Generation of Ir–Sn and Rh–Sn bonds from the oxidative addition of tin(IV) halides to [Ir(μ-Cl)(1,5-COD)]2 and [Rh(μ-Cl)(1,5-COD)]2

Facile oxidative addition of SnCl4, MeSnCl3, and SnBr4 across Ir(I) and Rh(I) cyclooctadiene complexes resulted in the formation of the corresponding Ir–Sn and Rh–Sn heterobimetallic complexes. Treatment of SnCl4 with [Ir(COD)(μ-Cl)]2 and [Rh(COD)(μ-Cl)]2 afforded [Ir(COD)(μ-Cl)Cl(SnCl3)]2 (1) and [Rh(COD)(μ-Cl)Cl(SnCl3)]2 (2), respectively. Reaction of the organotin halide MeSnCl3 with [Ir(COD)(μ-Cl)]2 led to the formation of [Ir(COD)(μ-Cl)Cl(MeSnCl2)]2 (3). The reaction of SnBr4 to IrI and RhI precursors gave [Ir(COD)(μ-Br)Br(SnBr3)]2 (4) and [Rh(COD)(μ-Br)Br(SnBr3)]2 (5) respectively, which indicates halide exchange at post-oxidative addition stage. The structures of complexes 1–5 were confirmed by X-ray crystallography. A cis-addition of Sn–X bond across IrI/RhI is proposed from the analysis of the geometrical features of “X–M–Sn” triangular units in 1–5.

Graphical abstractFacile oxidative addition of SnCl4, MeSnCl3, and SnBr4 across Ir(I) and Rh(I) cyclooctadiene complexes resulted in the formation of the corresponding Ir–Sn and Rh–Sn heterobimetallic complexes. A cis-addition of Sn–X bond across IrI/RhI is proposed from the analysis of the geometrical features of “X–M–Sn” triangular units in the complexes.Figure optionsDownload full-size imageDownload as PowerPoint slide