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H+/AuPPh3+ Exchange for the Hydride Complexes CpMoH(CO)2(L) (L = PMe3, PPh3, CO). Formation and Structure of [Cp(CO)2(PMe3)Mo(AuPPh3)2]+[BF4]-

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posted on 1997-07-02, 00:00 authored by Rossana Galassi, Rinaldo Poli, E. Alessandra Quadrelli, James C. Fettinger
The reaction of CpMoH(CO)2L with AuPPh3+BF4- in THF at −40 °C proceeds directly to the MoAu2 cluster compounds [CpMo(CO)2L(AuPPh3)2]+BF4- (L = PMe3 (1), PPh3 (2)) with release of protons. A 1:1 reaction leaves 50% of the starting hydride unreacted. At lower temperature, however, the formation of a [CpMo(CO)2(PMe3)(μ-H)(AuPPh3)]+ intermediate is observed. This compound evolves to the cation of 1 and CpMoH(CO)2(PMe3) upon warming and is deprotonated by 2,6-lutidine to afford CpMo(CO)2(PMe3)(AuPPh3). The X-ray structure of 1 can be described as a four-legged piano stool with the PMe3 and the “η2-(AuPPh3)2” ligands occupying relative trans positions. [Cp(CO)2(PMe3)Mo(AuPPh3)2]+[BF4]- (Mr = 1298.41):  monoclinic, space group P21/n, a = 18.1457(13) Å, b = 9.7811(7) Å, c = 26.096(2) Å, β = 105.086(5)°, V = 4472.0(5) Å3, Z = 4. The reaction of CpMoH(CO)2(PMe3) with 3 equiv of AuPPh3+ affords a MoAu3 cluster, [CpMo(CO)2(PMe3)(AuPPh3)3]2+ (3), in good yields under kinetically controlled conditions. Under thermodynamically controlled conditions, 3 dissociates extensively into 1 and free AuPPh3+. It is proposed that the hydride ligand helps build higher nuclearity Mo−Au clusters. The difference in reaction pathways for the interaction of AuPPh3+ with CpMoH(CO)2L when L = PR3 or CO and for the interaction of CpMoH(CO)2(PMe3) with E+ when E = H, Ph3C or AuPPh3 is discussed. The lower acidity and greater aurophilicity of the [CpMo(CO)2L(μ-H)(AuPPh3)]+ intermediate when L = PMe3 favor attack by AuPPh3+ before deprotonation.