A yellow coloured compound formed when methyl ketone is oxidised by sodium hypoiodite is:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \(CHI_3\)
A methyl ketone (\( \ce{CH3CO-R} \)) is oxidised by sodium hypoiodite (NaOI, from \( \ce{I2} \) + NaOH): the three hydrogens of the methyl group are replaced by iodine and the \( \ce{CI3} \) group then splits off as iodoform (triiodomethane, CHI3), a yellow solid with a typical smell. The rest of the molecule becomes the sodium salt of an acid with one carbon less:
\( \ce{R-CO-CH3} \) + 3NaOI → R–COONa + \( \ce{CHI3} \)↓ (yellow) + 2NaOH
Why the other options are incorrect
Option (1) \(CI_4\) — Incorrect. Tetraiodomethane has no hydrogen; the methyl carbon leaves as \( \ce{CHI3} \), keeping one H.
Option (3) \(CH_2I_2\) — Incorrect. di-iodomethane would need only two H atoms replaced; in the haloform reaction all three H atoms of \( \ce{CH3} \) are replaced.
Option (4) \(CH_3I\) — Incorrect. Methyl iodide is not formed in this oxidation; it is a colourless liquid, not the yellow precipitate.
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.”