CHAPTER 7 117

Solutions 7.1. a) 4-chloro-4-ethylheptane b) 1-bromo-1-methylcyclohexane c) 4,4-dibromo-1-chloropentane d) (S)-5-fluoro-2,2-dimethylhexane 7.2.

a)

Br

SH

SH+ Br

b)

I OMe + IO Me

7.3.

a)

Br O

O

O

O

+ Br

b)

I ClΙCl

+

7.4.

OBr

O

Br+

118 CHAPTER 7

7.5.

Br

Cl

Br

Cl

+

7.6. a) the rate of the reaction is tripled. b) the rate of the reaction is doubled. c) the rate of the reaction will be six times faster. 7.7.

a)

SH

b) Cl c)

OH

7.8.

BrH3C

HF

OMe

S

MeOCH3

HF

S

14

12

3

2

3

4

The reaction does proceed with inversion of configuration. However, the Cahn-Ingold-Prelog system for assigning a stereodescriptor (R or S) is based on a prioritization scheme. Specifically, the four groups connected to a chirality center are ranked (one through four). In the reactant (above left), the highest priority group is the leaving group (bromide) which is then replaced by a group that does not receive the highest priority. In the product, the fluorine atom has been promoted to the highest priority as a result of the reaction, and as such, the prioritization scheme has changed. In this way, the stereodescriptor (S) remains unchanged, despite the fact that chirality center undergoes inversion.

CHAPTER 7 119

7.9.

a) b)

c) d)

7.10.

7.11.

BrOδ+

HH δ+

Being formed

Beingbroken

This step is favorable (downhill in energy) because ring strain is alleviated when the three-membered ring is opened. 7.12.

a)

N

N

H

H

RS

R

CH3

N

N

H

CH3HN

N

H

CH3

− H+

Nicotine

120 CHAPTER 7

b)

H3CN

H3C OHR

SR

CH3

CH3

N

CH3 OHH3C

choline 7.13. a) The rate of the reaction will be doubled, because the change in concentration of sodium chloride will not affect the rate. b) The rate of the reaction will remain the same, because the change in concentration of sodium chloride will not affect the rate. 7.14. Draw the carbocation intermediate generated by each of the following substrates in an SN1 reaction:

(a)

(b)

(c)

(d)

7.15.

Br

The first compound will generate a tertiary carbocation, while the second compound will generate a tertiary benzylic carbocation that is resonance stabilized. The second compound leads to a more stable carbocation, so that compound will lose its leaving group more rapidly than the first compound. 7.16.

a) + NaΙ

Cl

b)

SH+

HS+ Br

c)

O

O

+

O

O

+ Cl

d)

ClΙ+

CHAPTER 7 121

7.17.

SH

+HS

Diastereomers 7.18. a) No b) Yes c) No d) Yes e) Yes f) No 7.19. a) No b) Yes c) Yes d) Yes e) No f) No g) No h) Yes i) No j) No k) Yes l) No 7.20. a) No b) Yes c) Yes d) No e) No f) No 7.21. a)

O H

H Br

O

H

HBr

Br+

b)

OH

H BrO

H H

Br

H BrBr

+

c)

Br

OH H

O

H

HOH

OH H

122 CHAPTER 7

d)

ΙEtOH

O

H

Et OEtEtOH

e)

OH H

H

MeOH

OHO

H

Me

O H

H

MeOH

OMe

f)

OMe

H

MeOH

MeOHMeO

OHH

OH H

H

OH

g)

Br

S H

SH h)

Ι

EtOH

O

Et

HEtOH

OEt

CHAPTER 7 123

7.22.

c)

LG Nuc Attack H+- -

d)

C+LG Nuc Attack H+- rearr. -

e)

LG Nuc Attack H+- -+ H+

f)

C+LG Nuc Attack H+- rearr. -

+ H+

g) LG Nuc Attack-

h)

LG Nuc Attack H+- -

Problem 7.20c and 7.20h exhibit the same pattern. Both problems are characterized by three mechanistic steps: 1) loss of a leaving group, 2) nucleophilic attack, and 3) proton transfer. 7.23.

H

HO

H

O H

H

OHH

OH H

H

OH

OH

HO

H

The chirality center at C2 is lost when the leaving group leaves to form a carbocation with trigonal planar geometry. The chirality center at C3 is lost during the hydride shift in the following step. Once again, the chirality center is converted into a trigonal planar sp2 hybridized center (which is no longer a chirality center).

124 CHAPTER 7

7.24. a)

ClMeOH

O

H

Me MeOHOMe

b)

BrEtOH

O

H

Et EtOHO

c)

ΙH

OH

OH

H

HO

HOH

d)

Br

OH

O

H

OHO

7.25.

N

H

HH

Me Ι

MeΙN

Me

Me

Me

Me

N

H

H

H

Me

N

Me

Me Me

NH3

H3NN

H

Me

Me

Me

N

H

MeH

Me Ι

MeΙ

NH3

N

H

H

Me

Me

N

H

Me Me

7.26. a) SN1 b) SN2 c) Neither d) SN1 e) Both f) Neither g) Both

CHAPTER 7 125

7.27. a) SN1 b) SN2 c) SN2 d) SN2 e) SN2 7.28.

a)

TsOCl

Br

NH2

OMe

F

Cl

b)

TsOCl

Br

NH2

OMe

F

Cl

7.29. a) SN1 b) SN2 c) SN1 d) SN2 e) SN1 f) SN2 g) SN2 h) SN1 7.30. Acetone is a polar aprotic solvent and will favor SN2 by raising the energy of the nucleophile, giving a smaller Ea. 7.31.

a)

MeOH

OMe

SN1

b)

BrCl

HMPA

Cl

SN2

c)

O H H Br BrSN1

Racemic

d) OTs

NaCN

DMFCN

SN2

126 CHAPTER 7

e)

IH2O OH SN1

Racemic

f)

BrNaCN

DMSOCN SN2

7.32. No. Preparation of this amine via the Gabriel synthesis would require the use of a tertiary alkyl halide, which will not undergo an SN2 process. 7.33.

(a)

INaOH

OH

(b)

BrOH

HBr

(c)

IOHHI

(d)

Br NaSH

DMSO

SH

(e)

Br

O

O

DMSO O

O

(f)

OH

Br

1) TsCl, pyridine

2) NaBr, DMSO

(g)

I OO

(h)

Br OH

H2O

(i) OH CN1) TsCl, pyridine

2) NaCN

7.34.

OH SH

(R)-2-butanol

(R)-2-butanethiol

1) TsCl, pyridine

2) NaI, DMSO3) NaSH, DMSO

CHAPTER 7 127

7.35.

N

Cl

O

HO

NH2 Nuc

N

Cl

O

HO

NH2

N

NucO

HO

NH2

N

Nuc

O

HO

NH2 Nuc

N

Cl

ClO

HO

NH2

melphalan

Nuc

Nuc

7.36. a) Systematic Name = 2-chloropropane Common Name = isopropyl chloride b) Systematic Name = 2-bromo-2-methylpropane Common Name = tert-butyl bromide c) Systematic Name = 1-iodopropane Common Name = propyl iodide d) Systematic Name = 2-chlorobutane Common Name = propyl iodide d) Systematic Name = (R)-2-bromobutane Common Name = (R)-sec-butyl bromide e) Systematic Name = 1-chloro-2,2-dimethylpropane Common Name = neopentyl chloride f) Systematic Name = chlorocyclohexane Common Name = cyclohexyl chloride

128 CHAPTER 7

7.37.

Ι

Ι

Ι Ι

Increasing reactivity (SN2)

7.38.

a)

Cl

Cl

secondary primary b)

Br Br

primarymore sterically

hindered

primaryless sterically

hindered

c)

Cl Cl

secondary tertiary d)

Br I

betterleaving group

7.39. No. Preparation of this compound via the process above would require the use of a tertiary alkyl halide, which will not undergo an SN2 process. 7.40. a) NaSH b) sodium hydroxide c) methoxide dissolved in DMSO 7.41.

a)

Cl

Cl

tertiary primary b)

Br

Br

primary tertiary

c)

Cl Cl

allylic d)

Cl OTs

betterleaving group

CHAPTER 7 129

7.42. a) The rate of the reaction is doubled.

b) The rate of the reaction is doubled. 7.43. a) The rate of the reaction is doubled b) The rate of the reaction will remain the same. 7.44. a) aprotic b) protic c) aprotic d) protic e) protic 7.45.

a)

BrO

2

1

3

4H

R

b)

HO

NCR

13

2 4

c) The reaction is an SN2 process, and it does proceed with inversion of configuration. However, the prioritization scheme changes when bromide (#1) is replaced with a cyano group (#2). As a result, the Cahn-Ingold-Prelog system assigns the same configuration to the reactant and the product. 7.46.

ΙO

OMe

H H

δ− δ−

7.47. Iodide functions as a nucleophile and attacks (S)-2-iodopentane, displacing iodide as a leaving group. The reaction is an SN2 process, and therefore proceeds via inversion of configuration. The product is (R)-2-iodopentane. The reaction continues until a racemic mixture is obtained.

130 CHAPTER 7

7.48.

O H

H BrO

H

H BrBr

− H2O

Racemic The chirality center is lost when the leaving group leaves to form a carbocation with trigonal planar geometry. The nucleophile can then attack either face of the planar carbocation, leading to a racemic mixture. 7.49.

OH

H O S

O

O

OH

OH

OH H

OS

O

O

HO

HO

H

OH H

E

Reaction coordinate

Racemic

7.50.

Increasing stability

7.51.

secondary tertiary primary secondary

CHAPTER 7 131

7.52.

OH H Cl O H

H

Cl

Cl

H

7.53.

Br

O

O

O

O

Br+

7.54. a)

ClMeOH O

H

Me MeOHOMe

3 Steps b)

ClSH

SH

2 Steps c)

OH H Ι O

H

H ΙΙ− H2O

3 Steps

132 CHAPTER 7

d)

O

H

Et

EtOH

EtOH

H

OEt

OTs

4 Steps 7.55.

a)

Br EtOH OEt

b)

NaBr

OTs Br

c)

OH HCl

Cl

d)

NaCN

DMSO CN 7.56.

O

O

7.57.

BrO H

OH

Although the substrate is primary, it is still sterically hindered. As a result, SN2 reactions at neopentyl halides do not occur at an appreciable rate.

CHAPTER 7 133

7.58. a)

BrH

OH

O

H

HOH

HO

H

b) The substrate is primary, and therefore, the reaction must proceed via an SN2 process. SN2 reactions are highly sensitive to the strength of the nucleophile, and the nucleophile (water) is a weak nucleophile. As a result, the reaction occurs slowly.

c) Br

O HOH Br+

Hydroxide is a strong nucleophile, which favors the SN2 process. 7.59.

a) OTs OHNaOH

b)

OH CN1) TsCl, py

2) NaCN

c) OH BrHBr

d)

Cl SHNaSH

e)

Br O

O

NaO

O

134 CHAPTER 7

7.60.

a) Ι + OH

b)

ΙO

O

+

c)

ΙCN+

d) Ι+ SH

e)

Ι+ H2O

f)

Ι+ H2S

7.61.

a)

SH

b)

SEt

c)

CN

7.62. The second method is more efficient because the alkyl halide (methyl iodide) is not sterically hindered. The first method is not efficient because it employs a tertiary alkyl halide, and SN2 reactions do not occur at tertiary substrates. 7.63.

a)

OH 1) TsCl, pyridine

2) NaBr

Br

b) OH Cl

HCl

c) ClNaOH

OH

CHAPTER 7 135

7.64.

a) BrS H

SH Br+

b) Br NaSHRate = k

c) The rate would be slower.

d)

E

Reaction coordinate

e)

BrHS

Et

H H

δ− δ−

7.65. a) SN1 (tertiary substrate)

b)

O HH Br

O H

H

Br Br

c) OHRate = k

d) No. The rate is not dependent on the concentration or strength of the nucleophile.

e)

E

Reaction coordinate

136 CHAPTER 7

7.66. a) SN2

b)

BrCN

CN

Br+

c)

BrNaCNRate = k

d) Yes. The reaction rate would double.

e)

E

Reaction coordinate 7.67.

H

I

OHOH H

HO

H

HO

H

CHAPTER 7 137

7.68.

a) O

O I

O

O

I+

b) This reaction occurs via an SN2 process. As such, the rate of the reaction is highly sensitive to the nature of the substrate. The reaction will be faster in this case, because the methyl ester is less sterically hindered than the ethyl ester. 7.69.

O

Br

HO

Br

OBase

7.70.

Br

Ι

Ι

EtOH

OEt

O

H

Et

EtOH

7.71. When the leaving group leaves, the carbocation formed is resonance stabilized:

O OTs O O

Resonance stabilized 7.72. Iodide is a very good nucleophile (because it is polarizable), and it is also a very good leaving group (because it can stabilize the negative charge by spreading the charge over a large volume of space). As such, iodide will function as a nucleophile to displace the chloride ion. Once installed, the iodide group is a better leaving group than chloride, thereby increasing the rate of the reaction.

138 CHAPTER 7

7.73.

OH

HO

H

H

HO

H

OH H

HO

H

OH

OHH

OH