The reaction of Pt(COD)2 (bis(1,5-cyclooctadiene)platinum) and tBu3SnH (tri-t-butyl tin hydride) in a 1:1 ratio yielded Pt(H)(COD)(tBu3Sn), 1  . Compound 1  , reacts with one equivalent each of tBu3SnH and tBuNC (tert-butyl isocyanide) to give the di-hydro Pt-Sn compound Pt(H)2(tBu3Sn)2(tBuNC)2, 2  . Compound 2  contains an 18 electron platinum atom in a distorted octahedral geometry. Compound 2  loses its hydride ligands in solution and in the solid state to give the complex Pt(tBu3Sn)2(tBuNC)2, 3  . The Pt atom in 3  has a 16 electron configuration with a formal oxidation state of +2. Addition of an equimolar mixture of H2 and D2 to a solution of 2  results in rapid formation of HD gas. When CO gas is purged through a hexane solution of 2  , at room temperature it forms Pt(CO)(tBu3Sn)2(tBuNC)2, 4  . Compounds 3  and 4  in the presence of H2 gas regenerate 2  . Compound 1  , upon reaction with another equivalent of tBu3SnH and subsequent CO gas purge affords Pt(SnBut3)2(CO)2, 5  . Compound 5  reacts with two equivalents of tBuNC and loses a CO molecule to form compound 4  . Compound 1  reacts with one equivalent of NHC carbene, ItBu (N, N'-di-tert-butylimidazol-2-ylidene)to give the dimer [Pt(SntBu3)(ItBu)(H)]2, 6  . Compound 6  activates H2, CO and C2H4 at room temperature and forms [Pt(SntBu3)(ItBu)(H)3], 7  , [Pt(SntBu3)(ItBu)(CO)(H)], 8  and [Pt(SntBu3)(ItBu)(C2H4)(H)], 9  respectively.

Activation of small molecules by Pt-Sn heterobimetallic complexes

247th National Meeting (2014)

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