The major products A and B from the following reactions are:

Solution

Chemistry ma’am · 3 steps

The acetyl group tames the ring to one bromine, and LiAlH4 strips the amide's oxygen right off.

1. Bromination gives A. In acetanilide the nitrogen lone pair is shared with the C=O group, so it pushes much less electron density into the ring than the -NH₂ of aniline does. The ring is still ortho/para directing but only moderately activated, so Br₂ in acetic acid adds just one bromine. The bulky -NHCOCH₃ group blocks the ortho positions, so the product is p-bromoacetanilide.
2. Reduction gives B. LiAlH₄ reduces an amide all the way down to an amine. The first hydride adds to the C=O, the oxygen leaves bound to aluminium to give an iminium ion, and a second hydride turns that into CH₂. No -OH is left behind.

   C₆H₅NHCOCH₃ →(LiAlH₄) C₆H₅NHCH₂CH₃

3. So. A is p-bromoacetanilide and B is N-ethylaniline, which is option (b).

The idea to keep. Acetylating -NH₂ cuts its activating power, so you get monosubstitution at para. And LiAlH₄ turns an amide C=O into CH₂, never into CH-OH.

Answer from the NTA final answer key. Solution by PrepWiser.