Reaction 1: Aniline \((\mathrm{C_6H_5NH_2})\) with \(\mathrm{Br_2}\) (excess) in \(\mathrm{H_2O}\) gives A (Major Product).
Reaction 2: Aniline with (i) \((\mathrm{CH_3CO})_2\mathrm{O}\); (ii) \(\mathrm{Br_2}\), \(\mathrm{CH_3COOH}\); (iii) HCl gives B (Major Product).
Consider the above reactions, the product A and product B respectively are
Reaction 2: Aniline with (i) \((\mathrm{CH_3CO})_2\mathrm{O}\); (ii) \(\mathrm{Br_2}\), \(\mathrm{CH_3COOH}\); (iii) HCl gives B (Major Product).
Consider the above reactions, the product A and product B respectively are
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The Correct Option is C
Solution and Explanation
Reaction 2: (i) Acetic anhydride converts \(-\mathrm{NH_2}\) into \(-\mathrm{NHCOCH_3}\) (acetanilide). The lone pair is now partly delocalised onto the carbonyl, so the ring is only moderately activated.
(ii) \(\mathrm{Br_2}\) in \(\mathrm{CH_3COOH}\) gives mono-bromination, mainly at the para position (the bulky acetamido group blocks ortho): p-bromoacetanilide.
(iii) HCl hydrolyses the amide back to the amine: 4-bromoaniline (B).
So A = 2,4,6-tribromoaniline and B = 4-bromoaniline, which is option (3).
Answer: (3) 2,4,6-tribromoaniline and 4-bromoaniline
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