The acid chloride shown is first treated with two equivalents of methylamine, and the isolated product is then reduced with excess lithium aluminium hydride followed by work-up. Which structure is the final product?

Ethanoyl chloride reacts with methylamine to give N-methylethanamide (the second equivalent of amine neutralises HCl). Lithium aluminium hydride reduces the amide C=O to CH2, giving N-methylethanamine, a secondary amine with an N-ethyl group. The alcohol would be the product only if the C-N bond were cleaved, which does not occur with amides.