首页> 外文期刊>Journal of the American Chemical Society >A THEORETICAL STUDY OF CHLORINE ATOM AND METHYL RADICAL ADDITION TO NITROGEN BASES - WHY DO CL ATOMS FORM TWO-CENTER-THREE-ELECTRON BONDS WHEREAS CH3 RADICALS FORM TWO-CENTER-TWO-ELECTRON BONDS
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A THEORETICAL STUDY OF CHLORINE ATOM AND METHYL RADICAL ADDITION TO NITROGEN BASES - WHY DO CL ATOMS FORM TWO-CENTER-THREE-ELECTRON BONDS WHEREAS CH3 RADICALS FORM TWO-CENTER-TWO-ELECTRON BONDS

机译:氯原子和甲基自由基与氮基的理论研究-为什么CL原子形成两中心三电子键,而CH3自由基形成两中心两电子键

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摘要

Ab initio molecular orbital calculations have been carried out on a series of adducts between chlorine atom and NH3, NMe(3), NCl3, HN=CH2, and pyridine, and between methyl radical and HN=CH2 and pyridine. A two-center-three-electron (2c-3e) bond is predicted for all the chlorine adducts, whereas the CH3 adducts with the unsaturated systems form two-center-two-electron (2c-2e) bonds following promotion of one of the nitrogen lone pair electrons into a pi* orbital. For chlorine adducts, the greater strength of the 2c-2e N-Cl bond compared with the 2c-3e N-Cl bond is not sufficient to compensate for the required promotion energy in both the saturated and unsaturated amines. On the other hand, for CH3 adducts of the unsaturated nitrogen bases, HN=CH2 and pyridine, the C-N and C-C bond energies are sufficiently high and the promotion energy is sufficiently low that adducts to both N and C with 2c-2e bonds can be formed. Adducts between CH3 and saturated nitrogen centers are less stable than the separated species because of the inability of CH3 to form effective 2c-3e bonds in neutral systems (due to its low electron affinity), and because of the high excitation energy required to promote an electron from the nitrogen lone pair (due to the absence of suitable low-lying empty orbitals in these systems).
机译:从头算分子轨道计算已对氯原子与NH3,NMe(3),NCl3,HN = CH2和吡啶以及甲基与HN = CH2和吡啶之间的一系列加合物进行了计算。预测所有氯加合物的二中心三电子(2c-3e)键,而具有不饱和体系的CH3加合物在促进其中一个加成后形成二中心二电子(2c-2e)键。氮孤对电子成pi *轨道。对于氯加合物,与2c-3e N-Cl键相比,2c-2e N-Cl键的强度更高,不足以补偿饱和胺和不饱和胺中所需的促进能。另一方面,对于不饱和氮碱的CH3加合物,HN = CH2和吡啶,CN和CC键的能量足够高,促进能足够低,以至于N和C都可以通过2c-2e键加成。形成。由于CH3无法在中性系统中形成有效的2c-3e键(由于其较低的电子亲和力),并且由于需要较高的激发能来促进氢原子的形成,因此CH3与饱和氮中心之间的加合物不如分离的物种稳定。来自氮孤对的电子(由于这些系统中没有合适的低空空轨道)。

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