Question

Find A,B,C,D?

Answer 1

In this reaction takes place when the $NaN{O_2}$ is added as A the following reaction takes place.
In place of B $PhC{H_2}N{H_2}$ is formed hydrochloric acid reacts to it. Whereas in place of C we will have $PhC{H_2}CON{H_2}$ which on further reaction with D that is none other than $CHC{l_3}/KOH$

Complete step-by-step answer: In this reaction takes place when the $NaN{O_2}$ is added as A the following reaction takes place.
In place of B $PhC{H_2}N{H_2}$ is formed hydrochloric acid reacts to it. Whereas in place of C we will have $PhC{H_2}CON{H_2}$ which on further reaction with D that is none other than $CHC{l_3}/KOH$
Where $NaN{O_2}$ is none other than sodium nitrite. Further we have seen that B is $PhC{H_2}N{H_2}$ that is benzylamine. Then, C is $PhC{H_2}CON{H_2}$ and d is what we call chloroform In our day to day life.
We will now discuss the reaction which is as follows:
We are given benzyl chloride which further changes to nitromethyl benzene on addition of A so,this A which converts benzyl chloride to nitromethylbenzene is none other than sodium nitrite.
This reaction further continues when hydrochloric acid is added to the nitromethylbenzene then it gives B which is the formation of none other than benzylamine which on further reverse reaction from C forms B so, the C must be $PhC{H_2}CON{H_2}$ then B that is benzylamine further reacts with D to give benzyl isocyanide.So D is also none other than chloroform.

Note: All the reactions in organic chemistry should be remembered very carefully. each and every reaction can be applied anywhere where the conversions are concerned.