Question

A) How will you convert benzene into
$p$-nitrobromobenzene?

B) How will you convert benzene into
$m$-nitrobromobenzene?

Answer 1

Electrophilic aromatic substitution of benzene with bromine in the presence of lewis acid $\left(FeB{r}_3\right)$ gives bromobenzene

Bromobenzene to
𝑝−nitrobromobenzene:

Nitration of bromobenzene with concentrated nitric acid and concentrated sulphuric acid gives a mixture of 𝑜-bromonitrobenzene and $p$-bromonitrobenzene, which can be separated by fractional distillation.



This is an electrophilic substitution reaction of aryl halide.

(ii) Halogens attached to the benzene ring are ortho and para directing whereas the nitro group is meta directing. So, the first step involved here is nitration of benzene, followed by halogenation.



Nitration of benzene occurs via heating in the presence of concentrated nitric acid and concentrated sulphuric acid, which gives nitrobenzene as the product.

Nitrobenzene to
m−nitrobromobenzene


Nitrobenzene undergoes halogenation in the presence of lewis acid $\left(FeB{r}_3\right)$ and bromine $\left(B{r}_2\right)$, which led to the formation of $m$-nitrobromobenzene because nitrobenzene is a meta-directing group