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

摘要

Reactions of Nb(NAr)(N=C^tBu2)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=C^tBu2)2(OAr′), Nb(NAr)(N=C^tBu2)(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=C^tBu₂)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=C^tBu₂)(O-2,4,6-Me₃C₆H₂)₂(HN=C^tBu₂). Similar reactions of >3a with 1.0 equiv of (CF₃)₃COH or 2.0 equiv of (CF₃)₂CHOH afforded Nb(NAr)(N=C^tBu₂)₂[OC(CF₃)₃](HN=C^tBu₂) or Nb(NAr)(N=C^tBu₂)[OCH(CF₃)₂]₂(HN=C^tBu₂), 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^tBu₂) ligand. The reaction of Nb(NAr)(N=C^tBu₂)₂(OAr′) with 1.0 equiv of 2,4,6-Me₃C₆H₂OH gave the disproportionation products Nb(NAr)(N=C^tBu₂)(OAr′)₂ and Nb(NAr)(N=C^tBu₂)(O-2,4,6-Me₃C₆H₂)₂(HN=C^tBu₂) 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=C^tBu₂)₂(OC₆F₅)₃(HN=C^tBu₂) as the sole isolated product, which was formed from once generated Nb(NAr)(N=C^tBu₂)₂(OC₆F₅)(HN=C^tBu₂) by treating with C₆F₅OH.