Abstract

At high temperatures in toluene, [2,5-Ph(2)-3,4-Tol(2)(eta(5)-C(4)COH)]Ru(CO)(2)H (3) undergoes hydrogen elimination in the presence of PPh(3) to produce the ruthenium phosphine complex [2,5-Ph(2)-3,4-Tol(2)-(eta(4)-C(4)CO)]Ru(PPh(3))(CO)(2) (6). In the absence of alcohols, the lack of RuH/OD exchange, a rate law first order in Ru and zero order in phosphine, and kinetic deuterium isotope effects all point to a mechanism involving irreversible formation of a transient dihydrogen ruthenium complex B, loss of H(2) to give unsaturated ruthenium complex A, and trapping by PPh(3) to give 6. DFT calculations showed that a mechanism involving direct transfer of a hydrogen from the CpOH group to form B had too high a barrier to be considered. DFT calculations also indicated that an alcohol or the CpOH group of 3 could provide a low energy pathway for formation of B. PGSE NMR measurements established that 3 is a hydrogen-bonded dimer in toluene, and the first-order kinetics indicate that two molecules of 3 are also involved in the transition state for hydrogen transfer to form B, which is the rate-limiting step. In the presence of ethanol, hydrogen loss from 3 is accelerated and RuD/OH exchange occurs 250 times faster than in its absence. Calculations indicate that the transition state for dihydrogen complex formation involves an ethanol bridge between the acidic CpOH and hydridic RuH of 3; the alcohol facilitates proton transfer and accelerates the reversible formation of dihydrogen complex B. In the presence of EtOH, the rate-limiting step shifts to the loss of hydrogen from B.

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