See image — Aldehydes Ketones and Carboxylic Acids Chemistry Question

💡 Solution & Explanation

Step 1 – Identify the substrates. Compound A and Compound B are enantiomers of trans-2-hydroxycyclopropyl methyl ketone (trans-1-acetyl-2-hydroxycyclopropane). They are a racemic pair (R,R and S,S, or the relevant trans enantiomers). Step 2 – Reaction with NaOH (base, heat). The hydroxyl group is deprotonated by NaOH to give an alkoxide. The alpha-carbon of the ketone (the cyclopropane carbon bearing the C=O) is also activated. Under basic conditions, the alkoxide on C2 of the cyclopropane can initiate a retro-cyclopropane ring-opening (or intramolecular aldol / elimination sequence). Step 3 – Mechanism. The key transformation is a base-mediated ring-opening of the cyclopropane. The alkoxide acts as an internal nucleophile/base: deprotonation alpha to the ketone generates a carbanion/enolate that is stabilized. With the strained cyclopropane and the beta-hydroxy ketone motif, treatment with base promotes a retro-aldol or elimination. Specifically, the trans-2-hydroxycyclopropyl methyl ketone undergoes base-promoted ring opening: the C–C bond of the cyclopropane breaks, generating a homoallylic system. The ring opening of the cyclopropane beta-hydroxy ketone under basic conditions gives a 1,4-dicarbonyl equivalent or an enone via elimination. The alkoxide leaves (as the driving force from ring strain relief) and the enolate collapses: the net result is that the three-membered ring opens to give a 3-carbon unit that, combined with intramolecular aldol condensation under heat, forms a five-membered ring enone. Step 4 – Product identification. Since both A and B are enantiomers (racemic mixture), they must give the same achiral product. The ring opening of the cyclopropane beta-hydroxy ketone and subsequent aldol condensation/dehydration under heat yields cyclopent-2-en-1-one (2-cyclopentenone). This is a symmetrical, achiral product, consistent with both enantiomers giving a single compound C. The molecular formula: starting material C5H8O2, loss of water gives C5H6O, which matches cyclopentenone. Step 5 – Why other options fail. (b) The cyclopropane diketone product would require oxidation and no ring opening — not consistent with base/heat conditions on a beta-hydroxy ketone. (c) Cyclohexenone is 6-membered and would require a C6 framework; starting material only has 5 carbons total — impossible. (d) Option (d) appears to show a different cyclopentenone isomer (double bond at a different position), but the conjugated cyclopent-2-en-1-one (option a) is the thermodynamically favored product under the heat condition. Both enantiomers A and B give the same achiral cyclopentenone, confirming a single compound C is formed. Therefore, the correct answer is A.