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

For the usual octahedral Ni(II) hydration-isomer series, distinguish coordinated water from water of crystallization:The first two coordination spheres have no cis/trans choice because they contain zero or one chloride. The last hydration class has two geometrical isomers: the chlorides occupy adjacent or opposite octahedral positions. Thus there are three hydration classes and four structures when cis/trans isomerism is included. Every formula retains two chlorides and six waters overall; the coordination number remains six. The schematic in B6(c) displays the distinct octahedral connectivities. This enumeration states the usual six-coordinate interpretation, rather than treating unspecified changes in coordination number as further members of the same elementary series.

The neutral product is cis-$[\mathrm{NiC}{\mathrm{l}}_2(\mathrm{dppe})]$, and the displaced material is six waters per nickel:The two phosphorus donors of dppe chelate the same metal, forcing adjacent donor positions. The diamagnetism identifies a square-planar, paired $d^8$ Ni(II) centre rather than a tetrahedral high-spin $d^8$ centre. Its valence count is $8+4_{\mathrm{P},\mathrm{P}}+4_{\mathrm{Cl},\mathrm{Cl}}=16$ electrons. The reaction is ligand substitution, not reduction of Ni(II).

The relevant chelate rings contain four atoms for dppm, five for dppe, and six for dppp, counting the nickel and both phosphorus atoms. The one-carbon bridge in dppm forces a compact ring and the small bite angle $75.62^{\circ }$. A two-carbon tether accommodates a five-membered chelate closer to the preferred square-planar cis angle, giving $86.99^{\circ }$. The more flexible three-carbon tether can open slightly beyond $90^{\circ }$, giving $91.77^{\circ }$.

For a simplified symmetric chelate with Ni–P distance $r$ and P–P donor separation $d$,Increasing the accessible donor separation therefore increases the bite angle $\beta$. The ligand is a three-dimensional tether, so ring puckering and substituent crowding matter as well as the number of methylene groups. This geometrical explanation does not require different formal oxidation states or electron counts for the three complexes.

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.