Question

Give the structures of $A,\,B$ and $C$ in the following reactions:
(i) ${C_6}{H_5}{N_2}^ + C{l^ - }\xrightarrow{{CuCN}}A\xrightarrow[{{H^ + }}]{{{H_2}O}}B\xrightarrow[\Delta ]{{N{H_3}}}C$
(ii) ${C_6}{H_5}N{O_2}\xrightarrow{{Sn\, + \,HCl}}A\xrightarrow[{273\,K}]{{NaN{O_2}\, + \,HCl}}B\xrightarrow[\Delta ]{{{H_2}O,\,{H^ + }}}C$

Answer 1

In the question a sequence of reactions are given. Identify the reactant and reagent in each step to identify the product formed in each step which will in turn be the reactant in the next step. In this way proceed in a stepwise manner and identify $A,\,B$ and $C$ in the given reaction sequences.

Complete step by step solution:
(i) The starting compound is ${C_6}{H_5}{N_2}^ + C{l^ - }$ which when treated with $CuCN$yields cyanobenzene, ${C_6}{H_5}CN$. $($Sandmeyer’s Reaction$)$
${C_6}{H_5}{N_2}^ + C{l^ - }\,\xrightarrow{{CuCN}}\,{C_6}{H_5}CN$
Hence the compound $A$ is ${C_6}{H_5}CN$, the starting compound for the next step. In the next step cyanobenzene undergoes hydrolysis to form benzoic acid, ${C_6}{H_5}COOH$.
${C_6}{H_5}CN\,\xrightarrow[{{H^ + }}]{{{H_2}O}}\,{C_6}{H_5}COOH$
Hence the compound $B$ is ${C_6}{H_5}COOH$, the starting compound for the next step where the reagent is ammonia. In the next step benzoic acid reacts with ammonia in presence of heat to form benzamide, ${C_6}{H_5}CON{H_2}$.
${C_6}{H_5}COOH\,\xrightarrow[\Delta ]{{N{H_3}}}\,{C_6}{H_5}CON{H_2}$
Hence the compound $C$ is ${C_6}{H_5}CON{H_2}$.
(ii) The starting compound is ${C_6}{H_5}N{O_2}$ which when treated with $Sn$ in presence of $HCl$ is reduced to yield aniline, ${C_6}{H_5}N{H_2}$.
${C_6}{H_5}N{O_2}\,\xrightarrow{{Sn\, + \,HCl}}\,{C_6}{H_5}N{H_2}$
Hence the compound $A$ is ${C_6}{H_5}N{H_2}$, the starting compound for the next step where the reagent is $NaN{O_2}\, + \,HCl$ which leads to the formation of nitrous acid, which basically reacts with aniline. In the next step aniline undergoes diazotization reaction in presence of nitrous acid to form benzene diazonium salt, ${C_6}{H_5}{N_2}^ + C{l^ - }$.
${C_6}{H_5}N{H_2}\,\xrightarrow[{273\,K}]{{NaN{O_2}\, + \,HCl}}\,{C_6}{H_5}{N_2}^ + C{l^ - }$
Hence the compound $B$ is ${C_6}{H_5}{N_2}^ + C{l^ - }$, the starting compound for the next step. In the next step ${C_6}{H_5}{N_2}^ + C{l^ - }$ undergoes hydrolysis to form phenol, ${C_6}{H_5}OH$.
${C_6}{H_5}{N_2}^ + C{l^ - }\,\xrightarrow[\Delta ]{{{H_2}O,\,{H^ + }}}\,{C_6}{H_5}OH$
Hence the compound $C$ is ${C_6}{H_5}OH$.

Note:
For the given questions you must have basic knowledge about aniline and its derivatives and the reactions they undergo in presence of different reagent. You must also know which compound acts as a reducing agent, when a compound undergoes hydrolysis, diazotization reaction and Sandmeyer’s reaction.