Dinuclear versus Mononuclear Zinc Compounds from Reduction of LZnCl2 (L = α-Diimine Ligands): Effects of the Ligand Substituent, Reducing Agent, and Solvent
posted on 2008-11-24, 00:00authored byJie Yu, Xiao-Juan Yang, Yanyan Liu, Zhifeng Pu, Qian-Shu Li, Yaoming Xie, Henry F. Schaefer, Biao Wu
A Zn−Zn-bonded compound, [K(THF)2]2[(LiPr)Zn−Zn(LiPr)] (2) and three mononuclear zinc compounds, [Zn(LMe)2Na2(Et2O)2] (3), [Zn(LEt)2Na2(THF)2] (4), and [Zn(LEt)2K2]n (5), with N-aryl substituted α-diimine ligands LiPr, LMe, and LEt (L = [(2,6-R2C6H3)N(Me)C]2, R = iPr, Me, Et, respectively) have been synthesized from the reduction of the LZnCl2 precursors by the alkali metal Na or K. X-ray structural analyses show that the compounds have a [Zn2L2]2− (2) or [ZnL2]2− (3−5) core incorporating Na+ or K+ ions solvated by THF or Et2O molecules, except for 5, which displays a 2D polymeric structure formed by intermolecular K−C bonds due to the lack of K−solvent interactions. In compound 2, the formal Zn2+ ion in the precursor is reduced to Zn+, while in 3−5 it remains unreduced. The neutral ligands in the precursor, however, are doubly reduced to a dianion, L2−, in all compounds, as evidenced by the bond lengths of the N–CC–N moiety of the ligands. Effects of the ligand substituent, reducing agent, and solvent on the products have been studied, which reveal that the steric bulk of the ligand is the most important factor that determines whether the Zn−Zn bond can be formed. DFT computations show that the Zn−Zn bond in 2 is formed mainly by the 4s orbitals of zinc. The results of natural charge and orbital analyses also confirm the reduction of the ligands to their dianionic forms.