Past exam of the chemistry course of the year 4 of the natural sciences course of the University of Cambridge / 2022 / Paper 2 / Section H / 33 / d

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.

Use the neutral-fragment skeletal count: each AlMe supplies two electrons, each CEt supplies three and the bridging H supplies one. HenceThere are 13 nonhydrogen skeletal vertices, eight Al plus five C, and $16=n+3$ pairs. In the usual PSEP classification this is therefore arachno, corresponding to a 15-vertex closo parent with two vertices absent. The μ${}_4$-H cap contributes an electron but is not counted as a carbon-like skeletal vertex. The aluminium cube is a convenient subframework; its eight corners alone are not the complete Wade skeleton. Calculations on capped cubic carbaalanes likewise identify 16 skeletal pairs for the five-carbon-cap/one-hydride-cap case.^[ref]

The requested equality is $32$ electrons if each unsaturated $C=CHPh$ unit is counted as a four-electron donor to the chosen extended cluster framework. It needs that additional convention; a count confined to the localized ${\mathrm{Al}}_7{\mathrm{C}}_5$ skeleton gives $28$ instead.

For $B$, an $\mathrm{AlR}$ fragment contributes two skeletal electrons, a singly substituted cap $\mathrm{CR}$ contributes three, and the bridging hydrogen contributes one:For the second formula, let $x$ be the contribution assigned to each unsaturated unit. The comparison isThus $N=N_B$ requires $x=4$. A bookkeeping partition that includes the two alkene $\pi$ pairs in the extended cluster electron pool givesIn contrast, assigning the $C=C$ double bond entirely to the exoskeletal substituent leaves each corresponding cage carbon with two electrons for the cage:The difference is exactly the two $\pi$ pairs, four electrons, rather than an arithmetic discrepancy. These are different electron partitions: do not compare $28$ from one partition with $32$ from the other as though both were the same skeletal count, and do not infer a structural classification merely by inserting the extended count into Wade's rules for twelve localized skeletal vertices.

Both the original and revised question pages print the same neutral molecular formula; the revised copy does not supply a charge, a drawing, or an explicit instruction about these $\pi$ electrons.^[ref] The primary study identifies the open cluster with the two genuine $C=C$ groups and reports their hydroalumination to give the capped aluminium cube. Its accessible abstract does not establish the orbital partition needed to select the four-electron convention.^[ref]

Conditional conclusion: the advertised $32$-electron equality follows under the explicit extended-framework convention above. Without evidence that those $\pi$ pairs belong in the cluster electron pool, retain the $28$-electron localized count and this qualification. The source checks do not justify asserting a typo or inventing a charge to force agreement.

TODO: confirm the intended orbital partition from a complete primary orbital analysis or an authoritative clarification of the question.