Iodoform test distinguishes methyl ketones from other ketones Steps:

• Alkaline aqueous iodine (iodoform reagent) reacts with methyl ketones (CH₃COR) to form yellow iodoform precipitate via haloform reaction.
• Examine structures: pentan-2-one is CH₃COCH₂CH₂CH₃ (methyl ketone, positive); pentan-3-one is CH₃CH₂COCH₂CH₃ (no CH₃CO group, negative).
• Pairs with differing results (precipitate vs. no precipitate) can be distinguished.
• Option B shows this difference, allowing distinction.

Why B is correct:

• Pentan-2-one undergoes haloform reaction due to CH₃C(O)- group, cleaving to CH₃I and forming iodoform; pentan-3-one lacks methyl group adjacent to carbonyl, so no reaction.

Why the others are wrong:

• A: Butanal is an aldehyde, not a ketone; test not applicable for ketone pair distinction.
• C: Propanone (methyl ketone) positive; propan-2-ol oxidizes to propanone, also positive—no difference.

Final answer: B