See image — Aldehydes Ketones and Carboxylic Acids Chemistry Question

💡 Solution & Explanation

Concept: Intramolecular aldol condensation. The starting material is a keto-aldehyde: CH3-C(=O)-CH2-CH2-CH2-CH2-C(=O)-H (hept-6-one-1-al, i.e., 6-oxoheptanal). Under basic conditions (HO-) with heat, an intramolecular aldol condensation occurs. Step 1 - Identify acidic alpha-hydrogens: The aldehyde end (CHO) has no alpha-H on its carbonyl carbon since it is terminal. However, the methyl ketone end (CH3-CO-) has alpha-H on the CH2 adjacent to the ketone, and the CH3 group. The aldehyde carbon itself is the electrophilic carbonyl. Step 2 - Enolate formation: Base abstracts an alpha-proton from the carbon alpha to the ketone. The most favorable intramolecular cyclization considers ring size. The alpha carbon of the ketone (the CH2 directly attached to C=O of ketone, i.e., C2) attacking the aldehyde carbonyl (C7) would form a 6-membered transition state ring... let us count: C2-C3-C4-C5-C6-C7 = 6 atoms in the ring forming a 5-membered carbocycle (5 carbons in ring with O outside). Wait, let us recount the chain: C1(CH3)-C2(C=O)-C3(CH2)-C4(CH2)-C5(CH2)-C6(CH2)-C7(C=O,H). Alpha to the ketone at C2: C1 (CH3) and C3. Alpha to aldehyde at C7: C6. Step 3 - Ring size analysis: If enolate forms at C6 (alpha to aldehyde C7) and attacks ketone C2: ring would contain C2-C3-C4-C5-C6 = 5-membered ring. This gives a cyclopentane ring. After aldol addition and dehydration, a cyclopentenone or cyclopentenyl product forms. Step 4 - Determining regiochemistry: Enolate at C6 attacks ketone carbonyl C2. The aldol product after cyclization gives a 5-membered ring with OH at C2 and the CH3 group exocyclic. After dehydration (elimination of water), the double bond forms between C2 and C3 (or C1-C2), placing the C=C conjugated with the exocyclic acetyl group (CH3CO-). This gives 1-(cyclopent-1-en-1-yl)ethan-1-one, i.e., 1-acetylcyclopent-1-ene, which matches option (b): a cyclopentenyl ring with the C(=O)-CH3 group at the vinylic position (C1 of the double bond), giving an enone system with extended conjugation. Step 5 - Why other options fail: (a) cyclohept-2-en-1-one would require a 7-membered ring, which is less favorable than 5-membered ring formation. (c) requires a different connectivity and does not reflect simple intramolecular aldol. (d) has the acetyl group at a saturated sp3 carbon, which would not be the thermodynamically favored conjugated product after dehydration. The 5-exo-trig cyclization (or equivalently the formation of the 5-membered ring with conjugated enone) at 73% yield is consistent with intramolecular aldol condensation giving the conjugated cyclopentenyl ketone, option (b). Therefore, the correct answer is B.