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Synthesis, structure and reactivity of high-valent early transition metal complexes containing pi coordinated carbocycles and hard pi donor ligands.

机译:含pi配位碳环和硬pi供体配体的高价早期过渡金属配合物的合成,结构和反应活性。

摘要

Ta(DIPP)Cl₄ (1) has been prepared from the reaction of TaCl₅ with 1 equiv of Me₃SiDIPP and is shown to be dimeric. While Ta(DIPP)Cl₄ reacts with Et₂O to form trans-Ta(DIPP)Cl₄(OEt₂) (2), TaCl₅ and Me₃SiDMP react in the presence of Et₂O to provide trans-Ta(DMP)Cl₄(OEt₂) (3) directly. TaCl₅ reacts with 2 equiv of Me₃SiDMP to form the bis phenoxide complex Ta(DMP)₂Cl₃(OEt₂) (6). All of these compounds cyclize 3-hexyne upon their two electron reduction, to form (η⁶-C₆Et₆)Ta(OR)ₓCl₃₋ₓ (OR = DIPP, DMP, x = 1; OR = DMP, x = 2), but do not readily undergo the analogous reaction with 2-butyne. However, (η⁶-C₆Me₆)Ta(DIPP)Cl₂ (10) can be prepared in essentially quantitative yield from the reaction of (η⁶-C₆Me₆)Ta(DIPP)₂Cl (9) with Ta(DIPP)₂Cl₃(OEt₂). The arene ligand in (η⁶-C₆Me₆)Ta(DIPP)Cl₂ (10) is characterized by a folded structure and considerable localization of the π electron density in a 1,4-diene fashion. The thermolysis of (η⁶-C₆Me₆)Ta(DIPP)₂Cl 9 produces free C₆M₆ along with the "tucked in" complex (η¹-C₆Me₅CH₂)Ta(DIPP)₂Cl₂ (11). Compound 11 can be prepared in higher yields from 9 by its thermal decomposition in the presence of Me₃SiCl. Alkylation of tantalum(III) arene complexes affords stable dialkyl (η⁶-C₆Me₆)Ta(DIPP)R₂ and monoalkyl halide species (η⁶-C₆Me₆)Ta(DIPP)RX. The alkyl hydride complexes (η⁶-C₆Me₆)Ta(DIPP)R(H) are also prepared from (η⁶-C₆Me₆)Ta(DIPP)RX and LiBEt₃H. The arene ring in (η⁶-C₆Me₆)Ta(DIPP)Et₂ (18) exhibits a structure consistent with a diene-diyl distortion. The first evidence for the formation of a d¹ arene species is presented in cyclic voltammetry experiments on these compounds. The addition of 4 equiv of LiNHAr to a solution of ZrCl₄(THF)₂ in THF/PY yields Zr(=NAr)(NHAr)₂(PY)₂ (26). Zr(=NAr)(NHAr)(PY)₂ reacts with 1 or 2 equiv of TMSCl in THF/PY to provide Zr(=NAr)(NHAr)Cl(PY)₂ (27) and Zr(=NAr)Cl₂(PY)₃ (28) respectively. MCl₄(THF)₂ (M = Zr or Hf) react with 4 equiv LiNHAr followed by the addition of 2 equiv TMSCl to provide M(=NAr)Cl₂(THF)₂ (24, M = Zr; 25 M = Hf). Both 24 and 25 react with K₂ (C₈H₈) to provide ((η⁸-C₈H₈)M(=NAr))₂ (31, M = Zr; 32, M = Hf). Zr(=NAr)Cl₂(THF)₂ reacts with 1 equiv Li(C₅H₄Me) to provide ((η⁵-C₅H₄Me)Zr(=NAr)Cl)₂ (30). An imprecise structure of this dimeric compound has been determined which exhibits bridging imido functionalities. (η⁵-C₅Me₅)ZrCl₃(THF) reacts with 3 equiv LiNHAr in THF/PY to provide ((η⁵-C₅Me₅)Zr(=NAr)(NHAr)(PY))₂ (33). Compound 33 reacts with 1 equiv TMSCl in THF/PY to provide monomeric (η⁵-C₅Me₅)Zr(=NAr)Cl(PY)₂ (35).
机译:Ta(DIPP)Cl 3(1)是由TaCl 3与1当量的Me 3 SiDIPP反应制得的,显示为二聚体。当Ta(DIPP)Cl 3与Et 2 O反应形成反式Ta(DIPP)Cl 3(OEt 2)(2)时,TaCl 3和Me 3 SiDMP在Et 2 O存在下反应,直接提供反式Ta(DMP)Cl 3(OEt 2)(3)。 。 TaCl 3与2当量的Me 3 SiDMP反应形成双酚盐配合物Ta(DMP)2 Cl 3(OEt 2)(6)。所有这些化合物在两个电子还原时都环化3-己炔,形成(η⁶-C₆Et₆)Ta(OR)ₓCl₃₋ₓ(OR = DIPP,DMP,x = 1; OR = DMP,x = 2),但是不容易与2-丁炔进行类似反应。但是,(η3 -C 4 Me 3)Ta(DIPP)Cl 2(10)可以由(η3 -C 4 Me 3)Ta(DIPP)2 Cl(9)与Ta(DIPP)2 Cl 3(OEt 2)反应以基本上定量的产率制备。 (η1 -C 4 Me 3)Ta(DIPP)Cl 2(10)中的芳烃配体的特征在于其折叠结构和π电子密度以1,4-二烯的形式相当局部化。 (η3 -C 4 Me 3)Ta(DIPP)2 Cl 9的热分解产生游离的C 3 M 4以及“嵌入”的配合物(η1 -C 4 Me 3 CH 2)Ta(DIPP)2 Cl 2(11)。在Me 9SiCl存在下,通过热分解可以从9制备更高产率的化合物11。钽(Ⅲ)芳烃配合物的烷基化得到稳定的二烷基(η1 -C 4 Me 3)Ta(DIPP)R 2和单烷基卤化物(η1 -C 4 Me 3)Ta(DIPP)RX。烷基氢化物配合物(η⁶-C₆Me₆)Ta(DIPP)R(H)也由(η⁶-C₆Me₆)Ta(DIPP)RX和LiBEt₃H制备。 (η1 -C 4 Me 3)Ta(DIPP)Et 2(18)中的芳烃环显示出与二烯-二基变形一致的结构。在这些化合物的循环伏安法实验中提供了形成d 1芳烃物质的第一个证据。在ZrCl 3(THF)2在THF / PY中的溶液中加入4当量的LiNHAr,得到Zr(= NAr)(NHAr)2(PY)2(26)。 Zr(= NAr)(NHAr)(PY)2在THF / PY中与1或2当量的TMSCl反应,得到Zr(= NAr)(NHAr)Cl(PY)2(27)和Zr(= NAr)Cl 2( PY)₃(28)。 MCl 3(THF)2(M = Zr或Hf)与4当量的LiNHAr反应,然后加入2当量的TMSC1,得到M(= NAr)Cl 2(THF)2(24,M = Zr; 25M = Hf)。 24和25都与K 2(C 3 H 6)反应,得到((η3 -C 4 H 5)M(= NAr))2(31,M = Zr; 32,M = Hf)。 Zr(= NAr)Cl 2(THF)2与1当量的Li(C 3 H 4 Me)反应生成((η3 -C 3 H 4 Me)Zr(= NAr)Cl)2(30)。已经确定该二聚化合物的不精确结构显示出桥接的酰亚胺基官能团。 (η3 -C 4 Me 3)ZrCl 3(THF)在THF / PY中与3当量的LiNHAr反应,得​​到((η3 -C 4 Me 5)Zr(= NAr)(NHAr)(PY))2(33)。化合物33与1当量的TMSC1在THF / PY中反应,得到单体(η3 -C 4 Me 3)Zr(= NAr)Cl(PY)2(35)。

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