See image — Aldehydes Ketones and Carboxylic Acids Chemistry Question

💡 Solution & Explanation

Step 1 – Identify the starting material: The starting material is 1-methyl-1-(hydroxymethyl)cyclopentane, a cyclopentane bearing both a methyl group and a –CH2OH group at the same carbon (C1). Step 2 – Reaction A (H+/heat, acid-catalyzed dehydration/ring expansion): Under acidic conditions with heat, the primary alcohol (–CH2OH) is protonated and loses water to generate a primary carbocation at the exocyclic carbon. This primary carbocation undergoes a 1,2-alkyl shift (ring expansion): the C1–C2 bond of the cyclopentane migrates to the exocyclic carbocation, expanding the five-membered ring to a six-membered ring and placing the positive charge at C1 of the new ring (which is tertiary). Elimination of a proton then gives methylenecyclohexane or, more precisely, after proton loss from the ring carbon adjacent to the cationic center, the thermodynamically favored product is 1-methylcyclohex-1-ene (methylcyclohexene). Actually, tracing carefully: starting cyclopentane C1 has methyl and CH2OH. Protonation of OH gives CH2+; the ring C1–C2 bond migrates to give a six-membered ring carbocation at what was C1 (now tertiary, bearing the methyl). Deprotonation of an adjacent CH gives 1-methylcyclohex-1-ene (A), a six-membered ring with a trisubstituted double bond between C1 (bearing methyl) and C2. Step 3 – Reaction B (O3/Zn, ozonolysis with reductive workup): Ozonolysis of 1-methylcyclohex-1-ene cleaves the double bond. Since the double bond is endocyclic in a six-membered ring, cleavage opens the ring and gives a single bifunctional compound: a six-carbon chain (actually a 1,6-dicarbonyl compound). The double bond between C1 (bearing methyl, so it becomes a ketone end) and C2 (CH, becomes an aldehyde end). Cleavage of the ring double bond in 1-methylcyclohex-1-ene gives a linear dialdehyde/keto-aldehyde: specifically 6-oxoheptanal (CH3–C(=O)–(CH2)3–CHO), i.e., a 7-carbon chain with a ketone at one end and an aldehyde at the other: OHC–CH2–CH2–CH2–C(=O)–CH3. Step 4 – Reaction C (HO-/heat, intramolecular aldol condensation): The keto-aldehyde (B) = 6-oxoheptanal undergoes intramolecular aldol condensation under basic conditions with heat. The base deprotonates the alpha carbon of the ketone (alpha to the methyl ketone), and the resulting enolate attacks the aldehyde carbonyl intramolecularly, forming a five-membered ring beta-hydroxy carbonyl intermediate (aldol product), which then undergoes dehydration (heat) to give the alpha,beta-unsaturated cyclic ketone. The ring formed has 5 members: the enolate carbon attacks the aldehyde carbon to close a 5-membered ring. The product after aldol and dehydration is 1-(cyclopent-1-en-1-yl)ethan-1-one, i.e., a cyclopentenone ring bearing an acetyl (COCH3) substituent on the double bond — which corresponds to option (c). Wait — re-examining: the ozonolysis of 1-methylcyclohex-1-ene (double bond between C1-methyl and C2) gives: C1 end → ketone (CH3CO–), C2 end → aldehyde (–CHO), with a –(CH2)3– tether between them → OHC–(CH2)3–CO–CH3 (6-oxoheptanal). Intramolecular aldol: alpha carbons of the methyl ketone (the CH3 or the CH2 next to C=O). Deprotonation of the CH2 alpha to the ketone (C5), enolate attacks the aldehyde (C1) to close a 5-membered ring (atoms: C1(aldehyde)–C2–C3–C4–C5(enolate) = 5 atoms in ring). Aldol adduct is a cyclopentanol-ketone; dehydration gives the cyclopentenone ring fused with the acetyl. Product: 2-acetylcyclopent-2-en-1-ol after aldol, then dehydration gives 2-acetylcyclopent-2-en-1-one? No — the aldehyde becomes C=O (ketone in ring if substituted, or aldehyde). The aldehyde carbon becomes part of the ring as C=O; combined with the enolate forming the C–C bond, after aldol and dehydration we get a cyclopentenone with a methyl ketone exocyclic substituent = 1-(cyclopent-1-en-1-yl)ethanone, option (c). Why other options fail: (a) cyclohexenone would require a 6-membered ring aldol which would need a 7-carbon dialdehyde, not formed here. (b) cyclopentenone without substituent doesn't account for the methyl group from the starting material. (d) has CHO and methyl on a cyclopentene, inconsistent with the aldol product. Therefore, the correct answer is C.