Question

Nitrobenzene reacts with ${Br}_2$ in the presence of $Fe{Br}_3$ 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 greater than those at the ortho and para positions
• B. Aromaticity is lost in the $\sigma$-complexes formed by the attack of ${Br}^{+}$ at the ortho and para positions but not at the meta position
• C. The $\sigma$-complex formed by the attack of ${Br}^{+}$ 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 by the loss of $H^{+}$ from the $\sigma$-complexes, the meta-oriented $\sigma$-complex loses $H^{+}$ most readily

Answer 1

Answer: (C)

The $\sigma$-complex formed by the attack of ${Br}^{+}$ at the meta position is the least destabilized and the most stable among the three $\sigma$-complexes

Nitrobenzene reacts with ${Br}_2$ in the presence of $Fe{Br}_3$ to give m-bromonitrobenzene as the major product.The best reason for the formation of m-bromonitrobenzene as the major product is as given below.
The $\sigma$-complex formed by the attack of ${Br}^{+}$ at the meta position is the least destabilized and the most stable among the three $\sigma$-complexes. Hence, the nitro group acts as meta directing group. It deactivates the ring.