6.40E
Question
Answer questions 6-39 taking all stereoisomers into account.
(a)
(b)
(c)
Step-by-Step Solution
Verified(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
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:
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:
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:
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
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
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