Synthesis and crystal structures of two (cyclopentadienyl)titanium(III) hydroborate complexes, [Cp∗TiCl(BH4)]2 and Cp2Ti(B3H8)

Treatment of Cp∗TiCl3 and Cp2TiCl2 with NaB3H8 affords the titanium(III) hydroborate compounds [Cp∗TiCl(BH4)]2 and Cp2Ti(B3H8), respectively. The former compound arises by means of a new reaction, the metal-induced fragmentation of the B3H8 anion, and can also be made by treating Cp∗TiCl3 with LiBH4. The latter compound has been previously described, but not characterized crystallographically. Both compounds have been studied by single crystal X-ray diffraction. Dimeric [Cp∗TiCl(BH4)]2 has bridging chloride ligands and terminal Cp∗ and BH4 ligands. The Ti–Ti distance is 3.452(1) Å, which indicates that there is no metal–metal bonding interaction. The Ti–Cl distances are 2.440(2) Å and the Ti–Cl–Ti and Cl–Ti–Cl angles of 89.97(8) and 90.03(8)° so that the Ti2Cl2 unit is nearly a perfect square. The BH4 groups are each tridentate, with a Ti–B distance of 2.220(9) Å and an average Ti–H distance of 1.98(5) Å. In Cp2Ti(B3H8), the B3H8 ligand is bidentate, as is usually seen, and the Ti–B and Ti–H distances are 2.600(3) and 1.96(2) Å. The dihedral angle between the Ti–B(1)–B(2) plane and the B(1)–B(2)–B(3) plane is 123.4°. The Ti–B distances are 0.04 Å longer than those in niobium analog, Cp2Nb(B3H8), despite the fact that the single bond metallic radius of Ti is 0.02 Å smaller than that of Nb. This lengthening of the bond is probably a consequence of the presence of one fewer skeletal bonding electron in Cp2Ti(B3H8).

Treatment of Cp∗TiCl3 and Cp2TiCl2 with NaB3H8 affords the compounds [Cp∗TiCl(BH4)]2 and Cp2Ti(B3H8), respectively. The former compound arises by means of a new reaction, the metal-induced fragmentation of the B3H8 anion. Dimeric [Cp∗TiCl(BH4)]2 has bridging chloride ligands and tridentate BH4 groups. The Ti–Ti distance of 3.452(1) Å indicates that there is no metal–metal bond. In Cp2Ti(B3H8), the Ti–B distances are 0.04 Å longer than those in niobium analog, Cp2Nb(B3H8), probably because it has one fewer skeletal bonding electron.Figure optionsDownload as PowerPoint slide