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Reaction of methoxide ion with the Z ‐ and E ‐isomers of O ‐methylbenzohydroximoyl cyanide. Theoretical calculations on the conformations of the tetrahedral intermediate
Author(s) -
Johnson James E.,
Jano Issam,
McAllister Michael A.
Publication year - 1999
Publication title -
journal of physical organic chemistry
Language(s) - English
Resource type - Journals
SCImago Journal Rank - 0.325
H-Index - 66
eISSN - 1099-1395
pISSN - 0894-3230
DOI - 10.1002/(sici)1099-1395(199903)12:3<240::aid-poc123>3.0.co;2-b
Subject(s) - chemistry , methoxide , cyanide , ion , crystallography , nucleophilic substitution , transition state , computational chemistry , substitution reaction , stereochemistry , medicinal chemistry , catalysis , inorganic chemistry , organic chemistry
Theoretical calculations (HF/6–31 + G*, MP2/6–31 + G*//HF/6–31 + G*, B3LYP/6–31 + G*//HF/6–31 + G*, HF‐SCIPCM/6–31 + G*//HF/6–31 + G*, and B3LYP‐SCIPCM/6–31 + G*//HF/6–31 + G*) were carried out on the tetrahedral intermediate RC(CN)(OCH 3 )—NOCH 3 − , where R = C 6 H 5 , CH 2 CH or H, generated by nucleophilic attack of methoxide ion on ( Z )‐ or ( E )‐ O ‐methylhydroximoyl cyanide [RC(CN)NOCH 3 ]. These calculations indicate that the staggered conformation 4C , which would lead to an E ‐substitution product, is approximately 7 (R = H) or 8 kcal mol −1 (R = CH 2 CH) less stable than conformation 4A , which leads to a Z ‐substitution product. Furthermore, the lowest energy transition state leading to a Z ‐substitution product is approximately 5 kcal mol −1 lower in energy than the lowest energy transition state leading to an E ‐substitution product. These results provide an explanation for the experimental results reported recently that ( Z )‐ O ‐methylbenzohydroximoyl cyanide gives only the less thermodynamically stable Z ‐substitution product under conditions where the purported tetrahedral intermediate should have a long enough lifetime to establish an equilibrium between all the staggered conformations. Copyright © 1999 John Wiley & Sons, Ltd.