86802-75-7Relevant articles and documents
METHYLDIPLATINUM(I) COMPLEXES
Azam, Kazi A.,Puddephatt, Richard J.,Brown, Michael P.,Yavari, Ahmad
, p. C31 - C34 (1982)
Methyldiplatinum(I) complexes of general formula +- (dppm = Ph2PCH2PPh2), have been prepared by oxidative addition of methyl iodide to or by reaction of (SbF6> with the ligand
A hydrido(μ-hydrido)methyldiplatinum(II) complex. Reductive elimination of methane induced by reaction with an alkyne
Azam, Kazi A.,Puddephatt, Richard J.
, p. 1396 - 1399 (2008/10/08)
Reaction of [Pt2Cl2(μ-CH2)(μ-dppm)2] with H[SbF6] gives [Pt2Cl(μ-Cl)Me(μ-dppm)2][SbF6] (II) and reduction of II with NaBH4 gives [Pt2H(μ-H)Me(μ-dppm)2][SbF6] (III), characterized by 1H and 31P{1H} NMR spectroscopy. Complex III undergoes A-frame inversion on the NMR time scale, and it is proposed that a transition state with a linear MePtHPtH unit is involved; the activation energy at -60°C is ΔG? = 41 kJ mol-1. Complex III reacts with CF3C≡CCF3 in MeCN solution to give 90% [Pt2H(MeCN)(μ-CF3C=CCF3)(μ-dppm) 2][SbF6] and methane and 10% [Pt2Me(MeCN)(μ-CF3C=CCF3)(μ-dppm) 2][SbF6] (VIII) and hydrogen. Pure VIII was prepared by the similar reaction of CF3C≡CCF3 with [Pt2(μ-H)-Me2(μ-dppm)2][SbF6], with elimination of methane. The alkyne has ability to induce reductive elimination of methane from III or VIII, whereas phosphines induce loss of H2 from III and react only slowly with VIII.