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Conversion of 13( S )‐hydroperoxy‐9( Z ),11( E )‐octadecadienoic acid to the corresponding hydroxy fatty acid by KOH: A kinetic study
Author(s) -
Simpson Thomas D.,
Gardner Harold W.
Publication year - 1993
Publication title -
lipids
Language(s) - English
Resource type - Journals
SCImago Journal Rank - 0.601
H-Index - 120
eISSN - 1558-9307
pISSN - 0024-4201
DOI - 10.1007/bf02536318
Subject(s) - chemistry , yield (engineering) , oxygen atom , oxygen , fatty acid , stereochemistry , organic chemistry , thermodynamics , molecule , physics
Transformation of 13( S )‐hydroperoxy‐9( Z ),11( E )‐octadecadienoic acid (13 S ‐HPOD) to 13( S )‐hydroxy‐9( Z ),11( E )‐octadecadienoic acid (13 S ‐HOD) under alkaline conditions (0.05 to 5 M KOH) occurred first‐order with respect to 13 S ‐HPOD concentration. Overall yield was about 80%. The energy of activation at higher concentrations (3.75 to 5 M KOH) was determined to be in the range of 15.3 to 15.6 kcal. Compared to the 13 S ‐HPOD conversion, 13( S )‐hydroperoxy‐9( Z ),11( E ),15( Z )‐octadecatrienoic acid (13 S ‐HPOT) was converted at a faster rate to the corresponding hydroxy fatty acid (13 S ‐HOT), with the reaction also being first‐order. Chiral phase high‐performance liquid chromatography demonstrated that in the transformation the stereochemistry of both the 13 S ‐HPOD and 13 S ‐HPOT reactants was preserved. Manometric analyses of the KOH/13 S ‐HPOD reaction showed an uptake of gas, which amounted to 11% of the mols of reactant 13 S ‐HPOD on the assumption that the gas was O 2 . As there is a theoretical loss of 1 oxygen atom in the reaction, the fate of this oxygen (possibly via active oxygen species) may involve reaction with 13 S ‐HPOD/13 S HOD to form the 20% by‐products.