question_answer

The number of possible enantiomeric pairs that can be produced during monochlorination of 2-methylbutane is:

A) 3                                 

B) 4

C) 1                                 

D) 2

Correct Answer: D

Solution :

2-methylbutane has 4 types of hydrogen atoms.

\[{{\operatorname{CH}}_{3}}-\underset{{}}{\overset{\overset{{}}{\mathop \underset{|}{\mathop C{{H}_{3}}}\,}\,}{\mathop{CH}}}\,-C{{H}_{2}}C{{H}_{3}}\]

Hence, on monochlorination, it gives 4 product.

(i)

It will exist in from of enantiomeric pair.

(ii)

It has no chiral carbon, so no enaniomerism possible.

(iii)

It has no chiral carbon atom, so no enantiomerism.