Which among the following compounds give yellow solid when reacted with NaOI / NaOH ?
Choose the correct answer from the options given below :

Solution

Chemistry ma’am · 3 steps

The yellow solid is iodoform: look for CH₃CO⁻, or CH₃CH(OH)⁻ that can become it.

1. What the test needs. NaOI / NaOH gives yellow iodoform, CHI₃, only with a methyl ketone, CH₃CO-R (R is H, alkyl or aryl), or with an alcohol CH₃CH(OH)-R. Hypoiodite first oxidises that alcohol to the methyl ketone.
2. Why acids fail. In a carboxylic acid the CH₃CO⁻ is joined to OH. In NaOH the acid becomes the carboxylate ion, and its α-hydrogens are not acidic enough to be replaced by iodine.

   Compound	Key group	Iodoform?
   (A) Butan-2-ol, CH₃CH(OH)CH₂CH₃	CH₃CH(OH)⁻, oxidised to butan-2-one	Yes
   (B) Propan-1-ol, CH₃CH₂CH₂OH	oxidises to CH₃CH₂CHO, no CH₃CO⁻	No
   (C) Butan-2-one, CH₃COCH₂CH₃	CH₃CO⁻, a methyl ketone	Yes
   (D) Acetic acid, CH₃COOH	CH₃CO⁻ on OH, a carboxylic acid	No
   (E) Propanal, CH₃CH₂CHO	CH₃CH₂CO⁻, the methyl is one carbon too far	No

3. So. Only (A) and (C) give the yellow CHI₃ precipitate, which is option (a).

The idea to keep. For the iodoform test, the CH₃ must sit directly on the carbonyl carbon, or on the carbinol carbon of an alcohol. Ethanal and ethanol pass; acids and esters do not.

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