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

Reactions of Nb(NAr)(N═CtBu2)3 (3a, Ar = 2,6-Me2C6H3) with 1.0, 2.0, or 3.0 equiv of Ar′OH (Ar′ = 2,6-iPr2C6H3) afforded Nb(NAr)(N═CtBu2)2(OAr′), Nb(NAr)(N═CtBu2)(OAr′)2, or Nb(NAr)(OAr′)3, respectively (at 25 °C), whereas the reaction with 2.0 equiv of 2,6-tBu2C6H3OH afforded Nb(NAr)(N═CtBu2)2(O-2,6-tBu2C6H3) upon heating (70 °C) without the formation of bis(phenoxide) and the reaction of 3a with 2.0 equiv of 2,4,6-Me3C6H2OH afforded Nb(NAr)(N═CtBu2)(O-2,4,6-Me3C6H2)2(HN═CtBu2). Similar reactions of 3a with 1.0 equiv of (CF3)3COH or 2.0 equiv of (CF3)2CHOH afforded Nb(NAr)(N═CtBu2)2[OC(CF3)3](HN═CtBu2) or Nb(NAr)(N═CtBu2)[OCH(CF3)2]2(HN═CtBu2), 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═CtBu2) ligand. The reaction of Nb(NAr)(N═CtBu2)2(OAr′) with 1.0 equiv of 2,4,6-Me3C6H2OH gave the disproportionation products Nb(NAr)(N═CtBu2)(OAr′)2 and Nb(NAr)(N═CtBu2)(O-2,4,6-Me3C6H2)2(HN═CtBu2) 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 C6F5OH afforded Nb(N═CtBu2)2(OC6F5)3(HN═CtBu2) as the sole isolated product, which was formed from once generated Nb(NAr)(N═CtBu2)2(OC6F5)(HN═CtBu2) by treating with C6F5OH.

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   Crystal structure refinement with SHELXL · 2015

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