See image — Aldehydes Ketones and Carboxylic Acids Chemistry Question

💡 Solution & Explanation

Step 1 - Identify the starting material: The starting material is a cyclopentanone with two substituents at C2: a carboxylic acid (COOH) group and a tertiary alcohol group C(OH)(CH3) [i.e., a 1-hydroxy-1-methyl group attached at C2 of the ring]. Step 2 - Reaction with KBrO / H+ (oxidative conditions, haloform-type or oxidative decarboxylation): KBrO under acidic conditions with heat acts as an oxidant. The tertiary alcohol C(OH)(CH3) adjacent to the carbonyl is oxidized. More specifically, KBrO (analogous to hypohalite) under these conditions performs an oxidative cleavage of the alpha-hydroxy ketone or effects a haloform-type reaction on the methyl ketone side. The C(OH)(CH3) group (which is effectively a tertiary alcohol with a methyl) undergoes oxidation: the C-C bond between the ring carbon and the C(OH)(CH3) group cleaves oxidatively, removing the C(OH)(CH3) fragment. This gives intermediate (X) which retains the COOH group at C2 of cyclopentanone — i.e., 2-carboxycyclopentan-1-one. Step 3 - Heating of (X): Compound (X) is 2-carboxycyclopentan-1-one (a beta-keto acid, since the COOH at C2 is beta to the ring carbonyl at C1). Beta-keto acids readily undergo decarboxylation upon heating (loss of CO2). Heating (X) causes loss of CO2, giving cyclopentan-1-one (cyclopentanone) as product (Y). Step 4 - Why other options fail: (a) still retains the hydroxy-methyl group — this would be the product if only the COOH were removed without oxidation of the alcohol; (b) is intermediate (X), not the final product (Y); (d) methylcyclopentane has no ketone and would require reduction, not consistent with these conditions. Therefore, the correct answer is C.