The halide shown is converted to its Grignard reagent with magnesium in dry ether; an excess of the reagent is allowed to react with ethyl ethanoate (ethyl acetate), and the mixture is then hydrolysed with $\mathrm{H}{}_3\mathrm{O}{}^{+}$. Which structure is the final product?

The first equivalent of propylmagnesium bromide adds to the ester and expels ethoxide, giving pentan-2-one. The ketone is more reactive than the ester, so a second equivalent adds to give a tertiary alkoxide with two propyl groups: 4-methylheptan-4-ol after hydrolysis. Pentan-2-one is the intermediate, 2-methylpentan-2-ol would need a methyl Grignard, and heptan-4-ol would need a formate ester.