Haloform reaction cleavage of methyl ketone

Steps:

• Butan-2-one (CH₃COCH₂CH₃) is a methyl ketone that undergoes haloform reaction with alkaline I₂, triiodinating the methyl group.
• This leads to cleavage, forming sodium propanoate (CH₃CH₂COONa) and iodoform (CHI₃).
• Adding excess dilute H₂SO₄ acidifies the mixture, protonating the carboxylate to propanoic acid (CH₃CH₂COOH).
• The organic products are iodoform (noted as iodomethane in the question) and propanoic acid as product M.

Why D is correct:

• In the haloform reaction, RCOCH₃ cleaves to RCOOH + CHI₃ (after acidification), where R = CH₂CH₃, so RCOOH is CH₃CH₂COOH (propanoic acid).

Why the others are wrong:

• A: Ethanoate ion forms from acetone (CH₃COCH₃), not butan-2-one.
• B: Ethanoic acid is the acidified form from acetone, not butan-2-one.
• C: Propanoate ion is the intermediate before acidification, but the final product is the acid.

Final answer: D