Oxidative Reactivity of (N2S2)PdRX Complexes (R = Me, Cl; X = Me, Cl, Br): Involvement of Palladium(III) and Palladium(IV) Intermediates
datasetposted on 19.02.2016, 06:45 by Jia Luo, Nigam P. Rath, Liviu M. Mirica
A series of (N2S2)PdRX complexes (N2S2 = 2,11-dithia[3.3](2,6)pyridinophane; R = X = Me, 1; R = Me, X = Cl, 2; R = Me, X = Br, 3; R = X = Cl, 4) were synthesized, and their structural and electronic properties were investigated. X-ray crystal structures show that for the corresponding Pd(II) complexes the N2S2 ligand adopts a κ2 conformation, with the pyridine N donors binding in the equatorial plane. Cyclic voltammetry (CV) studies suggest that the Pd(III) oxidation state is accessible at moderate redox potentials. In situ EPR, ESI-MS, UV–vis, and low-temperature electrochemical studies were employed to detect the formation of Pd(III) species during the oxidation of Pd(II) precursors. In addition, the [(N2S2)PdIVMe2](PF6)2 ([12+](PF6)2) complex was isolated by oxidation of 1 with 2 equiv of FcPF6, and its structural characterization reveals an octahedral Pd(IV) center. The reversible PdIV/III redox couple for the Pd(IV) species supports the observed formation of the Pd(III)–dimethyl species upon chemical reduction of 12+. In addition, reactivity studies reveal ethane, MeCl, and MeBr elimination upon one-electron oxidation of 1 (as well as the one-electron reduction of 12+), 2, and 3, respectively. Mechanistic studies suggest the initial formation of a Pd(III) species, followed by methyl group transfer/disproportionation and subsequent reductive elimination from a Pd(IV) intermediate, although a halogen radical pathway cannot be completely excluded during C–halide bond formation. Interestingly, computational results suggest that the N2S2 ligand stabilizes to a greater extent the Pd(IV) vs the Pd(III) oxidation state, likely due to steric rather than electronic effects.