Ethyl acetate is treated with (i) excess $\mathrm{CH}{}_3\mathrm{MgBr}$ in dry ether; (ii) $\mathrm{H}{}_3\mathrm{O}{}^{+}$. The final product is

The first equivalent adds to the ester and ethoxide is lost to give acetone. Acetone is more reactive than the ester and takes up a second equivalent of $\mathrm{CH}{}_3\mathrm{MgBr}$, giving $(\mathrm{CH}{}_3){}_3\mathrm{C}-\mathrm{OMgBr}$. Hydrolysis gives $(\mathrm{CH}{}_3){}_3\mathrm{C}-\mathrm{OH}$.