162053-75-0Relevant academic research and scientific papers
Complexes of the Bidentate Ligands Z-PPh2CH2(t-Bu)=NNR2 (R = Me or H) with Rhodium and Iridium
Perera, Sarath D.,Shaw, Bernard L.
, p. 633 - 640 (2007/10/02)
Treatment of 0.5 equivalent of 2> with the phosphino dimethylhydrazone Z-PPh2CH2C(t-Bu)=NNMe2(L1) or the phosphino hydrazone Z-PPh2CH2C(t-Bu)=NNH2 (L2) gave the chelate complexes (R = Me 1a or H 1b).Complex 1a reacts with another mole of L1 to give the bis(phosphine)rhodium(I) complex trans-2> 2a.The analogous iridium(I) complex 2b was prepared by treating with 2 mol equivalents of L1.Complex 2a reacts with (cod = cycloocta-1,5-diene) to give a 1:1 mixture of the chelate 1a and .Treatment of with 2 equivalents of L2 gave the hydridoiridium(III) complex 3a.Treatment of 2> (M = Rh or Ir) with two equivalents of L1 in methanol in the presence of NH4PF4 gave the cationic complexes (M = Rh 4a or Ir 4b) in which L1 is bidentate.When 2> was treated with 2 equivalents of L1 in benzene it yielded the neutral complexes > (M = Rh 5a or Ir 5b) in which L1 is monodentate through phosphorus.Treatment of 2> with 2 equivalents of L2 in CD2Cl2 gave the cationic chelate complexes (M = Rh 4c or Ir 4d).Treatment of 2> or 2> (C8H14 = cyclooctene) with 2 equivalents of L2 per rhodium atom gave the cationic bis(phosphine)rhodium(I) complex 6.Treatment of with 1 equivalent of L1 in C6D6 gave the Wilkinson-type complex 7, which readily reacts with dioxygen to give the adduct 8.The rhodium(III) complex 8 reacts with sulfur dioxide to give the rhodium(III) sulfate 9.Treatment of 2> with 2 equivalents of L1 per rhodium atom gave the bis(phosphine) complexes (M = Rh 10a or Ir 10b), containing one chelate and one monodentate phosphine ligand.These complexes reacted rapidly with dioxygen to give corresponding dioxygen adductc (M = Rh 11a or Ir 11b).Proton 31P- and some 13C- NMR data are given.
