Question

When benzene diazonium chloride is treated with cuprous chloride in HCl, chlorobenzene is formed. This reaction is called:
A) Perkin reaction
B) Hinsberg reaction
C) Gattermann reaction
D) Sandmeyer reaction

Answer 1

We know that benzene diazonium chloride can be used as a precursor for synthesis of different compounds based on the reagent or conditions it is introduced to.

Complete step by step answer:
We have a very significant intermediate, benzene diazonium salt $\left( {{C_6}{H_5}N_2^ + C{l^ - }} \right)$ which can be used to prepare different compounds. ${C_6}{H_5}N_2^ + C{l^ - }$can be prepared by treating aniline with $HN{O_2}$ (prepared in situ by $NaN{O_2}\;and\;HCl$) at low temperature $\left( {273 - 278\;K} \right)$ as follows:
${C_6}{H_5}N{H_2} \to {C_6}{H_5}N_2^ + C{l^ - } + NaCl + 2{H_2}O$
The diazonium group being a good leaving group can be replaced in different manners. One application of ${C_6}{H_5}N_2^ + C{l^ - }$ is to prepare aryl halides.

Sandmeyer reaction: In this reaction, the diazonium group can be replaced by $C{l^ - }, B{r^ - }\;or\;C{N^ - }$ by using cuprous ions. For example, chlorobenzene can be prepared by treating ${C_6}{H_5}N_2^ + C{l^ - }$ with CuCl in HCl as per the following reaction:
${C_6}{H_5}N_2^ + C{l^ - } \xrightarrow{CuCl/HCl} {C_6}{H_5}Cl + {N_2}$

Gattermann reaction: In this reaction, the diazonium group can be replaced by $C{l^ - }\;or\;B{r^ - }$ by using Cu powder. For example, chlorobenzene can be prepared by treating ${C_6}{H_5}N_2^ + C{l^ - }$with $C{u_2}C{l_2}$ in HCl as per the following reaction:
${C_6}{H_5}N_2^ + C{l^ - } \xrightarrow[HCl]{Cu} {C_6}{H_5}Cl + {N_2} + CuCl$
As we can see that in the question, we have CuCl in HCl.

Hence, from the given options, the correct one is D.

Note: We have to be careful with the name of the reactions. Gattermann reaction is similar to Sandmeyer but in this, Cu powder is used in place of CuCl. Hinsberg reaction involves benzene sulfonyl chloride whereas Perkin reaction is for aromatic aldehydes.