Halogeno, hydrido, and hydrocarbyl derivatives of (η-cyclo-octatetraene)(η-pentamethylcyclopentadienyl)zirconium(IV): X-ray crystal structure of [Zr(η-C5Me5)(η-C8H8)H]

Abstract

The sequential addition of K₂(C₈H₈) and Li(C₅Me₄R)(R = Me or Et) to ZrCl₄ affords [Zr(η-C₅Me₄R)(η-C₈H₈)Cl] which further reacts with MR′(M = MgCl or Li; R′= Me, Et, CH₂Ph, C₆H₄Me-4, CMeCH Me, C₂Me, or C₂Ph) to afford the corresponding hydrocarbyl derivatives [Zr(η-C₅Me₄R)(η-C₈H₈)R′]. Reaction of [Zr(η-C₅Me₄R)(η-C₈H₈)Cl] with LiAIH₄ or Mg(Cl)CHMe₂ affords [Zr(η-C₅Me₄R)(η-C₈H₈) H]. Both i.r. and n.m.r. data suggest that the compounds [Zr(η-C₅Me₄R)(η-C₈H₈)X](X = Cl, H, or R′) have a ‘planar’ 1–8-η-C₈H₈ ligand and should therefore be regarded as 18-electron complexes of Zr^IV. An X-ray diffraction study of [Zr(η-C₅Me₅)(η–C₈H₈) H] has shown that the molecule is monomeric and has one terminal hydride ligand. All eight carbon atoms of the C₈ ring are within bonding distance of the Zr atom, with Zr–C varying from 2.43(4) to 2.68(4)Å and the ring itself slightly folded into an envelope conformation. It is bonded to the metal atom on the convex face. Crystals of [Zr(η-C₅Me₅)(η-C₈H₈)H] are orthorhombic, space group Pbc2₁, with two crystallographically distinct molecules in the asymmetric unit. Surprisingly, although the two molecules appear to be related by a pseudo-glide plane, in fact one molecule is twisted in its crystal siting relative to the other, though their molecule symmetry is identical. The structure could not be very accurately determined; it has been refined to R 0.066 (R′ 0.055) for 1 019 reflections measured at 210 K to 2θ= 50°.