JEE Main · 2022 · 29 Jun 2022 · Shift 1 · Aldehydes Ketones and Carboxylic Acids

Fermentation of glucose by zymase gives ethanol, so $A$ is $\mathrm{C}{\mathrm{H}}_3\mathrm{C}{\mathrm{H}}_2\mathrm{OH}$. Treatment with sodium hypoiodite ($\mathrm{NaOI}$) is the iodoform reaction: the $\mathrm{C}{\mathrm{H}}_3\mathrm{CH}(\mathrm{OH})-$ unit of ethanol, equivalently its oxidation product acetaldehyde $\mathrm{C}{\mathrm{H}}_3\mathrm{CHO}$, is cleaved so that the methyl carbon leaves as iodoform $\mathrm{CH}{\mathrm{I}}_3$. The remaining carbon departs as the carboxylate, and because acetaldehyde carries only hydrogen on the other side, that carboxylate is formate, $\mathrm{HCO}{\mathrm{O}}^{-}\mathrm{N}{\mathrm{a}}^{+}$. Sodium formate holds a single carbon atom, so product $B$ contains $1$ carbon.