Synthesis of (Arylmido)niobium(V) Complexes Containing Ketimide, Phenoxide Ligands, and Some Reactions with Phenols and Alcohols

[Image: see text] Reactions of Nb(NAr)(N=C(t)Bu(2))(3) (3a, Ar = 2,6-Me(2)C(6)H(3)) with 1.0, 2.0, or 3.0 equiv of Ar′OH (Ar′ = 2,6-(i)Pr(2)C(6)H(3)) afforded Nb(NAr)(N=C(t)Bu(2))(2)(OAr′), Nb(NAr)(N=C(t)Bu(2))(OAr′)(2), or Nb(NAr)(OAr′)(3), respectively (at 25 °C), whereas the reaction with 2.0 equiv of 2,6-(t)Bu(2)C(6)H(3)OH afforded Nb(NAr)(N=C(t)Bu(2))(2)(O-2,6-(t)Bu(2)C(6)H(3)) upon heating (70 °C) without the formation of bis(phenoxide) and the reaction of 3a with 2.0 equiv of 2,4,6-Me(3)C(6)H(2)OH afforded Nb(NAr)(N=C(t)Bu(2))(O-2,4,6-Me(3)C(6)H(2))(2)(HN=C(t)Bu(2)). Similar reactions of 3a with 1.0 equiv of (CF(3))(3)COH or 2.0 equiv of (CF(3))(2)CHOH afforded Nb(NAr)(N=C(t)Bu(2))(2)[OC(CF(3))(3)](HN=C(t)Bu(2)) or Nb(NAr)(N=C(t)Bu(2))[OCH(CF(3))(2)](2)(HN=C(t)Bu(2)), respectively. On the basis of their structural analyses and the reaction chemistry, it was suggested that these reactions proceeded via coordination of phenol (alcohol) to Nb and the subsequent proton (hydrogen) transfer to the ketimide (N=C(t)Bu(2)) ligand. The reaction of Nb(NAr)(N=C(t)Bu(2))(2)(OAr′) with 1.0 equiv of 2,4,6-Me(3)C(6)H(2)OH gave the disproportionation products Nb(NAr)(N=C(t)Bu(2))(OAr′)(2) and Nb(NAr)(N=C(t)Bu(2))(O-2,4,6-Me(3)C(6)H(2))(2)(HN=C(t)Bu(2)) with 1:1 ratio, clearly indicating the presence of the above mechanism and the fast equilibrium (between the ketimide and the phenoxide). The reaction of 3a with 1.0 or 2.0 equiv of C(6)F(5)OH afforded Nb(N=C(t)Bu(2))(2)(OC(6)F(5))(3)(HN=C(t)Bu(2)) as the sole isolated product, which was formed from once generated Nb(NAr)(N=C(t)Bu(2))(2)(OC(6)F(5))(HN=C(t)Bu(2)) by treating with C(6)F(5)OH.