Question

Conversion of benzene diazonium chloride to chlorobenzene is an example of which of the following reactions?
(a) Claisen
(b) Friedel-Crafts
(c) Sandmeyer
(d) Wurtz

Answer 1

Hint: When sodium nitrite is treated with dilute hydrochloric acid, it forms nitrous acid. When this nitrous acid gets protonated, it releases one equivalent of water and forms the nitrosonium ion. This ion is an electrophile and when reacted with aniline at a temperature of 273-278 K gives the product benzene diazonium chloride. When this benzene diazonium chloride is treated with $CuCl$ dissolved in $HCl$, we get chlorobenzene.

Complete step by step solution:
In order to solve this question, first let us understand how benzene diazonium chloride is prepared. When sodium nitrite is treated with dilute hydrochloric acid, it forms nitrous acid. When this nitrous acid gets protonated, it releases one equivalent of water and forms the nitrosonium ion. This ion is an electrophile and when reacted with aniline at a temperature of 273-278 K gives the product benzene diazonium chloride. The reactions are given below:
$ { NaNO }_{ 2 }(s)+HCl(dil.)\rightarrow HONO(nitrous\quad acid)+NaCl(aq)$
$ ({ C }_{ 6 }{ H }_{ 5 })-{ NH }_{ 2 }+HONO+HCl\xrightarrow { 273-278\quad K } ({ C }_{ 6 }{ H }_{ 5 })-\overset { + }{ N } \equiv N{ Cl }^{ - }+2{ H }_{ 2 }O$
When this benzene diazonium chloride is treated with CuCl dissolved in HCl, we get chlorobenzene. This reaction is called the Sandmeyer reaction. It follows a free radical mechanism. The mechanism involves the transfer of a single electron from copper to the diazonium salt which results in the formation of a diazo radical and copper (II) chloride. This follows the evolution of nitrogen gas from the diazo radical to form aryl radical. This aryl radical reacts with copper (II) chloride to give chlorobenzene and copper (I) chloride (which is the catalyst). The reaction is given below:
$({ C }_{ 6 }{ H }_{ 5 })-\overset { + }{ N } \equiv N{ Cl }^{ - }\xrightarrow { CuCl/HCl } ({ C }_{ 6 }{ H }_{ 5 })-Cl+{ N }_{ 2 }$
Sandmeyer reaction is quite important since it can be used to produce substitution products that are sometimes not possible to achieve by direct substitution.

Hence the correct answer is (C) Sandmeyer

Note: There are a lot of variations of the Sandmeyer reaction. It can be used not only to produce chlorobenzene but also other aryl halides such as bromobenzene, fluorobenzene (Balz-Schiemann reaction) and other products such as phenols, aryl thioethers and aryl nitriles. For the formation of the phenol, the diazonium salt is simply hydroxylated with acidified water. For the formation of fluorobenzene, the benzene diazonium salt is treated with fluoroboric acid at 273-298 K.