 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. $A$ 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.

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 $1,1,2,2,1$, 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.