The final product D in the following reaction is
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 2-bromo-4-methylaniline
The –\( \ce{NH2} \) group of aniline is so strongly activating that bromination goes straight to the tribromo product. To get a monobromo product, the activating effect is reduced by acetylation (protecting the amino group), then the protecting group is removed by hydrolysis. The acetamido group –\( \ce{NHCOCH3} \) is still ortho/para directing, but less activating, because the nitrogen lone pair is also in resonance with the C=O group.
Following the sequence on p-nitrotoluene:
A: \( \ce{H2} \)/Pd reduces –\( \ce{NO2} \) to –\( \ce{NH2} \): p-toluidine (4-methylaniline).
B: \( \ce{Ac2O} \) acetylates –\( \ce{NH2} \): N-(4-methylphenyl)acetamide, \( \ce{CH3C6H4NHCOCH3} \).
C: \( \ce{Br2} \)/AcOH brominates once. The para position is already taken by \( \ce{CH3} \), and –\( \ce{NHCOCH3} \) is the stronger director, so Br enters ortho to the –NHCOCH3 group: 2-bromo-4-methylacetanilide.
D: \( \ce{H3O^{+}} \) hydrolyses the amide back to –\( \ce{NH2} \): 2-bromo-4-methylaniline.
\( \displaystyle \ce{CH3C6H4NO2 + 3H2} \xrightarrow[\ce{C2H5OH}]{\ce{Pd}} \ce{CH3C6H4NH2 + 2H2O} \)
\( \displaystyle \ce{CH3C6H4NH2 + (CH3CO)2O -> CH3C6H4NHCOCH3 + CH3COOH} \)
\( \displaystyle \ce{CH3C6H4NHCOCH3 + Br2} \xrightarrow{\ce{CH3COOH}} \ce{CH3C6H3(Br)NHCOCH3 + HBr} \)
\( \displaystyle \ce{CH3C6H3(Br)NHCOCH3 + H2O} \xrightarrow{\ce{H+}} \ce{CH3C6H3(Br)NH2 + CH3COOH} \)
In structure 3 the Br is next to the \( \ce{NH2} \) group and the \( \ce{CH3} \) is para to it.
Why the other options are incorrect
Option (1) 4-bromo-2-methylaniline — Incorrect. Here Br is para and \( \ce{CH3} \) ortho to \( \ce{NH2} \); but the \( \ce{CH3} \) was para to nitrogen from the start and cannot move.
Option (2) 3-bromo-4-methylaniline — Incorrect. Br is meta to the amino group, i.e. ortho to \( \ce{CH3} \); the stronger –\( \ce{NHCOCH3} \) director sends Br ortho to itself instead.
Option (4) 3,5-dibromo-4-methylaniline — Incorrect. Only one Br is introduced (the acetyl group prevents polybromination), and both Br atoms here are meta to nitrogen.
NCERT (2026-27 edition):
Electrophilic Substitution – …
