1313012-43-9Relevant articles and documents
Preparation of half-sandwich osmium complexes by deprotonation of aromatic and pro-aromatic acids with a hexahydride bronsted base
Esteruelas, Miguel A.,Garci-Raboso, Jorge,Olivan, Montserrat
, p. 3844 - 3852 (2011/08/09)
Figure Persented: Half-sandwich osmium(II) and osmium(IV) complexes have been prepared by reaction of the hexahydride complex OsH6(P iPr3)2 (1) with phenol, pyrrole, and methylcyclopentadienes. The reaction with phenol initially leads to OsH 3(OPh)(PiPr3)2 (2). In toluene, 2 undergoes reductive elimination of phenol, which tautomerizes to give OsH 2(η4-2,4-cyclohexadien-1-one)(PiPr 3)2 (3). The equilibrium mixture of 2 and 3 evolves into OsH(η5-PhO)(PiPr3)2 (4) with loss of molecular hydrogen. The addition of HBF4 to diethyl ether solutions of 4 leads to [OsH(η6-PhOH)(PiPr 3)2]BF4 (6). The reaction of 1 with pyrrole gives OsH(η5-C4H4N)(PiPr 3)2 (7), which by addition of HBF4 affords [OsH2(η5-C4H4N)(P iPr3)2]BF4 (8). Similarly, treatment of 1 with methylcyclopentadiene leads to OsH(η5-C 5H4Me)(PiPr3)2 (9), which reacts with HBF4 to give [OsH2(η5-C 5H4Me)(PiPr3)2]BF 4 (10). Treatment of toluene solutions of 1 with tetramethylcyclopentadiene gives a mixture of the trihydride OsH 3(η5-C5HMe4)(P iPr3) (11; 56%) and the dihydride-tolyl derivatives OsH2(m-tolyl)( η5-C5HMe4)(P iPr3) (12; 14%) and OsH2(p-tolyl)( η5-C5HMe4)(PiPr3) (13; 30%). However in n-octane the trihydride 11 is formed in 85% yield. In contrast to tetramethylcyclopentadiene, pentamethylcyclopentadiene reacts with 1 in toluene to give selectively the trihydride OsH3(η 5-C5Me5)(PiPr3) (14). Complexes 2 and 7 have been characterized by X-ray diffraction analysis.
