Question

$C_6{H}_{5}Cl\underset{{Cu}_{2}O}{\overset{{NH}_3}{\to }}A\underset{Alc.KOH}{\overset{{CHCl}_3}{\to }}B$.

In this reaction $B$ is:

• A. $C_6{H}_{5}CN$
• B. $C_6{H}_5{CH}_{2}CN$
• C. $C_6{H}_5{CH}_{2}NC$
• D. $C_6{H}_{5}NC$

Answer 1

The correct option is D $C_6{H}_{5}NC$

In the first reaction, nucleophilic aromatic substitution is occurring. It occurs under extreme conditions (high temperature and high pressure).

In this, the nucleophile (ammonia here) replaces the halogen from the benzene ring.

$C_6{H}_{5}Cl+N{H}_3+C{u}_{2}O$ (high pressure and temperature) $\to C_6{H}_{5}N{H}_2$ (aniline)
now the second reaction is the carbylamine reaction

$RN{H}_2+CHC{l}_3+KOH\to RNC$

Here, $R$ = phenyl group = $Ph$

So, $C_6{H}_{5}N{H}_2+CHC{l}_3+KOH\to C_6{H}_{5}NC$