Formation of the Cu(II)-phenoxyl radical by reaction of O₂ with the Cu(II)-phenolate complex via the Cu(I)-phenoxyl radical.

Abstract

Reaction of Cu(ClO₄)₂•6H₂O with a tripodal 2N2O ligand, H₂Me₂NL having a p-(dimethylamino)phenol moiety in CH₂Cl₂/MeOH (1:1 v/v) in basic conditions under inert gas atmosphere gave [Cu(Me₂NL)(H₂O)] (1). The same reaction carried out under aerobic condition gave [Cu(Me₂NL)(MeOH)]ClO₄ (2), which could be obtained also from the isolated complex 1 by reaction with O₂ in CH₂Cl₂/MeOH. The X-ray crystal structures of 1 and 2 revealed similar square-pyramidal structures, but 2 showed the (dimethylamino)phenoxyl radical features. Complex 1 exhibited characteristic Cu(II) EPR signals of the dx²-y² ground state in CH₂Cl₂/MeOH at 77 K, while 2 was EPR silent. The EPR and XAFS results suggest that 2 is assigned to the Cu(II)-(dimethylamino)phenoxyl radical. However, complex 1 showed the different features in the absence of MeOH. The EPR spectrum of the CH₂Cl₂ solution of 1 exhibited distortion from the dx²-y² ground state and the presence of the temperature-dependent equilibrium between the Cu(II)-(dimethylamino)phenolate and the Cu(I)-(dimethylamino)phenoxyl radical. From these results, the Cu(II)-phenoxyl radical complex 2 is concluded to be formed by the reaction of 1 with O₂ via the Cu(I)-phenoxyl radical species.