Iodoform test identifies methyl ketones or oxidizable alcohols

Steps:

• Identify the iodoform test: yellow CHI3 precipitate from CH3COR or CH3CH(OH)R with alkaline I2/NaI.
• Analyze each reaction's product for the required structural motif.
• Check A: 2-chloropropane + dilute NaOH(aq) → 2-propanol (CH3CH(OH)CH3), matches CH3CH(OH)-.
• Eliminate others as products lack the motif.

Why A is correct:

• 2-Propanol oxidizes to acetone (CH3COCH3), a methyl ketone that reacts with alkaline I2 to form CHI3 precipitate per iodoform reaction.

Why the others are wrong:

• B: Ethanol oxidizes to CH3COOH (reflux ensures full oxidation), which lacks CH3CO- or CH3CH(OH)- group.
• C: Methyl ethanoate hydrolyzes to CH3OH and CH3COOH, neither undergoes iodoform test.
• D: Propanal reduces to CH3CH2CH2OH, a primary alcohol not oxidizable to methyl ketone/aldehyde.

Final answer: A