Consider the above reactions, the product A and product B respectively are

Solution

Chemistry ma’am · 4 steps

Free -NH₂ fills every ortho and para spot with Br; acetylation calms it down to one para Br.

1. Product A. The free -NH₂ group pushes its lone pair into the ring, so the ring becomes very electron-rich. With bromine water, every free ortho and para position reacts at once, even at room temperature. That gives the white precipitate of 2,4,6-tribromoaniline.
2. Tame the group. Acetic anhydride turns aniline into acetanilide, C₆H₅NHCOCH₃. The nitrogen lone pair is now shared with the C=O group, so it activates the ring much less. The bulky -NHCOCH₃ group also crowds the ortho positions.
3. Product B. In CH₃COOH, bromine therefore adds only once, mainly at the para position, giving 4-bromoacetanilide. Acid hydrolysis with HCl then removes the acetyl group and gives back the free amine: 4-bromoaniline.

   C₆H₅NH₂ →((CH₃CO)₂O) C₆H₅NHCOCH₃ →(Br₂, CH₃COOH) p-BrC₆H₄NHCOCH₃ →(HCl, H₂O) p-BrC₆H₄NH₂

4. So. A is 2,4,6-tribromoaniline and B is 4-bromoaniline, in that order. That is option (c).

The idea to keep. Free -NH₂ means over-substitution at every ortho and para spot. To get just one substituent, protect it as an acetamide first and hydrolyse the acetyl group off at the end.

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