Abstract

The molecular mechanism of H(2) activation by two transition metal thiolate complexes [Cp*M(PMe(3))(SDmp)](BAr(F)(4)) (M = Ir, Rh) (Ohki, Y; Sakamoto, M; Tatsumi, K. J. Am. Chem. Soc., 2008, 130, 11610-11611) has been investigated using density functional theory calculations. According to our calculations, the reaction of the iridium thiolate complex with H(2) is likely to proceed through the following steps: (1) the oxidative addition of H(2) to the iridium center to generate a dihydride intermediate; (2) the reductive elimination of one Ir-bound hydrogen to produce the hydride thiol product. For the rhodium thiolate complex, its reaction with H(2) is to form the dihydrogen intermediate first, and then the H-H bond is heterolytically cleaved at the Rh-S bond via a four-center transition state to yield the hydride thiol product. The rate-determining step is the oxidative addition step (with a barrier of 18.0 kcal/mol in the solvent) for the iridium complex, and the formation of the dihydrogen complex (with a barrier of 13.9 kcal/mol in the solvent) for the rhodium complex. The calculated free energy profiles for both metal thiolate complexes can reasonably account for the observed reversible H(2) activation by two metal thiolate complexes under mild conditions.

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