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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?

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journal contribution
posted on 30.10.1996, 00:00 by Michael L. McKee, Athanassios Nicolaides, Leo Radom
Ab initio molecular orbital calculations have been carried out on a series of adducts between chlorine atom and NH3, NMe3, 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 π* 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).