See image — Hydrocarbons Chemistry Question

💡 Solution & Explanation

Step 1 – Reaction (a): The starting material is cyclopentylidene-cyclobutane with a CH3 group on the cyclobutane ring attached via a wedge bond. The double bond (cyclopentylidene = cyclobutane) is reduced by H2/Ni via syn addition. The cyclobutane ring already bears a defined stereocenter (CH3 on wedge). After syn addition of H2 across the double bond, a new stereocenter is created at the ring junction. Because the two newly formed stereocenters can be either syn or anti relative to each other, two stereoisomeric products result. These two products are diastereomers of each other (they are not mirror images because the starting material is chiral and the addition is syn). Therefore (a) → (q) Diastereomers. Step 2 – Reaction (b): The starting material is 1,2-dimethylcyclopentene (both methyl groups on the double-bond carbons). Syn addition of D2/Pt places both D atoms on the same face of the ring. The two carbons bearing D (and CH3) become stereocenters. Syn addition to a symmetric alkene like 1,2-dimethylcyclopentene gives the cis product (both D on same face, both CH3 on opposite face relative to each other in the ring), which is the meso compound because the molecule has an internal plane of symmetry. Therefore (b) → (p) Meso compound. Step 3 – Reaction (c): The starting material is 6-methoxynaphthalene with a CH2=C(CH3)- or similar prochiral exocyclic alkene substituent (naproxen precursor type structure). H2/Pt reduces the double bond. The reduction creates one new chiral center at the carbon bearing the naphthalene ring. Since no chiral catalyst is used, both faces of the double bond are attacked equally, giving equal amounts of R and S product — a racemic mixture. Therefore (c) → (r) Racemic. Step 4 – Reaction (d): The starting material is methylenecyclopentane (cyclopentane ring with exocyclic =CH2). H2/Pt reduces the double bond. The product is methylcyclopentane. No stereocenters are created because the exocyclic carbon becomes –CH3 (only one carbon type attached) and the ring carbon bearing CH3 has two identical ring-H environments making it not a stereocenter. The product has no chiral center at all. Therefore the product is optically inactive due to absence of a chiral center. Therefore (d) → (s). Summary: (a)→(q), (b)→(p), (c)→(r), (d)→(s). Therefore, the correct answer is {"a": "q", "b": "p", "c": "r", "d": "s"}.