Über unsymmetrische B-Fluorborazine

Zusammenfassung Unsymmetrische B-Butyl-B-fluorborazine wurden durch Umhalogenierung entsprechender Chlorborazine mit NaF dargestellt, während B-Methyl-B-fluorborazine so nicht rein erhalten wurden. Pentamethylmonofluorborazin wurde durch Methylierung von N-Trimethyl-B-trifluorborazin mit 2 CH₃Li erhalten.

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On unsymmetrical B-fluoro-borazines

Unsymmetrical B-butyl-B-fluoroborazines were obtained from the corresponding B-butyl-B-chloroborazines with NaF; however, B-methyl-B-fluoroborazines could not be obtained in this way in a pure state. Pentamethylmonofluoroborazine was prepared from N-trimethyl-B-trifluoroborazine and 2 moles methyllithium.

     

Der Dihydridoiridium(III)‐Komplex [IrH2Cl(PiPr3)2] als Molekülbaustein für unsymmetrische Rhodium–Iridiumund Iridium–Iridium‐Zweikernverbindungen

The Dihydridoiridium(III) Complex [IrH₂Cl(P i Pr₃)₂] as a Molecular Building Block for Unsymmetrical Binuclear Rhodium–Iridium and Iridium–Iridium Compounds The title compound [IrH₂Cl(PiPr₃)₂] ( 3 ) reacts with the chloro‐bridged dimers [RhCl(PiPr₃)₂]₂ ( 1 ) and [IrCl(C₈H₁₄)(PiPr₃)]₂ ( 5 ) by cleavage of the Cl‐bridges to give the unsymmetrical binuclear complexes 4 and 6 with Rh(μ‐Cl)₂Ir and Ir(μ‐Cl)₂Ir as the central building block. The reactions of 3 with the bis(cyclooctene) and (1,5‐cyclooctadiene) compounds [MCl(C₈H₁₄)₂]₂ ( 7 , 8 ) and [MCl(η⁴‐C₈H₁₂)]₂ ( 9 , 10 ) (M = Rh, Ir) occur analogously and afford the rhodium(I)‐iridium(III) and iridium(I)‐iridium(III) complexes 11 – 14 in 70–80% yield. Treatment of [(η⁴‐C₈H₁₂)M(μ‐Cl)₂IrH₂(PiPr₃)₂] ( 13 , 14 ) with phenylacetylene leads to the formation of the substitution products [(η⁴‐C₈H₁₂)M(μ‐Cl)₂IrH(C≡CPh)(PiPr₃)₂] ( 15 , 16 ) without changing the central molecular core. Similarly, the compound [(η⁴‐C₈H₁₂)Rh(μ‐Br)₂IrH(C≡CPh)(PiPr₃)₂] ( 18 ) has been prepared; it was characterized by X‐ray crystallography.