Speciation of Ferric Phenoxide Intermediates during the Reduction of Iron(III)−μ-Oxo Dimers by Hydroquinone
2014-11-03T00:00:00Z (GMT) by
The aqueous speciation of iron(III)–tris(pyridylmethyl)amine (TPA) complexes was determined from potentiometric titration data, and the overall formation constants (β) for relevant species were calculated. At pH < 3 the mononuclear complex [Fe(TPA)]+3(aq) predominates (log β = 10.75(15). Above pH 3 Fe3+–OH2 hydrolysis produces the μ-oxo dimer [Fe2(μ-O)(TPA)2(H2O)2]+4 (1a; log β = 19.91(12)). This species is a diprotic acid with the conjugate bases [Fe2(μ-O)(TPA)2(H2O)(OH)]+3 (1b; log β = 15.53(6)) and [Fe2(μ-O)(TPA)2(OH)2]+2 (1c; log β = 10.27(7)). The pKas of 1a are 4.38(14) and 5.26(9). Compounds 1a–c quantitatively oxidize hydroquinone to benzoquinone with concomitant formation of 2 equiv of Fe(II). Kinetic and spectroscopic data at pH 5.6 are consistent with rapid equilibrium formation of a diiron(III)–phenoxide intermediate followed by rate-controlling electron transfer. The equilibrium constant for the formation of the intermediate complex is 25(3) M–1, and the rate constant for its decomposition is 0.56(9) s–1. A kinetic isotope effect of kH/kD = 1.5 was determined from proton inventory experiments in mixed H/D media. The μ-oxo–diiron(III) phenoxide intermediate is hydrolyzed in a pH dependent process to form a mononuclear iron(III)–phenoxide, which complicates the kinetics by introducing a fractional dependence on total iron(III) concentration in the pH range 4.1–5.2. The pH-dependent cleavage of μ-oxo–diiron(III)–phenoxides was investigated with phenol, a redox-inert proxy for hydroquinone. The addition of phenol to 1 facilitates acidic cleavage of the μ-oxo dimer to form [Fe(TPA)(OPh)(H2O)]+2, which becomes the dominant iron(III)–phenoxide as the pH decreases to 4. The 2-naphtholate analogue of this intermediate, [Fe(TPA)(2-naphtholate)(OCH3)]ClO4 (6), was characterized by single-crystal X-ray diffraction (C29H28FeN4O2,ClO4; P21; a = 13.2646(2) Å, b = 15.2234(3) Å, c = 13.7942(3) Å; Z = 4).