An NMR spectroscopic study of the reactions of (2-(2-methoxyethoxy)ethyl)diphenylphosphine with rhodium(I) olefin complexes
作者:V.Vijay Sen Reddy、Ashima Varshney、Gary M. Gray
DOI:10.1016/0022-328x(90)80180-8
日期:1990.7
[(coe)2Rh(μ-Cl)]2 with Ph2PMe yields [(coe)(Ph2PMe)Rh(μ-Cl)]2 (VI) at a Rh/Ph2PMe ratio of 1/1, [(Ph2PMe)2Rh(μ-Cl)]2 (VII) at a Rh/Ph2PMe ratio of 1/2 and ClRh(Ph2PMe)3 (VIII) at a Rh/Ph2PMe ratio of 1/3. The reaction of [(coe)2Rh(μ-Cl)]2 with I gives a different product, [Ph2P(CH2CH2O)2CH3-P,ORh(μ-Cl)]2 (IX) at a Rh/I ratio of 1/1, but similar products at Rh/I ratios of 1/2, [Ph2P(CH2CH2O)2CH3-P2Rh(μ-Cl)]2 (XI)
Ph 2 P(CH 2 CH 2 O)2 CH 3(I)与[(cod)Rh(μ-Cl)] 2和[(coe)2 Rh(μ-Cl)] 2(cod = 1使用31 P和13 C NMR光谱研究了1,5-环辛二烯(coe =顺式环辛烯)和具有[[coe)2 Rh(μ-Cl)] 2的Ph 2 PMe 。这些研究表明两种Rh络合物给出了完全不同的反应产物。I与[(cod)Rh(μ-Cl)] 2在Rh / I比为1/1的反应中生成(cod)RhPh 2 P(CH 2 CH2 O)2 CH 3 - P氯(II)。在该溶液中将Rh / I比提高到1/2会导致形成ClRhPh 2 P(CH 2 CH 2 O)2 CH 3 - P 3(III),但不会形成任何两个与Rh配位的I的配合物。可以通过首先使II与AgBF 4反应形成[(cod)RhPh 2 P(CH 2 CH 2 O)2 CH 3 - P,O ] BF