Question

Write the structures of A, B and C in the following reaction:

Hint:

Acetylation is the standard "protection" strategy in aromatic synthesis to prevent over-substitution and oxidation of the sensitive \( -NH_{2} \) group.

Answer 1

Concept:

• Direct bromination of aniline is difficult to control because the \( -NH_{2} \) group is extremely activating, leading to 2,4,6-tribromoaniline.

• To obtain a mono-substituted product, the activating power of the amino group must be reduced by acetylation (protection).

• The acetyl group is electron-withdrawing, which reduces the electron density on Nitrogen and thus on the ring.
Step 1: Formation of A (Acetylation)
Aniline reacts with acetic anhydride \( (CH_{3}CO)_{2}O \) to form acetanilide.
Structure A: \( C_{6}H_{5}NHCOCH_{3} \)

Step 2: Formation of B (Bromination)
Acetanilide reacts with \( Br_{2} \) in acetic acid. The \( -NHCOCH_{3} \) group is ortho/para directing. Due to steric hindrance of the bulky acetyl group, the para product is the major one.
Structure B: \( p\text{-Br}-C_{6}H_{4}NHCOCH_{3} \) (4-bromoacetanilide)

Step 3: Formation of C (Hydrolysis)
Acidic hydrolysis (\( H^{+} \)) removes the acetyl protecting group to restore the primary amine.
Structure C: \( p\text{-Br}-C_{6}H_{4}NH_{2} \) (4-bromoaniline or p-bromoaniline) The final answer is A: Acetanilide, B: 4-bromoacetanilide, C: 4-bromoaniline.