Neutral and Cationic Group 4 Metal Compounds Containing Octamethyldibenzotetraazaannulene (Me<sub>8</sub>taa<sup>2-</sup>) Ligands. Synthesis and Reactivity of (Me<sub>8</sub>taa)MX<sub>2</sub> and (Me<sub>8</sub>taa)MX<sup>+</sup> Complexes (M = Zr, Hf; X = Cl, Hydrocarbyl, NR<sub>2</sub>, OR)
作者:Alfredo Martin、Roger Uhrhammer、Thomas G. Gardner、Richard F. Jordan、Robin D. Rogers
DOI:10.1021/om970814x
日期:1998.2.1
group from Zr to a Me8taa imine carbon in coordinating solvents. The reaction of (Me8taa)H2 with the appropriate ZrR4 compound yields (Me8taa)Zr(CH2Ph)2 (3c) and (Me8taa)Zr(CH2CMe3)2 (3d). The reaction of (Me8taa)H2 and Zr(NR2)4 yields (Me8taa)Zr(NR2)2 (6a, R = Me; 6b, R = Et). Spectroscopic data for (Me8taa)MX2 compounds 2, 3, 4, and 6 are consistent with cis, C2v-symmetric structures. Dialkyl complexes
含有双阴离子四氮杂大环的面外(Me 8 taa)MX 2和(Me 8 taa)MX +配合物(M = Zr,Hf; X = Cl,烃基,NR 2或OR)的合成和反应性描述了配体八甲基二苯并四氮杂壬烯(Me 8 taa 2-)。[Li(Et 2 O)] 2 [Me 8 taa](1)与MCl 4(THF)2的反应产生(Me 8 taa)MCl 2络合物(2a,M = Zr;2b,M = Hf)。2a的烷基化,b在烃溶剂中用LiCH 2 SiMe 3或LiMe生成(Me 8 taa)M(CH 2 SiMe 3)2(3a,M = Zr; 4a,M = Hf)或(Me 8 taa)MMe 2(3b, M = Zr;4b,M = Hf)络合物。通过在配位溶剂中将Me基团从Zr迁移到Me 8 taa亚胺碳上,化合物3b进行了重排。(Me 8 taa)H 2与适当的ZrR 4化合物反应生成(Me 8 taa)Zr(CH2