Abstract

Methanol solutions of rhodium(III) tetra(p-sulfonatophenyl) porphyrin [(TSPP)Rh(III)] have a hydrogen ion dependent equilibrium between bis-methanol, monomethoxy monomethanol, and bis-methoxy complexes. Reactions of dihydrogen (D(2)) with solutions of [(TSPP)Rh(III)] complexes in methanol produce equilibrium distributions of a rhodium hydride [(TSPP)Rh(III)-D(CD(3)OD)](-4) and rhodium(I) complex [(TSPP)Rh(I)(CD(3)OD)](-5). The rhodium hydride complex in methanol functions as a weak acid with an acid dissociation constant of 1.1(0.1) x 10(-9) at 298 K. Patterns of rhodium hydride substrate reactions in methanol are illustrated by addition with ethene, acetaldehyde, and carbon monoxide to form rhodium alkyl, alpha-hydroxyethyl, and formyl complexes, respectively. The free energy change for the addition reaction of [(TSPP)Rh(III)-D(CD(3)OD)](-4) with CO in methanol to produce a formyl complex (DeltaG(o)(298K) = -4.7(0.1) kcal mol(-1)) is remarkably close to DeltaG(o)(298K) values for analogous reactions in water and benzene. Addition reactions of the rhodium hydride ([(TSPP)Rh(III)-D(CD(3)OD)](-4)) with vinyl olefins invariably yield the anti-Markovnikov product which places the rhodium porphyrin on the less hindered terminal primary carbon center. Addition of the rhodium-methoxide unit in [(TSPP)Rh(III)-OCD(3)(CD(3)OD)](-4) with olefins to form beta-methoxyalkyl complexes places rhodium on the terminal carbon for alkene hydrocarbons and vinyl acetate, but vinyl olefins that have pi-electron withdrawing substituents have a thermodynamic preference for placing rhodium on the interior carbon where negative charge is better accommodated. Equilibrium thermodynamic values for addition of the Rh-OCD(3) unit to olefins in methanol are evaluated and compared with values for Rh-OH addition to olefins in water.

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