See image — Aldehydes Ketones and Carboxylic Acids Chemistry Question

💡 Solution & Explanation

Concept: The reaction shown is a base-catalyzed isomerization of one alpha,beta-unsaturated cyclopentenone to another. The key mechanistic question is: what open-chain diketone intermediate could undergo an intramolecular aldol condensation (with dehydration) to give both the starting material and the product? Step 1 – Identify the carbon skeletons. The reactant is 2-ethylidenecyclopentan-1-one and the product is 2-(dimethylmethylidene)cyclopentan-1-one (isopropylidenecyclopentenone). Both are five-membered ring enones. Numbering the carbons: the ring has 5 carbons plus the exocyclic alkene substituent. Step 2 – Retro-aldol analysis. Under base catalysis, an intramolecular aldol condensation of a diketone can form cyclopentenones. Working backward (retro-aldol + hydration of the double bond), the cyclopentenones can be opened to a linear diketone. The retro-aldol of 2-ethylidenecyclopentan-1-one and 2-isopropylidenecyclopentan-1-one should lead to the same diketone intermediate. Step 3 – Identify the diketone. For a cyclopentenone formed by intramolecular aldol condensation of a 1,6-diketone (heptanedione), the ring closure involves C1 and C5 (or equivalent). Heptane-2,5-dione (CH3-CO-CH2-CH2-CO-CH2CH3, i.e., 5-methylheptane-2,5-dione is not right; the correct compound is heptane-2,5-dione: CH3-C(=O)-CH2-CH2-C(=O)-CH2CH3) can undergo base-catalyzed intramolecular aldol condensation. The alpha carbon of one ketone attacks the other ketone carbonyl, forming a five-membered ring. Depending on which alpha carbon is deprotonated, one gets either the ethylidene product or the isopropylidene product — this accounts for the equilibrium between the two cyclopentenones. Step 4 – Confirm option (c). Option (c) is drawn as a 1,5-diketone (heptanedione): CH3-C(=O)-CH2-CH2-C(=O)-CH2CH3 (heptane-2,5-dione). This diketone has two different sets of alpha hydrogens. Base-catalyzed intramolecular aldol of the terminal methyl alpha proton gives the ethylidene cyclopentenone, and deprotonation at the other alpha position gives the isopropylidene (dimethyl) product. Thus option (c) is the common intermediate for both isomers. Step 5 – Eliminate other options. (a) is a linear enone, not a diketone, so it cannot undergo intramolecular aldol to form the cyclopentanone ring directly. (b) is also a linear enone (monoketone with alkene), not suitable as an aldol intermediate for ring formation in this context. (d) is excluded because (c) correctly fits. Therefore, the correct answer is C.