Which of the following will not give iodoform test?

The correct answer is Option (2) → Benzyl alcohol

The iodoform test uses iodine and sodium hydroxide (sodium hypoiodite). A compound gives yellow iodoform, \( \ce{CHI3} \), only if it has a \( \ce{CH3CO-} \) group, or a \( \ce{CH3CH(OH)-} \) group that the reagent first oxidises to \( \ce{CH3CO-} \).

Option (1) Ethanal — Gives the iodoform test. \( \ce{CH3CHO} \) has the \( \ce{CH3CO-} \) group (the carbonyl carbon carries H):

    \( \displaystyle \ce{CH3CHO + 3I2 + 4NaOH -> CHI3 + HCOONa + 3NaI + 3H2O} \)

Option (2) Benzyl alcohol — Does not give the iodoform test. In \( \ce{C6H5CH2OH} \) the carbon carrying –OH is joined to a phenyl group and two H atoms, not to a methyl group. Oxidation gives benzaldehyde, \( \ce{C6H5CHO} \), which has no \( \ce{CH3CO-} \) group, so no iodoform forms.

Option (3) Ethanol — Gives the iodoform test. \( \ce{CH3CH2OH} \) has the \( \ce{CH3CH(OH)-} \) unit (with H as the other group). It is oxidised to ethanal, which then reacts as in Option (1):

    \( \displaystyle \ce{CH3CH2OH + 4I2 + 6NaOH -> CHI3 + HCOONa + 5NaI + 5H2O} \)

Option (4) Isopropyl alcohol — Gives the iodoform test. \( \ce{(CH3)2CHOH} \) has the \( \ce{CH3CH(OH)-} \) unit. It is oxidised to acetone, a methyl ketone:

    \( \displaystyle \ce{(CH3)2CHOH + 4I2 + 6NaOH -> CHI3 + CH3COONa + 5NaI + 5H2O} \)

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.”

“Iodoform reaction with sodium hypoiodite is also used for detection of CH₃CO group or CH₃CH(OH) group which produces CH₃CO group on oxidation.”