Question

Nitrobenzene reacts with $B{r}_2$ in the presence of $FeB{r}_3$ to give m-bromonitrobenzene as the major product. Which of the following provides the best reason for the formation of m-bromonitrobenzene as the major product?

• A. The electron density at the meta position is lesser than those at the ortho and para positions
• B. Aromaticity is lost in the $\sigma$- complexes formed by the attack of $B{r}^{+}$ at the ortho and para positions but not at the meta position
• C. The $\sigma -$ complex formed by the attachment of $B{r}^{+}$ at the meta position is the least destabilized and the most stable among the three $\sigma -$ complexes
• D. In the final step of regeneration of benzene ring but the loss of $H^{+}$ from the $\sigma -$ complexes, the meta oriented $\sigma -$ complex loses $H^{+}$ most readily.

Answer 1

The correct option is C The $\sigma -$ complex formed by the attachment of $B{r}^{+}$ at the meta position is the least destabilized and the most stable among the three $\sigma -$ complexes

Nitrobenzene reacts with $B{r}_2$ in the presence of $FeB{r}_3$ to give m-brominitrobenzene as the major product. The best reason for the formation of m-bromonitrobenzene as the major product is as given below.

The groups which direct the incoming group to meta position are called meta directing groups. Hence, the nitro group acts as meta directing group. Nitro group reduces electron
density in ring due to strong -I effect. Overall electron density decreases making further substitution difficult. The electron density on ortho and para position is comparatively less than that at meta position.

The $\sigma -$ complex formed by the attack of $B{r}^{+}$ at the meta position is the least destabilized and the most stable among the three $\sigma -$ complexes. Hence, m-bromonitrobenzene is fromed as a major product.