See image — Aldehydes Ketones and Carboxylic Acids Chemistry Question

💡 Solution & Explanation

Step 1 - Identify the starting material: The compound is CH3-C(=O)-CH2-C(CH3)2-CHO, which is 2,2-dimethyl-4-oxopentanal (a keto-aldehyde). It has a ketone (CH3CO-) at one end and an aldehyde (-CHO) at the other end, separated by a CH2 and a quaternary carbon bearing two methyl groups. Step 2 - Reaction conditions: KOH/H2O is a base-catalyzed intramolecular aldol condensation condition. Step 3 - Intramolecular aldol mechanism: Under basic conditions, the alpha carbon of the ketone (the -CH2- between the ketone and the quaternary carbon) is deprotonated to form an enolate. This enolate carbon then attacks the aldehyde carbonyl intramolecularly. The aldehyde carbon is 3 carbons away from the alpha carbon of the ketone (counting: alpha-C, quaternary-C, aldehyde-C), forming a 5-membered ring upon cyclization (the ring includes: enolate carbon, quaternary carbon, aldehyde carbon, plus the new C-C bond and the oxygen becomes OH). Step 4 - Aldol addition product: The intramolecular attack of the enolate on the aldehyde gives a beta-hydroxy ketone in a 5-membered ring. The ring formed contains: C(ketone)-CH(-)-C(CH3)2-CH(OH)- closing back, giving a cyclopentane ring with a ketone and a hydroxyl group. Step 5 - Dehydration: Under the basic aqueous conditions with heat, the beta-hydroxy ketone undergoes dehydration (elimination of water) to give an alpha,beta-unsaturated ketone (enone). The OH on the former aldehyde carbon and the H on the adjacent carbon (the alpha carbon to ketone) are eliminated to form a double bond. Step 6 - Product structure: The product is a cyclopentenone ring. The ring is 5-membered with: the carbonyl (ketone) at C1, a double bond between C4 and C5 (alpha,beta to the carbonyl through the ring), and gem-dimethyl substituents at C3. This corresponds to 4,4-dimethylcyclopent-2-en-1-one or described as 3,3-dimethylcyclopent-4-en-1-one depending on numbering. Looking at option (b): it shows a 5-membered ring with the ketone at the top, a double bond on the lower-left portion, and gem-dimethyl at the bottom carbon - this matches the product of the intramolecular aldol condensation where the gem-dimethyl quaternary carbon is retained in the ring and the double bond forms between the former alpha-carbon of ketone and the former aldehyde carbon. Step 7 - Why other options fail: - Option (a): has gem-dimethyl adjacent to the carbonyl (C2 position), which would require the quaternary carbon to be alpha to the ketone - not consistent with the connectivity. - Option (c): is a 6-membered ring product, but the chain length only supports a 5-membered ring cyclization. - Option (d): is a 5-membered ring with only one methyl substituent, losing the gem-dimethyl pattern of the starting material. Therefore, the correct answer is B.