Carbonyl compound identified by reduction product and iodoform test

Steps:

• LiAlH4 reduces aldehydes to primary alcohols and ketones to secondary alcohols, so X must be a ketone.
• Alkaline I2 (iodoform test) gives yellow precipitate (CHI3) with methyl ketones (CH3COR) or acetaldehyde, so X is a methyl ketone.
• Among options, only butanone (CH3COCH2CH3) is a methyl ketone.
• Other ketones like pentan-3-one lack the CH3CO- group and fail iodoform test.

Why B is correct:

• Butanone (CH3COCH2CH3) is a methyl ketone that reduces with LiAlH4 to butan-2-ol (secondary alcohol) and gives iodoform precipitate.

Why the others are wrong:

• A. Butanal: Aldehyde reduces to primary alcohol (butan-1-ol), not secondary.
• C. Ethanal: Acetaldehyde reduces to primary alcohol (ethanol), not secondary.
• D. Pentan-3-one: Symmetric ketone (CH3CH2COCH2CH3) lacks methyl ketone structure, fails iodoform test.

Final answer: B