256952-80-4Relevant academic research and scientific papers
Synthesis of Group 4 [(RN-o-C6H4)2O]2- complexes where R is SiMe3 or 0.5 Me2SiCH2CH2SiMe2
Schrock, Richard R.,Liang, Lan-Chang,Baumann, Robert,Davis, William M.
, p. 163 - 173 (1999)
Complexes that contain the [(Me3SiN-o-C6H4)2O]2- ligand ([1]2-) of the type [1]M(NMe2)2, [1]MCl2, and [1]MMe2 have been prepared where M=Ti, Zr, or Hf. Although cations prepared by addition of [Ph3C][B(C6F5)4] or [PhNMe2H][B(C6F5)4] to [1]ZrMe2 or [1]HfMe2 could not be observed in NMR studies, addition of [(η5-C5H4Me)2Fe][B(C 6H5)4] to [1]HfMe2 in the presence of THF led to isolation of {[1]HfMe(THF)2}[B(C6H5)4]. An X-ray study showed the cation to be a distorted octahedron in which the [1]2- ligand is in the mer arrangement and is significantly twisted from a planar NC2OC2N arrangement. The THF ligands are trans to one another. No well-behaved activity for the polymerization of 1-hexene could be observed with activated [1]ZrMe2, while {[1]HfMe(THF)2}[B(C6H5)4] was inactive. The reaction between Li2[O(o-C6H4NH)2] and Me2ClSiCH2CH2SiMe2Cl in THF produced a cyclic diamido/ether ligand H2[2]. The reaction between H2[2] and Zr(NMe2)4 or ZrR4 (R=CH2Ph, CH2SiMe3) gave [2]Zr(NMe2)2(HNMe2) and Zr[2]2, respectively. The dimethylamine in [2]Zr(NMe2)2(HNMe2) could be replaced with pyridine or 2,4-lutidine to give [2]Zr(NMe2)2(L) (L=pyridine or 2,4-lutidine), which then could be converted into [2]ZrCl2(L) with excess Me3SiCl. The reaction between [2]ZrCl2(py) and two equivalents of Me3SiCH2MgCl gave a bimetallic complex in which one of the trimethylsilyl methyl groups has been doubly C-H activated, as confirmed by X-ray crystallography.
