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ISBN-13: 9781121180611

Contents 1. Structure and Bonding 1 2. Acids and Bases 33 3. Introduction to Organic Molecules and Functional Groups 57 4. Alkanes 75 5. Stereochemistry 111 6. Understanding Organic Reactions 139 7. Alkyl Halides and Nucleophilic Substitution 159 8. Alkyl Halides and Elimination Reactions 193 9. Alcohols, Ethers, and Epoxides 223 10. Alkenes 257 11. Alkynes 287 12. Oxidation and Reduction 309 13. Mass Spectrometry and Infrared Spectroscopy 337 14. Nuclear Magnetic Resonance Spectroscopy 351 15. Radical Reactions 373 16. Conjugation, Resonance, and Dienes 397 17. Benzene and Aromatic Compounds 421 18. Electrophilic and Aromatic Substitution 443 19. Carboxylic Acids and the Acidity of the O-H Bond 479 20. Introduction to Carbonyl Chemistry 501 21. Aldehydes and Ketones — Nucleophilic Addition 535 22. Carboxylic Acids and Their Derivatives — Nucleophilic Acyl Substitution 567 23. Substitution Reactions of Carbonyl Compounds at the a Carbon 603 24. Carbonyl Condensation Reactions 631 25. Amines 659 26. Carbon-Carbon Bonding-Forming Reactions in Organic Synthesis 693 27. Carbohydrates 715 28. Amino Acids and Proteins 751 29. Lipids 785 30. Synthetic Polymers 801

iii

Credits 1. Structure and Bonding: Chapter 1 from Study Guide/Solutions Manual to accompany Organic Chemistry, Third Edition by Smith 1

2. Acids and Bases: Chapter 2 from Study Guide/Solutions Manual to accompany Organic Chemistry, Third Edition by Smith 33

3. Introduction to Organic Molecules and Functional Groups: Chapter 3 from Study Guide/Solutions Manual to accompany Organic Chemistry, Third Edition by Smith 57

4. Alkanes: Chapter 4 from Study Guide/Solutions Manual to accompany Organic Chemistry, Third Edition by Smith 75 5. Stereochemistry: Chapter 5 from Study Guide/Solutions Manual to accompany Organic Chemistry, Third Edition by Smith 111

6. Understanding Organic Reactions: Chapter 6 from Study Guide/Solutions Manual to accompany Organic Chemistry, Third Edition by Smith 139

7. Alkyl Halides and Nucleophilic Substitution: Chapter 7 from Study Guide/Solutions Manual to accompany Organic Chemistry, Third Edition by Smith 159

8. Alkyl Halides and Elimination Reactions: Chapter 8 from Study Guide/Solutions Manual to accompany Organic Chemistry, Third Edition by Smith 193

9. Alcohols, Ethers, and Epoxides: Chapter 9 from Study Guide/Solutions Manual to accompany Organic Chemistry, Third Edition by Smith 223

10. Alkenes: Chapter 10 from Study Guide/Solutions Manual to accompany Organic Chemistry, Third Edition by Smith 257

11. Alkynes: Chapter 11 from Study Guide/Solutions Manual to accompany Organic Chemistry, Third Edition by Smith 287

12. Oxidation and Reduction: Chapter 12 from Study Guide/Solutions Manual to accompany Organic Chemistry, Third Edition by Smith 309

13. Mass Spectrometry and Infrared Spectroscopy: Chapter 13 from Study Guide/Solutions Manual to accompany Organic Chemistry, Third Edition by Smith 337

14. Nuclear Magnetic Resonance Spectroscopy: Chapter 14 from Study Guide/Solutions Manual to accompany Organic Chemistry, Third Edition by Smith 351

15. Radical Reactions: Chapter 15 from Study Guide/Solutions Manual to accompany Organic Chemistry, Third Edition by Smith 373

16. Conjugation, Resonance, and Dienes: Chapter 16 from Study Guide/Solutions Manual to accompany Organic Chemistry, Third Edition by Smith 397

17. Benzene and Aromatic Compounds: Chapter 17 from Study Guide/Solutions Manual to accompany Organic Chemistry, Third Edition by Smith 421

18. Electrophilic and Aromatic Substitution: Chapter 18 from Study Guide/Solutions Manual to accompany Organic Chemistry, Third Edition by Smith 443

19. Carboxylic Acids and the Acidity of the O-H Bond: Chapter 19 from Study Guide/Solutions Manual to accompany Organic Chemistry, Third Edition by Smith 479

iv

20. Introduction to Carbonyl Chemistry: Chapter 20 from Study Guide/Solutions Manual to accompany Organic Chemistry, Third Edition by Smith 501

21. Aldehydes and Ketones — Nucleophilic Addition: Chapter 21 from Study Guide/Solutions Manual to accompany Organic Chemistry, Third Edition by Smith 535

22. Carboxylic Acids and Their Derivatives — Nucleophilic Acyl Substitution: Chapter 22 from Study Guide/Solutions Manual to accompany Organic Chemistry, Third Edition by Smith 567

23. Substitution Reactions of Carbonyl Compounds at the a Carbon: Chapter 23 from Study Guide/Solutions Manual to accompany Organic Chemistry, Third Edition by Smith 603

24. Carbonyl Condensation Reactions: Chapter 24 from Study Guide/Solutions Manual to accompany Organic Chemistry, Third Edition by Smith 631

25. Amines: Chapter 25 from Study Guide/Solutions Manual to accompany Organic Chemistry, Third Edition by Smith 659 26. Carbon-Carbon Bonding-Forming Reactions in Organic Synthesis: Chapter 26 from Study Guide/Solutions Manual to accompany Organic Chemistry, Third Edition by Smith 693

27. Carbohydrates: Chapter 27 from Study Guide/Solutions Manual to accompany Organic Chemistry, Third Edition by Smith 715

28. Amino Acids and Proteins: Chapter 28 from Study Guide/Solutions Manual to accompany Organic Chemistry, Third Edition by Smith 751

29. Lipids: Chapter 29 from Study Guide/Solutions Manual to accompany Organic Chemistry, Third Edition by Smith 785 30. Synthetic Polymers: Chapter 30 from Study Guide/Solutions Manual to accompany Organic Chemistry, Third Edition by Smith 801

v

Smith: Study Guide/ Solutions Manual to accompany Organic Chemistry, Third Edition

1. Structure and Bonding

© The McGraw−Hill Companies, 2011

Text

Study Guide/Solutions Manual to accompany Organic Chemistry, Third Edition

Structure and Bonding 1–1 C Chhaapptteerr 11:: SSttrruuccttuurree aanndd B Boonnddiinngg

IIm mppoorrttaanntt ffaaccttss • The general rule of bonding: Atoms strive to attain a complete outer shell of valence electrons (Section 1.2). H “wants” 2 electrons. Second-row elements “want” 8 electrons. nonbonded electron pair

H

C

N

O

X

Usual number of bonds in neutral atoms

1

4

3

2

1

Number of nonbonded electron pairs

0

0

1

2

3

X = F, Cl, Br, I

The sum (# of bonds + # of lone pairs) = 4 for all elements except H.

•

Formal charge (FC) is the difference between the number of valence electrons of an atom and the number of electrons it “owns” (Section 1.3C). See Sample Problem 1.4 for a stepwise example.

Definition: Examples:

formal charge

number of valence electrons C

C

• C shares 8 electrons. • C "owns" 4 electrons. • FC = 0

•

=

–

number of electrons an atom "owns"

C

• Each C shares 6 electrons. • Each C "owns" 3 electrons. • FC = +1

• C shares 6 electrons. • C has 2 unshared electrons. • C "owns" 5 electrons. • FC =
1

Curved arrow notation shows the movement of an electron pair. The tail of the arrow always begins at an electron pair, either in a bond or a lone pair. The head points to where the electron pair “moves” (Section 1.5). Move an electron pair to O.

O H C N H

A

O H C N H

B

Use this electron pair to form a double bond.

•

Electrostatic potential plots are color-coded maps of electron density, indicating electron rich and electron deficient regions (Section 1.11).

1

2

Smith: Study Guide/ 1. Structure and Bonding Text Study Guide/Solutions Manual to accompany: Organic Chemistry, 3rd Edition Solutions Manual to accompany Organic Chemistry, Third Edition

© The McGraw−Hill Companies, 2011

Chapter 1–2   TThhee iim mppoorrttaannccee ooff LLeew wiiss ssttrruuccttuurreess ((SSeeccttiioonnss 11..33,, 11..44)) A properly drawn Lewis structure shows the number of bonds and lone pairs present around each atom in a molecule. In a valid Lewis structure, each H has two electrons, and each second-row element has no more than eight. This is the first step needed to determine many properties of a molecule. [linear, trigonal planar, or tetrahedral] (Section 1.6)

Geometry Hybridization

Lewis structure

[sp, sp2, or sp3] (Section 1.8)

Types of bonds

[single, double, or triple] (Sections 1.3, 1.9)

 R Reessoonnaannccee ((SSeeccttiioonn 11..55)) The basic principles: • Resonance occurs when a compound cannot be represented by a single Lewis structure. • Two resonance structures differ only in the position of nonbonded electrons and
bonds. • The resonance hybrid is the only accurate representation for a resonance-stabilized compound. A hybrid is more stable than any single resonance structure because electron density is delocalized. O

O

O

CH3CH2 C

CH3CH2 C O

delocalized charges

CH3CH2 C O



O



delocalized
bonds

resonance structures

hybrid

The difference between resonance structures and isomers: • Two isomers differ in the arrangement of both atoms and electrons. • Resonance structures differ only in the arrangement of electrons. O

O

CH3 C

O

CH3CH2 C O CH3

isomers

CH3CH2 C O H

O H

resonance structures

 G Geeoom meettrryy aanndd hhyybbrriiddiizzaattiioonn The number of groups around an atom determines both its geometry (Section 1.6) and hybridization (Section 1.8). Number of groups 2 3 4

Geometry

Bond angle (o)

Hybridization

Examples

linear trigonal planar tetrahedral

180 120 109.5

sp sp2 sp3

BeH2, HCCH BF3, CH2=CH2 CH4, NH3, H2O

Smith: Study Guide/ Solutions Manual to accompany Organic Chemistry, Third Edition

1. Structure and Bonding

© The McGraw−Hill Companies, 2011

Text

Study Guide/Solutions Manual to accompany Organic Chemistry, Third Edition

Structure and Bonding 1–3

D Drraaw wiinngg oorrggaanniicc m moolleeccuulleess ((SSeeccttiioonn 11..77)) • Shorthand methods are used to abbreviate the structure of organic molecules. CH3 H =

CH3 C

C

C

H

H

CH3

skeletal structure

•

CH3 CH3

(CH3)2CHCH2C(CH3)3

=

isooctane

condensed structure

A carbon bonded to four atoms is tetrahedral in shape. The best way to represent a tetrahedron is to draw two bonds in the plane, one in front, and one behind. Four equivalent drawings for CH4 H C H

H H H

H H

HH

C

C

C H

H

H

H H

HH

Each drawing has two solid lines, one wedge, and one dashed line.



B Boonndd lleennggtthh • Bond length decreases across a row and increases down a column of the periodic table (Section 1.6A). C H

>

N H

>

O H

H F

H Cl

<

<

H Br

Increasing bond length Increasing bond length

•

Bond length decreases as the number of electrons between two nuclei increases (Section 1.10A). CH3 CH3

<

CH2 CH2 < H C C H

Increasing bond length

•

Bond length increases as the percent s-character decreases (Section 1.10B). Csp H

Csp2 H

Csp3 H

Increasing bond length

•

Bond length and bond strength are inversely related. Shorter bonds are stronger bonds (Section 1.10). longest C–C bond weakest bond

C C

C C

Increasing bond strength

C C

shortest C–C bond strongest bond

3

4

Smith: Study Guide/ 1. Structure and Bonding Text Study Guide/Solutions Manual to accompany: Organic Chemistry, 3rd Edition Solutions Manual to accompany Organic Chemistry, Third Edition

© The McGraw−Hill Companies, 2011

Chapter 1–4 •

Sigma () bonds are generally stronger than
bonds (Section 1.9). C C

1 strong m bond

C C

1 stronger m bond 1 weaker / bond

C C

1 stronger m bond 2 weaker / bonds

  EElleeccttrroonneeggaattiivviittyy aanndd ppoollaarriittyy ((SSeeccttiioonnss 11..1111,, 11..1122)) • Electronegativity increases across a row and decreases down a column of the periodic table. • A polar bond results when two atoms of different electronegativity are bonded together. Whenever C or H is bonded to N, O, or any halogen, the bond is polar. • A polar molecule has either one polar bond, or two or more bond dipoles that reinforce.  D Drraaw wiinngg LLeew wiiss ssttrruuccttuurreess:: A A sshhoorrttccuutt Chapter 1 devotes a great deal of time to drawing valid Lewis structures. For molecules with many bonds, it may take quite awhile to find acceptable Lewis structures by using trial-and-error to place electrons. Fortunately, a shortcut can be used to figure out how many bonds are present in a molecule. Shortcut on drawing Lewis structures—Determining the number of bonds: [1] Count up the number of valence electrons. [2] Calculate how many electrons are needed if there were no bonds between atoms and every atom has a filled shell of valence electrons; i.e., hydrogen gets two electrons, and second-row elements get eight. [3] Subtract the number obtained in Step [2] from the sum obtained in Step [1]. This difference tells how many electrons must be shared to give every H two electrons and every second-row element eight. Since there are two electrons per bond, dividing this difference by two tells how many bonds are needed. To draw the Lewis structure: [1] Arrange the atoms as usual. [2] Count up the number of valence electrons. [3] Use the shortcut to determine how many bonds are present. [4] Draw in the two-electron bonds to all the H’s first. Then, draw the remaining bonds between other atoms making sure that no second-row element gets more than eight electrons and that you use the total number of bonds determined previously. [5] Finally, place unshared electron pairs on all atoms that do not have an octet of electrons, and calculate formal charge. You should have now used all the valence electrons determined in the first step. Example: Draw all valid Lewis structures for CH3NCO using the shortcut procedure. [1] Arrange the atoms. H • In this case the arrangement of atoms is implied by the way the structure is H C N C O drawn. H

Smith: Study Guide/ Solutions Manual to accompany Organic Chemistry, Third Edition

1. Structure and Bonding

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Text

Study Guide/Solutions Manual to accompany Organic Chemistry, Third Edition

Structure and Bonding 1–5 [2] Count up the number of valence electrons. 3H's 2C's 1N 1O

x x x x

1 electron per H 4 electrons per C 5 electrons per N 6 electrons per O

= = = =

3 electrons 8 electrons 5 electrons + 6 electrons 22 electrons total

[3] Use the shortcut to figure out how many bonds are needed. • Number of electrons needed if there were no bonds: 3 H's x 4 second-row elements x

•

2 electrons per H = 8 electrons per element =

6 electrons + 32 electrons 38 electrons needed if there were no bonds

Number of electrons that must be shared: 38 electrons – 22 electrons 16 electrons must be shared

•

Since every bond takes two electrons, 16/2 = 8 bonds are needed.

[4] Draw all possible Lewis structures. • Draw the bonds to the H’s first (three bonds). Then add five more bonds. Arrange them between the C’s, N, and O, making sure that no atom gets more than eight electrons. There are three possible arrangements of bonds; i.e., there are three resonance structures. • Add additional electron pairs to give each atom an octet and check that all 22 electrons are used. H

H

H C N C O

H C N C O

H H

H

H

H C N C O

H

H C N C O

H

Bonds to H's added.

H C N C O

H

H

H

H

H C N C O

H C N C O

H

All bonds drawn in. Three arrangements possible.

•

Calculate the formal charge on each atom. H

H

H C N C O H

H C N C O H

+1

•

H

Electron pairs drawn in. Every atom has an octet.

–1

H H C N C O H

–1

+1

You can evaluate the Lewis structures you have drawn. The middle structure is the best resonance structure, since it has no charged atoms.

Note: This method works for compounds that contain second-row elements in which every element gets an octet of electrons. It does NOT necessarily work for compounds with an atom that does not have an octet (such as BF3), or compounds that have elements located in the third row and later in the periodic table.

5

6

Smith: Study Guide/ 1. Structure and Bonding Text Study Guide/Solutions Manual to accompany: Organic Chemistry, 3rd Edition Solutions Manual to accompany Organic Chemistry, Third Edition

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Chapter 1–6 C Chhaapptteerr 11:: A Annssw weerrss ttoo PPrroobblleem mss 1.1 The mass number is the number of protons and neutrons. The atomic number is the number of protons and is the same for all isotopes. Nitrogen-14 Nitrogen-13 a. number of protons = atomic number for N = 7 7 7 b. number of neutrons = mass number – atomic number 7 6 c. number of electrons = number of protons 7 7 d. The group number is the same for all isotopes. 5A 5A 1.2 The atomic number is the number of protons. The total number of electrons in the neutral atom is equal to the number of protons. The number of valence electrons is equal to the group number for second-row elements. The group number is located above each column in the periodic table. a. atomic number [1] 31P 15 15

1.3

c. valence e– 5

d. group number 5A

[2] 19 F 9

9

9

7

7A

[3] 12H

1

1

1

1A

Ionic bonds form when an element on the far left side of the periodic table transfers an electron to an element on the far right side of the periodic table. Covalent bonds result when two atoms share electrons. a.

F

F

covalent

1.4

b. total number of e– 15

b. Li+ Br

ionic

H H c. H C C H H H

All C–H and C–C bonds are covalent.

d. Na+ N H H

ionic

Both N–H bonds are covalent.

a. Ionic bonding is observed in NaF since Na is in group 1A and has only one valence electron, and F is in group 7A and has seven valence electrons. When F gains one electron from Na, they form an ionic bond. b. Covalent bonding is observed in CFCl3 since carbon is a nonmetal in the middle of the periodic table and does not readily transfer electrons.

1.5 Atoms with one, two, three, or four valence electrons form one, two, three, or four bonds, respectively. Atoms with five or more valence electrons form [8 – (number of valence electrons)] bonds. a. O 8
6 valence e
= 2 bonds

c. Br 8
7 valence e
= 1 bond

b. Al 3 valence e
= 3 bonds

d. Si 4 valence e
= 4 bonds

Smith: Study Guide/ Solutions Manual to accompany Organic Chemistry, Third Edition

1. Structure and Bonding

© The McGraw−Hill Companies, 2011

Text

Study Guide/Solutions Manual to accompany Organic Chemistry, Third Edition

Structure and Bonding 1–7 1.6 [1] Arrange the atoms with the H’s on the periphery. [2] Count the valence electrons. [3] Arrange the electrons around the atoms. Give the H’s 2 electrons first, and then fill the octets of the other atoms. [4] Assign formal charges (Section 1.3C). a.

[1]

[2] Count valence e
. 2C x 4 e
= 8 6H x 1 e
= 6 total e
= 14 [2] Count valence e
. 1C x 4 e
= 4 5H x 1 e
= 5 1N x 5 e
= 5 total e
= 14

H H

H C C H H H

b.

[1]

H

H C N H H H

c.

[1]

[2] Count valence e
. 1C x 4 e
=4 =3 3H x 1 e
negative charge = 1 =8 total e


H

H C H

d.

[1]

H H H C C H H H

[3]

All 14 e
used. All second-row elements have an octet. H

H H C N H

H C N H

H H

H H

e


12 used. N needs 2 more electrons for an octet. [3]

H

H

H C

H C

H

H

[The –1 charge on C is 6 used. explained in Section 1.3C.] C needs 2 more electrons for an octet. e


H H [2] Count valence e
. [3] = 4 1C x 4 e
H C Cl H C Cl 3H x 1 e
= 3 H H = 7 1Cl x 7 e–
8 e used. Complete octet. total e
= 14 Cl needs 6 more electrons for an octet.

H

H C Cl H

1.7

[3]

Follow the directions from Answer 1.6. a. HCN

b. H2CO

H C N

H C O H

c. HOCH2CO2H

Count valence e
. 1C x 4 e
= 4 1H x 1 e
= 1 1N x 5 e
= 5 total e
= 10 Count valence e
. 1C x 4 e
= 4 2H x 1 e
= 2 1O x 6 e
= 6 = 12 total e


H O Count valence e
. H O C C O H 2C x 4 e
= 8 H 4H x 1 e
= 4

3O x 6 e
= 18 total e
= 30

H C N

4

e


used.

H C N

Complete N and C octets.

H C O

H C O

H

H

6 e
used.

Complete O and C octets.

H O H O C C O H H

16 e
used.

H O H O C C O H H

Complete octets.

7

8

Smith: Study Guide/ 1. Structure and Bonding Text Study Guide/Solutions Manual to accompany: Organic Chemistry, 3rd Edition Solutions Manual to accompany Organic Chemistry, Third Edition

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Chapter 1–8 1.8 Formal charge (FC) = number of valence electrons – [number of unshared electrons + 1/2 (number of shared electrons)] H

a.

6
[2 + 1/2(6)] = +1

5
[0 + 1/2(8)] = +1

+

c.

CH3 N C

b.

H N H

O O O

H

4
[0 + 1/2(8)] = 0

5
[0 + 1/2(8)] = +1

4
[2 + 1/2(6)] =
1

6
[4 + 1/2(4)] = 0

6
[6 + 1/2(2)] =
1

1.9 H

H

a. CH3O

[1] H C O H

[1] H C C

b. HC2

c. (CH3NH3)

[1] H H H C N H H H

d. (CH3NH)–

[1] H H C N H H

H

[2] Count valence e
. [3] H C O 1C x 4 e
= 4 H 3H x 1 e
= 3
8 e used. 1O x 6 e
= 6
= 13 total e Add 1 for (
) charge = 14

H C O

[2] Count valence e
. [3] H C C 2C x 4 e
= 8 1H x 1 e
= 1 4 e
used. total e
= 9 Add 1 for (
) charge = 10

H C C

H

[4] H C O

H

H

Assign charge.

[4] H C C Assign charge.

[2] Count valence e
. [3] H H 1C x 4 e
= 4 H C N H 6H x 1 e
= 6 H H 1N x 5 e
= 5 14 e
used. total e
= 15 Subtract 1 for (+) charge = 14

[4]

[2] Count valence e
. [3] H 1C x 4 e
= 4 H C N H 4H x 1 e
= 4 H 1N x 5 e
= 5
used. 10 e total e
= 13 Add 1 for (
) charge = 14

[4]

H H

Assign charge.

Count valence e
. 2C x 4 e
= 8 4H x 1 e
= 4 2Cl x 7 e
= 14 total e
= 26

H H C H

H C Cl Cl

H H C Cl

H

Complete octet and assign charge.

H C Cl H

H

b. C3H8O (three isomers) Count valence e
. 3C x 4 e
= 12 8H x 1 e
= 8 1O x 6 e
= 6 total e
= 26

H H H H C C C O H H H H

H O H H C C C H H H H

H H

H

H C C O C H H H

H

H C N H

1.10 a. C2H4Cl2 (two isomers)

H H

H C N H

H

Smith: Study Guide/ Solutions Manual to accompany Organic Chemistry, Third Edition

1. Structure and Bonding

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Text

Study Guide/Solutions Manual to accompany Organic Chemistry, Third Edition

Structure and Bonding 1–9 c. C3H6 (two isomers)

H

Count valence e
. 3C x 4 e
=12 6H x 1 e
= 6 total e
= 18

H H C C C H H H H

H

H C C C H H H

1.11 Two different definitions: • Isomers have the same molecular formula and a different arrangement of atoms. • Resonance structures have the same molecular formula and the same arrangement of atoms. 2 lone pairs

N in the middle

N at the end

3 lone pairs

O

a.

N C O

and

C N O

b.

O

HO C O

different arrangement of atoms = isomers

and

HO C O

same arrangement of atoms = resonance structures

1.12 Isomers have the same molecular formula and a different arrangement of atoms. Resonance structures have the same molecular formula and the same arrangement of atoms. 2 lone pairs 3 lone pairs CH3 bonded to C=O H bonded to C=O a.

O

O

CH3 C OH

CH3 C OH

O

c.

same arrangement of atoms = resonance structures (C2H5O2)– C2H4O2 O

CH3 C OH

A

C

D

different arrangement of atoms = isomers

O

CH3 C OH

H C CH2OH

A

B

A

b.

CH3 C OH

O

O

O

d. CH3 C OH

H

H C CH2OH

B

D

different arrangement of atoms = isomers

different molecular formulas = neither

1.13 Curved arrow notation shows the movement of an electron pair. The tail begins at an electron pair (a bond or a lone pair) and the head points to where the electron pair moves. a.

H C O

H C O

H

H

The net charge is the same in both resonance structures.

b.

CH3 C C CH2

CH3 C C CH2

H H

H H

The net charge is the same in both resonance structures.

9

10

Smith: Study Guide/ 1. Structure and Bonding Text Study Guide/Solutions Manual to accompany: Organic Chemistry, 3rd Edition Solutions Manual to accompany Organic Chemistry, Third Edition

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Chapter 1–10 1.14 Compare the resonance structures to see what electrons have “moved.” Use one curved arrow to show the movement of each electron pair. a.

CH2 C

C CH3

H

CH2 C

H

C CH3

H

b.

O C O

O C O

O

O

H

One electron pair moves: one curved arrow.

Two electron pairs move: two curved arrows.

1.15 To draw another resonance structure, move electrons only in multiple bonds and lone pairs and keep the number of unpaired electrons constant. a.

CH3 C C H H

b.

C CH3

CH3 C C

H

H H

CH3 C CH3

CH3 C CH3

Cl

Cl

C CH3

c.

H C C Cl

H

H C C Cl H H

H H

1.16 A “better” resonance structure is one that has more bonds and fewer charges. The better structure is the major contributor and all others are minor contributors. To draw the resonance hybrid, use dashed lines for bonds that are in only one resonance structure, and use partial charges when the charge is on different atoms in the resonance structures. a.

CH3 C N CH3

CH3 C N CH3

H H

H H

hybrid:

+ +

All atoms have octets. one more bond major contributor

CH3 C N CH3

b.

CH2

C

CH2

CH2

H

CH2

H

These two resonance structures are equivalent. They both have one charge and the same number of bonds. They are equal contributors to the hybrid. hybrid:





H H

C

CH2



C

CH2

H

1.17 Draw a second resonance structure for nitrous acid.
+ H O N O

H O N O

major contributor fewer charges

minor contributor

H O


– N

O

hybrid

1.18 All representations have a carbon with two bonds in the plane of the page, one in front of the page (solid wedge) and one behind the page (dashed line). Four possibilities:

H

H

H

Cl Cl

C

C

C

Cl Cl

Cl Cl

H

H

Cl

Cl C H

HH

Smith: Study Guide/ Solutions Manual to accompany Organic Chemistry, Third Edition

1. Structure and Bonding

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Text

Study Guide/Solutions Manual to accompany Organic Chemistry, Third Edition

Structure and Bonding 1–11 1.19 To predict the geometry around an atom, count the number of groups (atoms + lone pairs), making sure to draw in any needed lone pairs or hydrogens: 2 groups = linear, 3 groups = trigonal planar, 4 groups = tetrahedral. N has 2 atoms + 2 lone pairs 4 groups = tetrahedral (or bent molecular shape)

3 groups = trigonal planar 4 groups = tetrahedral

4 groups = tetrahedral O

CH3 C CH3

a.

c.

NH2

3 groups = trigonal planar

b.

4 groups = tetrahedral

4 groups = tetrahedral

4 groups = tetrahedral

CH3 C N

d.

CH3 O CH3

2 groups = linear

2 groups = linear

4 groups = tetrahedral (or bent molecular shape)

1.20 To predict the bond angle around an atom, count the number of groups (atoms + lone pairs), making sure to draw in any needed lone pairs or hydrogens: 2 groups = 180°, 3 groups = 120°, 4 groups = 109.5°. This C has 3 groups, so both angles are 120°.

2 groups = 180°

H

a. CH3 C C Cl

H

b. CH2 C Cl

c. CH3 C Cl H

This C has 4 groups, so both angles are 109.5°.

2 groups = 180°

1.21 To predict the geometry around an atom, use the rules in Answer 1.19.

3 groups H H trigonal planar H C HO C C C C C C C C C C H H H H H

4 groups 2 groups tetrahedral linear (or bent molecular shape)

4 groups tetrahedral (or bent molecular shape) H 4 groups H H O H H tetrahedral C C C C C CH3 H H H H H

3 groups trigonal planar

enanthotoxin

11

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Chapter 1–12 1.22 Reading from left to right, draw the molecule as a Lewis structure. Always check that carbon has four bonds and all heteroatoms have an octet by adding any needed lone pairs. (CH3)2CHCH(CH2CH3)2

(CH3)3CCH(OH)CH2CH3

H

(CH3)2CHCHO

H C H

H H

H C H H H H H H

a. H C C C C C C H

H

H C H

HH C H

H C H

b. H C

H H H H H H C H

c. H C

C H

H CH3(CH2)4CH(CH3)2

C

H C H

H H

O H H C C C H

HH C H

H C H H

H O

d. H C C

C H

H H C H

H H H

H

H

H C H

double bond needed to give C an octet

H

1.23 Simplify each condensed structure using parentheses. CH2CH3

a.

CH3CH2CH2CH2CH2Cl

CH2OH

b. CH3CH2CH2 C CH2CH3 H

CH3(CH2)4Cl

CH3

CH3

c. HOCH2 C CH2CH2CH2 C CH2 C CH3 H

CH3(CH2)2CH(CH2CH3)2

CH3

CH3

(HOCH2)2CH(CH2)3C(CH3)2CH2C(CH3)3

1.24 Draw the Lewis structure of lactic acid. H H O H C C

CH3CH(OH)CO2H

O C O H

H H

1.25 In shorthand or skeletal drawings, all line junctions or ends of lines represent carbon atoms. The carbons are all tetravalent. 1H

1H

1H

O

O

O

3 H's

0 H's

O

a.

b.

0 H's

O

0 H's

O

octinoxate (2-ethylhexyl 4-methoxycinnamate)

3 H's

avobenzone

1.26 In shorthand or skeletal drawings, all line junctions or ends of lines represent carbon atoms. Convert by writing in all carbons, and then adding hydrogen atoms to make the carbons tetravalent. H

a.

H C C

CH3

CH3 H C C H H H

b.

H H O H C C H H H C C C H H H H

CH3 H

c.

CH3

C

C H

C

H H C Cl C C H H H

H

HH

d.

CH3

N

C

C

N

CH3 H H

CH3

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Structure and Bonding 1–13 1.27 A charge on a carbon atom takes the place of one hydrogen atom. A negatively charged C has one lone pair, and a positively charged C has none. H

d.

c.

b.

a.

positive charge no lone pairs one H needed

negative charge one lone pair one H needed

positive charge no lone pairs no H's needed

O

H

H

negative charge one lone pair one H needed

1.28 Draw each indicated structure. Recall that in the skeletal drawings, a carbon atom is located at the intersection of any two lines and at the end of any line. O

a. (CH3)2C CH(CH2)4CH3 =

c.

N

= (CH3)2CH(CH2)2CONHCH3

H H

b.

H

CH3 C

C CH2CH2Cl

H2N C H

C H H

O

Cl =

d.

HO

O

= HO(CH2)2CH=CHCO2CH(CH3)2

H2N

1.29 To determine the orbitals used in bonding, count the number of groups (atoms + lone pairs): 4 groups = sp3, 3 groups = sp2, 2 groups = sp, H atom = 1s (no hybridization). All covalent single bonds are , and all double bonds contain one  and one
bond. Each H uses a 1s orbital.

H H H H C C C H

All single bonds are
bonds.

Each C–C bond is Csp3–Csp3. Each C–H bond is Csp3–H1s.

H H H

Total of 10
bonds.

Each C has 4 groups and is sp3 hybridized.

1.30 [1] Draw a valid Lewis structure for each molecule. [2] Count the number of groups around each atom: 4 groups = sp3, 3 groups = sp2, 2 groups = sp, H atom = 1s (no hybridization). Note: Be and B (Groups 2A and 3A) do not have enough valence e– to form an octet, and do not form an octet in neutral molecules. H

a. [1] H C Be

H

[2] Count groups around each atom: H

H

Be has 2 bonds.

H C Be

H

H

4 groups sp3

2 groups sp

[3] All C–H bonds: Csp3–H1s C–Be bond: Csp3–Besp Be–H bond: Besp–H1s

13

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Chapter 1–14 CH3

b. [1] CH3 B CH3

[2] Count groups around each atom:

[3] All C–H bonds: Csp3–H1s C–B bonds: Csp3–Bsp2

CH3 CH3 B CH3

B forms 3 bonds.

4 groups sp3 H

3 groups sp2

H

[2] Count groups around each atom:

c. [1] H C O C H H

H

H

[3] All C–H bonds: Csp3–H1s C–O bonds: Csp3–Osp3

H

H C O C H

4 groups sp3

H

H

4 groups sp3

1.31 To determine the hybridization, count the number of groups around each atom: 4 groups = sp3, 3 groups = sp2, 2 groups = sp, H atom = 1s (no hybridization). a.

b.

CH3 C CH

3 groups 3 groups sp2 sp2

2 groups sp

4 groups sp3

c. CH2 C CH2

N CH3

3 groups sp2

2 groups sp

1.32 All single bonds are . Multiple bonds contain one  bond, and all others are
bonds. All CH bonds are  bonds. O

a. CH3

C

H

 bond

one  bond, one
bond

 bond

b. CH3 C N one  + two
bonds

 bond  bond

c.

H

O C

 bond

one  bond, one
bond O CH3

 bond

1.33 Bond length and bond strength are inversely related: longer bonds are weaker bonds. Single bonds are weaker and longer than double bonds, which are weaker and longer than triple bonds. bond 1: single bond

a.

bond 3: double bond

C C

increasing bond strength: 1 < 3 < 2 increasing bond length: 2 < 3 < 1

bond 2: triple bond

b.

bond 1: single bond CH3

H N

N

bond 2: double bond C N

bond 3: triple bond

increasing bond strength: 1 < 2 < 3 increasing bond length: 3 < 2 < 1

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Structure and Bonding 1–15 1.34 Bond length and bond strength are inversely related: longer bonds are weaker bonds. Single bonds are weaker and longer than double bonds, which are weaker and longer than triple bonds. Increasing percent s-character increases bond strength and decreases bond length. CH3

a.

CH3 C C H

or H

Csp–H1s

CH2 N H

H CH3 N H

or

Nsp2–H1s 33% s-character shorter bond

Csp2–H1s 33% s-character

50% s-character shorter bond

Nsp3–H1s 25% s-character

H

H

b.

c.

C CH2

C O

H C OH

or

H

H

2

Csp –H1s 33% s-character shorter bond

Csp3–H1s 25% s-character

1.35 Electronegativity increases across a row of the periodic table and decreases down a column. Look at the relative position of the atoms to determine their relative electronegativity. most electropositive most electropositive most electronegative most electronegative a.

Se < S < O

Na < P < Cl

b.

increasing electronegativity

most electropositive most electropositive most electronegative most electronegative c.

increasing electronegativity

S < Cl < F

increasing electronegativity

d.

Pa Both O–H protons [(b) and (c)] are more acidic than the C–H proton (a) by the element effect. The most acidic proton has added resonance stabilization when it is removed, making its conjugate base the most stable.

OH H OH

c a

b O

conjugate base by loss of (c):

O O

O

H OH

H OH

resonance stabilization negative charge on O in both resonance structures This makes (c) most acidic.

O

conjugate base by loss of (b): H O

no resonance stabilization, but negative charge on O, an electronegative atom

OH

O

conjugate base by loss of (a):

2.62

OH

OH

OH

OH

This conjugate base has two resonance structures, but one places a negative charge on C.

Lewis bases are electron pair donors: they contain a lone pair or a
bond. Brønsted–Lowry bases are proton acceptors: to accept a proton they need a lone pair or a
bond. This means Lewis bases are also Brønsted–Lowry bases. lone pairs on O both

O

a. b.

2.63

O

H

C

CH3 Cl

lone pairs on Cl both

d.

c.

H

neither = no lone pairs or
bond


bonds both

A Lewis acid is an electron pair acceptor and usually contains a proton or an unfilled valence shell of electrons. A Brønsted–Lowry acid is a proton donor and must contain a hydrogen atom. All Brønsted–Lowry acids are Lewis acids, though the reverse may not be true. a. H3O+ both contains a H

b.

Cl3C+

Lewis acid unfilled valence shell on C

c. BCl3

d. BF4

Lewis acid unfilled valence shell on B

neither no H or unfilled valence shell

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Acids and Bases 2–21

2.64

Label the Lewis acid and Lewis base and then draw the products. Cl

a.

+ BCl3

Cl

Lewis base

O

c.

Cl B Cl

CH3

Cl

Lewis acid

new bond

b.

CH3

CH3

C

C

CH3

CH3 H OSO3H

+ CH3

Cl

OH

Lewis base

C C H CH3

+

HSO4

new bond

BF3

CH3CH2OH + BF3

nucleophile

electrophile

OH2

CH3CH2 O H

+ H2O

d. electrophile

b.

CH3 C O CH3

c.

CH3 S CH3

Br

electrophile

+

e. Br Br CH3

+ BF3

nucleophile

nucleophile

AlCl3

CH3SCH3 + AlCl3

nucleophile

nucleophile

FeBr3

Br Br Fe Br Br

electrophile

C O BF3 CH3

electrophile

Draw the product of each reaction. CH2CH3

CH2CH3 + H2O

a. CH3CH2 C

CH2CH3

CH2CH3

b. CH3CH2 C

CH3CH2 C

+ NH3

CH2CH3

CH2CH3 + (CH3)2NH

CH3CH2 CH3

e. CH3CH2 C

OH

CH2CH3

CH2CH3

CH2CH3 CH3CH2 C NH3

CH2CH3

CH3CH2 CH3

+ CH3OH

CH2CH3

c. CH3CH2 C

CH2CH3

d. CH3CH2 C

CH3CH2 C OH2

CH2CH3

CH3CH2 C

CH2CH3

CH3CH2 C

O CH3

CH2CH3

2.67 nucleophile OH H Br

proton transfer

OH2

Br + Br

electrophile

NHCH3

CH2CH3

CH3CH2 CH3

+ (CH3)2O

CH2CH3

a.

new bond

A Lewis acid is also called an electrophile. When a Lewis base reacts with an electrophile other than a proton, it is called a nucleophile. Label the electrophile and nucleophile in the starting materials and then draw the products.

a.

2.66

CH3 C Cl

Lewis acid

Lewis base

2.65

O

+ OH

CH3

CH3

Lewis acid

C

+ H2 O

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Chapter 2–22 Br

b.

proton transfer

H Br

+ Br

electrophile nucleophile H

H Br

+ Br

c. H

Br + Br

proton transfer

+ HBr

H

H

nucleophile electrophile

Draw the products of each reaction. In part (a), –OH pulls off a proton and thus acts as a Brønsted–Lowry base. In part (b), –OH attacks a carbon and thus acts as a Lewis base.

2.68

a.

CH2

+ C(CH3)2

b. OH + (CH3)3C+

H2O + CH2=C(CH3)2

(CH3)3COH

H

OH

Answer each question about esmolol.

2.69

most acidic second most acidic Esmolol contains C–H, N–H, and O–H bonds. H Since acidity increases across a row of the N periodic table, the OH bond is most acidic, followed by the NH bond.

OH

a.

O CH3O

esmolol

O

O H

b.

O

Na+ O

H N

O

Na+ H

CH3O

+ H2

CH3O O

O

O H

c.

O

OH H H

H N

O

H Cl

CH3O

O

OH

d, e, f.

O

* CH3O

* O

*

*

*

H N

*

N

+ Cl–

CH3O O

*

H N

All sp2 C are indicated with an arrow. The N is the only trigonal pyramidal atom. The
+ C's are indicated with a (*).

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Acids and Bases 2–23

2.70

Draw the product of protonation of either O or N and compare the conjugate acids. When acetamide reacts with an acid, the O atom is protonated because it results in a resonancestabilized conjugate acid. O

protonate O

CH3

O CH3

C

NH2

C

H

O

NH2

CH3

C

H

resonance stabilization of the + charge O is more readily protonated because the product is resonance stabilized.

NH2

O

protonate N

acetamide

CH3

C

no other resonance structure NH3

2.71 O

O

HO

O OH

pKa = 2.86

HO

O O

+

O

O

O O

O

O O

This group destabilizes the second negative charge.

O

HO

O

+ +  stabilizes the (
) charge of the conjugate base.

COO
now acts as an electron-donor group which destabilizes the conjugate base, making removal of the second proton more difficult and thus it is less acidic than CH3COOH.

The nearby COOH group serves as an electron-withdrawing group to stabilize the negative charge. This makes the first proton more acidic than CH3COOH.

2.72

O

pKa = 5.70

O

The COOH group of glycine gives up a proton to the basic NH2 group to form the zwitterion. O

a.

acts as a base

NH2CH2 C

proton transfer acts as an acid

OH

glycine b.

O

O + Cl


NH3CH2 C O

OH

most basic site O

c.

H

NH2

CH2 C O

most acidic site

Na+

OH

O

zwitterion form

H Cl

NH3CH2 C

O NH3CH2 C

O NH2CH2 C O

+ Na+ + H2O

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Chapter 2–24 2.73

Use curved arrows to show how the reaction occurs. O [1]

O

H

O

O H

H OH O

O [2]

+ H O H

OH

H

Protonate the negative charge on this carbon to form the product.

2.74

Compare the OH bonds in Vitamin C and decide which one is the most acidic. OH O

HO

O

Vitamin C ascorbic acid

This is the most acidic proton since the conjugate base is most resonance stabilized.

HO

OH

loss of H+ OH

OH O

HO

O

O

OH

O

OH

OH

O

OH O

HO

O

HO

O

O OH

The most delocalized anion with 3 resonance structures.

Removal of either of these H's does not give a resonancestabilized anion. HO

OH O

HO

O OH

loss of H+ HO

OH O

HO

O

O

HO

O

HO

only 2 resonance structures This proton is less acidic since its conjugate base is less resonance stabilized.

O O

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57

Introduction to Organic Molecules and Functional Groups 3–1 C Chhaapptteerr 33:: IInnttrroodduuccttiioonn ttoo O Orrggaanniicc M Moolleeccuulleess aanndd FFuunnccttiioonnaall G Grroouuppss

Increasing strength



TTyyppeess ooff iinntteerrm moolleeccuullaarr ffoorrcceess ((33..33)) Type of force van der Waals (VDW)

Cause Due to the interaction of temporary dipoles • Larger surface area, stronger forces • Larger, more polarizable atoms, stronger forces

Examples All organic compounds

dipole–dipole Due to the interaction of permanent dipoles (DD) hydrogen bonding Due to the electrostatic interaction of a H atom (HB or H-bonding) in an O–H, N–H, or H–F bond with another N, O, or F atom.

H2O

ion–ion

NaCl, LiF

Due to the interaction of two ions

(CH3)2C=O, H2O



PPhhyyssiiccaall pprrooppeerrttiieess Property Boiling point (3.4A)

•

Observation For compounds of comparable molecular weight, the stronger the forces the higher the bp. CH3CH2CH2CH2CH3

CH3CH2CH2CHO VDW, DD MW = 72 bp = 76 oC

VDW MW = 72 bp = 36 oC

CH3CH2CH2CH2OH VDW, DD, HB MW = 74 bp = 118 oC

Increasing strength of intermolecular forces Increasing boiling point

3.1 For compounds with similar functional groups, the larger the surface area, the higher the bp. CH3CH2CH2CH3 bp = 0 oC

CH3CH2CH2CH2CH3 bp = 36 oC

Increasing surface area Increasing boiling point

•

For compounds with similar functional groups, the more polarizable the atoms, the higher the bp. CH3I

CH3F bp =
78

oC

bp = 42 oC

Increasing polarizability Increasing boiling point

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Chapter 3–2 Melting point (3.4B)

•

For compounds of comparable molecular weight, the stronger the forces the higher the mp. CH3CH2CH2CH2CH3

CH3CH2CH2CHO VDW, DD MW = 72 mp =
96 oC

VDW MW = 72 mp =
130 oC

CH3CH2CH2CH2OH VDW, DD, HB MW = 74 mp =
90 oC

Increasing strength of intermolecular forces Increasing melting point

•

For compounds with similar functional groups, the more symmetrical the compound, the higher the mp. CH3CH2CH(CH3)2 mp =
160

oC

(CH3)4C mp =
17 oC

Increasing symmetry Increasing melting point

Solubility (3.4C)

Types of water-soluble compounds: • Ionic compounds • Organic compounds having
5 C’s, and an O or N atom for hydrogen bonding (for a compound with one functional group). Types of compounds soluble in organic solvents: • Organic compounds regardless of size or functional group. • Examples: CCl4

soluble

CCl4

soluble

H2O

soluble

O

CH3CH2CH2CH3 butane

CH3

H2O

insoluble

C

CH3

acetone

Key: VDW = van der Waals, DD = dipole–dipole, HB = hydrogen bonding MW = molecular weight

 R Reeaaccttiivviittyy ((33..88)) • Nucleophiles react with electrophiles. • Electronegative heteroatoms create electrophilic carbon atoms, which tend to react with nucleophiles. • Lone pairs and  bonds are nucleophilic sites that tend to react with electrophiles. CH3CH2 Cl


+

electrophilic site

OH
+

CH3 CH3 O CH3

CH3 N CH3

basic and nucleophilic site

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Introduction to Organic Molecules and Functional Groups 3–3 Chapter 3: A Annssw weerrss ttoo PPrroobblleem mss 3.1 CH3CH2 OH

CH3CH2 O H

H OSO3H

Na+ H

+ HSO4

CH3CH2 OH2

H2SO4

CH3CH3

CH3CH2 O Na+ + H2

no reaction

NaH

CH3CH3

no reaction

3.2 Identify the functional groups based on Tables 3.1, 3.2, and 3.3. alkene (double bond) HO

alkene (double bond)

CO2H

ether

O

CO2CH2CH3

O HO

carboxylic acid

N H

OH

shikimic acid

alcohols (hydroxy groups)

ester NH2

oseltamivir

amide amine

3.3 One possible structure for each functional group: O

a. aldehyde = R

b. ketone =

R

C

H

CH3CH2CH2

C

c. carboxylic acid =

H

O

O

O

C

C

C

CH3

R

d. ester =

CH2CH3

R

3.4 One possible structure for each description: O

a. C5H10O CH3CH2CH2CH2

C

O H

CH3CH2CH2

aldehyde ketone b. C6H10O

CH3CH2

O

H

C

C

C H

C

CH3

ketone ketone CH3

alkene

O

O

O

CH3CH2

O C

H C

C

H H

C H

alkene

H

R

C

CH3CH2CH2

OH

O O

R CH3CH2

C

O

CH3

C

OH

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Chapter 3–4 3.5 Summary of forces: • All compounds exhibit van der Waals forces (VDW). • Polar molecules have dipole–dipole forces (DD). • Hydrogen bonding (H-bonding) can occur only when a H is bonded to an O, N, or F. a. only nonpolar C–C and C–H bonds VDW only

e. CH3CH2CH2COOH

(CH3CH2)3N

c.

• VDW forces • polar C–N bonds - DD • no H on N so no H-bonding

• VDW forces • polar C–O bonds and a net dipole - DD • H bonded to O H-bonding

O

b.

d.

f.

CH2 CHCl

• VDW forces • polar C–Cl bond - DD

• VDW forces • 2 polar C–O bonds and a net dipole - DD • no H on O so no H-bonding

CH3 C C CH3

only nonpolar C–H and C–C bonds VDW only

3.6 One principle governs boiling point: • Stronger intermolecular forces = higher bp. Increasing intermolecular forces: van der Waals < dipole–dipole < hydrogen bonding Two factors affect the strength of van der Waals forces, and thus affect bp: • Increasing surface area = increasing bp. Longer molecules have a larger surface area. Any branching decreases the surface area of a molecule. • Increasing polarizability = increasing bp. a. (CH3)2C=CH2 or (CH3)2C=O

c. CH3(CH2)4CH3 or CH3(CH2)5CH3 longer molecule, more surface area higher boiling point

only VDW VDW and DD polar, stronger intermolecular forces higher boiling point b. CH3CH2COOH or CH3COOCH3

d.

CH2 CHCl

or

CH2 CHI

I is more polarizable. higher boiling point

no H-bonding VDW, DD, and H-bonding stronger intermolecular forces higher boiling point

3.7 Increasing intermolecular forces: van der Waals < dipole–dipole < hydrogen bonding O CH3CH2

C

O NH2

N–H bonds allow for hydrogen bonding. stronger intermolecular forces higher boiling point

H

C

N

CH3

CH3

no hydrogen bonding weaker intermolecular forces

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61

Introduction to Organic Molecules and Functional Groups 3–5 3.8 a.

or

b.

NH2

more polar stronger intermolecular forces (H-bonding) higher mp

3.9

or

more spherical packs better higher mp

Compare the intermolecular forces to explain why sodium acetate has a higher melting point than acetic acid. O CH3

C

O OH

CH3

acetic acid

C

O– Na+

sodium acetate

a. VDW, DD, and H-bonding b. not ionic, lower melting point

a. VDW, DD, ionic bonds b. Ionic bonds are the strongest: higher melting point.

3.10 In the more ordered solid phase, molecules are much closer together than in the less ordered liquid phase. The shape of a molecule determines how closely it can pack in the solid phase so symmetry is important. In the liquid phase, molecules are already farther apart, so symmetry is less important and thus it doesn’t affect boiling point. 3.11 A compound is water soluble if it is ionic or if it has an O or N atom and
5 C’s. a. CH3CH2OCH2CH3

b. CH3CH2CH2CH2CH3

an O atom that can H-bond with water
5 C's water soluble

c. (CH3CH2CH2CH2)3N

nonpolar not water soluble

an N atom that can H-bond to H2O, but > 5 C's not water soluble

3.12 Hydrophobic portions will primarily be hydrocarbon chains. Hydrophilic portions will be polar. Circled regions are hydrophilic because they are polar. All other regions are hydrophobic since they have only C and H. OH C C H

O COOH

a.

c.

b.

HO O

norethindrone

arachidonic acid

OH

benzo[a]pyrene derivative

62

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Chapter 3–6 3.13 Like dissolves like. • To be soluble in water, a molecule must be ionic, or have a polar functional group capable of Hbonding for every 5 C’s. • Organic compounds are generally soluble in organic solvents regardless of size or functional group. nonpolar long hydrocarbon chain

polar

O

O

OH

a. N

b.

polar

vitamin B3 (niacin) soluble in water due to two polar functional groups and only 6 C's in the molecule

nonpolar

O

vitamin K1 (phylloquinone) soluble in organic solvents two polar C–O bonds but the compound has > 10 C's water insoluble

3.14 A soap contains both a long hydrocarbon chain and a carboxylic acid salt. ionic salt a. CH3CO2–Na+

short chain carboxylic acid salt

ionic salt

b. CH3(CH2)14CO2–Na+

c. CH3(CH2)12COOH

long chain

no salt

This is a soap because it contains both a long chain and a carboxylic acid salt.

d. CH3(CH2)9CO2–Na+

long chain This is a soap because it contains both a long chain and a carboxylic acid salt.

3.15 Detergents have a polar head consisting of oppositely charged ions, and a nonpolar tail consisting of C–C and C–H bonds, just like soaps do. Detergents clean by having the hydrophobic ends of molecules surround grease, while the hydrophilic portion of the molecule interacts with the polar solvent (usually water). a detergent

SO3 Na+

nonpolar tail hydrophobic This end interacts with the grease to dissolve it.

polar head ionic - hydrophilic This end interacts with the water solvent to maintain the micelle's solubility in water.

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63

Introduction to Organic Molecules and Functional Groups 3–7 3.16 amide

ester

ester

ester

ether

ether

ester

H

O

ester

O

O

O

O

O

O O

O

HN

O

O

O

ether

N

O

O

H N

amide

ether

H

O

O

O

O

N

O O

ester amide

O

O

O

O

O

O

O

N

O O

N H

O

ester

ester

ester

ester

amide

amide

valinomycin

nonactin

3.17 Electronegative heteroatoms like N, O, or X make a carbon atom an electrophile. A lone pair on a heteroatom makes it basic and nucleophilic.
Bonds create nucleophilic sites and are more easily broken than  bonds. nucleophilic

a.

electrophilic b. H O H

Br

C bonded to Br electrophilic

nucleophilic

nucleophilic c.

d.

nucleophilic

N

CH3

electrophilic

3.18 Electrophiles and nucleophiles react with each other. O

a.

CH3CH2 Br

electrophile b.

+

YES

c.

+

Br

nucleophile

CH3

C

Cl

+

electrophile

nucleophile

CH3 C C CH3

nucleophile

OH

NO

d.

CH3 C C CH3

nucleophile

OCH3

YES

nucleophile + Br electrophile

YES

H

amide

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Chapter 3–8

3.19 Identify the functional groups based on Tables 3.1, 3.2, and 3.3. amide aromatic rings

aromatic ring

aromatic ring a.

CH2

HO

c.

CH3CH2CO2 C CH3 C H

e.

Darvon

O

N

carboxylic acid

sulfide CH3 CH3

COOH

penicillin G

ibuprofen

carboxylic acid

amine amine

NH2

alkene

amine N

CO2H

b.

S

O

CH2N(CH3)2

ester

amide

H N

O

d.

alkene

carboxylic acid pregabalin

f.

OH

H

O

alkyne

alkene

alkenes

O

O

alcohol

ester

ketone pyrethrin I

histrionicotoxin

3.20 CH3 OH

CH2OH

CH3 CH CH2 CH3

CH3CH2CH2CH2 OH

alcohol

alcohol

CH3 C OH

CH3 CH CH3

CH3

alcohol

alcohol

CH3 CH3CH2 O CH3CH2

CH3 O CH CH3

ether

ether

CH3 O CH2CH2CH3

ether

3.21 A cyclic ester is called a lactone. A cyclic amide is called a lactam. O

a.

N CH3

O

b.

amine

c.

ether

d.

O

NH O

ester lactone

amide lactam

3.22 Draw the constitutional isomers and identify the functional groups. O

ketone OH

carboxylic acid

H

O OH

ester

O

aldehyde O

ester O

H

alcohol

O OH

alcohol

O

aldehyde

O

H

O

aldehyde OH

alcohol

H

O

ether

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65

Introduction to Organic Molecules and Functional Groups 3–9 3.23 Use the rules from Answer 3.5. O

O OH

a.

OCH3

b.

c.

d. N

VDW VDW VDW no dipole–dipole dipole–dipole dipole–dipole no H-bonding (no O–H bond) no H-bonding (no N–H bond) (nonpolar C–C, C–H bonds) no H-bonding (no O, N, F)

VDW dipole–dipole H-bonding (O–H bond)

3.24 Increasing intermolecular forces: van der Waals < dipole–dipole < H-bonding a. increasing intermolecular forces: CH3CH3 < CH3Cl < CH3NH2

c. increasing intermolecular forces: (CH3)2C=C(CH3)2 < (CH3)2CHCOCH3 < (CH3)2CHCOOH

VDW VDW dipole–dipole dipole–dipole H-bonding b. increasing intermolecular forces: VDW

VDW dipole–dipole

VDW

d. increasing intermolecular forces:

CH3Cl < CH3Br < CH3I

VDW dipole–dipole H-bonding

CH3Cl < CH3OH < NaCl

Increasing polarizability stronger intermolecular forces

VDW VDW ionic dipole–dipole dipole–dipole H-bonding

3.25 O CH3

H O C CH3

C O H

O

hydrogen bonding between two acetic acid molecules

3.26 A = VDW forces; B = H-bonding; C = ion–ion interactions; D = H-bonding; E = H-bonding; F = VDW forces. 3.27 Use the principles from Answer 3.6. a.

I

CH3(CH2)4

CH3(CH2)5

I

CH3(CH2)6

I

Increasing size, increasing surface area, increasing boiling point

b.

CH3CH2CH2CH3 < (CH3)3N < CH3CH2CH2NH2

VDW

VDW dipole–dipole

VDW dipole–dipole H-bonding

Increasing boiling point

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Chapter 3–10 c.

(CH3)3COC(CH3)3 < CH3(CH2)3O(CH2)3CH3 < CH3(CH2)7OH VDW VDW VDW dipole–dipole dipole–dipole dipole–dipole smaller surface area larger surface area H-bonding highest bp

Increasing boiling point d.

<

Br <

VDW

VDW dipole–dipole

OH

VDW dipole–dipole H-bonding

OH

<

VDW dipole–dipole H-bonding larger surface area

Increasing boiling point

<

e.

<

smallest surface area most branching

largest surface area

Increasing boiling point OH

O

f.

< VDW

< VDW dipole–dipole

VDW dipole–dipole H-bonding

Increasing boiling point

3.28

In CH3CH2NHCH3, there is a N–H bond so the molecules exhibit intermolecular hydrogen bonding, whereas in (CH3)3N the N is bonded only to C, so there is no hydrogen bonding. The hydrogen bonding in CH3CH2NHCH3 makes it have much stronger intermolecular forces than (CH3)3N. As intermolecular forces increase, the boiling point of a molecule of the same molecular weight increases.

3.29

Stronger forces, higher mp. CH3

O CH(CH3)2

menthone VDW dipole–dipole lower melting point

CH3

OH CH(CH3)2

menthol VDW dipole–dipole H-bonding stronger forces higher melting point

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67

Introduction to Organic Molecules and Functional Groups 3–11 3.30

Stronger forces, higher mp. More symmetrical compounds, higher mp. CH3

a. (CH3)3CH < (CH3)2C=O < (CH3)2CHOH

VDW

VDW DD

c.

VDW DD H-bonding

< VDW

NH2

< VDW DD

Increasing intermolecular forces Increasing melting point

Cl

VDW DD H-bonding

Increasing intermolecular forces

b. CH3F < CH3Cl < CH3I

Increasing melting point

Increasing polarizability Increasing melting point

3.31


119 oC not symmetrical


118 oC not symmetrical

In both compounds the CH3 group dangling from the chain makes packing in the solid difficult, so the mp is low.

3.32


91 oC symmetrical higher mp


25 oC most spherical highest mp

This molecule can pack somewhat better since it has no CH3 group dangling from the chain, so the mp is somewhat higher. It also has the most surface area and this increases VDW forces compared to the first two compounds.

This compound packs the best since it is the most spherical in shape, increasing its mp.

Boiling point is determined solely by the strength of the intermolecular forces. Since benzene has a smaller size, it has less surface area and weaker VDW interactions and therefore a lower boiling point than toluene. The increased melting point for benzene can be explained by symmetry: benzene is much more symmetrical than toluene. More symmetrical molecules can pack more tightly together, increasing their melting point. Symmetry has no effect on boiling point. benzene bp = 80 oC mp = 5 oC very symmetrical closer packing in solid form higher mp

CH3

and

toluene bp = 111 oC mp =
93 oC

less symmetrical lower mp

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Chapter 3–12 3.33 Increasing polarity = increasing water solubility. Neither compound is very H2O soluble. a.

CH3CH2CH2CH3 <

(CH3)3CH

<

CH3OCH2CH3 < CH3CH2CH2OH

VDW

VDW VDW DD more spherical (This nonpolar, hydrophobic molecule is more compact, making it more water soluble than its straight-chain isomer, drawn to the left.)

b.

Br

polar no H-bonding

OH

O

polar H-bonding to H2O, not itself

VDW DD H-bonding

polar and H-bonding More opportunities for H-bonding with its O atom and its H on O.

3.34 Look for two things: • To H-bond to another molecule like itself, the molecule must contain a H bonded to O, N, or F. • To H-bond with water, a molecule need only contain an O, N, or F. These molecules can H-bond with water. All of these molecules have an O or N atom. b. CH3NH2, c. CH3OCH3, d. (CH3CH2)3N, e. CH3CH2CH2CONH2, g. CH3SOCH3, h. CH3CH2COOCH3

Each of these molecules can H-bond to another molecule like itself. Both compounds have N–H bonds. b. CH3NH2, e. CH3CH2CH2CONH2

3.35 Draw the molecules in question and look at the intermolecular forces involved. no H bonded to O O

diethyl ether

OH

H bonded to O: hydrogen bonding

1-butanol

VDW forces dipole–dipole forces H-bonding • Both have
5 C's and an electronegative O atom, so they can H-bond to water, making them soluble in water. • Only 1-butanol can H-bond to another molecule like itself, and this increases its boiling point. VDW forces dipole–dipole forces

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69

Introduction to Organic Molecules and Functional Groups 3–13 3.36

Use the solubility rule from Answer 3.13. OH CCl3

a.

Cl

C

Cl

H

O

CH3

b.

O

COOH

2 polar functional groups but > 10 C's not water soluble

O HO

d.

N

many polar bonds with N and O atoms many opportunities for H-bonding water soluble

OH OH

O HO

O

HO

O

CH3

caffeine

HO

OCH3

aspartame many polar bonds with N and O atoms many opportunities for H-bonding water soluble

OH

N N

N H

CH3O

CH3

N

H2N

mestranol

DDT no N or O not water soluble

O

e.

c.

OH

f. OH

sucrose many polar bonds with O 11 O's and 12 C's many opportunities for H-bonding with H2O water soluble

carotatoxin 1 polar functional group but > 10 C's not water soluble

3.37 (CH3)2CHCH(CH3)2 B 6 C's Branching makes less surface area, weaker VDW. lowest bp

CH3(CH2)4CH3 C 6 C's no branching

CH3(CH2)5CH3 D 7 C's

CH3(CH2)6CH3 A 8 C's highest bp

C, D, and A are all long chain hydrocarbons, but the size increases from C to D to A, increasing the VDW forces and increasing bp.

3.38

a.

Water solubility is determined by polarity. Polar molecules are soluble in water, while nonpolar molecules are soluble in organic solvents. Arrows indicate polar functional groups.

CH3

CH2NH2

b.

HO CH3 CH3

O CH3

CH3

CH3

CH3

CH3

vitamin E only 2 polar functional groups many nonpolar C–C and C–H bonds (29 C's) soluble in organic solvents insoluble in H2O

HOCH2

OH N

CH3

pyridoxine vitamin B6 many polar bonds and few nonpolar bonds soluble in H2O It is also soluble in organic solvents since it is organic, but is probably more soluble in H2O.

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Chapter 3–14 3.39

Compare the functional groups in the two components of sunscreen. Dioxybenzone will most likely be washed off in water because it contains two hydroxy groups and is more water soluble. O

OH

O

CH3O

C(CH3)3

avobenzone two ketones one ether

3.40

O

OCH3

dioxybenzone two hydroxy groups one ketone one ether more water soluble

Because of the O atoms, PEG is capable of hydrogen bonding with water, which makes PEG water soluble and suitable for a product like shampoo. PVC cannot hydrogen bond to water, so PVC is water insoluble, even though it has many polar bonds. Since PVC is water insoluble, it can be used to transport and hold water. O

H-bond

H

no H-bonding

H O

O

O

O Cl

poly(ethylene glycol) PEG water soluble

3.41

OH

Cl

Cl

Cl

poly(vinyl chloride) PVC water insoluble

Molecules that dissolve in water are readily excreted from the body in urine whereas less polar molecules that dissolve in organic solvents are soluble in fatty tissue and are retained for longer periods. Compare the solubility properties of THC and ethanol to determine why drug screenings can detect THC and not ethanol weeks after introduction to the body. CH3 OH CH3CH2 CH3

O CH3

(CH2)4CH3

OH

ethanol

tetrahydrocannabinol THC THC has relatively few polar bonds compared to the number of nonpolar bonds, making it soluble in organic solvents and therefore soluble in fatty tissue.

Ethanol has 1 O atom and only 2 C's, making it soluble in water.

Due to their solubilities, THC is retained much longer in the fatty tissue of the body, being slowly excreted over many weeks, while ethanol is excreted rapidly in urine after ingestion.

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71

Introduction to Organic Molecules and Functional Groups 3–15 3.42

Compare the intermolecular forces of crack and cocaine hydrochloride. Stronger intermolecular forces increase both the boiling point and the water solubility. CH3

ionic bond

H

N

CH3 N

Cl

COOCH3

COOCH3

O

O

O C

O C

H

H

cocaine hydrochloride a salt

cocaine (crack) neutral organic molecule

The molecules are identical except for the ionic bond in cocaine hydrochloride. Ionic forces are extremely strong forces, and therefore the cocaine hydrochloride salt has a much higher boiling point and is more water soluble. Since the salt is highly water soluble, it can be injected directly into the bloodstream where it dissolves. Crack is smoked because it can dissolve in the organic tissues of the nasal passage and lungs. 3.43

A laundry detergent must have both a highly polar end of the molecule and a nonpolar end of the molecule. The polar end will interact with water, while the nonpolar end surrounds the grease/organic material. H O H

H O H

a.

OCH2CH2O

CH2CH2O H

polar interacts with water by H-bonding at all O atoms, as well as H's bonded to O's.

nonpolar interacts with organic material

H O H H O H

CH2CH2O H N CH2CH2O

b. O

H O H H

H O H

H O H

nonpolar interacts with organic material

polar interacts with water by H-bonding at O and H atoms

H O H

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Chapter 3–16 3.44

An emulsifying agent is one that dissolves a compound in a solvent in which it is not normally soluble. In this case the phospholipids can dissolve the oil in its nonpolar tails and bring it into solution in the aqueous vinegar solution. Or, the nonpolar tails dissolve in the oil, and the polar head brings the water-soluble compounds into solution. In any case, the phospholipids make a uniform medium, mayonnaise, from two insoluble layers. vinegar oil aqueous organic hydrophilic hydrophobic These two ingredients will not mix. The emulsifying agent (egg yolk) has phospholipids that have both hydrophobic and hydrophilic portions, making the mayonnaise uniform.

3.45 O

a.

OH O

O

O H N

HCl

O

N H

OH O

O

N

O H N

OH HH O

O

N H

N H

five functional groups that have many opportunities for H-bonding water soluble

ionic salt more water soluble

c. Since the hydrochloride salt is ionic and therefore more water soluble, it is more readily transported in the bloodstream.

3.46

Use the rules from Answer 3.17. nucleophilic

a.

+


I 

nucleophilic

nucleophilic

+ c.

+ O

+

+

e. CH3OH

electrophilic

electrophilic

electrophilic O

b.

CH2

Cl–

N H O

b.

OH H H O

d.

f.

nucleophilic All the C=C's are nucleophilic.

CH3

nucleophilic

C

+ Cl

electrophilic (All lone pairs on O and Cl are nucleophilic.)

N

Cl–

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73

Introduction to Organic Molecules and Functional Groups 3–17 3.47 NO

+ Br

a.

d.

nucleophilic nucleophilic +

CH2Cl

b.

CN

YES

+

+

e. nucleophilic

nucleophilic

OH

NO

nucleophilic nucleophilic H3O

YES

electrophilic

electrophilic O

c.

CH3

C

+

CH3

CH3

YES

nucleophilic electrophilic

3.48

More rigid cell membranes have phospholipids with fewer C=C’s. Each C=C introduces a bend in the molecule, making the phospholipids pack less tightly. Phospholipids without C=C’s can pack very tightly, making the membrane less fluid, and more rigid. The double bonds introduce kinks in the chain, making packing of the hydrocarbon chains less efficient. This makes the cell membrane formed from them more fluid. O CH2O O (CH3)3NCH2CH2

H C O

O P O CH2

O

O

3.49 amine can H-bond NH2 HO

hydroxy group can H-bond OH

HO O

OH O

O O O

HO O O HN HO O most acidic proton HO

N H

O

Cl H N

O

O

C

Cl

N H H2N

H N O

O OH OH

OH O

vancomycin

N H

amide can H-bond

NHCH3

a. 7 amide groups [regular (unbolded) arrows] b. OH groups bonded to sp3 C's are circled. OH groups bonded to sp2 C's have a square. c. Despite its size, vancomycin is water soluble because it contains many polar groups and many N and O atoms that can H-bond to H2O. d. The most acidic proton is labeled (COOH group). e. Four functional groups capable of Hbonding are ROH, RCOOH, amides, and amines.

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Chapter 3–18 3.50 O OH

= C

CHO

O

OH

H

=

O

H

CHO

A

B

The OH and CHO groups are close enough that they can intramolecularly H-bond to each other. Since the two polar functional groups are involved in intramolecular Hbonding, they are less available for H-bonding to H2O. This makes A less H2O soluble than B, whose two functional groups are both available for H-bonding to the H2O solvent.

H

a. melting point H

Fumaric acid has its two larger COOH groups on opposite ends of the molecule, and in this way it can pack better in a lattice than maleic acid, giving it a higher mp. H COOH fumaric acid C C

b. solubility H

H C C

HOOC
+
+COOH

Maleic acid is more polar, giving it greater H2O solubility. The bond dipoles in fumaric acid cancel.

maleic acid

c. removal of the first proton (pKa1) H

H

HOOC

C C HOOC

H C C

COOH

H

H

COOH

loss of 1 proton

loss of 1 proton H C C O C

HOOC C O

O

H C C

O

H

COO

H

In maleic acid, intramolecular H-bonding stabilizes the conjugate base after one H is removed, making maleic acid more acidic than fumaric acid. d. removal of the second proton (pKa2) H

Intramolecular H-bonding is not possible here.

H C C

O C

C O O O

Now the dianion is held in close proximity in maleic acid, and this destabilizes the conjugate base. Thus, removing the second H in maleic acid is harder, making it a weaker acid than fumaric acid for removal of the second proton.

OOC

H C C

H

C

O

The OH and the CHO are too far apart to intramolecularly H-bond to each other, leaving more opportunity to H-bond with solvent.

3.51 HOOC

H

COO

The two negative charges are much farther apart. This makes the dianion from fumaric acid more stable and thus pKa2 is lower for fumaric acid than maleic acid.

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4. Alkanes

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Text

Study Guide/Solutions Manual to accompany Organic Chemistry, Third Edition

Alkanes 4–1 C Chhaapptteerr 44:: A Allkkaanneess  G Geenneerraall ffaaccttss aabboouutt aallkkaanneess ((44..11––44..33)) • Alkanes are composed of tetrahedral, sp3 hybridized C’s. • There are two types of alkanes: acyclic alkanes having molecular formula CnH2n + 2, and cycloalkanes having molecular formula CnH2n. • Alkanes have only nonpolar C
C and C
H bonds and no functional group so they undergo few reactions. • Alkanes are named with the suffix -ane.  C Cllaassssiiffyyiinngg C C’’ss aanndd H H’’ss ((44..11A A)) • Carbon atoms are classified by the number of C’s bonded to them; a 1o C is bonded to one other C, and so forth. C C

C

C

C C

C C

C

C

1o C

•

2o C

CH3 CH3

C C C

CH3CH2 C

C

3o C

H

1o C

4o C

C CH3 CH3

4o C

3o C

o

2 C

Hydrogen atoms are classified by the type of carbon atom to which they are bonded; a 1o H is bonded to a 1o C, and so forth. H C C

C

C

H C C

H C C

1o H

2o H

CH3

3o H

2o H

 N Naam meess ooff aallkkyyll ggrroouuppss ((44..44A A)) CH3

=

methyl CH3CH2

=

isopropyl

CH3CH2CHCH3

=

sec-butyl =

propyl (CH3)2CH

=

butyl

ethyl CH3CH2CH2

CH3CH2CH2CH2

(CH3)2CHCH2

=

isobutyl =

(CH3)3C

tert-butyl

3o H

CH3CH2 C CH3

C

1o H

H

=

75

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Chapter 4–2

C Coonnffoorrm maattiioonnss iinn aaccyycclliicc aallkkaanneess ((44..99,, 44..1100)) • Alkane conformations can be classified as staggered, eclipsed, anti, or gauche depending on the relative orientation of the groups on adjacent carbons. eclipsed

staggered

anti

gauche

HH

H

CH3

H

H H

H H

H

H

H H

• •

•

H

H

H

H

H CH3

H

CH3

H

•

H

CH3

•

Dihedral angle of • Dihedral angle of 2 2 CH3’s = 180o CH3’s = 60o A staggered conformation is lower in energy than an eclipsed conformation. An anti conformation is lower in energy than a gauche conformation. Dihedral angle = 0o

Dihedral angle = 60o



TTyyppeess ooff ssttrraaiinn • Torsional strain—an increase in energy due to eclipsing interactions (4.9). • Steric strain—an increase in energy when atoms are forced too close to each other (4.10). • Angle strain—an increase in energy when tetrahedral bond angles deviate from 109.5o (4.11).

TTw woo ttyyppeess ooff iissoom meerrss [1] Constitutional isomers—isomers that differ in the way the atoms are connected to each other (4.1A). [2] Stereoisomers—isomers that differ only in the way atoms are oriented in space (4.13B). cis

trans

CH3 CH3 CH3

constitutional isomers

CH3

CH3

CH3

stereoisomers



C Coonnffoorrm maattiioonnss iinn ccyycclloohheexxaannee ((44..1122,, 44..1133)) • Cyclohexane exists as two chair conformations in rapid equilibrium at room temperature. • Each carbon atom on a cyclohexane ring has one axial and one equatorial hydrogen. Ring-flipping converts axial to equatorial H’s, and vice versa. An axial H flips equatorial. Hax Heq

Ring-flip.

Heq Hax

An equatorial H flips axial.

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4. Alkanes

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Text

Study Guide/Solutions Manual to accompany Organic Chemistry, Third Edition

Alkanes 4–3 •

In substituted cyclohexanes, groups larger than hydrogen are more stable in the more roomy equatorial position. The larger CH3 group is equatorial. H CH3

H CH3

Conformation 1

Conformation 2

more stable 95%

•

axial

5%

Disubstituted cyclohexanes with substituents on different atoms exist as two possible stereoisomers. • The cis isomer has two groups on the same side of the ring, either both up or both down. • The trans isomer has two groups on opposite sides of the ring, one up and one down. CH3

CH3

H

CH3 H

H

trans isomer

CH3 H

cis isomer

 O Oxxiiddaattiioonn––rreedduuccttiioonn rreeaaccttiioonnss ((44..1144)) • Oxidation results in an increase in the number of C
Z bonds or a decrease in the number of C
H bonds. O CH3CH2 OH

ethanol

•

CH3

C

OH

Increase in C–O bonds = oxidation

acetic acid

Reduction results in a decrease in the number of C
Z bonds or an increase in the number of C
H bonds. H

H C C

H

H H H C C H

H

ethylene

H H

ethane

Increase in C–H bonds = reduction

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Chapter 4–4 C Chhaapptteerr 44:: A Annssw weerrss ttoo PPrroobblleem mss 4.1 The general molecular formula for an acyclic alkane is CnH2n + 2. Number of C atoms = n 23 25 27

2n + 2 2(23) + 2 = 2(25) + 2 = 2(27) + 2 =

Number of H atoms 48 52 56

4.2 Isopentane has 4 C’s in a row with a 1 C branch. H

H

a. CH3CH2 C CH3

b.

CH3

H C CH3 CH3 C CH3

H

c. CH3CH2CH(CH3)2 re-draw

H

isopentane

d.

isopentane

H C

H C H

e.

f. CH3CH(CH3)CH2CH3

CH3 C H CH3 H

re-draw isopentane

5 C's in a row pentane isopentane

isopentane

4.3 To classify a carbon atom as 1°, 2°, 3°, or 4° determine how many carbon atoms it is bonded to (1° C = bonded to one other C, 2° C = bonded to two other C’s, 3° C = bonded to three other C’s, 4° C = bonded to four other C’s). Re-draw if necessary to see each carbon clearly. To classify a hydrogen atom as 1°, 2°, or 3°, determine if it is bonded to a 1°, 2°, or 3° C (A 1° H is bonded to a 1° C; a 2° H is bonded to a 2° C; a 3° H is bonded to a 3° C). Re-draw if necessary. 1° C

a.

1° C's

1° C

[1] CH3CH2CH2CH3

[2] (CH3)3CH

[3]

[4]

4° C

CH3

1° C's

2° C's

1° C

CH3 C H CH3 3° C

4° C's All other C's are 1° C's.

All other C's are 2° C's. 3° C

re-draw

re-draw 1° H's 1° H's

HH

1° H's CH3

b.

[1] CH3CH2CH2CH3

[2]

[3] CH C 3

CH3 C H CH3

2° H's

CH3 CH3

1° H's 1° H's

3° H

C CH3

CH3 CH3

All 1° H's

[4]

H H

CH3 CH3

H H

1° H's

CH3 H H

H

3° H All others are 2° H's.

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4. Alkanes

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Alkanes 4–5

4.4 Use the definition of 1°, 2°, 3°, or 4° carbon atoms from Answer 4.3. 1° C

4° C O

3° C All other tetrahedral C's are 2° C's. HO

3° C

4.5 Constitutional isomers differ in the way the atoms are connected to each other. To draw all the constitutional isomers: [1] Draw all of the C’s in a long chain. [2] Take off one C and use it as a substituent. (Don’t add it to the end carbon: this re-makes the long chain.) [3] Take off two C’s and use these as substituents, etc. Five constitutional isomers of molecular formula C6H14: [1] long chain

[2] with one C as a substituent CH3

CH3CH2CH2CH2CH2CH3

[3] using two C's as substituents CH3

CH3

CH3CH2CH2 C CH3 CH3CH2 C CH2CH3 H

CH3CH2 C CH3

H

H CH3 C

CH3

H C CH3

CH3 CH3

4.6 Molecular formula C8H18 with one CH3 substituent: CH3

CH3

CH3CH2CH2CH2CH2 C CH3

CH3

CH3CH2CH2CH2 C CH2CH3

H

CH3CH2CH2 C CH2CH2CH3

H

H

4.7 Draw each alkane to satisfy the requirements. CH3

a.

b. 4° C

CH3

c. CH C CH CCH 3 2 3 1° C

1° C All other C's are 2° C's.

H

1° H

H

3° H

2° H

4.8 Draw each compound as a skeletal structure to compare the compounds. C3

C2

C3 CH3(CH2)3CH(CH3)2 =

CH3CH2CH(CH3)CH2CH2CH3 =

A

B

C

CH3 bonded to C3 identical to compound C

CH3 bonded to C2

CH3 bonded to C3 identical to compound A

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Chapter 4–6 4.9 Use the steps from Answer 4.5 to draw the constitutional isomers. Five constitutional isomers of molecular formula C5H10 having one ring: [2]

[1]

[3]

CH3

CH3

CH3

CH3

CH3

CH2CH3

Follow these steps to name an alkane: [1] Name the parent chain by finding the longest C chain. [2] Number the chain so that the first substituent gets the lower number. Then name and number all substituents, giving like substituents a prefix (di, tri, etc.). [3] Combine all parts, alphabetizing the substituents, ignoring all prefixes except iso.

4.10

4-tert-butyl

[1] CH3 a. CH3CH2CH2

[2]

CH3 C CH3 C CH2CH2CH2CH3 CH3

4

8 carbons = octane H [1] H C CH3 H C CH2 CHCH3 CH3 CH3

b.

[3] 4-tert-butyl-4-methyloctane

CH3 CH3 C CH3 CH3CH2CH2 C CH2CH2CH2CH3 CH3 6 7 8 1 2 3

4-methyl 5 H 6

[2]

[3] 2,4-dimethylhexane

H C CH3 1 H C CH2 CHCH3 4 CH3 2 CH3

6 carbons = hexane 4-methyl 2-methyl 6-isopropyl

[1]

H [2] CH3 C CH3 CH2CH3 CH3 CH3CH2CH2 C CH2 CH2 C CH3 CH3CH2CH2 H H 9 8 7 6

c.

H C CH3 C CH2 CH2 H 5 4

d.

[2]

CH3 CHCH2 CH3

5

CH2CH2CH3

1

C CH3

3

CH3 CHCH2

7

3-methyl

[3] 2,4-dimethylheptane

CH2CH2CH3

2 3

H

6

[3] 6-isopropyl-3-methylnonane

CH2CH3 C CH3 H

9 carbons = nonane

[1]

1

2

C CH3 H 4

CH3

4-methyl

7 carbons = heptane 2-methyl

4.11 Use the steps in Answer 4.10 to name each alkane. a. CH3CH2CH(CH3)CH2CH3 [1]

re-draw

[2]

1

2

3

4

5

CH3CH2 CH CH2CH3

CH3CH2 CH CH2CH3

CH3

CH3

5 carbons = pentane

[3] 3-methylpentane

3-methyl

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Alkanes 4–7

b. (CH3)3CCH2CH(CH2CH3)2 re-draw

[1]

[2] 2 CH3 1

CH3 CH3 C CH2 CH CH2CH3 CH3

[3] 4-ethyl-2,2-dimethylhexane 4

5

6

CH3 C CH2 CH CH2CH3

CH2CH3

CH3

CH2CH3

4-ethyl

6 carbons = hexane

2,2-dimethyl c. CH3(CH2)3CH(CH2CH2CH3)CH(CH3)2

re-draw [1]

4-isopropyl [3] 4-isopropyloctane

[2]

CH3 CH3CH2CH2CH2 CH CH CH3

CH3 8 7 6 5 4 CH3CH2CH2CH2 CH CH CH3

CH2CH2CH3

CH2CH2CH3

3

8 carbons = octane

2

1

2,2,4,4-tetramethyl [2]

d. [1]

[3] 2,2,4,4-tetramethylpentane 1 2 3 4 5

5 carbons = pentane

2-methyl [2]

[1]

1

[3] 3-ethyl-2,5-dimethylheptane

34 5 6 7

e. 2 or

3-ethyl 5-methyl longest chain = 7 carbons = heptane Number so there are more substituents. Pick the upper option. 2-methyl

f.

[2] 1

[1]

3

5

[3] 5-sec-butyl-3-ethyl-2,7-dimethyldecane 5-sec-butyl

2 10 carbons = decane

3-ethyl

6

8 9 10 7-methyl

4.12 To work backwards from a name to a structure: [1] Find the parent name and draw that number of C’s. Use the suffix to identify the functional group (-ane = alkane). [2] Arbitrarily number the C’s in the chain. Add the substituents to the appropriate C’s. [3] Re-draw with H’s to make C’s have four bonds. a. 3-methylhexane [1] 6 carbon alkane

[2]

[3] CH3

C

C C C C C

C

methyl on C3

C C C C C

CH3 CH3CH2 CH CH2CH2CH3

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Chapter 4–8 b. 3,3-dimethylpentane [1]

5 carbon alkane

methyl groups on C3

[2]

[3] CH3

CH3 C C C C C

C

C

CH3CH2 C CH2CH3

C C C

CH3

CH3

c. 3,5,5-trimethyloctane [1] 8 carbon alkane

[2]

C C C C C C C C

C C C C C C C C

[3]

methyl groups on C3 and C5 CH3

CH3

CH3

CH3

CH3CH2 CH CH2 C

CH3

CH2CH2CH3

CH3

d. 3-ethyl-4-methylhexane [1]

[2]

[3]

ethyl group on C3

6 carbon alkane

CH2CH3 CH3CH2 CH CH CH2CH3 CH3

CH2CH3 C

C C C C C

C C C C C C CH3

methyl group on C4

e. 3-ethyl-5-isobutylnonane [1]

[2]

[3]

isobutyl group on C5

9 carbon alkane C C C C C C C C C

CH3

CH3

CH2 CH CH3

CH2 CH CH3

C C C C C C C C C

CH3CH2 CH CH2 CH CH2CH2CH2CH3 CH2CH3

CH2CH3

ethyl group on C3

4.13 Use the steps in Answer 4.10 to name each alkane. [1]

[3] hexane

[2]

H H H H H H

H C C C C C C H

no substituents, skip [2]

H H H H H H

6 carbons = hexane

2-methyl [1]

[2]

H H H CH3 H

H C C C C H H H H

C H H

H H H CH3 H

H C C C C

5

H H H H

1

[3] 2-methylpentane

1

[3] 3-methylpentane

C H H

5 carbons = pentane 3-methyl [1]

H H CH3 H H

H C C C H H H

[2] H H CH3 H H

C C H H H

5 carbons = pentane

H C C C

5

H H H

C C H H H

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Alkanes 4–9 2,2-dimethyl [1]

[2]

H H CH3 H

H C C C

C

H

H H CH3 H

4

1

H H CH3 H H C C C

C

[3] 2,2-dimethylbutane

H

H H CH3 H

4 carbons = butane

[1]

H H

H

H

H C C

C

C H

H CH3 CH3 H

[2]

4

1

H H

H

H

H C C

C

C H

[3] 2,3-dimethylbutane

H CH3 CH3 H

4 carbons = butane 2,3-dimethyl

4.14 Follow these steps to name a cycloalkane: [1] Name the parent cycloalkane by counting the C’s in the ring and adding cyclo-. [2] Numbering: [2a] Number around the ring beginning at a substituent and giving the second substituent the lower number. [2b] Number to assign the lower number to the substituents alphabetically. [2c] Name and number all substituents, giving like substituents a prefix (di, tri, etc.). [3] Combine all parts, alphabetizing the substituents, ignoring all prefixes except iso. (Remember: If a carbon chain has more C’s than the ring, the chain is the parent, and the ring is a substituent.) 1 2 CH3 [2] 3 C C C CH 3

[1] a.

C

4

6 carbons in ring = cyclohexane

C

1,1-dimethyl [3] 1,1-dimethylcyclohexane

C 6

5 Number so the substituents are at C1.

1,2,3-trimethyl [1]

[2]

3 4C

b.

CH3 C 2 C CH3

[3] 1,2,3-trimethylcyclopentane

5C C CH3

5 carbons in ring = cyclopentane

1 Number so the first substituent is at C1, second at C2. 1 2

3 C C [2] C

[1] c.

C

CH3 4 C

5

6 carbons in ring = cyclohexane

C

[3] 1-butyl-4-methylcyclohexane 6

1-butyl 4-methyl Number so the earlier alphabetical substituent is at C1, butyl before methyl.

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Chapter 4–10 1

[1]

[2]

[3] 1-sec-butyl-2-isopropylcyclohexane

C

5C

d.

4C

C C

C

3

6 carbons in ring = cyclohexane

1-sec-butyl

6

2

2-isopropyl

Number so the earlier alphabetical substituent is at C1, butyl before isopropyl.

[1]

[2]

e.

1

3

5

C

C

C

C

C

[3] 1-cyclopropylpentane

4 2 1-cyclopropyl

longest chain = 5 carbons = pentane

Number so the cyclopropyl is at C1. 5

[1]

6

4 3

[2]

[3] 3-butyl-1,1-dimethylcyclohexane

2

f. 3-butyl

6 carbons in ring = cyclohexane

1,1-dimethyl

1

Number so the two methyls are at C1.

4.15 To draw the structures, use the steps in Answer 4.12. a. 1,2-dimethylcyclobutane [1] 4 carbon cycloalkane

[2]

CH3 1 C C 4

methyl groups on C1 and C2

C C C C

[3] CH3

C C 3 CH3 2

CH3

b. 1,1,2-trimethylcyclopropane [1] 3 carbon cycloalkane

c. 4-ethyl-1,2-dimethylcyclohexane [1] 6 carbon cycloalkane [2] C C

C

[1] 5 carbon cycloalkane C

[3]

CH3

C

C C

3

CH3CH2 4 C 2 CH3 C C

CH3

[2]

4C

CH

CH3

CH3

[3] CH CH 3 2

C C ethyl 5 C 1 CH3 6 on C4 2 CH3's

C

d. 1-sec-butyl-3-isopropylcyclopentane C

3 CH3's

3

C C

C

2C

C C CH3 1 CH 3

C

C

CH3

[2]

CH3

C3

isopropyl

[3]

C 2 C C 5 1 CH CH3

CH3 CH2

sec-butyl

CH3 CH3

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Alkanes 4–11

e. 1,1,2,3,4-pentamethylcycloheptane [1] 7 carbon cycloalkane

[2]

5 CH3's

CH3

[3]

CH3

CH3

CH3 3 C 4 C C5

C C C C C C C

2 CH3 C 1 C 6 C C7 CH3 CH3

CH3 CH3

CH3

4.16 To name the cycloalkanes, use the steps from Answer 4.14. [1]

5 carbons in ring = cyclopentane

[1]

[2]

C C

[3] methylcyclobutane

C C CH3

CH3

4 carbons in ring = cyclobutane

[1]

methyl

[2] CH3

CH3

CH3

3 carbons in ring = cyclopropane [1]

[2]

CH3

3 carbons in ring = cyclopropane

CH3

C C C

[3] ethylcyclopropane CH2CH3

ethyl

3 carbons in ring = cyclopropane CH3

[3] 1,2-dimethylcyclopropane

1,2-dimethyl

CH2CH3

[1]

C C C

[2]

CH3

[3] 1,1-dimethylcyclopropane

CH3

1,1-dimethyl

4.17 Compare the number of C’s and surface area to determine relative boiling points. Rules: [1] Increasing number of C’s = increasing boiling point. [2] Increasing surface area = increasing boiling point (branching decreases surface area). CH3(CH2)6CH3

8 C's linear largest number of C's no branching highest bp

CH3(CH2)5CH3

7 C's linear

CH3CH2CH2CH2CH(CH3)2

(CH3)3CCH(CH3)2

7 C's one branch 7 C's three branches increasing branching decreasing surface area decreasing bp

Increasing boiling point: (CH3)3CCH(CH3)2 < CH3CH2CH2CH2CH(CH3)2 < CH3(CH2)5CH3 < CH3(CH2)6CH3

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Chapter 4–12 4.18 To draw a Newman projection, visualize the carbons as one in front and one in back of each other. The C
C bond is not drawn. There is only one staggered and one eclipsed conformation. Br

rotation here H H H C C Br

C in front

H

H

H H

60o

H H

H Br

H

H H

H H

1 staggered

2 eclipsed

C behind

4.19 Staggered conformations are more stable than eclipsed conformations. H

H

H CH3 HH H H

HH

CH3 H

H

HH

H

H CH3 H H

H H

CH3 H

eclipsed energy maximum

H H H C C CH3

Energy

rotation here

H

H

CH3 CH3

H

H

H H H 60o

0o

CH3

H

H

H H

120o

H

H H

H H

H 180o

240o

300o

staggered energy minimum

360o = 0o

Dihedral angle

4.20 4.0 kJ/mol H

H

H CH3

To calculate H,CH3 destabilization: H,H eclipsing 4.0 kJ/mol of destabilization

H H

4.0 kJ/mol

14 kJ/mol (total)
8.0 kJ/mol for 2 H,H eclipsing interactions = 6 kJ/mol for one H,CH3 eclipsing interaction

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Alkanes 4–13

To determine the energy of conformations keep two things in mind: [1] Staggered conformations are more stable than eclipsed conformations. [2] Minimize steric interactions: keep large groups away from each other. The highest energy conformation is the eclipsed conformation in which the two largest groups are eclipsed. The lowest energy conformation is the staggered conformation in which the two largest groups are anti.

4.21

rotation here H H

CH3

CH3

CH3 C CH2CH3 CH3

CH3

60o

H

CH3

H H

H

1 staggered most stable

H

CH3 H

CH3 CH3

H

2 eclipsed

3 staggered most stable 60o

CH3 CH3

CH3

60o

H H

60o

CH3 CH3

CH3 H

H

H

6 eclipsed least stable

H

CH3

CH3

60o H CH3

CH3

CH3

60o

H

H

H

5 staggered

4 eclipsed least stable

4.22 To determine the most and least stable conformations, use the rules from Answer 4.21. Cl

1,2-dichloroethane

H

H

ClCH2 CH2Cl

60o

H

H

H

H

H

H Cl

Cl

H Cl

1 staggered, anti

2 eclipsed

rotation here

60o

H

H

Cl

H Cl

3 staggered, gauche 60o

60o H

Cl H

H H

H Cl

6 eclipsed

60o

H

H

H

Cl

60o

H

H

H H

Cl

Cl Cl

5 staggered, gauche

4 eclipsed

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Chapter 4–14 highest energy Cl groups eclipsed least stable 4

3

5

1 most stable

180o

Eclipsed forms are higher in energy.

6

Energy

2

Staggered forms are lower in energy.

1 most stable Cl groups anti

60o 120o 180o 0o 120o 60o Dihedral angle between 2 Cl's

4.23 Add the energy increase for each eclipsing interaction to determine the destabilization. HH

H CH

CH3 H

3

HCH3

4.24

H CH

3

b.

a.

H

1 H,H interaction = 2 H,CH3 interactions (2 x 6.0 kJ/mol) =

12.0 kJ/mol

Total destabilization =

16 kJ/mol

4.0 kJ/mol

CH3

3 H,CH3 interactions (3 x 6.0 kJ/mol) = 18 kJ/mol Total destabilization

Two points: • Axial bonds point up or down, while equatorial bonds point out. • An up carbon has an axial up bond, and a down carbon has an axial down bond. equatorial axial up

CH3 Br H H

HO H

equatorial

H

H Cl H OH

equatorial

CH3

axial up H

axial down

Br

HO Cl

OH

Up carbons are dark circles. Down carbons are clear circles.

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89

Alkanes 4–15

4.25

Draw the second chair conformation by flipping the ring. • The up carbons become down carbons, and the axial bonds become equatorial bonds. • Axial bonds become equatorial, but up bonds stay up; i.e., an axial up bond becomes an equatorial up bond. • The conformation with larger groups equatorial is the more stable conformation and is present in higher concentration at equilibrium. axial H eq Br

a.

more stable Br is equatorial.

Draw in the H

Draw second conformation.

Br

and label the C as up or down.

Axial bond is up = up carbon

eq

H

Up carbons switch to down carbons.

Br

axial

Axial bond is down = down carbon

axial eq Cl

Cl

Draw in the H

b.

and label the C as up or down.

Draw second conformation. Up carbons switch to down carbons.

H

Cl

axial

eq Axial bond is up = up carbon

H

Axial bond is down = down carbon

more stable Cl is equatorial.

Axial bond is up = up carbon

eq

H

c.

CH2CH3

more stable CH2CH3 is equatorial.

Draw second conformation.

Draw in the H

CH2CH3

and label the C as up or down.

eq

Up carbons switch to down carbons.

H CH2CH3

Axial bond is down = down carbon

4.26 Larger axial substituents create unfavorable diaxial interactions, whereas equatorial groups have more room and are favored. H H

H H

C C H

CH2CH3

H H

larger substituent more important to be equatorial

equatorial CH2CH3

H

C H

H

H

H C C H

more compact substituent less important to be equatorial

H

C C

The axial conformation containing the C CH group is not as unstable as the axial conformation containing the CH2CH3, so it is present in higher concentration at equilibrium.

The H's and CH3 of the sp3 hybridized C have severe 1,3-diaxial interactions with the two other axial H's.

H

C CH

equatorial C CH

CH3

H

H

The sp hybridized C's are linear and point down. The 1,3-diaxial interactions with the two other axial H's are less severe.

90

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Chapter 4–16 4.27 Wedges represent “up” groups in front of the page, and dashes are “down” groups in back of the page. Cis groups are on the same side of the ring, and trans groups are on opposite sides of the ring. a.

cis-1,2-dimethylcyclopropane CH3

trans-1-ethyl-2-methylcyclopentane

b.

or

CH3

CH3

or

CH3

CH3CH2

cis = same side of the ring both groups on wedges or both on dashes

CH3

CH3CH2

CH3

trans = opposite sides of the ring one group on a wedge, one group on a dash

4.28 Cis and trans isomers are stereoisomers. cis-1,3-diethylcyclobutane

a. trans-1,3-diethylcyclobutane

cis = same side of the ring both groups on wedges or both on dashes

b. cis-1,2-diethylcyclobutane

constitutional isomer different arrangement of atoms

trans = opposite sides of the ring one group on a wedge, one group on a dash

4.29 To classify a compound as a cis or trans isomer, classify each non-hydrogen group as up or down. Groups on the same side = cis isomer, groups on opposite sides = trans isomer. down bond (up) (equatorial) HH(up) HO

a.

down bond (up) (equatorial)

H

H

(up) Br

Cl

H

HO

up bond (axial)

OH OH

down bond (equatorial) both groups down = cis isomer

Cl Cl

b.

up bond H (down) (equatorial)

H H

c. H (down)

H

H Br

Br

down bond H (equatorial) one group up, one down = trans isomer

Cl

one group up, one down = trans isomer

Br

4.30 CH3

a.

trans: CH3

CH3

c. CH3

groups on same side cis isomer CH3

cis:

H CH3

b. H

CH3

groups on opposite sides trans isomer (one possibility) CH3 CH3 H

H

two chair conformations for the cis isomer Same stability since they both have one equatorial, one axial CH3 group.

H CH3

H

H CH3 H

CH3

both groups equatorial more stable two chair conformations for the trans isomer d. The trans isomer is more stable because it can have both methyl groups in the more roomy equatorial position.

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Alkanes 4–17

4.31 CH2CH3

up

axial CH2CH3

CH3CH2 H

CH3

a.

H

down (equatorial)

CH3

c.

1,1-disubstituted

trans-1,3-disubstituted CH3

up (axial) H

H CH2CH3

b.

up

CH3

d.

CH3CH2 up (equatorial)

H

cis-1,2-disubstituted

down

H

trans-1,4-disubstituted

4.32 Oxidation results in an increase in the number of C
Z bonds, or a decrease in the number of C
H bonds. Reduction results in a decrease in the number of C
Z bonds, or an increase in the number of C
H bonds. a.

CH3

O

O

C

C

H

CH3

O

c.

OH

Decrease in the number of C–H bonds. Increase in the number of C–O bonds. Oxidation

CH3

C

CH3

HO OH C CH3 CH3

No change in the number of C–O or C–H bonds. Neither

O

b.

CH3

C

CH3CH2CH3

CH3

d.

Decrease in the number of C–O bonds. Increase in the number of C–H bonds. Reduction

O

OH

Decrease in the number of C–O bonds. Increase in the number of C–H bonds. Reduction

4.33 The products of a combustion reaction of a hydrocarbon are always the same: CO2 and H2O. flame

a.

CH3CH2CH3

b.

+

+

5 O2

9 O2

flame

3 CO2 +

6 CO2 +

4 H2O + heat

6 H2O + heat

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Chapter 4–18 4.34 Lipids contain many nonpolar C–C and C–H bonds and few polar functional groups. NH2 HOOC

a. CH3(CH2)7CH=CH(CH2)7COOH

b.

oleic acid

H N

O CH3O

O

aspartame

only one polar functional group 18 carbons a lipid

many polar functional groups only 14 carbons not a lipid

4.35 “Like dissolves like.” Beeswax is a lipid, and therefore, it will be more soluble in nonpolar solvents. H2O is very polar, ethanol is slightly less polar, and chloroform is least polar. Beeswax is most soluble in the least polar solvent. Increasing polarity H2O

CH3CH2OH

CHCl3

Increasing solubility of beeswax

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Alkanes 4–19 4.36 Use the rules from Answer 4.3. 1°

4°

3°

CH3 CH3

2°

a.

[1]

2°

[2] 4°

[2]

CH3

CH3

3°

1°

CH2

CH2

C b. [1] CH3 CH

C

CH2

CH3 CH2

CH

All CH3's have 1° H's. All CH2's have 2° H's. All CH's have 3° H's.

2° 1°

CH3

CH2

CH2 CH

CH3

CH3

4.37 One possibility: CH3

a. CH3 C CH3

b.

d. (CH3)3CH

c. CH3CH2CH3

CH3

4.38 Use the rules from Answer 4.3. 1° O

O

O

O

2° O

4°

4°

O

OH OH C(CH3)3

2° 2°

2° 3°

4° 1°

4.39 a. Five constitutional isomers of molecular formula C4H8: CH2 CH3CH CHCH3

CH2 CHCH2CH3

CH3

CH3

C

CH3

b. Nine constitutional isomers of molecular formula C7H16: H

H

CH3 C CH2CH2CH2CH3

CH3CH2CH2CH2CH2CH2CH3

CH3CH2 C CH2CH2CH3

CH3 CH3

CH3

CH3 C CH2CH2CH3

CH3CH2 C CH2CH3

CH3

CH3 H

CH3CH2 C CH2CH3 CH2CH3

H CH3 C

CH3 C CH3

CH3 CH3

CH3 H CH3 C

H C CH2CH3

CH3 CH3

H CH3 C

H CH2 C CH3

CH3

CH3

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Chapter 4–20 c. Twelve constitutional isomers of molecular formula C6H12 containing one ring:

4.40 Use the steps in Answers 4.10 and 4.14 to name the alkanes. a. CH3CH2CHCH2CHCH2CH2CH3

[1]

CH2CH3

CH3

1 2 3 4 5 6 7 8 [2] CH3CH2CHCH2CHCH2CH2CH3 CH3

8 carbons = octane

CH2CH3

3-methyl

7-methyl

7

b. CH3CH2CCH2CH2CHCHCH2CH2CH3

[1]

5-ethyl

CH3

CH2CH3

CH3

2 3 CH2CH3

[2] 1

[3] 5-ethyl-3-methyloctane

[3] 3,3,6-triethyl-7-methyldecane

CH3CH2CCH2CH2CHCHCH2CH2CH3

CH2CH3 CH2CH3

CH2CH3 CH2CH3

10 carbons = decane

3,3,6-triethyl

c. CH3CH2CH2C(CH3)2C(CH3)2CH2CH3 [1]

[2]

re-draw CH3 CH3

CH3CH2CH2 C

[3] 3,3,4,4-tetramethylheptane

CH3 CH3 3

CH3CH2CH2 C

C CH2CH3

CH3 CH32

4

CH3 CH3

C CH2CH3

1

3,3,4,4-tetramethyl

7 carbons = heptane d. CH3CH2C(CH2CH3)2CH(CH3)CH(CH2CH2CH3)2

3,3-diethyl

re-draw [1]

CH3CH2 H

CH3CH2 C

C

4

[2]

CH3CH2 H

H

CH3CH2 C

C CH2CH2CH3

1

CH3CH2 CH3 CH2CH2CH3

CH3CH2 H CH3CH2 C

C CH2CH2CH3

CH3CH2 CH3

7 carbons = heptane

C CH2CH2CH3

4-methyl

5-propyl

3,3-diethyl

e. (CH3CH2)3CCH(CH3)CH2CH2CH3 re-draw

[3] 3,3-diethyl-4-methyl-5-propyloctane

CH3CH2 CH3 CH2CH2CH3

8 carbons = octane

[1]

C

5 H

CH3CH2 H

[2]

CH3CH2 C

1

4

[3] 3,3-diethyl-4-methylheptane

C CH2CH2CH3

CH3CH2 CH3

6

4-methyl

7

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Alkanes 4–21

f. CH3CH2CH(CH3)CH(CH3)CH(CH2CH2CH3)(CH2)3CH3 3 4 5 re-draw H H H [1] [2] CH3CH2 C

H

H

H

CH3CH2 C

C

C CH2CH2CH2CH3

1 2

C

CH3 CH3 CH2CH2CH3

3,4-dimethyl

CH3 CH3 CH2CH2CH3

[3] 3,4-dimethyl-5-propylnonane

C CH2CH2CH2CH3

5-propyl

9 carbons = nonane

g. (CH3CH2CH2)4C re-draw [1]

4

4,4-dipropyl

[2]

CH2CH2CH3

CH2CH2CH3

CH3CH2CH2 C CH2CH2CH3

[3] 4,4-dipropylheptane

CH3CH2CH2 C CH2CH2CH3

CH2CH2CH3

1

2

3

CH2CH2CH3

7 carbons = heptane

1

h. [1]

[2] 5

3

7 6-isopropyl

3-methyl

10 carbons = decane i.

[3] 6-isopropyl-3-methyldecane

6

[2]

[1]

10 carbons = decane

8

6

4-isopropyl

4

[3] 8-ethyl-4-isopropyl-2,6-dimethyldecane

1

8-ethyl

j.

2,6-dimethyl 4-isopropyl 4 [3]

[2]

[1]

4-isopropyloctane

1 8 carbons = octane

2,2,5-trimethyl k.

2,2,5-trimethylheptane 1 2

1 2 CH(CH2CH3)2

l.

=

3 3-cyclobutylpentane

4 5 3-cyclobutyl

5 m. 4 3

1 2

1-sec-butyl

2-isopropyl

1-sec-butyl-2-isopropylcyclopentane

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Chapter 4–22 5 6 1

n.

4

1-isobutyl-3-isopropylcyclohexane

3 2

1-isobutyl

3-isopropyl

4.41 2,2-dimethyl

CH3

3,3-dimethyl

CH3

4,4-dimethyl

CH3

CH3 C CH2CH2CH2CH2CH3

CH3CH2 C CH2CH2CH2CH3

CH3CH2CH2 C CH2CH2CH3

1

1

1

CH3

H

CH3

3

2 2,2-dimethylheptane

3,3-dimethylheptane

H

CH3 C

1

1

2

CH3

4

CH3

H

CH3 C CH2CH2CH2 C CH3

CH3CH2 C

1

1

2

6 CH3

3

2,6-dimethyl 2,6-dimethylheptane

1

3

H

C CH2CH2CH2CH3

CH3CH3

2,3-dimethyl 2,3-dimethylheptane

2,5-dimethylheptane

H

CH3

5

H

CH3 C

CH3

2,5-dimethyl

2,4-dimethyl 2,4-dimethylheptane H

CH2CH2 C CH2CH3

CH3

2

2

H

H

CH3 C CH2 C CH2CH2CH3

CH3

4 4,4-dimethylheptane

H

H

H

CH3CH2 C CH2 C CH2CH3

C CH2CH2CH3

1

CH3 CH3

4 3,4-dimethyl

3

CH3

CH3

5 3,5-dimethyl

3,5-dimethylheptane

3,4-dimethylheptane

4.42 Use the steps in Answer 4.12 to draw the structures. a. 3-ethyl-2-methylhexane [1]

6 C chain

C C

[2]

C C C C C C

C C C C

H

CH3 CH2CH3

methyl on C2

[3] CH3 C

H C CH2CH2CH3

CH3 CH2CH3

ethyl on C3

b. sec-butylcyclopentane [1]

5 C ring

[2]

isopropyl on C4 c. 4-isopropyl-2,4,5-trimethylheptane [1]

7 C chain

[2]

CH3

CH

C C C C CH3

C C C C C C C

CH3 C C C

CH3 CH3

methyls on C2, C4, and C5 d. cyclobutylcycloheptane [1] 7 C cycloalkane

[2]

CH3

[3]

CH3 CH

CH3 CH CH2 C CH3

CHCH2CH3

CH3 CH3

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Alkanes 4–23

e. 3-ethyl-1,1-dimethylcyclohexane [1] 6 C cycloalkane

[2]

[3]

CH3CH2 C

C C

C

C

ethyl on C3 C

C

2 ethyl groups

C

C

C

C

C

C C

g. 6-isopropyl-2,3-dimethylnonane [1] 9 C alkane

C

[2]

C

C C C

C

isopropyl

methyl

CH3

CH3 C

C

C

C

C

C

C

C

C

8 C alkane

C

C

C

C

C

C

C

C

C

C

CH3

[2] CH3 CH3 C

C

C

C

C

[1]

C

C

CH3 CH3 C C C C

CH2CH3

[3]

ethyl on C1

[2]

CH2CH3

or

j. trans-1-tert-butyl-4-ethylcyclohexane 6 C ring

C

C

methyl

CH3

[1]

[3]

CH3

CH3

i. cis-1-ethyl-3-methylcyclopentane 5 C ring

C

CH

5 methyl groups

h. 2,2,6,6,7-pentamethyloctane [1]

C

[3]

C C

C C

2 methyl groups on C1

CH3

C

C C C

C

CH3

C

C C

[2]

8 C cycloalkane

C

C

C

f. 4-butyl-1,1-diethylcyclooctane [1]

C

methyl on C3 C(CH3)3

[2] CH3CH2

CH3

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Chapter 4–24 4.43 Draw the compounds. a. 2,2-dimethyl-4-ethylheptane

CH3

e. 1-ethyl-2,6-dimethylcycloheptane

CH2CH3

alphabetized incorrectly ethyl before methyl

CH3

CH2CH3

2

CH3

CH3 C CH2 C CH2CH2CH3

Numbered incorrectly. Re-number so methyls are at C1 and C4.

H

1

4

CH3

4-ethyl-2,2-dimethylheptane 2-ethyl-1,4-dimethylcycloheptane b. 5-ethyl-2-methylhexane

f. 5,5,6-trimethyloctane

Longest chain was not chosen = heptane

3

Numbered incorrectly. Re-number so methyls are at C3 and C4.

4

2,5-dimethylheptane 3,4,4-trimethyloctane g. 3-butyl-2,2-dimethylhexane

H

c. 2-methyl-2-isopropylheptane

CH3C CH3

Longest chain was not chosen = octane

CH3 C CH2CH2CH2CH2CH3 CH3

2,3,3-trimethyloctane d. 1,5-dimethylcyclohexane

1

4-tert-butyloctane

CH3

1 Numbered incorrectly. Re-number so methyls are at C1 and C3.

4

Longest chain not chosen = octane

h. 1,3-dimethylbutane

H

Longest chain not chosen = pentane

3 CH3

CH3 CCH2 CH3

1,3-dimethylcyclohexane

CH2 CH3

2-methylpentane

4.44 CH3

a.

H

CH3 CH2CH2CH3

CH3

H CH2CH2CH3

b.

CH3 H

H CH2CH3

re-draw 4 4-isopropylheptane

CH3 CH2CH3

re-draw

3 3-ethyl-3-methylpentane

c.

CH3CH2

CH2CH2CH3

CH3CH2CH2

H CH2CH3

re-draw

4

5

4,4-diethyl-5-methyloctane

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Alkanes 4–25

4.45 Use the rules from Answer 4.17. a.

CH3CH2CH3 CH3CH2CH2CH3 CH3CH2CH2CH2CH3 4 C's 3C's 5 C's lowest boiling point highest boiling point

b. (CH3)2CHCH(CH3)2

CH3CH2CH2CH(CH3)2

most branching lowest boiling point

CH3(CH2)4CH3 least branching highest boiling point

4.46 a.

(CH3)3CC(CH3)3 branching = lower surface area lower boiling point more spherical, better packing = higher melting point

CH3(CH2)6CH3 no branching = higher surface area higher boiling point

b. There is a 159° difference in the melting points, but only a 20° difference in the boiling points because the symmetry in (CH3)3CC(CH3)3 allows it to pack more tightly in the solid, thus requiring more energy to melt. In contrast, once the compounds are in the liquid state, symmetry is no longer a factor, the compounds are isomeric alkanes, and the boiling points are closer together. 4.47 CH3

a.

CH3

H

or H

H

H

H CH3

H

1 gauche CH3,CH3 = 3.8 kJ/mol of destabilization

CH3

higher energy 2 gauche CH3,CH3 3.8 kJ/mol x 2 = 7.6 kJ/mol of destabilization

Energy difference = 7.6 kJ/mol – 3.8 kJ/mol =

CH3

H

CH3

b.

H

H

or

CH3 H

CH3

CH3

H CH3

CH3

2 gauche CH3,CH3 3.8 kJ/mol x 2 = 7.6 kJ/mol of destabilization

higher energy 3 eclipsed H,CH3 6 kJ/mol x 3 = 18 kJ/mol of destabilization

Energy difference = 3.8 kJ/mol

H CH3

18 kJ/mol – 7.6 kJ/mol = 10.4 kJ/mol

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Chapter 4–26 4.48 Use the rules from Answer 4.21 to determine the most and least stable conformations. a. CH3 CH2CH2CH2CH3

b. CH3CH2CH2 CH2CH2CH3 H CH CH CH 2 2 3

H H

CH2CH2CH3

H

H

HH

H

H

H H

eclipsed least stable

staggered most stable

CH3CH2 CH2CH3

CH2CH3 H

H

H CH2CH3

staggered ethyl groups anti most stable

All staggered conformations are equal in energy. All eclipsed conformations are equal in energy.

H H

HH

eclipsed ethyl groups eclipsed least stable

4.49

b.

a.

c.

Br Cl H

Cl

Cl

H

H

H

H

Cl

H

H

H CH3

H

H CH3

H

4.50 CH3 a.

C

H

H

H

C

H Br

C

b.

H

c.

H Br

Cl H

H Cl

CH3 H

C

Br Cl

CH3

Cl H

Cl

H

H Br Cl

H Br

Br CH3

H

CH2CH3 Cl Cl CH2CH3

4.51 CH3CH2 H H 60°

[1]

H

CH3CH2 CH2CH2CH3

H CH2CH3

H H

H CH3

H 60°

H H

CH3 H

1 60° H H

H

CH3 H

6

CH2CH3 CH3

3

2

CH3CH2

H

60° 60°

H

H H

60°

H H

H H

H

CH3CH2 CH3

5

CH3 CH2CH3

4

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Alkanes 4–27

least stable 4

Energy

2

Eclipsed forms are higher in energy.

6

Staggered forms are lower in energy.

5

3 1 most stable

1 most stable

60o

120o

180o

60o

0o

120o

180o

Dihedral angle between two alkyl groups

most stable CH3CH2 H

[2] CH3CH2 CHCH2CH3

H

H

CH3

H CH2CH3

H H

60°

H H

H

60° CH3

CH3

CH3 CH3

CH3

1

CH2CH3 CH3

3

2 60°

CH3CH2

60°

H CH3

H

H CH3 60°

CH3 H

60°

H

CH3CH2

CH3 H

H

CH3 CH2CH3

CH3

5

6

H H

least stable 4

least stable 4

Energy

2

3

120o

Staggered forms are lower in energy.

5

1 most stable

180o

Eclipsed forms are higher in energy.

6

1 most stable

60o

0o

60o

120o

180o

Dihedral angle (between CH3CH2 in back and CH3 in front)

101

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Chapter 4–28 4.52 Two types of strain: 4.1 Torsional strain is due to eclipsed groups on adjacent carbon atoms. 4.2 Steric strain is due to overlapping electron clouds of large groups (ex: gauche interactions). H

CH3 H

CH3

H

a.

H H

H CH3

b. CH3

H

c. H CH3

CH3

two sites three bulky methyl groups close = steric strain

HH

CH2CH3 CH3CH2

two bulky ethyl groups close = steric strain eclipsed conformation = torsional strain

eclipsed conformation = torsional strain

4.53 The barrier to rotation is equal to the difference in energy between the highest energy eclipsed and lowest energy staggered conformations of the molecule. a. CH3 CH(CH3)2 CH3

CH3

H

CH3

CH3 H

H

H

b. CH3 C(CH3)3

H

most stable

H CH3

H

H

CH3

CH3

H H

least stable Destabilization energy =

CH3 H

H CH3

H

CH3 H

most stable

least stable Destabilization energy =

2 H,CH3 eclipsing interactions 2(6.0 kJ/mol) = 12.0 kJ/mol 1 H,H eclipsing interaction = 4.0 kJ/mol Total destabilization = 16.0 kJ/mol 16.0 kJ/mol = rotation barrier

3 H,CH3 eclipsing interactions 3(6.0 kJ/mol) = 18.0 kJ/mol Total destabilization = 18.0 kJ/mol 18.0 kJ/mol = rotation barrier

4.54 Cl H

H

H

H H

most stable

HH

H Cl

H H

least stable

2 H,H eclipsing interactions = 2(4.0 kJ/mol) = 8.0 kJ/mol Since the barrier to rotation is 15 kJ/mol, the difference between this value and the destabilization due to H,H eclipsing is the destabilization due to H,Cl eclipsing. 15.0 kJ/mol – 8.0 kJ/mol = 7.0 kJ/mol destabilization due to H,Cl eclipsing

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Alkanes 4–29

4.55 The gauche conformation can intramolecularly hydrogen bond, making it the more stable conformation. H OH

hydrogen bonding

O

HOCH2 CH2OH

H

H

H

O

rotation here

H

H

H

H

OH

H

anti

gauche

H

Hydrogen bonding can occur only in the gauche conformation, making it more stable.

4.56 axial H OH axial

a. [1]

c. HO

d.

down HO

HO

eq

eq HO H

H

eq

Br

a. [2]

H

H

H

eq

H

H

ax

ax OH

c. down

up

CH eq Br 3 H eq

H eq CH3eq

CH3

d.

OH

HO

ax

CH3 up

H

b. OH eq

HO

OH ax H

ax

Br

d.

both up = cis

axial a.

eq HO

ax Br

c.

H

axial

H eq

eq

OH

Br

b.

H eq CH3 eq

H ax OH H

one up, one down = trans up

axial

[3]

ax

H up OH

b.

HO

ax

ax

H

OH OH eq

eq HO

eq H

H eq

H

H

one up, one down = trans

axial

H

ax

4.57 ax

ax H

H CH3 CH3

eq

eq

both groups equatorial more stable

eq H

H eq CH3

CH3

ax

ax

OH

ax

104

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Chapter 4–30 4.58 Axial/equatorial substituent location Disubstituted cyclohexane a. 1,2-cis disubstituted b. 1,2-trans disubstituted c. 1,3-cis disubstituted d. 1,3-trans disubstituted e. 1,4-cis disubstituted f. 1,4-trans disubstituted

Conformation 1 Axial/equatorial Axial/axial Axial/axial Axial/equatorial Axial/equatorial Axial/axial

Conformation 2 Equatorial/axial Equatorial/equatorial Equatorial/equatorial Equatorial/axial Equatorial/axial Equatorial/equatorial

4.59 A cis isomer has two groups on the same side of the ring. The two groups can be drawn both up or both down. Only one possibility is drawn. A trans isomer has one group on one side of the ring and one group on the other side. Either group can be drawn on either side. Only one possibility is drawn. [1]

a.

cis

[2]

[3]

a.

a.

trans

cis

trans

cis

b. cis isomer

b. cis isomer

trans

b. cis isomer ax

ax

ax

ax ax

ax eq

eq

eq

both groups equatorial more stable c. trans isomer ax

eq

eq

eq

larger group equatorial more stable c. trans isomer

larger group equatorial more stable c. trans isomer

ax

ax eq

eq eq

eq

eq

ax

eq ax

larger group equatorial more stable d.

ax both groups equatorial more stable d.

d. The cis isomer is more stable than the trans since one conformation has both groups equatorial.

both groups equatorial more stable

The trans isomer is more stable than the cis since one conformation has both groups equatorial.

The trans isomer is more stable than the cis since one conformation has both groups equatorial.

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Alkanes 4–31

4.60 Compare the isomers by drawing them in chair conformations. Equatorial substituents are more stable. See the definitions in Problem 4.59. a. cis

CH2CH3 ax

ax

CH2CH3 ax

CH2CH3

H CH2CH3 eq

trans H

CH2CH3 eq H

H

CH3CH2 H eq

CH2CH3

H

ax

b. 1-ethyl-3-isopropylcyclohexane

ax (CH3)2CH

cis (CH3)2CH

H

CH2CH3

eq

H

both groups equatorial most stable of all conformations trans isomer

ax

H

trans

CH2CH3

H

H

CH2CH3

H

(CH3)2CH

H

(CH3)2CH

eq

ax

both groups equatorial most stable of all conformations cis isomer

The cis isomer is more stable than the trans isomer because its more stable conformation has two groups equatorial.

4.61 Only the more stable conformation is drawn. CH3

CH3

CH3

CH3

or CH3

CH3

re-draw to see axial and equatorial CH3 CH3

CH3

CH3

CH3 CH3

more stable substituents on C1, C3, C5 = all equatorial

4.62

OH

re-draw to see axial and equatorial

OH CH3

CH3 HO

all equatorial menthol

HO

eq CH2CH3

The trans isomer is more stable than the cis isomer because its more stable conformation has two groups equatorial.

eq

H

H

eq

ax CH2CH3 H

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Chapter 4–32 4.63 a. HO

OH O OH

HO

HO

or HO

HO

HO

O

OH

O

b. HO

HO

or

HO

O OH

OH HO

most stable All groups are equatorial.

HO

OH HO

OH

OH

4.64 OH OH OH

a.

O

OH

HO

OH

OH

OH O

HO

c.

O

OH

OH OH

more stable More groups are equatorial.

O

HO

OH

HO

HO

constitutional isomer OH OH

OH O OH

OH OH

b.

OH

HO

HO HO

OH

galactose

O

d. OH

glucose All groups are equatorial. more stable

OH

OH

stereoisomer

4.65 CH2CH3

a.

and

d.

same molecular formula C4H8 different connectivity constitutional isomers

same molecular formula C10H20 different connectivity constitutional isomers

CH3 H

and H

CH3

CH3

H

H

=

CH2CH3

1 down, 1 up = 1 down, 1 up = trans trans same arrangement in three dimensions identical

and

f. H

CH3

molecular formula: C6H12

different molecular formulas not isomers CH3

CH3

CH3

molecular formula: C6H10

different arrangement in three dimensions stereoisomers

c.

and

e.

H

CH2CH3

CH2CH3

and

b.

CH2CH3

and

CH3CH2

CH2CH3

CH3CH2

1 down, 1 up = both down = cis trans different arrangement in three dimensions stereoisomers

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Alkanes 4–33 ax

ax

g. eq

eq

H

CH3CH2

CH2CH3

up

CH3

H

up

up both up = cis

h.

H

CH3

and

up

and

H

3,4-dimethylhexane

both up = cis

2,4-dimethylhexane

same molecular formula C8H18 different IUPAC names constitutional isomers same arrangement in three dimensions identical

4.66 CH3

a.

CH

H

CH3

CH3

CH3CH2

H

H

CH2CH3 CH3

CH3

H

CH3

CH3

b.

and H

H

H

and

CH3 CH2CH3

CH(CH3)2

re-draw

re-draw CH3

CH3 CH2CH3

CH3 CH CH

CH3

CH2

CH3 CH

CH

CH2CH3

CH2CH3

3-ethyl-2-methylpentane 3-ethyl-2-methylpentane same molecular formula same name identical molecules

same molecular formula different arrangement of atoms constitutional isomers

4.67 constitutional isomer

One possibility:

stereoisomer

a. trans

cis H

b.

H

H

H

OH OH

HO

cis

OH

H OH

c.

OH

trans

H Cl

cis Cl

Cl

Cl

Cl

Cl

trans

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Chapter 4–34 4.68 Three constitutional isomers of C7H14:

1,1-dimethylcyclopentane

1,2-dimethylcyclopentane

1,3-dimethylcyclopentane

or

or

trans

cis

cis

trans

4.69 Use the definitions from Answer 4.32 to classify the reactions. O

a.

=

CH3CHO

CH3

C

CH3CH2OH

H

d.

CH2 CH2

Decrease in the number of C–O bonds. Reduction

H C C H

Decrease in the number of C–H bonds. Oxidation CH3

b.

Increase in the number of C–Z bonds. Oxidation

Increase in the number of C–O bonds. Oxidation

c.

CH2 CH2

CH2Br

e.

O

HOCH2CH2OH

f.

CH3CH2OH

CH2 CH2

Loss of one C–O bond and one C–H bond. Neither

Two new C–O bonds. Oxidation

4.70 Use the rule from Answer 4.33. flame

a. CH3CH2CH2CH2CH(CH3)2

flame 7 CO2 + 8 H2O + heat

4 CO2 + 5 H2O + heat

b.

11 O2

(13/2) O2

4.71 1 C–O bond

2 C–O bonds [1]

H O

a.

[2] 2 C–H bonds

H

benzene

an arene oxide [1] increase in C–O bonds oxidation reaction

b.

OH H

1 C–H bond

phenol [2] loss of 1 C–O bond, loss of 1 C–H bond neither

Phenol is more water soluble than benzene because it is polar (contains an O–H group) and can hydrogen bond with water, whereas benzene is nonpolar and cannot hydrogen bond.

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Alkanes 4–35

4.72 Lipids contain many nonpolar C–C and C–H bonds and few polar functional groups. OH O

HO

HO

OH

a. HO OH mevalonic acid

O HO

c.

d. HO estradiol

O

HO

many polar functional groups not a lipid

OH O

HO

few polar functional groups a lipid

OH

OH

sucrose

many polar functional groups not a lipid

b. squalene no polar functional groups a lipid

4.73 O OH

O

COH

OH

CNHCH2CH2SO3
Na+

This polar part of the molecule interacts with water.

HO

OH

HO

cholic acid a bile acid

OH

a bile salt This nonpolar part of the molecule can interact with lipids to create micelles that allow for transport of lipids through aqueous environments.

4.74 The mineral oil can prevent the body’s absorption of important fat-soluble vitamins. The vitamins dissolve in the mineral oil, and are thus not absorbed. Instead, they are expelled with the mineral oil. 4.75 The amide in the four-membered ring has 90° bond angles giving it angle strain, and therefore making it more reactive. amide H N

penicillin G

S

O

N

CH3 CH3

O COOH

strained amide more reactive

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Chapter 4–36 4.76 Cl H

Example: HH

I

HH

HH

H

Cl C

C

H

H

H

I C

H

Although I is a much bigger atom than Cl, the C–I bond is also much longer than the C–Cl bond. As a result the eclipsing interaction of the H and I atoms is not very much different in magnitude from the H,Cl eclipsing interaction.

HH

H

C

H

H

H

longer bond

H

H

4.77 H

H

H

decalin

H

trans-decalin

cis-decalin H

H H

H H

H

trans The trans isomer is more stable since the carbon groups at the ring junction are both in the favorable equatorial position.

1,3-diaxial interaction cis This bond is axial, creating unfavorable 1,3-diaxial interactions.

4.78 4

2 1

5

3

1

1

(1-methylbutyl)cyclopentane

3

4

(1,1-dimethylpropyl)cyclopentane

4 3 2 1

2

3

3 pentylcyclopentane

1 2

2

(2-methylbutyl)cyclopentane

(2,2-dimethylpropyl)cyclopentane

3 2

2 1 (1-ethylpropyl)cyclopentane

2 3

1 (1,2-dimethylpropyl)cyclopentane

1

3 4

(3-methylbutyl)cyclopentane

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Stereochemistry 5–1 C Chhaapptteerr 55:: SStteerreeoocchheem miissttrryy

IIssoom meerrss aarree ddiiffffeerreenntt ccoom mppoouunnddss w wiitthh tthhee ssaam mee m moolleeccuullaarr ffoorrm muullaa ((55..22,, 55..1111)).. [1] Constitutional isomers—isomers that differ in the way the atoms are connected to each other. They have: • different IUPAC names • the same or different functional groups • different physical and chemical properties. [2] Stereoisomers—isomers that differ only in the way atoms are oriented in space. They have the same functional group and the same IUPAC name except for prefixes such as cis, trans, R, and S. • Enantiomers—stereoisomers that are nonsuperimposable mirror images of each other (5.4). • Diastereomers—stereoisomers that are not mirror images of each other (5.7). CH3

CH2CH3

C

CH3CH2

C

H Br

CH3

C H Br

H Br

A

CH3

C

CH2CH3

C H Br

Br

C H Br

C

enantiomers

CH3

C H Br

H

B

CH3CH2 C

H

Br

D enantiomers

A and B are diastereomers of C and D.



A Assssiiggnniinngg pprriioorriittyy ((55..66)) • Assign priorities (1, 2, 3, or 4) to the atoms directly bonded to the stereogenic center in order of decreasing atomic number. The atom of highest atomic number gets the highest priority (1). • If two atoms on a stereogenic center are the same, assign priority based on the atomic number of the atoms bonded to these atoms. One atom of higher atomic number determines a higher priority. • If two isotopes are bonded to the stereogenic center, assign priorities in order of decreasing mass number. • To assign a priority to an atom that is part of a multiple bond, consider a multiply bonded atom as an equivalent number of singly bonded atoms. highest atomic number = highest priority 1

3

Br

CH2CH2CH3

4 H C CH2I *

3

Cl

2 I is NOT bonded directly to the stereogenic center.

4 CH3 C CH2CH2CH2CH2CH3 *

1 OH

2

CH(CH3)2

1 This is the highest priority C since it is bonded to 2 other C's. * = stereogenic center

4 H C CH2OH 3 * COOH

2 This C is considered bonded to 3 O's.

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Chapter 5–2

SSoom mee bbaassiicc pprriinncciipplleess • When a compound and its mirror image are superimposable, they are identical achiral compounds. A plane of symmetry in one conformation makes a compound achiral (5.3). • When a compound and its mirror image are not superimposable, they are different chiral compounds called enantiomers. A chiral compound has no plane of symmetry in any conformation (5.3). • A tetrahedral stereogenic center is a carbon atom bonded to four different groups (5.4, 5.5). • For n stereogenic centers, the maximum number of stereoisomers is 2n (5.7). plane of symmetry CH3 C H

H

CH3

CH3

*C

C H

plane of symmetry

[* = stereogenic center]

CH3CH2

H

no stereogenic centers

H Cl

1 stereogenic center

CH3

*C Cl

H

CH3

CH3

C* H Cl

2 stereogenic centers

*C Cl

H

CH3 C* H

Cl

2 stereogenic centers

Chiral compounds generally contain stereogenic centers. A plane of symmetry makes these compounds achiral.



O Oppttiiccaall aaccttiivviittyy iiss tthhee aabbiilliittyy ooff aa ccoom mppoouunndd ttoo rroottaattee ppllaannee--ppoollaarriizzeedd lliigghhtt ((55..1122)).. • An optically active solution contains a chiral compound. • An optically inactive solution contains one of the following: • an achiral compound with no stereogenic centers. • a meso compound—an achiral compound with two or more stereogenic centers. • a racemic mixture—an equal amount of two enantiomers.

TThhee pprreeffiixxeess RR aanndd SS ccoom mppaarreedd w wiitthh dd aanndd ll The prefixes R and S are labels used in nomenclature. Rules on assigning R,S are found in Section 5.6. • An enantiomer has every stereogenic center opposite in configuration. If a compound with two stereogenic centers has the R,R configuration, its enantiomer has the S,S configuration. • A diastereomer of this same compound has either the R,S or S,R configuration; one stereogenic center has the same configuration and one is opposite. The prefixes d (or +) and l (or –) tell the direction a compound rotates plane-polarized light (5.12). • d (or +) stands for dextrorotatory, rotating polarized light clockwise. • l (or –) stands for levorotatory, rotating polarized light counterclockwise.

TThhee pphhyyssiiccaall pprrooppeerrttiieess ooff iissoom meerrss ccoom mppaarreedd ((55..1122)) Type of isomer Physical properties Constitutional isomers Different Enantiomers Identical except the direction of rotation of polarized light Diastereomers Different Racemic mixture Possibly different from either enantiomer

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Stereochemistry 5–3

EEqquuaattiioonnss • Specific rotation (5.12C): specific = rotation

•

[
]




=

l x c


= observed rotation (o) l = length of sample tube (dm) c = concentration (g/mL)

Enantiomeric excess (5.12D): ee

=

% of one enantiomer – % of other enantiomer

=

[
] mixture x 100% [
] pure enantiomer

dm = decimeter 1 dm = 10 cm

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Chapter 5–4 C Chhaapptteerr 55:: A Annssw weerrss ttoo PPrroobblleem mss 5.1 Cellulose consists of long chains held together by intermolecular hydrogen bonds forming sheets that stack in extensive three-dimensional arrays. Most of the OH groups in cellulose are in the interior of this three-dimensional network, unavailable for hydrogen bonding to water. Thus, even though cellulose has many OH groups, its three-dimensional structure prevents many of the OH groups from hydrogen bonding with the solvent and this makes it water insoluble. 5.2 Constitutional isomers have atoms bonded to different atoms. Stereoisomers differ only in the three-dimensional arrangement of atoms.

and

a. 2,3-dimethylpentane

2,4-dimethylpentane different connectivity of atoms different names constitutional isomers O

b.

and

c.

and

different connectivity of atoms constitutional isomers

OH

d.

four-membered ring three-membered ring

and trans isomer

different connectivity of atoms constitutional isomers

cis isomer

Both are 1,2-dimethylcyclobutane, but the CH3 groups are oriented differently. stereoisomers

5.3 Draw the mirror image of each molecule by drawing a mirror plane and then drawing the molecule’s reflection. A chiral molecule is one that is not superimposable on its mirror image. A molecule with one stereogenic center is always chiral. A molecule with zero stereogenic centers is not chiral (in general). CH3

a.

C Cl

CH3 CH3 C Br

CH3 Br

c. Cl

CH3

O

CH3

identical

CH3 Br H

C Cl

Cl

Br H

stereogenic center nonsuperimposable mirror images chiral molecules

CH3

achiral molecule

CH3

C

O

identical

achiral molecule

b.

CH3

H Br

d.

Br H

C F

C CH2CH3

CH3CH2

stereogenic center nonsuperimposable mirror images chiral molecules

F

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Stereochemistry 5–5 5.4 A plane of symmetry cuts the molecule into two identical halves. 2 H's are behind one another. H H

a.

CH3

C CH3

b.

CH3

H

H

H

CH3

c.

H

CH3 C

CH3

d.

H

C H Cl

H

CH3 C H Cl

one possible plane of symmetry plane of symmetry

plane of symmetry

plane of symmetry

5.5 Rotate around the middle C
C bond so that the Br atoms are eclipsed. rotate CH3 here H C Br

Br

Br

H

C

C CH3

H

H

Br

C

CH3

CH3

C2 C3 plane of symmetry

5.6 To locate a stereogenic center, omit: All C’s with 2 or more H’s, all sp and sp2 hybridized atoms, and all heteroatoms. (In Chapter 25, we will learn that the N atoms of ammonium salts [R4N+X–] can sometimes be stereogenic centers.) Then evaluate any remaining atoms: a tetrahedral stereogenic center has a carbon bonded to four different groups. H

a.

CH3CH2 C CH2CH3

d.

Cl

bonded to 2 identical ethyl groups 0 stereogenic centers

CH3CH2CH2OH

0 stereogenic centers CH3

e. (CH3)2CHCH2CH2 C CH2CH3 H

b.

(CH3)3CH

This C is bonded to 4 different groups. 1 stereogenic center

0 stereogenic centers c.

H CH3 C CH=CH2 OH

This C is bonded to 4 different groups. 1 stereogenic center

f.

H CH3CH2 C CH2CH2CH3 CH3

This C is bonded to 4 different groups. 1 stereogenic center

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Chapter 5–6 5.7 Use the directions from Answer 5.6 to locate the stereogenic centers. Br

H

a.

c.

CH3CH2CH2 C CH3

d.

OH

Br

stereogenic center

3 C's bonded to 4 different groups 3 stereogenic centers

both C's bonded to 4 different groups 2 stereogenic centers

H

b. (CH3)2CHCH2 C COOH NH2

stereogenic center

5.8 Use the directions from Answer 5.6 to locate the stereogenic centers. O

OH H2N CH3O

O

O

CH3O

NH2

H N

4 C's bonded to 4 different groups 4 stereogenic centers

aliskiren

5.9 Find the C bonded to four different groups in each molecule. At the stereogenic center, draw two bonds in the plane of the page, one in front (on a wedge), and one behind (on a dash). Then draw the mirror image (enantiomer). stereogenic center

stereogenic center

c. CH3SCH2CH2CH(NH2)COOH

a. CH3CH(Cl)CH2CH3 H CH3

C

H

Cl

Cl CH2CH3

CH3CH2

C

CH3

CH3SCH2CH2

mirror images nonsuperimposable enantiomers

CH3CH2CH(OH)CH2OH H CH3CH2

OH C

CH2OH

C

COOH

H

HO HOCH2

mirror images nonsuperimposable enantiomers

C

CH2CH3

H

H2N HOOC

mirror images nonsuperimposable enantiomers

stereogenic center b.

NH2

H

C

CH2CH2SCH3

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Stereochemistry 5–7 5.10 Draw the chiral molecule with only C and H atoms. CH2CH3

CH2CH3

CH3 C CH2CH2CH3

or

CH3 C CH(CH3)2

H

H

5.11 Use the directions from Answer 5.6 to locate the stereogenic centers. O

C bonded to H and 3 different C's: 1 stereogenic center

a.

Cl

b. Cl

c.

Each labeled C is bonded to: d. H, Cl, CH2, CHCl: 2 stereogenic centers

O

4 C's bonded to 4 different groups: 4 stereogenic centers

NH2

O O

e.

O

N H

CO2H

C bonded to H, 2 different O's and 1 C: 1 stereogenic center

no stereogenic centers

CO2H

5.12 O

a.

b.

O O O

cholesterol

HO

simvastatin

All stereogenic C's are circled. Each C is sp3 hybridized and bonded to 4 different groups.

5.13 Assign priority based on atomic number: atoms with a higher atomic number get a higher priority. If two atoms are the same, look at what they are bonded to and assign priority based on the atomic number of these atoms. a. –CH3, –CH2CH3

higher priority

c. –H, –D

higher mass higher priority

e. –CH2CH2Cl, –CH2CH(CH3)2

higher priority H

b. –I, –Br

higher priority

d. –CH2Br, –CH2CH2Br

higher priority

f. –CH2OH, –CHO

=

C O

H

=

C O O C

2 H's, 1 O

2 O's, 1 H

2 C–O bonds

C bonded to 2 O's has higher priority.

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Chapter 5–8 5.14 Rank by decreasing priority. Lower atomic number = lower priority. Highest priority = 1, Lowest priority = 4 priority priority 2 c. –CH2CH3 C bonded to 2 H's + 1 C a. –COOH C = second lowest 3 atomic number 3 –CH3 C bonded to 3 H's –H H = lowest atomic number 4 –NH2

4 H = lowest atomic number 1 –CH(CH3)2 C bonded to 1 H + 2 C's decreasing priority: –CH(CH3)2, –CH2CH3, –CH3, –H –H

N = second highest 2 atomic number O = highest atomic number 1

–OH

decreasing priority: –OH, –NH2, –COOH, –H

b. –H –CH3

priority 4

H = lowest atomic number C bonded to 3 H's

d. –CH=CH2 C bonded to 1 H + 2 C's

3

1 Cl = highest atomic number –CH2Cl C bonded to 2 H's + 1 Cl 2 decreasing priority: –Cl, –CH2Cl, –CH3, –H –Cl

priority 2

–CH3

C bonded to 3 H's

3

–C
CH

C bonded to 3 C's

1

4 H = lowest atomic number decreasing priority: –C
CH, –CH=CH2, –CH3, –H –H

5.15 To assign R or S to the molecule, first rank the groups. The lowest priority group must be oriented behind the page. If tracing a circle from (1)
(2)
(3) proceeds in the clockwise direction, the stereogenic center is labeled R; if the circle is counterclockwise, it is labeled S. 2

2 Cl

a.

C CH3

3

2

COOH

b.

H Br

C CH3

3

1

counterclockwise S isomer

CH2Br

c.

H OH

1

ClCH2 CH3

counterclockwise S isomer

CH2Br

rotate

C OH

CH3 HO

H 3

3

C CH2Cl

d. 1

counterclockwise S isomer

clockwise R isomer

lowest priority group now back

5.16 CH3O

O

2

Cl

CH3OOC

N

1

S

H Cl

N

Cl H

3

3

2

COOCH3 N

1

S

S

clopidogrel

clockwise R isomer

counterclockwise S isomer Plavix

5.17 The maximum number of stereoisomers = 2n where n = the number of stereogenic centers. a. 3 stereogenic centers 23 = 8 stereoisomers

b. 8 stereogenic centers 28 = 256 stereoisomers

2

1

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Stereochemistry 5–9 5.18 a. CH3CH2CH(Cl)CH(OH)CH2CH3

b. CH3CH(Br)CH2CH(Cl)CH3

2 stereogenic centers = 4 possible stereoisomers

CH3CH2

CH2CH3

C H Cl

C

CH3CH2

H HO

CH2CH3

C

CH2CH3

C H OH

A

H Cl

CH3CH2

H Cl H Br H Cl

CH3CH2 HO

H

Br H Cl H

A

B

H Cl Br H

H Br Cl H

C

D

CH2CH3

C H

C

C

B

C OH

2 stereogenic centers = 4 possible stereoisomers

C H Cl

D

5.19 a. CH3CH(OH)CH(OH)CH3

b. CH3CH(OH)CH(Cl)CH3

2 stereogenic centers = 4 possible stereoisomers CH3

CH3

C HO

CH3

C

H

H OH

A

CH3

H HO

CH3

C

C

B

C H OH

H HO

CH3

C H

CH3

C

H HO

CH3

C

C

OH

HO

H

CH3

H OH

H HO

C

C

CH3

CH3

H Cl

H Cl

C

A

CH3

C

identical C is a meso compound. A and B are enantiomers. Pairs of diastereomers: A and C, B and C.

5.20

2 stereogenic centers = 4 possible stereoisomers

C

C H

B

CH3

CH3

H Cl

H Cl

C

C

CH3 OH

CH3

C

C

D

H OH

Pairs of enantiomers: A and B, C and D. Pairs of diastereomers: A and C, A and D, B and C, B and D.

A meso compound must have at least two stereogenic centers. Usually a meso compound has a plane of symmetry. You may have to rotate around a C–C bond to see the plane of symmetry clearly. CH3CH2

a.

C HO

H

CH3

CH2CH3

b.

C H

OH

2 stereogenic centers plane of symmetry meso compound

C HO H

OH

H Br

H

C

rotate

H Br Br H

c. CH3

2 stereogenic centers no plane of symmetry not a meso compound

Br H

2 stereogenic centers plane of symmetry meso compound

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Chapter 5–10 5.21 Use the definition in Answer 5.20 to draw the meso compounds. a. BrCH2CH2CH(Cl)CH(Cl)CH2CH2Br

Cl

Br

H H

Cl

b.

HO

OH

HO

Br

c.

OH H H

plane of symmetry

NH2

H2N

H H

H2N

plane of symmetry

NH2

plane of symmetry

5.22 The enantiomer must have the exact opposite R,S designations. Diastereomers with two stereogenic centers have one center the same and one different. If a compound is R,S: Exact opposite: R and S interchanged.

Its enantiomer is: S,R

One designation remains the same, the other changes.

Its diastereomers are: R,R and S,S

5.23 The enantiomer must have the exact opposite R,S designations. For diastereomers, at least one of the R,S designations is the same, but not all of them. a. (2R,3S)-2,3-hexanediol and (2R,3R)-2,3-hexanediol One changes; one remains the same: diastereomers b. (2R,3R)-2,3-hexanediol and (2S,3S)-2,3-hexanediol Both R's change to S's: enantiomers c. (2R,3S,4R)-2,3,4-hexanetriol and (2S,3R,4R)-2,3,4-hexanetriol Two change; one remains the same: diastereomers

5.24 The enantiomer must have the exact opposite R,S designations. For diastereomers, at least one of the R,S designations is the same, but not all of them. HO H HO H

a.

R HO

R

R S

H OH HO H

sorbitol

HO H HO H

b. OH

HO

R

R

H OH

S

S

H OH H OH

c. OH

H OH

HO

S HO H

S

S

R

OH

H OH

A

B

One changes; three remain the same. diastereomer

All stereogenic centers change. enantiomers

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Stereochemistry 5–11 5.25 Meso compounds generally have a plane of symmetry. They cannot have just one stereogenic center. Cl

a.

b.

c. OH

plane of symmetry meso compound

no plane of symmetry not a meso compound

no plane of symmetry not a meso compound

5.26 Cl

2 stereogenic centers = 4 stereoisomers maximum

a.

c. Cl

Draw the cis and trans isomers: CH3

CH3

CH3

CH3

Draw the cis and trans isomers: Cl

Cl

cis A

Cl

Cl

identical

A

CH3

CH3

CH3

CH3

Cl

Cl

trans B

identical

C

Cl

Cl

B identical

Pair of enantiomers: B and C. Pairs of diastereomers: A and B, A and C.

Pair of diastereomers: A and B. Only 3 stereoisomers exist. Only 2 stereoisomers exist.

b.

2 stereogenic centers = 4 stereoisomers maximum

HO

Draw the cis and trans isomers: cis

trans OH

CH3

A

HO

CH3

B

Pairs of enantiomers: A and B, C and D. Pairs of diastereomers: A and C, A and D, B and C, B and D.

OH

CH3

HO

C All 4 stereoisomers exist.

CH3

D

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Chapter 5–12 5.27 Four facts: • Enantiomers are mirror image isomers. • Diastereomers are stereoisomers that are not mirror images. • Constitutional isomers have the same molecular formula but the atoms are bonded to different atoms. • Cis and trans isomers are always diastereomers. CH3

a.

C Br

S

Br

and H CH2OH

C

HOCH2

S

c.

H CH3

d.

C CH3

H NH2

(S)-alanine [
] = +8.5 mp = 297 oC

OH HO

same molecular formula, opposite configuration at one stereogenic center enantiomers

COOH

and

OH

1,3-isomer 1,4- isomer constitutional isomers

and

5.28

OH

HO

same molecular formula same R,S designation: identical b.

HO

and

OH HO

trans

cis

Both 1,3 isomers, cis and trans: diastereomers

a. Mp = same as the S isomer. b. The mp of a racemic mixture is often different from the melting point of the enantiomers. c. –8.5, same as S but opposite sign d. Zero. A racemic mixture is optically inactive. e. Solution of pure (S)-alanine: optically active Equal mixture of (R)- and (S)-alanine: optically inactive 75% (S)- and 25% (R)-alanine: optically active

5.29 [
] =


l xc


= observed rotation l = length of tube (dm) c = concentration (g/mL)

[
] =

10° = +100 = specific rotation 1 dm x (1 g/10 mL)

5.30 Enantiomeric excess = ee = % of one enantiomer
% of other enantiomer. a. 95%
5% = 90% ee

5.31

b. 85%
15% = 70% ee

a. 90% ee means 90% excess of A, and 10% racemic mixture of A and B (5% each); therefore, 95% A and 5% B. b. 99% ee means 99% excess of A, and 1% racemic mixture of A and B (0.5% each); therefore, 99.5% A and 0.5% B. c. 60% ee means 60% excess of A, and 40% racemic mixture of A and B (20% each); therefore, 80% A and 20% B.

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Stereochemistry 5–13 5.32 a. [
] mixture

ee =

+10

x 100% = 42% ee

+24 x 100%

[
] pure enantiomer

b.

[
] solution

x 100% = 80% ee

+24

[
] solution = +19.2

5.33 a.

[
] mixture x 100% = 60% ee +3.8

b. % one enantiomer – % other enantiomer = ee 80% – 20% = 60% ee

[
] mixture = +2.3

80% dextrorotatory (+) enantiomer 20% levorotatory (–) enantiomer

5.34 • Enantiomers have the same physical properties (mp, bp, solubility), and rotate the plane of polarized light to an equal but opposite extent. • Diastereomers have different physical properties. • A racemic mixture is optically inactive. cis isomer

trans isomers

CH3

CH3

A

CH3

enantiomers

CH3

CH3

B

CH3

C

A and B are diastereomers of C.

a. The bp's of A and B are the same. The bp's of A and C are different. b. Pure A: optically active Pure B: optically active Pure C: optically inactive Equal mixture of A and B: optically inactive Equal mixture of A and C: optically active c. There would be two fractions: one containing A and B (optically inactive), and one containing C (optically inactive).

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Chapter 5–14 5.35 Use the definitions from Answer 5.2. CH3

O

a.

c.

and

H CH3

and

O

b.

both up cis

Both compounds are 1,2-dimethylcyclohexane. one cis, one trans = stereoisomers

O

and

H

one up, one down trans

same molecular formula C4H8O different connectivity constitutional isomers

and

d.

O

same molecular formula C7H14 different connectivity constitutional isomers

C5H8O

C5H10O

different molecular formulas not isomers

5.36 Use the definitions from Answer 5.3.

CH3

a.

C CH3

OH

CH3

CH2OH H

HOCH2 H

O

c.

C

O

e. OHC

OH OH

CH3

HO

OH

threose identical achiral

identical achiral COOH

b.

C HSCH2

chiral

COOH

H NH2

H H2N

d.

C CH2SH

H

Br

Br

H

cysteine chiral

identical achiral

5.37 CHO C CH3

CH3

a. OH H

A R isomer

HO

C

CH3

b. H CHO

S enantiomer

C

OHC

HO H

c. H OH

R identical

CH3

C

CHO

S enantiomer

CHO OH

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Stereochemistry 5–15 5.38 A plane of symmetry cuts the molecule into two identical halves.

H

CH3CH2

a.

C Cl

b.

C

H

C

COOH

H HO

C

H

H

CH3

H and OH are aligned.

The plane of symmetry bisects the molecule.

plane of symmetry

CH2CH3

C Cl

d.

C

C HOOC

CH2CH3

CH3CH2

CH3

HO H HO H

Cl

CH3CH2

Cl

c.

C H Cl

H

Cl

e.

C CH2CH3

no plane of symmetry

no plane of symmetry

plane of symmetry

5.39 Use the directions from Answer 5.6 to locate the stereogenic centers. a. CH3CH2CH2CH2CH2CH3 All C's have 2 or more H's. 0 stereogenic centers

g. O

bonded to 4 different groups 1 stereogenic center

H

b.

CH3CH2O C CH2CH3 CH3

h.

1 stereogenic center All C's have 2 or more H's, or are sp2 hybridized. 0 stereogenic centers

c. (CH3)2CHCH(OH)CH(CH3)2 0 stereogenic centers H

H

d. (CH3)2CHCH2 C CH2 C

H

C

CH3

Cl CH2CH3

CH3 CH3

i.

3 stereogenic centers Each indicated C bonded to 4 different groups = 2 stereogenic centers

H

e.

CH3 C CH2CH3 D

OH

bonded to 4 different groups 1 stereogenic center

HO

j. OH

OH

OH

O

HO

OH OH

f. OH

OH

OH

Each indicated C bonded to 4 different groups = 6 stereogenic centers

Each indicated C bonded to 4 different groups = 5 stereogenic centers

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Chapter 5–16 5.40 Stereogenic centers are circled. Eight constitutional isomers: Cl Cl

Cl Cl Cl

Cl

Cl

Cl

5.41

a.

H2N H

NH2

H NH2

amphetamine O

O

CH3

COOH

b.

H COOH

H CH3

O

HOOC

ketoprofen

5.42 Draw a molecule to fit each description. a.

OH

O

c.

b. O

alcohol

ketone

cyclic ether

5.43 Assign priority based on the rules in Answer 5.13. a.
OH,
NH2

c.
CH(CH3)2,
CH2OH

higher atomic number higher priority b.
CD3,
CH3 D higher mass than H higher priority

e.
CHO,
COOH

C bonded to O higher priority d. –CH2Cl, –CH2CH2CH2Br

C has 3 bonds to O. higher priority f.
CH2NH2,
NHCH3 higher atomic number higher priority

C bonded to Cl higher priority

5.44 Assign priority based on the rules in Answer 5.13. a.
F >
OH >
NH2 >
CH3

d. –COOH > –CHO > –CH2OH > –H

b.
(CH2)3CH3 >
CH2CH2CH3 >
CH2CH3 >
CH3

e. –Cl > –SH > –OH > –CH3

c.
NH2 >
CH2NHCH3 >
CH2NH2 >
CH3

f. –CCH > –CH=CH2 > –CH(CH3)2 > –CH2CH3

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Stereochemistry 5–17 5.45 Use the rules in Answer 5.15 to assign R or S to each stereogenic center. 1

1 I

a.

H

c.

C

C

H CH3

CH3CH2

2

T

CH3

switch H and CH3

C

CH3 D

H D

T

2

3

3

counterclockwise It looks like an S isomer, but we must reverse the answer, S to R.

counterclockwise S isomer

R isomer 1

2

NH2

b.

CH3 3 H

Cl

C

d.

CH2CH3

Br ICH2

2

Cl switch H and Br

C

H ICH2

H

C

1 Br

3

clockwise, but H in front S isomer

counterclockwise It looks like an S isomer, but we must reverse the answer, S to R.

R isomer CH3

e.

CH(CH3)2

C

HOOC

C

f.

CH3 SH

H HO

NH2 H

C

CH3

C

H

g.

H

S S

CH3

HO

H

S, R

R, R

Cl

h. Cl

S

5.46 a. (3R)-3-methylhexane

c. (3R,5S,6R)-5-ethyl-3,6-dimethylnonane CH3 H R

CH3 H

S

CH3 H

b. (4R,5S)-4,5-diethyloctane 4R

R

H CH2CH3

d. (3S,6S)-6-isopropyl-3-methyldecane

H CH2CH3

H CH(CH3)2 S S

CH3CH2 H

5S

H CH3

5.47 5 a.

6

H

3

6R 9

1 b.

4

2

7

4R 5

6R 3

5S

3R

1

10

c.

9

7

1

S (3S)-3-methylhexane

(4R,6R)-4-ethyl-6-methyldecane

(3R,5S,6R)-5-isobutyl-3,6-dimethylnonane

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Chapter 5–18 5.48 Two enantiomers of the amino acid leucine. COOH C (CH3)2CHCH2

COOH

H NH2

H H2N

S isomer naturally occurring

C CH2CH(CH3)2

R isomer

5.49 S

a. OH

NH Cl

b.

H NH2

HO

CH3CH2O2C

S CH3

COOH

ketamine

L-dopa

S

enalapril

O

N

N H

c.

O

CO2H

S

5.50 CO2CH3 H

H N

methylphenidate

H N

H

CO2CH3 H H

R,R

S,S

5.51 a. 1R,2S C1

OH

d. NHCH3

OH

S

R NHCH3

e.

enantiomer of ephedrine

C2

ephedrine b. 1S,2S C1

OH

OH NHCH3

R

R NHCH3

diastereomer of ephedrine

C2

pseudoephedrine c. Ephedrine and pseudoephedrine are diastereomers (one stereogenic center is the same; one is different).

5.52 OH C C H

NH2 H N

a. HO

O

S

b.

O

c.

O O

N O

amoxicillin

COOH

O

N

O

norethindrone

CH3

O

heroin

S

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Stereochemistry 5–19 5.53 O a. CH3CH(OH)CH(OH)CH2CH3

c.

b. CH3CH2CH2CH(CH3)2

OH OH

HO OH

HO

0 stereogenic centers

4 stereogenic centers 24 = 16 possible stereoisomers

2 stereogenic centers 22 = 4 possible stereoisomers

5.54 a. CH3CH(OH)CH(OH)CH2CH3 CH3

CH2CH3

C

CH3CH2

C

H HO

CH3

C H OH

C

H HO

A

CH3

CH2CH3

C H OH

HO

C

CH3

C H OH

H

B

CH3CH2 C

H HO

C

H

OH

D

Pairs of enantiomers: A and B, C and D. Pairs of diastereomers: A and C, A and D, B and C, B and D. b. CH3CH(OH)CH2CH2CH(OH)CH3 OH

OH

OH

OH

OH

OH

A

C

B

OH

OH

identical meso compound

Pair of enantiomers: A and B. Pairs of diastereomers: A and C, B and C. c. CH3CH(Cl)CH2CH(Br)CH3 Cl

Br

Br

A

Cl

Cl

Br

Br

C

B

Cl

D

Pairs of enantiomers: A and B, C and D. Pairs of diastereomers: A and C, A and D, B and C, B and D. d. CH3CH(Br)CH(Br)CH(Br)CH3 Br

Br

Br

Br

Br

Br

A identical

Br

Br

Br

Br

Br

Br

Br

Br

Br

B

C

D

meso compound Pair of enantiomers: B and C. Pairs of diastereomers: A and B, A and C, A and D, B and D, C and D.

Br

Br

Br

identical meso compound

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Chapter 5–20 5.55 HOCH2

a.

C

CH3

CH3

H OH

H HO

C

CH2OH

C

H HO

CH3

C

C H OH

HO

H I

I H

H I

NH2

H

I H

I H

H2N

OH

d.

C

C

H HO

OH H

diastereomer

I H

H2N

H2N

or

HO

enantiomer

CH2OH

diastereomer

HO

CH3

or

C H OH

enantiomer

c.

CH3

diastereomer

enantiomer

b.

CH2OH

HO

diastereomer

CH3

diastereomer

CH3

CH3

or CH2CH3

CH3CH2

CH3CH2

CH3CH2

diastereomer

enantiomer

diastereomer

5.56 CH3

CH3

CH3

CH3

a. CH3

CH3

A

identical

CH3

CH3

B

C

meso compound Pair of enantiomers: B and C. Pairs of diastereomers: A and B, A and C.

b.

CH3

CH3

CH3

CH3

CH3

A

CH3

CH3

CH3

B

identical

identical

Pair of diastereomers: A and B. Meso compounds: A and B.

c.

Cl

Cl

Br

Cl

Br

Br

B

A

C

Cl Br

D

Pairs of enantiomers: A and B, C and D. Pairs of diastereomers: A and C, A and D, B and C, B and D.

5.57

achiral

achiral

chiral

chiral

achiral

achiral

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Stereochemistry 5–21 5.58 Explain each statement. All molecules have a mirror image, but only chiral molecules have enantiomers. A is not chiral, and therefore, does not have an enantiomer.

a. OH

B has one stereogenic center, and therefore, has an enantiomer. Only compounds with two or more stereogenic centers have diastereomers.

b. Cl

C is chiral and has two stereogenic centers, and therefore, has both an enantiomer and a diastereomer.

c. Cl OH

HO

rotate

OH

D has two stereogenic centers, but is a meso compound. Therefore, it has a diastereomer, but no enantiomer since it is achiral. plane of symmetry

d. OH

e. HO

E has two stereogenic centers, but is a meso compound. Therefore, it has a diastereomer, but no enantiomer since it is achiral.

OH

plane of symmetry

5.59 OHC H HO

C2 C3

OH H

C R

C R

OHC

a. CH2OH

HO

D-erythrose

C S H

HOCH2

CH2OH C R

b. HO

H OH

2S,3R diastereomer

2R,3R

H

C R

CHO

OHC

C R

H

C HO S H

OH

2R,3R identical

OHC

H OH

c.

C S

d. CH2OH

H OH

C H R HO

2S,3S enantiomer

C S

CH2OH

2R,3S diastereomer

5.60 Re-draw each Newman projection and determine the R,S configuration. Then determine how the molecules are related. CHO H

CHO CH2OH

H

OH

HO

A

OHC

re-draw

OH

OH H

H HO

CH2OH

R,R a. A and B are identical. b. A and C are enantiomers.

CHO OH

HO

H

H

CH2OH

HO

H

CH2OH

B re-draw OHC

CH2OH

R,R

OH

H

CH2OH

H

H HO

H

H

OH

CHO

OH

C re-draw

OHC

H OH

HO H

D re-draw OHC

H CH2OH

HO

CH2OH

S,S c. A and D are diastereomers. d. C and D are diastereomers.

H

OH

S,R

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Chapter 5–22 5.61 R R

trans

NH2

A

R NH2

NH2

S

cis

trans

trans

B

C

D

NH2

NH2

E NH2 group is in a different location.

A (trans, R) and B (cis, R) are diastereomers. A (trans, R) and C (trans, R) are identical molecules. A (trans, R) and D (trans, S) are enantiomers. A (trans, R) and E are constitutional isomers.

5.62 a.

CH3

and

g.

C

and

C

H Br

enantiomers

H

2S,3S CH3

b.

CH3

CHO

C H HO

OH

OH

OH

1,4-cis 1,4-trans diastereomers

i.

C

H HO

H OH

and 6 H's 12 H's different molecular formulas not isomers

2R,3R

one different configuration diastereomers

C

d.

CH3

H

j. BrCH2

and

and

Cl

k.

and H

Cl

Br on end HO CH3

Cl

Cl Br H

and

C H

enantiomers

H

1,3-trans diastereomers

1,3-cis

mirror images not superimposable enantiomers

C

C

CH3 CH3

and

I

HOCH2 BrCH2

CH2OH CH3

enantiomers

different molecular formulas not isomers

f.

H

HO H

H

C

2R,3S

e.

CH3

2S,3S

CH3

and H

C

and HO

OHC

C

C

H

h.

same molecular formula different connectivity constitutional isomers

c.

H Br

H Br

Br

identical

and

CH3

CH3

CH3

l. Br I

and

C H

CH2Br

CH3

CH2OH C Br H

Br in middle different connectivity constitutional isomers

H

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Stereochemistry 5–23 5.63 a. A and B are constitutional isomers. A and C are constitutional isomers. B and C are diastereomers (cis and trans). C and D are enantiomers. plane of symmetry b.

A A has two planes of symmetry. achiral

B

C

D

achiral

chiral

chiral mirror images and not superimposable enantiomers

c. Alone, C and D would be optically active. d. A and B have a plane of symmetry. e. A and B have different boiling points. B and C have different boiling points. C and D have the same boiling point. f. B is a meso compound. g. An equal mixture of C and D is optically inactive because it is a racemic mixture. An equal mixture of B and C would be optically active. 5.64 H HO H

N

ee =

CH3O

[
] mixture

x 100%

[
] pure enantiomer quinine = A quinine's enantiomer = B

N

quinine

a.
50 x 100% = 30% ee
165

b. 30% ee = 30% excess one compound (A) remaining 70% = mixture of 2 compounds (35% each A and B) Amount of A = 30 + 35 = 65% Amount of B = 35%


83 x 100% = 50% ee
165

50% ee = 50% excess one compound (A) remaining 50% = mixture of 2 compounds (25% each A and B) Amount of A = 50 + 25 = 75% Amount of B = 25%

73% ee = 73% excess of one compound (A) remaining 27% = mixture of 2 compounds (13.5% each A and B) Amount of A = 73 + 13.5 = 86.5% Amount of B = 13.5% [] mixture x 100% e. 60% = –165


120 x 100% = 73% ee
165 c. [] = +165 d. 80%
20% = 60% ee

[] mixture =
99

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Chapter 5–24 5.65 OH HO

OH

O

HO

O

O OH

O

HO

HCl, H2O

OH

amygdalin (laetrile)

COOH

only one of the products formed

CN

mandelic acid

OH

a. The 11 stereogenic centers are circled. Maximum number of stereoisomers = 211 = 2048 b. Enantiomers of mandelic acid: HO H

H OH

HOOC

COOH

R c. 60%
40% = 20% ee 20% = [] mixture/
154 x 100% [] mixture =
31

+50

d. ee =

+154

S

x 100% = 32% ee

[
] for (S)-mandelic acid = +154

32% excess of the S enantiomer 68% of racemic R and S = 34% S and 34% R S enantiomer: 32% + 34% = 66% R enantiomer = 34%

5.66 sp2

a. Each stereogenic center is circled. b. The stereogenic centers in mefloquine are labeled. c. Artemisinin has 7 stereogenic centers. 2n = 27 = 128 possible stereoisomers d. One N atom in mefloquine is sp2 and one is sp3. e. Two molecules of artemisinin cannot intermolecularly H-bond because there are no O–H or N–H bonds.

CF3

H

N

CF3

O O

O H

H

O

R HO

O

H H N

sp3

H S

H

artemisinin

mefloquine

CF3

CF3 N

f.

CF3

N

CF3

H Cl H H N

HO H

H HO H

HH N

+

Cl

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135

Stereochemistry 5–25 5.67 c. diastereomer

a. Each stereogenic center is circled.

O

H S N

N

N S H

O

N

OH CONH2 O

saquinavir Trade name: Invirase

N

S

S

O

H N

HS

R

H N H

O

N

OH CONH2 O

H NH

H NH

(CH3)3C

(CH3)3C

d. constitutional isomer

b. enantiomer

O

H N

N

N H

O

N

OH CONH2 O

N H

H N H

O

O H

NH (CH3)3C

new amine

NH

O

H N

H

new aldehyde

N OH O

H NH

(CH3)3C

5.68 HO

a. HO

O

HO

O

OH

re-draw

more stable ring all groups equatorial

salicin b. diastereomer O

HO

HO O

re-draw

OH O

HO HO

OH

All other groups on the ring are equatorial.

HO

OH

O

OH

axial HO

c. enantiomer HO

O

OH

OH OH

HO

OH O

HO HO

O

HO

O OH

OH

re-draw

HO HO

O

O OH

HO OH

In the enantiomer, all groups are still equatorial, but all down bonds are up bonds and all up bonds are down bonds.

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Chapter 5–26 5.69 Allenes contain an sp hybridized carbon atom doubly bonded to two other carbons. This makes the double bonds of an allene perpendicular to each other. When each end of the allene has two like substituents, the allene contains two planes of symmetry and it is achiral. When each end of the allene has two different groups, the allene has no plane of symmetry and it becomes chiral.

CH3

H

CH3

C C C

CH3

H

H

CH3

HC C C C CH C CH CH CH CH CHCH2CO2H

C C C

B

allene

H

no plane of symmetry

mycomycin re-draw

chiral

A

CH CH CH CHCH2CO2H

HC C C C

These two substituents are at 90o to these two substituents. Allene A contains two planes of symmetry, making it achiral.

C C C H

H

The substituents on each end of the allene in mycomycin are different. Therefore, mycomycin is chiral.

5.70 HO O

O

OH

O

OH

NH2 O

discodermolide

OH

a. The 13 tetrahedral stereogenic centers are circled. b. Because there is restricted rotation around a C–C double bond, groups on the end of the double bond cannot interconvert. Whenever the substituents on each end of the double bond are different from each other, the double bond is a stereogenic site. Thus, the following two double bonds are isomers: R

R

R

C C H

H C C

H

H

R

These compounds are isomers.

There are three stereogenic double bonds in discodermolide, labeled with arrows. c. The maximum number of stereoisomers for discodermolide must include the 13 tetrahedral stereogenic centers and the three double bonds. Maximum number of stereoisomers = 216 = 65,536.

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137

Stereochemistry 5–27 5.71 racemic mixture of 2-phenylpropanoic acid

salts formed by proton transfer COO

COOH C

H NH2 H CH3

S

C

+ (R)-sec-butylamine

diastereomers COO

COOH H NH2

C

+ R

R

S

enantiomers

H CH3

H CH3

+

H NH3

(R)-sec-butylamine

H CH3

+

H NH3

C

R R

These salts are diastereomers, and they are now separable by physical methods since they have different physical properties.

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Understanding Organic Reactions 6–1 C Chhaapptteerr 66:: U Unnddeerrssttaannddiinngg O Orrggaanniicc R Reeaaccttiioonnss

W Wrriittiinngg oorrggaanniicc rreeaaccttiioonnss ((66..11)) • Use curved arrows to show the movement of electrons. Full-headed arrows are used for electron pairs and half-headed arrows are used for single electrons. +

C

C Z

C Z

Z

+

Z

full-headed arrow

half-headed arrows

•

C

Reagents can be drawn either on the left side of an equation or over an arrow. Catalysts are drawn over or under an arrow.



TTyyppeess ooff rreeaaccttiioonnss ((66..22)) [1] Substitution

C Z

+ Y

C Y

+

Z

Z = H or a heteroatom

Y replaces Z

[2] Elimination

C

C

X

Y

+ reagent

Two  bonds are broken.

[3] Addition

C C

+

+

C C

X Y


bond

X Y

C C X Y

This
bond is broken.



IIm mppoorrttaanntt ttrreennddss Values compared Bond dissociation energy and bond strength

Two  bonds are formed.

Trend The higher the bond dissociation energy, the stronger the bond (6.4). Increasing size of the halogen CH3 F


Ho = 456 kJ/mol

CH3 Cl

CH3 Br

351 kJ/mol

293 kJ/mol

Increasing bond strength

CH3

I

234 kJ/mol

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Chapter 6–2 Ea and reaction rate

The larger the energy of activation, the slower the reaction (6.9A).

Energy

Ea

larger Ea
slower reaction slower reaction

Ea

faster reaction

Reaction coordinate

Ea and rate constant

The higher the energy of activation, the smaller the rate constant (6.9B).

Go products
Go > 0 Go reactants

Keq < 1

more stable reactants

Free energy

Free energy

Equilibrium always favors the species lower in energy.

Go reactants
Go < 0 Go products

Equilibrium favors the starting materials.

Keq > 1

more stable products

Equilibrium favors the products.



R Reeaaccttiivvee iinntteerrm meeddiiaatteess ((66..33)) • Breaking bonds generates reactive intermediates. • Homolysis generates radicals with unpaired electrons. • Heterolysis generates ions. Reactive intermediate

General structure

Reactive feature

Reactivity

radical

C

unpaired electron

electrophilic

carbocation

C

carbanion

C

positive charge; only six electrons around C net negative charge; lone electron pair on C

electrophilic nucleophilic

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Understanding Organic Reactions 6–3   EEnneerrggyy ddiiaaggrraam mss ((66..77,, 66..88))

Energy

transition state

Ea reactants

Ea determines the rate.
Ho is the difference in bonding energy between the reactants and products.


Ho products

Reaction coordinate

 C Coonnddiittiioonnss ffaavvoorriinngg pprroodduucctt ffoorrm maattiioonn ((66..55,, 66..66)) Variable Value Meaning Keq Keq > 1 More product than starting material is present at equilibrium.
Go


Go < 0

The energy of the products is lower than the energy of the reactants.


Ho


Ho < 0

Bonds in the products are stronger than bonds in the reactants.


So


So > 0

The product is more disordered than the reactant.

  EEqquuaattiioonnss ((66..55,, 66..66))
Go = 2.303RT log Keq Keq depends on the energy difference between reactants and products. R = 8.314 J/(K•mol), the gas constant T = Kelvin temperature (K)


Go free energy change

=


Ho



T
So

change in bonding energy

T = Kelvin temperature (K)

  FFaaccttoorrss aaffffeeccttiinngg rreeaaccttiioonn rraattee ((66..99)) Factor Effect energy of activation higher Ea  slower reaction concentration higher concentration  faster reaction temperature higher temperature  faster reaction

change in disorder

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Chapter 6–4 C Chhaapptteerr 66:: A Annssw weerrss ttoo PPrroobblleem mss 6.1 [1] In a substitution reaction, one group replaces another. [2] In an elimination reaction, elements of the starting material are lost and a
bond is formed. [3] In an addition reaction, elements are added to the starting material. OH

O

Br

c.

a.

CH3

C

Br replaces OH = substitution reaction

b.

CH3

CH3

C

CH2Cl

Cl replaces H = substitution reaction

H

O

O

d.

CH3CH CHCH3

CH3CH2CH(OH)CH3

OH


bond formed elements lost (H + OH) elimination reaction

addition of 2 H's addition reaction

6.2 Alkenes undergo addition reactions. HCl

CH2 CH2

CH3CH2Cl

Br2

CH2 CH2

BrCH2CH2Br

addition of 1 H and 1 Cl addition reaction

addition of 2 Br's addition reaction

6.3 Heterolysis means one atom gets both of the electrons when a bond is broken. A carbocation is a C with a positive charge, and a carbanion is a C with a negative charge. heterolysis

heterolysis

CH3

a.

CH3 C OH

b.

heterolysis

c. CH3CH2 Li

Br

CH3

Electrons go to the more electronegative atom, Br.

Electrons go to the more electronegative atom, O.

Electrons go to the more electronegative atom, C.

CH3 CH3 C

OH

Br

CH3CH2

CH3

carbocation

Li

carbanion

carbocation

6.4 Use full-headed arrows to show the movement of electron pairs, and half-headed arrows to show the movement of single electrons. CH3

a. (CH3)3C N N

(CH3)3C

+

N N

c.

CH3 C

CH3

+

Br

CH3 C Br

CH3

b.

CH3

+

CH3

CH3 CH3

d.

HO OH

CH3

2 HO

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Understanding Organic Reactions 6–5 6.5 Increasing number of electrons between atoms = increasing bond strength = increasing bond dissociation energy = decreasing bond length. Increasing size of an atom = increasing bond length = decreasing bond strength. a. H Cl

or H Br Br is larger than Cl. longer, weaker bond higher bond dissociation energy

b. CH3 OH

CH3 SH

or

S is larger than O. longer, higher bond weaker bond dissociation energy

c. (CH3)2C O

CH3 OCH3

or

higher bond dissociation energy

single bond fewer electrons

6.6 To determine
Ho for a reaction: [1] Add the bond dissociation energies for all bonds broken in the equation (+ values). [2] Add the bond dissociation energies for all of the bonds formed in the equation ( values). [3] Add the energies together to get the
Ho for the reaction. A positive
Ho means the reaction is endothermic. A negative
Ho means the reaction is exothermic. a. CH3CH2 Br + H2O

CH3CH2 OH

[1] Bonds broken

+ HBr

[2] Bonds formed


Ho (kJ/mol)


Ho (kJ/mol)

CH3CH2 Br

+ 285

CH3CH2 OH

 389

H OH

+ 498

H Br

 368

+ 783 kJ/mol

Total

 757 kJ/mol

Total

[3] Overall
Ho = sum in Step [1] + sum in Step [2] + 783 kJ/mol  757

kJ/mol

ANSWER: + 26 kJ/mol endothermic

b. CH4 + Cl2

CH3Cl + HCl

[1] Bonds broken

[2] Bonds formed


Ho (kJ/mol)


Ho (kJ/mol)

CH3 H

+ 435

CH3 Cl

 351

Cl Cl

+ 242

H Cl

 431

Total

+ 677 kJ/mol

Total

 782 kJ/mol

[3] Overall
Ho = sum in Step [1] + sum in Step [2] + 677 kJ/mol  782 kJ/mol ANSWER:  105 kJ/mol exothermic

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Chapter 6–6

6.7 Use the directions from Answer 6.6. In determining the number of bonds broken or formed, you must take into account the coefficients needed to balance an equation. a. CH4 + 2 O2

CO2 + 2 H2O

[1] Bonds broken
Ho (kJ/mol) CH3 H O O


Ho (kJ/mol) OC O  535 x 2 =  1070

+ 435 x 4 = + 1740 + 497 x 2 = + 994

sum in Step [1] + sum in Step [2]

HO H  498 x 4 =  1992

+ 2734 kJ/mol

Total

[3] Overall
Ho =

[2] Bonds formed

+ 2734 kJ/mol

 3062 kJ/mol

Total

 3062 kJ/mol ANSWER:

b. 2 CH3CH3 + 7 O2

4 CO2 + 6 H2O

[1] Bonds broken

[3] Overall
Ho =

[2] Bonds formed


Ho (kJ/mol) CH3CH2 H + 410 x 12 = + 4920 O O

+ 497 x 7 = + 3479

C C

+ 368 x 2 =

Total

 328 kJ/mol

+736


Ho (kJ/mol) OC O

 535 x 8 =  4280

sum in Step [1] + sum in Step [2]

HO H  498 x 12 =  5976 Total

 10256 kJ/mol

+ 9135 kJ/mol

+ 9135 kJ/mol  10256 kJ/mol ANSWER:  1121 kJ/mol

6.8 Use the following relationships to answer the questions: Keq = 1 then
G° = 0; Keq > 1 then
G° < 0; Keq < 1 then
G° > 0 a. A negative value of
G° means the equilibrium favors the product and Keq is > 1. Therefore Keq = 1000 is the answer. b. A lower value of
G° means a larger value of Keq, and the products are more favored. Keq = 10
2 is larger than Keq = 10
5, so
G° is lower. 6.9 Use the relationships from Answer 6.8. a. Keq = 5.5. Keq > 1 means that the equilibrium favors the product. b.
G° = 40 kJ/mol. A positive
G° means the equilibrium favors the starting material.

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Understanding Organic Reactions 6–7 6.10 When the product is lower in energy than the starting material, the equilibrium favors the product. When the starting material is lower in energy than the product, the equilibrium favors the starting material. a.
G° is positive so the equilibrium favors the starting material. Therefore the starting material is lower in energy than the product. b. Keq is > 1 so the equilibrium favors the product. Therefore the product is lower in energy than the starting material. c.
G° is negative so the equilibrium favors the product. Therefore the product is lower in energy than the starting material. d. Keq is < 1 so the equilibrium favors the starting material. Therefore the starting material is lower in energy than the product. 6.11 H H

OCH3

Keq = 2.7

OCH3

a. The Keq is > 1 and therefore the product (the conformation on the right) is favored at equilibrium. b. The
G° for this process must be negative since the product is favored. c.
G° is somewhere between 0 and
6 kJ/mol. 6.12 A positive
H° favors the starting material. A negative
H° favors the product. a.
H° is positive (80 kJ/mol). The starting material is favored. b.
H° is negative (–40 kJ/mol). The product is favored. 6.13 a. False. b. True. c. False. d. True. e. False.

The reaction is endothermic. This assumes that
G° is approximately equal to
H°. Keq < 1 The starting material is favored at equilibrium.

6.14 a. True. b. False.
G° for the reaction is negative. c. True. d. False. The bonds in the product are stronger than the bonds in the starting material. e. True.

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Chapter 6–8

6.15

Energy

transition state

Ea product


H° starting material Reaction coordinate

6.16 A transition state is drawn with dashed lines to indicate the partially broken and partially formed bonds. Any atom that gains or loses a charge contains a partial charge in the transition state. CH3

CH3

a. CH3 C OH2 CH3

b. CH3O H

H2O

CH3 C CH3 CH3

+

transition state: CH3  C

OH

CH3O





transition state:

+

CH3O

H2O

H

OH

OH2

CH3

6.17 transition state 2: E

Energy

transition state 1: D

B Ea(A–B) Ea(A–C)


H°A–B = endothermic
H°A–C = exothermic

A

a. Reaction A–C is exothermic. Reaction A–B is endothermic. b. Reaction A–C is faster. c. Reaction A–C generates a lower-energy product. d. See labels. e. See labels. f. See labels.

C Reaction coordinate

6.18 reactive intermediate

Energy

transition state 1

Ea1

a. b. c. d. e.

transition state 2 Ea2


H°1

Reaction coordinate


H°2
H°overall

Two steps since there are two energy barriers. See labels. See labels. One reactive intermediate is formed (see label). The first step is rate determining since its transition state is at higher energy. f. The overall reaction is endothermic since the energy of the products is higher than the energy of the reactants.

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Understanding Organic Reactions 6–9 6.19

Energy

Ea(B–C) B Ea(A–B)

relative energies: C < A < B B C is rate-determining.

A C Reaction coordinate

6.20 Ea, concentration, and temperature affect reaction rate.
H°,
G°, and Keq do not affect reaction rate. a. Ea = 4 kJ/mol corresponds to a faster reaction rate. b. A temperature of 25 °C will have a faster reaction rate since a higher temperature corresponds to a faster reaction. c. No change: Keq does not affect reaction rate. d. No change:
H° does not affect reaction rate.

Energy

6.21 The Ea of an endothermic reaction is at least as large as its
H° because the Ea essentially “includes” the
H° in its total. The Ea measures the difference between the energy of the starting material and the energy of the transition state, and in an endothermic reaction, the energy of the products is somewhere in between these two values.

Ea


H° = (+) value for an endothermic reaction

Reaction coordinate

6.22 a. b. c. d. e.

False. The reaction occurs at the same rate as a reaction with Keq = 8 and Ea = 80 kJ/mol. False. The reaction is slower than a reaction with Keq = 0.8 and Ea = 40 kJ/mol. True. True. False. The reaction is endothermic.

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Chapter 6–10

6.23 All reactants in the rate equation determine the rate of the reaction. [1] rate = k[CH3CH2Br][
OH]

[2] rate = k[(CH3)3COH]

a. Tripling the concentration of CH3CH2Br only  The rate is tripled. b. Tripling the concentration of
OH only  The rate is tripled. c. Tripling the concentration of both CH3CH2CH2Br and
OH  The rate increases by a factor of 9 (3  3 = 9).

a. Doubling the concentration of (CH3)3COH  The rate is doubled. b. Increasing the concentration of (CH3)3COH by a factor of 10  The rate increases by a factor of 10.

6.24 The rate equation is determined by the rate-determining step. a.

CH3CH2 Br

b.

(CH3)3C

+

CH2 CH2

OH

(CH3)3C +

Br slow

+

OH fast

Br

one step rate = k[CH3CH2Br][–OH]

+ H2O + Br

(CH3)2C

CH2

+ H2O

two steps The slow step determines the rate equation. rate = k[(CH3)3CBr]

6.25 A catalyst is not used up or changed in the reaction. It only speeds up the reaction rate. OH and H are added to the starting material. a.

H2O

CH2 CH2

H2 adds to the starting material.

I– not used up = catalyst.

CH3CH2OH

b.

I–

CH3Cl

CH3OH

–OH

H2SO4
OH

H2SO4 is not used up = catalyst.

substitutes for

O

H2

c.

Cl
.

Pt

Pt not used up = catalyst.

6.26 Use the directions from Answer 6.1. O

HO OH

O

H OH

c.

a.

addition of 2 H's addition reaction


bond formed elements lost (H + OH) elimination reaction

b.

H H

Cl replaces H = substitution reaction

H Cl

O

d.

C

OH

O Cl

H replaces Cl = substitution reaction

C

H

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Understanding Organic Reactions 6–11 6.27 H

a. homolysis of

c. heterolysis of CH3 MgBr

b. heterolysis of CH3 O H

CH3 C H H

H CH3 C

H

CH3 O

CH3

H

H

MgBr

carbanion

radical

6.28 Use the rules in Answer 6.4 to draw the arrows. Br

a.

Br O

b.

CH3

Cl CH3

Br

d.

+

Cl

e.

+

+ Br

Br Br

O

CH3 C CH3

c.

+

+

Cl

C

CH3

f.

CH3 Cl

CH3CH2 Br

+

H

CH3 C CH3

CH3CH2OH +

OH

CH3

C H +

H + H2O

C C

OH CH3

H

Br

H

6.29 a.

I

+

OH +

OH

H

OH

I

H C

c.

H O

H C

O

b. CH3 C CH2CH2CH3 OCH2CH3

CH3

C

d.

CH2CH2CH3

H

H

C H H

Br

+

H

H + H2O +

C C

Br

H H C

Cl

H

H

+ HCl

OCH2CH3

6.30 Draw the curved arrows to identify the product X.

O

+

H Br

[1]

O H

+

Br

Br

[2]

OH

A

B

X

6.31 Follow the curved arrows to identify the intermediate Y. CO2H [1] O

CO2H O

O

C

[2]

O

COOH

O O

D

Y

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Chapter 6–12

6.32 Use the rules from Answer 6.5. a.

I

CCl3

Br CCl3

largest halogen intermediate weakest bond bond strength

b. H2N NH2

Cl CCl3

smallest halogen strongest bond

single bond weakest bond

HN NH

N N

double bond intermediate bond strength

triple bond strongest bond

6.33 Use the directions from Answer 6.6. a. CH3CH2 H

+ Br2

CH3CH2 Br

[1] Bonds broken

+ HBr

[2] Bonds formed


Ho (kJ/mol)

[3] Overall
Ho =


Ho (kJ/mol)

CH3CH2 H

+ 410

CH3CH2 Br

 285

+ 602 kJ/mol

Br Br

+ 192

H Br

 368

 653 kJ/mol

+ 602 kJ/mol

Total

 653 kJ/mol

Total

b. OH + CH4

CH3 + H2O

[1] Bonds broken

[2] Bonds formed


Ho (kJ/mol) CH3 H

c.

ANSWER:  51 kJ/mol


Ho (kJ/mol) H OH

+ 435 kJ/mol

CH3 OH + HBr

CH3 Br

 498 kJ/mol

+ 435 kJ/mol  498 kJ/mol ANSWER:  63 kJ/mol

+ H2O

[2] Bonds formed

[1] Bonds broken

[3] Overall
Ho =


Ho (kJ/mol)

[3] Overall
Ho =


Ho (kJ/mol)

CH3 OH

+ 389

CH3 Br

 293

+ 757 kJ/mol

H Br

+ 368

H OH

 498

 791 kJ/mol

Total

+ 757 kJ/mol

d. Br + CH4

Total

ANSWER:  34 kJ/mol

H + CH3Br

[1] Bonds broken

[2] Bonds formed


Ho (kJ/mol) CH3 H

 791 kJ/mol

+ 435 kJ/mol


Ho (kJ/mol) CH3 Br

293 kJ/mol

[3] Overall
Ho = + 435 kJ/mol  293 kJ/mol ANSWER: + 142 kJ/mol

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151

Understanding Organic Reactions 6–13 6.34 propane

propene

CH3 CH2CH3

CH3

CH2CH3


Ho = 356 kJ/mol This bond is formed from two sp3 hybridized C's.

CH3 CH CH2

CH3

CH CH2


Ho = 385 kJ/mol This bond is formed from one sp2 and one sp3 hybridized C. The higher percent scharacter in one C makes a stronger bond; thus the bond dissociation energy is higher.

6.35 H

H

H

CH2 CH C H

CH2 CH C H

CH2 CH

C H

hybrid:




H

CH2 CH C H

H




6.36 The more stable radical is formed by a reaction with a smaller H°. H CH3 CH2

C H

CH3 CH2

C H

H


Ho = 410 kJ/mol = less stable radical

H

This C–H bond is stronger.

A

H CH3 C

CH3

CH3 C

H

CH3


Ho = 397 kJ/mol = more stable radical

H

This C–H bond is weaker.

B

Since the bond dissociation for cleavage of the C–H bond to form radical A is higher, more energy must be added to form it. This makes A higher in energy and therefore less stable than B. 6.37 Use the bond dissociation energy for the CC  bond in ethane as an estimate of the  bond strength in ethylene. Then you can estimate the
bond strength as well. CH3 CH3 o

H = 368 kJ/mol

CH2 CH2

635 – 368 = 267 kJ/mol =
bond

o

H = 635 kJ/mol

6.38 Use the rules from Answer 6.10. a. Keq = 0.5. Keq is less than one so the starting material is favored. b. Go = 100 kJ/mol. Go is less than 0 so the product is favored. c. Ho = 8.0 kJ/mol. Ho is positive, so the starting material is favored. d. Keq = 16. Keq is greater than one so the product is favored. e. Go = 2.0 kJ/mol. Go is greater than zero so the starting material is favored. f. Ho = 200 kJ/mol. Ho is positive so the starting material is favored. g. So = 8 J/(K•mol). So is greater than zero so the product is more disordered and favored. h. So = 8 J/(K•mol). So is less than zero so the starting material is more disordered and favored.

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Chapter 6–14

6.39

a. A negative
G° must have Keq > 1. Keq = 102. b. Keq = [products]/[reactants] = [1]/[5] = 0.2 = Keq.
G° is positive. c. A negative
G° has Keq > 1, and a positive
G° has Keq < 1.
G° =
8 kJ/mol will have a larger Keq.

6.40 R

axial

R

equatorial H

R

Keq

–CH3 –CH2CH3 –CH(CH3)2 –C(CH3)3

H

18 23 38 4000

a. The equatorial conformation is always present in the larger amount at equilibrium since the Keq for all R groups is greater than 1. b. The cyclohexane with the –C(CH3)3 group will have the greatest amount of equatorial conformation at equilibrium since this group has the highest Keq. c. The cyclohexane with the –CH3 group will have the greatest amount of axial conformation at equilibrium since this group has the lowest Keq. d. The cyclohexane with the –C(CH3)3 group will have the most negative
G° since it has the largest Keq. e. The larger the R group, the more favored the equatorial conformation. f. The Keq for tert-butylcyclohexane is much higher because the tert-butyl group is bulkier than the other groups. With a tert-butyl group, a CH3 group is always oriented over the ring when the group is axial, creating severe 1,3-diaxial interactions. With all other substituents, the larger CH3 groups can be oriented away from the ring, placing a H over the ring, making the 1,3-diaxial interactions less severe. Compare: tert-butylcyclohexane

isopropylcyclohexane

CH3 CH3 H C CH3 H

H

H H

severe 1,3-diaxial interactions with the CH3 group and the axial H's

C

CH3 CH3

less severe 1,3-diaxial interactions

6.41 Calculate Keq, and then find the percentage of axial and equatorial conformations present at equilibrium. F (axial)

a.

H

fluorocyclohexane 1 part

F (equatorial) H

1.5 parts

G° = –5.9log Keq G° = –1.0 kJ/mol –1.0 kJ/mol = –5.9log Keq Keq = 1.5

b. Keq = [products]/[reactants] 1.5 = [products]/[reactants] 1.5[reactants] = [products] [reactants] = 0.4 = 40% axial [products] = 0.6 = 60% equatorial

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Understanding Organic Reactions 6–15 6.42 Reactions resulting in an increase in entropy are favored. When a single molecule forms two molecules, there is an increase in entropy. increased number of molecules
S° is positive. products favored decreased number of molecules
S° is negative. starting material favored increased number of molecules H2O
S° is positive. products favored

+

a. b.

CH3

c.

(CH3)2C(OH)2

+

CH3CH3

CH3

(CH3)2C=O

d. CH3COOCH3

+

H2O

+

CH3COOH

+

CH3OH

no change in the number of molecules neither favored

6.43 Use the directions in Answer 6.16 to draw the transition state. Nonbonded electron pairs are drawn in at reacting sites. Br

+

a.

OH

+

c.

Br

O

+





+ Br

transition state:

+

BF3

transition state:

Cl

F B Cl

d.

C CH3

F

F

Cl





C H

+ H2O

transition state:

H

CH3 CH3 + C CH3

H

+

C C

H

F F B

CH3

H

CH3

NH3




O H NH2

transition state:

F

b.

+

NH2

H3O

+

H

+

C H OH2 H

6.44

A

B

Reaction coordinate • one step A B • exothermic since B lower than A • low Ea (small energy barrier)


H°

B

A Reaction coordinate • one step A B • endothermic since B higher than A • high Ea (large energy barrier)

A

B EaA–B

Energy

Ea

Energy

Ea
H°

d.

c.

b. Energy

Energy

a.

EaB–C C

Reaction coordinate
H°overall • two steps • A lowest energy • B highest energy • Ea(A-B) is ratedetermining, since the transition state for Step [1] is higher in energy.

Ea = 16 kJ/mol A


H° =
80 kJ/mol B

Reaction coordinate • one step A B • exothermic since B lower than A

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Chapter 6–16

6.45 a.

CH3 H

b.

Cl + CH4

CH3 + HCl

Cl

CH3 + HCl

[1] Bonds broken
Ho (kJ/mol)

 431 kJ/mol

H Cl

 431 kJ/mol ANSWER:

+ 4 kJ/mol

d. The Ea for the reverse reaction is the difference in energy between the products and the transition state, 12 kJ/mol.

c.

Ea = 16 kJ/mol

Energy

+ 435 kJ/mol


Ho (kJ/mol)

+ 435 kJ/mol

CH3 H

[3] Overall
Ho =

[2] Bonds formed

16 kJ/mol Energy


H° = 4 kJ/mol

12 kJ/mol 4 kJ/mol

Reaction coordinate

Reaction coordinate

6.46 D Energy

B

a. b. c. d.

B, D, and F are transition states. C and E are reactive intermediates. The overall reaction has three steps. A–C is endothermic. C–E is exothermic. E–G is exothermic. e. The overall reaction is exothermic.

F C

A

E G

Reaction coordinate

6.47 O CH3

C

O

CH3 CH3 C O

OH

CH3

CH3

C

CH3 CH3 C OH

O

CH3

Since pKa (CH3CO2H) = 4.8 and pKa [(CH3)3COH] = 18, the weaker acid is formed as product, and equilibrium favors the products. Thus,
H° is negative, and the products are lower in energy than the starting materials. transition state

O CH3


C  O


CH3 H

O C CH3 CH3

Energy

transition state: Ea starting materials

Reaction coordinate


H° products

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155

Understanding Organic Reactions 6–17 6.48 H H

H H

H Cl

+H

[1]

H

H

+ Cl

Cl

[2]

H

a. Step [1] breaks one
bond and the HCl bond, and one CH bond is formed. The
H° for this step should be positive since more bonds are broken than formed. b. Step [2] forms one bond. The
H° for this step should be negative since one bond is formed and none is broken. c. Step [1] is rate-determining since it is more difficult. d. Transition state for Step [1]: Transition state for Step [2]: 


H H

H H +

Cl

+ H


Cl 

H

e.

Energy

Ea2


H°1 is positive.

Ea1


H°2 is negative.
H°overall is negative.

Reaction coordinate

6.49 B

C

Energy

Ea Ea (CH3)3C +

A

+ H 2O + I

Ea (CH3)3C

+ HI

OH (CH3)3C

+

OH2


H°3
H°2


H°1 = 0

+ I


H°overall (CH3)3C

Reaction coordinate

a. The reaction has three steps, since there are three energy barriers. b. See above.

I

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Chapter 6–18

c. Transition state A (see graph for location): + (CH3)3C

OH H

Transition state B: (CH3)3C


+


I

Transition state C:
+  I

OH2


+

(CH3)3C

d. Step [2] is rate-determining since this step has the highest energy transition state. 6.50 Ea, concentration, catalysts, rate constant, and temperature affect reaction rate so (c), (d), (e), (g), and (h) affect rate. 6.51 a. b. c. d.

rate = k[CH3Br][NaCN] Double [CH3Br] = rate doubles. Halve [NaCN] = rate halved. Increase both [CH3Br] and [NaCN] by factor of 5 = [5][5] = rate increases by a factor of 25.

6.52 O

acetyl C Cl chloride CH3 CH3O

[1] slow

O CH3O

O

[2]

CH3 C Cl fast

CH3

C

+ Cl– OCH3

methyl acetate

a. Only the slow step is included in the rate equation: Rate = k[CH3O–][CH3COCl] b. CH3O– is in the rate equation. Increasing its concentration by 10 times would increase the rate by 10 times. c. When both reactant concentrations are increased by 10 times, the rate increases by 100 times (10
10 = 100). d. This is a substitution reaction (OCH3 substitutes for Cl). 6.53 a. b. c. d. e. f.

True: Increasing temperature increases reaction rate. True: If a reaction is fast, it has a large rate constant. False: Corrected - There is no relationship between
G° and reaction rate. False: Corrected - When the Ea is large, the rate constant is small. False: Corrected - There is no relationship between Keq and reaction rate. False: Corrected - Increasing the concentration of a reactant increases the rate of a reaction only if the reactant appears in the rate equation.

6.54 a. The first mechanism has one step: Rate = k[(CH3)3CI][–OH] b. The second mechanism has two steps, but only the first step would be in the rate equation since it is slow and therefore rate-determining: Rate = k[(CH3)3CI] c. Possibility [1] is second order; possibility [2] is first order.

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157

Understanding Organic Reactions 6–19 d. These rate equations can be used to show which mechanism is plausible by changing the concentration of –OH. If this affects the rate, possibility [1] is reasonable. If it does not affect the rate, possibility [2] is reasonable. CH3 e. CH3 C

CH2

I

Energy


–

H


– OH

Ea


H°

A = (CH3)3CI + –OH B = (CH3)2C=CH2 + I– + H2O

B

A Reaction coordinate transition state [1]

f.

transition state [2] Energy

B Ea[1]


H°[1]

(CH3)3C+ + I– + –OH

CH3

[1] CH3 C +CH3
I
–

Ea[2]


H°[2]
H°overall

(CH3)3CI

(CH3)2C=CH2 + H2O

CH3

[2] CH3 C CH2
+ H

OH


–

Reaction coordinate

6.55 The difference in both the acidity and the bond dissociation energy of CH3CH3 versus HC
CH is due to the same factor: percent s-character. The difference results because one process is based on homolysis and one is based on heterolysis. Bond dissociation energy: CH3CH2 H

sp3 hybridized 25% s-character

HC C H

sp hybridized 50% s-character Higher percent s-character makes this bond shorter and stronger.

Acidity: To compare acidity, we must compare the stability of the conjugate bases: CH3CH2

sp3 hybridized 25% s-character

HC C

sp hybridized 50% s-character Now a higher percent s-character stabilizes the conjugate base making the starting acid more acidic.

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Chapter 6–20

6.56 Ha Hb

a.

Hb

Hb

Hb

C C CH3

C C CH3

C C CH3

C C CH3

H H

H H

H H

H H

Hb

b.

Hb

C C CH3

C C CH3

H H

H H

Ha Hb

Ha

Ha

C C CH3

C C CH3

C C CH3

H H

H H

H H

c. C–Ha is weaker the C–Hb since the carbon radical formed when the C–Ha bond is broken is highly resonance stabilized. This means the bond dissociation energy for C–Ha is lower. 6.57 In Reaction [1], the number of molecules of reactants and products stays the same, so entropy is not a factor. In Reaction [2], a single molecule of starting material forms two molecules of products, so entropy increases. This makes
G° more favorable, thus increasing Keq. 6.58 O CH3

C

O

+ CH3CH2OH

OH

CH3

C

OCH2CH3

+ H2O

Keq = 4

ethyl acetate To increase the yield of ethyl acetate, H2O can be removed from the reaction mixture, or there can be a large excess of one of the starting materials.

6.59 a.

O H

O

O

CH3CH2 O H

ethanol phenol

O

resonance stabilized less energy for homolysis O O H

Csp2–O higher % s-character shorter bond

no resonance stabilization

Less energy is required for cleavage of C6H5O–H because homolysis forms the more stable radical. O

b.

CH3CH2 O

CH3CH2 O H

Csp3–O lower % s-character longer bond

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159

Alkyl Halides and Nucleophilic Substitution 7–1 C Chhaapptteerr 77:: A Allkkyyll H Haalliiddeess aanndd N Nuucclleeoopphhiilliicc SSuubbssttiittuuttiioonn  G Geenneerraall ffaaccttss aabboouutt aallkkyyll hhaalliiddeess • Alkyl halides contain a halogen atom X bonded to an sp3 hybridized carbon (7.1). • Alkyl halides are named as halo alkanes, with the halogen as a substituent (7.2). • Alkyl halides have a polar C–X bond, so they exhibit dipole–dipole interactions but are incapable of intermolecular hydrogen bonding (7.3). • The polar C–X bond containing an electrophilic carbon makes alkyl halides reactive towards nucleophiles and bases (7.5).   TThhee cceennttrraall tthheem mee ((77..66)) • Nucleophilic substitution is one of the two main reactions of alkyl halides. A nucleophile replaces a leaving group on an sp3 hybridized carbon. R X

+

Nu

R Nu

nucleophile

+

X

leaving group

The electron pair in the C
Nu bond comes from the nucleophile.

• •

One  bond is broken and one  bond is formed. There are two possible mechanisms: SN1 and SN2.

  SSNNN11 aanndd SSNNN22 m meecchhaanniissm mss ccoom mppaarreedd SN2 mechanism One step (7.11B)

•

[2] Alkyl halide

•

Order of reactivity: CH3X > RCH2X > R2CHX > R3CX (7.11D)

•

Order of reactivity: R3CX > R2CHX > RCH2X > CH3X (7.13D)

[3] Rate equation

• •

rate = k[RX][:Nu–] second-order kinetics (7.11A)

• •

rate = k[RX] first-order kinetics (7.13A)

[4] Stereochemistry

•

backside attack of the nucleophile (7.11C) inversion of configuration at a stereogenic center favored by stronger nucleophiles (7.17B) better leaving group
faster reaction (7.17C)

•

trigonal planar carbocation intermediate (7.13C) racemization at a stereogenic center favored by weaker nucleophiles (7.17B) better leaving group
faster reaction (7.17C)

favored by polar aprotic solvents (7.17D)

•

• [5] Nucleophile

•

[6] Leaving group

•

[7] Solvent

•

•

SN1 mechanism Two steps (7.13B)

[1] Mechanism

• • •

favored by polar protic solvents (7.17D)

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Chapter 7–2 Increasing rate of the SN1 reaction H

H

H

R

H C Br

R C Br

R C Br

R C Br

H

H

R

R

1o

2o

3o

both SN1 and SN2

SN 1

methyl SN2

Increasing rate of an SN2 reaction



IIm mppoorrttaanntt ttrreennddss • The best leaving group is the weakest base. Leaving group ability increases across a row and down a column of the periodic table (7.7). Increasing basicity NH3

Increasing basicity

H2O

F

Increasing leaving group ability

•

Cl

Br

I

Increasing leaving group ability

Nucleophilicity decreases across a row of the periodic table (7.8A). For 2nd row elements with the same charge:

CH3

NH2

OH

F

Increasing basicity Increasing nucleophilicity

•

Nucleophilicity decreases down a column of the periodic table in polar aprotic solvents (7.8C). Down a column of the periodic table

F

Cl

Br

I

Increasing nucleophilicity in polar aprotic solvents

•

Nucleophilicity increases down a column of the periodic table in polar protic solvents (7.8C). Down a column of the periodic table

F

Cl

Br

I

Increasing nucleophilicity in polar protic solvents

•

The stability of a carbocation increases as the number of R groups bonded to the positively charged carbon increases (7.14). + CH3

+ RCH2

+ R2CH

+ R3C

methyl

1o

2o

3o

Increasing carbocation stability

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Alkyl Halides and Nucleophilic Substitution 7–3

IIm mppoorrttaanntt pprriinncciipplleess

•

Principle Electron-donating groups (such as R groups) stabilize a positive charge (7.14A).

•

Example 3 Carbocations (R3C+) are more stable than 2o carbocations (R2CH+), which are more stable than 1o carbocations (RCH2+). o

•

Steric hindrance decreases nucleophilicity but not basicity (7.8B).

•

(CH3)3CO– is a stronger base but a weaker nucleophile than CH3CH2O–.

•

Hammond postulate: In an endothermic reaction, the more stable product is formed faster. In an exothermic reaction, this fact is not necessarily true (7.15).

•

SN1 reactions are faster when more stable (more substituted) carbocations are formed, because the rate-determining step is endothermic.

•

Planar, sp2 hybridized atoms react with reagents from both sides of the plane (7.13C).

•

A trigonal planar carbocation reacts with nucleophiles from both sides of the plane.

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Chapter 7–4 C Chhaapptteerr 77:: A Annssw weerrss ttoo PPrroobblleem mss 7.1 Classify the alkyl halide as 1°, 2°, or 3° by counting the number of carbons bonded directly to the carbon bonded to the halogen. C bonded to 2 C's 2° alkyl halide

C bonded to 1 C 1° alkyl halide

I

CH3

a.

CH3CH2CH2CH2CH2 Br

F

b.

c. CH3 C

CHCH3

d.

CH3 Cl

C bonded to 3 C's 3° alkyl halide

C bonded to 3 C's 3° alkyl halide

7.2 Use the directions from Answer 7.1. F This F is bonded to a C which is not bonded to any other C's. Therefore, it cannot be classified as 1°, 2°, or 3°. HO

S

O OCOC2H5

H F O F

H

bonded to C bonded to 3 C's 3° alkyl halide bonded to C bonded to 2 C's 2° alkyl halide

7.3 Draw a compound of molecular formula C6H13Br to fit each description. Br

a.

b.

Br

c. Br

1° alkyl halide one stereogenic center

2° alkyl halide two stereogenic centers

3° alkyl halide no stereogenic centers

7.4 To name a compound with the IUPAC system: [1] Name the parent chain by finding the longest carbon chain. [2] Number the chain so the first substituent gets the lower number. Then name and number all substituents, giving like substituents a prefix (di, tri, etc.). To name the halogen substituent, change the -ine ending to -o. [3] Combine all parts, alphabetizing substituents, and ignoring all prefixes except iso. a.

(CH3)2CHCH(Cl)CH2CH3

re-draw [1]

2-methyl [2]

Cl

5 carbon alkane = pentane

[3] 3-chloro-2-methylpentane 2 Cl

3-chloro

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Alkyl Halides and Nucleophilic Substitution 7–5 2-bromo

b. Br

[1]

[2]

Br

[3] 2-bromo-5,5-dimethylheptane

2 7 carbon alkane = heptane

5,5-dimethyl

c. 2-methyl

[2]

[1] Br

Br

[3] 1-bromo-2-methylcyclohexane

1-bromo

1 6 carbon cycloalkane = cyclohexane

2-fluoro

d. F

[1]

F

[2]

[3] 2-fluoro-5,5-dimethylheptane 5,5-dimethyl

1 2 3 4 7

7 carbon alkane = heptane

7.5 To work backwards from a name to a structure: [1] Find the parent name and draw that number of carbons. Use the suffix to identify the functional group (-ane = alkane). [2] Arbitrarily number the carbons in the chain. Add the substituents to the appropriate carbon. a. 3-chloro-2-methylhexane [1]

6 carbon alkane

[2]

methyl at C2

1 2 3 4 5 6 Cl

chloro at C3

b. 4-ethyl-5-iodo-2,2-dimethyloctane [1]

8 carbon alkane

[2] ethyl at C4

1 2 3 4 5 6 7 8

I

2 methyls at C2

iodo at C5

c. cis-1,3-dichlorocyclopentane [1]

5 carbon cycloalkane

[2]

chloro groups at C1 and C3, both on the same side Cl

Cl

C3

C1 d. 1,1,3-tribromocyclohexane [1]

6 carbon cycloalkane

3 Br groups

[2]

Br

C3

Br Br

C1

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Chapter 7–6 e. propyl chloride [1] 3 carbon alkyl group

[2] chloride on end

CH3CH2CH2

CH3CH2CH2 Cl

f. sec-butyl bromide [1] 4 carbon alkyl group

[2] bromide

CH3 CHCH2CH3

CH3 CHCH2CH3 Br

7.6 Boiling points of alkyl halides increase as the size (and polarizability) of X increases. Remember: stronger intermolecular forces = higher boiling point. a.

smallest halogen least polarizable lowest boiling point

b.

CH3CH2CH2I

middle size halogen intermediate boiling point

largest halogen most polarizable highest boiling point

CH3(CH2)5Br

CH3(CH2)4CH3

weakest forces nonpolar lowest boiling point

7.7

CH3CH3CH2Cl

CH3CH2CH2F

VDW, DD forces intermediate boiling point

CH3(CH2)5OH

OH is capable of hydrogen bonding. strongest forces highest boiling point

a. Because an sp2 hybridized C has a higher percent s-character than an sp3 hybridized C, it holds electron density closer to C. This pulls a little more electron density towards C, away from Cl, and thus a Csp2–Cl bond is less polar than a Csp3–Cl bond. Cl

b.

Cl

sp3 C–Cl bond intermediate boiling point

lowest boiling point

Br

larger halogen, sp3 C–Br bond highest boiling point

7.8 Since more polar molecules are more water soluble, look for polarity differences between methoxychlor and DDT. methoxychlor

DDT

H CH3O

C

H OCH3

CCl3

2 methoxy groups more polar The O atoms can hydrogen bond to H2O. more biodegradable

Cl

C

Cl

CCl3

2 chloro groups less polar readily soluble in an organic medium

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Alkyl Halides and Nucleophilic Substitution 7–7 7.9 To draw the products of a nucleophilic substitution reaction: [1] Find the sp3 hybridized electrophilic carbon with a leaving group. [2] Find the nucleophile with lone pairs or electrons in
bonds. [3] Substitute the nucleophile for the leaving group on the electrophilic carbon. +

a. Br

leaving group

+ Br

OCH2CH3 OCH2CH3

nonbonded e
pairs nucleophile

Cl

OH

b.

+ leaving group

nonbonded e
pairs nucleophile

+

I

c.

leaving group

+ Na+Cl–

Na+
OH

N3

+ I

N3

nonbonded e
pairs nucleophile

Br

CN

+

d. leaving group

+ Na+Br–

Na+
CN

nonbonded e
pairs nucleophile

7.10 Use the steps from Answer 7.9 and then draw the proton transfer reaction.

a.

+

Br

N(CH2CH3)3

substitution

+

N(CH2CH3)3 + Br

nucleophile leaving group

+

b. (CH3)3C Cl

H2O

substitution

(CH3)3C

proton

+ Cl

(CH3)3C

transfer

H

nucleophile

leaving group

O H

7.11 Draw the structure of CPC using the steps from Answer 7.9. N

nucleophile

+

Cl

substitution

+

N

leaving group +

CPC

Cl

O H

+ HCl

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Chapter 7–8 7.12 Compare the leaving groups based on these trends: • Better leaving groups are weaker bases. • A neutral leaving group is always better than its conjugate base. a. Cl–, I–

b. NH3, NH2–

farther down a column of the periodic table less basic better leaving group

c. H2O, H2S

neutral compound less basic better leaving group

farther down a column of the periodic table less basic better leaving group

7.13 Good leaving groups include Cl–, Br–, I–, H2O. a.

CH3CH2CH2 Br

c. CH3CH2CH2 OH2

b. CH3CH2CH2OH

Br
is a good leaving group.

No good leaving group. is too strong a base.

H2O is a good leaving group.


OH

d. CH3CH3

No good leaving group. H
is too strong a base.

7.14 To decide whether the equilibrium favors the starting material or the products, compare the nucleophile and the leaving group. The reaction proceeds towards the weaker base. a.

CH3CH2 NH2

+

Br

nucleophile better leaving group weaker base pKa (HBr) = –9

b.

I

+

CN

nucleophile pKa (HCN) = 9.1

+

CH3CH2 Br

NH2

leaving group

Reaction favors starting material.

pKa (NH3) = 38

CN

+

I

leaving group better leaving group weaker base pKa (HI) = –10

Reaction favors product.

7.15 It is not possible to convert CH3CH2CH2OH to CH3CH2CH2Cl by nucleophilic substitution with NaCl because –OH is a stronger base and poorer leaving group than Cl–. The equilibrium favors the reactants, not the products. CH3CH2CH2OH

Na Cl

weaker base

CH3CH2CH2Cl

Na

OH

stronger base

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Alkyl Halides and Nucleophilic Substitution 7–9 7.16 Use these three rules to find the stronger nucleophile in each pair: [1] Comparing two nucleophiles having the same attacking atom, the stronger base is a stronger nucleophile. [2] Negatively charged nucleophiles are always stronger than their conjugate acids. [3] Across a row of the periodic table, nucleophilicity decreases when comparing species of similar charge. O

a. NH3, NH2

b. CH3 , HO

c. CH3NH2, CH3OH

d.

Across a row of the periodic A negatively charged Across a row of the periodic nucleophile is stronger table, nucleophilicity decreases table, nucleophilicity decreases than its conjugate acid. with species of the same charge. with species of the same charge. stronger nucleophile stronger nucleophile stronger nucleophile

C CH3

O

CH3CH2O

same attacking atom (O) stronger base stronger nucleophile

7.17 Polar protic solvents are capable of H-bonding, and therefore must contain a H bonded to an electronegative O or N. Polar aprotic solvents are incapable of H-bonding, and therefore do not contain any O–H or N–H bonds. a. HOCH2CH2OH contains 2 O–H bonds polar protic

b. CH3CH2OCH2CH3 no O–H bonds polar aprotic

c. CH3COOCH2CH3 no O–H bonds polar aprotic

7.18 • In polar protic solvents, the trend in nucleophilicity is opposite to the trend in basicity down a column of the periodic table so that nucleophilicity increases. • In polar aprotic solvents, the trend is identical to basicity so that nucleophilicity decreases down a column. a. Br
and Cl
in polar protic solvent

farther down the column more nucleophilic in protic solvent b.
OH and Cl
in polar aprotic solvent

farther up the column and to the left in the row more basic more nucleophilic

In polar aprotic solvents: nucleophilicity increases O F nucleophilicity Cl increases

c. HS
and F
in polar protic solvent In polar protic solvents: nucleophilicity increases farther down the column O F nucleophilicity and left in the row S increases more nucleophilic in protic solvent

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Chapter 7–10 7.19 The stronger base is the stronger nucleophile except in polar protic solvents when nucleophilicity increases down a column. For other rules, see Answers 7.16 and 7.18. a.

H2O

OH

no charge weakest nucleophile

NH2

negatively charged intermediate nucleophile

negatively charged farther left in periodic table strongest nucleophile

b. Br F Basicity decreases down a Basicity decreases column in polar aprotic solvents. across a row. weakest nucleophile intermediate nucleophile c.

H2O

OH

strongest nucleophile

CH3COO

OH
OH

weakest nucleophile

strongest nucleophile

weaker base than intermediate nucleophile

7.20 To determine what nucleophile is needed to carry out each reaction, look at the product to see what has replaced the leaving group. a. (CH3)2CHCH2CH2 Br

(CH3)2CHCH2CH2

c. (CH3)2CHCH2CH2 Br

SH

SH replaces Br. HS
is needed.

b. (CH3)2CHCH2CH2 Br

(CH3)2CHCH2CH2

(CH3)2CHCH2CH2

OCOCH3

OCOCH3 replaces Br. CH3COO
is needed.

OCH2CH3

d. (CH3)2CHCH2CH2 Br

(CH3)2CHCH2CH2

C CH

CCH replaces Br. HCC
is needed.

OCH2CH3 replaces Br. CH3CH2O
is needed.

7.21 The general rate equation for an SN2 reaction is rate = k[RX][:Nu
]. a. [RX] is tripled, and [:Nu–] stays the same: rate triples. b. Both [RX] and [:Nu–] are tripled: rate increases by a factor of 9 (3  3 = 9). c. [RX] is halved, and [:Nu–] stays the same: rate halved. d. [RX] is halved, and [:Nu–] is doubled: rate stays the same (1/2  2 = 1). 7.22 The transition state in an SN2 reaction has dashed bonds to both the leaving group and the nucleophile, and must contain partial charges. a.

CH3CH2CH2 Cl

+

OCH3

CH3CH2CH2 OCH3

+

Cl

CH3CH2CH2 CH3O 


b.

Br

+

SH

SH

+

Br




SH 
Br

Cl 


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Alkyl Halides and Nucleophilic Substitution 7–11 7.23 All SN2 reactions have one step. Cl 


CH3CH2CH2

Energy

CH3O 


Ea CH3CH2CH2 Cl


H°

+ OCH3

CH3CH2CH2 OCH3 + Cl

Reaction coordinate

7.24 To draw the products of SN2 reactions, replace the leaving group by the nucleophile, and then draw the stereochemistry with inversion at the stereogenic center. CH3CH2 D

a.

D CH CH 2 3

C

+

Br

CH3CH2O

OCH2CH3

H

b.

C

+

I

CN

CN

H

7.25 Increasing the number of R groups increases crowding of the transition state and decreases the rate of an SN2 reaction. Cl

or

a. CH3CH2 Cl 1° alkyl halide

CH3 Cl

b.

methyl halide faster reaction

or 2° alkyl halide

c.

Cl

Br

2° alkyl halide faster reaction

1° alkyl halide faster reaction

or

Br

3° alkyl halide

7.26 These three methyl groups make the alkyl halide sterically hindered. This slows the rate of an SN2 reaction even though it is a 1° alkyl halide.

CH3 CH3 C CH2Br CH3

7.27 +

H N

N H

CH3 SR2 H N

A

+ SR2

N H

loss of

CH3

a proton

H N

N CH3

nicotine

7.28 In a first-order reaction, the rate changes with any change in [RX]. The rate is independent of any change in [nucleophile]. a. [RX] is tripled, and [:Nu–] stays the same: rate triples. b. Both [RX] and [:Nu–] are tripled: rate triples. c. [RX] is halved, and [:Nu–] stays the same: rate halved. d. [RX] is halved, and [:Nu–] is doubled: rate halved.

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Chapter 7–12 7.29 In SN1 reactions, racemization always occurs at a stereogenic center. Draw two products, with the two possible configurations at the stereogenic center. leaving group

nucleophile

CH3

a.

C (CH3)2CH

CH3

H2O

C

Br

(CH3)2CH

CH2CH3

CH3

H CH3CH2

Cl

CH2CH3

+

HBr

OH

enantiomers

CH3COO

CH3

(CH3)2CH

CH2CH3

nucleophile

b.

C

+

OH

H

CH3

CH3CH2

H

OOCCH3

+

OOCCH3

CH3CH2

+

Cl–

CH3

diastereomers leaving group

7.30 Carbocations are classified by the number of R groups bonded to the carbon: 0 R groups = methyl, 1 R group = 1°, 2 R groups = 2°, and 3 R groups = 3°. a.

+ b. (CH3)3CCH2

+

1 R group 1° carbocation

2 R groups 2° carbocation

c.

+

d. + 2 R groups 2° carbocation

3 R groups 3° carbocation

7.31 For carbocations: Increasing number of R groups = Increasing stability. +

+

CH3CH2CH2CH2

+

CH3CHCH2CH3

CH3 C CH3 CH3

1° carbocation least stable

2° carbocation intermediate stability

3° carbocation most stable

7.32 For carbocations: Increasing number of R groups = Increasing stability. + a.

+

+

(CH3)2CHCH2CH2

(CH3)2CHCHCH3

1° carbocation least stable

2° carbocation intermediate stability

CH2

+ (CH3)2CCH2CH3

3° carbocation most stable

+

b. + 1° carbocation least stable

2° carbocation intermediate stability

3° carbocation most stable

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Alkyl Halides and Nucleophilic Substitution 7–13 7.33 The rate of an SN1 reaction increases with increasing alkyl substitution. a.

or

(CH3)3CBr

(CH3)3CCH2Br

CH3

1° alkyl halide 3° alkyl halide faster SN1 reaction slower SN1 reaction b.

Br

or

2° alkyl halide slower SN1 reaction

Br

Br

c.

or

3° alkyl halide faster SN1 reaction

Br

2° alkyl halide slower SN1 reaction

3° alkyl halide faster SN1 reaction

7.34 • For methyl and 1° alkyl halides, only SN2 will occur. • For 2° alkyl halides, SN1 and SN2 will occur. • For 3° alkyl halides, only SN1 will occur. CH3 H

a. CH3 C

C

Br

b.

Br

CH3 CH3

Br

d.

2° alkyl halide SN1 and SN2

1° alkyl halide SN 2

2° alkyl halide SN1 and SN2

Br

c.

3° alkyl halide SN1

7.35 • Draw the product of nucleophilic substitution for each reaction. • For methyl and 1° alkyl halides, only SN2 will occur. • For 2° alkyl halides, SN1 and SN2 will occur and other factors determine which mechanism operates. • For 3° alkyl halides, only SN1 will occur. Strong nucleophile favors SN2. I

CH3OH Cl

a.

OCH3

c.

+ HCl

3° alkyl halide only SN1

Br

b.

OCH2CH3

CH3CH3O

+

I

+

HBr

2° alkyl halide Both SN1 and SN2 are possible. Weak nucleophile favors SN1. SH

SH

+ Br

CH3OH

d. Br

1° alkyl halide only SN2

2° alkyl halide Both SN1 and SN2 are possible.

OCH3

7.36 First decide whether the reaction will proceed via an SN1 or SN2 mechanism. Then draw the products with stereochemistry. +

a. H Br

2° alkyl halide SN1 and SN2

H2O

Weak nucleophile favors SN1.

+ HBr

+ H OH

HO H

enantiomers

SN1 = racemization at the stereogenic C

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Chapter 7–14 Cl

HC C

C C H

+

b.

+

Cl

SN2 = inversion at the stereogenic C

D H

H D

1° alkyl halide SN2 only

7.37 Compounds with better leaving groups react faster. Weaker bases are better leaving groups. a. CH3CH2CH2Cl

c. (CH3)3C OH or

CH3CH2CH2I

or

weaker base better leaving group b. (CH3)3CBr

d. CH3CH2CH2OH

or

weaker base better leaving group

7.38

+

OH2

weaker base better leaving group

(CH3)3CI

or

(CH3)3C

CH3CH2CH2 OCOCH3

weaker base better leaving group

• Polar protic solvents favor the SN1 mechanism by solvating the intermediate carbocation and halide. • Polar aprotic solvents favor the SN2 mechanism by making the nucleophile stronger. b. CH3CN polar aprotic solvent no O–H or N–H bond favors SN2

a. CH3CH2OH polar protic solvent contains an O–H bond favors SN1

c. CH3COOH polar protic solvent contains an O–H bond favors SN1

d. CH3CH2OCH2CH3 polar aprotic solvent no O–H or N–H bond favors SN2

7.39 Compare the solvents in the reactions below. For the solvent to increase the reaction rate of an SN1 reaction, the solvent must be polar protic. a.

(CH3)3CBr

3o RX

+

H2O

H2O

or

H2O (CH3)3COH

+

HBr

– SN1 reaction (CH3)2C=O + CH3OH

b. Cl

CH3OH

+

or DMSO

HCl

Br

+

OH

1o RX – SN2 reaction

d.

+

H Cl

CH3O

H2O

OH

+

Br

or

CH3OH

+

or

2o RX strong nucleophile SN2 reaction

DMF [HCON(CH3)2]

Polar aprotic solvent increases the rate of an SN2 reaction.

DMF

HMPA

CH3OH

Polar protic solvent increases the rate of an SN1 reaction.

OCH3

3o RX – SN1 reaction

c.

Polar protic solvent increases the rate of an SN1 reaction.

H OCH3

Cl

HMPA [(CH3)2N]3P=O

Polar aprotic solvent increases the rate of an SN2 reaction.

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Alkyl Halides and Nucleophilic Substitution 7–15 7.40 To predict whether the reaction follows an SN1 or SN2 mechanism: [1] Classify RX as a methyl, 1°, 2°, or 3° halide. (Methyl, 1° = SN2; 3° = SN1; 2° = either.) [2] Classify the nucleophile as strong or weak. (Strong favors SN2; weak favors SN1.) [3] Classify the solvent as polar protic or polar aprotic. (Polar protic favors SN1; polar aprotic favors SN2.) a.

+

CH2Br

CH3CH2O

+

CH2OCH2CH3

Br

SN2 reaction

1° alkyl halide SN2

b.

+

Br

+

Br

Strong nucleophile favors SN2.

2° alkyl halide SN1 or SN2 I

c.

N3

N3

3° alkyl halide SN1

OCH3

CH3OH

+

SN2 reaction = inversion at the stereogenic center The leaving group was "up." The nucleophile attacks from below.

+

SN1 reaction

HI

Weak nucleophile favors SN1.

d.

+

+

H2O

Weak nucleophile favors SN1.

Cl

3° alkyl halide SN1

OH

+ HCl

SN1 reaction forms two enantiomers.

HO

7.41 Vinyl carbocations are even less stable than 1° carbocations. CH3CH2CH2CH2CH=CH

CH3CH2CH2CH2CH2CH2

CH3CH2CH2CH2CHCH3

vinyl carbocation least stable

1° carbocation intermediate stability

2° carbocation most stable

7.42 Na+
CN

a.

CN

carbon framework

CN

nucleophile

b. (CH3)3CCH2CH2 carbon framework

Cl

SH

(CH3)3CCH2CH2

nucleophile

Cl

Na+
SH

(CH3)3CCH2CH2

SH

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Chapter 7–16 OH

Cl

OH

Na+
OH

c. nucleophile carbon framework

Na+
C

d.

CH3CH2 C C H

carbon framework

CH CH3CH2 C C H

CH3CH2 Cl

nucleophile

7.43 Cl CH2CH3

CH3O

CH3OCH2CH3

Cl CH3

CH3CH2O

CH3OCH2CH3

7.44 Use the directions from Answer 7.4 to name the compounds. [1] a.

[2]

CH3 CH3 C CH2CH2 F

[3] 1-fluoro-3,3-dimethylbutane

1

CH3 C CH2CH2 F

3 CH3

CH3

1-fluoro 3,3-dimethyl

4 carbon alkane = butane

b.[1]

CH3

[2]

3-ethyl 3

I

2

2-methyl [2]

3-ethyl-1-iodo-2-methylhexane

1-iodo

6 carbon alkane = hexane c. [1] (CH3)3CCH2Br

[3]

1 I

CH3

[3] 1-bromo-2,2-dimethylpropane

CH3 C CH2 Br

CH3

2 CH3 1

CH3 C CH2 Br

1-bromo 2,2-dimethyl

CH3

3 carbon alkane = propane d. [1]

[2] Br

6

2 Br

Cl

8 carbon alkane = octane

6-methyl

Br

[2]

2

I

I

5 carbon cycloalkane = cyclopentane

f.

[1]

Cl

1-bromo [3] cis-1-bromo-3-iodocyclopentane 3-iodo

1 Cl

[2]

trans-1,2-dichloro

Cl

6 carbon cycloalkane = cyclohexane

[3] 6-bromo-2-chloro-6-methyloctane

Cl 6-bromo 2-chloro

Br

e. [1]

1

Cl

2

[3] trans-1,2-dichlorocyclohexane

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Alkyl Halides and Nucleophilic Substitution 7–17 g. [1] (CH3)3CCH2CH(Cl)CH2Cl CH3

CH3

[2]

CH3

CH3 C CH2 C CH2Cl CH3

H

[3] 1,2-dichloro-4,4-dimethylpentane

CH3 C CH2 C CH2Cl

H Cl

Cl

4,4-dimethyl

1,2-dichloro

5 carbon alkane = pentane h.

[1]

[2]

4

I H

2 1

[3] (2R)-2-iodo-4,4-dimethylhexane

I H

6 carbon alkane = hexane (Indicate the R/S designation also)

4,4-dimethyl

2

3

(2R)-2-iodo

I H4

Clockwise R

1

7.45 To work backwards to a structure, use the directions in Answer 7.5. e. 1-bromo-4-ethyl-3-fluorooctane

a. isopropyl bromide Br CH3 CHCH3

Bromine on middle C makes it an isopropyl group.

b. 3-bromo-4-ethylheptane

3

4-ethyl 4 3-bromo

Br

c. 1,1-dichloro-2-methylcyclohexane 1 Cl

1,1-dichloro

Cl

2

2-methyl

d. trans-1-chloro-3-iodocyclobutane 1-chloro 3 Cl 1 I

3-iodo

4-ethyl Br

1 1-bromo

3 F

4 3-fluoro

f. (3S)-3-iodo-2-methylnonane 2-methyl 4 3 1 I H 3S 3-iodo g. (1R,2R)-trans-1-bromo-2-chlorocyclohexane 1R 1-bromo Br 1 Cl 2-chloro 2R h. (5R)-4,4,5-trichloro-3,3-dimethyldecane 5R 3,3-dimethyl Cl H 1 4 5 3 Cl Cl

4,4,5-trichloro

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Chapter 7–18 7.46 H H

CH3

a.

CH3 C CH2CH2F CH3

g.

d. Br

1° halide

I

h. I H

e.

2° halide

1° halide 2° halide

I

c.

1° halide

2° halide

2° halide

2° halide

Br

b.

C C Cl Cl H

Cl

3° halide

(CH3)2CCH2

(CH3)3CCH2Br Cl

1° halide

f. Cl

Both are 2° halides.

7.47 1-chloro

3-chloro

1

Cl 1

3 3-chloropentane

Cl

1-chloropentane 1-chloro

Cl

Cl

1 1-chloro-2,2-dimethylpropane

2 2-chloro-2-methylbutane

Two stereoisomers

2-chloro

2

2 * Cl

2-chloropentane [* denotes stereogenic center]

3

Cl H 4 1 Clockwise "4" in back = R

2

3

4 1 Cl H Clockwise "4" in front = S

Two stereoisomers 3-methyl 3

2 3 *

2-chloro Cl 2-chloro-3-methylbutane [* denotes stereogenic center]

4H

2

2

3 4H

Cl

1 Counterclockwise "4" in front = R

1Cl Counterclockwise "4" in back = S

Two stereoisomers 1-chloro 2-methyl

*

3

4 Cl

Cl

1-chloro-2-methylbutane [* denotes stereogenic center]

3

H

2

1

Clockwise "4" in front = S

3

1-chloro 1-chloro-3-methylbutane

2-chloro 2-methyl

2

3-methyl

Cl

4

H

Cl

2 1 Clockwise "4" in back = R

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Alkyl Halides and Nucleophilic Substitution 7–19 7.48 Use the directions from Answer 7.6. Br a. (CH3)3CBr

or

CH3CH2CH2CH2Br

b.

larger surface area = stronger intermolecular forces = higher boiling point

I

or

c.

Br

or

more polar = nonpolar larger halide = more polarizable = only VDW forces higher boiling point higher boiling point

7.49 a.

CH3CH2CH2CH2 Br

b.

CH3CH2CH2CH2 Br

SH

c.

CH3CH2CH2CH2 Br

CN

d.

CH3CH2CH2CH2 Br

OCH(CH3)2

e.

CH3CH2CH2CH2 Br

C CH

f.

CH3CH2CH2CH2 Br

g.

CH3CH2CH2CH2 Br

NH3

h.

CH3CH2CH2CH2 Br

Na+ I


i.

CH3CH2CH2CH2 Br

Na+ N3


CH3CH2CH2CH2OH + Br

OH

H2O

CH3CH2CH2CH2SH + Br CH3CH2CH2CH2CN + Br


CH3CH2CH2CH2OCH(CH3)2 + Br


CH3CH2CH2CH2C CH + Br
CH3CH2CH2CH2OH2 + Br


CH3CH2CH2CH2NH3 + Br


CH3CH2CH2CH2OH + HBr

CH3CH2CH2CH2NH2 + HBr

CH3CH2CH2CH2I + Na+ Br
CH3CH2CH2CH2N3 + Na+ Br


7.50 Use the steps from Answer 7.9 and then draw the proton transfer reaction, when necessary. O

a.

+ Cl

CH3

C

CH3

nucleophile leaving group I

b.

leaving group

+

I

c.

leaving group

d.

Cl

+

Cl

CN

+ NaI

O O

Na+ –CN

nucleophile

H2O

OH + HI

nucleophile

+ CH3CH2OH

leaving group

+ O

nucleophile

OCH2CH3

+ HCl

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Chapter 7–20 Br

OCH3

+ Na+ –OCH3

e.

nucleophile

leaving group Cl

f.

CH3

+

leaving group

+ NaBr

SCH3 + Cl

CH3SCH3

nucleophile

7.51 A good leaving group is a weak base. OH

c.

a. bad leaving group OH is a strong base.

b.

CH3CH2CH2CH2 Cl

e.

This has only C–C and C–H bonds. No good leaving group. OH2

d.

Cl good leaving group weak base

bad leaving group NH2 is a strong base.

f.

good leaving group H2O is a weak base.

CH3CH2NH2

CH3CH2CH2

I

I good leaving group weak base

7.52 Use the rules from Answer 7.12. a. increasing leaving group ability:
NH2 <
OH < F


c. increasing leaving group ability: Cl
< Br
< I


least basic best leaving group

most basic worst leaving group

most basic worst leaving group

b. increasing leaving group ability:
NH2 <
OH < H2O most basic worst leaving group

least basic best leaving group

d. increasing leaving group ability: NH3 < H2O < H2S most basic worst leaving group

least basic best leaving group

least basic best leaving group

7.53 Compare the nucleophile and the leaving group in each reaction. The reaction will occur if it proceeds towards the weaker base. Remember that the stronger the acid (lower pKa), the weaker the conjugate base. I

NH2

+ I

a.

weaker base pKa (HI) = –10 b.

CH3CH2I

+ CH3O

stronger base pKa (CH3OH) = 15.5

+

NH2

Reaction will not occur.

stronger base pKa (NH3) = 38 CH3CH2OCH3

+ I

weaker base pKa (HI) = –10

Reaction will occur.

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Alkyl Halides and Nucleophilic Substitution 7–21 c.

OH

F

+ F

weaker base pKa (HF) = 3.2 CN

d.

+

Reaction will not occur.

OH

stronger base pKa (H2O) = 15.7 I

+ I

+

CN

Reaction will not occur.

stronger base pKa (HCN) = 9.1

weaker base pKa (HI) = –10

7.54 Br

A SCH3 replaces Br.
SCH is needed. 3

SCH3 a.

b.

OCH(CH3)2 OCH(CH3)2 replaces Br.
OCH(CH ) 3 2

needed.

C

c.

C

CH3

N(CH3)3

d.

Br

C CCH3 replaces Br. C CCH3 is needed. N(CH3)3 replaces Br. N(CH3)3 is needed.

7.55 Use the directions in Answer 7.16. a. Across a row of the periodic table nucleophilicity decreases.
OH <
NH

2<

d.

CH3


b. • In a polar protic solvent (CH3OH), nucleophilicity increases down a column of the periodic table, so:
SH is more nucleophilic than
OH. • Negatively charged species are more nucleophilic than neutral species so
OH is more nucleophilic than H2O. H2O <
OH <
SH

c. • In a polar protic solvent (CH3OH), nucleophilicity increases down a column of the periodic table, so: CH3CH2S
is more nucleophilic than CH3CH2O
. • For two species with the same attacking atom, the more basic is the more nucleophilic so CH3CH2O
is more nucleophilic than CH3COO
. CH3COO
< CH3CH2O
< CH3CH2S


Compare the nucleophilicity of N, S, and O. In a polar aprotic solvent (acetone), nucleophilicity parallels basicity. CH3SH < CH3OH < CH3NH2

e.

In a polar aprotic solvent (acetone), nucleophilicity parallels basicity. Across a row and down a column of the periodic table nucleophilicity decreases. Cl
< F
<
OH

f.

Nucleophilicity decreases across a row so
SH is more nucleophilic than Cl
. In a polar protic solvent (CH3OH), nucleophilicity increases down a column so Cl
is more nucleophilic than F
. F
< Cl
<
SH

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Chapter 7–22 7.56 Polar protic solvents are capable of hydrogen bonding, and therefore must contain a H bonded to an electronegative O or N. Polar aprotic solvents are incapable of hydrogen bonding, and therefore do not contain any O–H or N–H bonds. a.

c.

(CH3)2CHOH

e.

no O–H or N–H bond aprotic

contains O–H bond protic b.

CH2Cl2

d.

CH3NO2

no O–H or N–H bond aprotic f. HCONH2 contains an N–H bond protic

NH3 contains N–H bond protic

no O–H or N–H bond aprotic

N(CH3)3

7.57 O CH3

C

O N

N

H

H

CH3

CH3

The amine N is more nucleophilic since the electron pair is localized on the N.

C

O N

N

H

H

CH3

CH3

C

N

N

H

H

The amide N is less nucleophilic since the electron pair is delocalized by resonance.

7.58 1° alkyl halide SN2 reaction Br

a. Mechanism:

+

CN

CN

Energy

b. Energy diagram:

c. Transition state:

Ea Br +

+ Br

acetone

CN

Br 



H° CN

+ Br


CN

Reaction coordinate

d. Rate equation: one step reaction with both nucleophile and alkyl halide in the only step: rate = k[R–Br][–CN] e. [1] The leaving group is changed from Br
to I
: Leaving group becomes less basic  a better leaving group  faster reaction. [2] The solvent is changed from acetone to CH3CH2OH: Solvent changed to polar protic  decreases reaction rate. [3] The alkyl halide is changed from CH3(CH2)4Br to CH3CH2CH2CH(Br)CH3: Changed from 1° to 2° alkyl halide  the alkyl halide gets more crowded and the reaction rate decreases. [4] The concentration of
CN is increased by a factor of 5. Reaction rate will increase by a factor of 5. [5] The concentration of both the alkyl halide and
CN are increased by a factor of 5: Reaction rate will increase by a factor of 25 (5 x 5 = 25).

CH3

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Alkyl Halides and Nucleophilic Substitution 7–23 7.59 Use the directions for Answer 7.25. Br

Br

Br

a. 2° alkyl halide intermediate reactivity

3° alkyl halide least reactive

1° alkyl halide most reactive

Br

Br

b.

Br

3° alkyl halide least reactive

c.

2° alkyl halide intermediate reactivity

1° alkyl halide most reactive

Br

Br Br

2° alkyl halide intermediate reactivity

vinyl halide least reactive

1° alkyl halide most reactive

7.60 better leaving group

a. CH3CH2Br CH3CH2Cl

+

OH

+

OH

faster reaction

stronger nucleophile

b.

Br

+

OH

Br

+

H2O

faster reaction

stronger nucleophile

c.

d.

Cl

+

NaOH

Cl

+

NaOCOCH3

I

+

OCH3

+

OCH3

I

faster reaction

CH3OH DMSO

faster reaction

polar aprotic solvent less steric hindrance

e.

Br

+

OCH2CH3

Br

+

OCH2CH3

faster reaction

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Chapter 7–24 7.61 All SN2 reactions proceed with backside attack of the nucleophile. When nucleophilic attack occurs at a stereogenic center, inversion of configuration occurs. CH3

CH3 a.

C H

+

Cl

* D

I

b.

+

Cl

*

c.

C

OCH3

+

H

OH

OH

*

*

+ I OCH2CH3

*

OCH2CH3

H

d.

inversion of configuration

+ Cl

D

* OCH3

Br

+

CN

CN

* *

+ Cl

No bond to the stereogenic center is broken, since the leaving group is not bonded to the stereogenic center.

H

inversion of configuration

+ Br

[* denotes a stereogenic center]

7.62 Follow the definitions from Answer 7.30. a.

CH3CH2CHCH2CH3

c.

2° carbocation

e.

(CH3)2CHCH2CH2

1° carbocation

2° carbocation

CH2CH3 b.

d.

f.

3° carbocation

CH2

3° carbocation

1° carbocation

7.63 For carbocations: Increasing number of R groups = Increasing stability.

a.

CH2

b. 1° carbocation least stable

2° carbocation intermediate stablity

3° carbocation most stable

1° carbocation least stable

2° carbocation intermediate stablity

7.64 


Cl H

H H


Cl  C C

H C C


Cl H

H H

+

3 Cl groups – electron-withdrawing destabilizing least stable

methyl group without added Cl's more stable

H

H

H C O C H

H

H

H

H C O C H

resonance stabilized most stable

H

3° carbocation most stable

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183

Alkyl Halides and Nucleophilic Substitution 7–25 7.65 Step [1]

CH3

a. Mechanism: SN1 only

CH3 C CH2CH3 I

+ H2O

CH3 CH3

A

CH3

Step [2] C CH2CH3 + H2O

B

CH3 C CH2CH3 OH2

+ I

C

Energy

b. Energy diagram:

Ea1

B
H°1

Ea2
H°2
H°overall = 0

C CH3

A CH3

CH3 C CH2CH3

CH3 C CH2CH3

OH2

I

Reaction coordinate c. Transition states:

CH3 CH3


+

CH3
+ C CH2CH3 CH3 OH2
+

C CH2CH3 I
–

d. rate equation: rate = k[(CH3)2CICH2CH3] e. [1] Leaving group changed from I
to Cl
: rate decreases since I
is a better leaving group. [2] Solvent changed from H2O (polar protic) to DMF (polar aprotic): rate decreases since polar protic solvent favors SN1. [3] Alkyl halide changed from 3° to 2°: rate decreases since 2° carbocations are less stable. [4] [H2O] increased by factor of five: no change in rate since H2O is not in rate equation. [5] [R–X] and [H2O] increased by factor of five: rate increases by a factor of five. (Only the concentration of R–X affects the rate.)

7.66 The rate of an SN1 reaction increases with increasing alkyl substitution. Br

a.

Br

Br

1° alkyl halide least reactive

3° alkyl halide most reactive

2° alkyl halide intermediate reactivity b.

Br

Br Br

1° alkyl halide least reactive Br

2° alkyl halide intermediate reactivity

3° alkyl halide most reactive

Br Br

c. aryl halide least reactive

2° alkyl halide intermediate reactivity

3° alkyl halide most reactive

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Chapter 7–26 7.67 The rate of an SN1 reaction increases with increasing alkyl substitution, polar protic solvents, and better leaving groups. a.

(CH3)3CCl

+ H2O

(CH3)3CI

+ H2O

b.

Br

Cl c.

better leaving group faster reaction

Br + CH OH 3

Cl

2° halide faster SN1 reaction

+ H2O

3° halide d. faster SN1 reaction

I

1° halide slower SN1 reaction

I

CH3OH

+

aryl halide slower reaction

+ H O 2

+ CH3CH2OH

CH3CH2OH

+ CH3CH2OH

DMSO

polar protic solvent faster reaction polar aprotic solvent slower reaction

7.68 CH3CH2 a.

Br

CH3

C

CH3 b.

+ H2O

HO

Cl

CH3

+ CH3OH

CH3CH2

CH3

+

C

CH3

CH3 C

CH3CH2

CH3CH2

CH2CH3 C

OH

CH3 C

+ HCl OCH3

Br

OCH2CH3

c.

+

+ CH3CH2OH Br

d.

+ HBr

CH3CH2O

+ HBr

OH +

H2 O

+

OH + HBr

7.69 The 1o alkyl halide is also allylic, so it forms a resonance-stabilized carbocation. Increasing the stability of the carbocation by resonance, increases the rate of the SN1 reaction. CH3OH CH3CH2CH2CH CHCH2Br

CH3CH2CH2CH CHCH2OCH3

CH3CH2CH2CHCH CH2

HBr

OCH3 CH3CH2CH2CH CHCH2

CH3CH2CH2CH CH CH2

Br

CH3CH2CH2CH CH CH2

Br

resonance-stabilized carbocation

Use each resonance structure individually to continue the mechanism: CH3CH2CH2CH CH CH2

CH3OH

CH3CH2CH2CH CHCH2 H OCH3

CH3CH2CH2CH CHCH2OCH3

Br CH3CH2CH2CH CH CH2

CH3CH2CH2CH CH CH2 H OCH3

CH3OH

Br

CH3CH2CH2CHCH CH2 OCH3

HBr

HBr

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Alkyl Halides and Nucleophilic Substitution 7–27 7.70 H

H Br

a.

+

CN + Br

CN acetone

1° alkyl halide SN2 only b.

+

OCH3

DMSO

+

Br H

strong nucleophile polar aprotic solvent Both favor SN2.

2° alkyl halide SN1 and SN2 Br

c.

Br

H OCH3

reaction at a stereogenic center inversion of configuration

OCH3

+ CH3OH

+ HBr CH2CH2CH3

CH2CH2CH3

3° alkyl halide SN1 only CH2CH3 d.

C H

CH2CH3 C

+ CH3COOH

I CH3

H

CH2CH3 C CH3COO CH3

OOCCH3 CH3

H

2° alkyl halide Weak nucleophile favors SN1. SN1 and SN2 e.

+

OCH2CH3

Br

2° alkyl halide SN1 and SN2

f.

OCH2CH3

DMF

+ Br

HI

reaction at a stereogenic center racemization of product

reaction at a stereogenic center inversion of configuration

strong nucleophile polar aprotic solvent Both favor SN2. OCH2CH3

OCH2CH3

Cl

+ CH3CH2OH

+

+ HCl

two products – diastereomers Nucleophile attacks from above and below.

Weak nucleophile favors SN1.

2° alkyl halide SN1 and SN2

7.71 H

CN (axial) CN

Br

a. (CH3)3C

(eq) acetone

H

(CH3)3C

H

inversion (equatorial to axial)

H

polar aprotic solvent SN2 reaction Large tert-butyl group in more roomy equatorial position. Br (axial) H

b. (CH3)3C

CN

acetone

H (CH3)3C

H

H

polar aprotic solvent SN2 reaction

inversion (axial to equatorial) CN (eq)

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Chapter 7–28 7.72 Br [2]

C

[1] Na+H (CH3)2NCH2CH2O H

OCH2CH2N(CH3)2

H Na+

(CH3)2NCH2CH2O

C

+ NaBr

+ H2 H

diphenhydramine

7.73 First decide whether the reaction will proceed via an SN1 or SN2 mechanism (Answer 7.40), and then draw the mechanism.

H

Br

OCH2CH3 Br

OCH2CH3 H

CH3CH2OH

can attack from above or below

3° alkyl halide SN1 only

+

Br OCH2CH3 OCH2CH3

+ HBr

+

7.74 nucleophile O

O

C O H

O

C O

C

OH

+ Br O

Br

Br

C7H10O2 leaving group

7.75 Br

H H N CH3

Br– N

Br

H N

CO32–

CH3

N H CH3

N

CO32–

N

+ HCO3–

Br–

N CH3 N

nicotine + NaHCO3

+ NaBr

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Alkyl Halides and Nucleophilic Substitution 7–29 7.76 a. Hexane is nonpolar and therefore few nucleophiles will dissolve in it. b. (CH3)3CO– is a stronger base than CH3CH2O–: The three electron-donating CH3 groups add electron density to the negative charge of the conjugate base, destabilizing it and making it a stronger base. c. By the Hammond postulate, the SN1 reaction is faster with RX that form more stable carbocations. (CH3)3C

(CH3)2C CF3

3° Carbocation is stabilized by three electron-donor CH3 groups.

Although this carbocation is also 3°, the three electron-withdrawing F atoms destabilize the positive charge. Since the carbocation is less stable, the reaction to form it is slower.

d. The identity of the nucleophile does not affect the rate of SN1 reactions since the nucleophile does not appear in the rate-determining step. Polar aprotic solvent favors SN2 reaction.

e. 2° alkyl halide SN1 or SN2

H

Br

H

Br

acetone

Br

Br

S

(2R)-2-bromobutane optically active Strong nucleophile favors SN2 reaction.

This compound reacts with Br
until a 50:50 mixture results, making the mixture optically inactive. Then either compound can react with Br
and the mixture remains optically inactive.

7.77 CH3NH2 Cl

N

Cl

+

CH3NH2

N

Cl

+

N

Cl

N CH3

+

H N H CH3

CH3NH3

N

Cl

N

H

+

CH3NH3

CH3

N

N

H CH3

CH3NH2

+

Cl

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Chapter 7–30 7.78 In the first reaction, substitution occurs at the stereogenic center. Since an achiral, planar carbocation is formed, the nucleophile can attack from either side, thus generating a racemic mixture. 3° alkyl halide CH3OH

OCH3

two steps

SN1

Br

(6R)-6-bromo-2,6-dimethylnonane

Br OCH3

achiral, planar carbocation

racemic mixture optically inactive

In the second reaction, the starting material contains a stereogenic center, but the nucleophile does not attack at that carbon. Since a bond to the stereogenic center is not broken, the configuration is retained and a chiral product is formed. 3° alkyl halide Br

two steps

CH3OH

SN1

OCH3

configuration retained

Br

optically active

Reaction does not occur at the stereogenic center.

(5R)-2-bromo-2,5-dimethylnonane

7.79 The nucleophile has replaced the leaving group. Missing reagent:

a.

O

I

O C CH

Cl

b.

The nucleophile has replaced the leaving group. Missing reagent: C CH

N3

c.

N3

SH SH

d.

The nucleophile has replaced the halide. Starting material: Cl

The nucleophile has replaced the halide. Starting material: Cl The leaving group must have the opposite orientation to the position of the nucleophile in the product.

7.80 To devise a synthesis, look for the carbon framework and the functional group in the product. The carbon framework is from the alkyl halide and the functional group is from the nucleophile. SH

a. carbon framework

functional group

Cl

Na

SH

SH

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189

Alkyl Halides and Nucleophilic Substitution 7–31 Na O

b.

O

Cl

O

functional carbon group framework Na

c.

CH3CH2CN

CN

CH3CH2Cl

CH3CH2CN

functional group

carbon framework

O

Cl

d. functional group carbon framework

Na

O

O

2o halide O Na

or

O

Cl

O

1o halide

This path is preferred. The strong nucleophile favors an SN2 reaction so an unhindered 1o alkyl halide reacts faster.

carbon functional framework group e.

CH3CH2 OCOCH3

carbon framework

Na

CH3CH2Cl

functional group

OCOCH3

CH3CH2OCOCH3

7.81 Work backwards to determine the alkyl chloride needed to prepare benzalkonium chloride A. CH3

a.

CH3(CH2)17Cl

CH2N(CH3)2

CH2

N

(CH2)17CH3

Cl–

CH3

B

b.

A CH3

CH3(CH2)17N(CH3)2

CH2Cl

CH2

N

(CH2)17CH3

CH3

A

C

7.82 I

B very crowded 3° halide O Na

D

OCH3

Na+ OCH3

C

E

CH3I

OCH3

A E unhindered methyl halide

preferred method The strong nucleophile favors SN2 reaction so the alkyl halide should be unhindered for a faster reaction.

Cl–

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Chapter 7–32 7.83 H C C H

CH3(CH2)7CH2Br

NaH

H C C

CH3(CH2)7CH2C CH

A

B

NaH

CH3(CH2)7CH2C C

+ H2

C

+ H2

CH3(CH2)11CH2Br H

H

addition of H2

C C CH3(CH2)7CH2

CH3(CH2)7CH2C CCH2(CH2)11CH3

(1 equiv)

CH2(CH2)11CH3

D

muscalure

7.84 O

a.

H

[1] Na+H

[2] CH3–Br O

O

CH3

+ H2

(Chapter 9)

b.

CH3CH2CH2 C C H

[1] Na+ NH2

[2] CH3CH2–Br CH3CH2CH2 C C

CH3CH2CH2 C C CH2CH3

(Chapter 11)

c. H CH(CO2CH2CH3)2

+ Na+ Br–

[1] Na+ –OCH2CH3

CH(CO2CH2CH3)2

[2] C6H5CH2–Br

(Chapter 23)

+ Na+ Br– + NH3

C6H5CH2 CH(CO2CH2CH3)2

+ Na+ Br– + CH3CH2OH

7.85 quinuclidine The three alkyl groups are "tied back" in a ring, making the electron pair more available.

N

triethylamine

CH3CH2

This electron pair is more hindered by the three CH2CH3 groups. N These bulky groups around the N CH2CH3 cause steric hindrance and this CH2CH3 decreases nucleophilicity.

This electron pair on quinuclidine is much more available than the one on triethylamine. less steric hindrance more nucleophilic

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Alkyl Halides and Nucleophilic Substitution 7–33 7.86 O

O H

O

[1]

H

+

H2

O

O CH3 Br

CH3

[2]

minor product O

+ NaBr

O CH3 Br

CH3

[2]

major product

7.87 Br

Br

a. OH

base

(CH3)3C

b.

O (CH3)3C

OH

O

intramolecular SN2

(CH3)3C

O

O

base (CH3)3C

c. (CH3)3C

(CH3)3C

Br Br

Br

(CH3)3C

Br

base OH

O (CH3)3C

O

3° alkyl halide harder reaction OH

d. Br (CH3)3C

intramolecular SN2

intramolecular SN2

(CH3)3C

O O

base

Br (CH3)3C

3° alkyl halide harder reaction

intramolecular SN2

(CH3)3C

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Chapter 7–34 7.88 Cl bonded to sp2 C cannot undergo SN1.

Cl O

Cl

Cl O

Cl

CH3OH

Cl bonded to sp3 C no resonance stabilization possible for the carbocation formed here

Cl

Cl bonded to sp3 C Resonance-stabilized carbocation forms. best for SN1 Cl

Cl

Cl

Cl

Cl

+

(1 equiv)

O

J

Cl

O

re-draw Cl

Cl O

Cl

Cl

+

CH3OH

O

O CH3

Cl

H

Cl

Cl

+ O

K

OCH3

HCl

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Text

Study Guide/Solutions Manual to accompany Organic Chemistry, Third Edition

193

Alkyl Halides and Elimination Reactions 8–1 C Chhaapptteerr 88:: A Allkkyyll H Haalliiddeess aanndd EElliim miinnaattiioonn R Reeaaccttiioonnss  A A ccoom mppaarriissoonn bbeettw weeeenn nnuucclleeoopphhiilliicc ssuubbssttiittuuttiioonn aanndd --eelliim miinnaattiioonn Nucleophilic substitutionA nucleophile attacks a carbon atom (7.6). Nu

H

H Nu +

C C

C C X

substitution product

X

good leaving group

-EliminationA base attacks a proton (8.1). B

H C C

C C

+ H B+ +

X

elimination product

• •

Similarities In both reactions RX acts as an electrophile, reacting with an electron-rich reagent. Both reactions require a good leaving group X:– willing to accept the electron density in the C–X bond.

X

good leaving group

• •

Differences In substitution, a nucleophile attacks a single carbon atom. In elimination, a Brønsted–Lowry base removes a proton to form a
bond, and two carbons are involved in the reaction.

  TThhee iim mppoorrttaannccee ooff tthhee bbaassee iinn EE22 aanndd EE11 rreeaaccttiioonnss ((88..99)) The strength of the base determines the mechanism of elimination. • Strong bases favor E2 reactions. • Weak bases favor E1 reactions. strong base OH CH3

E2

CH3 C CH2

+

H2O

+ Br

CH3

same product different mechanism

CH3 C CH3 Br H2O

weak base

E1

CH3 C CH2 CH3

+

H3O

+

+ Br

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Chapter 8–2   EE11 aanndd EE22 m meecchhaanniissm mss ccoom mppaarreedd E2 mechanism one step (8.4B)

•

E1 mechanism two steps (8.6B)

[1] Mechanism

•

[2] Alkyl halide

•

rate: R3CX > R2CHX > RCH2X (8.4C)

•

rate: R3CX > R2CHX > RCH2X (8.6C)

[3] Rate equation [4] Stereochemistry

• • •

• • •

[5] Base

•

rate = k[RX][B:] second-order kinetics (8.4A) anti periplanar arrangement of H and X (8.8) favored by strong bases (8.4B)

rate = k[RX] first-order kinetics (8.6A) trigonal planar carbocation intermediate (8.6B) favored by weak bases (8.6C)

[6] Leaving group

•

•

[7] Solvent

•

[8] Product

•

better leaving group
faster reaction (8.4B) favored by polar aprotic solvents (8.4B) more substituted alkene favored (Zaitsev rule, 8.5)

•

• •

better leaving group
faster reaction (Table 8.4) favored by polar protic solvents (Table 8.4) more substituted alkene favored (Zaitsev rule, 8.6C)

  SSuum mm maarryy cchhaarrtt oonn tthhee ffoouurr m meecchhaanniissm mss:: SSNNN11,, SSNNN22,, EE11,, oorr EE22 Alkyl halide type 1o RCH2X 2o R2CHX 3o R3CX

Conditions strong nucleophile strong bulky base strong base and nucleophile strong bulky base weak base and nucleophile weak base and nucleophile strong base

Mechanism SN2 E2 SN2 + E2 E2 SN1 + E1 SN1 + E1 E2

  ZZaaiittsseevv rruullee • -Elimination affords the more stable product having the more substituted double bond. • Zaitsev products predominate in E2 reactions except when a cyclohexane ring prevents trans diaxial arrangement.

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Alkyl Halides and Elimination Reactions 8–3 C Chhaapptteerr 88:: A Annssw weerrss ttoo PPrroobblleem mss 8.1 • The carbon bonded to the leaving group is the  carbon. Any carbon bonded to it is a  carbon. • To draw the products of an elimination reaction: Remove the leaving group from the  carbon and a H from the  carbon and form a
bond. a.

b.

  CH3CH2CH2CH2CH2 Cl

1



2

K+
OC(CH3)3

1

CH3CH2CH2CH CH2

(CH3CH2)2C

CH2

CH3CH=C(CH3)CH2CH3

Cl 1

c.

K+
OC(CH3)3

2 

Br

K+
OC(CH3)3

1

8.2 Alkenes are classified by the number of carbon atoms bonded to the double bond. A monosubstituted alkene has one carbon atom bonded to the double bond, a disubstituted alkene has two carbon atoms bonded to the double bond, etc. 4 C's bonded to C=C tetrasubstituted

2 C's bonded to each C=C disubstituted

OH

a.

b. vitamin D3

3 C's bonded to each C=C trisubstituted vitamin A

H CH2

3 C's bonded to each C=C trisubstituted 2 C's bonded to the C=C disubstituted

HO

8.3 To have stereoisomers at a C=C, the two groups on each end of the double bond must be different from each other. two different groups (CH3CH2 and H) a. two CH3 groups no stereoisomers possible

two different groups (H and CH3)

b. CH3CH2CH CHCH3 stereoisomers possible

two different groups two different groups (cyclohexyl and H) (cyclohexyl and H) c.

CH CH

stereoisomers possible

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Chapter 8–4 8.4 Two definitions: • Constitutional isomers differ in the connectivity of the atoms. • Stereoisomers differ only in the 3-D arrangement of atoms in space.

c.

and

C C H

and

d.

trans

H C C

and H

H

CH3

cis trans different arrangement of atoms in space stereoisomers

different connectivity of atoms constitutional isomers

b.

CH3CH2

CH3

CH3CH2

a.

trans

CH3CH2 CH3

identical

H

CH3CH2

C C

and

CH3

C C

H

H

H

different connectivity of atoms constitutional isomers

8.5 Two rules to predict the relative stability of alkenes: [1] Trans alkenes are generally more stable than cis alkenes. [2] The stability of an alkene increases as the number of R groups on the C=C increases. a.

or monosubstituted

b.

CH2CH3

CH3CH2 C C

CH3CH2

trisubstituted more stable

H

H

CH3

or

c.

C C

or H

H

CH3

disubstituted more stable

disubstituted

CH2CH3

trans more stable

cis

8.6 Use the rules from Answer 8.5 to explain the energy differences.

cis-2-butene

trans-2-butene

cis-2,2,5,5-tetramethyl-3-hexene

trans-2,2,5,5-tetramethyl-3-hexene

more steric interaction between larger tert-butyl groups in the cis isomer larger difference in stability

less steric interaction between smaller CH3 groups smaller energy difference

8.7

A

B

Alkene A is more stable than alkene B because the double bond in A is in a six-membered ring. The double bond in B is in a four-membered ring, which has considerable angle strain due to the small ring size.

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Alkyl Halides and Elimination Reactions 8–5 8.8 In an E2 mechanism, four bonds are involved in the single step. Use curved arrows to show these simultaneous actions: [1] The base attacks a hydrogen on a  carbon. [2] A
bond forms. [3] The leaving group comes off. OCH2CH3

CH3CH2 H CH3CH2 C

CHCH3

transition state: (CH3CH2)2C=CHCH3 +

Br

+

HOCH2CH3

CH3CH2 C

new
bond

 carbon

CH3CH2 H

Br



OCH2CH3

CHCH3

 Br

8.9 For E2 elimination to occur there must be at least one hydrogen on a  carbon.
carbon

CH3 H

CH3 C

C H

no H's on
carbon inert to E2 elimination

CH3 Br

8.10 In both cases, the rate of elimination decreases. stronger base faster reaction a. CH CH Br 3 2

+

OC(CH3)3

CH3CH2 Br

+

OH

better leaving group faster reaction b. CH3CH2 Br

+

OC(CH3)3

CH3CH2 Cl

+

OC(CH3)3

8.11 As the number of R groups on the carbon with the leaving group increases, the rate of an E2 reaction increases. a. (CH3)2CHCH2CH2CH2Br

(CH3)2CHCH2CH(Br)CH3

2° alkyl halide intermediate reactivity

1° alkyl halide least reactive

3° alkyl halide most reactive CH3

Cl

Cl

b. Cl

1° alkyl halide least reactive

(CH3)2C(Br)CH2CH2CH3

CH3

2° alkyl halide intermediate reactivity

3° alkyl halide most reactive

8.12 Use the following characteristics of an E2 reaction to answer the questions: [1] E2 reactions are second order and one step. [2] More substituted halides react faster. [3] Reactions with strong bases or better leaving groups are faster. [4] Reactions with polar aprotic solvents are faster.

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Chapter 8–6 Rate equation: rate = k[RX][Base] a. tripling the concentration of the alkyl halide = rate triples b. halving the concentration of the base = rate halved c. changing the solvent from CH3OH to DMSO = rate increases (Polar aprotic solvent is better for E2.) d. changing the leaving group from I to Br = rate decreases (I is a better leaving group.) e. changing the base from OH to H2O = rate decreases (weaker base) f. changing the alkyl halide from CH3CH2Br to (CH3)2CHBr = rate increases (More substituted halide reacts faster.) 8.13 The Zaitsev rule states: In a -elimination reaction, the major product has the more substituted double bond. CH3 H

a.




CH3 C

C

H

Br

CH2CH3

(CH3)2C

loss of H and Br

CH3

b.




CHCH2CH3

+

(CH3)2CHCH

trisubstituted major product

Br

CHCH3

disubstituted minor product

CH3

CH3

CH2

CH3

CH3

CH3

loss of H and Br CH3

trisubstituted minor product

tetrasubstituted major product

disubstituted minor product

Cl

c.

CH3

monosubstituted minor product

disubstituted major product

CH3


Cl

d.

CH3CH2CH2CH2CH=CHCH3

loss of H and Cl




trisubstituted

CH3

CH3 ONLY product

loss of H and Cl

CH3

CH3

8.14 An E1 mechanism has two steps: [1] The leaving group comes off, creating a carbocation. [2] A base pulls off a proton from a  carbon, and a
bond forms. CH3 CH3

C

CH2CH3

[1]

+ CH3OH

CH3 CH3

[2]

+

(CH3)2C=CHCH3 + CH3OH2 + Cl

H

Cl

transition state [1]:

C CH CH3 CH3OH + Cl

transition state [2]:

CH3 CH3 +C 

CH2CH3

Cl 


CH3 CH3 C

+

CH CH3 H

OCH3

+

H

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Alkyl Halides and Elimination Reactions 8–7 8.15 The Zaitsev rule states: In a -elimination reaction, the major product has the more substituted double bond. 2

CH3

CH3

1




C CHCH3 CH3CH2

CH3CH2 C CH2CH3 + H2O

a.

CH3CH2

 Cl 1

trisubstituted major product 1

CH2CH2CH3 CH3




b. 3

I

2

CH2CH2CH3 +

CH3

CH3OH

+

C CH2 CH3CH2

disubstituted

CH2CH2CH3 +

tetrasubstituted major product

CH2

CH2CH2CH3 CH3

+

disubstituted

trisubstituted

8.16 Use the following characteristics of an E1 reaction to answer the questions: [1] E1 reactions are first order and two steps. [2] More substituted halides react faster. [3] Weaker bases are preferred. [4] Reactions with better leaving groups are faster. [5] Reactions in polar protic solvents are faster. Rate equation: rate = k[RX]. The base doesn't affect rate. a. doubling the concentration of the alkyl halide = rate doubles b. doubling the concentration of the base = no change (Base is not in the rate equation.) c. changing the alkyl halide from (CH3)3CBr to CH3CH2CH2Br = rate decreases (More substituted halides react faster.) d. changing the leaving group from Cl to Br = rate increases (better leaving group) e. changing the solvent from DMSO to CH3OH = rate increases (Polar protic solvent favors E1.) 8.17 Both SN1 and E1 reactions occur by forming a carbocation. To draw the products: [1] For the SN1 reaction, substitute the nucleophile for the leaving group. [2] For the E1 reaction, remove a proton from a  carbon and create a new
bond. CH3

CH3

Br

a.

+ H2O leaving group nucleophile and base

SN1 product

CH3

b. CH3 C CH2CH2CH3

nucleophile and base

CH3

E1 products CH3

+ CH3CH2OH

Cl

leaving group

CH2

OH

CH3 C CH2CH2CH3 OCH2CH3

SN1 product

CH3

CH3 C CH2 CH3CH2CH2

C CHCH2CH3 CH3

E1 products

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Chapter 8–8 8.18 E2 reactions occur with anti periplanar geometry. The anti periplanar arrangement uses a staggered conformation and has the H and X on opposite sides of the C–C bond. H

H

HO

C

CH3 CH3

C

CH3

H

Br

C

base

H C

CH3

H

H and Br are on opposite sides = anti periplanar

8.19 The E2 elimination reactions will occur in the anti periplanar orientation as drawn. To draw the product of elimination, maintain the orientation of the remaining groups around the C=C. CH3CH2O H

CH3

a.

C6H5

C

CH3

C6H5 C H Br

C6H5 C C

C6H5

H

The two benzene rings remain on opposite sides of the newly formed C=C. This makes them trans.

The two benzene rings are anti in this conformation (one wedge, one dash). diastereomers CH3CH2O H

b.

C6H5

C6H5 C6H5 C C H CH3 Br

C6H5 C C

CH3

H

The two benzene rings are gauche in this conformation (both drawn on dashes, behind the plane).

The two benzene rings remain on the same side of the newly formed C=C. This makes them cis.

8.20 Note: The Zaitsev products predominate in E2 elimination except when substituents on a cyclohexane ring prevent a trans diaxial arrangement of H and X. axial H's H

CH(CH3)2

two conformations

a. CH3

Cl

CH3

H CH3

H H H

CH(CH3)2

Cl

A Use this conformation. It has Cl axial and two axial H's.

H

H

H H

B

H Cl CH(CH3)2

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Alkyl Halides and Elimination Reactions 8–9

H CH3

H H H

H

1

H

H

CH(CH3)2


OH

CH3

Cl

2

CH(CH3)2 H H

CH3

H H

H

[loss of H(2) + Cl]

A

re-draw

CH(CH3)2

CH(CH3)2

CH3

CH3

trisubstituted major product

disubstituted H CH(CH3)2 CH3

CH3

H

two conformations

b.

CH3

Cl

Cl H H

CH(CH3)2

H

H

1

2

H

H

B

CH(CH3)2

CH3
OH

H H

H H

H

Use this conformation. It has Cl axial and one axial H.

Cl

H

Cl

H

A

CH3

H

[loss of H(1) + Cl]

re-draw

two different axial H's

CH(CH3)2

CH(CH3)2

B only one axial H on a  carbon

CH(CH3)2

H

H H

H

CH(CH3)2

[loss of H(1) + Cl]

= CH3

disubstituted only product

8.21 Draw the chair conformations of cis-1-chloro-2-methylcyclohexane and its trans isomer. For E2 elimination reactions to occur, there must be a H and X trans diaxial to each other. Two conformations of the cis isomer: Cl CH3

H

H

H H

A reacting conformation (axial Cl)

This reacting conformation has only one group axial, making it more stable and present in a higher concentration than B. This makes a faster elimination reaction with the cis isomer.

H

CH3 H

H

H

Two conformations of the trans isomer: Cl

H H Cl

H H

CH3 H

Cl H

H H

CH3

B reacting conformation (axial Cl)

This conformation is less stable than A, since both CH3 and Cl are axial. This slows the rate of elimination from the trans isomer.

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Chapter 8–10 8.22 E2 reactions are favored by strong negatively charged bases and occur with 1°, 2°, and 3° halides, with 3° being the most reactive. E1 reactions are favored by weaker neutral bases and do not occur with 1° halides since they would have to form highly unstable carbocations. CH3

a.

C(CH3)3

CH3 C CH3

+

Cl

Cl

c.

OCH3

+

strong negatively charged base E2

I

b.

+

CH3OH

weak neutral base E1

H2O

d. CH3CH2Br

weak neutral base E1

+

OC(CH3)3

strong negatively charged base E2

8.23 Draw the alkynes that result from removal of two equivalents of HX. Br

Cl Cl

a.

C C CH2CH3

NH2

C C CH2CH3

c. CH3 C CH2CH3

NH2 CH3C CCH3

Br

H H

+ HC CCH2CH3 Br

KOC(CH3)3

b. CH3CH2CH2CHCl2

CH3CH2C

DMSO

CH

NH2

d.

C Br

8.24 1° halide SN2 or E2 H

b.

K+
OC(CH3)3

Cl

a.

CH3 C CH2CH3 Cl

2° halide any mechanism

strong bulky base E2

OH

strong base SN2 and E2

CH2CH3 I

c.

+

weak base SN1 and E1

CH3 CH CHCH3

+

SN2 product

disubstituted major E2 product

OCH2CH3

SN1 product

+

+ E1 product

CH3CH2O Cl

3° halide no SN2

CH3CH2OH

E2

major E2 product

monosubstituted minor E2 product CH2CH3

CHCH3

strong base

d.

CH2 CH CH2CH3

OH

CH2CH3 CH3CH2OH

3° halide no SN2

H CH3 C CH2CH3

minor E2 product

E1 product

C

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Alkyl Halides and Elimination Reactions 8–11 8.25 3° halide no SN2

weak base SN1 and E1

CH3

CH3

Br

CH3OH

CH3

overall reaction

The steps: CH3

SN1 CH3

CH3OH + Br

CH3 H CH3

CH3

Br

+

+

CH3 CH3

or E1

OCH3

O CH3 H CH3 Br

CH3

HBr

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Chapter 8–12 8.26 a. CH3CH2CH2CH2CH2CH2Br

CH3CH2CH2CH2CH CH2

Br

b.

CH3CH2CH2CH2CH CHCH2CH3 +

CH3CH2CH2CH2CH2CH CHCH3

CH3

c. CH3CH2CHCHCH3

CH3CH2CH C(CH3)2

+

CH3CH CHCH(CH3)2

Cl CH2

I

d.

8.27 To give only one product in an elimination reaction, the starting alkyl halide must have only one type of
carbon with H’s. a.

CH2 CHCH2CH2CH3








CH2

CH2CH2CH2CH3

Cl

 b. (CH3)2CHCH CH2

 Cl

C(CH3)3

e.

CH2Cl

c.

Two  carbons are identical.

CH2Cl




CH2



CH3

CH3




(CH3)2CHCH2


Cl

d.






C(CH3)3

Two  carbons are identical.

8.28 To have stereoisomers, the two groups on each end of the double bond must be different from each other. farnesene

geranial CHO

b.

a.

two methyl groups no stereoisomers

2 H's — no two different groups stereoisomers at each end can have stereoisomers

two methyl groups no stereoisomers

two different groups at each end can have stereoisomers

8.29 Use the definitions in Answer 8.4. CH3

a.

CH2

and

and

c. trans trans

different connectivity constitutional isomers

trans trans identical

H CH3CH2

b.

CH3

and

C C CH3

CH3CH2

CH2CH3

stereoisomers

CH2CH3 C C

CH3

CH3

C

d.

CH3 CH3

C

and

CH3

CH3

stereoisomers

H

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Alkyl Halides and Elimination Reactions 8–13 8.30 There are three different isomers. Cis and trans isomers are diastereomers. H

Cl

H

C C H

Br

H

C C Br

Cl

H C C

H

Cl

B

A constitutional isomer of B and C

Br

C

diastereomers

8.31 Double bond can be cis or trans. OH

a. five sp3 stereogenic centers (four circled, one labeled) b. Two double bonds can both be cis or trans. c. 27 = 128 stereoisomers possible

CH2CH CH(CH2)3COOH

PGF2
HO

CH CHCH(OH)(CH2)4CH3

sp3 stereogenic center

Double bond can be cis or trans.

8.32 Use the rules from Answer 8.5 to rank the alkenes. CH3CH2 H

monosubstituted

b.

CH3

CH3CH2

H

C C

C C

a. CH2 CHCH2CH2CH3

H

H

least stable

disubstituted cis intermediate stability

CH2 CHCH(CH3)2

CH2 C(CH3)CH2CH3

monosubstituted least stable

disubstituted intermediate stability

CH3

disubstituted trans most stable (CH3)2C

CHCH3

trisubstituted most stable

8.33 A larger negative value for
H° means the reaction is more exothermic. Since both 1-butene and cis-2-butene form the same product (butane), these data show that 1-butene was higher in energy to begin with, since more energy is released in the hydrogenation reaction. 1-butene

+ H2

CH3

CH3

+ H2

C C H

CH3CH2CH2CH3


Ho = 127 kJ/mol

1-butene

H

cis-2-butene

CH3CH2CH2CH3


Ho = 120 kJ/mol

Energy

CH2 CHCH2CH3

cis-2-butene

larger
H° for 1-butene higher in energy

butane smaller
H° for cis-2-butene lower in energy, more stable

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Chapter 8–14 8.34 Cl

a.


1

(CH3)3CO

(loss of 2 H) major product disubstituted

(loss of 1 H) monosubstituted

DBU

b. O




O

only product

Cl

O

O

3

I

c.

CH3CH=CHCH2CH2CH(CH3)2

2

CH3

1




OH CH3CH2C(CH3)=C(CH3)CH2CH2CH3

2

CH3CH2CH(CH3)C(CH3) CHCH2CH3

(loss of 1 H) major product tetrasubstituted

(loss of 2 H) trisubstituted

CH2

(loss of 3 H) disubstituted  d.

Cl

OC(CH3)3




only product

2 e. 1

2

f.




OH


I

Br

CH3CH2CH2CH2CH=CHCH3

(loss of 1 H) major product disubstituted

(loss of 2 H) monosubstituted

OH

1 (loss of 2 H) major product trisubstituted

(loss of 1 H) disubstituted

8.35 To give only one alkene as the product of elimination, the alkyl halide must have either: • only one
carbon with a hydrogen atom • all identical
carbons so the resulting elimination products are identical CH3 H

a. CH3 C H

C H Cl

CH3 C CH3

H

CH3 H CH3 C

C H

Cl

C H H Cl

b.

CH=CH2 Cl

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Alkyl Halides and Elimination Reactions 8–15 Cl Cl

c.

8.36 Draw the products of the E2 reaction and compare the number of C’s bonded to the C=C.

1 A




2

2 1
major product trisubstituted

Br Br

1

(CH3)2CHCH=CHCH3





disubstituted


2

B




(CH3)2CHCH=CHCH3

1 major product disubstituted

2

monosubstituted

A yields a trisubstituted alkene as the major product and a disubstituted alkene as minor product. B yields a disubstituted alkene as the major product and a monosubstituted alkene as minor product. Since the major and minor products formed from A have more alkyl groups on the C=C (making them more stable) than those formed from B, A reacts faster in an elimination reaction. 8.37 a. Mechanism: by-products Br H

OC(CH3)3

+ HOC(CH3)3 + Br


(CH3)3COH

b. Rate = k[R–Br][–OC(CH3)3] [1] Solvent changed to DMF (polar aprotic) = rate increases [2] [–OC(CH3)3] decreased = rate decreases [3] Base changed to –OH = rate decreases (weaker base) [4] Halide changed to 2° = rate increases (More substituted RX reacts faster.) [5] Leaving group changed to I– = rate increases (better leaving group) 8.38 K+ –OC(CH3)3

+

Cl

A 1-chloro-1-methylcyclopropane

B

The dehydrohalogenation of an alkyl halide usually forms the more stable alkene. In this case A is more stable than B even though A contains a disubstituted C=C whereas B contains a trisubstituted C=C. The double bond in B is part of a three-membered ring, and is less stable than A because of severe angle strain around both C’s of the double bond.

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Chapter 8–16 8.39 H

a.

KOH

CH3CH2CH2 C

NaOCH2CH3

b. Cl

Br

trans isomer more stable major product

trans isomer more stable major product

8.40 CH3

a.

Cl CH3

CH3

CH3

CH2

CH3

CH3

CH3

trisubstituted

disubstituted

tetrasubstituted major product Br

b. trisubstituted This isomer is more stable — large groups farther away. major product

disubstituted trisubstituted

Cl

c.

disubstituted

trisubstituted major product

8.41 Use the rules from Answer 8.22. a.

OCH3

CH3CH CHCH3

strong base E2

Br

2° halide

CH3OH

b.

weak base E1

Br

2° halide

CH2CH2CH3

H2O

Cl CH3

weak base 3° halide E1 OH Cl

e. 2° halide

strong base E2 OH

f. 2° halide

Cl

CH3CH2CH CH2

(cis and trans)

strong base E2

1° halide d.

CH3CH CHCH3

OC(CH3)3

I

c.

CH3CH2CH CH2

(cis and trans)

strong base E2

CHCH2CH3 CH3

CH2CH2CH3 CH3

CH2CH2CH3 CH3

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Alkyl Halides and Elimination Reactions 8–17 8.42 The order of reactivity is the same for both E2 and E1: 1° < 2° < 3° a.

Br

Br

Br

1° halide

2° halide

3° halide

Increasing reactivity in E1 and E2 CH3

b.

Br

CH3

Cl

CH3

Cl

3° halide

2° halide

3° halide + better leaving group

Increasing reactivity in E1 and E2

8.43 CH3

a.

Cl

Cl OH

3° halide – faster reaction CH3

Cl

H2O

b. CH3 Cl

c. (CH3)3CCl

H2O

strong base – E2

H2O (CH3)3CCl

OH

OH

OH DMSO

3° halide – faster reaction polar aprotic solvent faster reaction

8.44 In a ten-membered ring, the cis isomer is more stable and, therefore, the preferred elimination product. The trans isomer is less stable because strain is introduced when two ends of the double bond are connected in a trans arrangement in this medium-sized ring.

Br

bromocyclodecane

cis-cyclodecene

8.45 With the strong base –OCH2CH3, the mechanism is E2, whereas with dilute base, the mechanism is E1. E2 elimination proceeds with anti periplanar arrangement of H and X. In the E1 mechanism there is no requirement for elimination to proceed with anti periplanar geometry. In this case the major product is always the most stable, more substituted alkene. Thus, C is the major product under E1 conditions. (In Chapter 9, we will learn that additional elimination products may form in the E1 reaction due to carbocation rearrangement.) Cl OCH2CH3

A

strong base E2

B

Since this is an E2 mechanism, dehydrohalogenation needs an anti periplanar H to form the double bond. There is only one H trans to Cl, so the disubstituted alkene B must form.

Cl CH3OH

A

weak base E1

B

C

disubstituted alkene

trisubstituted alkene more stable

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Chapter 8–18 8.46 H and Br must be anti during the E2 elimination. Rotate if necessary to make them anti; then eliminate. CH3

a.

C6H5

C6H5

Br

H

CH3

E2

CH2CH3

CH3 CH2CH3

CH3 Br

b.

Br

H

CH3

C6H5

CH3

CH3

rotate

CH2CH3

C6H5

E2

CH3

CH2CH3

CH3

CH3

CH2CH3

C6H5

CH2CH3

H

C6H5

c.

C6H5

C6H5

CH3

H

CH3

Br

H

rotate

CH2CH3

CH3

CH2CH3

E2

Br

CH3

CH3

CH3

8.47 Cl

a.

two chair conformations H

CH3

Cl

H

H H

H H CH3

H

B

H

H CH3

Choose this conformation. axial Cl

H

H

Cl

B

(CH3)2CH

CH3 (CH3)2CH A H

CH(CH3)2

(CH3)2CH

H axial Cl

H H

H

one axial H

(CH3)2CH

=

CH3

H

only product CH3 CH(CH3)2

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211

Alkyl Halides and Elimination Reactions 8–19 Cl

b.

H

two chair conformations

H (CH3)2CH

CH(CH3)2 H

H

H

c.

1

re-draw

CH3

CH3

CH(CH3)2

CH(CH3)2

D

D

= D

D

H

(loss of
2 H)

(loss of 2 H)

D

Cl H Cl

D H H

(CH3)2CH CH3

(loss of
1 H) major product trisubstituted re-draw

D

H

H

H (CH3)2CH CH3

H
1 H two axial H's B

B

Choose this conformation. axial Cl

H

H H

A

H

Cl

Cl (CH3)2CH CH3


2

Cl

axial

H H Cl

H

CH3

(CH3)2CH CH3

CH3

H

2

D H

H

=

D

H H

D

enantiomers


OH

D

This conformation reacts. axial Cl

D

D = H

D

H

(loss of 1 H) D Cl

d. =

D

H

H

1

H H D

(loss of 2 D)

H

Cl

D

2

This conformation reacts. axial Cl

H

D Cl H

D

=

D

H

D

H

enantiomers


OH

D H

H H

D

(loss of 1 D)

=

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Chapter 8–20 8.48 enantiomers

a. H CH3

H

CH3 C C CH2CH3

CH3CH2 CH3

C3

C2

H

H H CH3 CH3

C C

Cl H

enantiomers

C C CH CH 2 3 CH3 Cl

Cl

–HCl

H and Cl are arranged anti in each stereoisomer, for anti periplanar elimination.

CH3CH2

Cl

H

CH2CH3

C C CH3 CH3

C C CH3

CH3 CH3 CH2CH3 CH3CH2

identical

–HCl H

H CH3

C C

Cl

D

C –HCl

H

CH3

C C

B

A

2-chloro-3-methylpentane

H

H

H CH3

CH3

–HCl CH3

H

C C C C CH3 CH2CH3 CH3CH2 CH3

identical

b. Two different alkenes are formed as products. c. The products are diastereomers: Two enantiomers (A and B) give identical products. A and B are diastereomers of C and D. Each pair of enantiomers gives a single alkene. Thus diastereomers give diastereomeric products. 8.49 The trans isomer reacts faster. During elimination, Br must be axial to give trans diaxial elimination. In the trans isomer, the more stable conformation has the bulky tert-butyl group in the more roomy equatorial position. In the cis isomer, elimination can occur only when both the tertbutyl and Br groups are axial, a conformation that is not energetically favorable. Br Br

H

= This conformation must react, but it contains two axial groups.

cis Br

Br

H

= trans

preferred conformation

8.50 CH2CHCl2

a.

C CH

NaNH2 (2 equiv)

CH3

b.

CH3CH2 C

CHCH2Br

CH3 Br

Cl

c.

CH3 C CH2CH3 Cl

NaNH2 (2 equiv)

NaNH2 (excess)

CH3 CH3CH2 C

C CH

CH3

HC C CH2CH3

CH3 C C CH3

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213

Alkyl Halides and Elimination Reactions 8–21 H H

d.

NaNH2

C C

C C

(2 equiv)

Cl Cl

8.51 H H

Br

a. CH3C CCH3

CH3 C CH2CH3

or

CH3 C C CH3 Br Br

Br CH3

CH3 Br

b. CH3 C C CH

CH3 C

CH3

CH3

CH3 Br

C CH3

or

CH3 C

CH3 Br

CH CH2Br

or

C C

CH2CHBr2

CH3

CH3 Br H

c.

CH3 C

Br Br

or

C C

C C

Br H

H H

8.52 H H CH3 C C CH3

CH3 C C CH3

CH3 CH=C=CH2

CH2=CH CH=CH2

C

Br Br

2,3-dibromobutane

sp sp A

sp B

8.53 Use the “Summary chart on the four mechanisms: SN1, SN2, E1, or E2” on p. 8–2 to answer the questions. a. b. c. d. e. f. g. h. i. j.

Both SN1 and E1 involve carbocation intermediates. Both SN1 and E1 have two steps. SN1, SN2, E1, and E2 have increased reaction rates with better leaving groups. Both SN2 and E2 have increased rates when changing from CH3OH (protic solvent) to (CH3)2SO (aprotic solvent). In SN1 and E1 reactions, the rate depends on only the alkyl halide concentration. Both SN2 and E2 are concerted reactions. CH3CH2Br and NaOH react by an SN2 mechanism. Racemization occurs in SN1 reactions. In SN1, E1, and E2 mechanisms, 3° alkyl halides react faster than 1° or 2° halides. E2 and SN2 reactions follow second-order rate equations.

8.54 Br

a.

OC(CH3)3

sterically 1° halide hindered base SN2 or E2 OCH2CH3

I

b. 1° halide SN2 or E2 Cl

c. CH3 C CH3 Cl

dihalide

strong nucleophile NH2 (2 equiv)

strong base

HC C CH3

E2

OCH2CH3

SN2

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Chapter 8–22 Br

d.

sterically hindered base

1° halide SN2 or E2

CH2CH3

2° halide SN1, SN2, E1, E2

CH2CH3

E2

sterically hindered base

Br

major product

OCH2CH3

Br

CH3CH2OH

CH2CH3

(CH3)2CH

SN1 product 2 NaNH2

CHCH2Br

CHCH3

CH2CH3

CH2CH3

weak base

3° halide no SN2 g.

CH2CH3

OC(CH3)3

e.

f.

E2

DBU

(CH3)2CH

E1 products

C CH

dihalide Br Cl Cl

h.

KOC(CH3)3

dihalide

CH3

OCH2CH3

I

CH3CH2OH

i. 2° halide SN1, SN2, E1, E2 j.

(2 equiv) DMSO

CH3 CH3 C C CH

CH3CH CHCH3

SN1 product

weak base

E1 product

H2O Cl

CH3CH2C(CH3) CHCH3 OH

weak base

3° halide no SN2

(cis and trans) E1 product

(cis and trans) E1 product

SN1 product

E1 product

8.55 [1] NaOCOCH3 is a good nucleophile and weak base, and substitution is favored. [3] KOC(CH3)3 is a strong, bulky base that reacts by E2 elimination when there is a
hydrogen in the alkyl halide. a.

CH3Cl

[1] NaOCOCH3

[2] NaOCH3

[3] KOC(CH3)3

Cl

CH3OCOCH3

CH3OCH3

CH3OC(CH3)3

b.

[1] NaOCOCH3

[2] NaOCH3

[3] KOC(CH3)3

OCOCH3

OCH3

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215

Alkyl Halides and Elimination Reactions 8–23

c.

[1] NaOCOCH3 Cl

Cl

d.

OCOCH3 [2] NaOCH3

[1] NaOCOCH3

OCOCH3

[2] NaOCH3 OCH3

+ E2

SN2 [3] KOC(CH3)3

[3] KOC(CH3)3

8.56 a. two enantiomers:

(CH3)3C

(CH3)3C

A

B

b. The bulky tert-butyl group anchors the cyclohexane ring and occupies the more roomy equatorial position. The cis isomer has the Br atom axial, while the trans isomer has the Br atom equatorial. For dehydrohalogenation to occur on a halo cyclohexane, the halogen must be axial to afford trans diaxial elimination of H and X. The cis isomer readily reacts since the Br atom is axial. The only way for the trans isomer to react is for the six-membered ring to flip into a highly unstable conformation having both (CH3)3C and Br axial. Thus, the trans isomer reacts much more slowly. Br

trans diaxial

Br

(CH3)3C

(CH3)3C H

H

cis-1-bromo-4-tert-butylcyclohexane

trans-1-bromo-4-tert-butylcyclohexane OCH3

c. two products: C=

(CH3)3C

OCH3

D = (CH3)3C

d. cis-1-Bromo-4-tert-butylcyclohexane reacts faster. With the strong nucleophile –OCH3, backside attack occurs by an SN2 reaction, and with the cis isomer, the nucleophile can approach from the equatorial direction, avoiding 1,3-diaxial interactions. 1,3-diaxial interactions Br

OCH3

H (CH3)3C

H

OCH3

equatorial approach preferred cis-1-bromo-4-tert-butylcyclohexane

(CH3)3C

Br

axial approach

trans-1-bromo-4-tert-butylcyclohexane

e. The bulky base –OC(CH3)3 favors elimination by an E2 mechanism, affording a mixture of two enantiomers A and B. The strong nucleophile –OCH3 favors nucleophilic substitution by an SN2 mechanism. Inversion of configuration results from backside attack of the nucleophile.

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Chapter 8–24 8.57 Cl

H

H

a.

strong base SN2 and E2

2° halide SN1, SN2, E1, E2 Cl

OH

OH

H

SN2 product major E2 product inversion at stereogenic center HO

H2O

b.

H

H

weak base SN1 and E1

2° halide SN1, SN2, E1, E2

c.

Cl C6H5

3° halide no SN2

minor E2 product

OH

SN1 products

major E1 product CH3 CH3

minor E2 product

minor E1 product

CH3 CH3

CH3OH

OCH3 C6H5

weak base SN1 and E1

minor E1 product

CH3 CH3

CH3 CH3

C6H5 OCH3

SN1 products

C6H5

E1 product

NaOH

d.

strong base SN2 and E2

Cl

OH

SN2 product

2° halide SN1, SN2, E1, E2 CH3 Br

e.

3° halide no SN2

weak base good nucleophile SN1

f. D

2° halide SN1, SN2, E1, E2

achiral SN1 product

OH

KOH strong base SN2 and E2

minor E2 product

CH3 OOCCH3

CH3COO

Br

major E2 product

(trans diaxial elimination of D, Br) D

SN2 product inversion at stereogenic center

E2 product

8.58 a.

b.

CH3OH Cl

weak base SN1 and E1

KOH Cl

strong base E2

OCH3 OCH3

SN1

SN1

E1

E1

E1

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Alkyl Halides and Elimination Reactions 8–25 8.59 CH3

CH3

Br

a.

OC(CH3)3

OC(CH3)3 CH3

strong bulky base E2

3° halide

CH2

major product more substituted alkene

No substitution occurs with a strong bulky base and a 3o RX. The C with the leaving group is too crowded for an SN2 substitution to occur. Elimination occurs instead by an E2 mechanism. Br

b. 1° halide

OCH3

strong nucleophile SN2

OCH3

All elimination reactions are slow with 1° halides. The strong nucleophile reacts by an SN2 mechanism instead. c.

CH3 Cl

3° halide

minor product only

OH

strong base E2 I

d. Cl

2° halide

good nucleophile, weak base SN2 favored

CH3

More substituted alkene is favored.

minor product only

I

major product

The 2o halide can react by an E2 or SN2 reaction with a negatively charged nucleophile or base. Since I– is a weak base, substitution by an SN2 mechanism is favored.

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Chapter 8–26 8.60 3° halide, weak base: SN1 and E1 CH3CH2OH

a.

+

overall reaction

Cl

+

+ HCl

OCH2CH3

The steps:

+ HCl

SN1

Any base (such as CH3CH2OH or Cl–) can be used to remove a proton to form an alkene. If Cl– is used, HCl is formed as a reaction by-product. If CH3CH2OH is used, (CH3CH2OH2)+ is formed instead.

OCH2CH3

CH3CH2OH

H

Cl

or E1 H

+ HCl

Cl

or

E1

+ HCl H

Cl CH3 Cl

b.

3° halide strong base E2

CH3

OH

CH2 + H2O + Cl

+

overall reaction CH3

CH3

Cl

Each product:

one step

H

OH

or

+ H2O + Cl

CH2 H

OH CH2

Cl

one step

8.61 Draw the products of each reaction with the 1° alkyl halide. Cl

a.

strong nucleophile SN2

H

Cl

b. H

NaOCH2CH3

OCH2CH3 H

KCN

strong nucleophile SN2

Cl

c. H

CN H

DBU

sterically hindered base E2

H

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Alkyl Halides and Elimination Reactions 8–27 8.62 H H

H Br

H

H2O

H

O H

OH

H2O Br

above

H

H

H

+

HBr

+

HBr

H

Br– H H

H

O H

re-draw

Br

below H

OH

H

H

H

+ H

H

HBr

H

Br

8.63 good nucleophile O CH3

C

O

CH3 CHCH3 O

C

CH3

CH3COO
is a good nucleophile and a weak base and so it favors substitution by SN2.

O CH3 CHCH3

(only)

Br CH3CH2O

strong base

CH3 CHCH3 + CH3CH CH2 The strong base gives both SN2 and E2 products, but since the 2° RX is OCH2CH3

20%

80%

somewhat hindered to substitution, the E2 product is favored.

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Chapter 8–28 8.64 Cl

OCH3 CH3OH

+

+

+

+

HCl

OCH3

3° halide weak base SN1 and E1

Cl

H

CH3OH

Cl

OCH3

OCH3

Cl–

HCl

Cl H

or

HCl

Cl–

CH3OH

HCl OCH3

Cl

or

OCH3

H

HCl

Cl H

8.65 E2 elimination needs a leaving group and a hydrogen in the trans diaxial position. Two different conformations:

Cl Cl

Cl

Cl

Cl Cl

This conformation has Cl's axial, but no H's axial.

Cl Cl Cl

Cl Cl Cl

This conformation has no Cl's axial.

For elimination to occur, a cyclohexane must have a H and Cl in the trans diaxial arrangement. Neither conformation of this isomer has both atoms—H and Cl—axial; thus, this isomer only slowly loses HCl by elimination.

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Alkyl Halides and Elimination Reactions 8–29 8.66 H Br CH 3

H and Br are anti periplanar. Elimination can occur.

CH3O


O O H H

H (in the ring) and Br are NOT anti periplanar. Elimination cannot occur using this H. Instead elimination must occur with the H on the CH3 group.

O O

–HBr

CH3 Br

H major product

Elimination can occur here. H CH3O


O O

O O

–HBr

H

H

Elimination cannot occur in the ring because the required anti periplanar geometry is not present.

8.67 leaving group H N

C6H5O O

S

N

C6H5O

DBN

N O

B

H

H H

H H N

O

overall reaction

S O

O

H C CH2Cl O

SN2

E2

H

A sequence of two reactions forms the final product: E2 elimination opens the five-membered ring. Then the sulfur nucleophile displaces the Cl– leaving group to form the six-membered ring.

H H

N

C6H5O

Cl

S

O

N

CH2 C

O

O

8.68 a.

H D




SeOC6H5

SeOC6H5 and H are on the same side of the ring. syn elimination

CH3

D

b.

C H Br

CH3

Br

rotate

C H Br

H C

CH3

H

C

CH3

Zn

CH3

Br

Both Br atoms are on the opposite sides of the C–C bond. anti elimination

H

C C H

CH3

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223

Alcohols, Ethers, and Epoxides 9–1 C Chhaapptteerr 99:: A Allccoohhoollss,, EEtthheerrss,, aanndd EEppooxxiiddeess

G Geenneerraall ffaaccttss aabboouutt R RO OH H,, R RO OR R,, aanndd eeppooxxiiddeess • All three compounds contain an O atom that is sp3 hybridized and tetrahedral (9.2). CH3

O

CH3

H

109o an alcohol

•

60o

O

O

CH3 H H

111o an ether

H H

an epoxide

All three compounds have polar C–O bonds, but only alcohols have an O–H bond for intermolecular hydrogen bonding (9.4). H C H

=

O

=

H

C H H

H

H

H

O

hydrogen bond

•

Alcohols and ethers do not contain a good leaving group. Nucleophilic substitution can occur only after the OH (or OR) group is converted to a better leaving group (9.7A). + R OH

+

R OH2

H Cl

+ Cl

strong acid weak base good leaving group

•

Epoxides have a leaving group located in a strained three-membered ring, making them reactive to strong nucleophiles and acids HZ that contain a nucleophilic atom Z (9.15). leaving group With strong nucleophiles,

O

O C C

Nu

OH

H OH

C C

[1]

C C

[2]

Nu

+ OH

Nu

Nu



A A nneew w rreeaaccttiioonn ooff ccaarrbbooccaattiioonnss ((99..99)) • Less stable carbocations rearrange to more stable carbocations by shift of a hydrogen atom or an alkyl group. Besides rearrangement, carbocations also react with nucleophiles (7.13) and bases (8.6). C C

1,2-shift

R (or H)

C C R (or H)



PPrreeppaarraattiioonn ooff aallccoohhoollss,, eetthheerrss,, aanndd eeppooxxiiddeess ((99..66)) [1] Preparation of alcohols R X

+

OH

R OH

+ X

• •

The mechanism is SN2. The reaction works best for CH3X and 1o RX.

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Chapter 9–2 [2] Preparation of alkoxides (a Brønsted–Lowry acid–base reaction) R O H + Na+ H

R O

Na+ + H2

alkoxide

[3] Preparation of ethers (Williamson ether synthesis) R X

+

OR'

R OR'

• •

+ X

The mechanism is SN2. The reaction works best for CH3X and 1o RX.

[4] Preparation of epoxides (Intramolecular SN2 reaction) •

B H O

[1]

C C

O C

X

+

halohydrin

O C C

[2]

C

+X

X

H B+

A two-step reaction sequence: [1] Removal of a proton with base forms an alkoxide. [2] Intramolecular SN2 reaction forms the epoxide.



R Reeaaccttiioonnss ooff aallccoohhoollss [1] Dehydration to form alkenes [a] Using strong acid (9.8, 9.9)

C C H OH

H2SO4 or TsOH

• C C

+

• •

Order of reactivity: R3COH > R2CHOH > RCH2OH. The mechanism for 2o and 3o ROH is E1; carbocations are intermediates and rearrangements occur. The mechanism for 1o ROH is E2. The Zaitsev rule is followed.

• •

The mechanism is E2. No carbocation rearrangements occur.

•

Order of reactivity: R3COH > R2CHOH > RCH2OH. The mechanism for 2o and 3o ROH is SN1; carbocations are intermediates and rearrangements occur. The mechanism for CH3OH and 1o ROH is SN2.

H2O

•

[b] Using POCl3 and pyridine (9.10)

C C H OH

POCl3

C C

pyridine

+ H2O

[2] Reaction with HX to form RX (9.11) R OH + H X

R X

+ H2O

• •

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225

Alcohols, Ethers, and Epoxides 9–3 [3] Reaction with other reagents to form RX (9.12) R OH R OH

+ SOCl2 +

R Cl

• •

pyridine

PBr3

R Br

Reactions occur with CH3OH and 1o and 2o ROH. The reactions follow an SN2 mechanism.

[4] Reaction with tosyl chloride to form alkyl tosylates (9.13A) O R OH + Cl S

CH3

pyridine

O R O S

O

CH3

O R OTs

• The C–O bond is not broken so the configuration at a stereogenic center is retained.



R Reeaaccttiioonnss ooff aallkkyyll ttoossyyllaatteess Alkyl tosylates undergo either substitution or elimination depending on the reagent (9.13B). Nu

C C

+

OTs

•

Substitution is carried out with strong :Nu– so the mechanism is SN2.

•

Elimination is carried out with strong bases so the mechanism is E2.

H Nu C C H OTs

B

+ HB+

C C

–OTs

+



R Reeaaccttiioonnss ooff eetthheerrss Only one reaction is useful: Cleavage with strong acids (9.14) R O R'

+

H X

R X

+

+

R' X

(2 equiv) (X = Br or I)

H2O

• •

With 2o and 3o R groups, the mechanism is SN1. With CH3 and 1o R groups the mechanism is SN2.



R Reeaaccttiioonnss ooff eeppooxxiiddeess Epoxide rings are opened with nucleophiles :Nu– and acids HZ (9.15). • OH

O C

C

[1] Nu

[2] H2O

or HZ

C

C

•

Nu

(Z)

•

The reaction occurs with backside attack, resulting in trans or anti products. With :Nu–, the mechanism is SN2, and nucleophilic attack occurs at the less substituted C. With HZ, the mechanism is between SN1 and SN2, and attack of Z– occurs at the more substituted C.

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Chapter 9–4 C Chhaapptteerr 99:: A Annssw weerrss ttoo PPrroobblleem mss 9.1 • Alcohols are classified as 1°, 2°, or 3°, depending on the number of carbon atoms bonded to the carbon with the OH group. • Symmetrical ethers have two identical R groups, and unsymmetrical ethers have R groups that are different. OH OH

CH3

O

1° alcohol

symmetrical ether

2° alcohol

OH

CH3

OH

3° alcohol

O

unsymmetrical ether

O

unsymmetrical ether

1° alcohol

9.2 Use the definition in Answer 9.1 to classify each OH group in cortisol. O

OH

HO

2° alcohol

1° alcohol

OH

H H

3° alcohol H

O

cortisol

9.3 To name an alcohol: [1] Find the longest chain that has the OH group as a substituent. Name the molecule as a derivative of that number of carbons by changing the -e ending of the alkane to the suffix -ol. [2] Number the carbon chain to give the OH group the lower number. When the OH group is bonded to a ring, the ring is numbered beginning with the OH group, and the “1” is usually omitted. [3] Apply the other rules of nomenclature to complete the name. 1 OH

a. [1]

OH

[2]

[3] 3,3-dimethyl-1-pentanol

5 carbons = pentanol b. [1]

CH3

CH3

[2]

OH

2-methyl

[3] cis-2-methylcyclohexanol

OH

1

6 carbon ring = cyclohexanol

6-methyl c.

[1]

OH

OH

[2]

3 9 carbons = nonanol

5-ethyl

[3] 5-ethyl-6-methyl-3-nonanol

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Alcohols, Ethers, and Epoxides 9–5 9.4 To work backwards from a name to a structure: [1] Find the parent name and draw its structure. [2] Add the substituents to the long chain. OH

7

3

2

c. 2-tert-butyl-3-methylcyclohexanol

a. 7,7-dimethyl-4-octanol 4

OH

1 5 b. 5-methyl-4-propyl-3-heptanol 3

OH

OH

d. trans-1,2-cyclohexanediol

or

OH

OH

OH

9.5 To name simple ethers: [1] Name both alkyl groups bonded to the oxygen. [2] Arrange these names alphabetically and add the word ether. For symmetrical ethers, name the alkyl group and add the prefix di. To name ethers using the IUPAC system: [1] Find the two alkyl groups bonded to the ether oxygen. The smaller chain becomes the substituent, named as an alkoxy group. [2] Number the chain to give the lower number to the first substituent. IUPAC name:

a. common name: CH3 O CH2CH2CH2CH3

butyl

methyl

butyl methyl ether

b. common name: OCH3

CH3 O CH2CH2CH2CH3

substituent: methoxy 1-methoxybutane IUPAC name: OCH3

substituent – methoxy

methyl larger group – 6 C's cyclohexane

cyclohexyl cyclohexyl methyl ether c. common name: CH3CH2CH2 O CH2CH2CH3

propyl

propyl

dipropyl ether

larger group – 4 C's butane

methoxycyclohexane IUPAC name: CH3CH2CH2 O CH2CH2CH3

propoxy

propane

1-propoxypropane

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Chapter 9–6 9.6 Name each ether using the rules from Answer 9.5. 1 CH3 a.

b.

CH3CH2 O CH2CH3

CH3 C

O CH3

CH3

ethoxy

ethane

2-methoxy

2-methyl

propane 2-methoxy-2-methylpropane

1-ethoxyethane

9.7 Three ways to name epoxides: [1] Epoxides are named as derivatives of oxirane, the simplest epoxide. [2] Epoxides can be named by considering the oxygen as a substituent called an epoxy group, bonded to a hydrocarbon chain or ring. Use two numbers to designate which two atoms the oxygen is bonded to. [3] Epoxides can be named as alkene oxides by mentally replacing the epoxide oxygen by a double bond. Name the alkene (Chapter 10) and add the word oxide. H

Three possibilities: [1] methyloxirane [2] 1,2-epoxypropane [3] propene oxide

O

a. CH3

c.

O H

Three possibilities: [1] cis-2-methyl-3-propyloxirane [2] cis-2,3-epoxyhexane [3] cis-2-hexene oxide

1 CH3

b.

O

1-methyl epoxy group

2

Two possibilities: [1] 6 carbons = cyclohexane 1,2-epoxy-1-methylcyclohexane [2] 1-methylcyclohexene oxide

9.8 Two rules for boiling point: [1] The stronger the forces the higher the bp. [2] Bp increases as the extent of the hydrogen bonding increases. For alcohols with the same number of carbon atoms: hydrogen bonding and bp’s increase: 3° ROH < 2° ROH < 1° ROH. CH3

a.

CH3

OH

b.

O

OH

OH OH

VDW lowest bp

VDW DD intermediate bp

VDW DD hydrogen bonding highest bp

3° ROH lowest bp

2° ROH intermediate bp

1° ROH highest bp

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Alcohols, Ethers, and Epoxides 9–7 9.9 Draw dimethyl ether and ethanol and analyze their intermolecular forces to explain the observed trend. dimethyl ether CH3

O

ethanol CH3CH2OH

CH3

VDW DD no HB much lower bp

VDW DD HB Two molecules of CH3CH2OH can hydrogen bond to each other. stronger forces = much higher bp

Both molecules contain an O atom and can hydrogen bond with water. They have fewer than 5 C's and are therefore water soluble. H

H

O H

CH3

O

O H

CH3

CH3CH2

O

H

9.10 Strong nucleophiles (like –CN) favor SN2 reactions. The use of crown ethers in nonpolar solvents increases the nucleophilicity of the anion, and this increases the rate of the SN2 reaction. The nucleophile does not appear in the rate equation for the SN1 reaction. Nonpolar solvents cannot solvate carbocations so this disfavors SN1 reactions as well. 9.11 Compare the number of carbons and functional groups to determine if each compound is soluble in water, an organic solvent, or both. CH3

O O

N

CH3

O

O H O

Br–

H O

H

O

S

CO2CH3

eplerenone 24 C's and 6 O's (from four functional groups) soluble in organic solvents probably borderline water solubility

HO

S

tiotropium bromide 19 C's, 4 O's a salt – water soluble Because it has many C's, it is also probably soluble in organic solvent.

9.12 Draw the products of substitution in the following reactions by substituting OH or OR for X in the starting material. a. CH3CH2CH2CH2 Br

+ OH

b.

Cl

+

OCH3

c.

CH2CH2–I

+

OCH(CH3)2

CH2CH2–OCH(CH3)2

d.

Br +

OCH2CH3

OCH2CH3

alcohol

CH3CH2CH2CH2 OH + Br

OCH3

+

Cl

+

unsymmetrical ether

+

Br

I

unsymmetrical ether

unsymmetrical ether

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Chapter 9–8 9.13 To synthesize an ether using a Williamson ether synthesis: [1] First find the two possible alkoxides and alkyl halides needed for nucleophilic substitution. [2] Classify the alkyl halides as 1°, 2°, or 3°. The favored path has the less hindered halide. Two possibilities:

Two possibilities:

CH3 CH3 O

a.

CH3 O

b.

CH3

CH3CH2 O C CH3

CH3CH2 O C CH3

H

H

CH3 CH3O

+ Br

CH3CH2 O

CH3Br + O

+

CH3CH2

H

methyl halide

2° halide

CH3

Br C CH3

2° halide

Br +

1° halide

O C CH3 H

less hindered RX preferred

less hindered RX preferred

9.14 NaH and NaNH2 are strong bases that will remove a proton from an alcohol, creating a nucleophile. a. CH3CH2CH2 O H + Na+ H CH3

b.

CH3CH2CH2 O Na+ + H2 CH3

+ + Na NH2

C O H

C O

H

Na+ + NH3

+

H

c.

O

H

Na+ H

CH3CH2CH2–Br + Na+ + H2

O

O

O

d.

H

CH2CH2CH3 + Br–

O

Na+ H

+ Na+ + H2

Br

O

+ Br–

Br C6H10O

9.15 Dehydration follows the Zaitsev rule, so the more stable, more substituted alkene is the major product. H

a.

CH3 C CH3

TsOH

CH2 CH CH3

+ H2O

OH CH3CH2

b.

TsOH OH

+ H2O

C CHCH3 CH3

trisubstituted major product

CH2

disubstituted minor product

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231

Alcohols, Ethers, and Epoxides 9–9 CH3 OH

c.

CH2

CH3

TsOH

+

+ H2O

trisubstituted disubstituted major product minor product

9.16 The rate of dehydration increases as the number of R groups increases. (CH3)2CHCH2CH2CH2OH

a.

1° alcohol slowest reaction

(CH3)2CHCH2CH(OH)CH3

(CH3)2C(OH)CH2CH2CH3

2° alcohol intermediate reactivity

3° alcohol fastest reaction CH3

OH

OH

b. HO

1° alcohol slowest reaction

CH3

3° alcohol fastest reaction

2° alcohol intermediate reactivity

9.17 There are three steps in the E1 mechanism for dehydration of alcohols and three transition states. transition state [1]:

transition state [2]:

CH3

CH3

+

CH3 C CH3

+ OH H

CH3 C

transition state [3]: CH3 CH3

+OH




+

C CH2

CH3

H

2




OSO3H

OSO3H

9.18 transition state [1]:

transition state [2]:

H

H

CH3CH2 C H

+ OH

CH3 C




H

H

OSO3H




OSO3H CH2 OH2

+

9.19 H H

HSO4

H

+

This alkene is also formed in addition to Y from the rearranged carbocation.




+

rearranged 3° carbocation

H2SO4

The initially formed 2° carbocation gives two alkenes: or HSO4

H

H +

H

H + CH 2 H

HSO4

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Chapter 9–10 9.20 CH3 H

a. CH3 C H

C CH2CH3

CH3 H

rearrangement

+

CH3 C

+

1,2-H shift

3° carbocation more stable

H

rearrangement 1,2-H shift

+

c.

+

rearrangement 1,2-methyl shift

H

2° carbocation

b.

+

C CH2CH3

2° carbocation

3° carbocation more stable

+ 3° carbocation more stable

2° carbocation

9.21 CH3 H CH3 C

C CH3

H

CH3 H

H2O

CH3 C

overall reaction

Cl

CH3 H

C CH3

H

CH3 C

OH

C CH3

HCl

OH H

The steps:

CH3 H CH3 C Cl

CH3 H

C CH3

H

CH3 C

H2O

and

O H H

CH3 H CH3 C

C CH3

H

Cl

CH3 H

C CH3

CH3 C

CH3 H

C CH3

H

CH3 C

H H2O

2° carbocation

H O

3° carbocation

C CH3 H

H

Rearrangement of H forms a more stable carbocation.

Cl

9.22 CH3

a. CH3 C CH2CH3

HCl

CH3 CH3 C CH2CH3

c.

OH

HBr

Br

+ H2O

Cl

OH HI

b.

+ H2O

I

OH

+ H2O

9.23 • CH3OH and 1° alcohols follow an SN2 mechanism, which results in inversion of configuration. • Secondary (2°) and 3° alcohols follow an SN1 mechanism, which results in racemization at a stereogenic center. H OH

a.

C

CH3CH2CH2

I H D

HI

C CH3CH2CH2

D

1° alcohol so inversion of configuration

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Alcohols, Ethers, and Epoxides 9–11 CH3

b.

Br

HBr

3° alcohol, so Br
attacks from above and below. The product is achiral.

CH3

OH

achiral starting material HO

achiral product Cl

HCl

c.

Cl

3° alcohol = racemization

9.24 OH

a.

HCl

Cl

HCl

c.

Cl

OH

(product formed after a 1,2-H shift) Cl HCl

b. OH

(product formed after a 1,2-CH3 shift)

9.25 Substitution reactions of alcohols using SOCl2 proceed by an SN2 mechanism. Therefore, there is inversion of configuration at a stereogenic center. H OH

SOCl2

Reactions using SOCl2 proceed by an SN2 mechanism = inversion of configuration.

Cl H

pyridine S

R

9.26 Substitution reactions of alcohols using PBr3 proceed by an SN2 mechanism. Therefore, there is inversion of configuration at a stereogenic center. PBr3

H OH

Reactions using PBr3 proceed by an SN2 mechanism = inversion of configuration.

Br H

S

R

9.27 Stereochemistry for conversion of ROH to RX by reagent: [1] HX—with 1°, SN2, so inversion of configuration; with 2° and 3°, SN1, so racemization. [2] SOCl2—SN2, so inversion of configuration. [3] PBr3—SN2, so inversion of configuration. OH

a.

SOCl2

c.

Cl

OH

PBr3

Br

pyridine OH

HI

I

b. 3° alcohol, SN1 = racemization

I

SN2 = inversion

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Chapter 9–12 9.28 To do a two-step synthesis with this starting material: [1] Convert the OH group into a good leaving group (by using either PBr3 or SOCl2). [2] Add the nucleophile for the SN2 reaction. H H

H

PBr3

CH3 C OH

N3

CH3

CH3 C Br

CH3

CH3 C N3

CH3CH2O

CH3

H CH3 C OCH2CH3 CH3

bad leaving group

good leaving group

9.29 O

+ CH3

a. CH3CH2CH2CH2 OH H OH

b. CH3CH2CH2

C

CH3CH2CH2CH2 O S

SO2Cl

pyridine

CH3 + Cl

O

H OTs TsCl

CH3

pyridine

CH3CH2CH2

C

+ Cl

CH3

9.30 OTs +

a.

1° tosylate

b.

CN

SN2 CN

+

CH3CH2CH2 OTs

1° tosylate

K+
OC(CH3)3

E2

CH3CH CH2 + K+
OTs + HOC(CH3)3

strong bulky base

H OTs

c.

CH3

HS H +

C

CH2CH2CH3

2° tosylate

+ OTs

strong nucleophile

SH

SN2 strong nucleophile

CH3

C

SN2 product (inversion of configuration)

CH2CH2CH3

(Substitution is favored over elimination.)

9.31 HO H

S

TsCl

TsO H

NaOH

H OH

SN2

pyridine

retention S enantiomers

inversion R

One inversion from starting material to product.

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Alcohols, Ethers, and Epoxides 9–13 9.32 These reagents can be classified as: [1] SOCl2, PBr3, HCl, and HBr replace OH with X by a substitution reaction. [2] Tosyl chloride (TsCl) makes OH a better leaving group by converting it to OTs. [3] Strong acids (H2SO4) and POCl3 (pyridine) result in elimination by dehydration. H

a.

CH3 C OH CH3

H

SOCl2

pyridine

CH3 C OH

d.

CH3

H

b.

H

CH3 C Cl

CH3

TsCl

H

CH3 C OH

CH3

CH3

H

c.

CH3

H

e.

CH3 C OTs

pyridine

CH3 C Br

CH3

H

CH3 C OH

H

HBr

[1] PBr3 [2] NaCN

CH3 C CN CH3

H H2SO4

CH3 C OH

CH2 CHCH3

CH3 C OH

f.

CH3

CH3

POCl3

CH2 CHCH3

pyridine

9.33 HBr

a. CH3CH2 O CH2CH3 CH3

b.

2 CH3CH2 Br

HBr

CH3 C O CH2CH3

c.

+ H2O

O CH3

HBr

Br

CH3 CH3 C Br

H

+ CH3Br +

+ H2O

CH3CH2Br + H2O

H

9.34 Ether cleavage can occur by either an SN1 or SN2 mechanism, but neither mechanism can occur when the ether O atom is bonded to an aromatic ring. An SN1 reaction would require formation of a highly unstable carbocation on a benzene ring, a process that does not occur. An SN2 reaction would require backside attack through the plane of the aromatic ring, which is also not possible. Thus cleavage of the Ph–OCH3 bond does not occur. HBr

OCH3

anisole

OH

+

CH3Br

phenol

Br

bromobenzene

NOT formed SN1:

SN2: O CH3

CH3OH

H

highly unstable carbocation

CH3 O

H

Br

9.35 Compare epoxides and cyclopropane. For a compound to be reactive towards nucleophiles, it must be electrophilic. 
O

+

+

epoxide

cyclopropane

O is electronegative and pulls electron density away from C's. This makes them electrophilic and reactive with nucleophiles.

Cyclopropane has all C's and H's, so all nonpolar bonds. There are no electrophilic C's so it will not react with nucleophiles.

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Chapter 9–14 9.36 Two rules for reaction of an epoxide: [1] Nucleophiles attack from the back side of the epoxide. [2] Negatively charged nucleophiles attack at the less substituted carbon. CH3

a.

O

CH3 OH

[1] CH3CH2O [2] H2O

O C C

b. CH3

OCH2CH3

H

Attack here: less substituted C backside attack

[1] H

H

H

HO

C C

H [2] H O 2

CH3 H

H

C C C CH

Attack here: less substituted C backside attack

9.37 In both isomers,
OH attacks from the back side at either C–O bond. cis-2,3-dimethyloxirane [1] –OH

O H CH3

C

C

H CH3

CH3

OH

H C

[2] H2O

C

HO

+ H CH3

HO

H

C H CH3

CH3

C OH

[1] –OH [2] H2O

O H CH3

C

C

H CH3

enantiomers HO

HO O H CH3

C

[1] –OH C

CH3 H

CH3

OH

H C

[2] H2O

HO

trans-2,3-dimethyloxirane

HO

HO

+

C H

CH3

CH3 H C

H CH3

C OH

identical Rotate around the C–C bond to see the plane of symmetry. HO

OH C

CH3

H

C H

CH3

meso compound

[1]

O

–OH

[2] H2O

H CH3

C

C

CH3 H

HO

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Alcohols, Ethers, and Epoxides 9–15 9.38 Remember the difference between negatively charged nucleophiles and neutral nucleophiles: • Negatively charged nucleophiles attack first, followed by protonation, and the nucleophile attacks at the less substituted carbon. • Neutral nucleophiles have protonation first, followed by nucleophilic attack at the more substituted carbon. BUT – trans or anti products are always formed regardless of the nucleophile. CH3

a.

O

HBr

Br CH3

c.

OH

CH3CH2 CH3CH2

H H

CH3

b.

[1] CN [2] H2O

negatively charged nucleophile: attack at less substituted C

CH3CH2OH H2SO4

neutral nucleophile: attack at more substituted C

neutral nucleophile: attack at more substituted C O

CH3CH2 CH3CH2

O

CH3 OH CN

O

d.

CH3CH2 CH3CH2

OH C

C

CH3CH2O

H

HO H H

[1] CH3O [2] CH3OH

H C

CH3CH2 CH3CH2

negatively charged nucleophile: attack at less substituted C

H

H

C OCH3

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Chapter 9–16 9.39 Draw the structure of each alcohol, using the definitions in Answer 9.1. OH

OH

a.

OH

1°

b.

OH

2°

3°

enol

9.40 Use the directions from Answer 9.3. a.

(CH3)2CHCH2CH2CH2OH

[1]

[2]

H CH3 C CH2CH2CH2OH

H

[3]

4-methyl-1-pentanol

[3]

4-ethyl-6-methyl-3-heptanol

[3]

4-ethyl-5-methyl-3-octanol

[3]

cis-1,4-cyclohexanediol

[3]

3,3-dimethylcyclohexanol

[3]

(2R,3R)-2,3-butanediol

[3]

5-methyl-2,3,4-heptanetriol

CH3 C CH2CH2CH2OH

CH3

CH3

5 carbons = pentanol

b.

1 4-methyl

(CH3)2CHCH2CH(CH2CH3)CH(OH)CH2CH3

[1]

H

[2]

H OH

CH3 C CH2 C CH CH3

CH2CH3

H

CH3 C CH2 C CH

CH2CH3

CH3

7 carbons = heptanol

c. [1]

CH2CH3

CH2CH3

6-methyl

4-ethyl

5-methyl [2] OH OH

8 carbons = octanol

d. [1]

3

H OH

3 4

HO

OH

4-ethyl

[2]

1

HO

OH

cis

cyclohexanediol

e. [1]

[2]

HO

HO

6 carbons = cyclohexanol

f.

[1]

HO H

3,3-dimethyl [2]

HO H

2 HO H

3

HO H

2R,3R

4 carbons = butanediol

g. OH

[1] OH

OH

7 carbons = heptanetriol

[2]

4

OH

OH

5-methyl

2

OH

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Alcohols, Ethers, and Epoxides 9–17 h. [1]

[2] HO

CH(CH3)2

[3]

1 CH(CH3)2

HO

trans-3isopropylcyclopentanol

5 carbons = cyclopentanol 3-isopropyl

9.41 Use the rules from Answers 9.5 and 9.7. a.

d.

O

O

1,2-epoxy-2-methylhexane or 2-butyl-2-methyloxirane or 2-methylhexene oxide

dicyclohexyl ether 4,4-dimethyl b.

CH2CH3

e.

OCH2CH2CH3

longest chain = heptane

substituent = 3-propoxy 4,4-dimethyl-3-propoxyheptane

c.

O epoxy 2 5 carbons = cyclopentane

1,2-epoxy-1-ethylcyclopentane or 1-ethylcyclopentene oxide CH3

O

f.

ethyl isobutyl ether or 1-ethoxy-2-methylpropane

CH3

CH3 C O C CH3 CH3

CH3

tert-butyl tert-butyl di-tert-butyl ether

9.42 Use the directions from Answer 9.4. a. 4-ethyl-3-heptanol

3

d. 6-sec-butyl-7,7-diethyl-4-decanol

4

OH

6

OH

4

1 OH

b. trans-2-methylcyclohexanol

OH

or CH3 HO

1

e. 3-chloro-1,2-propanediol HO

CH3

OH

3

c. 2,3,3-trimethyl-2-butanol

f. diisobutyl ether 2

2

O

7

3 Cl

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Chapter 9–18 1

g. 1,2-epoxy-1,3,3-trimethylcyclohexane

i. (2R,3S)-3-isopropyl-2-hexanol 3 O

2

1

OH

3 j. (2S)-2-ethoxy-1,1-dimethylcyclopentane 1

h. 1-ethoxy-3-ethylheptane O 1

3 2

O

9.43 Eight constitutional isomers of molecular formula C5H12O containing an OH group: OH 1-pentanol OH

1 OH 3-methyl-1-butanol 1° alcohol

3

1° alcohol

1 OH 2-methyl-1-butanol 1° alcohol

2-pentanol OH 2 1

2° alcohol

OH

OH

3-pentanol 2° alcohol

2,2-dimethyl-1-propanol 1° alcohol

2

3-methyl-2-butanol 2° alcohol

OH

2-methyl-2-butanol 3° alcohol

2

9.44 Use the boiling point rules from Answer 9.8. a.

CH3CH2OCH3

ether no hydrogen bonding lowest bp b.

(CH3)2CHOH

2° alcohol hydrogen bonding intermediate bp

CH3CH2CH2OH

1° alcohol hydrogen bonding highest bp

CH3CH2CH2CH2CH2CH3

CH3CH2CH2CH2CH2CH2OH

no OH group lowest water solubility

one OH group intermediate water solubility

HOCH2CH2CH2CH2CH2CH2OH

two OH groups highest water solubility

9.45 Melting points depend on intermolecular forces and symmetry. (CH3)2CHCH2OH has a lower melting point than CH3CH2CH2CH2OH because branching decreases surface area and makes (CH3)2CHCH2OH less symmetrical so it packs less well. Although (CH3)3COH has the most branching and least surface area, it is the most symmetrical so it packs best in a crystalline lattice, giving it the highest melting point. OH

–108 °C lowest melting point

OH

–90 °C intermediate melting point

OH

26 °C highest melting point

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Alcohols, Ethers, and Epoxides 9–19 9.46 Stronger intermolecular forces increase boiling point. All of the compounds can hydrogen bond, but both diols have more opportunity for hydrogen bonding since they have two OH groups, making their bp’s higher than the bp of 1-butanol. 1,2-Propanediol can also intramolecularly hydrogen bond. Intramolecular hydrogen bonding decreases the amount of intermolecular hydrogen bonding, so the bp of 1,2-propanediol is somewhat lower. Increasing boiling point OH

HO

OH

HO OH

1-butanol 118 °C

1,2-propanediol 187 °C

1,3-propanediol 215 °C

9.47 a. CH3CH2CH2OH b. CH3CH2CH2OH c. CH3CH2CH2OH

d. CH3CH2CH2OH e. CH3CH2CH2OH f. CH3CH2CH2OH g. CH3CH2CH2OH

h. CH3CH2CH2OH i.

CH3CH2CH2OH

H2SO4

CH3CH CH2

NaH

+ H2O

CH3CH2CH2O Na+

HCl

CH3CH2CH2Cl

ZnCl2 HBr

+ H2

+ H2O

CH3CH2CH2Br + H2O

SOCl2

CH3CH2CH2Cl

pyridine PBr3

CH3CH2CH2Br

TsCl

CH3CH2CH2OTs

pyridine [1] NaH

[2] CH3CH2Br

CH3CH2CH2O Na+

[1] TsCl

CH3CH2CH2OTs

[2] NaSH

CH3CH2CH2OCH2CH3

CH3CH2CH2SH

9.48 a.

OH

NaH

b.

OH

NaCl

c.

OH

HBr

d.

OH

HCl

e.

OH

H2SO4

f.

OH

NaHCO3

Br

g.

OH

[1] NaH

Cl

h.

OH

POCl3 pyridine

O

N.R.

Na+ + H2

+ H2O

N.R.

O

[2] CH3CH2Br

O CH2CH3

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Chapter 9–20 9.49 Dehydration follows the Zaitsev rule, so the more stable, more substituted alkene is the major product. TsOH

a. OH

tetrasubstituted major product

b.

CH2CH3 OH

c.

OH

disubstituted CH2CH3

CHCH3

TsOH

TsOH

trisubstituted disubstituted major product d.

TsOH

CH3CH2CH2CH2OH

CH3CH2CH CH2

OH TsOH

e. tetrasubstituted major product

disubstituted

Two products formed by carbocation rearrangement

9.50 The more stable alkene is the major product.

OH H2SO4

trans and disubstituted major product

monosubstituted

cis and disubstituted

9.51 OTs is a good leaving group and will easily be replaced by a nucleophile. Draw the products by substituting the nucleophile in the reagent for OTs in the starting material. a.

CH3CH2CH2CH2 OTs

b.

CH3CH2CH2CH2 OTs

c.

CH3CH2CH2CH2 OTs

d.

CH3CH2CH2CH2 OTs

CH3SH

SN2 NaOCH2CH3

SN2 NaOH

SN2 K+

CH3CH2CH2CH2 SCH3 + HOTs

CH3CH2CH2CH2 OCH2CH3

CH3CH2CH2CH2 OH

+ Na+

+ Na+

–OTs

–OTs

–

OC(CH3)3

E2

CH3CH2CH CH2 + (CH3)3COH + K+ –OTs

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Alcohols, Ethers, and Epoxides 9–21 9.52 HBr

a.

b.

+

HO H OH

H D

Br H

HCl ZnCl2

H Br

Cl

1° Alcohol will undergo SN2. inversion

H D

SOCl2

c.

pyridine

HO H

SOCl2 always implies SN2. inversion

H Cl

TsCl pyridine

d.

2° Alcohol will undergo SN1. racemization

KI

HO H

TsO H

SN2 inversion

H

I

Configuration is maintained. C–O bond is not broken.

9.53 NaH

(a) (b) (c) (2R)-2-hexanol

B=

O H TsCl pyridine

HO H

CH3I

A=

PBr3

CH3O H CH3O

C= TsO H

E=

D= H OCH3

CH3O

F= H Br

CH3O H

Routes (a) and (c) given identical products, labeled B and F.

9.54 Acid-catalyzed dehydration follows an E1 mechanism for 2o and 3o ROH with an added step to make a good leaving group. The three steps are: [1] Protonate the oxygen to make a good leaving group. [2] Break the CO bond to form a carbocation. [3] Remove a  hydrogen to form the
bond.

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Chapter 9–22 OH

a.

CH3 H OSO3H

CH3

CH3 +

overall reaction

The steps:

CH2 +

+ H2O

H O H CH3

H H

+ HSO4

CH3

+ H2SO4

CH3

+ H2O

H

HSO4

2° carbocation and

+ HSO4

H CH3

1,2-H shift

CH2

CH2

H

3° carbocation and

2° carbocation

H

+ HSO4 CH3

CH3

CH3

b.

CH3 OH

H OSO3H

overall reaction

+ H2SO4

+ H2SO4

CH3 + H2O CH3

The steps: CH3 CH3

CH3 + HSO4

CH3 + H O 2

1,2-CH3 shift

2° carbocation

CH3 + H2SO4

CH3 H

O H H

CH3

+ HSO4

3° carbocation

CH3

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Alcohols, Ethers, and Epoxides 9–23 9.55 a.

Br

OH

Br

C

B

A

(E1 with acid)

OH

(E2 with base)

D

(E2 with base)

(E1 with acid)

3,3-dimethylcyclopentene

major product

b. The best starting material to form 3,3-dimethylcyclopentene would be C since the alkene can be formed as a single product by an E2 elimination with base. 9.56 With POCl3 (pyridine), elimination occurs by an E2 mechanism. Since only one carbon has a
hydrogen, only one product is formed. With H2SO4, the mechanism of elimination is E1. A 2° carbocation rearranges to a 3° carbocation, which has three pathways for elimination.

O OH

Cl

P Cl

H Cl

O

POCl2

N

POCl2

N

O H

+ N H

V

+

W

Cl

+ –OPOCl2

+ N H

OH

H OSO3H

+ OH2

+ + H2O

V

+ HSO4–

2° carbocation

1,2–CH3 shift

+

3° carbocation

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Chapter 9–24 HSO4–

H

H

+

+

+ HSO4–

HSO4–

H

3° carbocation

3° carbocation

X

3° carbocation

Y

+ H2SO4

+

Z +

H2SO4

H2SO4

9.57 To draw the mechanism: [1] Protonate the oxygen to make a good leaving group. [2] Break the CO bond to form a carbocation. [3] Look for possible rearrangements to make a more stable carbocation. [4] Remove a  hydrogen to form the
bond. Dark and light circles are meant to show where the carbons in the starting material appear in the product.

OH

+ H2O

O H H

H OSO3H

H

2o carbocation

+ HSO4–

+ HSO4

–

3o carbocation

9.58 HO

H

HBr

a.

Br

H

H

S HO

H

b.

Br

R HO

H

HCl

c.

Cl

H

SOCl2

pyridine

PBr3 follows SN2 = inversion.

H

H

Cl

2° alcohol SN1 = racemization

R

S HO

d.

2° alcohol SN1 = racemization

R

H

PBr3

Br

H

R

Cl

SOCl2 follows SN2 = inversion.

+ H2SO4

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247

Alcohols, Ethers, and Epoxides 9–25 9.59 OH

3-methyl-2-butanol

[1] HBr

H

1,2-H shift

H

Br

Br

[2] –H2O

2° carbocation 2-methyl-1-propanol

HBr

HO

3° carbocation

H2O

Br Br

H Br

H2O

SN2 no carbocation

The 2° alcohol reacts by an SN1 mechanism to form a carbocation that rearranges.

The 1° alcohol reacts with HBr by an SN2 mechanism. no carbocation intermediate = no rearrangement possible

9.60 Conversion of a 1° alcohol into a 1° alkyl chloride occurs by an SN2 mechanism. SN2 mechanisms occur more readily in polar aprotic solvents by making the nucleophile stronger. No added ZnCl2 is necessary. HCl R OH

R Cl HMPA

polar aprotic solvent This makes Cl< a better nucleophile.

9.61 H Cl

A

Cl

OH

H2O

OH2

two resonance structures for the carbocation B

Cl

Cl Cl Cl

C

9.62 CH3CH2CH2CH2OH

H2SO4, NaBr

overall reaction

CH3CH2CH2CH2Br

CH3CH2CH CH2

CH3CH2CH2CH2OCH2CH2CH2CH3

Step [1] for all products: Formation of a good leaving group CH3CH2CH2CH2

OH

H OSO3H

CH3CH2CH2CH2

O H

+ HSO4

H

Formation of CH3CH2CH2CH2Br: CH3CH2CH2CH2

O H H

Br

CH3CH2CH2CH2Br + H2O

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Chapter 9–26 Formation of CH3CH2CH=CH2: CH3CH2CH CH2 O H H

CH3CH2CH CH2

+ H2O

H2SO4

H

HSO4–

Ether formation (from the protonated alcohol): CH3CH2CH2CH2

CH3CH2CH2CH2

OH2

CH3CH2CH2CH2

O CH2CH2CH2CH3 + H2O H

HSO4

O H

CH3CH2CH2CH2 O CH2CH2CH2CH3

H2SO4

9.63 H OSO3H

OH O

OH2

+

O

+ HSO4–

O

+ HSO4

–

H

1,2-shift O +

O + H2SO4

+ H2O

H O

+

9.64 HSO4– OH

H OSO3H

OH2

OH

O

OH + HSO4–

H

O

OH

+ H2O

+ H2SO4

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Alcohols, Ethers, and Epoxides 9–27 9.65 a.

c. O

Br

O

CH3CH2OCH2CH2CH3

O

2° halide

CH3CH2O + BrCH2CH2CH3

O

Br

1° halide

1° halide less hindered RX preferred path

OCH2CH2CH3

b.

CH3CH2OCH2CH2CH3

CH3CH2Br + OCH2CH2CH3

1° halide

Neither path preferred.

OCH2CH2CH3

Br O

OCH2CH2CH3

+ BrCH2CH2CH3

2° halide

1° halide less hindered RX preferred path

9.66 A tertiary halide is too hindered and an aryl halide too unreactive to undergo a Williamson ether synthesis. Two possible sets of starting materials:

CH3

CH3

CH3

O C CH3

Br C CH3

Br

O C CH3 CH3

CH3

aryl halide unreactive in SN2

O

CH3

3° alkyl halide too sterically hindered for SN2

9.67 a. b.

(CH3)3COCH2CH2CH3

O

HBr

(CH3)3CBr + BrCH2CH2CH3

(2 equiv) HBr

(2 equiv)

2

Br

H2O

H2O

c.

OCH3

HBr

(2 equiv)

Br

CH3Br H2O

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Chapter 9–28 9.68 overall reaction

CH3

a. O

I

CH3 H

I

+ H2O

I

The steps: I CH3 O

CH3

+ I

H

+

CH3

O

H

H

CH3

O

CH3

CH3

O

Cl

Na+ H

H

O

Cl

CH3

O

I CH3

H

I

H

b.

H

I

I

+ H2 + NaCl

+ H2 + Na+

O

9.69 OC(CH3)3

O C(CH3)3

H–OCOCF3

CH3

OH

H

CH2

+ CF3CO2–

C

OH

CH3

+

CH3

C CH 2

H

CH3

CF3CO2–

+ CF3CO2H

9.70 O

a. H

C

O C

H

H H

HBr

d. H C

BrCH2CH2OH

H

O

b. H

C

C

H

H H

O

c. H

C

C

H

H2O

e. H

H2SO4

[2] H2O

[1] HC C– H H

C

H

HOCH2CH2OH

H H

[2] H2O

O CH3CH2OCH2CH2OH

C

H

CH2 CH2OH

[1]
OH C

f. H C

HC C

[2] H2O

O HOCH2CH2OH

[1] CH3CH2O
H H

C

[1] CH3S


H H

CH3SCH2CH2OH

[2] H2O

9.71 O

a. CH 3 CH3

H H CH3

b.

O

CH3CH2OH H2SO4 [1] CH3CH2O
Na+ [2] H2O

O

CH3 OH CH3 C CH3CH2O

C H H CH3 OH OCH2CH3

OH HBr

c. O

d.

Br OH

[1] NaCN [2] H2O

CN

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Alcohols, Ethers, and Epoxides 9–29 9.72 CH3

H

Cl C

a.

CH3

C

H O

Cl

H C

H

CH3

H CH3 C

C

O

H

CH3

O C4H8O

Na+ H H

Cl

CH3 C

b. H O

H

Cl

CH3 H C

C CH3

CH3 H C

C

O

H

CH3

OH C

c. CH3

C

H

H

CH3 H

Cl

rotate

C

CH3

CH3

H

C

CH3 H

O C4H8O

Na+ H

Cl

C

The 2 CH3 groups are anti in the starting material, making them trans in the product.

CH3 H

CH3 H

Cl

C

C CH3

O H

The 2 CH3 groups are gauche in the starting material, making them cis in the product.

C

H

O

backside attack

CH3 H C

C

H CH3

O C4H8O

Na+ H

9.73 O

O

CH3CH2O

O

CH3CH2O

+

Cl

CH3CH2OCH2

Cl

9.74 a. (1R,2R)-2-isobutylcyclopentanol

f.

b. 2° alcohol

[1]

NaH HO

O

A

H2SO4

HO

A

[2] HO

c. stereoisomer

POCl3

[3]

pyridine HO

HO

Cl

HCl

(1R,2S)-2-isobutylcyclopentanol

[4] HO

SOCl2

d. constitutional isomer [5] HO

pyridine

Cl

OH

TsCl (1S,3S)-3-isobutylcyclopentanol

[6] HO

e. constitutional isomer with an ether O

butoxycyclopentane

pyridine

TsO

CH3

Cl–

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Chapter 9–30 9.75 KOC(CH3)3

OTs

a.

b.

OH

*

Bulky base favors E2.

HBr

H CH3

H CH3

O

c.

Keep the stereochemistry at the stereogenic center [*] the same here since no bond is broken to it.

Br

*

CH3CH2 H

[1]
CN

CH3CH2 C C H CH2CH3 H

[2] H2O

C

+

C

H CH3CH2 CH2CH3 H

NC

KCN (CH3)3C

H

SN2 inversion

Br

CH3

O

O

CH3CO2

OTs

pyridine

H D

CN

OH

H D Br

Br

OH

CH3CO2 H D

HBr

OCH3

OH

[1] NaOCH3 [2] H2O

OH

OH

OCH3 O

NaH

OCH2CH3

CH3CH2I

i. CH2CH3 CH3

j.

CH3CH2 C O CH3 CH3

CH2CH3

CH2CH3 HI

(2 equiv)

CH3 CH3CH2 C

I

+

I

CH3

+ H2O

CH3

9.76 a.

CH2CH3

SN2 inversion

TsCl

OH

h.

C

CH3

f.

g.

C

CN

(CH3)3C

PBr3

OH

e.

H

H

OTs

d.

HO

OH

OH

b.

OH

c.

OH

HBr or PBr3

Br

HCl, ZnCl2 or SOCl2, pyridine

Cl

[1] Na+ H
O

[2] CH3CH2Cl O

identical

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Alcohols, Ethers, and Epoxides 9–31 [1] TsCl, pyridine

d.

OH

OTs

[2] N3
N3

Make OH a good leaving group (use TsCl); then add N3
.

9.77 a.

OH

HCl or SOCl2, pyridine

b.

OH

H2SO4

c.

OH [1] Na+ H


d.

OH

Cl

[2] CH3Cl

O

OCH3

CN

OTs [2] –CN

[1] TsCl, pyridine

Make OH a good leaving group (use TsCl); then add
CN.

9.78 Br

(a) = HBr or PBr3

or other strong base

OH OTs

(c) = (e) =

(d) = KOC(CH3)3 or other strong base

TsCl, pyridine

H2SO4 or TsOH

(b) = KOC(CH3)3

Br (f) = KOC(CH3)3

NBS

or other strong base HOCl OH

(g) = NaH

(h) =

OH

+ enantiomer

O Cl


OH,

H2O

OH

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Chapter 9–32 9.79 OH

O

O

Cl

O

O O

NaH

(CH3)2CHNH2

N H H

O

proton transfer O OH

N H

propranolol

9.80 a. All 2° OH groups on stereogenic centers are circled (40 stereogenic centers). OH

OH

O O

HO

OH

O

OH

O

OH NH2

OH

OH

OH

HO

OH OH

OH

OH OH

OH O HO

O

N

N

H

H

HO OH

OH

O OH

OH

O

OH

OH

OH H

OH

HO

HO O

OH

O O

HO

OH O

palytoxin

OH

OH HO

OH

H

OH

OH

OH OH

OH

OH

OH

This carbocation is resonance stabilized, so loss of H2O to form it is easier than loss of H2O from a 2° alcohol, where the carbocation is not resonance stabilized. R

b.

R C OH OR'

hemiacetal

H+

R R C OH2 OR'

R

R

R C

R C

OR'

OR'

resonance-stabilized cation H2O

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255

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Text

Alcohols, Ethers, and Epoxides 9–33 9.81 If the base is not bulky, it can react as a nucleophile and open the epoxide ring. The bulky base cannot act as a nucleophile, and will only remove the proton. –N(CH

H

2CH3)2

Li+ O

+ H OH

O

+

HN(CH2CH3)2

LiOH

OH

9.82 First form the 2° carbocation. Then lose a proton to form each product. H

H CH3CH2 C CH2 OH

CH3CH2 C CH2

H OH2

H

OH2

H

1o alcohol

no 1o carbocation at this step

H

H

1,2-shift

CH3CH2 C CH2

+

H2O

H

+ H2O

2o carbocation

H

CH3CH2 C CH2 H

CH3CH2 C CH2

H2O

H

+

CH3

H

+

H

+

C C

CH3

H

H

CH3CH2CH=CH2

H3O

H

C C

CH3 CH C CH3

=

CH3

H3O

CH3

H2O

9.83 OH OH CH3 C

C CH3

OH

OH2 OH

H OSO3H

CH3 C

CH3 CH3

CH3 C

C CH3

pinacol

C CH3

CH3 C

shift

CH3 CH3

CH3 CH3

CH3

1,2-CH3

CH3

OH C CH3

+ H2O

+ HSO4

CH3 H2SO4 +

CH3 C CH3

O

CH3 O H

C

CH3 C

CH3

CH3

HSO4

C CH3

pinacolone

9.84 O O H

O C

I

O

H H O

Na+ OH

C I

O

O

O

O

H H O

C

I

O

H H O

O H

C

C

H I

O

H H

O

I

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10. Alkenes

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Text

Study Guide/Solutions Manual to accompany Organic Chemistry, Third Edition

257

Alkenes 10–1 C Chhaapptteerr 1100:: A Allkkeenneess  G Geenneerraall ffaaccttss aabboouutt aallkkeenneess • Alkenes contain a carbon–carbon double bond consisting of a stronger  bond and a weaker
bond. Each carbon is sp2 hybridized and trigonal planar (10.1). • Alkenes are named using the suffix -ene (10.3). • Alkenes with different groups on each end of the double bond exist as a pair of diastereomers, identified by the prefixes E and Z (10.3B). Two higher priority groups on the same side

Two higher priority groups on opposite sides CH3

CH3

CH3

C C H

CH2CH3

H

E isomer (2E)-3-methyl-2-pentene

•

CH3

Z isomer (2Z)-3-methyl-2-pentene

Alkenes have weak intermolecular forces, giving them low mp’s and bp’s, and making them water insoluble. A cis alkene is more polar than a trans alkene, giving it a slightly higher boiling point (10.4). cis-2-butene more polar isomer

trans-2-butene

CH3

CH3

H

CH3

C C

less polar isomer

C C H

H

H

CH3

no net dipole

a small net dipole higher bp

•

CH2CH3

C C

lower bp

Since a
bond is electron rich and much weaker than a  bond, alkenes undergo addition reactions with electrophiles (10.8).

  SStteerreeoocchheem miissttrryy ooff aallkkeennee aaddddiittiioonn rreeaaccttiioonnss ((1100..88)) A reagent XY adds to a double bond in one of three different ways: • Syn addition—X and Y add from the same side. H BH2

C C

•

BH2 C C

•

Syn addition occurs in hydroboration.

Anti addition—X and Y add from opposite sides. C C

•

H

X2 or X2, H2O

X

•

C C X(OH)

Anti addition occurs in halogenation and halohydrin formation.

Both syn and anti addition occur when carbocations are intermediates. C C

H X or H2O, H+

H

X(OH) C C

•

H

and

C C X(OH)

Syn and anti addition occur in hydrohalogenation and hydration.

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Chapter 10–2

A Addddiittiioonn rreeaaccttiioonnss ooff aallkkeenneess [1] Hydrohalogenation—Addition of HX (X = Cl, Br, I) (10.9–10.11) RCH CH2 + H X

R CH CH2 X

H

alkyl halide

• • • • •

The mechanism has two steps. Carbocations are formed as intermediates. Carbocation rearrangements are possible. Markovnikov’s rule is followed. H bonds to the less substituted C to form the more stable carbocation. Syn and anti addition occur.

[2] Hydration and related reactions—Addition of H2O or ROH (10.12) RCH CH2 + H OH

H2SO4

R CH CH2 OH H

alcohol

RCH CH2 + H OR

H2SO4

R CH CH2 OR H

ether

For both reactions: • The mechanism has three steps. • Carbocations are formed as intermediates. • Carbocation rearrangements are possible. • Markovnikov’s rule is followed. H bonds to the less substituted C to form the more stable carbocation. • Syn and anti addition occur.

[3] Halogenation—Addition of X2 (X = Cl or Br) (10.13–10.14) RCH CH2

+

X X

R CH CH2 X

X

vicinal dihalide

• • • •

The mechanism has two steps. Bridged halonium ions are formed as intermediates. No rearrangements occur. Anti addition occurs.

[4] Halohydrin formation—Addition of OH and X (X = Cl, Br) (10.15) RCH CH2 + X X

H2O

R CH CH2 OH X

halohydrin

• • • • • •

The mechanism has three steps. Bridged halonium ions are formed as intermediates. No rearrangements occur. X bonds to the less substituted C. Anti addition occurs. NBS in DMSO and H2O adds Br and OH in the same fashion.

[5] Hydroboration–oxidation—Addition of H2O (10.16) [1] BH3 or 9-BBN • Hydroboration has a one-step mechanism. RCH CH2 R CH CH2 • No rearrangements occur. [2] H2O2, HO H OH • OH bonds to the less substituted C. alcohol • Syn addition of H2O results.

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Alkenes 10–3 C Chhaapptteerr 1100:: A Annssw weerrss ttoo PPrroobblleem mss 10.1 Six alkenes of molecular formula C5H10:

trans cis diastereomers

10.2 To determine the number of degrees of unsaturation: [1] Calculate the maximum number of H’s (2n + 2). [2] Subtract the actual number of H’s from the maximum number. [3] Divide by two. a. C2H2 [1] maximum number of H's = 2n + 2 = 2(2) + 2 = 6 [2] subtract actual from maximum = 6 – 2 = 4 [3] divide by two = 4/2 = 2 degrees of unsaturation b. C6H6 [1] maximum number of H's = 2n + 2 = 2(6) + 2 = 14 [2] subtract actual from maximum = 14 – 6 = 8 [3] divide by two = 8/2 = 4 degrees of unsaturation c. C8H18 [1] maximum number of H's = 2n + 2 = 2(8) + 2 = 18 [2] subtract actual from maximum = 18 – 18 = 0 [3] divide by two = 0/2 = 0 degrees of unsaturation

d. C7H8O Ignore the O. [1] maximum number of H's = 2n + 2 = 2(7) + 2 = 16 [2] subtract actual from maximum = 16 – 8 = 8 [3] divide by two = 8/2 = 4 degrees of unsaturation e. C7H11Br Because of Br, add one more H (11 + 1 H = 12 H's). [1] maximum number of H's = 2n + 2 = 2(7) + 2 = 16 [2] subtract actual from maximum = 16 – 12 = 4 [3] divide by two = 4/2 = 2 degrees of unsaturation f. C5H9N Because of N, subtract one H (9 – 1 H = 8 H's). [1] maximum number of H's = 2n + 2 = 2(5) + 2 = 12 [2] subtract actual from maximum = 12 – 8 = 4 [3] divide by two = 4/2 = 2 degrees of unsaturation

10.3 One possibility for C6H10: a. a compound that has 2
bonds

c. a compound with 2 rings

b. a compound that has 1 ring and 1
bond

d. a compound with 1 triple bond

10.4 To name an alkene: [1] Find the longest chain that contains the double bond. Change the ending from -ane to -ene. [2] Number the chain to give the double bond the lower number. The alkene is named by the first number. [3] Apply all other rules of nomenclature. To name a cycloalkene: [1] When a double bond is located in a ring, it is always located between C1 and C2. Omit the “1” in the name. Change the ending from -ane to -ene. [2] Number the ring clockwise or counterclockwise to give the first substituent the lower number. [3] Apply all other rules of nomenclature.

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Chapter 10–4 3-methyl 1

a. [1] CH2 CHCH(CH3)CH2CH3

CH3 CH CHCHCH 2 [2] 2CH3

5 C chain with double bond pentene

[3] 3-methyl-1-pentene

1-pentene 3 CH3CH2

b. [1] (CH3CH2)2C CHCH2CH2CH3

C CHCH2CH2CH3

[2] CH3CH2

7 C chain with double bond heptene

[3] 3-ethyl-3-heptene

3-heptene 3-ethyl 2-ethyl

c. [1]

[2]

4-methyl

[3] 2-ethyl-4-methyl-1-pentene

1 1-pentene

5 C chain with double bond pentene

1

[2]

d. [1]

4

[3] 3,4-dimethylcyclopentene

3

3,4-dimethyl

5 C ring with a double bond cyclopentene

1-methyl

e. [1]

[2]

[3] 5-tert-butyl-1-methylcyclohexene 1

5-tert-butyl 6 C ring with a double bond cyclohexene

10.5 Use the rules from Answer 10.4 to name the compounds. Enols are named to give the OH the lower number. Compounds with two C=C’s are named with the suffix -adiene.

a.

1

3

[2]

[1]

[3] 4-ethyl-3-hexen-1-ol

OH

OH

4-ethyl 6 C chain with double bond hexene [1]

OH

1

4

[2]

[3] 5-ethyl-6-methyl-7-octen-4-ol

OH

b. 6-methyl

5-ethyl 8 C chain with double bond octene [1]

5 [2]

2

c.

[3] 2,6-dimethyl-2,5-heptadiene 2-methyl

6-methyl 7 C chain with two double bonds heptadiene

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Alkenes 10–5 10.6 To label an alkene as E or Z: [1] Assign priorities to the two substituents on each end using the rules for R,S nomenclature. [2] Assign E or Z depending on the location of the two higher priority groups. • The E prefix is used when the two higher priority groups are on opposite sides. • The Z prefix is used when the two higher priority groups are on the same side of the double bond. higher priority

Cl

CH3

a.

higher priority

C C

higher priority Two higher priority groups are on opposite sides: E isomer.

OCH3

Br

H

H

H

c. higher priority

higher priority

C6H5

O H

higher priority

CH3CH2

b.

CH2CH3

higher priority

kavain

C C H

higher priority

In both double bonds, the two higher priority groups are on opposite sides: E isomers.

CH3

Two higher priority groups are on the same side: Z isomer.

10.7 To name an alkene: First follow the rules from Answer 10.4. Then, when necessary, assign an E or Z prefix based on priority, as in 10.6. 2-hexene

Br

CH2CH2CH3

CH3

[2]

C C H

higher priority

CH2CH2CH3

CH3

a. [1]

[3] (2E)-3-bromo-2-hexene

C C

E alkene

6 C chain with double bond hexene

Br

H

higher priority

3-bromo

3-decene CH3CH2

b. [1]

CH2CH2CH2CH2C(CH3)3

CH3CH2

[2]

C C H

higher priority

CH2CH3

10 C chain with double bond decene

CH3

C C

Z alkene

H

CH3

CH2CH2CH2CH2CCH3

9,9-dimethyl higher priority 9,9-dimethyl

CH2CH3

4-ethyl

[3] (3Z)-4-ethyl-9,9-dimethyl-3-decene

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Chapter 10–6 10.8 To work backwards from a name to a structure: [1] Find the parent name and functional group and draw, remembering that the double bond is between C1 and C2 for cycloalkenes. [2] Add the substituents to the appropriate carbons. a. (3Z)-4-ethyl-3-heptene

The higher priority groups are on the same side = Z.

c. (1Z)-2-bromo-1-iodo-1-hexene 6 carbons

7 carbons

The double bond is between C1 and C2. I Br

4-ethyl The double bond is between C3 and C4. b. (2E)-3,5,6-trimethyl-2-octene

The higher priority groups are on the same side = Z.

The double bond is between C2 and C3.

8 carbons The higher priority groups are on opposite sides = E.

3,5,6-trimethyl

10.9 Draw all of the stereoisomers and then use the rules from Answer 10.6 to name each diene.

E

E

E

(2E,4E)-2,4-hexadiene

Z

Z

(2E,4Z)-2,4-hexadiene

Z

(2Z,4Z)-2,4-hexadiene

10.10 To rank the isomers by increasing boiling point: Look for polarity differences: small net dipoles make an alkene more polar, giving it a higher boiling point than an alkene with no net dipole. Cis isomers have a higher boiling point than their trans isomers. CH3

CH3

CH3CH2

C C CH3

H

CH3CH2

C C CH3

H

All dipoles cancel. smallest surface area no net dipole lowest bp

CH2CH3

C C CH2CH3

Two dipoles cancel. no net dipole trans isomer intermediate bp

H

H

Two dipoles reinforce. net dipole cis isomer highest bp

10.11 Recall from Section 5.13B that the odor of a molecule is determined more by shape than by functional groups. That is why the R and S isomers of limonene smell so differently.

H

(R)-limonene

H

(S)-limonene

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Alkenes 10–7 10.12 Increasing number of double bonds = decreasing melting point. O O

OH

stearic acid no double bonds highest melting point

OH

stearidonic acid

O

4 double bonds lowest melting point

OH

linolenic acid 3 double bonds intermediate melting point

10.13 Br

a.

H2SO4 OH

CH2=CHCH2CH2CH2CH3

NaOCH2CH3

b.

+ CH3CH=CHCH2CH2CH3

10.14 To draw the products of an addition reaction: [1] Locate the two bonds that will be broken in the reaction. Always break the
bond. [2] Draw the product by forming two new  bonds. HCl

H

CH3

two new
bonds

a.

c.

Cl

b.

H HCl

CH3 CH3

CH3

CH3CH2CH2CH CHCH2CH2CH3

two new
bonds

Cl

H H

HCl

CH3CH2CH2 C C CH2CH2CH3 H Cl

two new
bonds

10.15 Addition reactions of HX occur in two steps: [1] The double bond attacks the H atom of HX to form a carbocation. [2] X attacks the carbocation to form a CX bond. transition state step [1]: H + H

H H Cl

[1]

transition state step [2]:

H H H

H

[2] Cl

Cl

+ H

Cl 


+

Cl 


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Chapter 10–8 10.16 Addition to alkenes follows Markovnikov’s rule: When HX adds to an unsymmetrical alkene, the H bonds to the C that has more H’s to begin with. 2 H's H adds here.

no H's Cl adds here. CH3

CH3 Cl

HCl

a.

c.

H H

H

CH2

HCl

Cl CH3

no H's Cl adds here.

one H H adds here.

H

no H's Cl adds here CH3

b.

CH3

HCl

C CH2

Cl C

CH2

CH3 H

CH3

2 H's H adds here.

10.17 To determine which alkene will react faster, draw the carbocation that forms in the ratedetermining step. The more stable, more substituted carbocation, the lower the Ea to form it and the faster the reaction. CH3

CH3

H

C C CH3

CH3

C

H X

H

C C

H

CH3

H

CH2 H

CH3CHCH2 H X

3° carbocation faster reaction

H

H

2° carbocation slower reaction

10.18 Look for rearrangements of a carbocation intermediate to explain these results.

H Cl

CH3

Cl

H

H

H H

Cl CH3

1-chloro-3methylcyclohexane

H CH3

H

H

2° carbocation

CH3 H

+ Cl–

2° carbocation

Rearrangement would not further stabilize this carbocation.

H

1,2-H shift Cl CH3

3° carbocation

CH3 Cl

1-chloro-1methylcyclohexane

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Alkenes 10–9 10.19 Addition of HX to alkenes involves formation of carbocation intermediates. Rearrangement of the carbocation will occur if it forms a more stable carbocation. H

a.

H CH3

CH3 C C

H

H CH3

H C C CH2CH3 CH2CH3

H C C CH2CH3

H

H Br

3° carbocation no rearrangement b.

CH3

H H

H

C C H

CH3 C C CH2CH3 CH2CH3

CH3 C C CH2CH3

H Br

H CH3 CH3 C C CH(CH3)2

CH3 C C CH(CH3)2 CH(CH3)2

(cis or trans)

CH3 C C CH2CH3

H

H H

H

H

H H Br H

2° carbocation 2° carbocation Rearrangement would not further stabilize either carbocation. no rearrangement

C C

c.

H H

CH3 C C CH2CH3

H

(cis or trans)

CH3

H H

1,2-H shift

H CH3 CH3 CH3 C C

H

H

2° carbocation Rearrangement would not further stabilize the carbocation.

C CH3

H H

2° carbocation rearrangement

3° carbocation more stable

H H

H CH3 CH3

CH3 C C CH(CH3)2

CH3 C C

Br H

H H

C CH3 Br

10.20 To draw the products, remember that addition of HX proceeds via a carbocation intermediate. Addition of H+ (from HBr) from above and below gives an achiral, trigonal planar carbocation. a.

H H

CH3 CH3

CH3 CH3

Addition of Br– from above and below.

CH3

CH3

CH3

Br CH3

CH3 CH3

Br CH3

enantiomers

Addition of H+ (from HCl) from above and below by Markovnikov's rule forms an achiral 3° carbocation. b. CH3

CH3 CH3

CH3

Cl– attacks from above and below. CH3

H

CH3

CH3

CH3

Cl

H

achiral, trigonal planar 3° carbocation

diastereomers

CH3 Cl

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Chapter 10–10 10.21 The product of syn addition will have H and Cl both up or down (both on wedges or both dashes), while the product of anti addition will have one up and one down (one wedge, one dash). CH3 H

CH3 H

CH3 H

CH3 H

Cl CH3

Cl CH3

Cl CH3

Cl CH3

A syn addition

B anti addition

C anti addition

D syn addition

10.22 CH3

CH2CH2CH3

CH3

or

CH2 C

a. CH3 C CH2CH2CH3 OH

C CHCH2CH3

CH3

CH3

H+ would add here to form a 3° carbocation. CH3 OH

b.

CH3

CH2

or H+

H+ would add here to form a 3° carbocation.

would add here to form a 3° carbocation. or

c. H+

OH

H+ would add here to form a 3° carbocation.

CH3CH CHCH3

would add here to form a 2° carbocation.

(cis or trans)

10.23 H

H2O

OH

H2SO4

1-pentene

HO

H

enantiomers

10.24 The two steps in the mechanism for the halogenation of an alkene are: [1] Addition of X+ to the alkene to form a bridged halonium ion [2] Nucleophilic attack by X
transition state [1]: X X X

Step [1] C C

transition state [2]:


X

X +

+ X C C

X

Step [2]

C C

C

+

C

C

X

X X

C 


10.25 Halogenation of an alkene adds two elements of X in an anti fashion. a.

Br2

Br

Br

Cl2

CH3 Cl

CH3 Cl

Cl

Cl

b. Br

Br

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Alkenes 10–11 10.26 To draw the products of halogenation of an alkene, remember that the halogen adds to both ends of the double bond but only anti addition occurs. Cl

Cl

Cl CH3

Cl CH3

Cl2

a.

c.

Br2

CH3

CH3

CH3 Br

enantiomers H

b. C

Br

Br2 Br H

C C

H

C

H Br

achiral meso compound

10.27 The two steps in the mechanism for the halogenation of an alkene are: [1] Addition of X+ to the alkene to form a bridged halonium ion [2] Nucleophilic attack by X
CH3

trans-2-butene

C C

H

Addition of Br+ can occur from above or below:

Br Br

CH3

overall

H

Attack of Br– can occur from the left or right:

C

CH3

C C CH3

H

Br

Br Br

below CH3

CH3

H

Br

H

above Br

H

H C

H

enantiomers

C

C CH3

Br

CH3

right

left

left

right + Br

CH3 H

Br C

C

H CH3

CH3 H

Br C

C

+ Br CH3 H C Br

Br C

H CH3

H CH3

CH3

C

H

C Br

+ Br

H

CH3

CH3

C

H

C Br

H CH3

+ Br Br

H C

CH3 H

Br

diastereomers

C

Br CH3

Br

H C

CH3 H

C

CH3 Br

CH3 H C Br

All four compounds are identical—an achiral meso compound.

Br C

H CH3

Br

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Chapter 10–12 10.28 Halohydrin formation adds the elements of X and OH across the double bond in an anti fashion. The reaction is regioselective so X ends up on the carbon that had more H’s to begin with. Br

NBS

a.

DMSO, H2O

HO

CH3

Cl2

b. HO

H2O

Br

CH3 OH

CH3 OH

Cl

Cl

Cl bonds to the carbon with more H's to begin with.

10.29 H

a. (CH3)2S

H

H CH2CH3 H B N CH2CH3

CH3

H B S

b. (CH3CH2)3N CH3

H CH2CH2CH2CH3 H B P CH2CH2CH2CH3

c. (CH3CH2CH2CH2)3P

H CH2CH3

H CH2CH2CH2CH3

10.30 In hydroboration the boron atom is the electrophile and becomes bonded to the carbon atom that had more H’s to begin with. CH3

a.

C CH2

CH3

BH3

CH3 C CH2

CH3

BH3

CH2BH2

H BH2

C with more H's. B will add here.

C with more H's. B will add here.

BH3

b.

CH2

c.

BH2 H

C with more H's. B will add here.

10.31 The hydroboration–oxidation reaction occurs in two steps: [1] Syn addition of BH3, with the boron on the less substituted carbon atom [2] OH replaces the BH2 with retention of configuration. a. CH3CH2CH CH2

BH3

CH3CH2CH CH2 H

CH2CH3

b.

c. CH3

CH3

BH3

H2O2, –OH

CH3CH2CH CH2

BH2

H

OH

CH2CH3 H

CH2CH3 H2O2, –OH H

CH2CH3 H

CH2CH3 H

BH2

BH2

OH

OH

CH3

CH3 H

CH3

BH2

BH2 H2O2,

CH3

CH3 H OH

CH3 H

–OH

CH3

CH3

H OH

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Alkenes 10–13 10.32 Remember that hydroboration results in addition of OH on the less substituted C. OH HO

a.

c.

b.

(E or Z isomer can be used.)

OH

10.33 CH2

a.

CH2

H2O H2SO4

OH

Hydration places the OH on the more substituted carbon.

CH3

[1] BH3

CH2OH

[2] H2O2, HO

Hydroboration–oxidation places the OH on the less substituted carbon.

OH

Hydration places the OH on the more substituted carbon.

H2O H2SO4

b.

OH

[1] BH3 [2] H2O2, HO OH

H2O H2SO4

c.

[1] BH3 [2] H2O2, HO

Hydroboration–oxidation places the OH on the less substituted carbon.

Hydration places the OH on the more substituted carbon.

Hydroboration–oxidation places the OH on the less substituted carbon.

HO

10.34 There are always two steps in this kind of question: [1] Identify the functional group and decide what types of reactions it undergoes (for example, substitution, elimination, or addition). [2] Look at the reagent and determine if it is an electrophile, nucleophile, acid, or base. acid: catalyzes loss of H2O

acid

a.

CH2

HBr

CH3

c.

Br

alkene: addition reactions

H2SO4

OH

alcohol: substitution and elimination

nucleophile and base Cl

b.

CH2CH2CH3

OCH3 NaOCH3

2° alkyl halide: substitution and elimination

CH3CH2CH=CHCH3 (cis and trans)

CHCH2CH3

CH2CH2CH3

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Chapter 10–14 10.35 To devise a synthesis: [1] Look at the starting material and decide what reactions it can undergo. [2] Look at the product and decide what reactions could make it. CH3

? Br

a.

Cl OH

alkyl halide Can undergo substitution and elimination. K+ –OC(CH3)3

halohydrin: Can form from an alkene with Cl2 and H2O. Cl2

Cl

Br H2O

OH

OH is added to the more substituted C.

CH3

?

OH

b.

OH

alcohol Can undergo substitution and elimination.

alcohol Can be formed by substitution and addition.

CH3 OH

CH3 [1] BH 3

H2SO4

[2] H2O2, HO–

(major product)

CH3 OH

OH is added to the less substituted C.

10.36 Use the directions from Answer 10.2 to calculate degrees of unsaturation. a. C3H4 [1] maximum number of H's = 2n + 2 = 2(3) + 2 = 8 [2] subtract actual from maximum = 8 – 4 = 4 [3] divide by 2 = 4/2 = 2 degrees of unsaturation

f. C8H9Br Because of Br, add one H (9 + 1 = 10 H's). [1] maximum number of H's = 2n + 2 = 2(8) + 2 = 18 [2] subtract actual from maximum = 18 – 10 = 8 [3] divide by 2 = 8/2 = 4 degrees of unsaturation

b. C6H8 [1] maximum number of H's = 2n + 2 = 2(6) + 2 = 14 [2] subtract actual from maximum = 14 – 8 = 6 [3] divide by 2 = 6/2 = 3 degrees of unsaturation

g. C8H9ClO Ignore the O; count Cl as one more H (9 + 1 = 10 H's). [1] maximum number of H's = 2n + 2 = 2(8) + 2 = 18 [2] subtract actual from maximum = 18 – 10 = 8 [3] divide by 2 = 8/2 = 4 degrees of unsaturation

c. C40H56 [1] maximum number of H's = 2n + 2 = 2(40) + 2 = 82 [2] subtract actual from maximum = 82 – 56 = 26 [3] divide by 2 = 26/2 = 13 degrees of unsaturation d. C8H8O Ignore the O. [1] maximum number of H's = 2n + 2 = 2(8) + 2 = 18 [2] subtract actual from maximum = 18 – 8 = 10 [3] divide by 2 = 10/2 = 5 degrees of unsaturation e. C10H16O2 Ignore both O's. [1] maximum number of H's = 2n + 2 = 2(10) + 2 = 22 [2] subtract actual from maximum = 22 – 16 = 6 [3] divide by 2 = 6/2 = 3 degrees of unsaturation

h. C7H9Br Because of Br, add one H (9 + 1 = 10 H's). [1] maximum number of H's = 2n + 2 = 2(7) + 2 = 16 [2] subtract actual from maximum = 16 –10 = 6 [3] divide by 2 = 6/2 = 3 degrees of unsaturation i. C7H11N Because of N, subtract one H (11 – 1 = 10 H's). [1] maximum number of H's = 2n + 2 = 2(7) + 2 = 16 [2] subtract actual from maximum = 16 – 10 = 6 [3] divide by 2 = 6/2 = 3 degrees of unsaturation j. C4H8BrN Because of Br, add one H, but subtract one for N (8 + 1 – 1 = 8 H's). [1] maximum number of H's = 2n + 2 = 2(4) + 2 = 10 [2] subtract actual from maximum = 10 – 8 = 2 [3] divide by 2 = 2/2 = 1 degree of unsaturation

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Alkenes 10–15 10.37 First determine the number of degrees of unsaturation in the compound. Then decide which combinations of rings and
bonds could exist. C10H14 [1] maximum number of H's = 2n + 2 = 2(10) + 2 = 22 [2] subtract actual from maximum = 22 – 14 = 8 [3] divide by two = 8/2 = 4 degrees of unsaturation

possibilities: 4
bonds 3
bonds + 1 ring 2
bonds + 2 rings 1
bond + 3 rings 4 rings

10.38 The statement is incorrect because in naming isomers with more than two groups on a double bond, one must use an E/Z label, rather than a cis/trans label. higher priority

Cl

higher priority

Cl

higher priority

higher priority N(CH2CH3)2

O

O

enclomiphene E isomer

N(CH2CH3)2

zuclomiphene Z isomer

10.39 Name the alkenes using the rules in Answers 10.4 and 10.6. CH3

a.

CH2=CHCH2CH(CH3)CH2CH3

CH2 CHCH2CHCH2CH3

6 C chain with a double bond = hexene

1-hexene 4-methyl

4-methyl-1-hexene

5-ethyl

2-methyl b.

5-ethyl-2-methyl-2-octene 2-octene

8 C chain with a double bond = octene 2-isopropyl 2-isopropyl-4-methyl-1-pentene

c. 1-pentene 5 C chain with a double bond = pentene d.

CH3

CH3

4-methyl CH3

C C H

CH3 C C

CH2CH(CH3)2

CH3

higher 3-methyl priority 5-methyl

CH2CHCH3

H

e. 1-ethyl 5-isopropyl

CH2CH(CH3)2

Higher priority groups are on opposite sides = E alkene.

6 C chain with a double bond = hexene

6 C ring with a double bond = cyclohexene

C C H

2-hexene

CH3 (2E)-3,5-dimethyl-2-hexene

CH3

higher priority

1-ethyl-5-isopropylcyclohexene

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Chapter 10–16 1-sec-butyl-2-methylcyclopentene

2-methyl

f.

1-sec-butyl 5 C ring with a double bond = cyclopentene E double bond (higher priority groups on opposite sides with bold bonds)

4-isopropyl

g.

(4E)-4-isopropyl-4-hepten-3-ol

3-ol

OH

OH

7 C chain with a double bond = heptene

4-heptene

2-cyclohexene 5-sec-butyl-2-cyclohexenol 1-ol

h. OH

OH

5-sec-butyl

6 C ring with a double bond = cyclohexene

10.40 Use the directions from Answer 10.8. 2 4-ethyl

a. (3E)-4-ethyl-3-heptene 7 carbons

e. (2Z)-3-isopropyl-2-heptene 3-isopropyl 7 carbons

3 Higher priority groups on opposite sides = E. 3,3-dimethyl

b. 3,3-dimethylcyclopentene 5 carbon ring

1

Higher priority groups on the same side = Z. 1

f. cis-3,4-dimethylcyclopentene

or 3

4

5 carbon ring

3,4-dimethyl

c. cis-4-octene g. trans-2-heptene 8 carbons

4 Higher priority groups on the same side = cis.

7 carbons

2 Higher priority groups on opposite sides = trans. 4-propyl

2 d. 4-vinylcyclopentene 5 carbon ring

4

1-isopropyl

1 h. 1-isopropyl-4-propylcyclohexene 6 carbon ring

10.41 a.

(2E,4S)-4-methyl-2-nonene (2E,4R)-4-methyl-2-nonene (2Z,4S)-4-methyl-2-nonene A B C b. A and B are enantiomers. C and D are enantiomers. c. Pairs of diastereomers: A and C, A and D, B and C, B and D.

(2Z,4R)-4-methyl-2-nonene D

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Alkenes 10–17 10.42 CH3

CH3

H CH3

a.

c.

(1Z,4S)-1,4-dimethylcyclodecene diastereomer

(1E,4R)-1,4-dimethylcyclodecene CH3

CH3 CH3 H

b.

(1Z,4R)-1,4-dimethylcyclodecene diastereomer

(1E,4S)-1,4-dimethylcyclodecene enantiomer

10.43 Name the alkene from which the epoxide can be derived and add the word oxide. 1-ethyl H

derived a.

O

O

c.

from

H

derived from

H

6 carbon ring cyclohexene 1-ethylcyclohexene 1-ethylcyclohexene oxide

(3E)- 3-heptene oxide or trans-3-heptene oxide

H

7 carbon chain heptene (3E)- 3-heptene or trans-3-heptene

2-methyl O

b.

derived

derived

d.

C(CH3)3

C(CH3)3

from

from

2-methyl-2-hexene oxide

O

6 carbon chain hexene 2-methyl-2-hexene

5 carbon ring cyclopentene 4-tert-butylcyclopentene oxide 4-tert-butylcyclopentene

10.44

2-sec-butyl

a. 2-butyl-3-methyl-1-pentene As written, this is the parent chain, but there is another longer chain containing the double bond.

new name: 2-sec-butyl-1-hexene

b. (Z)-2-methyl-2-hexene new name: 2-methyl-2-hexene Two groups on one end of the C=C are the same (2 CH3's), so no E and Z isomers are possible.

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Chapter 10–18 1 c. (E)-1-isopropyl-1-butene new name: (3E)-2-methyl-3-hexene

As written, this is the parent chain, but there is another longer chain containing the double bond.

1

d. 5-methylcyclohexene 4 As written the methyl is at C5. Re-number to put it at C4. e. 4-isobutyl-2-methylcylohexene

new name: 4-methylcyclohexene 2

1

1

5 As written this methyl is at C2. Re-number to put it at C1.

f.

1-sec-butyl-2-cyclopentene

3 1

1 3

2

new name: 3-sec-butylcyclopentene

This has the double bond between C2 and C3. Cycloalkenes must have the double bond between C1 and C2. Re-number. g.

1-cyclohexen-4-ol

new name: 5-isobutyl-1-methylcyclohexene

OH

OH

3

1 The numbering is incorrect. When a compound contains both a double bond and an OH group, number the C skeleton to give the OH group the lower number.

3-cyclohexenol (The "1" can be omitted.)

OH

h.

6

OH

3-ethyl-3-octen-5-ol The numbering is incorrect. When a compound contains both a double bond and an OH group, number the C skeleton to give the OH group the lower number.

4 5 6-ethyl-5-octen-4-ol

10.45 COOH

a. and b. COOH CH3 H

CH3O

H

E

Z HOOC

E

S

E E bongkrekic acid

Z

R

Z

c. Since there are 7 double bonds and 2 tetrahedral stereogenic centers, 29 = 512 possible stereoisomers.

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Alkenes 10–19 10.46

H

2-methyl-1-pentene (2E)-4-methyl-2-pentene

4-methyl-1-pentene

(2E)-3-methyl-2-pentene

(3R)-3-methyl-1-pentene

2-methyl-2-pentene (2Z)-4-methyl-2-pentene

2-ethyl-1-butene

(2Z)-3-methyl-2-pentene

(3S)-3-methyl-1-pentene

H

10.47 O

stearic acid

OH

hIghest melting point no double bonds

OH

intermediate melting point one E double bond

O

elaidic acid O OH

oleic acid

lowest melting point one Z double bond

10.48 O

a.

O

OH

all trans double bonds higher melting point

OH

O

eleostearic acid

b.

OH

all cis double bonds lower melting point

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Chapter 10–20 10.49 O

a.

O

O

O O

O

O

*

O

O

O

O

O

A A has one tetrahedral stereogenic center, labeled with an asterisk [*].

B

O

O

O

b.

O O

O

R

S O

H

O H

O

O

O

O

10.50 transition state [1]: Br

[1]

+ Br  H C C H H

[2] Br

H C C H

H

Br H

H Br

H

H Br

C C H

Energy

Br2 H

CH2 CH2




A

transition state [2]: + Br

A

H

CH2 CH2

BrCH2CH2Br

+ Br2

H

H

C C Br 


H

Reaction coordinate

10.51 The more negative the
H°, the larger the Keq assuming entropy changes are comparable. Calculate the
H° for each reaction and compare. CH2 CH2 + HI

CH3CH2

[1] Bonds broken

I

[2] Bonds formed

Ho (kJ/mol) C–C
bond

+ 267

I

+ 297

H

Total

+ 564 kJ/mol

Ho (kJ/mol) CH2ICH2 H

 410

I

 222

C

Total

[3] Overall Ho = sum in Step [1] + sum in Step [2] + 564 kJ/mol  632

kJ/mol

– 68

kJ/mol

 632 kJ/mol

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Alkenes 10–21 CH2 CH2 + HCl

CH3CH2 Cl

[3] Overall Ho =

[2] Bonds formed

[1] Bonds broken

C–C
bond

+ 267

CH2ClCH2 H

 410

sum in Step [1] + sum in Step [2]

H Cl

+ 431

C Cl

 339

+ 698 kJ/mol

+ 698 kJ/mol

Total

 749 kJ/mol

Ho (kJ/mol)

Total

Ho (kJ/mol)

 749

kJ/mol

– 51 kJ/mol Compare the H°: Addition of HI: –68 kJ/mol more negative H°, larger Keq. Addition of HCl: –51 kJ/mol

10.52 H

HBr

a.

Br

Br2, H2O

f. Br H

HI

b.

NBS

g.

aqueous DMSO

I H

H2O H2SO4 CH3CH2OH

d.

H2SO4 Cl2

e.

[2] H2O2,

i.

OH

Br

Br +

OH

OH

HO– OH

OH H

OH

H

[1] BH3

h.

c.

Br +

H

[1] 9-BBN –

[2] H2O2, HO

OCH2CH3

OH

Cl +

Cl

Cl

Cl

10.53 CH3

a.

C CH2 CH3 CH3

b.

C CH2

HBr

c.

C CH2 CH3 CH3

d.

C CH2 CH3

HI

e.

CH3

f.

CH3 C CH3

C CH2

C CH2

H2SO4 CH3CH2OH H2SO4 Cl2

CH3 CH3 C CH3 OH

CH3

h.

CH3

CH3

CH3 CH3 C CH2Cl Cl

NBS aqueous DMSO [1] BH3 [2] H2O2, HO–

CH3

CH3 C CH3 OCH2CH3

C CH2

i.

C CH2 CH3

CH3 C CH2Br OH

CH3

g.

CH3 H2O

CH3 Br2, H2O

CH3

CH3 I

CH3 C CH2

CH3 C CH3 Br

CH3 CH3

CH3

CH3

[1] 9-BBN [2] H2O2, HO–

CH3 CH3 C CH2Br OH CH3 CH3 C CH2OH H CH3 CH3 C CH2OH H

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Chapter 10–22 10.54 Br Br2

Br

Halogenation

a. OH

Br2

b.

Halohydrin formation: Br adds to the C that had more H's to begin with.

Br

H2O OCH3

Br2

Same as halohydrin formation, except CH3OH in place of H2O.

Br

c.

CH3OH

10.55 Br

a.

Br

d.

CH3CH2CH2CH=CHCH3 Br

Cl

b.

e.

CH2 or

CH3

Cl

Cl CH3

CH3

c.

CH3

C CH3

C CH2

CH3CH2

f.

(CH3CH2)3CBr

Br

C CHCH3 CH3CH2

10.56 Hydroboration–oxidation results in addition of an OH group on the less substituted carbon, whereas acid-catalyzed addition of H2O results in the addition of an OH group on the more substituted carbon. a.

OH

c.

hydroboration–oxidation and acid-catalyzed addition

acid-catalyzed addition

OH

b. hydroboration–oxidation

e.

OH

OH

d. hydroboration–oxidation

or

OH

Both methods would give product mixtures.

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Alkenes 10–23 10.57 a.

Cl (CH3CH2)2C=CHCH2CH3 CH3CH2

H

CH3CH2

CH3CH2 H

HCl

C C

Cl2

d. CH3CH2 C Cl

CH2CH3

Cl

C H CH2CH3

e.

CH2

b.

CH2Br OH

Br adds here to less substituted C.

(CH3CH2)2C=CH2 CH3CH2

f.

CH3CH2 C CH3

H2SO4

CH3CH2

[1] 9-BBN

CH3CH2

H2O

C CH2

CH2

OH

H

[1] BH3

(CH3)2C=CHCH3

CH2OH

[2] H2O2, HO


OH adds here to less substituted C.

H adds here to less substituted C. c.

Br2 H2O

H adds here to less substituted C.

(CH3)2C

DMSO, H2O

CHCH3

H (CH3)2C

OH Br

Br adds here to less substituted C.

[2] H2O2, HO


BH3 adds here to less substituted C.

NBS

g.

BH2

OH

Br2

h.

CHCH3

Br Br

10.58 CH3CH2

H

or

C C CH3CH2CH2

CH3

CH2CH2CH3

or

CH3CH C

H

Cl

C CHCH2CH3

CH3

CH3CH2 C CH2CH2CH3 CH3

CH3CH2

10.59 H

a.

CH3

H

C C

C C

=

H

CH3

C

H

CH3

b.

CH3

CH3CH2

H

CH3CH2 CH3

CH3 C C H

CH3CH2

CH3 CH3CH2

=

C C

c.

CH3

=

C C

H

CH3 C C

CH3

H

Cl2

HO

H2O

C CH3 CH3CH2

NBS DMSO, H2O

H

C

Br

H

CH3 H C

C

CH3 H

b.

CH2

H2O H2SO4

(CH3)3C

CH3

(CH3)3C

OH

I H

H

I

C

HO CH3 H

C CH3CH2 CH3

CH3

CH3 H

HO

OH

HI

Cl

CH3 CH3CH2 C

10.60 a. (CH3)3C

C Br

Cl

HO

C

CH3

Br

CH3CH2 CH3 C

CH3 H

Br

Br

H

Br2

C Br

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Chapter 10–24 CH3

Cl Cl2

c. CH3

CH3

CH3

Cl

d.

Cl CH3

CH3

Cl

[1] BH3

CH3

only anti addition

CH3 H

[2] H2O2, HO
CH3

CH3

CH3 H

OH

Br CH2CH3

CH3

only syn addition

OH

Br CH2CH3

HBr

e.

Cl

Cl2

f.

OH

only anti addition

+

H2O OH

Cl

NBS

g.

Br

DMSO, H2O

h.

H2O

CH3CH=CHCH2CH3

Br HO CH3

HO CH3

H OH

HO H

H2SO4

OH

10.61 Cl CH3 CH2 C CH3

Br2

H

NaCl

H

C

CH2 C CH3

C Br

Cl– acts as the nucleophile.

Cl

CH3 CH3

Br

+ Br

Br acts as the electrophile and is therefore added to the C with more H's to begin with.

CH3

+

Br Br

10.62 Draw each reaction. (a) The cis isomer of 4-octene gives two enantiomers on addition of Br2. (b) The trans isomer gives a meso compound. H

H

Br2

a.

H

Br

Br

H

H Br

cis-4-octene

Br

(4R,5R)-4,5-dibromooctane

(4S,5S)-4,5-dibromooctane

enantiomers

H

Br2

b. H

trans-4-octene

H

Br

H

H Br

(4R,5S)-4,5-dibromooctane meso compound

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Alkenes 10–25 10.63 CH3CH2

CH3CH2

CH2CH3

C C H

H

H

H Cl CH3CH2

CH3CH2

CH2CH3

H C C H

H

H Cl–

CH3CH2 H

C Cl

Cl H

CH2CH3

By protonation of the alkene, the cis and trans isomers produce identical carbocation intermediates.

Cl–

CH3CH2

CH2CH2CH3

CH2CH3 H Cl

C C H H

H

C C

C

CH3CH2 C

CH2CH2CH3

H

CH3CH2

CH2CH2CH3

Cl

C Cl

CH2CH2CH3

H

Both cis- and trans-3-hexene give the same racemic mixture of products, so the reaction is not stereospecific.

10.64 The alkene that forms the more stable carbocation reacts faster, according to the Hammond postulate. H Br

a.

CH CH

CH CH

CH CH

H

H

CH CH

CH CH

CH CH

H

H

H

This 2° carbocation is resonance stabilized, making it more stable, so the starting alkene reacts faster with HBr. CH3 CH CH CH3 H Br

b.

CH3 CH CH CH3 H

2° carbocation no resonance stabilization

CH3

Markovnikov

CH3

addition

CH2 C

CH3 CH2OCH3 H Br

C(CH3)2

faster

H

3° carbocation

H Br CH2 C

CH2

Markovnikov addition

CH3 CH2 H

C CH2–OCH3

3° carbocation

This carbocation is still 3°, but the nearby electronegative O atom withdraws electron density from the carbocation, destabilizing it. Thus, the reaction to form this carbocation occurs more slowly.

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Chapter 10–26 10.65 Cl H

H

H Cl

a.

Cl

H

H

H

H O

H HO

H

C

O

O

O C H H

CH3

O C CH3

H H

H

H

H

H Br

b.

Cl

H

and

H

Br

Br

C H CH3

C CH3

2o carbocation

+ HCl

H

H

1,2-H shift

CH3

3o carbocation

+ Br–

10.66 a.

H CH3

CH3

CH3

1,2-shift

H2O

CH3

CH3

O H – CH3 HSO4

CH3

OH CH3

H2SO4 HSO4–

H OSO3H

2o carbocation

H2O

3o

+ H2SO4

carbocation

H OSO3H

b.

OH

OH H2SO4

+ H2SO4 O

HSO4–

CH3

H –OSO

3H

O

CH3

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Alkenes 10–27 10.67 H

H OH2

H + H3O+ O H

H2O

OH H2O

H

H O H

1,2-shift

H2O

H2O

H

H

OH + H3O+

H2O

O H

1,2-shift

H2O

OH

H

H + H3O+

10.68 The isomerization reaction occurs by protonation and deprotonation. H

CH2

OSO3H

C

CH3

CH3

CH3 C H

CH3

CH3

C

CH3 C H CH3

CH3 HSO4–

2,3-dimethyl-1-butene

CH3

C C CH3

CH3

2,3-dimethyl-2-butene

10.69 H

H C C

H

H Br H

H CH3 C

C C H

H C C

H

H

CH3CH CHCH2

Br

CH3CH=CH–CH2Br

H

Br Br CH3CHCH CH2

Since two resonance structures can be drawn for the intermediate carbocation, two different products result from attack by Br–.

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Chapter 10–28 10.70 Br HBr

A

OCH3

OCH3

B

H Br

COOCH3

C

COOCH3

D

H Br

Br

O OCH3

OCH3

H

Br

HBr

H

This carbocation is resonance stabilized by the O atom, and therefore preferentially forms and results in B.

C

H O

CH3

O C

CH3


+ O H This carbocation is destabilized by the
+ on the adjacent C, so it does not form. This carbocation is formed preferentially and results in product D. It is not destabilized by an adjacent electron-withdrawing COOCH3 group.

10.71 Br Br

H Br

OH

OH

O

Br

O

Br + HBr

+ Br + Br

10.72 OH PBr3

OH

K+ –OC(CH3)3

Br

H2O + H2SO4

a. OH adds to more substituted C.

POCl3, pyridine Br

b.

CH3 C CH3

K+ –OC(CH3)3

Br

Br2

CH3 CH CH2

CH3 C CH2Br

H OH

c.

H H2SO4

[1] BH3 [2] H2O2,

d.

CH3 CH CH2I

K+ –OC(CH3)3

OH

HO–

+ H2 CH3

(CH3)2C

CH2

HCl

CH3 C Cl

CH3

e.

CH3

Cl

Br

Br2

K+ –OC(CH3)3

H2O

f.

CH3CH CH2

O–

NaH

Br2

Br

OH

Br

CH3CH CH2

NaNH2 (2 equiv)

CH3C CH

CH3I

OCH3

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Alkenes 10–29 10.73 Cl

Cl2

a.

H2O

+ H2 Br

CN NaCN

Br

OH

O

O

NaH

NaOH

c.

O O

OH

HBr

b.

Cl

Na+ H–

CH3CH2CH2I

(from b.) [1] –SH

d.

O

[2] H2O

OH

+ enantiomer SH

(from a.)

e.

O

[1] –C
CH

OH

[2] H2O

C

+ enantiomer

(from a.)

CH

10.74 Br

Br K+ –OC(CH3)3

a.

d.

Br

NaNH2

C CH

(2 equiv) Br Br2

b.

Br

(from b.) OH

OH

Br2

c.

O

NaH

Br

e.

(from a.) Br

(from c.)

H2O

(from a.)

10.75 OH

Br OH POCl3

Br2

pyridine

H2O

A

B

major product

H2SO4 OH Br2

NaH O

H2O

O NaH

Br

C

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Chapter 10–30 10.76 TsO H

TsO–

H

+ TsOH + TsO–

A

isocomene

10.77 Na+ HCO3– O

H O

O

O

O

O

O

O I

I

I

OCH3 HO

OCH3

OCH3

B

HO

HO Na+

I

OCH3 HO

I–

C

H2CO3

10.78

TsO H OH

OH2

nerol

+ TsOH H2O O H H

H

OH

OH


-terpineol

OSO2Cl OH

OH

OH

OTs

H OSO2Cl

HSO3Cl


-cyclogeraniol

nerol OSO2Cl

10.79 O

H OSO3H

O H OH

+

HSO4–

H2O

H O H

HSO4–

OH

HO

OH

+ H2SO4

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Alkynes 11–1 C Chhaapptteerr 1111:: A Allkkyynneess  G Geenneerraall ffaaccttss aabboouutt aallkkyynneess • Alkynes contain a carbon–carbon triple bond consisting of a strong  bond and two weak
bonds. Each carbon is sp hybridized and linear (11.1). 180o H C C H

=

acetylene

sp hybridized

• • •

Alkynes are named using the suffix -yne (11.2). Alkynes have weak intermolecular forces, giving them low mp’s and low bp’s, and making them water insoluble (11.3). Since its weaker
bonds make an alkyne electron rich, alkynes undergo addition reactions with electrophiles (11.6).

 A Addddiittiioonn rreeaaccttiioonnss ooff aallkkyynneess [1] Hydrohalogenation—Addition of HX (X = Cl, Br, I) (11.7) •

X H

H X

R C C H

R C C H

(2 equiv)

X H

Markovnikov’s rule is followed. H bonds to the less substituted C in order to form the more stable carbocation.

geminal dihalide

[2] Halogenation—Addition of X2 (X = Cl or Br) (11.8) •

X X

X X

R C C H

R C C H

(2 equiv)

•

X X

Bridged halonium ions are formed as intermediates. Anti addition of X2 occurs.

tetrahalide

[3] Hydration—Addition of H2O (11.9)

R C C H

H2O H2SO4 HgSO4

H

R C C

H

HO

enol

•

O R

C

CH3

ketone

•

Markovnikov’s rule is followed. H bonds to the less substituted C in order to form the more stable carbocation. The unstable enol that is first formed rearranges to a carbonyl group.

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Chapter 11–2 [4] Hydroboration–oxidation—Addition of H2O (11.10)

R C C H

R

C C

[2] H2O2, HO

H

•

O

H

R

[1] BH3

OH

C

C

H

H H

enol

The unstable enol, first formed after oxidation, rearranges to a carbonyl group.

aldehyde



R Reeaaccttiioonnss iinnvvoollvviinngg aacceettyylliiddee aanniioonnss [1] Formation of acetylide anions from terminal alkynes (11.6B) +

R C C H

B

R C C

+

+

HB

•

Typical bases used for the reaction are NaNH2 and NaH.

[2] Reaction of acetylide anions with alkyl halides (11.11A) H C C

+

R X

H C C R + X

• •

The reaction follows an SN2 mechanism. The reaction works best with CH3X and RCH2X.

[3] Reaction of acetylide anions with epoxides (11.11B) [1] O H C C CH2CH2OH

H C C

[2] H2O

• •

The reaction follows an SN2 mechanism. Ring opening occurs from the back side at the less substituted end of the epoxide.

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Alkynes 11–3 C Chhaapptteerr 1111:: A Annssw weerrss ttoo PPrroobblleem mss 11.1 • An internal alkyne has the triple bond somewhere in the middle of the carbon chain. • A terminal alkyne has the triple bond at the end of the carbon chain. HC C CH2CH2CH3

CH3 C C CH2CH3

HC C CH CH3 CH3

terminal alkyne

internal alkyne

terminal alkyne

11.2 Csp–Csp3 (c)

Csp2–Csp (b)

OH

(a) Csp3–Csp2

Increasing bond strength: a < c < b

O

santalbic acid

11.3 Like alkenes, the larger the number of alkyl groups bonded to the sp hybridized C, the more stable the alkyne. This makes internal alkynes more stable than terminal alkynes. 11.4 To name an alkyne: [1] Find the longest chain that contains both atoms of the triple bond, change the -ane ending of the parent name to -yne, and number the chain to give the first carbon of the triple bond the lower number. [2] Name all substituents following the other rules of nomenclature. CH2CH2CH3

a.

H C C CH2CCH2CH2CH3

CH2CH3 CH3

c.

CH2=CHCH2CHC

CH2CH2CH3

CH3

4,4-dipropyl-1-heptyne b. CH3C CCClCH2CH3

CCCH2CH2CH3

(Number to give the lower number to the first site of unsaturation.)

4-ethyl-7,7-dimethyl-1-decen-5-yne d.

1 5 (The longest chain must contain both functional groups.)

CH3

3

4-chloro-4-methyl-2-hexyne 3-isopropyl-1,5-octadiyne

11.5 To work backwards from a name to a structure: [1] Find the parent name and the functional group. [2] Add the substituents to the appropriate carbon. OH

a. trans-2-ethynylcyclopentanol 5 C ring with OH at C1

OH

OH on C1 C CH

ethynyl at C2

or

C CH

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Chapter 11–4 tert-butyl at C4

b. 4-tert-butyl-5-decyne 10 C chain with a triple bond

triple bond at C5 c. 3-methylcyclononyne 9 C ring with a triple bond at C1

3-methyl

triple bond at C1

11.6 Two factors cause the boiling point increase. The linear sp hybridized C’s of the alkyne allow for more van der Waals attraction between alkyne molecules. Also, since a triple bond is more polarizable than a double bond, this increases the van der Waals forces between two molecules as well. 11.7 To convert an alkene to an alkyne: [1] Make a vicinal dihalide from the alkene by addition of X2. [2] Add base to remove two equivalents of HX and form the alkyne. a. Br2CH(CH2)4CH3

Na+ –NH2

BrCH CHCH2CH2CH2CH3

Na+ –NH2

HC CCH2CH2CH2CH3

not isolated b. CH2=CCl(CH2)3CH3 c. CH2=CH(CH2)3CH3

Na+ –NH2 Cl2

HC CCH2CH2CH2CH3 CH2CHCH2CH2CH2CH3 Cl Cl

Na+ –NH2

HC CCH2CH2CH2CH3

(2 equiv)

11.8 Acetylene has a pKa of 25, so bases having a conjugate acid with a pKa above 25 will be able to deprotonate it. a. CH3NH– [pKa (CH3NH2) = 40] pKa > 25 = Can deprotonate acetylene. b. CO32– [pKa (HCO3–) = 10.2] pKa < 25 = Cannot deprotonate acetylene.

c. CH2=CH– [pKa (CH2=CH2) = 44] pKa > 25 = Can deprotonate acetylene. d. (CH3)3CO– {pKa [(CH3)3COH] = 18} pKa < 25 = Cannot deprotonate acetylene.

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Alkynes 11–5

11.9 To draw the products of reactions with HX: • Add two moles of HX to the triple bond, following Markovnikov’s rule. • Both X’s end up on the more substituted C. Br

2 HBr

a. CH3CH2CH2CH2 C C H

CH3CH2CH2CH2 C CH3 Br

b. CH3 C C CH2CH3

Br

2 HBr

Br

CH3 CH2 C CH2CH3

CH3 C CH2 CH2CH3

Br

Br

Br

c.

C CH

2 HBr

C CH3 Br

11.10 a.

+

Cl

c.

Cl

+

+

CH3

b. CH3 O CH2

O

NH

NH

CH2

11.11 Addition of one equivalent of X2 to alkynes forms trans dihalides. Addition of two equivalents of X2 to alkynes forms tetrahalides. CH3CH2 C C CH2CH3

2 Br2

Br Br CH3CH2 C C CH2CH3 Br Br

CH3CH2 C C CH2CH3

Cl

Cl2

CH2CH3 C C

CH3CH2

trans dihalide Cl

11.12 Cl2

CH3 C C CH3

Cl

CH3 C C

CH3

Cl

The two Cl atoms are electron withdrawing, making the
bond less electron rich and therefore less reactive with an electrophile.

11.13 To draw the keto form of each enol: [1] Change the C–OH to a C=O at one end of the double bond. [2] At the other end of the double bond, add a proton. H H C H

CH2

a.

OH

O

new C–H bond OH

H H

OH H

O H

H

new C–H bond

O

b. H

c.

new C–H bond

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Chapter 11–6 11.14 Treatment of alkynes with H2O, H2SO4, and HgSO4 yields ketones. H2O

CH3CH=C(OH)CH2CH3

H2SO4, HgSO4

CH3C(OH)=CHCH2CH3

CH3 C C CH2CH3

O

O

Two enols form.

two ketones after tautomerization

11.15 OH

OH

OH

a.

OH

b.

enol tautomers

constitutional isomers, but not tautomers

11.16 Reaction with H2O, H2SO4, and HgSO4 adds the oxygen to the more substituted carbon. Reaction with [1] BH3, [2] H2O2, –OH adds the oxygen to the less substituted carbon. a. (CH3)2CHCH2 C C H

(CH3)2CHCH2

b.

C C H

CH3

H2O H2SO4, HgSO4

C CH

O CH2 C CH3

H

Forms a ketone. H2O is added with the O atom on the more substituted carbon.

CH3

[1] BH3 [2] H2O2, HO

C CH

CH3 C

CH3 C

O CH2 CH2 C H

H

Forms an aldehyde. H2O is added with the O atom on the less substituted carbon.

O

H2O

Forms a ketone. H2O is added with the O atom on the more substituted carbon.

C

H2SO4, HgSO4

CH3 H H C C O H

[1] BH3 [2] H2O2, HO

Forms an aldehyde. H2O is added with the O atom on the less substituted carbon.

11.17 a. H C C H

[1] NaH

[1] NaH

H C C

+ H2

C C

+ H2

[2] (CH3)2CHCH2–Cl

(CH3)2CHCH2 C C H

[2] CH3CH2–Br

C C CH2CH3 + NaBr

b.

C CH [1] NaNH2

C C

+ NH3

[2] (CH3)3CCl

CH3 C CH2

+ NaCl

CH3

+

C CH

+ NaCl

1° alkyl halide substitution product 3° alkyl halide elimination product

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Alkynes 11–7

11.18 a.

(CH3)2CHCH2C CH3 CH3 C CH2 H

CH CH3 CH3 C CH2Cl H

C C H

terminal alkyne only one possibility

C C H

1° RX

b. CH3C CCH2CH2CH2CH3 CH3

C C

[1]

[1]

CH2CH2CH2CH3

[2]

C C CH2CH2CH2CH3

CH3Cl

[2] CH3 C C

Cl CH2CH2CH2CH3

internal alkyne two possibilities

1° RX

c. (CH3)3CC CCH2CH3 CH3 CH3 C C C CH3

CH3 CH3 C C C CH3

CH2CH3

internal alkyne only one possibility

Cl CH2CH3

1° RX

CH3 CH3 C Cl CH3

The 3° alkyl halide would undergo elimination.

C CCH2CH3

3° RX too crowded for SN2 reaction

11.19 Na+ H

HC C H

HC C

Na+ H H C C CH2CH3

CH3CH2–Br

C C CH2CH3 + H2 H

+ H2

+ Na+Br–

+ Na+

(CH3)2CHCH2C

Br

H (CH3)2CHCH2CH2

– C C CH2CH3 + Br

11.20 CH3

CH3

CH3

CH3 C C C C CH3 CH3

CH3 C C C

CH3

CH3

+

X C CH3

CH3

CH3

The 3° alkyl halide is too crowded for nucleophilic substitution. Instead, it would undergo elimination with the acetylide anion.

2,2,5,5-tetramethyl-3-hexyne

11.21 CH3

a.

[1]

CH3 OH

C C H

O

[2] H2O

Epoxide is drawn up, so the acetylide anion attacks from below at less substituted C.

C CH

b.

O

[1]

C C H

[2] H2O

C

OH

CH

+ C

OH CH

enantiomers

Backside attack of the nucleophile (–C
CH) at either C since both ends are equally substituted.

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Chapter 11–8 11.22 a. CH3CH2CH2Br

CH3CH2C C–Na+

b. (CH3)2CHCH2CH2Cl

c. (CH3CH2)3CCl

CH3CH2C C–Na+

CH3CH2C C–Na+

d. BrCH2CH2CH2CH2OH

e. f.

CH3CH2CH2C CCH2CH3

(CH3CH2)2C

CH3CH2C C–Na+

O

CH3CH2C C–Na+

O

CH3CH2C C–Na+

(CH3)2CHCH2CH2C

CCH2CH3

CHCH3

CH3CH2C CH

BrCH2CH2CH2CH2O–Na+

CH3CH2C CH

H2O

CH3CH2C CCH2CH2OH

H2O

CH3CH2C CCH2CHCH3 OH

11.23 To use a retrosynthetic analysis: [1] Count the number of carbon atoms in the starting material and product. [2] Look at the functional groups in the starting material and product. Determine what types of reactions can form the product. Determine what types of reactions the starting material can undergo. [3] Work backwards from the product to make the starting material. [4] Write out the synthesis in the synthetic direction. ?

CH3CH2C CCH2CH3

HC CH

6 C's CH3CH2C CCH2CH3

HC C H

Na+ H

HC C

2 C's

CH3CH2C C

CH3CH2–Br

+ CH3CH2Br

CH3CH2C C H

Na+ H

HC C

+ CH3CH2Br

CH3CH2–Br

CH3CH2C C

CH3CH2C CCH2CH3

11.24 O HC CH

CH3CH2CH2 C H

product: 4 carbons, aldehyde functional group (can be made by hydroboration–oxidation of a terminal alkyne)

starting material: 2 carbons, C
C functional group (can form an acetylide anion by reaction with NaH)

O

Retrosynthetic analysis:

CH3CH2CH2 C

Forward direction: H C C H

CH3CH2C C H

H C C H

H Na+ H C C H

CH3CH2–Br

CH3CH2C C H

[1] BH3 [2] H2O2,

HO–

O CH3CH2CH2 C H

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Alkynes 11–9

11.25 a., b.

N

O

O O

N

O

c. d. * = sp hybridized C e. 3 < 1 < 2

H

HN

(1)

(2)

*

* *

O

*

C*

(3)

HO

erlotinib

most acidic C–H proton

phomallenic acid C most acidic proton

shortest C–C single bond Csp–Csp2

11.26 Use the rules from Answer 11.4 to name the alkynes.

3-hexyne

1-hexyne

2-hexyne

3-methyl-1-pentyne

4-methyl-1-pentyne

3,3-dimethyl-1-butyne

4-methyl-2-pentyne

11.27 Use the rules from Answer 11.4 to name the alkynes. 5

3

a. CH3CH2CH(CH3)C CCH2CH3

CH3

5-methyl-3-heptyne

3-heptyne

5-methyl

f.

CH3CHC CCHCH3 CH3

5 2

CH3

7-methyl

CH3CH2CHC CCHCHCH2CH3 CH2CH3

3,6-diethyl HC C CH(CH2CH3)CH2CH2CH3

1-hexyne

3-ethyl

6-methyl h.

3,6-diethyl-7-methyl-4-nonyne

d.

2,5-octadiyne

2 5-ethyl

6 4-ethyl

4-nonyne

CH3CH2

CCH3

g.

2,5-dimethyl

c.

CH2CH2CH3

CH3CH2C CCH2C

2,5-dimethyl-3-hexyne

CH3

3-methyl 3-ethyl-3-methyl-1-hexyne

CH3CH2 C C CH

3-ethyl

3-hexyne b.

e.

3-ethyl-1-hexyne

(2E)-4,5-diethyl-2-decen-6-yne

1 ethynyl 1-ethynyl-6-methylcyclohexene

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Chapter 11–10 11.28 Use the directions from Answer 11.5 to draw each structure. a. 5,6-dimethyl-2-heptyne

d. cis-1-ethynyl-2-methylcyclopentane 1-ethynyl

2-heptyne

or

C CH

C CH

2-methyl

b. 5-tert-butyl-6,6-dimethyl-3-nonyne 6,6-dimethyl

e. 3,4-dimethyl-1,5-octadiyne 3,4-dimethyl

3-nonyne 5-tert-butyl

diyne

c. (4S)-4-chloro-2-pentyne Cl H

diyne

f. (6Z)-6-methyl-6-octen-1-yne

4-chloro S configuration

6-methyl

2-pentyne

Z 1

11.29 Keto–enol tautomers are constitutional isomers in equilibrium that differ in the location of a double bond and a hydrogen. The OH in an enol must be bonded to a C=C. O

a.

CH3

C

OH CH3

and

CH2

C

CH3

• C=O • C=C • one more CH bond • OH on C=C keto–enol tautomers

b.

and OH is not bonded to the C=C. NOT keto–enol tautomers

H

and

O • C=O • one more CH bond

• C=C • OH on C=C

keto–enol tautomers O

OH

O

OH

c.

d.

OH

and NOT keto–enol tautomers

OH is not bonded to the C=C.

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Alkynes 11–11

11.30 To draw the enol form of each keto form: [1] Change the C=O to a C–OH. [2] Change one single C–C bond to a double bond, making sure the OH group is bonded to the C=C. O

OH

E/Z isomers possible

a. O

b. CH3CH2CHO =

O

OH

CH2CH3

CH2CH3

c.

OH +

CH2CH3

OH H

H

E/Z isomers possible

11.31 Use the directions from Answer 11.13 to draw each keto form. a. OH

O

OH

c.

OH

O

O

b. H

11.32 Tautomers are constitutional isomers that are in equilibrium and differ in the location of a double bond and a hydrogen atom. O

O

OH

O

O

a.

A

OH

OH

tautomer

O

O

c.

b. constitutional isomer

OH

d.

constitutional isomer

neither

11.33 O

O H

A H2O H

O

OH + H3O

H

B

H

+ H2O

+ H2 O

11.34 O H HO

O

O

H OH

HO

O

H OH

11.35 NHCH3

X H O H enamine 2

NHCH3 H

+ H2O

NCH3

NCH3

H + H2O

Y imine

+ H3O

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Chapter 11–12 11.36 The equilibrium always favors the formation of the weaker acid and the weaker base. a. HC C

+ CH3OH

c.

HC CH + CH3O

pKa = 15.5

pKa = 25 weaker acid Equilibrium favors products.

b. CH3C CH pKa = 25

+

CH3

+

CH3C C

Na+Br–

HC CH +

pKa = 25 weaker acid Equilibrium favors starting materials. d. CH3CH2C C

CH4

HC C– Na

+ HBr

pKa = –9

CH3CH2C CH + CH3COO

+ CH3COOH

pKa = 25

pKa = 4.8

pKa = 50 weaker acid Equilibrium favors products.

weaker acid Equilibrium favors products.

11.37 HC CCH2CH2CH2CH3

a.

e.

CH3 C CH2CH2CH2CH3

(2 equiv)

Cl Br

HBr

b.

O

Cl

HCl

Br

g.

H

[2] H2O2, HO– NaH

f.

CH3 CCH2CH2CH2CH3

(2 equiv)

[1] BH3

C

CH2CH2CH2CH2CH3

Na+ C CCH2CH2CH2CH3

[1] –NH2

CH3CH2C CCH2CH2CH2CH3

[2] CH3CH2Br

Cl Cl Cl2

c.

HC CCH2CH2CH2CH3 Cl Cl

(2 equiv) H2O

d.

h.

H

[2]

OH

O

C C

H2SO4, HgSO4 H

[1] –NH2

CH3

CH2CH2CH2CH3

C

O

HOCH2CH2 C CCH2CH2CH2CH3

[3] H2O CH2CH2CH2CH3

11.38 Br Br

HBr

a.

c.

(2 equiv)

O [1] BH3

d.

(2 equiv)

[2] H2O2, HO–

Br Br

11.39 O

a.

(CH3CH2)3C

C CH

H2O H2SO4, HgSO4

b.

(CH3CH2)3C

C CH

c.

(CH3CH2)3C

C CH

d.

(CH3CH2)3C

C CH

[1] BH3 [2] H2O2, HO–

HCl (2 equiv) [1] NaH [2] CH3CH2Br

O

H2SO4

Br Br

Br2

b.

H2O

(CH3CH2)3C

(CH3CH2)3C

C

CH3

CH2CHO

(CH3CH2)3CCCl2CH3

(CH3CH2)3C

C CCH2CH3

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Alkynes 11–13 11.40 Reaction rate (which is determined by Ea) and enthalpy (
H°) are not related. More exothermic reactions are not necessarily faster. Since the addition of HX to an alkene forms a more stable carbocation in an endothermic, rate-determining step, this carbocation is formed faster by the Hammond postulate. H R C C R'

HX

H X

R C C R'

H HX

R C C R'

R C C R'

H X

R C C R'

R C C R'

H H

H H

H H

sp2 hybridized carbocation more stable faster reaction

sp hybridized carbocation less stable slower reaction

11.41 O

OH C

a. OH

O

b.

C

CH3

CH3

CH3 O

C

c.

CH

C

CH3 C CH

CH2

OH C

CH3

d.

CH3CH2 O

CH

C

CH2

C C

CH2CH3

OH

11.42 To determine what two alkynes could yield the given ketone, work backwards by drawing the enols and then the alkynes. H HC C CH2CH3

OH C C

H

CH2CH3

HO

O CH3

C

H CH3 C C CH3

C C

CH2CH3

CH3

2-butanone

CH3

11.43 a.

CH2CHO

C CH

b.

(CH3)2CHC

CCH(CH3)2

O

11.44 Equilibrium favors the weaker acid. CH3CH2CH2CH2–Li+ is a strong enough base to remove the proton of an alkyne because its conjugate acid, CH3CH2CH2CH3, is weaker than a terminal alkyne.

RC C H

CH3CH2CH2CH2–Li+

RC C–Li+

pKa = 25

CH3CH2CH2CH3

pKa = 50 much weaker acid

11.45 a.

C CH

2 HBr

Br Br

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Chapter 11–14

b.

Cl

2 Cl2 (CH3)3CC

(CH3)3CC

CH

CHCl2

Cl

c.

CH

Cl Cl

[1] Cl2

CH

[2] NaNH2

C C

C C

(2 equiv)

H H [1] BH3

C CH

d.

C

[2] H2O2, HO

e.

O Cl

HCl

C CH

C

CH2

(1 equiv) HC C + D2O

f. g.

HC CD

C C CH3

h.

+ DO– O

H2O

C

CH2

H2SO4

CH3CH2C C + CH3CH2CH2–OTs

CH3

O C

CH2CH3

CH3

CH3CH2C CCH2CH2CH3

CH3

[1] HC C

+ OTs

CH3

[2] HO–H

O

O

i.

H

OH

C CH

C CH H

j.

CH3C C–H

[1] Na+ H–

[2] CH3C

Br

+

C–

CH3C C–H

2° halide E2 CH2–I

k. CH3CH2C C–H

[1]

Na+ –NH2

C C–H [1] Na+ H–

l.

CH3CH2C C

C C

[2] O

[2]

CH3CH2C C

CH2

C CCH2CH2O– [3] HO–H

C C CH2CH2OH

11.46 CH2CH2Br

KOC(CH3)3

CH=CH2

Br2

H H C C H Br Br

A

B

KOC(CH3)3 C CH

DMSO (2 equiv)

C

D

C CCH3

CH3I

NaNH2

C C E

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Alkynes 11–15

11.47 most acidic H [1] NaNH2 [2] CH3I H C C CH2CH2CH2OH

CH3 C C CH2CH2CH2OH

NaNH2 will remove the proton from the OH since it is more acidic.

A

H C C CH2CH2CH2OCH3 = B

H C C CH2CH2CH2O CH3

I

11.48 stereogenic center at the site of reaction

identical

Cl HC C

a.

C CH D H

H D

stereogenic center NOT at the site of reaction

O H CH3

[1] HC C H

CH3 [2] H2O

HO C

CH3

C

H CH3

C

OH

H C

O C CH

d.

CH3 H

H CH3

[1] HC C H CH3

[2] H2O

HO

H C

H CH3

CH3

CH3

11.49 TsCl

OH pyridine

H D

HC C

OTs

D H

SN2 B

A retention PBr3

SN 2

D H Br

A inversion

inversion HC C

SN2

H D

C inversion

OH

H

C

C C

C CH

OH

CH3

HC

HC

enantiomers

H D

H

Configuration is retained.

CH3 H

H D

C

C CH

inversion

Cl HC C

b.

c.

CH3 H

C H CH3

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Chapter 11–16 11.50 H–C CCH2OR'

new C–C bond

CH3 Li+ OH

a.

Br

PBr3

C CCH2OR'

C CCH2OR' B

OCOR

OCOR

OCOR

A

C OH

O

[1] D = HC C OR

b.

OH OR

H C C

[2] H2O

OH

H C C

E = TsCl, pyridine

These 2 C's are added. OCH2CH3

OCH2CH3 [1] NaH CH3 C C

[2] CH3I

OH

OH

[1] NaH

H C C

H C C

H C C [2] F = CH3CH2Br

G

OTs

new C–C bond

11.51 Br H CH3CH2 C C

CH3CH2 H

Br H H CH3–C H

H

C C Br

NH2

Br

1-butyne

H2N H

C CH3

CH3

CH3 C C CH3

C C CH3

Br

CH3CH2 C C H

NH2

H

Br

major product more substituted alkyne

2-butyne

(E and Z isomers possible) NH2

Reaction by-products: H

Br H CH3CH2 C C

NH2 H

H

Br H NH2

C H

2 NH3 + 2 Br–

CH3CH C CH2

CH3 C C Br H

1,2-butadiene

11.52 Draw two diagrams to show  and
bonds.


bond sp2 sp

sp3

+

CH2 C CH3

vinyl cation

sp3 sp2 sp sp3 H

H H

H C

H

C

H

sp2

C H

vacant p orbital for the carbocation

C

C

H

C H

H



sp3



All H's use 1s orbitals. All bonds above are  bonds.

The positive charge in a vinyl carbocation resides on a carbon that is sp hybridized, while in (CH3)2CH+, the positive charge is located on an sp2 hybridized carbon. The higher percent s-character on carbon destabilizes the positive charge in the vinyl cation. Moreover, the positively charged carbocation is now bonded to an sp2 hybridized carbon, which donates electrons less readily than an sp3 hybridized carbon.

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Alkynes 11–17

11.53 A carbanion is more stable when its lone pair is in an orbital with a higher percentage of the smaller s orbital. A carbocation is more stable when its positive charge is due to a vacant orbital with a lower percentage of the smaller s orbital. In HCC+, the positively charged C uses two p orbitals to form two
bonds. If the  bond is formed using an sp hybrid orbital, the second hybrid orbital would have to remain vacant, a highly unstable situation. See also Problem 11.52. HC C

CH2 CH

sp hybridized higher % s-character more stable

HC C

sp2 hybridized lower % s-character less stable

CH2 CH

sp hybridized sp hybridized Vacant orbital has 50% s-character. Vacant orbital is a p orbital. more stable less stable

11.54 CH3

H Cl

CH3

+

C

Cl– attack on the opposite side to the H yields the Z isomer.

C C

Cl CH3C CCH3

Cl

C CH3

H

CH3

CH3

CH3

C C

H H

Cl

Cl

Cl– attack on the same side as the H yields the E isomer.

11.55 a. C

C

C

C

C

C

C

C

H

H C O

H

H CH3CH2 Li+

C H O

OH H OH

+ CH3CH3

+

Li+

H OSO3H OH

HSO4

OH2

b. CH3 C C CH

CH3 C C CH

H

+ OH

CH3 C C CH

H2O C C CH

CH3

H

H

H

H O H

CH3 C C C H H

resonance structures

HSO4

HSO4 CH3 CH CH

O H C O

H2SO4

H

CH3 CH C C H

OH CH3 CH C C

H

H

O H CH3 CH C C H

H H OSO3H

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Chapter 11–18 11.56 H C C OCH3 CH3CH2CH2 H OH2 H CH3CH2CH2C C OCH3

H OH2

H2O

H

OH C C OCH3 CH3CH2CH2

OH C C OCH3 CH3CH2CH2

H

C C OCH3 CH3CH2CH2

H

H2O

O CH3CH2CH2CH2 C OCH3

O H H2O CH3CH2CH2CH2 C OCH3

OH CH3CH2CH2CH2 C OCH3 H2O

H OH2

11.57 a.

H

H CH3 C C CH2

CH3 C C CH2

NH2

CH3 C C CH2

CH3 C C CH H

H

2-butyne

NH3

NH2 H

H2O

NH2

NH2 CH3CH2 C C

NH2

NH3

CH3 C C CH

CH3CH2 C C H

acetylide anion

CH3 C C CH

H

1-butyne

NH3

H

b. A more stable internal alkyne can be isomerized to a less stable terminal alkyne under these reaction conditions because when CH3CH2C
CH is first formed, it contains an sp hybridized C–H bond, which is more acidic than any proton in CH3–C
C–CH3. Under the reaction conditions, this proton is removed with base. Formation of the resulting acetylide anion drives the equilibrium to favor its formation. Protonation of this acetylide anion gives the less stable terminal alkyne. 11.58 Cl

a.

2 –NH2 Cl Cl2

b.

KOC(CH3)3

Cl

(2 equiv) DMSO

Cl

c.

POCl3 OH

KOC(CH3)3

Cl2 Cl

pyridine

(2 equiv) DMSO

Cl

11.59 a. C6H5CH2CH2Br

KOC(CH3)3

C6H5CH=CH2

Br2

C6H5CHBrCH2Br

NaNH2 excess

C6H5C CH

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305

Alkynes 11–19

KOC(CH3)3

b. C6H5CHBrCH3

H2SO4

c. C6H5CH2CH2OH

Br2

C6H5CH=CH2

Br2

C6H5CH=CH2

NaNH2

C6H5CHBrCH2Br

excess NaNH2

C6H5CHBrCH2Br

excess

C6H5C CH

C6H5C CH

11.60 The alkyl halides must be methyl or 1°. a. HC C

CH2CH2CH(CH3)2

HC C

Cl CH2CH2CH(CH3)2

CH3

b. CH3

CH3

C C C CH2CH3

CH3 Cl

C C C CH2CH3

CH3

c.

C C

1° RX

CH3 CH2CH2CH3

C C

Cl CH2CH2CH3 1° RX

11.61 a.

HC C–H

b.

HC C–H

Na+ H–

(CH3)2CHCH2–Cl

HC C

Na+ H–

(CH3)2CHCH2C

NaH

CH3CH2CH2–Cl

HC C

CH

CH3CH2CH2C C

CH3CH2CH2C CH

CH3CH2CH2–Cl [1] BH3

c. CH3CH2CH2C CH (from b.) d.

[2] H2O2, HO–

H2SO4 HgSO4

(from b.)

e.

O

H2O

CH3CH2CH2C CH

CH3CH2CH2

2 HCl

CH3CH2CH2C CH

CH3CH2CH2C CCH2CH2CH3

CH3CH2CH2CH2CHO

C

CH3

CH3CH2CH2CCl2CH3

(from b.)

f.

CH3CH2CH2C CCH2CH2CH3

O

H2O H2SO4, HgSO4

CH3CH2CH2

C

CH2CH2CH2CH3

(from b.)

11.62 Cl2

a. CH3CH2CH CH2 b. CH3CH2C CH (from a.) c. CH3CH2C CH (from a.)

d. CH3CH2CH CH2

HBr (2 equiv) Cl2 (2 equiv)

Br2

Cl

Cl

CH3CH2CH CH2 CH3CH2CBr2CH3

CH3CH2CCl2CHCl2

CH3CH2CHBrCH2Br

2 –NH2

CH3CH2C CH

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Chapter 11–20 e.

NaH

CH3CH2C CH

CH3CH2CH2 Br

CH3CH2C C

CH3CH2C C CH2CH2CH3

(from a.) [1]

f.

CH3CH2C C

O CH3CH2C C CH2CH2OH

[2] H2O

(from e.) O

OH

[1]

g. CH3CH2C C

[2] H2O

(from e.)

C CCH2CH3

(+ enantiomer)

11.63 NaH

CH3CH2CH2CH2CH2CH2Br

a.

HC CH

b.

CH3CH2CH2CH2CH2CH2C CH

HC C

NaH

CH3CH2CH2CH2CH2CH2C CH

CH3CH2CH2CH2CH2CH2C C

CH3CH2Br

CH3CH2CH2CH2CH2CH2C CCH2CH3

(from a.)

c.

CH3CH2CH2CH2CH2CH2C C

[2] H2O

(from b.) d.

O

[1]

CH3CH2CH2CH2CH2CH2C CCH2CH2OH

CH3CH2CH2CH2CH2CH2C CCH2CH2OH

[1] NaH

CH3CH2CH2CH2CH2CH2C CCH2CH2OCH2CH3

[2] CH3CH2Br

(from c.)

11.64 K+ –OC(CH3)3

CH2 CH2

Br

Br

Br2

NaNH2

NaH

HC CH

HC C

Br (2 equiv) Br H2O

Br

C C

NaH

C C

HC C

H2SO4, HgSO4

O

11.65 a.

H2SO4

2 –NH2

Br2

OH

Br

CH3 C CH

NaH

CH3 C C

Br

CH3 C CCH2CH2CH3 SOCl2 OH

NaH

Cl2

b. (from a.)

[1] CH3 C C

Cl

H2O OH

O

(from a.) [2] H2O

Cl

OH CH3C CCH2CHCH3

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Alkynes 11–21

11.66 a.

H2SO4

OH

CH2 CH2

Br

Br2

NaNH2 (2 equiv)

Br

NaH

HC CH

HC C Br

PBr3

OH

[1] NaH C C

HC C

[2] O

HO

[3] H2O

OH

b.

H2SO4

Br2

CH2 CH2

C C

Br

H2O

NaH

O

OH

C C

NaH

HO

C C

CH3CH2Br

O

CH3CH2 O

(from a.)

(from a.)

11.67 Br Br

Br

CH3 C C H

CH3

Br

CH3

H

HO

C C

CH3 C C H

H O

H2O

CH3

Br C C

C C H H

H O

H

Br–

Br H

+ H2O

H OH2

H2O

CH3

O CH3

C

Br C C H

H O

CH2Br

H3O+

H

H2O

11.68 Only this carbocation forms because it is resonance stabilized. The positive charge is delocalized on oxygen. H TsOH O TsO H

H

O

O

re-draw

O

+ TsO–

H

not resonance stabilized

(not formed)

H

OCH3

NOT O

O CH3OH

O

OCH3 H

CH3OH

OCH3

X + CH3OH2

O

Y

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Chapter 11–22 11.69 CH3 C C

O H

C

O

H CH3

CH3 OH

OH2 C C

C C

O

+ H

C

CH3 H2O C

CH3 C C

O H2O

CH3

O

C OH H

O

C

O CH3

C O

CH3

C O H

H

C

O

CH3

C OH

CH3

H

HCOOH

resonance structures

C OH H

enol

O O

C

H

H

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Oxidation and Reduction 12–1 C Chhaapptteerr 1122:: O Oxxiiddaattiioonn aanndd R Reedduuccttiioonn

SSuum mm maarryy:: TTeerrm mss tthhaatt ddeessccrriibbee rreeaaccttiioonn sseelleeccttiivviittyy • A regioselective reaction forms predominately or exclusively one constitutional isomer (Section 8.5). CH3 I

CH3 OH

major product trisubstituted alkene

•

minor product disubstituted alkene

A stereoselective reaction forms predominately or exclusively one stereoisomer (Section 8.5).

H Br

H

Na+ –OCH2CH3

C C

+

C C

H H

C C

H

H

trans alkene major product

•

CH2

+

H

cis alkene minor product

An enantioselective reaction forms predominately or exclusively one enantiomer (Section 12.15). C C CH2OH

allylic alcohol

O C C

Sharpless reagent

or

CH2OH

C C O

CH2OH

One enantiomer is favored.



D Deeffiinniittiioonnss ooff ooxxiiddaattiioonn aanndd rreedduuccttiioonn Oxidation reactions result in: • an increase in the number of C–Z bonds, or • a decrease in the number of C–H bonds.

Reduction reactions result in: • a decrease in the number of C–Z bonds, or • an increase in the number of C–H bonds. [Z = an element more electronegative than C]



R Reedduuccttiioonn rreeaaccttiioonnss [1] Reduction of alkenes—Catalytic hydrogenation (12.3)

R

CH CH R

H2 Pd, Pt, or Ni

H H R C C R H H

alkane

• •

Syn addition of H2 occurs. Increasing alkyl substitution on the C=C decreases the rate of reaction.

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Chapter 12–2 [2] Reduction of alkynes H H

2 H2

R C C R

R C C R

Pd-C

[a]

•

Two equivalents of H2 are added and four new C–H bonds are formed (12.5A).

•

Syn addition of H2 occurs, forming a cis alkene (12.5B). The Lindlar catalyst is deactivated so that reaction stops after one equivalent of H2 has been added.

H H

alkane

[b]

R C C R

R

R

H2

C C

Lindlar catalyst

H

•

H

cis alkene

[c]

R C C R

R

Na

H

•

C C

NH3 H

R

Anti addition of H2 occurs, forming a trans alkene (12.5C).

trans alkene

[3] Reduction of alkyl halides (12.6) R X

[1] LiAlH4 [2] H2O

R H

alkane

• •

The reaction follows an SN2 mechanism. CH3X and RCH2X react faster than more substituted RX.

• •

The reaction follows an SN2 mechanism. In unsymmetrical epoxides, H– (from LiAlH4) attacks at the less substituted carbon.

• • •

The mechanism has one step. Syn addition of an O atom occurs. The reaction is stereospecific.

•

Ring opening of an epoxide intermediate with –OH or H2O forms a 1,2-diol with two OH groups added in an anti fashion.

[4] Reduction of epoxides (12.6) O C C

OH

[1] LiAlH4 [2] H2O

C C H

alcohol



O Oxxiiddaattiioonn rreeaaccttiioonnss [1] Oxidation of alkenes [a] Epoxidation (12.8) C C

+

O C C

RCO3H

epoxide

[b] Anti dihydroxylation (12.9A)

[1] RCO3H C C

HO C

C

[2] H2O ( H+ or HO–)

OH

1,2-diol

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Oxidation and Reduction 12–3 [c] Syn dihydroxylation (12.9B) HO

[1] OsO4; [2] NaHSO3, H2O C C

OH C

or [1] OsO4, NMO; [2] NaHSO3, H2O or KMnO4, H2O, HO–

C

•

Each reagent adds two new C–O bonds to the C=C in a syn fashion.

•

Both the  and
bonds of the alkene are cleaved to form two carbonyl groups.

1,2-diol

[d] Oxidative cleavage (12.10) R'

R R

C O

[2] Zn, H2O or CH3SCH3

H

R'

R

[1] O3

C C

O C

+

R

H

ketone

aldehyde

[2] Oxidative cleavage of alkynes (12.11) [a]

R

[1] O3

R C C R'

internal alkyne

R' C O

[2] H2O

+

•

O C

HO

OH

The  bond and both
bonds of the alkyne are cleaved.

carboxylic acids

[b]

[1] O3

R C C H

terminal alkyne

[2] H2O

R C O

+

CO2

HO

[3] Oxidation of alcohols (12.12, 12.13) H

[a]

[b]

PCC R C OH or HCrO4– H A-26 1o alcohol Amberlyst resin H R C OH H

•

Oxidation of a 1o alcohol with PCC or HCrO4– (Amberlyst A-26 resin) stops at the aldehyde stage. Only one C–H bond is replaced by a C–O bond.

•

Oxidation of a 1o alcohol under harsher reaction conditions—CrO3 (or Na2Cr2O7 or K2Cr2O7) + H2O + H2SO4—affords a RCOOH. Two C–H bonds are replaced by two C–O bonds. Since a 2o alcohol has only one C–H bond on the carbon bearing the OH group, all Cr6+ reagents—PCC, CrO3, Na2Cr2O7, K2Cr2O7, or HCrO4– (Amberlyst A-26 resin)—oxidize a 2o alcohol to a ketone.

C O H

aldehyde

R

CrO3

C O

H2SO4, H2O

1o alcohol

HO

carboxylic acid

H

[c]

R

PCC or CrO3 or R HCrO4– 2o alcohol Amberlyst A-26 resin

R C OH

•

R C O R

ketone

[4] Asymmetric epoxidation of allylic alcohols (12.15) H

CH2 OH C C

R

H

(CH3)3C

OOH

Ti[OCH(CH3)2]4

H R

O C C

CH2OH H

with (–)-DET

or

H CH2OH R C C H O

with (+)-DET

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Chapter 12–4 C Chhaapptteerr 1122:: A Annssw weerrss ttoo PPrroobblleem mss 12.1 Oxidation results in an increase in the number of C–Z bonds (usually C–O bonds) or a decrease in the number of C–H bonds. Reduction results in a decrease in the number of C–Z bonds (usually C–O bonds) or an increase in the number of C–H bonds. O

O

oxidation

a.

c.

reduction

b.

CH3

d.

C

O CH3

CH3

CH2 CH2

C

OCH3

CH3CH2Cl

oxidation

neither

1 new C–H bond and 1 new C–Cl bond

12.2 Hydrogenation is the addition of hydrogen. When alkenes are hydrogenated, they are reduced by the addition of H2 to the
bond. To draw the alkane product, add a H to each C of the double bond. CH3

CH2CH(CH3)2

CH3 CH2CH(CH3)2

C C

a. CH3

CH3 C

H

H

c.

C H H

b.

12.3 Draw the alkenes that form each alkane when hydrogenated. a.

or

H2 Pd-C

or

c.

or or

or or

or

H2 Pd-C

or

b.

H2 Pd-C

12.4 Cis alkenes are less stable than trans alkenes, so they have larger heats of hydrogenation. Increasing alkyl substitution increases the stability of a C=C, decreasing the heat of hydrogenation. CH3CH2 H

H

cis alkane less stable larger heat of hydrogenation

H

H

or

b.

C C

or

C C

a.

CH3CH2

CH2CH3

CH2CH3

trans alkane

trisubstituted

disubstituted less stable larger heat of hydrogenation

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Oxidation and Reduction 12–5 12.5 Hydrogenation products must be identical to use hydrogenation data to evaluate the relative stability of the starting materials. Different products are formed. Hydrogenation data can't be used to determine the relative stability of the starting materials.

2-methyl-2-pentene

3-methyl-1-pentene

12.6 Increasing alkyl substitution on the C=C decreases the rate of hydrogenation. With one equivalent of H2, only the more reactive of the double bonds will be reduced.

disubstituted more reactive

CH3

trisubstituted

CH2

H2, Pd-C

CH3

(1 equiv)

C

CH3 CH3

C H CH3

limonene

12.7 Two enantiomers are formed in equal amounts:

new stereogenic center H

a.

CH3 C CH2CH2CH3

C C H

b.

CH3

CH2CH3

CH2CH3

CH2CH3 CH2CH2CH3

CH2

H

CH3

CH3 H

=

C

CH3

+ CH3

CH2CH3

+ CH2CH2CH3 H

CH3CH2CH2 C CH 3 H

H CH3

diastereomers c.

C(CH3)3

C(CH3)3

C(CH3)3

+

diastereomers

12.8

A

Molecular formula before hydrogenation C10H12

Molecular formula after hydrogenation C10H16

Number of rings 3

Number of
bonds 2

B C

C4H8 C6H8

C4H10 C6H12

0 1

1 2

Compound

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Chapter 12–6 12.9 H2, Pd-C CH2OCO(CH2)16CH3 CHOCO(CH2)6(CH2CH=CH)2(CH2)4CH3 CH2OCO(CH2)16CH3

CHOCO(CH2)6(CH2CH2CH2)2(CH2)4CH3 B

excess

CH2OCO(CH2)16CH3

H2, Pd-C

CH2OCO(CH2)16CH3

1 equiv

A

CH2OCO(CH2)16CH3

CH2OCO(CH2)16CH3

CHOCO(CH2)6CH2CH=CH(CH2)7CH3 CHOCO(CH2)9CH2CH=CH(CH2)4CH3 CH2OCO(CH2)16CH3

CH2OCO(CH2)16CH3

or

C A has 2 double bonds. lowest melting point

C has 1 double bond. intermediate melting point

C

B has 0 double bonds. highest melting point

12.10 Hydrogenation of HC
CCH2CH2CH3 and CH3C
CCH2CH3 yields the same compound. The heat of hydrogenation is larger for HC
CCH2CH2CH3 than for CH3C
CCH2CH3 because internal alkynes are more stable (lower in energy) than terminal alkynes. 12.11 O CH2 C CH3

O

C CH2CH3

H2

H H C

CH2CH3 C C

Lindlar catalyst

A

CH3

cis-jasmone

H

(perfume component isolated from jasmine flowers)

H

12.12 To draw the products of catalytic hydrogenation remember: • H2 (excess), Pd will reduce alkenes and alkynes to alkanes. • H2 (excess), Lindlar catalyst will reduce only alkynes to cis alkenes. a.

b.

CH2=CHCH2CH2 C C CH3

CH2=CHCH2CH2 C C CH3

H2 (excess) Pd-C H2 (excess) Lindlar catalyst

CH3CH2CH2CH2CH2CH2CH3

CH2 CH CH2 CH2

CH3 C C

H

H

12.13 Use the directions from Answer 12.12. a. CH3OCH2CH2C CCH2CH(CH3)2

H2 (excess)

CH3OCH2CH2CH2CH2CH2CH(CH3)2

Pd-C

b. CH3OCH2CH2C CCH2CH(CH3)2

H2 (1 equiv)

CH3OCH2CH2 C C

Lindlar catalyst

c.

CH3OCH2CH2C CCH2CH(CH3)2

H2 (excess) Lindlar catalyst

H CH3OCH2CH2

H CH2CH(CH3)2

C C H

H

CH3OCH2CH2

d. CH3OCH2CH2C CCH2CH(CH3)2

CH2CH(CH3)2

Na, NH3

H

C C H

CH2CH(CH3)2

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Oxidation and Reduction 12–7 12.14 D2 CH3CD2CD2CH2CH2CH3

Pd-C

CH3

D2

CH3 C C CH2CH2CH3

C Lindlar catalyst

CH2CH2CH3 C

D

D

CH3

Na

C

ND3

D C

D

CH2CH2CH3

12.15 H2 H

R

Lindlar catalyst

A

B

H

12.16 LiAlH4 reduces alkyl halides to alkanes and epoxides to alcohols. CH3 [1] LiAlH4

Cl

a.

H

b.

[2] H2O

O

[1] LiAlH4

H

[2] H2O

H replaces Cl.

CH3 OH H H

12.17 To draw the product, add an O atom across the
bond of the C=C. O

mCPBA

a.

(CH3)2C

CH2

b.

(CH3)2C

C(CH3)2

(CH3)2C

c.

CH2

CH2

mCPBA

O mCPBA

(CH3)2C

C(CH3)2

12.18 For epoxidation reactions: • There are two possible products: O adds from above and below the double bond. • Substituents on the C=C retain their original configuration in the products. H

CH3

mCPBA

C C

a.

CH3 H

H

H CH3CH2

b.

H

mCPBA CH3CH2 H

H

CH3

c.

H

CH2CH3

C C H

O C C

mCPBA

CH3 H H C C H enantiomers O

O C C

CH3CH2 CH2CH3 H C C H CH2CH3 O H

identical

CH3

CH3

O

O

H

H

enantiomers

O CH2

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Chapter 12–8 12.19 Treatment of an alkene with a peroxyacid followed by H2O, HO adds two hydroxy groups in an anti fashion. cis-2-Butene and trans-2-butene yield different products of dihydroxylation. cis-2Butene gives a mixture of two enantiomers and trans-2-butene gives a meso compound. The reaction is stereospecific because two stereoisomeric starting materials give different products that are also stereoisomers of each other. CH3 C C H

HO

CH3

[1] RCO3H

H

[2] H2O, HO


C CH3

C

CH3 H

OH

H

cis-2-butene

CH3 C C H

CH3 H

OH C

C

HO

H

CH3

enantiomers HO

H

[1] RCO3H

CH3

[2] H2O, HO


H C

CH3

CH3 H

CH3

C

OH

H

OH C

C H CH3

HO

identical meso compound

trans-2-butene

12.20 Treatment of an alkene with OsO4 adds two hydroxy groups in a syn fashion. cis-2-Butene and trans-2-butene yield different stereoisomers in this dihydroxylation, so the reaction is stereospecific. HO CH3 C C H

CH3

[1] OsO4

H

[2] NaHSO3, H2O

OH C

CH3

C

H

H

[1] OsO4

HO

H CH3

[2] NaHSO3, H2O

CH3

C

C

HO

OH C

trans-2-butene

CH3

CH3 H

C H CH3

H

CH3 H OH

identical meso compound

cis-2-butene CH3 C C H

CH3 H

H C

CH3

C

HO

OH

enantiomers

12.21 To draw the oxidative cleavage products: • Locate all the
bonds in the molecule. • Replace all C=C’s with two C=O’s. Replace this
bond with two C=O's. [1] O3

a. (CH3)2C

CHCH2CH2CH2CH3

[2] Zn, H2O

[1] O3 [2] Zn, H2O

+

O CHCH2CH2CH2CH3

aldehyde

ketone +

O

One ketone and one aldehyde are formed.

Two aldehydes are formed.

O

[2] Zn, H2O

c.

O

H

[1] O3

b.

(CH3)2C

H

O O H

A dicarbonyl compound is formed.

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Oxidation and Reduction 12–9 12.22 To find the alkene that yields the oxidative cleavage products: • Find the two carbonyl groups in the products. • Join the two carbonyl carbons together with a double bond. This is the double bond that was broken during ozonolysis. CH3

a. (CH3)2C O

+ (CH3CH2)2C

O

(CH3)2C

c.

C(CH2CH3)2

CH3 O only

C CH3

CH3

b.

CHCH3 +

C CH3

With only one product, the alkene must be symmetrical around the double bond. Join this C to the same C in another identical molecule.

Join these two C's. O

CH3

C

CH3CHO

Join these two C's.

12.23 O

a.

[1] O3

O

O

[2] CH3SCH3

H

H

H

O

[1] O3

b.

[2] CH3SCH3 H

c.

[2] CH3SCH3

O O O

O

O

O H

O

H

H H

a.

[2] H2O

O

O

[1] O3 CH3CH2

C

OH

+

HO

C

CH2CH2CH3

internal alkyne O [1] O3

b.

C

C C [2] H2O

internal alkyne terminal alkyne c.

O

+ OH

HO

C

identical compounds

internal alkyne

H C C CH2 CH2 C C CH3

[1] O3 [2] H2O

O CO2 +

HO

C O

C

O OH

+ HO

C

H O

12.24 To draw the products of oxidative cleavage of alkynes: • Locate the triple bond. • For internal alkynes, convert the sp hybridized C to COOH. • For terminal alkynes, the sp hybridized C–H becomes CO2. CH3CH2 C C CH2CH2CH3

H

O O

[2] CH3SCH3

H

H

H

[1] O3

d. H

O

[1] O3

CH3

O

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Chapter 12–10 12.25 a.

c.

CH3CH2CH(CH3)CO2H

b.

CO2 + CH3(CH2)8CO2H

CH3CH2CH C C CHCH2CH3

CH3(CH2)8 C CH

CH3

CH3CH2CO2H, HO2CCH2CO2H, CH3CO2H CH3CH2 C C CH2 C C CH3

CH3

d.

HO2C(CH2)14CO2H

12.26 For oxidation of alcohols, remember: • 1° Alcohols are oxidized to aldehydes with PCC. • 1° Alcohols are oxidized to carboxylic acids with oxidizing agents like CrO3 or Na2Cr2O7. • 2° Alcohols are oxidized to ketones with all Cr6+ reagents. O PCC

OH

a.

OH

H

c.

CrO3

OH

H2SO4, H2O

O OH OH

O

O PCC

b.

CrO3

d.

H2SO4, H2O

12.27 Upon treatment with HCrO4–– Amberlyst A-26 resin: • 1° Alcohols are oxidized to aldehydes. • 2° Alcohols are oxidized to ketones. H OH

O

HCrO4–

a.

c.

Amberlyst A-26 resin

b. HO

OH

H

HCrO4–

OH

OH

O

HCrO4–

O

Amberlyst A-26 resin

O

O

Amberlyst A-26 resin

12.28 NaOCl

a. OH

b.

O

The by-products of the reaction with sodium hypochlorite are water and table salt (NaCl), as opposed to the by-products with HCrO4–– Amberlyst A-26 resin, which contain carcinogenic Cr3+ metal.

+ NaCl + H2O

Oxidation with NaOCl has at least two advantages over oxidation with CrO3, H2SO4 and H2O. Since no Cr6+ is used as oxidant, there are no Cr by-products that must be disposed of. Also, CrO3 oxidation is carried out in corrosive inorganic acids (H2SO4) and oxidation with NaOCl avoids this.

12.29 O HO

OH

ethylene glycol

O H

H O

OH

HO O

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Oxidation and Reduction 12–11 12.30 To draw the products of a Sharpless epoxidation: • With the C=C horizontal, draw the allylic alcohol with the OH on the top right of the alkene. • Add the new oxygen above the plane if (
)-DET is used and below the plane if (+)-DET is used. OH (CH3)3C

a.

OH

OOH

Ti[OCH(CH3)2]4 (+)-DET

O H

(+)-DET adds O below the plane.

OH

b.

re-draw

OH

(CH3)3C

H O

OOH

OH

Ti[OCH(CH3)2]4 (–)-DET

(
)-DET adds O above the plane.

12.31 Sharpless epoxidation needs an allylic alcohol as the starting material. Alkenes with no allylic OH group will not undergo reaction with the Sharpless reagent. This alkene is part of an allylic alcohol and will be epoxidized. geraniol

OH

This alkene is not part of an allylic alcohol and will not be epoxidized.

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Chapter 12–12 12.32 Use the rules from Answer 12.1. OH

a.

d.

CH2CH3 reduction

C CH

O

reduction

O

b. c.

CH3CH2Br

e.

OH oxidation

OH CH2 CH2

OH

HO

f.

neither 1 C–H and 1 C–Br bond are removed.

ClCH2CH2Cl oxidation (2 new C–Cl bonds)

CH2 CH2

O

O oxidation

12.33 Use the principles from Answer 12.2 and draw the products of syn addition of H2 from above and below the C=C. H2 Pd-C

a. CH3

Pd-C

CH3 CH2CH3

c. CH3

C CH2 (CH3)2CH

+ CH3

CH3 CH2CH3

H2

H2 Pd-C

CH2CH3

+

Pd-C

CH3CH2

d.

CH3

CH3

H2

b.

CH3

CH3 CH2CH3 H C (CH3)2CH

CH2CH3

+ CH3

C

(CH3)2CH H

CH3

12.34 Increasing alkyl substitution increases alkene stability, decreasing the heat of hydrogenation. 2-methyl-2-butene trisubstituted smallest
Ho = –112 kJ/mol

2-methyl-1-butene disubstituted intermediate
Ho = –119 kJ/mol

3-methyl-1-butene monosubstituted largest
Ho = –127 kJ/mol

12.35 A possible structure: a. Compound A: molecular formula C5H8: hydrogenated to C5H10. 2 degrees of unsaturation, 1 is hydrogenated. 1 ring and 1
bond b. Compound B: molecular formula C10H16: hydrogenated to C10H18. 3 degrees of unsaturation, 1 is hydrogenated. 2 rings and 1
bond c. Compound C: molecular formula C8H8: hydrogenated to C8H16. 5 degrees of unsaturation, 4 are hydrogenated. 1 ring and 4
bonds

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Oxidation and Reduction 12–13 12.36 c.

a. monosubstituted largest heat of hydrogenation b. fastest reaction rate

O

[1] O3

H

[2] Zn, H2O

A

c.

a. tetrasubstituted smallest heat of hydrogenation b. slowest reaction rate

+

H

C

H

O

[1] O3 O [2] Zn, H2O

B

+ O

identical c.

a. trisubstituted intermediate heat of hydrogenation b. intermediate reaction rate

O

[1] O3 [2] Zn, H2O

O

+

H

C

12.37 Work backwards to find the alkene that will be hydrogenated to form 3-methylpentane. 2 possible enantiomers: or H

R isomer

3-methylpentane

H

S isomer

12.38 O

O H2 (excess)

OH

a.

OH

Pd-C

stearidonic acid

stearic acid O

b.

H2 (1 equiv)

O OH

OH

Pd-C O

O OH

O OH

c. trans one possibility

OH

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Chapter 12–14 O

O

d.

OH

O OH

<

OH

<

stearidonic acid 4 cis C=C's

product in (b) 3 cis C=C's

product in (c) 2 cis and one trans C=C's

12.39 H2

a.

H2

b.

Lindlar catalyst Na NH3

c.

[1] LiAlH4

h.

Pd-C

no reaction no reaction

(CH3)3COOH

j.

[1] CH3CO3H

OH

k.

O

OH O

syn addition

H2O,

[2] H2O

OH OH

KMnO4

g.

OH

[1] LiAlH4

l.

OH

[2] NaHSO3, H2O

no reaction

mCPBA

OH

anti addition

[1] OsO4 + NMO

f.

H

Ti[OCH(CH3)2]4 (
)-DET

OH

[2] H2O, HO


O

H

[2] CH3SCH3

O

e.

O

[1] O3

i.

CH3CO3H

d.

no reaction

[2] H2O

syn addition

HO


OH

12.40 a. CH3CH2CH2 C C CH2CH2CH3

H2 (excess) Pd-C

CH3CH2CH2

H2

b. CH3CH2CH2 C C CH2CH2CH3

C C Lindlar catalyst

c. CH3CH2CH2 C C CH2CH2CH3 d. CH3CH2CH2 C C CH2CH2CH3

CH2CH2CH3

Na

H H

H

CH2CH2CH3

[1] O3

C C CH3CH2CH2 H O

[2] H2O

CH3CH2CH2

NH3

cis alkene

C

trans alkene O

+

OH

CH3CH2CH2

C

OH

identical

12.41 a.

CH3CH2 C CH3

H C

H2 Pd-C

CH2OH

H C * C CH2OH CH3 H

[* = new stereogenic center]

CH3

CH3

CH3CH2 H

C CH3CH2

CH2CH2OH H

+

HOCH2CH2 H

Two enantiomers are formed.

C CH2CH3

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Oxidation and Reduction 12–15 b.

CH3CH2

H

C

C

CH3

O

mCPBA CH3CH2 CH3

CH2OH

H CH2OH

CH3CH2 CH3

H CH2OH

e.

f.

CH3 C C CH3CH2

CH2OH

CH3 C C CH3CH2

CH2OH

(CH3)3COOH Ti[OCH(CH3)2]4 (+)-DET

H

O

c. CH3CH2 C

CH3CH2

CrO3

CH2OH

C

H

[1] PBr3

C CH2OH

CH3CH2 CH3

COOH CH3CH2 C

H2

HCrO4–

H C

C CH3

O OH

H2SO4, H2O O

d.

OH

PCC

OH

CF3CO3H

H

OH

O

O

H

e.

f.

OH

[1] OsO4 [2] NaHSO3, H2O

OH

OH

H OH

OH OH

OH

OH OH

[1] HCO3H [2] H2O, HO–

OH OH OH

OH

OH

OH H O

OH

g.

OH

(CH3)3COOH

re-draw

Ti[OCH(CH3)2]4 (+)-DET

h.

OH

KMnO4 H2O, HO


OH

OH

OH

OH OH OH

CH3

CH3CH2

Amberlyst A-26 CH2OH

OH

Na2Cr2O7

C

CH3

Pd-C

c.

H

C

12.42

OH

CH2Br

CH3CH2

H C C

CH3

b.

C

CH3

CHO

H2SO4, H2O

OH

H

[2] LiAlH4 [3] H2O

h.

a.

O CH3 CH3CH2

CH3CH2

C

CH3

C

CH3

H C

g.

H

C

CH2OH

d. CH3CH2 CH3

PCC

C

CH3

C

CH3CH2

H

CH2OH H O

(CH3)3COOH Ti[OCH(CH3)2]4 (
)-DET

H

CH3 CH3CH2

OH

H C CHO

CH2OH H

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Chapter 12–16 12.43 OH

O PCC

a.

OH

c.

CrO3

OH

H2SO4, H2O

O

O PCC

b. CH3CH2CH2CH2OH

CH3CH2CH2

C

H

12.44 OH

a.

b.

[1] SOCl2

CH2

[3] H2O

[1] OsO4

OH

[2] NaHSO3, H2O

CH2OH O

[1] mCPBA

c.

[2] LiAlH4

Cl

pyridine

CH2

[2] LiAlH4

OH

[3] H2O

CH3

H2

d.

Lindlar catalyst

12.45 a. H2, Pd-C

d. [1] LiAlH4

b. mCPBA O c. KMnO4 H2O, HO–

or [1] OsO4 [2] NaHSO3, H2O OH

OH f. PBr3

h. [1] LiAlH4 [2] H2O

or HBr

[2] H2O

g. CrO3, or PCC or HCrO4–, H2SO4, H2O Amberlyst A-26 resin O

e. H2O (–OH) OH

OH

Br

OH

(+ enantiomer)

12.46 Alkenes treated with [1] OsO4 followed by NaHSO3 in H2O will undergo syn addition, whereas alkenes treated with [2] CH3CO3H followed by –OH in H2O will undergo anti addition. a.

[1]

(CH3)2CH

CH(CH3)2 C C

H

[2] (CH3)2CH H

H

H C C CH(CH3)2

[1] OsO4 [2] NaHSO3, H2O

[1] CH3CO3H [2] –OH, H2O

anti addition

HO (CH3)2CH H

OH C

H

HO (CH3)2CH H

C CH(CH3)2

H CH(CH ) 3 2 C

C OH

HO

OH

rotate (CH3)2CH H

C

C H

CH(CH3)2

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325

Oxidation and Reduction 12–17 b.

C6H5

[1]

H

C C H

C6H5

C6H5

[2]

C6H5

[1] CH3CO3H

H

[2] –OH, H2O

rotate

C

H C6H5

syn addition

HO

OH C

[2] NaHSO3, H2O

C C H

HO

[1] OsO4

H

HO

H C

C6H5

H C6H5

C6H5

C

H C6H5

C6H5

C

H C

+ enantiomer OH

+ enantiomer

OH OH

CH3CH2CH2

c. [1]

CH2CH2CH3

C C H

HO [1] OsO4

OH C

CH3CH2CH2

[2] NaHSO3, H2O

H

CH3CH2CH2

H

C H

CH2CH2CH3

syn addition CH3CH2CH2

[2]

H

H

H

[1] CH3CO3H

C C CH2CH2CH3

[2] –OH, H2O

H

CH2CH2CH3 OH

12.47 OH O

H

A

O

OH H OH

+ AlH3 + Li+

H3Al H Li

D– (from LiAlD4) opens the epoxide ring from the back side, so it is oriented on a wedge in the final product. D

OH

H

CH2CH2CH3 OH

OH

CH3CH2CH2

H

rotate

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Chapter 12–18 12.48 Cl H OH [1] C

mCPBA

O

O

O O

O

[2] Br

OH Br

Br

OH

H

O

OH

H

Na+ H

Br

[3]

Cl

C

OH

OH

O

Br

Na+ + H2 + O

Br–

O

+ HO

12.49 Use the directions from Answer 12.21. a. (CH3CH2)2C CHCH2CH3

[1] O3 [2] CH3SCH3

(CH3CH2)2C

O CHCH2CH3

CH3

[1] O3

b.

+

O

O

[2] Zn, H2O

+ O C CH3

O C CH

c.

C

[1] O3

OH + CO2

[2] H2O O

d.

[1] O3

C C

C

[2] H2O

O OH

+

HO

C

identical

12.50 a. (CH3)2C

b.

O and CH2 O

O and

O

(CH3)2C

CH2

c.

d.

CH3CH2CH2CH CHCH2CH2CH3

CH3CH2CH2CHO only

Join this C to the same C in another identical molecule. O

O and 2 equivalents of

Join both of these C's to a C from formaldehyde.

CH2 O

formaldehyde C

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Oxidation and Reduction 12–19 12.51 Use the directions from Answer 12.22. Join these two C's. O

[1] O3

a. C10H18

H

[2] CH3SCH3 O

2 degrees of unsaturation

one ring + one
bond

O [1] O3

b. C10H16

[2] CH3SCH3

3 degrees of unsaturation

O

two rings + one
bond

12.52 Join these two C's.

a. CH3CH2CH2CH2COOH and

Join these two C's. COOH and

CH3CH2CH2CH2C CH c.

CO2

C CCH3 CH3COOH

Join these two C's. CH3CH2C CCH2CH2CH3

b. CH3CH2COOH and CH3CH2CH2COOH

12.53 a.

squalene A

B

B

C

B

B

A

[1] O3 [2] Zn, H2O H

+ O

O

+ O H

2 equiv (from portion A)

O

O H

4 equiv (from portion B)

1 equiv (from portion C)

COOH

b.

c.

linolenic acid

[1] O3 [2] Zn, H2O

[1] O3 [2] Zn, H2O O

zingiberene

H O

+

+ H

COOH

O

O

2 equiv

12.54 A

a. C8H12

O

[1] O3 [2] CH3SCH3

H

+

+ H H

O

O

O O H

H H

O

H

H O

O

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Chapter 12–20 [1] NaNH2

H2 (excess)

b. C6H10

[2] CH3I

Pd-C

B

C

C7H12

12.55 The hydrogenation reaction tells you that both oximene and myrcene have 3
bonds (and no rings). Use this carbon backbone and add in the double bonds based on the oxidative cleavage products.

H2

C10H16

Pd-C

2,6-dimethyloctane

3 degrees of unsaturation O

Oximene:

(CH3)2C

O

CH2 O

CH2(CHO)2

CH3 C CHO O

O

Myrcene:

(CH3)2C

O

CH2 O

H

C

(2 equiv)

C CH2CH2

CHO

12.56 a.

H2 (2 equiv)

H2

Pd-C

Lindlar catalyst

C7H12 A

A

oxidative cleavage

B

Na, NH3

O OH

+ other product(s)

C

b. A does not react with NaH because it is not a terminal alkyne.

12.57 OH H2SO4

H2 Pd-C

decalin

C10H18O A

C10H16 B

C10H16 C ozonolysis O

O

C10H16O2 E

CHO

D

O

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329

Oxidation and Reduction 12–21 12.58 Since hydrogenation of DHA forms CH3(CH2)20COOH, DHA is a 22-carbon fatty acid. The ozonolysis products show where the double bonds are located. DHA CO2H

A

B

B

B

B

C

B

[1] O3 [2] Zn, H2O

+

CH3CH2CHO

OHCCH2CHO

+

OHCCH2CH2CO2H

5 equiv (from portion B)

(from portion A)

(from portion C)

12.59 The stereogenic center (labeled with *) in both structures can be R or S. * O

O

H

O H

H

H2

ozonolysis

O O H

Pd-C

or

H

* butylcyclopheptane

O

possible structures for dictyopterene D'

H

12.60 OH OH

a.

re-draw

(CH3)3COOH Ti[OC(CH3)2]4 (
)-DET

b.

H C C (CH3)3C

CH2OH H

H (CH3)3C

(CH3)3COOH

C

Ti[OC(CH3)2]4 (+)-DET

C

H

O

OH H

CH2OH H

O

12.61 OH

re-draw OH

(CH3)3COOH Ti[OC(CH3)2]4 (
)-DET

O

CH2OH H

major product 87%

CH2OH

+ O

H

minor product 13%

enantiomeric excess = % one enantiomer – % second enantiomer ee = 87% – 13% = 74%

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Chapter 12–22 12.62 O

H

a.

Replace this O to make an alkene. H

H

H

CH2 H

HOCH2

HOCH2

CH3 CH3 C C CH3 O

H CH2OH

H

(+)-DET

CH2OH

CH3 C

(–)-DET Replace this O to make an alkene.

H

O

c.

(–)-DET HO

HO

b.

Replace this O to make an alkene.

C

CH3

CH3

12.63 Use retrosynthetic analysis to devise a synthesis of each hydrocarbon from acetylene. a.

CH3CH2CH CH2

CH3CH2CH CH

NaH HC CH

CH3CH2Cl

C CH

C CH

HC CH

H2

CH3CH2C CH

CH3CH2CH CH2

Lindlar catalyst

CH3

CH3

C C

b. H

CH3 C C CH3

C CH

CH3 C CH

HC CH

H NaH

HC CH

CH3Cl

C CH

NaH

CH3 C CH

CH3 C C

CH3Cl

CH3

H2

CH3 C C CH3

C C

Lindlar catalyst CH3

d.

CH3 C C CH3

NaH C CH

CH3Cl

C CH

CH3 C CH

CH3 C CH

(CH3)2CHCH2CH2CH2CH2CH(CH3)2

HC CH

H

HC CH

CH3

H

HC CH

H

H

C C

c.

CH3

NaH

Cl C CH

HC C

NaH

CH3 C C

CH3Cl

CH3 C C CH3

Na

CH3

NH3 H

HC CCH2CH(CH3)2 NaH

C C

C CH Cl C C

HC CH H2 (2 equiv) Pd-C

H

C C CH3

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331

Oxidation and Reduction 12–23 12.64 Cl

NaH

HC CH

HC C NaH

Cl H2 Lindlar catalyst

12.65 Br

–NH 2

Br2

Na NH3

(2 equiv)

cis

trans

Br

12.66 CH3

O

a.

CH3 C H

C

HC CH

C

CH3 H

NaH

CH3

H

CH3 C C CH3

C H CH3Cl

HC C

HC C CH3

HC C CH3

NaH

C C CH3

O CH3 H

C

C

CH3Cl

mCPBA CH3 H

HC CH

CH3 C C CH3

CH3 C H

H2 Lindlar catalyst CH3 C H

O

b. H C

C

CH3

H CH3 H

CH3

CH3

(+ enantiomer) CH3 C C CH3

c.

C

H

H

C

H CH3 H

H CH3

H

C C CH3 CH3

H2O, HO


C

C

CH3 H

+ enantiomer

H

CH3 KMnO4

HC CH

O

mCPBA

CH3 C C CH3

C C H CH3

(from a.)

CH3 C C

CH3

OH

HC C CH3

H H

Na, NH3

(from a.) HO

CH3 C C CH3

C C

CH3 HO C H CH3

OH C H CH3

HC C CH3

HC CH

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Chapter 12–24 HO

d.

OH C

H CH3

H

C H

CH3 CH3 C C CH3

C C

CH3

CH3

HC C CH3

HC CH

H

(+ enantiomer) H

CH3

HO

KMnO4

C C CH3 H

H2O, HO


(from b.)

OH

(+ enantiomer)

C

C H CH3

H

CH3

12.67 [1] BH3

a. C6H5CH CH2

PCC

C6H5CH2 CH2OH

[2] H2O, –OH

C6H5CH2 CHO

HO H2O

b. C6H5CH CH2

H2SO4

[1] BH3

c. C6H5CH CH2

[2] H2O,

O PCC

C6H5CH CH3

C5H5

CrO3

C6H5CH2 CH2OH

–OH

C

CH3

C6H5CH2 COOH

H2SO4, H2O

HC CH NaH

O mCPBA

d. C6H5CH CH2

C6H5

C C H H H

HO

[1] C CH

C6H5CH CH2C CH

[2] H2O

12.68 Br Br2

Br

NaH

2 NaNH2

1-pentene CH3Cl Na

(2E)-2-hexene

NH3

12.69 a.

b.

CH3CH2CH CH2

CH3CH2CH2CH2OH

[1] 9-BBN or BH3 [2] H2O2,

HO–

POCl3

pyridine

CrO3 CH3CH2CH2CH2OH

CH3CH2CH CH2

mCPBA

H2SO4, H2O

CH3CH2CH2COOH

HO O [1] LiAlH4 CH3CH2CH CH3 CH3CH2CH CH2 [2] H2O H2O, H2SO4 CrO3 H2SO4, H2O CH3CH2

C O

CH3

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Oxidation and Reduction 12–25 c.

OH

[1] OsO4 [2] NaHSO3, H2O

OH

[2] CH3I (2 equiv) O

mCPBA

d.

OCH3 OH

H2O

HC C

O

OCH3

[1] NaH (2 equiv)

C CH

NaH

O PCC

C CH

C CH

HC CH

12.70 NaH

a. HC CH

CH3CH2Br

HC C

NaH

HC C CH2CH3

C C CH2CH3 O

H2 Lindlar catalyst

OH

OH

b.

HO CH2CH2 C C CH2CH3

Na

c.

OH

O CH2CH2 C C CH2CH3

OH

NH3

(from a.)

H2O

PBr3

Br

(from b.) PCC

d.

CHO

OH (from b.)

12.71 H2

a.

[2] H2O2, HO–

Lindlar catalyst [1] LiAlH4

mCPBA

b.

OH

[1] 9-BBN or BH3

C CH

O

[2] H2O

OH

(from a.)

c.

C CH

HO NaH

C C O H

mCPBA

d. (from c.)

H

CH3

(+ enantiomer)

CH3Cl

C C CH3

Na NH3

KMnO4 H2O, HO–

OH

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Chapter 12–26 12.72 HC CH

NaH

Br

C CH

NaH CH3CH2 C CH

Br

CH3CH2 C C

Na, NH3 Cl

H

Cl2

anti addition

Cl H

(3R,4S)-3,4-dichlorohexane

12.73 a.

OH

HC CH

K+–OC(CH3)3

PBr3

NaH

CH2 CH2

Br Br

C CH

Br2

2 NaNH2

Br

NaH CH3CH2 C CH

HC CH

Br Br

CH3CH2 C C

Na, NH3

H2

b.

(from a.)

H

Lindlar catalyst

HO

[1] OsO4

c.

(from b.)

O

mCPBA

[2] NaHSO3, H2O

H C

CH3CH2 H

C

H

CH2CH3

OH

12.74 OCH2CH3

H OCH2CH3 H OCH3

Li

OCH3

OCH3

Li

OCH3

Li

H OCH2CH3

OCH3

OCH3

H OCH2CH3

Li

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Oxidation and Reduction 12–27 12.75 The favored conformation for both molecules places the tert-butyl group equatorial. OH (CH3)3C

OH

H H

= (CH ) C 3 3 A

A

OH

(CH3)3C

=

(CH3)3C

OH

B

This OH is axial and will react faster because the OH group is more hindered.

B

This OH is equatorial and will react more slowly because the OH group is less hindered.

12.76 O H O

Cl O

O

O C O

Cl

O

O

O

O H

O

C

O mCPBA

OSiR3

OSiR3

R = alkyl group O C

O

H

Cl O

O

O

H O

O

O OH

O

O C

Cl

O

O O

OH OSiR3

C O

O H O

O O

Cl

OSiR3

OSiR3

O

O

C O

Cl

HO

C

Cl

C

Cl

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Chapter 12–28 12.77 O

mCPBA

"down" bond

a.

O comes in from below.

X

"up" bond

OH Br2

b.

Br

H2O

Br

Br+ comes in from below. X H2O attacks from the back side at the more substituted C. This places the OH group axial, on an "up" bond.

NaH

O

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337

Mass Spectrometry and Infrared Spectroscopy 13–1 C Chhaapptteerr 1133:: M Maassss SSppeeccttrroom meettrryy aanndd IInnffrraarreedd SSppeeccttrroossccooppyy  M Maassss ssppeeccttrroom meettrryy ((M MSS)) • • • • • • • •

Mass spectrometry measures the molecular weight of a compound (13.1A). The mass of the molecular ion (M) = the molecular weight of a compound. Except for isotope peaks at M + 1 and M + 2, the molecular ion has the highest mass in a mass spectrum (13.1A). The base peak is the tallest peak in a mass spectrum (13.1A). A compound with an odd number of N atoms gives an odd molecular ion. A compound with an even number of N atoms (including zero) gives an even molecular ion (13.1B). Organic chlorides show two peaks for the molecular ion (M and M + 2) in a 3:1 ratio (13.2). Organic bromides show two peaks for the molecular ion (M and M + 2) in a 1:1 ratio (13.2). The fragmentation of radical cations formed in a mass spectrometer gives lower molecular weight fragments, often characteristic of a functional group (13.3). High-resolution mass spectrometry gives the molecular formula of a compound (13.4A).

  EElleeccttrroom maaggnneettiicc rraaddiiaattiioonn • The wavelength and frequency of electromagnetic radiation are inversely related by the following equations:  = c/ or  = c/ (13.5). • The energy of a photon is proportional to its frequency; the higher the frequency the higher the energy: E = h (13.5).   IInnffrraarreedd ssppeeccttrroossccooppyy ((IIR R,, 1133..66 aanndd 1133..77)) • Infrared spectroscopy identifies functional groups. • IR absorptions are reported in wavenumbers: wavenumber = ~ = 1/


• • •

The functional group region from 4000–1500 cm–1 is the most useful region of an IR spectrum. C–H, O–H, and N–H bonds absorb at high frequency,
2500 cm–1. As bond strength increases, the wavenumber of an absorption increases; thus triple bonds absorb at higher wavenumber than double bonds. C C

~ 1650 cm–1

C C

~ 2250 cm–1

Increasing bond strength Increasing
~

•

The higher the percent s-character, the stronger the bond, and the higher the wavenumber of an IR absorption. C

C H

C H H

Csp3–H 25% s-character 3000–2850 cm–1

Csp2–H 33% s-character 3150–3000 cm–1

Csp–H 50% s-character 3300 cm–1

Increasing percent s-character Increasing ~


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Chapter 13–2 C Chhaapptteerr 1133:: A Annssw weerrss ttoo PPrroobblleem mss 13.1

The molecular ion formed from each compound is equal to its molecular weight.

a. C3H6O molecular weight = 58 molecular ion (m/z) = 58

b. C10H20 molecular weight = 140 molecular ion (m/z) = 140

c. C8H8O2 molecular weight = 136 molecular ion (m/z) = 136

d. C10H15N molecular weight = 149 molecular ion (m/z) = 149

13.2

Some possible formulas for each molecular ion: a. Molecular ion at 72: C5H12, C4H8O, C3H4O2 b. Molecular ion at 100: C8H4, C7H16, C6H12O, C5H8O2 c. Molecular ion at 73: C4H11N, C2H7N3

13.3

To calculate the molecular ions you would expect for compounds with Cl, calculate the molecular weight using each of the two most common isotopes of Cl (35Cl and 37Cl). Do the same for Br, using 79Br and 81Br. a. C4H935Cl = 92 C4H937Cl = 94 Two peaks in 3:1 ratio at m/z 92 and 94

d. C4H11N = 73 One peak at m/z 73

b. C3H7F = 62 One peak at m/z 62

e. C4H4N2 = 80 One peak at m/z 80

c. C6H1179Br = 162 C6H1181Br = 164 Two peaks in a 1:1 ratio at m/z 162 and 164

13.4

After calculating the mass of the molecular ion, draw the structure and determine which C–C bond is broken to form fragments of the appropriate mass-to-charge ratio. e– Cleave bond [1].

[1] CH3 CH3 CH3 C

C

H

H

m/z = 100

CH3 C CH3

+

CH3CHCH2CH3

H

m/z = 43

CH2CH3

e– Cleave bond [1].

CH3 C CH2CH3 H

m/z = 57

+

(CH3)2CH

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Mass Spectrometry and Infrared Spectroscopy 13–3

13.5 Break this bond. CH3 H CH3 C

CH3

C

e–

C CH3 H

CH3 H

+

CH3 C CH3 CH3

m/z = 114

H

CH3

C

C CH3

H

H

m/z = 57 This 3° carbocation is more stable than others that can form, and is therefore the most abundant fragment.

13.6 a.

OH

Cleave bond [1].

+

CH3

C CH2 CH3

OH

H

m/z = 59

CH3 C CH2 CH3 H

[1]

OH

[2] Cleave bond [2].

CH2CH3

+

CH3 C H

m/z = 45 b. OH

– H2O

CH3 C CH2 CH3

CH2=CHCH2CH3

+

CH3CH=CHCH3

m/z = 56

H

m/z = 56

13.7 O

a.

C

O C

CH3CH2 C O

CH3CH2

(from cleavage of bond [2]) [1]

b.

13.8

(from cleavage of bond [1])

[2]

CH3CH2CH2CH2CH2CH2OH

+ CH2OH

c. CH3CH2CH2CHO

CH3CH2CH2C O

HC O

Use the exact mass values given in Table 13.1 to calculate the exact mass of each compound. C7H5NO3

mass: 151.0270

C8H9NO2

C10H17N

mass: 151.0634 compound X

mass: 151.1362

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Chapter 13–4 13.9 CH3

benzene C6H6 m/z = 78

toluene C7H8 m/z = 92

CH3

CH3

p-xylene C8H10 m/z = 106

GC–MS analysis: Three peaks in the gas chromatogram. Order of peaks: benzene, toluene, p-xylene, in order of increasing bp. Molecular ions observed in the three mass spectra: 78, 92, 106.

13.10 Wavelength and frequency are inversely proportional. The higher frequency light will have a shorter wavelength. a. Light having a
of 102 nm has a higher  than light with a
of 104 nm. b. Light having a
of 100 nm has a higher  than light with a
of 100 m. c. Blue light has a higher  than red light. 13.11 The energy of a photon is proportional to its frequency, and inversely proportional to its wavelength. a. Light having a  of 108 Hz is of higher energy than light having a  of 104 Hz. b. Light having a
of 10 nm is of higher energy than light having a
of 1000 nm. c. Blue light is of higher energy than red light. 13.12 The larger the energy difference between two states, the higher the frequency of radiation needed for absorption. The 400 kJ/mol transition requires a higher  of radiation than a 20 kJ/mol transition. 13.13 Higher wavenumbers are proportional to higher frequencies and higher energies. a. IR light with a wavenumber of 3000 cm–1 is higher in energy than IR light with a wavenumber of 1500 cm–1. b. IR light having a
of 10 m is higher in energy than IR light having a
of 20 m. 13.14 Stronger bonds absorb at a higher wavenumber. Bonds to lighter atoms (H versus D) absorb at higher wavenumber. a.

CH3 C C CH2CH3 or CH2 C(CH3)2

stronger bond higher wavenumber

b.

CH3 H

or CH3 D

lighter atom H higher wavenumber

13.15 Cyclopentane and 1-pentene are both composed of C–C and C–H bonds, but 1-pentene also has a C=C bond. This difference will give the IR of 1-pentene an additional peak at 1650 cm–1 (for the C=C). 1-Pentene will also show C–H absorptions for sp2 hybridized C–H bonds at 3150–3000 cm–1. 13.16 Look at the functional groups in each compound below to explain how each IR is different. O CH3

C

A

CH3

C=O peak at ~1700 cm–1

OH

CH3OCH CH2

B C=C peak at 1650 cm–1 Csp2–H at 3150–3000 cm–1

C O–H peak at 3200–3600 cm–1

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Mass Spectrometry and Infrared Spectroscopy 13–5

13.17 a. Compound A has peaks at ~3150 (sp2 hybridized C–H), 3000–2850 (sp3 hybridized C–H), and 1650 (C=C) cm–1. b. Compound B has a peak at 3000–2850 (sp3 hybridized C–H) cm–1. 13.18 All compounds show an absorption at 3000–2850 cm–1 due to the sp3 hybridized C–H bonds. Additional peaks in the functional group region for each compound are shown. O

e.

a. no additional peaks

b.

OH

Csp2–H at 3150–3000 cm–1 C=O at ~1700 cm–1 C=C at 1650 cm–1 O–H above 3000 cm–1 The OH of a COOH is much broader than the OH of an alcohol and occurs at 3500–2500 cm–1 (see Chapter 19).

OH

O–H bond at 3600–3200 cm–1

c.

O

Csp2–H at 3150–3000 cm–1 C=C bond at 1650 cm–1

d.

f.

CH3O HO

N H

O–H at 3600–3200 cm–1 N–H at 3500–3200 cm–1 Csp2–H at 3150–3000 cm–1 C=O at ~1700 cm–1 C=C at 1650 cm–1 aromatic ring at 1600, 1500 cm–1

O

C=O bond at ~1700 cm–1

13.19 Possible structures are (a) CH3COOCH2CH3 and (c) CH3CH2COOCH3. Compounds (b) and (d) also have an OH group that would give a strong absorption at ~3600–3200 cm–1, which is absent in the IR spectrum of X, thus excluding them as possibilities. 13.20 a. Hydrocarbon with a molecular ion at m/z = 68 IR absorptions at 3310 cm–1 = Csp–H bond 3000–2850 cm–1 = Csp3–H bonds 2120 cm–1 = C
C bond Molecular formula: C5H8 H C C CH2CH2CH3

or H C C CHCH3 CH3

b. Compound with C, H, and O with a molecular ion at m/z = 60 IR absorptions at 3600–3200 cm–1 = O–H bond 3000–2850 cm–1 = Csp3–H bonds Molecular formula: C3H8O CH3CH2CH2 O H

or

CH3CH O H CH3

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Chapter 13–6 13.21 c.

a.

e.

(CH3)3CCH(Br)CH(CH3)2

O

molecular formula: C5H10O molecular ion (m/z): 86

molecular formula: C6H6 molecular ion (m/z): 78

Cl

d.

b.

molecular formula: C10H16 molecular ion (m/z): 136

molecular formula: C8H17Br molecular ions (m/z): 192, 194

molecular formula: C5H11Cl molecular ions (m/z): 106, 108

13.22 O CH2CH2CH3

C9H12 molecular weight = 120

C

OCH2CH3

CH2CH3

C9H10O molecular weight = 134

C8H10O molecular weight = 122

13.23 Examples are given for each molecular ion. a. molecular ion 102: C8H6, C6H14O, C5H10O2, C5H14N2 b. molecular ion 98: C8H2, C7H14, C6H10O, C5H6O2 c. molecular ion 119: C8H9N, C6H5N3 d. molecular ion 74: C6H2, C4H10O, C3H6O2 13.24 Likely molecular formula, C8H16 (one degree of unsaturation—one ring or one
bond). Four structures with m/z = 112

13.25 CH3 CH3 CH CO2CH3 Cl

B C4H7O2Cl molecular weight: 122, 124 should show 2 peaks for the molecular ion with a 3:1 ratio Mass spectrum [1]

C

OCH3

C8H10O molecular weight: 122

Mass spectrum [2]

CH3CH2CH2Br

A C3H7Br molecular weight: 122, 124 should show 2 peaks for the molecular ion with a 1:1 ratio Mass spectrum [3]

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Mass Spectrometry and Infrared Spectroscopy 13–7

13.26 H2 or

or

or

or Pd-C

Possible structures C7H12 (exact mass 96.0940)

13.27 [1] [2]

a.

OH

OH

OH

(from cleavage of bond [1])

[1]

O

[2]

(from cleavage of bond [2])

O

O

b.

(from cleavage of bond [1])

(from cleavage of bond [2])

[1] [2]

c.

O

O

O

(from cleavage of bond [1])

(from cleavage of bond [2])

13.28 [1] a.

– H2O

C6H5 CH CH2

H H

Cleave bond [1].

H H

m/z = 104

H

e–

C6H5 C C OH

m/z = 122

(resonance-stabilized carbocation)

C6H5 C H

m/z = 91 e– CH2 C Cleave bond [1].

[3] CH3 H H

b.

CH2 C

C C H

m/z = 68

– H2O

[1]

CH3 H H CH2 C C C OH

[2]

H H

m/z = 86

H H C C OH H H

m/z = 71 e– Cleave bond [2]. e– Cleave bond [3].

CH2 C CH3

m/z = 41 CH2 OH

m/z = 31

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Chapter 13–8 13.29 Cleave bond [1].

[1] [2]

CH3

+

CH3

CH3CH2 C CH2CH2CH2CH3 m/z = 127 CH2CH3

CH3CH2 C CH2CH2CH2CH3

CH3

CH2CH3

[3]

Cleave bond [2].

m/z = 142

+

CH2CH2CH2CH3

m/z = 85

CH3CH2 C

CH2CH3 CH3

Cleave bond [3].

CH2CH3

+

C CH2CH2CH2CH3

m/z = 113

CH2CH3

13.30 Ketone A

Ketone B

O

O

m/z = 128

m/z = 128


cleavage


cleavage

O CH2CH3

+

O

+

CH3

+

+

m/z = 99

m/z = 113

This is ketone A since
cleavage gives a fragment with m/z of 99.

This is ketone B since
cleavage gives a fragment with m/z of 113.

13.31 One possible structure is drawn for each set of data: a. A compound that contains a benzene ring and has a molecular ion at m/z = 107

c. A compound that contains a carbonyl group and gives a molecular ion at m/z = 114 O

NH2 CH3

CH2CH2CH2CH2CH3

C7H14O

C7H9N

b. A hydrocarbon that contains only sp3 hybridized carbons and a molecular ion at m/z = 84

C

d. A compound that contains C, H, N, and O and has an exact mass for the molecular ion at 101.0841 O CH3

C6H12

C

NHCH2CH2CH3

C5H11NO

13.32 Use the values given in Table 13.1 to calculate the exact mass of each compound. C8H11NO2 (exact mass 153.0790) is the correct molecular formula. 13.33 Molecules with an odd number of N's have an odd number of H's, making the molecular ion odd as well.

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Mass Spectrometry and Infrared Spectroscopy 13–9

13.34 Two isomers such as CH2=CHCH2CH2CH2CH3 and (CH3)2C=CHCH2CH3 have the same molecular formulas and therefore give the same exact mass, so they are not distinguishable by their exact mass spectra. 13.35
Cleavage of a 1° alcohol (RCH2OH) forms an alkyl radical R• and a resonance-stabilized carbocation with m/z = 31. CH2OH

CH2 OH

m/z = 31

resonance-stabilized carbocation

13.36 An ether fragments by
cleavage because the resulting carbocation is resonance stabilized. [1] (CH3)2CH

[2]

CH2

O

CH2 CH3

Cleave bond [1]. Cleave bond [2].

(CH3)2CH

CH2

O

CH2 CH3

CH2 O

CH2 CH3

resonance-stabilized carbocations (CH3)2CH

CH3

CH2

O

CH2

(CH3)2CH

CH2

O

13.37 a. (CH3)2C O or (CH3)2CH OH stronger bond higher ~
absorption

b. (CH3)2C NCH3

or (CH3)2CH NCH3

stronger bond higher ~
absorption

c.

H or

H

stronger bond higher ~
absorption

13.38 Locate the functional groups in each compound. Use Table 13.2 to determine what IR absorptions each would have. a.

Csp3–H at 2850–3000 cm–1

d. O

b.

C CH

Csp–H at 3300 cm–1 Csp3–H at 2850–3000 cm–1 C–C triple bond at 2250 cm–1

e.

OH

O

OH

O–H at 3200–3600 cm–1 Csp3–H at 2850–3000 cm–1

f.

O–H at 3200–3600 cm–1 Csp2–H at 3000–3150 cm–1 Csp3–H at 2850–3000 cm–1 C=C at 1650 cm–1

O–H at > 3000 cm–1 Csp2–H at 3000–3150 cm–1 C=O at ~1700 cm–1 phenyl group at 1600, 1500 cm–1 The OH of the RCOOH is even broader than the OH of an alcohol (3500–2500 cm–1), as we will learn in Chapter 19.

C

c.

Csp3–H at 2850–3000 cm–1 C=O at 1700 cm–1

OH

CH2

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Chapter 13–10 13.39 a.

and

C=C bond 1650 cm–1 Csp2–H at 3150–3000 cm–1

HC CCH2CH2CH3 C
C bond 2250 cm–1 Csp–H at 3300 cm–1

O

b. CH CH 3 2

OCH3 OCH3

d.

O

and CH3(CH2)5

no C=O bond

C

OCH3

C=O bond ~1700 cm–1

O

C

and

OH

CH3

C

e.

OCH3

O–H bond no O–H bond > 3000 cm–1 [See note on OH in Answer 13.38f.]

and

CH3C CCH3

CH3CH2C CH

Csp–H bond 3300 cm–1 C
C bond at ~2250 cm–1

no C
C absorption due to symmetry

O

c. CH3CH2

C

and CH3

f.

CH3CH CHCH2OH

HC CCH2N(CH2CH3)2

O–H bond 3200–3600 cm–1 Csp2–H at 3150–3000 cm–1 C=C bond at 1650 cm–1

C=O bond 1700 cm–1

and

CH3(CH2)5C N

Csp–H bond 3300 cm–1

13.40 The IR absorptions above 1500 cm–1 are different for each of the narcotics. CH3

HO

C

O

CH3O

O O

O H

H

HO

O

N CH3

CH3

morphine

C

O

H

H

O

heroin • C=O bond at ~1700 cm–1 • no O–H bond

• O–H bond at ~3200–3600 cm–1 • no C=O bond

N CH3

H

N OH

O

CH3

oxycodone • C=O bond at ~1700 cm–1 • O–H bond at ~3200–3600 cm–1

13.41 Look for a change in functional groups from starting material to product to see how IR could be used to determine when the reaction is complete. Loss of the C=C will be visible in the IR by disappearance of the peak at 1650 cm–1.

H2

a.

Pd

OH

O PCC

b.

[1] O3

c.

CH3 O

[2] CH3SCH3

d.

OH

[1] NaH [2] CH3Br

Loss of the O–H group will be visible in the IR by disappearance of the peak at 3200–3600 cm–1 and appearance of the C=O at ~1700 cm–1.

O C CH3

O

Loss of the C=C will be visible in the IR by disappearance of the peak at 1650 cm–1 and appearance of the C=O at ~1700 cm–1.

Loss of the O–H will be visible in the IR by disappearance of the peak at 3200–3600 cm–1.

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Mass Spectrometry and Infrared Spectroscopy 13–11 13.42 In addition to Csp3–H at ~3000–2850 cm–1: Spectrum [1]:

Spectrum [2]:

CH2=C(CH3)CH2CH2CH2CH3 (B) C=C peak at 1650 cm–1 Csp2–H at ~3150 cm–1

(CH3CH2)3COH (F) OH at 3600–3200 cm–1

Spectrum [3]:

Spectrum [4]:

(CH3)2CHOCH(CH3)2 (D) No other peaks above 1500 cm–1

CH(CH3)2

(C)

Csp2–H at ~3150 cm–1 Phenyl peaks at 1600 and 1500 cm–1 Spectrum [5]:

Spectrum [6]:

CH3CH2CH2CH2COOH (A) OH at ~3500–2500 cm–1 C=O at ~1700 cm–1

CH3COOC(CH3)3 (E) C=O at ~1700 cm–1

13.43 In addition to Csp3–H at ~3000–2850 cm–1: O CH3

C

O CH3

~1700 cm–1

CH3CH2

C

H

CH2OH C

H

~1700 cm–1

H

OH

C

CH3 O CH=CH2

H

(OH) 3200–3600 cm–1 (OH) 3200–3600 cm–1 (C=C)1650 cm–1 (C=C)1650 cm–1 (Csp2–H) 3150–3000 cm–1

(Csp2–H) 3150–3000 cm–1 No enols (such as CH3CH=CHOH) are drawn since these compounds are not stable.

13.44 a. Compound with a molecular ion at m/z = 72 IR absorption at 1725 cm–1 = C=O bond Molecular formula: C4H8O O

O

O CH3

No additional peaks above 1500 cm–1

c. Compound with a molecular ion at m/z = 74 IR absorption at 3600–3200 cm–1 = O–H bond Molecular formula: C4H10O OH

or

b. Compound with a molecular ion at m/z = 55 The odd molecular ion means an odd number of N's present. Molecular formula: C3H5N IR absorption at 2250 cm–1 = C
N bond CH3CH2C N

OH OH

OH

or

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Chapter 13–12 13.45

Chiral hydrocarbon with a molecular ion at m/z = 82 Molecular formula: C6H10 IR absorptions at 3300 cm–1 = Csp–H bond 3000–2850 cm–1 = Csp3–H bonds 2250 cm–1 = C
C bond Two possible enantiomers: CH3

*

HC CCHCH2CH3

C

CH3

stereogenic center

CH3CH2

CH3

or C CH

C

HC C

H

H

CH2CH3

13.46 The chiral compound Y has a strong absorption at 2970–2840 cm–1 in its IR spectrum due to sp3 hybridized C–H bonds. The two peaks of equal intensity at 136 and 138 indicate the presence of a Br atom. The molecular formula is C4H9Br. Only one constitutional isomer of this molecular formula has a stereogenic center: Br

Br

and

Y=

two possible enantiomers

13.47 O H

CH3

Zn(Hg) HCl

m/z = 92; molecular formula C7H8 IR absorptions at: 3150–2950 cm–1 = Csp3–H and Csp2–H bonds 1605 cm–1 and 1496 cm–1 due to phenyl group

Z

13.48 H O

Br

O

H–Br

H

+ Br

H Br H

CH3

CH3 HBr

O

C CH2 CH3

CH3

H

C

H O CH2

H

Br

Br

H2O

13.49 O

fragments:

J C6H12O m/z = 100 IR absorption at 2962 cm–1 = Csp3–H bonds 1718 cm–1 = C=O bond

O

O


cleavage product
cleavage product m/z = 43 m/z = 85 The fragment at m/z = 57 could be due to (C4H9)+ or (C3H5O)+.

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Mass Spectrometry and Infrared Spectroscopy 13–13

13.50 O C

NH2 CHO

L C7H6O m/z = 106

K C7H9N m/z = 107 IR absorptions at 3373 and 3290 cm–1 = N–H 3062 cm–1 = Csp2–H bonds 2920 cm–1 = Csp3–H bonds 1600 cm–1 = benzene ring

m/z = 105

m/z = 77

IR absorption at 3068 cm–1 = Csp2–H bonds on ring 2850 cm–1 = Csp3–H bond 2820 cm–1 and 2736 cm–1 = C–H of RCHO (Appendix E) 1703 cm–1 = C=O bond 1600 cm–1 = aromatic ring

The odd molecular ion indicates the presence of a N atom.

13.51 Possible structures of P: Cl Cl2

CH3O

CH3O

Cl

or

Cl

or

CH3O

CH3O

FeCl3

C7H7ClO m/z = 142, 144 IR absorption at 3096–2837 cm–1 = Csp3–H bonds and Csp2–H bonds 1582 cm–1 and 1494 cm–1 = benzene ring The peak at M + 2 shows the presence of Cl or Br. Since Cl2 is a reactant, the compound presumably contains Cl.

13.52 The mass spectrum has a molecular ion at 71. The odd mass suggests the presence of an odd number of N atoms; likely formula, C4H9N. The IR absorption at ~3300 cm–1 is due to N–H and the 3000–2850 cm–1 is due to sp3 hybridized C–H bonds. H

N

Br

H

Br

N

N H

H

+

Na+ Br– + H2

W + Na+ + H2

Na+ H

13.53 The
,-unsaturated carbonyl compound has three resonance structures, two of which place a single bond between the C and O atoms. This means that the C–O bond has partial single bond character, making it weaker than a regular C=O bond, and moving the absorption to lower wavenumber. O

+

O

+

three resonance structures for 2-cyclohexenone

O

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Chapter 13–14 13.54 PCC

a. and b.

H

PCC

NaOCOCH3

O

O O Cr OH

OH H

A molecular ion at 154 C10H18O IR at 1730 cm–1 (C=O)

O

+

H B

PCC

+

O Cr OH

O H

OH

OH

O

+

B

+

OH

H

+ +H B

B

a. and c.

+ O

H

O

HO

isopulegone + + H B

+ Cr4+

B

citronellol

O

isopulegone + Cr4+ + H B+

O

+ H2O

O

O H OH

B

OH

B

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351

Nuclear Magnetic Resonance Spectroscopy 14–1 C Chhaapptteerr 1144:: N Nuucclleeaarr M Maaggnneettiicc R Reessoonnaannccee SSppeeccttrroossccooppyy 11   1H HN NM MR R ssppeeccttrroossccooppyy [1] The number of signals equals the number of different types of protons (14.2).

CH3 O CH3

Ha

CH3CH2

CH3 O CH2CH3

Cl

Ha

Ha Hb

Ha

all equivalent H's 1 NMR signal

2 types of H's 2 NMR signals

Hb Hc

3 types of H's 3 NMR signals

[2] The position of a signal (its chemical shift) is determined by shielding and deshielding effects. • Shielding shifts an absorption upfield; deshielding shifts an absorption downfield. • Electronegative atoms withdraw electron density, deshield a nucleus, and shift an absorption downfield (14.3). C H

•

This proton is shielded. Its absorption is upfield, 0.9–2 ppm.

C H X

This proton is deshielded. Its absorption is farther downfield, 2.5–4 ppm.

Loosely held
electrons can either shield or deshield a nucleus. Protons on benzene rings and double bonds are deshielded and absorb downfield, whereas protons on triple bonds are shielded and absorb upfield (14.4). H C

C H H

deshielded H downfield absorption

shielded H upfield absorption

[3] The area under an NMR signal is proportional to the number of absorbing protons (14.5). [4] Spin-spin splitting tells about nearby nonequivalent protons (14.6–14.8). • Equivalent protons do not split each other’s signals. • A set of n nonequivalent protons on the same carbon or adjacent carbons split an NMR signal into n + 1 peaks. • OH and NH protons do not cause splitting (14.9). • When an absorbing proton has two sets of nearby nonequivalent protons that are equivalent to each other, use the n + 1 rule to determine splitting. • When an absorbing proton has two sets of nearby nonequivalent protons that are not equivalent to each other, the number of peaks in the NMR signal = (n + 1)(m + 1). In flexible alkyl chains, peak overlap often occurs, resulting in n + m + 1 peaks in an NMR signal. 1133  CN NM MR R ssppeeccttrroossccooppyy ((1144..1111))  13C [1] The number of signals equals the number of different types of carbon atoms. All signals are single lines. [2] The relative position of 13C signals is determined by shielding and deshielding effects. • Carbons that are sp3 hybridized are shielded and absorb upfield. • Electronegative elements (N, O, and X) shift absorptions downfield. • The carbons of alkenes and benzene rings absorb downfield. • Carbonyl carbons are highly deshielded, and absorb farther downfield than other carbon types.

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Chapter 14–2 C Chhaapptteerr 1144:: A Annssw weerrss ttoo PPrroobblleem mss 14.1

Use the formula
= [observed chemical shift (Hz)/ of the NMR (MHz)] to calculate the chemical shifts.

a. CH3 protons:
= [1715 Hz] / [500 MHz] = 3.43 ppm

14.2

Calculate the chemical shifts as in Answer 14.1.

a. one signal:
= [1017 Hz] / [300 MHz] = 3.39 ppm

14.3

b. The positive direction of the
scale is downfield from TMS. The CH3 protons absorb upfield from the OH proton.

OH proton:
= [1830 Hz] / [500 MHz] = 3.66 ppm

second signal:
= [1065 Hz] / [300 MHz] = 3.55 ppm

b. one signal: 3.39 = [x Hz] / [500 MHz] x = 1695 Hz

second signal: 3.55 = [x Hz] / [500 MHz] x = 1775 Hz

To determine if two H’s are equivalent replace each by an atom X. If this yields the same compound or mirror images, the two H’s are equivalent. Each kind of H will give one NMR signal.

2 kinds of H's 2 NMR signals

e. CH3CH2CO2CH2CH3 4 kinds of H's 4 NMR signals

g. CH3CH2OCH2CH3 2 kinds of H's 2 NMR signals

d. (CH3)2CHCH(CH3)2 2 kinds of H's 2 NMR signals

f. CH3OCH2CH(CH3)2 4 kinds of H's 4 NMR signals

h. CH3CH2CH2OH 4 kinds of H's 4 NMR signals

c. CH3CH2CH2CH3

a. CH3CH3 1 kind of H 1 NMR signal b. CH3CH2CH3 2 kinds of H's 2 NMR signals

14.4 Each C is a different distance from the Cl. This makes each C different, and each set of H's different. There are 8 different kinds of protons.

CH3CH2CH2CH2CH2CH2CH2CH2Cl

14.5

Draw in all of the H’s and compare them. If two H’s are cis and trans to the same group, they are equivalent. H

H

CH3

Ha

a. 4 identical H's Hb

H

CH3 H

2 NMR signals

14.6

Ha

CH3

b.

Hc H

CH3

Ha

Ha

H

Ha

H

c. Hb

H

H H

Hd

4 NMR signals

Hb

Hb H

CH3 H

Hc Hb

CH3

Hc 3 NMR signals

If replacement of H with X yields enantiomers, the protons are enantiotopic. If replacement of H with X yields diastereomers, the protons are diastereotopic. In general, if the compound has one stereogenic center, the protons in a CH2 group are diastereotopic.

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Nuclear Magnetic Resonance Spectroscopy 14–3

CH3CH2CH2CH2CH2CH3

a.

c.

replacement of H with X CH3 C CH3CH2CH2CH2

Pick one configuration at the existing stereogenic center.

CH3 C

H X

CH3CH2CH2CH2

CH3

X H

H

C HO

enantiomers = enantiotopic H's b.

CH3CH(OH)CH2CH2CH3

X

replacement of H with X CH3

C

C CH2CH3

H

X

HO

H

C CH2CH3

H

diastereomers = diastereotopic H's

CH3CH2CH2CH2CH3

replacement of H with X no stereogenic center

CH3CH2CHCH2CH3

neither enantiotopic nor diastereotopic

X

14.7

The two protons of a CH2 group are different from each other if the compound has one stereogenic center. Replace one proton with X and compare the products.

a. The stereogenic center makes the H's in the CH2 group diastereotopic and therefore different from each other. stereogenic center Cl H Hb Hd

CH3 C C CH3

Hc

H H He

Ha

c.

b. Hc H H

Hd

Cl C C O CH3

He

Hb Hc stereogenic center H H H Hd Ha CH3 C C C CH3

stereogenic center Ha

H CH3 Hb

Hf Hg

5 NMR signals

7 NMR signals

5 NMR signals

14.8

He

Br H H

Decreased electron density deshields a nucleus and the absorption goes downfield. Absorption also shifts downfield with increasing alkyl substitution. a. FCH2CH2CH2Cl F is more electronegative than Cl. The CH2 group adjacent to the F is more deshielded and the H's will absorb farther downfield. b. CH3CH2CH2CH2OCH3 The CH2 group adjacent to the O will absorb farther downfield because it is closer to the electronegative O atom.

c. CH3OC(CH3)3 The CH3 group bonded to the O atom will absorb farther downfield.

14.9 O

a. ClCH2CH2CH2Br Ha Hb Hc 3 types of protons: Hb < Hc < Ha

b. CH3OCH2OC(CH3)3 Ha Hb Hc 3 types of protons: Hc < Ha < Hb

c. CH3

C

CH2CH3

Ha Hb Hc 3 types of protons: Hc < Ha < Hb

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Chapter 14–4

14.10 O

a.

CH3 C C H

CH3CH CH2

CH3CH2CH3

Hc Hb Ha Hc protons are shielded because they are bonded to an sp3 C. Ha is shielded because it is bonded to an sp C. Hb protons are deshielded because they are bonded to an sp2 C. Hc < Ha < Hb

b.

CH3

C

Ha

OCH2CH3

Hb Hc

Hc protons are shielded because they are bonded to an sp3 C. Ha protons are deshielded slightly because the CH3 group is bonded to a C=O. Hb protons are deshielded because the CH2 group is bonded to an O atom. Hc < Ha < Hb

14.11 An integration ratio of 2:3 means that there are two types of hydrogens in the compound, and that the ratio of one type to another type is 2:3. a. CH3CH2Cl 2 types of H's 3:2 - YES

b. CH3CH2CH3 2 types of H's 6:2 or 3:1 - no

c. CH3CH2OCH2CH3 2 types of H's 6:4 or 3:2 - YES

d. CH3OCH2CH2OCH3 2 types of H's 6:4 or 3:2 - YES

14.12 To determine how many protons give rise to each signal: • Divide the total number of integration units by the total number of protons to find the number of units per H. • Divide each integration value by this value and round to the nearest whole number. C8H14O2 total number of integration units = 14 + 12 + 44 = 70 units total number of protons = 14 H's 70 units/14 H's = 5 units per H

Signal [A] = 14/5 = 3 H Signal [B] = 12/5 = 2 H Signal [C] = 44/5 = 9 H

14.13 downfield absorption closer to O

downfield absorption closer to O

CH3O2CCH2CH2CO2CH3

CH3CO2CH2CH2O2CCH3

A ratio of absorbing signals 2:3 Signal [1] = 4 H = 2.64 Signal [2] = 6 H = 3.69 6 H's with downfield absorption

B ratio of absorbing signals 3:2 Signal [1] = 6 H = 2.09 Signal [2] = 4 H = 4.27 4 H's with downfield absorption

14.14 To determine the splitting pattern for a molecule: • Determine the number of different kinds of protons. • Nonequivalent protons on the same C or adjacent C’s split each other. • Apply the n + 1 rule.

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355

Nuclear Magnetic Resonance Spectroscopy 14–5

a.

O

O

C

C

CH3CH2

c.

Cl

Ha Hb Ha: 3 peaks - triplet Hb: 4 peaks - quartet

Ha

CH2CH2Br

e.

d.

CH3 C Br

H

Ha

H

Hb

Cl

Br

Ha Ha: 2 peaks - doublet Hb: 4 peaks - quartet

Ha

Ha: 2 peaks - doublet Hb: 2 peaks - doublet

f.

C C

Br

H C C

CH3

Hb Hc

Ha: 1 peak - singlet Hb: 3 peaks - triplet Hc: 3 peaks - triplet

Hb

H

b.

CH3

CH3CH2

ClCH2CH(OCH3)2

Hb

H

Ha Hb Ha: 2 peaks - doublet Hb: 3 peaks - triplet

Ha: 2 peaks - doublet Hb: 2 peaks - doublet

14.15 Identical protons do not split each other. Ha

Ha

H Cl

H Cl

Cl C C Cl

Br C C Cl

Cl H

Br H

Ha

Ha: 1 singlet All protons are equivalent.

Hb

Ha: 2 peaks - doublet Hb: 2 peaks - doublet

14.16 Use the directions from Answer 14.14. Ha

Ha

Ha O

a. O

Hb

Hb

c. CH3 C H

Ha: quartet Hb: triplet 2 NMR signals

Hb

Ha: doublet Hb: quartet 2 NMR signals

O

b. CH3

C

OCH2CH2OCH3 Ha and Hd are both singlets.

Ha

Hb Hc

Hd

Hb: triplet Hc: triplet 4 NMR signals

Ha: triplet Hb: doublet Hc Hc: singlet 3 NMR signals

d. Cl2CHCH2CO2CH3 Ha Hb

14.17 CH3CH2Cl 3 units 2 units

3 chemical shift (ppm)

1

There are two kinds of protons, and they can split each other. The CH3 signal will be split by the CH2 protons into 2 + 1 = 3 peaks. It will be upfield from the CH2 protons since it is farther from the Cl. The CH2 signal will be split by the CH3 protons into 3 + 1 = 4 peaks. It will be downfield from the CH3 protons since the CH2 protons are closer to the Cl. The ratio of integration units will be 3:2.

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Chapter 14–6

14.18 Cl

a.

(CH3)2CHCO2CH3

b.

CH3CH2CH2CH2CH3

c.

CH2Br

Ha

C C

d.

H Hb Ha H Ha Hb Hc Ha: split by 2 H's Ha: split by 1 H 3 peaks 2 peaks Hc: split by 4 equivalent H's Hb: split by 2 sets of H's 5 peaks (1 + 1)(2 + 1) = 6 peaks Hb: split by 2 sets of H's (3 + 1)(2 + 1) = 12 peaks (maximum) Since this is a flexible alkyl chain, the signal due to Hb will have peak overlap, and 3 + 2 + 1 = 6 peaks will likely be visible.

split by 6 equivalent H's 6 + 1 = 7 peaks

H

H

Hb (all H's) Br H Hc Ha: split by 2 different H's (1+1)(1+1) = 4 peaks Hb: split by 2 different H's (1+1)(1+1) = 4 peaks Hc: split by 2 different H's (1+1)(1+1) = 4 peaks C C

14.19 CH3CH2

O

b.

a. CH3OCH2CH3 Ha

CH3CH2

H b Hc

Ha: singlet at ~3 ppm Hb: quartet at ~3.5 ppm Hc: triplet at ~1 ppm

C

c. CH3OCH2CH2CH2OCH3 OCH(CH3)2

H a Hb

Hc Hd

Ha: triplet at ~1 ppm Hb: quartet at ~2 ppm Hc: septet at ~3.5 ppm Hd: doublet at ~1 ppm

Ha

d.

Hb Hc Hb Ha

Ha

CH2CH3

Ha

C C H

Ha: singlet at ~3 ppm Hb: triplet at ~3.5 ppm Hc: quintet at ~1.5 ppm

Hb

H

Hc Ha: triplet at ~1 ppm Hb: multiplet (8 peaks) at ~2.5 ppm Hc: triplet at ~5 ppm

14.20 Hb Splitting diagram for Hb Hb Cl

Jab = 13.1 Hz

Cl Ha

Hc

Hc

1 trans Ha proton splits Hb into 1 + 1 = 2 peaks a doublet

2 Hc protons 2 Hc protons split Hb into 2 + 1 = 3 peaks Now it's a doublet of triplets.

trans-1,3-dichloropropene Jbc = 7.2 Hz

14.21 Cl

H

Hb

ClCH2 C3H4Cl2

Cl

CH3 Ha A Ha: 1.75 ppm, doublet, 3 H, J = 6.9 Hz Hb: 5.89 ppm, quartet, 1 H, J = 6.9 Hz

singlet Cl

H

doublet

H doublet B signal at 4.16 ppm, singlet, 2 H signal at 5.42 ppm, doublet, 1 H, J = 1.9 Hz signal at 5.59 ppm, doublet, 1 H, J = 1.9 Hz

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Nuclear Magnetic Resonance Spectroscopy 14–7

14.22 Remember that OH (or NH) protons do not split other signals, and are not split by adjacent protons. triplet

singlet

doublet

triplet

b. CH3CH2CH2OH

a. (CH3)3CCH2OH

singlet

c. (CH3)2CHNH2

singlet singlet singlet 3 NMR signals

7 peaks 3 NMR signals

12 peaks (maximum) 6 peaks (more likely, resulting from peak overlap) 4 NMR signals

14.23 H

Hc

C CH3

Hd 5 H's on benzene ring

OH

Hb

A

Ha

Ha: doublet at ~1.4 due to the CH3 group, split into two peaks by one adjacent nonequivalent H (Hc). Hb: singlet at ~2.7 due to the OH group. OH protons are not split by nor do they split adjacent protons. Hc: quartet at ~4.7 due to the CH group, split into four peaks by the adjacent CH3 group. Hd: Five protons on the benzene ring.

14.24 Use these steps to propose a structure consistent with the molecular formula, IR, and NMR data. • Calculate the degrees of unsaturation. • Use the IR data to determine what types of functional groups are present. • Determine the number of different types of protons. • Calculate the number of H’s giving rise to each signal. • Analyze the splitting pattern and put the molecule together. • Use the chemical shift information to check the structure. • Molecular formula C7H14O2 2n + 2 = 2(7) + 2 = 16 16 – 14 = 2/2 = 1 degree of unsaturation 1
bond or 1 ring • IR peak at 1740 cm–1 C=O absorption is around 1700 cm–1 (causes the degree of unsaturation). No signal at 3200–3600 cm–1 means there is no O–H bond. • NMR data: absorption ppm integration singlet 1.2 26 26 units/3 units per H = 9 H's 1.3 10 triplet 10 units/3 units per H = 3 H's (probably a CH3 group) 4.1 6 quartet 6 units/3 units per H = 2 H's (probably a CH2 group) • 3 kinds of H's • number of H's per signal total integration units: 26 + 10 + 6 = 42 units 42 units / 14 H's = 3 units per H • look at the splitting pattern The singlet (9 H) is likely from a tert-butyl group: The CH3 and CH2 groups split each other: CH3

CH3 C CH3 CH3 CH2

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Chapter 14–8

• Join the pieces together.

O C

CH3CH2O

O

CH3 C CH3

or

CH3CH2

C

CH3 O C CH3

CH3

CH3

Pick this structure due to the chemical shift data. The CH2 group is shifted downfield (4 ppm), so it is close to the electron-withdrawing O.

14.25

•

Molecular formula: C3H8O

• •

IR peak at 3200–3600 cm–1 NMR data: • doublet at ~1.2 (6 H) • singlet at ~2.2 (1 H) • septet at ~4 (1 H)

•

Calculate degrees of unsaturation 2n + 2 = 2(3) + 2 = 8 8 – 8 = 0 degrees of unsaturation  Peak at 3200–3600 cm–1 is due to an O–H bond. 3 types of H's septet from 1 H singlet from 1 H doublet from 6 H's from the O–H proton

split by 6 H's split by 1 H

 Put information together: CH3 HO C CH3 H

14.26 a. Absorption [A]: Absorption [B]: Absorption [C]: Absorption [D]:

singlet at ~3.8 ppm multiplet at ~3.6 ppm triplet at ~2.9 ppm singlet at ~1.9 ppm

b. H N H

CH3 N

CH3O

split by 2 adjacent nonequivalent H's into a triplet

O H H HH

CH3O– CH2N CH2 adjacent to five-membered ring CH3C=O

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Nuclear Magnetic Resonance Spectroscopy 14–9 14.27 Identify each compound from the 1H NMR data. singlet at 2.5 a. CH2 CHCOCH3

O

HCl

H H

CH3

b.

triplet at 3.6

Cl

(CH3)2C

O

base

O

OH

H2O

H H

singlet A

B

singlet at 2.2

triplet at 3.05

singlet at 1.3 singlet at 3.8

14.28 Each different kind of carbon atom will give a different 13C NMR signal. O

b.

a. CH3CH2CH2CH3

CH3CH2

Ca Cb Cb Ca

C

c. CH3CH2CH2 O CH2CH2CH3 OCH3

Ca Cb Cc Cc Cb Ca same groups on both sides of O 3 kinds of C's 3 13C NMR signals

Each C is different. 4 kinds of C's 4 13C NMR signals

2 kinds of C's 2 13C NMR signals

CH3CH2

d.

H

C C H

H

Each C is different. 4 kinds of C's 4 13C NMR signals

14.29 Hb Ha

Ha Hb Ha Hd

a.

Ha

H H H

Hc

H H H

Hb Ha

H C C C H

H C C C H

H Cl Cl

Cl H Cl

Hc

Hb 4 1H NMR signals

b.

2 1H NMR signals

Ha

H Cl H

H C C C H

H C C C H

H H Cl

H Cl H

Hb 3 1H NMR signals

all H's identical 1 1H NMR signal

Hc

H H H

H H H

H H Cl

H Cl H

H C C C H

H C C C H

H C C C H

H C C C H

H Cl Cl

Ca Cl H Cl Ca Cb

Each C is different. 3 kinds of C's 3 13C NMR signals

c.

H H Cl

Hc

2 kinds of C's 2 13C NMR signals

H H Cl

Each C is different. 3 kinds of C's 3 13C NMR signals

Ca H Cl H Ca Cb 2 kinds of C's 2 13C NMR signals

Although the number of 13C signals cannot be used to distinguish these isomers, each isomer exhibits a different number of signals in its 1H NMR spectrum. As a result, the isomers are distinguishable by 1H NMR spectroscopy.

14.30 Electronegative elements shift absorptions downfield. The carbons of alkenes and benzene rings, and carbonyl carbons are also shifted downfield. O

a.

CH3CH2OCH2CH3

The CH2 group is closer to the electronegative O and will be farther downfield.

b.

BrCH2CHBr2

The C of the CHBr2 group has two bonds to electronegative Br atoms and will be farther downfield.

c.

H

C

OCH3

The carbonyl carbon is highly deshielded and will be farther downfield.

d. CH3CH CH2 The CH2 group is part of a double bond and will be farther downfield.

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Chapter 14–10

14.31 a. In order of lowest to highest chemical shift: Ca C b C c Cd CH3CHCH2CH3 OH

C d < C a < C c < Cb

b. In order of lowest to highest chemical shift: (CH3CH2)2C=O

Ca Cb

Cc

Ca < Cb < Cc

14.32 • molecular formula C4H8O2 2n + 2 = 2(4) + 2 = 10 10 – 8 = 2/2 = 1 degree of unsaturation

O

• no IR peaks at 3200–3600 or 1700 cm–1 no O–H or C=O

O

• 1H NMR spectrum at 3.69 ppm only one kind of proton •

13C

NMR spectrum at 67 ppm only one kind of carbon

This structure satisifies all the data. One ring is one degree of unsaturation. All carbons and protons are identical.

14.33 O OH

• molecular formula C4H8O 2n + 2 = 2(4) + 2 = 10 10 – 8 = 2/2 = 1 degree of unsaturation •

13C

NMR signal at > 160 ppm due to C=O

• molecular formula C4H8O 2n + 2 = 2(4) + 2 = 10 10 – 8 = 2/2 = 1 degree of unsaturation • all 13C NMR signals at < 160 ppm NO C=O

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Nuclear Magnetic Resonance Spectroscopy 14–11

14.34 Use the directions from Answer 14.3. a.

e.

(CH3)3CH

O

2 kinds of H's b.

h.

f.

CH3CH2

H

Hb

Br

Hc

H H

g.

Hd

CH3 C C CH3

Ha

HO H

Hf

He

Hc 6 kinds of H's

CH3CH2CH2OCH2CH2CH3

3 kinds of H's

C C H

i.

3 kinds of H's CH2CH3

H

Hc

H

H

C C

c. CH3CH2OCH2CH2CH2CH2CH3 7 kinds of H's

Hb

CH3

4 kinds of H's

(CH3)3CC(CH3)3

Ha CH3

Hd

CH2CH3

C C

Hb

5 kinds of H's

1 kind of H

d.

CH3

Ha

Hd

j.

4 kinds of H's

Ha

H

Hb

H

Hc

CH3 O

H

Hd

4 kinds of H's

14.35 CH3

H H

Hb

CH3

H

a.

H

Hc

Ha

H

Ha

Hb

b.

Hd

Hf

Hd

H

H

H

H

H

H

Hb CH2CH3

Hg

Hf

H

H

H

Hd

Hd H

Ha

H

H

H

Hb H

H

H H c

Ha

CH3

Hd H Ha CH3 d.

Hc

H CH3

c.

H

Hb 3 kinds of protons He

Hc

4 kinds of protons Hb

Hc H

Hb 4 kinds of protons

Hd CH3

Ha

Hc

7 kinds of protons

14.36 equivalent O O

a. CH3 O

b.

CH3

CHO

c.

d.

N

N

OH

H N

N

CH3

caffeine 4 NMR signals

OCH3 OH

vanillin 6 NMR signals

thymol 7 NMR signals

CH3O

N H

HO

capsaicin 15 NMR signals

equivalent

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Chapter 14–12

14.37 a. 2.5 = x Hz/300 MHz x = 750 Hz b. ppm = 1200 Hz/300 MHz = 4 ppm c. 2.0 = x Hz/300 MHz x = 600 Hz


(in ppm) = [observed chemical shift (Hz)] /  of the NMR (MHz)]

14.38 2.16 = x Hz/500 MHz x = 1080 Hz (chemical shift of acetone in Hz) 1080 Hz + 1570 Hz = 2650 Hz 2650 Hz/500 MHz = 5.3 ppm, chemical shift of the CH2Cl2 signal 14.39 Use the directions from Answer 14.8. a. CH3CH2CH2CH2CH3 or

c.

CH3CH2CH2OCH3

CH3OCH2CH3

or Increasing alkyl substitution farther downfield

Adjacent O deshields the H's. farther downfield b. CH3CH2CH2I

or CH3CH2CH2F

d. CH3CH2CHBr2 or

CH3CH2CH2Br

Two electronegative Br's deshield the H. farther downfield

More electronegative F deshields the H's. farther downfield

14.40 Use the directions from Answer 14.12. [total number of integration units] / [total number of protons] [13 + 33 + 73] / 10 = ~12 units per proton

Signal of 13 units is from 1 H. Signal of 33 units is from 3 H's. Signal of 73 units is from 6 H's.

14.41 Hb Hb

Hb

Ha

CH3

a. CH3CO2C(CH3)3 and CH3CO2CH3 c.

Ha

Hb

Ha

Hb

CH3

CH3

Ha

Ha

Ha : Hb = 1:3 Ha : Hb = 1:1 different ratio of peak areas Hb in CH3CO2CH3 is farther downfield than all H's in CH3CO2C(CH3)3.

and

Hb

CH3

Ha

Hb Hb

Hb

CH3

Ha : Hb = 3:1

Ha : Hb = 3:2

different ratio of peak areas

b. CH3OCH2CH2OCH3 and CH3OCH2OCH3

Ha

Hb Hb

Ha

Ha

Hb

Ha

Ha : Hb = 3:2 Ha : Hb = 3:1 different ratio of peak areas

14.42 The following compounds give one singlet in a 1H NMR spectrum: CH3 CH3CH3

CH3 C C CH3

Cl

Cl Br Br

CH3

O

C C CH3

CH3

(CH3)3C

C

C(CH3)3

Ha

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Nuclear Magnetic Resonance Spectroscopy 14–13 14.43 CH3

Ha

CH3CH2CH2CH2CH2CH3

CH3CHCH2CH2CH3

Ha Hb Hc Hc Hb Ha

Ha Hb Hc Hd He

3 signals: Ha: split by 2 Hb protons - triplet Hc: split by 2 Hb protons - triplet Hb: split by 3 Ha + 2 Hc protons - 12 peaks (maximum) Since Hb is located in a flexible alkyl chain, peak overlap occurs, so that only 3 + 2 + 1 = 6 peaks will likely be observed. Hc

5 signals: Ha: split by 1 Hb proton - doublet Hb: split by 6 Ha + 2 Hc protons - 21 peaks (maximum) Hc: split by 1 Hb + 2 Hd protons - 6 peaks (maximum) Hd: split by 2 Hc + 3 He protons - 12 peaks (maximum) He: split by 2 Hd protons - triplet Since Hb, Hc, and Hd are located in a flexible alkyl chain, it is likely that peak overlap occurs, so that the following is observed: Hb (6 + 2 + 1 = 9 peaks), Hc (1 + 2 + 1 = 4 peaks), and Hd (2 + 3 + 1 = 6 peaks). Ha Ha CH3 Hc

CH3CH2CHCH2CH3 CH3

Ha Hb

Hb Ha

Hd

4 signals: CH3 CH3 Ha: split by 2 Hb protons - triplet CH CH CHCH3 3 Hb: split by 3 Ha + 1 Hc protons - 8 peaks (maximum) Hc: split by 4 Hb + 3 Hd protons - 20 peaks (maximum) Ha Hb Hb Ha Hd: split by 1 Hc proton - doublet Since Hb and Hc are located in a flexible alkyl chain, it 2 signals: is likely that peak overlap occurs, so that the following Ha: split by 1 Hb proton - doublet is observed: Hb (3 + 1 + 1 = 5 peaks) and Hc (4 + 3 + H b: split by 6 Ha protons - septet 1 = 8 peaks).

CH3CH2 C CH3

Ha Hb

CH3

Hc Hc

3 signals: Ha: split by 2 Hb protons - triplet Hb: split by 3 Ha protons - quartet Hc: no splitting - singlet

14.44 O

a. CH3CH(OCH3)2

e. (CH3)2CH

CH3 protons split by 1 H = doublet CH proton split by 3 H's = quartet O

b. CH3OCH2CH2

C

OCH3

both CH2 groups split each other = triplets c.

CH2CH3

CH3 protons split by 2 H's = triplet CH2 protons split by 3 H's = quartet

d. CH3OCH2CHCl2 CH2 protons split by 1 H = doublet CH proton split by 2 H's = triplet

H a Hb

C

O OCH2CH3

CH3CH2CH2

Hc H d

Ha protons split by 1 H = doublet Hb proton split by 6 H's = septet Hc protons split by 3 H's = quartet Hd protons split by 2 H's = triplet f.

h.

HOCH2CH2CH2OH

g. CH3CH2CH2CH2OH Ha Hb Hc Hd

OH

Ha H b H c Ha protons split by 2 H's = triplet Hc protons split by 2 H's = triplet Hb protons split by CH3 + CH2 protons = 12 peaks (maximum) Since Hb is located in a flexible alkyl chain, it is likely that peak overlap occurs, so that only 3 + 2 + 1 = 6 peaks will be observed.

Ha Hb Ha protons split by 2 CH2 groups = quintet Hb protons split by 2 H's = triplet

C

O

i.

CH3CH2

Ha

C

H

Hb

Ha: split by CH3 group + Hb = 8 peaks (maximum) Hb: split by 2 H's = triplet

Ha protons split by 2 H's = triplet Hb protons split by CH3 + CH2 protons = 12 peaks (maximum) Hc protons split by 2 different CH2 groups = 9 peaks (maximum) Hd protons split by 2 H's = triplet Since Hb and Hc are located in a flexible alkyl chain, it is likely that peak overlap occurs, so that the following is observed: Hb (3 + 2 + 1 = 6 peaks), Hc (2 + 2 + 1 = 5 peaks).

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Chapter 14–14

H

CH3

j.

k.

C C CH3CH2

H

l.

C C Br

Hb

Ha

H

CH3

Ha

Hb

H

Hc

Ha: split by 1 H = doublet Hb: split by 1 H = doublet

Ha: split by 1 H = doublet Hb: split by 1 H = doublet

CH3

H

Ha

C C H

H

Hb

Ha: split by Hb + Hc doublet of doublets (4 peaks) Hb: split by Ha + Hc doublet of doublets (4 peaks) Hc: split by CH3, Ha + Hb - 16 peaks

14.45 Ha

H

Hb

H

Br

H

Br

C C

Ha

C C CO2CH3

Hb H

Ha: split by 1 H = doublet Hb: split by 1 H = doublet Ha and Hb are geminal.

CO2CH3

Ha: split by 1 H = doublet Hb: split by 1 H = doublet Ha and Hb are trans.

Both compounds exhibit two doublets for the H's on the C=C, but the coupling constants (Jgeminal and Jtrans) are different. Jgeminal is much smaller than Jtrans (0–3 Hz versus 11–18 Hz).

14.46 Hb

Ha C C

Hc

CN

Jab = 11.8 Hz Jbc = 0.9 Hz Jac = 18 Hz

Ha: doublet of doublets at 5.7 ppm. Two large J values are seen for the H’s cis (Jab = 11.8 Hz) and trans (Jac = 18 Hz) to Ha. Hb: doublet of doublets at ~6.2 ppm. One large J value is seen for the cis H (Jab = 11.8 Hz). The geminal coupling (Jbc = 0.9 Hz) is hard to see. Hc: doublet of doublets at ~6.6 ppm. One large J value is seen for the trans H (Jac = 18 Hz). The geminal coupling (Jbc = 0.9 Hz) is hard to see. Ha

Splitting diagram for Ha 1 trans Hc proton splits Ha into 1 + 1 = 2 peaks a doublet

Jac = the coupling constant between Ha and Hc

1 cis Hb proton splits Ha into 1 + 1 = 2 peaks Now it's a doublet of doublets. Jab = the coupling constant between Ha and Hb

14.47 Four constitutional isomers of C4H9Br: Br Br

Br Br

4 different C's

4 different C's

2 different C's

3 different C's

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Nuclear Magnetic Resonance Spectroscopy 14–15 14.48 Only two compounds in Problem 14.42 give one signal in their 13C NMR spectrum: CH3CH3

14.49 O R

C

O OR'

C

R

OR'

The O atom of an ester donates electron density so the carbonyl carbon has less
+, making it less deshielded than the carbonyl carbon of an aldyhyde or ketone. Therefore, the carbonyl carbon of an aldehyde or ketone is more deshielded and absorbs farther downfield.

14.50 d.

a. HC(CH3)3 2 signals

CH2

g. O

5 signals

7 signals e. CH3CH2

b.

CH2CH3

h.

O

C C

5 signals

H

4 signals

H

3 signals c. CH3OCH(CH3)2

f.

i.

OH

3 signals

7 signals

3 signals

14.51 O

a. CH3CH2 Ca Cb

O

OH

C

OH

Cc

Ca < Cb < Cc

b. CH3CH2CHCH2CH3

c.

C

Ca

Ca Cb Cc Ca < Cb < Cc

d. CH2 CHCH2CH2CH2Br

CH2CH3

Ca

Cb Cc

Cc < Cb < Ca

Cb Cc Cb < Cc < Ca

14.52 19 ppm

62 ppm

a. CH3CH2CH2CH2OH 14 ppm 35 ppm

16 ppm

205 ppm

b. (CH3)2CHCHO 41 ppm

143 ppm

23 ppm

c. CH2=CHCH(OH)CH3 113 ppm

69 ppm

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Chapter 14–16

14.53 Ca

Cb Ca

Cb

O

Ca

Cd a.

CH3O2C

H

H

H

b.

Cc

C C

Cc

CO2CH3

H

CH3O2C

Cb Ca

Ca

Cb

CO2CH3

Cb Ca

3 signals

3 different C's 3 signals

H

Cc

C C

Cc

Cd

CO2CH3 C C

H

O

Cc CO2CH3

Cb Ca 4 signals

O

Ce O

C b Cc Ca

5 signals

14.54 Use the directions from Answer 14.24. a. C4H8Br2: 0 degrees of unsaturation IR peak at 3000–2850 cm–1: Csp3–H bonds NMR: singlet at 1.87 ppm (6 H) (2 CH3 groups) singlet at 3.86 ppm (2 H) (CH2 group) CH3 CH3 C CH2Br Br

b. C3H6Br2: 0 degrees of unsaturation IR peak at 3000–2850 cm–1: Csp3–H bonds NMR: quintet at 2.4 ppm (split by 2 CH2 groups) triplet at 3.5 ppm (split by 2 H's) Br

Br

c. C5H10O2: 1 degree of unsaturation IR peak at 1740 cm–1: C=O NMR: triplet at 1.15 ppm (3 H) (CH3 split by 2 H's) triplet at 1.25 ppm (3 H) (CH3 split by 2 H's) quartet at 2.30 ppm (2 H) (CH2 split by 3 H's) quartet at 4.72 ppm (2 H) (CH2 split by 3 H's) O CH3CH2

C

O

CH2CH3

d. C6H14O: 0 degrees of unsaturation IR peak at 3600–3200 cm–1: O–H NMR: triplet at 0.8 ppm (6 H) (2 CH3 groups split by CH2 groups) singlet at 1.0 ppm (3 H) (CH3) quartet at 1.5 ppm (4 H) (2 CH2 groups split by CH3 groups) singlet at 1.6 ppm (1 H) (O–H proton) CH3 CH3CH2 C CH2CH3 OH

e. C6H14O: 0 degrees of unsaturation IR peak at 3000–2850 cm–1: Csp3–H bonds NMR: doublet at 1.10 ppm (integration = 30 units) (from 12 H's) septet at 3.60 ppm (integration = 5 units) (from 2 H's) H

H

CH3 C O C CH3 CH3

CH3

f. C3H6O: 1 degree of unsaturation IR peak at 1730 cm–1: C=O O NMR: triplet at 1.11 ppm C multiplet at 2.46 ppm CH3CH2 H triplet at 9.79 ppm

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Nuclear Magnetic Resonance Spectroscopy 14–17

14.55 Two isomers of C9H10O: 5 degrees of unsaturation (benzene ring likely) Compound B: IR absorption at 1688 cm–1: C=O NMR data: Absorptions: triplet at 1.22 (3 H) (CH3 group split by 2 H's) quartet at 2.98 (2 H) (CH2 group split by 3 H's) multiplet at 7.28–7.95 (5 H) (likely a monosubstituted benzene ring)

Compound A: IR absorption at 1742 cm–1: C=O NMR data: Absorptions: singlet at 2.15 (3 H) (CH3 group) singlet at 3.70 (2 H) (CH2 group) broad singlet at 7.20 (5 H) (likely a monosubstituted benzene ring) O CH2

C

O C

CH3

CH2CH3

14.56 Compound C: molecular ion 146 (molecular formula C6H10O4) IR absorption at 1762 cm–1: C=O 1H NMR data: Absorptions: Ha: doublet at 1.47 (3 H) (CH3 group adjacent to CH) Hb: singlet at 2.07 (6 H) (2 CH3 groups) Hc: quartet at 6.84 (1 H adjacent to CH3)

Hc O CH3

Hb

C

H O C O CH3

O C

CH3

Hb

Ha

14.57 Hb O

[1] LiC [2] H2O

CH

OH

Hc

CH3 C C CH

Ha

CH3

Ha D

Compound D: molecular ion 84 (molecular formula C5H8O) IR absorptions at 3600–3200 cm–1: OH 3303 cm–1: Csp–H 2938 cm–1: Csp3–H 2120 cm–1: C C 1H NMR data: Absorptions: Ha: singlet at 1.53 (6 H) (2 CH3 groups) Hb: singlet at 2.37 (1 H) alkynyl CH and OH Hc: singlet at 2.43 (1 H)

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Chapter 14–18

14.58 Compound E: C4H8O2: 1 degree of unsaturation IR absorption at 1743 cm–1: C=O NMR data:

Compound F: C4H8O2: 1 degree of unsaturation IR absorption at 1730 cm–1: C=O NMR data:

total integration units/# H's (23 + 29 + 30)/8 = ~10 units per H

total integration units/# H's (18 + 30 + 31)/8 = ~10 units per H

Ha: quartet at 4.1 (23 units - 2 H) Hb: singlet at 2.0 (29 units - 3 H) Hc: triplet at 1.4 (30 units - 3 H)

Ha: singlet at 4.1 (18 units - 2 H) Hb: singlet at 3.4 (30 units - 3 H) Hc: singlet at 2.1 (31 units - 3 H)

O CH3

C

Hb

O OCH2CH3

CH3

Ha Hc

Hc

C

CH2OCH3

Ha

Hb

14.59 Compound H: C8H11N: 4 degrees of unsaturation IR absorptions at 3365 cm–1: N–H 3284 cm–1: N–H 3026 cm–1: Csp2–H 2932 cm–1: Csp3–H 1603 cm–1: due to benzene 1497 cm–1: due to benzene

Compound I: C8H11N: 4 degrees of unsaturation IR absorptions at 3367 cm–1: N–H 3286 cm–1: N–H 3027 cm–1: Csp2–H 2962 cm–1: Csp3–H 1604 cm–1: due to benzene 1492 cm–1: due to benzene

NMR data:

NMR data:

multiplet at 7.2–7.4 ppm, 5 H on a benzene ring Ha: triplet at 2.9 ppm, 2 H, split by 2 H's Hb: triplet at 2.8 ppm, 2 H, split by 2 H's Hc: singlet at 1.1 ppm, 2 H, no splitting (on NH2) Ha

multiplet at 7.2–7.4 ppm, 5 H on a benzene ring Ha: quartet at 4.1 ppm, 1 H, split by 3H's Hb: singlet at 1.45 ppm, 2 H, no splitting (NH2) Hc: doublet at 1.4 ppm, 3 H, split by 1 H CH3

CH2CH2NH2

H

Hb

Hc

C NH2

Hc

Hb

Ha

14.60 a. C9H10O2: 5 degrees of unsaturation IR absorption at 1718 cm–1: C=O NMR data: multiplet at 7.4–8.1 ppm, 5 H on a benzene ring quartet at 4.4 ppm, 2 H, split by 3 H's triplet at 1.3 ppm, 3 H, split by 2 H's O C

OCH2CH3

downfield due to the O atom

b. C9H12: 4 degrees of unsaturation IR absorption at 2850–3150 cm–1: C–H bonds NMR data: singlet at 7.1–7.4 ppm, 5 H, benzene septet at 2.8 ppm, 1 H, split by 6 H's doublet at 1.3 ppm, 6 H, split by 1 H CH3 C CH3 H

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Nuclear Magnetic Resonance Spectroscopy 14–19 14.61 a. Compound J has a molecular ion at 72: molecular formula C4H8O 1 degree of unsaturation O IR spectrum at 1710 cm–1: C=O 1H NMR data (ppm): C CH3 CH2CH3 1.0 (triplet, 3 H), split by 2 H's 2.1 (singlet, 3 H) 2.4 (quartet, 2 H), split by 3 H's b. Compound K has a molecular ion at 88: molecular formula C5H12O 0 degrees of unsaturation IR spectrum at 3600–3200 cm–1: O–H bond 1 OH H NMR data (ppm): CH 0.9 (triplet, 3 H), split by 2 H's 3 C CH3 1.2 (singlet, 6 H), due to 2 CH3 groups CH2CH3 1.5 (quartet, 2 H), split by 3 H's 1.6 (singlet, 1 H), due to the OH proton

14.62 Compound L has a molecular ion at 90: molecular formula C4H10O2 0 degrees of unsaturation total integration units/# H's IR absorptions at 2992 and 2941 cm–1: Csp3–H (25 + 46 + 7)/10 = ~8 units per H 1H NMR data (ppm): O Ha: 1.2 (doublet, 3 H), split by 1 H H+ CH3 OH Hb: 3.3 (singlet, 6 H), due to 2 CH3 groups H Hc: 4.8 (quartet, 1 H), split by 3 adjacent H's

H

Hc

CH3 C OCH3

Ha

OCH3

L

14.63

TsOH

multiplet at ~2.0 singlet at ~1.7 multiplet at ~1.5 triplet at ~0.95 triplet at ~0.95

2 singlets at ~1.6 M

OH

OH2

H

N triplet at ~5.1

H OTs

two signals at ~4.6

triplet at ~2.0

OTs

1,2-H shift H

TsOH

N

H H H2O

OTs

or TsOH H OTs

M

Hb Hb

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Chapter 14–20

14.64 Compound O has a molecular formula C10H12O. 5 degrees of unsaturation IR absorption at 1687 cm–1 1H NMR data (ppm): Ha: 1.0 (triplet, 3 H), due to CH3 group, split by 2 adjacent H's Hb: 1.7 (sextet, 2 H), split by CH3 and CH2 groups Hc: 2.9 (triplet, 2 H), split by 2 H's 7.4–8.0 (multiplet, 5 H), benzene ring

O C CH2CH2CH3

Hc Hb Ha O

14.65 Compound P has a molecular formula C5H9ClO2. 1 degree of unsaturation 13C NMR shows 5 different C's, including a C=O. O 1H NMR data (ppm): C ClCH2CH2 OCH2CH3 Ha: 1.3 (triplet, 3 H), split by 2 H's Hc Hb Hd Ha Hb: 2.8 (triplet, 2 H), split by 2 H's Hc: 3.7 (triplet, 2 H), split by 2 H's P Hd: 4.2 (quartet, 2 H), split by CH3 group

14.66 Compound Q: Molecular ion at 86. Molecular formula: C5H10O: O 1 degree of unsaturation C CH3 CH2CH3 IR absorption at ~1700 cm–1: C=O NMR data: Ha: doublet at 1.1 ppm, 2 CH3 groups split by 1 H Hb: singlet at 2.1 ppm, CH3 group Hc: septet at 2.6 ppm, 1 H split by 6 H's

Hb O

[1] strong base [2] CH3I

CH3

C

Hc Ha = Q

CH(CH3)2

MW = 86

14.67 a. Compound R, the odor of banana: C7H14O2 1 degree of unsaturation 1H NMR (ppm): Ha: 0.93 (doublet, 6 H) Hb: 1.52 (multiplet, 2 H) Hc: 1.69 (multiplet, 1 H) Hd: 2.04 (singlet, 3 H) He: 4.10 (triplet, 2 H)

Hc

O CH3

C

b. Compound S, the odor of rum: C7H14O2 1 degree of unsaturation 1H NMR (ppm):

OCH2CH2CHCH3 CH3

Hd

He Hb Ha

Ha

Ha: 0.94 (doublet, 6 H) Hb: 1.15 (triplet, 3 H) Hc: 1.91 (multiplet, 1 H) Hd: 2.33 (quartet, 2 H) He: 3.86 (doublet, 2 H)

Hc

O CH3CH2

C

OCH2CHCH3 CH3

Hb Hd

He

Ha Ha

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Nuclear Magnetic Resonance Spectroscopy 14–21

14.68 C6H12: 1 degree of unsaturation K+ –OC(CH3)3

T

1H

Br

H+

U

1H

NMR of T (ppm): Ha: 1.01 (singlet, 9 H) Hb: 4.82 (doublet of doublets, 1 H, J = 10, 1.7 Hz) Hc: 4.93 (doublet of doublets, 1 H, J = 18, 1.7 Hz) Hd: 5.83 (doublet of doublets, 1 H, J = 18, 10 Hz) NMR of U: 1.60 (singlet) ppm.

CH3 C

OH

CH3

CH3 C CH3

Hc

Ha (CH3)3C

H C C

Hd

H

H

Hb

All H's are identical, so there is only one singlet in the NMR.

14.69 Both A and B have the same molecular ion—since they are isomers—and show a C=O peak in their IR spectra. 1H NMR spectroscopy is the best way to distinguish the two compounds. O CH3O

C

O

or

C(CH3)3

CH3

C

OC(CH3)3

B

A

Both A and B have two singlets in a 3:1 ratio in their 1H NMR spectra. But A has a peak at ~3 ppm due to the deshielded CH3 group bonded to the O atom. B has no proton that is so deshielded. Both of its singlets are in the 1–2.5 ppm region.

14.70 a. C6H12O2: 1 degree of unsaturation IR peak at 1740 cm–1: C=O 1H NMR 2 signals: 2 types of H's 13C NMR: 4 signals: 4 kinds of C's, including one at ~170 ppm due a C=O

O CH3

C

C

O

CH3

CH3 CH3

b. C6H10: 2 degrees of unsaturation IR peak at 3000 cm–1: Csp3–H bonds peak at 3300 cm–1: Csp–H bond peak at ~2150 cm–1: C
C bond 13 C NMR: 4 signals: 4 kinds of C's

CH3 HC C C CH3 CH3

14.71 O H

C

O N(CH3)2

N,N-dimethylformamide

H

C + CH3 N CH3

cis to the O atom cis to the H atom

A second resonance structure for N,N-dimethylformamide places the two CH3 groups in different environments. One CH3 group is cis to the O atom, and one is cis to the H atom. This gives rise to two different absorptions for the CH3 groups.

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Chapter 14–22

14.72 Ho

Ho

Ho Ho

Ho Hi Hi

Ho

Hi Hi

Ho

Hi Hi

Ho

Ho Ho

Ho

Ho

18-Annulene has 18
electrons that create an induced magnetic field similar to the 6
electrons of benzene. 18-Annulene has 12 protons that are oriented on the outside of the ring (labeled Ho), and 6 protons that are oriented inside the ring (labeled Hi). The induced magnetic field reinforces the external field in the vicinity of the protons on the outside of the ring. These protons are deshielded and so they absorb downfield (8.9 ppm). In contrast, the induced magnetic field is opposite in direction to the applied magnetic field in the vicinity of the protons on the inside of the ring. This shields the protons and the absorption is therefore very far upfield, even higher than TMS (–1.8 ppm).

14.73 Ca stereogenic center CH3 H CH3 C H

C CH3 Replace a CH3 group

with X.

OH

Cb

H HO C CH3 X C CH3 H

Replace Ca.

3-methyl-2-butanol

H

or

HO C CH3 CH3 C X H

Replace Cb.

The CH3 groups are not equivalent to each other, since replacement of each by X forms two diastereomers. Thus, every C in this compound is different and there are five 13C signals.

14.74 O CH3 P OCH3 OCH3

One P atom splits each nearby CH3 into a doublet by the n + 1 rule, making two doublets.

Ha

Hb Ha

All 6 Ha protons are equivalent.

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Radical Reactions 15–1 C Chhaapptteerr 1155:: R Raaddiiccaall R Reeaaccttiioonnss

G Geenneerraall ffeeaattuurreess ooff rraaddiiccaallss • A radical is a reactive intermediate with an unpaired electron (15.1). • A carbon radical is sp2 hybridized and trigonal planar (15.1). • The stability of a radical increases as the number of C’s bonded to the radical carbon increases (15.1). least stable

most stable RCH2

CH3

o

1

R2CH

R3C

o

3o

2

Increasing alkyl substitution Increasing radical stability

•

Allylic radicals are stabilized by resonance, making them more stable than 3o radicals (15.10). CH2 CH CH2

CH2 CH CH2

two resonance structures for the allyl radical



R Raaddiiccaall rreeaaccttiioonnss [1] Halogenation of alkanes (15.4) R H

X2 h or


X = Cl or Br

• •

R X

alkyl halide

• •

The reaction follows a radical chain mechanism. The weaker the C–H bond, the more readily the H is replaced by X. Chlorination is faster and less selective than bromination (15.6). Radical substitution results in racemization at a stereogenic center (15.8).

[2] Allylic halogenation (15.10) CH2 CH CH3

NBS h
or ROOR

CH2 CHCH2Br

allylic halide

•

The reaction follows a radical chain mechanism.

[3] Radical addition of HBr to an alkene (15.13) H H • A radical addition mechanism is followed. HBr RCH CH2 R C C H • Br bonds to the less substituted carbon atom to h,
, or H Br form the more substituted, more stable radical. ROOR alkyl bromide

[4] Radical polymerization of alkenes (15.14) CH2 CHZ

ROOR Z

Z

polymer

Z

•

A radical addition mechanism is followed.

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Chapter 15–2 C Chhaapptteerr 1155:: A Annssw weerrss ttoo PPrroobblleem mss 15.1

1° Radicals are on C’s bonded to one other C; 2° radicals are on C’s bonded to two other C’s; 3° radicals are on C’s bonded to three other C’s. a. CH3CH2 CHCH2CH3

c.

b.

2° radical

15.2

3° radical

1° radical

b. (CH3)3CCHCH3

c. (CH3)3CCH2

d.

Reaction of a radical with: • an alkane abstracts a hydrogen atom and creates a new carbon radical. • an alkene generates a new bond to one carbon and a new carbon radical. • another radical forms a bond. a. CH3 CH3 b. CH2 CH2

15.4

2° radical

The stability of a radical increases as the number of alkyl groups bonded to the radical carbon increases. Draw the most stable radical. a. (CH3)2CCH2CH3

15.3

d.

Cl CH3 CH2

H Cl

Cl

c.

Cl

d.

Cl

Cl Cl

Cl

Cl

CH2 CH2

+

O O

Cl O O

Monochlorination is a radical substitution reaction in which a Cl replaces a H generating an alkyl halide. Cl2

a.

Cl




b. CH3CH2CH2CH2CH2CH3

c. (CH3)3CH

H

Cl2

ClCH2CH2CH2CH2CH2CH3




Cl

H

CH3 C CH3




H CH3CH2 C CH2CH2CH3

Cl

Cl

Cl2

CH3 C CH2CH2CH2CH3

CH3 C CH2Cl

CH3

CH3

15.5 A

Cl2


Cl Cl2

B

Cl Cl


Cl

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Radical Reactions 15–3

15.6 Br

Initiation:

Propagation:

CH3

h or


Br

Br

H Br

CH3 Br

Br

+

Br

CH3

Br Br

CH3

Termination:

+

Br

+

H

Br

Br

Br Br

or CH3

CH3

CH3 CH3

or CH3

Br

CH3 Br

15.7 Step 1:

CH3 H

+

Br

CH3

1 bond broken +435 kJ/mol CH3

Step 2:

15.8

1 bond formed –368 kJ/mol

Br Br

CH3 Br +

1 bond broken +192 kJ/mol

1 bond formed –293 kJ/mol


H° = –101 kJ/mol

Br

The rate-determining step for halogenation reactions is formation of CH3• + HX. CH3 H

+

I

CH3

1 bond broken +435 kJ/mol

15.9


H° = +67 kJ/mol

H Br

H

I

1 bond formed –297 kJ/mol


H° = +138 kJ/mol This reaction is more endothermic and has a higher Ea than a similar reaction with Cl2 or Br2.

The weakest C–H bond in each alkane is the most readily cleaved during radical halogenation. a.

b.

H

H

3° most reactive

c.

CH3CHCH2CH3 H

3° most reactive

2° most reactive

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Chapter 15–4 15.10 To draw the product of bromination: • Draw out the starting material and find the most reactive C–H bond (on the most substituted C). • The major product is formed by cleavage of the weakest C–H bond. Br2

a.

Br2

c. Br







Br Br

Br2

b.

Br2

Br

d.







15.11 Cl2

Cl

Cl

h


Cl

Cl Cl

This is the desired product, 1-chloro-1-methylcyclohexane, but many other products are formed.

15.12 If 1o C–H and 3o C–H bonds were equally reactive there would be nine times as much (CH3)2CHCH2Cl as (CH3)3CCl since the ratio of 1o to 3o H’s is 9:1. The fact that the ratio is only 63:37 shows that the 1o C–H bond is less reactive than the 3o C–H bond. (CH3)2CHCH2Cl is still the major product, though, because there are nine 1o C–H bonds and only one 3o C–H bond. 15.13 CH3

Br2

a. CH3 C H




CH3

CH3

CH3 CH3 C Br

c.

CH3

H2O

C CH2

CH3 CH3 C OH

H2SO4

CH3

CH3

(from b.) CH3

b.

(CH3)3CO–K+

CH3 C Br CH3

CH3 C CH2

CH3

d.

CH3

(from a.)

Cl2

C CH2

CH2Cl CH3 C Cl

CCl4

CH3

CH3

(from b.)

15.14 Br2

a.

Br K+ –OC(CH ) 3 3

h
mCPBA

b. (from a.)

Br2

Br

+ enantiomer Br

O

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Radical Reactions 15–5

15.15 Since the reaction does not occur at the stereogenic center, leave it as is. C1

C4

H Br

CH3

H Br

Cl2

C

C

h


CH2CH3

H Br C CH2CH3

ClCH2

(2R)-2-bromobutane

CH3

CH2CH2Cl

R

S

15.16 Cl Cl

Cl2

a. CH3CH2CH2CH2CH3

CH3CH2CH2CH2CH2Cl




Cl

b.

Cl2

CH3

H

c.

H Cl

Cl2

d.

H Cl

CH3

CH3

C

CH3 Cl H

CH(CH2CH3)2

H Cl

H Cl

h

(Consider attack at C2 and C3 only.)

CH3 H

Cl H

CH2CH3

Cl Cl

CH3

ClCH2CH2CH(CH2CH3)2

CH3CH2 C CH2CH3

h

CH2CH3

C

Cl

CH3

Cl

Cl2

CH3CH2 C CH2CH3

Cl

CH3

CH3

H CH3 CH3CH2CH2

CH3CH2CH2

Cl

Cl

CH2Cl




C

CH3CH2CHCH2CH3

Cl

C

CH(CH2CH3)2

H Cl

15.17 O N

Chain propagation:

O N O

+

O3

+

O

O N O + O2 O N

+

O2

The radical is re-formed.

15.18 Draw the resonance structure by moving the
bond and the unpaired electron. The hybrid is drawn with dashed lines for bonds that are in one resonance structure but not another. The symbol • is used on any atom that has an unpaired electron in any resonance structure. a.

CH3 CH CH CH2

CH3 CH CH CH2


hybrid: CH3 CH CH

c.


CH2

hybrid:





b. d.





hybrid:

hybrid:







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Chapter 15–6 15.19 Reaction of an alkene with NBS or Br2 + h
yields allylic substitution products. Br NBS

a.

c. CH2 CH CH3

h


Br2

CH2CHCH3 Br Br

NBS

b. CH2 CH CH3

CH2 CH CH2Br

h


15.20 c. CH2 C(CH2CH3)2

CH3 CH CH CH2Br

h


+

h


b.

+ BrCH2C(CH2CH3) CHCH3

NBS h


CH3 CH3

Br

CH3 CH3

+ Br

15.21 CH3

CH3

CH3

NBS

CH3 Br

CH3

Br

h
Br

Br

15.22 Reaction of an alkene with NBS + h
yields allylic substitution products. a.

one possible product: high yield

Br

b.

CH3CH2CH CHCH2Br

CH2 C CH2CH3

CH3CH(Br)CH CH2 CH3 CH3

CHBrCH3

NBS

NBS

a. CH3 CH CH CH3

c.

Br

Cannot be made in high yield by allylic halogenation. Any alkene starting material would yield a mixture of allylic halides.

15.23 OOH

second resonance structure

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Radical Reactions 15–7

15.24 The weakest C–H bond is most readily cleaved. To draw the hydroperoxide products, add OOH to each carbon that bears a radical in one of the resonance structures. O

O

C H

C

OH

OH

linoleic acid

This allylic C–H bond is most readily cleaved. O

O

C

C

OH

OH

hydroperoxide products: HO

O

O

O

C

C

OH O

O C HO

OH

OH

(E/Z isomers are possible.)

OH

O

15.25 R

O H (CH3)3C

O

C(CH3)3

(CH3)3C

C(CH3)3

BHT H

H

H

H

CH3 O (CH3)3C

O C(CH3)3

H

CH3 O

(CH3)3C

H

C(CH3)3

H

CH3

H

(CH3)3C H

CH3

O C(CH3)3

(CH3)3C

H

H

CH3

CH2 CHCH2CH2CH2CH3

HBr

CH3CHCH2CH2CH2CH3 Br

CH2 CHCH2CH2CH2CH3

HBr

CH2BrCH2CH2CH2CH2CH3

ROOR

b.

HBr

Br

HBr ROOR

H CH3

15.26 a.

C(CH3)3

Br

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Chapter 15–8 HBr

c. CH3CH CHCH2CH2CH3

CH3CH CH2CH2CH2CH3

or HBr, ROOR

CH3CH2 CHCH2CH2CH3 Br

Br

15.27 In addition of HBr under radical conditions: • Br• adds first to form the more stable radical. • Then H• is added to the carbon radical. CH3 H

2 radical possibilities:

Br C

CH3

or

C

C

CH3 H

CH3

CH3 H

H C Br H

H

C

C Br

CH3 H

3° radical more stable This radical forms.

1° radical less stable

15.28 Br HBr ROOR

a.

c.

Br2

Br

Br

Br HBr

b.

15.29 Transition state 1: H CH3 C

Energy




Transition state 2:

CH2 Br

H




CH3 C CH2




H Br Br

CH3 CH CH2 +




Br

–18 kJ/mol

[Use the bond dissociation energies in Appendix C.]

H CH3 C CH2 Br

–29 kJ/mol H CH3 C CH2 + Br H Br

Reaction coordinate

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Radical Reactions 15–9

15.30 H

a.

H

CH2 C

H

H

CH2 C CH2 C CH2 C6H5

C6H5

b.

C

C6H5

CH2 CH OCOCH3

C6H5

O

O

O

COCH3 COCH3 COCH3

polystyrene

poly(vinyl acetate)

15.31 Initiation:

RO

OR

Cl

[1]

+ CH2 C

2 RO

Cl

[2] ROCH2 C

H

H

Cl

Cl

Propagation:

ROCH2 C

Cl

[3]

ROCH2 C

CH2 C H

Cl

Cl

H

H

CH2 C

Cl CH2 C H

H

H

Repeat Step [3] over and over.

Termination:

carbon radical

new C–C bond Cl Cl

[4] C CH2

CH2 C C CH2

[one possibility]

H H

15.32 a. increasing bond strength: 2 < 3 < 1 b. and c.

CH3 CH2 C CHCH3

CH3 H CH2 C CHCH3

H H

H

1° radical least stable

2° radical intermediate stability

CH3 H CH2 C CHCH3 H

3° radical most stable

d. increasing ease of H abstraction: 1 < 3 < 2

15.33 Use the directions from Answer 15.2 to rank the radicals. a.

(CH3)2CHCH2CH(CH3)CH2

1° radical least stable

(CH3)2CHCHCH(CH3)2

(CH3)2CCH2CH(CH3)2

2° radical intermediate stability

3° radical most stable

b. 2° radical least stable

3° radical intermediate stability

allylic radical most stable

382

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Chapter 15–10 15.34 Draw the radical formed by cleavage of the benzylic C–H bond. Then draw all of the resonance structures. Having more resonance structures (five in this case) makes the radical more stable, and the benzylic C–H bond weaker. CH2 H

benzylic C–H bond bond dissociation energy = 356 kJ/mol CH2

CH2

CH2

CH2

CH2

15.35 Hb

Hd

H H

CH2 CHCHCHC(CH3)CH2 H

Ha

H

Hc

Ha = bonded to an sp3 3° carbon Hb = bonded to an allylic carbon Hc = bonded to an sp3 1° carbon Hd = bonded to an sp3 2° carbon

Increasing ease of abstraction: Hc < Hd < Ha < Hb

15.36 Cl a.

Cl b.

(CH3)3CCH2CH2CH2CH3

(CH3)3CCH2CH2CH2CH2Cl

Cl

(CH3)3CCH2CH2CHCH3

(CH3)3CCH2CHCH2CH3

CH2Cl

Cl

(CH3)2CCH2CH2CH2CH3

(CH3)3CCHCH2CH2CH3

Cl

Cl

Cl

c.

Cl

Cl Cl d.

CH3

CH3

Cl CH3

CH3

CH2Cl

15.37 To draw the product of bromination: • Draw out the starting material and find the most reactive C–H bond (on the most substituted C). • The major product is formed by cleavage of the weakest C–H bond. Br

a.

c.

CH3

Br CH3 Br

b.

(CH3)3CCH2CH(CH3)2

(CH3)3CCH2C(CH3)2 Br

d.

(CH3)3CCH2CH3

(CH3)3CCHCH3

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Radical Reactions 15–11

15.38 Draw all of the alkane isomers of C6H14 and their products on chlorination. Then determine which letter corresponds to which alkane. Cl

*

*

h


A

Cl

Cl

Cl2

Cl

Cl Cl2

Cl

*

Cl

*

* *

h


Cl Cl

B Cl2

*

h


Cl Cl

C Cl

Cl2

Cl2

*

Cl

h


D

Cl Cl

h


*

Cl Cl

*

[* = stereogenic center]

E

15.39 Halogenation replaces a C–H bond with a C–X bond. To find the alkane needed to make each of the alkyl halides, replace the X with a H. Br

a.

Cl

c.

Br

b.

d.

(CH3)3CCH3

(CH3)3CCH2Cl

15.40 For an alkane to yield one major product on monohalogenation with Cl2, all of the hydrogens must be identical in the starting material. For an alkane to yield one major product on bromination, it must have a more substituted carbon in the starting material. a.

Cl

c.

d. Cl

b. Br

These two compounds can be formed in high yield from an alkane.

three different C–H bonds

Br

Br on 2° carbon The product with Br on 3° carbon will form predominantly.

These two compounds cannot be formed in high yield from an alkane.

384

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Chapter 15–12 15.41 Chlorination with two equivalents of Cl2 yields a variety of products. Cl

Cl2 (2 equiv)

Cl

Cl

Cl

Cl




Cl

Cl Cl

The desired product is only one of four products formed.

15.42 In bromination, the predominant (or exclusive) product is formed by cleavage of the weaker C–H bond to form the more stable radical intermediate.

CH3

Br2

CH2 H

CH3




CH2Br

CH3

weaker bond H o C stronger bond
H = 356 kJ/mol o As usual, more product is formed
H = 460 kJ/mol by homolysis of the weaker bond.

CH3 Br

D NOT formed

15.43 Chlorination is not selective so a mixture of products results. Bromination is selective, and the major product is formed by cleavage of the weakest C–H bond. Cl

Cl Cl

Cl2

a.




Cl

Y

Br2

b.


Br

Y

K+–OC(CH3)3

Br2

c.


Y

Br

15.44 Draw the resonance structures by moving the
bonds and the radical. CH2

a.

b.

Cl

Cl

CH2

c.

Z

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Radical Reactions 15–13 15.45 Reaction of an alkene with NBS + h
yields allylic substitution products. Br NBS

a.

h
Br

NBS

b. CH3CH2CH CHCH2CH3

c. (CH3)2C CHCH3

h


NBS

(CH3)2C

h


Br

CH3CH2CH CHCHCH3

CHCH2Br

(CH3)2C(Br)CH Br

CH3CH2CHCH CHCH3

CH2 C(CH3)CH(Br)CH3

BrCH2C(CH3) CHCH3

CH2 Br

NBS

d. h


Br Br

15.46 It is not possible to form 5-bromo-1-methylcyclopentene in good yield by allylic bromination because several other products are formed. 1-Methylcyclopentene has three different types of allylic hydrogens (labeled with *), all of which can be removed during radical bromination. Br

*

*

NBS

Br

Br

h


*

Br Br

5-bromo-1-methylcyclopentene

15.47 Br NBS h
Br

X

Br

Br

Br

Br

15.48 Cl + Cl2

a.

h

Cl Cl Cl

CH3 + Br2

b. CH3 CH3




CH3 Br CH3 CH3

(major product)

386

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Chapter 15–14 Br

c.

+ Br2

Br

Br

h


(minor product)

(two major products)

Br h


+ NBS

CH2

d.

CH2 Br Br Br2

HBr

e.

Br

h. HBr ROOR

f.

Br Br

Br

i. g.

h


HBr

CH2

ROOR

CH=CHCH2Br

NBS

Br

15.49 Br Br HBr

a.

Br

Br2

b.

NBS

c.

h


Br

+ enantiomer

15.50 KOC(CH3)3

Br2 h


O

Br

Br

A

ozonolysis cyclohexanone

B

C

D

15.51 CH3 a. CH3 b.

Br2 h
Cl2 h


CH3 Br CH2Cl

CH3

CH3

CH3

Cl

Cl

CH3

CH3

Cl

Cl

Cl

CH3

Cl

CH3

Cl

O acetone

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Radical Reactions 15–15 Cl Cl2

c.

Cl

Cl

Cl

h
Cl Cl

Cl Cl

CH3 d.

Br2 h

CH3

CH3

CH3 Br2


e. H

CH3

CH3

Br

Br

CH3

CH3

+ CH3 Br

CCl3 f.

C CH3CH2

CH3

Br CH3 CCl3

Cl2


F H

C ClCH2CH2

F H

CCl3

+

C

F Cl

CH3CH2

CCl3

+

F Cl

+

C CH2CH3 F H

F H CH3

C

C

CCl3

+

CH3

H Cl

C

C

CCl3

Cl H

15.52 Cl

Cl

Cl2

Cl

Cl

Cl

a. h


Cl

Cl

Cl

Cl

Cl

A

B

C

D

E Cl

(2R)-2-chloropentane Cl

Cl Cl

G

F

b. There would be seven fractions, since each molecule drawn has different physical properties. c. Fractions A, B, D, E, and G would show optical activity. 15.53 Cl2

a. H

Cl

Cl

Cl

h


Cl

1

2

3 Cl

4 Cl Cl

Cl

5

6

9

Cl

7

Cl

10

11

8

Cl

Cl

388

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Chapter 15–16 a. There would be 10 fractions, since 4 and 5 (two enantiomers) would be in one fraction. b. All fractions except the one that contains 4 and 5 would be optically active. 15.54 Br NBS

CH2

Br CH2

CH2

CH2

CH2

h


A

Br

CH2Br

Br

CH2Br

15.55 CH3

a.

CH3 h

CH3 C CH3 + Br2

CH3 C CH3 + HBr

H

Br

C–H bond broken +381 kJ/mol

Br–Br bond broken +192 kJ/mol

C–Br bond formed –272 kJ/mol

total bonds broken = +573 kJ/mol

Br

b. Initiation:

(CH3)3C

Propagation:

h or


Br

H

(CH3)3C

Termination: (one possibility)

Br

+

total bonds formed = –640 kJ/mol

Br

Br

+

H–Br bond formed –368 kJ/mol

+

Br

Br Br

(CH3)3C

c.
H° = (bonds broken) – (bonds formed) = (+381 kJ/mol) + (–368 kJ/mol) = +13 kJ/mol
H° = (bonds broken) – (bonds formed) = (+192 kJ/mol) + (–272 kJ/mol) = –80 kJ/mol

H Br

(CH3)3C

Br


H° = –67 kJ/mol

Br

Br

Br Br

d. and e. Transition state 1:

Transition state 2:

+13 kJ/mol Energy

Transition state 1: (CH3)3C

H

CH3

(CH3)3C CH3

+ Br


C CH

3

H

Br


–80 kJ/mol

(CH3)3C

Reaction coordinate

Br

Br

Transition state 2: CH3

CH3
C CH3 Br

Br




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Radical Reactions 15–17

15.56 O

Initiation:

O

N

Br

N

+

Br

h
O NBS

O

Propagation:

+

Br

H Br

Br

Br

+

Br (from NBS)

H Br

(from NBS) +

Br

+

Br

Br Br Termination: (one possibility)

+

Br Br

Br

15.57 Calculate the
H° for the propagation steps of the reaction of CH4 with I2 to show why it does not occur at an appreciable rate. CH3 H

+

I

CH3

+435 kJ/mol


Ho = +138 kJ/mol

I

–297 kJ/mol I

CH3

H

I

CH3

+151 kJ/mol

I

This step is highly endothermic, making it difficult for chain propagation to occur over and over again.


Ho = –83 kJ/mol

I

–234 kJ/mol

15.58 Calculate
H° for each of these steps, and use these values to explain why this alternate mechanism is unlikely. [1]

+ Cl C–H bond broken +435 kJ/mol

C–Cl bond formed –351 kJ/mol

+ Cl2 Cl–Cl bond broken +242 kJ/mol

H–Cl bond formed
H° = –189 kJ/mol –431 kJ/mol

CH4

CH3Cl + H

H

[2]

HCl +


H° = +84 kJ/mol

The
H° for Step [1] is very endothermic, making this mechanism unlikely.

Cl

15.59 H Br

Br–

1,2-CH3 shift

Br

Br–

3,3-dimethyl-1-butene

2o carbocation

3o carbocation

2-bromo-2,3-dimethylbutane

390

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Chapter 15–18 Br

3,3-dimethyl-1-butene

HBr

Br

HBr peroxide

The 2o radical does NOT rearrange.

Br

Br

1-bromo-3,3-dimethylbutane

Addition of HBr without added peroxide occurs by an ionic mechanism and forms a 2o carbocation, which rearranges to a more stable 3o carbocation. The addition of H+ occurs first, followed by Br–. Addition of HBr with added peroxide occurs by a radical mechanism and forms a 2o radical that does not rearrange. In the radical mechanism Br• adds first, followed by H•. 15.60 Cl

Initiation:

CH3 H

Propagation:

One possibility for termination:

h or


Cl

+

Cl

+

Cl

CH3

Cl Cl

CH3

CH3

Cl

CH3

H Cl

CH3 Cl

Cl

CH3 CH3

CH3 CH2 H CH3CH2

+

Cl Cl Cl

CH3CH2 CH3CH2 Cl

H Cl

Cl

15.61 H

Step 1:

CH3 CH CH2

+

Cl

CH3

C CH2


H1° = –72 kJ/mol

Cl

1 bond broken +267 kJ/mol H

Step 2:

1 bond formed –339 kJ/mol H

CH3 C CH2

+ H–Cl

Cl

CH3 C CH2 H Cl

1 bond broken +431 kJ/mol

1 bond formed –397 kJ/mol


H2° = +34 kJ/mol This step of propagation is endothermic. It prohibits chain propagation from occurring over and over.

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Radical Reactions 15–19

15.62 Cl2

a.

Cl




mCPBA

f.

O

(from b.) Cl

b.

K+ –OC(CH3)3

g. (from a.)

Br

Br2

Br

(from c.)

Br

Br

NBS

c.

Br

ROOR

h.

(from b.)

(from b.)

OH

i.

H2SO4

OH

mCPBA

OH

O

(from h.) Cl

Cl2

e.

OH

(from c.) H2O

d.

OH

Br

OH

j.

[1] NaCN

O

Cl

(from b.)

[2] H2O

(from f.)

CN

15.63 Br2

a.

Br

H2O

K+–OC(CH3)3

OH

H2SO4

h


major product mCPBA

b.

Cl2

c.

O

Cl

+ enantiomer

Cl

(from a.)

(from a.)

15.64 a. CH3CH2CH3

b.

Br

Br2 h


CH3CHCH3

K+ –OC(CH3)3

CH3CH CH2

Br Br

Br2

CH3 C C H

Br

H H

K+ –OC(CH3)3

HBr

K+ –OC(CH3)3

CH3C CH

(2 equiv) DMSO

Br

ROOR

c. HC CH

NaH

HC C

CH3CH2Br

H2

HBr

Lindlar catalyst

ROOR

Br

15.65 a. CH3 CH3

b.

HC CH

(from a.)

Br2 h
NaH

CH3 CH2Br

HC C

K+ –OC(CH3)3

CH3 CH2Br

(from a.)

CH2 CH2

HC CCH2CH3

Br2

BrCH2 CH2Br

2 NaNH2

HC CH

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Chapter 15–20 O

[1] HC

mCPBA

c. CH2 CH2

C

HC CCH2CH2OH

(from b.)

(from a.)

[2] H2O

d. HC CCH2CH3

NaH

C CCH2CH3

CH3 CH2Br

CH3CH2 C C CH2CH3

(from a.)

(from b.)

Na NH3

H2O

e. CH3CH2 C C CH2CH3

H2SO4 HgSO4

(from d.)

O

15.66 Cl

Cl2

K+ –OC(CH3)3

[1] O3

CHO

OHC

h


[2] (CH3)2S

15.67 O2 abstracts a H here.

O O

COOH HH

COOH

COOH

O O

C5H11

C5H11

C5H11

arachidonic acid

+ HOO R H OOH

COOH C5H11

O O COOH

+

COOH

R H

C5H11

C5H11

another molecule of arachidonic acid

5-HPETE

This process is repeated.

15.68 [1] O

H

O O

O O

[2]

H O

O

O O

O

[3]

+ HOO

O

O OH

+ O

Then, repeat Steps [2] and [3].

15.69 HOO

a. OOH

(cis and trans)

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Radical Reactions 15–21

O2 abstracts a H here. O O

b. O O

[1]

H H

+ HOO [2a]

[3a] R H

O O

O O

[2b]

OOH

[3b]

HOO

+ R

R H

O O

O O

[RH = 1-hexene]

+ R

Repeat Steps [2] and [3] again and again.

15.70 (CH3)3C H O

OCH3

Abstraction of the phenol H produces a resonance-stabilized radical.

BHA

(CH3)3C

(CH3)3C

O

OCH3

O

(CH3)3C OCH3

(CH3)3C

(CH3)3C

O

OCH3

O

O

OCH3

OCH3

15.71 Abstraction of the labeled H forms a highly resonance-stabilized radical. Four of the possible resonance structures are drawn. OH

HO

OH

OH O

HO

O OH

vitamin C

O

HO –O

O

OH O

HO O

OH

O

O

HO

O

O

O O

X OH

OH O

HO O

O

O

HO

O

O

O O

15.72 The monomers used in radical polymerization always contain double bonds. a.

b.

CH2 CHCO2Et COOEt COOEt COOEt

polyisobutylene

poly(ethyl acrylate) (used in Latex paints)

Et = CH2CH3

394

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Chapter 15–22 15.73 a.

CH3

CH3 CH2 C

CH2 C COOCH3

CH2

CH3

C CH2

C

COOCH3 COOCH3 COOCH3

methyl methacrylate

b.

CH3

PMMA

CH3 CH2 C HO

CO2CH2CH2OH

O

hydroxyethyl methacrylate

poly-HEMA

O O

O OH

15.74 Overall reaction:

CH2 CHCN

ROOR CN

Initiation:

RO

OR

[1]

CN CN

+ CH2 C

2 RO

CN CN

[2] ROCH2 C

H

H

CN

CN

Propagation:

ROCH2 C

CN

[3]

ROCH2 C

CH2 C H

CN

NC

CH2 C

[4]

C CH2 H

CN CH2 C H

H

H

Repeat Step [3] over and over.

Termination:

carbon radical

new C–C bond CN CN CH2 C C CH2

H

[one possibility]

H H

15.75 a. CH3O

CH=CH2

A OCH3

OCH3

OCH3

b. The OCH3 group stabilizes an intermediate carbocation by resonance. This makes A react faster than styrene in cationic polymerization.

OCH3

OCH3

OCH3

three of the possible resonance structures

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Radical Reactions 15–23

15.76 singlet singlet at 2.23 at 4.04 Cl2

Cl Cl

doublet at 1.69

Cl

Cl

h


Cl

Cl Cl

multiplet Cl B at 4.34

A

C

doublet at 5.85

15.77 molecular formula C3H6Cl2 Integration: (57 units + 29 units)/6 H's = 14 units per H one signal is 57 units/14 units per H = 4 H's second signal is 29 units/14 units per H = 2 H's 1H NMR data: quintet at 2.2 (2 H's) split by 4 H's triplet at 3.7 (4 H's) split by 2 H's

triplet Cl2 Cl

h


Cl

quintet minor product

15.78 a. CH3CH3

Cl2

equivalent H's singlet

doublet quartet

ClCH2CH2Cl

CH3CHCl2

h


Cl

b. Initiation: Propagation:

X h

Cl

Y

Cl

or


+

Cl

H

H +

CH3 C H

Cl

CH3 C

H

H Cl

H H

H +

CH3 C

Cl

Cl

CH3 C Cl

Cl

H

H H

Cl

Formation of Y: CH3 CH Cl Cl

Cl

Cl H CH2 CH2Cl Cl

Cl ClCH2CH2Cl

X Cl

+

Cl

H Cl

CH2 CH2Cl

CH2 CH2Cl

Termination:

Cl

CH3CHCl2 Y

CH3 CH Cl

Formation of X:

H Cl

CH3 CH Cl

Cl Cl

Cl

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Chapter 15–24 15.79 RO

OR

RO

O

O CH3

C

OR O

[1] H

OR

CH3

[2]

C

O

H

C

O

CH3 [3]

+

H OR

CH3

O C

(Repeat Steps [2] and [3].)

15.80 Initiation:

R3SnH

R3Sn

+ Z

Propagation:

+ HZ CH2

Br

+ R3SnBr

R3Sn R3SnH

CH2

R3SnH CH2

CH2

R3SnH

+

+

R3Sn

+

R3Sn

R3Sn

15.81 CO2H O

O

COOH

O

O

O

COOH

O O

A O

COOH

O

O

COOH

O

O OOH + R

O O R H

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397

Conjugation, Resonance, and Dienes 16–1 C Chhaapptteerr 1166:: C Coonnjjuuggaattiioonn,, R Reessoonnaannccee,, aanndd D Diieenneess

C Coonnjjuuggaattiioonn aanndd ddeellooccaalliizzaattiioonn ooff eelleeccttrroonn ddeennssiittyy • The overlap of p orbitals on three or more adjacent atoms allows electron density to delocalize, thus adding stability (16.1). • An allyl carbocation (CH2=CHCH2+) is more stable than a 1o carbocation because of p orbital overlap (16.2). • In any system X=Y–Z:, Z is sp2 hybridized to allow the lone pair to occupy a p orbital, making the system conjugated (16.5).

FFoouurr ccoom mm moonn eexxaam mpplleess ooff rreessoonnaannccee ((1166..33)) [1] The three atom “allyl” system: X

Y Z *

X Y Z *

* = +, –, •, or ••

[2] Conjugated double bonds:

[3] Cations having a positive charge adjacent to a lone pair:

X Y

[4] Double bonds having one atom more electronegative than the other:

X Y

+

+

X Y +

Electronegativity of Y > X

X Y



R Ruulleess oonn eevvaalluuaattiinngg tthhee rreellaattiivvee ““ssttaabbiilliittyy”” ooff rreessoonnaannccee ssttrruuccttuurreess ((1166..44)) [1] Structures with more bonds and fewer charges are more stable. CH3

more stable resonance structure

CH3 C O

+

C O

CH3

CH3

all neutral atoms one more bond

charge separation

[2] Structures in which every atom has an octet are more stable. +

+

CH3 O CH2

CH3 O CH2

more stable resonance structure

All 2nd row elements have an octet.

[3] Structures that place a negative charge on a more electronegative element are more stable. The (–) charge is on the more electronegative O atom. O CH3

C

O CH2

CH3

C

CH2

more stable resonance structure

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Chapter 16–2   TThhee uunnuussuuaall pprrooppeerrttiieess ooff ccoonnjjuuggaatteedd ddiieenneess [1] The C–C  bond joining the two double bonds is unusually short (16.8). [2] Conjugated dienes are more stable than similar isolated dienes. Ho of hydrogenation is smaller for a conjugated diene than for an isolated diene converted to the same product (16.9). [3] The reactions are unusual: • Electrophilic addition affords products of 1,2-addition and 1,4-addition (16.10, 16.11). • Conjugated dienes undergo the Diels–Alder reaction, a reaction that does not occur with isolated dienes (16.12–16.14). [4] Conjugated dienes absorb UV light in the 200–400 nm region. As the number of conjugated
bonds increases, the absorption shifts to longer wavelength (16.15).  R Reeaaccttiioonnss ooff ccoonnjjuuggaatteedd ddiieenneess [1] Electrophilic addition of HX (X = halogen) (16.10–16.11) CH2 CH CH CH2

HX

(1 equiv)

CH2 CH CH H

X

1,2-product kinetic product

• • • •

CH2

+

CH2 CH CH

CH2

H

X

1,4-product thermodynamic product

The mechanism has two steps. Markovnikov’s rule is followed. Addition of H+ forms the more stable allylic carbocation. The 1,2-product is the kinetic product. When H+ adds to the double bond, X– adds to the end of the allylic carbocation to which it is closer (C2 not C4). The kinetic product is formed faster at low temperature. The thermodynamic product has the more substituted, more stable double bond. The thermodynamic product predominates at equilibrium. With 1,3-butadiene, the thermodynamic product is the 1,4-product.

[2] Diels–Alder reaction (16.12–16.14) Z

1,3-diene

• • • • • • •




Z

The three new bonds are labeled in bold.

dienophile

The reaction forms two  and one
bond in a six-membered ring. The reaction is initiated by heat. The mechanism is concerted: all bonds are broken and formed in a single step. The diene must react in the s-cis conformation (16.13A). Electron-withdrawing groups in the dienophile increase the reaction rate (16.13B). The stereochemistry of the dienophile is retained in the product (16.13C). Endo products are preferred (16.13D).

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Conjugation, Resonance, and Dienes 16–3 C Chhaapptteerr 1166:: A Annssw weerrss ttoo PPrroobblleem mss 16.1 Isolated dienes have two double bonds separated by two or more  bonds. Conjugated dienes have two double bonds separated by only one  bond. a.

b.

One
bond separates two double bonds = conjugated diene

c.

Two
bonds separate two double bonds = isolated diene

d.

One
bond separates two double bonds = conjugated diene

Four
bonds separate two double bonds = isolated diene

16.2 isolated isolated isolated

O

OOH

C

OH

5-HPETE conjugated

16.3

a.

Conjugation occurs when there are overlapping p orbitals on three or more adjacent atoms. Double bonds separated by 2  bonds are not conjugated. CH2 CH CH CH CH CH2

All of the carbon atoms are sp2 hybridized. Each
bond is separated by only one  bond. conjugated b. The two
bonds are separated by three  bonds. NOT conjugated

c.

e. O

The two
bonds are separated by only one  bond. conjugated

d.

+ This carbon is not sp2 hybridized. NOT conjugated

+

Three adjacent carbon atoms are sp2 hybridized and have an unhybridized p orbital. conjugated

16.4 Two resonance structures differ only in the placement of electrons. All  bonds stay in the same place. Nonbonded electrons and
bonds can be moved. To draw the hybrid: • Use a dashed line between atoms that have a
bond in one resonance structure and not the other. • Use a  symbol for atoms with a charge or radical in one structure but not the other.

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Chapter 16–4 resonance hybrid: a.


+

+

+

The + charge is delocalized on two carbons.


+

resonance hybrid: b.

+ +


+

The + charge is delocalized on two carbons.


+

The + charge is delocalized on two carbons.


+ resonance hybrid:

c.

+


+

+

Each different kind of carbon atom will give a different 13C signal. When a carbocation is delocalized as in structure B, carbons become equivalent.

16.5


+ The two
+ carbons are identical.

CH3


+ B 4 different kinds of C 4 13C NMR signals

A 5 different kinds of C 5 13C NMR signals

SN1 reactions proceed via a carbocation intermediate. Draw the carbocation formed on loss of Cl and compare. The more stable the carbocation, the faster the SN1 reaction.

16.6

CH3CH2CH2Cl is a 1o halide, which does not react by an SN1 reaction because cleavage of the C–Cl bond forms a highly unstable 1o carbocation.

3-chloro-1-propene CH2 CHCH2Cl more reactive CH2

CH CH2

CH2 CH

CH2

1-chloropropane less reactive

resonance-stabilized carbocation Two resonance structures delocalize the positive charge on 2 C's making 3-chloro-1-propene more reactive.

CH3CH2CH2Cl

CH3CH2CH2

only one Lewis structure very unstable

16.7 a. CH2 CH CH CH CH2

CH2 CH CH CH

CH2

CH2

CH CH CH CH2

Move the charge and the double bond. O

b.

CH3CH2

C

O C

CH3

H

Move the charge and the double bond.

d. Move the charge and the double bond.

CH3CH2

C

+

C H

CH3

c. CH3 CH Cl Move the lone pair.

+

CH3 CH Cl

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Conjugation, Resonance, and Dienes 16–5 16.8 To compare the resonance structures remember: • Resonance structures with more bonds are better. • Resonance structures in which every atom has an octet are better. • Resonance structures with neutral atoms are better than those with charge separation. • Resonance structures that place a negative charge on a more electronegative atom are better. no octet

a. CH3

one more bond + NH2

+ C NH2

CH3 C

CH3

+

+

NH2 + CH3 C

c.

hybrid CH3 least stable most stable one more bond All atoms have an octet. better resonance structure intermediate stability CH3

b.

CH3

C

CH3

NH

least stable

C

O

O

+

one more bond better resonance structure intermediate stability

least stable

+ most stable




O

O

O

O

NH

CH3

negative charge on the more electronegative atom better resonance structure intermediate stability

C




d.

NH

hybrid most stable

N

least stable

–

+

N

one more bond better resonance structure

N

most stable

intermediate stability

16.9 OCH3

OCH3

OCH3

OCH3

largest contribution more bonds and all atoms have octets

16.10 O

a. CH3

C

O OH

A no charges All atoms have an octet. 4 bonds (C–C + C–O) most stable lowest energy resonance structure

CH3

C

O 


O OH

CH3

C

b. OH

+ + OH

C CH3

D

C B 2 charges 2 charges C does not have an octet. All atoms have an octet. 4 bonds (C–C + C–O) 3 bonds (C–C + C–O) intermediate stability least stable intermediate energy highest energy

resonance hybrid lower in energy than any single resonance form

c. In order of increasing energy: D < A < C < B

16.11 Remember that in any allyl system, there must be p orbitals to delocalize the lone pair. CH2

O

a.

O

b.

CH3 C

c. O

sp2 hybridized trigonal planar geometry

sp2 hybridized trigonal planar geometry

sp2 hybridized trigonal planar geometry

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Chapter 16–6 16.12 The s-cis conformation has two double bonds on the same side of the single bond. The s-trans conformation has two double bonds on opposite sides of the single bond. a. (2E,4E)-2,4-octadiene in the s-trans conformation

c. (3Z,5Z)-4,5-dimethyl-3,5-decadiene in both the s-cis and s-trans conformations Z

double bonds on opposite sides s-trans b. (3E,5Z)- 3,5-nonadiene in the s-cis conformation

s-cis

Z

s-trans Z

Z

Z E double bonds on the same side s-cis

16.13 s-cis conjugated OH

conjugated

s-trans

E E

conjugated

isolated

Z Z HO

Z

O

isolated OH

Z isolated

isolated

16.14 Bond length depends on hybridization and percent s-character. Bonds with a higher percent s-character have smaller orbitals and are shorter. HC C C CH

sp hybridized carbons 50% s-character shortest bond

CH2 CH CH CH2

CH3 CH3

sp2 hybridized carbons 33% s-character intermediate length

sp3 hybridized carbons 25% s-character longest bond

16.15 Two equivalent resonance structures delocalize the
bond and the negative charge. O CH3 C

O

CH3

CH3 C O

O 


hybrid:

O

C O 


These bond lengths are equal because they are identical.

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Conjugation, Resonance, and Dienes 16–7 16.16 The less stable (higher energy) diene has the larger heat of hydrogenation. Isolated dienes are higher in energy than conjugated dienes, so they will have a larger heat of hydrogenation. or

a. Double bonds separated by one
bond = conjugated diene smaller heat of hydrogenation

Double bonds separated by two
bonds = isolated diene larger heat of hydrogenation or

b. Double bonds separated by one
bond = conjugated diene smaller heat of hydrogenation

Double bonds separated by two
bonds = isolated diene larger heat of hydrogenation

16.17 Isolated dienes are higher in energy than conjugated dienes. Compare the location of the double bonds in the compounds below. *

*

* 3 conjugated double bonds most stable

* 2 conjugated double bonds intermediate stability

0 conjugated double bonds least stable

*

16.18 Conjugated dienes react with HX to form 1,2- and 1,4-products. a.

HCl

CH3 CH CH CH CH CH3

CH3 CH CH H

Cl

CH CH CH3

Cl

H

isolated diene c.

Cl

HCl

Cl

Cl d.

Cl

HCl

A

B

Cl

1,4-product

1,2-product

HCl

b.

CH3 CH CH CH CH CH3

C

This double bond is more reactive, so C is probably a minor product because it results from HCl addition to the less reactive double bond.

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Chapter 16–8 16.19 The mechanism for addition of DCl has two steps: [1] Addition of D+ forms a resonance-stabilized carbocation. [2] Nucleophilic attack of Cl
forms 1,2- and 1,4-products. [1]

D

D Cl

+ Cl

D

Cl

[2]

[2]

Cl

Cl D

D

16.20 Label the products as 1,2- or 1,4-products. The 1,2-product is the kinetic product, and the 1,4product, which has the more substituted double bond, is the thermodynamic product. This is C1.

The H added here.

The H added here.

CH3

CH3

HCl

CH3

Cl

+

Cl

This is C4.

1,2-product kinetic product

1,4-product thermodynamic product

16.21 To draw the products of a Diels–Alder reaction: [1] Find the 1,3-diene and the dienophile. [2] Arrange them so the diene is on the left and the dienophile is on the right. [3] Cleave three bonds and use arrows to show where the new bonds will be formed. COOH +

a.




re-draw

COOH

COOH

dienophile

diene

CH3

re-draw




+

b.

COOCH3 COOCH3

diene Rotate to make it s-cis.

CH3

+ O

dienophile

CH3

dienophile

re-draw

c.

COOCH3

diene Rotate to make it s-cis.

O




O

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Conjugation, Resonance, and Dienes 16–9 16.22 For a diene to be reactive in a Diels–Alder reaction, a diene must be able to adopt an s-cis conformation. rotate s-trans cannot rotate unreactive

s-cis most reactive The diene is always in the s-cis conformation.

s-cis reactive

16.23

(2Z,4Z)-2,4-hexadiene

(2E,4E)-2,4-hexadiene sterically unhindered more reactive

sterically hindered

Steric interactions between the two CH3 groups make it difficult for the diene to adopt the needed s-cis conformation.

16.24 Electron-withdrawing substituents in the dienophile increase the reaction rate. H

H CH2 CH2

HOOC

COOH

no electron-withdrawing groups least reactive

H C C

CH2 C

one electron-withdrawing group intermediate reactivity

COOH

two electron-withdrawing groups most reactive

16.25 A cis dienophile forms a cis-substituted cyclohexene. A trans dienophile forms a trans-substituted cyclohexene. CH3OOC

a.

COOCH3

+ H

COOCH3




C C

+

H

COOCH3

cis dienophile CH3OOC

b.

COOCH3




+ COOCH3

COOCH3

COOCH3

trans-substituted products

O




+

COOCH3 +

trans dienophile

c.

COOCH3

cis-substituted products

H

H

COOCH3

H

O

H

O

+ O

cis dienophile

H

O

identical

cis-substituted product

H

O

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Chapter 16–10 16.26 The endo product (with the substituents under the plane of the new six-membered ring) is the preferred product. +

a.




CH2 CHCOOCH3

COOCH3 COOCH3




+

b.

endo substituent

COOCH3 both groups endo COOCH3

COOCH3

16.27 To find the diene and dienophile needed to make each of the products: [1] Find the six-membered ring with a C–C double bond. [2] Draw three arrows to work backwards. [3] Follow the arrows to show the diene and dienophile. COOCH2CH3

COOCH2CH3

COOCH2CH3

+

a.

CH3O

CH3O

COOCH3

COOCH3

CH3O

COOCH3

+

b. COOCH3

COOCH3 Cl

Cl

c.

Cl

Cl

H H

H H

CH3OOC

O

O Cl

O

Cl

O

O

O O

O

O

+

16.28 CH3O

CH3O

+

CH3O

CH3O

NC O

O NC H

(+ enantiomer)

H O A

O

16.29 Conjugated molecules absorb light at a longer wavelength than molecules that are not conjugated. a.

or conjugated longer wavelength

b. not conjugated

or

all double bonds conjugated longer wavelength

one set of conjugated dienes

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Conjugation, Resonance, and Dienes 16–11 16.30 Sunscreens contain conjugated systems to absorb UV radiation from sunlight. Look for conjugated systems in the compounds below. O

a.

O

b. CH3O

c.

O

CH3O

OH

conjugated system could be a sunscreen

O

not a conjugated system

conjugated system could be a sunscreen

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Chapter 16–12 16.31 Use the definition from Answer 16.1. CH2=CHC N

3
bonds with only 1  bond between conjugated

1
bond with no adjacent sp2 hybridized atoms NOT conjugated

2 multiple bonds with only 1  bond between conjugated

CH2OCH3

CH2

3
bonds with 2 or more  bonds between NOT conjugated

This C is sp2. 1
bond with an adjacent sp2 hybridized atom

1
bond with no adjacent sp2 hybridized atoms NOT conjugated

The lone pair occupies a p orbital, so there are p orbitals on three adjacent atoms. conjugated

16.32 Use the definition from Answer 16.1. conjugated a.

conjugated

conjugated

isolated

conjugated isolated

c.

b. isolated

isolated

conjugated

isolated

conjugated

16.33 Although 2,3-di-tert-butyl-1,3-butadiene has four adjacent p orbitals, the bulky tert-butyl groups prevent the diene from adopting the s-cis conformation needed for the Diels–Alder reaction. Thus, this diene does not undergo a characteristic reaction of conjugated dienes. 16.34 a. (CH3)2C

CH CH2

(CH3)2C CH CH2

e.

CH3O

CH CH CH2

CH3O

b.

c. d.

CH CH CH2

CH3O CH CH CH2

CH2

CH2

f.

O

O

N(CH3)2

O

N(CH3)2

g.

O

h.

O

O

O

O

O

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Conjugation, Resonance, and Dienes 16–13 16.35 CH2

CH2

CH2

CH2

CH2

a.

OH

OH

OH

OH

OH

b.

CH2

c.

CH2

CH2

CH2

CH2

CH2

CH2

CH2

CH2

CH2

CH2

OCH3

OCH3

OCH3

OCH3

OCH3

d.

CH2 OCH3

16.36

resonance hybrid:





Five resonance structures delocalize the negative charge on five C's making them all equivalent.





All of the carbons are identical in the anion.




16.37 The N atom of C6H5CH2NH2 is surrounded by four groups and is sp3 hybridized. Although the N atom of C6H5NH2 is surrounded by four groups, it is also bonded to a benzene ring. To be conjugated with the benzene ring, the N atom must be sp2 hybridized and its lone pairs must occupy a p orbital. In this way the lone pair can be delocalized, as shown in one resonance structure. NH2

sp2 conjugated

NH2

[+ three other resonance structures]

NH2

sp3

The N atom is not conjugated with the benzene ring.

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Chapter 16–14 16.38 CH2

a.

CH CH2

H

CH3CH2CH2 H

more acidic CH2 CH CH2

CH2 CH

less acidic CH3CH2CH2

CH2

only one Lewis structure

Resonance stabilization delocalizes the negative charge on 2 C's after loss of a proton. This makes propene more acidic than propane. b. Draw the products of cleavage of the bond.

ethane CH3 CH3

CH3

+

1-butene CH3 CH2CH=CH2

CH3

CH3

+

CH2 CH CH2

CH2 CH CH2

One resonance-stabilized radical forms. This makes the bond dissociation energy lower because a more stable radical is formed.

Two unstable radicals form.

16.39 Use the directions from Answer 16.12. a. (3Z)-1,3-pentadiene in the s-trans conformation

c. (2E,4E,6E)-2,4,6-octatriene

d. (2E,4E)-3-methyl-2,4-hexadiene in the s-cis conformation

double bonds on opposite sides s-trans b. (2E,4Z)-1-bromo-3-methyl-2,4-hexadiene Br

s-cis

16.40 E

Z C

1,4-pentadiene

(3E)-1,3-pentadiene

(3Z)-1,3-pentadiene

2-methyl-1,3-butadiene C

C

3-methyl-1,2-butadiene

2,3-pentadiene

16.41

2E,4E

2E,4Z

2Z,4E

2Z,4Z

1,2-pentadiene

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Conjugation, Resonance, and Dienes 16–15 16.42 and

a.

and

c.

Z

E (3E)-1,3,5-hexatriene (3E)-1,3,5-hexatriene both s-cis both s-trans different conformations b.

(3E)-1,3,5-hexatriene

(3Z)-1,3,5-hexatriene

different stereoisomers

and (3Z)-1,3,5-hexatriene (3Z)-1,3,5-hexatriene both s-cis both s-trans different conformations

16.43 Use the directions from Answer 16.16 and recall that more substituted double bonds are more stable. Increasing heat of hydrogenation

conjugated diene one tetra-, one disubstituted double bond smallest heat of hydrogenation

conjugated diene one di-, one trisubstituted double bond smaller intermediate heat of hydrogenation

isolated diene one di-, one trisubstituted double bond larger intermediate heat of hydrogenation

isolated diene both disubstituted double bonds largest heat of hydrogenation

16.44 Conjugated dienes react with HX to form 1,2- and 1,4-products. Br

a.

HBr

major product, formed by addition of HBr to the more substituted C=C

(1 equiv) isolated diene b.

Br

HBr

Br

(E and Z isomers can form.)

(1 equiv) 1,2-product

1,4-product Br

Br

c.

Br

HBr

(1 equiv)

Br

1,2-product

1,4-product

1,2-product

(E and Z isomers) 1,4-product

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Chapter 16–16 16.45 This cation forms because it is benzylic and resonance stabilized.

CH CHCH3

A

CH CH2CH3

H Br

Br

CH CH2CH3

CH CH2CH3

CH CH2CH3

CH CH2CH3

Br CHCH2CH3

C H CH2CH CH2

C–CH

CH3

H

B

2o

H Br

CH CH2CH3

1,2H shift

Br

Br

CHCH2CH3

C

This 2o carbocation is also benzylic, making it resonance stabilized, as above.

carbocation

16.46 To draw the mechanism for reaction of a diene with HBr and ROOR, recall from Chapter 15 that when an alkene is treated with HBr under these radical conditions, the Br ends up on the carbon with more H’s to begin with. RO OR

RO

OR

HOR +

H Br

Br

Use each resonance structure to react with HBr.

Br

Br

H Br Br

Br Br

H Br

Br

Br Br

16.47 C2 CH2

a. and b.

HCl

X H adds here at C1.

CH3 Cl

Y Cl added at C2. 1,2-product kinetic product

CH3

C4 Cl

Z

Cl added at C4. 1,4-product thermodynamic product

Y is the kinetic product because of the proximity effect. H and Cl add across two adjacent atoms. Z is the thermodynamic product because it has a more stable trisubstituted double bond.

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Conjugation, Resonance, and Dienes 16–17 If addition occurred at the other C=C, the following allylic carbocation would form:

Addition occurs at the labeled double bond due to the stability of the carbocation intermediate. c.

CH2

CH3

CH3

CH2

The two resonance structures for this allylic cation are 3o and 2o carbocations. more stable intermediate Addition occurs here.

CH2

The two resonance structures for this allylic cation are 1o and 2o carbocations. less stable

16.48 Addition of HCl at the terminal double bond forms a carbocation that is highly resonance stabilized since it is both allylic and benzylic. Such stabilization does not occur when HCl is added to the other double bond. This gives rise to two products of electrophilic addition. Cl H Cl

Cl

1,2-product

Cl

1,4-product

Cl

(+ three more resonance structures that delocalize the positive charge onto the benzene ring)

16.49 There are two possible products: disubstituted C=C (CH3)2C H

CH CH C(CH3)2

(CH3)2C

Br

CH CH C(CH3)2

H

Br

trisubstituted C=C 1,2-product

1,4-product

The 1,2-product is always the kinetic product because of the proximity effect. In this case, it is also the thermodynamic (more stable) product because it contains a more highly substituted C=C (trisubstituted) than the 1,4-product (disubstituted). Thus, the 1,2-product is the major product at high and low temperature. 16.50 The electron pairs on O can be donated to the double bond through resonance. This increases the electron density of the double bond, making it less electrophilic and therefore less reactive in a Diels–Alder reaction. CH2 CH OCH3

CH2

+

CH OCH3

methyl vinyl ether This C now bears a net negative charge.

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Chapter 16–18 16.51 Use the directions from Answer 16.21.


a.

re-draw dienophile

diene

COOCH3


b. Cl

diene

COOCH3

Cl

Cl

trans-substituted products

trans dienophile COOCH3

c.

COOCH3

COOCH3

COOCH3

Cl

Cl


Cl

diene

cis dienophile

cis-substituted products

O H O




d.

H

diene

dienophile

endo ring

e.


O

re-draw

O

H

endo substituent dienophile

diene

O

O

f.

H

H




+

H

O

H

Both C=C's of the dienophile react so two Diels–Alder reactions take place.

or

excess

H O

diene

O

O

H

H

O

H

dienophile

16.52 Use the directions from Answer 16.27. CH3

a.

CH3

COOCH3

COOCH3

COOCH3

COOCH3

COOCH3

b.

+ CH3

O

c.

CH3

COOCH3

CH3

CH3

O H

H

COOCH3

COOCH3

O + COOCH3

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Conjugation, Resonance, and Dienes 16–19 +

d.

identical Cl

Cl O

e.

O

O C CH3

Cl

CH3

O

O

O

f.

O

+

O C

C

CH3

O

+

O

O

O

O

O O

16.53 O

O

This pathway is preferred because the dienophile has electronwithdrawing C=O groups that make it more reactive. O

O

O

O CH2

+

CH2 O

O

no electron-withdrawing groups less reactive

16.54 COOCH3

a. diene

COOCH3




+ HC C COOCH3

dienophile CO2CH3




+ CH3O2C C C CO2CH3

b. diene

dienophile

CO2CH3

CO2CH3 CO2CH3

416

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Chapter 16–20 16.55 OCH3 OCH3

+

CHO




CHO

OCH3

+ CHO

C

OCH3

H

1,2-disubstituted product major

O C

OCH3

1,3-disubstituted product minor

The major product is formed when the circled carbons with a + and – react.

H

O

resonance hybrids:

+OCH3



+OCH3

O

 

+



+ H

+H

+



For the 1,2-product, carbons with unlike charges would react. This is favored because the electron-rich and the electronpoor C's can bond.

O 

For the 1,3-product, there are no partial charges of opposite sign on reacting carbons. This arrangement is less attractive.

16.56 COOH




+

COOH COOH COOH

These are the only two double bonds that are conjugated and have the s-cis conformation needed for a Diels–Alder reaction.

16.57 H C +




Cl Cl

Cl

Cl

X

Cl

Cl

Y Cl

mCPBA

Cl Cl

1 equiv

Cl Cl

Cl

Cl Cl




C H

Cl

Cl

Cl

Cl Z

aldrin

O

dieldrin

This double bond is more electron rich, so it is epoxidized more readily.

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417

Conjugation, Resonance, and Dienes 16–21 16.58 In each problem, the synthesis must begin with the preparation of cyclopentadiene from dicyclopentadiene.




2

dicyclopentadiene

a.

HO

[1] OsO4

COOCH3

cyclopentadiene

HO

[2] NaHSO3, H2O




COOCH3

COOCH3

O




O

b.

O

mCPBA

O

O

O

c.

O

O

O

O

H2 (excess)

CHO




[1] NaH

Pd-C

CHO

[2] CH3CH2CH2CH2Br OH

O

16.59 O

re-draw




O

a.

diene

O

dienophile COOCH3




re-draw

b.

COOCH3

COOCH3

dienophile

diene

16.60 A transannular Diels–Alder reaction forms a tricyclic product from a monocyclic starting material.


O O

O O

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Chapter 16–22 16.61 CH3CH CHCH2OH

CH3CH CHCH2 OH2

+ Br

CH3CH CHCH2

two resonance structures

H Br

+ H2O

CH3CH CHCH2Br

+ Br CH3CHCH CH2

CH3CHCH CH2

Br + Br

16.62 a.

HCl

(CH3)2C=CHCH2CH2CH=CH2

(CH3)2C(Cl)CH2CH2CH2CH=CH2

isolated diene

(CH3)2C=CHCH2CH2CHClCH3

major product

minor product

I

b.

HI

I

conjugated diene

1,4-product

1,2-product O

O H


c.

O

+

O

H

O

O O

O

diene

dienophile COOCH3

+

d. diene

COOH




+ COOH

COOH

diene

cis dienophile HBr

f.

COOCH3 +

trans dienophile

+

e.

COOCH3




COOH

COOH

COOH

Br

Br

conjugated diene 1,2-product

1,4-product

16.63 The mechanism is E1, with formation of a resonance-stabilized carbocation.

OH

OH2 H

H A

A

H2O

A

H A

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Conjugation, Resonance, and Dienes 16–23 16.64 Cl


2

a.
1

loss of H (from
1 C ) + Cl conjugated more stable major product

loss of H (from
2 C) + Cl more substituted

b. Dehydrohalogentaion generally forms the more stable product. In this reaction, loss of H from the
1 carbon forms a more stable conjugated diene, so this product is preferred even though it does not contain the more substituted C=C.

16.65 singlet at 1.42 ppm H

O

mCPBA

Each H is a doublet of doublets in the 5.2–5.4 ppm region.

H

isoprene

H H

H

doublet of doublets at 5.5–5.7 ppm

two doublets at 2.7–2.9 ppm

16.66 1 H: doublet of doublets at 6.0 ppm The IR shows an OH absorption at 3200–3600 cm–1.

H Br

H2O

H

B H

Each H is a doublet of doublets in the 4.9–5.2 ppm region.

OH

OH peak at 1.5 ppm 6 H: singlet at 1.3 ppm

16.67

isolated diene shortest wavelength 1

2 conjugated bonds intermediate wavelength 2

3 conjugated bonds intermediate wavelength 3

4 conjugated bonds longest wavelength 4

16.68 O

The phenol makes ferulic acid an antioxidant. Loss of H forms a highly stabilized phenoxy radical that inhibits radical formation during oxidation.

C HO OCH3

ferulic acid

OH

The highly conjugated
system makes it a sunscreen.

420

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Chapter 16–24 16.69 O

+

H

H

O O

H

O3 cleaves the C=C.

O

[1] O3

H O




O O

[2] (CH3)2S

O

H

H

O

O

O

The Diels–Alder reaction establishes the stereochemistry of the four carbons on the sixmembered ring. All four carbon atoms bonded to the six-membered ring are on the same side.

Endo product formed.

16.70 O O

O

CH3 CH3




+ O




+

H

O

Diels–Alder reaction

O

H COOCH3

CH3O2C

COOCH3

A

O

loss of CO2

O C O

B

16.71 diene 1st

Diels–Alder reaction


C

CO2CH3 CO2CH3 CO2CH3

CO2CH3

+

CH3O2C C C CO2CH3

Either of these alkenes becomes the dienophile 2nd Diels–Alder for an intramolecular Diels– reaction Alder reaction in a second step.

D

CO2CH3

+

CO2CH3

two products C16H16O4

16.72 Retro Diels–Alder reaction forms a conjugated diene. Intramolecular Diels–Alder reaction then forms N. CO2CH3

CO2CH3 CO2CH3

NOCH3 HN




NOCH3 NOCH3

retro Diels–Alder HN

M

intramolecular Diels–Alder reaction

HN

N

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421

Benzene and Aromatic Compounds 17–1 C Chhaapptteerr 1177:: B Beennzzeennee aanndd A Arroom maattiicc C Coom mppoouunnddss  C Coom mppaarriinngg aarroom maattiicc,, aannttiiaarroom maattiicc,, aanndd nnoonnaarroom maattiicc ccoom mppoouunnddss ((1177..77)) •

Aromatic compound

•

A cyclic, planar, completely conjugated compound that contains 4n + 2
electrons (n = 0, 1, 2, 3, and so forth). An aromatic compound is more stable than a similar acyclic compound having the same number of
electrons.

• •

Antiaromatic compound

•

A cyclic, planar, completely conjugated compound that contains 4n
electrons (n = 0, 1, 2, 3, and so forth). An antiaromatic compound is less stable than a similar acyclic compound having the same number of
electrons.

• •

A compound that is not aromatic

•

A compound that lacks one (or more) of the requirements to be aromatic or antiaromatic.

  PPrrooppeerrttiieess ooff aarroom maattiicc ccoom mppoouunnddss • Every carbon has a p orbital to delocalize electron density (17.2). • They are unusually stable. Ho for hydrogenation is much less than expected, given the number of degrees of unsaturation (17.6). • They do not undergo the usual addition reactions of alkenes (17.6). • 1H NMR spectra show highly deshielded protons because of ring currents (17.4).   EExxaam mpplleess ooff aarroom maattiicc ccoom mppoouunnddss w wiitthh 66

eelleeccttrroonnss ((1177..88))

benzene

N

N H

pyridine

pyrrole

+

cyclopentadienyl anion

tropylium cation

  EExxaam mpplleess ooff ccoom mppoouunnddss tthhaatt aarree nnoott aarroom maattiicc ((1177..88))

not cyclic

not planar

not completely conjugated

422

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Chapter 17–2 C Chhaapptteerr 1177:: A Annssw weerrss ttoo PPrroobblleem mss 17.1

Move the electrons in the
bonds to draw all major resonance structures. N(CH3)2

O

N(CH3)2

O

O

N(CH3)2

O

N(CH3)2

diphenhydramine

17.2

Look at the hybridization of the atoms involved in each bond. Carbons in a benzene ring are surrounded by three groups and are sp2 hybridized. Csp2–Csp3

a.

b.

H

Csp2–Csp2

Csp2–H1s

Csp2–Csp2

Cp–Cp

Csp2–Csp2 Cp–Cp

shortest of all the indicated bonds in (a) and (b)

17.3 • • •

To name a benzene ring with one substituent, name the substituent and add the word benzene. To name a disubstituted ring, select the correct prefix (ortho = 1,2; meta = 1,3; para = 1,4) and alphabetize the substituents. Use a common name if it is a derivative of that monosubstituted benzene. To name a polysubstituted ring, number the ring to give the lowest possible numbers and then follow other rules of nomenclature. isopropyl group

a.

OH 1

3

isopropylbenzene

b. I 4

2

c.

PhCH(CH3)2

1 CH2CH3

Two groups are 1,3 = meta.

butyl group

phenol m-butylphenol CH3 1 2 Br

ethyl

2 3

iodo Two groups are 1,4 = para. p-ethyliodobenzene

2-bromo

d. Cl 5

5-chloro

toluene (CH3 group must be at the "1" position, if the molecule is named as a toluene derivative.)

2-bromo-5-chlorotoluene

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423

Benzene and Aromatic Compounds 17–3

17.4

Work backwards to draw the structures from the names. a. isobutylbenzene

c. cis-1,2-diphenylcyclohexane

e. 4-chloro-1,2-diethylbenzene

isobutyl group

Cl

b. o-dichlorobenzene

f. 3-tert-butyl-2-ethyltoluene

d. m-bromoaniline Br

Cl Cl NH2

aniline

17.5 Cl Cl Cl

Cl

Cl

Cl

Cl

Cl

1,2,3-trichlorobenzene

Cl

1,2,4-trichlorobenzene

1,3,5-trichlorobenzene

17.6 Molecular formula C10H14O2: 4 degrees of unsaturation IR absorption at 3150–2850 cm–1: sp2 and sp3 hybridized C–H bonds NMR absorptions (ppm): 1.4 (triplet, 6 H) O O 4.0 (quartet, 4 H) 6.8 (singlet, 4 H)

17.7

Count the different types of carbons to determine the number of 13C NMR signals. Ce Cf Ca

a.

Cb

CH2CH3

Cd Ca

Cc

Cb Cc

c. Cd

b. Cl

Cb 4 types of C's in the benzene ring 6 signals

17.8

Cc Cb

CH3

Cd Cc Cb

C C Ca b c 4 signals

All C's are different. 7 signals

Each of the three isomeric trichlorobenzenes exhibits a different number of 13C NMR signals. Cd

Cc

Cc

Ca Cl

Ca Cl

Cl

Cb 4 signals

Cc Cb

Cd Cl

Ca Cl

Cf

Cl

Ce 6 signals

Cl

Cb

Ca

Cb Ca

Cl

Ca

Cl

Cb 2 signals

424

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Chapter 17–4 17.9

The less stable compound has a larger heat of hydrogenation. CH3

CH2

A

B no benzene ring, less stable larger
H°

benzene ring, more stable smaller
H°

17.10 The protons on sp2 hybridized carbons in aromatic hydrocarbons are highly deshielded and absorb at 6.5–8 ppm whereas hydrocarbons that are not aromatic show an absorption at 4.5–6 ppm, typical of protons bonded to the C=C of an alkene. H

H

H

H

H

a. H

H H

not aromatic alkene H's ~ 4.5–6 ppm

H

H

c.

b.

H H

aromatic ring H's ~ 6.5–8 ppm

aromatic ring H's ~ 6.5–8 ppm

17.11 To be aromatic, a ring must have 4n + 2
electrons. 16
e 4n 4(4) = 16 antiaromatic

20
e 4n 4(5) = 20 antiaromatic

22
e 4n + 2 4(5) + 2 = 22 aromatic

17.12 Annulenes have alternating double and single bonds. An odd number of carbon atoms in the ring would mean there would be two adjacent single bonds. Therefore an annulene having an odd number of carbon atoms cannot exist. 17.13

17.14 In determining if a heterocycle is aromatic, count a nonbonded electron pair if it makes the ring aromatic in calculating 4n + 2. Lone pairs on atoms already part of a multiple bond cannot be delocalized in a ring, and so they are never counted in determining aromaticity. O

a.

b. O

Count one lone pair from O. 4n + 2 = 4(1) + 2 = 6 aromatic

+

O

no lone pair from O 4n + 2 = 4(1) + 2 = 6 aromatic

c.

d. O

N

N

With one lone Both N atoms are part of a double bond, so the lone pair from each O there would be 8 electrons. pairs cannot be counted: there are 6 electrons. If O's are sp3 hybridized, the 4n + 2 = 4(1) + 2 = 6 ring is not completely aromatic conjugated. not aromatic

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Benzene and Aromatic Compounds 17–5

17.15 H

quinine (antimalarial drug)

N

HO H

N is sp3 hybridized and the lone pair is in an sp3 hybrid orbital.

CH3O N

N is sp2 hybridized and the lone pair is not part of the aromatic ring. This means it occupies an sp2 hybrid orbital.

17.16 a. The five-membered ring is aromatic because it has 6
electrons, two from each
bond and two from the N atom that is not part of a double bond.

F

N

2 N

N F

CF3 N

CF3

N N

N

sp2 hybridized N lone pair in sp2 orbital

sp2 hybridized N lone pair in sp2 orbital

NH2 O

F

N

F

sp3 hybridized N lone pair in sp3 orbital

2

NH2 O

sp2 hybridized N lone pair in p orbital F

2 F

sp2 hybridized N lone pair in p orbital

b. and c.

sitagliptin

17.17

17.18 Compare the conjugate base of 1,3,5-cycloheptatriene with the conjugate base of cyclopentadiene. Remember that the compound with the more stable conjugate base will have a lower pKa. B

B

H H

1,3,5-cycloheptatriene pKa = 39

H

8
Electrons make this conjugate base especially unstable (antiaromatic).

Since the conjugate base is unstable, the pKa of 1,3,5-cycloheptatriene is high.

H H

cyclopentadiene

H

6
electrons aromatic conjugate base very stable anion

Since the conjugate base is very stable, the pKa of cyclopentadiene is much lower.

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Chapter 17–6 17.19 The compound with the most stable conjugate base is the most acidic.

Conjugate bases: no resonance delocalization

2 resonance structures

aromatic conjugate base most stable

most unstable base so least acidic acid

The acid is intermediate in acidity.

The acid is the most acidic.

17.20

17.21 To be aromatic, the ions must have 4n + 2
electrons. Ions in (b) and (c) do not have the right number of
electrons to be aromatic. 2
electrons 4(0) + 2 = 2 aromatic

a.

d.

10
electrons 4(2) + 2 = 10 aromatic

17.22 absorbs at 7.6 ppm H

A

=

The NMR indicates that A is aromatic. The C’s of the triple bond are sp hybridized. Each triple bond has one set of electrons in p orbitals that overlap with other p orbitals on adjacent atoms in the ring. This overlap allows electrons to delocalize. Each C of the triple bonds also has a p orbital in the plane of the ring. The electrons in these p orbitals are localized between the C’s of the triple bond, and not delocalized in the ring. Although A has 24
e– total, only 18 e– are delocalized around the ring.

2 antibonding MOs 1 bonding MO

+

17.23 In using the inscribed polygon method, always draw the vertex pointing down.

2
electrons All bonding MOs are filled. aromatic

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Benzene and Aromatic Compounds 17–7

17.24 Draw the inscribed pentagons with the vertex pointing down. Then draw the molecular orbitals (MOs) and add the electrons. Cation:

Radical:

2 antibonding MOs 3 bonding MOs 4
electrons Not all bonding MOs are filled. not aromatic

5
electrons Not all bonding MOs are filled. not aromatic

17.25 C60 would exhibit only one 13C NMR signal because all the carbons are identical. 17.26 a. If the Kekulé description of benzene was accurate, only one product would form in Reaction [1], but there would be four (not three) dibromobenzenes (A–D), because adjacent C–C bonds are different—one is single and one is double. Thus, compounds A and B would not be identical. A has two Br’s bonded to the same double bond, but B has two Br’s on different double bonds. b. In the resonance description, only one product would form in Reaction [1], since all C’s are identical, but only three dibromobenzenes (ortho, meta, and para isomers) are possible. A and B are identical because each C–C bond is identical and intermediate in bond length between a C–C single and C–C double bond. Br

[1]

Br

Br Br

[2]

Br

Br

Br

Br Br

A

B

C

D

17.27

propylbenzene

1,2,3-trimethylbenzene

isopropylbenzene

o-ethyltoluene

p-ethyltoluene

m-ethyltoluene

1,2,4-trimethylbenzene

1,3,5-trimethylbenzene

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Chapter 17–8 17.28 C8H10:

Br

Br

Br

C8H9Br:

Br

Br

Br

Br

Br

Br

3 isomers

2 isomers

3 isomers

1 isomer

17.29 To name the compounds use the directions from Answer 17.3. NH2

d.

a.

CH3CH2

aniline

CH2CH2CH3

g. Cl

sec-butylbenzene

Br

b.

e.

m-chloroethylbenzene Cl

toluene p-chlorotoluene

h.

Br NH2

CH2CH3

CH3

1-ethyl-3-isopropyl-5-propylbenzene

Br

Cl

c.

CH(CH3)2

o-chloroaniline

aniline

2,3-dibromoaniline OH

f.

NO2 NO2

2,5-dinitrophenol

phenol (OH at C1)

Ph

cis-1-bromo-2-phenylcyclohexane

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Benzene and Aromatic Compounds 17–9

17.30 a. p-dichlorobenzene

d. o-bromonitrobenzene

Cl

g. 2-phenyl-2-propen-1-ol

Br

Cl

OH

NO2

b. m-chlorophenol

e. 2,6-dimethoxytoluene

h. trans-1-benzyl-3-phenylcyclopentane

OCH3

Cl

CH3 OCH3

OH

c. p-iodoaniline

or

f. 2-phenyl-1-butene

I H2N

17.31 a. constitutional isomers of molecular formula C8H9Cl, and b. names of the trisubstituted benzenes Cl Cl

Cl Cl

Cl

stereoisomers for this isomer

Cl

Cl

Cl

Cl Cl

Cl

2-chloro-1,3-dimethylbenzene

1-chloro-2,3-dimethylbenzene

4-chloro-1,2-dimethylbenzene

Cl Cl Cl

1-chloro-2,4-dimethylbenzene c. stereoisomers Cl

1-chloro-3,5-dimethylbenzene

2-chloro-1,4-dimethylbenzene

Cl

17.32 Count the electrons in the
bonds. Each
bond holds two electrons.

a.

b.

10
electrons

7
electrons

c.

d.

10
electrons

e.

14
electrons

12
electrons

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Chapter 17–10 17.33 To be aromatic, the compounds must be cyclic, planar, completely conjugated, and have 4n + 2
electrons. Circled C's are not sp2. not completely conjugated not aromatic

a.

14
electrons in outer ring aromatic

b.

4 benzene rings joined together aromatic

c.

Circled C is not sp2. not completely conjugated not aromatic

d.

12
electrons does not have 4n + 2
electrons not aromatic

e.

12
electrons does not have 4n + 2
electrons not aromatic

f.

17.34 In determining if a heterocycle is aromatic, count a nonbonded electron pair if it makes the ring aromatic in calculating 4n +2. Lone pairs on atoms already part of a multiple bond cannot be delocalized in a ring, and so they are never counted in determining aromaticity. S

O

a.

c. O

6
electrons counting a lone pair from S 4(1) + 2 = 6 aromatic

not aromatic

not aromatic

N

N

b.

H N

f.

h.

10
electrons 4(2) + 2 = 10 aromatic

6
electrons, counting a lone pair from O 4(1) + 2 = 6 aromatic

17.35 Circled C's are not sp2. not aromatic

a.

c.

Circled C is not sp2. not aromatic

d.

4
electrons 4(1) = 4 antiaromatic

O

b.

10
electrons in 10-membered ring 4(2) + 2 = 10 aromatic

Count these 2e–.

N N

N

6
electrons counting a lone pair from O 4(1) + 2 = 6 aromatic

N

6
electrons counting the lone pair from N 4(1) + 2 = 6 aromatic

O

d. O

g.

e.

10
electrons 4(2) + 2 = 10 aromatic

These lone pairs are on doubly bonded N atoms, so they can't be counted.

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Benzene and Aromatic Compounds 17–11 17.36 6
electrons in this ring

6
electrons in this ring

+

A

A resonance structure can be drawn for A that places a negative charge in the five-membered ring and a positive charge in the seven-membered ring. This resonance structure shows that each ring has 6
electrons, making it aromatic. The molecule possesses a dipole such that the sevenmembered ring is electron deficient and the five-membered ring is electron rich. 17.37 Each compound is completely conjugated. A compound with 4n + 2
electrons is especially stable, while a compound with 4n
electrons is especially unstable.

pentalene

azulene

8
electrons 4(2) = 8 antiaromatic unstable

heptalene

10
electrons 4(2) + 2 = 10 aromatic very stable

12
electrons 6(2) = 12 antiaromatic unstable

17.38 Benzene has C–C bonds of equal length, intermediate between a C–C double and single bond. Cyclooctatetraene is not planar and not aromatic so its double bonds are localized. cyclooctatetraene

a

c b

6
electrons: aromatic all bonds of equal length intermediate

d not aromatic longer single bond localized double bond: shorter

17.39 N

N N

N H

purine

d 4.2

more acidic lower pKa pKa < 4.2

Increasing acidity

 O Otthheerr ffaaccttss • Extraction is a useful technique for separating compounds having different solubility properties. Carboxylic acids can be separated from other organic compounds by extraction, because aqueous base converts a carboxylic acid into a water-soluble carboxylate anion (19.12). • A sulfonic acid (RSO3H) is a strong acid because it forms a weak, resonance-stabilized conjugate base on deprotonation (19.13). • Amino acids have an amino group on the  carbon to the carboxy group [RCH(NH2)COOH]. Amino acids exist as zwitterions at pH
6. Adding acid forms a species with a net (+1) charge [RCH(NH3)COOH]+. Adding base forms a species with a net (–1) charge [RCH(NH2)COO]– (19.14).

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Carboxylic Acids and the Acidity of the O–H Bond 19–3 C Chhaapptteerr 1199:: A Annssw weerrss ttoo PPrroobblleem mss To name a carboxylic acid: [1] Find the longest chain containing the COOH group and change the -e ending to -oic acid. [2] Number the chain to put the COOH carbon at C1, but omit the number from the name. [3] Follow all other rules of nomenclature.

19.1

3 a.

2

4 H

CH3

c.

CH3CH2CH2 C CH2COOH CH3 2

H

CH3CH2 C CH2

1

CH3CH2

1

C COOH CH2CH3

Number the chain to put COOH at C1. 6 carbon chain = hexanoic acid 2,4-diethylhexanoic acid

Number the chain to put COOH at C1. 6 carbon chain = hexanoic acid 3,3-dimethylhexanoic acid

O

4 b.

6

H

CH3 C CH2CH2COOH

d.

1 Cl Number the chain to put COOH at C1. 5 carbon chain = pentanoic acid 4-chloropentanoic acid

4

1 OH

8 9

Number the chain to put COOH at C1. 9 carbon chain = nonanoic acid 4-isopropyl-6,8-dimethylnonanoic acid

19.2 a. 2-bromobutanoic acid

c. 3,3,4-trimethylheptanoic acid

O

O

e. 3,4-diethylcyclohexanecarboxylic acid O

HO

OH HO

Br

b. 2,3-dimethylpentanoic acid

d. 2-sec-butyl-4,4-diethylnonanoic acid

O

O

HO

f. 1-isopropylcyclobutanecarboxylic acid COOH

OH

19.3 O

a.
-methoxyvaleric acid




OH

c.
,-dimethylcaproic acid



OCH3

O O

b. -phenylpropionic acid



OH




d.
-chloro--methylbutyric acid OH

Cl



OH


O

481

482

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Chapter 19–4 19.4 C5 or  carbon OH

a.

OH

C2 or
carbon b.

CO2H

CO2H

HO

C3 or  carbon IUPAC: 2-hydroxypropanoic acid common:
-hydroxypropionic acid

IUPAC: 3,5-dihydroxy-3-methylpentanoic acid common: ,-dihydroxy--methylvaleric acid

19.5 O

O

a.

O Li+

lithium benzoate

b. Na+ O

O

O

c.

H

sodium formate or sodium methanoate

O

d.

K+

O Br

potassium 2-methylpropanoate

sodium 4-bromo-6-ethyloctanoate

19.6 COO–Na+

COOH

C2 2-propylpentanoic acid

19.7

19.8

sodium 2-propylpentanoate

More polar molecules have a higher boiling point and are more water soluble. COOCH3

CH2CH2CH2OH

least polar lowest boiling point least H2O soluble

intermediate polarity intermediate boiling point

CH2COOH

most polar highest boiling point most H2O soluble

Look for functional group differences to distinguish the compounds by IR. Besides sp3 hybridized C–H bonds at 3000–2850 cm–1 (which all three compounds have), the following functional group absorptions are seen: O CH3CH2CH2CH2

C

OH

O OH

Na+

CH3CH2CH2CH2

C

OCH3

carboxylic acid 2 strong absorptions ~1710 (C=O) ~2500–3500 (OH) cm–1

ester 1 strong absorption ~1700 (C=O) cm–1

Molecular formula: C4H8O2 one degree of unsaturation

1

O

alcohol 1 strong absorption ~3600–3200 (OH) cm–1

19.9 H NMR data (ppm): 0.95 (triplet, 3 H) 1.65 (multiplet, 2 H) 2.30 (triplet, 2 H) 11.8 (singlet, 1 H)

O HO

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Carboxylic Acids and the Acidity of the O–H Bond 19–5 19.10 O H

O

O OH

HO

OH H H

Hb

Ha

Hc

Hc Hd Hd

2 singlets 2 singlets 1:1 ratio 1:1 ratio Although both compounds have an absorption at 10–12 ppm in their 1H NMR spectra (due to Hb and Hc), Ha, which is bonded directly to the carbonyl carbon, is much farther downfield than Hd because it is more deshielded.

19.11 HO

HO COOH

HO

There are five tetrahedral stereogenic centers. Both double bonds can exhibit cis–trans isomerism. Therefore, there are 27 = 128 stereoisomers.

COOH

HO

OH

PGF2
a prostaglandin

OH

enantiomer

19.12 1° Alcohols are converted to carboxylic acids by oxidation reactions. O

a.

OH

c.

OH

COOH

CH2OH

O

b.

(CH3)2CH

C

(CH3)2CH

OH

CH2 OH

19.13 a.

CH2OH

Na2Cr2O7

A

b.

CH3C CCH3

B

c. O2N

COOH

H2SO4, H2O

KMnO4

CH3

O2N

COOH

C (Any R group with benzylic H's can be present para to NO2.) OH

[1] O3 [2] H2O

2° OH

d.

CH3COOH (2 equiv)

D

O

CrO3

OH

CO2H

H2SO4, H2O

1° OH

19.14 a.

b.

COOH

CH3

CH3

NaOH

OH

NaOCH3

COO

CH3

Na+ + H2 O

O

Na+

+ HOCH3

d.

CH3

NaH

c. CH3 C OH CH3

COOH

CH3

C O Na+ + H2 CH3

NaHCO3

COO

Na+

+ H2CO3

484

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Chapter 19–6 19.15 CH3COOH has a pKa of 4.8. Any base with a conjugate acid with a pKa higher than 4.8 can deprotonate it. a. F– pKa (HF) = 3.2 not strong enough b. (CH3)3CO– pKa [(CH3)3COH]= 18 strong enough c. CH3– pKa (CH4) = 50 strong enough

d. –NH2 pKa (NH3) = 38 strong enough e. Cl– pKa (HCl) = –7.0 not strong enough

19.16 Ha

H OHb

Increasing acidity: Ha < Hb < Hc OHc

H OHb – Ha

O

OHc

negative charge on C unstable conjugate base

O

mandelic acid Ha

H O

– Hb

OHc O Ha

Ha

H OHb

– Hc

negative charge on O more stable conjugate base H OHb

O O

O O

negative charge on O, resonance stabilized most stable conjugate base

19.17 Electron-withdrawing groups make an acid more acidic, lowering its pKa. CH3CH2 COOH

least acidic pKa = 4.9

ICH2

COOH

CF3 COOH

one electron-withdrawing group intermediate acidity pKa = 3.2

three electron-withdrawing F's most acidic pKa = 0.2

19.18 Acetic acid has an electron-donating methyl group bonded to the carboxy group. The CH3 group both stabilizes the acid and destabilizes the nearby negative charge on the conjugate base, making CH3COOH less acidic (with a higher pKa) than HCOOH. O

electron-donating CH3 group

CH3




+ C OH

acetic acid CH3 stabilizes the partial positive charge.

O CH3

C

O–

conjugate base CH3 destabilizes the negative charge. a less stable conjugate base

19.19 a. CH3COOH

least acidic

HSCH2COOH

HOCH2COOH

intermediate acidity

most acidic

b. ICH2CH2COOH

least acidic

ICH2COOH

I2CHCOOH

intermediate acidity

most acidic

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Carboxylic Acids and the Acidity of the O–H Bond 19–7 19.20 a.

CH3

COOH

least acidic

COOH

Cl

intermediate acidity

COOH

most acidic O

b. CH3O

COOH

CH3

COOH

C

COOH

CH3

least acidic

intermediate acidity

most acidic

19.21 OH

Phenol A has a higher pKa than phenol because of its substituents. Both the OH and CH3 are electron-donating groups, which make the conjugate base less stable. Therefore, the acid is less acidic.

HO

A

19.22 To separate compounds by an extraction procedure, they must have different solubility properties. a. CH3(CH2)6COOH and CH3CH2CH2CH2CH=CH2: YES. The acid can be extracted into aqueous base, while the alkene will remain in an organic layer. b. CH3CH2CH2CH2CH=CH2 and (CH3CH2CH2)2O: NO. Both compounds are soluble in organic solvents and insoluble in water. Neither is acidic enough to be extracted into aqueous base. c. CH3(CH2)6COOH and NaCl: one carboxylic acid, one salt: YES. The carboxylic acid is soluble in an organic solvent while the salt is soluble in water. d. NaCl and KCl: two salts: NO. 19.23 To separate compounds by an aqueous extraction technique, compounds must have different solubility properties. CH3CH2COOH and CH3CH2CH2OH are low molecular weight organic compounds that can hydrogen bond to water, so they are water soluble. They also both dissolve in organic solvents. As a result, they are inseparable because of their similar solubility properties. 19.24 weaker conjugate base better leaving group CF3SO3H

CF3 is electron withdrawing. stronger acid lower pKa

CF3SO3–

stronger conjugate base worse leaving group CH3SO3H

CH3 is electron donating. weaker acid higher pKa

CH3SO3–

485

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Chapter 19–8 19.25 phenylalanine COOH C H2N

methionine

COOH

H

H

COOH

C

C NH2

H2N

COOH

H CH2CH2SCH3

H CH3SCH2CH2

C

R

R

NH2

S

S

19.26 Since amino acids exist as zwitterions (i.e., salts), they are too polar to be soluble in organic solvents like diethyl ether. Thus, they are soluble in water. 19.27 COOH H3N C H

COO

COO

H 3N C H

H2N C H

H

H

H

pH = 1

glycine

pH = 11

neutral form

19.28 COO

pI =

pKa(COOH) + pKa(NH3+)

(2.58) + (9.24)

=

2

2

= 5.91

H3N C H CH2

19.29 +

+ H3N CH COOH H

electron-withdrawing group

H3N CH COO– H

The nearby (+) stabilizes the conjugate base by an electron-withdrawing inductive effect, thus making the starting acid more acidic.

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Carboxylic Acids and the Acidity of the O–H Bond 19–9 19.30 Use the directions from Answer 19.1 to name the compounds. a.

(CH3)2CHCH2CH2CO2H

b.

BrCH2COOH

COOH

4-methylpentanoic acid

g.

o-bromobenzoic acid

2-bromoacetic acid or 2-bromoethanoic acid

Br

O

h.

c.

CH3CH2

COOH p-ethylbenzoic acid

OH

4,4,5,5-tetramethyloctanoic acid d.

–

+

O

lithium butanoate

CH3CH2CH2COO Li

O– Na+

i.

sodium 2-methylhexanoate 1-ethylcyclopentanecarboxylic acid

e.

COOH

COOH

10

j.

3 7

2,4-dimethylcyclohexanecarboxylic acid

f.

5

7-ethyl-5-isopropyl-3-methyldecanoic acid

COOH

19.31 OH

f. o-chlorobenzoic acid

a. 3,3-dimethylpentanoic acid

COOH

O Cl OH

b. 4-chloro-3-phenylheptanoic acid

CH3

O

c. (2R)-2-chloropropanoic acid

Cl

O

K+

O

h. sodium
-bromobutyrate

Cl

Cl

e. m-hydroxybenzoic acid

C

O

OH

Na+

Br

O

d. ,-dichloropropionic acid

Cl

g. potassium acetate O

i. 2,2-dichloropentanedioic acid O

O

O

HO OH

HOOC

OH Cl Cl

j. 4-isopropyl-2-methyloctanedioic acid OH

O HO

OH O

19.32 O

O OH

pentanoic acid

OH

3-methylbutanoic acid

O– Na+

sodium pentanoate

O– Na+

sodium 3-methylbutanoate

OH

OH

2-methylbutanoic acid

2,2-dimethylpropanoic acid O

O

O

O

O

O

O– Na+

sodium 2-methylbutanoate

O– Na+

sodium 2,2-dimethylpropanoate

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Chapter 19–10 19.33 O

a.

OH

OH O

lowest boiling point

intermediate boiling point

O

highest boiling point O

OH

b. HO

lowest boiling point

intermediate boiling point

highest boiling point

19.34 O OH

a.

CrO3

OH

[1] O3

c.

KMnO4

CH3

+ CO2

[2] H2O

H2SO4, H2O

b. (CH3)2CH

COOH

C C H

HOOC

d. CH3(CH2)6CH2OH

COOH

Na2Cr2O7

CH3(CH2)6COOH

H2SO4, H2O

19.35 [1] BH3

a.

CH2

[2] H2O2, OH

B

[1] NaNH2

HC CH

HC CCH3

[2] CH3I

C

[2] CH3CH2I

CH(CH3)2

(CH3)2CHCl

c.

COOH

H2SO4, H2O

A

[1] NaNH2

b.

CrO3

CH2OH

–

CH3CH2C CCH3

[1] O3

CH3CH2COOH

[2] H2O

D

CH3COOH

E + F

COOH

KMnO4

AlCl3

G

H

19.36 Bases: [1] –OH pKa (H2O) = 15.7; [2] CH3CH2– pKa (CH3CH3) = 50; [3] –NH2 pKa (NH3) = 38; [4] NH3 pKa (NH4+) = 9.4; [5] HC
C– pKa (HC
CH) = 25.

a.

CH3

COOH

pKa = 4.3 All of the bases can deprotonate this.

b.

–OH,

Cl

c. (CH3)3COH

OH

pKa = 9.4 CH3CH2–, –NH2, and HC
C– can deprotonate this.

pKa = 18 CH3CH2–, –NH2, and HC
C– can deprotonate this.

19.37 a.

COOH

K+


OC(CH

COO
K+

3)3

+ HOC(CH3)3 pKa = 18

Reaction favors products.

pKa = 4.2 OH

b.

pKa
16

NH3

O

NH4+

pKa = 9.4

Reaction favors reactants.

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Carboxylic Acids and the Acidity of the O–H Bond 19–11 Na+ NH2

+

OH

c.

+ NH3 + Na+

O

pKa = 10 d.

Reaction favors products.

pKa = 38 CH3–Li+

COOH

COO Li+

CH3

CH4

Reaction favors products.

pKa = 50

CH3

pKa
4 e.

Na+H

OH

Na+

O

Reaction favors products.

pKa = 35

pKa
16 CH3

f.

H2

OH

Na2CO3

O– Na+

CH3

With the same pKa for the starting acid and the conjugate acid, an equal amount of starting materials and products is present.

Na+ HCO3

pKa = 10.2 pKa = 10.2

19.38 The stronger acid has a lower pKa and a weaker conjugate base. COOH

CH2OH

a.

or

c.

carboxylic acid stronger acid lower pKa weaker conjugate base

CH3

or

COOH

Cl

Cl is electron withdrawing. stronger acid lower pKa weaker conjugate base

CH3 is electron donating. weaker acid higher pKa stronger conjugate base

alcohol weaker acid higher pKa stronger conjugate base

or ClCH2COOH FCH2COOH F is more electronegative. weaker acid stronger acid higher pKa lower pKa stronger conjugate base weaker conjugate base b.

d.

or

NCCH2COOH

CN is electron withdrawing. stronger acid lower pKa weaker conjugate base

CH3COOH

weaker acid higher pKa stronger conjugate base

19.39 Br

a.

COOH

least acidic

COOH

OH

b.

CH3

COOH

Cl more electronegative most acidic

OH Cl

least acidic

OH O2N

most acidic

intermediate acidity

COOH

c.

Cl

Br is electronegative intermediate acidity

COOH

CH3

COOH CF3

least acidic

intermediate acidity

OH

most acidic

OH

OH

d. Br

least acidic

O2N

intermediate acidity

O2N

NO2

most acidic

COOH

489

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Chapter 19–12 19.40 O

a.

BrCH2COO–

BrCH2CH2

weakest base

COO–

(CH3)3

intermediate basicity

CCOO–

strongest base

O2N

weakest base b.

O

NH

weakest base

O

O

c. intermediate basicity

strongest base

CH2

intermediate basicity

strongest base

19.41 Increasing acidity ICH2COOH

pKa values

BrCH2COOH

least acidic 3.12

FCH2COOH

2.86

2.66

F2CHCOOH

F3CCOOH

most acidic 0.28

1.24

19.42 The OH of the phenol group in morphine is more acidic than the OH of the alcohol (pKa
10 versus pKa
16). KOH is basic enough to remove the phenolic OH, the most acidic proton. most acidic proton HO The OH is part of a phenol. Methylation occurs here.

O

O

[1] KOH

H

an alcohol

H

HO

N

[2] CH3I

H

H

HO

codeine

N CH3

HO

CH3

O

H

H

HO

N CH3

O

O

O H

H

N

Many resonance structures stabilize the conjugate base.

O

O

O

HO

morphine

CH3O

O H

CH3

O

H

N CH3

H HO

O H

N CH3

H HO

H

N CH3

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Carboxylic Acids and the Acidity of the O–H Bond 19–13 19.43 a. The negative charge on the conjugate base of p-nitrophenol is delocalized on the NO2 group, stabilizing the conjugate base, and making p-nitrophenol more acidic than phenol (where the negative charge is delocalized only around the benzene ring). OH O + N O

p-nitrophenol pKa = 7.2

O

O O + N

O +

O

O

OH

O

N

phenol pKa = 10

two of the possible resonance structures for the conjugate base (See part b. for all the possible resonance structures.)

b. In the para isomer, the negative charge of the conjugate base is delocalized over both the benzene ring and onto the NO2 group, whereas in the meta isomer it cannot be delocalized onto the NO2 group. This makes the conjugate base from the para isomer more highly resonance stabilized, and the para substituted phenol more acidic than its meta isomer. OH O2N

O

O

O2N

O O

O2N

pKa = 7.2 p-nitrophenol

O

N O

N

O O2N

O

negative charge on two O atoms very good resonance structure more stable conjugate base O2N stronger acid

OH

NO2

O

O

O

NO2

O

NO2

O

NO2

NO2

O

O

NO2

pKa = 8.3 m-nitrophenol

19.44 A CH3O group has an electron-withdrawing inductive effect and an electron-donating resonance effect. In 2-methoxyacetic acid, the OCH3 group is bonded to an sp3 hybridized C, so there is no way to donate electron density by resonance. The CH3O group withdraws electron density because of the electronegative O atom, stabilizing the conjugate base, and making CH3OCH2COOH a stronger acid than CH3COOH. – H+

O CH3O

OH

more acidic acid

O CH3O

O

more stable conjugate base

492

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Chapter 19–14 In p-methoxybenzoic acid, the CH3O group is bonded to an sp2 hybridized C, so it can donate electron density by a resonance effect. This destabilizes the conjugate base, making the starting material less acidic than C6H5COOH. O

– H+

CH3O

O

O CH3O

CH3O

OH

O

O

less acidic acid

like charges nearby less stable conjugate base

19.45 The O in A is more electronegative than the N in C so there is a stronger electron-withdrawing inductive effect. This stabilizes the conjugate base of A, making A more acidic than C. CO2H CO2H

O

A pKa = 3.2

O

N H

B pKa = 3.9

C

CO2H

pKa = 4.4

Since the O in A is closer to the COOH group than the O atom in B, there is a stronger electron-withdrawing inductive effect. This makes A more acidic than B.

19.46 CO2H

CO2H O2N

O2N

D

C

E –

H+

–

O

CO2H

H+

– H+

O O

O

O O2N

O2N

O

Since the benzene ring is bonded to the
carbon (not the carbonyl carbon), this compound is not much different than any alkylsubstituted carboxylic acid. least acidic

The electron withdrawingSince the NO2 group is bonded to a inductive effect of the NO2 benzene ring that is bonded directly to the group helps stabilize the COO– carbonyl group, inductive effects and resonance effects stabilize the conjugate group. base. For example, a resonance intermediate acidity structure can be drawn that places a (+) charge close to the COO– group. Two of the resonance structures for the conjugate base of C: most acidic O

O

O

N O

O N

O

O

O

unlike charges nearby stabilizing

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Carboxylic Acids and the Acidity of the O–H Bond 19–15 19.47 O CH3

C * OH

NaOH CH3

O

O

C * O

C * O

CH3 +

H3O+

H3O

labeled O atom

OH

O CH3

The resonance-stabilized carboxylate anion can now be protonated on either O atom, the one with the label and the one without the label.

C * OH

CH3

C * O

The label is now in two different locations.

19.48 O

a.

Ha

O

Hc

Hb

loss of Hb:

Ha

Hc

The most acidic proton forms the most stable conjugate base.

O O

1,3-cyclohexanedione increasing acidity: Hb < Ha < Hc O

one Lewis structure least stable conjugate base

O

loss of Ha:

Hc

Hb

O Hc

O

Hb

O

loss of Hc: H a

O

O

Ha

Hb

O

2 resonance structures intermediate stability

Ha

Hb

O

Hb

O

3 resonance structures most stable conjugate base

Hb N

N

b.

Hc O

Ha

N

Hc

loss of Hb: O

Ha

O

Ha

acetanilide N

N

Hc O

Ha

Ha

Hb

Hb

N

N

N

O

one Lewis structure least stable conjugate base

Ha

N

Hc O

Hb Hc

Hc O

7 resonance structures most stable conjugate base

increasing acidity: Ha < Hc < Hb

loss of Ha:

N

Hc

Ha N

loss of Hc: Ha

O

Ha

2 resonance structures that delocalize the negative charge intermediate stability

Hc O

O

Ha

Hc O

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Chapter 19–16 19.49 O

O

O

H

H

H

H H

O

O H

H

H

HO

OH

O

O H

H

H

H

HO

O

H

The conjugate base is resonance stabilized. Two of the structures place a negative charge on an O atom. weaker conjugate base stronger acid

The conjugate base has only one Lewis structure. stronger conjugate base weaker acid

O

19.50 As usual, compare the stability of the conjugate bases. With RSO3H, loss of a proton forms a conjugate base that has three resonance structures, all of which are equivalent and place a negative charge on a more electronegative O atom. With the conjugate base of RCOOH, there are only two of these resonance structures. Thus, the conjugate base RSO3– is more highly resonance stabilized than RCOO–, so RSO3H is a stronger acid than RCOOH. O

O

O

O

R S O

R S O

R S O

O

O

O

O

O

base

R S O H O O R

C

base O H

R

C

O

R

C

three resonance structures for the conjugate base

two resonance structures for the conjugate base

O

19.51 The negatively charged C is more nucleophilic than the negatively charged O atom.

CH3COOH

CH2COO–

strong base (2 equiv)

CH3CH2CH2CH2 Br

X

CH3CH2CH2CH2CH2COO– H3O+

Two equivalents of strong base remove both the O–H and C–H protons.

CH3CH2CH2CH2CH2COOH

hexanoic acid

19.52 CH3

O

O

C

C

NH2

acetamide

CH3

C

N

CH3

N

CH3

C

O O

H

O is more electronegative than N, making the conjugate base of CH3COOH more stable than the conjugate base of acetamide. Therefore, acetamide is less acidic.

O OH

C

somewhat less stable with the (–) charge on N

O CH3

O H

CH3

C

O

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Carboxylic Acids and the Acidity of the O–H Bond 19–17 19.53 COOH

A

• •

B

•

Dissolve both compounds in CH2Cl2. Add 10% NaHCO3 solution. This makes a carboxylate anion (C10H7COO–) from B, which dissolves in the aqueous layer. The other compound (A) remains in the CH2Cl2. Separate the layers.

19.54 OH

OH

and

• •

•

Dissolve both compounds in CH2Cl2. Add 10% NaOH solution. This converts C6H5OH into a phenoxide anion, C6H5O–, which dissolves in the aqueous solution. The alcohol remains in the organic layer (neutral) since it is not acidic enough to be deprotonated to any significant extent by NaOH. Separate the layers.

19.55 To separate two compounds in an aqueous extraction, one must be water soluble (or be able to be converted into a water-soluble ionic compound by an acid–base reaction), and the other insoluble. 1-Octanol has greater than 5 C’s, making it insoluble in water. Octane is an alkane, also insoluble in water. Neither compound is acidic enough to be deprotonated by a base in aqueous solution. Since their solubility properties are similar, they cannot be separated by an extraction procedure. 19.56 O C one double bond or ring a. Molecular formula: C3H5ClO2 ClCH2CH2 OH C=O and O–H IR: 3500–2500 cm–1, 1714 cm–1 NMR data: 2.87 (triplet, 2 H), 3.76 (triplet, 2 H), and 11.8 (singlet, 1 H) ppm

b. Molecular formula: C8H8O3 5 double bonds or rings CH3O COOH IR: 3500–2500 cm–1, 1688 cm–1 C=O and O–H NMR data: 3.8 (singlet, 3 H), 7.0 (doublet, 2 H), 7.9 (doublet, 2 H), and 12.7 (singlet, 1 H) ppm para disubstituted benzene ring 5 double bonds or rings c. Molecular formula: C8H8O3 OCH2COOH IR: 3500–2500 cm–1, 1710 cm–1 C=O and O–H NMR data: 4.7 (singlet, 2 H), 6.9–7.3 (multiplet, 5 H), and 11.3 (singlet, 1 H) ppm monosubstituted benzene ring

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Chapter 19–18 19.57 Compound A: Molecular formula C4H8O2 (one degree of unsaturation) IR absorptions at 3600–3200 (O–H), 3000–2800 (C–H), and 1700 (C=O) cm–1 1 H NMR data: Absorption singlet singlet triplet triplet

ppm 2.2 2.55 2.7 3.9

# of H’s 3 1 2 2

Explanation a CH3 group 1 H adjacent to none or OH 2 H’s adjacent to 2 H’s 2 H’s adjacent to 2 H’s

Structure: O CH3

C

A

CH2CH2OH

Compound B: Molecular formula C4H8O2 (one degree of unsaturation) IR absorptions at 3500–2500 (O–H) and 1700 (C=O) cm–1 1 H NMR data: Absorption doublet septet singlet (very broad)

ppm 1.6 2.3 10.7

# of H’s 6 1 1

Explanation 6 H’s adjacent to 1 H 1 H adjacent to 6 H’s OH of RCOOH

Structure: CH3 CH3 C COOH H

B

19.58 Compound C: Molecular formula C4H8O3 (one degree of unsaturation) IR absorptions at 3600–2500 (O–H) and 1734 (C=O) cm–1 1 H NMR data: Absorption triplet quartet singlet singlet

ppm 1.2 3.6 4.1 11.3

# of H’s 3 2 2 1

Explanation a CH3 group adjacent to 2 H’s 2 H’s adjacent to 3 H’s 2 H’s OH of COOH

Structure: O O

OH

C

19.59 Compound D: Molecular formula C9H9ClO2 (five degrees of unsaturation) C NMR data: 30, 36, 128, 130, 133, 139, 179 = 7 different types of C’s 1 H NMR data: 13

Absorption triplet triplet two signals singlet

ppm 2.7 2.9 7.2 11.7

# of H’s 2 2 4 1

Explanation 2 H’s adjacent to 2 H’s 2 H’s adjacent to 2 H’s on benzene ring OH of COOH

Structure: COOH Cl

19.60 Molecular formula C6H12O2 (1 double bond due to COOH) 1H

NMR: 1.1 (singlet), 2.2 (singlet), and 11.9 (singlet) ppm

CH3 CH3 C CH2COOH CH3

D

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Carboxylic Acids and the Acidity of the O–H Bond 19–19 19.61 Molecular formula: C8H6O4: 6 degrees of unsaturation IR 1692 cm–1 (C=O) 1H NMR 8.2 and 10.0 ppm (singlets)

O

O

HO

OH

COOH

aromatic H

19.62 A

COOH

COOH

B

O

C

OH

3 different C's Spectrum [2]: peaks at 27, 39, 186 ppm 5 different C's Spectrum [1]: peaks at 14, 22, 27, 34, 181 ppm 4 different C's Spectrum [3]: peaks at 22, 26, 43, 180 ppm

19.63 GBL: Molecular formula C4H6O2 (two degrees of unsaturation) IR absorption at 1770 (C=O) cm–1 1 H NMR data: Absorption multiplet triplet triplet

ppm 2.28 2.48 4.35

# of H’s 2 2 2

Explanation 2 H’s adjacent to several H’s 2 H’s adjacent to 2 H’s 2 H’s adjacent to 2 H’s

Structure: O O

GBL

19.64 HOOC

H C

threonine

OH

HOOC

H

C

H H2N

C H2N

CH3

OH

HOOC

C CH3

H

HOOC

C

H H2N

2S,3S

2R,3S

OH H C

C CH3

H2N

H

C

OH H CH3

2S,3R naturally occurring

2R,3R

19.65 NH

NH

COOH

NH2

COOH

proline

enantiomer

O

O

COO

zwitterion

19.66 a. methionine O H3N CH C OH

b. serine

O

O

H3N CH C O

H2N CH C O

CH2

CH2

CH2

CH2

CH2

CH2

CH2SCH3

CH2SCH3

CH2SCH3

OH

OH

OH

pH = 1

pH = 6 form at isoelectric point

pH = 11

H3N CH C OH

pH = 1

H3N CH C O

O H2N CH C O

pH = 6 form at isoelectric point

pH = 11

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Chapter 19–20 19.67 pKa(COOH) + pKa(NH3+)

a. cysteine pI =

= (2.05) + (10.25) / 2 = 6.15

2

b. methionine pI =

pKa(COOH) + pKa(NH3+)

= (2.28) + (9.21) / 2 = 5.75

2

19.68 O

O H2N

C

C OH

lysine This lone pair is localized on the N atom, making it a base.

H2N H

N H

H2N H

OH

tryptophan This lone pair is delocalized in the
system to give 10
electrons, making it aromatic. This is similar to pyrrole (Chapter 17). Since these electrons are delocalized in the aromatic system, this N atom in tryptophan is not basic.

19.69 The first equivalent of NH3 acts as a base to remove a proton from the carboxylic acid. A second equivalent then acts as a nucleophile to displace X to form the ammonium salt of the amino acid. O R

R

OH X

O

O

NH3

acid–base reaction

NH3

O– NH4+

SN2 reaction

X

R

O– NH4+ NH3

19.70 a. At pH = 1, the net charge is (+1). NH3

b. increasing pH: As base is added, the most acidic proton is removed first, then the next most acidic proton, and so forth. NH3

C

HOOCCH2CH2

COOH H

COOH H

base (1 equiv) NH3 C

HOOCCH2CH2

COO–

H

base (2nd equiv) NH3 C

OOCCH2CH2

COO–

H

base (3rd equiv) NH2 C

OOCCH2CH2 H

NH3 C

OOCCH2CH2 H

C

HOOCCH2CH2

c. monosodium glutamate

COO–

COO– Na+

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Carboxylic Acids and the Acidity of the O–H Bond 19–21 19.71 The first equivalent of NaH removes the most acidic proton; that is, the OH proton on the phenol. The resulting phenoxide can then act as a nucleophile to displace I to form a substitution product. With two equivalents, both OH protons are removed. In this case the more nucleophilic O atom is the stronger base; that is, the alkoxide derived from the alcohol (not the phenoxide), so this negatively charged O atom reacts first in a nucleophilic substitution reaction. nucleophile most acidic proton

NaH

–

O

(CH2)4OH

–O

(CH2)4O–

(1 equiv) HO

[1] CH3I

CH3O

(CH2)4OH

[2] H2O

(CH2)4OH NaH (2 equiv)

[1] CH3I

HO

(CH2)4OCH3

[2] H2O

nucleophile

19.72 O HO

COOH

HO

C O

p-hydroxybenzoic acid less acidic than benzoic acid

+

HO

O C O

like charges on nearby atoms destabilizing The OH group donates electron density by its resonance effect and this destabilizes the conjugate base, making the acid less acidic than benzoic acid. O H

OH

O COOH

C O

o-hydroxybenzoic acid more acidic than benzoic acid

Intramolecular hydrogen bonding stabilizes the conjugate base, making the acid more acidic than benzoic acid.

19.73

O

Hd OHe

HaO HbO Hc O

2-hydroxybutanedioic acid increasing acidity: Hd < Hc < Hb < He < Ha

Ha and He must be the two most acidic protons since they are part of carboxylic acids. Loss of a proton forms a resonance-stabilized carboxylate anion that has the negative charge delocalized on two O atoms. Ha is more acidic than He because the nearby OH group on the
carbon increases acidity by an electron-withdrawing inductive effect. Hb is the next most acidic proton because the conjugate base places a negative charge on the electronegative O atom, but it is not resonance stabilized. The least acidic H’s are Hc and Hd since these H’s are bonded to C atoms. The electronegative O atom further acidifies Hc by an electron-withdrawing inductive effect.

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501

Introduction to Carbonyl Chemistry 20–1 C Chhaapptteerr 2200:: IInnttrroodduuccttiioonn ttoo C Caarrbboonnyyll C Chheem miissttrryy  R Reedduuccttiioonn rreeaaccttiioonnss [1] Reduction of aldehydes and ketones to 1o and 2o alcohols (20.4) O R

C

OH

NaBH4, CH3OH H(R')

R C H(R')

or

H

[1] LiAlH4 [2] H2O

1o or 2o alcohol

or

H2, Pd-C

[2] Reduction of
,-unsaturated aldehydes and ketones (20.4C) OH NaBH4 CH3OH

• reduction of the C=O only R

O

O H2 (1 equiv)

R

Pd-C

• reduction of the C=C only

R

OH H2 (excess) Pd-C

• reduction of both
bonds

R

[3] Enantioselective ketone reduction (20.6) [1] (S)- or (R)CBS reagent

O C

R

HO H C

[2] H2O

H OH

R

(R) 2o alcohol

C

or

R

•

A single enantiomer is formed.

(S) 2o alcohol

[4] Reduction of acid chlorides (20.7A) [1] LiAlH4 [2] H2O O R

C

RCH2OH

1o alcohol

•

LiAlH4, a strong reducing agent, reduces an acid chloride to a 1o alcohol.

•

With LiAlH[OC(CH3)3]3, a milder reducing agent, reduction stops at the aldehyde stage.

Cl O

[1] LiAlH[OC(CH3)3]3 [2] H2O

R

C

H

aldehyde

502

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Chapter 20–2 [5] Reduction of esters (20.7A) [1] LiAlH4

RCH2OH

[2] H2O

1o alcohol

O R

C

•

LiAlH4, a strong reducing agent, reduces an ester to a 1o alcohol.

•

With DIBAL-H, a milder reducing agent, reduction stops at the aldehyde stage.

OR' O

[1] DIBAL-H

C

R

[2] H2O

H

aldehyde

[6] Reduction of carboxylic acids to 1o alcohols (20.7B) O R

C

[1] LiAlH4 OH

[2] H2O

RCH2OH

1o alcohol

[7] Reduction of amides to amines (20.7B) O R

C

[1] LiAlH4 [2] H2O

N

RCH2 N

amine



O Oxxiiddaattiioonn rreeaaccttiioonnss Oxidation of aldehydes to carboxylic acids (20.8) O R

C

O

CrO3, Na2Cr2O7, K2Cr2O7, KMnO4 H

or Ag2O, NH4OH

C

R

OH

All Cr6+ reagents except PCC oxidize RCHO to RCOOH. Tollens reagent (Ag2O + NH4OH) oxidizes RCHO only. Primary (1°) and secondary (2°) alcohols do not react with Tollens reagent.

•

carboxylic acid


PPrreeppaarraattiioonn ooff oorrggaannoom meettaalllliicc rreeaaggeennttss ((2200..99)) [1] Organolithium reagents: R X [2] Grignard reagents:

•

R X

+

2 Li

R Li

+ Mg

(CH3CH2)2O

[3] Organocuprate reagents:

R X 2 R Li

+

2 Li

+ CuI

+

LiX

R Mg X

R Li + LiX R2Cu Li+

+

LiI

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503

Introduction to Carbonyl Chemistry 20–3 [4] Lithium and sodium acetylides:

Na+ –NH2

R C C H

R C C

Na+

+

NH3

a sodium acetylide R Li

R C C H

R C C Li

+ R

H

a lithium acetylide



R Reeaaccttiioonnss w wiitthh oorrggaannoom meettaalllliicc rreeaaggeennttss [1] Reaction as a base (20.9C) R M

+

H O R

R H

M+

+

• •

O R

RM = RLi, RMgX, R2CuLi This acid–base reaction occurs with H2O, ROH, RNH2, R2NH, RSH, RCOOH, RCONH2, and RCONHR.

[2] Reaction with aldehydes and ketones to form 1o, 2o, and 3o alcohols (20.10) O R

C

OH

[1] R"MgX or R"Li

R C H (R')

[2] H2O

H (R')

R"

1o, 2o, or 3o alcohol

[3] Reaction with esters to form 3o alcohols (20.13A) [1] R"Li or R"MgX (2 equiv)

O R

C

OR'

OH R C R"

[2] H2O

R"

3o alcohol

[4] Reaction with acid chlorides (20.13) [1] R"Li or R"MgX (2 equiv) [2] H2O O R

C

OH

•

More reactive organometallic reagents—R"Li and R"MgX—add two equivalents of R" to an acid chloride to form a 3o alcohol with two identical R" groups.

•

Less reactive organometallic reagents— R'2CuLi—add only one equivalent of R' to an acid chloride to form a ketone.

R C R" R"

3o alcohol Cl O [1] R'2CuLi [2] H2O

R

C

R'

ketone

504

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Chapter 20–4 [5] Reaction with carbon dioxide—Carboxylation (20.14A) O

[1] CO2

R MgX

R C

[2] H3O+

OH

carboxylic acid

[6] Reaction with epoxides (20.14B) OH

O C

[1] RLi, RMgX, or R2CuLi

C

C

[2] H2O

C

R

alcohol

[7] Reaction with
,-unsaturated aldehydes and ketones (20.15B)

[1] R'Li or R'MgX

OH R

[2] H2O C

C

C

C

•

More reactive organometallic reagents— R'Li and R'MgX—react with
,unsaturated carbonyls by 1,2-addition.

•

Less reactive organometallic reagents— R'2CuLi— react with
,-unsaturated carbonyls by 1,4-addition.

R'

O R

C

allylic alcohol

C

O H C C R

[1] R'2CuLi [2] H2O

C R'

ketone

  PPrrootteeccttiinngg ggrroouuppss ((2200..1122)) [1] Protecting an alcohol as a tert-butyldimethylsilyl ether CH3 R O H

+ Cl

CH3

Si C(CH3)3 CH3

R O Si C(CH3)3 N

CH3

NH

R O TBDMS

Cl TBDMS

tert-butyldimethylsilyl ether

[2] Deprotecting a tert-butyldimethylsilyl ether to re-form an alcohol CH3 R O Si C(CH3)3

CH3

(CH3CH2CH2CH2)4N+ F– R O H

+

F Si C(CH3)3

CH3 R O TBDMS

CH3 F

TBDMS

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505

Introduction to Carbonyl Chemistry 20–5 C Chhaapptteerr 2200:: A Annssw weerrss ttoo PPrroobblleem mss 20.1 [1] Csp2–Csp2

a. [3] Csp2–Csp2

20.2

[2] : C –O
: Cp–Op sp2

O

-sinensal

sp2

A carbonyl compound with a reasonable leaving group undergoes substitution reactions. Those without good leaving groups undergo addition. O

a. CH3

C

O

b. CH3

CH3CH2CH2

no good leaving group addition reactions

20.3

b. The O is sp2 hybridized. Both lone pairs occupy sp2 hybrid orbitals.

H

C

O

O

c. Cl

CH3

Cl–good leaving group substitution reactions

C

C

d. OCH3

OCH3–reasonable leaving group substitution reactions

H

no good leaving group addition reactions

A carbonyl compound with a reasonable leaving group (NR2 or OR bonded to the C=O) undergoes substitution reactions. Those without good leaving groups undergo addition. O

no good leaving group addition reactions

O O

O

O

OH N

O

H

All other C=O's have leaving groups. substitution reactions

OH HO O

O

H O

O O

20.4

Aldehydes are more reactive than ketones. In carbonyl compounds with leaving groups, the better the leaving group, the more reactive the carbonyl compound. O

a.

CH3CH2CH2

C

O H

or

CH3CH2CH2

C

O CH3

c.

O CH3CH2

C

O

or CH3

less hindered carbonyl more reactive

C

Cl

or

CH3

C

OCH3

better leaving group more reactive

less hindered carbonyl more reactive b.

CH3CH2

O

CH3CH(CH3)

C

O

d. CH2CH3

CH3

C

O

or OCH3

better leaving group more reactive

CH3

C

NHCH3

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Chapter 20–6 NaBH4 reduces aldehydes to 1° alcohols, and ketones to 2° alcohols.

20.5

O a.

CH3CH2CH2

OH

NaBH4

C

H

CH3CH2CH2 C H

CH3OH

NaBH4

c.

H

CH3OH

O

OH

NaBH4 O

b.

OH

CH3OH

1° Alcohols are prepared from aldehydes and 2° alcohols are from ketones.

20.6

H OH

O

a.

OH

b.

O

OH

O

c.

20.7 OH

3° Alcohols cannot be made by reduction of a carbonyl group, because they do not contain a H on the C with the OH.

1-methylcyclohexanol

20.8 O

a.

O

OH

[1] LiAlH4

d.

[2] H2O O

Pd-C

OH

NaBH4

O

e.

b.

O

O

H2 (1 equiv)

c.

NaBH4 (excess)

NaBD4

f.

CH3OH

Pd-C

20.9 O

OH

NaBH4

a.

OH

CH3OH NaBH4

CHO

b.

CH3OH

c. (CH3)3C

O

OH NaBH4 CH3OH

(CH3)3C

OH

(CH3)3C

20.10 O

HO H Cl

A

Cl

[1] (S)-CBS reagent [2] H2O

OH

CH3OH

CH3OH O

OH H2 (excess)

B

OH

D OH

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Introduction to Carbonyl Chemistry 20–7 20.11 Part [1]: Nucleophilic substitution of H for Cl O

O

C

CH3

Cl

[1]

O

CH3 C Cl

[2]

H

H3Al H

C

CH3

H replaces Cl.

aldehyde

+ AlH3

Cl–

+

H

Part [2]: Nucleophilic addition of H– to form an alcohol O

O CH3

C

H

[3]

H3Al H

OH

H OH

CH3 C H

[4]

H + AlH3

+ –OH

CH3 C H H

1o alcohol

20.12 Acid chlorides and esters can be reduced to 1° alcohols. Keep the carbon skeleton the same in drawing an ester and acid chloride precursor. O O CH2OH

a.

C

or Cl

OH

CH2OH

OCH3

Cl

b.

C

O

C

O

CH3O

or O

c.

OCH3

CH3O

or

Cl

C

O

OCH3

CH3O

20.13 O

a.

O

[1] LiAlH4 OH

C

c.

OH

[2] H2O

N(CH3)2

[1] LiAlH4

CH2N(CH3)2

[2] H2O O

O

b.

[1] LiAlH4 NH2

NH2

[2] H2O

d.

NH

[1] LiAlH4

NH

[2] H2O

20.14 O CH2NH2

a.

C

O NH2

c.

O N

b.

CH2CH3

CH2CH3

C

O N

CH2CH3

CH2CH3

or

N CH2CH3

N H

N H

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Chapter 20–8 20.15 O COOCH3

a.

[1] LiAlH4

O

b. CH3O

[1] LiAlH4

CH3O

OH

CH3O

OH

[2] H2O

+ HOCH3 OH

NaBH4

COOCH3

CH3OH O

O

c.

OH

[2] H2O NaBH4

CH3O

OH

CH3OH

[1] LiAlH4 OH [2] H2O NaBH4 CH3OH

HO

OH

+ HOCH3

Neither functional group reduced

20.16 CO2CH2CH3

a. O

b.

CO2CH2CH3 O

CO2CH2CH3

c.

H2

CO2CH2CH3

(1 equiv) Pd-C

O

H2

CO2CH2CH3

(2 equiv) Pd-C

OH

[1] LiAlH4 OH

[2] H2O

O

d.

+ CH3CH2OH

OH

CO2CH2CH3

NaBH4

CO2CH2CH3

CH3OH O

OH

20.17 Tollens reagent reacts only with aldehydes. Ag2O, NH4OH

a.

b.

CH2OH Na2Cr2O7 H2SO4, H2O

COOH

Ag2O, NH4OH

OH

No reaction

OH

O C

CHO O Na2Cr2O7 H2SO4, H2O

OH

O C

OH

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Introduction to Carbonyl Chemistry 20–9 20.18 O OH CHO

c. B

CHO

PCC O

HO

B

H

OH

OH CH2OH

NaBH4

a. B

d. B

O

Ag2O, NH4OH

C HO

CH3OH HO

O

O

OH [1] LiAlH4

b. B

[2] H2O

OH

CH2OH

e. B

C

CrO3 H2SO4, H2O

HO

OH

O OH

20.19 a. CH3CH2Br + 2 Li

c. CH3CH2Br + 2 Li

CH3CH2Li + LiBr

b. CH3CH2Br + Mg

CH3CH2MgBr

CH3CH2Li

2 CH3CH2Li + CuI

+ LiBr

LiCu(CH2CH3)2 + LiI

20.20 HC CCH2CH2CH2CH2CH2CH3 + NaH

Na+ C CCH2CH2CH2CH2CH2CH3 + H2

hydrogen gas

HC CCH2CH2CH2CH2CH2CH3

BrMgC CCH2CH2CH2CH2CH2CH3 + CH4

methane gas

+ CH3MgBr

20.21 + LiOH

Li + H2O

a.

CH3 H2O CH 3 C MgBr + b. CH3

MgBr

c.

CH3 CH3 C H

+ HOMgBr

+ HOMgBr

+ H2O

d. CH3CH2C C Li

+ H2O

CH3CH2C CH + LiOH

CH3

20.22 To draw the product, add the benzene ring to the carbonyl carbon and protonate the oxygen. OH O

a.

H

C

OH

H C H [1] H

O

MgBr

c.

[2] H2O

CH3CH2

C

[1] H

MgBr CH3CH2 C H

[2] H2O

OH O

b. CH3CH2

C

[1] CH2CH3

MgBr CH3CH2 C CH2CH3

d. [2] H2O

O

[1] [2] H2O

MgBr

OH

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Chapter 20–10 20.23 To draw the products, add the alkyl or phenyl group to the carbonyl carbon and protonate the oxygen. [1] CH3CH2CH2Li

a.

[2] H2O

O

b.

H

C

Li

[1] H

O [2] H2O

HO CH2CH2CH3

OH

O

OH

[1] C6H5Li

c.

d.

C C

Na+

[1] CH2=O

C C CH2OH

[2] H2O

H C H

[2] H2O

20.24 Addition of RM always occurs from above and below the plane of the molecule. H OH

O

a.

CH3

C

H

CH3

b.

H OH

[1] CH3CH2MgBr

+

[2] H2O O

[1] CH3CH2Li

CH2CH3

CH3

+

OH

CH3

OH

[2] H2O

CH2CH3

20.25 OH OH

a.

O

CH3 C CH3

CH3MgBr +

H

b.

c.

CH2OH

H

MgBr

C

CH3

+

C

+ CH3CH2MgBr

H

+ H

C

O

OH

O

MgBr

O

OH

O

C CH2CH3

MgBr

+

d.

H

OH

O

+ CH3MgBr

CH2CH3

H OH

or O

C CH2CH3

+ CH3CH2MgBr H

H

20.26 OH

O

a.

+ CH3Li

linalool (three methods)

OH

Li

O Li

lavandulol O

+ CH2=O Li

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Introduction to Carbonyl Chemistry 20–11 b.

c. Linalool is a 3° ROH. Therefore, it has no H on the carbon with the OH group, and cannot be prepared by reduction of a carbonyl compound.

CHO

OH NaBH4 CH3OH

20.27 N(CH3)2

N(CH3)2

OH

O

BrMg

OCH3

OCH3

venlafaxine

20.28 CH3 O

CH3 O TBDMS–Cl

CH3OH C CH [1] Li C CH

imidazole [2] H2O HO

TBDMSO

TBDMSO

estrone

(CH3CH2CH2CH2)4NF CH3OH C CH

ethynylestradiol HO

20.29 O

a.

CH3CH2

C

Cl

b.

CH3CH2 C CH2CH2CH2CH3 CH2CH2CH2CH3

[2] H2O

O C

OH

[1] CH3CH2CH2CH2MgBr (2 equiv)

OCH3

[1] CH3CH2CH2CH2MgBr (2 equiv)

OH CH3CH2CH2CH2 C CH2CH2CH2CH3

[2] H2O O

c.

OCH2CH3

[1] CH3CH2CH2CH2MgBr (2 equiv)

OH CH3CH2CH2CH2 C CH2CH2CH2CH3

[2] H2O

CH2CH2CH2CH2CH3

20.30 O

OH

a.

C CH3

CH3O

C

+

MgBr

CH3 (2 equiv)

O

b. (CH3CH2CH2)3COH

CH3O

C

CH2CH2CH3 + CH3CH2CH2MgBr (2 equiv)

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Chapter 20–12 O

OH

c. CH3 C CH2CH(CH3)2

C

CH3O

CH2CH(CH3)2

CH3

+ CH3MgBr (2 equiv)

20.31 The R group of the organocuprate has replaced the Cl on the acid chloride. O

a. CH3CH2

C

Cl

CH3CH2

[2] H2O

O

O

C

C

Cl

c. CH3CH2

CH3

O

[1] (Ph)2CuLi Cl

[2] H2O

Ph

O

[1] [CH3CH2CH(CH3)]2CuLi

O

b.

[1] (CH3)2CuLi

[2] H2O

20.32 O

a. (CH3)2CHCH2

C

O

O

[1] LiAlH[OC(CH3)3]3 Cl [2] H O 2

C

(CH3)2CHCH2

H

c. (CH3)2CHCH2

O

b.

(CH3)2CHCH2

C

C

[2] H2O O

Cl

HO

[1] CH3CH2Li (2 equiv)

d.

(CH3)2CHCH2

O

[1] (CH3CH2)2CuLi Cl

C

[1] LiAlH4 Cl

[2] H2O

[2] H2O

(CH3)2CHCH2CH2OH

20.33 O C

a.

O 2CuLi

CH3

Cl

C

CH3

Cl + [(CH3)3C]2CuLi

b. O

or O C

O

or O C

CH3

(CH3)2CuLi Cl

Cl O

+ [(CH3)2CH]2CuLi O

20.34 O [1] Mg

Br

MgBr

[2] CO2

a.

b. c. CH3O

Cl

[1] Mg

CH2Br

MgCl [1] Mg

CH3O

C

O O

[2] CO2

[3] H3O+

COO–

CH2MgBr

[2] CO2

C

OH

[3] H3O+

CH3O

COOH

CH2COO– [3] H3O+

CH3O

CH2COOH

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Introduction to Carbonyl Chemistry 20–13 20.35 OH

a.

O

O

c.

b.

OH

OH

O

OH

O

d.

(+ enantiomer)

20.36 The characteristic reaction of
,-unsaturated carbonyl compounds is nucleophilic addition. Grignard and organolithium reagents react by 1,2-addition and organocuprate reagents react by 1,4-addition. O

O CH3

a. CH3

[1] (CH3)2CuLi

O [1] (CH3)2CuLi

c.

[2] H2O

[1] H C C Li [2] H2O

O CH3

[2] H2O

CH3 CH3 HO C CH CH3

HO C CH [1] H C C Li [2] H2O

CH3

[1] (CH3)2CuLi

b.

[2] H2O

O

O

CH3

[1] H C C Li [2] H2O HO C CH

20.37 [1] (CH2=CH)2CuLi

a. O

[2] H2O

O

O

(from a.)

O

b.

[1] CH2=CHLi

c.

[1] CH2=CHLi [2] H2O

OH

[2] H2O

OH

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Chapter 20–14 20.38 a. CH3CH2OH

HBr or PBr3 PCC

OH

Mg

CH3CH2Br

O

CH3CH2MgBr

[1] CH3CH2MgBr

OH

[2] H2O HBr

OH

b.

Br

(from a.) OH

c.

H2SO4

H2

+

Pd-C

(from a.) OH

d.

HBr or PBr3

Mg

Br

H2O CH3CH2OH

O

e.

MgBr

PCC

O

OH

CH3CHO OH

H2O

MgBr

PCC

(from d.) H2SO4

CH3CH2OH

CH2=CH2

O

mCPBA

20.39 O

a.

O

NaBH4 H

CH3OH

OH

[1] LiAlH4

OH

h.

[1] C6H5Li

H

[2] H2O O

H2 OH

[1] (CH3)2CuLi

i. H

Pd-C

H

PCC

O Na2Cr2O7

e.

H

H

O

No reaction

Ag2O NH4OH

j.

[1] HC
CNa H

O O OH

H2SO4, H2O

O

f.

No reaction [2] H2O

O

d.

OH

H

[2] H2O

O

c.

H O

H

OH

[2] H2O

O

b.

[1] CH3MgBr

g.

k.

OH

l.

[2] H2O [1] CH3C
CLi

H

O

OH

OH

[2] H2O TBDMSCl

OH

imidazole

O–TBDMS

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Introduction to Carbonyl Chemistry 20–15 20.40 O

a.

O

OH

NaBH4 CH3OH

[2] H2O

O

O

OH

[1] LiAlH4

b.

OH [1] CH3MgBr

g.

[2] H2O

[2] H2O O

O

OH

H2

c.

[1] (CH3)2CuLi

i.

Pd-C O

PCC

O

No reaction

Na2Cr2O7

e.

No reaction

[2] H2O

O

d.

C6H5 OH

[1] C6H5Li

h.

OH

[1] HC
CNa

j.

[2] H2O O

No reaction

H2SO4, H2O

OH

[1] CH3C
CLi

k.

[2] H2O

f.

O

OH

Ag2O

No reaction

NH4OH

O–TBDMS

TBDMSCl

l.

imidazole

20.41 Li (2 equiv)

a.

Br

b.

Br

c.

Br

d.

Li

MgBr

e.

MgBr

(CH3CH2CH2CH2)2CuLi

f.

Li

+

Li Mg

LiBr

[1] Li (2 equiv) [2] CuI (0.5 equiv)

H2O

+ LiOH

D2O

D

CH3C
CH

+

+ DOMgBr LiC
CCH3

20.42 a.

MgBr

b.

MgBr

CH2 O

H2O OH

O

HO

H2O

OH

c.

MgBr

CH3CH2COCl

H2O

g.

MgBr CH3COOH

h.

MgBr

HC CH

i.

MgBr

CO2

+ CH3COO HC C O

H3O+

OH O

d.

MgBr CH3CH2COOCH3

e.

MgBr

f.

MgBr

H2O

OH H2O

+

j.

MgBr

k.

MgBr

l.

MgBr

OH

H2O

D2O

D +

OD

OH O

CH3CH2OH + CH3CH2O

H2O

OH

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Chapter 20–16 20.43 O C

O Cl

a.

C

(CH3CH2CH2CH2)2CuLi

CH2CH2CH2CH3

O C

OCH3

b.

(CH3CH2CH2CH2)2CuLi

No reaction

O

O CH3

c.

CH3

[1] (CH3CH2CH2CH2)2CuLi [2] H2O

O

d.

CH3

[1] (CH3CH2CH2CH2)2CuLi

CH3

[2] H2O

CH2CH2CH2CH3 HO

20.44 Arrange the larger group [(CH3)3C–] on the left side of the carbonyl. HO H

H OH

a.

NaBH4, CH3OH S

O

R HO H

b.

[1] (S)-CBS reagent; [2] H2O R H OH

c.

[1] (R)-CBS reagent; [2] H2O S

20.45 HO

OH CH3

a. NaBH4, CH3OH

C

d. [1] CH3Li; [2] H2O

O

O

CH3 C

CH3 HO

CH2CH3 C

CH3 b. H2 (1 equiv), Pd-C

CH3

e. [1] CH3CH2MgBr; [2] H2O

OH

O

A c. H2 (excess), Pd-C

CH3

CH3

C

f.

[1] (CH2=CH)2CuLi; [2] H2O

CH3

CH=CH2

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Introduction to Carbonyl Chemistry 20–17 20.46 O C

a. (CH3)2CHCH2CH2

Cl

b. (CH3)2CHCH2CH2

[1] (CH2=CH)2CuLi Cl

c. (CH3)2CHCH2CH2

[1] C6H5MgBr (2 equiv) Cl

O C

d. (CH ) CHCH CH 3 2 2 2

(CH3)2CHCH2CH2

[2] H2O

O C

(CH3)2CHCH2CH2

[2] H2O

C

H

O

O C

O

[1] LiAlH[OC(CH3)3]3

CH=CH2

OH (CH3)2CHCH2CH2

[2] H2O

C C6H5 C6H5 OH

[1] LiAlH4 Cl

C

(CH3)2CHCH2CH2

[2] H2O

C H H

20.47 O

a.

CH3CH2

C

[1] LiAlH4 OCH2CH2CH3

O

b.

CH3CH2

C

[1] CH3CH2CH2MgCl (2 equiv) OCH2CH2CH3

O

c.

CH3CH2

C

CH3CH2CH2OH

[2] H2O OH CH3CH2 C CH2CH2CH3

[2] H2O

CH2CH2CH3 O

[1] DIBAL-H OCH2CH2CH3

[2] H2O

CH3CH2

C

H

20.48 OH

a. HO

CHO

O

CrO3

CHO

PCC

O

CHO

COOH

HO

NH4OH

H

b. HO

Ag2O

CHO

OH c. HO

H2SO4, H2O O

Na2Cr2O7

CHO

d. HO

HOOC

COOH

H2SO4, H2O

20.49 a.

NaBH4

O COOCH3

OH

O

[1] LiAlH4

c.

CH3OH

COOCH3

OH

(CH3)2N

[2] H2O

OH

(CH3)2N

O

b. O

[1] LiAlH4 COOCH3 [2] H2O

OH

d. CH2OH

[1] LiAlH[OC(CH3)3]3

O OH

Cl O

[2] H2O

O OH

H O

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Chapter 20–18 20.50 CH3

a. CH3

CH3 [1] CO2

MgBr

CH3

COOH

[1] CH2=O

MgBr

f.

[2] H3O+ CH3

CH3

O OH

O [1] CH3CH2MgBr

b.

CH2OH

[2] H2O O

[1] (CH3)2CuLi

g.

[2] H2O

[2] H2O

OH O

c.

OH

[1] C6H5Li

CHO

[1] CH3MgBr

h.

[2] H2O

COCl

d.

O

i.

[1] C6H5MgBr (excess)

OH [1] C6H5Li [2] H2O

OH

OH

C

O

[2] H2O

[1] (CH3)2CuLi

j. C6H5

e.

COOCH2CH3

[1] CH3MgCl (excess)

CH3

[2] H2O

OH

[2] H2O

C CH3

20.51 O

[1] C6H5MgBr

a.

HO C6H5

+

HO C6H5

[2] H2O

b. (CH3)3C

O

[1] CH3Li [2] H2O

c.

(CH3)3C

OH

[1] CH3CH2MgBr

O

OH

+

CH2CH3

[2] H2O

CH3

O

[1] (CH2=CH)2CuLi

OH

[1] Mg COOH [2] CO2

H

[3] H3O+

O

OH

CH=CH2

CH=CH2

Br H

CH2CH3

+

[2] H2O

e.

+ (CH3)3C

CH3

OH

d.

[1] (S)-CBS reagent

H OH

f. [2] H2O O

H OH [1] (R)-CBS reagent

g. [2] H2O

CH3

[2] H2O

OH

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Introduction to Carbonyl Chemistry 20–19 O

h.

[1] LiAlH4

OCH2CH3

CH2OH + HOCH2CH3

[2] H2O

H

H

20.52 Both ketones are chiral molecules with carbonyl groups that have one side more sterically hindered than the other. In both reductions, hydride approaches from the less hindered side. The CH3 groups on the bridgehead carbon make the top more hindered. H– attacks from below to afford an exo OH group. Attack comes from below.

2 H's less hindered Attack comes from above. H

H CH3

CH3

[1] LiAlH4 H

CH3

H

[2] H2O O

[1] LiAlH4

from above

OH

O

OH

[2] H2O H

endo OH group

from

exo OH group below

The concave shape of the six-membered ring makes the bottom face of the C=O more sterically hindered. Addition of the H– occurs from above to place the new C–H bond exo, making the OH endo.

20.53 Since a Grignard reagent contains a carbon atom with a partial negative charge, it acts as a base and reacts with the OH of the starting halide, BrCH2CH2CH2CH2OH. This acid–base reaction destroys the Grignard reagent so that addition cannot occur. To get around this problem, the OH group can be protected as a tert-butyldimethylsilyl ether, from which a Grignard reagent can be made. basic site Br

OH

Mg

acidic site




OH

BrMg

proton transfer

CH3

O

+ HOMgBr

These will react. INSTEAD: Use a protecting group. Br

OH

TBDMS–Cl imidazole

OTBDMS

Br

Mg ether

BrMg

OTBDMS O [1]

protected OH

[2] H2O OH

(CH3CH2CH2CH2)4NF OH

A

OH OTBDMS

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Chapter 20–20 20.54 Compounds F, G, and K are all alcohols with aromatic rings so there will be many similarities in their proton NMR spectra. These compounds will, however, show differences in absorptions due to the CH protons on the carbon bearing the OH group. F has a CH2OH group, which will give a singlet in the 3–4 ppm region of the spectrum. G is a 3o alcohol that has no protons on the C bonded to the OH group so it will have no peak in the 3–4 ppm region of the spectrum. K is a 2o alcohol that will give a doublet in the 3–4 ppm region of the spectrum for the CH proton on the carbon with the OH group. [1] O3 [2] CH3SCH3

O

C CHO

O

OH

[1] Mg H2SO4

[1] C6H5MgBr

HCl

Cl

D

[2] H2O

[2] CH2=O

CH2OH F

[3] H2O

B

A

mCPBA

[1] LiAlH4

[1] (CH3)2CuLi

O

E

[2] H2O

OH

Br

OH

MgBr

PBr3

G OH CH3

[2] H2O

[1] C6H5CHO

Mg

[2] H2O

I

H

J

K

20.55 O O

R

a.

Cl

Si R

H OH

[1] (R)-CBS reagent

R Si

[2] H2O

R

O

O

H OH Cl

O

NaI

I

O

R Si

A

R

O

B (CH3CH2)3SiCl imidazole

H OH HO

H N

KF

O

R

H OSi(CH2CH3)3 NHCH(CH3)2

Si HO

R

O

D

O

R (CH3)2CHNH2

Si R

O

H OSi(CH2CH3)3 I C

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Introduction to Carbonyl Chemistry 20–21

O

b.

O

N(CH3)2

O

N(CH3)2

[1] C6H5MgBr

N(CH3)2

H2SO4

[2] H2O

E

HO

O

F (Z and E isomers) O

O

[1] LiCu(CH CH CHC5H11)2 COOCH3

c.

COOCH3

OR' RO

[2] H2O

RO

G

OR'

O

several steps COOH

HO OH

PGE1

20.56 H OH

O

O O

a.

O

O

O

CH3 MgBr

O

+ + MgBr

O

H OH

+

+ MgBr

CH3 MgBr

OH HOMgBr + OH H OH O

b.

O

O

O

O CH2CH2CH2CH2 BrMg CH2CH2CH2CH2 MgBr

MgBr

BrMg

H OH

O O

O

+ + MgBr

+ + MgBr

OH HOMgBr +

OH

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Chapter 20–22 20.57 O

O CH3 MgBr

O OCH3

O O OCH3

O

OCH3

+

O O

(CH3)3COH + OH

O CH3 MgBr

O

O

OCH3

O CH3 MgBr

H OH CH3 C CH3 CH3

O

H OH

OH

O

O

OH

CH3 MgBr

O

OH

O

+ OH

+ OH

+ OCH3

H OH H OH

CH3OH + OH

20.58 MgBr

+ CH O 3

O

O

C

CH3O C OCH3

OCH3

O CH3O

C

+ CH3O

MgBr MgBr HO H

O O

O

C

C

CH3O C

MgBr

MgBr

+ CH3O

MgBr

H2O OH C

+ CH3OH + HOMgBr

20.59 CH2OH

a.

MgBr

O H

C

b. H

H

MgBr OH

or H

O MgBr

O

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Introduction to Carbonyl Chemistry 20–23 OH

c.

BrMg

(C6H5)2C=O

(C6H5)3COH

O MgBr

e.

BrMg

d. HO

O

or

O MgBr

or

O

CH3MgBr

20.60 O

OH

a.

C

CH3O

O

C

MgBr

c. (CH3CH2CH2CH2)2C(OH)CH3

(2 equiv)

CH3O

b. CH3 C CH2CH2CH(CH3)2

CH3O

C

CH3

BrMg

(2 equiv)

O

OH

C

CH2CH2CH(CH3)2

+ CH3–MgBr

CH3

(2 equiv)

20.61 OH Li

O

c.

H

(two ways)

C CH2CH3

O

or

(three ways) H

+

Li

O

OH

b.

OH

+

a.

O

C +

or

Li

+

O

or (three ways)

+

or

Li CH2CH3

O Li CH2CH3 +

Li

O

or +

Li

CH3CH2

O C

Li OCH3

+

(2 equiv)

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Chapter 20–24 20.62 OH

a.

O

+

MgBr

HO

H

H

c.

O

C6H5

b.

H

+ BrMg C6H5

O

H MgBr

+

OH

20.63 C6H5

C6H5

Br

Mg

Br

KOC(CH3)3

C6H5 H2

C6H5

C6H5

Pd-C

LiAlH4

Br

C6H5

H2O

MgBr

C6H5

C6H5

20.64 PBr3

a.

OH

Mg Br O

PCC H

[1] BrMg CH2CH3 [2] H2O

OH

CH3OH PBr3 CH3Br Mg O

b.

[1] CH3MgBr H

OH

O

PCC

[2] H2O

[1] CH3MgBr

OH

[2] H2O

(from a.) OH

PBr3

Br

c. CH3OH

OH

OH

[2] H2C=O [3] H2O

(from b.)

d.

[1] Mg

H2SO4

PCC

CH2 CH2

mCPBA

O

[1] BrMg CH2CH3 OH

(from a.) [2] H2O

20.65 Br

Mg

O

MgBr

OH

H2O

H CHO

MgBr Mg Br

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525

Introduction to Carbonyl Chemistry 20–25 20.66 OH

Br

PBr3

a.

MgBr

Mg

CH2OH

[1] CH2=O [2] H2O

MgBr

b.

COOH

[1] CO2 [2] H3O+

(from a.) OH

c.

O

CrO3

[1] CH3MgBr

H2SO4, H2O

H2SO4

OH

[2] H2O

mCPBA

O

major product OH

d.

mCPBA

H2SO4

O

OH

[1] C6H5MgBr [2] H2O

O [1] CH3CH2CH2MgBr

e.

OH

Pd-C

+

OH

[1] CH3MgBr

H2SO4

[1] O3

CH3

[2] H2O

[2] (CH3)2S

(from c.)

O CHO

major product OH

[1]

O

MgBr

OH

O

(from a.) H

PCC

g.

PCC [2] H2O

(from a.) [1]

MgBr

(from a.)

O

OH H2SO4

h. [2] H2O

(from c.)

C6H5

H2

(from c.)

f.

C6H5

H2SO4

[2] H2O

O

O

PCC

major product

526

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Chapter 20–26 20.67 OH

OH

f. (from c.)

O

PCC

b.

MgBr

Cl

SOCl2

a.

c.

PBr3

[1] CH3CHO

(CH3)2CHD

[1]

O

MgBr

h. MgBr

(from b.)

[2] H2O

MgBr

OH

[2] H2O

[1]

O

i. [1] O

MgBr

CHO

(from c.) Mg

OH

D2O

g.

Br

PCC

[2] H2O

MgBr OH

OH

[1] CH2=O

(from c.) OH

d.

[2] H2O

[2] H2O

(from c.)

[1] 2 Li

j.

OH

[(CH3)2CH]2CuLi

Br [2] CuI (0.5 equiv)

MgBr

[1] CO2

e.

+

[2] H3O

(from c.)

(from c.)

COOH

[1] O [2] H2O O

20.68 OH H

O

H

H

H

PCC H HO

O

[1] NaH

H

H HO

H

H

[2] CH3I

H

CH3O

estradiol [1] HC CLi [2] H2O OH C CH H H

H

CH3O

mestranol

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Introduction to Carbonyl Chemistry 20–27 20.69 Br

a.

CH3CH2Cl

Br2

AlCl3

h


Br

Br2

K+ –OC(CH3)3

[2] H2O2, –OH MgBr

Mg

O

OH

[1]

FeBr3

[2] H2O

CH3Cl

Br2

AlCl3

h


OH MgBr [1] H2C=O

Br Mg

[2] H2O O

OH MgBr [1] CH3CH=O

b.

OH

[1] BH3

PCC

[2] H2O

(from a.)

O

CH3

O Cl

CH3Br

AlCl3

O

Br NaOH

OH PCC

H

O

OH

Mg [1] CH3MgBr

PCC

[2] H2O

(from a.)

20.70 HBr

a.

[1] Mg

ROOR

Br

[2] CO2 [3] H3O+

[1] Mg

b. Br

(from a.)

OH

c.

Br2

Br

[2] H2C=O [3] H2O

h


O

[1] Mg

OH

[2] O [3] H2O

20.71 Br

Br2

a.

MgBr

Mg

[2] H3O+

FeBr3 MgBr

b.

[1] CH2=O [2] H2O

(from a.)

COOH

[1] CO2

CH2OH

PCC

CHO CH3OH

PCC

CH2=O

OH

528

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Chapter 20–28 OH

O

[1] PhMgBr

CHO

PCC

(from a.)

c.

[2] H2O

(from b.) PBr3

d. CH3CH2CH2OH

Mg CH3CH2CH2Br

PCC

CH3CH2CH2OH

CH3CH2CH2MgBr [1] CH3CH2CH2MgBr

CH3CH2CHO HO

[1] PhMgBr

PCC

[2] H2O

OH

O

(from a.) [2] H2O

O OH

e.

PCC

H

OH

O

O

H2O

O Br

Br2

PCC

FeBr3

MgBr

(from a.)

20.72 PCC

a. HO

[1] CH3CH2MgBr

H O

OH

O

CH3CH2MgBr

CH3CH2Br

O

PCC

b.

OH

Mg

PBr3

CH3CH2OH

PCC

[2] H2O

OH

[1] CH3CH2CH2MgBr [2] H2O

OH

c.

Mg

PBr3

CH3CH2CH2OH

CH3CH2CH2MgBr

CH3CH2CH2Br

PCC

H

[1] CH3CH2CH2CH2MgBr

PCC

[2] H2O O CH3CH2CH2CH2OH

H

d.

OH

O

Mg

PBr3

CH3CH2CH2CH2MgBr

CH3CH2CH2CH2Br

[1] (CH3)2CHCH2MgBr

PBr3

Mg

[2] H2O O

OH

Br

(from c.) (CH3)2CHCH2OH

PBr3

(CH3)2CHCH2Br

MgBr [1] CO2

Mg (CH3)2CHCH2MgBr

[2] H3O+

CO2H

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Introduction to Carbonyl Chemistry 20–29 e.

HBr

OH

Br

Mg

MgBr

H2O

K+ –OC(CH3)3

PBr3

+ OH

H

(from c.)

Br

O

20.73 a.

[1] (CH3)2CuLi

O

[1] CH3MgBr [2] H2O O

[2] H2O

H2SO4

OH

H2, Pd-C

+

b.

PCC

CHO

OH

OH

[1] A

OTs TsCl

[2] H2O

pyridine

PBr3

–CN

CN Mg

Br

MgBr

A O

c.

OCH3

[1] LiAlH4

OH

Br2 FeBr3

[2] H2O

O CH3CH2I

OH NaH Br

Br

O

Br

(+ ortho isomer) MgBr

[1] Br2, FeBr3

d.

H2O

[2] Mg

OH mCPBA [1] NaH

e. HC CH

O HO CH3CHC CH

[2] CH3CH=O [3] H2O

TBDMS–Cl imidazole

OTBDMS CH3CHC CH

NaH

OTBDMS CH3CHC C

[1] O [2] H2O

OTBDMS

OH

CH3CHC C OH

(CH3CH2CH2CH2)4NF

OH

20.74 Hb Hb

O

CH3 Ha

H

CH3 Hb

IR peak: 1716 cm–1 (C=O) 1H NMR: 2 signals (ppm) CH3 doublet 1.2 (Hb) H CH3 Ha septet 2.7 (Ha) Hb

C7H14O

A

Hd

Hd

Hb

NaBH4

H CH3

CH3OH

H Hc

OHa CH3

H CH3 CH3 Hd

Hd

C7H16O

B

Hc

IR peak: 3600–3200 cm–1 (OH) 1H NMR: 4 signals (ppm) doublet 0.9 (Hd) singlet 1.5 (Ha) multiplet 1.7 (Hc) triplet 3.0 (Hb)

530

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Chapter 20–30 20.75

Hb

H

Hb

H

O

CH3 CH3

C4H8O

Ha Ha

Hc

Hc

H

H

IR peak 3600–3200 cm–1 (OH) NMR: 4 signals (ppm) singlet (6 H) 1.2 (Ha) singlet (1 H) 1.6 (Hb) singlet (2 H) 2.7 (Hc) multiplet (5 H) 7.2 (benzene ring)

OHb

[1] C6H5MgBr

1H

CH3 CH3

[2] H2O

Ha Ha C10H14O

C NMR: 2 signals (ppm) singlet (6 H) 1.3 (Ha) singlet (2 H) 2.4 (Hb)

D

1H

20.76 O CH3

C

Hc OCH2CH3

Ha Hb C4H8O2

E

IR peak 1743 cm–1 (C=O) 1H NMR: 2 signals (ppm) triplet (3 H) 1.2 (Hc) singlet (3 H) 2.0 (Ha) quartet (2 H) 4.1 (Hb)

[1] CH3CH2MgBr (excess) [2] H2O

Hb OH Hb d CH3CH2 C CH2CH3 Ha

CH3 Hc C6H14O

F

Ha

IR peak 3600–3200 cm–1 (OH) 1H NMR: 4 signals (ppm) triplet (6 H) 0.9 (Ha) singlet (3 H) 1.1 (Hc) quartet (4 H) 1.5 (Hb) singlet (1 H) 1.55 (Hd)

20.77 Molecular ion at m/z = 86: C5H10O (possible molecular formula). Determine the number of integration units per H: Total number of integration units: 25 + 17 + 24 + 17 = 83 83 units/10 H's = 8.3 units per H Divide each integration value by 8.3 to determine the number of H's per signal: 25 units/ 8.3 = 3 H's 24 units/ 8.3 = 3 H's 17 units/ 8.3 = 2 H's H Hb

CH3CH2CH2C N

O

[1] CH3MgBr

H H

CH3

[2] H3O+

Hc

b

CH3 H H Ha Hd Hd

G

IR peak 1721 cm–1 (C=O) 1H NMR: 4 signals (ppm) triplet (3 H) 0.9 (Ha) sextet (2 H) 1.6 (Hb) singlet (3 H) 2.1 (Hc) triplet (2 H) 2.4 (Hd)

20.78 Molecular ion at m/z = 86: C5H10O (possible molecular formula). Hb Hd Hd [1] (CH3)3CLi [2] CH2=O [3] H2O

CH3 H H H

OH

He H Hf

H H Hc Hc

H

Ha

IR peaks: 3600–3200 cm–1 (OH) 1651cm–1 (C=C) 1H NMR: 6 signals (ppm) singlet (1 H) 1.7 (Ha) singlet (3 H) 1.8 (Hb) triplet (2 H) 2.2 (Hc) triplet (2 H) 3.8 (Hd) two signals at 4.8 and 4.9 due to 2 H's: He and Hf

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531

Introduction to Carbonyl Chemistry 20–31 20.79 Hb

He

[1] CH3MgBr

O

[2] H2O

IR peak 3600–3200 cm–1 (OH) 1H NMR: 5 signals (ppm) triplet (3 H) 0.94 (Ha) multiplet (2 H) 1.39 (Hb) multiplet (2 H) 1.53 (Hc) singlet (1 H) 2.24 (Hd) triplet (2 H) 3.63 (He)

OH Ha

Hc

Hd

CH3 MgBr

O

HOMgBr

OH

O H OH

20.80 O

OH Br

[1] (R)-CBS reagent [2] H2O

HO

OTBDMS Br imidazole (2 equiv)

HO

COOCH3

COOCH3

[1] NaH

HO

C6H5

Br TBDMSCl

[2] Br

A

TBDMS O COOCH3

Br

O

C6H5

Br NH3 H2N

O

B

A

+

OTBDMS H N

B TBDMS O

COOCH3 OTBDMS H N TBDMS O

O

C6H5

[1] LiAlH4 [2] H2O

O

C6H5

(CH3CH2CH2CH2)4NF OH

(R)-salmeterol

C6H5

532

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Chapter 20–32 20.81 larger group equatorial axial

OH [1] L-selectride (CH3)3C

O

(CH3)3C

[2] H2O

H

= (CH3)3C

OH

equatorial

cis-4-tert-butylcyclohexanol major product

4-tert-butylcyclohexanone

L-Selectride adds H– to a C=O group. There are two possible reduction products—cis and trans isomers— but the cis isomer is favored. The key element is that the three sec-butyl groups make L-selectride a large, bulky reducing agent that attacks the carbonyl group from the less hindered direction. O– O (CH3)3C

OH axial H2O

H

(CH3)3C R3B H

H equatorial

(CH3)3C

equatorial

cis product

When H– adds from the equatorial direction, the product has an axial OH and a new equatorial H. Since the equatorial direction is less hindered, this mode of attack is favored with large bulky reducing agents like L-selectride. In this case, the product is cis. R3B H

Axial H's hinder H axial attack. H (CH3)3C

H O (CH3)3C

H axial

axial O–

H2O

OH equatorial

(CH3)3C

trans product The axial H's hinder H– attack from the axial direction. As a result, this mode of attack is more difficult with larger reducing agents. In this case the product is trans. This product is not formed to any appreciable extent.

20.82 The  carbon of an
,-unsaturated carbonyl compound absorbs farther downfield in the 13C NMR spectrum than the
carbon, because the  carbon is deshielded and bears a partial positive charge as a result of resonance. Since three resonance structures can be drawn for an
,unsaturated carbonyl compound, one of which places a positive charge on the  carbon, the decrease of electron density at this carbon deshields it, shifting the 13C absorption downfield. This is not the case for the
carbon. 122.5 ppm O

150.5 ppm   mesityl oxide

O

O

O 


hybrid:

+

+

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Text

Introduction to Carbonyl Chemistry 20–33 20.83 CH2CN

OH

CN

OH

[1] Li+ –N[CH(CH3)2]2 [2]

N(CH3)2

CH2NH2

OH

[1] LiAlH4

O

OCH3

[2] H2O

OCH3

OCH3

OCH3

[3] H2O

W

H H C

CN

–

N[CH(CH3)2]2

X C15H19NO2

H C CN

O

H OH

O

OCH3

venlafaxine

Y

OCH3

HN[CH(CH3)2]2

CN

OH

CN

OCH3

OCH3 OH

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535

Aldehydes and Ketones 21–1 C Chhaapptteerr 2211:: A Allddeehhyyddeess aanndd K Keettoonneess— —N Nuucclleeoopphhiilliicc A Addddiittiioonn  G Geenneerraall ffaaccttss • Aldehydes and ketones contain a carbonyl group bonded to only H atoms or R groups. The carbonyl carbon is sp2 hybridized and trigonal planar (21.1). • Aldehydes are identified by the suffix -al, while ketones are identified by the suffix -one (21.2). • Aldehydes and ketones are polar compounds that exhibit dipole–dipole interactions (21.3).  CO O ((2211..44))  SSuum mm maarryy ooff ssppeeccttrroossccooppiicc aabbssoorrppttiioonnss ooff R RC CH HO O aanndd R R222C –1 C=O ~1715 cm for ketones IR absorptions • increasing frequency with decreasing ring size ~1730 cm–1 for aldehydes • For both RCHO and R2CO, the frequency decreases with conjugation. Csp –H of CHO

~2700–2830 cm–1 (one or two peaks)

CHO C–H
to C=O

9–10 ppm (highly deshielded proton) 2–2.5 ppm (somewhat deshielded Csp –H)

C=O

190–215 ppm

2

1

H NMR absorptions

13

C NMR absorption

3

 N Nuucclleeoopphhiilliicc aaddddiittiioonn rreeaaccttiioonnss [1] Addition of hydride (H–) (21.8) O R

C

OH

NaBH4, CH3OH

R C H(R')

or

H(R')

H

[1] LiAlH4 [2] H2O

1o

• •

The mechanism has two steps. H:– adds to the planar C=O from both sides.

• •

The mechanism has two steps. R:– adds to the planar C=O from both sides.

or 2o alcohol

[2] Addition of organometallic reagents (R–) (21.8) O R

C

OH

[1] R"MgX or R"Li

R C H(R')

[2] H2O

H(R')

R"

1o,

2o,

or 3o alcohol

[3] Addition of cyanide (–CN) (21.9) O R

C

NaCN H(R')

HCl

OH R C H(R') CN

cyanohydrin

• •

The mechanism has two steps. CN adds to the planar C=O from both sides.

–

536

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Chapter 21–2 [4] Wittig reaction (21.10) R

R C O

+ Ph3P

C C

•

alkene

•

R"

• •

The reaction is fastest at pH 4–5. The intermediate carbinolamine is unstable, and loses H2O to form the C=N.

• •

The reaction is fastest at pH 4–5. The intermediate carbinolamine is unstable, and loses H2O to form the C=C.

•

The reaction is reversible. Equilibrium favors the product only with less stable carbonyl compounds (e.g., H2CO and Cl3CCHO). The reaction is catalyzed with either H+ or –OH.

C

(R')H

(R')H

Wittig reagent

The reaction forms a new C–C  bond and a new C–C
bond. Ph3P=O is formed as by-product.

[5] Addition of 1o amines (21.11) R

R

R"NH2

C O

C N

mild acid

(R')H

(R')H

imine

[6] Addition of 2o amines (21.12) O

NR2 R2NH

H

(R')H

(R')H

mild acid

enamine

[7] Addition of H2O—Hydration (21.13) H2O H+ or –OH

O R

C

H(R')

OH R C H(R') OH

•

gem-diol

[8] Addition of alcohols (21.14) O R

C

H+ H(R')

OR" R C H(R')

+ R"OH (2 equiv)

OR''

acetal

+

H2O

• • •

The reaction is reversible. The reaction is catalyzed with acid. Removal of H2O drives the equilibrium to favor the products.

 O Otthheerr rreeaaccttiioonnss [1] Synthesis of Wittig reagents (21.10A) RCH2X

[1] Ph3P [2] Bu

Li

Ph3P CHR

• •

Step [1] is best with CH3X and RCH2X since the reaction follows an SN2 mechanism. A strong base is needed for proton removal in Step [2].

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537

Aldehydes and Ketones 21–3 [2] Conversion of cyanohydrins to aldehydes and ketones (21.9) OH

O

–OH

R C H(R') CN

C

H(R')

+ H2O

aldehyde or ketone

+ –CN

R

•

This reaction is the reverse of cyanohydrin formation.

[3] Hydrolysis of nitriles (21.9) OH R C H(R') CN

OH

H2O H+ or –OH


R C H(R') COOH

-hydroxy carboxylic acid

[4] Hydrolysis of imines and enamines (21.12) NR

O

NR2 H

(R')H

or

H2O, H

(R')H

imine

+

(R')H

H

+ RNH2 or R2NH

aldehyde or ketone

enamine

[5] Hydrolysis of acetals (21.14) OR" R C H(R') OR"

•

O

H+ + H2O

R

C

H(R')

aldehyde or ketone

+ R"OH (2 equiv)

•

The reaction is acid catalyzed and is the reverse of acetal synthesis. A large excess of H2O drives the equilibrium to favor the products.

538

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Chapter 21–4 C Chhaapptteerr 2211:: A Annssw weerrss ttoo PPrroobblleem mss As the number of R groups bonded to the carbonyl C increases, reactivity towards nucleophilic attack decreases.

21.1

a.

(CH3)2C

O

CH3CH O

2 R groups

CH2 O

b.

O

O

O

1 R group 0 R groups

Increasing reactivity decreasing alkyl substitution

Increasing reactivity decreasing steric hindrance

More stable aldehydes are less reactive towards nucleophilic attack.

21.2

CHO

CHO

cyclohexanecarbaldehyde This aldehyde has no added resonance stabilization.

benzaldehyde Several resonance structures delocalize the partial positive charge on the carbonyl carbon, making it more stable and less reactive towards nucleophilic attack. O

O

O

C

C

C

H

H

O

O

C

C

H

O H

C

H

O C

H

H

• To name an aldehyde with a chain of atoms: [1] Find the longest chain with the CHO group and change the -e ending to -al. [2] Number the carbon chain to put the CHO at C1, but omit this number from the name. Apply all other nomenclature rules. • To name an aldehyde with the CHO bonded to a ring: [1] Name the ring and add the suffix -carbaldehyde. [2] Number the ring to put the CHO group at C1, but omit this number from the name. Apply all other nomenclature rules.

21.3

CHO

a.

(CH3)3CC(CH3)2CH2CHO

c.

4 3 2 1 H

5

H O

5 C chain = pentanal

O

3,3,4,4-tetramethylpentanal 8

CHO

b.

6 8 C chain = octanal

1

7

CHO

5

4

3

2

2,5,6-trimethyloctanal

Cl Cl

2 1

CHO

Cl 3 4 Cl

3,3-dichlorocyclobutane4 C ring = carbaldehyde cyclobutanecarbaldehyde

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Aldehydes and Ketones 21–5 Work backwards from the name to the structure, referring to the nomenclature rules in Answer 21.3.

21.4

a. 2-isobutyl-3-isopropylhexanal

c. 1-methylcyclopropanecarbaldehyde CHO

3 carbon ring

O

6 C chain

CH3

H

b. trans-3-methylcyclopentanecarbaldehyde 5 carbon ring

CHO

d. 3,6-diethylnonanal

CHO

9 C chain H

or O CH3

CH3

21.5 • To name an acyclic ketone: [1] Find the longest chain with the carbonyl group and change the -e ending to -one. [2] Number the carbon chain to give the carbonyl C the lower number. Apply all other nomenclature rules. • To name a cyclic ketone: [1] Name the ring and change the -e ending to -one. [2] Number the C’s to put the carbonyl C at C1 and give the next substituent the lower number. Apply all other nomenclature rules. a.

1 O

4

c.

5

O

8 C chain = octanone (CH3)3C

3

b.

1

8 O

5-ethyl-4-methyl-3-octanone

CH3

(CH3)3C O

5 C ring = cyclopentanone

(CH3)3CCOC(CH3)3

5 C chain = pentanone

CH3

3

2

3

4

5

O

2,2,4,4-tetramethyl-3-pentanone

2 1

O

3-tert-butyl-2-methylcyclopentanone

Most common names are formed by naming both alkyl groups on the carbonyl C, arranging them alphabetically, and adding the word ketone.

21.6

a. sec-butyl ethyl ketone

d. 3-benzoyl-2-benzylcyclopentanone

e. 6,6-dimethyl-2-cyclohexenone 6 C ketone

benzyl group: benzoyl group: 5 C ketone O

b. methyl vinyl ketone

O

O CH2

C

f. 3-ethyl-5-hexenal O

O

c. p-ethylacetophenone O

CHO O

CH2

540

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Chapter 21–6 Compounds with both a C–C double bond and an aldehyde are named as enals.

21.7

a. (2Z)-3,7-dimethyl-2,6-octadienal 3

7 6

b. (2E,6Z)-2,6-nonadienal

2

1

3

7

CHO

6

2 cucumber aldehyde

1 CHO neral

Even though both compounds have polar C–O bonds, the electron pairs around the sp3 hybridized O atom of diethyl ether are more crowded and less able to interact with electrondeficient sites in other diethyl ether molecules. The O atom of the carbonyl group of 2-butanone extends out from the carbon chain making it less crowded. The lone pairs of electrons on the O atom can more readily interact with the electron-deficient sites in the other molecules, resulting in stronger forces.

21.8

O O

diethyl ether

2-butanone

For cyclic ketones, the carbonyl absorption shifts to higher wavenumber as the size of the ring decreases and the ring strain increases. Conjugation of the carbonyl group with a C=C or a benzene ring shifts the absorption to lower wavenumber.

21.9

a.

CHO

or

conjugated C=O lower wavenumber

CHO

b.

or

O

higher wavenumber

O

smaller ring higher wavenumber

21.10 Since a charge-separated resonance structure can be drawn for a carbonyl group, more electron donor R groups stabilize the (+) charge on this resonance form. The two R groups on the ketone C=O thus help to stabilize it. O R

C

O R

With an aldehyde, only one electron-donor R: O CH3CH2CH2CH2CH2

C

R

CH3CH2CH2CH2CH2

C

R

With a ketone, two electron-donor R groups:

O H

C

O H

The H of RCHO does not stabilize the charge-separated resonance structure, so it contributes less to the hybrid. The C=O has more double bond character. higher wavenumber

CH3CH2

C

O CH2CH2CH3

CH3CH2

C

CH2CH2CH3

The 2 R groups stabilize the charge-separated resonance structure, so it contributes more to the hybrid. The C=O has more single bond character. weaker bond lower wavenumber

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Aldehydes and Ketones 21–7 21.11 The number of lines in their 13C NMR spectra can distinguish the constitutional isomers. O

O

2-pentanone 5 lines

O

3-pentanone 3 lines

3-methyl-2-butanone 4 lines

21.12 [1] DIBAL-H

a. CH3CH2CH2COOCH3

CH3CH2CH2CHO

[2] H2O PCC

b. CH3CH2CH2CH2OH

CH3CH2CH2CHO

[1] BH3

c.

HC CCH2CH3

CH3CH2CH2CHO

[2] H2O2, HO–

d. CH3CH2CH2CH CHCH2CH2CH3

[1] O3 [2] Zn, H2O

CH3CH2CH2CHO

21.13 O Cl

a.

O

O

CH3

c.

AlCl3

H2O

C CH

H2SO4 HgSO4 O

O [1] (CH3)2CuLi

Cl

b.

[2] H2O

21.14 OCH3

CH3O

CH3O OCH3 [1] O3

OHC

CHO

[2] (CH3)2S

Cleave this C=C with O3.

21.15

LiAlH4 or NaBH4

O

O

C

C H

H

weaker base Equilibrium favors the weaker base. The H– nucleophile is a much stronger base than the alkoxide product.

stronger base

21.16 Addition of hydride or R–M occurs at a planar carbonyl C, so two different configurations at a new stereogenic center are possible. O

a.

H OH

NaBH4 CH3OH

new stereogenic center O

b. (CH3)3C

H OH

H OH

Add

OH CH=CH2 stereochemistry:

[1] CH2 CHMgBr [2] H2O

Add stereochemistry:

CH=CH2 OH

(CH3)3C

(CH3)3C

(CH3)3C

CH=CH2 OH

542

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© The McGraw−Hill Companies, 2011

Chapter 21–8 21.17 Treatment of an aldehyde or ketone with NaCN, HCl adds HCN across the double bond. Cyano groups are hydrolyzed by H3O+ to replace the 3 C–N bonds with 3 C–O bonds. OH CHO

a.

CHCN

NaCN

OH

b.

HCl

OH

H3O+,


CN

COOH

21.18 HO HO HO

O O OH HO HO

O O

CN

HO

enzyme

OH

C

O

CN

enzyme H

C

H

+

HCN

toxic by-product

amygdalin

21.19 CH3

a.

CH3 +

C O

Ph3P=CH2

C CH2

CH3

b.

CH3 O +

Ph3P=CHCH2CH2CH2CH3

CHCH2CH2CH2CH3

21.20 BuLi

a. Ph3P + Br
CH2CH3

Ph3P
CH2CH3 Br–

b. Ph3P + Br
CH(CH3)2

Ph3P
CH(CH3)2

Ph3P=CHCH3 BuLi Ph3P=C(CH3)2

Br–

c. Ph3P + Br
CH2C6H5

BuLi Ph3P
CH2C6H5

Ph3P=CHC6H5

Br–

21.21 CHO

a.

CH2CH3 +

Ph3P CHCH2CH3 CH2CH3

CHO +

b.

c.

CHO +

Ph3P CHC6H5

COOCH3 Ph3P CHCOOCH3

COOCH3

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Aldehydes and Ketones 21–9 21.22 To draw the starting materials of the Wittig reactions, find the C=C and cleave it. Replace it with a C=O in one half of the molecule and a C=PPh3 in the other half. The preferred pathway uses a Wittig reagent derived from a less hindered alkyl halide. CH3 CH2CH3 C C CH3 H

a.

CH3 C CH3

CH2CH3

+

PPh3

O C

or H

CH3 C O + Ph3P CH3

2° halide precursor (CH3)2CHX

b.

C C H

CH2CH3

+

C PPh3 H

H

1° halide precursor XCH2CH2CH3 preferred pathway

CH3CH2

CH2CH3

CH3CH2

CH2CH3 C

(only one route possible)

O C

H

H

(cis or trans) C6H5

c.

CH3

C6H5

C C H

C PPh3 H

C6H5

CH3

+

H

or

O C H

(cis or trans) 1° halide precursor C6H5CH2X

CH3 C O

+

Ph3P

C

H

H

(both routes possible) 1° halide precursor XCH2CH3

21.23 a.

Two-step sequence: O

HO CH3

[1] CH3MgBr

H2SO4

[2] H2O

One-step sequence: O

b.

minor product

Ph3P=CH2

tetrasubstituted major product

(E and Z isomers)

only product

Two-step sequence: O

[1] C6H5CH2MgBr

OH

[2] H2O

CH2C6H5

H2SO4

trisubstituted conjugated C=C

One-step sequence: O

CHC6H5

Ph3P=CHC6H5

CH2C6H5

trisubstituted

CHC6H5 only product

21.24 When a 1o amine reacts with an aldehyde or ketone, the C=O is replaced by C=NR. a.

CHO

O

b.

CH3CH2CH2CH2NH2

CH3CH2CH2CH2NH2

CH

NCH2CH2CH2CH3

NCH2CH2CH2CH3

c.

O

CH3CH2CH2CH2NH2

NCH2CH2CH2CH3

544

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Chapter 21–10 21.25 Remember that the C=NR is formed from a C=O and an NH2 group of a 1° amine. CH3

a.

CH3 C

C NCH2CH2CH3

b.

O + NH2CH2CH2CH3

H

H CH3

N

CH3

O

NH2

21.26 overall reaction H2O H H

O

H

rhodopsin

N

H

OPSIN

N

OPSIN

11-cis-retinal NH2 OPSIN

H3O

nucleophilic attack proton transfer OPSIN

OPSIN

NH2 C O

NH

H

OPSIN

C OH H

21.27 N(CH3)2

N(CH3)2

CH3 N

H

+

CH3

21.28 O H

This carbon has four bonds to C's. To make an enamine, it needs a H atom, which is lost as H2O when the enamine is formed.

21.29 • Imines are hydrolyzed to 1° amines and a carbonyl compound. • Enamines are hydrolyzed to 2° amines and a carbonyl compound. O

H2O

a.

CH N

+

C H+

H2N

H

imine

1° amine H2O

b.

CH2 N CH3

enamine

H+

CH2 N H CH3

2° amine

C OH2 H

H OH2

O

NH

+

O

H2O

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545

Aldehydes and Ketones 21–11 c.

(CH3)2NCH

C(CH3)2

O

H2O H+

enamine

+

(CH3)2NH

C CH(CH3)2 H

2° amine

21.30 • A substituent that donates electron density to the carbonyl C stabilizes it, decreasing the percentage of hydrate at equilibrium. • A substituent that withdraws electron density from the carbonyl C destabilizes it, increasing the percentage of hydrate at equilibrium. a. CH3CH2CH2CHO

or

b.

CH3CH2COCH3

one R group on C=O higher percentage of hydrate

2 R groups on C=O

CH3CF2CHO

or

CH3CH2CHO

F atoms are electron withdrawing. higher percentage of hydrate

21.31 H O H

H OH2

O H

O H

O H

O H +

O

+ H2O

+ H3O

(+ 1 resonance structure)

H2O

21.32 Treatment of an aldehyde or ketone with two equivalents of alcohol results in the formation of an acetal (a C bonded to 2 OR groups). O

a.

O

+

2 CH3OH

OCH3

TsOH

+ HO

b.

OH

TsOH

OCH3

21.33 OCH3

OCH3

a.

b.

OCH3

O

OCH3

2 OR groups on different C's 2 ethers

O

c.

2 OR groups on same C acetal

CH3 CH3

2 OR groups on same C acetal

d.

OCH3 OH

1 OR group and 1 OH group on same C hemiacetal

O

O

546

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Chapter 21–12 21.34 The mechanism has two parts: [1] nucleophilic addition of ROH to form a hemiacetal; [2] conversion of the hemiacetal to an acetal. O O

O

TsO
H

+ HOCH CH OH 2 2

+ H2O

overall reaction

O H

TsO H HO O CH2CH2OH

O H

HO O CH2CH2OH + TsO
H

+ TsO

hemiacetal

HOCH2CH2OH

H HO O CH2CH2OH

H O

O CH2CH2OH

O CH2CH2OH

O CH2CH2OH

TsO
H + H2O

+ TsO–

O CH2CH2OH

O

TsO

O H

O

O + TsO
H

carbocation re-drawn

acetal

21.35 CH3O OCH3

a.

b.

+

H2O

O

H2SO4 CH3

O

O

O CH3

+ 2 CH3OH

+

c.

H2O

H2SO4 CH3

O

O

OH

+ C CH3

OH

O

H2SO4

+ H2O

C

CH3

C

CH3

+ HOCH2CH2OH

21.36 HO

O

O H2O O

H2SO4

safrole

+ H

H

HO

formaldehyde

21.37 Use an acetal protecting group to carry out the reaction. O

O

O

O

H3

O+

[2] H2O

TsOH COOCH2CH3

O

O

[1] CH3Li (2 equiv)

HOCH2CH2OH COOCH2CH3

OH

OH

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Aldehydes and Ketones 21–13 21.38 OH

O

C4

C1

OH O

a. HO

O

H

b.

C1 O

H

C4

OH

C5

C1

C5

C1

21.39 The hemiacetal OH is replaced by an OR group to form an acetal. OH

a.

OCH2CH3

O

H+

+ CH3CH2OH

O

OH HO

OCH2CH3

O

b.

H+

+ CH3CH2OH

HO

HO

O

HO

21.40 acetal

O

HO H O

O

H

OH

O O

H OCH3

H

acetal HO

O

H O

HO

COOH HO

monensin

O

H

O

HO

O

H O

OH

acetal

OH

acetal

O

OH

H

digoxin

hemiacetal Ether O atoms are indicated in bold.

21.41 HO OH

a.

HO OH O

*

*

*

HO

*

OH * OH

5 stereogenic centers (labeled with *)

OH

d. HO

CHO OH

HO OH

b.

HO OH

O HO

hemiacetal C

OH OH

HO OH O OH

HO OH

-D-galactose

O OCH3

HO OH


-D-galactose

c.

e.

HO OH

+

O HO OH OCH3

O

548

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Chapter 21–14 21.42 Use the rules from Answers 21.3 and 21.5 to name the aldehydes and ketones. O

a. (CH3)3CCH2CHO

O

O

re-draw H

4 C = butanal 3,3-dimethylbutanal

h. (CH ) C 3 3

re-draw

C CH(CH3)2

5 C = pentanone 2,2,4-trimethyl3-pentanone (common name: tert-butyl isopropyl ketone)

O

5 C = pentanone 2-chloro-3-pentanone

b. Cl

O

8 C = octanone 8-phenyl-3-octanone

c. Ph d. CHO

5 C ring 2-methylcyclopentanecarbaldehyde

O

O

j.

f. (CH3)2CH

CH3 O

CHO

O

6 C ring = cyclohexanone 5-isopropyl-2-methylcyclohexanone

k.

CHO

6 C = hexanal 3,4-diethylhexanal

E 8 C = octenone (5E)-2,5-dimethyl-5-octen-4-one

CHO

trans-2-benzylcyclohexanecarbaldehyde l.

CH2Ph

o-nitroacetophenone

NO2

6 C ring = cyclohexanone 5-ethyl-2-methylcyclohexanone

e.

g.

CH3

i.

6 C = hexenal 3,4-diethyl-2-methyl-3-hexenal

21.43 f. 2-formylcyclopentanone

a. 2-methyl-3-phenylbutanal

O

O

CHO

H O

O

b. dipropyl ketone CHO

g. (3R)-3-methyl-2-heptanone

c. 3,3-dimethylcyclohexanecarbaldehyde h. m-acetylbenzaldehyde CHO

O

d.
-methoxypropionaldehyde

O H

i. 2-sec-butyl-3-cyclopentenone

O

OCH3 O

j. 5,6-dimethyl-1-cyclohexenecarbaldehyde

e. 3-benzoylcyclopentanone

CHO O

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549

Aldehydes and Ketones 21–15 21.44 O

O

H

O

H

H

2-ethylbutanal

hexanal

O

O

H

2,2-dimethylbutanal

O

H

(2S)-2,3-dimethylbutanal

O

3,3-dimethylbutanal

O

H

(2R)-2,3-dimethylbutanal

H

4-methylpentanal

O

H

O

H

(2S)-2-methylpentanal

(2R)-2-methylpentanal

O

H

H

(3R)-3-methylpentanal

(3S)-3-methylpentanal

21.45 H

= C6H5CH2CHO

phenylacetaldehyde

O

a. NaBH4, CH3OH b. [1] LiAlH4; [2] H2O

g.

C6H5CH2CH2OH C6H5CH2CH2OH

h.

C6H5CH2CH(OH)CH3 NaCN, HCl

e. Ph P CHCH 3 3

OCH2CH3

CH3CH2OH (excess), H+

OCH2CH3

C6H5CH2CH(OH)CN

H NH

C6H5CH2CH=CHCH3

mild acid

H

(CH3)2CHNH2

j.

O

OH

HO

NCH(CH3)2

mild acid

(E and Z isomers)

N

i.

(E and Z isomers) f.

(E and Z isomers)

N(CH2CH3)2

c. [1] CH3MgBr; [2] H2O d.

H

(CH3CH2)2NH, mild acid

O

H+

21.46 O

2-heptanone a. NaBH4, CH3OH

OH

d.

[1] LiAlH4; [2] H2O

e. OH

c. [1] CH3MgBr; [2] H2O

NC

OH

CHCH3

OH

b.

NaCN, HCl

Ph3P CHCH3

+ f. (CH3)2CHNH2, mild H

(E and Z isomers) NCH(CH3)2

550

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Chapter 21–16 N(CH2CH3)2

(CH3CH2)2NH, mild H+

g.

N(CH2CH3)2

(E and Z isomers) CH3CH2O OCH2CH3

CH3CH2OH (excess), H+

h.

NH N

i.

N

mild H+

(E and Z isomers) OH

HO

j.

O

O

H+

21.47 H Ph3P CHCH2CH3

O

a.

CHO Ph3P CHCOOCH3

c.

CHCH2CH3

COOCH3

H

+ COOCH3 H CHO

b.

Ph3P

O

d.

Ph3P CH(CH2)5COOCH3

H CH(CH2)5COOCH3

H

21.48 a. CH3CH2Cl

[1] Ph3P

[1] Ph3P CH3CH=C(CH3)2

c.

CH2Cl

[2] BuLi [3] (CH3)2C=O

CH=CHCH2CH2CH3 [2] BuLi [3] CH3CH2CH2CHO

(E and Z isomers)

c. Ph3P CHCH=CH2

BrCH2CH=CH2

[1] Ph3P

b.

CH2Br

[2] BuLi [3] C6H5CH2CH2CHO

CH=CHCH2CH2C6H5

(E and Z isomers)

21.49 a.

Ph3P CHCH2CH2CH3

b.

Ph3P C(CH2CH2CH3)2

BrCH2CH2CH2CH3 BrCH(CH2CH2CH3)2

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Aldehydes and Ketones 21–17 21.50 HO

CHO

O

[2] H2O2, –OH

CH HO C CH

[1] LiC A

HO

COOH

C Ag2O

B [1] BH3

NH4OH O

O

HO

[2] H2O

TBDMSO E TBDMS–Cl

D H2O

TBDMSO

OH

F [1] CH3Li [2] H2O

H2SO4 HgSO4

N

NH (CH3CH2CH2CH2)4N+F
HO

G OH

21.51 a. CH3CH2CHO + b.

mild

H2N

O

CH3CH2CH N

acid

HOCH2CH2OH

O

H+

O

H3O+

c.

H2N

N

O

O

d.

mild

+

C6H5

N H

N

acid

(E and Z isomers)

C6H5 HO

CN C

e. C6H5

H3O+,


HO

COOH

C6H5

C6H5

C6H5

CH3CH2OH OH

f.

O H3O+

N

g.

OCH2CH3

H+

O

OCH3

h. CH3O

O

H3O+

OCH3

+

HN

O + HOCH3

CH3O

21.52 CH3CH2O

OCH2CH3

O + HOCH2CH3

a. CH3O

O

OCH3 OCH3

OCH3

b.

HO

O + HOCH3

OCH2CH3 OCH3

OCH3

H

c. O + HOCH2CH3

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Chapter 21–18 21.53 Consider para product only, when an ortho, para mixture can result. Br Br2

O

CH3COCl

FeBr3

O

HOCH2CH2OH Br

AlCl3

O

Br

B

A

O

Mg

O

H+

MgBr D

C

[1] CH3CHO [2] H2O O

O

O

H2O

O

PCC

O

H+ G

F

O

OH

O E

21.54 Ph3P=CHCH2CH2CH3

a. CH3CH2CH2CHO

CH3CH2CH2

CH2CH2CH3

H O

HO CN

NaCN

b.

+

+

C C H

CH3CH2CH2 H C C H CH2CH2CH3

NC OH

HCl NaBH4

O

c.

OH

CH3OH

CH3CH2

d. HO

+

OH CH3CH2

O

CH3OH

HO

CH3CH2

+

O

O

HO

HCl OCH3

OH

OCH3

21.55 new stereogenic center O

OH

O

OH

O

OH

CHO

HO

A achiral S

new stereogenic center H

CHO

chiral

H O

S HO H B

An equal mixture of enantiomers results, so the product is optically inactive.

H O

OH

O OH

A mixture of diastereomers results. Both compounds are chiral and OH they are not enantiomers, so the mixture is optically active.

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Aldehydes and Ketones 21–19 21.56 OH H

HO

a.

acetal O

O

O

H

H

acetal

acetal O

O

etoposide

O O H

CH3O

O

OCH3 OH

b. Lines of cleavage are drawn in. OH H

HO O

H

O O

O

H

O

H+

O O H

CH3O

OH H

HO H2O

O

OCH3

O

+

HO

H

CH3O

OH

HO

OH H

O CH HO H

O

OH OH

+ CH3CHO + CH2=O

OCH3 OH

21.57 Use the rule from Answer 21.1. O

O H

a. O

O

O

b.

O

Increasing reactivity decreasing steric hindrance

Increasing reactivity decreasing steric hindrance

21.58 O

H2O

HO OH

Less stable carbonyl compounds give a higher percentage of hydrate. Cyclopropanone is an unstable carbonyl compound because the bond angles around the carbonyl carbon deviate considerably from the desired angle. Since the carbonyl carbon is sp2 hybridized, the optimum bond angle is 120°, but the three-membered ring makes the C–C–C bond angles only 60°. This destabilizes the ketone, giving a high concentration of hydrate when dissolved in H2O.

21.59 Electron-donating groups decrease the amount of hydrate at equilibrium by stabilizing the carbonyl starting material. Electron-withdrawing groups increase the amount of hydrate at equilibrium by destabilizing the carbonyl starting material. Electron-donating groups make the IR absorption of the C=O shift to lower wavenumber because they stabilize the charge-separated resonance form, giving the C=O more single bond character.

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Chapter 21–20 O 2N

COCH3

CH3O

COCH3

a.

p-nitroacetophenone NO2 withdrawing group less stable

b.

higher percentage of hydrate

p-methoxyacetophenone CH3O donating group more stable lower percentage of hydrate

c.

higher wavenumber

lower wavenumber

21.60 Use the principles from Answer 21.22. CH3CH2

a.

CH2CH2CH3

CH3CH2

H

CH3CH2

C C CH3CH2

CH3CH2

CH2CH2CH3

C O

Ph3P C

C PPh3

or H

CH2CH3

CH3

C C

b. H

CH3

or

Ph3P C

H

CH2CH3

C PPh3

H

O C

H

H

(Both routes possible.)

1° alkyl halide precursor (XCH2CH3)

1° alkyl halide precursor (XCH2CH2CH3) O

PPh3

c.

Ph3P CH2

or

methyl halide precursor (CH3X) preferred pathway C6H5

d.

H

2° alkyl halide precursor [(CH3CH2)2CHX]

CH2CH3

C O H

O C

CH3CH2

1° alkyl halide precursor (XCH2CH2CH2CH3) preferred pathway CH3

CH2CH2CH3

C6H5 O

Ph3P

CH2 O

2° alkyl halide precursor (CH3CH2CH2CHXCH3) or

C6H5




PPh3

H

1° alkyl halide precursor (C6H5CH2X) preferred pathway

2° alkyl halide precursor X

21.61 a.

N

O

O

O

H

+ HOCH2CH2OH

c.

O

+ NH2

b.

N

O

+

HN

d. CH3O

OCH3 OCH3

CH3O + HOCH3

H O

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Aldehydes and Ketones 21–21 21.62 PCC

a. C6H5–CH2OH b. C6H5–COCl

C6H5–CHO

[1] LiAlH[OC(CH3)3]3

f. C6H5–CH=CH2

[2] Zn, H2O C6H5–CHO

g. C6H5–CH=NCH2CH2CH3

[2] H2O c. C6H5–COOCH3

[1] DIBAL-H

[1] O3

h. C6H5–CH(OCH2CH3)2

C6H5–CHO

[2] H2O [1] LiAlH4 d. C6H5–COOH

e. C6H5–CH3

[2] H2O KMnO4

PCC

C6H5–CH2OH

C6H5–COOH

H2O

C6H5–CHO

H+ H2O

C6H5–CHO

H+

C6H5–CHO

[1] LiAlH4 [2] H2O

C6H5–CHO

PCC

C6H5–CH2OH

C6H5–CHO

21.63 PCC

a. OH Cl

b.

c. CH3COCl (CH3CH2)2CuLi

e. CH3C CCH3

O

O

(CH3)2CuLi

d. CH3CH2C CH

O

O

H2O H2SO4 HgSO4

O

H2O H2SO4 HgSO4

O

21.64 O

a. One-step sequence:

preferred route only one product formed

Ph3P

or O

Two-step sequence:

OH H2SO4

[1] CH3CH2CH2MgBr [2] H2O

b. One-step sequence:

CHO

Ph3P

preferred route only one product formed

or

OH CHO [1] (CH ) CHMgBr 3 2

H2SO4

Two-step sequence: [2] H2O

+ other alkenes that result from carbocation rearrangement

21.65 a.

CH3CH2CH2CH CHCH3 PCC

One possibility:

CHO

OH PBr3 OH

Br

[1] Ph3P [2] BuLi

CHO Ph3P CHCH3

CH3CH2CH2CH CHCH3

(E and Z)

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Chapter 21–22 b.

C6H5CH CHCH2CH2CH3 CH3OH

One possibility:

SOCl2 CH3Cl

Br2

AlCl3

h


CHO

Br [1] Ph3P [2] BuLi

C6H5CH PPh3

C6H5CH CHCH2CH2CH3

(from a.)

(E and Z)

c. PCC

OH OH

O Br

PBr3

O

[1] Ph3P

Ph3P CHCH(CH3)2

[2] BuLi

21.66 H2O

a.

PCC OH

H2SO4

Ph3P CHCH2CH3

H+

O

CHCH2CH3

O

b.

O

HOCH2CH2OH

O

(from a.) PBr3

CH3CH2CH2OH O

c.

CH3CH2CH2Br

H2NCH2CH2CH3

[1] Ph3P [2] BuLi

Ph3P CHCH2CH3

NCH2CH2CH3

mild acid NH3 (excess)

(from a.)

BrCH2CH2CH3 PBr3 HOCH2CH2CH3

OH

d.

Br

HBr

MgBr

Mg

CH3OH PCC [1] H2C=O

OH

O H

PCC

MgBr

[2] H2O [1]

(from a.)

C(OCH2CH3)2

O

CH3CH2OH (2 equiv)

PCC

TsOH 2

e.

[1] OsO4 [2] NaHSO3, H2O

OH OH

O TsOH PCC OH

OH

O O

[2] H2O

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Aldehydes and Ketones 21–23 21.67 CH3OH SOCl2 CH3Cl

a.

KMnO4 CH3

AlCl3

OH

COOH

Br

PBr3

OH

[1] LiAlH4

PCC CHO

[2] H2O

PPh3

[1] PPh3 [2] BuLi

PPh3 CH CH

CHO

(E and Z isomers) OH

b.

CH3Cl (from a.) AlCl3

OH

OCH3

[1] NaH [2] CH3Br

CH3

OCH3

P(C6H5)3

h


CH3

PBr3

(+ ortho isomer)

OCH3

Br2

Br

P(C6H5)3

CH3OH

Br

BuLi OCH3

CH3O

CH CH

(CH3)3COH

C(CH3)3 P(C6H5)3

(E and Z isomers)

HCl

CH3Cl (from a.)

C(CH3)3

(CH3)3CCl AlCl3

AlCl3

C(CH3)3

C(CH3)3

KMnO4

CH3

C(CH3)3

[1] LiAlH4 [2] H2O [3] PCC OHC

HOOC

(+ ortho isomer)

21.68 O

a.

OH

HOCH2CH2OH

O

O OH

H+

PBr3

O

[1] Mg

O

PCC OH

O

O

Br [2] (CH3)2CO [3] H2O

OH H2O, H+

O [1] LiAlH4

OH

O

[2] H2O OH

OH CH3CH2CH2OH PCC

b.

O

[1] Ph3P

O Br

(from a.)

[2] BuLi

O

O

PPh3

CH3CH2CHO O

H3O+

O CH=CHCH2CH3

(E/Z mixture)

O CH=CHCH2CH3

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Chapter 21–24 21.69 O

a. CH3CH2OH

PCC CH3

C

H

PBr3 Mg

CH3CH2Br

CH3CH2MgBr

H2SO4

b. CH3CH2OH

O [1] C CH3 H [2] H2O

Br2

CH2=CH2

H OH PCC C CH3 CH2CH3

Br

2 NaNH2

Br

O CH3

CH2CH3

[1] CH3

O

O

OH

O C

[2] H2O H

(from a.)

OH

PCC

CH2CH3

HC C Na+

[2] NaHSO3, H2O HO

CH3 C OCH2CH3

TsOH

NaH

HC CH

[1] OsO4

O

OCH2CH3

CH3CH2OH

C

TsOH

21.70 O

O O

mCPBA

O

O

OH NHC(CH3)3

O

(CH3)3CNH2

NHC(CH3)3

O

H2O

O

O

O H3O+

X

OH NHC(CH3)3

HO

+ (CH3)2C=O

HO

albuterol

21.71 Br

Mg

O

Br

CHO

O

[1]

HOCH2CH2OH TsOH

BrMg

O

O

O

OH

O

O

[2] H2O

H2O H+ OH A

21.72 a.

O H

Na+H–

O

Cl

+ H2 + Na+

b.

O O

acetal

O

CH3

O

+ Cl– O

methoxy methyl ether

CHO

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Aldehydes and Ketones 21–25 H O

O

c.

O

H OH2

H H2O

O

O

+ H2O

O

O

O

O

H2O H

+ HOCH3

(The three organic products are boxed in.)

O CH2

H

+ H3O H O

O

H

H OH2

O

O

OH

H

CH2 OH

H2O CH2 O H

CH2 O

H2O

+ H3O

21.73 H

H

N

N

H2O

H2O

H

OH

H

HO

N

H N

HO

H OH2

H H N

a. H H OH2

H2O

(+ 1 resonance structure) O

O H2O

NH3

O NH2 H OH2

H

H2O NH2

(+ 1 resonance structure)

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Chapter 21–26

NH2

H

O

NH2

H

O

H

H

H H N

O

b. NH2 O

N

proton transfer

N

H OH2

OH

NH2 O

NH2 O

H2O N

N

N

N

OH2

N

H

OH2 H2O

H

N

N

NH2 O

NH2 O

proton OH

transfer H OH2

H

N H H

O

H3O+

H2O N

N

+ H3O+

N H2O

N

H

21.74 The OH groups react with the C=O in an intramolecular reaction, first to form a hemiacetal, and then to form an acetal. O

O OH

HO

OH

O O

OH

OH adds here to form a hemiacetal. Then, the acetal is formed by a second intramolecular reaction.

hemiacetal

acetal C9H16O2

21.75 O

H218O

18O

H+

overall O

OH

H+

OH 18

OH

+

H218O

H H218O

+ OH H 18

OH

+ H318O

OH2 18

OH

18O

+ H2O

H

H218O

+ H318O 18

O

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Aldehydes and Ketones 21–27 21.76 H H OH

a.

O

OCH2CH3

O

OCH2CH3

O

H

H OH2

+

OH

O

H

H2O

H OH

H OH2

+ H OCH2CH3 H2O

O H3O+

O

O

HO

H O

HO

H

OH + H2O

O

H

H

O

b.

H OH2 H

OH

OH

+

H

H

O

O

OH

H

H OH2

O

OH

H OH

OH2

H

OH

O

O

O

H

H OH

H OH

H O

H2O

H

O

O

O

O

H OH

+ H3O+

O

21.77 OCH3

OCH3 +

CH3O O HO

H OTs

enol ether

CH3O

OH

H O

O

O

acetal

O

OH

OH

+ OTs H OTs +

H OTs

OTs

H

O

O

H

CH3O

O

OH OTs

+ CH3OH

OH

+ OTs

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Chapter 21–28 21.78 H O HO

O

NH2

+ CH3

HO

C

N C H H CH3

H

HO

H OH2

H OH

proton

HO

CH3

transfer

HO

N C H

HO

dopamine

H OH2

HO

HO

HO NH

HO

NH

HO

H

CH3 H3O

HO

CH3 N H

HO

CH3

CH3

H2O

salsolinol

N C H

(+ 3 more resonance structures)

C

HO H2O

H

H2O

21.79 O O

O (CH3)2S

CH2 H

Bu Li

(CH3)2S

CH2

+ (CH3)2S

S(CH3)2

sulfur ylide

X

sulfonium salt

X

+ Bu H + LiX

21.80 Hemiacetal A is in equilibrium with its acyclic hydroxy aldehyde. The aldehyde can undergo hydride reduction to form 1,4-butanediol and Wittig reaction to form an alkene. OH

O

a.

OH O

OH

C

A

OH H

This can now be reduced with NaBH4. OH O

O

b.

OH

A

C

Ph3P=CHCH2CH(CH3)2 H

reacts with the Wittig reagent

1,4-butanediol

(CH3)2CHCH2CH=CHCH2CH2CH2OH

(E and Z isomers)

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Aldehydes and Ketones 21–29 21.81 O

H OH2

CH3–MgI

O

HO

OCH3

HO

OCH3

OCH3

OCH3

OCH3

OCH3

H2O

5,5-dimethoxy-2-pentanone

OCH3 OCH3

H2O

H H OH2

H

HO

H OH2

HO

OCH3

HO

HO

OCH3

OH

OCH3

OH

H

O

OCH3

(+ 1 resonance structure)

H

H2O

H2O

H2O

HO

H O CH3 O OH

OH

O

H3O

OH

H

(+ 1 resonance structure)

Y

H2O

H O CH3

21.82 cyclopropenone (1640 cm–1) O

O

2-cyclohexenone (1685 cm–1) O

O

These three resonance structures include an aromatic ring; 4n + 2 = 2
electrons. Although they are charge separated, the stabilized aromatic ring makes these three structures contribute to the hybrid more than usual. Since these three resonance contributors have a C–O single bond, the absorption is shifted to a lower wavenumber.

O

O

O

There are three resonance structures for 2cyclohexenone, but the charge-separated resonance structures are not aromatic so they contribute less to the resonance hybrid. The C=O absorbs in the usual region for a conjugated carbonyl.

21.83 CHO

a.

and

c. O

aldehyde ketone The sp2 hybridized C–H bond of the aldehyde absorbs at 2700–2830 cm–1.

b.

and O

higher wavenumber for C=O

O

conjugated with a benzene ring lower wavenumber

O

and

O

smaller ring higher wavenumber for C=O

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Chapter 21–30 21.84 A. Molecular formula C5H10O IR absorptions at 1728, 2791, 2700 cm–1 NMR data: singlet at 1.08 (9 H) singlet at 9.48 (1 H) ppm

B. Molecular formula C5H10O IR absorption at 1718 cm–1 NMR data: doublet at 1.10 (6 H) singlet at 2.14 (3 H) septet at 2.58 (1 H) ppm C. Molecular formula C10H12O IR absorption at 1686 cm–1 NMR data: triplet at 1.21 (3 H) singlet at 2.39 (3 H) quartet at 2.95 (2 H) doublet at 7.24 (2 H) doublet at 7.85 (2 H) ppm D. Molecular formula C10H12O IR absorption at 1719 cm–1 NMR data: triplet at 1.02 (3 H) quartet at 2.45 (2 H) singlet at 3.67 (2 H) multiplet at 7.06–7.48 (5 H) ppm

1 degree of unsaturation C=O, CHO 3 CH3 groups CHO

CHO CH3 C CH3 CH3

1 degree of unsaturation C=O 2 CH3's adjacent to H CH3 CH adjacent to 2 CH3's

O

5 degrees of unsaturation (4 due to a benzene ring) C=O CH3 adjacent to 2 H's CH3 CH2 adjacent to 3 H's 2 H's on benzene ring 2 H's on benzene ring

O

CH3

5 degrees of unsaturation (4 due to a benzene ring) C=O CH3 adjacent 2 H's 2 H's adjacent to 3 H's CH2 a monosubstituted benzene ring

O

21.85 C7H16O2: 0 degrees of unsaturation IR: 3000 cm–1: C–H bonds NMR data (ppm): Ha: quartet at 3.5 (4 H), split by 3 H's Hb: singlet at 1.4 (6 H) Hc: triplet at 1.2 (6 H), split by 2 H's

CH3

Hb

CH3CH2 O C O CH2CH3

Hc Ha

CH3

Ha Hc

Hb

21.86 B. Molecular formula C9H10O A. Molecular formula C9H10O 5 degrees of unsaturation 5 degrees of unsaturation IR absorption at 1720 cm–1
C=O IR absorption at 1700 cm–1
C=O –1 IR absorption at ~2700 cm–1
CH of RCHO IR absorption at ~2700 cm
CH of RCHO NMR data (ppm): NMR data (ppm): 2 triplets at 2.85 and 2.95 (suggests –CH2CH2–) triplet at 1.2 (2 H's adjacent) multiplet at 7.2 (benzene H's) quartet at 2.7 (3 H's adjacent) signal at 9.8 (CHO) doublet at 7.3 (2 H's on benzene) doublet at 7.7 (2 H's on benzene) O singlet at 9.9 (CHO) CH2 CH2 C

H

O C H

CH2 CH3

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Aldehydes and Ketones 21–31 21.87 C. Molecular formula C6H12O3 1 degree of unsaturation IR absorption at 1718 cm–1
C=O To determine the number of H's that give rise to each signal, first find the number of integration units per H by dividing the total number of integration units (7 + 40 + 14 + 21 = 82) by the number of H's (12); 82/12 = 6.8. Then divide each integration unit by this number (6.8). O OCH3 NMR data (ppm): singlet at 2.2 (3 H's) OCH3 doublet at 2.7 (2 H's) singlet at 3.2 ( 6 H's – 2 OCH3 groups) triplet at 4.8 (1 H)

21.88 D. Molecular ion at m/z = 150: C9H10O2 (possible molecular formula) 5 degrees of unsaturation IR absorption at 1692 cm–1
C=O O NMR data (ppm): triplet at 1.5 (3 H's – CH3CH2) quartet at 4.1 (2 H's – CH3CH2) doublet at 7.0 (2 H's – on benzene ring) doublet at 7.8 (2 H's – on benzene ring) singlet at 9.9 (1 H – on aldehyde)

CHO

21.89 OH OH OH

a. HO

b.

CHO

HO HO HO

OH

OH CHO OH

21.90 OH

OH O

HO HO

OH

H Cl

OH O

HO HO

OH

OH


-D-glucose

CH3OH

HO HO

above

OH

O OH

OH H

O

below

OH

Cl

+

HCl

acetal

OH O

HO HO

OCH3

CH3

OH CH3OH

O

HO HO

OH O CH3 H

The carbocation is trigonal planar, so CH3OH attacks from two different directions, and two different acetals are formed.

HO HO

O OH

+ acetal OCH3

O OH

H2O Cl

O

HO HO

HO HO

OH

Cl OH

OH

O

HO HO

OH2

OH

HCl

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Chapter 21–32 21.91 H+

H O

H O

OH

O

OH

O OH

O

O

OH

O

OH

O H

O

O

H2O H3O

21.92 O

a.

O

brevicomin O

HO

b.

O

OH

Br

O

[1] Ph3P

O

O

Br [2] BuLi

H+

PPh3 CH3CH2CHO

brevicomin

O

H3O+

OH

H3O+ O

OH

O

[1] OsO4

O

PCC

O CH=CHCH2CH3

[2] NaHSO3, H2O OH

CH3CH2CH2OH

(E and Z isomers)

OH

21.93 OH

a.

HO HO

acetal carbon

OCH3

O

c.

OH HO O HO

O

CH3O CH3O

CH3O

OH O HO

(OH can be up or down.)

OH OH

[2] CH3OH, HCl

HO HO

O OH HO O HO

O HO

OCH3

(OCH3 can be up or down.)

OCH3 [3] NaH (excess) CH3I (excess)

O CH3O

OH

(OH can be up or down in both products.)

hemiacetal carbon

HO

OH

HO CH3O

OH HO

+ b. [1] H3O HO

OCH3

O

CH3O CH3O

O CH3O O CH3O

OCH3 O OCH3 OCH3

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567

Carboxylic Acids and Their Derivatives 22–1 C Chhaapptteerr 2222:: C Caarrbbooxxyylliicc A Acciiddss aanndd TThheeiirr D Deerriivvaattiivveess— —N Nuucclleeoopphhiilliicc A Accyyll SSuubbssttiittuuttiioonn  SSuum mm maarryy ooff ssppeeccttrroossccooppiicc aabbssoorrppttiioonnss ooff R RC CO OZZ ((2222..55)) IR absorptions • All RCOZ compounds have a C=O absorption in the region 1600–1850 cm–1. • RCOCl: 1800 cm–1 • (RCO)2O: 1820 and 1760 cm–1 (two peaks) • RCOOR': 1735–1745 cm–1 • RCONR'2: 1630–1680 cm–1 • Additional amide absorptions occur at 3200–3400 cm–1 (N–H stretch) and 1640 cm–1 (N–H bending). • Decreasing the ring size of a cyclic lactone, lactam, or anhydride increases the frequency of the C=O absorption. • Conjugation shifts the C=O to lower wavenumber. 1

H NMR absorptions

13

C NMR absorption

• •

C–H  to the C=O absorbs at 2–2.5 ppm. N–H of an amide absorbs at 7.5–8.5 ppm.

•

C=O absorbs at 160–180 ppm.

 SSuum mm maarryy ooff ssppeeccttrroossccooppiicc aabbssoorrppttiioonnss ooff R RC CN N ((2222..55)) • •

IR absorption 13 C NMR absorption

C
N absorption at 2250 cm–1 C
N absorbs at 115–120 ppm.

––   SSuum mm maarryy:: TThhee rreellaattiioonnsshhiipp bbeettw weeeenn tthhee bbaassiicciittyy ooff ZZ– aanndd tthhee pprrooppeerrttiieess ooff R RC CO OZZ

• Increasing basicity of the leaving group (22.2) • Increasing resonance stabilization (22.2)

R

O

O

O

C

C

C

Cl

R

acid chloride

O

anhydride

O R

R

O OH

carboxylic acid

~

R

C

O OR'

ester

R

C

NR'2

amide

• Increasing leaving group ability (22.7B) • Increasing reactivity (22.7B) • Increasing frequency of the C=O absorption in the IR (22.5)

 G Geenneerraall ffeeaattuurreess ooff nnuucclleeoopphhiilliicc aaccyyll ssuubbssttiittuuttiioonn • The characteristic reaction of compounds having the general structure RCOZ is nucleophilic acyl substitution (22.1). • The mechanism consists of two basic steps (22.7A): [1] Addition of a nucleophile to form a tetrahedral intermediate [2] Elimination of a leaving group • More reactive acyl compounds can be used to prepare less reactive acyl compounds. The reverse is not necessarily true (22.7B).

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Chapter 22–2

N Nuucclleeoopphhiilliicc aaccyyll ssuubbssttiittuuttiioonn rreeaaccttiioonnss [1] Reactions that synthesize acid chlorides (RCOCl) [a] From RCOOH (22.10A):

O R

O +

C

SOCl2

OH

R

+

C

+

SO2

Cl

HCl

[2] Reactions that synthesize anhydrides [(RCO)2O] [a] From RCOCl (22.8):

O R

C

+–

Cl

O

O

O

C

C

R'

R

O

O

C

Cl–

+

R'

O


OH

[b] From dicarboxylic acids (22.10B):

O

O

+

OH

H2O

O

O

cyclic anhydride

[3] Reactions that synthesize carboxylic acids (RCOOH) O

O

[a] From RCOCl (22.8): [b] From (RCO)2O (22.9):

R

R

C

+

Cl

O

O

C

C

O

H2O

pyridine

R

R

C

+

OH

R

H2O

2

R

C

Cl–

OH

O +

OR'

H2O (H+ or –OH)

R

C

O OH

(with acid)

or R

R

C

C

O–

+

(with base)

O

H2O, H+

[d] From RCONR'2 (R' = H or alkyl, 22.13):

+

N H

O +

O

[c] From RCOOR' (22.11):

C

+

+

R'2NH2

+

R'2NH

+

+ N H

OH

O R

C

NR'2

R' = H or alkyl

O

H2O, –OH R

C

O–

[4] Reactions that synthesize esters (RCOOR') O

[a] From RCOCl (22.8):

R

C

O +

Cl

R'OH

pyridine

R

C

OR'

Cl–

R'OH

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569

Carboxylic Acids and Their Derivatives 22–3 [b] From (RCO)2O (22.9):

O R

C

O

O O

C

+

R'OH

R

O

[c] From RCOOH (22.10C):

R

C

OH

+

+

RCOOH

OR'

O

H2SO4

R'OH

C

R

R

C

+ H2O

OR'

[5] Reactions that synthesize amides (RCONH2) [The reactions are written with NH3 as the nucleophile to form RCONH2. Similar reactions occur with R'NH2 to form RCONHR', and with R'2NH to form RCONR'2.] O

[a] From RCOCl (22.8):

R

C

O +

NH3

Cl

R

(2 equiv)

[b] From (RCO)2O (22.9):

O R

C

O O

C

[c] From RCOOH (22.10D):

R

R

NH3

R

(2 equiv)

R

OH

R

[2]


C

NH2

R

C

C

NH2

+

RCOO– NH4+

+

H2O

O OH

R'NH2

+

C

R

DCC

O

[d] From RCOOR' (22.11):

NH4+Cl–

O

[1] NH3

O C

+ NH2

O +

O C

C

+ H2O

NHR'

O +

NH3

OR'

R

C

NH2

+

R'OH



N Niittrriillee ssyynntthheessiiss ((2222..1188)) Nitriles are prepared by SN2 substitution using unhindered alkyl halides as starting materials. R X

+

–CN

R C N

X–

+

SN2

R = CH3, 1o



R Reeaaccttiioonnss ooff nniittrriilleess [1] Hydrolysis (22.18A) R C N

O

H2O (H+ or –OH)

R

C

O OH

(with acid)

or

R

C

O–

(with base)

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Chapter 22–4 [2] Reduction (22.18B) [1] LiAlH4 [2] H2O

R CH2NH2

1o amine

R C N O

[1] DIBAL-H [2] H2O

R

C

H

aldehyde

[3] Reaction with organometallic reagents (22.18C) R C N

O

[1] R'MgX or R'Li [2] H2O

R

C

R'

ketone

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571

Carboxylic Acids and Their Derivatives 22–5 C Chhaapptteerr 2222:: A Annssw weerrss ttoo PPrroobblleem mss 22.1

The number of C–N bonds determines the classification as a 1o, 2o, or 3o amide. NH2 1° amide O

O

O

1° amide H2N O

NH

oxytocin All seven others are 2° amides.

NH O

N H

O

O

O

HN

N

H N

O N H H2N

NH2 S

H N

O

1° amide

S

3° amide

O HO

22.2

As the basicity of Z increases, the stability of RCOZ increases because of added resonance stabilization. R

O

O

C

C

O

R + Br

Br

R

C

+

Br

The basicity of Z determines how much this structure contributes to the hybrid. Br– is less basic than –OH, so RCOBr is less stable than RCOOH.

22.3 O CH3 Cl

CH3

CH3 NH2

H

C

O Cl

CH3

O

O

C

C

NH2

H

C

Cl

NH2

This resonance structure contributes little to the hybrid since Cl– is a weak base. Thus, the C–Cl bond has little double bond character, making it similar in length to the C–Cl bond in CH3Cl. This resonance structure contributes more to the hybrid since –NH2 is more basic. Thus, the C–N bond in HCONH2 has more double bond character, making it shorter than the C–N bond in CH3NH2.

22.4 a. (CH3CH2)2CH

COCl

b.

C6H5COOCH3

re-draw

re-draw

O

O Cl 2-ethylbutanoyl chloride

2-ethyl

OCH3

methyl benzoate

alkyl group = methyl acyl group = benzoate

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Chapter 22–6 O

c. CH3CH2CON(CH3)CH2CH3 e.

re-draw

CH3CH2

O

C

O

C

benzoic propanoic anhydride

O N

acyl group = propanoic

N-ethyl-N-methyl N-ethyl-N-methylpropanamide

acyl group = propanamide

1

O

d.

C

H

acyl group = benzoic

OCH2CH3

CN

f.

3

alkyl group = ethyl acyl group = formate

3-ethylhexanenitrile

6 carbon chain = hexanenitrile

ethyl formate

22.5 a. 5-methylheptanoyl chloride

d. N-isobutyl-N-methylbutanamide

g. sec-butyl 2-methylhexanoate

O

O

O N

Cl

b. isopropyl propanoate

O

e. 3-methylpentanenitrile

h. N-ethylhexanamide

O O

CN O

O H

O

O O

N H

f. o-cyanobenzoic acid

c. acetic formic anhydride

OH

CH3

CN

CH3CONH2 has two H’s bonded to N that can hydrogen bond. CH3CON(CH3)2 does not have any H’s capable of hydrogen bonding. This means CH3CONH2 has much stronger intermolecular forces, which leads to a higher boiling point.

22.6

22.7 a. CH3

O

O

O

C

C

C

OCH2CH3

and CH3

N(CH2CH3)2

c. CH3CH2CH2

amide: C=O at lower wavenumber

O

b.

and

O

O

d.

O O

smaller ring: C=O at a higher wavenumber

NHCH3

and

CH3CH2CH2

2° amide: 1 N–H absorption at 3200–3400 cm–1

O O

O

anhydride: 2 C=O peaks

NH2

1° amide: 2 N–H absorptions O

and

C

Cl

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573

Carboxylic Acids and Their Derivatives 22–7 22.8 Hb

O

Hc signal from the 5 H's on the aromatic ring

Ha

O

H H

CH3

Ha

A

Molecular formula C9H10O2 5 degrees of unsaturation IR: 1743 cm–1 from ester C=O 3091–2895 cm–1 from sp2 and sp3 C–H 1 H NMR: Ha = 2.06 ppm (singlet, 3 H) – CH3 Hb = 5.08 ppm (singlet, 2 H) – CH2 Hc = 7.33 ppm (broad singlet, 5 H)

O

H H O

Hb

Hb signal from the 10 H's on the two aromatic rings

B Molecular formula C14H12O2 9 degrees of unsaturation IR: 1718 cm–1 from conjugated ester C=O 3091–2953 cm–1 from sp2 and sp3 C–H 1H NMR: Ha = 5.35 ppm (singlet, 2 H) Hb = 7.26–8.15 ppm (multiplets, 10 H)

22.9 NH2

H2N

H H H N S O

HO

CH3

N

H H H N S

O

CH3

N O

O

COOH

H COOH

cephalexin (Trade name: Keflex)

amoxicillin a. 4 stereogenic centers b. 24 = 16 possible stereoisomers c. enantiomer

a. 3 stereogenic centers b. 23 = 8 possible stereoisomers c. enantiomer

H2N

NH2

H H H N S O

HO

H H H N S

CH3

N

CH3

O

N

O

O H COOH

COOH

22.10 To draw the products of these nucleophilic acyl substitution reactions, find the nucleophile and the leaving group. Then replace the leaving group with the nucleophile and draw a neutral product. nucleophile O

a.

CH3

C

nucleophile

CH3OH Cl

CH3

O

O

C

C

OCH3

+ HCl

b. CH3

O NH3 OCH2CH3

CH3

C

NH2 + HOCH2CH3

leaving group

leaving group

22.11 More reactive acyl compounds can be converted to less reactive acyl compounds. a.

CH3COCl

CH3COOH

more reactive

YES

b. CH3CONHCH3 less reactive

NO

c.

less reactive CH3COOCH3

more reactive

CH3COOCH3

less reactive d.

(CH3CO)2O

more reactive

CH3COCl

NO

more reactive

YES

less reactive

CH3CONH2

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Chapter 22–8 22.12 The better the leaving group is, the more reactive the carboxylic acid derivative. The weakest base is the best leaving group. O

O

C

C


OCH 3

strongest base least reactive

b.

2

CH3CH2

C

OCH3

NH2

a.


NH

O


Cl

weakest base most reactive

intermediate

O

O

O

C

C

C

CH3CH2

NHCH3


NHCH 3

OH

CH3CH2


OH intermediate

strongest base least reactive

Cl

O O

C

CH2CH3


OOCCH

2CH3 weakest base most reactive

22.13 CH3

O

O

O

O

C

C

C

C

O

CH3

Cl3C

O

The Cl atoms are electron withdrawing, which makes the conjugate base (the leaving group, CCl3COO–) weaker and more stable.

CCl3

acetic anhydride trichloroacetic anhydride

22.14 O H2O

a.

O

O OH

+

pyridine Cl

O

b.

CH3COO


N – H Cl

c.

NH2 + NH4+Cl–

NH3 excess

O O

O CH3 + Cl–

(CH3)2NH

N(CH3)2

d.

+ (CH3)2NH2 Cl–

excess

22.15 The mechanism has three steps: [1] nucleophilic attack by O; [2] proton transfer; [3] elimination of the Cl
leaving group to form the product. OCH3

O OH

+

OCH3

O Cl

OCH3

O

Cl

O Cl O H N

H OCH3

OCH3

N

OCH3

OCH3

O O

OCH3

A

Cl

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575

Carboxylic Acids and Their Derivatives 22–9 22.16 O

O O

O

O OH

H2O

O NH3

HO

a.

c. O

b.

CH3OH

NH2

O

O

NH4

excess

O CH3

O

O (CH3)2NH

HO

d.

O N(CH3)2

excess

O (CH3)2NH2

22.17 Nucleophilicity decreases across a row of the periodic table so an NH2 group is more nucleophilic than an OH group. more nucleophilic Acetylation occurs here. H N

NH2 HO

C

CH3

O

HO

acetaminophen (active ingredient in Tylenol)

less nucleophilic

22.18 Reaction of a carboxylic acid with thionyl chloride converts it to an acid chloride. O

a. CH3CH2

C

O

SOCl2 OH

CH3CH2

C

O C

b.

Cl

O [1] SOCl2

OH

O [2] (CH3CH2)2NH (excess)

C

Cl

C

N(CH2CH3)2

(CH3CH2)2NH2 Cl–

22.19 COOH + CH3CH2OH

a.

H2SO4 C

OCH2CH3 + H2O

O O

COOH

b.

+

OH

C

H2SO4

+ H2 O

O

O COOH

c.

+

C NaOCH3

O

Na+

+ CH3OH

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Chapter 22–10 O

O

H2SO4

d. HO

+ H2O

OH

O

22.20 O C

O CH318OH

OH

C

18

OCH3 + H2O

22.21 O C

HO

OH O

H
OSO3H

H

C

HO

HO OH

O

H

O

HO OH H

HSO4–

+

O –OSO

O + H2SO4

3H

HSO4–

O H O

HO OH2 O

O

H2O +

H
OSO3H

+ HSO4–

22.22 O

a. O

CH3NH2

O H3NCH3

OH

b.

c.

+

O

O CH3NH2 DCC

O

O

[1] CH3NH2 OH

NHCH3

OH

NHCH3 + H2O

[2]


22.23 R

OH

OH

OH

O H

O H

C

C

C

C

C

OR'

R

OR'

R

OR'

R

product of step [1]

OH

R

O H

OH

R

product of step [5]

22.24 O [1] H3O+

OH

+

HO

O

+

HO

O

a.

O

CH3O

CH3O [2] H2O, –OH

octinoxate CH3O

O

C

OH

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Carboxylic Acids and Their Derivatives 22–11 O [1] H3O+

O

b.

OH

O

+

HO

+

HO

OH O

OH [2] H2

octyl salicylate

O, –OH

O OH

22.25 O

a. CH3CH2

–OH,

H2O

C 18 OCH3

18

O CH3CH2

C

O

+ H18OCH3

b.

O–

CH3CH2

–OH,

C

18

O

H2O CH3CH2

OCH3

C

O O

CH3CH2

C 18 O

+ HOCH3

22.26 HOCH2 O

OH HOCH2

HO

RCOOH, H2SO4

O O HO

RCOOCH2

OH

OH

RCOO

CH2OH

a long chain fatty acid

RCOO OOCR OOCR CH2 O O CH2OOCR OOCR

O

RCOO

sucrose

olestra

22.27 CH2OCO(CH2)15CH3

CH2 OH

Na+
OOC(CH2)15CH3

CH OH

Na+
OOC(CH2)15CH3

CH2 OH

Na+
OOC(CH2)7CH=CH(CH2)7CH3

hydrolysis

CHOCO(CH2)15CH3 CH2OCO(CH2)7CH=CH(CH2)7CH3 cis

glycerol

soap

cis

22.28 O

O NH

H2O, OH

OH

proton transfer

O

N H H

O

NH

O O

NH2

22.29 Aspirin has an ester, a more reactive acyl group, but acetaminophen has an amide, a less reactive acyl group. a. The ester makes aspirin more easily hydrolyzed with water from the air than acetaminophen. Therefore, Tylenol can be kept for many years, whereas aspirin decomposes. b. Similarly, aspirin will be hydrolyzed and decompose in the aqueous medium of a liquid medication, but acetaminophen is stable due to the less reactive amide group, allowing it to remain unchanged while dissolved in H2O.

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Chapter 22–12 ester O

C

amide H N

CH3

O COOH

CH3

O

HO

acetylsalicylic acid

C

acetaminophen

22.30 "Regular" amide is not hydrolyzed. H H H N S

R O

H3

H H N

R

O+

O

N O

H S HN OH

O

COOH

COOH

22.31 O

H N

O

H N

N H

N H

O

O

nylon 6,10

O Cl

Cl

H2N

+

NH2

O

22.32 OH

+

HOOC

COOH

HO

1,4-dihydroxymethylcyclohexane

In the polyester Kodel, most of the bonds in the polymer backbone are part of a ring, so there are fewer degrees of freedom. Fabrics made from Kodel are stiff and crease resistant, due to these less flexible polyester fibers.

terephthalic acid

O

O

O

O

O

O O

O

Kodel

22.33 O O

OH OH

+

OH

OH

Reaction occurs here (–H2O).

O

O O

O O

O

O O

PLA polyl(lactic acid)

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Carboxylic Acids and Their Derivatives 22–13 22.34 Acetyl CoA acetylates the NH2 group of glucosamine, since the NH2 group is the most nucleophilic site. HO

HO O

O HO

OH

+ CH3

C

O HO

NH2

HO

OH

SCoA HN

HO

glucosamine

O

NAG

22.35 a. CH3CH2CH2 Br

NaCN

CH3CH2CH2 CN

H2O, –OH

c. CN

CN

H2O, H+

b.

COO

COOH

CN

COOH

22.36 a.

CH3

O

OH

NH

C

C

C

NH2

CH3

O

b.

C

c.

NH

CH3CH2

NH2

OH

CH3CH2

C

O

OH C

NHCH3

NCH3

22.37 a. CH3CH2 Br

[1] NaCN [2] LiAlH4 [3] H2O

[1] DIBAL-H

b. CH3CH2CH2 CN

CH3CH2 CH2NH2

O CH3CH2CH2 C

[2] H2O

H

22.38 OCH3

a.

[1] CH3CH2MgCl CN

[2] H2O

OCH3 O C CH2CH3

O

CN

C

[1] C6H5Li

b.

[2] H2O

22.39 O

O CH2 CN

a.

[1] CH3MgBr

CH2 CN

CH2 C CH3

[2] H2O

c.

[1] DIBAL-H

CH2 C H

[2] H2O O

CH2 CN

b.

[1] (CH3)3CMgBr [2] H2O

O

CH2 C

CH2 CN C(CH3)3

d.

CH2 C H3O+

OH

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Chapter 22–14 22.40 CH3 C N

O

O

[1] CH3CH2MgBr

[1] CH3MgBr

CH3CH2 C N

[2] H2O

[2] H2O

22.41 O Cl

a.

g.

2,2-dimethylpropanoyl chloride

N

O

N-cyclohexylbenzamide

H O

b.

h.

cyclohexyl pentanoate

m-chlorobenzonitrile

O

Cl

CN

c.

O O O

i.

Cl 3-phenylpropanoyl chloride

isobutyl 2,2-dimethylpropanoate

d.

2-ethylhexanenitrile

O

j.

C

CN O

e.

Cl

O

cis-2-bromocyclohexanecarbonyl chloride

Br O

O

k. N

cyclohexanecarboxylic anhydride

N,N-diethylcyclohexanecarboxamide O

O

f.

l.

phenyl phenylacetate O

cyclopentyl cyclohexanecarboxylate

O

22.42 a. propanoic anhydride O

e. isopropyl formate

H

b.
-chlorobutyryl chloride

O

O

O O

i. benzoic propanoic anhydride O O

O

f. N-cyclopentylpentanamide

O

j. 3-methylhexanoyl chloride

NH

O

Cl Cl

O

Cl

c. cyclohexyl propanoate

g. 4-methylheptanenitrile

k. octyl butanoate

O

O CN

O

d. cyclohexanecarboxamide C

l. N,N-dibenzylformamide

h. vinyl acetate

O

O O

O NH2

O

CH3

H

N

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Carboxylic Acids and Their Derivatives 22–15 22.43 Rank the compounds using the rules from Answer 22.12. O

a.

O

O

NH2

OCH2CH2CH3


NH


OCH CH CH 2 2 3

2 strongest base least reactive


Cl

weakest base most reactive

intermediate

O

b.

Cl

O

O

O

O

ester least reactive

anhydride intermediate

O

O F3C

O O

CF3

anhydride with electronwithdrawing F's most reactive

O

O

c. OH

SH


OH

Cl


SH intermediate

strongest base least reactive


Cl

weakest base most reactive

22.44 a. Better leaving groups make acyl compounds more reactive. A has an electron-withdrawing NO2 group, which stabilizes the negative charge of the leaving group, whereas B has an electron-donating OCH3 group, which destabilizes the leaving group. O CH3

C

O O

NO2

CH3

A

C

OCH3

O B

an electron-withdrawing substituent an electron-donating substituent O O

O

N

O

N

O

O

OCH3

O

OCH3

O

one possible resonance structure leaving group from B Adjacent negative charges destabilize the leaving group.

one possible resonance structure

leaving group from A

Delocalizing the negative charge on the NO2 stabilizes the leaving group making A more reactive than B.

resonance structures for the leaving group b.

O

O C N

N

R

imidazolide

Nu

R C N Nu

O N

R

C

Nu

N

N

N

N

N

N

N

N

N

The leaving group is both resonance stabilized and aromatic (6
electrons), making it a much better leaving group than exists in a regular amide.

N

582

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© The McGraw−Hill Companies, 2011

Chapter 22–16 22.45 Reaction as an acid: O CH3

C

O

B NH2

CH3

O

C

NH

CH3

C

CH3CH2 NH2 NH

B

CH3CH2 NH

no resonance stabilization of the conjugate base

These two resonance structures make the conjugate base more stable, and therefore CH3CONH2 a stronger acid. Reaction as a base: O CH3

C

O NH2

CH3

C

This electron pair is localized on N.

CH3CH2 NH2

NH2

This electron pair is delocalized by resonance, making it less available for electron donation. Thus CH3CONH2 is a much weaker base.

22.46 CH3CH2CH2CH2COCl NH3

H2O

a.

CH3CH2CH2CH2COOH

pyridine

b.

d. N Cl– H

CH3CH2OH

c.

O

CH3COO
CH3CH2CH2CH2

O

CH3

CH3CH2CH2CH2CON(CH2CH3)2

excess N Cl– H

O

(CH3CH2)2NH2 Cl– C6H5NH2

f.

Cl–

excess

CH3CH2CH2CH2CONHC6H5 C6H5NH3 Cl–

22.47 O

O O

a.

SOCl2

d.

no reaction

H2O

2

NaCl

no reaction O

O

b.

(CH3CH2)2NH OH

e.

excess

N(CH2CH3)2 O

O CH3OH

c.

O H2N(CH2CH3)2 OCH3

O

O

f. OH

NH4 Cl–

(CH3CH2)2NH

e.

CH3CH2CH2CH2COOCH2CH3

pyridine

CH3CH2CH2CH2CONH2

excess

CH3CH2NH2 excess

NHCH2CH3 O O

H3NCH2CH3

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583

Carboxylic Acids and Their Derivatives 22–17 22.48 C6H5CH2COOH

a.

O

NaHCO3

O Na

b.

NaOH

O

c.

SOCl2

O

O

CH3OH H2SO4

+ H2CO3 g.

OCH3 O

CH3OH

h. + H2 O

–OH

O Na

i.

O

[1] NaOH [2] CH3COCl

O

Cl

d.

NaCl

no reaction

j.

O

CH3NH2 DCC

e.

NH3

O

k.

(1 equiv) O NH4

f.

O

[1] NH3 [2]


l.

NH2

O

O

NHCH3

[1] SOCl2

O

[2] CH3CH2CH2NH2

NHCH2CH2CH3 O

[1] SOCl2 [2] (CH3)2CHOH

OCH(CH3)2

22.49 c.

CH3CH2CH2CO2CH2CH3

O

H2O, –OH

HO

O O

a.

SOCl2

no reaction

d.

NH3

NH2

O

b.

HO

O

H3O+

HO

OH

e. CH3CH2NH2

NHCH2CH3

HO

22.50 NH2

a.

H3O+

CH2COOH

b.

CH2COO

H2O, –OH

O

22.51 a.

H3O+

COOH

d.

[1] CH3CH2Li [2] H2O

C6H5CH2CN

b.

H2O, –OH

COO

[2] H2O

C

[2] H2O

O

[1] LiAlH4

CH2NH2

f. O

H

e. [1] DIBAL-H

CH3

c. [1] CH3MgBr

O

[2] H2O

584

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© The McGraw−Hill Companies, 2011

Chapter 22–18 22.52 O COOH

C

SOCl2

a. OH

g.

O

h. C6H5CH2COOH

+

+ C6H5

N H (excess)

N

C6H5

[2] H2O

d. (CH3)2CHCOOH +

H3O+

C6H5CH2CH2CH2COOH


HOOC

O

COOH

(CH3CO)2O +

NH2

O

H2O –OH

NHCOCH3

(excess)

+

NH2

O

O

+ CH3COO– H3N

O O

f.

C6H5CH2CH2CH2CN

O

k.

NHCOCH3

O

C6H5(CH2)2NH(CH2)3CH3

OCH(CH3)CH2CH3

CH3

e.

CH3CH2CH2CH2

O

j.

C

(CH3)2CH

[1] SOCl2

CH2CH2CH3

O

H2SO4

CH3CH2CHOH

i.

C

[2] H2O, –OH

[2] CH3CH2CH2CH2NH2 [3] LiAlH4 [4] H2O

N Cl– H H O

[1] CH3CH2CH2MgBr

c. C6H5CN

CH3CH2CH2CH2Br

OH

b. C6H5COCl

O

[1] NaCN Cl

H3O+

OH

l.

C6H5CH2CH2COOCH2CH3

OH O

H2O –OH

C6H5CH2CH2COO– +

O

HOCH2CH3

22.53 Both lactones and acetals are hydrolyzed with aqueous acid, but only lactones react with aqueous base. O

a.

O O

O

X

H3O+

O

OH O OH

COOH OH

NaOH

b.

O

O O

O

X

H2O

O

CO2 Na+ OH

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Carboxylic Acids and Their Derivatives 22–19 22.54 Br NaCN

CN

H3O+

COOH SOCl2

A

COCl [1] (CH3)2CuLi

C

[2] H2O

O

D

B [1] LiAlH4

CH3OH,

[2] H2O

CH3

E

H+ PCC

CH2NH2

CO2CH3 [1] DIBAL-H

CHO [1] CH3Li

[2] H2O C

G

F

OH

[2] H2O

H

[1] LiAlH4

(CH3CO)2O

[2] H2O

CH2OTs NaCN

CH2OH

CH2NHCOCH3

CH2CN

TsCl, pyridine I

K

J

L [1] CH3MgBr [2] H2O O

M

22.55 O

H D

C CH3

a.

CH3COCl

OH

O

c.

pyridine CH3

CH3 Br

b.

CH3

NaCN

H+

COOH

O CN

H D

C

H D CH3CH2OH

D H

d.

CH3

C

+ Cl C H 6 5

CH3

C

CH3

CH3

C

C

H NH2 (2 equiv)

C6H5

CH3

H NH C

O

COOCH2CH3

+

CH3

22.56 Hydrolyze the amide and ester bonds in both starting materials to draw the products.

a.

O

O

CO2CH2CH3

H2O

O

CO2H

O HO

N H

OH

H2N NH2

oseltamivir

NH2

C6H5

C

H NH3 + Cl–

586

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© The McGraw−Hill Companies, 2011

Chapter 22–20 NH2

H N

HOOC

b.

NH2

H2O

O CH3O

HOOC

H2N

+ CH3OH

OH COOH

O

O

aspartame

phenylalanine

22.57 O C

Cl

a.

O

C Cl

C Cl

H

O

HO O

b.

NHNH2 + Cl

+ HN

N O

C

NH2NH

NH2NH

NH2NH2

O

O

O O

O

proton

O OH

transfer HO

OH

H+

O

OH

22.58 H+

O CH3

C

OH

OH CH3

C

OH

OH

18

O H

H

18

CH3 C OH2

18

18

O H

+ H2 O

OH

CH3 C OH

CH3 C OH

O H

OH

CH3 C 18

+ H2O

O H

O H CH3

C

O

H2O

18

CH3

OH

C

18

+ H3O+

OH

A OH CH3 C 18

O H

A

OH CH3

18

O H

A

+ H3O+

+ H2 O

Two possibilities for A:

OH CH3 C

C 18 O H

OH H2O

CH3

C

18

O

+ H3O+

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Carboxylic Acids and Their Derivatives 22–21 22.59 H2O

H

H2O

O H O O

OH

OH

OH

H
OH2

O

O

OH

O

OH

O

H2O

OH

H

-butyrolactone

H
OH2 O H3O

O

HO

HO

OH

H

H2O

OH

4-hydroxybutanoic acid GHB

22.60 enzyme CH2OH

CH2 O H

O O CO2H

H A

aspirin

O

O

OH CO2H

OH CO2H

CH2 A

H A

O O OH CO2H

A

CH2 H O O HA

CH2O C

CH2O C CH3

inactive enzyme

A

O H

OH

O

CO2H

OH CO2H

+ CH3

salicylic acid

A

588

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Chapter 22–22 22.61 O R

O

C

O R' +

OH

This bond is not broken. H CH3

O CH3

C

O R CH2CH3 X

C

mechanism

CH3

C

+ R'OH

O

H CH3

O

accepted

C

R

H2O SN2

O

H CH3

O

HO

C

CH2CH3

(2R)-2-butanol

CH3

C

O X

C

CH2CH3

According to the accepted mechanism, the stereochemistry around the stereogenic center is retained in the product.

This bond is cleaved. H CH3

O CH3

C

C

O

CH2CH3

CH3

C

H2 O SN2 alternative

X

CH3 H

O

OH

O

CH3CH2

C

OH

(2S)-2-butanol

This SN2 mechanism would form the product of inversion leading to (2S)-2butanol. Since (2R)-2-butanol is the only product formed, the SN2 mechanism does not occur during ester hydrolysis.

22.62 CH3O

CH3O O

C

O

O

OH

OH O

O

proton

O

OH

O O

OCH3

transfer

HO

OCH3

CO2H HO

OCH3

D

H OH2

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Carboxylic Acids and Their Derivatives 22–23 22.63 H OCH2CH3 O

OH

O

O

H–Cl

A

Cl

H

O

O CH2CH3

OCH2CH3

OH

OH

O

Cl

H–Cl

OCH2CH3 OH2

OCH2CH3

O

OH

OCH2CH3

O

O

OH

OCH2CH3 OH

O

H

Cl

H

H–Cl Cl Cl CO2CH2CH3 OCH2CH3

OCH2CH3

O

Cl

+ Cl

Cl

=

O

H2 O

B

22.64 The mechanism is composed of two parts: hydrolysis of the acetal and intramolecular Fischer esterification of the hydroxy carboxylic acid. H2O

H O

O OH

H OH2

O

O

H2O O

HO

O

O

OH O

O

H2O

H

O

O

OH

OH

O

+ H2O

HO

H HO HO

H OH2

O

HO

HO

HO HO

O

OH

O

OH

HO

OH

O

+ H3O+

O

+ H2O

H

H OH2 H2O HO

H O

OH OH

OH

HO

HO

O

OH

H OH2 OH OH

HO

+ H2O O

O

O O

HO

O H

HO

HO

+ H3O+

H2O

OH2 OH

+ H2O

590

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© The McGraw−Hill Companies, 2011

Chapter 22–24 22.65 (any base) B O

NH2 OH

+

CH3CH2O

X

C

H N

OCH2CH3

O C

OCH2CH3

O

OCH2CH3

C

H N

OH

OCH2CH3

OCH2CH3

OH

+ HB+

diethyl carbonate O

O HN

H

HN

+ CH3CH2O–

O Y

O

OCH2CH3 H N

O

O

H N H

C

OCH2CH3

OH B

+ HB+

HB+

O

OCH2CH3

+ CH3CH2O–

CH3CH2OH + B

22.66 sp3 C

sp2 C

O RCH2 Cl

R

less electrophilic C more crowded C since it is surrounded by four atoms

C

Cl

O R

C

Cl

more electrophilic C due to electron-withdrawing O more reactive

This resonance structure illustrates how the electronegative O atom withdraws more electron density from C. The sp2 hybridized C of RCOCl is much less crowded, and this makes nucleophilic attack easier as well.

22.67 O H3 HO

CN

O+

Ha

A C6H10O2: 2 degrees of unsaturation O

Hb H c A

IR: 1770 cm–1 from ester C=O in a five-membered ring 1H NMR: H = 1.27 ppm (singlet, 6 H) – 2 CH groups a 3 Hb = 2.12 ppm (triplet, 2 H) – CH2 bonded to CH2 Hc = 4.26 ppm (triplet, 2 H) – CH2 bonded to CH2

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591

Carboxylic Acids and Their Derivatives 22–25 proton HO

C N

HO

C N OH2

HO

transfer

C NH

H OH2

OH

imidic acid H2O

C NH2

HO

C NH2

HO

OH

C NH2

HO

O H OH2

O H

amide

H2O

H2O

H2O HO

NH2

HO

NH2

O H

H O

H OH2

O

H O

NH3 O

O

NH3

O

H3O

H2O

H2O

O

22.68 Fischer esterification is treatment of a carboxylic acid with an alcohol in the presence of an acid catalyst to form an ester. a. (CH3)3CCO2CH2CH3

(CH3)3CCOOH

O

c.

+ HOCH2CH3

O

b.

OH

O

HO

O

OH O

O

d.

HO

OH

HO

O

O

O

22.69 NH3 Br excess

a.

NH2 H3O+

NaCN Br

H N

OH NH2

CN O

DCC

O

NaOH

b.

Br

OH H2SO4 OH

OH

(from a.)

(from a.)

O

O [1]

c.

O

CN

[2] H2O

MgBr

Mg Br O

MgBr

592

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© The McGraw−Hill Companies, 2011

Chapter 22–26 O

d.

OH

(from b.)

CrO3

OH

H2SO4, H2O

H2SO4

OH

O O

(from c.) [1]

MgBr

e.

OH

(2 equiv)

O

[2] H2O O

(from d.) (from c.) [1]

f.

MgBr

OH

(2 equiv)

O

[2] H2O O

(from b.)

22.70 a.

Br

–CN

H2O,

CN

b.

CN H+ COOH

(from a.) c.

COOH

SOCl2

COCl

(from b.) d.

COOH

(from b.)

e.

HOCH2CH3

CO2CH2CH3

H2SO4

[1] CH3MgBr CN

(from a.)

C

[2] H2O

CH3

O

[1] DIBAL-H

f.

CN

(from a.)

g.

CHO

[2] H2O [1] LiAlH4

CN

(from a.) h.

CH2NH2

(from g.)

CH2NH2

[2] H2O CH3COCl

CH2NHCOCH3

22.71 a. CH3Cl

+ NaCN

CH3 CN

H3O+

CH3 COOH

[1] CO2 CH3 Cl + Mg

Br + NaCN

b.

sp2 Br + Mg

CH3 MgCl

[2] H3O+

CH3 COOH

This method can't be used because an SN2 reaction can't be done on an sp2 hybridized C. MgBr

[1] CO2 [2] H3O+

COOH

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Carboxylic Acids and Their Derivatives 22–27 This method can't be used because an SN2 reaction can't be done on a 3° C.

c. (CH3)3CCl + NaCN

(CH3)3C

(CH3)3C

Cl + Mg

d. HOCH2CH2CH2CH2Br

[1] CO2

MgCl

(CH3)3C

[2] H3O+

H3O+

HOCH2CH2CH2CH2 CN

+ NaCN

COOH

HOCH2CH2CH2CH2 COOH

This method can't be used because you can't make a Grignard reagent with an acidic OH group.

HOCH2CH2CH2CH2 Br + Mg

22.72 CH3OH O

SOCl2

CH3

CH3Cl

HNO3 H2SO4

AlCl3

CH3

COOH

KMnO4

O2N

H2SO4

O 2N

C

CH3CH2OH

OCH2CH3

O2N H2 Pd-C

(+ ortho isomer)

O C

OCH2CH3

H2N

22.73 O NH2

HO N H

N H

HO

CH3COCl

NaH

N H

O

N H

SOCl2

serotonin

O

N H CH3Cl

CH3COOH

SOCl2

CH3OH

O

CrO3 H2SO4, H2O

N H

CH3O

CH3CH2OH

N H

melatonin

22.74 O

O CH3CH2Cl

a. OH

COH

KMnO4

AlCl3

OH

CNH2

NH3

OH

OH

OH

salicylamide NO2

HNO3

Cl

(2 equiv)

OH

(+ para isomer)

b.

SOCl2

O

H2, Pd-C

H2SO4 HO

(+ ortho isomer)

HO

NH2

H N

CH3COCl HO

(More nucleophilic NH2 reacts first.)

C

CH3

O

acetaminophen

594

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Chapter 22–28 H N

c.

C

CH3

NaH

O

HO

H N

C

CH3

H N

CH3CH2Br

O

O

CH3

O

CH3CH2O

acetaminophen (from b.)

C

p-acetophenetidin

22.75 HO [1] CrO3, H2SO4, H2O [2] SOCl2 O

a.

O

O Cl

Cl2

Cl AlCl3

FeCl3

CH3OH O

SOCl2

b.

CH3Cl

CH3

AlCl3

CH3

HNO3 H2SO4

COOH

KMnO4

C

CH3OH

O2N

H

O2N

O2N

(+ ortho isomer)

H2, Pd-C O C

O

c.

FeBr3

Br

HNO3 H2SO4

O

O OCH3

N H

Br2

OCH3

+

C Cl [1] CrO3, H2SO4, H2O [2] SOCl2

Br O2N (+ ortho isomer)

H2N

CH3CH2CH2OH O Br

H2, Pd-C

Br Cl

H2N

OCH3

[1] CrO3, H2SO4, H2O [2] SOCl2 CH3CH2OH

H

N O

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Carboxylic Acids and Their Derivatives 22–29 (CH3)3COH

CH3OH

HCl

SOCl2

(CH3)3CCl

d.

CH3Cl

AlCl3

KMnO4

SOCl2 HO

AlCl3

Cl

(+ ortho isomer) Br

Br2 FeBr3

Br

CH3Cl

O

O Br

KMnO4

Br

NaOH

HO

AlCl3 (+ ortho isomer)

O O

O O

O O

(CH3)3C

Br

22.76 CH3Cl

a.

CH3

AlCl3 CH3OH

SOCl2

Br2

CH2Br

h


NaCN

CH2CN

H2O H+

CH2COOH HOCH2CH3 H2SO4

CH3Cl

CH2COOCH2CH3

ethyl phenylacetate NO2

(from a.) b.

CH3

CH3Cl

NO2 CH3

HNO3

NH2 COOH

KMnO4

H2SO4

AlCl3

NH2 COOH

H2

HOCH3

Pd-C

H2SO4

COOCH3

methyl anthranilate

(+ para isomer)

(from a.)

O CH3 KMnO 4

CH3Cl

c.

COOH

AlCl3 CH3CH2OH

OH CH3COOH H2SO4

[1] LiAlH4 [2] H2O

CrO3 H2SO4, H2O

O

benzyl acetate

CH3COOH

22.77 [1] LiAlH4 [2] H2O Cl

a. Cl

–CN

NC

excess

CN

O

H2O H2SO4

NH2

H2N

OH

HO O

b.

Br2 h


Br

KOC(CH3)3

[1] O3

CHO

CrO3

COOH

[2] Zn, H2O

CHO

H2SO4, H2O

COOH

596

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Chapter 22–30 22.78 O

a. CH3

C

O

CH313CH2OH OH

d. CH3 CH2OH

–

13

CH3 CH2Br

13C

CH3

H2SO4

OH

O CH3

18

OH

OCH2CH3

CH3COCl

CH3CH218OH

+ base (H18O–) PBr3

C

CH3

H218O

13

O CH3CH2OH

13

H2SO4, H2O

c. CH3CH2Br

13

OCH2CH3

O

CrO3

b. CH313CH2OH

C

CH3

H2SO4

CH3

CH218OH

C

18

OCH2CH3 18O

CrO3

13

H2SO4, H218O

18O

13 C 18 CH3 OH

CH3CH2OH H2SO4

13 C

CH3

+

OCH2CH3

H218O

22.79

a. HO

OH

HO

and

C

C

O

O

O

O

O

OH

O

O O

O O

O NH

b. ClOC

COCl

and

NH2

H2N

O

HN

NH

O

HN

22.80 O

a.

O

O O

O

O

O

O

OH

HO

O O O

O O

b.

O

O

HO

OH

O O O

O

O

HO

OH

HO

OH

O

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597

Carboxylic Acids and Their Derivatives 22–31 22.81 a. Docetaxel has fewer C’s and one more OH group than taxol. This makes docetaxel more water soluble than taxol. OH O

b.

O

O

O

OH O

N

O

H

carbamate

=

RO

OH HO O

docetaxel

carbamate

H

O

NHR

O

O O

O RO

O NHR

RO

1 most stable

O NHR

O

RO

4 least stable

NHR

RO

3

2 More basic N atom allows N to donate electron density more than O, so this structure contributes more than 3 to the hybrid.

Increasing stability: 4 < 3 < 2 < 1

H OH2

O

c.

OH O

O

NHR

O

NHR

H

O

C

H

H2O

NHR

H2O

H2O

NHR

H OH2

O C

+ NH2R

H OH2

O H3O H2O

H3NR OH O

d.

O

O

OH O

N H

O H3O+

OH

docetaxel

HO O

O

H O

O

O

CO2 COOH

OH

H3N OH

OH

O

O OH CH3CO2H

HO HO

H HO HO

O

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Chapter 22–32 22.82 O

a. CH3

C

O

and OCH3

CH3CH2

C

c.

OH

contains a broad, strong OH absorption at 3500–2500 cm–1 O

N(CH3)2

and

C=O at < 1700 cm–1 due to the stabilized amide

NH2

2 NH absorptions at 3200–3400 cm–1 C=O absorption higher wavenumber O

O

and

b.

O

O

d. O

Cl

OH

O

and CH3

Cl

Acid chloride CO absorbs at much higher wavenumber.

OH absorption at 3500–3200 cm–1 + C=O

ketone

only C=O

22.83 O

a. C6H5COOCH2CH3

CH3CH2COOCH2CH3

O

b.

most resonance stabilized

least resonance stabilized CH3COOCH3

CH3CONH2

Increasing wavenumber

CH3COCl

Increasing wavenumber

22.84 a.

C6H12O2
one degree of unsaturation IR: 1738 cm–1
C=O NMR: 1.12 (triplet, 3 H), 1.23 (doublet, 6 H), 2.28 (quartet, 2 H), 5.00 (septet, 1 H) ppm

d. C4H7ClO
one degree of unsaturation IR: 1802 cm–1
C=O (high wavenumber, RCOCl) NMR: 0.95 (triplet, 3 H), 1.07 (multiplet, 2 H), 2.90 (triplet, 2 H) ppm O

O

Cl O

b.

C4H7N IR: 2250 cm–1
triple bond NMR: 1.08 (triplet, 3 H), 1.70 (multiplet, 2 H), 2.34 (triplet, 2 H) ppm CH3CH2CH2C N

c. C8H9NO IR: 3328 (NH), 1639 (conjugated amide C=O) cm–1 NMR: 2.95 (singlet, 3 H), 6.95 (singlet, 1 H), 7.3–7.7 (multiplet, 5 H) ppm

e. C5H10O2
one degree of unsaturation IR: 1750 cm–1
C=O NMR: 1.20 (doublet, 6 H), 2.00 (singlet, 3 H), 4.95 (septet, 1 H) ppm O

O

f. C10H12O2
five degrees of unsaturation IR: 1740 cm–1
C=O NMR: 1.2 (triplet, 3 H), 2.4 (quartet, 2 H), 5.1 (singlet, 2 H), 7.1–7.5 (multiplet, 5 H) ppm O O

O N H

CH3

g. C8H14O3
two degrees of unsaturation IR: 1810, 1770 cm–1
2 absorptions due to C=O (anhydride) NMR: 1.25 (doublet, 12 H), 2.65 (septet, 2 H) ppm O

O O

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599

Carboxylic Acids and Their Derivatives 22–33 22.85 A. Molecular formula C10H12O2
five degrees of unsaturation IR absorption at 1718 cm–1
C=O NMR data (ppm): triplet at 1.4 (CH3 adjacent to 2 H's) singlet at 2.4 (CH3) quartet at 4.4 (CH2 adjacent to CH3) doublet at 7.2 (2 H's on benzene ring) doublet at 7.9 (2 H's on benzene ring)

O O

B. IR absorption at 1740 cm–1
C=O NMR data (ppm): singlet at 2.0 (CH3) triplet at 2.9 (CH2 adjacent to CH2) triplet at 4.4 (CH2 adjacent to CH2) multiplet at 7.3 (5 H's, monosubstituted benzene)

O O

22.86 Molecular formula C10H13NO2
five degrees of unsaturation IR absorptions at 3300 (NH) and 1680 (C=O, amide or conjugated) cm–1 NMR data (ppm): triplet at 1.4 (CH3 adjacent to CH2) singlet at 2.2 (CH3C=O) quartet at 3.9 (CH2 adjacent to CH3) CH3CH2O doublet at 6.8 (2 H's on benzene ring) singlet at 7.2 (NH) doublet at 7.4 (2 H's on benzene ring)

H N

CH3 O

phenacetin

22.87 Molecular formula C11H15NO2
five degrees of unsaturation IR absorption 1699 (C=O, amide or conjugated) cm–1 NMR data (ppm): triplet at 1.3 (3 H) (CH3 adjacent to CH2) singlet at 3.0 (6 H) (2 CH3 groups on N) quartet at 4.3 (2 H) (CH2 adjacent to CH3) doublet at 6.6 (2 H) (2 H's on benzene ring) doublet at 7.9 (2 H) (2 H's on benzene ring)

CH3

O N

CH3

OCH2CH3

C

22.88 a. Molecular formula C6H12O2
one degree of unsaturation IR absorption at 1743 cm–1
C=O 1 O H NMR data (ppm): triplet at 0.9 (3 H) – CH3 adjacent to CH2 O multiplet at 1.35 (2 H) – CH2 D multiplet at 1.60 (2 H) – CH2 singlet at 2.1 (3 H – from CH3 bonded to C=O) triplet at 4.1 (2 H) – CH2 adjacent to the electronegative O atom and another CH2

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Chapter 22–34 b. Molecular formula C6H12O2
one degree of unsaturation IR absorption at 1746 cm–1
C=O O 1 H NMR data (ppm): doublet at 0.9 (6 H) – 2 CH3's adjacent to CH O E multiplet at 1.9 (1 H) singlet at 2.1 (3 H) – CH3 bonded to C=O doublet at 3.85 (2 H) – CH2 bonded to electronegative O and CH

22.89 O

There is restricted rotation around the amide C–N bond. The 2 H's are in different environments (one is cis to an O atom, and one is cis to CH2Cl), so they give different NMR signals.

O

ClCH2 C

ClCH2 C NH2

N H

4.02 ppm

H

different environments

7.35 and 7.60 ppm

This resonance structure gives a significant contribution to the resonance hybrid.

22.90 18 18

O C6H5

C

–OH

OCH2CH3

ethyl benzoate

18

O C6H5 C OCH2CH3 OH

H2O

OH

C6H5 C OCH2CH3

18 –OH

OH

OH

O

C6H5 C OCH2CH3 O

C6H5

C

OCH2CH3 18

–

Two OH groups are now equivalent and either can lose H2O to form labeled or unlabeled ethyl benzoate.

OH

Unlabeled starting material was recovered.

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601

Carboxylic Acids and Their Derivatives 22–35 22.91 NH2

abbreviate as: R

N H

CH3O

NH2

R H

NH2

C

RCH2CH2NH2 H C O

O CH3OOC

CH3OOC CH3OOC

OCOCH3

CH3OOC

OCOCH3 OCH3

OCH3

proton transfer H RCH2CH2N

H2O

RCH2CH2NH OH2 H C

CH3OOC

any proton CH3OOC source

CH3OOC

CH3OOC

OCOCH3

CH3OOC

OCH3

RCH2CH2NH OH H C

OCOCH3

CH3OOC

OCH3

OCOCH3 OCH3

R

RCH2CH2N

H BH3

RCH2CH2N CH3O

CH3OOC CH3OOC

OCOCH3 OCH3 H3O

CH3O

CH3O

N

N O H

O C O OCOCH3

CH3OOC OCH3

CH3OOC

H H

OCOCH3 OCH3

N

CH3OOC

OCOCH3 OCH3

BH3 CH3O

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603

Substitution Reactions of Carbonyl Compounds 23–1 C Chhaapptteerr 2233:: SSuubbssttiittuuttiioonn R Reeaaccttiioonnss ooff C Caarrbboonnyyll C Coom mppoouunnddss aatt tthhee

C Caarrbboonn  K Kiinneettiicc vveerrssuuss tthheerrm mooddyynnaam miicc eennoollaatteess ((2233..44)) O

Kinetic enolate • The less substituted enolate • Favored by strong base, polar aprotic solvent, low temperature: LDA, THF, –78 oC

R

kinetic enolate O

Thermodynamic enolate • The more substituted enolate • Favored by strong base, protic solvent, higher temperature: NaOCH2CH3, CH3CH2OH, room temperature

R

thermodynamic enolate

 H Haallooggeennaattiioonn aatt tthhee

ccaarrbboonn [1] Halogenation in acid (23.7A) O R

C

O C

H

X2 R

CH3COOH

C

X

C

• •

The reaction occurs via enol intermediates. Monosubstitution of X for H occurs on the
carbon.

•

The reaction occurs via enolate intermediates. Polysubstitution of X for H occurs on the
carbon.


-halo aldehyde or ketone

X2 = Cl2, Br2, or I2

[2] Halogenation in base (23.7B) O R

C

C

R

O

X2 (excess) –OH

R

C

C

R

•

X X

H H X2 = Cl2, Br2, or I2

[3] Halogenation of methyl ketones in base—The haloform reaction (23.7B) O R

C

CH3

–OH

X2 = Cl2, Br2, or I2

•

O

X2 (excess) R

C

O–

+ HCX3

haloform

The reaction occurs with methyl ketones, and results in cleavage of a carbon–carbon  bond.

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Chapter 23–2  R Reeaaccttiioonnss ooff --hhaalloo ccaarrbboonnyyll ccoom mppoouunnddss ((2233..77C C)) [1] Elimination to form ,-unsaturated carbonyl compounds O R




Li2CO3

Br

LiBr DMF

O R

•

Elimination of the elements of Br and H forms a new
bond, giving an ,unsaturated carbonyl compound.

•

The reaction follows an SN2 mechanism, generating an -substituted carbonyl compound.






[2] Nucleophilic substitution O

O

Nu


Br

R

R


Nu

 A Allkkyyllaattiioonn rreeaaccttiioonnss aatt tthhee  ccaarrbboonn [1] Direct alkylation at the  carbon (23.8) O

•

O

C
H C

[1] Base

C
R C

[2] RX

+

X–

The reaction forms a new C–C bond to the  carbon. LDA is a common base used to form an intermediate enolate. The alkylation in Step [2] follows an SN2 mechanism.

• •

[2] Malonic ester synthesis (23.9) [1] NaOEt



[2] RX [3] H3O+,


H

•

R CH2COOH

•

H C COOEt COOEt

diethyl malonate

[1] NaOEt

[1] NaOEt

[2] RX

[2] R'X [3] H3O+,


The reaction is used to prepare carboxylic acids with one or two alkyl groups on the  carbon. The alkylation in Step [2] follows an SN2 mechanism.



R CHCOOH R'

[3] Acetoacetic ester synthesis (23.10)

O CH3

C  H C H

[1] NaOEt [2] RX [3] H3O+,


•

O CH3

C

CH2 R

•

COOEt

ethyl acetoacetate

O [1] NaOEt [2] RX

[1] NaOEt [2] R'X [3] H3O+,


CH3

C

CH R R'

The reaction is used to prepare ketones with one or two alkyl groups on the  carbon. The alkylation in Step [2] follows an SN2 mechanism.

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Substitution Reactions of Carbonyl Compounds 23–3 C Chhaapptteerr 2233:: A Annssw weerrss ttoo PPrroobblleem mss 23.1 • To convert a ketone to its enol tautomer, change the C=O to C–OH, make a new double bond to an
carbon, and remove a proton at the other end of the C=C. • To convert an enol to its keto form, find the C=C bonded to the OH. Change the C–OH to a C=O, add a proton to the other end of the C=C, and delete the double bond. [In cases where E and Z isomers are possible, only one stereoisomer is drawn.] HO

a. O

OH O

b.

O

d.

C6H5

OH H

C6H5

C6H5

H

c.

C6H5

C6H5

O O

O

OH

O

e. O

O O

OH

OH

f.

C6H5

Draw mono enol tautomers only.

OH C6H5

O

OH

OH

(Conjugated enols are preferred.)

23.2 O

OH

OH

(E and Z)

2-butanone

23.3

C=C has one C bonded to it.

C=C has two C's bonded to it. The more substituted double bond is more stable.

The mechanism has two steps: protonation followed by deprotonation. OH

OH H H OH2

O H H H

H2O H H

O H H

H3O+

O

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Chapter 23–4 23.4 Cl

H

H

H

O

O D

D

D Cl H

H

O D

D Cl

Cl

D H

D Cl H

O D

O D H Cl

H Cl

D D

D D

H

H D Cl

O D

O

Cl

23.5 a. CH3CH2O

O

O

O

O

O

C

C

C

C

C

C

OCH2CH3

CH3CH2O

H

b.

CH3

C

OCH2CH3

CH3CH2O

H

O

O

O

O

C

C

C

C

C

OCH2CH3

CH3

C

c.

CH3

O

C

C

CHC N

CH3

C

OCH2CH3

OCH2CH3

CH3

O C C

H O

C H

O CH

O

OCH2CH3

H O CH3

CHC N

C

C C N H

23.6

The indicated H’s are  to a C=O or C
N group, making them more acidic because their removal forms conjugate bases that are resonance stabilized. O

a.

CH3

C

O

b.

CH2CH2CH3

c.

CH3CH2CH2 CN

CH3CH2CH2

C

OCH2CH3

d. O

O CH3

23.7 O

O

O

–H+ O

O

–H+ O

O

–H+

O O

O

no resonance stabilization least acidic

O

O

O

O

O

O

Two resonance structures stabilize the conjugate base. intermediate acidity

Three resonance structures stabilize the conjugate base. most acidic

O

O

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Substitution Reactions of Carbonyl Compounds 23–5 23.8

In each of the reactions, the LDA pulls off the most acidic proton. O

O O LDA

a.

c.

THF

CHO

b.

O LDA

CH3 C OCH2CH3

CHO

LDA

CN

d.

THF

CH2 C

THF

OCH2CH3

CN

LDA THF

23.9 the gas O

O

O

O

O CH4

OCH2CH3

H3

OCH2CH3

O

O+

OCH2CH3

+

+ MgBr

H CH3–MgBr

The CH2 between the two C=O’s contains acidic H’s, so CH3MgBr reacts as a base to remove a proton. Thus, proton transfer (not nucleophilic addition) occurs. 23.10 • LDA, THF forms the kinetic enolate by removing a proton from the less substituted C. • Treatment with NaOCH3, CH3OH forms the thermodynamic enolate by removing a proton from the more substituted C. O

O

LDA, THF O

a.

O

O

NaOCH3

LDA, THF O

c.

CH3OH

NaOCH3

O

CH3OH

LDA, THF

O

b.

O

NaOCH3 CH3OH

23.11 a. This acidic H is removed with base to form an achiral enolate. O

O H CH3

NaOH

(2R)-2-methylcyclohexanone

O H2O

O CH3

CH3

H

H

achiral Protonation of the planar achiral enolate occurs with equal probability from two sides so a racemic mixture is formed. The racemic mixture is optically inactive.

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Chapter 23–6 b.

O


O


O

NaOH

or

H

H CH3

O

CH3




H2O

H CH3


H CH3

(3R)-3-methylcyclohexanone This stereogenic center is not located at the
carbon, so it is not deprotonated with base. Its configuration is retained in the product, and the product remains optically active.

23.12 O

Cl

a.

O

H2O, HCl O

b.

CH3CH2CH2

C

Br

Br2, CH3CO2H

Cl2

O

O

c. O

Br2 H

CH3CO2H

CH3CH2CH

C

H

Br

23.13 O

O

O Br Br

Br2, –OH

a.

O I2, –OH

b.

I I

I I

O

O I2, –OH

c.

O

+ HCI3

23.14 O

O

O

a. Br

Li2CO3 LiBr DMF

O CH3SH

c. Br

O

SCH3

O CH3CH2NH2

b. Br

NHCH2CH3

23.15 Bromination takes place on the
carbon to the carbonyl, followed by SN2 reaction with the nitrogen nucleophile. CH3

Br O

O

O

N O

O

O

O

NHCH3

Br2

LSD

CH3CO2H N COC6H5

N COC6H5

N

M

COC6H5

(Section 18.5)

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Substitution Reactions of Carbonyl Compounds 23–7 23.16 O

a. CH3CH2

C

CH2CH3

O

O

[1] LDA, THF CH3CH2

[2] CH3CH2I

C

O [1] LDA, THF

CHCH3

O

c.

O [2] CH3CH2I

CH2CH3 O

O [1] LDA, THF

[1] LDA, THF

b.

CH2CH2CN

d.

CH2CHCN

[2] CH3CH2I

[2] CH3CH2I

CH2CH3

23.17 O

O [1] LDA, THF

a.

[2] CH3I

O

O

O +

[1] LDA, THF

b.

[2] CH3I

COOCH3

c.

[1] LDA, THF

C

[2] CH3I

O

CH3 +

C

O

O

CH3

O

23.18 Three steps are needed: [1] formation of an enolate; [2] alkylation; [3] hydrolysis of the ester.

CO2CH2CH3

CO2CH2CH3

[1] LDA

CH3O

CO2CH2CH3

[2] CH3I

CH3O

CH3O

A [3] H3O+

The product is racemic because the new stereogenic center is formed by alkylation of a planar enolate with equal probability from above and below. CH3O

CO2H

naproxen

23.19 LDA

O

a.

O

O

CH3CH2Br

THF

O

b.

NaOCH2CH3

O

CH3Br

O

NaOCH2CH3

CH3CH2OH

O

O LDA

c. (from a.)

CH3CH2OH

THF

O CH3I

O

CH3Br

O

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Chapter 23–8 O

d.

O

O

LDA

CH3Br

THF

(from b.)

23.20 LDA THF

O O

Br

CH3I

O

O O

O

A

Br2

O

CH3CO2H

O

C

B

CH2 Li2CO3 LiBr DMF

O O


-methylene-butyrolactone

23.21 Decarboxylation occurs only when a carboxy group is bonded to the
C of another carbonyl group. O COOH

a.

COOH




b.

COOH

O

c.

COOH




d.

COOH




COOH


YES

NO

YES

NO

23.22 H3O+

[1] NaOEt

a. CH2(CO2Et)2

[2] Br




CH2 CH2COOH

b. CH2(CO2Et)2 [1] NaOEt

[1] NaOEt

H3O+

[2] CH3Br

[2] CH3Br




CH3 CH3 C COOH H

23.23 COOH

a.

Cl

b.

Cl

Br

O

Br

O

COOH

23.24 Locate the
C to the COOH group, and identify all of the alkyl groups bonded to it. These groups are from alkyl halides, and the remainder of the molecule is from diethyl malonate. a. (CH3)2CHCH2CH2CH2CH2 CH2COOH 

H3O+

[1] NaOEt CH2(CO2Et)2 [2] (CH3)2CHCH2CH2CH2CH2Br

b.




(CH3)2CHCH2CH2CH2CH2CH2COOH

COOH

 [1] NaOEt

[1] NaOEt

[2] CH3CH2CH2Br

[2] CH3CH2CH(CH3)CH2CH2Br

CH2(CO2Et)2

H3O+ COOH


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Substitution Reactions of Carbonyl Compounds 23–9 c.

(CH3CH2CH2CH2)2 CHCOOH

 H3O+

[1] NaOEt

[1] NaOEt

[2] CH3CH2CH2CH2Br

[2] CH3CH2CH2CH2Br

(CH3CH2CH2CH2)2CHCOOH

CH2(CO2Et)2




23.25 The reaction works best when the alkyl halide is 1° or CH 3X, since this is an SN2 reaction. a.

(CH3)3C–CH2COOH




COOH

b.

(CH3)3CX 3° alkyl halide (too crowded)

c.

(CH3)3C–COOH

This compound has 3 CH3 groups on the
carbon to the COOH. The malonic ester synthesis can be used to prepare mono- and disubstituted carboxylic acids only: RCH2COOH and R2CHCOOH, but not R3CCOOH.

X

aryl halide (leaving group on an sp2 hybridized C)

Aryl halides are unreactive in SN2 reactions.

23.26 O

O

a.

CH3

C

[1] NaOEt CH2CO2Et

[2] CH3I [3] H3O+,


C

CH3

O

O CH2CH3

b.

C

CH3

[1] NaOEt CH2CO2Et

CH3

[2] CH3CH2CH2Br [3] NaOEt [4] C6H5CH2I [5] H3O+,


C

CH CH2 CH2CH2CH3

23.27 Locate the
C. All alkyl groups on the
C come from alkyl halides, and the remainder of the molecule comes from ethyl acetoacetate. O

a.

CH3

C

O CH2 CH2CH3

CH3

C

CH2

COOEt

O

H3O+

[1] NaOEt [2] CH3CH2Br




CH3

C

CH2CH2CH3

 O

O

b. CH3

C

C

CH3

CH(CH2CH3)2

CH2

COOEt

[1] NaOEt

[1] NaOEt

[2] CH3CH2Br

[2] CH3CH2Br

O

H3O+


CH3

C

CH(CH2CH3)2

 

O

O

O

c.

CH3

C

CH2

[1] NaOEt

[1] NaOEt

[2] CH3CH2Br

[2] CH3(CH2)3Br

COOEt

23.28 O CH3

C

CH2CO2Et

+

Br

H3O+

Br

O

NaOEt (2 equiv)

CH3

X

CO2Et




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Chapter 23–10 23.29 O

O O

a.

CH3

C

H3

[1] NaOEt CH2CO2Et

[2] CH3O Br

CO2Et

b. CH3

C

O

[1] LDA, THF CH3

[2] CH3O




CH3O

nabumetone

CH3O O

O+

Br CH3O

nabumetone

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Substitution Reactions of Carbonyl Compounds 23–11 23.30 Use the directions from Answer 23.1 to draw the enol tautomer(s). In cases where E and Z isomers can form, only one isomer is drawn. O

a.

OH H

O

O

HO

O

HO

O

d.

H

(mono enol form) O

OH

conjugated enol (more stable)

e.

b.

O

conjugated enol (more stable)

O OH

O

c.

OH

OH

O

OH

O

OH

O

f. OEt

OH O

OH

OEt

OEt

unconjugated enol (less stable)

O

O

OH

OH

OH

OEt

OEt

23.31 O

O

O OCH2CH3

O

The ester C=O is resonance stabilized, and is therefore less available for tautomerization. Since the carbonyl form of the ester group is stabilized by electron delocalization, less enol is present at equilibrium.

OCH2CH3

ethyl acetoacetate

23.32 a. CH3CH2CH2CO2CH(CH3)2

c.

e. NC

O OH

O C CH2CH3

O

b. O

d. CH3O

CH2CN

f. HOOC

23.33 O CH2

a. CH3CH2 C OH

Ha Hb

Hc

Ha is part of a CH3 group = least acidic. Hb is bonded to an
C = intermediate acidity. Hc is bonded to O = most acidic.

b.

C

H

Hb O

CH3

Ha

Hc

Hc is bonded to an sp2 hybridized C = least acidic. Ha is bonded to an
C = intermediate acidity. Hb is bonded to an
C, and is adjacent to a benzene ring = most acidic.

O

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Chapter 23–12 Hc

O O

O

c.

d.

H

CH3 H

H

Hc

e. HO

Hb

Ha

Ha

H

Ha Hb Hc is bonded to an
C = least acidic. Ha is bonded to an
C, and is adjacent to a benzene ring = intermediate acidity. Hb is bonded to an
C between two C=O groups = most acidic.

O

COOH

sp3

H

H

Hb

Hc

sp3

hybridized Ha is bonded to an C = least acidic. Hb is bonded to an
C = intermediate acidity. Hc is bonded to O = most acidic.

hybridized Hb is bonded to an C = least acidic. Hc is bonded to an
C = intermediate acidity. Ha is bonded to O = most acidic.

23.34 CN

LDA

a.

d.

THF

O

CN

LDA THF

O O

OCH3

b.

LDA THF

O O

O

LDA

e.

THF

O

f.

LDA

c.

O

OCH3

LDA O

O

THF

O

O

THF

23.35 Enol tautomers have OH groups that give a broad OH absorption at 3600–3200 cm–1, which could be detected readily in the IR. 23.36 O

O Hb

Ha Ha

remove Ha

Hb

Ha Ha

Hb

Ha

O

O

Hb

Ha

O Hb Hb

remove Hb

Ha Ha

Hb Hb

Removal of Ha gives two resonance structures. The negative charge is never on O.

O

Hb

Ha Ha

O

Hb

Ha Ha

Removal of Hb gives three resonance structures. The negative charge is on O in one resonance structure, making the conjugate base more stable and Hb more acidic (lower pKa).

Hb

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Substitution Reactions of Carbonyl Compounds 23–13 23.37 O

O

HO

O

5,5-Dimethyl-1,3-cyclohexanedione exists predominantly in its enol form because the C=C of the enol is conjugated with the other C=O of the dicarbonyl compound. Conjugation stabilizes this enol.

5,5-dimethyl-1,3cyclohexanedione O

O

HO

The enol of 2,2-dimethyl-1,3-cyclohexanedione is not conjugated with the other carbonyl group. In this way it resembles the enol of any other carbonyl compound, and thus it is present in low concentration.

O

2,2-dimethyl-1,3cyclohexanedione

23.38 In the presence of acid, (R)-
-methylbutyrophenone enolizes to form an achiral enol. Protonation of the enol from either face forms an equal mixture of two enantiomers, making the solution optically inactive. O

OH

O

H3O+

O

H2O

H

H

H

achiral (E and Z isomers)

(R)-
-methylbutyrophenone

In the presence of base, (R)-
-methylbutyrophenone is deprotonated to form an achiral enolate, which can then be protonated from either face to form an optically inactive mixture of two enantiomers. O

O

O

O

H2O

–OH

H

H

H

achiral

(R)-
-methylbutyrophenone

23.39 Protonation in Step [3] can occur from below (to re-form the R isomer) or from above to form the S isomer as shown. H A

A H

H O

H O

[1]

H

O

O

H

OH

R isomer inactive enantiomer

O

[2]

OH

H

OH

achiral enol

H A A

[3] H

H O

HA

H

+ O

S isomer active enantiomer

H O

H

[4] O

H

H A

(+ one more resonance structure)

O OH

H

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Chapter 23–14 23.40 O CH3

C

O O

CH3

CH2

O

C

O

CH3

CH2

C

O O

CH3

CH2

C

O

CH3

ester The O atom of the ester OR group donates electron density by a resonance effect. The resulting resonance structure keeps a negative charge on the less electronegative C end of the enolate. This destabilizes the resonance hybrid of the conjugate base, and makes the
H's of the ester less acidic.

CH3

O

O

C

C

CH3

CH2

O CH3

CH2

C

CH3

no additional resonance structures

ketone

This structure, which places a negative charge on the O atom, is the major contributor to the hybrid, stabilizing it, and making the
H's of the ketone more acidic.

23.41 LDA reacts with the most acidic proton. If there is any H2O present, the water would immediately react with the base: [(CH3)2CH]2N

Li

[(CH3)2CH]2NH

H OH

Li

OH

LDA

23.42 O

O

2,4-pentanedione base (1 equiv) O

O

[1] CH3I

O

O

[2] H2O A

One equivalent of base removes the most acidic proton between the two C=O's, to form A on alkylation with CH3I.

base (2nd equiv)

O

O

[1] CH3I

O

O

[2] H2O

more nucleophilic site

B

With a second equivalent of base a dianion is formed. Since the second enolate is less resonance stabilized, it is more nucleophilic and reacts first in an alkylation with CH3I, forming B after protonation with H2O.

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Substitution Reactions of Carbonyl Compounds 23–15 23.43 O

NaOH

(CH3)3C

H2O

O (CH3)3C

A

(CH3)3C

B COCH3

O

H

H

(CH3)3C

C

COCH3

(CH3)3C

one axial and one equatorial group, less stable

Both groups are equatorial. more stable

This isomerization will occur since it makes a more stable compound.

(CH3)3C

COCH3

Both groups are equatorial. Compound C will not isomerize since it already has the more stable arrangement of substituents.

Isomerization occurs by way of an intermediate enolate, which can be protonated to either reform A, or give B. Since B has two large groups equatorial, it is favored at equilibrium. COCH3

COCH3

H

(CH3)3C

OH

(CH3)3C

planar enolate

23.44 Protons on the  carbon of an
,-unsaturated carbonyl compound are acidic because of resonance. There is no H on this C, so a planar enolate cannot form and this stereogenic center cannot change.  O








X

Remove the H on this  C.

O

O H

OH H OH

Removal of this proton forms a resonance-stabilized anion. One resonance structure places a negative charge on O. O

O

Protonation of the O planar enolate can occur from below (to re-form starting material), or from above to form Y.

H

Y

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Chapter 23–16 23.45 The mechanism of acid-catalyzed halogenation consists of two parts: tautomerization of the carbonyl compound to the enol form, and reaction of the enol with halogen. A higher percentage of the more stable enol is present. O

OH

CH3COOH

OH

Br2

O

O Br

2-pentanone

C=C has 1 bond to C

A

C=C has 2 bonds to C more stable (E and Z isomers)

Br

B major product formed from the more stable enol

23.46 • The mechanism of acid-catalyzed halogenation [Part (a)] consists of two parts: tautomerization of the carbonyl compound to the enol form, and reaction of the enol with halogen. • In the haloform reaction [Part (b)], the three H’s of the CH3 group are successively replaced by X, to form an intermediate that is oxidatively cleaved with base. O

O O

H

O H

O

O H

O

H H

a.

O H H

O H

O

O Br

O

H Br

O C

b.

C

H H

H

–

OH

+

Br H

H

I

I

H

I

+ I–

H

Repeat [1] and [2] two times.

C –CI

+

H

C

O

O O

Br

[2]

H

O O

O H

H

+ H2O

O

CI3

CI3

OH

3

CHI3

–OH

23.47 Use the directions from Answer 23.24. a.

CH3OCH2 CH2COOH




CH3OCH2Br




b.

C6H5

c.

COOH


C6H5

H

O C

H

Br Br

Br

O

[1]

H

Br

O H H

O H

Br

Br

COOH

and Br

O

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Substitution Reactions of Carbonyl Compounds 23–17 23.48 a. CH3CH2CH2CH2CH2 CH2COOH

H3O+

[1] NaOEt CH2(CO2Et)2 [2] CH3CH2CH2CH2CH2Br

CH3CH2CH2CH2CH2CH2COOH




 b.

[1] NaOEt

[1] NaOEt

[2] CH3CH2CH2CH2CH2Br

[2] CH3Br

CH2(CO2Et)2

COOH

H3O+


COOH

 c.

COOH

[1] NaOEt

[1] NaOEt

[2] (CH3)2CHCH2CH2CH2CH2Br

[2] CH3Br

CH2(CO2Et)2



H3O+

COOH




23.49 O O

O

O

[1] NaOEt O

O

O

[2] CH3CH2CH2Br

O

[1] NaOEt O

[2] CH3CH2CH2Br

O

O

O

H3O+
O O

valproic acid

23.50 H3O+

[1] NaOEt

a.

CH2(CO2Et)2 [2] BrCH2CH2CH2CH2CH2Br [3] NaOEt

b.

COOH

(from a.)

c.

COOH

[1] LiAlH4

COOH




CH2OH

[2] H2O

CH3OH H2SO4

CO2CH3

CH3

[1] CH3MgBr (2 equiv)

C OH

[2] H2O

CH3

(from a.)

d.

COOH

CH3CH2OH H2SO4

(from a.)

COOCH2CH3

[1] LDA [2] CH3I

CH3 COOCH2CH3

H

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Chapter 23–18 23.51 a.

O

[1] Na+ –CH(COOEt)2 HO [2] H2O CH3CHCH2 CH(COOEt)2

O CH3

c.

C

CH3

O [1] Na+ –CH(COOEt)2 C [2] H2O CH3 CH(COOEt)2

Cl

nucleophilic attack here b.

[1] Na+ –CH(COOEt)2 [2] H2O

CH2 O

d.

HOCH2CH(COOEt)2

CH3

O

O

C

C

O

[1] Na+ –CH(COOEt)2 [2] H2O CH

CH3

3

O C

CH(COOEt)2 + CH3COOH

23.52 Use the directions from Answer 23.27. O

O



a.

CH3 O

 CH3



COOEt

C

CH2

CH3

C




[2] Br

[1] NaOEt

[1] NaOEt

[2] CH3CH2Br

[2] Br

COOEt

O

c. O

CH2

O

b.

O

C

CH2

H3O+


H3O+

[1] NaOEt COOEt

O

H3O+

[1] NaOEt

O




[2] Br

 O

d. CH3

C

[1] NaOEt CH2

COOEt

O

H3O+

[2]




Br

Br [3] NaOEt

23.53 O

a.

CH3

C

[1] NaOEt CH2

COOEt

b.

CH3

C

CH2

c.

CH3

CH3

[1] NaOEt

[1] NaOEt

H3O+

[2] CH3Br

[2] CH3Br

COOEt

O C




[2] CH3CH2Br

O

O

H3O+

O

LDA CH(CH3)2

THF




CH2

C

C

CH2CH2CH3 O CH3

C

CH(CH3)2 O

CH3I CH3CH2

CH(CH3)2

C

CH(CH3)2

(from b.) O

d.

CH3

C

CH(CH3)2

(from b.)

NaOCH3 CH3OH

O CH3

C

O

CH3I C(CH3)2

CH3

C

C(CH3)3

O

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Substitution Reactions of Carbonyl Compounds 23–19 23.54 O

O

O

Li2CO3

a.

LiBr DMF

Br O

f.

(E and Z)

[2] Li2CO3, LiBr, DMF

O O

COOH

b.

O [1] Br2, CH3CO2H




O I2 (excess)

g.

O

+ CHI3

–OH

COOH [1] LDA

c. CH3CH2CH2CO2Et

COOH CH3CH2

h.

CH3CH2CHCO2Et

Cl

NaH

CN

C N

[2] CH3CH2I O

O Br

d.

O NHCH(CH3)2

(CH3)2CHNH2

O Br2 (excess)

i.

–OH

O

O

O

Br Br O

[1] LDA

e.

NaI

j.

[2] CH3CH2I

I

Cl

23.55 O

O

O

O

[1] LDA

O [1] LDA

a.

c.

Cl

[2]

[2] CH3I

H

H

O D H

O

[1] LDA

C

b.

CH3

H D [2]

C

CH3

I

23.56 CHO

a.

OH

NaBH4 CH3OH

p-isobutylbenzaldehyde

[1] PBr3

CN

[2] CH3I

[2] NaCN A

CN

[1] LDA

B

C H3O+


COOH

ibuprofen

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Chapter 23–20 COOH

b.

removal of the most acidic H

D

[2] CH3I

COO

[1] LDA

COOCH3

substitution reaction

+ I–

Removal of the most acidic proton with LDA forms a carboxylate anion that reacts as a nucleophile with CH3I to form an ester as substitution product. 23.57 CO2CH3

a.

CO2CH3

Cl

NH

+

N

K2CO3

S

Cl

S

Cl

sp2

Cl on hybridized C does not react. Cl on sp3 hybridized C reacts. H COOH

The N atom acts as a nucleophile to displace Cl–.

H COOCH3

CH3OH

b. HO

H COOCH3

TsCl

HO

H2SO4

Cl

NH

B

N

inversion of configuration SN2

Cl

C

CO2CH3

S isomer

S

TsO

pyridine

Cl

A

clopidogrel (racemic)

S

Cl

clopidogrel (single enantiomer)

23.58 O

O

O

O

LDA Br

NaOCH2CH3

THF

Br

–78 oC

A

B

CH3CH2OH

room temperature

A

C

23.59 O

O

 a.

O




+

CO2

COOH

In order for decarboxylation to occur readily, the COOH group must be bonded to the  C of another carbonyl group. In this case, it is bonded to the  carbon. [1] NaOEt (CH3CH2)3CCH(CO2Et)2

b. CH2(CO2Et)2 [2] (CH3CH2)3CBr

The 3° alkyl halide is too crowded to react with the strong nucleophile by an SN2 mechanism.

c.

H

O [1] LDA [2] CH3CH2I

LDA removes a H from the less substituted C, forming the kinetic enolate. This product is from the thermodynamic enolate, which gives substitution on the more substituted  C.

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623

Substitution Reactions of Carbonyl Compounds 23–21 23.60 O

O

O CH3

LDA

O

I

+

O

I–

O CH3

I

I–

+

23.61 O O

O

O

O

O

O

EtO O

O

OEt EtO

OEt

EtO

OEt

H H

H

EtO

O

OEt

O

+ HOEt

–

OEt –

OEt

O CO2Et H3O+

O

O

O

H

OEt

O

O OEt

O

O + HOEt

23.62 LDA = B O

C6H5

H

Br

B

Br

C6H5

C6H5

O

+

Br

+ HB+

O Br

+ HB+

C6H5 H

B

O

H

Br C6H5

Br –OC(CH

Br C6H5

3)3

C6H5

+ HB+ C6H5

O

H C6H5

O

Br C6H5

O

Br

O

Br C6H5

+

C6H5

B O

O

O

+ HOC(CH3)3

O

+ Br

This reaction occurs with both bases [LDA and KOC(CH3)3]. + Br

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Chapter 23–22 23.63 1st new C–C bond O H H

OCH2CH3

O

LDA THF –78 oC




Remove proton here.

H

OCH2CH3

[3]

Cl Cl

B

O OCH2CH3 LDA THF –78 oC

[2]

H

[1]

A

O

OCH2CH3

Cl

Cl

[4] O

Cl

OCH2CH3

C 2nd new C–C bond Cl

23.64 O COCH3

a.

O

Cl2

Cl

NHC(CH3)3

H2NC(CH3)3

H2O, HCl COO–

COCH3 –OH, I 2 excess

b.

H3O+

COOH

+ CHI3 O COCH3

[1] LiAlH4 [2] H2O

[1] LDA, THF

c.

OH

[2] CH3CH2CH2Br O COCH3

Br2

[1] LDA, THF

d.

Li2CO3

CH3COOH

[2] CH3Br

Br

23.65 O

O Br

Br2

a.

CH3COOH O

O Br

b.

OCH3

NaOCH3

(from a.) O

OH

O [1] LDA, THF

c.

d.

[1] LiAlH4

[2] CH2=CHCH2Br

[2] H2O O

O

O

O

O

[1] LDA, THF

[1] LDA, THF

[2] CH3Br

[2] CH3CH2Br

LiBr DMF

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Substitution Reactions of Carbonyl Compounds 23–23

O

O Br

e. (from a.)

CH3CH2Br [1] Li (2 equiv) [2] CuI (0.5 equiv)

Li2CO3 LiBr DMF

[2] H2O

O

O

O Br

[1] LDA, THF

f.

O

[1] (CH3CH2)2CuLi

NaOCH2CH3

[2] BrCH2CH2CH2CH2Br O

O

O [1] LDA, THF

g.

[1] LDA, THF

[2] CH3CH2Br

O

Li2CO3

CH3COOH

[2] CH3CH2Br

O

Br

O

[1] NaOCH2CH3

h.

O

O Br2

O

Br

Br2

Li2CO3 LiBr DMF

CH3COOH

[2] CH3CH2Br

LiBr DMF

(from g.)

23.66 O

O

O

Cl AlCl3

O

Cl2

Br2

FeCl3

CH3CO2H

Br

Cl

Cl H2NC(CH3)3 O NHC(CH3)3

Cl

bupropion

23.67 O

a.

OH

PBr3

Br

O

O [1] NaOEt

COOEt

COOEt

H3O+


[2]

mCPBA

Br

O O

O

O

b.

COOEt

O

[1] NaOEt [2] HC CCH2Br

H3O+

HOCH2CH2OH




TsOH

O

COOEt O

O

[1] NaH [2] CH3I H3O+

O

O

H2 Lindlar catalyst

O

O

CH3

626

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Chapter 23–24 O

O COOEt

c.

[1] NaOEt

O COOEt

[2]

OH

H3O+

NaBH4




CH3OH

CH2Br

CH3

Br2 h

23.68 O

O

Br2

a.

O

CH3CH2NH2

CH3COOH

NH3 (excess)

Br

CH3CH2Br

NHCH2CH3

PBr3 CH3CH2OH

O

b.

O

Li2CO3 LiBr DMF

Br

OH

NaBH4 CH3OH

(from a.) Br2

c.

Br

FeBr3 [1] Li (2 equiv) [2] CuI (0.5 equiv) O

(from b.)

O

[1] (C6H5)2CuLi [2] H2O

C6H5

SOCl2

d.

CH2Br

Br2 h


CH3

CH3OH

CH3Cl AlCl3

[1] Li (2 equiv) [2] CuI (0.5 equiv) O

(from b.)

[1] (C6H5CH2)2CuLi [2] H2O

O C6H5

NaBH4 CH3OH

OH C6H5

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627

Substitution Reactions of Carbonyl Compounds 23–25 23.69 O

O

O

CH3CH2I

LDA THF HO

–78 oC

A

O

CH3CH2O

most acidic H To synthesize the desired product, a protecting group is needed: O

O

O LDA

TBDMS–Cl imidazole HO

THF TBDMSO

A

TBDMSO

CH3CH2I

O O (CH3CH2CH2CH2)4N+F– HO

B

TBDMSO

23.70 CH2(COOEt)2 NaOEt OH

PBr3

Br

–

CH(COOEt)2

CH(COOEt)2 H3O+,


Cl

Y

SOCl2 COOH

O

23.71 O

a.

O

O

Y =

[1] CH3Li W

[2] CH3I

(CH3)2CH

Ha O

Y

Hb

C

Hc Hd

CH2CH2CH3

He

C7H14O
one degree of unsaturation IR peak at 1713 cm–1
C=O 1 H NMR signals at (ppm) He: triplet at 0.8 (3 H) Ha: doublet at 0.9 (6 H) Hd: sextet at 1.4 (2 H) Hc: triplet at 1.9 (2 H) Hb: septet at 2.1 (1 H)

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Chapter 23–26 O

O

b. CH3 Li

O

O

O

O

O

O

+ I–

Y

CH3

I

23.72 Removal of Ha with base does not generate an anion that can delocalize onto the carbonyl O atom, whereas removal of Hb generates an enolate that is delocalized on O. CO2CH3

CO2CH3

CO2CH3

Delocalization of this sort can't occur by removal of Ha, making Ha less acidic. H

H

H

B

O

H O

O

Ha Hb

B H CO2CH3

CO2CH3

H

CO2CH3

H

O

CO2CH3

H

O

H

O

O

Removal of Hb gives an anion that is resonance stabilized so Hb is more acidic.

Mechanism: CO2CH3

H

OCH3

CO2CH3

CO2CH3

O

O

CO2CH3

Br O

O

+ CH3OH CO2CH3

O

CO2CH3

CO2CH3 HO H

+ O

B

+ Br–

CO2CH3

O

–OH

H

O

+ HB+

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23. Substitution Reactions of Carbonyl Compounds at the a Carbon

629

© The McGraw−Hill Companies, 2011

Text

Substitution Reactions of Carbonyl Compounds 23–27 23.73 CH3 Li O

OCH2CH3

H OH2 O CH3

OCH2CH3

OH H3O+

OH2

OCH2CH3

CH3

+ Li+

OCH2CH3

CH3

+ H2O

CH3

OCH2CH3

+ H2O

+ H2O

X H CH3

OH OCH2CH3

H OH2 OH OCH2CH3

CH3

CH3

+ H2O

OH OCH2CH3

CH3

OCH2CH3

H + H2O O H

+ H2O O

+ H2O

H3O+

HOCH2CH3

23.74 O

O

O

O

O

O R X

e

H

e

H NH2

or O

H NH2

OH

H

proton transfer OH O R

+ X– + NH2

e

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631

Carbonyl Condensation Reactions 24–1 C Chhaapptteerr 2244:: C Caarrbboonnyyll C Coonnddeennssaattiioonn R Reeaaccttiioonnss

TThhee ffoouurr m maajjoorr ccaarrbboonnyyll ccoonnddeennssaattiioonn rreeaaccttiioonnss Reaction type

Reaction

[1] Aldol reaction (24.1)

O

2

RCH2

OH

–OH

C

H

H2O

[2] Claisen reaction (24.5)

RCH2

C

C

or

R

OR'

[1] NaOR'

O

[2] H3O+

C

RCH2

H

H3O+

CHO C R

(E and Z)
,-unsaturated carbonyl compound

-hydroxy carbonyl compound

O

2

CHCHO

H

aldehyde (or ketone)

RCH2

–OH

RCH2 C

O CH

ester

C

OR'

R


-keto ester O

O

[3] Michael reaction (24.8)

–OR'

+

R

–


,-unsaturated carbonyl compound

[4] Robinson annulation (24.9)

R

1,5-dicarbonyl compound

carbonyl compound

–OH

H2O

O

O


,-unsaturated carbonyl carbonyl compound compound

2-cyclohexenone



U Usseeffuull vvaarriiaattiioonnss [1] Directed aldol reaction (24.3) O

O R'CH2

C

[1] LDA R"

R" = H or alkyl

[2] RCHO [3] H2O

HO

O

R C CH C H R'

O

H2O

OH

+

O

O or

R"

-hydroxy carbonyl compound

–

OH

or H3O+

C

R C H

R"

C R'

(E and Z)
,-unsaturated carbonyl compound

632

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Chapter 24–2 [2] Intramolecular aldol reaction (24.4) [a] With 1,4-dicarbonyl compounds:

O

O NaOEt O

[b] With 1,5-dicarbonyl compounds:

EtOH

O

O NaOEt O

EtOH

[3] Dieckmann reaction (24.7) [a] With 1,6-diesters:

OEt

O [1] NaOEt

O O

O C

[2] H3O+

OEt

OEt

[b] With 1,7-diesters:

OEt O

O O

[1] NaOEt [2] H3

OEt

O+

O C

OEt

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Carbonyl Condensation Reactions 24–3 C Chhaapptteerr 2244:: A Annssw weerrss ttoo PPrroobblleem mss 24.1 O

CH2CHO

H

a. OH

c. CH3

C

OH

O

CH3 C CH2 C CH3

CH3

CH3

O

O

(CH3)3CCH2CHO

b.

HO

C(CH3)3

(CH3)3CCH2C

C CHO

H

O

HO

d.

H

24.2 O O

O

CHO

a.

c.

b.

(CH3)3C


H no
H no aldol reaction

C

d. (CH3)3C

H

no
H no aldol reaction

yes

C

CHO

e.


H

CH3


H yes

yes

24.3 base

O

a.

O

O

OH

base

O

c. HO HO

CHO

CHO

base

b.

(E and Z isomers)

24.4 H OSO3H O C C C H H H

HO H CH3

OH2 H CH3 C H

H

O

C C H

CH3 C H

C C

H

+

+ HSO4

O

H

O

HSO4

CH3CH CH C H

H +

H2O

H2SO4

Locate the
and  C’s to the carbonyl group, and break the molecule into two halves at this bond. The
C and all of the atoms bonded to it belong to one carbonyl component. The  C and all of the atoms bonded to it belong to the other carbonyl component.

24.5



O




 OH


CHO

a.

b. C6H5

c.

C6H5

OH H C6H5 O

O

CHO H

CHO CHO




O

C6H5 O

634

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Chapter 24–4 24.6 –OH

O

H

H H H

C

H OH

O

C

O

H C

C

H

H

C H

O

O

+ H2O OH

OH

H C

+

H C

–OH

C H

CHO

+ H2O

C H

H

O

O

–OH

24.7 CHO

a. CH3CH2CH2CHO and CH2=O

CHO

or

O

c. C6H5CHO

OH

and

O

O

b. C6H5COCH3 and CH2=O

(E and Z isomers)

24.8 CO2Et H

a.

CO2Et

CH2(CO2Et)2

O

H

c.

O

H CH3COCH2CN NC

COCH3 H

b.

COCH3

(E and Z isomers)

COCH3

CH2(COCH3)2 O

24.9 2

–

CHO

OH

CHO

a.

H2O

NaBH4

OH

CH3OH

OH

OH CHO

b.

CHO [1] (CH3)2CuLi

–OH

(E and Z mixture)

[2] H2O

CHO NaBH4

c. (from b.)

CH3OH

CH3CH2CH2CH C(CH2CH3)CH2OH

CHO [1] CH3MgBr

d. (from b.)

[2] H2O

CH3CH2CH2CH C(CH2CH3)CH(CH3)OH

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Carbonyl Condensation Reactions 24–5 24.10 Find the
and  C’s to the carbonyl group and break the bond between them. O

OH

O

a.

b. O

OH

O

c.

O

O

H

H

O

O H

O

24.11 O

O H N

CH3O CH3O O

O H2

N

CH3O

N

CH3O

Pd-C

X CH3O

donepezil

CH3O

24.12 All enolates have a second resonance structure with a negative charge on O. O

O H

O

EtO

O

O

O

H

H

H

EtO O

OH

O

+ OH

H

H OH

O

O

O

EtO H EtOH

24.13 O

O –

a. CHO

O O

OH

O

b.

H2O

[1] O3 [2] (CH3)2S

CHO O

–OH

H2O

–

OH

H2O

24.14

1-methylcyclopentene

EtOH

EtOH

H H

O

2-cyclohexenone

636

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Chapter 24–6 24.15 Join the
C of one ester to the carbonyl C of the other ester to form the -keto ester. O

a.




OCH3

O

O

O

b.

O




OCH2CH3

OCH3

O







OCH2CH3

24.16 In a crossed Claisen reaction between an ester and a ketone, the enolate is formed from the ketone, and the product is a -dicarbonyl compound. O

a. CH3CH2CO2Et and HCO2Et

H

O OEt

Only this compound can form an enolate. b.

O

and

CO2Et

HCO2Et

H C

OEt

O

Only this compound can form an enolate. O CH3

c.

C

O

and CH3

CH3

C

O

O

OEt

The ketone forms the enolate. d.

O

and

O

O

C

C

OEt

O

The ketone forms the enolate.

24.17 A -dicarbonyl compound like avobenzone is prepared by a crossed Claisen reaction between a ketone and an ester. O

O O

O

O CH3O

CH3O

O

C(CH3)3

Break the molecule into two components at either dashed line. avobenzone

C(CH3)3

or O O

CH3O

C(CH3)3

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637

Carbonyl Condensation Reactions 24–7 24.18 O

O

O

[1] NaOEt

a.

OEt

[2] (EtO)2C=O

O

b.

C

C6H5CH2

O

[1] NaOEt

C

[2] ClCO2Et

OEt

EtO

OEt

O

24.19 O

O

CO2Et

OEt

OEt CH3 O

[1] NaOEt

[1] NaOEt

[2] CH3I

O

[2] (EtO)2C=O A

EtO

EtO

B

H3O+
CO2H

ibuprofen

24.20 O CH3O2C

CO2CH3

base

CH3O

CH3O2C

CO2CH3

O OR

OR X

1,6-Diester forms a five-membered ring.

24.21 O

O

O

O

O

OEt

EtO

EtO

OEt O

24.22 A Michael acceptor is an
,-unsaturated carbonyl compound. O

a.
,-unsaturated yes - Michael acceptor

O

O

O

d.

c.

b.

O

not
,-unsaturated

not
,-unsaturated

CH3O


,-unsaturated yes - Michael acceptor

638

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Chapter 24–8 24.23 O

a.

+

CH2 CHCO2Et

CH3

C

[1] NaOEt [2] H2O

CH2CO2Et

CH3

O

O

C

C

OEt

COOEt [1] NaOEt

+ CH2(CO2Et)2

b.

[2] H2O

O

O EtO2C

O

O

c.

+ CH2

CH3

C

CH2CN

O

[1] NaOEt [2] H2O

CO2Et CN

O

24.24 O

O

O

O EtO2C

a. CO2Et

b.

O

O

O

O

24.25 The Robinson annulation forms a six-membered ring and three new carbon–carbon bonds: two  bonds and one
bond. new C–C bond O

O O

OH

re-draw

+

a.

O

O

O

H2O

O

O

new  and
bonds new C–C bond O

O COOEt +

b.

COOEt re-draw O

COOEt

OH

H2O

O

O

new  and
bonds

new C–C bond

O

c.

O +

re-draw

OH

O

O

H2O

O

new  and
bonds

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Carbonyl Condensation Reactions 24–9 new C–C bond EtO2C

O +

d. O

COOEt

EtOOC

re-draw O

OH

H2O

O

O

new  and
bonds

24.26 a.

O

O

O

b.

c.

O

O

O

O

O

O

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Chapter 24–10 24.27 The product of an aldol reaction is a -hydroxy carbonyl compound or an
,-unsaturated carbonyl compound. The latter type of compound is drawn as product unless elimination of H2O cannot form a conjugated system. a. (CH3)2CHCHO only

O

OH

–

OH H2O

–

(CH3)2CHCHC(CH3)2

d.

(CH3CH2)2C=O only

CHO

OH H2O O

CH3

–OH

b. (CH3)2CHCHO + CH2=O

e. (CH3CH2)2C=O + CH2=O

CH3 C CHO

H2O

H2O

CH2OH O

CHO

–OH

c. C6H5CHO + CH3CH2CH2CHO

–OH

O –

H2O

+ C6H5CHO

f.

OH H2O

(E and Z isomers) (E and Z isomers)

24.28 HO

CHO

O H

OH

CHO

H O

OH

OH

CHO

CHO

24.29 O

O

a. CH3

C

OH

[1] LDA

O

CH3CH2CH2 C CH2 C

CH3 [2] CH3CH2CH2CHO

H

[3] H2O

b. CH CH 3 2

CH3

C

OH

[1] LDA OEt

[2] O

O

[3] H2O

CHO

O

C

CH C

H

CH3

O O

24.30 O O CHO

a.

O

c.

O

b. OHC

CHO CHO

O

OEt

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Carbonyl Condensation Reactions 24–11 24.31 Locate the
and  C’s to the carbonyl group, and break the molecule into two halves at this bond. The
C and all of the atoms bonded to it belong to one carbonyl component. The  C and all the atoms bonded to it belong to the other carbonyl component. O O

a.

OH



b.



O



d.

c.

C6H5





O

O

CH CHCN


C6H5

O

H

e.

C6H5


O

CH3

O

O

O

CH3

C6H5

CH2=O O

C6H5






CH3CN

CHO

C6H5

24.32 O CHO (CH3)2C=O NaOEt EtOH

CH3O

CH3O

X

24.33 O

O

O

O

c.

b.

a.

d. HO

O

O

O H

H

O

O O

CH3

CH3

H

O

O

24.34 Ozonolysis cleaves the C=C, and base catalyzes an intramolecular aldol reaction. O

O

NaOH

[1] O3

H2O

[2] (CH3)2S O C

D C10H14O

24.35 O a. C6H5CH2CH2CH2CO2Et

O

C6H5CH2CH2CH2

OEt CH2CH2C6H5

O

b. (CH3)2CHCH2CH2CH2CO2Et

(CH3)2CHCH2CH2CH2

O OEt CH2CH2CH(CH3)2

642

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Chapter 24–12 CH3O

c. CH3O

O

O

CH2COOEt

OEt

OCH3

24.36 O

O

CH3CH2CH2CH2CO2Et + CH3CH2CO2Et

O

O

OEt O

OEt O

O

O OEt

OEt

24.37 O

O

a. CH3CH2CH2CO2Et only

O OEt

f. CH3CH2CO2Et + (EtO)2C=O

CH2CH3 O

O

b. CH3CH2CH2CO2Et + C6H5CO2Et

g.

O

c. CH3CH2CH2CO2Et + (CH3)2C=O

O h.

O

O

H

O

O + Cl

C

OEt

OEt

O

O O + HCO2Et

CH2CH3 O

EtO

O OEt

O

O

EtO

O OEt

d. EtO2CC(CH3)2CH2CH2CH2CO2Et O

O

e. C6H5COCH2CH3 + C6H5CO2Et

24.38 O

O

CH3O

CH3O

a. O

OEt

(EtO)2C=O

or O CH3O

O OEt

EtO

OEt

C6H5

b. O

O

O

C6H5 O

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Carbonyl Condensation Reactions 24–13 O CHO

c.

d. C6H5CH(COOEt)2

H

CH3

O

O

O

O

C6H5

OEt

OEt

EtO

OEt

or

O

CH3CHO

OEt

24.39 O

To form bond (b):

To form bond (a):

O

O

O

O

EtO

bond (a)

EtO

bond (b)

CH3

O

24.40 Only esters with 2 H’s or 3 H’s on the
carbon form enolates that undergo Claisen reaction to form resonance-stabilized enolates of the product -keto ester. Thus, the enolate forms on the CH2
to one ester carbonyl, and cyclization yields a five-membered ring. This is the only
carbon with 2 H's. O

O NaOCH3

OCH3

CH3O CH3O

O

OCH3 [1] nucleophilic attack

CH3O

CH3OH

O

CH3O

O

O

CO2CH3

CH3O2C

[2] loss of CH3O– [3] deprotonation

O

B O H3

CO2CH3

CH3O2C

O+

CH3O2C

OCH3

CO2CH3

CH3O2C

H O

O

O

highly resonance-stabilized enolate Formation of this enolate drives the reaction to completion.

acidic H between 2 C=O's

24.41 O

a.

O +

C6H5

C6H5

–OEt,

O

O

EtOH

C6H5 C6H5

O

O CO2Et

b.

–OEt,

+

O EtOOC

EtOH

O

O

O

c. + CH2(CN)2

–OEt,

EtOH CN CN

644

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Chapter 24–14 O

O O

d.

+

–OEt,

CO2Et

EtOH O EtO2C

24.42 O

O

O

a.

c.

b. O O

CO2Et

O EtO2C

O

d. CN

CO2Et

O O

O

O

O

O

O

C6H5

CO2Et EtO2C

CN

CO2Et

C6H5

24.43 O

O H

a.

Michael reaction

CH3 O

A

O

(E or Z isomer can be used.)

O O

b.

H OH O

O H

OH

O

O H2O

OH O

O

OH

H2O

H OH

OH O

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645

Carbonyl Condensation Reactions 24–15 24.44 O

O
OH

a.

+

H2O

O

O

O O

O +

b.


OH

re-draw C6H5

+

O

O

H2O

C6H5

O

C6H5

O O re-draw

c.


OH

+

+

H2O

O

O

O

O O


OH

re-draw

d.

+

O

H2O

O

O

24.45 O

O

c.

a. O

O

O

O

O

O

O

O

d. O

b. O

O

O

24.46 CH3CH2CH2CHO

a.

–OH

CH3CH2CH2

H2O

b.

–OH

CHO

(E and Z) H CH2

CH2=O, H2O

CH2CH3 CHO C CH2CH3

g. h.

CHO

c.

i.

[1] LDA [2] CH3CHO; [3] H2O

d.

e.

H

NaOEt, EtOH

CO2Et

j.

[1] CH3Li

k.

NaBH4 CH3OH

HOCH2CH2OH TsOH CH3NH2 mild acid

O

(CH3)2NH mild acid CrO3

CH3CH2CH2CH2OH

l.

O H

CH3CH2CH2 C

CH3 N

H CH3CH2CH=CHN(CH3)2

(E and Z) CH3CH2CH2COOH

H2SO4

[2] H2O

f.

CH3CH2CH2CH2OH

OH

CH3CH2CH2 CH2(CO2Et)2 EtO2C OH

H2 Pd-C

O

Br2 H

CH3COOH Br

646

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Chapter 24–16

m.

Ph3P=CH2

CH3CH2CH2

O

OH

n.

C CH2

NaCN, HCl

o.

CH3CH2CH2 C H

H

[1] LDA; [2] CH3I H

CN

24.47 O

O

O

–OH

a.

e.

H2O

C

O

O

NaOEt, EtOH

f.

b. O

(CH3)2C=O

O +

CHO

CH3

C

C6H5 NaOCH3

C6H5 +

CH3OH

O

O O

CO2Et

OH

H2O

O

C6H5

+

–OH

C6H5

H2O

g.

–

CH3

CHO

CN

O

d.

[2] CH3CH2CHO

O

NaOEt, EtOH

c. NCCH2CO2Et

CH3

O

[3] H2O

O

O

CH3

OH

[1] LDA

CH2CO2Et

[1] NaOEt, EtOH

CH2CO2Et

[2] H3O+

(E and Z)

h.

O

(E and Z) O

O

O

OEt

24.48 O

A [1] NaOEt

O CO2Et

[2](EtO)2CO [3] H3O+ G

[1] LDA [2] CH3CH2CHO [3] H2O, –OH H H2, Pd-C O

(1 equiv)

O

B [1] NaOEt [2] CH3CH2I

C H3O+
CO2Et

O

HO D [1] CH3CH2MgBr [2] H2O E [1] LDA [2] CH3I

O

I [1] LDA [2] HCO2Et [3] H3O+

O

O

F H2SO4

O

H

major product J [1] O3 O O K –OH

H2O O

[2] (CH3)2S

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Carbonyl Condensation Reactions 24–17 24.49 H

H

B

B

CH CHCl

C CHCl H

CH2 CH2Cl

CH2 CH2Cl

sp2

sp3

Vinyl halides undergo elimination by an E1cB mechanism more readily than alkyl halides because the carbanion intermediate formed from a vinyl halide has an sp2 hybridized C, while the carbanion derived from CH3CH2Cl is sp3 hybridized. The higher percent s-character of the sp2 hybridized anion makes it more stable, and therefore, it is formed more readily. 24.50 The final step in the reaction sequence involves an intramolecular crossed Claisen reaction between a ketone and an ester to form a
-dicarbonyl compound.

[1] O3

CrO3

O

[2] (CH3)2S CHO

EtOH

O

H2SO4 H2O

O

H2SO4 COOH

A

COOCH2CH3

B

C [1] NaOEt, EtOH [2] H3O+ O

O

D C13H20O2

24.51 O

[1] O3

O

O

[2] (CH3)2S

O O

O

NaOH EtOH

H

B

O H

O O OH

Form the enolate here to generate a five-membered ring in the product.

A

new C–C bond

The RCHO has the more accessible carbonyl.

24.52 O H

C

C

H H CO32–

[1] H

H

C

O

O

C

C

H

H + HCO3–

+ H

O

HH H

[2] O

C

C

C

H H H OCO2

H

HO

C

O

O

HH

[3]

C

C

H H

HOCH2 H

C

C

H

HOCH2 CH2OH

+ CO32–

Repeat steps [1]–[3] with these two H's and CO32–.

648

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Chapter 24–18 24.53 Enolate A is more substituted (and more stable) than either of the other two possible enolates and attacks an aldehyde carbonyl group, which is sterically less hindered than a ketone carbonyl. The resulting ring size (five-membered) is also quite stable. That is why 1-acetylcyclopentene is the major product. O

O H

H OH

O

H

O H H –

–OH

OH H O

OH O

A O most stable enolate less hindered carbonyl

O

H2O

H

OH

O

H2O

–

OH

1-acetylcyclopentene major product O

H H

–OH

O

–

H

O

H

O

H

O

O

OH H O

H OH

H

B The more hindered ketone carbonyl makes nucleophilic attack more difficult.

OH

OH O

H

H H2O CHO –OH

H2O

O H

H

H H –

O

H OH

O

O H

OH H2O

H

H H

–OH

OH

O

O

O

O

OH

C less stable enolate H2O

These two reacting functional groups are farther away than the reacting groups in the first two reactions, making it harder for them to find each other. Also, the product contains a less stable seven-membered ring.

O

–OH

24.54 All enolates have a second resonance structure with a negative charge on O. O

O O

O O

a. HH Na+ –OEt

O

O

O

O

O H +

O

H

+ EtOH

O H2O

+

O

O H3

O+

EtO

O

+ EtOH

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Carbonyl Condensation Reactions 24–19 Na+ –OCH3 H H

H

CH3O

b.

CH3O O O

O

O

O

O

O

O

COOCH3

H OH

+ CH3OH –OH

O O COOCH3

24.55 Removal of a proton from CH3NO2 forms an anion for which three resonance structures can be drawn. O

O

H CH2 N

O

CH2 N O

O

CH2 N O

CH2 N O

O

–OH

O C6H5

C

O

O

CH2 N

H

H OH

C6H5 C CH2

O

OH

NO2

OH

C6H5 C CH NO2

H

C6H5CH CHNO2

C6H5 C CH NO2

H H

H

+

–OH

+ H2O

–OH

24.56 All enolates have a second resonance structure with a negative charge on O.

O O

O

O

H H

O

+ H2O

OH

H OH H OH

O

O

O

O H H OH H OH

HO OH O

OH

O

HO

O

H O

H OH

O

650

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Chapter 24–20 24.57 Polymerization occurs by repeated Michael reactions. B H

B

H [1]

B

[2]

O

O

O

O

O

O

O O

O

O

O

O

Repeat step [2].

new C–C bonds B O

O

O

O

O

O

polytulipalin

24.58 O

O H

O H

O H

CH3

C

O

O O

C

O H

CH3

CH3

O

O

CH3COO–

+

C

H O

O O

O

H

O CH3COO

CH2

H CH3COO–

CH3COO– OH

OH

O

H OCOCH3

H

+ CH3COOH O

O

O

O

H

O

O

O

O

CH2 O

+ CH3COOH O

O

+ –OH

coumarin

24.59 C6 CO2Et

a.

O

+ EtOH

H

ethyl 2,4-hexadienoate CO2Et

OEt

EtO O

OEt

CO2Et

CO2Et

O EtO

O

OEt

diethyl oxalate

O CO2Et

EtO2C O OEt

CO2Et

+

OEt

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651

Carbonyl Condensation Reactions 24–21 b. The protons on C6 are more acidic than other sp3 hybridized C–H bonds because a highly resonance-stabilized carbanion is formed when a proton is removed. One resonance structure places a negative charge on the carbonyl O atom. This makes the protons on C6 similar in acidity to the
H’s to a carbonyl. c. This is a crossed Claisen because it involves the enolate of a conjugated ester reacting with the carbonyl group of a second ester. 24.60 O

O –OH,

a.

H2O C6H5

C6H5CHO

O

b.

O

–

OH, H2O

O CH2OH

O

PCC

CHO

H2C=O O

O –OH,

c.

HCO2Et

–OEt,

CH3CH2OH

O

H2O O

O

O

OH [1] LiAlH4

[1] LDA, THF

d.

[2] CH3CH2CH2CH2Br

[2] H2O

or

O

[1] LDA, THF

H2 (excess)

[2] CH3CH2CH2CHO

Pd-C

[3] H2O, –OH

(E and Z isomers)

O

e.

O

OH

–OH

H2 (excess)

H2O

Pd-C

24.61 CHO [1] O3

CHO

[2] (CH3)2S

CHO

NaOEt EtOH

A

B

24.62 O

O [1] LDA, THF

a.

[2] O H [3] H2O, –OH O

O –OH

b.

O

H2O

O

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Chapter 24–22 O

O

O [1] NaOEt

[1] Br2, CH3COOH

c.

O

[2] Li2CO3, LiBr, DMF

O

H3O+

O

O




O OEt

COOEt

[2] H2O O CN

NaOEt, EtOH

d.

CH2(CN)2

CN

O –

e.

[1] (CH3)2CuLi

OH, H2O

O

O

O

[2] H2O

24.63 O

OH

O [1] LDA

a. [2] CH3CH2CHO [3] H2O PCC OH O

O

O

O

O

[1] LDA

[1] NaOEt

[2] CH3CO2Et

[2] CH3Br

b. OH H2SO4 O

CrO3 OH

PBr3

H2SO4 H2O

CH3OH OH O

O

OH

[1] LDA

c.

CH3OH

O

C6H5

[2] C6H5CHO

SOCl2

– H2O

unstable

PCC [1] Br2, h


CH3Cl

C6H5

OH

[2] –OH

AlCl3

HO

O C6H5

d.

H2 (excess)

C6H5

Pd-C

(from c.) O [1] CH3MgBr

e.

O

OH H2SO4

[1] O3

[2] H2O

[2] (CH3)2S Mg

CH3OH

PBr3

CH3Br

major product

O CHO

–OH,

H2O

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653

Carbonyl Condensation Reactions 24–23 24.64 [1] NaOEt

OEt

a.

CO2Et

[2] H3O+

O

b.

O

[1] NaOEt

CO2Et

[2]

O

CO2Et

Br

(from a.)

O

PBr3 OH

C6H5

[1] NaOEt

c. CO2Et

[2]

O

CO2Et

Br

C6H5

C6H5

[1] LiAlH4

OH

[2] H2O

O

OH

(from a.) CH3OH + SOCl2 [1] CH3Cl, AlCl3

Br

[2] Br2, h


C6H5

OEt

d.

[1] NaOEt

O

O

C6H5

[2] H3O+

O

O

AlCl3 O Cl SOCl2 O

CrO3

OH

OH H2SO4 H2O

24.65 a.

[1] O3

CHO

–

[2] (CH3)2S

CHO

H2O

CHO

OH

NaBH4 CHO

b.

OH

CH3OH

(from a.) O

c.

CHO

CrO3

COOH

CHO

H2SO4

COOH

(from a.)

H2SO4

O CO2Et [1] NaOEt [2] CH3I

(from c.)

CO2Et

[1] NaOEt

CO2Et

[2] H3O+

H2O

O

d.

OH

O CO2Et

H3O+


CO2Et

654

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Chapter 24–24 24.66 O

O CHO

O

a. CH3O

+

O

CH3O

octinoxate [1] NaH

b.

[2] CH3Cl

HO

Br

Br2 CH3O

[2] H2C=O

CH3O

SOCl2

CH2OH

[1] Mg

(+ ortho isomer)

CH3O

PCC PCC

CH3OH

CH3OH O O

H2SO4

HO

HO

CHO O CH3O

CrO3 H2SO4, H2O

H2O

HO

CH2=O [1] PBr3

BrMg

NaOR, ROH

HO

octinoxate

[2] Mg H2O O

BrMg

Mg Br

PCC

PBr3

HO

HO

24.67 O

O

[1] HCO2Et

a.

[2]

CHO

H3O+

NaOEt, EtOH

O

O CHO

b.

O

NaOEt, EtOH

CHO

O

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Carbonyl Condensation Reactions 24–25 O O

O O

O

H

OEt H

OEt

c.

O

H OEt O

O

+

O

O

+

O

HCO2Et

OEt

+

O

H OH2

d.

=

O

O

O

O

+OH

+ H OH

H2O

H2O

+ OH

H OH2

OH

no
H

OH

O

OH

OH

+ H OH2

H2O

+O H

O

-hydroxy ketone + OH2

H

+

H2O

H2O

H2O H3O+

O

O

O

H2O

This reaction is an acid-catalyzed aldol that proceeds by way of enols not enolates. The -hydroxy ketone initially formed cannot dehydrate to form an
,-unsaturated carbonyl because there is no H on the
carbon. Thus, dehydration occurs but the resulting C=C is not conjugated with the C=O. O

e.

H CHO

This H is now more acidic because it is located between two carbonyl groups. As a result, it is the most readily removed proton for the Michael reaction in the next step.

656

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Chapter 24–26 24.68 Rearrangement generates a highly resonance-stabilized enolate between two carbonyl groups. OEt O

O

O OEt O

O

OEt

O

OEt

O

OEt

O

H OCH2CH3

OCH2CH3

OCH2CH3

OCH2CH3 CH3CH2O

H OCH2CH3 OCH2CH3 O

O

H

CO2CH2CH3

O OEt

O OEt CO2CH2CH3

CO2CH2CH3

O OCH2CH3

(+ 2 resonance structures)

H OCH2CH3

This product is a highly resonance stabilized enolate. This drives the reaction. H3O+ O

H

CO2CH2CH3

24.69 All enolates have a second resonance structure with a negative charge on O. O O

O

O

B

[2]

[1] +

H OH O

O H OH

[3]

B

O

[4]

OH

[5]

+ OH

HB+ + HB+

+ –OH

O OH

O

O HO

+

O

Repeat steps [1]–[5] by deprotonating the indicated CH3. O

H

isophorone + HB+

(+ 2 resonance structures)

H

B

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24. Carbonyl Condensation Reactions

657

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Text

Carbonyl Condensation Reactions 24–27 24.70 All enolates have a second resonance structure with a negative charge on O. B O

O

COOEt

O

H H

O CO2CH3

CO2CH3 CO2CH3

new bond

new bond H2O

+ –OH

O

CO2CH3

24.71 a.

–

H CH2

OH

N

CH2

CH2

N

(+ other resonance structures)

one possible resonance structure The negative charge is delocalized on the electronegative N atom. This factor is what makes the CH3 group bonded to the pyridine ring more H OH acidic, and allows the condensation to occur.

O C6H5

N

O H

CH2

OH

C6H5 C CH2

N

N

OH

C6H5 C CH

H

N

C6H5 C CH

H H

N

H

H2O

OH

CH CH

N

A OH

b. The condensation reaction can occur only if the CH3 group bonded to the pyridine ring has acidic hydrogens that can be removed with –OH. OH

N CH2

N

H

CH2

OH

H CH2

3-methylpyridine

Since the negative charge is delocalized on the N, the CH3 contains acidic H's and reaction will occur.

CH2

(+ other resonance structures)

2-methylpyridine N

N

N CH2

N CH2

N CH2

N CH2

No resonance structure places the negative charge on the N so the CH3 is not acidic and condensation does not occur.

N CH2

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25. Amines

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Text

Study Guide/Solutions Manual to accompany Organic Chemistry, Third Edition

659

Amines 25–1 C Chhaapptteerr 2255:: A Am miinneess

G Geenneerraall ffaaccttss • Amines are organic nitrogen compounds having the general structure RNH2, R2NH, or R3N, with a lone pair of electrons on N (25.1). • Amines are named using the suffix -amine (25.3). • All amines have polar C–N bonds. Primary (1°) and 2° amines have polar N–H bonds and are capable of intermolecular hydrogen bonding (25.4). • The lone pair on N makes amines strong organic bases and nucleophiles (25.8).

SSuum mm maarryy ooff ssppeeccttrroossccooppiicc aabbssoorrppttiioonnss ((2255..55)) Mass spectra Molecular ion Amines with an odd number of N atoms give an odd molecular ion. IR absorptions

N–H

3300–3500 cm–1 (two peaks for RNH2, one peak for R2NH)

1

NH CH–N

0.5–5 ppm (no splitting with adjacent protons) 2.3–3.0 ppm (deshielded Csp3–H)

C–N

30–50 ppm

H NMR absorptions

13

C NMR absorption



C Coom mppaarriinngg tthhee bbaassiicciittyy ooff aam miinneess aanndd ootthheerr ccoom mppoouunnddss ((2255..1100)) • Alkylamines (RNH2, R2NH, and R3N) are more basic than NH3 because of the electron-donating R groups (25.10A). • Alkylamines (RNH2) are more basic than arylamines (C6H5NH2), which have a delocalized lone pair from the N atom (25.10B). • Arylamines with electron-donor groups are more basic than arylamines with electron-withdrawing groups (25.10B). • Alkylamines (RNH2) are more basic than amides (RCONH2), which have a delocalized lone pair from the N atom (25.10C). • Aromatic heterocycles with a localized electron pair on N are more basic than those with a delocalized lone pair from the N atom (25.10D). • Alkylamines with a lone pair in an sp3 hybrid orbital are more basic than those with a lone pair in an sp2 hybrid orbital (25.10E).

PPrreeppaarraattiioonn ooff aam miinneess ((2255..77)) [1] Direct nucleophilic substitution with NH3 and amines (25.7A) R X

+

NH3

excess

R NH2

+ NH4+ X–

1o amine

• •

R' R X + R' N R' R'

+

R N R' X– R'

ammonium salt

•

The mechanism is SN2. The reaction works best for CH3X or RCH2X. The reaction works best to prepare 1o amines and ammonium salts.

660

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Chapter 25–2 [2] Gabriel synthesis (25.7A) • •

O R X

+

CO2–

–OH

N

R NH2

H2O

+

CO2–

1o amine

O

The mechanism is SN2. The reaction works best for CH3X or RCH2X. Only 1o amines can be prepared.

•

[3] Reduction methods (25.7B) [a] From nitro compounds

H2, Pd-C or

R NO2

[b] From nitriles

R NH2

Fe, HCl or Sn, HCl

1o amine

[1] LiAlH4 [2] H2O

R C N

R CH2NH2

1o amine O

[c] From amides

C

R

[1] LiAlH4 [2] H2O

NR'2

RCH2 N R' R'

R' = H or alkyl

1o, 2o, and 3o amines

[4] Reductive amination (25.7C) R C O

R2"NH

+

•

R

NaBH3CN

R' C N R"

R'

H R"

•

R', R" = H or alkyl 1o, 2o, and 3o amines

Reductive amination adds one alkyl group (from an aldehyde or ketone) to a nitrogen nucleophile. Primary (1°), 2°, and 3° amines can be prepared.



R Reeaaccttiioonnss ooff aam miinneess [1] Reaction as a base (25.9) R NH2

+

+

H A

R NH3

+

A

[2] Nucleophilic addition to aldehydes and ketones (25.11) With 1o amines:

With 2o amines:

O R

C

O

NR' C

H

R = H or alkyl

R'NH2

R

C

C

imine

H

R

C

C

H

R = H or alkyl

NR'2

R'2NH R

C

C

enamine

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25. Amines

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Text

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661

Amines 25–3 [3] Nucleophilic substitution with acid chlorides and anhydrides (25.11) O R

C

O +

R'2NH (2 equiv)

Z

R

Z = Cl or OCOR R' = H or alkyl

C

NR'2

1o, 2o, and 3o amides

[4] Hofmann elimination (25.12)

C C H NH2

[1] CH3I (excess)

•

C C

[2] Ag2O [3]


The less substituted alkene is the major product.

alkene

[5] Reaction with nitrous acid (25.13) With 1o amines: R NH2

With 2o amines:

NaNO2

+

R N N

Cl–

NaNO2

R N H

HCl

alkyl diazonium salt

R N N O

HCl

R

R

N-nitrosamine



R Reeaaccttiioonnss ooff ddiiaazzoonniiuum m ssaallttss [1] Substitution reactions (25.14) N2

Cl–

F

X

OH

phenol

With HBF4:

With CuX:

With H2O: +

aryl chloride or aryl bromide

aryl fluoride

X = Cl or Br With NaI or KI:

With CuCN:

I

aryl iodide

With H3PO2:

CN

H

benzonitrile

benzene

[2] Coupling to form azo compounds (25.15) N2+ Cl– +

Y

Y = NH2, NHR, NR2, OH (a strong electrondonor group)

N N

azo compound

Y

+

HCl

662

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Chapter 25–4 C Chhaapptteerr 2255:: A Annssw weerrss ttoo PPrroobblleem mss 25.1

Amines are classified as 1o, 2o, or 3o by the number of alkyl groups bonded to the nitrogen atom. 1o amine

2o amine a. H N 2

C6H5

H N

N H

NH2

b. CH3CH2O

2o amine

N CH3 O

1o amine

3o amine

25.2 3o amine CH3

a.

b.

HO C CH2NH2

CH3 N CH2CH2OH

CH3

3o alcohol

25.3

CH3

1o amine

1o alcohol

The N atom of a quaternary ammonium salt is a stereogenic center when the N is surrounded by four different groups. All stereogenic centers are circled. OH +

CH3

HO

CH3 +

a. CH3 N CH2CH2 N CH2CH3 CH3

H N

b. HO

H

N has 3 similar groups.

25.4 CH3 NH2

NH2

NH2

+ 3 more resonance structures

140 pm partial double Because the lone pair on N can be delocalized bond character on the benzene ring, the C–N bond has partial double bond character, making it shorter. Both the C and N atoms must be sp2 hybridized (+ have a p orbital) for delocalization to occur. The higher percent s-character of the orbitals of both C and N shortens the bond as well.

147 pm The C–N bond is formed from two sp3 hybridized atoms and the lone pair is localized on N.

25.5 NHCH2CH3

a.

CH3CH2CH(NH2)CH3

2-butanamine or sec-butylamine

c.

N(CH3)2

N,N-dimethylcyclohexanamine

e. N-ethyl-3-hexanamine CH3

b.

(CH3CH2CH2CH2)2NH

dibutylamine

d.

f.

NHCH2CH2CH3

NH2

2-methyl-5-nonanamine

2-methyl-N-propylcyclopentanamine

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Amines 25–5 25.6

An NH2 group named as a substituent is called an amino group.

a. 2,4-dimethyl-3-hexanamine

c. N-isopropyl-p-nitroaniline

g. N-methylaniline

e. N,N-dimethylethylamine

NHCH(CH3)2

NHCH3 N

O2N

NH2

b. N-methylpentylamine

d. N-methylpiperidine

f. 2-aminocyclohexanone

h. m-ethylaniline

O

NH2 NH2

NHCH3

N CH2CH3

25.7

Primary (1°) and 2o amines have higher bp’s than similar compounds (like ethers) incapable of hydrogen bonding, but lower bp’s than alcohols that have stronger intermolecular hydrogen bonds. Tertiary amines (3°) have lower boiling points than 1o and 2o amines of comparable molecular weight because they have no N–H bonds. O

a. (CH ) CHCH CH CH C CH CH 3 2 2 3 3 2 3 alkane lowest boiling point

25.8

ketone intermediate boiling point

(CH3)2CHCH2NH2

amine N–H can hydrogen bond. highest boiling point

CH3

alkane lowest boiling point

O

CH3

ether intermediate boiling point

NH2

amine N–H can hydrogen bond. highest boiling point

1o Amines show two N–H absorptions at 3300–3500 cm–1. 2o Amines show one N–H absorption at 3300–3500 cm–1. molecular weight = 59 one IR peak = 2° amine

25.9

b.

CH3 N CH2CH3 H

The NH signal occurs between 0.5 and 5.0 ppm. The protons on the carbon bonded to the amine nitrogen are deshielded and typically absorb at 2.3–3.0 ppm. The NH protons are not split. molecular formula C6H15N 1H NMR absorptions (ppm): 0.9 (singlet, 1 H) 1.10 (triplet, 3 H) 1.15 (singlet, 9 H) 2.6 (quartet, 2 H)

NH CH3 adjacent to CH2 (CH3)3C CH2 adjacent to CH3

N H

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Chapter 25–6 25.10 The atoms of 2-phenylethylamine are in bold. O N

(CH3CH2)2N

a.

CH3

b.

H

CH3O

O H

N H LSD lysergic acid diethyl amide

H

HO

N CH3

codeine

25.11 SN2 reaction of an alkyl halide with NH3 or an amine forms an amine or an ammonium salt. NH3

Cl

a.

NH2

excess

NH2

b.

CH3CH2Br

N(CH2CH3)3 Br–

excess

25.12 The Gabriel synthesis converts an alkyl halide into a 1o amine by a two-step process: nucleophilic substitution followed by hydrolysis. CH3O

NH2

a.

b.

Br

(CH3)2CHCH2CH2NH2

CH3O

NH2

Br

c.

(CH3)2CHCH2CH2Br

25.13 The Gabriel synthesis prepares 1° amines from alkyl halides. Since the reaction proceeds by an SN2 mechanism, the halide must be CH3 or 1°, and X can’t be bonded to an sp2 hybridized C. NH2

a.

NH2

b.

aromatic An SN2 does not occur on an aryl halide. cannot be made by Gabriel synthesis

can be made by Gabriel synthesis

NH2

N H

c.

d.

2° amine cannot be made by Gabriel synthesis

N on 3° C An SN2 does not occur on a 3° RX. cannot be made by Gabriel synthesis

25.14 Nitriles are reduced to 1o amines with LiAlH4. Nitro groups are reduced to 1o amines using a variety of reducing agents. Primary (1°), 2o, and 3o amides are reduced to 1o, 2o, and 3o amines respectively, using LiAlH4. O

a.

CH3CHCH2NH2

CH3CHCH2NO2

CH3

CH3

CH3CHC N

CH3CHCNH2

CH3

CH3 O

b.

CH2NH2

CH2NO2

C N

C NH2

c.

NH2

NO2

C

N

O NH2

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Amines 25–7 25.15 Primary (1°), 2o, and 3o amides are reduced to 1o, 2o, and 3o amines respectively, using LiAlH4. CONH2

O

CH2NH2

a.

NHCH3

c.

NHCH3

O

b.

N

N

25.16 General reaction:

(CH3)2CHNH2

[H]

R C N

isopropylamine

RCH2NH2

The amine needs 2 H's here. The C bonded to the N must have 2 H's for the amine to be formed by reduction of a nitrile.

25.17 Only amines with a CH2 or CH3 bonded to the N can be made by reduction of an amide. NH2

a.

N bonded to benzene cannot be made by reduction of an amide

NH2

NH2

b.

c.

N on 2° C on both sides cannot be made by reduction of an amide

N bonded to a 3° C cannot be made by reduction of an amide

N bonded to CH2 can be made by reduction of an amide

N H

d.

25.18 Reductive amination is a two-step method that converts aldehydes and ketones into 1o, 2o, and 3o amines. Reductive amination replaces a C=O by a C–H and C–N bond. a.

CHO

CH3NH2

NHCH3

c.

NaBH3CN O

b.

NH3 NaBH3CN

O

(CH3CH2)2NH NaBH3CN

NH2

d.

O

+

25.19 NH2

a.

O NH3

OH

b.

OH NHCH3

O NH2CH3

or

OH NH2

CH2=O

NH2

NaBH3CN

N(CH2CH3)2

NHCH(CH3)2

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Chapter 25–8 25.20 Only amines that have a C bonded to a H and N atom can be made by reductive amination; that is, an amine must have the following structural feature: H

a. NH2

C N

phentermine In phentermine, the C bonded to N is not bonded to a H, so it cannot be made by reductive amination. b. systematic name: 2-methyl-1-phenyl-2-propanamine

25.21 The pKa of many protonated amines is 10–11, so the pKa of the starting acid must be less than 10 for equilibrium to favor the products. Amines are thus readily protonated by strong inorganic acids like HCl and H2SO4, and by carboxylic acids as well. a. CH3CH2CH2CH2 NH2

CH3CH2CH2CH2 NH3 + Cl–

+ HCl

b. C6H5COOH

+

c.

pKa
10 weaker acid products favored

pKa = –7

pKa = 15.7 pKa
10 weaker acid reactants favored

pKa = 10.7 weaker acid products favored

pKa = 4.2

N H H

N H

(CH3)2NH2 + C6H5COO–

(CH3)2NH

+ HO–

+ H2O

25.22 An amine can be separated from other organic compounds by converting it to a water-soluble ammonium salt by an acid–base reaction. In each case, the extraction procedure would employ the following steps: • Dissolve the amine and either X or Y in CH2Cl2. • Add a solution of 10% HCl. The amine will be protonated and dissolve in the aqueous layer, while X or Y will remain in the organic layer as a neutral compound. • Separate the layers. a.

NH2

and X

b.

(CH3CH2CH2CH2)3N

and

+

H Cl

CH3

NH3 Cl–

CH3

X • insoluble in H2O • soluble in CH2Cl2

• soluble in H2O • insoluble in CH2Cl2 (CH3CH2CH2CH2)2O

+

H Cl

Y

(CH3CH2CH2CH2)3NH Cl–

• soluble in H2O • insoluble in CH2Cl2

(CH3CH2CH2CH2)2O

Y • insoluble in H2O • soluble in CH2Cl2

25.23 Primary (1°), 2o, and 3o alkylamines are more basic than NH3 because of the electron-donating inductive effect of the R groups. a. (CH3)2NH

and

NH3

2° alkylamine CH3 groups are electron donating. stronger base

b. CH3CH2NH2 1° alkylamine stronger base

and

ClCH2CH2NH2

1° alkylamine Cl is electron withdrawing. weaker base

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Amines 25–9 25.24 Arylamines are less basic than alkylamines because the electron pair on N is delocalized. Electron-donor groups add electron density to the benzene ring making the arylamine more basic than aniline. Electron-withdrawing groups remove electron density from the benzene ring, making the arylamine less basic than aniline. NH2

NH2

NH2

NH2

a.

NH2

NH2

b.

CH3OOC

O2N

CH3O

electronwithdrawing group least basic

arylamine intermediate basicity

electrondonating group most basic

electronwithdrawing group least basic

arylamine intermediate basicity

alkylamine most basic

25.25 Amides are much less basic than amines because the electron pair on N is highly delocalized. CONH2

NH2

NH2

amide least basic

arylamine intermediate basicity

alkylamine most basic

25.26 sp2 hybridized more basic

This N is also sp2 hybridized but the electron pair CH3 occupies a p orpital so it can N N delocalize onto the aromatic a. Electron pair on N CH3 occupies an sp2 ring. Delocalization makes hybrid orbital. this N less basic. DMAP 4-(N,N-dimethylamino)pyridine

b.

sp3 hybridized N stronger base

N

H

CH3

N

nicotine sp2 hybridized N higher percent s-character weaker base

25.27 This electron pair is delocalized, making it a weaker base.

Br 2

H

N

N

stronger base sp3 hybridized N 25% s-character

HO H

NH2 CH3O

a. N

stronger base This compound is similar to DMAP in Problem 25.26a.

b.

sp2 hybridized N 33% s-character

sp hybridized N 33% s-character N

c.

N(CH3)2

stronger base sp3 hybridized N 25% s-character

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Chapter 25–10 25.28 Amines attack carbonyl groups to form products of nucleophilic addition or substitution. CH3CH2CH2NH2

O

a.

b. CH3

O

O

C

C

O

NCH2CH2CH3

CH3CH2CH2NH2 CH3

CH3

COCl

c.

O

O

O

C

C

C

NHCH2CH2CH3

CH3

O

CONHCH2CH2CH3

CH3CH2CH2NH2

(CH3CH2)2NH

O

N(CH2CH3)2 O

(CH3CH2)2NH CH3

COCl

CH3

C

N(CH2CH3)2 CON(CH2CH3)2

(CH3CH2)2NH

25.29 [1] Convert the amine (aniline) into an amide (acetanilide). [2] Carry out the Friedel–Crafts reaction. [3] Hydrolyze the amide to generate the free amino group. O

a.

C

CH3

NH2

O C CH3 (CH3)3CCl (CH ) C NH 3 3 AlCl3

Cl

(+ ortho isomer)

O NH2

b.

C

CH3

O C CH3 H3O+ NH (CH3)3C

H N

Cl

Cl C

CH3

H N O AlCl3

O

C

CH3

NH2

H3O+

O CCH2CH3

O

O

(+ para isomer)

25.30 transition state

+

Energy

transition state:

N(CH3)3 H

Ea

OH




starting materials

(no 3-D geometry shown here)


H° products

Reaction coordinate

25.31 a. CH3CH2CH2CH2 NH2

b.

(CH3)2CHNH2

[1] CH3I (excess) [2] Ag2O [3]


[1] CH3I (excess) [2] Ag2O [3]


CH3CH2CH=CH2

CH3CH=CH2

c.

NH2

[1] CH3I (excess) [2] Ag2O [3]


NH2

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Amines 25–11 25.32 In a Hofmann elimination, the base removes a proton from the less substituted, more accessible  carbon atom, because of the bulky leaving group on the nearby
carbon. [1] CH3I (excess)

a.

CH2CHCH3

CH=CHCH3

[2] Ag2O [3]


NH2

major product

[1] CH3I (excess)

b. H2N

 major product +

 [1] CH3I (excess)



c.

[2] Ag2O [3]


CH2CH=CH2

N H

[2] Ag2O [3]


(3  C's) least substituted  carbon

CH3CH=CH(CH2)3N(CH3)2 + CH2=CH(CH2)2CHN(CH3)2 + CH3 CH2=CH(CH2)4N(CH3)2

major product, formed by removal of a H from the least substituted  C

25.33 K+ –OC(CH3)3

a.

K+ –OC(CH3)3

c. Cl

Br [1] CH3I (excess)

b. NH2

[1] CH3I (excess)

d.

[2] Ag2O [3]


(E and Z)

[2] Ag2O [3]


NH2

25.34 N2+ Cl–

NH2 NaNO2

a.

HCl

CH3

b.

H CH3CH2 N CH3

HCl

HCl

N H

CH3

NaNO2

NaNO2

c.

N NO

NaNO2

CH3CH2 N CH3

d.

HCl

NO

N2 Cl–

NH2

25.35 a.

NH2 [1] NaNO2, HCl [2] CuBr

b.

NH2

Br

[1] NaNO2, HCl

OH

c. CH3O

NH2

d.

N2+ Cl–

[2] H2O O2N

O2N

Cl

[1] NaNO2, HCl

[1] CuCN

CH2NH2

[2] LiAlH4 [3] H2O

Cl

25.36 NH2

NO2

a.

HNO3

H2

H2SO4

Pd-C

F [1] NaNO2, HCl [2] HBF4

CH3O

[2] HBF4

F

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Chapter 25–12 NO2

b.

NO2 HNO3

H2

H2SO4

Pd-C

(from a.)

OH

NH2 [1] NaNO2, HCl [2] H2O

NO2

OH

NH2 CH3

[1] NaNO2, HCl

c.

CH3Cl

[2] NaI

NH2

I

(+ para isomer)

AlCl3

I

(from a.) NH2

NH2 Cl

Cl

Cl2 (excess)

d.

Cl

Cl

[1] NaNO2, HCl [2] H3PO2

FeCl3

(from a.)

Cl

Cl

25.37 OH

NH2

N N

a.

NH2

c.

HO

OH

N N HO

b.

HO

N N

OH

25.38 To determine what starting materials are needed to synthesize a particular azo compound, always divide the molecule into two components: one has a benzene ring with a diazonium ion, and one has a benzene ring with a very strong electron-donor group. O2N

a. H2N

Cl

b. HO

N N

N N

CH3

O2N H2N

Cl HO

Cl– N2

Cl– N2

CH3

25.39 a. N N OH

b. O2N

NO2

alizarine yellow R

para red

+

N N Cl–

N N

OH COOH

+

NO2

O2N

N N Cl–

OH COOH

OH

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Amines 25–13 25.40 O

CH3

O

O

O O

O

–

O3S

N N

O

Dacron

O

methyl orange

To bind to fabric, methyl orange (an anion) needs to interact with positively charged sites. Since Dacron is a neutral compound with no cationic sites on the chain, it does not bind methyl orange well.

N CH3

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Chapter 25–14 25.41 a. CH3NHCH2CH2CH2CH3 N-methyl-1-butanamine (N-methylbutylamine) b.

i. N H

tripropylamine

CH2CH3

2-ethylpyrrolidine f. (C6H5)2NH diphenylamine

NH2

1-octanamine (octylamine) c.

(CH3CH2CH2)3N

e.

CH3CH2CH2CH(NH2)CH(CH3)2

2-methyl-3-hexanamine

N C(CH3)3

g.

NH2

j.

NH2

k.

CH2CH3

4,6-dimethyl-1-heptanamine

3-ethyl-2-methylcyclohexanamine

N-tert-butyl-N-ethylaniline

CH3 N

d.

h. O

l.

NH2

N(CH2CH3)2

CH2CH2CH3

4-aminocyclohexanone

N-methyl-N-propylcyclohexanamine

N,N-diethylcycloheptanamine

25.42 a. cyclobutylamine

h. 3-methyl-2-hexanamine

e. N-methylpyrrole

NH2

NH2 N

b. N-isobutylcyclopentylamine

CH3

i. 2-sec-butylpiperidine

f. N-methylcyclopentylamine N

NHCH3

H

c. tri-tert-butylamine

N H

g. cis-2-aminocyclohexanol

N[C(CH3)3]3

NH2

NH2

or

d. N,N-diisopropylaniline OH

N[CH(CH3)2]2

j. (2S)-2-heptanamine H NH2

OH

25.43 NH2

1-butanamine H N

N-methyl-1-propanamine

NH2

2-butanamine

N H

diethylamine

NH2

2-methyl-1-propanamine

N H

N-methyl-2-propanamine

NH2

2-methyl-2-propanamine

N

N,N-dimethylethanamine

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Amines 25–15 25.44 [* denotes a stereogenic center.] a.

*

N

CH3

b.

CH2CH3 * * CH3CH2CHCH CH CH 2 2 2 N CH2CH2CH2CH3

CH2CH3

CH3

1 stereogenic center 2 stereoisomers

N H

H CH3

CH3 CH3CH2

CH2CH2CH2 H CH3

N

Cl

CH3

N Cl

CH2CH2CH2

N Cl

CH3 CH2CH3

C CH3CH2

CH2CH2CH2CH3

CH2CH2CH2 H CH3

CH3 CH2CH3

C CH3CH2

CH2CH3

H CH3

CH3 CH2CH3

C

CH2CH3

H

CH3

2 stereogenic centers 4 stereoisomers

CH2CH2CH2CH3

CH3 CH2CH3

C CH3CH2

CH2CH2CH2CH3

N Cl

CH2CH2CH2

N Cl

CH2CH2CH2CH3

25.45 a.

c. (CH3CH2)2NH

(CH3CH2)2NH or N

sp3 hybridized N stronger base b.

2° alkylamine stronger base

sp2 hybridized N weaker base

HCON(CH3)2 or (CH3)3N

or

(ClCH2CH2)2NH

2° alkylamine Cl is electron withdrawing. weaker base or

d.

NH

N H

amide alkylamine weaker base stronger base

weaker base (delocalized electron pair on N)

stronger base

25.46 a.

NH2

NH2

NH3

arylamine intermediate least basic basicity

alkylamine most basic

O2N

b. N H

delocalized electron pair on N least basic

d. N

sp2 hybridized N intermediate basicity

NH2

c.

N H

sp3 hybridized N most basic

NH2 Cl

CH3

electronwithdrawing group least basic

intermediate basicity

(C6H5)2NH

C6H5NH2

diarylamine least basic

NH2

arylamine intermediate basicity

electrondonating group most basic NH2

alkylamine most basic

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Chapter 25–16 25.47 The electron-withdrawing inductive effect of the phenyl group stabilizes benzylamine, making its conjugate acid more acidic than the conjugate acid of cyclohexanamine. The conjugate acid of aniline is more acidic than the conjugate acid of benzylamine, since loss of a proton generates a resonance-stabilized amine, C6H5NH2. NH3

CH2NH3

pKa = 10.7

NH3

pKa intermediate

NH2

alkylamine cyclohexanamine

pKa = 4.6

CH2NH2

electron-withdrawing inductive effect of the sp2 hybridized C's benzylamine

NH2

resonance-stabilized aromatic amine aniline

25.48 The most basic N atom is protonated on treatment with acid. O

O

O

O NH

a.

NH2

CH3CO2H N

N

more basic

O CH2COOH

+ CH3CO2–

O CH2COOH

benazepril N

N

b.

NH

CH3CO2H

N

NH2 + CH3CO2– N

varenicline

most basic

25.49 Nc

O C

a.

NHNH2

N

Nb < Na < Nc

Nb Na

Na

Nc

N

NH2

b. N

Nb H

N b < Na < N c

Order of basicity: Nb < Na < Nc Nb – The electron pair on this N atom is delocalized on the O atom; least basic. Na – The electron pair on this N atom is not delocalized, but is on an sp2 hybridized atom. Nc – The electron pair on this N atom is on an sp3 hybridized N; most basic.

Order of basicity: Nb < Na < Nc Nb – The electron pair on this N atom is delocalized on the aromatic five-membered ring; least basic. Na – The electron pair on this N atom is not delocalized, but is on an sp2 hybridized atom. Nc – The electron pair on this N atom is on an sp3 hybridized N; most basic.

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Amines 25–17 25.50 The most basic N atom is protonated on treatment with acid. a 3° alkylamine with an sp3 hybridized N most basic O

O Cl

a. N

Cl

Cl

b. HCl

N

NH

N H Cl–

NH

aripiprazole a delocalized electron part of an amide pair on the benzene ring least basic intermediate O

Cl

N

O

25.51 The para isomer is the weaker base because the electron pair on its NH2 group can be delocalized onto the NO2 group. In the meta isomer, no resonance structure places the electron pair on the NO2 group, and fewer resonance structures can be drawn: O2N

NH2

O2N

NH2

O2N

NH2

O2N

NH2

O2N

NH2

meta NH2

NH2

O2N

O2N

NH2 O

para

N

NH2 O2N

NH2 O2N

O NH2 O

N

NH2 O

O

N O

25.52 N A

N

This two-carbon bridge makes it difficult for the lone pair on N to delocalize on the aromatic ring.

B

pKa of the conjugate acid = 5.2 pKa of the conjugate acid = 7.29 Resonance structures that place a double bond between the N atom and the benzene ring are stronger conjugate acid weaker conjugate acid destabilized. Since the electron pair is more weaker base stronger base localized on N, compound B is more basic. The electron pair of this arylamine is delocalized on the benzene ring, decreasing its basicity. N

B

N

Geometry makes it difficult to have a double bond here.

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Chapter 25–18 25.53 B

NH

N

N

pyrrole pKa = 23 stronger acid

N

N

N

B

NH

pyrrolidine pKa = 44 weaker acid

weaker conjugate base The electron pair is delocalized, decreasing the basicity. The N atom is sp2 hybridized.

N

stronger conjugate base The electron pair is not delocalized on the ring. The N atom is sp3 hybridized.

25.54 a. C6H5CH2CH2CH2Br

b. C6H5CH2CH2Br

NH3

excess

NaCN

c. C6H5CH2CH2CH2NO2

[1] LiAlH4 C6H5CH2CH2CN H2 Pd-C

d. C6H5CH2CH2CONH2

C6H5CH2CH2CH2NH2

[2] H2O

[1] LiAlH4 [2] H2O

C6H5CH2CH2CH2NH2 e. C6H5CH2CH2CHO

NH3 NaBH3CN

C6H5CH2CH2CH2NH2 C6H5CH2CH2CH2NH2

C6H5CH2CH2CH2NH2

25.55 a. (CH3CH2)2NH

NH2

b.

N(CH3)2

c.

H N

d.

O CH3

C

NHCH2CH3

H N

N(CH3)2

NH2 O

or

O

O CH3 N

C

H

O

25.56 In reductive amination, one alkyl group on N comes from the carbonyl compound. The remainder of the molecule comes from NH3 or an amine. NH2

a.

H

+ NH3

O

b.

H N

C6H5

H

C6H5

or

NH2

O (CH3CH2CH2)2NH

H

N H

H

or

H

C

O

CH2CH3

O

d.

NH2

C6H5

O

O

c. (CH3CH2CH2)2N(CH2)2CH(CH3)2

H2N

H

CH3CH2CH2

N H

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Text

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Amines 25–19 25.57 a.

NH2

C6H5 O

b.

O

C6H5

c.

C6H5

NH3

CHO

NH

NaBH3CN

C6H5

CH2NH2

NaBH3CN NH2

(CH3)2NH

d.

N(CH3)2

NaBH3CN

NaBH3CN

O

NH

25.58 a.

CH2NH2

Br

NH3 excess

CN

CH2NH2

[1] LiAlH4

b.

[2] H2O CONH2

CH2NH2 [1] LiAlH4

c.

[2] H2O CHO

CH2NH2

NH3

d.

NaBH3CN CH3

CH2Br

e.

NH3

h


excess

COOH

CONH2

[1] SOCl2

f.

[2] NH3 NH2

g.

CH2NH2

[1] LiAlH4 [2] H2O

CN

[1] NaNO2, HCl

CH2NH2

[1] LiAlH4 [2] H2O

[2] CuCN

h.

CH2NH2

Br2

NO2

HNO3

H2 Pd-C

H2SO4

NH2

Then as in (g).

25.59 Use the directions from Answer 25.22. Separation can be achieved because benzoic acid reacts with aqueous base and aniline reacts with aqueous acid according to the following equations: COOH NaOH

benzoic acid • soluble in CH2Cl2 • insoluble in H2O

COO–Na+ + H2O

(10% aqueous) • soluble in H2O • insoluble in CH2Cl2 +

NH2

NH3 Cl– H Cl

aniline • soluble in CH2Cl2 • insoluble in H2O

(10% aqueous) • soluble in H2O • insoluble in CH2Cl2

678

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Chapter 25–20 Toluene (C6H5CH3), on the other hand, is not protonated or deprotonated in aqueous solution so it is always soluble in CH2Cl2 and insoluble in H2O. The following flow chart illustrates the process. CH3

COOH

NH2

CH2Cl2 10% NaOH CH2Cl2

H2O COO–Na+

CH3

NH2

10% HCl H2O

CH2Cl2 +

Cl–

NH3

CH3

25.60 H N

N-ethylaniline

a.

b.

HCl

H H N

f.

CH3I (excess)




Cl–

+ CH2=CH2 O

CH3CH2COCl CH3COOH

N

g.

H H N CH3COO–

O

h. The product in (g) c.

N(CH3)2

Ag2O

N

N

H2SO4

N

(CH3)2C=O

HNO3

O

O2N

NO2

CH3 N

d.

e.

CH2O

i. The product in (g)

[1] LiAlH4

NaBH3CN CH3I

CH3 CH3 N

(excess)

I–

N

[2] H2O O

j. The product in (h)

H2

O N

Pd-C NH2 H2N

N

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25. Amines

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Text

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679

Amines 25–21 25.61 NH2 p-methylaniline

CH3

CH3COCl

f. HCl

a.

CH3

NH3

CH3COCl

CH3COOH

g.

h.

NaNO2

i. Step (b), then excess

e.

N2+Cl–

CH3

HCl

C CH3 NH

c. (CH3CO)2O CH 3 CH3I

NH3 CH3COO–

CH3

C CH3 NH

CH3

O

d.

NH2

Cl–

O

b.

AlCl3

AlCl3 CH3

O CH3COCl

C CH3 NH

CH3

AlCl3 N(CH3)3 I–

CH3

(CH3)2C=O

CH3

C O CH3

N C(CH3)2

j.

CH3CHO NaBH3CN

CH3

25.62 a. CH3CH2CH2CH2NH2 b. CH3CH2CH2CH2NH2 c. CH3CH2CH2CH2NH2

d. CH3CH2CH2CH2NH2

e. CH3CH2CH2CH2NH2 f. CH3CH2CH2CH2NH2

ClCOC6H5

CH3CH2CH2CH2NHCOC6H5

O=C(CH2CH3)2 [1] CH3I (excess) [2] Ag2O [3]
C6H5CHO NaBH3CN NaBH3CN

CH3CHO CH3I excess

CH3CH2CH2CH2N=C(CH2CH3)2 CH3CH2CH=CH2

CH3CH2CH2CH2NHCH2C6H5

CH3CH2CH2CH2NHCH2CH3 [CH3CH2CH2CH2N(CH3)3]+I–

25.63 a. CH3(CH2)6NH2

[1] CH3I (excess) [2] Ag2O [3]


CH3(CH2)4CH=CH2

[1] CH3I (excess)

b. NH2

[2] Ag2O [3]


(E + Z) major product

NHCH2CH3

680

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Chapter 25–22 c.

[1] CH3I (excess) N H

CH2=CH2

e.

major product + (CH3)2CHN(CH3)2

[2] Ag2O [3]


1

2 N H

[2] Ag2O [3]


3

CH2=CHCH3 + CH3CH2N(CH3)2 CH3

[2] Ag2O [3]


NH2 CH3

N(CH3)2

+

major product 2

(CH3)2N

[1] CH3I (excess)

d.

1

[1] CH3I (excess)

CH3

CH3

+

(E + Z)

(CH3)2N CH3

CH3

(E + Z)

CH2

major product CH3

25.64 N(CH3)2

O HN(CH3)2

a.

mild acid O NH2CH2CH2CH3

b.

NCH2CH2CH3

mild acid NH2

O NH3

c.

NaBH3CN

NH2

O

d.

[1] NaBH4, CH3OH

[1] CH3I (excess)

[2] H2SO4

[2] Ag2O [3]


O NH3 NaBH3CN

OH

O CH3NH2

mCPBA

NHCH3

e.

(from d.) O

O

O Br2

f.

CH3COOH

25.65 CH3 N

a. one stereogenic center benzphetamine

Br

NH2CH2CH2CH2CH3

NHCH2CH2CH2CH3

3

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681

Amines 25–23 b. Amides that can be reduced to benzphetamine: CHO N

N

and O

c. Amines + carbonyl compounds that form benzphetamine by reductive amination: NHCH3

O

or

H O

or

1

H N

NH CH2=O

CH3

 N

[1] CH3I

2

[2] Ag2O [3]


d.

(CH3)2NCH2

major product elimination across , 1 (elimination across , 2)

25.66 a.

CH2CH2Cl

NH3

g.

CH2CH2NH2

NH

NaNO2

N N O

HCl

excess O CO2

[1] KOH

b.

NH O

c. Br

NH2 +

[2] (CH3)2CHCH2Cl [3] –OH, H2O

NO2

Sn

NH + C6H5CHO

NaBH3CN

Br

NH2

O

+

i.

N

N H

[1] LiAlH4

d. CN

[2] H2O CONHCH2CH3

N CH2C6H5

CO2

HCl

e.

h.

j. CH3CH2CH2 N CH(CH3)2

CH2NH2 [1] LiAlH4

H CH2NHCH2CH3

[2] H2O

f. C6H5CH2CH2NH2 + (C6H5CO)2O C6H5CH2CH2NHCOC6H5 + C6H5CH2CH2NH3+ C6H5COO–

[1] CH3I (excess) [2] Ag2O [3]


CH3CH=CH2

(CH3)2NCH(CH3)2

CH3CH2CH2N(CH3)2

682

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Chapter 25–24 25.67 NH2 and H must be anti for the Hofmann elimination. Rotate around the C–C bond so the NH2 and H are anti. C6H5

a.

NH2

H

C6H5

NH2

CH3

b.

H

CH3

C6H5

CH2CH3

CH3

c.

H

(CH3)3C

C(CH3)3

NH2 CH2CH3

CH2CH3

C(CH3)3

rotate 120° counterclockwise

three steps (CH3)3C

rotate 120° counterclockwise C6H5

NH2 CH3

CH2CH3

H

C6H5

C(CH3)3

(CH3)3C

CH2CH3 NH2 CH3

H

three steps

three steps

(CH3)3C

(CH3)3C

25.68 N2+ Cl– Cl HO

Cl

Br

a. H2O

b. H3PO2

A Cl

h. C6H5NH2

d. CuBr

H

Cl

Cl

NH2

N N

OH

Cl

NC

Cl

F

Cl

e. CuCN i. C6H5OH

c. CuCl

N N

Cl

Cl

f. HBF4 I

I Cl

Cl

j. KI

g. NaI

25.69 Under the acidic conditions of the reaction, aniline is first protonated to form an ammonium salt that has a positive charge on the atom bonded to the benzene ring. The –NH3+ is now an electron-withdrawing meta director, so a significant amount of meta substitution occurs. NH2

H OSO3H

NH3

+ HSO4–

This group is now a meta director.

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Amines 25–25 25.70 CH3

R

C

CH3CH2CH2

[1] CH3I (excess)

H NH2

H C C

[2] Ag2O [3]


A

H

B

O

[1] O3

H

[2] CH3SCH3

CH2CH2CH3

H

C

O H

H

C

CH2CH2CH3

25.71

a.

Br

Br

H H Na+ N

Br

OH

H N

Br

Br

Na+

+ CH3CH2NH2

Br

N H CH2CH3

+ H2O

Na+

OH

+ H2O + Na+

N

CH2CH3

H A

H O

O

H N

O

H

b.

N

proton transfer

NH2

OH2

H

N

H3B–H

H

H

N + H2O

proton source + A

H N + BH3

25.72 OH

O

OH HN

H N

H2C=O

H A

OH

OH N

proton transfer A

OH

OH

OH H N

OH2

OH H2C N

N

N

A

H2O

H A OH H

OH H N

OH H N

N

684

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Chapter 25–26 25.73 N N

N N

N N

N N

N N

X

R N N

X

alkyl diazonium salt The N2+ group on an aromatic ring is stabilized by resonance, whereas the alkyl diazonium salt is not.

aryl diazonium salt

25.74 NaNO2

Overall reaction:

HCl, H2O

NH2

The steps:

OH

+

+

OH

NaNO2 + HCl

NH2

N N O

Cl

H

+

+ HCl

+

N O

N N O

N N O

H

H

N N O H

H

+

H Cl

+

H Cl

– N N OH2 + Cl

N N O H

H Cl

+

O H

OH

H

+ HCl

H +

B

A

O H H

H OH Cl

H

B

+ HCl

Cl

+

+N

A H OH

Cl

+

1,2-H shift

N N +

OH

+ HCl

N

+ H2O

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Amines 25–27 25.75 A nitrosonium ion (+NO) is a weak electrophile so electrophilic aromatic substitution occurs only with a strong electron-donor group that stabilizes the intermediate carbocation. H Cl

O N O

NaCl

HO N O

H Cl

H O N O H

Na N(CH3)2

O N H

N(CH3)2

O N H

Cl

N(CH3)2

O N H

O N H O N H

N(CH3)2

O N

N(CH3)2

especially good resonance structure All atoms have octets.

resonance-stabilized carbocation

N O

H2O

N O

N(CH3)2

N(CH3)2 H2O

OH

25.76 O

a.

NH2 NH3

[1] CH3I

NaBH3CN

[2] Ag2O [3]


O

OH NaBH4

b.

(Hofmann elimination, less substituted C=C favored)

H2SO4

(more substituted C=C favored)

CH3OH

(E + Z isomers formed)

25.77 NO2

HNO3 H2SO4

a.

b. NO2

Br2 FeBr3

NH2

H2 Pd-C

ClCOCH3

H2 Br

NO2

Pd-C

Br

NH2

Br

NHCOCH3

(from a.) CH3Cl AlCl3

c.

CH3

HNO3 O2N H2SO4

CH3

H2 Pd-C

H2N

CH3

[1] NaNO2, HCl [2] NaI

I

(+ ortho isomer) H2N

[1] NaNO2, HCl

d.

NC Br2 FeBr3

[2] CuCN

NC

Br

(from a.)

e.

Br2 FeBr3

HNO3 Br

H2SO4

NO2 Br

(+ para isomer)

H2 Pd-C

NH2 Br

[1] NaNO2, HCl [2] H2O

OH Br

CH3

686

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Chapter 25–28 f.

O2N

CH3

KMnO4

O2N

COOH

H2

H2N

Pd-C

COOH

[1] NaNO2, HCl HO [2] H2O

COOH

(from c.) O

O

ClCOCH3

g.

HNO3 H2SO4

AlCl3

h.

NH2

[1] NaNO2, HCl

O

O2N

O

H2N

H2

[1] NaNO2, HCl [2] NaI

Pd-C C6H5N2+Cl–

OH

I

N N

OH

[2] H2O

(from a.)

(from d., Step [1])

25.78 NH2

a.

CN

[1] NaNO2, HCl

COOH

H3O+

[2] NaCN

[2] NH2CH3 Br

Br

Br2

CH3COCl NH2

b.

CONHCH3

[1] SOCl2

Br –

CH3Cl

NHCOCH3 FeBr3

OH NHCOCH3 H2O

NHCOCH3 AlCl3

(+ ortho isomer)

NH2 [1] NaNO2, HCl [2] H3PO2 CH3

Br

COOH

COOCH2CH3 HOCH2CH3

c.

H2SO4 Br

(from a.) d.

COOH

[1] LiAlH4 [2] H2O

CH2OH

Br2

Br

FeBr3

CH2OH

(3x)

(from a.)

Br

HO H2N

e.

HO

HO [1] NaNO2, HCl [2] H2O

CH3Cl AlCl3

C6H5N2 CH3

(+ ortho isomer)

+Cl–

(from a., Step [1])

N N CH3

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Amines 25–29 25.79 CH3Cl

[1]

Br2 h


AlCl3

H2C=O

NH2

(excess)

–OH

Br

[2]

NH3

Br

OH

N H

NaBH3CN CHO CH NH 3 2

PCC

N H

NaBH3CN

(from [1])

[1] LiAlH4 [2] H2O

O

COOH [1] SOCl 2

KMnO4

[2] CH3NH2

[3]

NCH3

[1] LiAlH4

H

[2] H2O

N H

(from [1])

[4]

CHO

CN [1] DIBAL-H

[1] HNO3, H2SO4

Then route [2].

[2] H2O

[2] H2, Pd-C [3] NaNO2, HCl

[1] LiAlH4

[4] CuCN

[2] H2O NH2

Br

Br2

[5]

Then route [1]. COOH

MgBr [1] CO 2

Mg

Then route [3].

[2] H3O+

FeBr3 [1] H2C=O [2] H2O OH

Then route [2].

25.80 O

O

mCPBA

O

O

O

[1] LiAlH4

O

[2] H2O

O

PBr3 OH

O

PCC O NHCH3

O

CH3NH2

O

NaBH3CN

O

CH3NH2 O O

Br

a. CH3O

MDMA

MDMA

[part (b)]

[part (a)]

CH3O

CN [1] LiAlH 4

NaCN

[2] H2O

CH3O CH3O

NHCH3

O

25.81 CH3O

Br

O

CH3O

CH3O

NH2

CH3O CH3O

mescaline

688

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Chapter 25–30 CH3O

Br

CH3O

CH3O

HBr

b. CH3O

CH3O

ROOR

CH3O

excess

CH3O

CH3O

CH3O

Br

CH3O

Mg

MgBr

c. CH3O

NH2

NH3

O

[1]

CH3O

[2] H2O

CH3O CH3O

CH3O

OH

PBr3

CH3O

CH3O

CH3O

Br

CH3O CH3O

CH3O NH3 excess CH3O

NH2

CH3O CH3O

25.82 NO2

HNO3

a.

H2SO4

NH2

H2

CN

[1] NaNO2, HCl

COOH

H3O+

[2] CuCN

Pd-C

COOH

Cl2 FeCl3 Cl

Cl2

HNO3

b.

NO2

CH3Cl

NH2

CH3

OH

H2

NO2

CH3

Pd-C

CH3CH2Cl

d.

AlCl3

HNO3

CH3CH2

H2SO4

CH2NH2

[1] LiAlH4 [2] H2O H2 NO2 Pd-C

CH3CH2

NH2 [1] NaNO2, HCl [2] NaCN

(+ ortho isomer) CH3

CH3

CH3CH2 CH3

Br2

O2N

CH3

(from c.)

Br

Cl [1] NaNO2, HCl

N N

[2] NH2

(from b.) (from a.)

Cl

NH2

H3O+

CH3 Br

Cl NH2

Cl

H2N

NH2 [1] NaNO2, HCl [2] CuCN

COOH

Pd-C

FeBr3

f.

H2

CN

CH3CH2

(+ ortho isomer)

e. O2N

Cl

[2] H2O

Pd-C

H2SO4

Cl [1] NaNO2, HCl

NO2

HNO3

CH3

AlCl3

Cl

H2

FeCl3 (2X)

H2SO4

c.

Cl

Cl

Cl

CH3CH2

[1] NaNO2, HCl [2] H2O

CN

HO

CH3 Br

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Amines 25–31 25.83 O NO2

HNO3

a.

NO2 H 2

Cl2 FeCl3

H2SO4

NH2

O

H N

O

O

Pd-C Cl

Cl2 FeCl3 H N

Cl

Cl

O

Cl

(+ isomer)

Cl H2

b.

Pd-C H2N

O2N

NO2

[1] NaNO2, HCl

HNO3

[2] H2O

H2SO4 HO

HO

NH2

H2 Pd-C HO O

(+ ortho isomer)

(from a.)

O O H N O

HO O ClCOCH2CH3

c.

O

O

Cl2

AlCl3

OH NHCH3 NaBH4

Cl NH2CH3

NHCH3

CH3OH

H2O, HCl

25.84 CH3OH OH

NH2

a.

[1] HNO3, H2SO4

[1] NaNO2, HCl

[2] H2, Pd-C

[2] H2O

SOCl2

OH

OH

CH3Cl

Br2

AlCl3

FeBr3 (2X)

Br

OCH3 Br

[1] NaH

CH3

CH3

CH3 [1] Br2, h


Br NH2 CH3O Br

CH3CH2OH CrO3, H2SO4, H2O CH3COOH SOCl2 CH3COCl

b.

O

H N O

(from a.)

H N

Cl

–OH,

O

AlCl3 O

(+ ortho isomer)

H2O

Br

[2] CH3Cl

(+ ortho isomer)

NH2

Br

O NH2

[2] NH3 (excess)

690

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Chapter 25–32 CH3OH SOCl2 [1] CH3Cl, AlCl3

Cl2

c.

Cl

FeCl3

HOOC

[2] KMnO4

[1] LiAlH4 [2] H2O

Cl

HO Cl

PCC

O Cl H

(+ ortho isomer) NH2

(from a.) mild acid N CH

Cl

CH3OH SOCl2

CH3

CH3Cl

d.

H2SO4

AlCl3

CH3

HNO3

COOCH3

[1] KMnO4 [2] CH3OH, H+

O2N

COOCH3

H2 Pd-C

O2N

H2N PCC

(+ ortho isomer)

O (CH3)2CHNH

CO2CH3

OH

NaBH3CN

NH2

e. O2N

CH3

(from d.)

[1] H2, Pd-C

F

CH3

[2] NaNO2, HCl [3] HBF4

O

O

[1] KMnO4

F

F

C

[2] SOCl2

Cl

HN

(from a.)

Probably a strong enough activator that the Friedel–Crafts reaction will still occur. O NHCOCH3

O

Make two parts:

AlCl3

Cl

CH3

C

NH

O

f. I

I

CN CN NHCOCH3

NHCOCH3

NHCOCH3

[1] HNO3, H2SO4

[1] NaNO2, HCl

[2] H2, Pd-C

[2] CuCN

(from b.)

NH2

CN

(+ ortho isomer) CH3 O2N

[2] NaNO2, HCl

(from d.)

CH3

[1] H2, Pd-C [3] NaI

I

O [1] KMnO4

Cl

[2] SOCl2 I

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25. Amines

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691

Amines 25–33 25.85 molecular weight = 87 C5H13N two IR peaks = 1° amine

H

H

H

H

H

N

N

N

N

H

H H

H

H

N

H

H

H

N

H

H

N

H

N

25.86 H2N NHCH2CH3

CH2CH3

CH2NHCH3

Compound A: C8H11N Compound B: C8H11N IR absorption at 3400 cm–1
2° amine IR absorption at 3310 cm–1
2° amine 1H NMR signals at (ppm): 1H NMR signals at (ppm): 1.3 (triplet, 3 H) CH3 adjacent to 2 H's 1.4 (singlet, 1 H) amine H 2.4 (singlet, 3 H) CH3 3.1 (quartet, 2 H) CH2 adjacent to 3 H's 3.8 (singlet, 2 H) CH2 3.6 (singlet, 1 H) amine H 6.8–7.2 (multiplet, 5 H) benzene ring 7.2 (multiplet, 5 H) benzene ring

Compound C: C8H11N IR absorption at 3430 and 3350 cm–1
1° amine 1H NMR signals at (ppm): 1.3 (triplet, 3 H) CH3 near CH2 2.5 (quartet, 2 H) CH2 near CH3 3.6 (singlet, 2 H) amine H's 6.7 (doublet, 2 H) para disubstituted 7.0 (doublet, 2 H) benzene ring

25.87 HO

C N

HO

Compound D: Molecular ion at m/z = 71: C3H5NO (possible formula) IR absorption at 3600–3200 cm–1
OH 2263 cm–1
CN Use integration values and the molecular formula to determine the number of H's that give rise to each signal. 1H NMR signals at (ppm): 2.6 (triplet, 2 H) CH2 adjacent to 2 H's 3.2 (singlet, 1 H) OH 3.9 (triplet, 2 H) CH2 adjacent to 2 H's

NH2

Compound E: Molecular ion at m/z = 75: C3H9NO (possible formula) IR absorption at 3600–3200 cm–1
OH 3636 cm–1
N–H of amine 1H NMR signals at (ppm): 1.6 (quintet, 2 H) CH2 split by 2 CH2's 2.5 (singlet, 3 H) NH2 and OH 2.8 (triplet, 2 H) CH2 split by CH2 3.7 (triplet, 2 H) CH2 split by CH2

25.88 Guanidine is a strong base because its conjugate acid is stabilized by resonance. This resonance delocalization makes guanidine easily donate its electron pair; thus it's a strong base. NH H2N

C

HA NH2

guanidine

H2N

NH2

NH2

NH2

C

C

C

NH2

pKa = 13.6

H2N

NH2

H2N

NH2

NH2 H2N

C

NH2

692

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Chapter 25–34 25.89 CH3

N

[1] CH3I (excess)

CH3

N

CH3 I

[2] Ag2O

[2] Ag2O

[1] CH3I (excess)

[3]


[3]
N(CH3)2

C8H10

N(CH3)3

I

Y

25.90 One possibility: O

O

O

CH3COCl

a. O

O

AlCl3

O

Br2

H2NC(CH3)3

CH3CO2H

O

O

Br

O

O

O

(+ isomer)

NaBH4 CH3OH HO

O H3O+

HO OH

albuterol

b.

O

+

HN(CH2CH3)2

H2SO4 O N 2

OH

N H

A [1] NaNO2, HCl

H2 Pd-C

O

N H

N(CH2CH3)2

HO

HNO3

N H

[2] H2O

H2N

[1] NaH HO

O

Cl

[2]

Br2 O2N

COOH

HNO3

O SOCl2 O2N O

Br

COOH

H2SO4

H3O+ CO2

O

O O

COCl

A

O2N

O O

N(CH2CH3)2

H2

H2N

Pd-C O

Mg

O

O

N(CH2CH3)2

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26. Carbon−Carbon Bonding−Forming Reactions in Organic Synthesis

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C–C Bond-Forming Reactions 26–1 C Chhaapptteerr 2266:: C Caarrbboonn––C Caarrbboonn B Boonndd--FFoorrm miinngg R Reeaaccttiioonnss iinn O Orrggaanniicc SSyynntthheessiiss

C Coouupplliinngg rreeaaccttiioonnss [1] Coupling reactions of organocuprate reagents (26.1) • R'X can be CH3X, RCH2X, 2o cyclic + RCu R' X + R2CuLi R' R halides, vinyl halides, and aryl halides. + LiX X = Cl, Br, I • X may be Cl, Br, or I. • With vinyl halides, coupling is stereospecific. [2] Suzuki reaction (26.2) Y R' X

+

Pd(PPh3)4

R B

NaOH

Y

X = Br, I

R' R

+

HO BY2

+

NaX

•

[3] Heck reaction (26.3) Pd(OAc)2

R' X

Z

•

R'

Z

P(o-tolyl)3

X = Br or I

•

• •

(CH3CH2)3N (CH3CH2)3NH

X–

•

C Cyycclloopprrooppaannee ssyynntthheessiiss [1] Addition of dihalocarbenes to alkenes (26.4) C

X X C

CHX3

C

KOC(CH3)3

C

CH2I2

C C

Zn(Cu)

H H C C C

+ ZnI2

R'X is a vinyl halide or aryl halide. Z = H, Ph, COOR, or CN With vinyl halides, coupling is stereospecific. The reaction forms trans alkenes.

• •

The reaction occurs with syn addition. The position of substituents in the alkene is retained in the cyclopropane.

• •

The reaction occurs with syn addition. The position of substituents in the alkene is retained in the cyclopropane.

•

Metathesis works best when CH2=CH2, a gas that escapes from the reaction mixture, is formed as one product.

C

[2] Simmons–Smith reaction (26.5)

R'X is most often a vinyl halide or aryl halide. With vinyl halides, coupling is stereospecific.



M Meettaatthheessiiss ((2266..66)) 2 RCH CH2

Grubbs RCH CHR

catalyst Grubbs catalyst diene

+ CH2 CH2

+ CH2 CH2

693

694

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Chapter 26–2 C Chhaapptteerr 2266:: A Annssw weerrss ttoo PPrroobblleem mss 26.1 A new C–C bond is formed in each coupling reaction. a.

(CH2=CH)2CuLi

Cl

new C–C bond

b.

(CH3)2CuLi

Br

CH3

new C–C bond

c.

CH3O

CH3O

I (CH3CH2CH2CH2)2CuLi

CH3O

new C–C bond

CH3O

Br

d.

2

new C–C bond

CuLi

26.2 I OH

OH

A=

B=

(CH3CH2)2CuLi

several steps

I

(CH3)2CuLi

OH

several steps OH

O

CO2CH3 C18 juvenile hormone

26.3 Br

a. (CH3)2CHCH2CH2I

[1] Li

[(CH3)2CHCH2CH2]2CuLi

CH2CH2CH(CH3)2

[2] CuI

or Br

b.

[1] Li [2] CuI [1] Li

Cl

(CH3)2CHCH2CH2I

CuLi

CH2CH2CH(CH3)2

2

CuLi

Br

[2] CuI 2

Cl

c.

Cl

[1] Li [2] CuI

CuLi 2

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26. Carbon−Carbon Bonding−Forming Reactions in Organic Synthesis

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C–C Bond-Forming Reactions 26–3 26.4 I +

a.

b.

O

Pd(PPh3)4

O

NaOH

B

B(OCH3)2

Pd(PPh3)4

+

Br NaOH

c.

[1] Li

Br

B(OCH3)2 + LiOCH3

[2] B(OCH3)3

O

O

B

C C H + H B

d.

O

O

26.5 O

O

H B O

a. CH3CH2CH2 C C H

b.

I

Li

Li

CH3O

O

B(OCH3)3

Br

Br2

c.

I

B

H

CH3O

Pd(PPh3)4, NaOH

B(OCH3)2 Li

Li

I

NaOH

B(OCH3)2

B(OCH3)3

Br Pd(PPh3)4

CH3O

CH3O

Pd(PPh3)4

NaOH

(+ ortho isomer)

CH3O

26.6 Br

Heck

a.

+

b.

I

reaction OCH3

+

OCH3

Heck reaction

Br

c.

CN

+

OCH3

d.

Heck

CN

reaction Br

+

O

Heck

OCH3

reaction

O

26.7 Locate the double bond with the aryl, COOR, or CN substituent, and break the molecule into two components at the end of the C=C not bonded to one of these substituents. O

a.

Br OCH3

O OCH3

695

696

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Chapter 26–4 Br

b.

O

c.

O CO2CH3

CO2CH3

Br

26.8 Add the carbene carbon from either side of the alkene. H

CH3

a.

CHCl3

C C

CH3 C H

KOC(CH3)3

H

H

Cl Cl C C

H

+

CH3 H

C

H enantiomers

H H

C C

Cl Cl CH3CH2

b.

CH2CH3

CHCl3

C KOC(CH3)3 CH3CH2 H

C C H

H

Cl Cl C C

+

CH2CH3 H

CH3CH2 H

c.

CH3

CHCl3

CCl2

KOC(CH3)3

C

CH2CH3 H

C

identical CH3

C

Cl Cl

CH3

+

CCl2

H

H

enantiomers

26.9 CHCl3

a.

Cl Cl

KOC(CH3)3

Br Br

(from b.)

CHBr3

b.

LiCu(CH3)2

c.

Br Br

KOC(CH3)3

26.10 a.

CH2I2

+

ZnI2

Zn(Cu) H

b.

CH2I2

+

Zn(Cu) H

ZnI2

c.

CH2I2 Zn(Cu)

+

ZnI2

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C–C Bond-Forming Reactions 26–5 26.11 The relative position of substituents in the reactant is retained in the product.

CH3CH2 C C H

H H C

H

CH2I2

CH2CH3

Zn(Cu)

CH3CH2 C H

trans-3-hexene

C

CH3CH2 H C

C

H CH2CH3

C H H

H CH2CH3

two enantiomers of trans-1,2-diethylcyclopropane

26.12 Grubbs catalyst

a.

c. (E and Z)

Grubbs catalyst

OCH3

Grubbs catalyst

b.

(CH2=CH2 is also formed in each reaction.) (E and Z)

OCH3

OCH3

26.13 + cis-2-pentene

+

There are four products formed in this reaction including stereoisomers, and therefore, it is not a practical method to synthesize 1,2-disubstituted alkenes.

26.14 O

a.

Ph

+

O

O

Ph

CO2CH3

O

CO2CH3 CO2CH3

b. CO2CH3

26.15 High dilution conditions favor intramolecular metathesis.

O

Y =

O

O

Under high dilution conditions, an intramolecular reaction is favored, resulting in Y. Two different molecules do not often come into contact, so two double bonds in the same molecule react.

697

698

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Chapter 26–6 long chain 1 O

Alkenes 1 and 2 differ in proximity to the ether side chain. Three products are possible because alkene 1 can react with alkene 1 in another molecule (forming Z); alkene 1 can react with alkene 2 in another molecule (forming Z'); and alkene 2 can react with alkene 2 in another molecule (forming Z"). At usual reaction concentrations, the probability of two molecules approaching each other is greater than the probability of two sites (connected by a long chain) in the same molecule reacting together, so the intermolecular reaction is favored.

O

O

2

long chain

O

Z

O

=

O

O

O

O

O

Z' =

O

O

O

O

O

O

Z" =

O

O

O

O

O

26.16 Cleave the C=C bond in the product, and then bond each carbon of the original alkene to a CH2 group using a double bond.

a.

O

O

c. O

b. CH3O

CH3O OH

OH

CO2CH3

CO2CH3

CHO

CHO O

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C–C Bond-Forming Reactions 26–7 26.17 Cl

CH2CH2CH3

+

a.

(CH3CH2CH2)2CuLi

N

B(OCH3)2

b.

Pd(PPh3)4

Br

+

N

NaOH O O

Pd(OAc)2

Br

+

c.

P(o-tolyl)3 (CH3CH2)3N

N

d.

Cl

[1] Li [2] CuI [3]

Br

O

e.

Br

Pd(PPh3)4

B

+

O

f.

NaOH

Br +

CH3O

N

CO2CH3

Pd(OAc)2

CH3O

P(o-tolyl)3 (CH3CH2)3N

g.

Br

CO2CH3

[1] Li [2] B(OCH3)3 Br + Pd catalyst [3] O [1] H B

h. (CH3)3C C C H

O [2] C6H5Br, Pd(PPh3)4, NaOH

26.18 Br

a.

c. CO2CH3

Br

CO2CH3

Br

d. b.

Br

CH3O

CH3O

699

700

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Chapter 26–8 26.19 Each coupling reaction uses Pd(PPh3)4 and NaOH to form the conjugated diene.

I

A I O

H

H B

O

O

C C H

B

B O

H

ethynylcyclohexane

I

C It is not possible to synthesize diene D using a Suzuki reaction with ethynylcyclohexane as starting material. Hydroboration of ethynylcyclohexane adds the elements of H and B in a syn fashion affording a trans vinylborane. Since the Suzuki reaction is stereospecific, one of the double bonds in the product must therefore be trans.

D

26.20 Locate the styrene part of the molecule, and break the molecule into two components. The second component in each reaction is styrene, C6H5CH=CH2.

a.

Br

b.

c.

Br

Br

styrene part

styrene part

styrene part

26.21 Inversion of configuration occurs with substitution of the methyl group for the tosylate. a.

H

(CH3)2CuLi

OTs

H

(CH3)2CuLi

CH3

d.

CH3 OTs (CH3)2CuLi

26.22 HBr

1-butene

ROOR

Br

+

CuLi 2

[1] Li [2] CuI CuLi 2

H CH3

H

H (CH3)2CuLi

OTs H

H

OTs

b.

c.

CH3

octane

H

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C–C Bond-Forming Reactions 26–9 26.23 Add the carbene carbon from either side of the alkene. H Cl

CHCl3

a.

CHCl3

e.

KOC(CH3)3

Cl

KOC(CH3)3

Cl Cl C

H H

H

CH2I2

b.

H

C

H CH3 Br

CHBr3

Br

KOC(CH3)3

H

Cl Cl

CH3 Br

+

Br

Cl Cl C

H

H

H

CH2I2

C C

H

H

d.

C

H

Zn(Cu)

c.

H

+

C

CHCl3

f.

KOC(CH3)3

+

H

C

C

H

Zn(Cu) H

H

26.24 Since the new three-membered ring has a stereogenic center on the C bonded to the phenyl group, the phenyl group can be oriented in two different ways to afford two stereoisomers. These products are diastereomers of each other. CHI2 H

H H

+ Zn(Cu)

H H

H

26.25 High dilution conditions favor intramolecular metathesis. H

OH

H

Grubbs catalyst

N

a.

OH

N

OH

c.

Grubbs catalyst

CO2CH3

OH

O

O O

b.

CO2CH3

O

Grubbs catalyst

26.26 Retrosynthetically break the double bond in the cyclic compound and add a new =CH2 at each end to find the starting material. O

a.

O

O O

O

b.

O

CO2CH3 O

c. O

CO2CH3

702

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Chapter 26–10 26.27 Alkene metathesis with two different alkenes is synthetically useful only when both alkenes are symmetrically substituted; that is, the two groups on each end of the double bond are identical to the two groups on the other end of the double bond. a.

CH3CH2CH CH2

CH3CH2CH2CH CHCH2CH3 + CH3CH2CH CHCH2CH3 + CH3CH2CH2CH CHCH2CH2CH3

+

+ CH2 CH2 (Z + E)

CH3CH2CH2CH CH2

(Z + E)

b.

(Z + E)

This reaction is synthetically useful since it yields only one product.

+ c.

+

+

+

+

+

+ CH3CH CHCH3 + CH3CH2CH CHCH2CH3 (Z + E) (Z + E)

+

26.28 O O

O

M

O

O

26.29 All double bonds can have either the E or Z configuration. a.

c.

b.

26.30 Br Br CHBr3

a.

KOC(CH3)3 Br

COOH

+

b.

CO2CH3

COOH

CH3O2C

Pd(OAc)2 P(o-tolyl)3 (CH3CH2)3N

Br

+

c.

NHCOCH3

Pd(OAc)2 P(o-tolyl)3 (CH3CH2)3N

NHCOCH3

CH3

CH3

d.

B(OCH3)2

O

e.

B

Br

+ O

CH3O

+

Br

Pd(PPh3)4 NaOH Pd(PPh3)4 NaOH

CH3O

O

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C–C Bond-Forming Reactions 26–11 O

O

Grubbs

f.

catalyst

O

O CH2I2

g.

Zn(Cu) Br Br

h. (CH3)2CuLi

Product in (a) [1] Li

Cl

i.

[2] CuI Br

[3]

O

[1]

j.

H B

H

O CH3CH2 C C H

[2]

Br

H

Pd(PPh3)4 NaOH

26.31 O

Cl




Cl3C C

Cl

O Na+

C Na+ +

O C O

Cl

Cl

C

+

CO2

+

NaCl

Cl

26.32 This reaction follows the Simmons–Smith reaction mechanism illustrated in Mechanism 26.5. Br CHBr2

Zn(Cu)

[1]

CH Zn

[2]

Br

+

ZnBr2

26.33 Ph2S

O OCH3 Ph2S

a sulfur ylide

1,4-addition of the sulfur ylide to the
carbon

O OCH3

OCH3

a resonance-stabilized enolate

O

methyl trans-chrysanthemate

Ph2S

704

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Chapter 26–12 26.34 + N2

A N

N

CO2CH3

N

N

CO2CH3

CO2CH3

diazo compound

CO2CH3

B

26.35 I

PAr3

[1]

a.

Ar3P Pd

Pd(PAr3)2 oxidative addition

I

PAr3

[2]

+

I

Ar3P Pd syn addition

PAr3 Ar3P Pd

I

PAr3

[3]

H H

H

The H cis to Pd is eliminated.

+

syn elimination

H

E

This H is trans to Pd.

Ar3P Pd

[4] I

reductive elimination

Pd(PAr3)2 + HI

b. This suggests that the stereochemistry in Step [3] must occur with syn elimination of H and Pd to form E. Product F cannot form because the only H on the C bonded to the benzene ring is trans to the Pd species, and therefore it cannot be removed if elimination occurs in a syn fashion. 26.36 O H B

Br NaNH2

C CH

O

Br

O

(– HBr)

B O

(Z)-2-bromostyrene

Pd(PPh3)4 NaOH

A

26.37

Br2 FeBr3 O

O

H

H B O

B O

Br Pd(PPh3)4, NaOH

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C–C Bond-Forming Reactions 26–13 26.38 HO

SOCl2

OH

HO

Cl TBDMS–Cl imidazole

HC CH

NaH HC C

Cl

TBDMSO

O B

OH

O

O

Bu4NF

BH O

O HC C

B

OTBDMS

OTBDMS

O

26.39 a. CH3O

CH3O HNO3

NO2

H2SO4

H2

NH2

Pd-C Br2

[1] NaH OH [2] CH3Br PBr3

Synthesize these two components, and then use a Heck reaction to synthesize the product.

Br

CH3O

NaNO2

N2+Cl–

HCl

CH3O

H2O

OH

Br

CH3OH CH3CH2OH SOCl2 CH3CH2Cl

CH2CH3

AlCl3

h
Pd(OAc)2

Br

CH3O

Br2

CHCH3

KOC(CH3)3

Br CH3O

P(o-tolyl)3 (CH3CH2)3N

Br

CH3CH2OH SOCl2

Br

CH3CH2Cl

b. CH3O

AlCl3

(from a.) Br CO2Et

CH3O

CH3O CO2Et

Pd(OAc)2 P(o-tolyl)3 (CH3CH2)3N

CO2Et H2 Pd-C

CH3O

Br2

Br

CH3O MgBr

Mg

[1] (2 equiv) [2] H2O

FeBr3 OH

CH3O

706

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Chapter 26–14 26.40 CH2I2

a.

Zn(Cu) Br Br

b.

OH

OH Br Br

NBS

Br

NaOH

CHBr3

OH

h


OH

KOC(CH3)3

Br

c.

Br

CH2I2

(CH3CH2CH2CH2)2CuLi

Zn(Cu)

(from b.)

Br

d.

CuLi

CH2I2

2

Br

Zn(Cu)

(from b.)

26.41 Br

Br2

a.

CH2I2

CH2=CH2 Pd(OAc)2 P(o-tolyl)3 (CH3CH2)3N

FeBr3

Zn(Cu)

styrene

CHCl3

b. styrene (from a.)

Cl

KOC(CH3)3

Br

Cl

CO2CH3

CO2CH3

c. Pd(OAc)2 P(o-tolyl)3 (CH3CH2)3N

(from a.) Br

d.

[1] Li

Br CuLi

[2] CuI

(from a.)

2

For an alternate synthesis of styrene see Problem 26.39.

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26. Carbon−Carbon Bonding−Forming Reactions in Organic Synthesis

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Text

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C–C Bond-Forming Reactions 26–15 26.42 [1] NaH

a.

C CH

HC CH Cl

[2]

[1] NaH

H2 Lindlar catalyst

[2] Cl

CH2I2 Zn(Cu)

+ enantiomer

[1] NaH

b.

[1] NaH

C CH

HC CH [2]

Na Cl

[2]

NH3

Cl

CH2I2 Zn(Cu)

+ enantiomer

c.

O

CH2I2

CH2

Ph3P=CH2

Zn(Cu)

Br

d.

O

Br2 CH3CO2H

O

LiBr

O

Li2CO3, DMF

[1] (CH3)2CuLi

O

[2] H2O Ph3P=CH2

[1] (CH3)2CuLi

CHCl3

[2] H2O Cl

Cl

KOC(CH3)3

707

708

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Chapter 26–16 26.43 Either compound can be used to

a. CH3

OCH3

CH3

Br

Br

OCH3 synthesize the organoborane, so

two routes are possible. Possibility [1]: CH3Cl

Br2 CH3

AlCl3 Br2

[1] Li CH3

FeBr3 HNO3

Br

FeBr3

Br [2] B(OCH3)3

Br

NO2

H2

CH3

Br

B(OCH3)2

NH2

Pd-C

H2SO4

[1] NaNO2, HCl

Br

OH

[2] H2O [1] NaH [2] CH3I Br

Pd(PPh3)4 CH3

B(OCH3)2

Br

OCH3

NaOH

CH3

OCH3

OCH3

Possibility [2]: Br

OCH3

Pd(PPh3)4

[1] Li

(CH3O)2B

OCH3

[2] B(OCH3)3

CH3

Br

NaOH

(from Possibility [1])

(from Possibility [1])

CH3

HO

OCH3

HO

b.

Br

Br

The acidic OH makes it impossible to prepare an organolithium reagent from this aryl halide, so this compound must be used as the aryl halide that couples with the organoborane from bromobenzene. O2N

O2N

HNO3

Br2

H2SO4

FeBr3

Br2 FeBr3

Br

H2N Br

[1] Li

(CH3O)2B

[2] B(OCH3)3 HO Pd(PPh3)4 Br NaOH

[1] NaNO2, HCl Br

Pd-C

HO (CH3O)2B

HO

H2

Br [2] H2O

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C–C Bond-Forming Reactions 26–17 Br

O

c.

Br

O O

O

This can't be converted to an organoborane reagent via an organolithium reagent.

three B(OCH3)2

(as in 26.43b)

steps Br2 FeBr3

HNO3 Br

H2SO4

[1] NaNO2, HCl

H2 Br

NO2

Pd-C

NH2

Br

[2] H2O

Br

OH CH3COCl pyridine

B(OCH3)2

Br

O

Br

Pd(PPh3)4 NaOH

O O

O

26.44 EtO2C CO2Et EtO2C CO2Et

Grubbs catalyst

a.

Synthesis of starting material: [1] NaOEt

CH2(CO2Et)2

[2]

CH(COEt)2

EtO2C CO2Et

[1] NaOEt [2]

Cl

Cl SOCl2 OH

SOCl2 OH

OTBDMS

b.

OTBDMS

Grubbs catalyst

Synthesis of starting material: OH

PCC

H

OH

O OH

PBr3

H2O Br

Mg

O

O

MgBr

TBDMS–Cl imidazole OTBDMS

710

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Chapter 26–18 26.45 O

O

Grubbs catalyst

a.

Synthesis of starting material: O

SOCl2

CrO3 OH

H2SO4 H2O

OH

O

O Cl AlCl3

OH

O

NaBH4

[1] NaOEt

CH3OH

[2]

Cl SOCl2 OH

b.

O O

Grubbs catalyst

O O

Synthesis of starting material:

Br2

Br

MgBr

Mg

OH

FeBr3

OH

H2SO4

O CH2 CH2

PCC

CHO

H2O

mCPBA

OH TsOH O O

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C–C Bond-Forming Reactions 26–19 26.46 Cl

CH3

CH3O

N

Cl

CH3

CH3O

CO2CH3

N

[1] (CH3CH2CH2CH2)4N+ F—

CO2CH3 [2] PCC

Cl

CH3

CH3O

N

CO2CH3

OTBDMS

LiCu

OTBDMS

I

CHO

2

directed aldol [1] CH3CHO + LDA 
CH2CHO [2] H3O+

O N O

O

Cl CH3O

O

Cl

O

N

CH3

CH3O

N

several steps

CO2CH3

O

CH3O

OH

N H

O

CHO

A

maytansine

26.47 NH2

NO2 Br

Br H2

HNO3

a.

OH Br [1] NaNO2, HCl

Pd-C

H2SO4

OCH3 Br [1] NaH

[2] H2O

OCH3 CuLi

Br [1] Li

[2] CH3I

[2] CuI

2

Br OCH3

OCH3

OH H2O

(2 enantiomers) OH

b.

[1] OsO4

OTBDMS TBDMS–Cl

HO

Br

H2SO4

Br

[2] NaHSO3, H2O

TBDMSO

imidazole

[1] Li [2] CuI

Br

OTBDMS

Br

OTBDMS

TBDMSO TBDMSO

CuLi 2

(CH3CH2CH2CH2)4N+F– OH HO

(2 enantiomers) O H

HB

c.

O

O B O

Br Pd(PPh3)4, NaOH

712

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Chapter 26–20 Cl

Br

d.

Br

AlCl3

CO2CH3

CO2CH3 Pd(OAc)2 P(o-tolyl)3 (CH3CH2)3N

CO2CH3 Br

e.

CO2CH3

Pd(OAc)2 P(o-tolyl)3 (CH3CH2)3N

Br

CO2CH3 H

(+ enantiomer)

[1] Li

f.

H

Zn(Cu) CH2I2

O

mCPBA CuLi

[2] CuI

Br

2

(2 enantiomers)

26.48 O

a.

O

O

O

b. O O

O

O




[1] LiAlH4




2

CO2Et EtO2C

CO2Et

CO2Et

OH

[2] H2O

OH [1] NaH (2 equiv) Cl (2 equiv) [2]

O O

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C–C Bond-Forming Reactions 26–21 26.49 I

O

O

+ A

Pd(PPh3)4

B O

N H

OCH2CH3

N

NaOH

B

C

OCH2CH3

OCH2CH3 H3O+

H

D C11H11NO

N O

O

OCH2CH3

H OH2

H OH2 NH

OCH2CH3

NH

O

OCH2CH3

N H + H2O

O

N + H2O

O

O

C H

H

H2O N

N

D

N O

O

OCH2CH3 O

+ H2O

resonance-stabilized

H3O+

H OCH2CH3

26.50 O CH3O

CH3O

H O

X

O PPh3

PPh3

CH3O H O

Y

O Ph3P CH3O

CH3O O

+ PPh3

O Ph3P

+ Ph3P=O

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27. Carbohydrates

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715

Carbohydrates 27–1 C Chhaapptteerr 2277:: C Caarrbboohhyyddrraatteess   IIm mppoorrttaanntt tteerrm mss • Aldose A monosaccharide containing an aldehyde (27.2) • Ketose A monosaccharide containing a ketone (27.2) A monosaccharide with the O bonded to the stereogenic center farthest from the • D-Sugar carbonyl group drawn on the right in the Fischer projection (27.2C) • Epimers Two diastereomers that differ in configuration around one stereogenic center only (27.3) • Anomers Monosaccharides that differ in configuration at only the hemiacetal OH group (27.6) • Glycoside An acetal derived from a monosaccharide hemiacetal (27.7)  A Accyycclliicc,, H Haaw woorrtthh,, aanndd 33--D D rreepprreesseennttaattiioonnss ffoorr DDD--gglluuccoossee ((2277..66)) CHO

Haworth projection

CH2OH O H H OH

H

H H

HO

OH

HO H

OH

CH2OH O H H OH

OH H

H

H

OH

H

OH

H

HO H

CH2OH


anomer

CHO H

3-D representation

O

HO HO H

HO HO

H C OH

OH OH

OH H

HO C H H

H

H

H C OH

H

OH

 anomer

acyclic form

OH

OH

OH H

H C OH

O

H

OH

H

CH2OH

 R Reeaaccttiioonnss ooff m moonnoossaacccchhaarriiddeess iinnvvoollvviinngg tthhee hheem miiaacceettaall [1] Glycoside formation (27.7A) OH HO HO

OH O

OH OH

ROH HCl

OH

O

HO HO

+ HO

OR

HO

OH OR


-D-glucose

O

• •

Only the hemiacetal OH reacts. A mixture of
and  glycosides forms.

•

A mixture of
and  anomers forms.

OH

 glycoside


glycoside

[2] Glycoside hydrolysis (27.7B) OH

OH HO HO

O OH OR

H3O+

HO HO

OH O

HO + HO

OH OH

OH OH

 anomer


anomer

+

O

ROH

716

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Chapter 27–2

R Reeaaccttiioonnss ooff m moonnoossaacccchhaarriiddeess aatt tthhee O OH H ggrroouuppss [1] Ether formation (27.8) OR

OH O

HO HO

OH OH

O

RO RO

Ag2O RX

• •

All OH groups react. The stereochemistry at all stereogenic centers is retained.

• •

All OH groups react. The stereochemistry at all stereogenic centers is retained.

OR OR

[2] Ester formation (27.8) OH HO HO

O

Ac2O or OH

AcCl

OH

OAc O

AcO AcO

OAc AcO

pyridine



R Reeaaccttiioonnss ooff m moonnoossaacccchhaarriiddeess aatt tthhee ccaarrbboonnyyll ggrroouupp [1] Oxidation of aldoses (27.9B) CHO

COOH

COOH

•

[O]

or

• CH2OH

CH2OH

aldose

aldonic acid

Aldonic acids are formed using: • Ag2O, NH4OH • Cu2+ • Br2, H2O Aldaric acids are formed with HNO3, H2O.

COOH

aldaric acid

[2] Reduction of aldoses to alditols (27.9A) CHO

CH2OH NaBH4 CH3OH

CH2OH

CH2OH

aldose

alditol

[3] Wohl degradation (27.10A) This C–C bond is cleaved.

CHO H

OH

CHO [1] NH2OH

•

[2] Ac2O, NaOAc [3] NaOCH3 CH2OH

•

CH2OH

•

The C1–C2 bond is cleaved to shorten an aldose chain by one carbon. The stereochemistry at all other stereogenic centers is retained. Two epimers at C2 form the same product.

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27. Carbohydrates

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717

Carbohydrates 27–3 [4] Kiliani–Fischer synthesis (27.10B) CHO CHO

H

CHO

OH

HO

[1] NaCN, HCl

H

•

+

[2] H2, Pd-BaSO4

•

[3] H3O+ CH2OH

CH2OH

One carbon is added to the aldehyde end of an aldose. Two epimers at C2 are formed.

CH2OH



O Otthheerr rreeaaccttiioonnss [1] Hydrolysis of disaccharides (27.12) O

H3O+

This bond is cleaved.

O

+

O

O

O OH

OH

OH

A mixture of anomers is formed.

[2] Formation of N-glycosides (27.14B) CH2OH O H H OH

H H OH OH

RNH2 mild H+

CH2OH O H H OH

H H

+

NHR OH

CH2OH O H H OH

NHR H H OH

•

Two anomers are formed.

718

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Chapter 27–4 C Chhaapptteerr 2277:: A Annssw weerrss ttoo PPrroobblleem mss 27.1

A ketose is a monosaccharide containing a ketone. An aldose is a monosaccharide containing an aldehyde. A monosaccharide is called: a triose if it has three C’s; a tetrose if it has four C’s; a pentose if it has five C’s; a hexose if it has six C’s, and so forth. CHO

CH2OH

a. a ketotetrose

C O

b. an aldopentose

H C OH

COOH

=

CH3

CHO HO CH3

OHC

OH

H C OH CH2OH

HO H

C

CH3

CHO

CH2CH3

re-draw

H

HOCH2 H

CHO

re-draw

C

C

c.

CH2CH2OH

CH2CH2OH

CHO

= HO

CH3 H

H

CHO

CH2OH = H

CHO

re-draw HO

CH3

C

CH3

CH2OH CH2CH3

CH2CH3

OH

d. H C CHO

C

CHO

=

HO

CH3 H

H

For each molecule: [1] Convert the Fischer projection formula to a representation with wedges and dashes. [2] Assign priorities (Section 5.6). [3] Determine R or S in the usual manner. Reverse the answer if priority group [4] is oriented forward (on a wedge). CH2NH2

a. Cl

[1]

H

[1]

Cl C H

CH2NH2

[2]

H

[1]

CHO

CH2NH2

[3] 1

H

H

CHO

CHO

[3]

1 Cl C H 4

1

3

2

2

CHO

CHO

[3]

1 Cl C H 4

1 Cl C H CH2OH

CH2OH

COOH CH2Br H

[1]

Cl C CH2Br H

[2]

1 Cl C CH2Br 2 H

4

S configuration

H forward S configuration

3 3

3 COOH

H forward

3

3

d. Cl

Cl C H CH2NH2

CH2NH2

[2]

S configuration

2

2

[2]

CH2OH

COOH

Cl C CH2Br 2

4

Cl C H

CH2OH

CH2NH2

1 Cl C CH2Br 2

CH2NH2

CHO Cl

CHO

3

3

H

CHO

b. Cl

CH2NH2 Cl C CH2Br

CH2Br H

c.

H C OH

CH2OH

COOH

27.3

c. an aldotetrose

Rotate and re-draw each molecule to place the horizontal bonds in front of the plane and the vertical bonds behind the plane. Then use a cross to represent the stereogenic center in a Fischer projection formula.

a. CH3 C OH

b.

H C OH H C OH

CH2OH

27.2

CHO

H C OH

COOH

[3] 1

Cl C CH2Br 2 H

S configuration

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719

Carbohydrates 27–5 27.4 R

CHO

S

H C OH

R

HO C H H C OH

R

H C OH CH2OH D-glucose

27.5 a. aldotetrose: 2 stereogenic centers

b. a ketohexose: 3 stereogenic centers CH2OH

CHO

C O

H C * OH

H C* OH

H C * OH

[• = stereogenic center]

H C* OH H C* OH

CH2OH

CH2OH

A D sugar has the OH group on the stereogenic center farthest from the carbonyl on the right. An L sugar has the OH group on the stereogenic center farthest from the carbonyl on the left.

27.6

CHO

a.

H

OH

H

OH

HO

CHO HO H

H

HO

CH2OH

CHO

H

HO

OH

HO

H

OH CH2OH

B

A

H

H

CH2OH

OH group on the left: L sugar

H

C

OH group on the left: L sugar

OH group on the right: D sugar

b. A and B are diastereomers. A and C are enantiomers. B and C are diastereomers.

27.7

The D- notation signifies the position of the OH group on the stereogenic carbon farthest from the carbonyl group, and does not correlate with dextrorotatory or levorotatory. The latter terms describe a physical phenomenon, the direction of rotation of plane-polarized light.

27.8

There are 32 aldoheptoses; 16 are D sugars. C2 R

CHO

CHO

CHO

CHO

C3 R H

OH

H

OH

H

OH

H

OH

H

OH

H

OH

H

OH

H

OH

H

OH

HO

H

H

OH

H

OH

HO

H

OH

H

CH2OH

H

HO

H

H

OH

HO

OH

H

OH

H

CH2OH

CH2OH

H OH CH2OH

720

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Chapter 27–6 27.9

Epimers are two diastereomers that differ in the configuration around only one stereogenic center. epimers CHO H H

CHO

OH

HO

OH

H

H OH

CH2OH

CH2OH

D-erythrose

D-threose

CHO

and

H HO

OH H CH2OH

L-threose

27.10 a. D-allose and L-allose: enantiomers b. D-altrose and D-gulose: diastereomers but not epimers c. D-galactose and D-talose: epimers d. D-mannose and D-fructose: constitutional isomers e. D-fructose and D-sorbose: diastereomers but not epimers f. L-sorbose and L-tagatose: epimers 27.11 a.

HO

CH2OH

CH2OH

CH2OH

C O

C O

C O

H

H H

H

OH

HO

OH

HO

OH

CH2OH D-fructose

HO

H

H

HO

H

H

H

CH2OH L-fructose

enantiomers

27.12 CH2OH

S

C O

CH2OH

S

C O

H

HO

H

OH

HO

H

OH

H

HO

CH2OH D-fructose

b.

H H OH CH2OH

D-tagatose

OH CH2OH

D-tagatose

CH2OH C O

c.

HO H HO

H OH H CH2OH

L-sorbose

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721

Carbohydrates 27–7 27.13

Step [1]: Place the O atom in the upper right corner of a hexagon, and add the CH2OH group on the first carbon counterclockwise from the O atom. Step [2]: Place the anomeric carbon on the first carbon clockwise from the O atom. Step [3]: Add the substituents at the three remaining stereogenic centers, clockwise around the ring. a. Draw the
anomer of:

CH2OH O H

[1]

CHO

[2]

OH

H

OH

D sugar, CH2OH

H

OH

is drawn up.

H

OH

H H

H HO

[1]

O CH2OH

[2]

is drawn down.

H

farthest away C, OH on left = L sugar c. Draw the  anomer of:

OH

H

OH OH CH2OH

O CH2OH

OH

[2]

D sugar, CH2OH

is drawn up.

farthest away C, OH on right = D sugar

H

[3]

HO

O OH CH2OH OH OH H H H

The
anomer The first two substituents are has the OH and on the left so they are CH2OH trans. In drawn up. The third is on an L sugar, the the right, drawn down. OH must be drawn up.

CH2OH O H

OH H

H

H H

[1]

CH2OH O H

OH

OH

H

H L sugar, CH2OH

H

First three substituents are on the right so they are drawn down.

H

CH2OH

HO


anomer OH is down for a D sugar.

H

OH

CHO

HO

farthest away C, OH on right = D sugar

CHO HO

H

[3]

CH2OH O H H H OH

b. Draw the
anomer of: HO

H OH

H

CH2OH

CH2OH O H

 anomer OH is up for a D sugar.

[3]

H HO

CH2OH O OH H OH H H OH H

The first substituent is on the left so it is drawn up. The other two are on the right, drawn down.

722

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Chapter 27–8 27.14 To convert each Haworth projection into its acyclic form: [1] Draw the C skeleton with the CHO on the top and the CH2OH on the bottom. [2] Draw in the OH group farthest from the C=O. A CH2OH group drawn up means a D sugar; a CH2OH group drawn down means an L sugar. [3] Add the three other stereogenic centers, counterclockwise around the ring. “Up” groups go on the left, and “down” groups go on the right. "up" group on left

CH2OH

CH2OH is up = D sugar O

HO

H H

a.

OH

CHO

[1]

CHO

[2]

H H

H OH

CHO

[3] H

OH

H

OH

HO

OH H CH2OH

"down" groups on right

OH

H

H

CH2OH

OH CH2OH

OH on right = D sugar

CH2OH is down = L sugar H O H CH2OH OH H OH

H

b. HO

OH

"down" groups on right

CHO

[1]

CHO

[2]

CHO

[3] HO

H HO

"up" group on left

H

CH2OH

CH2OH

H

H

OH

H

OH

HO

H CH2OH

OH on left = L sugar

27.15 To convert a Haworth projection into a 3-D representation with a chair cyclohexane: [1] Draw the pyranose ring as a chair with the O as an “up” atom. [2] Add the substituents around the ring. O is an "up" atom. CH2OH

HO O

HO

H H

a.

OH

[1]

O

H

[2]

OH H

H

OH

H

H OH

HO

OH

H H

b.

O is an "up" atom. O

H

H

CH2OH OH H

H

[1]

O

[2] HO

OH

H

H OH

H

OH O

H

H

OH

HO

O

H

H HO

OH

OH

With so many axial groups, this is not the more stable conformation of this sugar.

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723

Carbohydrates 27–9 27.16 Cyclization always forms a new stereogenic center at the anomeric carbon, so two different anomers are possible. Two anomers of D-erythrose: CHO H

H

OH

H

OH

H

O H

H

OH

OH OH

H

CH2OH

H

OH

O H

H

OH

OH

H

H

D-erythrose

27.17 HO

HO CH3CH2OH HO OH HO HCl

HO O

a. HO HO

HO

HO O H

H

HO O

+ HO

OCH2CH3

HO

OCH2CH3

H

-D-mannose OH

OH

OH CH3CH2OH

O H

b. OH

OH

OH

OH O

OH O

+

H

HCl OH

OH

HO

OCH2CH3

OCH2CH3 OH HO H


-D-gulose

c.

CH2OH O OH CH3CH2OH HO H HCl H CH2OH H OH

CH2OH O OCH2CH3 + HO H H CH2OH H OH

CH2OH O CH2OH HO H H OCH2CH3 H OH

-D-fructose

27.18 H OH2

H

+

HOCH2 H H OH

OCH2CH3

O

HOCH2

H H OH

+

OCH2CH3

O

H

H

OH

H OH

H

+ H2O

HOCH2 O H H OH

H

+

H

+ CH3CH2OH

OH

resonance-stabilized carbocation HOCH2

+

O

H

H

H

OH

OH

H

724

Smith: Study Guide/ 27. Carbohydrates Text Study Guide/Solutions Manual to accompany: Organic Chemistry, 3rd Edition Solutions Manual to accompany Organic Chemistry, Third Edition

© The McGraw−Hill Companies, 2011

Chapter 27–10 H HOCH2

H2O

above HOCH2

O

+

H

H

OH

OH

O

H

H

OH

H OH

H

H

H2O

+

O H

HOCH2

OH

O

+ H3O+

H

H

OH

H OH

H

H

planar carbocation

HOCH2

H2O

H

below

H

H OH

HOCH2

H

O

HO

H

O

+

H

O H

HO

OH

H

+ H3O+

H

H

OH

H2O

27.19 a. All circled O atoms are part of a glycoside.

b. Hydrolysis of rebaudioside A breaks each bond indicated with a dashed line and forms four molecules of glucose and the aglycon drawn.

OH

HO

OH

OH

OH

OH O

O O

HO

O OH

OH

O

O HO

OH O

HO

O

OH

+

OH

HO

OH HO

+

OH

HO

OH

OH

OH HO

H

OH

rebaudioside A Trade name: Truvia

H O

O HO

O

OH

HO

+

OH

O

H

+

H O

OH HO

OH

HO

OH O

HO

OH

aglycon

HO

Both anomers of glucose are formed, but only the
anomer is drawn.

27.20 OH

CH3O

OH Ag2O

O

a.

OH

HO OH H OH

c.

H OCH2C6H5

NaH OH

C6H5CH2O

OCH3

CH3O

C6H5CH2O

OH

HO

C6H5CH2O

CH3I

O CH3O

O

b.

OCH3

C6H5CH2Cl

OH H OCH2C6H5 O OCH2C6H5

C6H5CH2O

H

O C6H5CH2O

OCH2C6H5 C6H5CH2O

C6H5CH2O

H OCH2C6H5

O

H3O+

OH

C6H5CH2O C6H5CH2O

H

+
anomer + HOCH2C6H5

Smith: Study Guide/ Solutions Manual to accompany Organic Chemistry, Third Edition

27. Carbohydrates

© The McGraw−Hill Companies, 2011

Text

Study Guide/Solutions Manual to accompany Organic Chemistry, Third Edition

725

Carbohydrates 27–11 OH

OH

d.

Ac2O OH

HO

OH

H C6H5COO

O +

C6H5

OH

HO

C

Cl

pyridine

OOCC6H5 C6H5COO

O

product in (c)

OOCC6H5 O

C6H5COO

OH H

f.

OAc

AcO AcO

O

e.

O

pyridine

OH H OH

OAc

OAc O

+

C6H5

C

C6H5CH2O

H

OCH2C6H5 O

Cl

pyridine

C6H5CH2O

+
anomer OCOC6H5

C6H5CH2O H

27.21 CH2OH HO HO H

CH2OH

CH2OH

O

H

H

NaBH4

HO

H

CH3OH

HO

OH

H

CH2OH

OH

HO

H

H

HO

H

H

HO

OH CH2OH

CH2OH

D-tagatose

H

H

OH

D-galactitol

D-talitol

27.22 Carbohydrates containing a hemiacetal are in equilibrium with an acyclic aldehyde, making them reducing sugars. Glycosides are acetals, so they are not in equilibrium with any acyclic aldehyde, making them nonreducing sugars. hemiacetal

hemiacetal HO

a.

CH2OH O H H OH

OH

b. H

H H

OH

reducing sugar

CH2OH O H H OH

H H OH OH

reducing sugar

acetal

CH2OH O H c. H OH

hemiacetal HO

HO

H H

O

O

d. HO

OCH2CH3 OH

O HO

nonreducing sugar

OH

OH HO

lactose reducing sugar

27.23 CHO

a.

HO

H

H

OH

H

OH

COOH Ag2O NH4OH

HO

b.

H

OH

H

OH

H

OH

CH2OH

H

OH

H

OH CH2OH

COOH Br2 H2O

H

OH

CHO H

CHO

c. HO

H

CH2OH

HO

H

HO

H

H

OH

H

OH CH2OH

CH2OH

COOH HNO3 H2O

HO H H

H OH OH COOH

OH

726

Smith: Study Guide/ 27. Carbohydrates Text Study Guide/Solutions Manual to accompany: Organic Chemistry, 3rd Edition Solutions Manual to accompany Organic Chemistry, Third Edition

© The McGraw−Hill Companies, 2011

Chapter 27–12 27.24 Molecules with a plane of symmetry are optically inactive. CHO CHO

a.

COOH

H

OH

H

OH

H

OH

H

CH2OH D-erythrose

H

HO

H

HO

OH

OH

HO

H

OH

HO

H

HO

H

COOH

H

OH

H

CH2OH

OH COOH

optically inactive

D-galactose

COOH

HO

H

H

optically inactive

HO H

c.

COOH

OH

HO

CHO

b.

H

H H

H

OH

CH2OH

COOH

D-lyxose

optically active

27.25 CHO HO H HO

H

OH

OH

H

OH

H

H

CHO

H

OH

or

HO H

CH2OH D-idose

CHO H

H

HO

OH

H

OH H OH

CH2OH

CH2OH

D-gulose

D-xylose

27.26 CHO HO

a.

H

H

OH

CHO HO

H

H

HO

H

HO

H

CH2OH

OH

H

CH2OH

b.

H

OH

H

OH

H

OH CH2OH

D-ribose

H

OH H OH CH2OH

D-threose

CHO

CHO

CHO

c.

OH

HO

H

HO

H

H

OH CH2OH

D-galactose

CHO HO

CHO

H

H

OH

H

OH

H

OH

H

OH

H

OH

OH

H

H

CH2OH

OH CH2OH

CHO HO H HO HO H

CHO

H

H

OH

OH

H

OH

H

HO

H

HO

OH CH2OH

H

H H OH CH2OH

Smith: Study Guide/ Solutions Manual to accompany Organic Chemistry, Third Edition

27. Carbohydrates

727

© The McGraw−Hill Companies, 2011

Text

Study Guide/Solutions Manual to accompany Organic Chemistry, Third Edition

Carbohydrates 27–13 27.27 Possible optically inactive D-aldaric acids: CHO

COOH

H

OH

H H

H

OH

OH

H

OH

OH

H

OH

CH2OH

COOH H

plane of symmetry

HO

H

H

H

COOH

A'

CHO

OH

OH

HO

OH

H

H

OH

COOH

CH2OH

A"

This OH is on the right for a D sugar.

There are two possible structures for the D-aldopentose (A' and A''), and the Wohl degradation determines which structure corresponds to A. This is A.

Product of Wohl degradation:

CHO

CHO

H

OH

H

OH

H

OH

CHO

Wohl

H

OH

H

OH

CH2OH

COOH

[O]

COOH

H

OH

HO

H

OH

H

CH2OH

COOH

A'

optically inactive

CHO

[O]

H

HO

OH

H

COOH

Wohl

H OH

H

H

H

CH2OH

optically active

OH

HO

OH CH2OH

A"

B

Since this compound has no plane of symmetry, its precursor is B, and thus A'' = A.

27.28 CHO HO

CH2OH

H

H

HO

OH

HO

H

H

H

H

HO

OH

CHO

H

HO

OH

rotate

H

H

180°

HO

OH

CH2OH

CHO

H OH D-idose

H

H

OH CH2OH

identical

27.29 Optically inactive alditols formed from NaBH4 reduction of a D-aldohexose. CHO

CH2OH

CH2OH

H

OH

H

OH

H

H

OH

H

OH

HO

H

H

OH

H

OH

HO

H

H

OH

H

OH

H

NaBH4

CH2OH

CH2OH

optically inactive

A'

H

HO

H

H

CH2OH

OH CH2OH

optically inactive

CHO K–F

COOH

A''

COOH

H

OH

H

OH

H

OH

HO

H

H

OH

H

OH

H

OH

HO

H

H

OH

H

OH

H

OH

H

A'

NaBH4

OH

HO

This is A.

This is B.

OH

CH2OH

H

OH

CHO H

CHO

OH

CH2OH

B'

[O]

COOH

optically inactive

CHO

CHO [O]

OH COOH

optically active

HO

H

HO

H

H

OH

H K–F

OH

HO

H

HO H

H OH

CH2OH

CH2OH

B''

A''

728

Smith: Study Guide/ 27. Carbohydrates Text Study Guide/Solutions Manual to accompany: Organic Chemistry, 3rd Edition Solutions Manual to accompany Organic Chemistry, Third Edition

© The McGraw−Hill Companies, 2011

Chapter 27–14 Two D-aldohexoses (A' and A'') give optically inactive alditols on reduction. A'' is formed from B'' by Kiliani–Fischer synthesis. Since B'' affords an optically active aldaric acid on oxidation, B'' is B and A'' is A. The alternate possibility (A') is formed from an aldopentose B' that gives an optically inactive aldaric acid on oxidation. 27.30 OH

OH

O

HO HO

HO HO

OH

OH O HO

OH HO

H OH2

O+ OH

+

OH

OH

HO

+ H2O

HO HO

OH +

O

H O HO

OH

O

HO HO

OH

OH

O

+

OH

O

HO

O OH

HO HO


-D-glucose OH H2O

above

OH

H

O

HO HO

H2O

+

OH HO

OH

O

HO HO

OH HO

O

HO HO

-D-glucose +

OH

+ H3O+

OH

planar carbocation

below

OH

O

HO HO

H2O

HO

O

HO HO

+

OH

HO

H

OH


-D-glucose

H2O

27.31 OH O

HO HO

OH O HO

HO HO

O

O OH OH

HO OH

OH

OH

H3O+

OH


-D-glucose

+

HO HO

O OH HO

-D-glucose


anomer

27.32
glycoside bond OH O HO

HO

4 1 O

OH O

HO

OH

cellobiose

OH

OH

Two possible anomers here.
OH is drawn.

The same products are formed on hydrolysis of the
and  anomers of maltose.

Smith: Study Guide/ Solutions Manual to accompany Organic Chemistry, Third Edition

27. Carbohydrates

© The McGraw−Hill Companies, 2011

Text

Study Guide/Solutions Manual to accompany Organic Chemistry, Third Edition

729

Carbohydrates 27–15 27.33 a. 4

OH HO

4

O

O HO

OH HO

4

O

O HO

OH HO

O O

O HO

b.

O HO

1

1

dextran

1

HO HO

O

1

6

O HO

HO

1

HO O 6

O HO

HO HO

27.34 OH O

OH O

O

HO

O HO NHAc

HO

NHAc

OH O

OH O

OHO NHAc

O

NHAc

chitin—a polysaccharide composed of NAG units OH O

OH O HO

O

NH2

HO

NH2

O HO

OH O

OH O

chitosan

OHO

O

NH2

NH2

27.35 OH

a.

CH2OH O H H OH

H OH

H H

OH

CH3NH2 mild H+

CH2OH O H H OH H

OH

b.

C6H5NH2 OH

OH

+

OH

CH2OH O H H OH

OH

H

mild

H

OH OH

NHC6H5 +

H+ OH

+ H2O

H

O

HO

NHCH3

H

OH O

HO

OH

NHCH3

H

OH

H

OH

H

O

HO

H OH

OH

+ H2O

NHC6H5

730

Smith: Study Guide/ 27. Carbohydrates Text Study Guide/Solutions Manual to accompany: Organic Chemistry, 3rd Edition Solutions Manual to accompany Organic Chemistry, Third Edition

© The McGraw−Hill Companies, 2011

Chapter 27–16 27.36 OH

OH

OH O

HO HO

O

HO HO

OH OH

+

OH2

OH

OH

H OH2

OH O

HO HO

HO HO

+

O+

+

H2O

OH

+ H2O

OH

OH

CH3CH2NH2

HO HO

above

H

H2O + NHCH2CH3

O

O

HO HO

NHCH2CH3 + OH

HO

OH

H3O+

O

HO HO

+

OH

OH

OH HO HO

below CH3CH2NH2

O HO

O

HO HO

+

OH

+

NHCH2CH3 H

H3O+

NHCH2CH3

H2O

27.37 O

O NH

a. N

HO CH2

O

N HO CH2

O

N

N

NH2

O H

H OH

OH

NH

b.

OH

27.38 a. Two purine bases (A and G) are both bicyclic bases. Therefore they are too big to hydrogen bond to each other on the inside of the DNA double helix. b. Hydrogen bonding between guanine and cytosine has three hydrogen bonds, whereas between guanine and thymine there are only two. This makes hydrogen bonding between guanine and cytosine more favorable. H H O N N H

H O N H

guanine G

H

cytosine C

H

O

NH

N N

N

O

N

N N H

N N

H N H

guanine G

O H

CH3

N N H

thymine T

Smith: Study Guide/ Solutions Manual to accompany Organic Chemistry, Third Edition

27. Carbohydrates

731

© The McGraw−Hill Companies, 2011

Text

Study Guide/Solutions Manual to accompany Organic Chemistry, Third Edition

Carbohydrates 27–17 27.39 Label the compounds with R or S and then classify. CHO H

H

OH

b.

a. CH3CH2 C OH

CH2CH3

c.

A R

R identical

d.

C CH3CH2

H OH

CHO

CHO

OH

H

CHO

R identical

C

CH2CH3 H

HO

CHO

S enantiomer

S enantiomer

27.40 Use the directions from Answer 27.2 to draw each Fischer projection. COOH

=

a. CH3 C Br

CH3 S

Br

Br

b. CH3 C Cl

H

C Br = H S

re-draw CH3

CH3

CH2CH3

C

C

CH3CH2

f.

S

CH2CH3 OCH2CH3

OCH2CH3

Cl S

Br H

Cl

S

R C

g. H

S

CH2CH3

S

Cl

Cl C

H Br H

C

R

Br

=

Br Cl

S R

re-draw

CH3

H

re-draw

CH3CH2

H Br

C

H Cl = CH3CH2 S

CH3

h. HO

Br

Br

CHO

CHO

H

27.41 Epimers are two diastereomers that differ in the configuration around only one stereogenic center. CHO

CHO

OH

HO H

H

H

HO

OH

HO

CH2OH

C4

OH H H CH2OH

D-xylose

L-arabinose

27.42 CHO HO

a.

H H

H

CHO H

OH

HO

OH

HO

CH2OH D-arabinose

CHO

OH H H CH2OH

enantiomer

HO

b. HO C3

H

CHO

H

H

H

c. HO

OH CH2OH

epimer

H

OH H OH

OH

O

d. HO

OH OH

CH2OH

diastereomer (but not epimer)

H H OH

CH2OH

CH2OH

H

H

S HO = S HO C H HO R H C OH

HO H

Br

CH3

CH3

HO H H OH re-draw HO C H Cl

Br

H C Br = H S Br C H Br S CHO

H

H

H

CH3 Br

C

Br

Cl

re-draw Br C H

C

CH3

OCH3

H

CH2CH3

Br Cl

CH2CH3 = CH3

Cl

d.

Br

OCH3

CH3O OCH2CH3 C

H

CH3 Cl

Cl

H

=

Cl C H

CH3

re-draw

CH3

H C Br

e.

H

H

c.

CH3

COOH

constitutional isomer

732

Smith: Study Guide/ 27. Carbohydrates Text Study Guide/Solutions Manual to accompany: Organic Chemistry, 3rd Edition Solutions Manual to accompany Organic Chemistry, Third Edition

© The McGraw−Hill Companies, 2011

Chapter 27–18 27.43 CHO HO

CHO

H

HO

CHO

H

H

CH2OH C O

OH

H

OH

H

OH

HO

H

H

OH

H

OH

H

OH

HO

H

H

OH

H

OH

HO

OH

H

OH

H

H

CH2OH

CH2OH

CH2OH

A

B

C

a. A and B epimers b. A and C diastereomers

CH2OH O

H H H

OH H

OH

CH2OH

OH

D

HO

OH

OH HO

O OH

HO

F

H

E

c. B and C enantiomers d. A and D constitutional isomers

e. E and F diastereomers

27.44

OH H

H

OH

OH

H

a. anomers, epimers, diastereomers, reducing sugars b. CHO O

H

H

H

A

H HO

OH

OH

OH

O

H

H

OH

H

OH

B

OH H

This is the acyclic form of both A and B.

OH CH2OH

D-xylose

27.45 Use the directions from Answer 27.13. a.
-D-talopyranose

[1]

CHO HO

H

HO

H

HO

H

H

OH CH2OH

CH2OH O H

[2]

CH2OH O H

OH

[3]

OH H

H

CH2OH O OH H OH OH H H H


anomer OH is up for a D sugar.

D sugar, CH2OH

is drawn up.

farthest away C, OH on right = D sugar

D-talose

b.
-D-mannopyranose CHO HO HO H H

[1]

CH2OH O H

[2]

CH2OH O H

H

H H OH OH CH2OH

D-mannose

OH

D sugar, CH2OH

is drawn up.

farthest away C, OH on right = D sugar


anomer OH is up for a D sugar.

[3]

CH2OH O OH H OH OH H OH H H H

Smith: Study Guide/ Solutions Manual to accompany Organic Chemistry, Third Edition

27. Carbohydrates

© The McGraw−Hill Companies, 2011

Text

Study Guide/Solutions Manual to accompany Organic Chemistry, Third Edition

733

Carbohydrates 27–19 c.
-D-galactopyranose

CH2OH O H

[1]

CHO H

CH2OH O H

[2]

H OH

OH

HO

H

D sugar, CH2OH

HO

H

is drawn up.

H

OH CH2OH


anomer OH is down.

CH2OH O H [3] OH H OH H OH H H OH

farthest away C, OH on right = D sugar

D-galactose

d.
-D-ribofuranose

CH2OH O

[1]

[2]

H

CHO H

OH

H

OH

H

OH CH2OH

CH2OH O

H

H

OH

[3]

CH2OH H O H OH


anomer OH is down.

D sugar, CH2OH

is drawn up.

H H OH OH

farthest away C, OH on right = D sugar

D-ribose

e.
-D-tagatofuranose

CH2OH O

[1]

CH2OH

H

C O HO

H

HO

H

H

OH CH2OH

[2]

CH2OH O

CH2OH

H

OH

[3]

D sugar, CH2OH

is drawn up.


anomer OH is down.

farthest away C, OH on right = D sugar

27.46 CHO

a. HO

H

HO

H

HO

H

HO

OH


anomer OH

D sugar HOOH OH

D sugar

HOOH HO

O

O HO

OH

 anomer

CH2OH

farthest away C, OH on right = D sugar CHO

b.

H H HO H

OH OH H OH CH2OH

farthest away C, OH on right = D sugar

D sugar

OH HO O

OH OHOH


anomer

D sugar

OH HO O

OH OH OH

 anomer

CH2OH OH

H

OH H

D-tagatose

H

CH2OH OH O

H

734

Smith: Study Guide/ 27. Carbohydrates Text Study Guide/Solutions Manual to accompany: Organic Chemistry, 3rd Edition Solutions Manual to accompany Organic Chemistry, Third Edition

© The McGraw−Hill Companies, 2011

Chapter 27–20 c.

CHO HO

D sugar HOOH OH

H

H

O

OH

HO

H

H

OH

OH

D sugar

HOOH HO O

OH

OH

OH


anomer

 anomer

CH2OH

farthest away C, OH on right = D sugar

27.47 Use the directions from Answer 27.14. "up" group on left

CH2OH

OH

a.

CH2OH is up = D sugar O

H OH

CHO

[2]

"up" group on left

H

OH

b. H

H HO

OH

HO

"down" group on right

[1]

CHO

H

H H

OH =

H

CH2OH

"up" group on left

H

H OH

HO

H

HO

H

CH2OH

CH2OH

OH on left = L sugar CH2OH is up = D sugar O [1] OH

CHO

CHO

[2]

CHO

[3] HO

OH H

OH OH

H OH

CHO

[3] HO

CH2OH

H

CH2OH

CHO

[2]

HO

OH

H

CH2OH

H

"down" group on right

OH

CH2OH

OH

OH

OH on right = D sugar

O H CH2OH OH H OH

HO O

H

"up" group on left CH2OH is down = L sugar

OH

CHO

[3]

H H

HO

CHO

H

H

c.

[1]

OH

H

H CH2OH

OH CH2OH

OH on right = D sugar

H

H

OH

H

OH

H

OH CH2OH

Smith: Study Guide/ Solutions Manual to accompany Organic Chemistry, Third Edition

27. Carbohydrates

© The McGraw−Hill Companies, 2011

Text

Study Guide/Solutions Manual to accompany Organic Chemistry, Third Edition

735

Carbohydrates 27–21 CHO

d.

OH

H

OH

O

H

OH

HO

OH OH

H

OH

OH

H

OH

CHO

=

HO HO

CH2OH OH

O

H

H

OH

= HO

H

H

OH CH2OH

OH

C O

CH2OH OH

OH

H

OH

OH

OH

H

H

OH

CH2OH

D sugar

e. HOCH2

H

O

f. HO

O

H

OH

H

OH

H

OH CH2OH

27.48 CHO HO H H

H

a.

OH OH

H

CH2OH

b.

CH2OH OH CH2OH H H O HO H O HO H OH

H

OH OH

 anomer

D-arabinose

H

H H

O OH OH

OH

H

OH

H


anomer

H

H H

O H OH

OH

H

OH

OH

two anomers in the pyranose form

27.49 Two anomers of D-idose, as well as two conformations of each anomer:


anomer

axial O

HO OH

OH OH OH O

OH OH

HO

OH

4 axial substituents

 anomer

OH

equatorial CH2OH group

OH OH OH O

4 equatorial OH groups More stable conformation for the
anomer—the CH2OH is axial, but all other groups are equatorial.

equatorial CH2OH group OH

OH

3 axial substituents

OH O

HO

axial

OH

axial

OH

HO

3 equatorial OH groups The more stable conformation for the  anomer—the CH2OH is axial, as is the anomeric OH, but three other OH groups are equatorial.

736

Smith: Study Guide/ 27. Carbohydrates Text Study Guide/Solutions Manual to accompany: Organic Chemistry, 3rd Edition Solutions Manual to accompany Organic Chemistry, Third Edition

© The McGraw−Hill Companies, 2011

Chapter 27–22 27.50 CH3O

HO

O

D-gulose

g. The product in (a), then H3O+

O

HO C6H5CH2O

OCH3 OCH2C6H5 O

c. C6H5CH2Cl, Ag2O

OOCC6H5 O O OOCC H 6 5 COC6H5 OCH3 OCH 3

O

+
anomer

+
anomer

HO OH

C6H5COO

OCH3

OH

b. CH3OH, HCl

HO OH

f. C6H5COCl, pyridine

CH3O

CH3O HO

OH

C6H5COO

OCH3 O

a. CH3I, Ag2O

CH3O

CH3O

OH

OAc

h. The product in (b), then Ac2O, pyridine

OAc O

+
anomer

O

OCH2C6H5 CH2C6H5

C6H5CH2O HO

OAc

i. The product in (g), then C6H5CH2Cl, Ag2O

OH O

CH3O

d. C6H5CH2OH, HCl

OCH3 O

+
anomer HO OH OAc

OCH2C6H5 CH3O

OAc

CH3O

CH3O

OCH3 O

OAc OAc OAc CH3O

CH3O

27.51 OH OH O

a. CH3OH, HCl HO CHO HO

OH

COOH

d. Br2, H2O

+
anomer OCH3

H

H

OH

H

OH

H

OH

b. (CH3)2CHOH, HCl HO

CH2OH

OH OH O OH

D-altrose

c. NaBH4, CH3OH

HO

HO

H

H

OH

H

OH

H

OH CH2OH

+
anomer

OCH(CH3)2

COOH

e. HNO3, H2O

HO

H

H

OH

CH2OH

H

OH

H

H

H

OH

H

OH

H

+
anomer

OCH2C6H5

j. The product in (d), then CH3I, Ag2O

O

e. Ac2O, pyridine

OAc OCH3

OH CH2OH

OH COOH

+
anomer OCH2C6H5

Smith: Study Guide/ Solutions Manual to accompany Organic Chemistry, Third Edition

27. Carbohydrates

737

© The McGraw−Hill Companies, 2011

Text

Study Guide/Solutions Manual to accompany Organic Chemistry, Third Edition

Carbohydrates 27–23 CHO

f. [1] NH2OH [2] (CH3CO)2O, NaOCOCH3 [3] NaOCH3

H

OH

H

OH

H

OH

OCH3 OCH3 O

h. CH3I, Ag2O CH3O

OCH3 OCH3

CH2OH CHO HO

H

H OH

+

AcO

OH

HO

OAc OAc O

i. Ac2O, pyridine

CHO

H

g. [1] NaCN, HCl [2] H2, Pd-BaSO4 HO H [3] H3O+

H OH

H

OH

H

OH

H

OH

H

OH

+
anomer

OAc

OAc

H

CH2OH

+
anomer

j. C6H5CH2NH2, mild H+ OH

OH O

HO

+
anomer

CH2OH

NHCH2C6H5

OH

27.52 a. CH3OH, HCl

CH2OH

OCH3 OH

H

CHO H HO H

OH

H

O

OH

H

H

OCH(CH3)2 OH

OH CH2OH

g. [1] NaCN, HCl [2] H2, Pd-BaSO4 [3] H3O+

HO H

OH

HO H

+

H

H

H

OH

H

OH

HO

OH

H

H

CH2OH

CH2OH H

CHO

H

HO

+
anomer

OH

c. NaBH4, CH3OH

H

H

CHO CH2OH

H

D-xylose

HO

+
anomer

H

CH2OH

CHO

f. [1] NH2OH [2] (CH3CO)2O, NaOCOCH3 [3] NaOCH3

H

H

b. (CH3)2CHOH, HCl

H

O

OH

OH CH2OH

OH H

h. CH3I, Ag2O CH2OCH3

H

O

OH

OCH3

H

H

OCH3 OCH3

H

CH2OH

+
anomer COOH

d. Br2, H2O

H HO H

OH H OH

i. Ac2O, pyridine

CH2OAc O OAc H

H H OAc OAc

H

+
anomer

CH2OH COOH

e. HNO3, H2O

H HO H

OH H OH COOH

j. C6H5CH2NH2, mild H+

CH2OH

H

O

OH

H

H

NHCH2C6H5 OH

H

+
anomer

738

Smith: Study Guide/ 27. Carbohydrates Text Study Guide/Solutions Manual to accompany: Organic Chemistry, 3rd Edition Solutions Manual to accompany Organic Chemistry, Third Edition

© The McGraw−Hill Companies, 2011

Chapter 27–24 27.53 OH

H

H

OH

H

O H

OH

H

H H

H

O

OH

OH

H

N H

O

O

H

H

HO

H3O+

OH OH

O

solanine HO

H H HO

HO

OH

OH

O

HO

+ HO

H

aglycon

monosaccharide (both anomers)

OH

N

H

+

O

HO

O O

H

OH H

OH

aglycon

H

OH

HO

HO

HO

+

monosaccharide (both anomers) H

HO

O

HO HO

H3O+

salicin

OH

OH

H

O

HO HO

OH

OH

O

HO

OH

OH

OH

monosaccharide (both anomers)

monosaccharide (both anomers)

27.54 CHO H

CHO

OH

HO

H

H H

CHO

HO

H

HO

HO

OH

H

OH

H

CH2OH

OH

H

OH

H

CH2OH

D-glucose

H H OH OH CH2OH

D-arabinose

D-mannose

27.55 CHO CHO

HO

a. H

OH

H

OH CH2OH

H

H

H

OH

H

OH CH2OH

H

OH

H

OH CH2OH

HO HO H

CHO

CHO

HO

H

H

H

HO

H

HO

H

HO

H

HO

OH CH2OH

H

OH CH2OH

H

OH H H OH CH2OH

c.

CHO

CHO

HO

H

H

HO

H

HO

H

HO

H

HO

H

HO

H

HO

H

H

HO

H

HO

OH

CHO

b.

CHO

CHO

HO H

OH CH2OH

H

OH CH2OH

H

OH

H OH CH2OH

Smith: Study Guide/ Solutions Manual to accompany Organic Chemistry, Third Edition

27. Carbohydrates

© The McGraw−Hill Companies, 2011

Text

Study Guide/Solutions Manual to accompany Organic Chemistry, Third Edition

739

Carbohydrates 27–25 27.56 OH

OH CH3OH

O

a. HO

OH

HO

HO HO

HCl

OH

+
anomer H

H NaBH4

HO

H

OH CH3OH

H

OH

H

CH3I

H

OH

Ag2O

H

OH

c.

H

H

OCH3

H

OCH3 CH2OCH3

COOH H

COOH

OH

H

Br2

HO

H

OH

H2O

H

OH

H

OH

H

OH

HO

OCH3

CH3O

CH2OH

OH

H Ac2O

H

CH2OH

OCH3 OCH2CH3

+
anomer

H

H

CHO

O

CH3CH2O CH3CH2O

CH2OCH3

OH

CH2OH

H

Ag2O

OH

CH2OH

OH

HO

OCH3

OCH2CH3

+
anomer

CHO

b.

CH3CH2I

O

OAc

AcO

pyridine

H

H

OAc

H

OAc

CH2OH

CH2OAc

27.57 Molecules with a plane of symmetry are optically inactive. CHO H H

NaBH4

OH

H

CHO

CH2OH

OH

CH3OH

OH

H H H

CH2OH

OH

H

OH

HO

NaBH4

H

H

OH

CH2OH

OH

CH3OH

OH

H

H

H

CH2OH

CH2OH

OH

HO

OH CH2OH

D-xylose

D-ribose

27.58 OH

OH O

a. HO

+

OCH3

H3O

HO

OH

OH

OH HOCH2

c. OH

O

OH

OH

OCH3 HO O

HO

O

HO

OH

OH

b.

OH O

H3O+

OCH3 HO O

HO

OCH2CH3 NHCH2CH3

OH

OH

H3O+

HOCH2

OH

O

HO

OH

OH

OH

OCH3 HO O

OH

OH

+ CH3OH

OH

+ CH3CH2OH

OH HOCH2

OH

O

+ NH2CH2CH3 OH OH

740

Smith: Study Guide/ 27. Carbohydrates Text Study Guide/Solutions Manual to accompany: Organic Chemistry, 3rd Edition Solutions Manual to accompany Organic Chemistry, Third Edition

© The McGraw−Hill Companies, 2011

Chapter 27–26 27.59 OH

OH O

HO HO

HO HO

OH OH

OH O

+

HO HO

OH2 OH

OH

OH H OH

H OH2 OH

OH O

HO HO

O

HO HO

resonance-stabilized carbocation

H OH

+

OH O

HO HO

OH

OH O

O

HO HO

OH OH

OH + O H

H OH2

H

H OH

H OH

27.60

H

O C6H5

C

H

OH C6H5

C

H

C6H5 CH2OH HO HO

O

H OH

B (any base)

OH H O HO HO

C6H5

O OCH3

OCH3

O HO HO

OH

O

+ HB+ OCH3

OH

OH OH2 C6H5

O H O HO

OH

B

C6H5

O OCH3

O O HO

+ H2O

O OH

+ HB+ OCH3

C6H5

O OH

(any base) C6H5

O HO HO

OCH3

O HO HO

O OCH3 OH

Smith: Study Guide/ Solutions Manual to accompany Organic Chemistry, Third Edition

27. Carbohydrates

741

© The McGraw−Hill Companies, 2011

Text

Study Guide/Solutions Manual to accompany Organic Chemistry, Third Edition

Carbohydrates 27–27 27.61 OH

OH O

HO HO

OH OH

OH HO H O HO

O

O HO

OH

O

HO HO OCH3

HO HO

O

+

OH

OCH3

OH

OH

OH

O +

A

OH +

+ H2O

H OH2

O+

HO HO

OH O

HO HO

OCH3

B HO OH

OH

O

HO HO

OCH3

OH

O

HO HO

HO

HO

B

HO HO

OCH3 H

OH

H OH2

+

O+

HO HO

OH

A

+ H2 O

+

OH

O

+ CH3OH OH

H2O

above

OH

H

O

HO HO

H2O

+

OH HO

OH

O

HO HO

OH HO

O

HO HO

+ H3O+

+

A OH

OH

below

OH

O

HO HO

H2O

HO

O

HO HO

+

HO

OH H

OH

H2O

27.62 COOH H

OH

H

OH

H

OH

H

OH

H

=

CH2OH OH O H H H

OH OH

H A

H

OH

OH

OH

CH2OH + OH OH H H H OH OH

HO

H

O OH

HO OH

A

=

H H

H

OH

O

H

O H H

H

OH

OH

O OH

OH +

H A

OH OH

OH

OH

A

H

H

H A

OH

OH

OH OH

OH

H H OH

OH

O

A

O+

OH

OH +

CH2OH

H

OH H

OH O

H H

H

H O H

A

O

H H

H

OH

+ H2O

OH

OH +

OH2 OH

OH

OH

+

A

742

Smith: Study Guide/ 27. Carbohydrates Text Study Guide/Solutions Manual to accompany: Organic Chemistry, 3rd Edition Solutions Manual to accompany Organic Chemistry, Third Edition

© The McGraw−Hill Companies, 2011

Chapter 27–28 27.63 OH

CHO H

OH

HO

H

H

H

C O

C O

C OH

C OH

HO

OH

H

H

H

H

H2O

OH

H

CH2OH

HO

OH

H

OH

H

CH2OH

Protonation of this enolate can occur from two directions.

D-glucose

H C OH

H OH

C OH

H

HO

OH

H

OH

H

H OH

+

OH

OH

CH2OH

CH2OH

enediol A

Protonation on O forms an enediol.

H2O

two protonation products

CHO HO

H

HO

H

H

OH

H

OH CH2OH H OH

H

H

C OH

C OH

C OH

C O

C O

C O H HO

OH

H

H H

HO

OH

H

OH

H

H

CH2OH

H

OH

H

CH2OH

enediol A Deprotonation of the OH at C2 of the enediol forms a new enolate that goes on to form the ketohexose.

HO

OH

H

H H C OH C O

H

HO

OH

H

OH

H

CH2OH

H OH

+

OH CH2OH

+ H2O

27.64 Two D-aldopentoses (A' and A'') yield optically active aldaric acids when oxidized. Optically active D-aldaric acids: CHO HO

H

H

OH

H

OH CH2OH

A'

COOH

COOH

[O]

HO H H

H

HO

H

OH

HO

H

OH

H

COOH

optically active

OH COOH

optically active

CHO

[O]

HO

H

HO

H

H

OH CH2OH

A" D-lyxose

OH

Smith: Study Guide/ Solutions Manual to accompany Organic Chemistry, Third Edition

27. Carbohydrates

743

© The McGraw−Hill Companies, 2011

Text

Study Guide/Solutions Manual to accompany Organic Chemistry, Third Edition

Carbohydrates 27–29

CHO HO

CHO

H

H

OH

H

OH

Wohl

H

OH

H

OH

[O]

H

OH

HO

H

OH

H

CH2OH

CH2OH

COOH

COOH

A'

[O] HO

H OH

Wohl

H

H

COOH

COOH

optically inactive

CHO

CHO OH

HO

H

HO

H

H

CH2OH

OH CH2OH

no plane of symmetry optically active

A"

Only A" undergoes Wohl degradation to an aldotetrose that is oxidized to an optically active aldaric acid, so A'' is the structure of the D-aldopentose in question. 27.65 CHO HO

CH2OH

H

HO

CH2OH

H

HO

H

OH

H

OH

HO

H

OH

H

OH

H

CH2OH

CH2OH

CHO

H

HO

H

HO

OH

H

H

CH2OH

OH CH2OH

identical (by rotating 180°)

D-arabinose

H

D-lyxose

27.66 Only two D-aldopentoses (A' and A'') yield optically inactive aldaric acids (B' and B''). CHO

COOH

H

OH

H

OH

H

OH

[O]

COOH

H

OH

H

H

OH

HO

H

OH

H

CHO

OH H

H

[O]

OH

CH2OH

COOH

A'

B'

B''

optically inactive

optically inactive

HO H

COOH

OH H OH CH2OH

A"

Product of Kiliani–Fischer synthesis: CHO

CHO H

OH

H

OH

H

OH

K–F

CH2OH

CHO

OH

HO

H

OH

H

OH

H

H

OH

H

OH

HO

OH

H

OH

H

H

CH2OH

A'

D' [O] COOH

plane of symmetry

H

CHO

H

HO

CHO

H

H

OH

OH

H

OH

H OH

HO H

H OH

CH2OH

CH2OH

CH2OH

C'

C"

D"

[O]

[O]

[O]

COOH

COOH

COOH

H

OH

HO

H

OH

H

OH

H

H

H

OH

H

OH

HO

H

OH

H

OH

H

HO

H

H

OH

OH

H

OH

H OH

HO H

H OH

COOH

COOH

COOH

COOH

optically inactive

optically active

optically active

optically active

CHO H

K–F HO H

OH H OH CH2OH

A"

744

Smith: Study Guide/ 27. Carbohydrates Text Study Guide/Solutions Manual to accompany: Organic Chemistry, 3rd Edition Solutions Manual to accompany Organic Chemistry, Third Edition

© The McGraw−Hill Companies, 2011

Chapter 27–30 Only A' fits the criteria. Kiliani–Fischer synthesis of A' forms C' and D' which are oxidized to one optically active and one optically inactive aldaric acid. A similar procedure with A'' forms two optically active aldaric acids. Thus, the structures of A–D correspond to the structures of A'–D'. 27.67 Only two D-aldopentoses (A' and A'') are reduced to optically active alditols. CHO HO

CH2OH

H

H

OH

H

OH

HO

[H]

CH2OH

H

HO

H

OH

HO

H

OH

H

CH2OH

HO

[H]

H

H

H

CH2OH

optically active

H

HO

OH

CH2OH

A'

CHO

H

OH CH2OH

A"

optically active

Product of Kiliani–Fischer synthesis: CHO

CHO HO

H

H

H

OH K–F

H

OH

HO H H

CH2OH

CHO H

HO

H

H

H

HO

H

HO

H

HO

H

OH

HO

H

HO

H

OH

H

OH

H

OH

H

CH2OH

C'

C"

[O]

[O]

COOH HO

OH

CH2OH

B'

H

CHO

HO

CH2OH

A'

CHO

OH

H

COOH HO

H

H

H

HO

H

HO

H

HO

H

HO

OH

HO

H

OH

H

OH

H

COOH

COOH

optically active

OH

OH

CH2OH

A"

OH H H

H

COOH

optically active

OH

COOH

H

H

H

H

[O]

HO

OH

H

K–F HO

B''

OH

H

HO

CH2OH

[O]

COOH

CHO

OH

OH COOH

optically active

optically inactive

Only A'' fits the criteria. Kiliani–Fischer synthesis of A'' forms B'' and C'', which are oxidized to one optically inactive and one optically active diacid. A similar procedure with A' forms two optically active diacids. Thus, the structures of A–C correspond to A''–C''. 27.68 D-gulose

CHO

CH2OH

CHO H

OH

H

OH

H

H

OH

H

OH

HO

HO

H

H

OH CH2OH

A

HO

H

H

OH CH2OH

B

OH H

H

OH CH2OH

C

CHO

CH2OH H

OH

HO

H

H

OH CH2OH

D

HO

H

H

OH CH2OH

E

COOH HO

CH2OH

H

H

OH

OH

H

OH

COOH

HO

H

F

H

H

OH CHO

G

Smith: Study Guide/ Solutions Manual to accompany Organic Chemistry, Third Edition

27. Carbohydrates

© The McGraw−Hill Companies, 2011

Text

Study Guide/Solutions Manual to accompany Organic Chemistry, Third Edition

745

Carbohydrates 27–31 27.69 A disaccharide formed from two galactose units in a 1
4--glycosidic linkage:
glycoside bond OH O

HO HO

OH O

1 O

OH 4

OH

HO OH

27.70 A disaccharide formed from two mannose units in a 1
4--glycosidic linkage:
glycoside bond HO

HO O

HO HO

1

O

HO 4

OH O

HO

OH

27.71 OH

a.

OH O

OH O

HO

OH

HO OH

OH

OCH3 OCH 3 O

CH3I

O

Ag2O

CH3O

OCH3 OCH3

(Both anomers of B and C are formed, but only one is drawn.)

O CH3O

OCH3

OH O

CH3I

b. HO

OH O HO HO

HO

Ag2O

H3O+ OCH3 O OCH3

O OH

+

CH3O OCH3 OH E

OCH3 O

HO CH3O CH3O +

OCH3 O OH F + CH3OH

(Both anomers of E and F are formed, but only one is drawn.)

OCH3

D

OH OCH3 + CH3OH

C

OCH3 OCH 3 O

H3O+

OCH3 O

HO CH3O

OH

CH3O

OCH3 O

CH3O OCH3 O CH3O CH3O

OCH3 OCH3

A

B OH

OCH3 O

27.72 OH

a.

 glycoside bond

OH O

1

HO OH

b. c.

4

O

OH

OH

OH O

O OH

HO OH

1 4--glycoside bond reducing sugar (hemiacetal)

HO

HO

1

O

HO HO

reducing sugar (hemiacetal)

 glycoside bond 6

OH O

1 6-
glycoside bond OH

746

Smith: Study Guide/ 27. Carbohydrates Text Study Guide/Solutions Manual to accompany: Organic Chemistry, 3rd Edition Solutions Manual to accompany Organic Chemistry, Third Edition

© The McGraw−Hill Companies, 2011

Chapter 27–32 27.73 a and b.

OH

OH

CH2OH O

HO

O CH2 O OH OH OH

1 6-
-glycoside bond

HO

OH

stachyose

1 2-
-glycoside bond

OH HOCH2 O

OH O

c.

CH2OH O

CH2OH O

OH

O OH

O OH OH

HO

OH

OH HOCH2

O

HO OH

OH

OH

O

O

CH2OH

OH

OH

OH

HO OH

OH

OH

O

2

HO

OH CH2OH OH O

O

=

O

OH

OH

O

HO O CH2

H3O+

OH

1 6-
-glycoside bond

OH

OH

CH2OH

OH

OH

OH

identical Two anomers of each monosaccharide are formed, but only one anomer is drawn.

d. Stachyose is not a reducing sugar since it contains no hemiacetal. e.

CH3I Ag2O

OCH3 CH3O

OCH3 O

CH3O O CH3O

O

CH3O

OCH3 O

O OCH3 CH3O

CH3O

f.

product in (e)

H3O+

O

O O OCH3 OCH3 CH3

CH2OCH3 O

CH3O OH

OCH3 OCH3

OCH3 OCH3 CH2OH O

CH2OH O OH

OCH3 OCH3

OH

OCH3 CH3O

OCH3

Two anomers of each monosaccharide are formed.

CH3OCH2 CH3O OH O CH3O

CH2OCH3

Smith: Study Guide/ Solutions Manual to accompany Organic Chemistry, Third Edition

27. Carbohydrates

© The McGraw−Hill Companies, 2011

Text

Study Guide/Solutions Manual to accompany Organic Chemistry, Third Edition

747

Carbohydrates 27–33 27.74 OH O

HO HO

isomaltose +
anomer OH O

Isomaltose must be composed of two glucose units in an
-glycosidic linkage. Since it is a reducing sugar it contains a hemiacetal. The free OH groups in the hydrolysis products show where the two monosaccharides are joined.

O

HO HO

OH

the hemiacetal

OH [1] CH3I, Ag2O [2] H3O+ OCH3 O

CH3O CH3O

1

CH3O

OH

6

O

CH3O CH3O

OH

CH3O

(Both anomers are present.) OH

27.75 OH O

HO HO

CH3O CH3O

CH3I

trehalose

OH O

Ag2O

OH

OCH3 O CH3O

H3O+ O

OH OH

O HO

OCH3

CH3O CH3O

OCH3 OCH3

O CH3O

OCH3 O CH3O

OH

(both anomers)

Trehalose must be composed of D-glucose units only, joined in an
-glycosidic linkage. Since trehalose is nonreducing it contains no hemiacetal. Since there is only one product formed after methylation and hydrolysis, the two anomeric C's must be joined. 27.76 HO a. HO

O H2N

CH2OH

1

O HO HO




c.

6

O

H

O

H

HO

H H

OH

NH2 O

NHCH2C6H5

O CH3

b. HO HO

OH OH O

4 1

d.

OH

O HO

OH

mannose

glucose

N HO CH2

O

NH

OH OH

O H OH

O

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Chapter 27–34 27.77 Ignoring stereochemistry along the way:

A HO

OH

OH H

H A O

O

[1]

OH OH

OH

A

OH A O O

OH

HO

OH

[4] OH

OH HO H A HO

OH

HO

C

[5]

O

CH3

OH

O

CH3 O

OH

C

CH3

[8]

HO

OH

H

[7]

HO

HO

O

[6]

OH2

HO

OH

HO

[3] OH

OH

OH

O

OH

[2]

OH

OH

H

OH

OH

HO

H A

HO

OH + HA

CH3

– H2O

O

O O

HO HO

[9]

OH

O O

O H

OH

HO

A

[10]

OH

OH O

O HO

OH

C

OH

CH3

OH

+ H2O

[11] O

O O

O

[14]

OH

O

O

– H2O HO

O

[13]

OH

O O

O

H2O HO

O

OH

O

O

OH

HO HO

O

HO

O

H A

A

[15]

[12]

HO

O H A

A

O

H

O

O

O HO

O

O

[16]

O O

O

O HO

O + HA

=

O

O

H O

HO

H

O

CH3

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Text

Carbohydrates 27–35 27.78 The hydrolysis data suggest that the trisaccharide has D-galactose on one end and D-fructose on the other. D-Galactose must be joined to its adjacent sugar by a -glycosidic linkage. D-Fructose must be joined to its adjacent sugar by an
-glycosidic linkage. HO

CH3O

OH O

HO OH

galactose

 O HO HO

HO

O HO


O

CH3I Ag2O

CH3O

OCH3 O O CH3O CH3O CH3O

O CH3O

O OH OH

glucose

HO

CH3O

O OCH3

(Hydrolysis cleaves all acetals, indicated with arrows.)

OCH3 O CH3O

O

CH3O CH3O

H3O+

fructose 2 anomers CH3O

CH3O

CH3O

OH OH

CH3O CH3O

CH3O O

O CH3O

OH

OCH3 OH

OH

2,3,4,6-tetra-O-methyl2,3,4-tri-O-methyl-D-glucose 1,3,6-tri-O-methyl-D-fructose D-galactose (Both anomers of each compound are formed.)

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Amino Acids and Proteins 28–1 C Chhaapptteerr 2288:: A Am miinnoo A Acciiddss aanndd PPrrootteeiinnss   SSyynntthheessiiss ooff aam miinnoo aacciiddss ((2288..22)) [1] From
-halo carboxylic acids by SN2 reaction NH3

R CHCOOH

R CHCOO–NH4+

(large excess) SN2

Br

+

NH4+ Br–

•

Alkylation works best with unhindered alkyl halides—that is, CH3X and RCH2X.

NH2

[2] By alkylation of diethyl acetamidomalonate O

H H C N C COOEt

CH3

COOEt

R

[1] NaOEt

H2N C COOH

[2] RX

H

[3] H3O+,


[3] Strecker synthesis O R

C

NH4Cl H

NaCN

NH2

NH2

H3O+

R C CN

R C COOH

H

H


-amino nitrile

  PPrreeppaarraattiioonn ooff ooppttiiccaallllyy aaccttiivvee aam miinnoo aacciiddss [1] Resolution of enantiomers by forming diastereomers (28.3A) • Convert a racemic mixture of amino acids into a racemic mixture of N-acetyl amino acids [(S)and (R)-CH3CONHCH(R)COOH]. • React the enantiomers with a chiral amine to form a mixture of diastereomers. • Separate the diastereomers. • Regenerate the amino acids by protonation of the carboxylate salt and hydrolysis of the N-acetyl group. [2] Kinetic resolution using enzymes (28.3B) H2N

C

COOH

(CH3CO)2O

AcNH

H R

C

COOH

acylase

H R

H2N

C

COOH

H R

(S)-isomer

(S)-isomer separate

H2N

C

COOH

R H

(CH3CO)2O

AcNH

C

COOH

R H

acylase

AcNH

C

COOH

R H

(R)-isomer enantiomers

enantiomers

NO REACTION

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Chapter 28–2 [3] By enantioselective hydrogenation (28.4) R

NHAc

H

AcNH

H2

C C

Rh*

COOH

C

H2O, –OH

COOH

H CH2R

C

COOH

H CH2R

S enantiomer Rh*

H2N

S amino acid

= chiral Rh hydrogenation catalyst



SSuum mm maarryy ooff m meetthhooddss uusseedd ffoorr ppeeppttiiddee sseeqquueenncciinngg ((2288..66)) • Complete hydrolysis of all amide bonds in a peptide gives the identity and amount of the individual amino acids. • Edman degradation identifies the N-terminal amino acid. Repeated Edman degradations can be used to sequence a peptide from the N-terminal end. • Cleavage with carboxypeptidase identifies the C-terminal amino acid. • Partial hydrolysis of a peptide forms smaller fragments that can be sequenced. Amino acid sequences common to smaller fragments can be used to determine the sequence of the complete peptide. • Selective cleavage of a peptide occurs with trypsin and chymotrypsin to identify the location of specific amino acids (Table 28.2).

A Addddiinngg aanndd rreem moovviinngg pprrootteeccttiinngg ggrroouuppss ffoorr aam miinnoo aacciiddss ((2288..77)) [1] Protection of an amino group as a Boc derivative R H2N

H C

R

[(CH3)3COCO]2O (CH3CH2)3N

CO2H

Boc N H

H C

CO2H

[2] Deprotection of a Boc-protected amino acid R

H C

Boc N H

CO2H

CF3CO2H or HCl or HBr

R H2N

H C

CO2H

[3] Protection of an amino group as an Fmoc derivative

H2N

C

C

R H

O

R H OH

+

CH2O

C

O Fmoc Cl

Na2CO3 Cl

H2O

Fmoc N H

C

C O

OH

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Amino Acids and Proteins 28–3 [4] Deprotection of an Fmoc-protected amino acid R H C

Fmoc N H

R H C

OH H2N

C

O

C

OH

O

N H

[5] Protection of a carboxy group as an ester O

O H2N

C

C

OH

CH3OH,

H2N

H+

C

C

H2N

OCH3

H R

H R

O

O C

C

C6H5CH2OH, H+

OH

H2N

C

C

OCH2C6H5

H R

H R

methyl ester

benzyl ester

[6] Deprotection of an ester group O H2N

C

C

O

O –OH

OCH3

H2N

C

H2O

H R

C

OH

H R

H2N

C

C

O H2O, –OH

OCH2C6H5

H2N

or H2, Pd-C

H R

C H R

benzyl ester

methyl ester



SSyynntthheessiiss ooff ddiippeeppttiiddeess ((2288..77)) [1] Amide formation with DCC R H Boc N H

C

OH + C O

O H2N

C H R

C

R H OCH2C6H5

DCC Boc N H

C

C O

H N

O C

C

OCH2C6H5

H R

[2] Four steps are needed to synthesize a dipeptide: a. Protect the amino group of one amino acid using a Boc or Fmoc group. b. Protect the carboxy group of the second amino acid using an ester. c. Form the amide bond with DCC. d. Remove both protecting groups in one or two reactions.

SSuum mm maarryy ooff tthhee M Meerrrriiffiieelldd m meetthhoodd ooff ppeeppttiiddee ssyynntthheessiiss ((2288..88)) [1] Attach an Fmoc-protected amino acid to a polymer derived from polystyrene. [2] Remove the Fmoc protecting group. [3] Form the amide bond with a second Fmoc-protected amino acid using DCC. [4] Repeat steps [2] and [3]. [5] Remove the protecting group and detach the peptide from the polymer.

C

OH

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Chapter 28–4 C Chhaapptteerr 2288:: A Annssw weerrss ttoo PPrroobblleem mss 28.1 H CH3 O

S C

S

OH

R

H2N H

O

H CH3 O

S C

R C

H

CH3

OH

S

H2N H

CH3 OH

H O

R

H NH2

R C

OH

H NH2

L-isoleucine

28.2 a.

b.

NH3+ C COO–

(CH3)2CH

c.

NH3+

(CH3)2CHCH2

C

COO–

+

HOOCCH2CH2 C COO–

N

H

H

NH3+

d.

COO–

H

H H

28.3

In an amino acid, the electron-withdrawing carboxy group destabilizes the ammonium ion (–NH3+), making it more readily donate a proton; that is, it makes it a stronger acid. Also, the electron-withdrawing carboxy group removes electron density from the amino group (–NH2) of the conjugate base, making it a weaker base than a 1o amine, which has no electron-withdrawing group.

28.4

The most direct way to synthesize an
-amino acid is by SN2 reaction of an
-halo carboxylic acid with a large excess of NH3. a.

NH3

Br CH COOH

large excess

H

b. Br CH COOH CH CH3

NH3

large excess

H2N CH COO– NH4+ H

c. Br CH COOH CH2

glycine

NH3

large excess

H2N CH COO– NH4+ CH2

H2N CH COO– NH4+ CH CH3

CH2

phenylalanine

CH2

CH3

CH3

isoleucine

28.5 CH3I

a.

CH3 H2N C COOH

alanine

c.

H

(CH3)2CHCH2Cl

b.

CH2CH(CH3)2 H2N C COOH

leucine

28.6 H H C N C COOEt

CH3

COOEt

CH(CH3)CH2CH3 H2N C COOH H

H

O

CH3CH2CH(CH3)Br

[1] NaOEt [2] CH2=O [3] H3O+,


CH2OH H2N C COOH H

serine

isoleucine

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Amino Acids and Proteins 28–5 28.7 O

a. H2N CHCOOH (CH3)2CH

CH CH3

O

b. H2N CHCOOH

C

H

C

(CH3)2CHCH2

CH2

O

c. H2N CHCOOH H

C6H5CH2

CH2

C

H

CH CH3

CH3

valine

CH3

leucine phenylalanine

28.8 a. BrCH2COOH

NH3 large excess

H

b. CH3CONH C COOEt COOEt

NH2CH2COO– NH4+

H

[1] NaOEt [2] (CH3)2CHCl

CH(CH3)CH2CH3 H

d. CH3CONH C COOEt

CH(CH3)2

H2N CHCOOH

[2] H3O+

H

H2N C COOH

[3] H3O+,


[1] NH4Cl, NaCN

c. CH3CH2CH(CH3)CHO

COOEt

[1] NaOEt [2] BrCH2CO2Et

H2N C COOH CH2CO2H

[3] H3O+,


A chiral amine must be used to resolve a racemic mixture of amino acids.

28.9

N CH3 H

a. C6H5CH2CH2NH2 achiral

c.

b.

d.

C

N

CH3CH2

achiral

H

NH2

N

chiral (can be used)

H

O

H HO

chiral (can be used)

28.10 NH2

C

NH2

COOH +

R

S Ac2O AcNH

C

AcNH

COOH +

H CH2CH(CH3)2

C

proton transfer

H CH2CH(CH3)2

S

C

C6H5

(R isomer only)

CH3 H COO–

enantiomers

R H2N

Step [1]: React both enantiomers with the R isomer of the chiral amine.

COOH

C

(CH3)2CHCH2 H

S

AcNH

enantiomers

(CH3)2CHCH2 H

H CH2CH(CH3)2

To begin: Convert the amino acids into N-acetyl amino acids (two enantiomers).

COOH

C

+

H3N

C

CH3 H

R

C6H5

AcNH

C

(CH3)2CHCH2 H

R

COO–

+

H3N

C

CH3 H

C6H5

diastereomers

R

These salts have the same configuration around one stereogenic center, but the opposite configuration about the other stereogenic center.

756

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Chapter 28–6 separate

Step [2]: Separate the diastereomers.

AcNH

+

COO–

C

H3N

H CH2CH(CH3)2

C

C6H5

AcNH

CH3 H

S

C

C

C6H5

C

CH3 H

R

R

H2O, –OH NH2

H3N

(CH3)2CHCH2 H

R

Step [3]: Regenerate the amino acid by hydrolysis of the amide.

+

COO–

H2O, –OH

COOH

NH2

H CH2CH(CH3)2

COOH

C

H2N

+

(CH3)2CHCH2 H

(S)-leucine

C

C6H5

CH3 H

(R)-leucine

The chiral amine is also regenerated.

The amino acids are now separated.

28.11 COOH

H2N

[1] (CH3CO)2O

H2N C H

[2] acylase

CH2CH(CH3)2

C

H COOH N C C O (CH3)2CHCH2 H

COOH

CH3

+

H CH2CH(CH3)2

(S)-leucine

(mixture of enantiomers)

N-acetyl-(R)-leucine

28.12 H

a. H2N CHCOOH H

CH3

(CH3)2CH

NHAc

b. H2N CHCOOH

C C COOH

H2NCOCH2

NHAc C C

H

CH2

c. H2N CHCOOH COOH

CH2

CH CH3

CH2

CH3

CONH2

NHAc

C C H

COOH

28.13 Draw the peptide by joining adjacent COOH and NH2 groups in amide bonds. amide O

a.

H2N CH C OH

(CH3)2CH H

O H2N CH C OH

CH CH3

CH2

CH3

CH2

Val

COOH

H2N

C

N-terminal

O

O

H2N CH C OH H2N CH C OH

H

CH2

Gly

C

C

C-terminal OH

H CH2CH2COOH

amide

O H2N CH C OH

O

O

Val–Glu

Glu

b.

C

H N

NH N

CH2

H2N

CH CH3 CH3

Leu His

H H C

C O

N-terminal

(CH3)2CHCH2 H O H N C C OH C N C H O H CH2 C-terminal

amide

NH

Gly–His–Leu N

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Amino Acids and Proteins 28–7 O

O

O

O

c. H2N CH C OH H2N CH C OH H2N CH C OH H2N CH C OH CH2

CH3

CH2SCH3

M

A

CH OH

CH OH

CH3

CH3

T

T

N-terminal

CH2SCH3 amide

CH3

CH2 H

O

CHOH H

C

C

C

H2N

H N

C O

C

CH3

H

N H

amide

C

O

H N

O

C

H

OH

C-terminal

CHOH CH3

M–A–T–T

28.14 N HN

CONH2

a.

H2N

C

O

CH2 H

C

C

N H

C

H N

O

H CH2

O C

C

H2N

b.

OH

H CH(CH3)2

C

Arg–Asn–Val R–N–V

NH

C

N H

C O

O

H N

C

C

OH

H CH2

CH2

CH2

CH2

CONH2

CH2

C

NH2

NH2

HN

CH2 H

C

H CH2

CH2 CH2

O

Lys–His–Gln K–H–Q

28.15 There are six different tripeptides that can be formed from three amino acids (A, B, C): A–B–C, A–C–B, B–A–C, B–C–A, C–A–B, and C–B–A. 28.16 The s-trans conformation has the two C’s oriented on opposite sides of the C–N bond. The s-cis conformation has the two C’s oriented on the same side of the C–N bond. O H2N

C

C

H H

O

H H C OH N C H

H2N

C

C HH

O

H

s-trans

N C H

H C

OH

O

s-cis

28.17 O H2N

HO

N H

H N O

O N H

leu-enkephalin

H N O

O OH

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Chapter 28–8 28.18 O HO HS O

a. H2N

H

H

N O

O

H2N

H2N


COOH

H

O

N

N COOH

H2N

H

S

H N

N COOH

HS

O

NH2 O

O

glutathione b.

H

OH

COOH

N S

N

N COOH

O

H

H

O OH

O

The peptide bond beween glutamic acid and its adjacent OH amino acid (cysteine) is formed from the COOH in the R

group of glutamic acid, not the
COOH.

O

This comes from the amino acid glutamic acid. O




OH

This carboxy group is used to form the amide bond in the peptide, not the
COOH, as is usual. That's what makes glutathione's structure unusual.

COOH

glutamic acid

28.19 O C6H5

a. S

O C6H5

N

CH3 N H

b. S

from Ala

N N H

from Val

28.20 Determine the sequence of the octapeptide as in Sample Problem 28.2. Look for overlapping sequences in the fragments. common amino acids Answer: Ala–Leu–Tyr Tyr–Leu–Val–Cys

Ala–Leu–Tyr–Leu–Val–Cys–Gly–Glu

Val–Cys–Gly–Glu

28.21 Trypsin cleaves peptides at amide bonds with a carbonyl group from Arg and Lys. Chymotrypsin cleaves at amide bonds with a carbonyl group from Phe, Tyr, and Trp. a. [1] Gly–Ala–Phe–Leu–Lys + Ala [2] Phe–Tyr–Gly–Cys–Arg + Ser [3] Thr–Pro–Lys + Glu–His–Gly–Phe–Cys–Trp–Val–Val–Phe b. [1] Gly–Ala–Phe + Leu–Lys–Ala [2] Phe + Tyr + Gly–Cys–Arg–Ser [3] Thr–Pro–Lys–Glu–His–Gly–Phe + Cys–Trp + Val–Val–Phe

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Amino Acids and Proteins 28–9 28.22 Edman degradation gives N-terminal amino acid: Carboxypeptidase identifies the C-terminal amino acid:

Leu–___–___–___–___–___–___ Leu–___–___–___–___–___–Glu

Partial hydrolysis common amino acids Ala–Ser–Arg Gly–Ala–Ser Leu–Gly–Ala–Ser–Arg–___–Glu or

Gly–Ala–Ser–Arg

Leu–___–Gly–Ala–Ser–Arg–Glu Cleavage by trypsin is after Arg and yields a dipeptide; therefore, this must be the peptide:

Leu–Gly–Ala–Ser–Arg–Phe–Glu

28.23 a.

O

(CH3)2CHCH2 H H2N

C

Leu

OH

C O

(CH3)2CHCH2 H [(CH3)3COCO]2O C OH Boc N C (CH3CH2)3N H O

(CH3)2CHCH2 H Boc N H

C

OH + C O

O H2N

C

C

H2N

C

O

C

C6H5CH2OH, H+

OH

H CH(CH3)2

H CH(CH3)2

C

Val new amide bond

DCC Boc N H

C

C O

O

H N

C

C

OCH2C6H5

H CH(CH3)2 HBr, CH3COOH

(CH3)2CHCH2 H H2N

C

C O

H N

C

OCH2C6H5

H CH(CH3)2

(CH3)2CHCH2 H OCH2C6H5

H2N

O C

C

OH

H CH(CH3)2

Leu–Val

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Chapter 28–10 O CH3 H

b.

C

H2N

Ala

C

CH3 H [(CH3)3COCO]2O

OH

Boc N H

(CH3CH2)3N

O

C

H2N

C

OH

C6H5CH2OH, H+

H2N

H CH(CH3)CH2CH3

O

A

C

OH

O

C

C

OCH2C6H5

CH(CH3)CH2CH3

H

Ile

C

B

new amide bond A

+

CH3 H

DCC

B

Boc N H

C

C O

O H2N

C

C H H

H N

O C

C

CH3 H

H2 Pd-C

OCH2C6H5

Boc N H

C

OH

H2N

C

C

C

C

OH

H CH(CH3)CH2CH3

C

O C6H5CH2OH, H+

C O

H CH(CH3)CH2CH3

O

H N

OCH2C6H5

H H

Gly

new amide bond O

C

+ H2N

C H H

C

CH3 H OCH2C6H5

DCC Boc N H

C

C O

O

H N

C

H

C

N H

C

OCH2C6H5

HBr CH3COOH

O CH(CH3)CH2CH3

CH3 H

Ala–Ile–Gly

H2N

C

C O

H N H

O C

C

N H

C

OH

O CH(CH3)CH2CH3

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Amino Acids and Proteins 28–11 O CH3 H

c.

H2N

C

CH3 H

C

OH [(CH3)3COCO]2O (CH3CH2)3N

Ala O

C

Boc N H

H2N OH

C

C

C H H

O

O C6H5CH2OH, H+

OH

H H

Gly

new amide bond +

DCC

B

Boc N H

CH3 H H2N

C

C

C

CH3 H

O

H N

O

C

C6H5CH2OH,

OH

H2N

+

H2N

C

C

H2 OCH2C6H5

Boc N H

Pd-C

C

C O

H H

O

H N

C

C

OH

H H

C

C

C

OCH2C6H5

new amide bond

C

C

CH3 H

DCC

OCH2C6H5

BOC N H

O

C

C

H N

O

O C

C

CH3 H N H

H H

C

C

H2

OCH2C6H5

Pd-C

O CH3 H Boc N H

C

CH3 H

B

DCC Boc N H

C

C O

H N

O C

C

H H

CH3 H N H

C

C O

H N

C O

new amide bond +

B

O CH3 H

D

OCH2C6H5

CH3 H H+

Ala O

C

C

C

A

CH3 H

A

H2N

O

H N

C

C

C

C

N H

C

OH

O

H H

D

O C

CH3 H

HBr OCH2C6H5

H H

CH3COOH

CH3 H H2N

C

C O

H N

O C

C

H H

CH3 H N H

C

C O

Ala–Gly–Ala–Gly

H N

O C H H

C

OH

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Chapter 28–12 28.24 All Fmoc-protected amino acids are made by the following general reaction: R H C

H2N

C

R H Fmoc–Cl

OH

Na2CO3, H2O

O

H H

The steps:

Fmoc N H

C

OH

C

Fmoc N H

C

C

OH

O

[1] base [2] Cl CH2 POLYMER

O

H H Fmoc N H

C

C

O CH2 POLYMER

O [1]

[2] DCC

N H H Fmoc N

O C

C

OH

H CH(CH3)CH2CH3 (CH3)2CHCH2 H Fmoc N H

C

C

H N

O H

C

O

H H

C

C

N H

C

[1] O CH2

O CH(CH3)CH2CH3

[1] N H CH3 H [2] DCC C OH Fmoc N C H O

N H POLYMER [2] DCC

H Fmoc N H

(CH3)2CHCH2 H Fmoc N H

C

C

H H

O C

C

C N H

C

O CH2 POLYMER

O CH(CH3)CH2CH3

OH

O

[1]

N (CH3)2CHCH2 H H H O O H H [2] HF H Fmoc N C C C C O CH2 POLYMER N C N N C C C H H O O H CH3 H CH(CH3)CH2CH3

(CH3)2CHCH2 O H H O H H H2N C C C C OH N C N N C C C H H O O H CH3 H CH(CH )CH CH 3

2

3

Ala–Leu–Ile–Gly

+ F CH2 POLYMER

28.25 Antiparallel
-pleated sheets are more stable then parallel
-pleated sheets because of geometry. The N–H and C=O of one chain are directly aligned with the N–H and C=O of an adjacent chain in the antiparallel
-pleated sheet, whereas they are not in the parallel
-pleated sheet. This makes the latter set of hydrogen bonds weaker.

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763

Amino Acids and Proteins 28–13 28.26 In a parallel
-pleated sheet, the strands run in the same direction from the N- to C-terminal amino acid. In an antiparallel
-pleated sheet, the strands run in the opposite direction.

H

H

O CH3 H

N

C

C

C

N

H CH3 H

H

C

H N

O

O CH3 H C

C

N

H CH3 H

C

C

HO

OH

O

H CH3 O H H CH3 O H C C N C C N C N C C N C H H O H CH H O H CH 3

H

H

N

N

O CH3 H H O CH3 H C C N C C OH C N C C N C H H CH O H CH3 H 3 O

H

C

O C

H CH3

3

CH3 H

H

C

N

N H

C O

O CH3 H C

C

H CH3

N H

C

C

OH

O

antiparallel

parallel

28.27 a. Ser and Tyr

b. Val and Leu

H2N CH COOH

H2N CH COOH

CH2

CH2

OH

side chains with OH groups

H2N CH COOH

c. 2 Phe residues

H2N CH COOH

CH CH3

CH2

CH3

CH CH3

H2N CH COOH

H2N CH COOH

CH2

CH2

CH3 OH

hydrogen bonding

side chains with only C–C and C–H bonds

van der Waals forces

van der Waals forces

28.28 a. The R group for glycine is a hydrogen. The R groups must be small to allow the
-pleated sheets to stack on top of each other. With large R groups, steric hindrance prevents stacking. b. Silk fibers are water insoluble because most of the polar functional groups are in the interior of the stacked sheets. The
-pleated sheets are stacked one on top of another so few polar functional groups are available for hydrogen bonding to water.

764

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Chapter 28–14 28.29 All L-amino acids except cysteine have the S configuration. L-Cysteine has the R configuration because the R group contains a sulfur atom, which has higher priority. 1 2

The S atom gives the R group a higher priority than COOH, resulting in the R configuration.

1

NH2

NH2

C

HSCH2

3

COOH

H

2

C

R

COOH

H

3

With all other R groups, the COOH has a higher priority than R, giving it the S configuration.

28.30 a.

R COOH

S

H2N C H

COOH

CH3 C SH

COOH

b.

H C NH2

H2N C H

CH3 C SH

CH3

H2N C H

CH3 C S

CH3

(R)-penicillamine

COOH

CH3

(S)-penicillamine

S C CH3 CH3

28.31 Amino acids are insoluble in diethyl ether because amino acids are highly polar; they exist as salts in their neutral form. Diethyl ether is weakly polar, so amino acids are not soluble in it. N-Acetyl amino acids are soluble because they are polar but not salts. NH3+ C

R H

NHCOCH3 R H

COO–

amino acid, a salt H2O soluble and ether insoluble

C COOH

N-acetyl amino acid ether soluble

28.32 The electron pair on the N atom not part of a double bond is delocalized on the five-membered ring, making it less basic. H2N CH COOH CH2

HA

H2N CH COOH CH2

+

NH

NH2

N

N

When this N is protonated... H2N CH COOH CH2 NH N

sp3 hybridized N atom

...the ring is no longer aromatic. H2N CH COOH CH2

HA

preferred path

When this N is protonated...

H2N CH COOH CH2

+

NH

NH

+N H

N H

...the ring is still aromatic.

6
electrons

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Amino Acids and Proteins 28–15 28.33 H2N CH COOH

H2N CH COOH

CH2

CH2

The ring structure on tryptophan is aromatic since each atom contains a p orbital. Protonation of the N atom would disrupt the aromaticity, making this a less favorable reaction.

H2N

HN

+

no p orbital on N This electron pair is delocalized on the bicyclic ring system (giving it 10
electrons), making it less available for donation, and thus less basic.

28.34 At its isoelectric point, each amino acid is neutral. a. +

COO–

H3N C H

b. +

COO–

H3N C H

CH3

CH2CH2SCH3

alanine

methionine

c.

+

COO–

H3N C H CH2COOH

aspartic acid

d.

COO– H2N C H

+

CH2CH2CH2CH2NH3

lysine

28.35 [1] glutamic acid: use the pKa’s 2.10 + 4.07 [2] lysine: use the pKa’s 8.95 + 10.53 [3] arginine: use the pKa’s 9.04 + 12.48 b. In general, the pI of an acidic amino acid is lower than that of a neutral amino acid. c. In general, the pI of a basic amino acid is higher than that of a neutral amino acid.

a.

28.36 a. threonine pI = 5.06 (+1) charge at pH = 1 H3N CH COOH

b. methionine pI = 5.74 (+1) charge at pH = 1 H3N CH COOH

c. aspartic acid pI = 2.98 (+1) charge at pH = 1 H3N CH COOH

d. arginine pI = 5.41 (+2) charge at pH = 1 H3N CH COOH

CH OH

CH2

CH2

CH2

CH3

CH2

COOH

CH2

S

CH2

CH3

NH C NH2 NH2

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Chapter 28–16 28.37 a. valine pI = 6.00 (–1) charge at pH = 11

b. proline pI = 6.30 (–1) charge at pH = 11

c. glutamic acid pI = 3.08 (–2) charge at pH = 11

COO–

H2N CH COO–

H2N CH COO–

CH2

CH2

CH2

CH2

COO–

CH2

H2N CH COO– CH CH3 HN

CH3

d. lysine pI = 9.74 (–1) charge at pH = 11

CH2 NH2

28.38 The terminal NH2 and COOH groups are ionizable functional groups, so they can gain or lose protons in aqueous solution. O

a.

CH3 H

H2N CH C OH CH3

H2N

Ala b. At pH = 1

C

C O

CH3 H H3N

C

C O

H N

O CH3 C

C

H CH3

N H

H N

O CH3 C

C

H CH3

N H

H C

C

OH

A–A–A

O

H C

C

OH

O

c. The pKa of the COOH of the tripeptide is higher than the pKa of the COOH group of alanine, making it less acidic. This occurs because the COOH group in the tripeptide is farther away from the –NH3+ group. The positively charged –NH3+ group stabilizes the negatively charged carboxylate anion of alanine more than the carboxylate anion of the tripeptide because it is so much closer in alanine. The opposite effect is observed with the ionization of the –NH3+ group. In alanine, the –NH3+ is closer to the COO– group, so it is more difficult to lose a proton, resulting in a higher pKa. In the tripeptide, the –NH3+ is farther away from the COO–, so it is less affected by its presence.

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Amino Acids and Proteins 28–17 28.39

H CH2CH(CH3)2 H2N

C

COOH

leucine

C

H2N

g. C6H5COCl, pyridine

C6H5

COOCH3

CH3

C

N H

C

C

OH

C

H2N

N H

C

C O

H CH2CH(CH3)2

h. [(CH3)3COCO]2O

Boc N H

(CH3CH2)3N

C

COOH

i. The product in (d), then NH2CH2COOCH3 + DCC

H CH2CH(CH3)2 H C N COOCH3 AcNH C C

COOCH2C6H5

O H CH2CH(CH3)2

d. Ac2O, pyridine AcNH

e.

C

H3N

C

H2N

C

H CH2CH(CH3)2 H C N COOCH Boc N C C 3 H O H H

COO

H CH2CH(CH3)2

k. Fmoc–Cl, Na2CO3, H2O Fmoc N H O

COOH

H CH2CH(CH3)2

f. NaOH (1 equiv)

H H

j. The product in (h), then NH2CH2COOCH3 + DCC

COOH

H CH2CH(CH3)2

HCl (1 equiv)

OH

O

H CH2CH(CH3)2

c. C6H5CH2OH, H+

C

H CH2CH(CH3)2

O

b. CH3COCl, pyridine

H CH2CH(CH3)2

O

H CH2CH(CH3)2

+ a. CH3OH, H

l. C6H5N=C=S

C6H5 S

N

C

COOH

CH2CH(CH3)2 N H

768

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Chapter 28–18 28.40

H CH2C6H5 H2N

C

COOH

phenylalanine

C

H2N

g. C6H5COCl, pyridine

COOCH3

CH3

C

N H

C

C

OH

C

H2N

h.

N H

[(CH3)3COCO]2O

C

O

(CH3CH2)3N

Boc N H

COOCH2C6H5

C

COOH

i. The product in (d), then NH2CH2COOCH3 + DCC

H CH2C6H5 H C N COOCH3 AcNH C C O

H CH2C6H5

d. Ac2O, pyridine

C

AcNH

C

H3N

f.

H CH2C6H5 H C N COOCH3 Boc N C C H O H H

COOH

H CH2C6H5

k. Fmoc–Cl, Na2CO3, H2O

H2N

C

Fmoc N H O

COOH

l. C6H5N=C=S

H CH2C6H5

NaOH (1 equiv)

C6H5

COO S

28.41 a. (CH3)2CHCH2CHCOOH Br

b. CH3CONHCH(COOEt)2

NH3

(CH3)2CHCH2CHCOO– NH4+ + NH4+Br–

excess

NH3 NH2

[1] NaOEt [2] O C O CH3

HO

CH2 CHCOOH

CH2Br

[3] H3O+,
O

NH2

O

c.

[1] NH4Cl, NaCN H

N H

[2] H3O+

HOOC

O

NH2

O [1] NH4Cl, NaCN

d. CH3O

CHO

e. CH3CONHCH(COOEt)2

[2] H3O+

H H

j. The product in (h), then NH2CH2COOCH3 + DCC

H CH2C6H5

e. HCl (1 equiv)

OH

C

H CH2C6H5

O

H CH2C6H5

c. C6H5CH2OH, H+

C6H5

C

H CH2C6H5

O

b. CH3COCl, pyridine

H CH2C6H5

O

H CH2C6H5

a. CH3OH, H+

COOH

HO

[1] NaOEt [2] ClCH2CH2CH2CH2NHAc [3] H3O+,


NH2 CH2CH2CH2CH2NH2 H2N C COOH H

N

C

COOH

CH2C6H5 N H

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769

Amino Acids and Proteins 28–19 28.42 a. Asn

b. His

O

H2N CH C OH

c. Trp

O

H2N CH C OH

CH2

CH2Br

C O

C O

NH2

NH2

CH2Br

CH2

O CH2Br

H2N CH C OH CH2

NH

NH

HN

N

N

HN

28.43 OH O

H H C N C COOEt

CH3

CHCH3

[1] NaOEt

H2N C COOH

[2] CH3CHO [3] H3O+,


COOEt

H

threonine

28.44 Br2

a. CH3CHO

CH3

C

+

NH3

H3NCH2COO–

excess

glycine

NH2

[1] NH4Cl, NaCN H

BrCH2COOH

H2SO4, H2O

CH3COOH

O

b.

CrO3

BrCH2CHO

CH3 C COOH

[2] H3O+

H

alanine

28.45 O N–K+ CH2(COOEt)2

Br2 Br CH(COOEt)2 CH3COOH

O

O

N

CH(COOEt)2

A B

O [1] NaOEt [2] ClCH2CH2SCH3 O

H H2N C COOH CH2CH2SCH3 D

[1] NaOH, H2O

COOEt N

[2] H3O+,


C COOEt CH2CH2SCH3

O

C

770

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Chapter 28–20 28.46 O

OEt H H C N C COOEt

CH3

COOEt

[1]

O

NaOEt

CH2 CHCOOEt H EtOH C N C COOEt +

CH3

O

[2]

COOEt H C N C COOEt

CH3

COOEt

CH2–CHCOOEt H OH2 [3]

H2N CHCOOH

H3O+

CH2




O

COOEt H C N C COOEt

CH3

CH2COOH

CH2CH2COOEt

glutamic acid + H O 2

28.47 CH3

CH3

COOH

C

R

CH3

H2N

C6H5

C

(R isomer only)

CH3 H

C

+

H3N

C

C6H5

CH3 H

H OH

R

enantiomers

S

proton transfer

COO–

COOH

HO H

H OH

Step [1]: React both enantiomers with the R isomer of the chiral amine.

C

+ CH3

COO– H3N

C

C6H5

diastereomers

CH3 H

HO H

R

C

R

S

These salts have the same configuration around one stereogenic center, but the opposite configuration about the other stereogenic center.

separate

Step [2]: Separate the diastereomers.

CH3

Step [3]: Regenerate lactic acid by protonation.

+

COO– H3N

C

C

H OH

CH3 H

R

R

C6H5

CH3

COO–

C

C

COOH

H OH

R

C

C6H5

CH3 H

HO H

R

S

H3O+ CH3

+

H3N

H3O+ CH3

C

HO H

S

COOH

+

H3N

C

C6H5

CH3 H

The chiral amine is also regenerated as an ammonium salt.

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771

Amino Acids and Proteins 28–21 28.48 H CH3 NH2

C

CH3

COOH

H2N

S To begin: Convert the amino acids into amino acid esters (two enantiomers).

C

enantiomers

COOH

R

CH3OH, H+ CH3

H CH3 H2N

C

S

COOCH3

H C

H2N

R

enantiomers

COOCH3

H OH

Step [1]: React both enantiomers with the R isomer of mandelic acid.

proton transfer C6H5

H OH C6H5

H

C

R

H3N

C

(R)-mandelic acid COOH

CH3

H OH

H CH3

COO–

C

COOCH3

C

C6H5

COO– H3N

R

S

H C

R

COOCH3 diastereomers

These salts have the same configuration around one stereogenic center, but the opposite configuration about the other stereogenic center. separate

Step [2]: Separate the diastereomers.

H OH C6H5

C

R

Step [3]: Regenerate the amino acids by hydrolysis of the esters.

H CH3 –

COO

H3N

C

COOCH3

S

C6H5

C

COO–

H2N

C

S

COOH

H3N

R

H2O, –OH

H CH3

CH3

H OH

H C

R

COOCH3

H2O, –OH

CH3 H2N

H C

R

COOH

+

H OH C6H5

C

COOH

The chiral acid is regenerated.

772

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Chapter 28–22 28.49 NH2

COOH

C

NH2

+

COOH

enantiomers

C6H5CH2 H

H CH2C6H5

R

S

To begin: Convert the amino acids into N-acetyl amino acids (two enantiomers).

C

Ac2O AcNH

COOH

C

AcNH

+

COOH

C

enantiomers

C6H5CH2 H

H CH2C6H5

R

S

N

proton transfer

Step [1]: React both enantiomers with the R isomer of the chiral amine.

CH3O CH3O

N

brucine H

O

O

H

H

N AcNH

C

COO–

H CH2C6H5

S

AcNH

CH3O CH3O

N

COO–

C

C6H5CH2 H

O

CH3O CH3O

R

H

N

N

H

O

O

O

diastereomers separate

Step [2]: Separate the diastereomers. H AcNH

C

N

COO– CH3O

H CH2C6H5

S

H AcNH

N

COO–

C

CH3O

C6H5CH2 H CH3O

N O

Step [3]: Regenerate the amino acid by hydrolysis of the amide.

R

H

CH3O

H2O, –OH

C

COOH

H CH2C6H5

(S)-phenylalanine

H

O

O

NH2

N

O

H2O, –OH NH2

C

N

COOH

C6H5CH2 H

(R)-phenylalanine

The amino acids are now separated.

+

CH3O CH3O

N O

H O

The chiral amine is also regenerated.

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Amino Acids and Proteins 28–23 28.50 NH2

Ac2O

a. (CH3)2CH CH COOH racemic mixture

H2N

acylase

C

AcNH

COOH

H CH(CH3)2

H CH(CH3)2

CH3CONH NHCOCH3

b.

COOH

H2

–OH

chiral Rh catalyst

H2O

H2N

COOH

C

+

COOH

C

CH2CH2CH2CH2NH2

H

(S)-isomer

COOH

c.

NHAc

OH

chiral Rh catalyst

N H

H2N

–

H2

H2O

C

H

COOH

(S)-isomer

CH2

N H

28.51 CH3O

COOH

O CH3

H

N

CH3O

H2

chiral Rh catalyst

CH3

CH3

H N H

O

HO

strong acid

COOH H NH2

HO

O

O

O

COOH

CH3

A

L-dopa

O

28.52 a.

C6H5CH2 H H2N

C

O

H N

C O

C

C

H

c.

H2NCH2CH2CH2CH2 H

OH

H2N

H H H2N

C

C

H N

O

O C

C

OH

H H

Lys–Gly d.

O C

C O

CH3

Phe–Ala b.

C

H N

C

NH2C(NH)NHCH2CH2CH2 H

OH

H2N

C

C O

H CH2

H N

O C

C

OH

CH2

H

R–H

CH2CONH2

NH

Gly–Gln

N

28.53 Amide bonds are bold lines (not wedges). C-terminal HSCH2 O

[1] CH 3 H H2N

C

C O

H N

C

C

H C

N H

C

O H N

C C

OH

O H CH2OH H CH2CH2CONH2

N-terminal Ala–Gln–Cys–Ser A–Q–C–S

C-terminal

[2] HO2CCH2 H H2N

C

C O

(CH3)2CH O H O H H C N C OH C N C C C N H H CH2C6H4OH O H CH2CH2CH2NHC(NH)NH2

N-terminal Asp–Arg–Val–Tyr D–R–V–Y

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Chapter 28–24 28.54 Name a peptide from the N-terminal to the C-terminal end. C-terminal

CH2COOH H H

a.

H2N

C

C

H N

O

C

O

CH2 H

C

C

N H CH H 2 COOH

C

b.

OH

HOOC

C

H N

H CH2

O

H H C O

C

O N H

C

N-terminal C

NH2

H CH3

CH2 CH2

Gly–Asp–Glu G–D–E

NH

Ala–Gly–Arg A–G–R

C HN

NH2

28.55 A peptide C–N bond is stronger than an ester C–O bond because the C–N bond has more double bond character due to resonance. Since N is more basic than O, an amide C–N bond is more stabilized by delocalization of the lone pair on N. O R

C

A

O R

NH2

C

B

+

NH

Structure B contributes greatly to the resonance hybrid and this shortens and strengthens the C–N bond.

28.56 Use the principles from Answer 28.16. O CH3 H H2N

C

C

N

C

H CH3 H

C O

s-trans

O OH

H2N

C

C

N

H

H CH3 C OH C s-cis HCH3 O

28.57 a. b. c. d.

A–P–F + L–K–W + S–G–R–G A–P–F–L–K + W–S–G–R + G A–P–F–L–K–W–S–G–R + G A + P–F–L–K–W–S–G–R–G

28.58 common amino acids

a.

Answer:

Gly–Ala Ala–His His–Tyr

Gly–Ala–His–Tyr

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775

Amino Acids and Proteins 28–25 common amino acids

b.

Answer:

Lys–His His–Gly–Glu

Lys–His–Gly–Glu–Phe

Gly–Glu–Phe

28.59 common amino acids

Arg–Arg–Val

Answer:

Val–Tyr Tyr–Ile–His

Arg–Arg–Val–Tyr–Ile–His–Pro–Phe

Ile–His–Pro–Phe

28.60 Gly is the N-terminal amino acid (from Edman degradation), and Leu is the C-terminal amino acid (from treatment with carboxypeptidase). Partial hydrolysis gives the rest of the sequence. common amino acids Answer:

Gly–Ala–Phe–His

Gly–Gly–Ala–Phe–His–Ile–His–Leu

Phe–His–Ile Ile–His–Leu

28.61 Edman degradation data give the N-terminal amino acid for the octapeptide and all smaller peptides. A

B A: Glu–Arg–Val–Tyr B: Ile–Leu–His–Phe C: Glu–Arg D: Val–Tyr

C D Glu–Arg–Val–Tyr–Ile–Leu–His–Phe

octapeptide

cleavage with trypsin

cleavage with chymotrypsin

carboxypeptidase

Glu–Arg–Val–Tyr–Ile–Leu–His + Phe

28.62 A and B can react to form an amide, or two molecules of B can form an amide. H H Boc N H

C

C O

A

H

O OH + H2N

C

C

DCC OH

H CH(CH3)2

B

Boc N H

H C

C O

H N

(CH3)2CH

O C

+ OH

H CH(CH3)2

H2N

H C

C O

H N

O C

OH

H CH(CH3)2

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Chapter 28–26 28.63 O

a. H2N

C

O

C

CH3OH, H+

OH

H2N

H CH(CH3)2 O

b.

H2N

C

C

C

C

OCH3

H CH(CH3)2 O C6H5CH2OH, H+

OH

H2N

H CH2CH(CH3)2 [(CH3)3COCO]2O

c. NH2CH2COOH

C

C

H Boc N

OCH2C6H5

H CH2CH(CH3)2 O C

(CH3CH2)3N

C

OH

H H

H Boc N

DCC

d. product in (b) + product in (c)

H

O C

C

N H

CH2CH(CH3)2 C

O

H H O

H

e. (CH3)3CO

N

C

C

O

C

OCH2C6H5

C

H H2

OCH2C6H5

(CH3)3CO

Pd-C

H CH(CH3)2

O

N

C O

C

C

OH +

C6H5CH3

H CH(CH3)2

O HBr

f. starting material in (e)

H2N

CH3COOH

C

C

OH

H CH(CH3)2 O

CF3COOH

g. product in (e)

H2N

C

C

OH

H CH(CH3)2 O C

h.

O OH

Na2CO3

+ Fmoc–Cl

C

H2 O

H2N H

OH

HN H Fmoc

28.64 O H H

a.

H2N

C

Gly

C

H H

[(CH3)3COCO]2O OH

(CH3CH2)3N

Boc N H

O

A

DCC Boc N H

C

C O

H N

C

OH

O

A H H

+ B

C

H2N H

C

C

O OH

C

H CH3

H2N

CH3

B H H

OCH2C6H5

HBr CH3COOH

C

C

H CH3

Ala

O C

C6H5CH2OH, H+

H2N

C

C O

H N

O C

C

H CH3

Gly–Ala

OH

OCH2C6H5

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Amino Acids and Proteins 28–27 O

C6H5CH2 H

b.

C

H2N

C6H5CH2 H

[(CH3)3COCO]2O OH

C

(CH3CH2)3N

Boc N H

O

Phe

C

C

DCC

+ B

Boc N H

C

O

O C6H5CH2OH, H+

OH

C

OCH2C6H5

H CH2CH(CH3)2

B C6H5CH2 H

HBr OCH2C6H5

C

C

Leu

O C

H2N

H CH2CH(CH3)2

O

H N

C

C

C

OH

A C6H5CH2 H

A

H2N

CH3COOH

C

H2N

H CH2CH(CH3)2

O

H N

C O

C

C

OH

H CH2CH(CH3)2

Phe–Leu CH3 CH3CH2CH

c.

C

H2N

CH3 H

Ile

[(CH3)3COCO]2O OH

C

(CH3CH2)3N

O

Boc N H

C6H5CH2 H2N

C

C

C

OH

O OH

C6H5CH2OH, H+

H CH3

O

C

Phe

C

H N

C O

O C

C

O

C

H CH3

CH3CH2CH H

H2 Pd-C

OCH2C6H5

C

Boc N H

H CH 3

C6H5CH2 H

OH

C

OCH2C6H5

B

Ala

C

C6H5CH2OH, H+

H2N

C

C O

C DCC

Boc N H

C

H N

O

C

C

C

OH

H CH3

C6H5CH2 H O

CH3CH2CH H C

OCH2C6H5

O

H N

O

CH3 H

H2N

CH3

DCC C

H2N

A

CH3

+ B

C

Boc N H

CH3CH2CH H

A

O

CH3CH2CH H

C

C

N H

C

C

OCH2C6H5

O

H CH3 HBr CH3COOH

CH3 C6H5CH2 H O

CH3CH2CH H H2N

C

C O

H N

C

C

H CH3

Ile–Ala–Phe

N H

C

C O

OH

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Chapter 28–28 28.65 Make all the Fmoc derivatives as described in Problem 28.24. C6H5CH2 H

C6H5CH2 H

[1] base C OH Fmoc N C [2] Cl CH2 POLYMER H

a.

C

Fmoc N H

O

C

O CH2 POLYMER

O

[1]

(Fmoc-Phe)

N H

H Fmoc N

[2] DCC

O C

C

OH

H CH2C6H5 C6H5CH2 H O

(CH3)2CHCH2 H Fmoc N H

C

C O

H N

N H C O CH2 POLYMER N C [2] DCC + H (CH3)2CHCH2 H H CH2C6H5 O C

C6H5CH2 H O

[1]

C

Fmoc N H

[1] N H

C

C

H Fmoc N

(Fmoc-Phe)

C

C O CH2 POLYMER N C H H CH2C6H5 O C

OH

O

(Fmoc-Leu)

[2] DCC CH3 H

+

Fmoc N H

C

C O

OH

(Fmoc-Ala)

(CH3)2CHCH2 C6H5CH2 O H H O H H Fmoc N C C N C C O CH2 POLYMER C N C C N C H H H CH3 O H CH2C6H5 O

[1] N H [2] HF

(CH3)2CHCH2 C6H5CH2 H O H O H H2N C C N C C OH C N C C N C H H H CH3 O H CH2C6H5 O

Ala–Leu–Phe–Phe

+

F CH2 POLYMER

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Amino Acids and Proteins 28–29 CH3

CH3

CH3CH2CH H

b.

C

Fmoc N H

OH

C

CH3CH2CH H

[1] base [2] Cl CH2 POLYMER

C

Fmoc N H

O

C O

(Fmoc-Ile)

O CH2 POLYMER

[1] N H

[2] DCC

CH3CH2CH H O H C N C C O CH2 POLYMER Fmoc N C C N C H H O O H CH3 H H

[1] N H

H Fmoc N

N H

C

C

C

N H

C

H CH3

H H Fmoc N H

[1]

C

C

O C

H H

O

OH

O

(Fmoc-Gly)

C6H5CH2 H

C N C H H O CH2C6H5 C

CH3CH2CH O

H N

C

C

H CH3

N H

CH3 [1]

H C

OH

O CH2 POLYMER

[2] DCC C OH + Fmoc N C H (Fmoc-Phe) O CH3 H Fmoc N

C

(Fmoc-Ala)

CH3CH2CH H O

+

[2] DCC

C

H CH3

CH3

CH3

O

H Fmoc N

C

N H

O CH2 POLYMER

H2N

[2] HF

O

H H

O C

C N C H H CH2C6H5 O C

H N

CH3CH2CH H O C

C

N H

C

H CH3

C

OH

O

Phe–Gly–Ala–Ile

+

F CH2 POLYMER

28.66 An acetyl group on the NH2 forms an amide. Although this amide does block an amino group from reaction, this amide is no different in reactivity than any of the peptide amide bonds. To remove the acetyl group after the peptide bond is formed would require harsh reaction conditions that would also cleave the amide bonds of the peptide. amide R H2N C COOH H

O

R H C N C COOH

CH3

H

N-acetyl amino acid

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Chapter 28–30 28.67 a. A p-nitrophenyl ester activates the carboxy group of the first amino acid to amide formation by converting the OH group into a good leaving group, the p-nitrophenoxide group, which is highly resonance stabilized. In this case the electron-withdrawing NO2 group further stabilizes the leaving group. O

NO2

O

NO2

O

NO2

O

NO2

O

O N+ O

p-nitrophenoxide O N+ O

O

O O

N O

The negative charge is delocalized on the O atom of the NO2 group.

b. The p-methoxyphenyl ester contains an electron-donating OCH3 group, making CH3OC6H4O– a poorer leaving group than NO2C6H4O–, so this ester does not activate the amino acid to amide formation as much. 28.68 R H

a. H2N

C C

= RNH 2

OH

O O

O CH2O

C

O

O

O O

N

CH2O

C

O O N

CH2O

HNR RNH2

O CO3

2–

HO N

O H OCO2–

O

O +

CH2O

O N

O

H

O

C HNR

C

O NHR

CH2O

C

O N

HNR H O

N-hydroxysuccinimide

O

Fmoc-protected amino acid + CO3

2–

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Amino Acids and Proteins 28–31

N H H CH2 O

b.

O

R H

C

C

N H

C

OH

CH2 O

O

R H

C

C

O

N H

R H C

OH

+ CO2 +

DMF

HN

C C

O

OH

O +

+

Fmoc-protected amino acid

N H2

proton transfer R H H2N

C

C

OH

+

N H

O

28.69 Reaction of the OH groups of the Wang resin with the COOH group of the Fmoc-protected amino acids would form esters by Fischer esterification. After the peptide has been synthesized, the esters can be hydrolyzed with aqueous acid or base, but the conditions cannot be too harsh to break the amide bond or cause epimerization.

H Fmoc N CHCOOH

H2SO4

R O

O

O

O

Fischer esterification

Wang resin

OH

OH

O

new ester

C O

new ester

O C O

CH R

CHR

NHFmoc

NHFmoc

28.70 Amino acids commonly found in the interior of a globular protein have nonpolar or weakly polar side chains: isoleucine and phenylalanine. Amino acids commonly found on the surface have COOH, NH2, and other groups that can hydrogen bond to water: aspartic acid, lysine, arginine, and glutamic acid. 28.71 The proline residues on collagen are hydroxylated to increase hydrogen bonding interactions. O N

O [O]

N

OH

The new OH group allows more hydrogen bonding interactions between the chains of the triple helix, thus stabilizing it.

782

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Chapter 28–32 28.72 O C

(CH3)2CHCH2

H

(CH3)2CHCH2

(CH3)2CHCHCHO

CH3COOH

[1] NH4Cl, NaCN

O C

Br

Br2

+ H [2] H3O ,


Br

CrO3 H2SO4, H2O NH2

(CH3)2CHCH2

(CH3)2CHCHCOOH

NH3+

NH3

excess

(CH3)2CHCHCOO–

valine

(Racemic valine and leucine are formed as products, but the synthesis of the tripeptide is drawn with one enantiomer only.)

C COOH H

leucine O H

CH(CH3)2 C

H2N

C

H

[(CH3)3COCO]2O

OH

(CH3CH2)3N

Boc N H

O

valine

CH(CH3)2 C

A

C

H2N

OH

C

C

H

O

O OH

C6H5CH2OH, H+

H2N

CH2CH(CH3)2

C

C

OCH2C6H5

H CH2CH(CH3)2

leucine

B

+

C6H5CH2OH, H H H2N

CH(CH3)2 C

C

OCH2C6H5

C

O

DCC

A + B

H CH(CH3)2 O H C N C C C Boc N OCH2C6H5 H O H CH2CH(CH3)2

H CH(CH3)2 O H C N C C C Boc N OH H O H CH2CH(CH3)2

H CH(CH3)2 H C N C C Boc–NH

H2 Pd-C

O

O C

OH

H CH2CH(CH3)2

H CH(CH3)2 O H CH(CH3)2 H C N C C C C Boc N N COOCH2C6H5 H H O H CH2CH(CH3)2

DCC

C

HBr CH3COOH H CH(CH3)2 H C N C C H2N O

O

H CH(CH3)2 C N COOH H H CH2CH(CH3)2 C

Val–Leu–Val

28.73 Perhaps using a chiral amine R*NH2 (or related chiral nitrogen-containing compound) to make a chiral imine, will now favor formation of one of the amino nitriles in the Strecker synthesis. Hydrolysis of the CN group and removal of R* would then form the amino acid. O R

NR*

NH2R* H

R

H

chiral imine chiral amine

CN

NR* NR* R C H Hydrolyze R C H CN

nitrile

amino nitrile Perhaps a large excess of one stereoisomer will be formed.

COOH

NH2

Remove R*

R C H COOH

amino acid enantiomerically enriched (?)

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Amino Acids and Proteins 28–33 28.74

H OH2

S C6H5

OH

S

R C6H5

N H

N

H

H2O

O

N

H O OH S

R N H

C6H5

N

HO OH

proton R N H

S

transfer C6H5

R N H

N H

thiazolinone H2O OH2 OH N R N S H

O H C6H5 H2O

N

C6H5

R N H

S

O C6H5 S

N

R N H

N-phenylthiohydantoin

H3O

H

proton C6H5 N transfer S

OH OH R N H

HO

OH

S C6H5

N H

N H

R

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785

Lipids 29–1 C Chhaapptteerr 2299:: LLiippiiddss

H Hyyddrroollyyzzaabbllee lliippiiddss [1] Waxes (29.2)—Esters formed from a long-chain alcohol and a long-chain carboxylic acid O R

R, R' = long chains of C's

C

OR'

[2] Triacylglycerols (29.3)—Triesters of glycerol with three fatty acids O O

R O

R, R', R" = alkyl groups with 11–19 C's

O R' O

R" O

[3] Phospholipids (29.4) [a] Phosphatidylethanolamine [b] Phosphatidylcholine (cephalin) (lecithin) O O

[c] Sphingomyelin (CH2)12CH3

O R

O

HO

R

O

O NH

O

O

O R'

O

+

O P O CH2CH2NH3 O–

O

O R' +

O P O CH2CH2N(CH3)3 O–

R, R' = long carbon chain

R, R' = long carbon chain

O

R +

P O CH2CH2NR'3 O–

R = long carbon chain R' = H or CH3



N Noonnhhyyddrroollyyzzaabbllee lliippiiddss [1] Fat-soluble vitamins (29.5)—Vitamins A, D, E, and K [2] Eicosanoids (29.6)—Compounds containing 20 carbons derived from arachidonic acid. There are four types: prostaglandins, thromboxanes, prostacyclins, and leukotrienes. [3] Terpenes (29.7)—Lipids composed of repeating five-carbon units called isoprene units Isoprene unit

Types of terpenes [1] monoterpene 10 C’s

[4] sesterterpene

25 C’s

[2] sesquiterpene

15 C’s

[5] triterpene

30 C’s

[3] diterpene

20 C’s

[6] tetraterpene

40 C’s

C C C C

C

786

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Chapter 29–2 [4] Steroids (29.8)—Tetracyclic lipids composed of three six-membered and one five-membered ring 18

12

19 1 2

A 3

CH3 10

C

9

B

8 7

5 4

CH3

17

11

6

13

D

14

15

16

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787

Lipids 29–3 C Chhaapptteerr 2299:: A Annssw weerrss ttoo PPrroobblleem mss Waxes are esters (RCOOR') formed from a high molecular weight alcohol (R'OH) and a fatty acid (RCOOH).

29.1

a.

O CH3(CH2)29CH2

C

b. O

O O

O(CH2)17 C

CH2(CH2)32CH3

O(CH2)17

C

Eicosapentaenoic acid has 20 C’s and 5 C=C’s. Since an increasing number of double bonds decreases the melting point, eicosapentaenoic acid should have a melting point lower than arachidonic acid; that is, < 4–9°C.

29.2

29.3 O O

HO

O

a.

O

O

OH H2O

O

HO OH

H+

O

HO

OH

O

A

O

b. O

H2 (excess)

c.

H2 (1 equiv)

O

Pd-C

O O

Pd-C O

O

B

two possible products: O

O

O

O O

O

O

O

or

O

O O

O

C

< A C 2 double bonds 1 double bond lowest melting point intermediate melting point

<

C

B 0 double bonds highest melting point

29.4 O CH3(CH2)9CH2

C

OH

lauric acid

Lauric acid is a saturated fatty acid but has only 12 C’s. The carbon chain is much shorter than palmitic acid (16 C’s) and stearic acid (18 C’s), making coconut oil a liquid at room temperature.

788

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Chapter 29–4 29.5 O

O

O

O O

O

O

O

O

O O

29.6

O

A lecithin is a type of phosphoacylglycerol. Two of the hydroxy groups of glycerol are esterified with fatty acids. The third OH group is part of a phosphodiester, which is also bonded to another low molecular weight alcohol. O O

O

one possibility: R

O

O

palmitic acid

(CH2)14CH3 O

O O R'

O

O

+

O P O CH2CH2N(CH3)3

(CH2)7CH=CH(CH2)7CH3

oleic acid

+

O P O CH2CH2N(CH3)3

O–

O–

general structure of a lecithin

29.7

Soaps and phosphoacylglycerols have hydrophilic and hydrophobic components. Both compounds have an ionic “head” that is attracted to polar solvents like H2O. This head is small in size compared to the hydrophobic region, which consists of one or two long hydrocarbon chains. These nonpolar chains consist of only C–C and C–H bonds and exhibit only van der Waals forces.

O

R O O O O–

Na+ – O

The R groups are hydrophobic and nonpolar.

O R

nonpolar, hydrophobic region

polar head

R'

O P O CH2CH2N(CH3)3

29.8

soap

phosphoacylglycerol

O

+

polar head

Phospholipids have a polar (ionic) head and two nonpolar tails. These two regions, which exhibit very different forces of attraction, allow the phospholipids to form a bilayer with a central hydrophobic region that serves as a barrier to agents crossing a cell membrane, while still possessing an ionic head to interact with the aqueous environment inside and outside the cell. Two different regions are needed in the molecule. Triacylglycerols have three polar, uncharged ester groups, but they are not nearly as polar as phospholipids. They do not have an ionic head with nonpolar tails and so they do not form bilayers. They are largely nonpolar C–C and C–H bonds so they are not attracted to an aqueous medium, making them H2O insoluble.

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Lipids 29–5 Fat-soluble vitamins are hydrophobic and therefore are readily stored in the fatty tissues of the body. Water-soluble vitamins, on the other hand, are readily excreted in the urine and large concentrations cannot build up in the body.

29.9

29.10 O

O COOCH3 OH

HO

COOCH3 HO

+

HO

R

S misoprostol diastereomers Only one tetrahedral stereogenic center is different in these two compounds.

29.11 Isoprene units are shown in bold. OH

a.

OH

grandisol

c.

geraniol

OH

b.

camphor

d.

vitamin A

O

29.12

HO

O

manoalide

O HO

O

29.13 OPP

OPP

farnesyl diphosphate

resonance-stabilized carbocation

isopentenyl diphosphate

–

OPP

OPP + H B+

+

OPP H B

790

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Chapter 29–6 29.14 single bond free rotation

OPP

resonance-stabilized carbocation B H

1,2-H shift

+

+

H

+ –OPP

H


-terpinene

resonance-stabilized carbocation

29.15 S a.

O

methyl group at C13

carbonyl at C17

b.

O

R H

methyl group at C10 H O

O

double bond between C4 and C5 carbonyl at C3

29.16 H

H H

H

H H

H H

H

HO

cholesterol equatorial OH

enantiomer H

H H

H

different here

H H

different here

H

HO

H H

HO

H

H

HO

diastereomer

diastereomer

H

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791

Lipids 29–7 29.17 H

H H

H

H H HO

H

O

H H A

H HO

H

O

H B

All four rings are in the same plane. The bulky CH3 groups (arrows) are located above the plane. Epoxide A is favored, because it results from epoxidation below the plane, on the opposite side from the CH3 groups that shield the top of the molecule somewhat to attack by reagents. In B, the epoxide ring is above the plane on the same side as the CH3 groups. Formation of B would require epoxidation of the planar C=C from the less accessible, more sterically hindered side of the double bond. This path is thus disfavored.

792

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Chapter 29–8 29.18 H H H O CH3(CH2)16

C

H

H

O

29.19 Each compound has one tetrahedral stereogenic center (circled), so there are two stereoisomers (two enantiomers) possible. All C=C’s have the Z configuration. O O

(CH2)14CH3 O O

O

O

O O

(CH2)7CH=CH(CH2)7CH3 (CH2)6(CH2CH=CH)2(CH2)4CH3

O

O

(CH2)7CH=CH(CH2)7CH3 O O

O

(CH2)14CH3 O O

(CH2)14CH3 (CH2)6(CH2CH=CH)2(CH2)4CH3

O

(CH2)6(CH2CH=CH)2(CH2)4CH3 (CH2)7CH=CH(CH2)7CH3

O

O

29.20 O O

(CH2)7CH=CH(CH2)7CH3 O O

O

(CH2)14CH3 (CH2)7CH=CH(CH2)7CH3

There is no stereogenic center in this triacylglycerol since these R groups are identical, making this triacylglycerol optically inactive.

O

29.21 O

O

O

O O

O

O

O

O

CHO

O O

M

CH3(CH2)4CHO = O CH2(CHO)2 = P

O

N [1] O3 [2] CH3SCH3

H2, Pd-C O O

The C=C's are assumed to be Z, since that is the naturally occurring configuration.

O O O O L

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793

Lipids 29–9 29.22 When R" = CH2CH2NH3+, the compound is called a phosphatidylethanolamine or cephalin. O O

(CH2)12CH3

(CH2)16CH3 O

HO

O NH

O O

(CH2)16CH3

(CH2)14CH3

O

+

O P O CH2CH2NH3

O

O–

+

P

O CH2CH2NH3

O–

cephalin

sphingomyelin

29.23 a. HO

O

b.

O

c.

O

O

COOH O HO

OH

PGF2


C15 S

[1] (R)-CBS reagent O [2] H2O

O

A

O

X

[1] Zn(BH4)2

B and C diastereomers

O

O

O

O

O O

HO H

O

R

a.

b.

CHO

d. O

lycopene

carvone

e.

c.

-carotene
-pinene

H OH

S

29.24

neral

OH

Use a chiral reducing agent to add hydride from one side only to form a single diastereomer.

[2] H2O O

O

794

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Chapter 29–10 O

O

f.

O

h.

j.

periplanone B

humulene OH

HO


-amyrin

i.

g.

patchouli alcohol

COOH

dextropimaric acid

29.25 A monoterpene contains 10 carbons and two isoprene units; a sesquiterpene contains 15 carbons and three isoprene units, etc. See Table 29.5.

a.

f.

i.

monoterpene CHO

diterpene

sesquiterpene COOH

b.

OH O

g.

monoterpene sesquiterpene j. O

O

c.

HO

h.

O

monoterpene sesquiterpene d. tetraterpene

e. tetraterpene

triterpene

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795

Lipids 29–11 29.26 tail

tail head

tail

head tail

lycopene tail

head

tail

head

tail head

head tail

head

head tail

tail

squalene tail

head

tail

head tail

tail

29.27 B H

+ =

+ –OPP OPP +


-pinene

resonance-stabilized carbocation

29.28 The unusual feature in the cyclization that forms flexibilene is that a 2° carbocation rather than a 3° carbocation is generated. Cyclization at the other end of the C=C would have given a 3° carbocation and formed a 14-membered ring. In addition, the 2° carbocation does not rearrange to form a 3° carbocation.

OPP

OPP

OPP

isopentenyl diphosphate

farnesyl diphosphate

OPP

OPP H

HB

B

=

2° carbocation

B

H

OPP OPP

flexibilene HB

796

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Chapter 29–12 29.29 a.

OH

+ H2O

+ H2O

OH2 H OH2

H H2O

X [1] O3 [2] Zn, H2O

b.

O O –OH,

H2O

CHO O O

O

16,17-dehydroprogesterone

29.30 CH3

CH3

a.

H

b. H

H

H

29.31 H

H

a.

HO HO

H

c. HO

H

OH

H

OH

H

equatorial OH

H

HO

H

H

H

HO

H

d.

b.

HO H

H

H

H

axial OH

H

equatorial OH

axial OH

29.32 H

a. HO

H

OH HO

H

(eq)

H

H

OH (ax)

Axial reacts faster.

H

b.

HO HO

H

OH

(eq)

H

OH

(ax)

Axial reacts faster.

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797

Lipids 29–13 29.33 O OH

a. and b. S

H

H CH3(CH2)5COCl

R O

O

c.

S

H

H

H

H

pyridine

O

H

methenolone

H

Primobolan

29.34 O O

H

H

= H HO

H

H OH

H

H

H

29.35 CH3 groups make this face more sterically hindered. a.

H

b.

H

= H O

H

H O

H

H

H2, Pd-C H

H

The bottom face is more accessible so the H2 is added from this face to form an equatorial OH.

O H H

H

H

H

H2 added from below

798

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Chapter 29–14 29.36 H H H H HO

H

cholesterol H

H

H

H

a. CH3COCl

H

d. oleic acid, H+

H

O

O

H

H

H

O

H

b. H2, Pd-C

CH3(CH2)7CH=CH(CH2)7

O

H

H

H

H H HO c.

H

H H

e. [1] BH3, THF

H

[2] H2O2, –OH

H

PCC

H HO

H

H

H

OH

H H

H

O

29.37 H

1,2-shift OPP + –OPP

resonancestabilized carbocation

1,2-shift

B =

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Lipids 29–15 29.38 Re-draw the starting material in a conformation that suggests the structure of the product. = OH

OH

OH2 H OH2 H2O

H2O

H2 O H2O O H H

H2O

OH H OH2

O H

O H

H

O

H OH2

800

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Chapter 29–16 29.39 (any base) B OPP

H

=

farnesyl diphosphate

OPP + –OPP

resonance-stabilized carbocation CH3

CH3

CH3

H A (any acid)

1,2-H shift H CH3

CH3

A + A

1,2-CH3 shift H

B

+ HB

CH3 CH3

epi-aristolochene

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Synthetic Polymers 30–1 C Chhaapptteerr 3300:: SSyynntthheettiicc PPoollyym meerrss

C Chhaaiinn--ggrroow wtthh ppoollyym meerrss— —A Addddiittiioonn ppoollyym meerrss [1] Chain-growth polymers with alkene starting materials (30.2) •

General reaction: Z

initiator Z

•

Z

Z

Z

Mechanism—three possibilities, depending on the identity of Z: Type

Identity of Z

Initiator

Comments

[1] radical polymerization

Z stabilizes a radical. Z = R, Ph, Cl, etc.

A source of Termination occurs by radicals (ROOR) radical coupling or disproportionation. Chain branching occurs.

[2] cationic polymerization

Z stabilizes a carbocation. Z = R, Ph, OR, etc.

H–A or a Lewis acid (BF3 + H2O)

[3] anionic polymerization

Z stabilizes a carbanion. An Z = Ph, COOR, COR, CN, organolithium etc. reagent (R–Li)

Termination occurs by loss of a proton. Termination occurs only when an acid or other electrophile is added.

[2] Chain-growth polymers with epoxide starting materials (30.3) O

–OH

R

R O

O R

• •

The mechanism is SN2. Ring opening occurs at the less substituted carbon of the epoxide.

801

802

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Chapter 30–2

EExxaam mpplleess ooff sstteepp--ggrroow wtthh ppoollyym meerrss— —C Coonnddeennssaattiioonn ppoollyym meerrss ((3300..66)) Polyamides

Polyesters

H

O

O

O

N

O

O

nylon 6

polyethylene terephthalate O

H O

N

O

N H

O

Polyurethanes

O

H

N

N

O

copolymer of glycolic and lactic acids

Kevlar

H

O

Polycarbonates O

O

O

a polyurethane

O

O

C

O

Lexan



SSttrruuccttuurree aanndd pprrooppeerrttiieess • • • • • •

Polymers prepared from monomers having the general structure CH2=CHZ can be isotactic, syndiotactic, or atactic depending on the identity of Z and the method of preparation (30.4). Ziegler–Natta catalysts form polymers without significant branching. Polymers can be isotactic, syndiotactic, or atactic depending on the catalyst. Polymers prepared from 1,3-dienes have the E or Z configuration depending on the monomer (30.4, 30.5). Most polymers contain ordered crystalline regions and less ordered amorphous regions (30.7). The greater the crystallinity, the harder the polymer. Elastomers are polymers that stretch and can return to their original shape (30.5). Thermoplastics are polymers that can be molded, shaped, and cooled such that the new form is preserved (30.7). Thermosetting polymers are composed of complex networks of covalent bonds so they cannot be melted to form a liquid phase (30.7).

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Synthetic Polymers 30–3 C Chhaapptteerr 3300:: A Annssw weerrss ttoo PPrroobblleem mss Place brackets around the repeating unit that creates the polymer.

30.1

Cl

Cl

Cl

Cl

Cl

poly(vinyl chloride) O

H

O

N O

n

H N

N H

nylon 6,6

O

N H

O

H N

N H

O

n

Draw each polymer formed by chain-growth polymerization.

30.2

Cl

Cl Cl Cl Cl Cl Cl

OCH3 OCH3 OCH3

c.

a. Cl

OCH3 CH3 CH3 CH3 O

OCH3

b.

d.

OCH3

CO2CH3

OCH3 OCH3

Draw each polymer formed by radical polymerization.

30.3

CH3 CH3 CH3 O

a.

CO2CH3

O

CN

O

O C O C O C

b.

CN

O

O

O C O C O C

CN

CN

803

804

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Chapter 30–4 Use Mechanism 30.1 as a model of radical polymerization.

30.4

Initiation: (CH3)3CO

[1]

OC(CH3)3

Propagation:

[3]

O

+

O

O

O

O

(CH3)3CO

[2]

O

2 (CH3)3CO

(CH3)3CO

Repeat Step [3] over and over.

(CH3)3CO

O

O

O

O

O

O

Termination: O

O

O

O

O

O

O

O

[4]

O

OAc OAc O

O

O

(Ac = CH3CO–)

Radical polymerization forms a long chain of polystyrene with phenyl groups bonded to every other carbon. To form branches on this polystyrene chain, a radical on a second polymer chain abstracts a H atom. Abstraction of Ha forms a resonance-stabilized radical A'. The 2° radical B' (without added resonance stabilization) is formed by abstraction of Hb. Abstraction of Ha is favored, therefore, and this radical goes on to form products with 4° C’s (A).

30.5

Ph

3° C Ph Hb RO Ph

Ph

abstraction of Hb Ha Ph

B' Ph

Ph

Ph Ph

Ph

2° radical no resonance stabilization

Ph Ph

abstraction of Ha

Ph

30.6

Ph

Ph

Ph

Ph

A Ph

resonance-stabilized benzylic radical

Ph

Ph

4° C

A' Ph

B

Ph

Ph

favored

Ph

Cationic polymerization proceeds via a carbocation intermediate. Substrates that form more stable 3° carbocations react more readily in these polymerization reactions than substrates that form less stable 1° carbocations. CH2=C(CH3)2 will form a more substituted carbocation than CH2=CH2. 3° carbocation

1° carbocation

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Synthetic Polymers 30–5 Cationic polymerization occurs with alkene monomers having substituents that can stabilize carbocations, such as alkyl groups and other electron-donor groups. Anionic polymerization occurs with alkene monomers having substituents that can stabilize a negative charge, such as COR, COOR, or CN.

30.7

a. CH2=C(CH3)COOCH3

b. CH2=CHCH3

electron-withdrawing group anion polymerization

c. CH2=CHOC(CH3)3

alkyl group cationic polymerization

an electron-donating resonance effect cationic polymerization

d. CH2=CHCOCH3 electron-withdrawing group anion polymerization

Use Mechanism 30.4 as a model of anion polymerization.

30.8

[1]

+

Bu Li

Initiation:

Propagation:

Li+

Bu

CN

CN Repeat Step [2]

[2]

Bu

Bu CN

CN

CN

over and over.

CN O

[3]

Termination:

C

O C O CN

CN

CN

O

CN

Styrene (CH2=CHPh) can by polymerized by all three methods of chain-growth polymerization because a benzene ring can stabilize a radical, a carbocation, and also a carbanion by resonance delocalization.

30.9

*

* *

* = •, +, or –

* *

30.10 Draw the copolymers formed in each reaction. +

Ph

a.

CN Ph

CN n

F

b.

F

F CF3

F

+

F

F

F F CF3 F F

n

805

806

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Chapter 30–6 30.11 RO

2 RO

OR

+

RO

RO

RO

30.12 O

a.

O

HO

O

OH

O

HO

c.

O

O

O

HO

OH

O n

30.13 Cl

Cl

Cl E

Cl

1

Cl

CH2

2

Cl E

1

CH2 H

Two higher priority groups (1's) are on the same side of the double bond - - - > Z configuration.

2

30.14 RO

2 RO

OR

+

RO

RO

RO

RO

A

The resonance-stabilized radical can react at two carbons.

RO

B

Cl E

E configuration of each double bond

neoprene All double bonds have the Z configuration.

OH n

n

b.

O

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Synthetic Polymers 30–7 30.15 O HOOC

+

a.

O

b.

O

COOH

H2N

H

O

O

HO

NH

+

O

NH2

O

NH2

c.

N

N

H2N(CH2)6NH2

H2O

H

H

H

N

N

OH

HO

N

N

H

H O

O

O

30.16 H OH2

O C

+

OH

[1]

C

OCH3

CH3O2C

CH3O2C

H

OCH3

O

OH

[2] CH3O2C

C OCH3 +

O

+ H O 2

H

H2O

HO

OH

[3] OH CH3O2C

H OH2

C OCH3 OCH2CH2OH

OH CH3O2C

C OCH3

[4]

OCH2CH2OH

H H O 2 O+

OH H

H OH2 CH3O2C

+

C OCH3 OCH2CH2OH

CH3O2C

[5]

O CH2CH2OH + CH3OH [6]

O

O

O

O

Repeat for all ester bonds. CH3O2C O CH2CH2OH

O +

H OH2

30.17 O OH

HO O

HO

O

OH

O

O O

This compound is less suitable than either nylon 6,6 or PET for use in consumer products because esters are more easily hydrolyzed than amides, so this polyester is less stable than the polyamide nylon. This polyester has more flexible chains than PET, and this translates into a less strong fiber.

808

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Chapter 30–8 30.18 O C6H5O

C

OH

H A OC6H5

C6H5O

C

OH

OC6H5

OH C6H5O C OC6H5 H A

C6H5O C OC6H5 OH

O

OH

O

H + A A HO

OH A O H

C6H5O

C

OH H

C6H5O C

O

C6H5O C OC6H5 OH

O O

OH

O

OH

+ HA

+ C6H5OH

+ A

Repeat this process for all other CO bonds. O O

O

C

O O

O

+

2 C6H5OH

30.19 O HO

OH

1,4-dihydroxybenzene

+

O

O

O

O

Cl O

epichlorohydrin (excess)

O OH

n

A H2N

NH2

OH

OH O

NH

O O

HN

O

n

OH OH O

O

NH

HN O

O OH

B

n

OH

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Synthetic Polymers 30–9 30.20 HO AlCl3

O

H

O

OH

OH

OH

OH

OH

H

OH

H

AlCl3

+ 3 resonance structures

resonance-stabilized carbocation

AlCl3

OH

HO AlCl3

OH + AlCl3 + H2O

30.21 HO OH OH

OH

R H2C=O

=

R

H+

R

CH2

cardinol

OH

R

30.22 Chemical recycling of HDPE and LDPE is not easily done because these polymers are both long chains of CH2 groups joined together in a linear fashion. Since there are only C–C bonds and no functional groups in the polymer chain, there are no easy methods to convert the polymers to their monomers. This process is readily accomplished only when the polymer backbone contains hydrolyzable functional groups. 30.23 a. Combustion of polyethylene forms CO2 + H2O. b. Combustion of polyethylene terephthalate forms CO2 + H2O. c. These reactions are exothermic. d. HDPE and PET must be separated from poly(vinyl chloride) prior to incineration because combustion of hydrocarbons (like HDPE) and oxygen-containing organics (like PET) releases only CO2 + H2O into the atmosphere. Poly(vinyl chloride) also contains Cl atoms bonded to a hydrocarbon chain. On combustion this forms HCl, which cannot be released directly into the atmosphere, making incineration of halogen-containing polymers more laborious and more expensive. 30.24 OH

a.

O

O OH

O

H

O O

b.

OH

H2N O

N

N H

O

O

810

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Chapter 30–10 30.25 Draw the polymer formed by chain-growth polymerization as in Answer 30.2. COCH3 COCH3 COCH3 F

F

F F

F F

F

O

a.

c.

F

OCH3

CH3O

b.

C CH3CH2O2C

NH

O

d.

C CH2

NH

N H

OCH3 CH2

NH

O

O

CO2CH2CH3 CO2CH2CH3

30.26 Draw the copolymers. a.

CN

and

d. CN

and

n

n

Cl

b.

Cl

and Cl Cl

c.

e. Ph

and

n

n

CN and CN n

30.27 a.

CO2Et CO2Et CO2Et

CO2Et CN

b. CN

c.

CN O

H N

N H

O H2N

OH

O C(CH3)3

d.

and

O

O

O

C(CH3)3

C(CH3)3 O

e.

O

O

O HO

O

OH

or

O

O O

f.

O

O O

O

O O

Cl

Cl

and

HO

OH

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Synthetic Polymers 30–11 30.28 =

a. CO2Et CO2Et

CO2Et

=

b. CN

c.

n

CO2Et

CN

CN

O

H N

N H

n O

H N

=

O

n

C(CH3)3 O

d.

O

C(CH3)3 O

=

n

C(CH3)3

O

e.

O

O

=

O

O

O O

f.

O

n

O O

O

O

=

O

O

O

n

30.29 An isotactic polymer has all Z groups on the same side of the carbon backbone. A syndiotactic polymer has the Z groups alternating from one side of the carbon chain to the other. An atactic polymer has the Z groups oriented randomly along the polymer chain. a.

c.

b. Cl

Cl

Cl

CN

CN

CN

CN

Ph

Ph

Ph

30.30 OH

CCl3

from ethylene oxide

n

30.31 O H2N

a.

NH2

and

HO2C

CO2H

H N

H N O

Ph

Ph

811

812

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Chapter 30–12 H N

O

b. O C N

and

N C O

HO

OH

O

O

O

N H O O O

COCl

c.

and

HO

OH

COCl O O

H N

O

d. N H

O

e.

HO

OH

and Cl

O

Cl

O O

O

f.

HO

COOH O

30.32 CN

Ph

ABS

30.33 H2N

NH2

O

a. N H

+

N H

O

HO

O

Quiana OH

O

ClOC

b.

COCl

H2N

+

NH2

O

O N H

Nomex

N H

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30. Synthetic Polymers

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Synthetic Polymers 30–13 30.34 O H N

O

O

O

H N

N

H

H

N O

O

O

N

H

H

N

N H

Kevlar

O

N H

30.35 O O O

O

O

O

H N

O

O

O

n

polyester A

O

n

nylon 6,6

PET

T g = < 0 oC Tm = 50 oC

Tg = 70 oC Tm = 265 oC

N H n

Tg = 53 oC Tm = 265 oC

a. Polyester A has a lower Tg and Tm than PET because its polymer chain is more flexible. There are no rigid benzene rings so the polymer is less ordered. b. Polyester A has a lower Tg and Tm than nylon 6,6 because the N–H bonds of nylon 6,6 allow chains to hydrogen bond to each other, which makes the polymer more ordered. c. The Tm for Kevlar would be higher than that of nylon 6,6 because in addition to extensive hydrogen bonding between chains, each chain contains rigid benzene rings. This results in a more ordered polymer. 30.36 O

O O

O

O

O

A

dibutyl phthalate

O

O

Diester A is often used as a plasticizer in place of dibutyl phthalate because it has a higher molecular weight, giving it a higher boiling point. A should therefore be less volatile than dibutyl phthalate, so it should evaporate from a polymer less readily.

814

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Chapter 30–14 30.37 Initiation: (CH3)3CO

[1]

OC(CH3)3

[2]

2 (CH3)3CO

(CH3)3CO

+

Propagation: (CH3)3CO

[3]

(CH3)3CO

(CH3)3CO

new C–C bond Repeat Step [3] over and over to form gutta-percha.

Termination:

[4]

30.38 H

F

[1]

O

F3B

H

H

F

[2]

Ph

+

F B O

Ph

+ F3B OH

H

[3]

1,2-H shift

H

a highly resonance-stabilized carbocation Ph Ph

[4]

Ph

+

Repeat Steps [3] and [4].

A

Ph

new C–C bond Ph

Ph

Ph

A is the major product formed due to the 1,2-H shift (Step [3]) that occurs to form a resonance-stabilized carbocation. B is the product that would form without this shift.

Ph

A major product

B

30.39 H F3B O

CN

C N


+

H H F3B O

CN

C

This carbocation is unstable because it is located next to an electron-withdrawing CN group that bears a
+ on its C atom. This carbocation is difficult to form, so CH2=CHCN is only slowly polymerized under cationic conditions.

N

H H

This 2° carbocation is more stable because it is not directly bonded to the electron-withdrawing CN group. As a result, it is more readily formed. Thus, cationic polymerization can occur more readily.

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Synthetic Polymers 30–15 30.40 Initiation:

[1]

+

Bu Li

Li+

Bu

Ph

Ph

Repeat Step [2]

[2]

Propagation: Bu

Bu Ph

Ph

Ph

over and over.

Ph

O [3]

Termination:

O

O C O Ph

Ph

Ph

Ph

30.41 The substituent on styrene determines whether cationic or anionic polymerization is preferred. When the substituent stabilizes a carbocation, cationic polymerization will occur. When the substituent stabilizes a carbanion, anionic polymerization will occur. OCH3

a.

cationic polymerization

NO2

b.

CF3

c.

anionic polymerization

CH2CH3

d.

anionic polymerization

cationic polymerization

30.42 The rate of anionic polymerization depends on the ability of the substituents on the alkene to stabilize an intermediate carbanion: the better a substituent stabilizes a carbanion, the faster anionic polymerization occurs. O

CN OCH3

O

OCH3

O

O

least

most

increasing ability to undergo anionic polymerization

30.43 The reason for this selectivity is explained in Figure 9.9. In the ring opening of an unsymmetrical epoxide under acidic conditions, nucleophilic attack occurs at the carbon atom that is more able to accept a
+ in the transition state; that is, nucleophilic attack occurs at the more substituted carbon. The transition state having a
+ on a C with an electron-donating CH3 group is more stabilized (lower in energy), permitting a faster reaction. [1]

O

O

H OH2

H O

+

HO

H

H

OH

+

[4]

[2]

H2O

HO

O H

H2O

Repeat Steps [4] and [5] over and over.

O

HO

OH

[3] H

+ H2O

[5]

HO

HO

O

OH

OH +

+ H3O+

H3O+

816

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Chapter 30–16 30.44 AlCl3

H ClCH2

C Cl

H ClCH2

H

C Cl AlCl3

H

ClCH2

+ HO

C

H

H

1,4-cyclohexanediol

resonance-stabilized carbocation

A

OH

+ AlCl4–

Cl AlCl3 H H

ClCH2

OH + AlCl3 + HCl

CH2 O

ClCH2

C O H

Repeat. CH2

CH2 O

O n

B

30.45 H N N C O

+ CH3 OH

O

C

H N

CH3

O

C

O

CH3 OH CH3

O H CH3 O H H N

CH3 OH

+

OCH3 O

30.46 (CH3)3CO–OC(CH3)3

a. OCH3

OCH3 OCH3 OCH3 Ziegler–Natta catalyst

CH3 BuLi (initiator)

c. O

BF3 + H2O

d.

CH2 C

CH3 CH2 C

CH3 CH2

C O

a urethane

b.

N

C

CO

CO

CO

CH2CH3

CH2CH3

CH2CH3

O

CH3

OH

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Synthetic Polymers 30–17

O

e.

–OH

O

O

OH

+

f. OH O

OH

O

O

O

Cl

O

O

O

(excess)

n

O

O

g.

–

HO

OH

OH

H2O

O

O

H N

h.

O

O

O

O

O

O

O –OH

NH2

O

H2O

or

NH

H N

i.

OCN

NCO + HO

OH

O

O

N H

O

O

O

j.

HO

Cl2C=O OH

O

O

30.47 Polyethylene bottles are resistant to NaOH because they are hydrocarbons with no reactive sites. Polyester shirts and nylon stockings both contain functional groups. Nylon contains amides and polyester contains esters, two functional groups that are susceptible to hydrolysis with aqueous NaOH. Thus, the polymers are converted to their monomer starting materials, creating a hole in the garment.

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Chapter 30–18 30.48 O HO

CH2

OH

Cl

+

OH

O

O

CH2

CH2

O

O

n

prepolymer H2N

O

O

NH2 OH

OH

OH O

CH2

CH2

O

O

O NH

n

HN

NH

HN O

CH2

O

O

OH

CH2

O

OH

OH

n

30.49 a.

Poly(vinyl alcohol) cannot be prepared from vinyl alcohol because vinyl alcohol is not a stable monomer. It is the enol of acetaldehyde (CH3CHO), and thus it can't be converted to poly(vinyl alcohol).

OH OH

vinyl alcohol

OH

poly(vinyl alcohol) b.

OAc

vinyl acetate

ROOR radical polymerization

–OH

OAc OAc

poly(vinyl acetate)

H2 O hydrolysis

+ CH3CO2– OH

OH

poly(vinyl alcohol)

O C

c. OH

H

H+

OH

O

O

poly(vinyl alcohol)

an acetal poly(vinyl butyral)

30.50 NBS h


Br

–

OH

OH

[1] BH3 HO [2] H2O2, –OH

OH

1,3-propanediol

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30. Synthetic Polymers

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Synthetic Polymers 30–19 30.51 CH3 CH3Cl

CH3Cl

AlCl3

AlCl3

KMnO4 HOOC

CH3

CH3

[1] OsO4 CH2 CH2

HO

[2] NaHSO3, H2O

COOH

terephthalic acid

OH

ethylene glycol or mCPBA

O

H2O, –OH

OH

HO

30.52 OH

OH

OH OH

+ CH2 O

OH

phenol OH

Since phenol has no substituents at any ortho or para position, an extensive network of covalent bonds can join the benzene rings together at all ortho and para positions to the OH groups.

Bakelite OH

CH3

OH

OH

OH

OH

+ CH2 O

p-cresol Since p-cresol has a CH3 group at the para position to the OH group, new bonds can be formed only at two ortho positions so that a less extensive three-dimensional network can form.

30.53 O

a.

O O


-caprolactone

O O

polycaprolactone

O

b.

O O

p-dioxanone

O

O

polydioxanone

819

820

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Chapter 30–20 30.54 O

O

H N

O O

O

O

H N

N H

O

H N

O

O

O

poly(ester amide) A

O

H2N

O

H2N

OH

NH2

OH

HO

O

OH

HO

O

O

the benzyl ester of lysine

leucine

30.55 CO2CH2Ph

O

O

OH

O

O

O

salicylic acid (2 equiv) +

n O

+

O

OH

O

O

benzyl salicylate (2 equiv)

CO2H

PolyAspirin

OH

HO

Cl

Cl

O

sebacic acid

O

sebacoyl chloride

30.56 H H N

O H2N

NH2

N

H2N H

N

N

H

N

N

[1]

NH2

N

O

H2N

proton transfer

N

N

OH H A

[2]

NH2

H N

N

NH2

[3]

melamine A H2N

H N

N N

N

N

N NH2

H2N N

N

H N

NH2

NH2

N N

N NH2

N

H A

H N

N

NH2 H2N

N

[5]

NH2

[6] H N

N

NH2

N

N

NH2

H2N

H H N

N

CH2

N NH2

[4]

N H2O

H N

N

H2N

N NH2

OH2 A

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Synthetic Polymers 30–21 30.57 [1]

a. H

re-draw [2]

intramolecular H abstraction

[3] Repeat Step [2].

b. Abstraction of the H is more facile than abstraction of the other H's because the H atom that is removed is six atoms from the radical. The transition state for this intramolecular reaction is cyclic, and resembles a six-membered ring, the most stable ring size. Other H's are too far away or the transition state would resemble a smaller, less stable ring.

n

butyl substituent

30.58 O O H2N

O

O NH2

H

H2N

H

O N H

OH

H2N

H2N

O

NH2 H2N

O N H

N H

formaldehyde

urea

O

N

N

N

H2N

O

N O

O N

H N

N O

N

O NH2

repeat

H2N

O N

N

N

NH N

NH2

repeat O

N

N

N H

CH2

N HN

O

O

N H NH

NH2

O

NH2

NH2