19F NMR Investigations of the Reaction of B(C6F5)3 with Different Tri(alkyl)aluminum Compounds

Tris(pentafluorophenyl)borane, B(C₆F₅)₃ reacts with triethylaluminum, AlEt₃ to a mixture of Al(C₆F₅)_3−nEt_n and Al₂(C₆F₅)_6−nEt_n compounds depending on the B/Al ratio. From excess borane to excess AlEt₃ the species Al(C₆F₅)₃ → Al(C₆F₅)₂Et Al₂(C₆F₅)₄Et₂ → Al₂(C₆F₅)₃Et₃ → Al₂(C₆F₅)₂Et₄ → Al₂(C₆F₅)Et₅ are formed and differentiated by their para-F signal in ¹⁹F NMR. The reaction between B(C₆F₅)₃ and the higher aluminum alkyls, tri(iso-butyl)aluminum and tri(n-hexyl)aluminum AlR₃ (R = i-Bu, n-C₆H₁₃) is slower and requires AlR₃ excess to shift the C₆F₅ R exchange equilibria to almost complete formation of Al(C₆F₅)R₂ and BR₃. At equimolar ratio the equilibrium lies on the side of the unchanged borane together with its boranate [B(C₆F₅)₃R]⁻ anion. For tri(n-octyl)aluminum even at large Al(n-C₈H₁₇)₃ excess no C₆F₅ alkyl exchange can be observed, but boranate anions form.