Iodoform Test for Methyl Carbinols

Steps:

• Yellow precipitate with alkaline I₂(aq) indicates positive iodoform reaction.
• This test detects alcohols with CH₃CH(OH)- group, oxidizable to CH₃CO- methyl ketones.
• Compare structures: identify option with CH₃CH(OH)R where R is H, alkyl, or aryl.
• Option D matches CH₃CH(OH)- structure.

Why D is correct:

• D has CH₃CH(OH)- group, which oxidizes to methyl ketone and forms CHI₃ precipitate per iodoform mechanism.

Why the others are wrong:

• A lacks CH₃CH(OH)-; it's a primary alcohol without methyl ketone potential.
• B is a tertiary alcohol; cannot oxidize to carbonyl.
• C has CH₃CH₂CH(OH)-; oxidizes to ketone without CH₃CO- group.

Final answer: D