JEE Main · 2024 · 09 Apr 2024 · Shift 2 · Aldehydes Ketones and Carboxylic Acids

A positive iodoform test (aq. KOH then KOI) requires either a methyl ketone (or an acetaldehyde-type group) or a compound that is converted into one.

The compound in option (3), ${\text{CH}}_3{\text{CH}}_2{\text{CH}}_2\text{CHO}$, is an aldehyde with no methyl group attached to the carbonyl carbon, so it cannot give a positive iodoform test.

The compound in option (2) is a secondary chloride (${\text{CH}}_3{\text{CH}}_2{\text{CHCl CH}}_3$-type) that on treatment with aq. KOH is hydrolysed to the corresponding secondary alcohol (${\text{CH}}_3{\text{CH}}_2{\text{CH(OH)CH}}_3$), which has a ${\text{CH}}_3$–CH(OH)– unit; potassium hypoiodite then oxidises this to the methyl ketone in situ and cleaves it to give a carboxylate salt and a yellow precipitate of iodoform:

So option (2) gives the positive iodoform test.