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

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.