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p-Methylaniline (NH2 group on benzene ring with CH3 at para position) undergoes the following sequence: (i) Ac2O to give A, (ii) Br2/CH3COOH to give B, (iii) H2O/H+ to give C. Identify product C.

1. A benzene ring with NH2, CH3 and COCH3 groups, where COCH3 is ortho to NH2
2. A benzene ring with COCH3 and Br substituents, and CH3 at para position
3. A benzene ring with NH2, Br and CH3 substituents
4. A benzene ring with NHCOCH3 and Br substituents, and CH3 at para position

Correct answer: A benzene ring with NH2, Br and CH3 substituents

Solution

Step 1: Ac2O converts -NH2 to -NHCOCH3 (A = 4-methylacetanilide). Step 2: Br2/CH3COOH brominates ortho to -NHCOCH3; since para is blocked by CH3, bromination occurs at the ortho position (B = 2-bromo-4-methyl acetanilide). Step 3: H2O/H+ hydrolyses acetamide to give -NH2 (C = 2-bromo-4-methylaniline). This is a benzene ring with NH2, Br (ortho to NH2), and CH3 (para to NH2) substituents.

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