186340-23-8Relevant articles and documents
Synthesis of the complexes [PdClR(cod)] (R = benzyl, ethyl; cod = 1,5-cyolooctadiene). β-Elimination from [PdClEt(cod)] to give the η1,η2, and η3 isomers of [Pd2(μ-Cl)2(C8H13)2]
Stockland Jr., Robert A.,Anderson, Gordon K.,Rath, Nigam P.,Braddock-Wilking, Janet,Ellegood, J. Christopher
, p. 1990 - 1997 (1996)
Treatment of [PdCl2(cod)] with tetrabenzyltin gives the benzylpalladium complex [PdCl(CH2Ph)(cod)] (cod = 1,5-cyclooctadiene), 1a, whose structure has been determined by X-ray crystallography. It adopts approximate square-planar geometry, with the double bonds perpendicular to the square plane. The corresponding ethylpalladium derivative 1b has been prepared by a similar method, but it is considerably less stable. It decomposes by β-elimination to produce ethene and a transient hydride complex, which either undergoes migratory insertion to give [Pd2(μ-Cl)2(η1,η2-C 8H13)2], 2a, or dinuclear reductive elimination with a second molecule of 1b to produce ethane, [PdCl2(cod)], free cyclooctadiene, and palladium metal. Complex 2a has also been prepared by reaction of [PdCl2(cod)] with NaBH4. At higher temperatures 2a converts to an equilibrium mixture with its η3-allyl isomer, 2b. Reactions of [PdCl2(cod)] or K2PdCl4 in the presence of cyclooctadiene in aqueous solution to produce 2a or 2b have also been investigated.