A New Method for Introducing Tin Ligands into Tetrairidium Dodecacarbonyl

The reaction of Ph3SnOH with Ir4(CO)12 in the presence of [Bu4N]OH in methanol solvent gave two products, [Bu4N][Ir4(CO)11(SnPh3)] (1; 45% yield) and [Bu4N][Ir4(CO)10(SnPh3)2(μ-H)] (2; 5.5% yield), and the reaction of Ir4(CO)11(PPh3) with Ph3SnOH in the presence of [Bu4N]OH gave the complex Ir4(CO)10(SnPh3)(PPh3)(μ-H) (3) in 44% yield. It is proposed that the reactions occur by the addition of the anion [OSnPh3]− generated in situ to a CO ligand of the Ir4(CO)12 to form a stannyl-substituted metallocarboxylate ligand that subsequently loses CO2 and transfers the SnPh3 group to a metal atom.