Which of the following compound can give positive iodoform test when treated with aqueous KOH solution followed by potassium hypoiodite.

Solution

Chemistry ma’am · 3 steps

Let aqueous KOH act first, then look for a CH₃CO⁻ group in what it makes.

1. What iodoform needs. Hypoiodite (OI⁻) gives the yellow CHI₃ precipitate only with a CH₃CO⁻ group, or with a CH₃CH(OH)⁻ group that it can oxidise to one. So decide what each compound becomes in aqueous KOH, then check that product.
2. Gem-dihalide hydrolysis. In (c), both Cl atoms sit on the same carbon. Aqueous KOH swaps them for two OH groups. Two OH groups on one carbon are unstable, so this gem-diol loses water and gives a ketone.

   CH₃CH₂CCl₂CH₃ →(aq. KOH) CH₃CH₂C(OH)₂CH₃ →(-H₂O) CH₃CH₂COCH₃

   Compound	After aq. KOH	CH₃CO⁻ or CH₃CH(OH)⁻?	Iodoform
   (a) pentan-3-one	unchanged, C₂H₅COC₂H₅	no, ethyl on both sides	negative
   (b) ethyloxirane	ring opens to HOCH₂CH(OH)CH₂CH₃	no, the CH(OH) carries CH₂OH, not CH₃	negative
   (c) 2,2-dichlorobutane	butan-2-one, CH₃COC₂H₅	yes, a methyl ketone	positive
   (d) butanal	unchanged, CH₃CH₂CH₂CHO	no, the CO carries H, not CH₃	negative

3. So. Only (c) turns into a methyl ketone, butan-2-one, which hypoiodite converts to yellow CHI₃. The answer is (c).

The idea to keep. When a reagent comes first, test the product, not the starting compound. A gem-dihalide on a carbon next to CH₃ hydrolyses to a methyl ketone, so it passes the iodoform test.

Answer from the NTA final answer key. Solution by PrepWiser.