2012
DOI: 10.1021/om3001026
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Iridium–Gold Cluster Compounds: Syntheses, Structures, and an Unusual Ligand-Induced Skeletal Rearrangement

Abstract: The new iridium−gold complex Ir 4 (CO) 11 (Ph)(μ-AuPPh 3 ), 1 was obtained from the reaction of the tetrairidium anion [Ir 4 (CO) 11 (Ph)] − with [Au(PPh 3 )] [NO 3 ]. Two new iridium−gold complexes Ir 4 (CO) 10 (AuPPh 3 ) 2 , 2, and Ir 4 (CO) 11 (AuPPh 3 ) 2 , 3, were obtained from the reaction of [HIr 4 (CO) 11 ] − with [Au(PPh 3 )][NO 3 ]. Compounds 1−3 were structurally characterized by single crystal X-ray diffraction analyses. Compound 2 adds CO reversibly to form compound 3. In this process, the octahe… Show more

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Cited by 19 publications
(26 citation statements)
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References 55 publications
(58 reference statements)
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“…In addition, compound 3 contains two SnPh 3 ligands that occupy equatorial coordination sites, one on each of the two Ir(CO) 2 groups. The Ir−Sn bond distances are similar in length, Ir3−Sn1 = 2.6415(18) Å, Ir2−Sn2 = 2.6413 (19) Å, and are similar to those found in the compounds Ir 3 (CO) 6 (SnPh 3 ) 3 -(μ-H) 3 (μ 3 -Bi) 5 and Ir 3 (CO) 6 (SnPh 3 ) 3 (μ-SnPh 2 ) 3 . 27 There is a hydride ligand bridging the Ir−Ir bond between the two tinsubstituted iridium atoms.…”
Section: Inorganic Chemistrysupporting
confidence: 75%
“…In addition, compound 3 contains two SnPh 3 ligands that occupy equatorial coordination sites, one on each of the two Ir(CO) 2 groups. The Ir−Sn bond distances are similar in length, Ir3−Sn1 = 2.6415(18) Å, Ir2−Sn2 = 2.6413 (19) Å, and are similar to those found in the compounds Ir 3 (CO) 6 (SnPh 3 ) 3 -(μ-H) 3 (μ 3 -Bi) 5 and Ir 3 (CO) 6 (SnPh 3 ) 3 (μ-SnPh 2 ) 3 . 27 There is a hydride ligand bridging the Ir−Ir bond between the two tinsubstituted iridium atoms.…”
Section: Inorganic Chemistrysupporting
confidence: 75%
“…These CO ligands must come from the decarbonylation of glyceraldehyde. The geometry of 2 is a “bow tie” or edge‐sharing bitetrahedron,6 previously unreported for Ir complexes, other than in a Au–Ir cluster 7. In spite of its high‐temperature formation,8 it retains 14 H ligands (according to 1 H NMR spectroscopic data and DFT calculations).…”
Section: Interatomic Distances In 2 As Determined By X‐ray Diffractiomentioning
confidence: 90%
“…9,10 Iridium is well-known for its ability to produce highly active homogeneous 11 and heterogeneous 12 catalysts. Recently, gold nanoparticles have been shown to exhibit significant activity for the catalytic oxidation of CO and for the oxidation and transformations of unsaturated hydrocarbons.…”
Section: And Withmentioning
confidence: 99%