A compound C${}_5$H${}_{12}$O does not evolve hydrogen with sodium. When treated with one equivalent of $\mathrm{HI}$ at room temperature it gives methanol and a tertiary alkyl iodide. The compound is

No reaction with Na means there is no $\mathrm{O}-\mathrm{H}$, so the compound is an ether. A tertiary iodide is formed with the alcohol part being methanol, so the ether cleaves by SN1 at the tertiary carbon, i.e. tert-butyl methyl ether (C${}_5$H${}_{12}$O). The other isomers would give $\mathrm{CH}{}_3\mathrm{I}$ or secondary/primary products by SN2.