Propanone is treated with one equivalent of $\mathrm{LDA}$ in THF at $-78{}^{\circ }\mathrm{C}$ and then with iodomethane. What is the major product?

$\mathrm{LDA}$ converts propanone completely into its enolate $\mathrm{CH}{}_2=\mathrm{C}(\mathrm{CH}{}_3)-\mathrm{O}{}^{-}$. This ambident nucleophile reacts with a soft electrophile like $\mathrm{CH}{}_3\mathrm{I}$ at carbon (C-alkylation), giving $\mathrm{CH}{}_3\mathrm{COCH}{}_2\mathrm{CH}{}_3$. O-alkylation would give the enol ether, which is minor. Propan-2-ol would need a reducing agent.