Stepwise oxidation of anilines by cis-[RuIV(bpy)2(py)(O)]2+

The net six-electron oxidation of aniline to nitrobenzene or azoxybenzene by cis-Ru(IV)(bpy)(2)(py)(O)](2+) (bpy is 2,2'-bipyridine; py is pyridine) occurs in a series of discrete stages. In the first, initial two-electron oxidation is followed by competition between oxidative coupling with aniline to give 1,2-diphenylhydrazine and capture by H(2)O to give N-phenylhydroxylamine. The kinetics are first order in aniline and first order in Ru(IV) with k(25.1 degrees C, CH(3)CN) = (2.05 +/- 0.18) x 10(2) M(-1) s(-1) (DeltaH(++) = 5.0 +/- 0.7 kcal/mol; DeltaS(++) = -31 +/- 2 eu). On the basis of competition experiments, k(H)2(O)/k(D)2(O) kinetic isotope effects, and the results of an (18)O labeling study, it is concluded that the initial redox step probably involves proton-coupled two-electron transfer from aniline to cis-Ru(IV)(bpy)(2)(py)(O)](2+) (Ru(IV)=O(2+)). The product is an intermediate nitrene (PhN) or a protonated nitrene (PhNH(+)) which is captured by water to give PhNHOH or aniline to give PhNHNHPh. In the following stages, PhNHOH, once formed, is rapidly oxidized by Ru(IV)=O(2+) to PhNO and PhNHNHPh to PhN=NPh. The rate laws for these reactions are first order in Ru(IV)=O(2+) and first order in reductant with k(14.4 degrees C, H(2)O/(CH(3))(2)CO) = (4.35 +/- 0.24) x 10(6) M(-1) s(-1) for PhNHOH and k(25.1 degrees C, CH(3)CN) = (1.79 +/- 0.14) x 10(4) M(-1) s(-1) for PhNHNHPh. In the final stages of the six-electron reactions, PhNO is oxidized to PhNO(2) and PhN=NPh to PhN(O)=NPh. The oxidation of PhNO is first order in PhNO and in Ru(IV)=O(2+) with k(25.1 degrees C, CH(3)CN) = 6.32 +/- 0.33 M(-1) s(-1) (DeltaH(++) = 4.6 +/- 0.8 kcal/mol; DeltaS(++) = -39 +/- 3 eu). The reaction occurs by O-atom transfer, as shown by an (18)O labeling study and by the appearance of a nitrobenzene-bound intermediate at low temperature.