Iodoform test is given by which one of the following compounds?
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → Acetophenone
The iodoform test is given by compounds that contain the CH3CO– group (methyl ketones and ethanal), or the CH3CH(OH)– group, which is first oxidised to \( \ce{CH3CO} \)– by the reagent. With iodine and sodium hydroxide, the three hydrogens of the methyl group are replaced by iodine, and the \( \ce{CI3} \) group is then split off as iodoform (CHI3), a yellow precipitate.
Acetophenone, \( \ce{C6H5COCH3} \), is a methyl ketone, so it gives the test:
\( \displaystyle \ce{C6H5COCH3 + 3I2 + 4NaOH -> C6H5COONa + CHI3 v + 3NaI + 3H2O} \)
Why the other options are incorrect
Option (1) Benzophenone — Incorrect. \( \ce{C6H5COC6H5} \) has no methyl group on the carbonyl carbon.
Option (3) Pentan-3-one — Incorrect. \( \ce{CH3CH2COCH2CH3} \) has ethyl groups, not a methyl group, on the carbonyl carbon.
Option (4) 2-Methylpropan-2-ol — Incorrect. \( \ce{(CH3)3COH} \) is a tertiary alcohol with no hydrogen on the carbon carrying –OH, so it cannot be oxidised to a methyl ketone.
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. This oxidation does not affect a carbon-carbon double bond, if present in the molecule. Iodoform reaction with sodium hypoiodite is also used for detection of CH3CO group or CH3CH(OH) group which produces CH3CO group on oxidation.”