...
首页> 外文期刊>Organometallics >Density Functional Study of the Mecahnism of C ident to C, O-H, and N-H Bond Activation at the Pd=X (X = Sn, Si, C) Bonds of the (PH_3)_2Pd=XH_2 Complexes
【24h】

Density Functional Study of the Mecahnism of C ident to C, O-H, and N-H Bond Activation at the Pd=X (X = Sn, Si, C) Bonds of the (PH_3)_2Pd=XH_2 Complexes

机译:在(PH_3)_2Pd = XH_2配合物的Pd = X(X = Sn,Si,C)键处C识别C,O-H和N-H键活化机理的密度泛函研究

获取原文
获取原文并翻译 | 示例
           

摘要

The mechanism of the activation of the C=C .7r bond of ethyne, the 0:.- H bond of water, and the N-H bond of ammonia at the Pd=X (X = Sn, Si, C) bonds of the model (PH3)2Pd= XH2 complexes 1 to produce the products (PH3)2Pd(CH=CH)XH2 '(Pd-X) 4, (PH3)2Pd(H)X- (OH)H2 7, (PH3)2Pd(H)X(NH2)H2 9, respectively, has been theoretically investigated using a density functional method (B3I.. YP). The reaction mechanism significantly depends on both substrate and atom X of the' complex I. In the activation of the C=C .7r bond of ethyne at the Pd=Sn bond, a weak rJ2-interaction orthe C=C .7r bond not with Pd but with Sn initiates the reaction. The rotation of the (PH3)2Pd group around the Pd-Sn axis makes the reaction facile so that a small energy barrier of only 6.2kcal/mol is required. Although the reaction at the Pd=Sibond also takes the same mechanism) the reaction is downhill without energy barrier. On the other hand, when X is C, the reaction proceeds by a completely different mechanism without the coordination of the C=C .7r bond to the C atom. In the activation of the 0- H bond of water at the Pd=Sn bond, after the coordination of the H2O oxygen to the Snatom, the O-H bond is heterolytically broken and the hydrogen is transferred to the Pd atom as a proton with an energy barrier of 16.9 kcal/mol. The same heterolytic cleavage takes place at the Pd=Si bond, requiring a smaner energy barrier, while at the Pd=C bond the homolytic cleavage takes place without the coordination of the H2O oxygen to the C atom. In the case of NH3, independent orthe atom X, heterolytic cleavage of the N- H bond occurs after coordination of the NH3 nitrogen to the X atom and the transfer ora hydrogen as a proton to the Pdatom. The large energy barrier of 33.2 kcal/mol required in the reaction system with X =Sn for which no activation activity has been experimentally observed was reduced to half in the reaction system with X= C, suggesting an activation activity. The factors that determine the activation mechanism are discussed.
机译:在模型的Pd = X(X = Sn,Si,C)键处,乙炔的C = C .7r键,水的0:.- H键和氨的NH键的激活机理(PH3)2Pd = XH2络合物1以产生产物(PH3)2Pd(CH = CH)XH2'(Pd-X)4,(PH3)2Pd(H)X-(OH)H2 7,(PH3)2Pd(理论上,已经使用密度泛函方法(B3I..YP)研究了H)X(NH2)H2 9。反应机理很大程度上取决于底物和配合物I的原子X。在乙炔的C = C .7r键在Pd = Sn键处活化时,弱的rJ2-相互作用或C = C.7r键不起作用与钯,但与锡引发反应。 (PH3)2Pd基团绕Pd-Sn轴的旋转使反应容易进行,因此只需要6.2kcal / mol的小能垒。尽管在Pd = Sibond下的反应也采用相同的机理),但该反应是下坡的,没有能量垒。另一方面,当X为C时,反应通过完全不同的机理进行,而没有与C原子的C = C.7r键配位。在Pd = Sn键上水的0- H键的活化中,在H2O氧与Snatom配位后,OH键被杂化断裂,氢作为质子转移到Pd原子上,具有能量16.9 kcal / mol的势垒。在Pd = Si键上发生相同的杂化裂解,需要较大的能垒,而在Pd = C键上,发生均质裂解,而H2O氧不与C原子配位。在独立于原子X的NH3的情况下,在NH3氮与X原子配位以及氢作为质子转移到Pdatom之后,NH键的杂合裂解发生。在X = Sn的反应系统中所需的33.2kcal / mol的大能量势垒在X = C的反应系统中减少了一半,这在实验上未观察到活化能,这表明没有活化活性。讨论了决定激活机制的因素。

著录项

相似文献

  • 外文文献
  • 中文文献
  • 专利
获取原文

客服邮箱:kefu@zhangqiaokeyan.com

京公网安备:11010802029741号 ICP备案号:京ICP备15016152号-6 六维联合信息科技 (北京) 有限公司©版权所有
  • 客服微信

  • 服务号