Which of the following on reaction with NaOI will give yellow precipitates ?
(A) Phenylethanone
(B) Sec-Butyl alcohol
(C) Phenylethanal
(D) Methyl n-propylketone
Choose the correct answer from the options given below :
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → (A), (B) and (D) only
Sodium hypoiodite (iodine + NaOH) gives a yellow precipitate of iodoform, \( \ce{CHI3} \), with compounds that have a CH3CO– group (methyl ketones and ethanal) or a CH3CH(OH)– group, which the hypoiodite first oxidises to \( \ce{CH3CO} \)–. This is the iodoform test:
\( \displaystyle \ce{R-CO-CH3 + 3NaOI -> R-COONa + CHI3 v + 2NaOH} \)
(A) Phenylethanone — Correct. Acetophenone, \( \ce{C6H5COCH3} \), is a methyl ketone.
(B) Sec-Butyl alcohol — Correct. Butan-2-ol, \( \ce{CH3CH(OH)CH2CH3} \), has the \( \ce{CH3CH(OH)} \)– group and is oxidised to the methyl ketone butan-2-one, which then gives iodoform.
(C) Phenylethanal — Incorrect. \( \ce{C6H5CH2CHO} \) has no \( \ce{CH3CO} \)– group; its α-carbon carries the phenyl group, not three hydrogens.
(D) Methyl n-propylketone — Correct. Pentan-2-one, \( \ce{CH3COCH2CH2CH3} \), is a methyl ketone.
NCERT (2026-27 edition):
Oxidation of Methyl Ketones by Haloform Reaction
