Past exam of the chemistry B course of the University of Cambridge / 2023 / Paper 2 / B7 / c

zinc reduction is required in A, C and F. Reading the drawn ligands with consistent ionic bookkeeping givesIn A the two chlorides of the precursor are removed and neutral solvent donors occupy the reduced centre. In C the bromide is removed while the aryl ligand remains and the neutral alkyne coordinates; the drawn neutral intermediate has Ni(I). In F an anionic alkoxide-derived product is released and the neutral solvated Ni(0) complex is regenerated. zinc supplies electrons by oxidation to Zn(II); halide and alkoxide ions can be taken up in zinc-containing products.

A is precatalyst activation. The two one-electron reductions C and F together require two electrons per catalytic turnover, corresponding to one Zn atom if its oxidation ends at $+2$. The diagram does not establish whether each formal reduction is a single elementary electron-transfer event or a sequence of surface reactions and ligand transfers.

B is oxidative addition; D is alkyne insertion into Ni–aryl followed by carbonyl-oxygen coordination; E is carbonyl insertion into Ni–alkenyl. The distinction is visible from the bonds made and from consistent electron counting.

For B the depicted Ni(0) starting complex has two phosphine and two MeCN donors, giving $10+4+4=18$ electrons. Addition across the aryl C–Br bond produces two X-type ligands and Ni(II). The drawn product has lost the two solvent donors, giving $8+4+2+2=16$ electrons. Intrinsic oxidative addition to the unsolvated 14-electron $\mathrm{N}{\mathrm{i}}^0(\mathrm{dppe})$ fragment raises its count to 16; the overall drawn $18\to 16$ change also includes loss of four solvent-donor electrons. A falling displayed count does not turn this oxidative addition into a reductive elimination.

Before D, the aryl–alkyne intermediate is Ni(I), $d^9$, with $9+4_{\mathrm{P},\mathrm{P}}+2_{\mathrm{aryl}}+2_{\mathrm{alkyne}}=17$ electrons. Syn migration couples aryl carbon to one alkyne carbon and leaves a Ni–alkenyl bond. Insertion alone removes the separate two-electron alkyne donor and would give a 15-electron alkenyl fragment. The ester carbonyl oxygen then binds, as drawn, restoring the count to $9+4+2_{\mathrm{alkenyl}}+2_{\mathrm{O}}=17$. Ni remains $+1$.

In E the Ni-bound alkenyl carbon attacks the coordinated ester carbonyl carbon, forming the new ring-closing C–C bond. The C=O pi electrons move to oxygen, and an Ni–O alkoxide bond replaces Ni–C. This is carbonyl migratory insertion, equivalently intramolecular addition to a coordinated carbonyl. A neutral O donor plus an alkenyl X ligand become one alkoxide X ligand, so the drawn count falls $17\to 15$, while Ni remains $+1$. A solvating molecule could fill the resulting vacancy, but it is not shown and should not be silently added to the stated count.

The favoured insertion leaves nickel attached to the methyl-substituted alkyne carbon; the aryl group bonds to the SiMe${}_3$-substituted carbon. This is the connectivity leading to the illustrated ring-closed product: the carbon carrying methyl subsequently attacks the ester carbonyl.

The bulky trimethylsilyl substituent is kept away from the crowded phosphine-ligated nickel centre in the insertion arrangement. The alternative orientation would place Ni next to the more demanding SiMe${}_3$ end. regioselectivity can also reflect electronic polarization of a silyl-substituted alkyne and stabilization of the resulting alkenyl complex. The ortho ester helps capture the insertion product by oxygen coordination and subsequent ring closure. Steric discrimination at insertion, reinforced by the accessible intramolecular cyclization, accounts for the observed direction; the product drawing alone is not a quantitative separation of steric and electronic contributions.