.15A and 7.6): An elimination reaction that results in the loss of HX from adjacent carbons of the substrate and the formation of a p bond.

(Sections 3.11A and 6.11B): The dispersal of elec­ trons (or of electrical charge). Delocalization of charge always sta­ bilizes a system. D elocalization

(Sections 25.1 and 25.4): One of the two molecules (the other is RNA) that carry genetic informa­ tion in cells. Two molecular strands held together by hydrogen bonds give DNA a “twisted ladder”-like structure, with four types of heterocyclic bases (adenine, cytosine, thymine, and guanine) making up the “rungs” of the ladder. D eoxyribonucleic acid (DNA)

D extrorotatory (Section 5.8B): A compound that rotates planepolarized light clockwise.

(Section 5.2C): Stereoisomers that are not mirror images of each other.

D iastereom ers

Diastereoselective reaction

(See Stereoselective reaction and

Sections 5.10B and 12.3C) Diastereotopic hydrogens (or ligands) (Section 9.8B): If replace­ ment of each of two hydrogens (or ligands) by the same groups yields compounds that are diastereomers, the two hydrogen atoms (or ligands) are said to be diastereotopic. 1,2-D iaxial interaction (Section 4.12): The interaction between two axial groups that are on adjacent carbon atoms.

(Sections 20.6A, 20.6B, 20.7, and 20.8): Salts synthesized from the reaction of primary amines with nitrous acid. Diazonium salts have the structure [R-N=N]+ X—. Diazonium salts of primary aliphatic amines are unstable and decompose rapidly; those from primary aromatic amines decompose slowly when cold, and are useful in the synthesis of substituted aromatics and compounds. D iazonium salts

azo

(Section 19.2A): An intramolecular Claisen condensation of a diester; the enolate from one ester group attacks the carbonyl of another ester function in the same molecule, leading to a cyclic product. D ieckm ann condensation

(Section 6.13D): A measure of a solvent’s ability to insulate opposite charges from each other. The dielectric constant of a solvent roughly measures its polarity. Solvents with high dielectric constants are better solvents for ions than are sol­ vents with low dielectric constants. D ielectric constant

Gl-7

Glossary D iels-A lder reaction (Section 13.11): In general terms, a reaction between a conjugated diene (a 4-p-electron system) and a com­ pound containing a double bond (a 2 -p-electron system), called a dienophile, to form a cyclohexene ring.

(di

(Section 13.11): A molecule containing two double bonds two, alkene or double bonds). In a Diels-Alder reaction, a diene in the conformation reacts with a dienophile.

Diene

= ene= conjugated

s-cis

(Section 13.11): The diene-seeking component of a Diels-Alder reaction. Dienophile

D ihedral angle (Sections 4.8 and 9.9D): See Fig. 4.4. The angle between two atoms (or groups) bonded to adjacent atoms, when viewed as a projection down the bond between the adjacent atoms. Dihydroxylation (Section 8.16): A process by which a starting material is converted into a product containing adjacent alcohol functionalities (called a “1 ,2 -diol” or “glycol”). D ipeptide

(Section 24.4): A peptide comprised of two amino

acids. (Section 24.2C): The charge-separated form of an amino acid that results from the transfer of a proton from a car­ boxyl group to a basic group. D ipolar ion

Dipole m om ent, m (Section 2.2): A physical property associated with a polar molecule that can be measured experimentally. It is defined as the product of the charge in electrostatic units (esu) and the distance that separates them in centimeters: m = d.

eX

(Section 2.13B): An interaction between mol­ ecules having permanent dipole moments. D ipole-dipole force

(Section 18.4C): A synthetic process in which the a-hydrogen of an ester is removed by a strong, bulky base such as LDA, creating a resonance-stabilized anion which will act as a nucleophile in an Sn2 reaction. D irect alkylation

D irected aldol reaction (Section 19.5B): A crossed aldol reaction in which the desired enolate anion is generated first and rapidly using a strong base (e.g., LDA) after which the carbonyl reactant to be attacked by the enolate is added. If both a and a are possible, this process favors generation of the kinetic enolate anion.

thermodynamicenolateanion

(Section 9.2C): An NMR signal comprised of two peaks with equal intensity, caused by signal splitting from one neighbor­ ing NMR-active nucleus. Doublet

Downfield (Section 9.2A): Any area or signal in an NMR spectrum that is to the left relative to another. (See “Upfield” for compari­ son.) A signal that is downfield of another occurs at higher fre­ quency (and higher S and ppm values) than the other signal. E (E)-(Z) system (Section 7.2): A system for designating the stereo­ chemistry of alkene diastereomers based on the priorities of groups in the Cahn-Ingold-Prelog convention. An isomer has the highest priority groups on opposites sides of the double bond, a isomer has the highest priority groups on the same side of the double bond.

E

(Sections 22.1A and 22.12): A carbohydrate that, on a molecular basis, undergoes hydrolytic cleavage to yield two mol­ ecules of a monosaccharide. Dispersion force (or London force) (Sections 2.13B, 4.9, and 4.11): Weak forces that act between nonpolar molecules or between parts of the same molecule. Bringing two groups (or mol­ ecules) together first results in an attractive force between them because a temporary unsymmetrical distribution of electrons in one group induces an opposite polarity in the other. When groups are brought closer than their , the force between them becomes repulsive because their electron clouds begin to interpenetrate each other.

vanderWaalsradii

Distortionless enhanced p olarization tran sfer (D E PT ) spectra

(Section 9.11E): A technique in 13C NMR spectroscopy by which the number of hydrogens at each carbon, e.g., C, CH, CH2 , and CH3 can be determined. (Section 24.2A): A sulfur-sulfur single bond in a peptide or protein formed by an oxidative reaction between the thiol groups of two cysteine amino acid residues.

Z

(Sections 6.15C, 6.17, and 6.18B): A unimolecular elimination in which, in a slow, rate-determining step, a leaving group departs from the substrate to form a carbocation. The carbo­ cation then in a fast step loses a proton with the resulting formation of a p bond. E 1 reaction

E2 reaction (Sections 6.15C, 6.16, and 6.18B): A bimolecular 1,2 elimination in which, in a single step, a base removes a proton and a leaving group departs from the substrate, resulting in the forma­ tion of a p bond. E clipsed conform ation (Section 4.8A): A temporary orientation of groups around two atoms joined by a single bond such that the groups directly oppose each other.

kineticenolateanion

D isaccharide

Disulfide linkage

(Section 1.2 and 1.13): Bonds composed of four electrons: two electrons in a sigma (s) bond and two electrons in a pi (p) bond. D ouble bonds

A n

e c lip s e d

c o n f o r m a t io n

(Section 24.5A): A method for determining the -terminal amino acid in a peptide. The peptide is treated with phenylisothiocyanate (C6 H5—N =C =S ), which reacts with the -terminal residue to form a derivative that is then cleaved from the peptide with acid and identified. Automated sequencers use the Edman degradation method. E dm an degradation

N N

(Section 13.9A): The full range of ener­ gies propagated by wave fluctuations in an electromagnetic field. Electrom agnetic spectrum

(Section 1.12B): An electron density surface shows points in space that happen to have the same elec­ tron density. An electron density surface can be calculated for any chosen value of electron density. A “high” electron density surface (also called a “bond” electron density surface) shows the of electron density around each atomic nucleus and regions where neighboring atoms share electrons (bonding regions). A “low” electron density surface roughly shows the of a molecule’s electron cloud. This surface gives information about molecular shape and volume, and usually looks the same as a van der Waals or space-filling model of the molecule. (Contributed by Alan Shusterman, Reed College, and Warren Hehre, Wavefunction, Inc.) Electron density surface

core

outline

Gl-8

Glossary

Electron impact (EI) (Sections 9.14 and 9.18A): A method of ion formation in mass spectrometry whereby the sample to be analyzed (analyte) is placed in a high vacuum and, when in the gas phase, bombarded with a beam of high-energy electrons. A valence elec­ tron is displaced by the impact of the electron beam, yielding a species called the (if there has been no fragmenta­ tion), with a + 1 charge and an unshared electron (a radical cation).

molecularion

Electronegativity (Sections 1.4A and 2.2): A measure of the abil­ ity of an atom to attract electrons it is sharing with another and thereby polarize the bond. Electrophile (Sections 3.4A and 8.1A): A Lewis acid, an electronpair acceptor, an electron-seeking reagent. Electrophilic aromatic substitutions (Sections 15.1, 15.2, and 21.8): A reaction of aromatic compounds in which an (“electron-seeker” - a positive ion or other electron-deficient species with a full or large partial positive charge) replaces a hydrogen bonded to the carbon of an aromatic ring.

electrophile

Electrophoresis (Section 25.6A): A technique for separating charged molecules based on their different mobilities in an electric field. Electrospray ionization (ESI) (Section 9.18A): A method of ion formation in mass spectrometry whereby a solution of the sample to be analyzed (analyte) is sprayed into the vacuum chamber of the mass spectrometer from the tip of a high-voltage needle, imparting charge to the mixture. Evaporation of the solvent in the vacuum chamber yields charged species of the analyte; some of which may have charges greater than + 1. A family of peaks unique to the formula weight of the analyte results, from which the formula weight itself can be calculated by computer.

m/z

Electrostatic potential map (maps of electrostatic potential, MEP) (Sections 1.8, 2.2A, and 3.3A): Electrostatic potential maps are models calculated by a computer that show the relative distrib­ ution of electron density at some surface of a molecule or ion. They are very useful for understanding interactions between molecules that are based on attraction of opposite charges. Usually we choose the van der Waals surface (approximately the outermost region of electron density) of a molecule to depict the electrostatic potential map because this is where the electron density of one molecule would first interact with another. In an electrostatic potential map, color trending toward red indicates a region with more negative charge, and color trending toward blue indicates a region with less negative charge (or more positive charge). An electrostatic poten­ tial map is generated by calculating the extent of charge interaction (electrostatic potential) between an imaginary positive charge and the electron density at a particular point or surface in a molecule. (Contributed by Alan Shusterman, Reed College, and Warren Hehre, Wavefunction, Inc.) Elimination reaction (Sections 3.1, 6.15-6.17, 7.5, 7.7): A reac­ tion that results in the loss of two groups from the substrate and the formation of a p bond. The most common elimination is a 1,2 elimination or b elimination, in which the two groups are lost from adjacent atoms. Elimination-addition (via benzyne) (Section 21.11B): A substi­ tution reaction in which a base, under highly forcing conditions, deprotonates an aromatic carbon that is adjacent to a carbon bear­ ing a leaving group. Loss of the leaving group and overlap of the adjacent orbitals creates a species, called with a p-bond in the plane of the ring (separate from the aromatic p-system).

p

benzyne,

Attack by a nucleophile on this p-bond followed by protonation yields a substituted aromatic compound.

enamine

Enamines (Sections 16.8 and 18.9): An group consists of an amine function bonded to the 2 carbon of an alkene.

sp

Enantiomeric excess or enantiomeric purity (Section 5.9B): A percentage calculated for a mixture of enantiomers by dividing the moles of one enantiomer minus the moles of the other enantiomer by the moles of both enantiomers and multiplying by 100. The enantiomeric excess equals the percentage optical purity. Enantiomers (Sections 5.2C, 5.3, 5.7, 5.8, and 5.16): Stereisomers that are mirror images of each other. Enantioselective reaction (See Stereoselective reaction and Sections 5.10B and 12.3C) Enantiotopic hydrogens (or ligands) (Section 9.8B): If replace­ ment of each of two hydrogens (or ligands) by the same group yields compounds that are enantiomers, the two hydrogen atoms (or ligands) are said to be enantiotopic. Endo group (Section 13.11B): A group on a bicyclic compound that is on the same side (syn) as the longest bridge in the com­ pound. Endergonic reaction (Section 6.7): A reaction that proceeds with a positive free-energy change. Endothermic reaction (Section 3.9A): A reaction that absorbs heat. For an endothermic reaction AH° is positive. Energy (Section 3.9): Energy is the capacity to do work. Energy of activation, Eact (Section 10.5B): A measure of the dif­ ference in potential energy between the reactants and the transition state of a reaction. It is related to, but not the same as, the free ener­ gy of activation, AG*. Enolate (Sections 18.1, 18.3, and 18.4): The delocalized anion formed when an enol loses its hydroxylic proton or when the car­ bonyl tautomer that is in equilibrium with the enol loses an a proton. Enthalpy change (Sections 3.9A and 3.10): Also called the heat of reaction. The AH°, is the change in enthalpy after a system in its standard state has undergone a trans­ formation to another system, also in its standard state. For a reac­ tion, AH° is a measure of the difference in the total bond energy of the reactants and products. It is one way of expressing the change in potential energy of molecules as they undergo reaction. The enthalpy change is related to the free-energy change, AG°, and to the entropy change, A °, through the expression:

standardenthalpychange,

S

AH° = AG° + TAS°

Entropy change (Section 3.10): The standard entropy change, A °, is the change in entropy between two systems in their stan­ dard states. Entropy changes have to do with changes in the rela­ tive order of a system. The more random a system is, the greater is its entropy. When a system becomes more disorderly its entropy change is positive.

S

Enzyme (Section 24.9): A protein or polypeptide that is a catalyst for biochemical reactions. Enzyme-substrate complex (Section 24.9): The species formed when a substrate (reactant) binds at the active site of an enzyme. Epimers, epimerization (Sections 18.3A and 22.8): Diastereomers that differ in configuration at only a single tetrahedral chirality cen­ ter. Epimerization is the interconversion of epimers.

Gl-9

Glossary Epoxidation (Section 11.13): The process of synthesizing an expoxide. Peroxycarboxylic acids (RCO3 H) are reagents com­ monly used for epoxidation.

groups eclipsed. Vertical lines represent bonds that project behind the plane of the page (or that lie in it). Horizontal lines represent bonds that project out of the plane of the page.

(Sections 11.13 and 11.14): An oxirane. A three-membered ring containing one oxygen and two carbon atoms.

Epoxide

zz = ±

E quatorial bond (Section 4.12): The six bonds of a cyclohexane ring that lie generally around the “equator” of the molecule:

Keq

E qu ilibriu m constant, (Section 3.6 A): A constant that expresses the position of an equilibrium. The equilibrium constant is calculated by multiplying the molar concentrations of the prod­ ucts together and then dividing this number by the number obtained by multiplying together the molar concentrations of the reactants.

Equilibrium control (See Thermodynamic control) Essential am ino acid (Section 24.2B) An amino acid that cannot be synthesized by the body and must be ingested as part of the diet. For adult humans there are eight essential amino acids (RCH(NH2 )CO2 H): valine (R = isopropyl), Leucine (R = isobutyl), isoleucine (R = -butyl), phenylalanine (R = benzyl), threonine (R = 1 -hydroxyethyl), methionine (R = 2 (methylthio)ethyl), lysine (R = 4-aminobutyl), and tryptophen (R = 3-methyleneindole).

sec

Essential oil (Section 23.3): A volatile odoriferous compound obtained by steam distillation of plant material. Esterification (Section 17.7A): The synthesis of an ester, usually involving reactions of carboxylic acids, acid chlorides or acid anhydrides with alcohols.

(Section 9.10): Protons that can be trans­ ferred rapidly from one molecule to another. These protons are often attached to electronegative elements such as oxygen or nitrogen.

E xch an geable protons

(Section 6.7): A reaction that proceeds with a negative free-energy change. Exergonic reaction

(Section 13.11B): A group on a bicyclic compound that is on the opposite side (anti) to the longest bridge in the compound. E xo group

Exon (Section 25.5A): Short for “expressed sequence,” an exon is a segment of DNA that is used when a protein is expressed. (See Intron). E xotherm ic reaction (Section 3.9A): A reaction that evolves heat. For an exothermic reaction, AH° is negative. F Fat (Section 23.2): A triacylglycerol. The triester of glycerol with carboxylic acids. F atty acid (Section 23.2): A long-chained carboxylic acid (usual­ ly with an even number of carbon atoms) that is isolated by the hydrolysis of a fat.

(Sections 5.13 and 22.2C): A two­ dimensional formula for representing the configuration of a chiral molecule. By convention, Fischer projection formulas are written with the main carbon chain extending from top to bottom with all Fischer projection form ula

F is c h e r

W e d g e -d a s h e d

p r o je c tio n

w e d g e fo r m u la

(Section 10.5C): A reaction in which fluorine atoms are introduced into a molecule.

Fluorination

(Section 1.7): The difference between the number of electrons assigned to an atom in a molecule and the number of electrons it has in its outer shell in its elemental state. Formal charge can be calculated using the formula: = — / 2 — , where is the formal charge, is the group number of the atom (i.e., the number of electrons the atom has in its outer shell in its elemental state), is the number of electrons the atom is sharing with other atoms, and is the number of unshared electrons the atom possesses.

F orm al charge

F

Z

S

FZS U

U

F ourier transform N M R (Section 9.5): An NMR method in which a pulse of energy in the radiofrequency region of the elec­ tromagnetic spectrum is applied to nuclei whose nuclear magnetic moment is precessing about the axis of a magnetic field. This pulse of energy causes the nuclear magnetic moment to “tip” toward the xy plane. The component of the nuclear magnetic moment in the x-y plane generates (“induces”) a radiofrequency signal, which is detected by the instrument. As nuclei relax to their ground states this signal decays over time; this time-dependent signal is called a “Free Induction Decay” (FID) curve. A mathematical operation (a Fourier transform) converts time-dependent data into frequencydependent data - the NMR signal. F ragm entation (Section 9.16): Cleavage of a chemical species by the breaking of covalent bonds, as in the formation of fragments during mass spectrometric analysis.

AG* (Section 6.7): The difference in free energy between the transition state and the reactants. F ree energy o f activation,

(Section 3.10): The s tandardfree-energy change, AG°, is the change in free energy between two systems in their standard states. At constant temperature, AG° = AH° —T AS° = — RTln Kgq,where AH° is the standard enthalpy change, SS° is the standard entropy change, and K eqis the equilibrium constant. A nega­ tive value of AG ° for a reaction means that the formation of products Free-energy change

is favored when the reaction reaches equilibrium.

F ree-energy diagram (Section 6.7): A plot of free-energy changes that take place during a reaction versus the reaction coordinate. It displays free-energy changes as a function of changes in bond orders and distances as reactants proceed through the transition state to become products. Freon

(Section 10.11D): A chlorofluorocarbon or CFC.

v

(Sections 2.15 and 13.9A): The number of full cycles of a wave that pass a given point in each second. F requency,

(Section 14.8C): Cagelike aromatic molecules with the geometry of a truncated icosahedron (or geodesic dome). The structures are composed of a network of pentagons and hexagons.

Fullerenes

Gl-10

G lossary

Each carbon is sp2 hybridized; the remaining electron at each car­ bon is delocalized into a system of molecular orbitals that gives the aromatic character.

wholemolecule

Functional class nomenclature (Section 4.3E): A system for naming compounds that uses two or more words to describe the compound. The final word corresponds to the functional group pre­ sent; the preceding words, usually listed in alphabetical order, describe the remainder of the molecule. Examples are methyl alco­ hol, ethyl methyl ether, and ethyl bromide. Functional group (Section 2.4): The particular group of atoms in a molecule that primarily determines how the molecule reacts.

Glycolipids (Section 22.16): Carbohydrates joined through glycosidic linkages to lipids. Glycoproteins (Section 22.16): Carbohydrates joined through glycosidic linkages to proteins. Glycoside (Section 22.4): A cyclic mixed acetal of a sugar with an alcohol. Grignard reagent (Section 12.6B): An organomagnesium halide, usually written RMgX. Ground state (Section 1.12): The lowest electronic energy state of an atom or molecule.

Functional group interconversion (Section 6.14): A process that converts one functional group into another.

H

Furanose (Section 22.2C): A sugar in which the cyclic acetal or hemiacetal ring is five membered.

1 H—1H correlation spectroscopy (COSY) (Section 9.12): A two­ dimensional NMR method used to display the coupling between hydrogen atoms.

G Gauche conformation (Section 4.9): A gauche conformation of butane, for example, has the methyl groups at an angle of 60° to each other:

Haloform reaction (Section 18.3C): A reaction specific to methyl ketones. In the presence of base multiple halogenations occur at the carbon of the methyl group; excess base leads to acyl substitution of the trihalomethyl group, resulting in a carboxylate anion and a (CHX3 ).

haloform

Halogenation (Sections 10.3-10.6 and 10.8): A reaction in which one or more halogen atoms are introduced into a molecule. Halohydrin (Section 8.14): A compound bearing a halogen atom and a hydroxyl group on adjacent (vicinal) carbons. Halonium ion (Section 8.12A): An ion containing a positive halo­ gen atom bonded to two carbon atoms.

conformation of butane

GC/MS analysis (Section 9.19): An analytical method that cou­ ples a gas chromatograph (GC) with a mass spectrometer (MS). The GC separates the components of a mixture to be analyzed by sweeping the compounds, in the gas phase, through a column con­ taining an adsorbant called a . The gaseous mole­ cules will cling to the surface of the stationary phase (be ) with different strengths. Those molecules that cling (adsorb) weak­ ly will pass through the column quickly; those that more strongly will pass through the column more slowly. The separated components of the mixture are then introduced into the mass spec­ trometer, where they are analyzed.

stationaryphase

adsorbed adsorb

(gem-)

Geminal substituents (Section 7.10A): Substituents that are on the same atom. Gene (Section 25.1): A section of DNA that codes for a given protein. Genetic code (Sections 25.5C and 25.5D): The correspondence of specific three-base sequences in mRNA (codons) that each code for a specific amino acid. Each codon pairs with the anticodon of a specific tRNA, which in turn carries the corresponding amino acid. Genome (Sections 25.1 and 25.9): The set of all genetic informa­ tion coded by DNA in an organism. Genomics (Section 24.14): The study of the complete set of genet­ ic instructions in an organism. Glycan (see Polysaccharide and Section 22.13): An alternate term for a polysaccharide; monosaccharies joined together by glycosidic linkages. Glycol (Sections 4.3F and 8.16): A diol.

Hammond-Leffler postulate (Section 6.13A): A postulate stating that the structure and geometry of the transition state of a given step will show a greater resemblance to the reactants or products of that step depending on which is closer to the transition state in energy. This means that the transition state of an endothermic step will resemble the products of that step more than the reactants, whereas the transition state of an exothermic step will resemble the reactants of that step more than the products. Heat of hydrogenation (Section 7.3A): The standard enthalpy change that accompanies the hydrogenation of 1 mol of a com­ pound to form a particular product. Heisenberg uncertainty principle (Section 1.11): A fundamen­ tal principle that states that both the position and momentum of an electron (or of any object) cannot be exactly measured simultaneously. Hemiacetal (Sections 16.7A and 22.2C): A functional group, con­ sisting of an 3 carbon atom bearing both an alkoxyl group and a hydroxyl group [i.e., RCH(OH)(OR') or R2 C(OH)(OR')].

sp

Hemiketal (See Hemiacetal and Section 16.7A) Henderson-Hasselbalch equation (Section 24.2): The Henderson-Hasselbalch equation = pH + log[HA]/[A~]) shows that when the concentration of an acid and its conjugate base are equal, the pH of the solution equals the p a of the acid.

(pKa

K

Hertz (Hz) (Sections 9.7A, 9.9C, and 13.9A): The frequency of a wave. Now used instead of the equivalent cycles per second (cps). Heteroatom (Section 2.3): Atoms such as oxygen, nitrogen, sulfur and the halogens that form bonds to carbon and have unshared pairs of electrons.

Gl-11

G lossary

Heterocyclic amines (Section 20.1B): A secondary or tertiary

amine in which the nitrogen group is part of a carbon-based ring.

Hydride (Section 7.8): A hydrogen anion, H:

Hydrogen with a filled 1 s shell (containing two electrons) and negative charge.

H eterocyclic compound (Sections 14.9): A compound whose molecules have a ring containing an element other than carbon.

Hydroboration (Sections 8.7,

H eterogeneous catalysis (Sections 7.13 and 7.14): Catalytic reac­

Hydrocarbon (Section 2.2): A molecular containing only carbon

tions in which the catalyst is insoluble in the reaction mixture.

and hydrogen atoms.

H eterolysis (Section 3.1A): The cleavage of a covalent bond so

Hydrogen abstraction (Section 10.1B): The process by which a

that one fragment departs with both of the electrons of the covalent bond that joined them. Heterolysis of a bond normally produces positive and negative ions.

species with an unshared electron (a radical) removes a hydrogen atom from another species, breaking the bond to the hydrogen homolytically.

H eteronuclear correlation spectroscopy (HETCOR or C-H HETCOR) (Section 9.12): A two-dimensional NMR method used

Hydrogen bond (Sections 2.13B, 2.13E, and 2.13F): A strong

to display the coupling between hydrogens and the carbons to which they are attached. Heterotopic (chem ically nonequivalent atoms) (Section 9.8A): Atoms in a molecule where replacement of one or the other leads to a new compound. Heterotopic atoms are not chemical shift equivalent in NMR spectroscopy. Hofmann rule (Sections 7.6C and 20.12A): When an elimination yields the alkene with the less substituted double bond, it is said to follow the Hofmann rule. HOM O (Sections 3.3A and 13.9C): The highest occupied molec­

ular orbital. Homogeneous catalysis (Sections 7.13 and 7.14A): Catalytic reactions in which the catalyst is soluble in the reaction mixture. Hom ologous series (Section 4.7): A series of compounds in which

each member differs from the next member by a constant unit. Hom olysis (Sections 3.1A and 10.1): The cleavage of a covalent

bond so that each fragment departs with one of the electrons of the covalent bond that joined them. Hom olytic bond dissociation energy, DH° (Section 10.2): The enthalpy change that accompanies the homolytic cleavage of a covalent bond. H om otopic (chem ically equivalent) atom s (Section 9.8A):

Atoms in a molecule where replacement of one or another results in the same compound. Homotopic atoms are chemical shift equiv­ alent in NMR spectroscopy. Huckel’s rule (Section 14.7): A rule stating that planar monocyclic rings with (4n + 2) delocalized p electrons (i.e., with 2, 6 , 10, 14, . . . , delocalized p electrons) will be aromatic. H und’s rule (Section 1.10A): A rule used in applying the aufbau

principle. When orbitals are of equal energy (i.e., when they are degenerate), electrons are added to each orbital with their spins unpaired, until each degenerate orbital contains one electron. Then electrons are added to the orbitals so that the spins are paired. Hybridization o f atomic orbitals (Sections 1.12 and 1.15): A

mathematical (and theoretical) mixing of two or more atomic orbitals to give the same number of new orbitals, called , each of which has some of the character of the original atomic orbitals.

orbitals

hybrid

Hydration (Sections 8.5-8.10 and 11.4): The addition of water to

8 .8 , and 11.4): The addition of a boron hydride (either BH3 or an alkylborane) to a multiple bond.

dipole-dipole interaction (4-38 kJ mol—1) that occurs between hydrogen atoms bonded to small strongly electronegative atoms (O, N, or F) and the nonbonding electron pairs on other such elec­ tronegative atoms. H ydrogenation (Sections 4.16A, 7.3A, and 7.13-7.15): A reaction

in which hydrogen adds to a double or triple bond. Hydrogenation is often accomplished through the use of a metal catalyst such as platinum, palladium, rhodium, or ruthenium. Hydrophilic group (Sections 2.13D and 23.2C): A polar group that seeks an aqueous environment.

Hydrophobic group (or lipophilic group) (Sections 2.13D and 23.2C): A nonpolar group that avoids an aqueous surrounding and seeks a nonpolar environment. H ydroxylation (Sections 8.16 and 11.15): The addition of hydrox­

yl groups to each carbon or atom of a double bond. Hyperconjugation (Sections 4.8 and 6.11B): Electron delocaliza­

tion (via orbital overlap) from a filled bonding orbital to an adja­ cent unfilled orbital. Hyperconjugation generally has a stabilizing effect. I Im ines (Section 16.8): A structure with a carbon-nitrogen double

E Z

bond. If the groups bonded to carbon are not the same, ( ) and ( ) isomers are possible. Index o f hydrogen deficiency (Section 4.17): The index of hydro­

gen deficiency (or IHD) equals the number of pairs of hydrogen atoms that must be subtracted from the molecular formula of the corresponding alkane to give the molecular formula of the com­ pound under consideration. Induced fit hypothesis (Section 24.9): An hypothesis regarding

enzyme reactivity whereby formation of the enzyme-substrate complex causes conformational changes in the enzyme that facili­ tate conversion of the substrate to product. Inductive effect (Sections 3.8B, 3.11B, and 15.11B): An intrinsic

electron-attracting or -releasing effect that results from a nearby dipole in the molecule and that is transmitted through space and through the bonds of a molecule. Infrared (IR) spectroscopy (Section 2.15): A type of optical spec­

troscopy that measures the absorption of infrared radiation. Infrared spectroscopy provides structural information about func­ tional groups present in the compound being analyzed.

a molecule, such as the addition of water to an alkene to form an alcohol.

Inhibitor (Section 24.9): A compound that can negatively alter the

Hydrazone (Section 16.8B): An imine in which an amino group

Integration (Section 9.2B): A numerical value representing the

( —NH2 , —NHR, —NR2 ) is bonded to the nitrogen atom.

relative area under a signal in an NMR spectrum. In 1H NMR, the

activity of an enzyme.

Gl-12

G lossary

integration value is proportional to the number of hydrogens pro­ ducing a given signal.

Isotactic polymer (Special Topic B.1): A polymer in which the configuration at each stereogenic center along the chain is the same.

Intermediate (Sections 3.1, 6.10, and 6.11): A transient species that exists between reactants and products in a state corresponding to a local energy minimum on a potential energy diagram.

Isotopes (Section 1.2A): Atoms that have the same number of pro­ tons in their nuclei but have differing atomic masses because their nuclei have different numbers of neutrons.

Intermolecular forces (Sections 2.13B and 2.13F): Also known as van der Waals forces. Forces that act between molecules because of permanent (or temporary) electron distributions. Intermolecular forces can be attractive or repulsive. Dipole-dipole forces (including hydrogen bonds) and dispersion forces (also called London forces), are intermolecular forces of the van der Waal type.

IUPAC system (Section 4.3): (also called the “systematic nomen­ clature”) A set of nomenclature rules overseen by the International Union of Pure and Applied Chemistry (IUPAC) that allows every compound to be assigned an unambiguous name.

Intron (Section 25.5A): Short for “intervening sequence,” an intron is a segments of DNA that is not actually used when a pro­ tein is expressed, even though it is transcripted into the initial mRNA. Inversion of configuration (Sections 6 .6 and 6.14): At a tetrahe­ dral atom, the process whereby one group is replaced by another bonded 180° opposite to the original group. The other groups at the tetrahedral atom “turn inside out” (shift) in the same way that an umbrella “turns inside out.” When a chirality center undergoes configuration inversion, its (R,S) designation may switch, depend­ ing on the relative Chan-Ingold-Prelog priorities of the groups before and after the reaction. Iodination (Section 10.5C): A reaction in which one or more iodine atoms are introduced into a molecule. Ion (Sections 1.4A and 3.1A): A chemical species that bears an electrical charge. Ion-dipole force (Section 2.13D): The interaction of an ion with a permanent dipole. Such interactions (resulting in solvation) occur between ions and the molecules of polar solvents. Ion-ion forces (Section 2.14A): Strong electrostatic forces of attraction between ions of opposite charges. These forces hold ions together in a crystal lattice. Ion sorting (Section 9.18B): Sorting of ions in a mass spectrome­ ter by . Ions are presented on the x-axis of the mass spectrum in order of increasing If z = +1, is equivalent to the molec­ ular mass of the molecule.

m/z

m/z.

m/z

Ionic bond (Section 1.4A): A bond formed by the transfer of elec­ trons from one atom to another resulting in the creation of oppo­ sitely charged ions. Ionic reaction (Sections 3.1A and 10.1): A reaction involving ions as reactants, intermediates, or products. Ionic reactions occur through the heterolysis of covalent bonds. Ionization (Section 9.14): Conversion of neutral molecules to ions (charged species). Isoelectric point (pT) (Section 24.2C): The pH at which the num­ ber of positive and negative charges on an amino acid or protein are equal. Isomers (Sections 1.3A and 5.2): Different molecules that have the same molecular formula. Isoprene unit (Section 23.3): A name for the structural unit found in all terpenes:

K Karplus correlation (Section 9.9D): An empirical correlation between the magnitude of an NMR coupling constant and the dihe­ dral angle between two coupled protons. The dihedral angles derived in this manner can provide information about molecular geometries. Kekule structure (Sections 2.1D and 14.4): A structure in which lines are used to represent bonds. The Kekule structure for benzene is a hexagon of carbon atoms with alternating single and double bonds around the ring, and with one hydrogen atom attached to each carbon. Ketal (See Acetal and Section 16.7B) Keto and enol forms (Sections 18.1-18.3): Tautomeric forms of a compound related by a common resonance-stabilized intermediate. An structure consists of an alcohol functionality bonded to the carbon of an alkene. Shifting the hydroxyl proton to the alkene and creation of a carbon-oxygen p-bond results in the form of the species.

enol sp2

keto

Ketose (Section 22.2A): A monosaccharide containing a ketone group or a hemiacetal or acetal derived from it. Kinetic control (Sections 7.6B, 13.10A): A principle stating that when the ratio of products of a reaction is determined by relative rates of reaction, the most abundant product will be the one that is formed fastest. Also called rate control. Kinetic energy (Section 3.9): Energy that results from the motion of an object. Kinetic energy = 1/ 2 where is the mass of the object and is its velocity.

v

(KE) mv2,

m

Kinetic enolate (Section 18.4A): In a situation in which more than one enolate anion can be formed, the is that which is formed most rapidly. This is usually the enolate anion with the less substituted double bond; the decrease in steric hin­ drance permits more rapid deprotonation by the base. A kinetic enolate anion is formed predominantly under conditions that do not permit the establishment of an equilibrium.

kineticenolateanion

Kinetic resolution (Section 5.10B): A process in which the rate of a reaction with one enantiomer is different than with the other, leading to a preponderance of one product stereoisomer. This process is said to be “stereoselective” in that it leads to the prefer­ ential formation of one stereoisomer over other stereoisomers that could possibly be formed. Kinetics (Section 6.5): A term that refers to rates of reactions. L Lactam (Section 17.8I): A cyclic amide. Lactone (Section 17.7C): A cyclic ester. LCAO (linear combination of atomic orbitals, Section 1.11): A mathematical method for arriving at wave functions for molecular

G lossary

Gl-13

obitals that involves adding or subtracting wave functions for atomic orbitals.

ondary amine with formaldehyde) to yield a b-aminoalkyl car­ bonyl compound.

L ea vin g group (Sections 6.2, 6.4, and 6.13E): The substituent that departs from the substrate in a nucleophilic substitution reaction.

(Sections 8.2 and 8.19): A rule for predicting the regiochemistry of electrophilic additions to alkenes and alkynes that can be stated in various ways. As originally stated (in 1870) by Vladimir Markovnikov, the rule provides that “if an unsymmetrical alkene combines with a hydrogen halide, the halide ion adds to the carbon with the fewer hydrogen atoms.” More commonly the rule has been stated in reverse: that in the addition of HX to an alkene or alkyne the hydrogen atom adds to the carbon atom that already has the greater number of hydrogen atoms. A modern expression of Markovnikov’s rule is:

(Section 3.15): An effect that restricts the use of certain solvents with strong acids and bases. In principle, no acid stronger than the conjugate acid of a particular solvent can exist to an appreciable extent in that solvent, and no base stronger than the conjugate base of the solvent can exist to an appreciable extent in that solvent. Leveling effect o f a solvent

(Section 5.8B): A compound that rotates planepolarized light in a counterclockwise direction. L evorotatory

Lewis structure (or electron-dot structure) (Sections 1.4B and 1.5): A representation of a molecule showing electron pairs as a pair of dots or as a dash. (Section 3.3): An acid is an electron pair acceptor, and a base is an electron pair donor.

Lew is acid -b ase theory

(Section 23.1): A substance of biological origin that is solu­ ble in nonpolar solvents. Lipids include fatty acids, triacylglycerols (fats and oils), steroids, prostaglandins, terpenes and terpenoids, and waxes.

L ip id

L ip id bilayers (Section 23.6A): A two-layer noncovalent molecu­ lar assembly comprised primarily of phospholipids. The hydropho­ bic phospholipid “tail” groups of each layer orient toward each other in the center of the two-layered structure due to attractive dis­ persion forces. The hydrophilic “head” groups of the lipids orient toward the aqueous exterior of the bilayer. Lipid bilayers are important in biological systems such as cell membranes.

M ark o v n ik o v’s ru le

Intheionicadditionofanunsymmetrical reagenttoamultiplebond,thepositiveportionofthereagent(the electrophile)attachesitselftoacarbonatomofthereagentinthe waythatleadstotheformationofthemorestableintermediate carbocation.

(Section 9.13): A technique, useful in structure elucidation, that involves the generation of ions from a molecule, the sorting and detecting of the ions, and the display of the result in terms of the mass/charge ratio and relative amount of each ion. M ass spectrom etry (M S)

Matrix-assisted laser desorption-ionization (MALDI) (Section 9.18A): A method in mass spectrometry for ionizing analytes that do not ionize well by electrospray ionization. The analyte is mixed with low molecular weight organic molecules that can absorb ener­ gy from a laser and then transfer this energy to the analyte, pro­ ducing + 1 ions which are then analyzed by the mass spectrometer. Mechanism (See Reaction mechanism)

Lipophilic group (or hydrophobic group) (Sections 2.13D and 23.2C): A nonpolar group that avoids an aqueous surrounding and seeks a nonpolar environment.

(Section 2.14A): The temperature at which an equi­ librium exists between a well-ordered crystalline substance and the more random liquid state. It reflects the energy needed to overcome the attractive forces between the units (ions, molecules) that com­ prise the crystal lattice.

(Section 18.4): (i-C3 H7 )2 N~Li+ The lithium salt of diisopropylamine. A strong base used to form from carbonyl compounds.

M eso com pound (Section 5.12A): An optically inactive com­ pound whose molecules are achiral even though they contain tetra­ hedral atoms with four different attached groups.

Lock-and-key hypothesis (Section 24.9): An hypothesis that explains enzyme specificity on the basis of complementary geom­ etry between the enzyme (the “lock”) and the substrate (the “key”), such that their shapes “fit together” correctly for a reaction to occur.

(Section 11.10): A methanesulfonate ester. Methanesulfonate esters are compounds that contain the CH3 SO3— group, i.e., CH3 SO3 R.

Lithium diisopropylam ide (LD A)

lithiumenolates

L U M O (Sections 3.3A and 13.9C): The lowest unoccupied mole­ cular orbital.

M elting Point

M esylate

(Section 15.10B): An electron-withdrawing group on an aromatic ring. The major product of electrophilic aromatic substitution on a ring bearing a meta-directing group will have the newly substituted electrophile located meta to the substituent. M eta directors

(Section 7.8): A methyl anion, :CH3 , or methyl species that reacts as though it were a methyl anion.

M ethanide M M acrom olecule

(Section 10.10): A very large molecule.

M ethylene

(Section 9.12): A technique based on NMR spectroscopy that is used in medicine. M agnetic resonance im aging (M R I)

a-

(Section 18.7): A reaction in which the hydrogen of diethyl propanedioate (diethyl malonate, also called “malonic ester”) is removed, creating a resonance-stabilized anion which can serve as a nucleophile in an Sn2 reaction. The a-carbon can be substituted twice; the ester functionalities can be converted into a carboxylic acid which, after decarboxylation, will yield a substituted ketone. M alon ic ester synthesis

(Section 19.8): The reaction of an enol with an iminium cation (formed from the reaction of a primary or sec­

M annich reaction

(Section 8.15A): The carbene with the formula :CH2 .

M ethylene group

(Section 2.4B): The —CH2— group.

(Section 23.2C): A spherical cluster of ions in aqueous solution (such as those from a soap) in which the nonpolar groups are in the interior and the ionic (or polar) groups are at the surface. M icelle

(See C on jugate addition and Sections 18.9 and 19.7): A reaction between an active hydrogen compound and an a,b-unsaturated carbonyl compound. The attack by the anion of the active hydrogen compound takes place at the b-carbon of the a,b-unsaturated carbonyl compound. A Michael addition is a type of conjugate addition. M ichael addition

Gl-14

G lossary

Molar absorptivity, e (Section 13.9B): A proportionality con­ stant that relates the observed absorbance ( ) at a particular wavelength (l) to the molar concentration of the sample (C) and the length ( ) (in centimeters) of the path of the light beam through the sample cell:

A

l

e=

A/CX l

Molecular formula (Section 1.3A): A formula that gives the total number of each kind of atom in a molecule. The molecular formula is a whole number multiple of the empirical formula. For example the molecular formula for benzene is CgHg; the empiri­ cal formula is CH.

Nitrogen rule (Section 9.17B): A rule that states that if the mass of the molecular ion in a mass spectrum is an even number, the par­ ent compound contains an even number of nitrogen atoms, and conversely. N-nitrosoamines (Section 20.6C): Amines bearing an N= O on the nitrogen, such as R— NH— N= O or Ar— NH— N= O. Often referred to as “nitrosamines” in the popular press. Nnitrosoamines are very powerful carcinogens. Node (Section 1.9): A place where a wave function (C) is equal to zero. The greater the number of nodes in an orbital, the greater is the energy of the orbital.

Molecular ion (Sections 9.14, 9.15, and 9.17): The cation pro­ duced in a mass spectrometer when one electron is dislodged from the parent molecule, symbolized M ..

Nonbenzenoid aromatic compound (Section 14.8B): An aro­ matic compound, such as azulene, that does not contain benzene rings.

Molecular orbital (MO) (Sections 1.11 and 1.15): Orbitals that encompass more than one atom of a molecule. When atomic orbitals combine to form molecular orbitals, the number of molec­ ular orbitals that results always equals the number of atomic orbitals that combine.

Nuclear magnetic resonance (NMR) spectroscopy (Sections 9.2 and 9.11 A): A spectroscopic method for measuring the absorption of radio frequency radiation by certain nuclei when the nuclei are in a strong magnetic field. The most important NMR spectra for organic chemists are 1H NMR spectra and 13C NMR spectra. These two types of spectra provide structural infor­ mation about the carbon framework of the molecule, and about the number and environment of hydrogen atoms attached to each carbon atom.

Molecularity (Section 6.5): The number of species involved in a single step of a reaction (usually the rate-determining step). Molecule (Section 1.4B): An electrically neutral chemical entity that consists of two or more bonded atoms. Monomer (Section 10.10): The simple starting compound from which a polymer is made. For example, the polymer polyethylene is made from the monomer ethylene. Monosaccharide (Sections 22.1A and 22.2): The simplest type of carbohydrate, one that does not undergo hydrolytic cleavage to a simpler carbohydrate. Mutarotation (Section 22.3): The spontaneous change that takes place in the optical rotation of a and b anomers of a sugar when they are dissolved in water. The optical rotations of the sugars change until they reach the same value. N Nanotube (Section 14.8C): A tubular structure with walls resem­ bling fused benzene rings, capped by half of a “buckyball” (buckminsterfullerene) at each end. The entire structure exhibits aromat­ ic character. Neighboring-group participation (Problem 6.49): The effect on the course or rate of a reaction brought about by another group near the functional group undergoing reaction. Newman projection formula (Section 4.8A): A means of repre­ senting the spatial relationships of groups attached to two atoms of a molecule. In writing a Newman projection formula we imagine ourselves viewing the molecule from one end directly along the bond axis joining the two atoms. Bonds that are attached to the front atom are shown as radiating from the center of a circle; those attached to the rear atom are shown as radiating from the edge of the circle:

Nucleic acids (Sections 25.1, 25.4, and 25.5): Biological polymers of nucleotides. DNA and RNA are, respectively, nucleic acids that preserve and transcribe hereditary information within cells. Nucleophile (Sections 3.4A, 6.2, 6.3, and 6.13B): A Lewis base, an electron pair donor that seeks a positive center in a molecule. Nucleophilic addition-elimination (Section 17.4): Addition of a nucleophile to a carbonyl (or other trigonal) carbon, yielding a tetrahedral intermediate, followed by elimination of a leaving group to yield a trigonal planar product. Nucleophilic addition to the carbonyl carbon (Section 16.6): A reaction in which a (an electron-pair donor) forms a bond to the carbon of a (C = O) group. To avoid violat­ ing the octet rule, the electrons of the carbon-oxygen p-bond shift to the oxygen, resulting in a four-coordinate (tetrahedral) carbon.

nucleophile carbonyl

Nucleophilic aromatic substitution (Section 21.11A): A substitu­ tion reaction in which a nucleophile attacks an aromatic ring bear­ ing strongly electron-withdrawing groups in ortho and/or para positions relative to the site of attack and the leaving group. This step is an addition reaction that yields and aryl carbanion (called a Meisenheimer Complex) which is stabilized by the electron-with­ drawing groups on the ring. Loss of the leaving group in an elimi­ nation step regenerates the aromatic system, yielding a substituted aromatic compound by what was, overall, an addition-elimination process. Nucleophilic substitution reaction (Section 6.2): A reaction initi­ ated by a nucleophile (a species with an unshared electron pair) in which the nucleophile reacts with a substrate to replace a sub­ stituent (called the leaving group) that departs with an unshared electron pair. Nucleophilicity (Section 6.13B): The relative reactivity of a nucle­ ophile in an Sn2 reaction as measured by relative rates of reaction. Nucleoside (Sections 22.15A, 25.2, and 25.3): A five-carbon monosaccharide bonded at the 1 ' position to a purine or pyrimidine.

Gl-15

G lossary

Nucleotide (Sections 25.2 and 25.3): A five-carbon monosaccha­ ride bonded at the 1 ' position to a purine or pyrimidine and at the 3' or 5' position to a phosphate group. O

(Sections 1.4 and 1.6): An empirical rule stating that atoms not having the electronic configuration of a noble gas tend to react by either transferring electrons or sharing electrons so as to achieve the valence electron configuration (i. e., eight electrons) of a noble gas. O ctet rule

(Section 9.11D): An NMR method for investigating the number of protons attached to a carbon atom by which each carbon signal is split into (n + 1 ) signals, where n = the number of protons on the carbon under observation. O ff-resonance decoupling

O il (Section 23.2): A triacylglycerol (see below) that is liquid at room temperature. O lefin

(Section 7.1): An old name for an alkene.

(Section 25.7): Synthesis of specific sequence of nucleotides, often by automated solid-phase techniques, in which the nucleotide chain is built up by adding a protected nucleotide in the form of a phosphoramidite to a protected nucleotide linked to a solid phase, (usually a “controlled pore glass”) in the pres­ ence of a coupling agent. The phosphite triester product is oxidized to a phosphate triester with iodine, producing a chain that has been lengthened by one nucleotide. The protecting group is then removed, and the steps (coupling, oxidation, deprotection) are repeated. After the desired oligonucleotide has been synthesized it is cleaved from the solid support and the remaining protecting groups removed. O ligonucleotide synthesis

O ligopeptide

(Section 24.4): A peptide comprised of 3 - 10 amino

acids. O lig o sa cch a rid es (Section 22.1 A): A carbohydrate that hydrolyzes to yield 2 - 1 0 monosaccharide molecules. O ptical p urity (Section 5.9B): A percentage calculated for a mix­ ture of enantiomers by dividing the observed specific rotation for the mixture by the specific rotation of the pure enantiomer and multiplying by 100. The optical purity equals the enantiomeric purity or enantiomeric excess.

(Sections 5.8 and 5.9): A compound that rotates the plane of polarization of plane-polarized light. O ptically active com pound

O rb ital (Section 1.10): A volume of space in which there is a high probability of finding an electron. Orbitals are described mathe­ matically by the squaring of wave functions, and each orbital has a characteristic energy. An orbital can hold two electrons when their spins are paired. O rgan om etallic com pound

(Section 12.5): A compound that con­

tains a carbon-metal bond. (Section 24.7D): Protecting groups in which one set of protecting groups is stable under condi­ tions for removal of the other, and vice versa.

O rth o g o n al p ro tectin g gro up s

(Section 15.10B): An electron-donating group on an aromatic ring. The major product of electrophilic aro­ matic substitution on a ring bearing such a group will have the newly substituted electrophile located ortho and/or para to the ortho-para-directing group. O rth o -p ara directors

(Section 22.8): A 1,2-bisarylhydrazone formed by reac­ tion of an aldose or ketose with three molar equivalents of an arylO sazone

hydrazone. Most common are phenylosazones, formed by reaction with phenylhydrazine, and 2,4-dinitrophenylhydrazones. O xidation (Sections 12.2 and 12.4): A reaction that increases the oxidation state of atoms in a molecule or ion. For an organic sub­ strate, oxidation usually involves increasing its oxygen content or decreasing its hydrogen content. Oxidation also accompanies any reaction in which a less electronegative substituent is replaced by a more electronegative one. O xidative cleavage (Sections 8.17 and 8.20): A reaction in which the carbon-carbon double bond of an alkene or alkyne is both cleaved and oxidized, yielding compounds with carbon-oxygen double bonds.

(Section 12.2): A chemical species that causes another chemical species to become oxidized (lose electrons, or gain bonds to more electronegative elements, often losing bonds to hydrogen in the process). The oxidizing agent is reduced in this process. O xidizin g agent

O xim e (Section 16.8B): An imine in which a hydroxyl group is bonded to the nitrogen atom.

(Sections 3.13 and 11.12): A chemical species with an oxygen atom that bears a formal positive charge.

O xonium ion

(Section 11.12): A salt in which the cation is a species containing a positively charged oxygen.

O xonium salt

(Sections 8 .6 and 11.4): The addition of —OH and —HgO2 CR to a multiple bond.

O xym ercu ration

O x y m e rcu ra tio n -d e m ercu ra tio n (Section 8 .6 ): A two-step process for adding the elements of water (H and OH) to a double bond in a Markovnikov orientation without rearrangements. An alkene reacts with mercuric acetate (or trifluoroacetate), forming a bridged mercurinium ion. Water preferentially attacks the more substituted side of the bridged ion, breaking the bridge and result­ ing, after loss of a proton, in an alcohol. Reduction with NaBH4 replaces the mercury group with a hydrogen atom, yielding the final product. O zonolysis (Sections 8.17B and 8.20): The oxidative cleavage of a multiple bond using O3 (ozone). The reaction leads to the forma­ tion of a cyclic compound called an which is then reduced to carbonyl compounds by treatment with dimethyl sulfide (Me2S) or zinc and acetic acid.

ozonide,

P

p orbitals (Section 1.10): A set of three degenerate (equal energy)

atomic orbitals shaped like two tangent spheres with a nodal plane at the nucleus. For orbitals of second row elements, the principal quantum number, (see Atomic orbital), is 2; the azimuthal quan­ tum number, l, is 1 ; and the magnetic quantum numbers, m, are + 1 , 0 , or —1 .

p n

Paraffin

(Section 4.15): An old name for an alkane.

P a rtia l hydrolysis (Section 24.5D): Random cleavage of a polypeptide with dilute acid, resulting in a family of peptides of varying lengths that can be more easily sequenced than the parent polypeptide. Once each fragment peptide is sequenced, the areas of overlap indicate the sequence of the initial peptide.

(Section 1.10A): A principle that states that no two electrons of an atom or molecule may have the same set of four quantum numbers. It means that only two electrons can Pauli exclusion principle

Gl-16

G lossary

occupy the same orbital, and then only when their spin quantum numbers are opposite. When this is true, we say that the spins of the electrons are paired.

Polar covalent bond (Section 2.2): A covalent bond in which the electrons are not equally shared because of differing electronega­ tivities of the bonded atoms.

Peptide (Section 24.4): A molecule comprised of amino acids bonded via amide linkages.

P olar m olecule

Peptide bond, peptide linkage (Section 24.4): The amide linkage between amino acids in a peptide. Peracid (See Peroxy acid, Section 11.13A) Periplanar (See Coplanar, Section 7.6D) Peroxide (Section 10.1A): A compound with an oxygen-oxygen single bond. Peroxy acid (Section 11.13A): An acid with the general formula RCO3 H, containing an oxygen-oxygen single bond. Phase sign (Section 1.9): Signs, either + or —, that are character­ istic of all equations that describe the amplitudes of waves. Phase transfer catalysis (Section 11.16): A reaction using a reagent that transports an ion from an aqueous phase into a nonpo­ lar phase where reaction takes place more rapidly. Tetraalkylammonium ions and crown ethers are phase-transfer catalysts. Phospholipid (Section 23.6): Compound that is structurally derived from Phosphatidic acids are derivatives of glycerol in which two hydroxyl groups are joined to fatty acids, and one terminal hydroxyl group is joined in an ester linkage to phosphoric acid. In a phospholipid the phosphate group of the phosphatidic acid isjoined in ester linkage to a nitrogen-containing compound such as choline, 2-aminoethanol, or L-serine.

phosphatidicacid.

Physical property (Section 2.14): Properties of a substance, such as melting point and boiling point, that relate to physical (as opposed to chemical) changes in the substance. Pi (p) bond (Section 1.13): A bond formed when electrons occu­ py a bonding p molecular orbital (i.e., the lower energy molecular orbital that results from overlap of parallel orbitals on adjacent atoms).

p

Pi (p) molecular orbital (Section 1.13): A molecular orbital formed when parallel orbitals on adjacent atoms overlap. Pi molecular orbitals may be lobes of the same phase sign overlap) or orbitals of opposite phase sign overlap).

p bonding(p antibonding(p

pKa(Section 3.6B): The pKais the negative logarithm of the acid­

Polarim eter

(Section 2.3): A molecule with a dipole moment.

(Section 5.8B): A device used for measuring optical

activity. (Section 6.13C): The susceptibility of the electron cloud of an uncharged molecule to distortion by the influence of an electric charge. P olarizability

Polym er (Section 10.10): A large molecule made up of many repeating subunits. For example, the polymer polyethylene is made up of the repeating subunit —(CH2 CH2 )n—. P olym erase chain reaction (P C R ) (Section 25.8): A method for multiplying (amplifying) the number of copies of a DNA mole­ cule. The reaction uses DNA polymerase enzymes to attach addi­ tional nucleotides to a short oligonucleotide “primer” that is bound to a complementary strand of DNA called a “template.” The nucleotide that the polymerases attach are those that are com­ plementary to the base in the adjacent position on the template strand. Each cycle doubles the amount of target DNA that existed prior to the reaction step, yielding an exponential increase in the amount of DNA over time. Polym erizations

(Section 10.10): Reactions in which individual

monomers)arejoined together to form long-chain

subunits (called macromolecules. Polypeptide

(Section 24.4): A peptide comprised of many (>10)

amino acids. (Sections 22.1A and 22.13): A carbohydrate that, on a molecular basis, undergoes hydrolytic cleavage to yield many molecules of a monosaccharide. Also called a glycan. Polysaccharide

Polyunsaturated fatty acid/ester (Section 23.2): A fatty acid or ester of a fatty acid whose carbon chain contain two or more dou­ ble bonds.

(Section 3.9): Potential energy is stored ener­ gy; it exists when attractive or repulsive forces exist between objects. P otential en ergy

Potential energy diagram (Section 4.8); A graphical plot of the potential energy changes that occurs as molecules (or atoms) react (or interact). Potential energy is plotted on the vertical axis, and the progress of the reaction on the horizontal axis

ity constant, Ka. pKa = —log Ka.

P rim a ry carbon (Section 2.5): A carbon atom that has only one other carbon atom attached to it.

Plane of symmetry (Sections 5.6 and 5.12A): An imaginary plane that bisects a molecule in a way such that the two halves of the molecule are mirror images of each other. Any molecule with a plane of symmetry will be achiral.

(Sections 24.1, 24.5, and 24.6): The covalent structure of a polypeptide or protein. This structure is determined, in large part, by determining the sequence of amino acids in the protein.

Plane-polarized light (Section 5.8A): Light in which the oscilla­ tions of the electrical field occur only in one plane.

P ro ch iral center (Section 12.3C): A group is prochiral if replace­ ment of one of two identical groups at a tetrahedral atom, or if addition of a group to a trigonal planar atom, leads to a new chirality center. At a tetrahedral atom where there are two identical groups, the identical groups can be designated pro-R and pro-5 depending on what configuration would result when it is imagined that each is replaced by a group of next higher priority (but not higher than another existing group).

Polar aprotic solvent (Section 6.13C): A polar solvent that does not have a hydrogen atom attached to an electronegative element. Polar aprotic solvents do hydrogen bond with a Lewis base (e.g., a nucleophile).

not

Polar protic solvent (Section 6.13D): A polar solvent that has at least one hydrogen atom bonded to an electronegative element. These hydrogen atoms of the solvent can form hydrogen bonds with a Lewis base (e.g., a nucleophile).

P rim a ry structu re

(Section 23.5): Natural C20 carboxylic acids that contain a five-membered ring, at least one double bond, and several

Prostaglandins

G lossary

oxygen-containing functional groups. Prostaglandins mediate a vari­ ety of physiological processes. Prosthetic group (Sections 24.9 and 24.12): An enzyme cofactor that is permanently bound to the enzyme. Protecting group (Sections 11.11D, 11.11E, 12.9, 15.14A, 16.7C, and 24.7A): A group that is introduced into a molecule to protect a sensitive group from reaction while a reaction is carried out at some other location in the molecule. Later, the protecting group is removed. Also called blocking group. (See also orthogonal pro­ tecting group.) Protein (Section 24.4): A large biological polymer of a-amino acids joined by amide linkages. Proteome Proteome (Sections 25.1 and 25.9): The set of all pro­ teins encoded within the genome of an organism and expressed at any given time. Proteomics (Section 24.14): The study of all proteins that are expressed in a cell at a given time. Protic solvent (Sections 3.12, 6.13C, and 6.13D): A solvent whose molecules have a hydrogen atom attached to a strongly electroneg­ ative element such as oxygen or nitrogen. Molecules of a protic solvent can therefore form hydrogen bonds to unshared electron pairs of oxygen or nitrogen atoms of solute molecules or ions, thereby stabilizing them. Water, methanol, ethanol, formic acid, and acetic acid are typical protic solvents. Proton decoupling (Section 9.11B): An electronic technique used in 13C NMR spectroscopy that allows decoupling of spin-spin interactions between 13C nuclei and 1 H nuclei. In spec­ tra obtained in this mode of operation all carbon resonances appear as singlets. Proton off-resonance decoupling (Section 9.11D): A technique used in 13C NMR spectroscopy that allows one-bond couplings between 13C nuclei and 1H nuclei. In spectra obtained in this mode of operation, CH3 groups appear as quartets, CH2 groups appear as triplets, CH groups appear as doublets, and carbon atoms with no attached hydrogen atoms appear as singlets. Psi (C) function (See Wave function and Section 1.9) Pyranose (Section 22.2C): A sugar in which the cyclic acetal or hemiacetal ring is six membered. Q Quartet (Section 9.2C): An NMR signal comprised of four peaks in a 1:3:3:1 area ratio, caused by signal splitting from three neigh­ boring NMR-active spin 1/2 nuclei. Quaternary ammonium salt (Sections 20.2B and 20.3D): Ionic compounds in which a nitrogen bears four organic groups and a positive charge, paired with a counterion. Quaternary structure (Sections 24.1 and 24.8C): The overall structure of a protein having multiple subunits (non-covalent aggregates of more than one polypeptide chain). Each subunit has a primary, secondary, and tertiary structure of its own. R R (Sections 2.4A and 4.3A): A symbol used to designate an alkyl group. Oftentimes it is taken to symbolize any organic group.

R,S-System(Section 5.7): A method for designating the configu­ ration of tetrahedral chirality centers.

Gl-17

R acem ic form (racem ate or racem ic m ixture) (Sections 5.9A, 5.9B, and 5.10A): An equimolar mixture of enantiomers. A racemic form is optically inactive.

(Section 6.12A): A reaction that transforms an optically active compound into a racemic form is said to pro­ ceed with racemization. Racemization takes place whenever a reaction causes chiral molecules to be converted to an achiral intermediate. R acem iza tio n

Radical (or free radical) (Sections 3.1A, 10.1, 10.6, and 10.7): An uncharged chemical species that contains an unpaired electron. R a d ica l addition to alken es (Sections 10.9 and 10.10): A process by which an atom with an unshared electron, such as a bromine atom, adds to an alkene with homolytic cleavage of the p-bond and formation of a s-bond from the radical to the car­ bon; the resulting carbon radical then continues the chain reac­ tion to product the final product plus another species with an unshared electron. R adical cation (Section 9.14): A chemical species containing an unshared electron and a positive charge.

(Section 10.3): Substitution of a hydrogen by a halogen through a radical reaction mechanism. R adical halogenation

R adical reaction (Section 10.1B): A reaction involving radicals. Homolysis of covalent bonds occurs in radical reactions.

Rate control (See Kinetic control) (Section 6.9A): If a reaction takes place in a series of steps, and if the first step is intrinsically slower than all of the others, then the rate of the overall reaction will be the same as (will be determined by) the rate of this slow step.

R ate-determ ining step

(Section 6.7): The abscissa in a potential energy diagram that represents the progress of the reaction. It rep­ resents the changes in bond orders and bond distances that must take place as reactants are converted to products. R eaction coordinate

(Sections 3.1 and 3.14): A step-by-step description of the events that are postulated to take place at the molecular level as reactants are converted to products. A mecha­ nism will include a description of all intermediates and transition states. Any mechanism proposed for a reaction must be consistent with all experimental data obtained for the reaction. R eaction m echanism

R earrangem en t (Sections 3.1, 7.8A, and 7.8B): A reaction that results in a product with the same atoms present but a different car­ bon skeleton from the reactant. The type of rearrangement called a 1 ,2 shift involves the migration of an organic group (with its elec­ trons) from one atom to the atom next to it.

(Sections 12.2 and 12.3): A chemical species that causes another chemical species to become reduced (to gain elec­ trons, or to lose bonds to electronegative elements, often gaining bonds to hydrogen in the process). The reducing agent is oxidized in this process. R educing agent

(Section 22.6A): Sugars that reduce Tollens’ or Benedict’sreagents. All sugars that contain hemiacetal or hemiketal groups (and therefore are in equilibrium with aldehydes or a-hydroxyketones) are reducing sugars. Sugars in which only acetal or ketal groups are present are nonreducing sugars. R educing sugar

(Sections 12.2 and 12.3): A reaction that lowers the oxidation state of atoms in a molecule or ion. Reduction of an organic compound usually involves increasing its hydrogen content R eduction

Gl-18

G lossary

or decreasing its oxygen content. Reduction also accompanies any reaction that results in replacement of a more electronegative sub­ stituent by a less electronegative one. (Section 20.4C): A method for synthesizing primary, secondary, or tertiary amines in which an aldehyde or ketone is treated with a primary or secondary amine to produce an imine (when primary amines are used) or an iminium ion (when secondary amines are used), followed by reduction to produce an amine product. R eductive am ination

Regioselective reaction (Sections 8.2C and 8.19): A reaction that yields only one (or a predominance of one) constitutional isomer as the product when two or more constitutional isomers are possible products.

(Section 5.15A): The relationship between the configurations of two chiral molecules. Molecules are said to have the same relative configuration when similar or identical groups in each occupy the same position in space. The configura­ tions of molecules can be related to each other through reactions of known stereochemistry, for example, through reactions that cause no bonds to a stereogenic center to be broken. Relative configuration

(Section 25.4C): A process in which DNA unwinds, allowing each chain to act as a template for the formation of its complement, producing two identical DNA molecules from one original molecule. Replication

Resolution (Sections 5.16B and 20.3F): The process by which the enantiomers of a racemic form are separated. Resonance (Sections 3.11 A, 13.5, and 15.11B): An effect by which a substituent exerts either an electron-releasing or electronwithdrawing effect through the p system of the molecule.

(Section 14.5): An energy of stabilization that represents the difference in energy between the actual compound and that calculated for a single resonance structure. The resonance ener­ gy arises from delocalization of electrons in a conjugated system. Resonance energy

Resonance structures (or resonance contributors) (Sections 1.8, 1.8A, 13.3B, and 13.5A): Lewis structures that differ from one another only in the position of their electrons. A single resonance structure will not adequately represent a molecule. The molecule is better represented as a of all of the resonance structures.

hybrid

Restriction endonucleases (Section 25.6): Enzymes that cleave double-stranded DNA at specific base sequences.

(Section 19.4B): Aldol reactions are reversible; under certain conditions an aldol product will revert to its aldol reaction precursors. This process is called a R etro -a ld o l rea ctio n

reaction.

retro-aldol

(Section 7.16B): A method for planning syntheses that involves reasoning backward from the target mole­ cule through various levels of precursors and thus finally to the starting materials. R etrosynthetic analysis

conformation to another. A chair-chair ring flip converts any equa­ torial substitutent to an axial substituent and vice versa. R ing strain (Section 4.10): The increased potential energy of the cyclic form of a molecule (usually measured by heats of combus­ tion) when compared to its acyclic form. S

sorbital (Section 1.10): A spherical atomic orbital. For sorbitals the azimuthal quantum number l= 0 (see Atomic orbital). (Section 1.4A): The product of a reaction between an acid and a base. Salts are ionic compounds composed of oppositely charged ions. Salt

S anger ^ -term in al analysis (Section 24.5B): A method for deter­ mining the ^-terminal amino acid residue of a peptide by its SNAr (nucleophilic aromatic substitution) reaction with dinitrofluorobenzene, followed by peptide hydrolysis and comparison of the prod­ uct with known standards. Sapon ification (Sections 17.7B and 23.2C): Base-promoted hydrolysis of an ester.

(Sections 2.1, 7.13, and 23.2): A compound that does not contain any multiple bonds. Saturated com pound

(Section 4.8): A chemical formula that depicts the spatial relationships of groups in a molecule in a way similar to dash-wedge formulas. Saw horse form ula

(Section 20.1): A derivative of ammonia in which there are two carbons bonded to a nitrogen atom. Secondary amines have a formula R2 NH, where the R groups can be the same or different. S econdary am ine

(Section 2.5): A carbon atom that has two other carbon atoms attached to it. S econdary carbon

(Sections 24.1 and 24.8A): The local con­ formation of a polypeptide backbone. These local conformations are specified in terms of regular folding patterns such as pleated sheets, a helixes, and turns. S econdary structure

(Section 9.6): Effects observed in NMR spectra caused by the circulation of sigma and pi electrons within the molecule. Shielding causes signals to appear at lower frequencies (upfield), deshielding causes signals to appear at high­ er frequencies (downfield). Shielding and deshielding

Sigm a ( s ) bond (Section 1.12A): A single bond. A bond formed when electrons occupy the bonding s orbital formed by the end-on overlap of atomic orbitals (or hybrid orbitals) on adjacent atoms. In a sigma bond the electron density has circular symmetry when viewed along the bond axis.

Ribonucleic acid (RN A)

(Section 1.12A): A molecular orbital formed by end-on overlap of orbitals (or lobes of orbitals) on adjacent atoms. Sigma orbitals may be (orbitals or lobes of the same phase sign overlap) or (orbitals or lobes of opposite phase sign overlap).

(Section 25.5B): A ribonucleic acid that acts as a reac­ tion catalyst.

(Sections 9.2C and 9.9): Splitting of an NMR sig­ nal into multiple peaks, in patterns such as doublets, triplets, quar­ tets, etc., caused by interactions of the energy levels of the mag­ netic nucleus under observation with the energy levels of nearby magnetic nuclei.

(Sections 25.1 and 25.5): One of the two classes of molecules (the other is DNA) that carry genetic infor­ mation in cells. RNA molecules transcribe and translate the infor­ mation from DNA for the mechanics of protein synthesis. R ibozym e

R ing flip (Sections 4.11 and 4.12): The change in a cyclohexane ring (resulting from partial bond rotations) that converts one ring

Sigm a ( s ) orbital

bonding antibonding

Sign al splitting

(Sections 11.11E and 17.7C): Conversion of an alcohol, R—OH, to a silyl ether (usually of the form R—O—SiR'3 , where

Silylation

Gl-19

G lossary

the groups on silicon may be the same or different). Silyl ethers are used as protecting groups for the alcohol functionality. Singlet (Section 9.2C): An NMR signal with only a single, unsplit peak. Site-specific cleavage (Section 24.5D): A method of cleaving peptides at specific, known sites using enzymes and specialized reagents. For example, the enzyme trypsin preferentially cat­ alyzes hydrolysis of peptide bonds on the C-terminal side of argi­ nine and lysine. Other bonds in the peptide are not cleaved by this reagent. Sn 1 reaction (Sections 6.9, 6.10, 6.12, 6.13, and 6.18B): Literally, substitution nucleophilic unimolecular. A multistep nucleophilic substitution in which the leaving group departs in a unimolecular step before the attack of the nucleophile. The rate equation is first order in substrate but zero order in the attacking nucleophile. Sn2 reaction (Sections 6.5B, 6 .6 - 6 .8 , 6.13, and 6.18A): Literally, substitution nucleophilic bimolecular. A bimolecular nucleophilic substitution reaction that takes place in a single step. A nucleophile attacks a carbon bearing a leaving group from the back side, caus­ ing an inversion of configuration at this carbon and displacement of the leaving group. Solid-phase peptide synthesis (SPPS) (Section 24.7D): A method of peptide synthesis in which the peptide is synthesized on a solid support, one amino acid residue at a time. The first amino acid of the peptide is bonded as an ester between its carboxylic acid group and a hydroxyl of the solid support (a polymer bead). This is then treated with a solution of the second amino acid and appropriate coupling reagents, creating a dipeptide. Excess reagents, byprod­ ucts, etc. are washed away. Further linkages are synthesized in the same manner. The last step of the synthesis is cleavage of the pep­ tide from the solid support and purification. Solubility (Section 2.13D): The extent to which a given solute dis­ solves in a given solvent, usually expressed as a weight per unit volume (e.g., grams per 100 mL).

Specific rotation (Section 5.8C): A physical constant calculated from the observed rotation of a compound using the following equation: [«I d

a cXl

where a is the observed rotation using the D line of a sodium lamp, is the concentration of the solution or the density of a neat liquid in grams per milliliter, and is the length of the tube in decimeters.

c

l

Spectroscopy (Section 9.1): The study of the interaction of ener­ gy with matter. Energy can be absorbed, transmitted, emitted or cause a chemical change (break bonds) when applied to matter. Among other uses, spectroscopy can be used to probe molecular structure. Spin decoupling (Section 9.10): An effect that causes spin-spin splitting not to be observed in NMR spectra. Spin-spin splitting (Section 9.9): An effect observed in NMR spectra. Spin-spin splittings result in a signal appearing as a mul­ tiplet (i.e., doublet, triplet, quartet, etc.) and are caused by magnet­ ic couplings of the nucleus being observed with nuclei of nearby atoms. Splitting tree diagrams (Section 9.9B): A method of illustrating the NMR signal splittings in a molecule by drawing “branches” from the original signal. The distance between the branches is pro­ portional to the magnitude of the coupling constant. This type of analysis is especially useful when multiple splittings (splitting of already split signals) occur due to coupling with non-equivalent protons. Staggered conformation (Section 4.8A): A temporary orientation of groups around two atomsjoined by a single bond such that the bonds of the back atom exactly bisect the angles formed by the bonds of the front atom when shown in a Newman projection formula: F ront atom

Solvent effect (Sections 6.13C and 6.13D): An effect on relative rates of reaction caused by the solvent. For example, the use of a polar solvent will increase the rate of reaction of an alkyl halide in an Sn1 reaction. Solvolysis (Section 6.12B): Literally, cleavage by the solvent. A nucleophilic substitution reaction in which the nucleophile is a molecule of the solvent.

sporbital (Section 1.14): A hybrid orbital that is derived by math­ ematically combining one s atomic orbital and one patomic orbital. Two s phybrid orbitals are obtained by this process, and they are oriented in opposite directions with an angle of 180° between them. sp2 orbital (Section 1.13): A hybrid orbital that is derived by math­ ematically combining one atomic orbital and two atomic orbitals. Three sp2 hybrid orbitals are obtained by this process, and they are directed toward the corners of an equilateral triangle with angles of 1 2 0 ° between them.

s

p

sp3 orbital (Section 1.12A): A hybrid orbital that is derived by mathematically combining one atomic orbital and three atomic orbitals. Four sp3 hybrid orbitals are obtained by this process, and they are directed toward the corners of a regular tetrahedron with angles of 109.5° between them.

s

p

Rear atom A s ta g g e re d c o n fo rm a tio n

Step-growth polymer (See also Condendsation polymer, Section 17.12 and Special Topic C): A polymer produced when bifunctional monomers (or potentially bifunctional monomers) react with each other through the intermolecular elimination of water or an alcohol. Polyesters, polyamides, and polyurethanes are all step-growth (condensation) polymers Stereochemistry (Sections 5.2, 6 .8 , and 6.14): Chemical studies that take into account the spatial aspects of molecules. Stereogenic carbon (Section 5.3): A single tetrahedral carbon with four different groups attached to it. Also called an The last usage is preferred.

riccarbon,astereocenter, orachiralitycenter.

asymmet­

Stereogenic center (Sections 5.3, 5.18):When the exchange of two groups bonded to the same atom produces stereoisomers, the atom is said to be a stereogenic atom, or stereogenic center.

Gl-20

G lossary

(Sections 1.13B, 4.9A, 4.13, 5.2B, and 5.14): Compounds with the same molecular formula that differ in the arrangement of their atoms in space. Stereoisomers have the same connectivity and, therefore, are not constitutional isomers. Stereoisomers are classified further as being either enantiomers or diastereomers.

Stereoisom ers

only

(Sections 5.10B, 8.21C, and 12.3C): In reactions where chirality centers are altered or created, a stereosel­ ective reaction produces a preponderance of one stereoisomer. Furthermore, a stereoselective reaction can be either enantioselective, in which case the reaction produces a preponderance of one enantiomer, or diastereoselective, in which case the reaction pro­ duces a preponderance of one diastereomer. Stereoselective reaction

(Section 8.13): A reaction in which a par­ ticular stereoisomeric form of the reactant reacts in such a way that it leads to a specific stereoisomeric form of the product. Stereospecific reaction

(Section 6.13A): An effect on relative reaction rates caused by the space-filling properties of those parts of a molecule attached at or near the reacting site.

Steric effect

S teric hindrance (Sections 4.8B and 6.13A): An effect on relative reaction rates caused when the spatial arrangement of atoms or groups at or near the reacting site hinders or retards a reaction.

(Section 23.4): Steroids are lipids that are derived from the following perhydrocyclopentanophenanthrene ring system: Steroid

(Section 20.9): An amide derivative of a sulfonic acid, usually made by. the reaction of ammonia, or a primary or secondary amine, with a sulfonyl chloride, resulting in compounds having the general formulas R'SO2 NH2 , R'SO 2 NHR, or R'SO2 NR2 , respectively. Sulfonam ides

(Section 11.10): A compound with the formula ROSO2 R' and considered to be derivatives of sulfonic acids, HOSO2 R'. Sulfonate esters are used in organic synthesis because of the excellent leaving group ability of the fragment —OSO2 R'. S ulfon ate ester

Superposable (Sections 1.13B and 5.1): Two objects are superposable if, when one object is placed on top of the other, all parts of each coincide. To be superposable is different than to be super­ imposable. Any two objects can be superimposed simply by putting one object on top of the other, whether or not all parts coin­ cide. The condition of superposability must be met for two things to be identical.

(Sections 7.14A and 7.15A): An addition that places both parts of the adding reagent on the same face of the reactant. Syn addition

(Section 8.16A): An oxidation reaction in which an alkene reacts to become a 1 ,2 -diol (also called a ) with the newly bonded hydroxyl groups added to the same face of the alkene.

Syn dihydroxylation

glycol

Syndiotactic polym er (Special Topic B.1): A polymer in which the configuration at the stereogenic centers along the chain alter­ nate regularly: ( ), ( ), ( ), ( ), etc.

RS RS

Synthetic equivalent (Sections 8.21, 18.6, and 18.7): A compound that functions as the equivalent of a molecular fragment needed in a synthesis.

(Sections 1.3A and 1.17): A formula that shows how the atoms of a molecule are attached to each other.

Stru ctu ral form ula

(Sections 3.11D and 15.10): An effect on the rate of reaction (or on the equilibrium constant) caused by the replacement of a hydrogen atom by another atom or group. Substituent effects include those effects caused by the size of the atom or group, called steric effects, and those effects caused by the ability of the group to release or withdraw electrons, called elec­ tronic effects. Electronic effects are further classified as being inductive effects or resonance effects.

Synthon (Sections 8.21B, 18.6, and 18.7): The fragments that result (on paper) from the disconnection of a bond. The actual reagent that will, in a synthetic step, provide the synthon is called the .

syntheticequivalent

Substituent effect

Substitution reaction (Sections 3.1, 6.2, 10.3, 15.1, and 17.4): A reaction in which one group replaces another in a molecule.

(Section 4.3F): A system for naming compounds in which each atom or group, called a substituent, is cited as a prefix or suffix to a parent compound. In the IUPAC sys­ tem only one group may be cited as a suffix. Locants (usually num­ bers) are used to tell where the group occurs. Substitutive nom enclature

(Sections 6.2 and 24.9): The molecule or ion that under­ goes reaction. Substrate Su g ar

(Section 22.1A): A carbohydrate.

(Sections 20.9 and 20.10): Sulfonamide antibacteri­ al agents, most of which have the general structure P-H2 NC6 H4 SO2 NHR. Sulfa drugs act as (they inhibit the growth of microbes) by inhibiting the enzymatic steps that are involved in the synthesis of folic acid; when deprived of folic acid, the microorganism dies. Su lfa d ru gs

antimetabolites

T

(Section 18.2): An isomerization by which tautomers are rapidly interconverted, as in keto-enol tautomerization. T autom erization

(Section 18.2): Constitutional isomers that are easi­ ly interconverted. Keto and enol tautomers, for example, are rapidly interconverted in the presence of acids and bases. T au tom ers

T erm inal residue analysis (Section 24.5): Methods used to deter­ mine the sequence of amino acids in a peptide by reactions involv­ ing the - and -terminal residues.

N C

Terpene (Section 23.3): Terpenes are lipids that have a structure that can be derived on paper by linking isoprene units. T erpenoids

(Section 23.3): Oxygen-containing derivatives of

terpenes. T ertia ry am ine (Section 20.1): A derivative of ammonia in which there are three carbons bonded to a nitrogen atom. Tertiary amines have a formula R3N where the R groups can be the same or different. T ertiary carbon (Section 2.5): A carbon atom that has three other carbon atoms attached to it.

(Sections 24.1 and 24.8B): The three dimen­ sional shape of a protein that arises from folding of its polypeptide chains superimposed on its a helixes and pleated sheets.

T ertiary structure

Gl-21

G lossary

(Section 17.4): A species created by the attack of a nucleophile on a trigonal carbon atom. In the case of a carbonyl group, as the electrons of the nucleophile form a bond to the carbonyl carbon the electrons of the carbon-oxygen p-bond shift to the oxygen. The carbon of the carbonyl group becomes four-coordinate (tetrahedral), while the oxygen gains an electronpair and becomes negatively charged. T etrahedral interm ediate

T herm odyn am ic control (Sections 13.10A and 18.4A): A princi­ ple stating that the ratio of products of a reaction that reaches equi­ librium is determined by the relative stabilities of the products (as measured by their standard free energies, AG°). The most abundant product will be the one that is the most stable. Also called equilib­ rium control.

(Section 18.4A): In a situation in which more than one enolate anion can be formed, the is the more stable of the possible enolate anions—usually the enolate with the more substituted double bond. A thermody­ namic enolate is formed predominantly under conditions that per­ mit the establishment of an equilibrium. T herm odyn am ic enolate

enolate

thermodynamic

Torsional b a rrie r (Section 4.8B): The barrier to rotation of groups joined by a single bond caused by repulsions between the aligned electron pairs in the eclipsed form.

(Sections 4.8B, 4.9, and 4.10): The strain associ­ ated with an eclipsed conformation of a molecule; it is caused by repulsions between the aligned electron pairs of the eclipsed bonds. Torsional strain

Two-dimensional (2D) NMR (Section 9.12): NMR techniques such as COSY and HETCOR that correlate one property (e.g., coupling), or type of nucleus, with another. (See COSY and HETCOR.) U Ultraviolet-visible (UV-Vis) spectroscopy (Section 13.9): A type of optical spectroscopy that measures the absorption of light in the visible and ultraviolet regions of the spectrum. Visible-UV spectra primarily provide structural information about the kind and extent of conjugation of multiple bonds in the compound being analyzed. Unimolecular reaction (Section 6.9): A reaction whose rate-deter­ mining step involves only one species. Unsaturated compound (Sections 2.1, 7.13, and 23.2): A com­ pound that contains multiple bonds. Upfield (Section 9.2A): Any area or signal in an NMR spectrum that is to the right relative to another. (See Downfield for compar­ ison.) A signal that is upfield of another occurs at lower frequency (and lower S and ppm values) than the other signal. V

Tosylate

Vicinal coupling (Section 9.9): The splitting of an NMR signal caused by hydrogen atoms on adjacent carbons. (See also Coupling and Signal Splitting.)

p

Vicinal (vie-) substituents (Section 7.10): Substituents that are on adjacent atoms.

(Section 25.5): Synthesis of a messenger RNA (mRNA) molecule that is complimentary to a section of DNA that carries genetic information.

Vinyl group (Sections 4.5 and 6.1): The CH2 — CH — group.

(Section 17.7A): A reaction involving the exchange of the alkoxyl portion of an ester for a different alkoxyl group, resulting in a new ester.

Vinylic substituent (Section 6.1): Refers to a substituent on a car­ bon atom that participates in a carbon-carbon double bond.

(Section 11.10): A p-toluenesulfonate ester, which is a compound that contains the P-CH3 C6 H4 SO3— group, i.e., -CH3 C6 H4 SO3R

T ranscription

Transesterification

(Sections 6 .6 , 6.7, and 6.10): A state on a potential energy diagram corresponding to an energy maximum (i.e., characterized by having higher potential energy than imme­ diately adjacent states). The term transition state is also used to refer to the species that occurs at this state of maximum poten­ tial energy; another term used for this species is T ran sition state

theactivated

complex.

T ranslation (Section 25.5E): The ribosomal synthesis of a polypeptide using an mRNA template. T riacylglycerols (Section 23.2): An ester of glycerol (glycerin) in which all three of the hydroxyl groups are esterified. Triflate (Section 11.10): A methanesulfonate ester, which is a compound that contains the CH3 SO3— group, i.e., -CH3 SO3R

p

Tripeptide

(Section 24.4): A peptide comprised of three amino

acids (Sections 1.2 and 1.14): Bonds comprised of one sigma (s) bond and two pi (p) bonds.

Triple bonds

Vinylic halide (Section 6.1): An organic halide in which the halo­ gen atom is attached to a carbon atom of a double bond.

VSEPR model (valence shell electron pair replusion) (Section 1.16): A method of predicting the geometry at a covalently bond­ ed atom by considering the optimum geometric separation between groups of bonding and non-bonding electrons around the atom W Wave function (or C function) (Section 1.9): A mathematical expression derived from corresponding to an energy state for an electron, i.e., for an orbital. The square of the C function, C2, gives the probability of finding the electron in a par­ ticular place in space.

quantummechanics

Wavelength, l (Sections 2.15 and 13.9A): The distance between consecutive crests (or troughs) of a wave. Wavenumber, v (Section 2.15): A way to express the frequency of a wave. The wavenumber is the number of waves per centimeter, expressed as cm—1.

Triplet

Waxes (Section 23.7): Esters of long-chain fatty acids and longchain alcohols.

(Section 22.1A): A carbohydrate that, when hydrolyzed, yields three monosaccharide molecules.

Williamson synthesis (Section 11.11B): The synthesis of an ether by the SN2 reaction of an alkoxide ion with a substrate bearing a suitable leaving group (often a halide, sulfonate, or sulfate).

(Section 9.2C): An NMR signal comprised of three peaks in a 1 :2 :1 area ratio, caused by signal splitting from two neighbor­ ing NMR-active spin 1/2 nuclei. T risacch arid es

Gl-22

G lossary

Y

(Section 16.10): An electrically neutral molecule that has a negative carbon with an unshared electron pair adjacent to a positive heteroatom.

Y lid e

Z

(Sections 7.6B and 7.8A): A rule stating that an elimination will give as the major product the most stable alkene (i.e., the alkene with the most highly substituted double bond). Z a itsev’s ru le

(See D ipolar ion and Section 24.2C): Another name for a dipolar ion.

Z w itterion

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P h o to C re d its

Chapter 21

Chapter 24

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Index A

Absolute configuration, 220-221 Absorption maxima for nonconjugated and conjugated dienes, 609 Absorption spectrum, 607 Acetaldehyde, 75, 729 Acetaldehyde enolate, 51 physical properties of, 75 Acetals, 747-750 acid-catalyzed formation, 748 cyclic, 748-749 hemiacetals, 744-747 as protecting groups, 749-750 thioacetals, 750 Acetanilide, nitration of, 716 Acetic acid, 70-71, 142 physical properties of, 75 substituted, synthesis of, 850-853 Acetoacetic ester synthesis, 845-850 acylation, 849 dialkylation, 845-846 substituted methyl ketones, 846-847 Acetone, 6 , 75, 729 Acetonides, 1016 Acetonitrile, 72 Acetyl-coenzyme A, 792 Acetyl group, 731 Acetylcholine, 923-924 Acetylcholinesterase, 924, 1122 Acetylenes, 34, 55, 154, 286, 321 Acetylenic hydrogen atom, 154, 307 of terminal alkynes, substitution of, 310-312 Achiral molecules, 192 Acid anhydrides, reactions of, 819-820 Acid-catalyzed aldol condensations, 880-881 Acid-catalyzed aldol enolization, 835 Acid-catalyzed halogenation, of aldehydes and ketones, 837 Acid-catalyzed hemiacetal formation, 745 Acid-catalyzed hydration, of alkenes, 340-342, 353 Acid chlorides, Acyl chlorides Acid derivatives, synthesis of, 794 Acid strength, 109 Acid-base reactions, 115-118 acids and bases in water, 1 0 2 Br0 nsted-Lowry acids and bases,

See

1 0 1 -1 0 2

opposite charges attract, 103-104 predicting the outcome of, 113-114 and the synthesis of deuterium and tri­ tium-labeled compounds, 130 water solubility as the result of salt for­ mation, 114-115 Acidic hydrolysis of a nitrile, 810 Acidity: effect of the solvent on, 125-126 hybridization, 117-118 inductive effects of, 118

relationships between structure and, 115-116 Acidity constant (Ka), 109-110 Acids: acetic, 70-71, 142 alcohols as, 513 aldaric, 1018-1019 alkanedioic, 783 a-amino, 1086, 1088, 1092-1094 amino, 1084-1094 aspartic, 1088, 1113 benzoic, 70-71 b-dicarboxylic, 816 Br0 nsted-Lowry, 101-102 butanoic, 780 carbolic acids, Phenols carboxylic acids, 70-71, 779-830 cholic acid, 1071 conjugate acid, 101 D-glucaric, 1017 deoxyribonucleic (DNA): 1100-1101, 1132-1133, 1155-1157 dicarboxylic, 783-784 diprotic, 1 0 2 ethanoic, 142, 780 fatty, 71, 313, 1052-1053, 1060-1073 folic, 946 formic, 70-71, 780 fumaric, 380 hexanoic, 780 glutamic, 1088, 1113 L-amino, 1086-1088 Lewis, 102-104 linoleic, 493 maleic, 380 malonic, 816 meta-chloroperoxybenzoic (MCPBA), 528 methanoic, 780 N-acetylmuramic, 1039, 1118 N-acylamino, 1093-1094 niacin (nicotinic acid), 1116 nitric, 715 nitrous, 935-937 in nonaqueous solutions, 128-130 nucleic, 81, 1132 octadecanoic, 780 omega-3 fatty, 1052-1054 pantothenic, 1116 pentanoic, 780 peroxy (peracid), 528 phosphatidic, 1074 phosphoric, 1074 pictric, 971 ribonucleic (RNA), 1132, 1146-1154 shikimic acid, 1048 sialyl Lewisx acids, 1000 strong acids, 743 sulfuric acid, 1 0 1 - 1 0 2 uronic acids, 1037-1038 zaragozic acid A (squalestatin S1), 530

See

Acrylonitrile, anionic polymerization of, 489-490 Actin, 162 Activating groups, 691 ortho-para directors, 692-693 Activation energies, 472-475 Active hydrogen compounds, 853-854 Active methylene compounds, 853-854 Active site, 1115 Acyclovir, 1139 Acyl chlorides (acid chlorides), 6 8 6 , 785, 794-796 aldehydes by reduction of, 734-736 reactions of, 795-796, 819 synthesis of, 794-796 using thionyl chloride, 795 Acyl compounds: relative reactivity of, 793-794 spectroscopic properties of, 787-789 Acyl groups, 685 Acyl halide, 685 Acyl substitution, 779, 792-793, 831 by nucleophilic addition-elimination, 792-794 Acyl transfer reactions, 792 Acylation, 849 Acylation reaction, 685 Acylium ions, 432 Adamantane, 175 Addition polymers, 486, 817 Addition reaction, 331-384 of alkenes, 332-333 Additions, 99 Adduct, 617 Adenosine diphosphate (ADP), 426 Adenosine triphosphate (ATP), 266-267, 426, 521 Adenylate cyclase, 1136 Adipocytes, 1055 Adrenaline, 922 Adrenocortical hormones, 1070 Adriamycin, Doxorubicin Aggregation compounds, 156 Aglycone, 1011 Aklavinone, 976 Alanine, 1086 isolectric point of, 1090 titration curve for, 1091 Albuterol, 505 Alcohol dehydrogenase, 554 Alcohols, 53, 65, 126, Primary alcohols; Secondary alcohols; Tertiary alcohols as acids, 513 addition of: acetals, 747-750 hemiacetals, 744-747 thioacetals, 750 alcohol carbon atom, 503 from alkenes: through hydroboration-oxidation, 347

See

Seealso

I-1

I-2

In d e x

through oxymercuration-demercuration, 344-347 from alkyl hydrogen sulfates, 340 and alkyl phosphates, 521 boiling points, 505 from carbonyl compounds, 548-584 conversion of, into alkyl halides, 514 dehydration of, 291, 297-303 carbocation stability and the transi­ tion state, 300-302 ethanol, 502, 506, 507-508 as a biofuel, 508 ethylene, 508 polymerization of, 487 hydrogen bonding, 506 infrared (IR) spectra of, 90 intermolecular dehydration, ethers by, 522-523 mesylates, 518-521 methanol, 258, 506, 507, 513 nomenclature of, 148-149, 503-504 oxidation of, 557-561 physical properties of, 505-507 primary, 65 propylene glycols, 506, 508 reactions of, 502-547 with hydrogen halides, alkyl halides from, 514-517 with PBr3 or SOCI2, alkyl halides from, 517-518 by reduction of carbonyl compounds, 552-561 spectroscopic evidence for, 561 structure of, 503-505 synthesis of, from alkenes, 509-511 --butyl ethers by alkylation of, 525 tosylates, 518-519 triflates, 518-519 Aldaric acids, 1018-1019 Aldehyde hydrates, 557 Aldehydes, 69-70 a,y8 -unsaturated, additions to, 889-894 acid-catalyzed halogenation of, 837 base-promoted halogenation of, 837 carbonyl group, 730 chemical analyses for, 761 derivatives of, 761 IR spectra of, 762-763 mass spectra of, 764 NMR spectra of, 763-764 nomenclature of, 730-732 nucleophilic addition to the carbon-oxygen double bond, 741-744 by oxidation of 1° alcohols, 733-734 oxidation of primary alcohols to, 557 by ozonolysis of alkenes, 734 in perfumes, 733 physical properties of, 732-733 preparation of carboxylic acids by oxi­ dation of, 789-790 reduction by hydride transfer, 554 by reduction of acyl chlorides, esters, and nitriles, 734-735 relative reactivity, 743 spectroscopic properties of, 762-764

tert

summary of addition reactions, 765-766 synthesis of, 733-738 Tollens’ test (silver mirror test), 761 UV spectra, 764 Alder, Kurt, 616, 620 Alditols, 1022 Aldol addition product, dehydration of, 879 Aldol addition reactions, 877 Aldol additions, 876-877 Aldol condensation reactions, 879 Aldol condensations, 870, 877 acid-catalyzed, 880-881 crossed, 882-888 cyclizations via, 888-889 Aldol reactions, synthetic applications of, 881-882 Aldose, 1004, 1016 Aldotetrose, 1004 Aliphatic aldehydes, 730 nomenclature of, 730 Aliphatic compounds, 633 Aliphatic ketones, nomenclature of, 731 Alkadienes, 599-600 Alkaloids, 908, 922 Alkanedioic acids, 783 Alkanes, 54, 138 bicyclic, 175 branched-chain, nomenclature of, 143-145 chemical reactions of, 177 chlorination of, 467, 477-479 combustion of, 491-492 defined, 138 IUPAC nomenclature of, 142-145 multiple halogen substitution, 466 no functional group, cause of, 63 nomenclature and conformations of, 137-185 petroleum as source of, 138 physical properties, 154-156 polycyclic, 175 reactions of, with halogens, 465-467 reactions with halogens, 465-467 shapes of, 140-141 sources of, 138-139 “straight-chain,” 140 synthesis of, 177-178 Alkatrienes, 599 Alkene diastereomers, (E)-(Z) system for designating, 286-287 Alkenes, 30, 54, 55, 138 addition of sulfuric acid to, 340 addition of water to, 340-342 mechanism, 341-342 addition reaction, 332-333 alcohols from, through oxymercuration-demercuration, 344-347 aldehydes by ozonolysis of, 734 anti 1,2-dihydroxylation of, 535 dipole moments in, 62 electrophilic addition: of bromine and chlorine, 354-355 defined, 333 of hydrogen halides, 334-339

functional group, 63 halohydrin formation from, 360 heat of reaction, 288-289 hydrogenation of, 178-179, 313-314 ionic addition to, 339 ketones from, 739-740 Markovnikov additions, 334 regioselective reactions, 338 Markovnikov’s rule, 334-339 defined, 334 theoretical explanation of, 336-337 mechanism for syn dihydroxylation of, 363-364 oxidation of, 363-365 environmentally friendly methods, 537 oxidative cleavage of, 365-368 physical properties of, 286 preparation of carboxylic acids by oxi­ dation of, 789 properties/synthesis, 285-330 radical addition to, 484-486 radical polymerization of, 486-490 rearrangements, 342-343 relative stabilities of, 288-290 stereochemistry of the ionic addition to, 339 stereospecific reactions, 358-359 synthesis of alcohols from, 509-511 use in synthesis, 540-541 Alkenylbenzenes, 706, 712-713 additions to the double bond of, 712 conjugated, stability of, 712 oxidation of the benzene ring, 713 oxidation of the side chain, 713 Alkenyne, 599 Alkoxide ions, 129 Alkoxides, 269 Alkoxyl group, 71 Alkoxyl radicals, 460 Alkoxymercuration-demercuration, syn­ thesis of ethers by, 525 Alkyl alcohols, 504 Alkyl aryl ethers, cleavage of, 973 Alkyl chlorides, 264-265 Alkyl chloroformates, 812-813 Alkyl groups, 142 branched, nomenclature of, 145-146 and the symbol R, 63 Alkyl halides, 64-67 alcohol reactions with hydrogen halides, 514-517 alcohol reactions with PBr3 or SOCI2, 517-518 conversion of alcohols into, 514 dehydrohalogenation of, 268-269, 291-297 bases used in, 269 defined, 268 favoring an E2 mechanism, 291-292 less substituted alkene, formation of, using bulky base, 294-295 mechanism for, 269, 296-297 orientation of groups in the transition state, 295-296 Zaitsev rule, 292-294

In d e x

elimination reactions of, 268-269 nomenclature of, 147 Alkyl hydrogen sulfates, alcohols from, 340 Alkyl phosphates, 521 Alkyl radicals, geometry of, 480 Alkylation of alkynide anions, 312 Alkylbenzenes: preparation of carboxylic acids by oxi­ dation of, 790 reactions of the side chain of, 706-711 reactivity of, and ortho-para direction, 705-706 Alkylboranes: oxidation/hydrolysis of, 350-352 regiochemistry and stereochemistry, 351-352 protonolysis of, 353-354 Alkylcycloalkanes, 149 Alkylcycloalkanols, 149 Alkyllithium, 129-130 Alkyloxonium ion, 126, 235 Alkylpotassium compounds, 562 Alkylsodium compounds, 562 Alkynes, 34, 54, 55-56, 138, 286 addition of hydrogen halides to, 369-370 addition reaction, 331-384 functional group, 63 hydrogenation of, 178-179, 315-317 nomenclature of, 154 oxidative cleavage of, 370 physical properties of, 286 synthesis of, 285-330 by elimination reactions, 308-310 laboratory application, 308-309 terminal: acidity of, 307-308 substitution of the acetylenic hydro­ gen atom of, 310-312 Alkynide anions, alkylation of, 312 Allenes, 224-225, 599 Allotropes, 176 Allyl cation, 594-595 Allyl group, 152-153 Allyl radical, 586-590 molecular orbital description of, 591-592 resonance description of, 592-593 stability of, 590-593 Allylic bromination: chemistry of, 590 with N-bromosuccinimide, 589-590 Allylic carbocations, 718 Allylic chlorination (high temperature), 587-589 Allylic halides, 256 Allylic hydrogen atom, 587 Allylic substitution, 586-590 a-amino acids, 1086, 1088 synthesis of, 1092-1094 from potassium phthalimide, 1092 resolution of DL-amino acids, 1093-1094 Strecker synthesis, 1093

a-aminonitrile, formation of, during Strecker synthesis, 1093 anomer, 1008 carbon, 832 a carbon atom, 268 helices, 1 1 1 0 , 1 1 1 2 hydrogens, 832 -keratin, 1 1 1 2 substituents, 1065 Altman, Sidney, 1115 Amides, 72, 91, 785-786, 804-809 from acyl chlorides, 804-805 amines vs., 918-919 from carboxylic acids and ammonium carboxylates, 806 from carboxylic anhydrides, 805 DCC-promoted amide synthesis, 807 from esters, 806 hydrolysis of, 807-809 by enzymes, 808 nitriles from the dehydration of, 809 reactions of, 821 reducing to amines, 927-929 synthesis of, 804 Amine salts, 915-933 Amines, 68-70, 793, 911-963 amides vs., 918-919 amine salts, 915-933 aminium salts, 919-920 analysis of, 947-949 in aqueous acids, solubility of, 915-933 arenediazonium salts: coupling reactions of, 941-942 replacement reactions of, 937-940 aromatic, 916 preparation of, through reduction of nitro compounds, 927 basicity of, 915-933 biologically important, 922-924 2-phenylethylamines, 922 antihistamines, 922-923 neurotransmitters, 923-924 tranquilizers, 923-924 vitamins, 922-923 chemical analysis, 947 conjugate addition of, 892 diazotization, 935-936 heterocyclic, 913 basicity of, 918 infrared (IR) spectra of, 91 nomenclature, 912-913 physical properties of, 913-914 preparation of, 924-933 through Curtius rearrangement, 932 through Hofmann rearrangement, 931-932 through nucleophilic substitution reactions, 924-927 through reduction of nitriles, oximes, and amides, 929-931 through reduction of nitro com­ pounds, 927 through reductive amination, 927-929 primary, 912 oxidation of, 934

aa aa aa

I-3

preparation of, through Curtius rearrangement, 932-933 preparation of, through reduction of nitriles, oximes, and amides, 929-931 preparation of, through reductive amination, 927-929 preparation of, through Hofmann rearrangement, 931-932 quaternary ammonium salts, 919-920 reactions of, 933-937 oxidation, 934-935 primary aliphatic amines with nitrous acid, 935 primary arylamines with nitrous acid, 935 secondary amines with nitrous acid, 937 tertiary amines with nitrous acid, 937 reactions with sulfonyl chlorides, 943-944 as resolving agents, 920-921 secondary, 912 oxidation of, 934 preparation of, through reduction of nitriles, oximes, and amides, 929-931 preparation of, through reductive amination, 927-929 spectroscopic analysis, 948-949 structure of, 913-915 summary of preparations and reactions of, 950-953 tertiary, 912 oxidation of, 934 preparation of, through reduction of nitriles, oximes, and amides, 929-931 preparation of, through reductive amination, 927-929 Aminium salts, 919-920 Amino acid sequencers, 1097 Amino acids, 1084-1094 a-amino acids, 1086, 1088 synthesis of, 1092-1094 as dipolar ions, 1089-1092 essential, 1088 L-amino acids, 1086-1088 structures and nomenclature, 1086-1089 Amino cyclitol, 1042 Amino sugars, 1039 Aminobenzene, 634 Ammonia, 793 reaction of, with alkyl halide, 235 shape of a molecule of, 38-39 Ammonium compounds, eliminations involving: Cope elimination, 950 Hofmann elimination, 949-950 Ammonium cyanate, 2 Ammonium ion, 14 Ammonium salts, 915 Ammonolysis, 803, 806 Amphetamine, 6 8 , 922 Amylopectin, 1034-1035

I-4

In d e x

Amylose, 1034 Anderson, C. D., 1139 Androsterone, 1068 Aneshansley, D., 979 Anet, F. A. L., 416 Anethole, 502 Angle strain, 162 Angular methyl groups, 1065 Aniline, 634, 694, 715-716 Anionic polymerization, 534 Annulenes, 644-646 Anomeric carbon atom, 1009 Anomeric effect, 1010 Anomers, 1009 Anthracene, 653 Anti 1,2-dihydroxylation, of alkenes, 535 Anti addition: defined, 315 of hydrogen, 316-317 Anti conformation, 160 Anti coplanar conformation, 295 Anti-Markovnikov addition, 339 of hydrogen bromide, 484-486 of water to an alkene, 351 Anti-Markovnikov hydration of a double bond, 347 Anti-Markovnikov syn hydration, 347 Antiaromatic compounds, 650-651 Antibodies, 1084, 1123 Antibonding molecular orbital, 24, 37 Anticodon, 1150 Antigens, 1040-1041 Antihistamines, 922-923 Antimetabolites, 946 Antioxidants, 494 Antisense oligonucleotides, 1157 Aprotic solvents, 260 Arbutin, 1045 Arbuzov reaction, 760 Arenediazonium salts: coupling reactions of, 941-942 replacement: by —F, 939 by —I, 939 by —OH, 939 by hydrogen, 939-940 replacement reactions of, 937-940 Arenes, 706 ketones from, 739-740 Arenium ion, 678, 700-701 Arginine, 1088, 1113 Arly halides: 1H NMR spectra, 988 infrared spectra, 988 as insecticides, 989 mass spectra, 988 and nucleophilic aromatic substitution, 980-988 by addition-elimination (SnA t mechanism), 981-982 through an elimination-addition mechanism (benzyne), 984-987 spectroscopic analysis of, 988 Aromatic amines, preparation of, through reduction of nitro compounds, 927 Aromatic compounds, 54, 632-675

13C NMR spectra, 660-663 benzene: discovery of, 633 halogenation of, 680-681 Kekulé structure for, 638-639 modern theories of the structure of, 640-643 nitration of, 681-682 nomenclature of benzene derivatives, 634-636 reactions of, 637 sulfonation of, 682-683 thermodynamic stability of, 639-640 benzenoid, 651-654 in biochemistry, 657-660 Birch reduction, 719-720 defined, 649 electrophilic aromatic substitution reac­ tions, 677 general mechanism for, 678-680 Friedel-Crafts acylation, 685-687 Clemmensen reduction, 690-691 synthetic applications of, 690-691 Friedel-Crafts alkylation, 684-685 Friedel-Crafts reactions, limitations of, 687 fullerenes, 654-655 1H NMR spectra, 660 heterocyclic, 655-657 Hückel’s rule, 643-651 infrared spectra of substituted benzenes, 663-664 mass spectra of, 665 nonbenzenoid, 654 nucleophilic substitution reactions, allylic and benzylic halides in, 717-719 reactions of, 676-728 reduction of, 719-720 spectroscopy of, 660-665 synthetic applications: orientation in disubstituted benzenes, 716-717 protecting and blocking groups, use of, 715-716 Aromatic cyclodehydration, 721 Aromatic ions, 647-648 Artificial sweeteners, 1032-1033 Aryl halides, 232, 267, 980-991 properties of, 964 Arylamines, basicity of, 916-918 Ashworth, Linda, 1132 Asparagine, 1088, 1113 Aspartic acid, 1088, 1113 Asymmetric atoms, Chirality centers Atomic force microscopy (AFM), 655 Atomic number (Z), 4, 197 Atomic orbitals (AOs), 22, 24, 36 hybrid, 26 Atomic structure, 2-3 and quantum mechanics, 2 0 - 2 1 Atoms, 4 Atropisomers, 219, 224 Attractive electric forces, summary of, 82 Aufbau principle, 23, 46

See

Aureomycin, 976 Automated peptide synthesis, 1108-1110 Autoxidation, 493-494, 509 Axial bonds, of cyclohexane, 167 B

B chains, 1102-1103 Back-bonding, G-12 Bacterial dehalogenation of a PCB deriva­ tive, 983 Baker, B. R., 1139 Baker, J. T., 530 Ball-and-stick models, 41 Balzani, V., 166 Barger, G., 707 Barton, Derek H. R., 167 Base-catalyzed hemiacetal formation, 746 Base peak, 426 Base-promoted halogenation, of aldehydes and ketones, 837 Bases: acid-base reactions, 115-118 Br0 nsted-Lowry, 101-102 conjugate base, 101 heterocyclic bases, 1132-1133 Lewis, 102-104, 106 Mannich bases, 894 in nonaqueous solutions, 128-130 used in dehydrohalogenation of alkyl halides, 269 Basic hydrolysis of a nitrile, 810-811 Basic principles, applications of, 45-46, 92, 131, 181 Basicity: nucleophilicity vs., 258 and polarizability, 275 Bask peak, 426 Bathorhodopsin, 609 Beer’s law, 608 Benedict’s reagents, 1016-1017, 1042 Benzaldehyde, 731 Benzene, 56-57, 632, 650 discovery of, 633 halogenation of, 680-681 Kekule structure for, 638-639 meta-disubstituted, 663 modern theories of the structure of, 640-643 molecular orbital explanation of the structure of, 641-643 monosubstituted, 663 nitration of, 681-682 nomenclature of benzene derivatives, 634-636 ortho-disubstituted, 663 para-disubstituted, 663 reactions of, 637 resonance explanation of the structure of, 640-641 sulfonation of, 682-683 thermodynamic stability of, 639-640 Benzene ring, 632 preparation of carboxylic acids by oxi­ dation of, 790 Benzene substitution, 636 Benzenoid, 651-654

I-5

In d e x

Benzenoid aromatic compounds, 651-654 Benzoic acid, 70-71 Benzyl, 636 Benzyl alcohol, 506 Benzyl chloroformates, 813 Benzyl group, 64 Benzylic carbocations, 718 Benzylic cations, 708-709 Benzylic halides, 256 Benzylic halogenation, 710 Benzylic hydrogen atoms, 708 Benzylic radicals, 708-711 halogenation of the side chain, 709-710 Benzylic substituent, 708 Benzyne, 984-987 Benzyne elimination-addition mechanism, 985 Berg, Paul, 1155 Bergman cycloaromatization, 894 Bergman, R. G., 894 Bernal, J. D., 1067fn Bertrand, J. A., 1120 Beryllium hydride, linear geometry of, 40 Beta (b) carbon atom, 268 b eliminations, 268 b hydrogen atom, 268 b-anomer, 1008 b bends, 1113 b-carotene, 865 b configuration, 1 1 1 2 b-dicarbonyl compounds: by acylation of ketone enolates, 875-876 enolates of, 844-845 b-dicarboxylic acids, 816 b-keto acids, 815 b-pleated sheets, 1 1 1 0 , 1 1 1 2 b substituents, 1065 Bhopal, India, methyl isocyanate accident, 814 BHT (butylated hydroxytoluene), 494 Bicyclic alkanes, 175 Bicycloalkanes, 150 Bijvoet, J. M., 222 Bimolecular reaction, 238 BINAP ligands, 210, 224 Biochemistry, aromatic compounds in, 657-660 Biological methylation, 266 Biologically active natural products, 357 Biologically important amines, 922-924 2-phenylethylamines, 922 antihistamines, 922-923 neurotransmitters, 923-924 tranquilizers, 923-924 vitamins, 922-923 Biomolecules, mass spectrometry (MS) of, 443 Biphenyl, 724 Birch, A. J., 719 Birch reduction, 719-720 Bloch, Felix, 386 Boat conformation, 164-165 Boduszek, B., 1120 Boekelheide, V., 653 Boiling points, 78-79

of ionic compounds, 74-75 Bombardier beetle, 979 Bond breaking, as endothermic process, 462 Bond dissociation energies, 461-465 Bond length, 24 Bond-line formula, 41, 43-45 Bonding molecular orbital, 24, 27, 31-32, 37 Bonding pairs, 38 Born, Max, 21 Borneol, 502 Boron trifluoride: dipole moment, 61 trigonal planar structure of, 39-40 Bovine chymotrypsinogen, 1104 Bovine ribonuclease, 1103 Bovine trypsinogen, 1104 Boyer, Paul D., 539 Bradsher, C. K., 618 Bradsher reaction, 721 Branched alkyl groups, nomenclature of, 145-146 Branched-chain alkanes, nomenclature of, 143-145 Bromides, 264 Bromine, 467 addition to and trans-2-butene, 359 electrophilic addition of bromine to alkenes, 354-356 reaction with methane, 475 selectivity of, 479-480 2-Bromobutane, 373 Bromoform, 838 Bromohydrin, 360 Bromonium ion, 355 Br0 nsted-Lowry acids and bases, 101-102, 312 strength of, 109-113 acidity and pKa, 1 1 0 - 1 1 1 acidity constant (Ka), 109-110 predicting the strength of bases, 1 1 2 Br0 nsted-Lowry theory, 103 Brown, Herbert C., 347 Buckminsterfullerene, 137, 176, 654 1,3-Butadiene, 600-602 bond lengths o, 600-601 conformations of, 601 molecular orbitals of, 601-602 Butane, 140, 154 conformational analysis of, 160-162 Butanoic acid, 780 Butanone, synthesis of 2-butanol by the nickel-catalyzed hydrogenation of, 208 Butenandt, Adolf, 1068 Butlerov, Alexander M., 5 Butyl alcohol, 506

cis-

C

13C NMR (carbon-13) NMR Spectroscopy, 417-422 broadband (BB) proton decoupled, 417 chemical shifts, 418-420 DEPT 13C NMR spectra, 420-422 interpretation of, 417

off-resonance decoupling, 420 one peak for each magnetically distinct carbon atom, 417-418 C-terminal residues, 1094, 1099 Cahn, R. S., 197 Cahn-Ingold-Prelog system of naming enantiomers, 196-201, 225, 286 Calicheamicin g1l, 492 Calicheamicin g1l activation for cleavage of DNA, 894 Camphene, 305 Cannizzaro reaction, 864 Cantharidin, 827 Capillary electrophoresis, 1156 Capillin, 55 Carbaldehyde, 730 Carbamates (urethanes), 812-813 Carbanions, 3, 104-106, 550, 986 Carbenes, 361-362 Carbenoids, 362 Carbocations, 104-106, 248-251 relative stabilities of, 249-251 structure of, 249 Carbohydrates, 1000-1049, Disaccharides; Monosaccharides; Polysaccharides amino sugars, 1039 carbohydrate antibiotics, 1042 classification of, 1 0 0 1 - 1 0 0 2 defined, 1 0 0 1 disaccharides, 1001, 1029-1033 Fischer’s proof of the configuration of D-(+)-glucose, 1027-1029 glycolipids and glycoproteins of the cell surface, 1040-1041 glycoside formation, 1010-1013 glycosylamines, 1038-1039 as a major chemical repository for solar energy, 1 0 0 2 monosaccharides, 1001, 1004-1010 mutarotation, 1009-1010 oligosaccharides, 1 0 0 1 photosynthesis and carbohydrate metab­ olism, 1002-1004 polysaccharides, 1001, 1033-1037 summary of reactions of, 1042 trisaccharides, 1 0 0 1 Carbolic acids, Phenols Carbon, 10 chemistry of, 176 Carbon compounds, 1 alkyl halides (haloalkanes), 64-70 amides, 72 carboxylic acids, 70-71 esters, 71 families of, 53-97 functional groups, 62-64 hydrocarbons, 54-57 nitriles, 72 polar and nonpolar molecules, 60-62 polar covalent bonds, 57-59 Carbon dating, 4 Carbon dioxide, 40-41 Carbon tetrachloride, 232 Carbon-carbon double bond, 30 Carbonic acid, derivatives of, 812-814

Seealso

See

I-6

In d e x

Carbonic anhydrase, 1115, 1118-1119 Carbon-oxygen double bond: nucleophilic addition of ketones to, 741-744 reversibility of nucleophilic additions to, 743 Carbonyl compounds, 831-868 acetoacetic ester synthesis, 845-850 acidity of the hydrogens of, 832-833 active hydrogen compounds, 853-854 alcohols by reduction of, 552-561 alcohols from, 548-584 condensation and conjugate addition reactions of, 869-910 defined, 549 enamines, synthesis of, 854-858 haloform reaction, 838-839 halogenation at the carbon, 834-836 Hell-Volhard-Zelinski (HVZ) reaction, 839-841 lithium enolates, 841-844 oxidation and reduction of, 550 racemization via enols and enolates, 834-836 reactions at the carbon of, 831-868 reactions of Grignard reagents with, 565-566 reactions with nucleophile, 550 substituted acetic acids, synthesis of, 850-853 Carbonyl dichloride, 812 Carbonyl functional groups, 548 infrared (IR) spectra of, 89-90 Carbonyl groups, 69-70 nucleophilic addition to, 729-778 stereoselective reductions of, 555-556 Carboxyl group, 70 activation of, 1107 Carboxyl radicals, decarboxylation of, 816 Carboxylate anion, 793 Carboxylate salts, 781 Carboxylic acid anhydrides, 796-797 reactions of, 797 synthesis of, 796 Carboxylic acid derivatives, 780 Carboxylic acids, 70-71, 779-830 -halo, 839-841 acidity of, 121-125, 781-782 acyl chlorides, 785 acyl compounds: chemical tests for, 816 spectroscopic properties of, 787-789 acyl substitution, 792-794 amides, 785-786, 804-809 carboxylic anhydrides, 785 decarboxylation of, 814-816 dicarboxylic acids, 783-784 esterification, 797-802 esters, 784-785 infrared (IR) spectra of, 90-91 lactams, 811 lactones, 802-804 nitriles, 786-787 nomenclature, 780-781 nucleophilic addition-elimination at the acyl carbon, 792-794

a

a

a

a

oxidation of primary alcohols tooxidation of primary alcohols to, 557-558 physical properties, 780-781 polyamides, 817-818 polyesters, 817-818 preparation of, 789-792 reactions of, 818-819 Carboxylic anhydrides, 785 Carboxypeptidase A, 1115 Carboxypeptidases, 1099 Carcinogenic compounds, 660 Carcinogens, and epoxides, 533-534 Carotenes, 1063-1064 Carrier ionophore, 539 Carvone, 194, 205 Catalytic antibodies, 1123-1124 Catalytic asymmetric dihydroxylation, 365 Catalytic cracking, 139 Catalytic hydrogrenation, 314 Catalytic triad, 1120 Catenanes, 166, 167 Cation-exchange resins, 1095 Cationic oxygen atom, 969 Celera Genomics Company, 1162 Cellobiose, 1031-1032 Cellulose, 1036-1037 Cellulose derivatives, 1037 Cellulose trinitrate, 1037 Cephalins, 1075 Chain-growth polymers, 486, 817 Chain-initiating step, in fluorination, 475 Chain-propagating steps, 475-476 Chain reaction, 469 Chain-terminating (dideoxynucleotide) method, 1155 Chair conformation, 163-164 Chair conformational structures, drawing, 168 Chargaff, Erwin, 1141 Chauvin, Yves, G- 6 Chemical Abstracts Service (CAS), 142, 175 Chemical bonds, 7-9 Chemical energy, 119, 120 Chemical exchange, 415 Chemical shift, 387-388, 400-401 parts per million (ppm) and the S scale, 401 Chemotherapy, 944-945 Chiral drugs, 209-211 Chiral molecules, 190 Fischer projections, 215-216 not possessing chirality center, 224 racemic forms (racemic mixture), 207-208 stereoselective reactions, 208-209, 374 synthesis of, 207-211 Chirality: biological significance of, 187-188, 194-196 importance of, 188 in molecules, 187-188 and stereochemistry, 186-188 testing for, 195-196 Chirality centers, 191-193, 225

compounds other than carbon with, 224 molecules with multiple, 211-215 meso compounds, 213-214 naming compounds with, 214-215 proceeding with retention of configura­ tion, 219-222 Chitin, 1039 Chloracne, 990 Chloral hydrate, 746 Chlordiazepoxide, 923 Chloride ion, 793, 794 Chlorination: of alkanes, 467, 477-479 of methane: activation energies, 472-475 energy changes, 470-480 mechanism of reaction, 467-470 overall free-energy change, 471-472 reaction of methane with other halo­ gens, 475-476 Chlorine, 479 electrophilic addition of bromine to alkenes, 354-355 reaction with methane, 475 Chlorine selectivity, lack of, 467-468 Chlorobenzene, 980 electrophilic substitutions of (table), 695 Chloroethane, 402-403 physical properties, 75 Chloroethene, 404 Chloroform, 232, 838 dipole moment, 62 in drinking water, 839 Chlorohydrin, 360 Chloromethane molecule, dipole moment, 61 Chloromethane, physical properties, 75 Chloromethylation, 829 Chloroplasts, 1002 Chlorpheniramine, 923 Cholesterol, 211 Cholic acid, 1071 Choline, 266 Cholinergic synapses, 923 Chromate ester, formation of, 559 Chromate oxidations, mechanism of, 558-560 Chromatography using chiral media, 223 Chromic oxide, 560 Chylomicrons, 1066 Chymotrypsin, 1116, 1120-1121, 1130 Chymotrypsinogen, 1120 Cialis, 459 Circulation of p electrons, shielding/deshielding by, 400 Cis, 286 -1-chloro-3-methylcyclopentane, 244 Cis-trans isomers, of cycloalkanes, 189 Cis-trans isomerism, 171-174 cis 1,2-disubstituted cyclohexanes, 174 cis 1,3-disubstituted cyclohexanes, 174 cis 1,4-disubstituted cyclohexanes, 172-173 and conformational structures of cyclohexanes, 171-174

cis

In d e x

trans 1,2-disubstituted cyclohexanes, 174

trans 1,3-disubstituted cyclohexanes, 173-174 trans 1,4-disubstituted cyclohexanes, 171-172 Cis-trans isomers, 33-34 physical properties, 62 Claisen condensation: crossed, 874-875 defined, 870 examples of, 870-871 intramolecular, 873 mechanism for, 871-872 synthesis of b-keto esters, 870-875 Claisen rearrangement, 977-978 Claisen-Schmidt condensations, 883 Cleavage: of ethers, 527-528 with hot basic potassium permanganate, 365-366 with ozone, 366-368 Clemmensen reduction, 690-691 Codon, 1150 Coenzymes, 1116 Coenzymes Q (CoQ), 978-979 Cofactor, 1116 Coil conformations, 1110, 1113 Collagen, 83 Collision-induced dissociation, CID), 1100 Combination bands, 8 6 Combustion of alkanes, 491-492 Competitive inhibitor, 1116 Complete sequence analysis, 1099-1100 Compounds, 2-3 Concept maps, 47, 52 Concerted reaction, 240 Condensation reactions, 817, 870 Condensations: acid-catalyzed aldol, 880-881 aldol, 870, 877, 880-889 Claisen, 870-875 Claisen-Schmidt, 883 crossed aldol, 882-888 crossed Claisen, 874-875 Darzens, 906 Dieckmann, 873 intramolecular Claisen, 873 Condensed structural formulas, 42-43 Configurations: inversion of, 265 (R) and (S), 197 relative and absolute, 2 2 0 - 2 2 1 Conformational analysis, 138, 157 of butane, 160-162 hyperconjugation, 158-159 of methylcyclohexane, 168-170 performing, 158-160 Conformational stereoisomers, 161, 219 Conformations, 157 eclipsed, 158 staggered, 158 Conformer, 157 Conjugate acid, 101 Conjugate acid-base strengths, summary and comparison of, 124

Conjugate addition, 889 reactions, 870, 882 Conjugate base, 101 Conjugated dienes: electrophilic attack on, 612-616 stability of, 602-604 Conjugated double bonds, 599-600 Conjugated proteins, 1122 Conjugated unsaturated systems, 585-631 alkadienes, 599-600 allyl cation, 594-595 allyl radical, 586-590 allylic substitution, 586-590 1,3-Butadiene/z0, 600-602 conjugated dienes: electrophilic attack on, 612-616 stability of, 602-604 defined, 585 Diels-Alder reaction, 616-624 electron delocalization, 600-602 polyunsaturated hydrocarbons, 599-600 resonance theory, 595-599 ultraviolet-visible spectroscopy, 604-612 Connectivity, 42 Constitutional isomers, 6 , 140, 188, 189 Constructive interference, 20 Cope elimination, 950 Cope rearrangement, 978 Corey, E. J., 319, 530, 620, 674, 844 Corey, Robert B., 1111 Corey-Posner, Whitesides-House reac­ tion, G-1, G- 8 Corpus luteum, 1069 Corrin ring, G-17-G-18 Coulson, C. A., 644 Couper, Archibald Scott, 5 Coupling constants, 406, 410-412 dependence on dihedral angle, 411-412 reciprocity of, 410 Coupling reactions, of arenediazonium salts, 941-942 Coupling (signal splitting), 390-392 Covalent bonds, 7 formation of, 8 homolysis and heterolysis of, 100, 460 and Lewis structures, 8-9 multiple, 9 polar, 57-59 and potential energy (PE), 120 Cracking, 139 Cram, Donald J., 538, 987 Cresols, 965 Crick, Francis, 1140-1142, 1146 Crixivan, 317 Cross peaks, 424 Crossed aldol condensations, 882-888 using strong bases, 8 8 6 - 8 8 8 using weak bases, 883-886 Crossed aldol reaction, 882 Crossed Claisen condensation, 874-875 Crown ethers, 537-539 defined, 537 as phase transfer catalysts, 538 and transport antibiotics, 539 Crutzen, P. J., 495

I-7

Cumulated double bonds, 599-600 Cumulenes, 599 Curl, R. F., 654-655 Curtius rearrangement, preparation of amines through, 932 alkylation of ammonia, 924-925 alkylation of azide ion and reduction, 925 alkylation of tertiary amines, 927 Gabriel synthesis, 925-926 Curved arrows, 16 illustrating reactions with, 106-107 Cyanohydrins, 755-756 preparation of carboxylic acids by hydrolysis of, 790, 790-791 Cycles per second (cps), 605 3’, 5’-Cyclic adenylic acid (cyclic AMP), 1136 Cyclic anhydrides, 796 Cyclic compounds, stereoisomerism of, 217-219 Cyclizations, via aldol condensations, 888-889 Cycloalkanes: angle strain, 162 cis-trans isomers of, 189 disubstituted, 171-174 higher, conformations of, 167 naming, 149-151 nomenclature of, 151-153 physical properties, 154-156 ring strain, 162-163 synthesis of, 177-178 torsional strain, 162 Cycloalkenes, 290 retro-Diels-Alder reaction, 435 Cycloalkyalkanes, 150 Cyclobutadiene, 57, 650 Cyclobutane, 163 Cyclocymopol enantiomers, 357 Cycloheptane, 167 Cycloheptatriene, 648 Cyclohexane, 181 conformations of, 163-165 substituted, 167-168 Cyclohexane derivatives, 217-219 1.2-dimethylcyclohexanes, 218-219 1.3-dimethylcyclohexanes, 218 1.4-dimethylcyclohexanes, 217-218 Cyclohexanol, 506 Cyclohexene, 362, 639 Cyclononane, 167 Cyclooctadecane, 167 Cyclooctane, 167 Cyclooctatetraene, 644, 651 Cyclooxygenase, 1074 Cyclopentadiene, 619, 647 Cyclopentadienyl anion, 651 Cyclopentane, 163, 217-218 Cyclopentanol, 506 Cyclopropane, 162-163 Cysteine, 1087-1088 Cytosine methylation, 1164 Cytosine-guanine base pair, 1131

I-8

In d e x

D

D-Glucaric acid, 1017 D-Glucosamine, 1039 d-Tubocurarine, 924 D vitamins, 1070-1071 Dacron, 817 Dactylyne, 55, 331, 357 D’Amico, Derin C., 865 Darvon (dextropropoxyphene), 186, 209 Darzens condensation, 906 Dash structural formulas, 41-42, 47 Daunomycin, 975-976 Daunorubicin, 211 de Broglie, Louis, 21 Debye, Peter J.W., 58 Deactivating groups, 691, 695 Deacylases, 1093-1094 Decalin, 175 Decarboxylation, 754 of carboxylic acids, 814-816 Deconvolution, 1125 Decyl alcohol, 81 Degenerate orbitals, 23 Degrees of freedom, 471 Dehydration, 744 of alcohols, 291, 297-303, 342 carbocation stability and the transi­ tion state, 300-302 of aldol addition product, 879 defined, 297 intermolecular, 522-523 of primary alcohols, 297-298 mechanism for, 302 rearrangement after, 306-307 of secondary alcohols, 297-298, 300 mechanism for, 299-302 rearrangements during, 303-305 of tertiary alcohols, 297-298, 300 mechanism for, 299-302 Dehydrobenzene, Benyzne Dehydrohalogen, ation of v/c-dibromides to form alkynes, 309-310 Dehydrohalogenation, 99, 268-269 of alkyl halides, 268-269, 291-297 bases used in, 269 defined, 268 favoring an E2 mechanism, 291-292 less substituted alkene, formation of, using bulky base, 294-295 mechanism for, 269, 296-297 orientation of groups in the transition state, 295-296 Zaitsev rule, 292-294 bases used in, 269 defined, 268 mechanisms of, 269 Deinsertion, G-12 Delocalization, and acidity of carboxylic acids, 122-123 Delocalization effect, 123 Delocalization: of a charge, 301 of electrons, 57 Deoxy sugars, 1038 Deoxyribonucleic acid (DNA): defined, 1132

See

determining the base sequence of, 1155-1157 DNA sequencing, 1100-1101, 1132 by the chain-terminating method, 1155-1157 heterocyclic bases, 1132-1133 microchips, 1162 primary structure of, 1139-1140 replication of, 1144-1146 secondary structure of, 1140-1144 DEPT spectra, 661 DEPT 13C NMR spectra, 420-422 Deshielding, protons, 399-400 Designer catalysts, 1084 Deuterium atoms, 4 Dextrorotary, use of term, 203 Diacyl peroxide, 487 Dialkyl carbonate, 812 Dialkyl ethers, 527 Dialkylation, 845-846 Dialkylcarbodiimides, 806 Diamond, 176 Diamox, 1119 Dianeackerone, 770 Diastereomers, 189, 482, 920 Diastereoselective reactions, 208, 556 Diatomic molecules, 60 1,3-Diaxial interaction, 169 of a group, 170-171 Diazo coupling reaction, 941 Diazonium salts, 935 syntheses using, 938 Diazotization, 935-936 deamination by, 939-940 Diborane, 347 Dibromobenzenes, 634 Dibromopentane, 211-212 Dibutyl ether, 507 Dicarboxylic acids, 783-784 Dicyclohexano-18-crown-6, 538 Dicyclohexylcarbodiimide, 806, 1107 Dieckmann condensation, 873 Dielectric constants, 261 Diels, Otto, 616, 620 Diels-Alder reaction, 586, 616-624, 978, 1084 factors favoring, 618 molecular orbital considerations favor­ ing an endo transition state, 621-622 stereochemistry of, 618-620 Diene, 617 Dienophile, 617 Diethyl ether, 507, 509, 522 physical properties, 75 Difference bands, 8 6 Digitoxigenin, 1071 Dihalocarbenes, 362 Dihedral angle, 158, 411 1,2-Dihydroxylation, 363 Diisobutylaluminum hydride (DIBAL-H), 735 Diisopropyl ether, 507 Diisopropylcarbodiimide, 806, 1107 Diisopropylphosphofluoridate (DIPF),

tert-butyl

1122

1,2-Dimethoxyethane (DME), 507 Dimethoxytrityl (DMTr) group, 1158 Dimethyl ether, 67, 507 intermolecular forces, 76 Dimethylbenzenes, 635 Dimethylcyclohexane, 219 2.4-Dinitrofluorobenzene, 1099 2.4-Dinitrophenylhydrazones, 752, 761 Diols, 149 Diosgenin, 1071 1.4-Dioxane, 504-505, 507 Dipeptides, 1094 Dipolar ions: amino acids as, 1089-1092 defined, 1089 Dipole, 58 Dipole moments, 58, 60, 92 in alkenes, 62 permanent, 75 Dipole-dipole forces, 75-76 Dipropyl ether, 507 Diprotic acid, 102 Dirac, Paul, 20 Direct alkylation: of esters, 843 of ketones, via lithium enolates, 842-843 Directed aldol reactions, 883 and lithium enolates, 8 8 6 - 8 8 8 Directive effect, 701 Disaccharides, 1001, 1029-1033 artificial sweeteners, 1032-1033 cellobiose, 1031-1032 lactose, 1033 maltose, 1001-1002, 1030-1031 sucrose, 1001-1002, 1029-1030 Dispersion forces, 77-78, 92, 160-161 Dispersive IR spectrometers, 84 Dissolving metal reduction, 316 Disubstituted benzenes, orientation in, 716-717 Disubstituted cycloalkanes, 171-174 Divalent carbon compounds, 361-362 Divalent, use of term, 5 DL-amino acids, resolution of, 1093-1094 Deoxyribonucleic acid (DNA) DNA, DNA sequence, 1100-1101, 1132 Doisy, Edward, 1068 Dopamine, 6 8 , 922 Dot structure, 41 Double-headed arrows (4 ), 17 Doublets, 392 Downfield, use of term, 388 Doxorubicin, 975-976

See

E

E1 reactions, 269, 271-273 mechanism for, 272-273 N1 reactions vs., 275 E2 elimination, 311 E2 reactions, 269-271 mechanism for, 270-271 Sn2 reactions vs., 273-275 Eclipsed conformations, 158 Edman degradation, 1097-1098 Edman, Pehr, 1097

S

I-9

In d e x

Eisner, T., 979 Electromagnetic spectrum, 604-606 Electron deficient, 464 Electron delocalization, 600-602 Electron density surfaces, 29 Electron-donating substituents, 693 Electron impact (EI) ionization, 427, 440-441 Electron probability density, 21 Electron-withdrawing effect of a phenyl group, 917 Electron-withdrawing substituents, 693 Electronegative groups, deshielding, 400 Electronegativity, 8 , 47 Electronegativity differences polarize bonds (principle), 131 Electronic spectra, 608 Electrons, 4 configurations, 22-23 delocalization of electrons, 57 energy of, 37 sharing, 8-9 Electrophiles, 105-106, 312, 357 as Lewis acids, 333 Electrophilic addition: of bromine and chlorine to alkenes, 354-355 defined, 333 of hydrogen halides to alkenes, 334-339 Electrophilic aromatic substitution: effect of substituents on, 697-706 electron-releasing and electron-with­ drawing groups, 697-698 inductive and resonance effects, 698-699 meta-directing groups, 700-701 ortho-para-directing groups, 701-705 ortho-para direction and reactivity of alkylbenzenes, 705-706 table, 696 and thyroxine biosynthesis, 707 Electrophilic aromatic substitutions, 973 Electrospray ionization (ESI), 440, 441 mass spectrometry (MS) with (ESIMS), 1126 with quadrupole mass analysis, 443 Electrostatic potential, 104 map, 17 Elemental carbon, chemistry of, 176 Elements, 2-4 defined, 4 electronegativities of, 7 periodic table of, 3 Eleutherobin, 357 Elimination reactions, 269, 274, 744 of alkyl halides, 268-269 defined, 291 synthesis of alkenes by, 291 synthesis of alkynes by, 308-310 Eliminations, 99 b eliminations, 268 1 ,2 eliminations, 268 Elion, Gertrude, 1139 Enal, 879

Enamines, 751, 754-755 synthesis of, 854-857 Enantiomeric excess, 207 Enantiomerically pure, use of term, 207 Enantiomerism, 223 Enantiomers, 188, 190 naming, 196-201 optical activity, 201-205 origin of, 205-207 plane-polarized light, 2 0 2 polarimeter, 203 specific rotation, 204-205 Pasteur’s method for separating, 223 properties of, 2 0 1 - 2 0 2 resolution, methods for, 223 selective binding of drug enantiomers to left- and right-handed coiled DNA, 211

separation of, 223 Enantioselective reactions, 208, 555 Endergonic reactions, 240 Endothermic reactions, 120, 461-462 Energies of activation, 472-475 Energy, 20 defined, 119 Energy state, 20 Enol form, 833 Enol tautomers, 833-834 Enolate anions, 832-833 Enolate chemistry, summary of, 857-858 Enolates: of b-dicarbonyl compounds, 844-845 conjugate addition of, 892-893 defined, 832 racemization via, 834-836 reactions via, 834-841 regioselective formation of, 842 Enols (alkene alcohols), 832-833 racemization via, 834-836 reactions via, 834-841 Enone, 879 Enthalpies, 120 Enthalpy change, 120 Environmentally friendly alkene oxidation methods, 537 Enzyme-substrate complex, 1115 Enzymes, 208-209, 521, 1115 resolution by, 223 Epichlorohydrin (1-(chloromethyl)oxirane), 533 Epimerization, 836 Epimers, 836, 1023 Epoxidation: alkene epoxidation, 529 Sharpless asymmetric epoxidation, 529-530 stereochemistry of, 530 Epoxides, 528-535 acid-catalyzed ring opening of, 531 anti 1 ,2 -dihydroxylation of alkenes via, 535 base-catalyzed ring opening of, 531-532 carcinogens and biological oxidation, 533-534 defined, 528

epoxidation, 528-530 polyethers from, 534-535 reactions of, 531-535 synthesis of, 528-529 Equatorial bonds, of cyclohexane, 167 Equilibrium, 17 Equilibrium constant (Keq), 109 Erythromycin, 976 Eschenmoser, A., 318, 620, G-17 Essential amino acids, 1088 Essential nutrients, 946 Essential oils, 1061 Esterifications, 797-802 Fischer, 797 transesterification, 800 Esters, 71, 784-785 from acyl chlorides, 799 aldehydes by reduction of, 734-737 amides from, 806 from carboxylic acid anhydrides, 799-800 direct alkylation of esters, 843 esterification, 797-802 acid-catalyzed, 798-800 reactions of, 820 saponification, 800-802 synthesis of, 797-802 Estradiol, 966, 1068, 1069 Estrogens, 1068 synthetic, 1069 Ethanal, 729 Ethane: bond length, 36 carbon-carbon bond of, 57 physical properties, 75 radical halogenation of, 477 sp2 hybridization, 30 structure of, 28-29 Ethanoic acid, 142, 780 Ethanol, 502, 506, 507-508 as a biofuel, 508 Ethanoyl group, 731 Ethene, 55 anionic polymerization of, 489 physical properties, 75 radical polymerization of, 487-488 Ethene (ethylene), 30 bond length, 36 Ethers, 67, Epoxides alkyl aryl, cleavage of, 973 boiling points, 505 cleavage of, 527-528 crown, 537-539 cyclic, naming, 504 dialkyl, 527 diethyl, 75, 507, 509, 522 diisopropyl, 507 dimethyl, 67, 76, 507 dipropyl, 507 as general anesthetics, 67 hydrogen bonding, 506 by intermolecular dehydration of alco­ hols, 522-523 nomenclature, 503-504 oxygen atom, 503 physical properties of, 505-507

Seealso

I-10

In d e x

polybromodiphenyl (PBDEs), 991 protecting groups, 525-526 silyl, 526 reactions of, 527-528 synthesis of, 522-526 by alkoxymercuration-demercuration, 525 synthesis/reactions of, 502-547 by alkylation of alcohols, 525 Williamson synthesis of, 523-524, 1014 Ethinyl estradiol, 55 Ethyl acetate, physical properties, 75 Ethyl alcohol, 65 physical properties, 75 Ethyl bromide, 372 Ethyl group, 63 Ethyl methyl ether, 507 Ethylamine, physical properties, 75 Ethylbenzene, 706, 709 Ethylene, 508 polymerization of, 487 Ethylene oxide, 505 Ethyllithium, 129 Ethyne: bond length, 36 physical properties, 75 hybridization, 34-35 structure of, 34-35 Ethyne (acetylene), 34, 55 Ethynylestradiol, 1069 Eucalyptol, 502 Eugenol, 966 Exchangeable protons, 415 Excited states, 25 Exergonic reactions, 240 Exhaustive methylation, 1014 Exons, 1147 Exothermic reactions, 120, 461 Extremozymes, 556 (E)-(Z) system for designating, 286-287

tert-butyl,

sp

F

Faraday, Michael, 633 Farnesene, 378 Fat substitutes, 1055-1056 Fats, 1052, 1054 Fatty acids, 71, 313, 1052-1053, 1060-1073 composition, 1054 omega-3, 1052-1054 reactions of the carboxyl group of, 1059 saturated, 1052 unsaturated, 1052 reactions of the alkenyl chain of, 1059 Fehling’s solution, 1016 Fibrous tertiary structures, 1114 First-order spectra, 415 Fischer, Emil, 1005-1006, 1027-1029, 1115 Fischer esterifications, 797 Fischer projections, 1006 defined, 215-216

drawing/using, 216-217 Fleet, G.W. J., 1048 Fleming, Alexander, 1116-1117 Floss, H., 1048 Fluoride anion, 259 Fluorination, chain-initiating step in, 475 Fluorine, 681 electronegativity of, 8 reaction with methane, 475 Fluorobenzene, 681 Fluorocarbons, boiling point, 78 Folic acid, 946 Formal charges, 47 calculating, 13-14 summary of, 15 Formaldehyde, 18, 730 bond angles, 70 Formic acid, 70-71, 780 Formyl group, 731 Fourier transform, 398 Fourier transform infrared (FTIR) spec­ trometer, 84 Fourier transform NMR spectrometers, 397-398 Franklin, Rosalind, 1140 Free-energy change: for the reaction, 240 relationship between the equilibrium constant and, 1 2 0 - 1 2 1 Free energy of activation, 240 Free induction decay (FID), 398 Free radicals, Radicals Freons, 495 Frequency (n), 604 Frequency of radiation, 84 Friedel, Charles, 684 Fructose, 1001 Fructosides, 1010 Fullerenes, 176, 654-655 Fumaric acid, 380 Functional group interconversion (func­ tional group transformation), 262-264 Functional groups, 59, 62-64 defined, 62 Furan, 657 Furanose, 1008

See

G

Gabriel synthesis of amines, 925-926, 1092 Galactan, 1033 Gamma globulin, 1104 Gas chromatography (GC), 386 Gates, M., 620 Gauche conformations, 160 Gauche interaction, 169 GC/MS (gas chromatography with mass spectrometry), 441, 442-443 analysis, 442-443 Gel electrophoresis, 1125 Gelb, M. H., 1104fn 746 Geminal dihalide (gem-dihalide), 310 Geminal diol ( -diol), 746 Genes:

gem-diols,

gem

defined, 1146 location of, for diseases on chromo­ some 19 (schematic map), 1133 Genetic code, 1150-1152 Genetics, basics of, 1132 Genomics, 1126-1128 Gentamicins, 1042 Geometric specificity, 1115 Geranial, 379 Gibbs free-energy change, 120fn Gibbs, J. Willard, 120fn Gilbert, G., 1155 Gilman reagents, G-1 using in coupling reactions, G-8 Globular tertiary structures, 1114 Glucan, 1033 Glucoside, 1010 Glutamic acid, 1088, 1113 Glutamine, 1088, 1113 Glutathione, 1101 Glycans, Polysaccharides: Glyceraldehyde-3-phosphate dehydroge­ nase (GAPDH), 659 Glyceraldehyde-3-phosphate (GAP), 831 Glycerol, 506 Glycidic ester, 906 Glycine, 1086 Glycogen, 1035-1036 Glycolipids, 1040-1041 Glycols, 149, 363 Glycolysis, 831 Glycoproteins, 1040-1041 Glycoside formation, 1010-1013 Glycosides, 1010 Glycosylamines, 1038-1039 Glycylvalylphenylalanine, 1095 Goodman, L., 158, 1139 Gramicidin, 539 Graphite, 176 Grignard reagents, 563 alcohols from, 566-568 Grignard synthesis, planning, 569-570 preparation of carboxylic acids by car­ bonation of, 791 reactions with carbonyl compounds, 565-566 reactions with epoxides (oxiranes), 565 restrictions on the use of, 572-573 Grignard, Victor, 563 Ground state, 26 Group numbers, atoms, 5 Grubbs’ catalysts, G-1, G-5-G-7 Grubbs, Robert, G-6

See

H

Half-chair conformations, 165 Halo alcohol, 359 a-Haloalcohols, 545 Haloalkanes, 64-67 Halocycloalkanes, 149 Haloform reaction, 839 Halogen addition, mechanism of, 355-358 Halogen atoms, 460 Halogen molecules, 460 Halogen substituents, 695 Halogens:

I-11

In d e x

compounds containing, 180 reactions of alkanes with, 465-467 Halohydrin: defined, 359 formation, 359-361 mechanism for, 360 Halomon, 357 Halonium ions, 356 Haloperoxidases, 357 Halothane, 509 Hammond-Leffler postulate, 256-257, 293, 301 Haptens, 1123 Harington, C., 707 Harpp, D. N., 840 Hassel, Odd, 167 Haworth formulas, 1008 Haworth, W. N., 1008fn HCN, conjugate addition of, 891 Heat contents, 120 Heat of hydrogenation, 288-289 Heats of reaction, 288-289 using homolytic bond dissociation ener­ gies to calculate, 462-463 Heck-Mizokori reaction, G-1, G-2, G16-G-17 Heisenberg uncertainty principle, 24 Heisenberg, Werner, 20 Hell-Volhard-Zelinski (HVZ) reaction, 839-841 Heme, 1122 Hemiacetals, 744-747, 1008, 1042 acid-catalyzed formation, 745 base-catalyzed formation, 746 Hemicarcerand, 987 Hemoglobin, 1122-1124 Henderson-Hasselbalch equation, 1090 Heparin, 1039 Heptyl alcohol, 506 Hertz, H. R., 605fn Hertz (Hz), 410, 605 Heteroatoms, 59 Heterocyclic amines, 913 basicity of, 918 Heterocyclic aromatic compounds, 655-657 Heterogeneous catalysis, 313 Heterolysis, 100, 131 Heterolytically, use of term, 299 Heteropolysaccharides, 1033 Heterotopic atoms, 402 Hexanoic acid, 780 Hexose, 1004 Hexyl alcohol, 506 High-density lipoproteins (HDLs), 1066 High-performance liquid chromatography (HPLC), 1096-1097 High-resolution mass spectrometry, 439-440 Highest occupied molecular orbital (HOMO), 105, 609 Hinsberg test, 943-944 Hirst, E. L., 1008fn Histamine, 923 Histidine, 1088 Histrionicotoxin, 911, 924

Hitchings, George, 1139 Hodgkin, Dorothy, G-17 Hofmann elimination, 920, 949-950 Hofmann product, 296 Hofmann rearrangement (Hofman degra­ dation), preparation of amines through, 931-932 Hofmann rule, 294, 950 Homogeneous asymmetric catalytic hydro­ genation, G-15-G-16 Homogeneous catalysis, 313 Homologous series, 155 Homologues, 155 Homolysis, 100, 460 Homolytic bond dissociation energies (DH°), 461-465 defined, 462 using to calculate heats of reaction, 462-463 using to determine the relative stabili­ ties of radicals, 463-465 Homopolysaccharides, 1033 Homotopic hydrogens, 401-402 Hooke’s law, 85 Horner-Wadsworth-Emmons reaction, 760-761, 771 Host-guest relationship, 538 Hot basic potassium permanganate, cleav­ age with, 365-366 Huckel’s rule, 643-651, 653, 667 annulenes, 644-646 aromatic ions, 647-648 diagramming the relative energies of orbitals in monocyclic conjugated systems based on, 643-644 NMR spectroscopy, 646-647 Huffman, D., 654 Hughes, Edward D., 238 Huheey, J. E., 200 Human Genome Project, 1157, 1162 Human genome, sequencing of, 1162 Human hemoglobin, 1103 Hund’s rule, 23, 46, 644 Hybrid atomic orbitals, 26, 37 Hybrid of resonance structures, 16 Hybridization, and acidity, 117-118 Hydrate formation, 746-747 Hydrating ions, 80 Hydration, of alkenes, 340-342, 353 Hydrazones, 752-753 Hydride ions, 550 Hydroboration: defined, 347, 353 mechanism of, 348-349 stereochemistry of, 349-350 synthesis of alkylboranes, 347-350 Hydroboration-oxidation: alcohols from alkenes through, 347 as regioselective reactions, 351-352 Hydrocarbons, 138, 286 IR spectra of, 87-89 Hydrogen, 8 anti addition of, 316-317 syn addition of, 315-316 Hydrogen abstraction, 460-461 Hydrogen atoms, classification of, 147

Hydrogen bonds, 75-76, 77fn, 92 formation of, 81-82 Hydrogen bromide, 101 anti-Markovnikov addition of, 484-486 Hydrogen chloride, 101 Hydrogen halides: addition to alkynes, 369-370 electrophilic addition to alkenes, 334-338 Hydrogenases, 209 Hydrogenation, 288-289 of alkenes, 178-179, 313-314, 332 of alkynes, 178-179, 315-317 in the food industry, 313 function of the catalyst, 314-315 Hydrogenolysis, 773 Hydrolysis, 209, 253 acetals, 747-748 of alkylboranes, 350-352 regiochemistry and stereochemistry, 351-352 of amides, 807-809 by enzymes, 808 Hydronium ion, 102 Hydrophilic, use of term, 81, 1057 Hydrophobic effect, 81 Hydrophobic, use of term, 81, 1057 Hydroxide ion, 102 Hydroxybenzene, 634 3-Hydroxybutanal, 876 Hydroxyl group, alcohols, 65 Hydroxyproline, 1096 4-Hydroxyproline, 1087 Hyperconjugation, 158-159, 249-250, 256 I

Ibuprofen, 209 Imines, 751-752 Index of hydrogen deficiency (IHD): defined, 178 gaining structural information from, 178-180 Indole system, 658 Induced field, 399 Induced fit, 1115 Inductive effects, 131, 608-609, 702 and acidity of carboxylic acids, 123 of other groups, 124-125 Industrial styrene synthesis, 709 Infrared (IR) spectra, 90-91 alcohols, 90 amines, 91 carbonyl functional groups, 89-90 carboxylic acids, 90-91 functional groups containing het­ eroatoms, 89-90 hydrocarbons, 87-89 phenols, 90 Infrared (IR) spectroscopy, 54, 83-87, 386 defined, 83 dispersive IR spectrometers, 84 Fourier transform infrared (FTIR) spec­ trometer, 84 interpreting IR spectra, 87-91 Infrared spectra, of substituted benzenes, 663-664

I-12

In d e x

Ingold, Sir Christopher, 238 Inhibitors, 1116 Initial ozonides, 367 Initial rates, 238 Insertion-deinsertion, G-12 Insulin, 1102-1103 Integration of signal areas, 390 Interferogram, 84 Intermediates, 98, 247 Intermolecular dehydration: of alcohols, ethers by, 522-523 complications of, 522-523 Intermolecular forces (van der Waals forces), 75-76 in biochemistry, 81-83 dipole-dipole forces, 75-76 dispersion forces, 77-78 hydrogen bonding, 76 organic templates engineered to mimic bone growth, 82-83 solubilities, 79-81 International Union of Pure and Applied Chemistry (IUPAC), 6 fn, 142 for naming alkanes, 142-145 Intramolecular Claisen condensation, 873 Introns, 1147 Inversion, 239, 754 Inversion of configuration, 265 Iodide, 259 Iodination, of methane, 476 Iodine, reaction with methane, 475 Iodomethane, 232 Ion sorting and detection, 442 Ion trap mass analyzers, 442 Ion-dipole forces, 80 Ionic bonds, 7-8 Ionic compounds, 8 ion-ion forces, 74-75 Ionic reactions, 100, 230-284, 460 carbocations, 248-251 relative stabilities of, 249-251 structure of, 249 E1 reaction, 271-273 E2 reaction, 269-271 free-energy diagrams, 240-243 leaving groups, 237 nucleophiles, 234-237 organic halides, 231-234 5^1 reaction, 246-248 mechanism for, 247-248 rate-determining step, 246-247 5^2 reaction, 237-240 measuring, 237-238 mechanism for, 238-240 stereochemistry of, 243-246 transition state, 239-243 temperature, reaction rate, and the equilibrium constant, 242-243 Ionization, 427 Ionophores, 539 Ions, 7 Ipatiew, W., 865 Iron(III) halides (ferric halides), 103 Isoborneol, 305 Isobutane, 140 Isobutyl alcohol, 506

Isobutylene, polymerization of, 489 Isoelectric focusing, 1127 Isoelectric point, 1089-1090 Isolable stereoisomers, 219 Isolated double bonds, 599-600 Isoleucine, 1087, 1113 Isomaltose, 1045 Isomers, 5-6 subdivision of, 189 Isooctane, 139 Isopentane, 140 Isoprene units, 1061-1062 Isopropyl alcohol, 42, 506 equivalent dash formulas for, 42 Isopropyl group, 63 Isopropylamine, 6 8 Isopropylbenzene, 706 Isotope-coded affinity tags (ICAT), 1127 Isotopes, 4 J

Jones reagent, 558 Joule (J), 120fn Jung, Michael E., 865 K

Kam, C. M., 1120 Kanamycins, 1042 Karplus correlation, 411 Karplus, Martin, 411 Katz, T., 667 Kekule, August, 5, 56, 638 Kekule structures, 56 for benzene, 638-639 Kekule-Couper-Butlerov theory of valence, 633 Ketene, 829 Keto form, 832-833 Keto tautomers, 833-834 Ketone enolates, b-dicarbonyl compounds by acylation of, 875-876 Ketones, 53-54, 69-70, 310 a,b-unsaturated, additions to, 889-894 acid-catalyzed halogenation of, 837 from alkenes, arenes, and 2 ° alcohols, 738-739 base-promoted halogenation of, 837 carbonyl group, 730 chemical analyses for, 761 derivatives of, 761 direct alkylation of, via lithium enolates, 842-843 IR spectra of, 762-763 mass spectra of, 764 from nitriles, 739-740 NMR spectra of, 763-764 nomenclature of, 730-732 nucleophilic addition to the carbon-oxygen double bond, 741-744 oxidation of secondary alcohols to, 558 in perfumes, 733 physical properties, 732-733 relative reactivity, 743 spectroscopic properties of, 762-764 summary of addition reactions, 765-766

synthesis of, 738-740, G-4 Tollens’ test (silver mirror test), 761 UV spectra, 764 Ketopentose, 1004 Ketose, 1004, 1016 Kharasch, M. S., 484 Kiliani-Fischer synthesis, 1023-1024, 1042 Kilocalorie of energy, 120 Kinetic control, 293 defined, 614 thermodynamic control of a chemical reaction vs., 614-616 Kinetic energy (KE), 119 Kinetic enolate, formation of, 842 Kinetic products, 614, 616 Kinetic resolution, 209 Kinetics, defined, 237 Knowles, William S., 210, 365, G-15 Kolbe reaction, 974-975 Kossel, W., 7 Kratschmer, W., 654 Kroto, H. W., 654-655 Kumepaloxane, 357 L

L-amino acids, 1086-1088 Lactams, 811 Lactones, 96, 802-804 Lactose, 1033 Ladder sequencing, 1100 Langmuir-Blodgett (LB) films, 1060 Laqueur, Ernest, 1068 (3£)-Laureatin, 331, 357 LCAO (linear combination of atomic orbitals) method, 25 Le Bel, J. A., 6-7, 1027 Leaving groups, 233, 236, 237 ionization of, 257 nature of, 262-264 Lecithins, 1075 Lehn, Jean-Marie, 538 Lerner, Richard A., 1084 Less substituted alkene: defined, 294 formation of, using bulky base, 294-295 Leucine, 1086, 1113 Leveling effect, 110, 308 Levitra, 459 Levorotatory, use of term, 203 Lewis acid-base reactions, 131 Lewis acid-base theory, 102-103 Lewis acids, 102-104 as electrophiles, 105, 333 Lewis bases, 102-104 as nucleophiles, 106 Lewis, G. N., 7, 102-104 Lewis structures, 18, 47 and covalent bonds, 8-9 defined, 9 rules for writing, 9-10 Ligands, G-8-G-9 BINAP, 210, 224 ligand exchange, G-11 in transition metal complexes, G-10

In d e x

Like charges repel (principle), 46, 181 Limonene, 194, 379 Linalool, 863 Lindlar’s catalyst, 316 Linear polymers, 1094 Linoleic acid, 493 Lipase, 209 Lipid bilayers, 1075 Lipids, 82, 1050-1083 defined, 1050-1051 fatty acids, 71, 313, 1052-1053, 1060-1073 glycolipids, 1078 in materials science and bioengineering, 1060 phosphatides, 1074-1077 phospholipids, 1074-1077 prostaglandins, 1073-1074 sphingosine, derivatives of, 1077-1078 steroids, 1064-1073 terpenes, 1061-1062 terpenoids, 1061-1062 triacylglycerols, 1052-1058 waxes, 1078 Lithium aluminum hydride, 553 overall summary of, 554-555 Lithium dialkyl cuprate reagents, G-1 using in coupling reactions, G-8 Lithium diisopropylamide (LDA), 841, 875 Lithium, electronegativity of, 8 Lithium enolates, 841-844 direct alkylation of ketones via, 842-843 and directed aldol reactions, 8 8 6 - 8 8 8 regioselective formation of enolates, 842 Lithium tri-terf-butoxyaluminum hydride, 735 Lobry de Bruyn-Alberda van Ekenstein transformation, 1013 Lock-and-key hypothesis, 1115 London forces, Dispersion forces Lone pairs, 38 Loop conformations, 1110, 1113 Loschmidt, Johann Josef, 638fn Low-density lipoproteins (LDLs), 1066 “Low-resolution” mass spectrometers, 439 Lowest unoccupied molecular orbital (LUMO), 105, 609 Lucas, H. J., 545 Lucite, 488 Lycopene, 610 Lycopodine, 908 Lysine, 1088, 1113 isolectric point of, 1091-1092 Lysozyme, 1085, 1116-1120

See

M Macrocyclic lactones, 803 Macromolecules, 486 Magnetic focusing, 442 Magnetic resonance, 386 Magnetic resonance imaging (MRI), in medicine, 425-426

MALDI (matrix-assisted laser desorption ionization) mass spectrometry, 441-442, 1126 Maleic acid, 380 Malonic acids, 816 Malonic ester synthesis, 976 of substituted acetic acids, 850-853 Maltose, 1001-1002, 1030-1031 Mannich bases, 894 Mannich reaction, 894-895 Mannosides, 1010 Map of electrostatic potential (MEP), 59 Markovnikov additions, 334 anti-, 339 exception to, 338-339 regioselective reactions, 338 Markovnikov’s rule, 334-341 defined, 334 modern statement of, 337-338 theoretical explanation of, 336-337 Mass spectrometry (MS), 426-443, 1127 bask peak, 426 of biomolecules, 443 determining molecular formulas and molecular weights using, 435-436 electron impact (EI) ionization, 427, 440-441 with electrospray ionization (ESI), 440, 441 mass spectrometry (MS) with (ESIMS), 1126 fragmentation by cleavage of two bonds, 434-435 GC/MS (gas chromatography with mass spectrometry), 441, 442-443 high-resolution, 439-440 instrument designs, 440-442 ion sorting and detection, 442 ion trap mass analyzers, 442 magnetic focusing, 442 matrix-assisted laser desorption-ionization (MALDI), 440, 441-442 molecular formula, determining, 436-439 molecular ion, 426 and isotopic peaks, 435-436 peptide sequencing using, 1 1 0 0 - 1 1 0 1 polypetides/proteins, 1125-1126 quadrupole mass analyzer, 442 time-of-flight (TOF) mass analyzer, 442 Matrix-assisted laser desorption-ionization (MALDI), 440, 441-442 Maxam, A., 1155 Mayo, F. R., 484 McLafferty rearrangement, 435, 764 Meisenheimer intermediate, 982-983 Melting point, 74-75, 77 Menthol, 502 Mercapto group, 658 6 -Mercaptopurine, 1139 Merrifield, R. B., 830, 1108-1109 Mescaline, 922 Meso compounds, 213-214 Messenger RNA (mRNA) synthesis, 1147 Mesylates, 518-521

I-13

Meta-chloroperoxybenzoic acid (MCPBA), 528 Meta directors, 692, 695 Meta-disubstituted benzenes, 663 Methane, 54 chlorination of: activation energies, 472-475 energy changes, 470-480 mechanism of reaction, 467-470 overall free-energy change, 471-472 reaction of methane with other halo­ gens, 475-476 iodination, 476 orbital hybridization, 25-26 physical properties of, 75 structure of, 26-28 tetrahedral shape of, 6-7 valance shell of, 38-39 Methanide ion, 304 Methanogens, 54 Methanoic acid, 780 Methanol, 258, 506, 507, 513 Methanolysis, 253 Methanoyl group, 731 Methionine, 266, 1088, 1113 Methoxide anion, 258 Methyl alcohol (methanol), 65 Methyl carbocation, 250 Methyl cyanoacrylate, 490 Methyl group, 63 Methyl halides, 255 Methyl ketones, synthesis of, 845-850 Methyl salicylate, 966 Methyl transfer reaction, 230 Methylaminium ion, 112 Methylbenzene, 634 Methylcyclohexane, 181 conformational analysis of, 168-170 Methyldopa, 209 Methylene chloride, 232 Methylene group, 64 Methylene, structure and reactions of, 361 2-Methylhexane, retrosynthetic analysis for, 320 Methyloxirane, 532 2-Methylpropene, addition of HBr to, 337 Micelles, 1057 Michael additions, 855, 870, 892-893 Michael, Arthur, 892 Micrometers, 84 492 Micron, 605 Millimicron, 605 Mirror planes of symmetry, 195-196, 225 Mitomycin, 1163 Mitscherlich, Eilhardt, 633 Mixed triacylglycerol, 1052 Miyaura-Suzuki coupling, G-1, G-2-G-3 Modern statement of Markovnikov’s rule, 337-338 Molar absorptivity, 608 Molecular formulas, 5 determining, 436-439 gaining structural information from, 178-180 Molecular handedness, 193

Micromonosporaechinospora,

I-14

In d e x

Molecular ion, 426 depicting, 427-435 and isotopic peaks, 435-436 Molecular orbitals (MOs), 23-24, 36-37 antibonding, 37 bonding, 24, 27, 31-32, 37 number of, 37 theory, 47 Molecular oxygen, 490-491 Molecular recognition, 538 Molecular structure determines properties (principle), 92 Molecularity, 238 Molecules, composition of, 8 Molina, M. J., 495 Molozonides, 367 Monensin, 539 Monomers, 486 Mononitrotoluenes, 693 Monosaccharides, 1001, 1004-1010 aldaric acids, 1018-1019 alditols, 1 0 2 2 aldonic acids, synthesis of, 1017-1018 bromine water, 1017-1018 carbohydrate synthesis, use of protect­ ing groups in, 1014 classification of, 1004 conversion to cyclic acetals, 1016 conversion to esters, 1015-1016 D and L designations of, 1004-1005 deoxy sugars, 1038 enolization, 1013 ethers, formation of, 1014-1015 isomerization, 1013 Kiliani-Fischer synthesis, 1023-1024 nitric acid oxidation, 1018-1019 oxidation reactions of, 1016-1021 Benedict’s reagents, 1016-1017 Tollens’ reagents, 1016-1017 oxidative cleavage of polyhydroxy compounds, 1 0 2 0 - 1 0 2 1 periodate oxidations, 1 0 2 0 - 1 0 2 1 reactions with phenylhydrazine, 1022-1023 reducing sugars, 1017 reduction of, 1 0 2 2 Ruff degradation, 1025 structural formulas for, 1005-1009 synthesis and degradation of, 1023-1025 tautomerization, 1013 uronic acids, 1037-1038 Monosubstituted benzenes, 663 Monovalent, use of term, 5 Montreal Protocol, 495 Moore, S., 1096 Morphine, 922 MRI (magnetic resonance imaging) scan, 385 MudPIT (multidimensional protein identi­ fication technology), 1127 Multiple covalent bonds, 9 Murchison meteorite, 193 Muscle action, chemistry of, 162 Mutagens, 1146 Mutarotation, 1009-1010

Mycomycin, 381 Myelin, 1078 Myelin sheath, 1050 Mylar, 817 Mylotarg, 492 Myoglobin, 1114 Myosin, 137, 1112 Myrcene, 378 N N-Acetyl-D-glucosamine, 1039 N-Acetylglucosamine, 1118 N-Acetylmuramic acid, 1039, 1118 N-Acylamino acids, 1093-1094 -Formylmethionine, 1151 N-Methylmorpholine N-oxide (NMO), 364 -Nitrosoamines, 936-937 N-terminal, 1094 NAD+, 658-659 NADH, 658-659 Naming enantiomers, 196-201 Nanoscale motors and molecular switches, 166 Nanotubes, 655 Naphthalene, 653 Naphthols, 965 Natural products chemistry, 2 Natural rubber, 1064 Naturally occurring phenols, 966 Nature prefers disorder to order (princi­ ple), 131 Nature prefers states of lower potential energy (principle), 131, 181 Nature tends toward states of lower poten­ tial energy (principle), 46 Neighboring group effects, 545 Neighboring-group participation, 282 Neomycins, 1042 Neopentane, 140, 147, 467 boiling point, 78 Neurotransmitters, 923-924 Neutrons, 4 Newman projection formula, 157 Newman projections, 157-158 Niacin (nicotinic acid), 1116 Nicolaou, K. C., 492, 530, 617, 620 Nicotinamide adenine dinucleotide, 658 Nicotine, 6 8 , 923 Ninhydrin, 1095-1096 Nitrate ion, 14 Nitric acid, 715 oxidation, 1018-1019 Nitric oxide, 491 Nitriles, 72, 786-787 acidic hydrolysis of, 810 aldehydes by reduction of, 734-737 basic hydrolysis of, 810-811 hydrolysis of, 809-810 ketones from, 739-740 preparation of carboxylic acids by hydrolysis of, 790 reactions of, 821 reducing to amines, 927-929 Nitrogen, compounds containing, 180 Nitrogen inversion, 915 Nitrous acid:

N N

reactions of amines with, 935-937 primary aliphatic amines, 935 primary arylamines, 935 secondary amines, 937 tertiary amines, 937 Nitrous oxide (laughing gas), 67 Noble gas structure, 19 Nodes, 20, 37 Nodes of Ranvier, 1051 Nonactin, 539 Nonaqueous solutions, acids and bases in, 128-130 Nonaromatic compounds, 649-651 Nonaromatic cyclohexadienyl carbocation, 678 Nonbenzenoid aromatic compounds, 654 Nonbonding pairs, 38 Nonpolar compounds, boiling point, 79 Nonpolar molecules, 60 Nonreducing sugars, 1017 Noradrenaline, 922 Norethindrone, 1069 Novestrol, 1069 Novrad (levopropoxyphene), 186, 209 Noyori, Ryoji, 210, 365, 529-530, G-15 Nuclear magnetic resonance (NMR) spec­ trometry, 385-426, 1127 13C NMR (carbon-13) NMR spec­ troscopy, 403, 417-422 broadband (BB) proton decoupled, 417 chemical shifts, 418-420 DEPT 13C NMR spectra, 420-422 interpretation of, 417 off-resonance decoupling, 420 one peak for each magnetically dis­ tinct carbon atom, 417-418 chemical shift, 387-388, 400-401 parts per million (ppm) and the S scale, 401 chemical shift equivalent, 401-405 heterotopic atoms, 402 homotopic hydrogens, 401-402 complex interactions, analysis of, 412-414 conformational changes, 416 coupling (signal splitting), 390-392 defined, 386 diastereotopic hydrogen atoms, 404 enantiotopic hydrogen atoms, 403-404 first-order spectra, 415 Fourier transform NMR spectrometers, 397-398 1H NMR spectra, 412, 416, 417, 423, 425 magnetic resonance imaging (MRI), 425-426 multidimensional NMR spectroscopy, 422 nuclear spin, 395-397 proton NMR spectra: complicating features, 412 interpreting, 392-395 and rate processes, 415-417 protons, shielding/deshielding, 399-400 second-order spectra, 415

I-15

In d e x

signal areas, integration of, 390 signal splitting, 405-414 spin decoupling, chemical exchange as cause of, 415-416 spin-spin coupling, 405-414 coupling constants, 410-411 origin of, 406-410 splitting tree diagrams, 406-410 vicinal coupling, 405-406 splitting patterns, recognizing, 410-411 two-dimensional NMR (2D NMR) techniques, 422-426 COSY spectrum, 423-424 heteronuclear correlation spec­ troscopy (HETCOR or C-H HETCOR), 423, 424-425 Nuclear magnetic resonance (NMR) spec­ trum, 386 Nuclear spin, 395-397 Nucleic acids, 1132 water solubility, 81 Nucleophiles, 105-106, 233, 234-237, 312 defined, 234 reactions of carbonyl compounds with, 550 Nucleophilic addition, 550 Nucleophilic addition-elimination, 792-794 Nucleophilic substitution, 233 reactions, 266-267 allylic and benzylic halides in, 717-719 Nucleophilicity, 258-259 basicity vs., 258 Nucleotides/nucleosides, 1039, 1133-1137 laboratory synthesis, 1137-1139 medical applications, 1139 silyl-Hilbert-Johnson nucleosidation, 1137 Nylon, 817 O

Octadecanoic acid, 780 Octet rule, 7 exceptions to, 1 1 - 1 2 Octyl alcohol, 506 Off-resonance decoupling, 420 Oils, 1052, 1054 Olah, George A., 249 Olefiant gas, 286 Olefins, 286 metathesis, G-5-G-7 Oleksyszyn, J., 1120 Olestra, 1055-1056, 1082-1083 Oligopeptides, 1094 Oligosaccharides, 1001 Olympiadane, 167 Omega-3 fatty acids, 1052-1054 Opposite charges attract (principle), 46, 92, 131 and acid-base reactions, 103-104 Optical activity: origin of, 205-207 plane-polarized light, 202, 225 polarimeter, 203

racemic forms (racemic mixture), 206-207 and enantiomeric excess, 207 specific rotation, 204-205 Optical purity, 207 Optical rotatory dispersion, 205 Optically active compounds, 201 Orbital hybridization, 25-26 Orbital overlap stabilized molecules (prin­ ciple), 46 Orbitals, 22 Organic chemistry: defined, 1 development of the science of, 2-3 oxidation-reduction reactions in, 550-552 structural theory of, 5-7 Organic compounds: as bases, 126-127 families of, 72-74 ion-ion forces, 74-75 molecular structure, 73 physical properties, 73 Organic halides: defined, 232 as herbicides, 990 physical properties of, 232-233 Organic molecules, 1 Organic reactions, 98-136 acid-base reactions, 115-118 predicting the outcome of, 113-114 and the synthesis of deuterium and tritium-labeled compounds, 130 acidity, effect of the solvent on, 125-126 acids and bases in nonaqueous solu­ tions, 128-130 additions, 99 Br0 nsted-Lowry acids and bases, 109-115 carbanions, 104-106 carbocations, 104-106 carboxylic acids, acidity of, 121-125 covalent bonds, homolysis and heterol­ ysis of, 1 0 0 electrophiles, 105-106 eliminations, 99 energy changes, 119-120 illustrating using curved arrows, 106-107 intermediates, 98 Lewis acids and bases, 102-104 mechanisms, 99-102, 107, 127-128 nucleophiles, 105-106 organic compounds as bases, 126-127 reaction mechanism, 98 rearrangements, 1 0 0 relationship between the equilibrium constant and the standard freeenergy change, 1 2 0 - 1 2 1 substitutions, 99 Organic synthesis, 317-323 defined, 317 from inorganic to organic, 321 planning, 318-319 retrosynthetic analysis, 318-319

Organic vitamin, 2 Organohalogen compounds, 232 Organolithium compounds, 562-563 reactions of, 563-566 Organomagnesium compounds, 562-563 reactions of, 563-566 Organometallic compounds, 561-562 Orientation, 608-609, 703 Ortho-disubstituted benzenes, 663 Orthogonal protecting groups, 1109-1110 Ortho-para direction, and reactivity of alkylbenzenes, 705-706 Ortho-para directors, 692-693, 933 Osazones, 1022-1023 Osmium tetroxide, 363, 365 Oxetane, 504 Oxidation: of alcohols, 557-561 of alkenes, 363-365 of alkylboranes, 350-352 regiochemistry and stereochemistry, 351-352 defined, 551 oxidation states in organic chemistry, 551-552 Oxidation-reduction reaction, 269 Oxidative addition-reductive elimination, G-12-G-13 Oxidative cleavage, 365-368 of alkenes, 365-368 of alkynes, 370 Oxidizing agents, 551, 715 Oximes, 752-753 reducing to amines, 927-929 Oxirane, 504, 507 Oxonium cation, 743 Oxonium ion, 126 Oxonium salts, 527 Oxygen, compounds containing, 180 Oxymercuration-demercuration: alcohols from alkenes from, 344-347 defined, 344, 353 mechanism of oxymercuration, 345-346 rearrangements, 345 regioselectivity of, 344-345 Oxytocin, 1101-1102 Ozone, cleavage with, 366-368 Ozone depletion and chlorofluorocarbons (CFCs), 495 Ozonides, 367 P

P-2 catalyst, 315 p53 (anticancer protein), 1104 P450 cytochromes, 533 Paclitaxel (Taxol), 365 Palindromes, 1155 Pantothenic acid, 1116 Paquette, Leo A., 176 Para-disubstituted benzenes, 663 Paraffins, 177 Partial hydrolysis, 1099 and sequence comparison, 1 1 0 0 - 1 1 0 1 Pasteur, Louis, 223

I-16

In d e x

Pasteur’s method for separating enantiomers, 223 Pauli exclusion principle, 23-24, 46 Pauling, Linus, 1111 Pedersen, Charles J., 538 Penicillamine, 209 Penicillinase, 812 Penicillins, 811-812 Pentane, 140, 506 Pentanoic acid, 780 Pentose, 1004 Pentyl alcohol, 506 Peptide bonds, 1094 Peptide linkages, 1094 Peptide synthesizers, 830 Peptides, 1094 chemical synthesis of, 830 synthesis of, 1107-1108 Perfumes, aldehydes in, 733 Pericyclic reactions, 978 Periodic table of the elements, 3 Perkin, Jr., William, 861-862 Permanent dipole moment, 75 Peroxides, 460 Peroxy acid (peracid), 528 Perspex, 488 Pettit, R., 645 Petroleum, 138 refining, 139 Phase sign, 21 Phase transfer catalysts, 538 Phenacetin, 827 Phenanthrene, 653 Phenanthrols, 965 Phenol, 634, 694 Phenols, 503, 964-978, 980-991 as acids, reactions of, 969-972 bromination of, 973 distinguishing/separating from alcohols and carboxylic acids, 971 1H NMR spectra, 988 industrial synthesis, 967-969 infrared (IR) spectra of, 90 infrared spectra, 988 Kolbe reaction, 974-975 laboratory synthesis, 967 mass spectra, 988 monobromination of, 973 naturally occurring, 966 nitration of, 974 nomenclature of, 965 physical properties of, 966 properties of, 964 reactions of the benzene ring of, 973-974 reactions with carboxylic acid anhy­ drides and acid chlorides, 972 spectroscopic analysis of, 988 strength of, as acids, 969-971 structure of, 965 sulfonation of, 974 synthesis of, 967-969 in the Williamson synthesis, 972 Phenyl groups, 64, 635 Phenyl halides, 232 unreactivity of, 267

Phenylalanine, 1087, 1113 Phenylalanine hydroxylase, 658 Phenylation, 987 Phenylethanal, infrared spectrum of, 763 2-Phenylethylamines, 922 Phenylhydrazones, 752 Phenylosazones, 1022-1023 Pheromones, 156-157, 379 Phillips, David C., 1117 Phosgene, 812 Phosphatides, 1074-1077 Phosphatidic acid, 1074 Phosphatidylserines, 1075 Phospholipids, 1074-1077 Phosphoramidite, 1157-1158 Phosphoranes, 757 Phosphoric acid, 1074 Phosphorus pentoxide, 809 Phosphorus tribromide, 514 Phosphorus ylides, 757-758 Photons, 604 Photosynthesis and carbohydrate metabo­ lism, 1002-1004 Phthalimide, 926 Phytostanols, 1067 Phytosterols, 1067 Pi (p) bond, 31, 37 Pi (p) complex, G-12 Pictric acid, 971 Pitsch, S., 1137 Plane of symmetry (mirror plane), 195-196 Plane-polarized light, 202, 225 Plaskon, R. R., 1120 Plasmalogens, 1075 Plexiglas, 488 Polar aprotic solvents, 260-261 Polar bonds, electronegativity differences as causes of, 92 Polar covalent bonds, 57-59 as part of functional groups, 59 Polar molecules, 60 Polarimeter, 201, 203 Polarizability, 259, 275 Polarized bonds underlie inductive effects (principle), 131 Polyacrylonitrile, 488 Polyamides, 817-818, 1085 Polybrominated biphenyls and biphenyl ethers (PBBs and PBDEs), 990-991 Polybromodiphenyl ethers (PBDEs), 991 Polychlorinated biphenyls (PCBs), 983, 990 Polycyclic alkanes, 175 Polycyclic aromatic hydrocarbons (PAH), 651 Polyesters, 817-818 Polyethers, from epoxides, 534-535 Polyethylene, 486-488 Polyethylene glycol (PEG), 505-506 Polyethylene oxide (PEO), 505-506 Polyketide anticancer antibiotic biosynthe­ sis, 975-976 Polymer polypropylene, 55

Polymerase chain reaction (PCR), 1131, 1158-1161 Polymerizations, 486-488 Polypeptides, 1094-1110, 1125-1126 analysis of, 1125-1126 hydrolysis, 1095-1097 as linear polymers, 1094 primary structure of, 1097-1101 C-terminal residues, 1099 complete sequence analysis, 1099-1100 Edman degradation, 1097-1098 examples of, 1101-1104 peptide sequencing using mass spec­ trometry and sequence databases, 1 1 0 0 -1 1 0 1

Sanger N-terminal analysis, 1098-1099 purification of, 1125 synthesis of, 1104-1110 activation of the carboxyl group, 1107 automated peptide synthesis, 1108-1110 peptide synthesis, 1107-1108 protecting groups, 1105-1107 Polysaccharides, 1001, 1033-1037 cellulose, 1036-1037 cellulose derivatives, 1037 glycogen, 1035-1036 heteropolysaccharides, 1033 homopolysaccharides, 1033 starch, 1034-1035 water solubility, 81 Polystyrene, 488 Poly(tetrafluoroethene), 488 Polyunsaturated fats/oils, 493, 1052 Polyunsaturated hydrocarbons, 599-600 Pophristic, V. T., 158 Positive entropy change, 131 Potassium permanganate, 363, 365 Potential energy diagram, 159 Potential energy (PE): and covalent bonds, 1 2 0 defined, 119 Powers, J. C., 1120 Prenylated proteins, 1104 Presnell, S., 1120 Primary alcohols, 65 chemical test for, 560-561 dehydration of, 297-298 mechanism for, 302 rearrangement after, 306-307 preparation of carboxylic acids by oxi­ dation of, 789-790 Primary alkyl halide, 64 Primary amines: addition of, 751-755 preparation of: through Curtius rearrangement, 932-933 through Hofmann rearrangement, 931-932 through reduction of nitriles, oximes, and amides, 929-931

In d e x

through reductive amination, 927-929 Primary carbocations, 250, 256 Primary carbon, 65, 140 atom, 64 Primary halide, 274 Primary structure: of polypeptides and proteins, 1097-1101 of a protein, 1085 Primer, 1155-1156 Prochirality, 556 Progesterone, 1069 Progestins, 1068-1069 Proline, 1087, 1096 Propene (propylene), 30, 55, 340 Propyl alcohol, 506 structural formulas for, 41 Propyl group, 63 Propylene glycols, 506, 508 Propylene oxide alginates, 6 Prostaglandins, 1073-1074 Prosthetic groups, 1116, 1122 Protecting groups, 575 acetals, 749-750 amino acids, 1105-1107 ethers, 525-526 orthogonal, 1109-1110 --butyl ethers, 525 Proteins, 1085, 1094-1126 analysis of, 1125-1126 conjugated, 1 1 2 2 defined, 1094 prenylated, 1104 primary structure of, 1085, 1097-1101 C-terminal residues, 1099 complete sequence analysis, 1099-1100 Edman degradation, 1097-1098 examples of, 1101-1104 peptide sequencing using mass spec­ trometry and sequence databases,

tert

1 1 0 0 -1 1 0 1

Sanger N-terminal analysis, 1098-1099 proteomics, 1126-1128 purification of, 1125 quaternary structure, 1085 secondary structure, 1085 synthesis of, 1104-1110 activation of the carboxyl group, 1107 protecting groups, 1105-1107 tertiary structure, 1085 water solubility, 81 Proteome, 1133 Proteomics, 1126-1128 Protic solvent, 125, 259 Proton NMR spectra: complicating features, 412 interpreting, 392-395 and rate processes, 415-417 Protonated alcohol, 126, 299 Protonolysis, of alkylboranes, 353-354 Protons, 4 Protons, shielding/deshielding, 399-400

Purcell, Edward M., 386 Purine-purine base pairs, 1141 Pyramidal inversion, 915 Pyranose, 1008 Pyrene, 653 Pyridoxal phosphate (PLP), 753 Pyridoxine (vitamin B6), 753-754 Pyrimidine-pyrimidine base pairs, 1141 Pyrolysis, 1064 Pyrrole, 656 Q

Quadrupole mass analyzer, 442 Quanta, 604 Quantum mechanics, and atomic structure, 2 0 -2 1

Quaternary ammonium hydroxides, 919, 949 Quaternary ammonium salts, 919-920 Quaternary structure, of a protein, 1085, 1114-1115 Quinine, 922 Quinones, 978-980 R

Racemic forms (racemic mixture), 206-207 and enantiomeric excess, 207 and synthesis of chiral molecules, 208-209 Racemization, 251-252 via enols and enolates, 834-836 Radical addition of a p bond, 461 Radical addition, to alkenes, 484-486 Radical anion, 317 Radical cation, 427 Radical halogenation, 465-467 Radical polymerization, of alkenes, 486-490 Radical reactions, homolytic bond dissoci­ ation energies (DH°), 461-465 Radicals, 100, 459-501 alkanes: chlorination of, 477-478 combustion of, 491-492 alkyl radicals, geometry of, 480 antioxidants, 494 autoxidation, 493 bromine, selectivity of, 479-480 chain reaction, 469, 484 chlorination: of alkanes, 467 of methane, 467-468 chlorine selectivity, lack of, 467 formation/production of, 460 heats of reaction, using homolytic bond association energies to calculate, 462-463 methane chlorination, 475-476 activation energies, 472-475 overall free-energy change, 471-472 reaction of methane with other halo­ gens, 475-476 molecular oxygen and superoxide, 490-491 multiple halogen substitution, 466 nitric oxide, 491

I-17

radical halogénation, 465-467 radical polymerization of alkenes, 486-490 reactions of, 460-501 tetrahedral chirality centers, reactions that generate, 481-484 using homolytic bond dissociation ener­ gies to determine the relative sta­ bilities of, 463-465 Random coil arrangement, 1113 Raney nickel, defined, 750 proteins, 1104 Rate constant, 238 Rate-determining step, 246-247, 335 Rate-limiting step, 246-247 Reaction coordinate, 240 Reaction mechanism, 98 Reagents: Benedict’s, 1016-1017, 1042 Gilman, G-1, G-8 Grignard, 563, 565-570, 791 Jones, 558 lithium dialkyl cuprate, G-1, G-8 Tollens’ reagents, 761, 1016-1017, 1042 Rearrangements, 100 alkenes, 342-343 during dehydration of primary alcohols, 306-307 during dehydration of secondary alco­ hols, 303-305 McLafferty rearrangement, 435 organic reactions, 1 0 0 oxymercuration-demercuration, 345 Receiver coil, 397 Reducing agent, 551 Reducing sugars, 1017 Reduction, 314 defined, 550 dissolving metal reduction, 316 Reductive amination, preparation of pri­ mary, secondary, and tertiary amines through, 927-929 Reductive elimination, G-13 Regioselectivity, of oxymercuration-demercuration, 344-345 Relative configuration, 220-221 Relative potential energy, 119 Relative probability, 20 Relative reactivity, aldehydes vs. ketones, 743 Relative stability, 119 Relaxation process, 426 Relaxation times, 426 Replacement nomenclature, defined, 504 Replacement reactions, of arenediazonium salts, 937-940 Resolution, 920 by enzymes, 223 kinetic, 209 Resonance, 17 Resonance effects, 608-609, 701-702 Resonance effects can stabilize molecules and ions (principle), 131 Resonance energy, 639, 641

Ras

I-18

In d e x

Resonance stabilization, 18 Resonance structures: estimating the relative stability of, 597-598 rules for writing, 595-597 Resonance structures (resonance contribu­ tors), 16 rules for writing, 17-19 Resonance theory, 15-20, 56, 595-599 Restricted rotation, and the double bond, 32-33 Restriction endonucleases, 1155 Retention times, 443 Retinal (compound), 70, 609 Retro-aldol reaction, 877-879 in glycolysis, 878 Retrosynthetic analysis, 318-320, 371-372 disconnections/synthons/synthetic equivalents, 372-374 key to, 371 stereochemical considerations, 373-375 Retrosynthetic arrow, 319 Reverse turns, 1113 Rhodium, G-9 Ribonucleic acid (RNA): defined, 1132 genetic code, 1150-1152 messenger RNA (mRNA) synthesis, 1147 and protein synthesis, 1146-1154 ribosomal rRNA, 1148-1149 RNA polymerase, 1147 transcription, 1147 transfer RNAs (tRNAs), 1148, 1149-1150 translation, 1152-1154 Ribosomes, 1148-1149 Ribozymes, 1115, 1148 Ring flip, 167 Ring fusion, 652 Ring-opening olefin metathesis polymer­ ization (ROMP), 1006 RNA, Ribonucleic acid (RNA) RNA polymerase, 1147 Roberts, J. D., 415-417, 985 Robertson, A., 674 Robinson annulation, 893 Robinson, Robert, 674, 893 Rotaxanes, 166 Rowland, F. S., 495 R,S-system of naming enantiomers, 196-201, 225 assigning (R) and (S) configurations, 197-198 Ruff degradation, 1025 Ruff, Otto, 1025fn Ruh-Pohlenz, C.,, 1137 Ruthenium carbene complexes, G-5-G-7

See

sorbitals, 22, 36

S

Salts, 8 amine, 915-933 aminium, 919-920 arenediazonium, 937-942

carboxylate salts, 781 diazonium, 935, 938 oxonium, 527 quaternary ammonium, 919-920 Sample matrix, 442 Sandmeyer reaction, 938 Sanger, Frederick, 1098, 1155 Sanger N-terminal analysis, 1098-1099 Saponification, 800-802 of triacylglycerols, 1056-1058 Saturated compounds, 54, 314 Saturated fatty acids, 1052-1053 Sawhorse formulas, 157 (S)-BINAP ligand, 210, 224 Schardinger dextrins, 1045 Schrock, Richard, G-6 Schrödinger, Erwin, 20 Schultz, Peter G., 1084 Schwann cells, 1050-1051 SDS-PAGE (sodium dodecyl sulfate-polyacrylamide gel electrophoresis), 1125 sec-butyl alcohol, 506 Second chirality center, in a radical halogenation, generation of, 482-484 Second-order spectra, 415 Secondary alcohols, 6 6 chemical test for, 560-561 dehydration of, 297-298, 300 mechanism for, 299-302 rearrangements during, 303-305 Secondary alkyl halide, 64 Secondary amines: addition of, 751-755 preparation of: through reduction of nitriles, oximes, and amides, 929-931 through reductive amination, 927-929 Secondary carbocations, 250, 256 Secondary carbon, 64, 6 6 , 140 Secondary halides, 274 Secondary orbital interaction, 621 Secondary structure, of a protein, 1085 Self-assembled monolayers (SAMs), 1060 Semicarbazone, 771 Sequence databases, peptide sequencing using, 1 1 0 0 - 1 1 0 1 Serine, 1087, 1113 Serine proteases, 1120-1122 Serotonin, 922 Sevin, 813 Sex hormones, 1068-1070 Sharing electrons, 8-9 Sharpless asymmetric epoxidation, 529-530 Sharpless, Barry, 210 Sharpless, K. B., 365, 529, G-15 Shell process, 587 Shells, 4 Shielding, protons, 399-400 1,2 shift, 304 Shikimic acid, 1048 Sialyl Lewisx acids, 1000 Sickle-cell anemia, 1103 Side chain:

defined, 706 halogenation of, 709 Sigma bonds (sbonds), 37 and bond rotation, 157-160 Sigma (s) bonds, 28 Signal splitting, 390-392 silyl-Hilbert-Johnson nucleosidation, 1137, 1163 Simmons, H. E., 362 Simmons-Smith cyclopropane synthesis, 362 Simple addition, 889 Simple triacylglycerols, 1052 Single-barbed curved arrows, 460 Single bonds, 28 Singlets, 392 Site-specific cleavage, of peptide bonds, 1100

Skou, Jens, 539 Smalley, R. E., 654-655 Smith, D.C.C., 1047 Smith, R. D., 362 SN1 reactions, 246-248 E1 reactions vs., 275 effect of the concentration and strength of the nucleophile, 258 effect of the structure of the substrate, 256 factors affecting the rates of, 254-264 mechanism for, 247-248 rate-determining step, 246-247 reactions involving racemization, 251-252 solvent effects on, 261 solvolysis, 253 stereochemistry of, 251-254 Sn2 reactions, 237-240, 311 E2 reactions vs., 273-275 effect of the structure of the substrate, 254-256 functional group interconversion using, 264-266 measuring, 237-238 mechanism for, 238-240 reactions involving racemization, 251-252 solvent effects on, 259 stereochemistry of, 243-246 SNAr mechanism), 982 Sodioacetoacetic ester, 845 Sodium acetate, physical properties, 75 Sodium alkynides, 310, 573-574 Sodium amide, 308 Sodium borohydride, 553-554 overall summary of, 554-555 Sodium ethynide, 310 Sodium hydride, 849 Sodium nitrite, 936 Solid-phase peptide synthesis (SPPS), 1108-1109 Solubilities, 79-81 water solubility guidelines, 81 Solvating ions, 80 Solvent effects, 259 Solvolysis, 253

I-19

In d e x

Solvomercuration-demercuration, 346, 525 Sonogashira coupling, G-1, G-4-G-5 orbitals, 36, 37 hybridization: ethane, 30 ethyne, 34-35 2 orbitals, 30, 37 3 orbitals, 37 Spackman, D. H., 1096 Spectator ions, 102, 236 Spectroscopy, defined, 386 Sphingolipids, 1069 sphingolipid storage diseases, 1051 Sphingosine, derivatives of, 1077-1078 Spin decoupling, chemical exchange as cause, 415-416 Spin-lattice relaxation, 426 Spin-spin coupling, 405-414 coupling constants, 410-412 dependence on dihedral angle, 411-412 reciprocity of, 410 origin of, 406-410 splitting tree diagrams, 406-410 vicinal coupling, 405-406 Spin-spin relaxation, 426 Spiranes, 182 Splitting patterns, recognizing, 410-411 Splitting tree diagrams, 406-410 splitting analysis for a doublet, 406 splitting analysis for a quartet, 407-408 splitting analysis for a triplet, 406-407 Square planar configuration, G-9 Stability, 119 Stachyose, 1045 Staggered conformations, 158 Starch, 1034-1035 STEALTH® liposomes, 1077 Stein, W. H., 1096 Step-growth polymers, 817 Stephens-Castro coupling, 1003 Stereocenters, Chirality centers Stereochemistry, 186-229, 265 and chirality, 186-188 constitutional isomers, 188 defined, 188 diastereomers, 189 enantiomers, 188, 190 of epoxidation, 530 of hydroboration, 349-350 of the ionic addition, to alkenes, 339 of Sn1 reaction, 251-254 of Sn2 reaction, 243-246 stereoisomers, defined, 188 Stereogenic atoms, Chirality centers Stereogenic carbon, 192 Stereogenic center, 192 Stereoisomerism, of cyclic compounds, 217-219 Stereoisomers, 33, 161, 171, 189 defined, 188 Stereoselective reactions, 208-210, 374 Stereoselective reductions, of carbonyl groups, 555-556

sspp2 sspp

See

See

Stereospecific reactions, 358-359, 530, 1115 alkenes, 358-359 Steric effect, 255 Steric factors, 181 Steric hindrance, 159-160, 181, 255 Steroids, 909, 1064-1073 adrenocortical hormones, 1070 cholesterol, 1066-1068 cholic acid, 1071 D vitamins, 1070-1071 defined, 1064 digitoxigenin, 1071 diosgenin, 1071 reactions of, 1072-1073 sex hormones, 1068-1070 stigmasterol, 1071 structure and systematic nomenclature of, 1065-1066 Stigmasterol, 1071 Stille coupling, G-1, G-3-G-4 Stoddart, J. F., 166, 167 Stork enamine reactions, 854-857 Stork, Gilbert, 855, 8 6 6 , 908 Strecker synthesis, 1093 Streptomycin, 1042 Strong acids, 743 Structural formulas, 5 condensed, 42-43 interpreting/writing, 41 Structural isomers, 6 fn Structural theory of organic chemistry, 5-7 Stupp, S. I., 83 Styrene, 488, 706 Substituent effect, 124 Substituents: classification of, 696 effect on electrophilic aromatic substi­ tution, 697 Substituted acetic acids, synthesis of, 850-853 Substituted benzenes, infrared spectra of, 663-664 Substituted cyclohexanes, 167-168 Substituted methyl ketones, 846-847 Substitution reactions, 586 Substitutions, 99 Substrate, 233, 236 Subtractive effect, 21 Sucrose, 1001-1002, 1029-1030 Suddath, F. L., 1120 Suicide enzyme substrate, 895-898 Sulfa drugs, 945-946 synthesis of, 947 Sulfacetamide, 946 Sulfapyridine, 946 Sulfonamides, 943 Sulfonate ester derivative, 518 Sulfonyl chlorides, reactions of amines with, 943-944 Sulfur dioxide, dipole moment, 61 Sulfuric acid, 101-102 addition to alkenes, 340-342 Sugars: amino, 1039 deoxy, 1038

nonreducing, 1017 reducing, 1017 Sunscreens, 664-665 Superacids, 110 Superglue, 490 Supernovae, 2 Superoxide, 491 Superposable, use of term, 33, 186 Syn 1,2-dihydroxylation, 363 Syn addition, 363 defined, 315 of hydrogen, 315-316 Syn coplanar transition state, 295 Synapses, 911 Synthesis, planning, 370-375 Synthetic detergents, 1057-1058 Synthetic equivalent, 847 Synthetic estrogens, 1069 Synthons, 372 T

Tandem mass spectrometry (MS/MS), 1100

Tautomerization, 833-834 Tautomers, 833-834 Teflon, 488 boiling point, 78 Terelene, 817 Terminal alkynes: acidity of, 307-308 substitution of the acetylenic hydrogen atom of, 310-312 Terminal residue analysis, 1097 Terpenes, 1061-1062 Terpenoids, 1061-1062 Terramycin, 976 alcohol, 506 ethers: by alkylation of alcohols, 525 protecting group, 525 Tertiary alcohols: dehydration of, 297-298, 300 mechanism for, 299-302 Tertiary alkyl halide, 64 Tertiary amine oxides, 950 Tertiary amines, 912 oxidation of, 934 preparation of: through reduction of nitriles, oximes, and amides, 929-931 through reductive amination, 927-929 preparation of, through reduction of nitriles, oximes, and amides, 929-931 preparation of, through reductive ami­ nation, 927-929 reactions of, with nitrous acid, 937 Tertiary carbocations, 250, 256 Tertiary carbon, 64, 6 6 , 140 Tertiary halides, 274, 275 Tertiary structure, of a protein, 1085, 1114 Testosterone, 1068-1069 Tetrachloroethene, dipole moment, 61 Tetrachloromertensene, 357 Tetracyclines, 966

te terrt-b t-buuty tyll

I-20

In d e x

Tetraethyllead, 562 Tetrahedral carbon atoms, 37 Tetrahedral chirality centers, reactions that generate, 481-484 Tetrahedral intermediate, 792 Tetrahedral vs. trigonal stereogenic cen­ ters, 193 Tetrahydrofuran (THF), 504-505, 507 Tetramethylsilane, 400-401 Tetravalent, use of term, 5 Tetrose, 1004 Thalidomide, 195 Thermal cracking, 139, 474 Thermodynamic enolate, 842 formation of, 842 Thermodynamic control, 614 Thermodynamic products, 614, 616 Thioacetals, 750 Thiols, 1088 Thionyl chloride, 514 Thiophene, 657 Three-dimensional formulas, 45-46 Threonine, 1087, 1113 Thymol, 966 Thyroxine, 676 Thyroxine biosynthesis, 681 iodine incorporation in, 707 Time-of-flight (TOF) mass analyzer, 442 Toliprolol, 992 Tollens’ reagents, 761, 1016-1017, 1042 Tollens’ test (silver mirror test), 761, 772 Toluene, 87, 634, 693, 706 Tomasz, Maria, 1163 Tool Kit for Organic Synthesis, 371 Torsional barrier, 159 Torsional strain, 159, 162 Tranquilizers, 923-924 Trans, 286 -Cycloheptene, 290 -Cyclohexene, 290 -Cyclooctene, 290 Transaminations, 754 Transannular strain, 167 Transcription, 1147 Transesterification, 800 Transfer RNAs (tRNAs), 1148, 1149-1150 Transition metal complexes, G-8-G-9 electron counting in, G-9-G-11 Transition metal-catalyzed carbon-carbon bond-forming reactions, G-1 Transition metals, defined, G-8 Transition state, 239-243 Translation, 1152-1154 Triacylglycerols, 1052-1058 biological functions of, 1055 hydrogenation of, 1054-1055 saponification of, 1056-1058 Trialkylboranes, oxidation of, 351 Trichloromethane, dipole moment, 62 Triflate ion, 262 Trigonal planar carbon atoms, 37 Trigonal pyramid, 38 Trigonal stereogenic centers, tetrahedral stereogenicceters vs., 193 Trimethylene glycol, 506 2,4,6-Trinitrophenol, 971

traannss tr trans

Triose, 1004 Tripeptides, 1094 Triplets, 392 Trisaccharides, 1001 Tritium, 4 Tropylium bromide, 649 Tryptophan, 1087, 1113 Tscherning, Kurt, 1068 d-Tubocurarine chloride, 924 Tumor suppressor, 1104 Two-dimensional NMR (2D NMR) tech­ niques, 422-426 COSY spectrum, 423-424 heteronuclear correlation spectroscopy (HETCOR or C-H HETCOR), 423, 424-425 Two-dimensional polyacrylamide gel elec­ trophoresis (2D PAGE), 1127 Tyrosine, 657, 966, 998, 1087, 1113 U

Ubiquinones, 978-979 Ultraviolet-visible (UV-Vis) spec­ troscopy, 604-612 absorption maxima for nonconjugated and conjugated dienes, 608-611, 609 analytical uses of, 611 electromagnetic spectrum, 604-606 UV-Vis spectrophotometer, 606-608 Unbranched alkanes, 140, 142 boiling points, 155 density, 156 melting points, 155-156 solubility, 156 Unfavorable entropy change, 81 Unimolecular reactions, 246 Unsaturated compounds, 54, 314 Unsaturated fatty acids, 1052 reactions of the alkenyl chain of, 1059 Unshared pairs, 38 Upfield, use of term, 388 Urea, 2 Urethanes, 812-813 Uronic acids, 1037-1038 Urushiols, 966 UV-A, UV-B, and UV-C regions, 664 V

Valence electrons, 4-5, 13 Valence shell, 4 Valence shell electron pair repulsion (VSEPR) model, 38, 47 Valeric acid, 780 Valine, 1086, 1113 Valinomycin, 539 Valium, 923 van der Waals forces, 75-76 van der Waals radii, 161 van der Waals surface, 29, 59 Vanillin, 502, 505 van’t Hoff, J. H., 6-7, 159, 1027 Vasopressin, 1101-1102 Vedejs, E., 758 Viagra, 459 Vibrational absorption, 85 Vicinal coupling, 405-406

Vicinal dihalide (vic-dihalide), 308, 354 Vinyl chloride, 232 anionic polymerization of, 489 Vinyl group, 152-153, 232 Vinylic anion, 317 Vinylic halides, 232 unreactivity of, 267 Vision, photochemistry of, 609 Vitalism, 2 Vitamin A, 1064 Vitamin B12, G-17-G-18 Vitamin C, 494 Vitamin D, 1070-1071 Vitamin E, 494 Vitamin K1, 979 Vitamins, 922-923 organic, 2 Voet, D., 707, 1050fn, 1067 Voet, J. G., 707, 1050fn, 1067 Volume, atoms, 4 von Hofmann, August W., 949 Vorbruggen, H., 1137 Vulcanization, natural rubber, 1064 W

Walden inversions, 239fn Walden, Paul, 239fn Walker, John E., 539 Warmuth, R., 987 Water: acid-catalyzed addition of, to alkenes, 340-342 tetrahedral structure for the electron pairs of a molecule of, 39 Water solubility: guidelines, 81 as the result of salt formation, 114-115 Watson, James, 1140-1142, 1146 Wave function (y), 20-21 - and + signs of, 22 Wave mechanics, 20 Wavelength (l), 84, 604 Wavenumbers, 84 Waxes, 1078 Weak nucleophiles, 743 Whitmore, F., 299 Wieland, Heinrich, 1067 Wilkins, Maurice, 1140 Wilkinson’s catalyst, G-11, G-13-G-16 Wilkinson’s catalyst tris(triphenylphosphine)rhodium chloride), 313 Williams, L. D., 1120 Williamson ether synthesis, 523-524, 1014 Williamson synthesis, phenols in, 972 Willstatter, Richard, 639 Windaus, Adolf, 1067 Winstein, S., 545 Withers, Stephen, 1118 Wittig reaction, 757-758 Horner-Wadsworth-Emmons reaction, 760-761 Wittig synthesis, how to plan, 758-759 Wohler, Friedrich, 2, 321 Wood alcohol, Methanol Woods, D. D., 946

See

In d e x

Woodward, R. B., 318, 620, 909, G-17 X

X-ray crystallography, 1127 X-rays, 605 Xylenes, 635

Y

Yates, John, 1127 Ylides, addition of, 757-758 Z

Z-Ala, 830 Zaitsev, A. N., 293

I-21

Zaitsev rule, 292-294, 304 Zaragozic acid A (squalestatin S1), 530 Ziegler-Natta catalysts, 488, 1064 Zinc, 103 Zingiberene, 180 Zwitterions, 1089

SPECIAL

IBond-Forming Carbon-Carbon and Other TO ~

I

G

Reactions of Transition Metal Organometallic Compounds

A number of transition metal-catalyzed carbon-carbon bond-forming reactions have been developed into highly useful tools for organic synthesis. The great power of many transi­ tion metal-catalyzed reactions is that they provide ways to form bonds between groups for which there are very limited or perhaps no other carbon-carbon bond-forming reactions available. For example, using certain transition metal catalysts we can form bonds between alkenyl (vinyl) or aryl substrates and sp2- or sp-hybridized carbons of other reactants. We shall provide examples of a few of these methods here, including the Heck-Mizokori reac­ tion, the Miyaura-Suzuki coupling, the Stille coupling, and the Sonogashira coupling. These reactions are types of cross-coupling reactions, whereby two reactants of appro­ priate structure are coupled by a new carbon-carbon bond. Olefin metathesis is another reaction type, whereby the groups of two alkene reactants exchange position with each other. We shall discuss olefin metathesis reactions that are pro­ moted by Grubbs’ catalyst. Another transition metal-catalyzed carbon-carbon bond-forming reaction we shall dis­ cuss is the Corey-Posner, Whitesides-House reaction. Using this reaction an alkyl halide can be coupled with the alkyl group from a lithium dialkyl cuprate reagent (often called a Gilman reagent). This reaction does not have a catalytic mechanism. All of these reactions involve transition metals such as palladium, copper, and ruthe­ nium, usually in complex with certain types of ligands. After we see the practical applica­ tions of these reactions for carbon-carbon bond formation, we shall consider some general aspects of transition metal complex structure and representative steps in the mechanisms of transition metal-catalyzed reactions. We shall consider as specific examples the mech­ anism for a transition metal-catalyzed hydrogenation using a rhodium complex called Wilkinson’s catalyst, and the mechanism for the Heck-Mizokori reaction.

s

G-1

G .1

C ro s s -C o u p lin g R e a c tio n s C a ta ly z e d b y T ra n s itio n M e ta ls

G-2

G.1 Cross-Coupling Reactions C atalyzed b y Transition M etals G .1 A

The Heck-Mizokori Reaction

The Heck-Mizokori reaction involves palladium-catalyzed coupling of an alkene with an alkenyl or aryl halide, leading to a substituted alkene. The alkene product is generally trans due to a 1 ,2 -elimination step in the mechanism. General Reaction

X

Pd catalyst Base (amine), heat

+

R

X = I, Br, Cl (in order of relative reactivity) Specific Example

NO2

+

Pd(OAc)2 (1 mol %) Bu3N,90°C

What product would you expect from each of the following reactions? (a)

ReviewProblemG.1

CO2H Pd catalyst Base (amine), heat Br

(b) H3CO. +

Pd catalyst Base (amine), heat

Cl What starting materials could be used to synthesize each of the following compounds by a Heck-Mizokori reaction? (a )

G .1 B

O

The Miyaura-Suzuki Coupling

The Miyaura-Suzuki coupling joins an alkenyl or aryl borate with an alkenyl or aryl halide in the presence of a palladium catalyst. The stereochemistry of alkenyl reactants is preserved in the coupling.

ReviewProblemG.2

G-3

S p e c ia l T o p ic G

C a r b o n -C a r b o n B o n d -F o r m in g a n d O t h e r R e a c tio n s

General Reactions

R

1

^

Pd catalyst Base

+

/ B( ° H)2

X Alkenyl borate

R

Aryl halide (or alkenyl halide)

B(OH)2

R2

+

Aryl borate

Pd catalyst Base

Alkenyl halide (or aryl halide)

Specific Example

HO

Br HO.

B(OH)2

Pd(Ph3P)4

+

EtONa, heat

9

Bombykol (a sex hormone released by the female silkworm)

ReviewProblemG.3

What is the product of the following Miyaura-Suzuki coupling? +

ReviewProblemG.4

f-

Br

i

Pd catalyst Base

What starting materials could be used to synthesize the following compound by a Miyaura-Suzuki coupling? O H

The Stille Coupling and Carbonylation

G .1 C

The Stille coupling is a cross-coupling reaction that involves an organotin reagent as one reactant. In the presence of appropriate palladium catalysts, alkenyl and aryl tin reactants can be coupled with alkenyl triflates, iodides, and bromides, as well as allylic chlorides and acid chlorides. General Reaction

R1

X

X = triflate, I, or Br

+

R2

R

Pd catalyst

R2

R

G .1

C ro s s -C o u p lin g R e a c tio n s C a ta ly z e d b y T ra n s itio n M e ta ls

G-4

Specific Example

CO2Et J

+

f SnBu3

CO2Et PdCl2(MeCN)2 DMF, 25°C

Ketones can be synthesized by a variation of the Stille coupling that involves coupling in the presence of carbon monoxide. The following reaction is an example. CO (50 psi) Pd(Ph3P)2Cl2 (1-2 mol %)

I +

Bu3Sn

O

THF, 50°C

ReviewProblemG.5

What is the product of each of the following reactions? (a )

O

Pd catalyst

SnBu3 (b )

(c)

(d )

O Cl +

B u3Sn\^ÿÿi ' \

O2N

Pd catalyst CO2Et

What starting materials could be used to synthesize each of the following compounds by a Stille coupling reaction? (a )

(b )

i-Bu

G .1 D

The Sonogashira Coupling

The Sonogashira coupling joins an alkyne with an alkenyl or aryl halide in the presence of catalytic palladium and copper. A copper alkynide is formed as an intermediate in the reac­ tion. (When palladium is not used, the reaction is called the Stephens-Castro coupling, and it is not catalytic.) In addition to providing a method for joining an alkyne directly to an aromatic ring, the Sonogashira coupling provides a way to synthesize enynes.

ReviewProblemG.6

G-5

S p e c ia l T o p ic G

C a r b o n -C a r b o n B o n d -F o r m in g a n d O t h e r R e a c tio n s

General Reactions

Cul, Pd catalyst Base (an amine)

A lk e n y l h alid e

R

Cul, Pd catalyst Base (an amine)

R

A ryl h alid e Specific Example

HO■ -----=

— H

+

Cul, Pd(Ph3P)4 Et2NH, 25°C '

I

HO

4

ReviewProblemG.7

Provide the products of each of the following reactions. (a)

OH Br-H

(b )

H3CO +

ReviewProblemG.8

Cul, Pd catalyst Base (an amine)

+

/

I

V

c

o

2c h 3

Cu|, Pd catalyst> Base (an amine)

What starting materials could be used to synthesize each of the following compounds by a Sonogashira coupling reaction? (a)

(b)

O

X!

G.2 O lefin M etathesis: Ruthenium Carbene Com plexes and G rubbs' Catalysts Pairs of alkene double bonds can trade ends with each other in a remarkable molecular “dance” called olefin metathesis Greek: to change; Greek: position). The overall reaction is the following.

(meta,

thesis,

R1 [ M ] ^ = r e p r e s e n t s a r u th e n iu m a lk y lid e n e c o m p le x

R2

G-6

G .2 O le fin M e ta th e s is : R u th e n iu m C a rb e n e C o m p le x e s a n d G ru b b s ' C a taly s ts

The generally accepted catalytic cycle for this “change partners” dance was proposed by Yves Chauvin and is believed to involve metallocyclobutane intermediates that result from reaction of metal alkylidenes (also called metal carbenes) with alkenes. The catalysts them­ selves are metal alkylidenes, in fact. Chauvin’s catalytic cycle for olefin metathesis is sum­ marized here.

A lk e n e 2

Richard Schrock investigated the properties of some of the first catalysts for olefin metathesis. His work included catalysts prepared from tantalum, titanium, and molybde­ num. The catalysts predominantly in use today, however, are ruthenium catalysts developed by Robert Grubbs, Grubbs’ so-called first generation and second generation catalysts are shown here.

PCy3 Cl.... R u ^ "'"H I Ph cr PCy3

y

Cl

Ru'

H

Cl ^ I ^ P h PCy3

Cy = c y clo h e xy l G ru b b s, 1995 G ru b b s, 1999 F irs t g e n e ra tio n G ru b b s c a ta ly s t S eco nd g e n e ra tio n G ru b b s c a ta ly s t (c o m m e rc ia lly a va ila ble) (c o m m e rc ia lly a va ila ble) Olefin metathesis has proved to be such a powerful tool for synthesis that the 2005 Nobel Prize in Chemistry was awarded to Chauvin, Grubbs, and Shrock for their work in this area. One example is ring-closing olefin metathesis as applied to synthesis of the anticancer agent epothilone B by Sinha, shown below.

©

2005 Nobel Prize

G-7

S p e c ia l T o p ic G

TBS. O O

C a r b o n -C a r b o n B o n d -F o r m in g a n d O t h e r R e a c tio n s

O

Grubbs 1999 second generation catalyst 75%

*

TBS O X)

O

O

OH

O

[mixture of (Z) and (E, Z) dienes]

Another example is ring-opening olefin metathesis polymerization (ROMP), as can be used for synthesis of polybutadiene from 1,5-cyclooctadiene.

n ReviewProblemG.9

What products would form when each of the following compounds is treated with (PCy3)2C!2Ru= CHPh, one of Grubbs’ catalysts? (a )

„a

O^

.C 6H5

(b)

OTBDMS A

N

xh

O

(C)

O''''"""'"'

,,"""\O

(d)

O

N

O

OH

G .4 S o m e B a c k g ro u n d o n T ra n s itio n M e t a l E le m e n ts a n d C o m p le x e s

G-8

G.3 The Corey-Posner, W hitesides-House Reaction: Use o f Lithium Dialkyl Cuprates (Gilman Reagents) in Coupling Reactions The Corey-Posner, Whitesides-House reaction involves the coupling of a lithium dialkylcuprate (called a Gilman reagent) with an alkyl, alkenyl, or aryl halide. The alkyl group of the lithium dialkylcuprate reagent may be primary, secondary, or tertiary. However, the halide with which the Gilman reagent couples must be a primary or cyclic secondary alkyl halide if it is not alkenyl or aryl. General Reaction

R2CuLi

R' — X

A lith iu m d ia lk y l c u p ra te (a G ilm an rea ge n t)

A lk e n y l, aryl, o r 1° o r c y c lic 2° a lk y l h a lid e

R— R'

-

RCu

LiX

Specific Example

3 (CH3)2CuLi h

CH3Cu

L ith iu m d im e th y lc u p ra te

Lil

75%

The required lithium dimethylcuprate (Gilman) reagent must be synthesized by a two-step process from the corresponding alkyl halide, as follows. Synthesis of an organolithium compound Synthesis of the lithium dialkylcuprate (Gilman) reagent

R— X

2

R— Li

2 Li

Cul

R— Li

R2CuLi

LiX

Lil

All of the reagents in a Corey-Posner, Whitesides-House reaction are consumed stoichiometrically. The mechanism does not involve a catalyst, as in the other reactions of transi­ tion metals that we have studied.

Show how 1-bromobutane could be converted to the Gilman reagent lithium dibutylcuprate, and how you could use it to synthesize each of the following compounds.

ReviewProblemG.10

G.4 Som e Background on Transition M e ta l Elem ents and Complexes Now that we have seen examples of some important reactions involving transition metals, we consider aspects of the electronic structure of the metals and their complexes. Transition metals are defined as those elements that have partly filled (or f ) shells, either in the elemental state or in their important compounds. The transition metals that are of most concern to organic chemists are those shown in the green and yellow portion of the periodic table given in Fig. G.1, which include those whose reactions we have just discussed. Transition metals react with a variety of molecules or groups, called to form In forming a complex, the ligands donate electrons to vacant

d

transitionmetalcomplexes.

ligands,

G-9

S p e c ia l T o p ic G

C a r b o n -C a r b o n B o n d -F o r m in g a n d O t h e r R e a c tio n s

1/IA

H 2

3

o 4

5

6

1.00797

2/IIA

3

4

Li

Be

6.941

9.01218

11

12

Na Mg 22.98977

24.305

3/IIIB

4/IVB

5/VB

19

20

21

22

23

6

/VIB

7/VIIB

24

25

8

/VIIIB 26

9/VIIIB 10/VIIIB 28

27

11/IB

12/IIB

29

30

Cr Mn Fe Co Ni Cu Zn

Ti

V

39.098

40.08

44.9559

47.90

50.9414

51.996

54.9380

55.847

58.9332

58.71

63.546

65.38

37

38

39

40

41

42

43

44

45

46

47

48

K

Ca Sc

Rb Sr

Y

Zr Nb Mo Tc Ru Rh Pd Ag Cd

85.4678

87.62

89.9059

91.22

92.9064

95.94

98.9062

101.07

102.9055

106.4

107.868

112.40

55

56

57

72

73

74

75

76

77

78

79

80

Hf

Cs Ba La 132.9054 1

137.34

Ta W Re Os

138.9055

178.49

180.9479

3

4

5

2

183.85

186.2

190.2

7 Valence electrons 6

8

Ir 192.22

Pt Au Hg 195.09

9

196.9665

1 0

200.59

11

1 2

Figure G.1 Im p o rta n t tra n sitio n elem ents are shown in th e green and ye llo w p o rtio n o f th e p e rio d ic ta b le . Given across th e b o tto m is th e to ta l num ber o f valence electrons (s and d) o f each elem ent.

orbitals of the metal. The bonds between the ligand and the metal range from very weak to very strong. The bonds are covalent but often have considerable polar character. Transition metal complexes can assume a variety of geometries depending on the metal and on the number of ligands around it. Rhodium, for example, can form complexes with four ligands in a configuration called On the other hand, rhodium can form complexes with five or six ligands that are trigonal bipyramidal or octahedral. These typi­ cal shapes are shown below, with the letter L used to indicate a ligand.

squareplanar.

LK \T

L

4«¡&L

L

K,:Rh— I L

L

L

I L

Square planar rhodium complex

Trigonal bipyramidal rhodium complex

L

L

/i

L — Rh— L

Octahedral rhodium complex

G.5 Electron Counting in M e ta l Com plexes Transition metals are like the elements that we have studied earlier in that they are most stable when they have the electronic configuration of a noble gas. In addition to s and orbitals, transition metals have five orbitals (which can hold a total of 1 0 electrons). Therefore, the noble gas configuration for a transition metal is not 8 as with carbon, nitrogen, oxygen, and so on. When the metal of a transition metal complex has 18 valence electrons, it is said to be .*

d

p

18electrons,

coordinativelysaturated

*We do not usually show the unshared electron pairs of a metal complex in our structures, because to do so would make the structure unnecessarily complicated.

G .5 E le c tro n C o u n tin g in M e t a l C o m p le x e s

To determine the valence electron count of a transition metal in a complex, we take the total number of valence electrons of the metal in the elemental state (see Fig. G.1) and sub­ tract from this number the oxidation state of the metal in the complex. This gives us what is called the electron count, The oxidation state of the metal is the charge that would be left on the metal if all the ligands (Table G.1) were removed.

d dn. dn=total number of valence electrons _ of the elemental metal

oxidation state of the metal in the complex

inthecomplex,

dn

Then to get the total valence electron count of the metal we add to the number of electrons donated by all of the ligands. Table G.1 gives the number of electrons donated by several of the most common ligands. total number of valence electrons = of the metal in the complex

dn+electrons donated by ligands

Let us now work out the valence electron count of two examples.

C om m on

L ig a n d s in T r a n s it io n M e t a l C o m p le x e s 3

Ligand

Number of Electrons Donated

Count as

Negatively charged ligands

Hydride, H Alkanide, R Halide, X Allyl anion

4

Cyclopentadienyl anion, Cp

6

Electrically neutral ligands

Carbonyl (carbon monoxide)

:C # O:

Phosphine

R3 P: or Ph3 P: \ /

Alkene

C=C / \

Diene Benzene

aUsed w ith perm ission from the Journal of Chemical c o p y rig h t © 1980, Division o f Chemical Education.

Education, Vol. 57, No. 1, 1980, pp. 170-175,

E x a m p le A Consider iron pentacarbonyl, Fe(CO)5, a toxic liquid that forms when finely divided iron reacts with carbon monoxide.

CO

OCV Fe

+

5 CO

-------»

F e (C O )5

or

I

^ F e — CO O C ^

CO

Iron pentacarbonyl

From Fig. G.1 we find that an iron atom in the elemental state has 8 valence electrons. We arrive at the oxidation state of iron in iron pentacarbonyl by noting that the charge on the complex as a whole is zero (it is not an ion), and that the charge on each CO ligand is also zero. Therefore, the iron is in the zero oxidation state.

G-10

G-11

S p e c ia l T o p ic G

C a r b o n -C a r b o n B o n d -F o r m in g a n d O t h e r R e a c tio n s

Using these numbers, we can now calculate d n and, from it, the total number of valence electrons of the iron in the complex.

dn= - = = d n+5(CQ) = 8

total number of valence electrons

0

8

8

+

5 (2

) =

18

We find that the iron of Fe(CO) 5 has 18 valence electrons and is, therefore, coordinatively saturated. E x a m p le B Consider the rhodium complex Rh[(C6 H5 )3 P]3 H2 Cl, a complex that, as we shall see later, is an intermediate in certain alkene hydrogenations.

H

1Rh

^ %, I

L = PhgP [i.e., (C6H5)3P]

L ^ l NH L Cl H The oxidation state of rhodium in the complex is +3. [The two hydrogen atoms and the chlorine are each counted as - 1 (hydride and chloride, respectively), and the charge on each of the triphenylphosphine ligands is zero. Removing all the ligands would leave a Rh3+ ion.] From Fig. G.1 we find that, in the elemental state, rhodium has 9 valence electrons. We can now calculate d n for the rhodium of the complex.

dn=9 -

3 =

6

Each of the six ligands of the complex donates two electrons to the rhodium in the com­ plex, and, therefore, the total number of valence electrons of the rhodium is 18. The rhodium of Rh[(C6 H5 )3 P]3 H2Cl is coordinatively saturated. total number of valence = ^ + electrons rhodium

6 (2

) =

6

+

12

= 18

G.6 M echanistic Steps in the Reactions o f Som e Transition M e ta l Complexes Much of the chemistry of organic transition metal compounds becomes more understand­ able if we are able to follow the mechanisms of the reactions that occur. These mechanisms, in most cases, amount to nothing more than a sequence of reactions, each of which repre­ sents a fundamental reaction type that is characteristic o f a transition metal complex. Let us examine three of the fundamental reaction types now. In each instance we shall use steps that occur when an alkene is hydrogenated using a catalyst called Wilkinson’s catalyst. In Section G.7 we shall examine the entire hydrogenation mechanism. In Section G .8 we shall see how similar types of steps are involved in the Heck-Mizokori reaction. A transition metal complex can lose a ligand (by dissociation) and combine with another ligand (by association). In the process it undergoes ligand exchange. For example, the rhodium complex that we encountered in Example B above can react with an alkene (in this example, with ethene) as follows:

1. L ig a n d D is s o c ia tio n - A s s o c ia tio n ( L ig a n d E x c h a n g e ).

H 2C = i= C H2

H

L

,''v

L

Rh L ^ l ^ H

H h 2c

=

c h

2

:Rh' Cl

Cl L = PhgP [i.e., (C 6 H 5 ) 3 P]

G .6 M e c h a n is tic S te p s in t h e R e a c tio n s o f S o m e T ra n s itio n M e t a l C o m p le x e s

Two steps are actually involved. In the first step, one of the triphenylphosphine lig­ ands dissociates. This leads to a complex in which the rhodium has only 16 elec­ trons and is, therefore, coordinatively

unsaturated.

H L

H

I '"^R h"

L

L

Rh-

L

L

H

L

H Cl

Cl

(18 e le ctro n s)

(16 e le ctro n s)

L _ Ph3P

In the second step, the rhodium associates with the alkene to become coordinatively saturated again. h 2c

H L

^ =

c h

L R h—

H

2

H Rh

H 2C = C H 2

*Cl H

Cl

(16 e le ctro n s)

(18 e le ctro n s)

complex.

The complex between the rhodium and the alkene is called a p In it, two electrons are donated by the alkene to the rhodium. Alkenes are often called p donors to distinguish them from s donors such as Ph3 P:, CP, and so on. In a p complex such as the one just given, there is also a donation of electrons from a populated orbital of the metal back to the vacant p * orbital of the alkene. This kind of donation is called “back-bonding.”

d

An unsaturated ligand such as an alkene can undergo in ser­ tioninto a bond between the metal of a complex and a hydrogen or a carbon. These reactions are reversible, and the reverse reaction is called d einsertion. I n s e r tio n - D e in s e r tio n .

The following is an example of insertion-deinsertion. H 2C = i= C H

I^

L

Cl

H

L

.Rh I I

insertion

:R h —

, -------------------------------

Cl

deinsertion

CH3

L

/ c h

2

,

L

H

H

(18 e le ctro n s)

(16 e le ctro n s)

In this process, a p bond (between the rhodium and the alkene) and a s bond (between the rhodium and the hydrogen) are exchanged for two new s bonds (between rhodium and carbon, and between carbon and hydrogen). The valence electron count of the rhodium decreases from 18 to 16. This insertion-deinsertion occurs in a stereospecific way, as a of the M— H unit to the alkene.

synaddition

I> C = |= C ^ ^ ^

C—

i

/

M— H

M

C

\ H

Coordinatively unsaturated metal com­ plexes can undergo oxidative addition of a variety of substrates in the following way.*

3. O x id a tiv e A d d itio n - R e d u c tiv e E lim in a tio n .

A

;m :

A— B

oxidative addition

;m : B

*Coordinatively saturated complexes also undergo oxidative addition.

G-12

G-13

S p e c ia l T o p ic G

C a r b o n -C a r b o n B o n d -F o r m in g a n d O t h e r R e a c tio n s

The substrate, A— B, can be H — H, H — X, R— X, RCO— H, RCO— X, and a number of other compounds. In this type of oxidative addition, the metal of the complex undergoes an increase in the number of its valence electrons Consider, as an example, the oxidative addition of hydrogen to the rhodium complex that follows (L = Ph3P).

andinitsoxidationstate. H

L Rh

oxidative addition H— H

Rh

reductive elimination

Cl

L Cl

H

(18 electrons) Rh is in +3 oxidation state.

(16 electrons) Rh is in +1 oxidation state.

Reductiveeliminationis the reverse of oxidative addition. With this background, we are now in a position to examine the mechanisms of two applications of transi­ tion metal complexes in organic synthesis.

G.7 The Mechanism fo r a Hom ogeneous H ydrogenation: Wilkinson's Catalyst The catalytic hydrogenations that we have examined in prior chapters have been hetero­ geneous processes. Two phases were involved: the solid phase of the catalyst (Pt, Pd, Ni, etc.), containing the adsorbed hydrogen, and the liquid phase of the solution, containing the unsaturated compound. In homogeneous hydrogenation using a transition metal com­ plex such as Rh[(C6H5)3 P]3Cl (Wilkinson’s catalyst), hydrogenation takes place , i.e., in solution. When Wilkinson’s catalyst is used to carry out the hydrogenation of an alkene, the fol­ lowing steps take place (L = Ph3P).

inasin­

glephase Step1

H L

La ' Rhr^

H— H

->

Cl

l

16 valence electrons

Oxidative addition

.Rh Cl

H H

18 valence electrons

Step 2 H L ///„,, I Ä*\\L ^Rh" L ^ l ^ Cl

H

18 valence electrons

H L Rh— H L Cl 16 valence electrons

L

Ligand dissociation

G .7 T h e M e c h a n is m f o r a H o m o g e n e o u s H y d r o g e n a tio n : W ilk in s o n 's C a ta ly s t

G-14

Step 3 H2 C =i=C H

H

L Rh— H +

H,C=

=c h ,

H

L ig an d a s s o c ia tio n

Rh L ^ l X Cl H

Cl

16 v a le n c e e le c tro n s

18 v a le n c e e le ctro n s

Step4 h 2c = = c h 2 L

L

H :Rh H

Cl / H3 Rh— Ch 2

^

In s e rtio n

L

^ C l

H

18 v a le n ce e le c tro n s 16 v a le n c e e le ctro n s

Step5 Cl

CH3

,

,

:Rh — CH2

Rh— Cl

H3C

CH

, H

16 v a le n c e e le c tro n s

R e d u ctive e lim in a tio n

3

14 v a le n c e e le ctro n s

Step6 H ,

, Rh

Cl

H2

,

O x id a tiv e a d d itio n

Rh— H , Cl

14 v a le n c e e le c tro n s 16 v a le n c e e le ctro n s (C ycle re p e a ts fro m ste p 3.) Step 6 regenerates the hydrogen-bearing rhodium complex and reaction with another molecule of the alkene begins at step 3. Because the insertion step 4 and the reductive elimination step 5 are stereospecific, the net result of the hydrogenation using Wilkinson’s catalyst is a of hydrogen to the alkene. The following example (with D2 in place of H2) illustrates this aspect.

synaddition

H

H D2

EtO2 C^

^C ü2Et

A c/s-alken e (d ie th y l m aleate)

Rh(Ph3P)3a

H H EtO2 C ^ ___^ *C O 2Et D

D

A m eso c o m p o u n d

cis-

What product (or products) would be formed if the trans-alkene corresponding to the alkene (see the previous reaction) had been hydrogenated with D2 and Wilkinson’s catalyst?

ReviewProblemG.11

G-15

S p e c ia l T o p ic G

C a r b o n -C a r b o n B o n d -F o r m in g a n d O t h e r R e a c tio n s

THE CHEMISTRY OF . . . H o m o g e n e o u s A s y m m e t r i c C a t a l y t i c H y d r o g e n a t i o n : E x a m p l e s I n v o lv in g l- D O P A , ( S ) - N a p r o x e n , a n d A s p a r t a m e

Development by Geoffrey Wilkinson of a soluble catalyst for hydrogenation [tris(triphenylphosphine)rhodium chloride, Section 7.13 and Special Topic G] led to Wilkinson's earn­ ing a share of the 1973 Nobel Prize in Chemistry. His initial discovery, while at Imperial College, University of London, inspired many other researchers to create novel catalysts based on the Wilkinson catalyst. Some of these researchers were themselves recognized by the 2001 Nobel Prize in Chemistry, 50% of which was awarded to William S. Knowles (Monsanto Corporation, retired) and Ryoji Noyori (Nagoya

University). (The other half of the 2001 prize was awarded to K. B. Sharpless for asymmetric oxidation reactions. See Chapter 8 .) Knowles, Noyori, and others developed chiral catalysts for homogeneous hydrogenation that have proved extraordinarily useful for enantioselective syntheses ranging from small laboratory-scale reactions to industrial- (ton-) scale reactions. An important example is the method devel­ oped by Knowles and co-workers at Monsanto Corporation for synthesis of l-DOPA, a compound used in the treatment of Parkinson's disease:

A s ym m etric Synthesis o f l-D O PA H 3 CO

_COOH NHAc

H 3 CO

COOH

H2 (100%) [(Rh(R,R)-DIPAMP)COD]+BF4 ~(cat.)

AcO

HO

COOH

H3 O+ H NHAc AcO

H NH, HO

(100 %yield, 95% ee [enantiomeric excess])

O

L-DOPA

II

(R,R)-DIPAMP (Chiral ligand for rhodium)

COD = 1,5-Cyclooctadiene

Another example is synthesis of the over-the-counter analgesic naproxen using a BINAP rhodium catalyst developed by Noyori (Sections 5.11 and 5.18). A s ym m etric Synthesis o f (S)-Naproxen CH2 m nu COOH

H

+

(S)-BINAP-Ru(OCOCH3 ) 2 (0.5 mol%) Ho — MeOH

H3CO

CH3 COOH

' H3CO

(S)-Naproxen (an anti-inflammatory agent) (92%yield, 97% ee)

P(Ph ) 2

(Ph)2P

P(Ph ) 2

(Ph)2P

(R)-BINAP (S)-BINAP (S)-BINAP and (R)-BINAP are chiral atropisomers (see Section 5.18).

G-16

G .8 T h e M e c h a n is m f o r an E x a m p le o f C ro s s -C o u p lin g : T h e H e c k -M iz o k o r i R e a c tio n

Catalysts like these are important for asymmetric chem­ ical synthesis of amino acids (Section 24.3D), as well. A final example is the synthesis of (S)-phenylalanine methyl ester,

a compound used in the synthesis of the artificial sweetener aspartame. This preparation employs yet a different chiral ligand for the rhodium catalyst.

A s ym m etric Synthesis o f A sp artam e COOH NHAc

(1 ) (KH)-PNNP-Rh(I) (cat.), H2 (83% ee) (catalytic asymmetric hydrogenation) (2) MeOH, HA

Ph

H NH

(S)-phenylalanine methyl ester (97% ee after recrystallization)

Ph >----- N '

C >C

(Ph)2P:

CO O CH3

N— =P(Ph)2

CH,

(R,R)-PNNP (Chiral ligand for rhodium)

H

/

/ z

HOOC.

COOH

H2 N H (S )-a s p a rtic acid

NH

COOH

COOCH 3

Aspartame

The mechanism of homogeneous catalytic hydrogenation involves reactions characteristic of transition metal organometallic compounds. A general scheme for hydro­

genation using Wilkinson's catalyst is shown here. We have seen structural details of the mechanism in Section G.7.

Cl[(C6H5)3P]2Rh

2 Cl[(C6H5)3P]3Rh

A general mechanism fo r th e W ilkinson catalytic h yd rogenation m e th o d , adapted w ith perm ission o f John W ile y & Sons, Inc. from N oyori, Asymmetric Catalysis in Organic Synthesis, p. 17. C o p y rig h t 1994.

(Cft)3P > Cl[(C6H5)3P]2Rh

Cl[(C6H5)3P]2RhH

H

G.8 The Mechanism fo r an Exam ple o f Cross-Coupling: The H e ck -M izo ko ri Reaction Having seen steps such as oxidative addition, insertion, and reductive elimination in the context of transition metal-catalyzed hydrogenation using Wilkinson’s catalyst, we can now see how these same types of mechanistic steps are involved in a mechanism proposed for the Heck-Mizokori reaction. Aspects of the Heck-Mizokori mechanism are similar to steps proposed for other cross-coupling reactions as well, although there are variations and cer­ tain steps that are specific to each, and not all of the steps below are involved or serve the same purpose in other cross-coupling reactions.

G-17

S p e c ia l T o p ic G

C a r b o n -C a r b o n B o n d -F o r m in g a n d O t h e r R e a c tio n s

A MECHANISM FOR THE REACTION T h e H e c k - M i z o k o r i R e a c t i o n U s in g a n A ry l H a l i d e S u b s t r a t e G E N E R A L R E A C T IO N

Ar— X

+

^

R

B Pd,cala'ysl > Base (an amine)

^

M E C H A N IS M

Pd(L)4 - 21-

(L = ligand, e.g., Ph3 P) Ar— X

base — HX Pd(L)2 Coordinatively unsaturated catalyst

Reductive elimination (regenerates catalyst) base

Oxidative addition (incorporates halide reactant)

H— Pd(L)2— X

Ar— Pd(L)2— X

f Ar R

Alkene insertion (incorporates alkenyl reactant, forms new C—C bond)

1 ,2 -syn elimination (forms the product as a trans alkene)

Ar Ar H

H R

—C bond rotation

H h

G.9 Vitam in B ^ : A Transition M e ta l Biomolecule The discovery (in 1926) that pernicious anemia can be overcome by the ingestion of large amounts of liver led ultimately to the isolation (in 1948) of the curative factor, called vit­ amin B 12. The complete three-dimensional structure of vitamin B 12 [Fig. G.2(a)] was elu­ cidated in 1956 through the X-ray studies of Dorothy Hodgkin (Nobel Prize, 1964), and in 1972 the synthesis of this complicated molecule was announced by R. B. Woodward (Harvard University) and A. Eschenmoser (Swiss Federal Institute of Technology). The synthesis took 11 years and involved more than 90 separate reactions. One hundred co­ workers took part in the project. Vitamin B 12 is the only known biomolecule that possesses a carbon-metal bond. In the stable commercial form of the vitamin, a cyano group is bonded to the cobalt, and the cobalt is in the +3 oxidation state. The core of the vitamin B 12 molecule is a [Fig. G.2( )] with various attached side groups. The corrin ring consists of four pyrrole

b

corrinring

G-18

G .9 V ita m in B 12: A T ra n s itio n M e t a l B io m o le c u le

subunits, the nitrogen of each of which is coordinated to the central cobalt. The sixth lig­ and [(below the corrin ring in Fig. G.2(a)] is a nitrogen of a heterocyclic group derived from 5,6-dimethylbenzimidazole. The cobalt of vitamin B 12 can be reduced to a + 2 or a +1 oxidation state. When the cobalt is in the +1 oxidation state, vitamin B 12 (called B 12s) becomes one of the most pow­ erful nucleophiles known, being more nucleophilic than methanol by a factor of 1 0 14. Acting as a nucleophile, vitamin B 12s reacts with adenosine triphosphate (Fig. 22.2) to yield the biologically active form of the vitamin [Fig. G.2(c)].

A carbon-cobalt s bond

2

OH

OH

(c)

(a)

Figure G.2 (a) The structu re o f vitam in B1 2 . In th e com m ercial fo rm o f th e vitam in (cyanocobalamin), R = CN. (b) The corrin ring system. (c) In th e b io lo g ica lly active fo rm o f th e vitam in (5'-deoxyadenosylcobalam in), th e 5 ' carbon atom o f 5 '-d e oxyadenosine is coord in a te d to th e cobalt atom . For th e structure o f adenine, see Section 25.2.

fw iL E ’T 'to

See Special Topic H in WileyPLUS

S e e Table 2.7 for a Table o f IR frequencies

Frequency (cm 40 0 0

3600

3200

2800

2400

2000

1800

v-

Alkane Alkene Alkyne Aromatic 1° alcohol 2° alcohol 3° alcohol Phenol Ether Ester Carboxyllc acid Ketone Aldehyde Amide 1° amine 2° amine 3° amine Alkyl chloride Alkyl bromide N itrile

1600

1400

800

1000

5=

■H

5= -H

=C

ertor

•-H vO-

free

600

sk

vC=

=C

1200

H

i>0- H

H

sociated

broad

'-H

C=

vO=

-H

XI-

2.5

N-H

N

vN-H

r

3.0

3.5

v=

déisr 'I

I1 'I

I

I

4.0

I

vC-B

W

-W-id I

5

ll'l'i

I

I

I

I

(Microns)

T y p ic a l IR a b s o r p t io n fr e q u e n c ie s fo r c o m m o n fu n c tio n a l g r o u p s .

Absorptions are as follows: stretching; 8 = bending; w = weak; m = medium; s = strong; sk = skeletal From Multiscale Organic Chemistry: A Problem-Solving Approach by John W. Lehman © 2002. Reprinted by permission of Pearson Education, Inc., Upper Saddle River, NJ.

I I I

10

I I

11 12 13 14

16

2 2 0

2 0 0

1S 0

1 6 0

1 4 0

1 2 0

1 0 0

8 0

6 0

4 0

2 0

5c (p p m )

A p p r o x im a t e C a r b o n - 1 3 C h e m ic a l S h ifts

r TABLE 9 . 2 ^

Type of Carbon Atom

Chemical Shift (S, ppm)

1° Alkyl, RCH3 2° Alkyl, RCH2R 3° Alkyl, RCHR2

G-4G 1G-5G 15-5G

i

Alkyl halide or amine, — C—X I X= Cl, Br, or N —

1G-65

Alcohol or ether, — C— O—

5G-9G

Alkyne, — ç =

6G-9G

\ Alkene,

Aryl, Nitrile,

_ C—

1GG-17G

1 GG-17G

C

1 2 G-1 SG

C= N O

Amide,

C

15G-18G

N O

Carboxylic acid or ester,

C

O

16G-1S5

O Aldehyde or ketone,

C

182-215

0

A p p r o x im a te

Type of Proton

Chemical Shift (5, ppm)

1° Alkyl, RCH3 2° Alkyl, RCH2R 3° Alkyl, R3 CH Allylic, r 2c = C— CH3 2

I

R

Ketone, RCCH3 II O

P r o to n C h e m ic a l S h ifts

0 .8 - 1 .2

1.2-1.5 1.4—1 .8 1.6-1.9

3 2 .1 - 2 .6

3

Benzylic, ArCH3 Acetylenic, RC # CH Alkyl iodide, RCH2 I Ether, ROCH2R Alcohol, HOCH2R

2.2-2.5 2.5-S.1 S.1-S.S 5.5-S.9 3.3-4.0

Type of Proton

Chemical Shift (S, ppm)

Alkyl bromide, RC^Br Alkyl chloride, RCH2 CI Vinylic, R2 C " CH2 Vinylic, R2 C = C H I R

S.4-S.6 5.6 -S .8 4.6-5.0 5.2-5.7

Aromatic, ArH Aldehyde, RCH II O

6.0-8.5 9.5-10.5

Alcohol hydroxyl, ROH Amino, R9 NH2 Phenolic, ArOH Carboxylic, RCOH II O

0.5-6.0 1.0-5.0a 4.5-7.7a 1 0 - 1 Sa

aThe chemical shifts o f these protons vary in d iffe re n t solvents and w ith te m p e ra tu re and concentration.

a