Abstract
The H−R cleavage upon reaction MH + HR → MH···HR → M(η2-H2)R, where MH represents 18-e trans-dihydrides Ru(H)(H)(PH2CH2PH2)2 (1), Ru(H)(H)(PH3)4 (2), Ru(H)(H)(NH3)4 (3); HR are HX (X = F, Cl) and HOR (R = H, CH3) is studied using the DFT B3PW91/LANL2DZ level of theoretical calculations. The H−R bond splits upon interaction of the HR with 1 and 3 which possess a hydride H of high proton attracting power and significantly electropositive H of PH2 and NH3 groups. The basicity of the transition metal plays only a minor role in H−R splitting. The H−R cleavage proceeds via transfer of the H atom from R to hydride H in Ru−H···H−R···H−P(N), as an exothermic process without barrier or H···H intermediate. The less acidic HOR yields a multi-H-bonded intermediate Ru−H···H−O···(H−P(N))2, where the H−O bond cleaves with a low barrier. Such an energetically facile mechanism of H−R splitting was not found for 2, where H of PH3 is too inert to interact with R and a multi-H-bonded complex is not formed. The computed relative energies and barriers are in agreement with available experimental data.
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