Facile oxidation of phenylphosphinic acid to phenylphosphonic acid using [Cu 2(μ-O 2CCH 3) 4(H 2O) 2]: X-ray crystal structures of monomeric [Cu(PhPO 3H) 2(C 5H 5N) 4]·2CH 3OH and [Cu(PhPO 3H) 2(C 5H 5N) 4

Abstract

Phenylphosphinic acid (HPhPO 2H) is oxidized to phenylphosphonic acid (PhPO 3H 2) at room temperature using a solution of [Cu 2(μ-O 2CCH 3) 4(H 2O) 2] in pyridine. The phenylphosphonic acid was recovered as the monomeric copper(II) complex [Cu(PhPO 3H) 2(C 5H 5N) 4]·H 2O ( 1a), and the reaction thought to proceed via a copper(I) intermediate. Recrystallization of 1a from methanol gave [Cu(PhPO 3H) 2(C 5H 5N) 4]·2CH 3OH ( 1b). The unsolvated complex [Cu(PhPO 3H) 2(C 5H 5N) 4] ( 1c) was prepared by refluxing polymeric [Cu(PhPO 3)(H 2O)] ( 2) in pyridine. The X-ray crystal structures of 1b and 1c show that in each of these monomeric complexes the copper(II) ion is ligated by four equatorial pyridine molecules and two axial monoanionic phenylphosphonate groups. A cyclic voltammetric study of 1a revealed a quasi-reversible Cu 2+/Cu + couple with E 1/2 = +228 mV (vs Ag/AgCl).