Carbene and carbon dioxide chemistry of binuclear iridium complexes

Mark Kent Reinking, Purdue University

Abstract

The reaction of (Ir(COD)$\sb2$) (PF$\sb6$) with CNBu$\sp{\rm t}$ and dmpm results in $\rm \lbrack Ir\sb2 (\mu$-CNBu$\sp{\rm t})\sb4$(dmpm)$\sb2$) (PF$\sb6 \rbrack \sb2$, 1. Complex 1 has a trans-trans dmpm framework with one bridging isocyanide and four terminally bound isocyanide ligands. Hydrogenation of 1 leads to a displacement of one isocyanide ligand and formation of $\rm \lbrack Ir\sb2 (\mu$-H)$\sb2$(CNBu$\sp{\rm t})\sb4$(dmpm)$\sb2$) (PF$\sb6\rbrack \sb2$ 2. Reacting (IrCODCl) $\sb2$ with dmpm and CO results in $\rm \lbrack Ir\sb2 (\mu$-CO)(CO)$\sb3$Cl(dmpm)$\sb2$) (Cl) which can be reduced in situ resulting in Ir$\sb2$(CO)$\sb4$(dmpm)$\sb2$, 3. The X-ray structure of 3 reveals a cis, trans diphosphine arrangement and all terminally bound carbonyl ligands. Complex 3 equilibrates the phosphorus environments in solution by rotation of the CO ligands about the bimetallic core. Complex 3 is easily oxidized by two electrons to give $\rm \lbrack Ir\sb2 (\mu$-CO)(CO)$\sb3$(dmpm)$\sb2\rbrack \sp{2+}$, 4. Complex 4 rapidly reacts with carbon monoxide to give the symmetrical, $\rm \lbrack Ir\sb2 (\mu$-CO)(CO)$\sb4$(dmpm)$\sb2\rbrack \sp{2+}$, 5, which is isoelectronic and isostructural with compound 1. Compound 3 is reactive towards CH$\sb2$Cl$\sb2$ in the presence of Me$\sb3$SiO$\sb2$SCF$\sb3$ to give $\rm \lbrack Ir\sb2 (\mu$-CH$\sb2$)(CO)$\sb4$(dmpm)$\sb2\rbrack \sp{2+}$, 6. The carbene carbon of 6 is electrophilic and is reactive towards nucleophiles. When compound 6 is treated with H$\sp{-}$, $\rm \lbrack Ir\sb2 (\mu$-CH$\sb3$)(CO)$\sb2$(dmpm)$\sb2\rbrack \sp{2+}$, 8, is formed. X-ray analysis of 8 shows a "boat" conformation with close, nonbonding contact between the $\mu$-CH$\sb3$ fragment and the methylene groups of the diphosphine ligand. Removal of one CO ligand from 3 results in Ir$\sb2$(CO)$\sb3$(dmpm)$\sb2$, 10. Compound 10 is reactive towards liquid CO$\sb2$ resulting in disproportionation of the CO$\sb2$ to give Ir$\sb2$(CO)$\sb3$(CO)$\sb2$(dmpm)$\sb2$, 11, and CO. Compound 11 was found to undergo oxygen atom transfer to CO to generate CO$\sb2$.

Degree

Ph.D.

Advisors

Kubiak, Purdue University.

Subject Area

Chemistry

COinS