Preparation and Thermal Reaction of Tetrastannapalladacyclopentane. Sn−Sn Bond Formation and Cleavage

Reaction of H2SnPh2 with [Pd(dmpe)2]n (n = 1 or 2; dmpe
= 1,2-bis(dimethylphosphino)ethane) in 4:1 ratio produces a tetrastannapalladacyclopentane,
[Pd(SnPh2SnPh2SnPh2SnPh2)(dmpe)] (1), via dehydrogenative Sn−Sn
bond formation. Heating of 1 in toluene at 70 °C
cleaves the Sn−Sn bonds, which is accompanied by migration
of Ph groups, to form a bis(triphenylstannyl)palladium complex, [Pd(SnPh3)2(dmpe)] (2). A similar reaction
of H2SnPh2 with [Pd(PCy3)2] yields a dipalladium(I) complex with bridging diphenylstannyl ligands,
[{Pd(PCy3)}2(μ-η2-HSnPh2)2] (3).