Q.Ethanol gives a positive iodoform test although it is not a methyl ketone. Explain, with the two reactions involved, why this happens, and predict whether propan-1-ol would behave the same way.

$\text{I}_2$ dissolved in $\text{NaOH}$ generates sodium hypoiodite in situ ($\text{I}_2 + 2\text{NaOH} \rightarrow \text{NaOI} + \text{NaI} + \text{H}_2\text{O}$), a mild oxidant that first converts ethanol, $\text{CH}_3\text{CH}_2\text{OH}$, into acetaldehyde, $\text{CH}_3\text{CHO}$. Acetaldehyde is exactly the required substrate for the haloform reaction (§ haloform), so the excess $\text{I}_2$/$\text{NaOH}$ then iodinates its methyl group exhaustively and cleaves it to give iodoform, $\text{CHI}_3$ (the positive yellow precipitate), and sodium formate, $\text{HCOONa}$. Propan-1-ol, on oxidation by the same $\text{NaOI}$, gives propanal, $\text{CH}_3\text{CH}_2\text{CHO}$ -- whose $\alpha$-carbon carries only two hydrogens and an ethyl-type substituent, not a full methyl group -- so it lacks the r …

Recognise that the test reagent doubles as a mild oxidant (via in-situ NaOI); check whether the resulting oxidation product actually contains the required $\t …

Assuming every primary alcohol gives a positive iodoform test -- only an alcohol whose oxidation product carries a terminal METHYL group directly on the resulting carbonyl …