Compound identification via %C and iodoform test

Steps:

• Compute %C for options: C4H10O isomers (A, B) yield 48/74 × 100 ≈ 64.9%; C4H8O isomers (C, D) yield 48/72 × 100 = 66.7%.
• Verify one O atom per molecule: all options (C4H10O or C4H8O) satisfy this.
• Recognize alkaline I2(aq) reaction producing yellow CHI3 precipitate as positive iodoform test, specific to methyl ketones (CH3COR) or certain alcohols/aldehydes.
• Match criteria: only option with 66.7% C and methyl ketone structure is D.

Why D is correct:

• Butanone (CH3COCH2CH3, C4H8O) has %C = 66.7% and CH3CO– group, which undergoes haloform reaction to form iodoform.

Why the others are wrong:

• A: %C ≈ 64.9%, fails composition; tertiary alcohol, negative iodoform.
• B: %C ≈ 64.9%, fails composition; secondary alcohol gives iodoform but %C mismatch.
• C: %C = 66.7% matches, but butanal (CH3CH2CH2CHO) lacks CH3C=O, negative iodoform.

Final answer: D