The iodoform test is used for the detection of
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \(COCH_3\)
In the iodoform test the compound is warmed with iodine and sodium hydroxide (sodium hypoiodite). A compound containing the \( \ce{CH3CO-} \) group (methyl ketones and acetaldehyde) has its methyl group converted into iodoform, a yellow precipitate, while the rest of the molecule becomes the salt of a carboxylic acid with one carbon atom less:
\( \displaystyle \ce{CH3COCH3 + 3I2 + 4NaOH -> CHI3 v + CH3COONa + 3NaI + 3H2O} \)
Alcohols with the \( \ce{CH3CH(OH)-} \) group also respond, because hypoiodite first oxidises them to the \( \ce{CH3CO-} \) group. So the test detects the \( \ce{COCH3} \) group.
Why the other options are incorrect
Option (2) \(COO^-\) — Incorrect. The carboxylate ion is the product left after the methyl group is removed; it does not give iodoform itself.
Option (3) \(CHO\) — Incorrect. An aldehyde group in general is detected by Tollens’ or Fehling’s reagent; among aldehydes only acetaldehyde, which has a \( \ce{CH3CO} \)– group, gives iodoform.
Option (4) \(OH\) — Incorrect. Most alcohols do not give the test; only those with the \( \ce{CH3CH(OH)} \)– group, such as ethanol and propan-2-ol, do.
NCERT (2026-27 edition):
Oxidation – “… (iii) 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.”