Question

In the following sequence of reactions,

$C{H}_{3}Br\stackrel{KCN}{\to }A\stackrel{H_3{O}^{+}}{\to }B\stackrel{LiAl{H}_4}{\to }C$

The end product C is:

Answer 1

The end product C is ethyl alcohol.
$C{H}_{3}Br\stackrel{KCN}{\to }\underset{A}{C{H}_3-CN}\stackrel{H_3{O}^{+}}{\to }\underset{B}{C{H}_3-COOH}\stackrel{LiAl{H}_4}{\to }\underset{C}{C{H}_3-C{H}_2-OH}$
Methyl bromide reacts with KCN to form acetonitrile. Br atom is replaced with CN group. Hydrolysis of cyano group gives carboxylic group. Reduction of carboxylic group gives hydroxyl group.