In the following molecules, which will not give positive Iodoform tes…

The correct answer is Option (3) → Methanol

The iodoform test uses iodine and sodium hydroxide (NaOI). It gives a yellow precipitate of iodoform, CHI₃, with:

• compounds containing the CH₃CO– group (methyl ketones and acetaldehyde), and

• alcohols containing the CH₃CH(OH)– group, because NaOI first oxidises them to \( \ce{CH3CO} \)– compounds.

    \( \displaystyle \ce{R-CH(OH)-CH3} \xrightarrow{\ce{NaOI}} \ce{R-CO-CH3} \xrightarrow{\ce{NaOI}} \ce{R-COO^-Na^+ + CHI3 v} \)

For propan-2-ol (R = \( \ce{CH3} \)) the complete balanced equation is:

    \( \displaystyle \ce{CH3CH(OH)CH3 + 4I2 + 6NaOH -> CHI3 v + CH3COONa + 5NaI + 5H2O} \)

• Option (1) Propan-2-ol — Gives a positive iodoform test. \( \ce{CH3CH(OH)CH3} \) has the \( \ce{CH3CH(OH)} \)– group (R = \( \ce{CH3} \)).

• Option (2) Ethanol — Gives a positive iodoform test. \( \ce{CH3CH2OH} \) has the \( \ce{CH3CH(OH)} \)– group (R = H); it is the only primary alcohol that gives the test.

• Option (3) Methanol — Does not give the iodoform test. \( \ce{CH3OH} \) has only one carbon, so it has no \( \ce{CH3-C(OH)} \) unit; it is oxidised to formaldehyde, which has no \( \ce{CH3CO} \)– group.

• Option (4) Butan-2-ol — Gives a positive iodoform test. \( \ce{CH3CH(OH)CH2CH3} \) has the \( \ce{CH3CH(OH)} \)– group (R = \( \ce{C2H5} \)).

NCERT (2026-27 edition):
Oxidation of Methyl Ketones by Haloform Reaction