Question

Statement-I: Benzamide on reaction with \( \text{Br}_2 + \text{NaOH} \) gives benzyl amine.
Statement-II: On nitration of aniline, the meta product is formed more than the ortho product.

• A. Both Statement I and Statement II are correct
• B. Statement I is correct but Statement II is incorrect
• C. Statement I is incorrect but Statement II is correct
• D. Both Statement I and Statement II are incorrect

Hint:

In Hofmann degradation of amides, the product is always an amine with one less carbon than the starting amide. In nitration of aniline, the meta product is preferred due to steric hindrance at the ortho position.

Answer 1

The Correct Option is C

Step 1: Analyze Statement I.
Benzamide, on reaction with \( \text{Br}_2 + \text{NaOH} \), undergoes Hofmann degradation. This reaction leads to the formation of an amine with one fewer carbon atom than the starting amide, but it does not produce benzylamine. Instead, it yields aniline (a primary aromatic amine). Therefore, Statement I is incorrect.

Step 2: Analyze Statement II.
In the nitration of aniline, the nitronium ion (\(\text{NO}_2^+\)) predominantly attacks the meta position because the amino group is an electron-donating group, which activates the ortho and para positions. However, steric hindrance at the ortho position (due to the large size of the amino group) makes the meta product more favored. Hence, Statement II is correct.

Step 3: Conclusion.
Since Statement I is incorrect and Statement II is correct, the correct answer is option (C).
Final Answer: (C) Statement I is incorrect but Statement II is correct.