Abstract

Influence of fluorine, chlorine, and bromine atoms on conformational preferences of free non-rigid molecules is analyzed on the basis of quantum chemical calculations. Thirty-six compounds are considered: 35 with the general formula X(CH2)nY (where X = Y = CH3, F, Cl, Br or X = F, Cl, Br, Y = CH3 and even n from 8 to 16) and n-octane. According to the previous computational and experimental results obtained by other researches, free n-alkane molecules with the number of carbon atoms (nC) greater than approximately 16–18 prefer to form hairpin conformers (where the chain is folded in two) instead of fully extended (all-trans) conformers that are more favorable for the compounds with lower nC. For halogenated n-alkanes, such investigation is made for the first time. Calculations at the ωB97X-D/6–31++G(d,p) level of theory were performed for all compounds and at the MP2/6–31++G(d,p) level for n-alkanes, 1,n-difluoro-, 1,n-dichloroalkanes with n from 8 to 16, 1,8-dibromooctane and 1,9-fluoro-, 1,9-chloro-, 1,9-bromononane. Conclusions about influence of X, Y, or n on conformational preferences drawn from the calculations on the different theoretical levels are in qualitative agreement. The most prominent distinctions are revealed for tetradecane and 1,14-dichlorotetradecane. It is found that differences in the electronic (ΔEel) and electronic plus zero-point (ΔE0) energies between the hairpin and all-trans conformers have similar tendencies by variations of X, Y, or n. When the ΔEel and ΔE0 values for the particular compound are close to zero and have opposite signs, the distance between X and Y equal to ~ 4.0 A or less is an indicator for the preference of the hairpin conformer. The most strong effect on the advantageousness of the hairpin conformers is caused by two bromine atoms placing at the ends of molecules: for all considered 1,n-dibromoalkanes, the hairpin conformers are more favorable than the all-trans ones.

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