See image — Alcohols Phenols and Ethers Chemistry Question

💡 Solution & Explanation

The iodoform test (reaction with I2/NaOH) gives a positive result for compounds that have the structural feature CH3CH(OH)- (methyl carbinol) or CH3CO- (methyl ketone), or ethanol, or acetaldehyde — essentially any compound that can generate a CH3CO- group upon oxidation or directly contains it. Step 1 - Concept: The iodoform reaction is positive for: • Methyl ketones (CH3COR) • Secondary alcohols of the type CH3CH(OH)R (which are oxidized to methyl ketones by NaOH/I2) • Ethanol and acetaldehyde Step 2 - Analyze option (a): CH3CH2CH(OH)CH3 is 2-butanol. It has the structural unit CH3CH(OH)-, meaning the OH-bearing carbon carries a methyl group. Under NaOH/I2 conditions, it is oxidized to methyl ethyl ketone (CH3COCH2CH3), which is a methyl ketone. Methyl ketones undergo iodoform reaction to give CHI3. So option (a) is POSITIVE. Step 3 - Analyze option (b): (CH3)2CHCOC2H5 is 3-methyl-2-pentanone — wait, more precisely it is isopropyl ethyl ketone: the carbonyl is flanked by isopropyl and ethyl groups. There is no CH3CO- group (the carbonyl carbon is not attached to a methyl group); it is (CH3)2CH-CO-C2H5. Neither side of the carbonyl is a simple methyl, so this does NOT give iodoform. NEGATIVE. Step 4 - Analyze option (c): CH3-C(=O)-OCH3 is methyl acetate (an ester). Esters generally do not give a positive iodoform test under mild warming conditions with NaOH/I2; NaOH would saponify it to acetate and methanol, neither of which gives iodoform efficiently. NEGATIVE. Step 5 - Analyze option (d): (CH3)2CHCH2OH is isobutanol (2-methyl-1-propanol), a primary alcohol where the carbon bearing OH has no methyl group directly — it is -CH2OH type. Oxidation gives isobutyraldehyde (CH3)2CHCHO, which also lacks a CH3CO- unit. NEGATIVE. Step 6 - Conclusion: Only option (a) satisfies the structural requirement for a positive iodoform test. Therefore, the correct answer is A.