Q 24P
Question
Draw and name all of the monochlorination products that you might obtain from the radical chlorination of the compounds below. Which of the products is chiral? Are any of the products optically active?
(a) 2-methylbutane
(b) Methylcyclopropane
(c) 2,2-dimethylpentane
Step-by-Step Solution
Verifieda)
Chlorination of 2-methyl butane
b)
Chlorination of methyl cyclopropane
c)
Chlorination of 2,2-dimethylpentane
A chiral molecule is not superimposable on its mirror image, and a chiral molecule has at least one carbon atom, which is bonded to four different groups.
The achiral molecule is superimposable on its mirror image, and an achiral molecule does not contain a carbon atom bonded to four different groups.
Chlorination of 2-methyl butane in the presence of light gives the possible four mono chlorination products. Thus, the reaction is as follows:
Chlorination of 2-methyl butane
Here, the compound 2-chloro-3-methyl butane is a chiral molecule as it has a carbon atom that is bonded with four different groups. Therefore, due to the presence of a chiral center, this molecule is also optically active.
Chlorination of methyl cyclopropane in the presence of light gives the possible three products given as follows:
Chlorination of methyl cyclopropane
All the formed compounds are achiral as they do not contain carbon atom which is bonded to four different groups. Therefore, all the formed compounds are optically inactive.
Chlorination of 2,2-dimethylpentane in the presence of light gives the possible 4 monochlorinated products as follows:
Chlorination of 2,2-dimethylpentane
Here, the compounds 3-chloro-2,2-dimethylpentane and 4-chloro-2,2-dimethylpentane are chiral molecules because they contain chiral carbon atoms. Due to the presence of a chiral center, those compounds are also optically active.