Synthesis of (arylmido)niobium(V) complexes containing ketimide, phenoxide ligands, and some reactions with phenols, alcohols

Synthesis of (arylmido)niobium(V) complexes containing ketimide, phenoxide ligands, and some reactions with phenols, alcohols
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含有酮亚胺、酚盐配体的(芳基氨基)铌(V)络合物的合成以及与酚、醇的一些反应

DOI:
10.1021/acsomega.8b01065
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发表时间:
2018
期刊:
影响因子:
4.1
通讯作者:
K. Nomura
K. Nomura
中科院分区:
化学3区
文献类型:
--
作者:
N. Srisupap;K. Wised;K. Tsutsumi;K. Nomura

文献摘要

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Nb(NAr)(N <$CtBu2)3的反应(3a,Ar = 2,6-Me 2C 6 H3),具有1.0、2.0或3.0当量的Ar′OH(Ar′ = 2,6-iPr 2C 6 H3)分别得到Nb(NAr)(N <$CtBu2)2(OAr′)、Nb(NAr)(N <$CtBu2)(OAr′)2或Nb(NAr)(OAr′)3(在25 ℃下),而与2.0当量的2,6-tBu 2C 6 H3 OH反应得到Nb(NAr)(N = CtBu 2)2(O-2,6-tBu 2C 6 H3)加热(70 °C)时,不会形成双(苯氧化物)和3a与2.0当量的2,4,6-Me 3C 6 H2OH反应得到Nb(NAr)(N-CtBu 2)(O-2,4,6-Me 3C 6 H2)2(HN-CtBu 2)。3a与1.0当量(CF 3)3COH或2.0当量(CF 3)2CHOH的类似反应分别得到Nb(NAr)(N = CtBu 2)2[OC(CF 3)3](HN = CtBu 2)或Nb(NAr)(N = CtBu 2)[OCH(CF 3)2]2(HN = CtBu 2)。基于它们的结构分析和反应化学,有人建议,这些反应进行通过苯酚(醇)的Nb和随后的质子(氢)转移到酮酰亚胺(N <$CtBu2)配体的配位。Nb(NAr)(N <$CtBu2)2(OAr′)与1.0当量的2,4,6-Me 3C 6 H2OH反应,得到比例为1:1的反硝化产物Nb(NAr)(N <$CtBu2)(OAr′)2和Nb(NAr)(N <$CtBu2)(O-2,4,6-Me 3C 6 H2)2(HN <$CtBu2),这清楚地表明上述机理的存在和快速平衡(在酮亚胺和酚盐之间)。3a与1.0或2.0当量的C6 F5 OH的反应得到作为唯一分离产物的Nb(N = CtBu 2)2(OC 6 F5)3(HN = CtBu 2),其通过用C6 F5 OH处理由一次生成的Nb(NAr)(N = CtBu 2)2(OC 6 F5)(HN = CtBu 2)形成。
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 of3awith 2.0 equiv of 2,4,6-Me3C6H2OH afforded Nb(NAr)(N═CtBu2)(O-2,4,6-Me3C6H2)2(HN═CtBu2). Similar reactions of3awith 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′)2and 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 of3awith 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.