= Solution
The methyl <nucleophile> attacks an electrophilic ring boron. That boron becomes tetrahedral, bearing its original methyl, the incoming methyl, and its two ring <nitrogen> neighbors. This interrupts its p-orbital participation, so the <metal> attaches to the remaining five-atom conjugated segment. \b[$A$ is the anionic $\eta^5$ heptamethylated-ring Cr(CO)$_3$ complex], paired formally with $[\mathrm{MgBr}]^+$; solvent ligation and ion pairing at magnesium are not specified by the spectrum. The original connectivity, with the attacked boron labeled $B_1$, is shown in the first panel below.
\Image[/past-exam-of-the-chemistry-b-course-of-the-university-of-cambridge/2024/paper-2-borazine.png]
{title=Structures inferred from nucleophilic addition, Lewis-acid binding and B–Br oxidative addition}
There are seven methyl groups. The two attached to tetrahedral $B_1$ are inequivalent because the <chromium>-bound face distinguishes their environments: one signal of relative integral one for each. The other two B–Me groups are equivalent to one another, giving a signal of integral two. The two N–Me groups adjacent to $B_1$ form another equivalent pair, while the N–Me opposite $B_1$ is unique. The predicted methyl-group multiplicities are therefore \b[$1,1,2,2,1$], the same unordered set as the observed ratio. Multiplying by three gives proton integrals totaling 21. This assignment assumes no fast process interchanging the two $B_1$ methyl environments.
The six-<electron> neutral $\eta^6$ ring donor has become an anionic six-<electron> $\eta^5$ donor, so Cr(0) retains an 18-<electron> count: $6_d+6_{\mathrm{ring}}+3(2)_{\mathrm{CO}}=18$. This <borazine dearomatization> explains both the changed <hapticity> and the extra methyl without requiring attack at <chromium> or loss of a carbonyl.
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