 Solution

Eight AlMe vertices form a cube, five faces are capped by ${\mu }_4$-C(Et) units, and the sixth face is capped by a bridging hydride. Choosing the hydride face as the bottom gives four equivalent side carbon caps and one distinct opposite carbon cap. Their ethyl groups give the 4:1 integrations: four methyl groups contribute 12 H and their methylenes 8 H; the remaining ethyl gives 3 H and 2 H.

The eight aluminium-bound methyl groups divide into two sets of four: the four aluminium atoms bordering the hydride face and the four on the opposite face. Each set gives a 12-proton singlet. The single bridging hydride gives the remaining one-proton signal. This matches the complete integration pattern without postulating five inequivalent ethyl caps. It specifies the symmetry-compatible connectivity; NMR alone does not measure every bond length or rule out solution averaging of small distortions.