See image — Aromatic Hydrocarbons Chemistry Question

💡 Solution & Explanation

Step 1 - Nitration of 4-hydroxybenzoic acid with HNO3: The hydroxyl group is a strong ortho/para director, and the carboxylic acid is a meta director. The incoming nitro group goes ortho to OH (and ortho to CO2H), giving 2-nitro-4-hydroxybenzoic acid (i.e., NO2 ortho to COOH, OH para to COOH). Product (1) = 2-nitro-4-hydroxybenzoic acid. Step 2 - Base/n-PrCl (O-alkylation): The phenolic OH is alkylated under basic conditions with n-propyl chloride (n-PrCl) to give the n-propyl ether. Product (2) = 2-nitro-4-(n-propoxy)benzoic acid. Step 3 - SOCl2 (acid chloride formation): The carboxylic acid is converted to the acid chloride. Product (3) = 2-nitro-4-(n-propoxy)benzoyl chloride. Step 4 - (1) Et2NH: The acid chloride reacts with diethylamine to form the diethylamide. This gives 2-nitro-4-(n-propoxy)-N,N-diethylbenzamide. (2) H2/Pd: Catalytic hydrogenation reduces the nitro group to an amino group (-NO2 -> -NH2). Final product (4) = 2-amino-4-(n-propoxy)-N,N-diethylbenzamide. This matches option (b): a benzene ring bearing NH2 ortho to the C(=O)NEt2 group and an n-propoxy group para to the amide. Why other options fail: - (a) still has NO2 (not reduced), so H2/Pd step is ignored. - (c) has a monoethylamide (NHEt) instead of diethylamide (NEt2), and an unexplained Cl substituent. - (d) still has NO2 (not reduced), so H2/Pd step is ignored. Therefore, the correct answer is B.