Question

Which one of the following exhibits enantiomerism?

• A. $BrC{H}_2-C{H}_2-C{H}_2-C{H}_{2}Br$
• B. $H_{3}C-\underset{\underset{Br}{|}}{C}H-C{H}_2-C{H}_3$
• C. $H_{3}C-C{H}_2-C{H}_2-C{H}_{2}Br$
• D. $H_{3}C-C{H}_2-C\left(B{r}_2\right)-C{H}_3$

Answer 1

Answer: (B)

$H_{3}C-\underset{\underset{Br}{|}}{C}H-C{H}_2-C{H}_3$

Here, B compound is the enantiomer.

An enantiomer, also known as an optical isomer, is one of two stereoisomers that are mirror images of each other that are non-superimposable (not identical), much as one's left and right hands are the same except for being reversed along one axis (the hands cannot be made to appear identical simply by reorientation).

Organic compounds that contain a chiral carbon usually have two non-superposable structures. These two structures are mirror images of each other and are, thus, commonly called enantiomorphs (enantio = opposite ; morph = form), hence this structural property is now commonly referred to as enantiomerism.

So as only option B have chiral carbon centre, hence it will show enantiomerism.