Complexes with Platinum−Ruthenium Bonds: The Stepwise Formation of Complexes Containing PtRu and PtRu2 Units

Abstract

The reaction of [Pt(dppm)2]Cl2, dppm = Ph2PCH2PPh2, with [HRu(CO)4]- in a 1:2 ratio leads in a multistep reaction to the new heteronuclear cluster [PtRu2(CO)5(μ-CO)(μ-dppm)2], which contains a triangle of metal atoms having both Pt−Ru edges bridged by dppm ligands. The initial step in the reaction is the formation of the heterobimetallic complex [PtRuH(CO)3(μ-dppm)2]+, which then reacts with additional [HRu(CO)4]- to form the PtRu2 cluster. In the absence of excess [HRu(CO)4]-, the complex [PtRuH(CO)3(μ-dppm)2]+ undergoes chloride for carbonyl substitution at ruthenium to give [PtRuHCl(CO)(μ-CO)(μ-dppm)2]. This complex then reacts further with chlorinated solvent to give [PtRu(Cl)2(CO)2(μ-dppm)2], which exists in two isomeric forms, one of which contains a semibridging carbonyl ligand. The structures of [PtRu2(CO)5(μ-CO)(μ-dppm)2] and one isomer of [PtRu(Cl)2(CO)2(μ-dppm)2] have been determined crystallographically.