Carbonyl compound positive for both DNP and iodoform tests

Steps:

• 2,4-DNP reagent forms a yellow precipitate with aldehydes and ketones, confirming carbonyl presence.
• Alkaline aqueous iodine (iodoform test) gives a yellow precipitate with methyl ketones (CH3COR) or acetaldehyde.
• Eliminate A (butan-2-ol), an alcohol without a carbonyl group, so no DNP reaction.
• Among carbonyls (B, C, D), identify methyl ketone structure for iodoform: butanone (CH3COCH2CH3) qualifies, while butanal (CH3CH2CH2CHO) and pentan-3-one (CH3CH2COCH2CH3) do not.

Why C is correct:

• Butanone is a methyl ketone (CH3CO- group), forming hydrazone with DNP and iodoform with alkaline iodine per standard organic test definitions.

Why the others are wrong:

• A: Secondary alcohol lacks carbonyl, fails DNP test.
• B: Straight-chain aldehyde (not acetaldehyde) lacks CH3CO- for iodoform.
• D: Symmetrical ketone without CH3CO- group fails iodoform.

Final answer: C