Pushing electrons - A guide for students of organic chemistry

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Puslvng Electrons

1°ANO Q.ORG.I 3,36€

A Guide for Students of Organic Chemistry Third EdItkni

Daniál R Weeks, Ph.D. Northwesterfl University

THOMSON BROOKS/COLE

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tfrl

Copyrighl O 1998 by’tbomson Leaming, me. Thomson Leaniingw ís a lrademark used herein under license. ALL RIGIITS RESERVED. No part of (his work covered by Lhe copyright hereon may be reproduced or used 10 any form or by any means — graphic, eleclronic, or mechanical, including pholocopying. recording, laping, Web distribution, or inforinauon storagc or relrieval sysenis — without lhe wrilten permission o( lhe publisber. For infonuation about our producis, conlaci us: Thomson Lcarning Academie Resource Cenler 1.800-423-0563 htlpj/wWW.wadsworth.coni For penflission Lo use material from lhis texi, conlact us by Web: http://www.lhoinsonrighls.com Fax: 1-800-730-2215 Phone: 1-800-730-2214 Prinled o Lhe Uniled SLales of America ~3 ~4

ISBN 0-03-020693-6

‘~3j34J/~e To my sons, John and Timothy

Postcard trom a Stuilenl

Dear Dan, Last night 1 was sitting on Lhe shore of San Francisco Bay watching lhe fog roIl IR and suddenly 1 underslood what organic chemistry was ali about. Ric

Preface

Among Lhe faciors that set organic chernislry apart from Lhe oLher Lradilional divisions ofchernisiry are Lhe endless varieLy of sLructure, lhe sLrong dependence on lhe resonance Lheory, and Lhe ernpha sis on mechanisms. As sLudents become more facile wiLh sLnicLure and more adroit at wriling reso nance slruclures and mechanisms, organic chernisLry changes from a bewiidering array of facLs lo a unified science. Newly encounLered reactions can be undersLood from previousiy Iearned principies. AL lhe inLroducLion of resonance or rnechanisrns, lhe professor begins using iiLlie arrows LhaL resembie Iishhooks (r-’. ,r-’~). Using Lhese arrows, eiecLrons are moved around moiecuies and appear—as if by wizardry—precisely where lhe professor needs Lhern. The penchant for rnoving electrons has eamed organic chemisLs Lhe pejoralive electron pu.sher LhaL, like Yankee Doodie, we wear proudiy. 1 have observed LhaL sLudents avoid pushing electrons. They find iL difficuit, Lherefore, Lo wrile resonance strucLures and lose a powerfui Lool Lo expiam reacLiviLy. Many sludenls never become comfortabie wilh bond-making and bond-breaking steps in organic mechanisms. WiLh 1h15 iow-lech, pencii-and-paper program, a sLudenL can ieam Lo push electrons wiLh Lhe sarne wizardry as Lhe professor. Whiie working lhrough Lhis book, lhe sLudenL wili Iearn lo conslruct Lewis sLrucLures of organic molecuies and functionai groups. The functionai groups wiIi be iearned early and in a way LhaL minimizes Lhe necessiLy for memorizalion. Then a sysLematic melhod is presenLed whereby ali Lhe possible resonance slrucLures for a moiecuie, ion, or radical can be generaLed by beginning wiLh one Lewis sLrucLure and moving eleclrons. Chapler 3 demonslraLes lhe individuai bond rnaking and bond breaking steps in reacLion mechanisms. Chaplers 4 and 5 are new Lo this ediLion. In Chapter 4 sludenls newiy acquired abiiily Lo push eiecLrons is used lo consirucL cornpieLe rnechanisms. The book conciudes wilh a chapler in which eiecLron pLlshing skiiis are applied Lo Lhe reaclions of biochemisLsy. The programrned approach of this book ernphasizes repeliLion and acLive parlicipaLion in iearning how lo push eieclrons. In Lhe producLion of three edilions of lhe book, i have received help from Dr. Joyce C. Brockweii and Professor Ernerilus Charies D. Hurd (Norlhweslern Universily), Dr. Rebecca Dickslein (Drexel UniversiLy), Dr. Kenneth Turnbuli (Wrighl Slale Universily), and Dr. Darreil J. Wcodman (IiniversiLy of Washington). Whiie relirement Lo rural westem Virginia has iLs advan tages, 1 needed access lo a chemisLry iibrary and vigorous chemisLry faduiLy in order lo wrile Lhe lhird edition. 1 found both aL Virginia Polylechnic insLiLule and SIate liniversity (Virginia Tech). vii

viU

Preface

Professor Richard Gandour, ChemisLly Department Head, provided a generous welcome. As always, Iam grateful to John Vondeling and Sandi Kiselica of Saunders Coliege Publishin for Iheir encouragernenL and paLience. Daniel R Weeks June 1997

o t e ro essor ___________________________________

Students wili be belter ai organic chemisLry if Lhey masier lhe skiils in Pushüzg Electrons. The book is a suppiemenL. IL knows lIs place. IL is short, melhodical, and repeiitious. A student wiil complete Pushüig Electrons in 12 Lo 16 hours and have Lhe rest of Lhe Lime for importam Lhings such as lis Lening lo you and sLudying Lhe main LexI. The approach is melhodical, Lhe pace is siow. Strong siudenLs wiIl race Lhrough certain sec tions, buL even Lhey wilI benefiL. 1 teil (bem in lhe LexL, “you can never be 100 skilIe4J aI pushing elec Lrons’ Weaker sLudenLs wiii appreciaie LhaL Lhe author sLiil holds a grudge againsi bis calculus pro fessors who skipped sLeps. The repetiiion used in teaching sLudenLs how Lo wriLe Lewis slmcLures and how• properly to push electrons has an addiLional benelit. SLudenls become comforLable writing and rewiiling organ ic siructures. The book may noL eliminaLe penLavalenL carbon buL it should reduce Lhe frequency of iLs appeai-ance. The stmcLures in Lhis book, except for rings and larger sLructures in Lhe lasL lwo chapters, show ali carbons and hydrogens. 1 assert tha lhe early iniroducLion of line-angle simcLures does sLuderits greaL hann. Assign Chapter 1, Lewis SlrucLures, right away. Chapters 2 and 3 are relatively independent of one anolher; assign lhem as you see lit. IL is noL necessary for sLudents lo complete ChapLer 2 before beginning ChapLer 3. AlLhough a chapLer may be inLemipLed, ii should be done from begin ning lo end. There are two significanL addilions lo Lhe Lhird ediLion. 1 have expanded Lhe ireatmenL of bio chemistry in Chapter 5. Consider Lhis as optional depending on wheLher or nol your course contains biochemistry. Chapter 4 describes straLegies for deaiing wiLh mechanism problems. 1 suggest Lhat you assign parIs of II as your sLudenls encounler mechanism problems of increasing difficulty. Once again, you may intenupt Chapier4 but assign ii from beginning Lo end.

To the Student

have Lhree instructions. Fira supply an answer wherever a une appears under a blank space. AcLualIy, ia Chapter 4 only blank spaces are provided, buL by Lhat Lime you wiII know whaL is required. The conect answer wiIl be a word, a number, a sLnjcLure, some arrows, or—in ChapLer 4—a mechanism. Second, don’t look up Lhe correct answers until you have made a serious try aL doing it yourself. Third, if you plan Lojust Iook up Lhe righL answers and transcribe Lhem, reLurn Lhe book and get your money back. This program uses Lwo efíecLive Iearning devices: active involvemeaL and repelition. You wifl participate acLively in Lhe Iearning process. Because só much of Lhe academic experience consists of receiving informaLion, it should be refreshing Lo work through a program using your own wils. You wiII see an example of an operation asid Lhen carry it ouL several Limes as Lhe supporting matedai is gradually removei The approach is meLhodical. Some of you wilI find LhaL you can acceler ate your trip through certain secLions. BuL lhe program has been written ia the hope ±at none of you will ever feel abandonei Expect to spend a toLal of 12 Lo 16 hours Lo complele Lhe program. 1 hope your professor wilI assign appropriate sections of lhe book as Lhe course proceeds. Failing LhaL, do Chapter 1 righL away. Then, slart ChapLers 2 and 3 (Lhey are relaLively independent of each other). If you sLart Lo feel losI, puL it aside for awhile and Lhen pick up where you IefL off. Begin ChapLer 4 when you need some help with solving mechanism problems. Chapter 5 is for courses where some biochemistry is inLroduced Loward Lhe end. From time Lo Lime during lhe course, 1 Lhink you wiII find Pushing Electrons useful for review. This is especially Lrue of ChapLer 3, lhe backbone of lhe book.

Contents

Preface Giossary

Lewis Structures

1

Molecuies and Functionai Groups

1

Saturated Groups Unsaturated Groups Larger Molecules Formal Charge lons

2

vii xv

1 6 14 20 25

Cations Anions

25 31

Free Radicais Answers

36 39

Resonance Structures

49

Simpie ions

50

Cations Anions Free Radicais Benzene and Benzenoid Aromatic Compounds More Compkcated lons Moiecuies Having Resonance Siructures with Charge Separation Answers

50 55 56 60 65 72 77

xiv

3

Contents

Mechanisms

88

Sigma Bond Breaking Sigma Bond Making Simultaneous Bond Making and Breaking

88 92 95

Sigma Bond Makíng and Sreaking Homolytic Sigma Bond Making and Bond Breaking Sigma Bond Making and Pi Bond Breaking Pi Bond Making and Sigma Bond Breaking Homolytic Sigma Bond Making and Pi Bond Breaking

4

5

________________________________________________________

95 102 105 110 118 120 127 135 138

Complex Mechanisms Rearrangements Ring Closures Answers

Glossary

Boldtaced numbers reter lo lhe page on which lhe term tirsl appears. Aliphatic One of two broad classes of organic compounds. The other is aromaric. Aliphacic compounds are composed of chains of carbon atoms. A

O fl o~vang ‘ k iviecnanlsm ro j ems

162

subgroupaloms wichinbutLhis is Lhe compounds ibaL inconlain ringsand of carbon areclass similar to alicyclic open-chain compounds chemical

Theme l—The Intercepted Intermediate Theme lI—Rearrangement Theme lil—An Unexpected Produci Theme IV—Stereochemistry

166

physical properties. Actually, the most sensible definition of aliphatic com pounds is “compounds Lhat are not aromatic’ 16 Aromatic One of two broad classes of organic compounds. The oLher is aliphafic. Benzene, compounds containing a benzene ring, and compounds

•

•

169 170

Answers

175

similar Lo benzene in chemical properties constiwte the aromaLic com pounds. 16

Some Reactions from Biochemistry

182

Carbanion An ou ia which a negative charge resides on a carbon aLom. 34

Bioorganic Reactions Enzymes Coenzymes Answers

182 187 193 198

Carbocation atom. 34

An ion in which a posiLive charge resides on a carbon

Decarboxylation A reactiori in which a carboxylic acid loses carbon dioxide. 164 Double bond The combination of a sigma bond and a pi bond between Lwo atoms. The carbon—carbon double bond is written C=C. In this notation Lhere is no convention regarding which une represents lhe sigma and which, lhe pi bond. The presence of a double bond makes a compound unsaturated. 7 Electron attracting and electron releasing groups Various groups of atoms found ia organic compounds are electron attracting or releasing, relative to Lhe carbon aLom to which they are atlached. The presence of lhese groups can have a marked effecL on chemical reactivity. Some elecLron releasing groups are: —NH2, —OCH3, and —CH3. Some electron attracting groups are: —NO2, —COOH, —F, —CI, Br, and CN. 74

xvi

Giossary

Glossary

Electrophile The word comes from lhe Greek meaning “electron lover.” Eiectrophiles are chemical entities (atoms, ions, radicais, or molecuies) lhat seek elecu-ons. Electrophiles are electron-poor. They are designated as E or E+. When an electrophiie reacts with some electron-rich center on au organic molecule, Lhe process is called electrophilic attack. 72

Nncleophiie The word comes from lhe Greek, meaning “nucleus lover.” Nucieophiles are chemical entiúes (ions or molecuies) (hat seek an electron poor center. Nucieophiies are eieclron-rich. They are designated as Nu or Nu. When a nucleophile reacis wilh some electron-poor center of an organic molecule, lhe process is cailed nucleophilic altack 72

Formal charge A Lewis structure is, after ali, only a graphic representa tion of a molecular strncture. As useful as Lhey are, Lewis sinictures are nol enlirely descriptive of a molecule. The formal charge is lhe charge that wouid be on an atom in a molecule if lhe Lewis structure were an entirely accurate representation of lhe molecule. ln one seuse, formal charge is sim piy electron bookkeeping. Rowever, lhe location of fonnal charge on a Lewis stnicture is always revealing oí Lhe chemistry of lhe molecule. 20

Pi bond and pi electrons Bonds that are lhe resuir of lateral (side Lo side) overiap of atomic p-orbitals. Pi bonds are weaicer ffian sigma bonds. Pi bonds are always found in combination with a sigma bond. Therefore, single bonds are never pi bonds. Pi electrons are lhe electrons in pi bonds. 50 Carbocation Carbocations are classified as primar~’, secondary, or ler tiary according Lo lhe following scheme.

Free radical A cheniical species having at least one unpaired valence electron. 36, 58, 102

Primary The carbon atom with lhe positive charge is direcily attached ro only mie olher carbon atom and two hydrogen atoms. For exampie,

Functionai group One can think of organic compounds as being com posed of two parIs: a hydrocarbon backbone and a functionai group. The lalter is au atom or group of atoms Lhat confer some distinct chemical or physicai property. The functional group is lhe site of chemical reactions. Thus, molecules with lhe sarne functional group undergo similar chemical reactions despite having quite different hydrocarbon backbones. 22

CH3—CH2

Grignard reagent A reagent formed by reaction of an alkyl halide wilh magnesium iii au elher solvent. These reagents are extremely usefui and versatile intermediates for synthesizing organic compounds. The actual siructure of lhe Grignard reagent is rather complex. lt is expedient, how ever, lo consider it au ion pair, as this text does. 97 Heterolytic cleavage (heterolysis) The breaking of a bond so that Lhe electrons Lhat formed Lhe bond are distributed uneveniy between the Lwo fragments. i-Jeteroiysis results in an anion and a cation. 27 »

+

Secondary The carbon atom with lhe positive charge is directly auached Lo Lwo other carbon atoms and one hydrogen atom. For example, +

CH3—CR--CR3 Tertiary The carbon atom with lhe positive charge is directiy attached Lo Lhree other carbon atoms and rio hydrogen aloms. For example,

CR3 CR3—C4 CR3 The order of stability of carbocations is tertiary > secondary 123, 124

>

primary.

Af1-:B Protonation The combination of a proton (hydrogen ion) with some organic base Lo give lhe conjugale acid of thaL base. 90

or A:—

+

The counterpart of heterolysis is homolysis. in which lhe electrons are dis tributed eveniy between Lhe two fragments. The result is a pair of radicais. ~

A’

+

Metabolisni The sum total of ali biochemical reaccions that take place in au organism. 185

li.

+

:8

b

R—B

Regiochemistry The study of reactions in which one direction of bond maldng or bond brealdng is preferred over ali olhers. 166 Saturated Compounds Lhat contam oniy singie bonds cannol add olher reagents. Thus, they are cailed salw-ated 1 Secondary carbocalion (See Carbocation.)

xvii

xviii

Glossary

Sigma bond and sigma electrons Bonds Lhat are lhe resuil of axial over Iap of atomic orbiLais. Ali singie bonds are sigma bonds. Sigma electrons are lhe elecirons in sigma bonds. 88, 96 SN1 and SN2 reactions Organic reacLions can be classified according Lo ibeir mechanisms. The 5N 1 and 5N2 reacLions are probably the besL known of Ibese. This classilication gives: lhe result, Lhe type of reagenL, and Lhe moleculariLy of lhe reaction. Thus, 5N’ sLands for “subsLituLion, nucle ophiiic, unimolecular,” and 8N2 siands for “substiLution, nucleophilic, bimoiecular’ 88,95

1

Lewis Siruclures

Stereochemisíry The study of lhe arrangemenL of aLoms in three-dimen sional space and how Lhe arrangemenL affects Lhe chemical and physical properties of malecules. 95, 170 Steric strain Sirain (instabiliLy) presem in a molecule because of lhe arrangement of lhe acoms. Steric sLrain usuaily arises because bond angies are forced Lo depart from their ideal value, or because nonbonded atams are required to be Loo dose Lo one another. 49 Tertiary carbocation

(See Carbocation.)

Unsaturated Compounds lhat conLain double or Lripie bonds are capable of adding olher reagenis. Thus, ihey are called unsaturated. 6 Unshared electrons Valence elecuans in a molecule LhaL are not involved ia binding ane aLom Lo anolher. la a Lewis sLrucLure ihey are Lhe exclusive property of one aLom. They are also called nonbonding eiecLrons. 4

In a perfect worid there wouid be no needfor this chapter ai lhe beginning of a course in oJganic chemistry. The sldlls described here are taught iii general chemis:ry. Skip ihis chapter if you feel enurely confideni that you are able lo write Lewis siruclures correctly and are co,nforiable with lhe concepis o! ions, radicais, multipie bonding, andformal charge. On the other hand, ibese sl 52. (7) 4,

(26)

(27)

o +

—

72

Chapter 2 Resonance Structures

53. (3)

H Cl

Molecules Having Resonance Structures with Charge $eparation

55. (4)

CH3

1-1

:0:

4

.

í~Ni +

~

~ —

7’

ci

c \•,

73

li Nogood! is not considered because ii makes no sense Lo push elecLrons away from Lhe more electronegaLive atom.

Neither is ii acceplabie lo propose a diradical resonance structure since lhal violares lhe principie thal ali resonance slnicrures for a single inolecule musI contaín lhe sarne nuinber of paired and unpaired eleclrons.

N. ‘N

:0:

II

Mo~ecuIes Having Resonance Structures with Charge Separation So far in Lhis chapLer ali Lhe resonance slnictures for a given molecuie, ion, or radical have possessed Lhe sarne formal charge distribuLion. These resonance stnicLures are signiílcant because Lhe energ:es of Lhe vabous stnicwres are sirnilar and, Lhus, each sLrucLure makes an importanL contribuLion Lo lhe hbrid 1 ln many cases, however, molecules for which there is a Lewis slnicture wiLh no formal charge separaLion also have resonance stnicLures with charge separation. These sLructures make smaller, albeit finite, contribuLions Lo Lhe hybrid. The Lewis slructure for fonnaldehyde is

/C=9• 4

~

H

H No good!

56. The Lewis sltucture for acrolein (propenal) is H CH2CIJ—C /

Pushing lhe pi eiectrons of lhe carbonyl bond toward lhe oxygen atom wiiil generate a second resonance sLructure.

H

H

/

c=o:

H

..

/ CH2=CH~C\

Of Lhe Lwo aLoms, carbon and oxygen, thaL are sharing Lhe pi elecLrons of Lhe carbonyi bond, oxygen is Lhe more elecLronegative. Therefore, iL can act as a receptor. Pushing Lhe pi eiectrons Lo lhe oxygen atom generaLes a new resonance strucLure.

4-11

c=o:

generaLes

C—O:

II

generates ________________________

supply arrow The second sLnicliire has a païr of pushable eleclrons (namely, Lhe pi eiecLmns of Lhe carbon— carbon double bond) nexL to a receptor, Lhe carbonyl carbon atom, which possesses a formal

H /

O ________________________

posiLive charge.

/

lhe inLroduction of charge separalion into a resonance structure usua]ly increases Lhe energyoflhat strucLure relaLive Lo one in which noseparaLion of charge appeaisllie uncharged sLructure malces a subsLanLially greaterconLribuLion to Lhe hybrid. However,proposing Lhal lhe charged sLruclure rnakes a smaller buL limite contribuLion emphasizes Lhe electrophiiic nature of a carbonyl carbon atom and lhe nudeophilic nature of Lhe carbonyl oxygen alom.An allemaLive sLnictLlre ia which Lhe pi elecLrons are pushed toward the carbon atom, i,e.,

H +/ CH2 = CH — C

—

g

supply anow

74

Chapter 2 Resonance structures

Molecules Having Resonance Structures with Charge Separation

Don’t refer back to compleLe Lhe following. The resonance method notation for acrolein is

75

of pi elecLrons ia Lhe 11—0 double bond are pushed Loward Lhe oxygen atom. That is,

:pç~~~.q:

1~%) 57. The Lewis sLnicLure for N-methylacetamide is shown. Pushing lhe pi elecirons of Lhe carboayl

gives

stipply arroWs (301

boad toward Lhe oxygen abro wiII generale a new resonance structure.

(31)

la (31) a pair of pi electrons ia Lhe ring can be pushed toward Lhe positive charge. .0

11/

CH3

CH3—C—N

“II

gives

________________________

________________________

+

supply agoW

generates

The unshared elecLrons on Lhe nilrogen alom ia the second stnicLure can be pushed Lo lhe posiuvely charged carbonyl carbon and

.0.

supply arrow (31)

32

The process can be repealed

CM

o

/ CH3~~C~N\

..

\+/

H gives supply azrow generaLing Don’t refer back Lo compleLe the following. The resonance meLhod notation for N melhylaceLamide is

supply arrow (32)

(33)

FIO iii Lhe complete resonasice melhod nolalion for niuobenzene. jWalch out for Lhree 0111cr slmctures that can be generaLed from (30).] 4

~

4

)

58. Once again Lhe presence of an aromaLic iing greatly enhances Lhe opportunities Lo push electrons and generaLe new resonance sLnictures, The fiLio group ia niLrobeazene is a powerful elecíron attractor. A pair of pi electrons from Lhe ring can be pushed LOWard Lhe niLrogen atom as Lhe pair

76

Chapter 2 Resonance Structures

Answers

Answers 59. Hypervalence (p. 12) resulis ia au interesting resonance structure wiLh separation of charge. A phosphonium ylide is ao inLermediaLe in lhe WitLig synLhesis of alkenes. The phosphorous atom in Lhe yiide is hypervalenL. By pushing pi electroas away from Lhe phosphorous one generates resonance sLnicture with charge separaLion buL with every aLom obeying Lhe octet rule.

Chapter 2 —%- CH2+ i. cii3— CHCH~

2.

+ CH3— CH—CH= CH2

H

(Ph)3P phosphoriium ylide 3. pi. oxygen. CH3

Exercises

CH3

Doe Lewis sLnjcwre is provided along with the letal number of resonance sLnicLures (including lhe one provided). Write Lhe others. 60. 2-cyclohexenone (3)

+

)c—o—H

~O—H

:O—H

.5 CH3—C—CH2—CH3

6!. Ethyl aceLate (3)

.

CH3—C—CH2—C113

~ CH3—C—O—CH2—CH3 6. oxygen. carbon. recepLor,

62. Propionic acid (3) H—Õ—~—CH2—CH3

CH3—6~ CH2

CH3—~CH2

63. Benzonitrile (7) 7. 64. Anylide(2)

H (Ph)3P=C”

caitcn,CH3C~

8.

+

CH3—CH—C6

CH3—CH—C~O C113

77

78

Chapter 2 Resonance Structures

9. pusbable electrons,

Answers 20.

receptor

4~

lO. CH3\ + ,CCH—CH—CH3 eH3

CH~~ 4

~

—

tH2—CKCH2

,C—CHCH—CH3 CH3

)

— ~

2I.,r-~

CI-12CH—ëH2

4

~

4

~

CH ,CH—CHQ—H CH3

4

~

~CH—~H—~—H CH3

22.

1

CH2

4

o

~

CH3

O’

+

o

‘4/..

12. no additional stn,cwres 13. ~CH2

çp’.

~CH2—CHO—H

21

CH3

~,

+

—CH--O-—H

4

14. no additional sinictures 15.

CH3

CH3

CH3—C—CH~5~CH2~CH,

4

~

CH3

24,fl*2— :N~C—CH CH2 CH

CH2

___

~

4

-..

N

C

CH

—

CH2

—

CH3 —C--CHO—CH2—CH3 + 25. no additional

CH3

26. no additional stnjctures 16. +

cX2~H2~~O 17.

~

,,

o~c—’Ç,J

4

27.

~j’EH2

crCH2~CH2~C~O

+

1~-”-~

4

W

28. no additional struclures

o=c—Ç~J 29. CH3—CHCHCH2

4

19.

30. CH~çj_~~ /CCH=CHCH3 CH3

:h.

(,

—

4

H2CC”

31.

o.

4

P

CH3 )C=C[I—CH—CH,

CH

=CH,

79

80

Chapter 2 Resonance Structures

Answers

32.

:01] ~

______

N~C—C

CK3

OH

:~=c=c—cH3

4

CK3

CH3

:OH :0K

CH3 H

1

/

C

CH3

~

CH3 o:: \ c=c / / H CK3

()—CH—() :oii

4

:01-]

~

N

36.

~ ~

~

~

Q

‘~..

41. ~

3LQ 38.

CH3 CH3

c≤Zb4

CH3

~

dEb4

39. :OH

:0K

~/

:OH

4

~/

~

SI

82

Chapter 2 Resonance Structures

42.

Answers

CH2—CH3

Ô 43.

(15)

CH3

‘I’L

f”Z~j~-_ ~;

(16)

CH2

(18)

(19)

CH3

~x~()

44,

~

L~,Ç~__1~~,JJ

4

012

(20)

(19) 45.

~~‘~=o:

:4~=0:

(9)

C112

(10)

4

CH2

+ (21)

(20)

Ç=o:

00:

(10)

(II)

48

NH2

H

4

W

I4’~»O~

4

NH2

Br (22)

H

Br (23)

NH2

H

Br (23)

NH2

II

Br (24)

~ +

______ (13)

(13)

s~cIures (22), (23), (24), ~d (25) (14)

H Br (23)

II Br (25),

83

84

Chapter 2 Resonance Structures

H>~3_

Br

Answers

H 4

—

~

CHCH2

+

)fr~;r~4

4

~

85

p+~

8r0 —

(26)

(27) 53. 012

4

HCI

UCI

4~j] ~ ~

HCI

~

(27)

:ij

54. 012

~

(27)

:üi

II

CH—

)

(29)

4

~

4

~

1%

50.

~ H NO2

H NO2

55.

CH3 :0:

4

~

CH3

tíi~i? •~ :0:

~

~

+1~~5j A

fourth stiucwre,

56.

/

o

:0+ ~

IICI

+/

HCI

H

+/

CH2CH—C

C112

CH—C J

11

4

~

CK2=

+[i~~\

57.

•1 ~

~ .

H

li —

4

4):

+

CH2—CHC

P

/

4).

:~yCH, CH,—C—N\

H

+

C112—CHC

.p~.

CH—C

4):

II

CH2CH~-C

—,

4)..

is not included because lhe sp4iybridized niLrogen requires lhat ii and lhe Lwo adjaceaL carbon acoms be linear. Titia adds excessive ring suam LO lhe stniclure. +

C113

o—. 4

HCI

KCI

C113 ~

+

52.

:0

~

II NO2 SI. ((.t)H_C—H CH3

CH3

CN

CN

Y

Iji

H3C—C—CH—C—H

~1

1

Ph—C—CH—Ph

Pb—C—CH—Ph

CH3 CH3

~1

132.

713 CH3

Y

~CH2

CH3

CH3

+

138.

~ + CH2

~

159

160

Chapter 3 Mechanlsms

139.

Answers

+

147.

\__J

:O—CH2—CH2—C112—c142—gr: 0—H 148.

140.

+CH2 II

—

COOR

‘Li 141.

COOR +

:ci:

149, CH3 1+.. C—OH Q

But you needri’l worry about Lhese poinls. You were given Lhe product, and no stereochemistxy was indicated iii lis siniclural formula. Your lask was lo gei tolhe producL süpulaled in lhe piobiem. Don’L answer questions you are not asked.

Theme I—The intercepted Intermediate

LÍCH3

167

Lj~~i:

mosL precious commodity, answer oniy Lhe questions you are asked. +

NaBr

5. Propose a reasonable mechanism for Lhis reaction, using curved artow nocation.

Br: The big picture: An alkene has losL its double bom!. Halogen atoms have appeared ai lhe ter mmi. This looks lilce eleccrophilic addition lo lhe alkene, which is one of those mechanisms you must know. flui when 8r2 is used, bromine atoms appear aL each end of lhe erstwhile dou bie bond. The only source of chiorine is NaCI.

CH2=CH—CH2—CH2—CH2—CH2---.Q< 6 5 4 3 2 1

H

H2S04 1120

CH3~Q

What is going on here? Where did thal ring come from? There are six carbons and one oxy gen in both starting malerial and producL. When a ring forms from a single molecule, one end

WiLh thal background, wriLe lhe first step

‘ti

lhe mechanism of bromine’s addicion lo the

must biLe Lhe oLher. Taking Lhe oxygen as a Iandmark, iL will have lo become attached to C

doubie bond. (lf you can’L, you’re noL ready for this chapter). Lhe C=C are lhe functiorial groups and you expect lhe chemisLry to happen there. Another comforling observation: The four connected —CH2—-- groups appear in boLh sLarting maLeri ai and produci. So lhe CH3 in the product comes from C in lhe starting material

_______

Enough anaiysis. LeL’s ~ry some chemisLry. One could prolonate Lhe hydroxyl and gen erate a carbocation. li is quile a reasonabie thing to uy, buL aL some point you wouid see lhaL lhe oxygen you need in Lhe product is gone. You must rejeci this good idea because iL is not working. What else could become protonated? The double bond? Try iL. Now comes the mntercepted mnlermediate part. Cleariy, finishing off Lhis reaction in Lhe usual way wilI lead toLhe 1,2-dibromo compound. BuL, inslead of using 8r as Lhe nucieophile, Lry using C1.

lf you have produced lhe more slable, secondary carbocaLion, you should see a way Lo dose lhe ring se that lhe oxygen becomes altached Lo C5. Once again, an intennediaLe cation has beeri inlercepted, this time by a nucleophile at lhe olher end of the molecule.

(b) You have compleled Lhe lask. There are some other facLors LhaL could be considered. FirsL, lhe lcss nucieophilic chioride ion competes successfuily because iL is presenl in excess. Second, lhe regiochemistry is correcL for opening lhe bromonium ion and, lhird, lhe stereochemistry is anti.

168

Chapter 4 On Solving Mechanism Problems

On Solving Mechanism Problems

Theme lil—An Unexpected Product

Now IL’s just a matler of getLing rid aí Lhat proton. l’ll leave lhat Lo you.

7. Propose a mechanism for ibis reaction. CH3

Theme II—Rearrangement

1 C113—C=C112

6. Propose a mechanism for this reacLion.

112s04

CH, —C—CH=C 7CH3 NCH

C113 +

1 C113_C_CH2_C%c

CH3

CH3—C_CH2—Õ:

CH3N

~043

CH37

“H

H2S04

H

+

H20

CH3 la the presence aí a sLrong acid, an alcohal has become ao dehydraLion. Both Lhe sLarting material and product contam

169

CH3

major minar WhaL’s going on? A sLarting maLerial wiLh ________ carbons becomes a praduct with ________ There are Lwo, isomerie products. Further observaLion reveals Lwo units aí Lhe starting material in the product. Circie Lhem ia Lhe major product above.

pitas waLer. Sounds like . carbon aLoms, but the

longesL chain mn Lhe starting material is _______ lhe longest chain in Lhe product is _______. lt musL be a rearrangemenL! In Lhe first course ia organic, carbacations are almosL always responsible for rearrangements. lo two sleps, generaLe a carbocaLioii.

Keep it simple, concentraLe an one Lhing a~ a Lime. Don’t let Lwo producls worry you; leL’s do some chemislry. The nucieophilic dauble bond and Lhe strong acid suggest proLonation. This time, Lhink abauL which carbon accepta Lhe proton and which becomes Lhe carbocation.

CH3 (a)

(a)

CH3—C=CH2 H—6—S03H supply arrows

producis

Naw do lhe rearrangemenL part.

The carbocation is electrophilic and caia aLtack another molecule aí Lhe starting material. Again, choose to generate Lhe more stable cation.

(b)

(b)

CH3 C113—C

+

CH2=C

7CF{3 NCH

CH3 supply arrows And lhe loss aí a prolon completes this El reaction.

(c)

products

Doing Lhe chemistry without worrying about lhe two products has worked! You ought La be able Lo arrive at both products from Lhe single carbocaLion.

170

Chapter 4 On Solving Mechanism Problems

On Solving Mechanism Problems

171

CI

Theme IV—Stereochemistry 8. The two reactions below are carried out under identical condilions but yield different products. Give mechanisms lhaL account for formalion of Lhe products. Here is a problem in which you wiIl have Lo grappie with lhe stereochemistry. You wiil aiways know Lhis when Lhe slructures

11

andor names include siereochemical designaLions. OH

HO

011

II

arrows

product (1)

~OH~

:ó—Fj HH

(1) cis-4-chiorocyclohexanoi OH

arrows

WhaL aboul product (3)? AnoLher ring has formed, so somehow lhe head must bite lhe tail

NaOH EIOH

CI irans-4-chiorocyclohexanol

producl (2)

(2)

(3)

(Lhe properjargon is lhat this is an inlramoiecular reacLion). Try lhe trick of sLudying Lhe rela Lionship between lhe sLnicLures of starLing material and producl.

ti.

9+

Keep it simple. Do one ihing aLa lime. FirsL, determine Lhe simple slructural chemistry. Then address lhe stereochemisLry and Lhen, lhe unexpected producL (3).

draw sLarlnig materiais, supply arrows *

*

*

*

*

The simple chemistry. In Lhe firsi reacLion —OH subsLltules for —Cito form (1). In Lhe second reaction we see lhe a, j3 eliminalion of HCI lo form lhe alkene (2). Sodium hydroxide in elhanol are ciassic SN2-E2 condiiions. This sounds Iike familiar LerriLory. lf Lhe firsL reacLion

Here are some probiems invoiving mechanisms LradiLionaiiy studied laler in Lhe course. 9. Using curved arrow notation, show a mechanism for lhe foilowing.

is SN2, lhe configuration wiIl be ___________________. lf Lhe second reaclion is E2, Lhe elim inverted/relainewracemízcd inalion wilI be -

O:

sywania

The person who crafLed Lhis problem has added a lask by lhe way in which Lhe problem is presented. AlLhough lhe strucLures designaLe lhe sLereochemislry unambiguously, they are noL sufílcienl for showing mechanisms. You musL rewriLe Lhem iii order lo show lhe Lhree dimensional shape of lhe six-membered ring. 1 have done lhaL for you.

—

CH2

—

CH2

—

—

AlCl1tanhy)

-Y 172

Chapter 4 On Solvlng Mechanlsm Problema

O,, Solvlng Mechanlsm Problema

173

What is going on? You have noc lost or gained any carbon atoms. Tbe beazene ring lias remained intacL. lt looks iike au intramolecular cyclization. Tbe chemistry is an acid chioride lreated wiLh anhydrous aluminum chioride forming a ketone aL lhe aromatic ring. Ii must be a Friedei-Crafts acylacion. Write Lhe mechanism ia three steps.

lo.

O:

I

I

CH3—C—CH2—C—CH3 1

23

CH3

O: 45

÷

H2M—f~H2

—,.

1120

CH3 Big piclure: Lhe producL has five carbons and lwo nitrogens. Conclude: lhe two slarting moi ecules have condensed. This is just a twist on problem 7. Circie lhe remnanis of lhe slarting molecuies ia Lhe product above. Conciude that Lhe lwo niLrogens have become attached Lo C2 and C4, respeclively. Do you know a reaction ia wbich a nucieophuic amino nitrogen attacks au eiectrophiiic caxbonyi carbon? You shouid. There are other puzzles: Lhe acid caialysis, Lhe uing ciosure, and lhe appearance of doubie bonds ia Lhe ring. Try lhe whole Lhing.

Here are three mechanism probiems where you receive no guidance at ali. Try attack irlg Iliem aiong Lhe lines suggested ia Lhis chapter. My versions of lhe answers cmi be found ia lhe answer section. After you have finished, see how ciosely we agree. You wiili find pienty of mechanism probiems ia your Lext to LeSt your newiy acquired skiiis. ii. This reacLion was first reported by Emii Fischer.

174

Chapter 4 On Solving Mechanlsm Problanis

Answers

175

Answers C=O: ~—C—OH HO—C—-H

HO’

+

Chapter 4

CH30112 CH3OH1

+

H—C—ON H—C

H

H

OH

ChH2—OH

C(CH3)3

12. Your only hinl is Ibal a nirrile group is considered a derivalive of a carboxylic acid.

+

CH3ÕH

2.

scribbling, whal’s going oW?, 2, CI, 5, Y

HCk~,

•O—CH3

lf

13.

Are your arrows going clockwise? li gets lhe job done, bul this proton is on a carboxylic acid, and It makes chemical sense that lhe carboxyl Lakes lhe electrons of lhe O—H bond.

O: O:

:~

o:

I

I

I

Ph—CH2—C—CH2—Ph

+

Ph—C—C—Ph

3.

5, oxygen. carbon

NaOH(~)

Ph

ïV

HN

CH3—CH2—CH=C—CH3

4. (a)

CH3

r\

CH3 +

(b)

CI:

:Br:

-T 176

Chapter 4 On Solving Mechanism Problems

Answers

5. C5,C6

CH3

(a)

CH2==CH—CH2-— cH2~~~cH2_CH2_Q:

(b) CH3—C+

—+

CH2=C

~CH3

)

+

CH3—C—-CH2—C

N CH3

CH3

CH3

H

177

:o—s0311

H—O—SO3H

CH3 H (b) CH3_CH_CH2_CH2~CH2~~CH2~Q

(a)

CH3—C—C1-12 ~CN 0CH2 CH3 L—I H t.-76—S03H

5, 3, 4

CH3

CH3—cf—CHz—OÇ~

CH3

CH3

~H—Ü—SO3H —*

~

(b) CH3

+

CH~—C—CH2 CH3

CH3

cH3

1

—>

major product

NCf~

CH3

H

H alkene,

0 CH3H 1

CH3~Q

CH3

6.

)

CH3—C—C-—C

—$

8. inverted, anti CI:

CH3

÷1

CH3—C—CH2

—*

CH3—C—CH2

0-13

HO

(a) HO—~,j..—

CH3

H CH3

(e) +

1

CH3

~

—*

CH3—C rcC —H ~“_

product

SO3H H%~NÉi:

7. 4,8, (b)

(a)

mirior produet

)

CH3

CH3 ) CH3CtH2

14~— 503H

CH3—C—CH3

(e)

OH

178

Chapter 4 On Solving Mechanism Problema

9.

Answers

0:

179

0:

r~Nt_CR2_CH2_CR2 —l!_a: AIO3

-*

-~

CH3—1j—CH2—C—CH3 N: “-

0:

RH/~ N cH3_c7~~v9R2

-~÷

/

N

-H

)

NCH

product

N Ml

A1C13

lo. There are many steps in this mechanism, but most of (bem are jusI prolon managemeni. Severai parts of lhis pracess—lhe nucieophihc addiLions (o dose Lhe ring and lhe subsequent dehydraiions lo form Lhe doubie bonds—are acid-catalyzed. JL seems in lhis answer ibaL pro tons go on and off oxygen andor nitrogen with careless case. IL’s ali right. Proon transfer Lo and from atoms of lhese elemenis has practicafly no energy banier. In situations iike this, proper proton management is necessary Lo avoid high-energy spedies such as Lwo posiiive charges on Lhe sarne molecule ora negative charge in an acidic medium.

ImpaLience with the iength of this mechanism tempts one to have lhe two additions occur simui taneously. Doing that is unrealisLic foraL ieas two reasons: ir wouid require both carbonyls to be proonaed at the sarne time, and Lhe necessaTy alignmeni of lhe two molecules is entrop ically unfavorabie. There is a question (unanswered,I think) of whether the first dehydration occurs before or after lhe iing doses. You couid rnake Lhe second double bond a C~N ar a CC. BuL iL basto be CC Lo get the right product. There is a reason why h goes thai way. li has to do with aromaicity. You try lo figure ir out. Ii.

The big picture. IL’s one of those intramolecular ring ciosures. Anaiysis indicaes that Cl becomes bound to lhe oxygen on (3. The OCH3 from the rnehanol also becomes bound to Cl. There are Lwo stereoisorneric products bui we are no going Lo bother with that just ye. The chemisuy iooks Iike acid-cataiyzed nucieophihc addition toLhe C=0.

+

O:

I

5: I

CH3—C—CR2-—C—CR3

5: I

I

______________________

______________________

CR3—C—CH2—C—CR3

OH R

HN

CR011 57H

~‘

I I CI-13—C—CH, —C—CH3 +H~ jjr~ 1+ [12N—NH2

fl0112

1 I C113—C—CR2—C—CH3 ~“-

1:

c~R0R

O:

NH2

_______

—s

HNI

0H

~

C—O—CH3

CHOH

CR011

CR011

-~

CROR

CROR

CROH

CR011

H—C—Õ—-R

H—C—0—If

H—C—b—H

CH2OH

CR2OR

CH2OH

180

Chaptcr4 On Solving Mechanism Problems

Answers

181

in one have become bound Lo Lhe carboriyl carbons of Lhe oLher. NaOH strongiy suggests aldoi-type chemistry.

CHOR

c!HOK

Ph

Ph

H—C—õ—CH3 CH2OH

CHOH

FIO:

tH—C—H

o=C

±~

-~

_-*N4 —*

o==C

&IOK

CR2

CH2

H—C-—6—H

Ph

Ph

CH2OH PhN

slereochemisLry?

.vPh

O=C l4~i

or H

HO

12.

OCH3

Tire hmnL should be ali you needed. This is just an unusual nucleophiiic acyl subsüwLion.

)

4) :O_C[13

H7

jj*

prcducc

Lr74NflU

‘-n3

13.

C=õ:

This ring is fonned by Lhe condensalion of Lwo moiecuies reminiscent of problem lo. By encirthng lhe staiting materiais iii lhe sLnicture of lhe product one concludes that ihe—CH2—gmups

Ph

1 Ph

_~~* S(CpS

product

Ph

ILk =

1

Ph

Ph

I~l4 Ph ~jN~I7OH o=C

1

CH2 Ph

1

c =0: Ph

______

-lizO sponianeOtis

Bioorganic Reactions

5

183

2. A step ia Lhe laboratoiy synihesis of an amino acid

:3

COOCR3CH3

CH3 — ~—~H— c~ :_\____7~N& CH2__f~ COOCH,CH3

Some Reactions from Biochemistry

—*

+

:Br:

:Br:

3a. Two conseculive steps lhai prolecl an amino acid for peplide synlhesis

In many academic programs, the chemislry course Lhat follows organic is hiochemisLry. Indeed, most modera lexis for lhe first course in organic chemisiry conclude wiLh severa! chaplers on Lhe chemislry of biochemically importam classes of compounds such as proteins, carbohydrates, lípids, and nucleic acids.

:~

I

Ii

(CH3)3C—O—C—O—C—O-—C(iH3)3

lt ís not an entirely foolish conceil Lo say Lhal biochemistry is organic chemistry occurring ia more complex syslems. Certainly, a lhorough biowledge of organic is crucial for lhose who would Icnow and use biochemistry. la Lhis section 1 shall inlroduce examples from hioorganic chemisiry and elemenLary No chemistry. 1 hope lo make lwo poiaIs: biochemical reaction mechanisms are noL fundamenlally dif fereni from organic mechanisms, and lhe skills ihat you learned in earlier seclions of Lhis book can be applied in biochemislry.

:3:

:~ ..

(CH3HC—õ—C—õ—C—õ—c(cR3)3 H,N~

-p

H,N

-I

-I

CH3~~CH~~CN..

supply arrõws

3b.

:3

Bioorganic Reactions

(CH3)3—C—Õ—~,6—)~—õ—C(dH3)3

We begin wilh a selection of reacLions thal are noi sLriclly biochemical; rather lhey are organic reac Lions used in lhe service of sludying biochemisLry. You musi be prelly far along in lhe course by now, sol feel safe ia iniLiaiing some shorlhand noialion lhal you have no doubi seen in your lext or in lecture. Also, 1 um omiiLing some unshared pairs of electrons from aioms lhaL are remole Lo lhe sue of Lhe aciton. Where an-ows are supplied, draw Lhe producis. Where Lhe producis are given, sup ply arrows. (In lhis chapter lhe inolecules are larger and more complex; ia some cases, you may have Lo squeeze your answers unto lhe space provided.)

CH3—CH—C00-

4. A sLep in Lhe laboraLory synLhesis of a peptide

1. A slep ia the laboraLory synlhesis of a sleroid

CH~

0 CH

I

:0

I

C—0—C—NH—Cl1,—C-—O: -p

o

0:

:0: supply arrows

182

:3

184

Chapter 5 Some Reactions from Blochemistry

Bloorganic Reactions

185

(e)

5. A step in Lhe synlhesis of a dinucleoLide

+

—CH2—CN

—

+

ribose 8. A step inche synthesis of a disaccharide

6. A deprotecLion step subsequenL Lo 5 ribose

:õ—H

:0:

ribose

H

õ=~Õ~C[IcCH~CN

o

:0: ~

+

H2C=CH—CN H

:0: CH2

ribose

CH2 — ribose

supply arrows

7. Three consecuLive sLeps ia lhe preparaLion of a sugar derivative (a) ff001, +

And now some reacLions thaL occur—with Lhe help of enzymes (see below)—in Iiving celis. The first four reactions are from glycolysis, Lhe metabolism of blood sugar.

H20 9.

(b) HOCH2 HO~\ HO

OH

suppiy arrOWS

CH2 —OPOr

C=O:

C=O: )

H6—CH

HOCH2

-~Si~ HO

H4~CH2—CH3 H~°~CH2—CH3

CH2 —0PO~’

HL—ÕH HC—OH —0P07

Hb—~D—H + HC=OH HC—OH CH2 —OPO~

supply arrows The elemeaL phosphonts is a big player ia biochemisLry. la Lhe exercises LhaL follow, phos

186

Chapter 5 Some Reactions (tom Biochemistry

Enzymes

phorus is in lhe phosphaLe sLaLe of oxidation. Treat iL as you would a carboxylate carbon. Somecimes phosphorus ends up wiih ten valence elecirons (see page 12). w.

o

~5: p~.

o

-~

0~0~Ç\v~Q:

adenine

H~—c?_4iH2

-~

(c?H2)3

4

-~

CH2

.~5r 1

13.

a&nine

d

0

187

H3Z—~H C00

+

7 p07

¶—C—i412 (c?H2)3

-

+

P07

H3Z_?H C00

HO HO

OH

suppiy OH

14.

II (a)

H~+7H -

....

~A B~—H

CHOH

-

fl:

NH

~d~H

HN—C—NH—CH

+

H—B

(CH,)3

CHOH

+

1

-

CH2—0—POï

1

H3N—CH

-

CH2—O—P0j~

p~

H~-CH

..

B

C00

C00

supply arrows (b)

>b~

..

H ~j% C~;; OH:0: CHOH ~‘

BH+ _..

+

And now three from the urna cycle

6 H2N: ~ C—Nl-12

o

(CH,) +

1

1

-

H3N—CH COO

—

P0~ —

I

Please notice that, although Lhe molecules are larger and more complex, you have used Lhe sarne symbolism you have been using ali along Lo complete Ihese exercises correctly.

~

:0:

-

CH2—O—PtÇ

12

:0

Enzymes Reactions Lhat occur in living cells are almost always caialyzed by enzymes. To chemists of the author’s age, i.e. pre-war, enzymes are marvelous. Thirty years ago, 1 heard a friend of mine . descnbe biochemisLry as foliows, First, the enzyme and Lhe substraLe combine lo form an enzyme substraLe complex, and Lhen a miracle occurs.” Since Lhen, much has been learned aboul 1mw Lhese wonderful natural catalysls work. Our knowledge of organic mechanisms has been a major player in these discoveries. Enzymes are eflicienL and, in mosi cases, remarkably specific. Being proteins, enzymes are large, complex molecules Lhat are produced wiLhin the celi (via reactions Lhat are catalyzed by other enzymes). Like alI catalysts, they lower Lhe acLivation energy of lhe reaction. They do so by using many triclcs such as lowering entropy by bringing reacLanLS logeLher in perfeci alignmenc, inducing steric sLrain in a reactant, providing an acid and/or base atjusL Lhe righL place, providing an oxidant andior reductant atjust Lhe righL place, using an easier paLhway from reacLant lo product, and so on.

188

Enzymes

Chapter 5 Some Reactions Irom Biochemistry

The examples and probiems LhaL follow are from enzyme-calaiyzed reactions. 1 use symbol ism Lhal abbreviales for lhe enzyme and shows oniy lhe reactive siles, a common pracLice in bio chemisLry.

Amide (peplide) bonds are hard Lo hydroiyze. Chymotrypsin overcomes Lhe difflculty by firsL con verting lhe amide Lo au ester and Lhen hydroiyzing Lhe esier, a much easier lask. An iliustraLion: Converting Lhe amide lo an esLer.

Chymoirypsin is a proLease, i.e., an enzyme lhaL caLalyzes Lhe hydrolysis of proleins. There are many proleases. This one is a digesLive enzyme. Siudy lhis Iengthy iliuslraLion Lo prepare for Lhe exercises lhaL foliow.

•0• CH2 —~~~-——~ ‘? ~

The Lask of chymoLrypsin is Lo cleave a protein mIo smaller fragmenLs by caLaiyzing lhe hydroiysis of some of Lhe peplide (amide) bonds ia Lhe chain of amino acids lhaL consLilute lhe pro tem. As wiLh ali caLaIysLs, Lhe enzyme iLseif emerges unchanged from lhe process. Nevertheiess, il is very busy during Lhe process. peplide bond H20 -,

R—C

NH

chycnotrypsin

R

:0

II..

R—C—O—H

÷

H2N—W

II

R’ The enzyme uses lhe serine —OH Lo bind Lhe subslrate. As Lhe nucleophilic aLLaCk occurs, Lhe —OH prolon is Lransferred, malcing lhe sedne oxygen more nucle ophilic wiLhout ever having Lo gain a fuli negalive charge.

“cH2 —0

\

H

/

~CH

HC

N

/

:0

/

H /

/ :0

Na H

\

~1

The astonishing efficiency occurs because lhe enzyme provides proton donors and proLon acceplors in just lhe righL piaces ai just Lhe iight Limes, lhus avoiding lhe appearance of naked + or — charge. The acid and base cenLers are functional groups from Lhe side chains of certain of Lhe ainino acids in Lhe proLein.

N

I

HC

1

This is an oversimplified drawing of Lhe active sile of lhe enzyme. The shaded arca represenls Lhe bulk of lhe enzyme. Ser stands for lhe amino acid, serine, lhe side chain of which concribules a hydroxymelhyl group. Asp is for aspar Lic acid, which conlribuLes lhe carboxy laLe anion. lis is for hisLidine, which conLribules iLs imidazole side chain. la lhe firsL drawing, dotted iines for Lhe hydrogen bonds are shown. The hydro gen bonds are cn’cial ia holding Lhe acLive siLe Logelher, buL Lo avoid unnec essa,3’ 011111cr Lhey wiII noL be shown ia lhe iiiuslraLions.

‘There is a wonderful oppoitunity for crealing confusion bem. 1101h lhe subsirale (lhe enzyme’s largel) and lhe caialysl (lhe enzyme ilseIf) are prolerns!

CH,—O—C—R

\

1-1 Now Lhe esLer forms, bound Lo lhe enzyme, and most imporlantly, Lhe pep lide C—N bond breaks. Once again, Lhe simuilaneous cransfer of a prolon eiimi nates lhe appearance of a flui negative charge on Lhe niLrogen.

R

:f’4—f{

two smalier polypeptide fragments

Iong protein chain (with many peptide bonds)

189

:N—H

(N:~cHw HC I

190

Chapter 5 Some Reactions froin Biochemistry

Enzymes

191

15. An exercise: Using arrows, show how Lhe serine ester is converted to the C-terminal polypep Lide fragment and lhe restored enzyme.

Cl-12—O—C—R

/N:~

0:

(a)

CH2—6—C---R

HC

:6 :0

—

H

1-1 A water molecule has (aken lhe place of the N terminal fragment (enzymes can do Lhat). k almosi lhe exact reverse of Lhe iliustration above, push elec trons lo complete lhe enzymatic process.

/ The N-terminal polypepcide fragmeni is now fuliy formed and is released for furiher digestion and metabolism. supply arrows

(b)

1~ CH2 —O:

\

H

1-1

—*

N / ~CH HC I

\

~C.

supply arrows

192

Chapter 5 Some Reactions trom Blochemlstry

Coenzymes

Exercise 9 (page 185) is a useful example of learning LO push electrons. However, iL is an egregious oversimplification aí a melabolic reaccion ia which a six-carbon sugar is cleaved into two three carbon fragmenis. That producL in exercise 9, Lhe free enolate anion, just would noL form in a ceil nor would Lhe other producL, the protonaLed aldehyde. This important reaction is catalyzed by Lhe enzyme aldolase ia a way that avoids boLh these high-energy intermediates. A!dolase uses Lhe side chain of a lysine (lys) to bind Lhe subsLrale.

193

CN2—OPO~

17.

+1

~.w.wHNC HO—CH la Lhe next step, a proton Lransfer neuLralizes Lhe anion.

HC ~‘

I

ló(a)

+

CH2—OPO~

Ç\CIl

~1

H

HO—CH

I-IC

\

~1~

We join Lhe action afier the enzyme-subsLrate complex has fonned. A cysteine anion and a his Lidine cation stand ready Lo act as base and acid, respecLively.

Hc!—à—H

CH2—OPO~

\‘,

4,1120

CH,—OPO~ Lys ~.NH2

supply arows

HO—CH2

1

(b)

o=c

HC

I

CH2—OPO!

+1

4

Hydrolysis yields the producL and reLurns Lhe enzyme lo iLs original state.

*4 CH

l~f

14S—CH2l9~\

N=C HO—CH

Coenzymes HC= HC

\ CH

c

~ ~N

2

+ 14

1

C1JOH CH2—OPO~

resonance sLnicture The slrucLure lo the IefL is a varia Lion oyt Lhe enolate anion referred Lo above. You ought Lo be able Lo draw a resonance slrucwre thaL shows how Lhe enzyme sLabilizes It.

Inibe two previous examples of caLalysis by an enzyme, we have seen Lhe enzyme bind lo the sub sLrale in some producLive way and then provide acid-base catalysis. Now we encounLer au enzyme lhat uses a coenzyme. A coenzyme is a small molecule, originaling from a vitamin, that is assoei aLed wiLh, buL not part of, Lhe enzyme. The coenzyme is alLered during lhe reaclion, but lhe celi finds a way Lo return iL lo iLs original sLaLe SO that iL can be used again. One of Lhe products of lhe aldolase-caLalyzed reacLion is glyceraldehyde-3-phosphaLe. The cdl next oxidizes (dehydrogenates) Lhe aldehyde to the carboxylic acid sLaLe of oxidaLion and releases it as lhe phosphaLe ester, 1 ,3-diphosphoglyceraie. The enzyme, glyceraldehyde-3-

194

Chapter 5 Some Reactions trem Bicchemistry

Ccenzymes CH2—OP0~

phosphate dehydrogenasc, uses ao —SH group from Lhe side chain of cysteine, and iL uses the coenzyme, nicotinamide adenine dinucleotide, (NAD~, which originales as niacin) as an oxidizing agent.

CHOH

cH2—oPo! glyceraldehyde-3-phosphate

18.

195

+

1v~

H

H

H—B

CR011

t~:

H-B~ H

H

NAÍY’

The cysteine side chain makes a nucle ophilic alLack Lo fonn lhe eniyme-sub straLe complex. The process is assisLed by proLon transfers. The side chains Lhat serve as lhe base and acid are noL known and, Lherefore, are wrilLen as —B and —B—H~. The coenzyme is nol involved

CH0R

+

enzyme-SH

R 1, 3-diphosphoglyceraLe Note lhaL Lhe erstwhile aldehyde carbon has become a thiol ester carbon, i.e., It has been oxidized.

yeL.

~1~ 20. lo Exercise 19, show a way Lhat the enzyme might promoLe Lhe release of lhe product, 1,3diphosphoglyceraLe. You may assume that HOPO3= is available and that side chain acid and CH2— 0P0~ (enzymelsubsLrale intermediale)

19.

base centers are present. (Use chymoLrypsin as a model, and use your own paper.)

/

CHOH

~

*

\ —C-—0..,.,

~—H

1

H

H Now the coenzyme becomes reduced by accepting an hydride ion. A simultaneous proLon Lrans fer from the alcohol Lo a base on Lhe aclive siLe of lhe enzyme drives Lhe process.

R

1~

*

*

*

Pyridoxal phosphate, from vitamin B6, serves as a coenzyme in a number of differenL reac tions that manage amino acid metabolism. The transaminases use pyridoxal phosphate te catalyze reacLions IA which Lhe —NH2 group from an amino acid is transferred lo a-ketoglutarate. The amino acid is converted Lo a keto acid which is metabolized, and a-ketogluLarate is converted Lo glutaniaLe and sent inLo lhe urea cycle.

R—CH---C00

+

-OOC—CH2—C1-12—C—C00

+NH3

O

animo acid

supply arrows

*

a—ketoglutaraLe enz —s

R—C—C00 O a—keloacid

+

-OOC—CH2—CK2—CF1—C00 +NH3 glutamate

196

Chapter 5 Some Reactions from Blochemistry

Coenzymes

21

197

CHO B4~

=O3PO—CH2yÇ~.OH H

+

B—H R—C—COO

+

R—~COcr

B~H

~N

G13 II” ~CH

H

//

CII) OH

pyridoxal phosphate

=03P0

We join Lhe action after the enzyme-substrae complex is formed. Tbis time the coeazyme is the culprit in binding Lhe sub sLrate.

CH3

The erstwhile aldehyde carbon of pyridoxal phosphate has been reduced. CH3

1-1

,11420

R—C—-COO

The a-carbon of Lhe erstwhile amino acid has been oxidized.

O NH2

/

CH2 O3PO _CH2%tt~xOK ii+

CH3

As Lhis point, Lhe enzyme expeis Lhe a-keoacid and replaces it with a-ketoglutaric acid. The coenzyme is i~i ics reduced form, pyridoxamine phosphate.

H B~—H R—C—COO II N

/

23. Propose a way in which Lhe enzyme can complete lhe process. ThaL is, from the enzyme substraLe complex below, provide a mechanism Lo produce gluLamate and rescore Lhe coen zyme to its original, oxidized state.

CH O3PO_C[l2%~z1Ç{OH

Ç :~CH3

~—H

OOC

§

C—CH2—CH2—COO-

H 03P0—CH2 supply arrows

‘Ir

H

198

Chapter 5 Some Reactions trem Biochemistry

Answers ribose

Answers

\:0: \.

P~—O—CH,—CH2 —0’4

/

Chapter 5

H~±O_CH2 ~

6.

:õ

2.

ribose

COOCH2CH3

CH3—C—NH—C

CH2

:64_6cH2fbH_cN ~

/

:0:

COOCH2CH3 3.

(a)

CH2 —tibose

~ :6

(CH3)3C— õ—c—õ— C—6-—C(CH3)3 7. (a)

HOCH2

HOCH2 HO

________

HO

4

HO

OH

C113—CH—C HOCH2 (b)

(b)

(CH3)3C—O—Ç

+

TO—C—O—C(CH3)3

HO

NH

OH(~

CH3—CH—C00 H~~CH2 —CH3

4.

I (CH3)3C—0—-C—-NH—CH2

I

HOCH2

(c)

d

HO CH2 —CH3 +~

—CH3

199

200

Chapter 5 Some Reactions from Biochemistry

8.

Answers 12.

13.

:0: +

:~

1

H2N—C—NH2 (CH2)3

1

+

H3N—CH

:0:

H2N—C—NH2

°

(CH2)3 :0: +

P07

1

H3N—CH

coo-

coo-.

14. HN—C—NH 9.

CH,—OP0~

(CH2)3

C=O:

+

1

H3N—CH

RO—CH

oo CH

CH

coo-

[IC— OH

15.

[IC— OH

(a)

‘CH2—0—C—R

—OPor lo.

1 NØ:

II. (a)

pfl

d

JQ 0—P07

~ ~c~H CHOH CH2—O—P07

(b) CHOH

CH2 —0— P07

P07

201

202

Chapter 5 Some Reactions trom Biochemistry

16. (a)

Cri2

1

+

OPO~

(b)

~w~-fNC NO

Cri

—

203

Cri2 — 0PO~ / OlOR

OPO7

~ —~c-6

HO—CH —Cri2

resonance siructure +R B

CHOH

\ CII

I

yri2

~wfJNC

HC-~o~-H I1C

Answers

~_~øNf_CONH2

Cri2— OPO3

CHv~N R

20.

‘9, CH2—0P0J 1

+

1

N

HN = C FIO—Cri

N

%

HC

1

CII

CI1~~N

-s

1-1 Cri2— OPO~’ CHOR /CH2 — 0P03 :0:

CHOR

_~s~

“e O ~ ri

/

liA

B

B

/

Ol2~ opor CFIOH

Li R

\

ri

/

c==6 / Por

B IIÇN

204

Chapter 5 Some Reactions from Blochemistry

Answers 23. Conrinued

21.

00C~~~~ k~7c—CH2—cF1,

R—C—C00

C00

I

+

P4: CII

»

03I~0 —

1

011

çf

1 22.

ii

~1

g!

+

B~Ff

_0o~ H—C—CH2—CH2—C00

1-1

U1 .011 ‘CH3

4’

1 H30~

23.

21

-00c~ C—C[T,—CH2 —C00

9

~~00C HC—CH2—C112 —C00

HN

0

205