The major product formed when acetanilide undergoes bromination in presence of acetic acid is
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → p-bromoacetanilide
The –\( \ce{NH2} \) group of aniline is a powerful ortho/para-directing, activating group; with bromine water aniline gives 2,4,6-tribromoaniline at once. To get a monosubstituted product, the –\( \ce{NH2} \) group is first protected by acetylation with acetic anhydride. In acetanilide the lone pair on nitrogen is also delocalised towards the carbonyl oxygen by resonance, so it is less available to the ring, and the activating effect of the –\( \ce{NHCOCH3} \) group is much smaller.
The –\( \ce{NHCOCH3} \) group is still ortho/para directing, but it is bulky, so attack at the less hindered para position is preferred. Bromination in acetic acid therefore gives mainly the para product:
\( \displaystyle \ce{C6H5NHCOCH3 + Br2} \xrightarrow{\ce{CH3COOH}} p\text{-}\ce{BrC6H4NHCOCH3}\,(\text{major}) + \ce{HBr} \)
So the major product is p-bromoacetanilide. Hydrolysis of this amide then gives p-bromoaniline.
Why the other options are incorrect
Option (1) 2,4,6-Tribromoacetanilide — Incorrect. Trisubstitution happens with free aniline and bromine water; acetylation lowers the activation so far that only one bromine enters.
Option (2) m-bromoacetanilide — Incorrect. The –\( \ce{NHCOCH3} \) group directs to the ortho and para positions; the meta isomer is not formed.
Option (4) o-bromoacetanilide — Incorrect. It forms only as the minor product, because the bulky –\( \ce{NHCOCH3} \) group crowds the ortho positions.
NCERT (2026-27 edition):
Electrophilic Substitution – …
