Past exam of the chemistry B course of the University of Cambridge / 2024 / Paper 2 / B9 / c

A ring-retaining bromination replaces the three hydridic B–H positions while preserving N–H. Its net stoichiometry isUse controlled bromination and conditions allowing elimination of hydrogen bromide from addition intermediates, rather than identifying the first bromine adduct as the final ring product. Boron-bound hydrogen has hydride-like character because boron is less electronegative than hydrogen; its replacement is consistent with electrophilic attack by bromine. The primary preparative work reports B-tribromoborazines obtained by bromination of borazine and its N-methyl analogue.^[ref]

The product must retain the unsaturated B${}_3$N${}_3$ ring. Merely adding three HBr molecules to borazine initially gives a saturated hydrogen-halide addition product, not the required formula. Addition followed by suitable elimination is distinct from simply asserting direct B–H protonolysis by HBr.

oxidative addition inserts Pt into one B–Br bond. All three B–Br positions of the starting ring are equivalent, so there is no distinct choice among them before the first substitution. The product is the square-planar Pt(II) boryl bromide $\mathrm{trans}-[\mathrm{Pt}(\mathrm{PC}{\mathrm{y}}_3{)}_2(\mathrm{Br})({\mathrm{B}}_3\mathrm{B}{\mathrm{r}}_2{\mathrm{N}}_3{\mathrm{H}}_3)]$, with the two phosphines trans, the Pt–B bond trans to Pt–Br, and the six-membered ring intact. The third panel shows this connectivity. This is a metal–boron sigma bond, not attachment of the whole ring as an ${\eta }^6$ pi ligand. Formal electron counting gives Pt(II), $d^8$, with $8+4_{\mathrm{P}}+2_{\mathrm{Br}}+2_{\mathrm{boryl}}=16$ electrons, appropriate for square-planar coordination. The primary synthesis establishes this trans-bromo(borazinyl) structure.^[ref]

The electron-rich Pt(0) reacts at the polarized B–Br bond, whose electrophilic boron and departing bromide provide a favorable bond-activation pathway. This accounts for B functionalization without needing cleavage of a ring B–N bond or selection of a nitrogen-bound product. It does not imply that arbitrary N–H activation is impossible under every condition.

Replacing an electronegative bromine substituent by a metal fragment changes the boron's electron demand. Relative to the B–Br sites, the metal-bound boron is less effective at accepting N-to-B pi donation, so its adjacent B–N bonds are expected to have less multiple-bond character and be longer; donation toward the remaining B–Br sites produces compensating unequal ring bonds. The original threefold equivalence is lost. The detailed lengths depend on the full electronic structure; a localized drawing with three rigid alternating double bonds would conceal the redistribution that the unequal lengths reveal.