Complexes with Platinum−Iridium Bonds: Stepwise Formation of a PtIr2Cluster Complex

The reaction of [Pt(dppm) 2 ]Cl 2 (dppm = Ph 2 PCH 2 PPh 2 ) with [Ir(CO) 4 ] - and dppm in a 1:2:1 ratio leads in a multistep reaction to the new heteronuclear cluster complex [PtIr 2 (CO) 2 (μ-CO)(μ-dppm) 3 ], which contains a triangle of metal atoms with each edge bridged by a dppm ligand and in which only the platinum atom is coordinatively unsaturated. The initial step in the reaction leads to formation of the neutral and cationic binuclear complexes [PtIrCl(CO) 2 (μ-dppm) 2 ] and [PtIr(CO) 3 (μ-dppm) 2 ] +, respectively. The binuclear complexes then react with additional [Ir(CO) 4 ] - to form the cluster [PtIr 2 (CO) 4 (μ-CO)(μ-dppm) 2 ] by insertion into a Pt−P linkage; this cluster can exist in two isomeric forms, each containing a triangular PtIr 2 group with the Ir−Ir and one Pt−Ir edge bridged by dppm ligands but differing in stereochemistry. These isomers equilibrate slowly at room temperature, and each reacts easily with dppm to form the final cluster complex [PtIr 2 (CO) 2 (μ-CO)(μ-dppm) 3 ]. The structures of the complexes [PtIr 2 (CO) 2 (μ-CO)(μ-dppm) 3 ], [PtIr(CO) 3 (μ-dppm) 2 ][PF 6 ], and one isomer of [PtIr 2 (CO) 4 (μ-CO)(μ-dppm) 2 ] have been determined crystallographically, and the reaction sequence has been determined by monitoring reactions through NMR methods.