Abstract
A free pendant 2-amino group in a benzoquinolinate ligand bound to Ir(III) can cause heterolytic dissociation of an adjacent Ir-H-2, depending on the nature of the phosphine, L. For L=PMePh2, heterolysis does not occur and an H-2 complex, [IrH(H-2)(bq-NH2)L-2]BF4, is seen, but if L=PPh3 or PCy3, heterolysis does occur and a hydride product, [IrH2(bq-NH3)L-2]BF4, is formed by proton transfer from the bound H-2 to the pendant NH2 group. The electronic effect of L is not dominant. Theoretical studies (DFT calculations) show that the H-2 complex is predicted to be more stable for all the phosphines used, if the anion is ignored. We propose that the hydride isomer, formed when L is bulky, depends on ion pairing effects for its stability. The calculated electrostatic potentials for the two isomers suggest that the counter anion has to be located much closer to the metal in the H-2 complex than in the hydride where the anion is much farther from the metal. The bulky phosphines PPh3 and PCy3 favor remote ion pairing and therefore favor the hydride isomer because steric effects disfavor close ion pairing as confirmed by ONIOM (B3PW91/UFF) calculations of the ion pair geometry. An improved synthesis of [IrH5(PCy3)(2)] is reported.