Question

Which of the following will give $m$-bromonitrobenzene as the final product?

• A.
• B.
• C.
• D. Both (a) and (b)

Answer 1

The correct option is A

In (a), benzene reacts with $N{O}_2^{+}$ electrophile to give nitrobenzene.
In nitrobenzene, $-N{O}_2$ is a -R group which will reduce the electron density at ortho and para position. Thus, electrophilic attack takes place at meta position.
Therefore, in bromination of nitrobenzene with $FeB{r}_3$, the $B{r}^{+}$ will attack the meta position and give
m-bromonitrobenzene product.

In (b), benzene undergoes mono bromination with $FeB{r}_3$ to form bromobenzene.
In bromonbenzene, bromine is a ortho/para directing group due to its +R effect. Hence, in nitration of bromobenzene $N{O}^{+}$ electrophile will attack at ortho and para position to give
o- bromonitrobenzene and p-bromonitrobenzene.

In (c), benzene reacts with $N{O}_2^{+}$ electrophile to give nitrobenzene.
$N{O}_2^{+}$ shows -R effect which will reduce the electron density at ortho and para position. Thus, electrophilic ($C{l}^{+}$) attack takes place at meta position to give m-chloronitrobenzene.

Thus, option (a) is correct.