See image — Aromatic Hydrocarbons Chemistry Question

💡 Solution & Explanation

Step 1 - Starting material and nitration (A): Ethylbenzene is treated with HNO3/H2SO4 (mixed acid, nitration conditions) to give a di-nitro product (A). The ethyl group is an ortho/para director. With two equivalents of nitrating agent, the two nitro groups enter at the ortho and para positions relative to the ethyl group, giving 1-ethyl-2,4-dinitrobenzene as the major di-nitro product (A). (The 1,2,4-substitution pattern is strongly favored over 1,2,3 because the second nitration is directed para to the first nitro group's deactivating effect, landing at position 4 relative to the ring, consistent with the ethyl group directing.) Step 2 - Oxidation with KMnO4/heat (B): Hot concentrated KMnO4 oxidizes the alkyl side chain (ethyl group) on the benzene ring to a carboxylic acid (-COOH), regardless of chain length. Thus 1-ethyl-2,4-dinitrobenzene is oxidized to 2,4-dinitrobenzoic acid (B): benzene ring with -COOH at C1, -NO2 at C2, -NO2 at C4. Step 3 - Treatment with SOCl2 (C): SOCl2 (thionyl chloride) converts a carboxylic acid (-COOH) to an acid chloride (-COCl). Therefore 2,4-dinitrobenzoic acid (B) is converted to 2,4-dinitrobenzoyl chloride (C): benzene ring with -C(=O)Cl at C1, -NO2 at C2, -NO2 at C4. This corresponds to option (b), drawn as O=C-Cl on the ring with NO2 at ortho and para positions. Why other options fail: - Option (a) shows the same connectivity as (b) but uses 'COCl' notation ambiguously placed; option (b) more clearly and correctly depicts the acid chloride as O=C-Cl, which is the standard representation of an acyl chloride product. - Options (c) and (d) show the ethyl group still intact, meaning no oxidation by KMnO4 occurred, which is incorrect since KMnO4/heat oxidizes alkyl chains to -COOH. - Option (d) shows a 2,3-dinitro pattern which is not the expected regiochemistry from nitration of ethylbenzene. Therefore, the correct answer is B.