The major products A and B formed in the following reaction sequence are :

Solution

Chemistry ma’am · 4 steps

Acetic anhydride caps the nitrogen first, and then the tamed ring takes just one bromine, at para.

1. Acylation on nitrogen. The lone pair on the -NH₂ nitrogen is the most nucleophilic site in aniline, so (CH₃CO)₂O acylates the nitrogen, not the ring. A is acetanilide, C₆H₅NHCOCH₃.
2. Activation is reduced. In -NHCOCH₃ the nitrogen lone pair is shared with the ring and also with the C=O group. The ring stays activated and ortho/para directing, but much less than in aniline, so it does not take three bromines the way aniline does.
3. Para wins. With Br₂ in CH₃COOH at room temperature, only one bromine goes in. The bulky acetamido group blocks the ortho positions, so the major product is the para isomer.
4. So. A is acetanilide and B is p-bromoacetanilide, 4-BrC₆H₄NHCOCH₃, which is option (c).

The idea to keep. Acetylating -NH₂ is the standard way to calm aniline down: you get monosubstitution, mostly para. The -NHCOCH₃ group is still an ortho/para director.

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