See image — Aldehydes Ketones and Carboxylic Acids Chemistry Question

💡 Solution & Explanation

The iodoform test is positive for compounds that have the structural motif CH3C(=O)- (methyl ketone) or CH3CH(OH)- (secondary alcohol where OH is on a carbon bearing a methyl group), because NaOI (I2/NaOH) can oxidize such secondary alcohols to methyl ketones and then perform trihalogenation/cleavage to give CHI3 (iodoform). Step 1 - Identify the iodoform-positive structural requirement: A compound gives a positive iodoform test if it contains: (i) a methyl ketone (CH3-C=O), or (ii) a secondary alcohol of the type CH3-CH(OH)-R (which gets oxidized to a methyl ketone under the reaction conditions). Step 2 - Analyze each option: (a) 1-acetyl-1-methylcyclohexane: contains a -C(=O)-CH3 (methyl ketone) group directly. This is a methyl ketone, so it GIVES a positive iodoform test. (b) Ph-CH=CH-CH(OH)-CH3: This is a secondary alcohol with the OH on a carbon that also bears a methyl group (CH3-CH(OH)-). Under basic iodine conditions, it can be oxidized to Ph-CH=CH-C(=O)-CH3 (a methyl ketone via enone). Thus it CAN give iodoform test (the CH3-CH(OH) unit qualifies). Positive iodoform test. (c) H2C=CH-C(OH)(CH3)(Ph): This compound has an OH group on a carbon bearing methyl, phenyl, and a vinyl group. The carbon with OH is tertiary (connected to vinyl, methyl, phenyl, and OH). For a positive iodoform test, we need CH3-CH(OH)- (a methethine unit) or a methyl ketone. Here, the OH-bearing carbon has no H on it (tertiary alcohol) and the methyl group is directly on the same carbon as OH but there is no CH3-CH(OH) arrangement—it is CH3-C(OH) where C is also bonded to Ph and vinyl, giving a tertiary alcohol. This cannot be oxidized to a methyl ketone under mild iodoform conditions because tertiary alcohols are not oxidized by NaOI. The molecule does not contain a pre-existing methyl ketone either. Therefore, this compound is INCAPABLE of giving the iodoform test. (d) PhCHD-CH2-C(=O)-CH3: contains a -C(=O)-CH3 (methyl ketone) group. This is a methyl ketone and GIVES a positive iodoform test. Step 3 - Conclusion: Only option (c) lacks both a methyl ketone and a CH3-CH(OH)- secondary alcohol motif (it has a tertiary alcohol that cannot be oxidized), making it incapable of showing the iodoform test. Therefore, the correct answer is C.