ChemInform Abstract: CATIONIC COMPLEXES OF PLATINUM AND PALLADIUM WITH P‐TOLYL ISOCYANIDE

Abstract

Cationic Complexes of Platinum and Palladium with p‐Tolylisocyanide.

Synthesis of the complexes [PdCIR(cod)] (R = benzyl, ethyl; cod = 1,5-cyclooctadiene). β-Elimination from [PdCIEt(cod)] to give the η1,η2 and η3 isomers of [Pd2(μ-Cl)2(C8H13)2]

“…It had been reported (15) that treatment of PdCl, with 2.3 equivalents of 1,5-cyclooctadiene in aqueous acetone for 20 h at ambient temperature produced 2a in high yield. This was shown to occur by Wacker oxidation of one equivalent of cyclooctadiene to cycloocten-5-one, and addition of the HPdCl moiety thus formed to a second cyclooctadiene was proposed to account for the formation of 2a (15 (16). The corresponding bromide-bridged complex was also produced by passing CO through an ethanol solution of PdC1, and NaCl in the presence of excess 3-bromocyclooctene (17).…”

Synthesis and Characterisation of Palladium(II) Complexes with 1,2-bis(N-methylbenzimidazolyl)Benzene

“…22 The far-IR spectra of the parent complexes [PdX(cotl)] 2 displayed bands at 270 and 157 cm -1 for n(Pd-Cl-Pd) and n(Pd-Br-Pd) respectively. 21 The products, [Pd 2 Cl 2 (cotl)(N-N)]Cl and [{PdBr(cotl)} 2 (N-N)] displayed peaks corresponding to bridged palladium-chloride stretches at 274 cm -1 and terminal palladiumbromide stretches at 179 cm -1 indicated the former has not undergone chlorobridge cleavage and latter has undergone bromobridge cleavage. 13,23 A comparison of the 1 H NMR spectra of the heterocycle, N-N, [Pd(cod)(cotl)]ClO 4 , [PdX(cotl)] 2 (X = Cl or Br) with those of their reaction products revealed the presence of coordinated N-N in the latter.…”

“…and [PdX(cotl)] 2 (X = Cl or Br) were prepared as described earlier. 21 The solvents used were distilled prior to their use.…”

“…and [PdX(cotl)] 2 (X = Cl or Br) were prepared as described earlier. 21 The solvents used were distilled prior to their use. (1 mmol) in THF (20 cm 3 ) was added the heterocycle, N-N (1 mmol) dissolved in THF (10 cm 3 ).…”