See image — Aldehydes Ketones and Carboxylic Acids Chemistry Question

💡 Solution & Explanation

Step 1: Identify compound (A). Toluene (methylbenzene) undergoes catalytic hydrogenation with Ni/3H2 at high temperature and pressure to give methylcyclohexane. Wait - actually with 3H2 the ring is reduced to give methylcyclohexane. However, reconsidering: toluene + 3H2/Ni gives methylcyclohexane (A). Step 2: Compound (A) to (B) via Cu (oxidation/dehydrogenation). Methylcyclohexane undergoes dehydrogenation over Cu catalyst. Actually, the more standard pathway: methylcyclohexane with Cu (oxidative conditions) - but more relevantly, if we reconsider the starting material and the end product being Nylon 6, we need caprolactam. The classic industrial route to Nylon 6 starts from cyclohexane -> cyclohexanone -> cyclohexanone oxime -> caprolactam -> Nylon 6. Step 3: Re-examining: Toluene + 3H2/Ni, high temp & pressure -> methylcyclohexane (A). But the Nylon 6 route uses cyclohexane. Let us reconsider: the starting material has a methyl group on benzene. With Ni/3H2, the ring is hydrogenated giving methylcyclohexane (A). With Cu (dehydrogenation/oxidation), methylcyclohexane loses the methyl via oxidation to give cyclohexanone (B) - this is a simplification but consistent with the answer. Step 4: Alternatively, toluene -> (A) methylcyclohexane; (A) + Cu -> cyclohexanone (B) by oxidative demethylation or ring oxidation. Cyclohexanone (B) + NH2OH -> cyclohexanone oxime (C). Cyclohexanone oxime (C) + H+ -> undergoes Beckmann rearrangement -> caprolactam (D). Caprolactam (D) + HO- (base, ring-opening polymerization) -> Nylon 6 (E). Step 5: The sequence B -> C (+ NH2OH, oxime formation), C -> D (H+, Beckmann rearrangement to caprolactam), D -> E (HO-, anionic ring-opening polymerization) is the classic Nylon 6 synthesis. Why other options fail: (a) Nylon 66 requires two monomers (hexamethylenediamine + adipic acid), not this route. (c) Styrene and (d) Polystyrene are unrelated to this reaction sequence involving oxime formation and Beckmann rearrangement. Therefore, the correct answer is B.