Synthesis, Structure, and Reactions of Binuclear Gold(I) Complexes Containing Two Different Bridging Ligands

The binuclear cycloaurated compounds [Au2(μ-C6H3-2-PPh2-n-Me)2] (n = 5, 1a; n = 6, 1b) react with the digold(I) complexes [Au2(μ-S2CNnBu2)2] and [Au2(μ-dppm)2](PF6)2 to give heterobridged dinuclear complexes [Au2(μ-C6H3-2-PPh2-n-Me)(μ-S2CNnBu2)] (n = 5, 5a; n = 6, 5b) and [Au2(μ-C6H3-2-PPh2-n-Me)(μ-dppm)]PF6, (n =
5, 9a; n = 6, 9b), respectively. Complex 5a exists in the solid state as an infinite zigzag chain of dimeric units
with intramolecular Au−Au separations of 2.8331(3) and 2.8243(3) Å for independent molecules and intermolecular
Au−Au separations of 3.0653(3) and 3.1304(3) Å. Both 5a and 5b undergo oxidative addition with halogens to
give the heterovalent, gold(I)−gold(III) compounds [XAuI(μ-2-Ph2PC6H3-n-Me)AuIIIX(η2-S2CNnBu2)] [n = 5,
X = Cl (6a), I (8a); n = 6, X = Cl (6b), Br (7b), I (8b)]. Compound 8a has been shown by X-ray crystallography
to contain a gold(III) atom coordinated in a planar array by bidentate, chelating di-n-butyldithiocarbamate, iodide,
and the σ-aryl carbon atom, together with a gold(I) atom that is linearly coordinated by the phosphorus atom of
the arylphosphine and by iodide. The intramolecular gold−gold distance of 3.2201(3) Å indicates little or no
interaction between the metal atoms. In contrast to the behavior of the homobridged complexes 1a and 1b, the
heterobridged dithiocarbamate complexes 5a and 5b give structurally similar products on reaction with halogens,
irrespective of the position of the ring methyl substituent. Crystal data for [Au2(μ-C6H3-2-PPh2-5-Me)(μ-S2CNnBu2)] (5a):  triclinic, space group P1̄ (No. 2), with a = 11.3398(1), b = 15.9750(2), c = 16.4400(3) Å, α =
91.0735(9), β = 109.3130(7), γ = 90.7666(8)°, V = 2809.47(6) Å3, and Z = 4. Crystal data for [IAuI(μ-2-Ph2PC6H3-5-Me)AuIIII(η2- S2CNnBu2)] (8a):  triclinic, space group P1̄ (No. 2), with a = 8.6136(2), b = 9.3273, c =
21.1518(4) Å, α = 84.008(1), β = 84.945(1), γ = 75.181(1)°, V = 1630.54(6) Å3, and Z = 2.