Chapter 7

Substitution Reactions

Review of Concepts Fill in the blanks below. To verify that your answers are correct, look in your textbook at

the end of Chapter 7. Each of the sentences below appears verbatim in the section

entitled Review of Concepts and Vocabulary.

• Substitution reactions exchange one _____________ for another.

• Evidence for the concerted mechanism, called SN2, includes the observation of a

__________-order rate equation. The reaction proceeds with ____________ of

configuration.

• SN2 reactions are said to be _________________ because the configuration of the

product is determined by the configuration of the substrate.

• Evidence for the stepwise mechanism, called SN1, includes the observation of a

__________-order rate equation.

• The __________ step of an SN1 process is the rate-determining step.

• An SN1 reaction is a stepwise process with a first-order rate equation.

• There are four factors that impact the competition between the SN2 mechanism

and SN1: 1) the _____________, 2) the ________________, 3) the ___________

_______________, and 4) the _______________.

• ______________________ solvents favor SN2.

Review of Skills Follow the instructions below. To verify that your answers are correct, look in your

textbook at the end of Chapter 7. The answers appear in the section entitled SkillBuilder

Review.

SkillBuilder 7.1 Drawing the Curved Arrows of a Substitution Reaction

LG Nuc LG NucNuc

DRAW CURVED ARROWS, SHOWING NUCLEOPHILIC ATTACK ACCOMPANIED BY SIMULTANEOUS LOSS OF A LEAVING GROUP

A CONCERTED MECHANISM

DRAW A CURVED ARROW SHOWING THE LOSS OF THE LEAVING GROUP TO FORM A CARBOCATION INTERMEDIATE, FOLLOWED BY ANOTHER CURVED ARROW SHOWING THE NUCLEOPHILIC ATTACK

A STEPWISE MECHANISM

- LGNuc- LG

SkillBuilder 7.2 Drawing the Product of an SN2 Process

Br

+ OH + Br

DRAW THE MAJOR PRODUCT OF THE FOLLOWING REACTION

CHAPTER 7 115

SkillBuilder 7.3 Drawing the Transition State of an SN2 Process

SHNaSH

Cl

DRAW THE TRANSITION STATE OF THE FOLLOWING REACTION

TRANSITION STATE

SkillBuilder 7.4 Drawing the Carbocation Intermediate of an SN1 Process

Cl

DRAW THE CARBOCATION THAT WOULD BE FORMED IF A CHLORIDE ION IS EXPELLED FROM THE FOLLOWING COMPOUND

- Cl

SkillBuilder 7.5 Drawing the Products of an SN1 Process

Br

NaCN

PREDICT THE PRODUCTS OF THE FOLLOWING SN1 REACTION

+

SkillBuilder 7.6 Drawing the Complete Mechanism of an SN1 Process

IDENTIFY THE TWO CORE STEPS AND THREE POSSIBLE ADDITIONAL STEPS OF AN SN1 PROCESS

TWO CORE STEPS

THREE POSSIBLE ADDITIONAL STEPS

SkillBuilder 7.7 Drawing the Complete Mechanism of an SN2 Process

IDENTIFY THE ONE CORE STEP (CONCERTED) AND TWO POSSIBLE ADDITIONAL STEPS OF AN SN2 PROCESS

CORE STEP

TWO POSSIBLE ADDITIONAL STEPS

116 CHAPTER 7

SkillBuilder 7.8 Determining whether a Reaction Proceeds via an SN1 Mechanism or an SN2

Mechanism

SN2

NUC

SUBSTRATE

SN1

LG

SOLVENT

FILL IN THE TABLE BELOW, SHOWING THE FEATURES

THAT FAVOR SN2 OR SN1 REACTIONS

SkillBuilder 7.9 Identifying the Reagents Necessary for a Substitution Reaction

OH CN

IDENTIFY THE REAGENTS NECESSARY TO ACHIEVE THE FOLLOWING TRANSFORMATION

1)

2)

Review of Reactions Follow the instructions below. To verify that your answers are correct, look in your

textbook at the end of Chapter 7. The answers appear in the section entitled Review of

Reactions.

H

HH LG

Nuc H

HHNuc LG

SN2

+DRAW THE CURVED ARROWS THAT SHOW

THE FLOW OF ELECTRON DENSITY DURING

THE FOLLOWING SN2 REACTION

LG Nuc

SN1

DRAW THE CURVED ARROWS THAT SHOW

THE FLOW OF ELECTRON DENSITY DURING

THE FOLLOWING SN1 REACTION

Nuc- LG

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 + IOMe

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.

Br

H3C

HF

OMe

S

MeO

CH3

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)

H3C

N

H3C OHR

SR

CH3

CH3

N

CH3 OH

H3C

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 Br

OH H

Br

H BrBr

+

c)

Br

OH H

O

H

H

OH

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

H

EtOH

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)

TsO

Cl

Br

I

NH2

OMe

F

Cl

b)

TsO

Cl

Br

I

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)

I

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)

Br

O2

1

3

4

HR

b)

H

O

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 Br

Br

− 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 MeOH OMe

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) I

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)

OH1) 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