See image — Alcohols Phenols and Ethers Chemistry Question

💡 Solution & Explanation

Concept: The iodoform test is positive for methyl ketones (CH3-CO-R), acetaldehyde (CH3CHO), and secondary alcohols of the type CH3-CH(OH)-R. For the test to be positive after hydrolysis followed by heating, the compound must yield one of these upon those treatments. Key reaction: Beta-ketoesters and malonic ester derivatives undergo hydrolysis (saponification of ester groups) followed by decarboxylation upon heating to give simpler carbonyl compounds. Option (a): CH3-CH(CO2Et)-C(=O)-CH2-CH3 is a beta-ketoester. On hydrolysis and heating (decarboxylation), the CO2Et group is lost as CO2, giving CH3-CH2-C(=O)-CH2-CH3 (3-pentanone / pentan-3-one). This is a diethyl ketone with no methyl ketone group, so it does NOT give a positive iodoform test. Option (b): The compound is a cyclobutane ring bearing both a CO2Et group and a -C(=O)-CH3 (acetyl) group on the same carbon. This is a beta-ketoester (the ester and ketone are on adjacent carbons in the cyclobutane framework, or on the same carbon). On hydrolysis, the ester is converted to carboxylic acid. On heating, decarboxylation occurs, leaving a cyclobutyl methyl ketone: cyclobutane ring with a -C(=O)-CH3 group. This product is a METHYL KETONE (CH3CO- group), which gives a POSITIVE iodoform test. Option (c): CH3-CH(CO2Et)2 is a malonic ester derivative. On hydrolysis and heating (decarboxylation), it gives CH3-CH2-CO2H (propanoic acid) or after loss of CO2 gives CH3CH2COOH. This is a carboxylic acid, not a methyl ketone, so it does NOT give iodoform test. Option (d): The compound appears to be an ethyl ester of cyclobutanone carboxylic acid (cyclobutanone ring with -C(=O)-O-Et). On hydrolysis, the ester becomes a carboxylic acid. On heating, decarboxylation gives cyclobutanone. Cyclobutanone is a cyclic ketone but NOT a methyl ketone, so it does NOT give a positive iodoform test. Therefore, only option (b) gives a methyl ketone (acetyl group on cyclobutane) after hydrolysis and decarboxylation, which gives a positive iodoform test. Therefore, the correct answer is B.