Iodoform Test for Methyl Carbinol Alcohols

Steps:

• Yellow precipitate with alkaline I2(aq) indicates positive iodoform reaction.
• This test is specific to alcohols with CH3CH(OH)R structure (secondary) or CH3CH2OH (primary), oxidizable to methyl ketones.
• Examine structures: A, B, C lack the required CH3CH(OH)- or CH3CH2OH group.
• D contains CH3CH(OH)R, confirming it as X.

Why D is correct:

• D has the CH3CH(OH)- group, which oxidizes to CH3COR (methyl ketone), reacting with I2/OH- to form CHI3 (yellow ppt) per iodoform mechanism.

Why the others are wrong:

• A: Primary alcohol without CH3CH2OH; no methyl ketone formation.
• B: Tertiary alcohol; cannot oxidize to carbonyl.
• C: Secondary alcohol without adjacent CH3 group; yields non-methyl ketone.

Final answer: D