Synthesis of Higher Nuclearity Iridium Clusters: Reaction of [Ir4(CO)11Ph]- with [Ir(COD)Cl]2

The reaction of the a-phenyl tetrairidium carbonyl anion [Et4N][Ir-4(CO)(11)Ph] with [Ir(COD)Cl] (COD = 1,5-cyclooctadiene) yielded the two known tetrairidium compounds Ir-4(CO)(10)(COD) and Ir-4(CO)(7)(COD)(mu(4)-C8H10) (1) and the three new higher nudearity complexes Ir-5(CO)(11)(Ph)(COD) (2) Ir-5(CO)(9)(Ph)(COD)(2) (3), and Ir-9(CO)(15)(Ph)(mu(3)-C8H10)(COD) (4), containing sigma-coordinated phenyl ligands. Compounds 2 and 3 contain trigonal-bipyramidal Ir-5 clusters. Compound 4 contains nine iridium atoms in the form of a tricapped octahedron. Compound 4 was shown to be formed by the condensation of 1 and 2. Compound 3 reacts with COD to yield the compound Ir-5(CO)(7)(COD)(2)(mu(4)-eta(2):eta(1)-C8H11) (5) in a cluster-opening process that cleaves two hydrogen atoms from one of the COD C-C double bonds, eliminates the sigma-phenyl ligand, and transfers one of the hydrogen atoms the other C-C double bond to form a metalated mu(4)-eta(2):eta(1)-C8H11 cyclooctyne ligand.