In the following reaction sequence,
The major product B is :

Solution

Chemistry ma’am · 4 steps

First tame the -NH₂ by acetylation, then let the -NHCOCH₃ group decide where Br goes.

1. Acetylation gives A. Ac₂O acetylates the amino group of p-toluidine, so A is 4-methylacetanilide. Its ring now carries -NHCOCH₃ at C-1 and -CH₃ at C-4. The ring itself is not touched in this step.
2. Compare the directors. Both groups are ortho, para directors, and their para positions are already taken by each other. The lone pair on N still pushes electron density into the ring by resonance, so -NHCOCH₃ is a stronger activator than -CH₃, which only gives hyperconjugation.
3. Bromination gives B. Br₂ in AcOH is ionic electrophilic substitution, so the stronger activator wins and Br enters ortho to -NHCOCH₃. Acetylation also tones down the activation, so only one Br goes in, not three as with free aniline.
4. So. B is 2-bromo-4-methylacetanilide, with Br next to the -NHCOCH₃ group. That is option (b).

The idea to keep. When two ortho, para directors compete, the stronger resonance donor sets the position. -NHCOCH₃ beats -CH₃, and acetylation keeps bromination to a single substitution.

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