Abstract
Using pulse radiolysis technique, the one-electron oxidation of vanillin (V-OH) with azide radicals, at pH 6 and 9 resulted in the formation of vanillin phenoxyl radical with k = 6.7 × 107 and 2.5 × 109 dm3 mol-1 s-1, respectively. The transient absorption spectra of the vanillin phenoxyl radical (V-O•) formed either at pH 6 or 9, showed a λmax at 410 nm. At pH 5, the •OH radicals seem to form an adduct with vanillin, λmax at 430 nm and k(•OH + V-OH) = 3.3 × 109 dm3 mol-1 s-1, while at pH 9, the •OH radical reaction resulted in the formation of vanillin phenoxyl radical with λmax at 410 nm and k(•OH + V-O-) = 6.6 × 109 dm3 mol-1 s-1. The reactivity of NO2•radicals with vanillin is lower by orders of magnitude signifying an incomplete reaction. In general, the rate constants for the reaction of •OH, N•, NO• radicals with vanillin were higher at pH 9 than at the lower pH. Its reactivity with other one-electron oxidants like CCl3OO•, CHCl2OO• and CH• radicals and the ability to chemically repair tryptophanyl and guanosyl radicals with k = 1.5 - 4 × 107 dm3 mol-1 s-1 indicate its antioxidative behaviour.
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