 Solution

$B$ is a 1:1 nitrogen-bound adduct, with $\mathrm{N}\to \mathrm{AlB}{\mathrm{r}}_3$ at one ring nitrogen. Its lone pair is diverted from N-to-B pi donation into the new coordinate bond. That nitrogen becomes approximately tetrahedral and the ring can pucker. The two B–N bonds adjacent to it lose pi character and lengthen; the neighboring unbound nitrogen atoms can donate more strongly into those borons, shortening the next pair of B–N bonds. The remaining pair forms a third class. The second structural panel marks these three two-bond sets; it represents connectivity and bond classes, not a measured geometry.

In the slow-exchange structure, two boron atoms flank the bound nitrogen and are equivalent to one another, while the third boron is unique. This gives two ${}^{11}$B environments in a $2:1$ ratio. At room temperature, rapid exchange of the Lewis acid among the three equivalent nitrogen sites averages every boron over the same environments, giving one boron-11 nuclear magnetic resonance signal. Cooling slows exchange relative to the resonance-frequency difference and resolves the two environments.

This is Chemical exchange in NMR, not restoration of equal instantaneous B–N bonds. A spectrum alone cannot distinguish intramolecular migration from reversible dissociation and reassociation; concentration dependence and crossover experiments would test that distinction. Related N-bound borazine adducts have been characterized by diffraction and variable-temperature NMR, supporting the structural logic while not supplying the exact temperature dependence of this particular hexamethyl compound.^[ref]