The
ruthenium(0) complex [Ru{κ2
C
2-MeIm(CH2)3ImMe}(CO)3] (1), MeIm(CH2)3ImMe = 1,3-bis(3-methylimidazol-2-yliden-1-yl)propane,
which contains a chelating bis(N-heterocyclic carbene) ligand, reacts
with MeOTf at room temperature to give the ionic ruthenium(II) methyl
derivative [RuMe{κ2
C
2-MeIm(CH2)3ImMe}(CO)3]OTf ([2]OTf), whereas an analogous reaction of 1 with
MeI renders the neutral ruthenium(II) acetyl derivative [RuI{C(O)Me}{κ2
C
2-MeIm(CH2)3ImMe}(CO)2] (3), in which the iodide and
acetyl ligands occupy mutually trans coordination
sites. The fact that [2]OTf reacts with [Et4N]I to give 3 evidences the participation of the cationic
species 2
+ in the synthesis of 3. The mechanisms of these reactions in THF solution have been modeled
by DFT calculations. They have shown that 2
+ can be made from complex 1 and either MeOTf or MeI.
In both cases, two plausible reaction pathways have been identified.
They start with a rate-determining SN2 substitution process
in which the metal atom of 1 attacks the C atom of MeI
or MeOTf, displacing the corresponding anion to give, depending on
how compound 1 approaches MeI or MeOTf, 2
+ or a less stable isomeric species 2′+ that is easily transformed into 2
+. A subsequent CO migratory insertion in 2
+ leads to an unsaturated (pentacoordinated) acetyl intermediate that
easily adds the iodide anion to end in 3. DFT calculations
have also shown that the reaction of 1 with MeOTf to
give [2]OTf is thermodynamically more favorable than
that of 1 with MeI to give [2]I due to the
resonance stabilization and greater solvation energy of the triflate
anion. These two effects are also responsible for the fact that the
incorporation of the triflate anion with 2
+ to give a putative triflate complex analogous to 3 is
thermodynamically disfavored, whereas the incorporation of the iodide
anion with 2
+ to give 3 is thermodynamically
favored.