Iodoform test for methyl ketones from alcohol oxidation

Steps:

• Yellow precipitate with alkaline I2(aq) indicates iodoform test, positive for methyl ketones (RCOCH3) or acetaldehyde.
• Compound X is an alcohol oxidized to Y, a carbonyl compound; for Y to be a methyl ketone, X must be a secondary alcohol of form RCH(OH)CH3.
• Primary alcohols oxidize to aldehydes (then carboxylic acids), and tertiary do not oxidize to carbonyls.
• Among options, only butan-2-ol (CH3CH(OH)CH2CH3) yields butan-2-one (CH3COCH2CH3), a methyl ketone.

Why B is correct:

• Butan-2-ol oxidizes to butan-2-one, which has the CH3CO- group required for iodoform reaction.

Why the others are wrong:

• A: Butan-1-ol oxidizes to butanal (CH3CH2CH2CHO), which lacks CH3CO- or CH3CH(OH)- and fails iodoform test.
• C: Methylpropan-1-ol (2-methylpropan-1-ol) oxidizes to 2-methylpropanal ((CH3)2CHCHO), not a methyl ketone.
• D: Methylpropan-2-ol (2-methylpropan-2-ol) is tertiary and resists oxidation to any carbonyl compound.

Final answer: B