Past exam of the chemistry B course of the University of Cambridge / 2024 / Paper 2 / B8 / b / ii / Solution

The trityl cation acts as an electrophile and removes one metal-bound methyl by methyl abstraction. The balanced conversion isThus $C=[\mathrm{C}{\mathrm{p}}_2\mathrm{ZrMe}{]}^{+}[\mathrm{B}({\mathrm{C}}_6{\mathrm{F}}_5{)}_4{]}^{-}$ and $D=\mathrm{P}{\mathrm{h}}_3\mathrm{CMe}$, 1,1,1-triphenylethane. Transfer is formally of $\mathrm{M}{\mathrm{e}}^{-}$ to carbon, not transfer of a hydride or a methyl radical. zirconium remains Zr(IV), $d^0$. Ionic electron counting changes from $12+4=16$ in the dimethyl to $12+2=14$ at the cation, creating an accessible coordination site. The borate is a weakly coordinating anion, although ion pairing can still occur in solution.