Which of the following compounds gives a positive iodoform test?

The correct answer is Option (4) → 1-Phenylethanol

• The iodoform test (iodine and sodium hydroxide, that is, sodium hypoiodite) is given by compounds with a \( \ce{CH3CO-} \) group or a \( \ce{CH3CH(OH)-} \) group, which is first oxidised to \( \ce{CH3CO-} \).

• 1-Phenylethanol, \( \ce{C6H5CH(OH)CH3} \), has the \( \ce{CH3CH(OH)-} \) group. It is oxidised to acetophenone, whose methyl group ends up as iodoform:

      \( \displaystyle \ce{C6H5CH(OH)CH3 + 4I2 + 6NaOH -> C6H5COONa + CHI3 + 5NaI + 5H2O} \)

Why the other options are incorrect

Option (1) 2-Phenylethanol — Incorrect. In \( \ce{C6H5CH2CH2OH} \) the –OH is on a terminal \( \ce{CH2} \); there is no methyl group on the carbinol carbon.

Option (2) Pentanal — Incorrect. In \( \ce{CH3CH2CH2CH2CHO} \) the carbonyl carbon carries a hydrogen and a butyl group, not a methyl group.

Option (3) Pentan-3-ol — Incorrect. In \( \ce{CH3CH2CH(OH)CH2CH3} \) the carbinol carbon carries two ethyl groups; oxidation gives pentan-3-one, not a methyl ketone.

NCERT (2026-27 edition):
8.4 Chemical Reactions – “Iodoform reaction with sodium hypoiodite is also used for detection of CH₃CO group or CH₃CH(OH) group which produces CH₃CO group on oxidation.”