Abstract
The reaction of Rh4(CO)12 with Pd(PBu
t
3)2 yielded the high nuclearity bimetallic hexarhodium-tripalladium cluster complex Rh6(CO)16[Pd(PBu
t
3)]3, 10, in 11% yield. Compound 10 was converted to the hexarhodium-tetrapalladium cluster Rh6(CO)16[Pd(PBu
t
3)]4, 11, in 62% yield by reaction with an additional quantity of Pd(PBu
t
3)2. Both compounds were characterized crystallographically. Structurally, both compounds consist of an octahedral cluster of six rhodium atoms with sixteen carbonyl ligands analogous to that of the known compound Rh6(CO)16. Compound 10 also contains three Pd(PBu
t
3) groups that bridge three Rh–Rh bonds along edges of the Rh6 octahedron to give an overall D3 symmetry to the Rh6Pd3 cluster. Compound 11 contains four edge bridging Pd(PBu
t
3) groups distributed across the Rh6 octahedron to give an overall D2d symmetry to the Rh6Pd4 cluster. Each Rh–Pd connection in both compounds contains a bridging carbonyl ligand that helps to stabilize the bond between the Pd(PBu
t
3) groups and the Rh atoms. Both compounds can be regarded as Pd(PBu
t
3) adducts of Rh6(CO)16.
