Question

The compound $C_7{H}_8$ undergoes the following reactions :
$C_7{H}_8\stackrel{3C{l}_2/hv}{\to }A\stackrel{B{r}_2/Fe}{\to }B\stackrel{Zn/HCl}{\to }$
The product 'C' is

(NEET - 2018 )

• A. 3-bromo-2,4,6-trichlorotoluene
• B. o-bromotoluene
• C. p-bromotoluene
• D. m-bromotoluene

Answer 1

The correct option is D m-bromotoluene

The compound $C_7{H}_8$ is toluene.
Toluene in presence of light with 3 mole of $C{l}_2$ gives benzotrichloride. It is a free radical reaction.
Benzotrichloride undergoes bromination (nucleophilic aromatic substitution) in presence of $FeB{r}_3$ at meta position since $CC{l}_3$ is a meta directing group. In last step $CC{l}_3$ get reduced to $C{H}_3$ in presence of Zn/HCl.

So the correct option is (a).