Which of the following gives iodoform test?

The correct answer is Option (2) → \(CH_3CH_2OH\)

The iodoform test (\( \ce{I2} \)/NaOH, yellow \( \ce{CHI3} \)) is given by compounds with a CH₃CO– group, and by alcohols with a CH₃CH(OH)– group, which hypoiodite first oxidises to \( \ce{CH3CO} \)–. Ethanol is oxidised to ethanal, \( \ce{CH3CHO} \), which then gives iodoform; the net change is:

    \( \displaystyle \ce{CH3CH2OH + 4I2 + 6NaOH -> CHI3 v + HCOONa + 5NaI + 5H2O} \)

Why the other options are incorrect

• Option (1) \(CH_3OH\) — Incorrect. Methanol has no carbon next to the carbinol carbon; it is oxidised to methanal, which cannot form iodoform.

• Option (3) \(C_2H_5COC_2H_5\) — Incorrect. Pentan-3-one has ethyl groups on both sides of C=O and no \( \ce{CH3CO} \)– group.

• Option (4) \(HCHO\) — Incorrect. Methanal has only hydrogen atoms on the carbonyl carbon, no methyl group.

NCERT (2026-27 edition):
Oxidation of Methyl Ketones by Haloform Reaction – “Aldehydes and ketones having at least one methyl group linked to the carbonyl carbon atom (methyl ketones) are oxidised by sodium hypohalite to sodium salts of corresponding carboxylic acids having one carbon atom less than that of carbonyl compound. The methyl group is converted to haloform.”

Note: The same/similar question has been asked in different years by NTA as per the details given below: