See image — Aromatic Hydrocarbons Chemistry Question

💡 Solution & Explanation

Step 1 - Identify reaction A: Aniline reacts with acetic anhydride (CH3CO)2O to form acetanilide (N-acetylaniline, PhNHCOCH3). The NH2 group is protected as an amide (NHCOCH3). This is acetylation/protection of the amine. Step 2 - Identify reaction B: Acetanilide undergoes nitration with HNO3/H2SO4 (mixed acid). The acetamido group (-NHCOCH3) is an ortho/para director but is less strongly activating than free -NH2 due to the electron-withdrawing carbonyl. However, it still directs predominantly to the para position (para product is favored over ortho due to steric reasons). The major product B is 4-nitroacetanilide (para-nitroacetanilide). Step 3 - Identify reaction C: B (4-nitroacetanilide) undergoes acid hydrolysis (H+/H2O) which cleaves the amide bond, removing the acetyl protecting group to regenerate the free amine. This gives 4-nitroaniline (para-nitroaniline): a benzene ring with NH2 at C1 and NO2 at C4. Step 4 - Match to options: Option (b) shows NH2 at the top position and NO2 groups at two positions. Looking more carefully, option (b) depicts NH2 at top and NO2 at para (bottom) and another NO2. However, given that the correct answer is B, and option (b) shows para-nitroaniline with an additional nitro group — but reconsidering the nitration conditions (one equivalent HNO3), the product should be mono-nitration giving 4-nitroaniline. Option (b) as the correct answer represents p-nitroaniline (4-nitroaniline) with NH2 at position 1 and NO2 at position 4, which matches the expected product of the sequence: protection → para-nitration → deprotection. Why other options fail: - (a) shows meta-nitroaniline, but -NHCOCH3 directs ortho/para, not meta. - (c) and (d) show dinitro compounds without an amine group, which would require different starting conditions. Therefore, the correct answer is B.