Answers are grounded in this paper and cite their sources.
Reactions of Ln[N(SiMe 3 ) 2 ] 2 (thf) 2 (Ln = Sm, Yb) with 1 equiv of (C 5 Me 4 H)SiMe 2 NHPh afforded the first linked cyclopentadienyl−anilido (or amido) lanthanide(II) complexes Me 2 Si(C 5 Me 4 )(NPh)Ln(thf) x ( 1, Ln = Yb, x = 3; 2, Ln = Sm, x = 0−1) in 75−84% isolated yields. Recrystallization of 1 from toluene/hexane yielded the less solvated complex [Me 2 Si(C 5 Me 4 )(NPh)Yb(thf)] 2 ( 1 ‘). Complex 1 adopts a monomeric structure containing one chelating Cp−anilido ligand and three thf ligands, while 1 ‘ forms a dimeric structure through an “intermolecular” Yb−Ph π-interaction. Reaction of 1 or 1 ‘ with azobenzene gave the binuclear Yb(III) complex Me 2 Si(C 5 Me 4 )(NPh)Yb(thf)(μ,η 2: η 3 -N 2 Ph 2 )Yb(NPh)(C 5 Me 4 )SiMe 2 ( 3 ), which contains a cis -oriented azobenzene dianion unit bonding in a η 3 fashion to one Yb atom and in a η 2 fashion to the other Yb atom. Reaction of 1 with 1 equiv of fluorenone gave the corresponding Yb(III) ketyl complex Me 2 Si(C 5 Me 4 )(NPh)Yb(thf) 2 (OC 13 H 8 ) ( 4 ) in 75% isolated yield. Treatment of 4 with hexane/ether led to formation of the pinacolate complex [Me 2 Si(C 5 Me 4 )(NPh)Yb(thf)] 2 (μ-O 2 C 26 H 16 ) ( 5 ), in which the pinacolate unit is arranged in a gauche conformation. Dissolving of 5 in THF cleaved the central C−C bond of the pinacolate unit in 5 and regenerated 4 quantitatively. The Sm(II) complex 2 showed moderate activity (44.8 kg of polymer (mol of Sm) - 1 h - 1 ) for the polymerization of ethylene under 1 atm at room temperature, yielding linear polyethylene with M n = 7.26 × 10 5 and M w / M n = 1.58.
No takes yet. Share an insight, caveat, or question.
Hou et al. (2001) studied this question.
Synapse has enriched 10 related papers. Consider them for comparative context: