1,2-Addition of Dihydrogen across Rhodium(III)–OMe Bonds
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The Rh(III) complexes [(tbpy)2Rh(OMe)(L)][X]n (tbpy = 4,4′-di-tert-butyl-2,2′-bipyridyl; L = MeOH, n = 2, X = OTf (OTf = trifluoromethanesulfonate), TFA (TFA = trifluoroacetate); L = TFA, n = 1, X = OTf) have been shown to activate dihydrogen via net 1,2-addition of the H–H bond across the RhIII–OMe bond. The bis(methoxide) complex [(tbpy)2Rh(OMe)2][OTf] was synthesized by addition of CsOH·H2O in methanol to [(tbpy)2Rh(OTf)2][OTf] in CH3CN. The addition of HTFA to [(tbpy)2Rh(OMe)2][OTf] leads to the formation of [(tbpy)2Rh(OMe)(MeOH)][OTf][TFA], which exists in equilibrium with [(tbpy)2Rh(OMe)(TFA)][OTf]. The mixture of [(tbpy)2Rh(OMe)(MeOH)][OTf][TFA] and [(tbpy)2Rh(OMe)(TFA)][OTf] activates dihydrogen at 68 °C to give methanol and [(tbpy)2Rh(H)(TFA)][OTf]. Studies indicate that the activation of dihydrogen has a first-order dependence on the Rh(III) methoxide complex and a dependence on hydrogen that is between zero and first order. Combined experimental and computational studies have led to a proposed mechanism for hydrogen activation by [(tbpy)2Rh(OMe)(MeOH)][OTf][TFA] that involves dissociation of MeOH, coordination of hydrogen, and 1,2-addition of hydrogen across the Rh–OMe bond. DFT calculations indicate that there is a substantial energy penalty for MeOH dissociation and a relatively flat energy surface for subsequent hydrogen coordination and activation.
