Question

Answer questions 6-39 taking all stereoisomers into account.

(a)

(b)

(c)

Short answer

(a)

(S)-2-chlorobutane

(R)-2-chlorobutane

1-chlorobutane

(b)

(R)-1-chloro-2-methylbutane

(S)-1-chloro-2-methylbutane

2-chloro-2-methylbutane

(R)-2-chloro-3-methylbutane

(S)-2-chloro-3-methylbutane

1-chloro-3methylbutane

(c)

(chloromethyl)cyclopentane

1-chloro-1-methylcyclopentane

(1R)-1-chloro-2-methylcyclopentane

(1S)-1-chlro-2-methylcyclopentane

(1R)-1-chloro-3-methylccylopentane

(1S)-1-chloro-3-methylcyclopentane

Step-by-step solution

n-Butane

In n-butane, there are two types of hydrogen. One type is attached with terminal carbon and the second type is attached with non-terminal carbon. Therefore, without considering the stereochemistry for radical chlorination only two monochlorinated products are possible which are as shown:

Isopentane

In isopentane, there are four types of hydrogen. Therefore, without considering the stereochemistry for radical chlorination four monochlorinated products are possible which are as follows:

Given compound

In the given compound there are four types of hydrogen. Therefore, without considering the stereochemistry for radical chlorination four types of monochlorinated products are possible as shown:

a) Stereochemistry

If we consider stereochemistry, compound A is optically active as the marked carbon is chiral. Therefore, two possible isomers are there. Compound B is optically inactive. Thus, considering the stereochemistry for radical chlorination total of three monochlorinated products are possible which are as shown:

(S)-2-chlorobutane

(R)-2-chlorobutane

1-chlorobutane

b) Stereochemistry

If we consider the stereochemistry, compounds A and C are optically active as the marked carbons are chiral. Therefore, two possible isomers are there for each of A and C. Compounds B and D are optically inactive. Therefore, considering the stereochemistry for radical chlorination total of six monochlorinated products are possible as shown:

(R)-1-chloro-2-methylbutane

(S)-1-chloro-2-methylbutane

2-chloro-2-methylbutane

(R)-2-chloro-3-methylbutane

(S)-2-chloro-3-methylbutane

1-chloro-3methylbutane

c) Stereochemistry

If we consider the stereochemistry, compounds C and D are optically active as the marked carbons are chiral. Therefore, two possible isomers are there for each of C and D. Compounds A and B are optically inactive. Therefore, considering the stereochemistry for radical chlorination total of six monochlorinated products are possible which are as shown:

(chloromethyl)cyclopentane

1-chloro-1-methylcyclopentane

(1R)-1-chloro-2-methylcyclopentane

(1S)-1-chlro-2-methylcyclopentane

(1R)-1-chloro-3-methylccylopentane

(1S)-1-chloro-3-methylcyclopentane