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.