Cationic and neutral complexes of ruthenium-(II) and -(III) containing tertiary phosphines or arsines and nitrogen-donor ligands

Abstract

In contrast to the reaction of [RuX₃(EPh₃)₂(HOMe)](X = Cl or Br; E = P or As) with N–N (N–N = 2,2′-bipyridyl or 1,10-phenanthroline) in CH₂Cl₂ which gives [RuX₃(EPh₃)(N–N)], the reaction in methanol gives [RuX(EPh₃)(N–N)₂]X (X = Cl or Br; E = P). For E = As these cations are only formed if [RuCl₃(AsPh₃)(N–N)] and N–N in methanol are reacted in the presence of Na[BPh₄]. However, reaction of [RuCl₃(PPh₃)₂(O₂NMe)], bipy, and Na[BPh₄] in methanol gives both [RuCl₂(PPh₃)₂(bipy)][BPh₄] and [RuCl(PPh₃)(bipy)₂][BPh₄] whereas [RuCl₃(PPh₃)(bipy)] and PhCN give [RuCI₂(PPh₃)(NCPh)(bipy)]Cl·H₂O. Reaction of mer-[RuCl₃(PMe₂Ph)₃] and excess of bipy in methanol followed by recrystallisation from acetone–light petroleum gives [RuCl(PMe₂Ph)₃(bipy)]Cl·2H₂O and [RuCl(PMe₂Ph)₂(bipy)(OCMe₂)]Cl, whereas in CH₂Cl₂[RuCl₂(PMe₂Ph)₂(bipy)] and a small amount of [Hbipy][RuCI₃(PMe₂Ph)(bipy)] is formed. With phen the product from CH₂Cl₂–hexane is [RuCl₂(PMe₂Ph)₂(phen)], but from methanol, followed by recrystallisation from CH₂Cl₂–light petroleum, the main product is [RuCl(PMe₂Ph)₂(phen)(CH₂Cl₂)]Cl together with small amounts of [Hphen][RuCl₃(PMe₂Ph)(phen)]·H₂O. With 3,4,7,8-tetramethyl-1,10-phenanthroline (Me₄phen), reaction with [RuCl₃(PMe₂Ph)₃] in methanol gives [RuCl(PMe₂Ph)₃(Me₄phen)]Cl which on recrystallisation from CH₂Cl₂–pentane gives [RuCl(PMe₂Ph)₂(Me₄phen)(CH₂Cl₂)]Cl. In contrast, reaction with 2,9-dimethyl-1,10-phenanthroline (Me₂phen) gives [Ru₂Cl₂(Me₂phen)₄]Cl₂ and [Ru₂Cl₃(PMe₂Ph)₆]Cl. All the complexes have been characterised by elemental analysis, ¹H and ³¹P n.m.r. spectra, and heteronuclear-decoupling studies and attempts have been made to rationalise the observed differences in product composition.