Ethyl methanoate ($\mathrm{HCOOC}{}_2\mathrm{H}{}_5$) is treated with excess $\mathrm{CH}{}_3\mathrm{MgBr}$ in dry ether and then with $\mathrm{H}{}_3\mathrm{O}{}^{+}$. Which of the following are formed?

The first $\mathrm{CH}{}_3\mathrm{MgBr}$ displaces ethoxide to give ethanal, which immediately adds a second $\mathrm{CH}{}_3{}^{-}$ to give $\mathrm{CH}{}_3-\mathrm{CH}(\mathrm{OMgBr})-\mathrm{CH}{}_3$, hydrolysed to propan-2-ol (secondary). The leaving ethoxide gives ethanol on work-up. No tertiary alcohol forms because the intermediate is an aldehyde, not a ketone.