Question

Which of the following reactions will best convert nitrobenzene into 2, 4, 6-trichlorobromobenzene?

A) $\text{ SnC}{{\text{l}}_{\text{2}}}\text{ / HCl , C}{{\text{l}}_{\text{2}}}-{{\text{H}}_{\text{2}}}\text{O , NaN}{{\text{O}}_{\text{2}}}\text{ / HCl }-{{\text{0}}^{\text{0}}}\text{C , CuBr }$
B) $\text{ C}{{\text{l}}_{\text{2}}}\text{ / AlC}{{\text{l}}_{\text{3}}}\text{ , SnC}{{\text{l}}_{\text{2}}}\text{/HCl , NaN}{{\text{O}}_{\text{2}}}\text{ / HCl }-{{\text{0}}^{\text{0}}}\text{C , CuBr }$
C) $\text{ SnC}{{\text{l}}_{\text{2}}}\text{ / HCl , B}{{\text{r}}_{\text{2}}}-{{\text{H}}_{\text{2}}}\text{O , NaN}{{\text{O}}_{\text{2}}}\text{ / HCl }-{{\text{0}}^{\text{0}}}\text{C , CuCl }$
D) $\text{ SnC}{{\text{l}}_{\text{2}}}\text{ / HCl , NaN}{{\text{O}}_{\text{2}}}\text{ / HCl }-{{\text{0}}^{\text{0}}}\text{C , }{{\text{H}}_{\text{2}}}\text{O , C}{{\text{l}}_{\text{2}}}\text{(aq) , HBr }$

Answer 1

The aromatic nitro compounds are reduced by the active metals. The nitro group is reduced to $\text{ }-\text{N}{{\text{H}}_{\text{2}}}\text{ }$ groups. The exposure to the excess of halogen results in the 2, 4, 6-tri halo substituted benzene ring. The reaction of the conversion of aromatic primary amines to the diazonium salt $\text{ ArN}_{2}^{+}{{\text{X}}^{-}}\text{ }$is a diazotization reaction. The diazonium group in the diazonium salts is replaced by the groups to get the desired product.

Complete answer:
The nitro compounds can be reduced with active metals such as $\text{ Fe }$ , $\text{ Sn }$ , $\text{ Zn }$ , etc. And conc. hydrochloric acid. The mixture of $\text{ SnC}{{\text{l}}_{\text{2}}}\text{ }$ and conc.$\text{ HCl }$ has been used for the reduction of aromatic nitro compounds.

Aniline when reacted with the chlorine molecules, the aniline is converted into 2, 4, 6-trichloroaniline. The reaction of conversion of aniline to the 2, 4, 6-trichloroaniline is as follows,

The diazonium salts have the general formula of $\text{ ArN}_{2}^{+}{{\text{X}}^{-}}\text{ }$ , where $\text{ }{{\text{X}}^{-}}\text{ }$ maybe anion like chloride, bromide, etc. and the group $\text{ N}_{2}^{+\text{ }}$ ($\text{ }-\text{N }\equiv \text{ N}-\text{ }$) is called as the diazonium group. These are obtained when the aromatic primary amines react with nitrous acid.
The diazonium salt is prepared by heating the ice-cold solution of aromatic primary amines in excess of the mineral acid like $\text{ HCl }$ or $\text{ }{{\text{H}}_{\text{2}}}\text{S}{{\text{O}}_{\text{4}}}\text{ }$ with the ice-cold solution of sodium nitrite dissolved in water. The temperature is maintained between the $\text{ 273 K }-278\text{ K }$ because most of the diazonium salts decompose at a higher temperature.
The benzene diazonium chloride of The 2, 4, 6-trichloroaniline is prepared by treating an ice-cold solution of aniline in hydrochloric acid with an ice-cold solution of sodium nitrite at about $\text{ }{{\text{0}}^{\text{0}}}\text{C }$ . The reaction of converting aromatic primary amine to diazonium salt is called diazotization. The 2, 4, 6-trichloroaniline is converted into diazonium salt as follows,
$\text{ NaN}{{\text{O}}_{\text{2}}}\text{ + HCl }\to \text{ NaCl + HONO }$

The diazonium salt of The 2, 4, 6-trichloroaniline when warmed with the cuprous bromide $\text{ CuBr }$ in hydrochloric acid the corresponding bromide is formed. The resultant compound is 2, 4, 6-trichlorobromobenzene. The diazonium salt is converted to the halide group. The reaction is as follows,

Thus, the nitrobenzene is converted into the 2, 4, 6-trichlorobromobenzene by the action of the following reagent : $\text{ SnC}{{\text{l}}_{\text{2}}}\text{ / HCl , C}{{\text{l}}_{\text{2}}}-{{\text{H}}_{\text{2}}}\text{O , NaN}{{\text{O}}_{\text{2}}}\text{ / HCl }-{{\text{0}}^{\text{0}}}\text{C , CuBr }$

Hence, (A) is the correct option.

Note:
Note that the preparation of aryl halide from diazonium salt is a good method and has many advantages. Aryl chlorides and bromides when obtained by direct halogenation of aromatic compounds give a mixture of products that are difficult to separate. However, in the diazonium salt replacement method, a pure single product is formed. This also provides a good synthetic route for the preparation of aryl iodides and fluorides which are not obtained by direct halogenation.