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

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posted on 09.01.2012, 00:00 by Richard D. Adams, Mingwei Chen
The reaction of the σ-phenyl tetrairidium carbonyl anion [Et4N]­[Ir4(CO)11Ph] with [Ir­(COD)­Cl] (COD = 1,5-cyclooctadiene) yielded the two known tetrairidium compounds Ir4(CO)10(COD) and Ir4(CO)7(COD)­(μ4-C8H10) (1) and the three new higher nuclearity complexes Ir5(CO)11(Ph)­(COD) (2) Ir5(CO)9(Ph)­(COD)2 (3), and Ir9(CO)15(Ph)­(μ3-C8H10)­(COD) (4), containing σ-coordinated phenyl ligands. Compounds 2 and 3 contain trigonal-bipyramidal Ir5 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 Ir5(CO)7(COD)2421-C8H11) (5) in a cluster-opening process that cleaves two hydrogen atoms from one of the COD C–C double bonds, eliminates the σ-phenyl ligand, and transfers one of the hydrogen atoms the other C–C double bond to form a metalated μ421-C8H11 cyclooctyne ligand.