Which of the following will give yellow precipitate with \(NaOH\) and \(I_2\)?
(A) \(C_6H_5-CO-CH_3\)
(B) \(C_6H_5-CH_2-CHO\)
(C) \(C_6H_5-CH_2-CH(OH)-CH_3\)
(D) \(C_6H_5-CH_2-CH_2-OH\)
(E) \(C_6H_5-CO-CH_2-CHO\)
Choose the correct answer from the options given below :
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → (A) and (C) only
NaOH and \( \ce{I2} \) together form sodium hypoiodite (NaOI). In the iodoform test, a compound with a CH3CO– group (a methyl ketone, or acetaldehyde) is oxidised: the methyl group is converted to iodoform, CHI3, a yellow precipitate, and the rest of the molecule becomes the sodium salt of an acid with one carbon less. A compound with a CH3CH(OH)– group also gives the test, because hypoiodite first oxidises it to \( \ce{CH3CO} \)–.
\( \ce{C6H5COCH3 + 3NaOI -> C6H5COONa + CHI3 v + 2NaOH} \)
(A) \( \ce{C6H5-CO-CH3} \) (acetophenone) — Correct. A methyl ketone: it has the \( \ce{CH3CO} \)– group and gives iodoform and sodium benzoate.
(B) \( \ce{C6H5-CH2-CHO} \) (phenylacetaldehyde) — Incorrect. The carbonyl carbon carries H and \( \ce{C6H5CH2} \)–, not \( \ce{CH3} \); there is no \( \ce{CH3CO} \)– group.
(C) \( \ce{C6H5-CH2-CH(OH)-CH3} \) (1-phenylpropan-2-ol) — Correct. It has the \( \ce{CH3CH(OH)} \)– group; it is oxidised to the methyl ketone \( \ce{C6H5CH2COCH3} \), which gives iodoform.
\( \ce{C6H5CH2CH(OH)CH3 + NaOI -> C6H5CH2COCH3 + NaI + H2O} \)
\( \ce{C6H5CH2COCH3 + 3NaOI -> C6H5CH2COONa + CHI3 v + 2NaOH} \)
(D) \( \ce{C6H5-CH2-CH2-OH} \) (2-phenylethanol) — Incorrect. A primary alcohol with no \( \ce{CH3} \) group; oxidation gives \( \ce{C6H5CH2CHO} \), not a methyl ketone.
(E) \( \ce{C6H5-CO-CH2-CHO} \) — Incorrect. Neither carbonyl group carries a \( \ce{CH3} \) group, so no iodoform is formed.
NCERT (2026-27 edition):
Oxidation of Methyl Ketones by Haloform Reaction
