Question

Which is the correct sequence of reactions for preparation of m-bromoaniline from benzene?

• A. $B{r}_2,FeB{r}_3\to Fe,HCl\to HN{O}_3,H_{2}S{O}_4$
• B. $B{r}_2,FeB{r}_3\to HN{O}_3,H_{2}S{O}_4\to Fe,HCl$
• C. $HN{O}_3,H_{2}S{O}_4\to B{r}_2,FeB{r}_3\to Fe,HCl$
• D. $HN{O}_3,H_{2}S{O}_4\to Fe,HCl\to B{r}_2,FeB{r}_3$

Answer 1

The correct option is C $HN{O}_3,H_{2}S{O}_4\to B{r}_2,FeB{r}_3\to Fe,HCl$

Benzene on reacting with $HN{O}_3,H_{2}S{O}_4$ undergoes aromatic electrophilic substitution reaction to form nitrobenzene. It reacts with $B{r}_2$ to form meta product because $N{O}_2$ is a meta directing group. Fe, HCl is a reducing agent it will reduce the nitro group in m-bromo nitrobenzene to amino group. Thus, it gives m-bromoaniline as the final product.

Hence, option (c) is the correct answer.