A Mechanistic Study on Reactions of Group 13 Diyls LM with Cp*SbX ₂: From Stibanyl Radicals to Antimony Hydrides

Abstract

Oxidative addition of Cp*SbX ₂ (X=Cl, Br, I; Cp*=C ₅Me ₅) to group 13 diyls LM (M=Al, Ga, In; L=HC[C(Me)N (Dip)] ₂, Dip=2,6‐ iPr ₂C ₆H ₃) yields elemental antimony (M=Al) or the corresponding stibanylgallanes [L(X)Ga]Sb(X)Cp* (X=Br 1, I 2) and ‐indanes [L(X)In]Sb(X)Cp* (X=Cl 5, Br 6, I 7). 1 and 2 react with a second equivalent of LGa to eliminate decamethyl‐1,1’‐dihydrofulvalene (Cp* ₂) and form stibanyl radicals [L(X)Ga] ₂Sb ^. (X=Br 3, I 4), whereas analogous reactions of 5 and 6 with LIn selectively yield stibanes [L(X)In] ₂SbH (X=Cl 8, Br 9) by elimination of 1,2,3,4‐tetramethylfulvene. The reactions are proposed to proceed via formation of [L(X)M] ₂SbCp* as reaction intermediate, which is supported by the isolation of [L(Cl)Ga] ₂SbCp ( 11, Cp=C ₅H ₅). The reaction mechanism was further studied by computational calculations using two different models. The energy values for the Ga‐ and the In‐substituted model systems showing methyl groups instead of the very bulky Dip units are very similar, and in both cases the same products are expected. Homolytic Sb−C bond cleavage yields van der Waals complexes from the as‐formed radicals ([L(Cl)M] ₂Sb ^. and Cp* ^. ), which can be stabilized by hydrogen atom abstraction to give the corresponding hydrides, whereas the direct formation of Sb hydrides starting from [L(Cl)M] ₂SbCp* via concerted β‐H elimination is unlikely. The consideration of the bulky Dip units reveals that the amount of the steric overload in the intermediate I determines the product formation (radical vs. hydride).

Abstract

It's radical: Reactions of Cp*SbX ₂ with LM (M=Ga, In) either selectively yielded Sb‐centered radicals [L(X)Ga] ₂Sb ^. or Sb hydrides [L(X)In] ₂SbH. The unexpected formation of Sb hydrides is the result of different steric congestion from the [L(X)M] ligands as was proven by computational calculations.