Abstract

The 1,3-dipolar cycloaddition of methyl azide to C 50 and the subsequent nitrogen elimination from the formed triazoline intermediate to yield the aziridine adduct have been studied using semiempirical AM1 methods. The results show that the cycloaddition of methyl azide to a [5,5] bond of C 50 leading to a closed [5,5]-triazoline intermediate takes place through two stepwise pathways a and b: both of them starting from the formation of one C–N bond followed by the formation of the other C–N bond. The two pathways are competitive and the activation energy of each path is about 22 kcal mol −1. The subsequent thermal loss of N 2 from the closed [5,5]-triazoline takes place through two paths, I and II: path I includes three steps while path II includes two steps. And path I is the preferred one with a total activation energy of approximately 41 kcal mol −1. There is also a pathway without the formation of triazoline, in which the formation of C–N bond between N 3CH 3 and C 50 to give an intermediate is followed by the elimination of N 2 together with the formation of the other C–N bond to form the closed [5,6]-C 50NCH 3, which is the preferred pathway among all the pathways with a total activation energy of about 14.5 kcal mol −1. It is expected that the closed [5,6]-C 50NCH 3 is the main product from addition of N 3CH 3 to C 50. The closed [5,6]-C 50NCH 3 isomer is more stable than its closed [5,5] isomer. The isomerization activation energy from the closed [5,6] isomer to the [5,5] isomer is 53.6 kcal mol −1, and that for the reverse isomerization is 35.7 kcal mol −1. This result indicates that the conversion of the two isomers can hardly take place at room temperature and some day both of them can have the chance to be actually isolated experimentally.

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