See image — Hydrocarbons Chemistry Question

💡 Solution & Explanation

Step 1 - Identify the starting material: The compound shown is 2-benzylidenecyclopentan-1-one (a cyclopentenone with a benzylidene substituent at C2, making an exocyclic C=C conjugated with the ring carbonyl). The molecule has an exocyclic alkene (C=C between ring C2 and the benzylic carbon) and a ring endocyclic C=C (the enone double bond within the five-membered ring). Actually, re-examining: the structure is a cyclopent-2-en-1-one where C2 bears a benzylidene group — i.e., the ring double bond is between C2 and C3, and there is also a C=C exocyclic at C2 connecting to phenyl. More precisely, the compound is (E/Z)-2-benzylidenecyclopentan-1-one: a five-membered ring ketone with an exocyclic double bond at C2 bearing a phenyl group. Step 2 - Reaction conditions: 1 mole of H2 at 2-3 atm with 10% Pd/C (heterogeneous hydrogenation). 1 mole of H2 reduces one degree of unsaturation. The exocyclic C=C (benzylidene double bond) is more reactive toward catalytic hydrogenation than the isolated enone system under mild conditions, so 1 mole H2 selectively reduces the exocyclic alkene. Step 3 - Product analysis: Reduction of the exocyclic C=C (CH=Ph → CH2-Ph) gives 2-benzylcyclopentan-1-one. In this product, C2 of the cyclopentanone ring becomes a new stereocenter (it bears H, benzyl group, and is part of the ring with the adjacent carbonyl). The carbonyl carbon (C1) is not a stereocenter. C2 is the only stereocenter generated. Step 4 - Stereochemistry of hydrogenation: Catalytic hydrogenation on a heterogeneous Pd/C surface delivers H2 from one face (syn addition), but since the substrate (2-benzylidenecyclopentan-1-one) is planar/achiral at the reacting center, the two faces of the exocyclic double bond are enantiotopic. Adsorption can occur from either face with equal probability, generating the (R) and (S) configurations at C2 in equal amounts. Step 5 - Optical activity conclusion: The product 2-benzylcyclopentan-1-one has one stereocenter (C2). Since both enantiomers are formed in equal amounts (the two faces are equally accessible on the achiral catalyst surface with an achiral substrate), the product is a racemic mixture (equal amounts of R and S enantiomers). A racemic mixture is optically inactive due to external compensation. Step 6 - Why other options fail: - (a) Diastereomers: Diastereomers require at least two stereocenters; the product has only one stereocenter, so diastereomers cannot form. - (c) Meso: A meso compound requires at least two stereocenters with an internal plane of symmetry; not applicable here. - (d) Optically pure enantiomer: This would require asymmetric induction (chiral catalyst or chiral substrate), which is absent here. Therefore, the correct answer is B.