Cannizzaro Reaction

Ethanol is not itself a methyl ketone, yet it gives a positive iodoform test -- and understanding why requires recognising that the test conditions ($\text{I}_2$ in $\text{NaOH}$) actually generate sodium hypoiodite, $\text{NaOI}$, in situ (from $\text{I}_2 + 2\text{NaOH} \rightarrow \text{NaOI} + \text{NaI} + \text{H}_2\text{O}$), and hypoiodite is itself a mild oxidising agent.

The two reactions involved. First, sodium hypoiodite oxidises ethanol, $\text{CH}_3\text{CH}_2\text{OH}$, to acetaldehyde, $\text{CH}_3\text{CHO}$ -- this is possible because ethanol has the required structural pattern for hypoiodite oxidation, a $\text{CH}_3\text{-CH(OH)-}$ unit (a secondary-alcohol-like arrangement where a methyl group sits on the same carbon as the $\text{-OH}$, even though ethanol's carbinol carbon is technically primary here it still bears the necessary adjacent methyl). Once this acetaldehyde has formed, it is precisely the substrate required for the haloform reaction described in §8.9: it possesses the $\text{CH}_3\text{-CHO}$ methyl-ketone-like pattern, so the remaining excess of $\text{I}_2$/$\text{NaOH}$ carries it through exhaustive iodination at the methyl carbon and then cleavage by hydroxide, giving iodoform, $\text{CHI}_3$ (the positive yellow precipitate), and sodium formate, $\text{HCOONa}$, as the two final products.

$\text{CH}_3\text{CH}_2\text{OH} \xrightarrow{\text{NaOI}} \text{CH}_3\text{CHO} \xrightarrow{\text{I}_2/\text{NaOH}} \text{CHI}_3\downarrow + \text{HCOONa}$ …