JEE Main · 2026 · 23 Jan 2026 · Shift 2 · Aldehydes Ketones and Carboxylic Acids

Hypochlorite is a stronger oxidant than hypoiodite, so sodium hypochlorite oxidises iodide to hypoiodite, $\mathrm{NaOCl}+\mathrm{KI}\to \mathrm{NaCl}+\mathrm{KOI}$, and that statement is true; KOI itself acts as an oxidising agent, not a reducing agent. An $\alpha ,\beta$-unsaturated methyl ketone such as $\mathrm{C}{\mathrm{H}}_3\mathrm{COCH}=\mathrm{C}{\mathrm{H}}_2$ keeps the $\mathrm{C}{\mathrm{H}}_3\mathrm{CO}-$ group and therefore gives a positive iodoform test. Isopropyl alcohol, $\mathrm{C}{\mathrm{H}}_3\mathrm{CH}(\mathrm{OH})\mathrm{C}{\mathrm{H}}_3$, carries the $\mathrm{C}{\mathrm{H}}_3\mathrm{CH}(\mathrm{OH})-$ unit and does give the test, so the claim that it does not is false, while methanoic acid has neither a methyl ketone nor a methyl carbinol group and gives a negative test. The true statements are the formation of KOI and the response of the $\alpha ,\beta$-unsaturated methyl ketone.