The Addition of Gold and Tin to Bismuth-Triiridium Carbonyl Complexes

The reaction of Ir3(CO)9(μ3-Bi) with PhAu(NHC) (1), where NHC = 1,3-bis(2,6-diisopropylphenylimidazol-2-ylidene), has yielded the compound Ir3(CO)8(Ph)(μ3-Bi)[μ-Au(NHC)] (2) by the loss of one CO ligand and the oxidative addition of the Au-C (phenyl) bond of 1 to one of the iridium atoms. The Au(NHC) group bridges one of the Ir-Bi bonds of the cluster. On the basis of X-ray crystal structural analysis and molecular orbital and quantum theory of atoms in molecules calculations, the Au-Bi interaction was determined to be substantial and is comparable in character to the Ir-Bi and Ir-Ir bonds in this cluster. Compound 2 reacts with 2 equiv of HSnPh3 to yield the compound Ir3(CO)7(SnPh3)2(μ3-Bi)[μ-Au(NHC)](μ-H) (3), which contains two terminally coordinated SnPh3 ligands. Compound 3 reacts with H2O to yield the compound Ir3(μ3-Bi)(CO)7[μ-Ph2Sn(OH)SnPh2][μ-Au(NHC)] (4) by cleavage of a phenyl ring from each of the SnPh3 ligands and formation of a bridging OH group between the two tin atoms to form a chelating Ph2Sn(OH)SnPh2 ligand.