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Palladium N-Heterocyclic Carbene Complexes: Synthesis from Benzimidazolium Salts and Catalytic Activity in Carbon-carbon Bond-forming Reactions
Published on: July 30, 2017
Hydrocarbon-Soluble Mercuracarborands: Syntheses, Halide Complexes, and Supramolecular Chemistry
Zhiping Zheng1, Carolyn B. Knobler, Mark D. Mortimer
1Department of Chemistry and Biochemistry, 405 Hilgard Avenue, University of California, Los Angeles, California 90095-1569.
Abstract:
The syntheses of macrocyclic species composed of carborane derivatives joined via their carbon vertices by electrophilic mercury atoms are described. The reaction of closo-1,2-Li(2)[C(2)B(10)H(10)(-)(x)()R(x)()] with HgI(2) gives Li(2)[(1,2-C(2)B(10)H(10)(-)(x)()R(x)()Hg)(4)I(2)] [R = Et, x = 2 (5.I(2)Li(2)); R = Me, x = 2 (6.I(2)Li(2)); R = Me, x = 4 (7.I(2)Li(2))]. 6.I(2)(K.[18]dibenzocrown-6)(2) crystallizes in the monoclinic space group C2/m [a = 28.99(2) Å, b = 18.19(1) Å, c = 13.61(1) Å, beta = 113.74(2) degrees, V = 6568 Å(3), Z = 4, R = 0.060, R(w) = 0.070]; 7.I(2)(NBu(4))(2) crystallizes in the monoclinic space group P2(1)/c [a = 12.77(1) Å, b = 21.12(2) Å, c = 20.96(2) Å, beta = 97.87(2) degrees, V = 5600 Å(3), Z = 2, R = 0.072, R(w) = 0.082]. The precursor to 7, closo-8,9,10,12-Me(4)-1,2-C(2)B(10)H(8) (4), is made in a single step by reaction of closo-1,2-C(2)B(10)H(12) with MeI in trifluoromethanesulfonic acid. The free hosts 5, 6, and 7 are obtained by reaction of the iodide complexes with stoichiometric quantities of AgOAc. A (199)Hg NMR study indicates that sequential removal of iodide from 5.I(2)Li(2) and 6.I(2)Li(2) with aliquots of AgOAc solution leads to formation of two intermediate host-guest complexes in solution, presumed to be 5(6)ILi and 5(2)(6)(2).ILi. Crystals grown from a solution of 6.I(2)Li(2) to which 1 equiv of AgOAc solution had been added proved to be an unusual stack structure with the formula 6(3).I(4)Li(4) [tetragonal, I4/m, a = 21.589(2) Å, c = 21.666(2) Å, V = 10098 Å(3), Z = 2, R = 0.058, R(w) = 0.084]. Addition of 2 equiv of NBu(4)Br ion to 5 or 6 gives 5.Br(2)(NBu(4))(2) and 6.Br(2)(NBu(4))(2), respectively, while addition of 1 equiv of KBr to 6 forms 6.BrK. 5.Br(2)(NBu(4))(2) crystallizes in the triclinic space group P&onemacr;, [a = 10.433(1) Å, b = 13.013(1) Å, c = 15.867(2) Å, alpha = 91.638(2) degrees, beta = 97.186(3) degrees, gamma = 114.202(2) degrees, V = 1492 Å(3), Z = 1, R = 0.078, R(w) = 0.104]. The hosts 5 and 6 form 1:1 supramolecular adducts with the polyhedral anions B(10)I(10)(2)(-) and B(12)I(12)(2)(-) in solution.
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