Synthesis and structure of a binuclear ruthenium 4-chlorobenzamidato complex

Abstract

The compound Ru 2Cl(4-Cl-C 6H 4CONH) 4 was prepared by reaction of Ru 2Cl(O 2CCH 3) 4 with 4-Cl-C 6H 4CONH 2 at 180°C. Crystals of the composition Ru 2Cl(4-Cl-C 6H 4CONH) 4CH 3OH were obtained by slow diffusion of CH 3OH containing Et 4NCl into a Me 2SO solution of the compound. The structure of the crystalline product, which loses solvent of crystallization on removal from the mother liquor, was solved by X-ray crystallography by mounting a single crystal in a capillary containing the mother liquor. The crystals belong to the space group P1̄ (triclinic crystal system) with a = 12.731(3) Å, b = 14.389(3) Å, c = 12.604(3) Å, α = 103.41(2)°, β = 106.43(2)°, γ = 64.90(2)°, V = 1988.6(8) Å 3 and Z = 2. There are two half ruthenium dimers linked by a Cl atom and an uncoordinated solvent CH 3OH molecule per asymmetric unit. The ruthenium dimers lie on two centers of inversion at 0, 0, 0 and 1/2, 0, 0. The chloride ions bridge dinuclear cations in the crystal, forming infinite zigzag chains. The average Ru-Ru distance is 2.296[1] Å and each ruthenium atom has a RuClN 2O 2 coordination sphere where the average Ru′-Ru-Cl angle is virtually linear (175.68[6]°). The metal oxidation states in the complex are + 2 and + 3, giving an average value of + 2.5. The arrangement of four bridging 4-Cl-benzamidato ligands is of the 2 : 2 type. The average Ru-N, Ru-O, Ru-Cl distances and Ru(1)-Cl(1)-Ru(2) angle are 2.036[6] Å, 2.044[5] Å, 2.583[2] Å and 117.26(8)°, respectively. The IR spectrum of the compound shows two N-H stretches at 3380 and 3340 cm −1. The electronic spectrum of the compound in Me 2SO exhibits bands at 558 nm (ε = 340 M −1 cm −1), 425 nm (1000) and 320 nm (22,700).