19 F NMR Investigations of the Reaction of B(C 6 F 5 ) 3 with Different Tri(alkyl)aluminum Compounds

Tris(pentafluorophenyl)borane, B(C 6 F 5 ) 3 reacts with triethylaluminum, AlEt 3 to a mixture of Al(C 6 F 5 ) 3−n Et n and Al 2 (C 6 F 5 ) 6−n Et n compounds depending on the B/Al ratio. From excess borane to excess AlEt 3 the species Al(C 6 F 5 ) 3  → Al(C 6 F 5 ) 2 Et  $\mathbin{\lower.3ex\hbox{$\buildrel\textstyle\leftarrow\over{\smash{\rightarrow}\vphantom{_{\vbox to.5ex{\vss}}}}$}}$  Al 2 (C 6 F 5 ) 4 Et 2  → Al 2 (C 6 F 5 ) 3 Et 3  → Al 2 (C 6 F 5 ) 2 Et 4  → Al 2 (C 6 F 5 )Et 5 are formed and differentiated by their para ‐F signal in 19 F NMR. The reaction between B(C 6 F 5 ) 3 and the higher aluminum alkyls, tri( iso ‐butyl)aluminum and tri( n ‐hexyl)aluminum AlR 3 (R =  i ‐Bu, n ‐C 6 H 13 ) is slower and requires AlR 3 excess to shift the C 6 F 5   $\leftrightarrow$  R exchange equilibria to almost complete formation of Al(C 6 F 5 )R 2 and BR 3 . At equimolar ratio the equilibrium lies on the side of the unchanged borane together with its boranate [B(C 6 F 5 ) 3 R] − anion. For tri( n ‐octyl)aluminum even at large Al( n ‐C 8 H 17 ) 3 excess no C 6 F 5   $\leftrightarrow$  alkyl exchange can be observed, but boranate anions form.