Diruthenium−Tin Complexes from the Reaction of Ph2SnH2 with Ru(CO)5 and Their Reactions with Bis(tri-tert-butylphosphine)platinum

Abstract

Ph2SnH2 reacts with 2 equiv of Ru(CO)5 to give the compound [Ru(CO)4H]2(μ-SnPh2) (1) in 57% yield by loss of CO from each molecule of Ru(CO)5 and by an oxidative addition of an Sn−H bond to each ruthenium atom. When compound 1 was irradiated with visible radiation, the compound Ru2(CO)8(μ-SnPh2) (2) was obtained by loss of hydrogen. A mechanism involving loss of CO followed by loss of H2 and readdition of CO is supported by isotopic labeling studies. Compound 1 reacts with Pt(PBut3)2 to yield the new trimetallic compound Ru2(CO)7(μ-SnPh2)(μ-H)2(μ-PtPBut3) (3). Compound 3 contains a Pt(CO)(PBut3) group bridging the Ru−Ru bond and two bridging hydrido ligands. Compound 2 reacts with Pt(PBut3)2 to yield the two products PtRu2(CO)8)(PBut3)(μ-SnPh2) (4; 78% yield) and Pt2Ru2(CO)8(PBut3)2(μ-SnPh2) (5; 15% yield) by the addition of one and two Pt(PBut3) groups to the metal−metal bonds of 2. The first Pt(PBut3) addition occurs at the Ru−Ru bond to form 4. The second Pt(PBut3) addition occurs at one of the Ru−Sn b...