Question:

In the following reaction sequence,


The major product B is :

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Acetylation converts NH2 into the weaker but still ortho/para-directing NHCOCH3 group; bromination in AcOH is electrophilic substitution ortho to the strongest director (para is blocked).
Updated On: Oct 31, 2026
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The Correct Option is B

Solution and Explanation

Step 1: p-Toluidine + Ac2O gives A = N-(4-methylphenyl)acetamide (p-methylacetanilide); the amine is protected as the amide.
Step 2: Br2/AcOH is an electrophilic aromatic substitution. Ring substituents: NHCOCH3 (strong activator, o/p) and CH3 (weak activator, o/p).
Step 3: The stronger activating group (NHCOCH3) controls the position. Its para position is occupied by CH3, so Br enters ortho to NHCOCH3 (position 2).
Step 4: B = N-(2-bromo-4-methylphenyl)acetamide. Meta bromination (option A) is not expected; acylation (option C) needs Friedel-Crafts conditions; benzylic bromination (option D) needs radical conditions (light/NBS).
Answer: (B)
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