Which of the following compounds will respond to iodoform test?
(A) Acetone
(B) Ethyl methyl ketone
(C) Acetophenone
(D) Pentan-2-one
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → (A), (B), (C) and (D)
The iodoform test is given by compounds that contain the \( \ce{CH3CO-} \) group (methyl ketones and ethanal), or the \( \ce{CH3CH(OH)-} \) group, which is oxidised to it. With iodine and sodium hydroxide (sodium hypoiodite), the methyl group is converted into iodoform, a yellow solid with a typical smell, and the rest of the molecule becomes the sodium salt of an acid with one carbon atom less:
\( \displaystyle \ce{R-CO-CH3 + 3NaOI -> R-COONa + CHI3 v + 2NaOH} \)
(A) Acetone, \( \ce{CH3COCH3} \) — Correct. It is a methyl ketone (R = \( \ce{CH3} \)); it gives iodoform and sodium acetate.
(B) Ethyl methyl ketone, \( \ce{CH3COCH2CH3} \) — Correct. The \( \ce{CH3CO-} \) group is present; it gives iodoform and sodium propanoate.
(C) Acetophenone, \( \ce{C6H5COCH3} \) — Correct. An aryl methyl ketone also has the \( \ce{CH3CO-} \) group; it gives iodoform and sodium benzoate.
(D) Pentan-2-one, \( \ce{CH3COCH2CH2CH3} \) — Correct. The carbonyl is at C2, next to a methyl group; it gives iodoform and sodium butanoate.
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.”