JEE Main · 2026 · 23 Jan 2026 · Shift 2 · Alcohols Phenols and Ethers

Hot concentrated HI cleaves both C--O bonds of the mixed ether, so its two alkyl groups appear as two different alkyl iodides, and aqueous NaOH converts these to the two parent alcohols. A yellow precipitate with NaOI (the iodoform reaction) requires each alcohol to be ethanol or a methyl carbinol $\mathrm{C}{\mathrm{H}}_3\mathrm{CH}(\mathrm{OH})-$. Ethyl sec-butyl ether satisfies this: it gives ethyl iodide and sec-butyl iodide, which hydrolyse to ethanol and butan-2-ol, and both $\mathrm{C}{\mathrm{H}}_3\mathrm{C}{\mathrm{H}}_2\mathrm{OH}$ and $\mathrm{C}{\mathrm{H}}_3\mathrm{CH}(\mathrm{OH})\mathrm{C}{\mathrm{H}}_2\mathrm{C}{\mathrm{H}}_3$ give the test. The phenyl-bearing ether would give 1-phenyl-1-propanol and the branched ether a tertiary alcohol, neither of which responds. The mixed ether is ethyl sec-butyl ether.