Polarographic and optical pulse radiolysis study of the radicals formed by hydroxyl radical attack on imidazole and related compounds in aqueous solutions

Abstract

The disproportionation-combination ratio for several allylic type radicals has been measured in the condensed phase at 90 K. Pairs of allylic radicals react exclusively by combination, with no disproportionation found within the limits of the analytical technique. Cross combination and disproportionation occurs between an allylic radical and an alkyl radical. In the cross disproportionation, the allylic radical acts exclusively as a hydrogen donor. This effect is attributed to the delocalization of the free spin, the hydrogen acceptor function being completely inhibited. Because of this, two allylic type radicals (or any radicals in which the density of the free spin is sufficiently delocalized) can only combine. The contribution of the resonant components of the allylic radicals to the combination products is found to be greatly dependent on steric effects.

OH-Initiated Oxidation of Imidazoles in Tropospheric Aqueous-Phase Chemistry

“…Bansal and Sellers investigated the ·OH reaction of imidazole, 1-methylimidazole, and 2-methylimidazole via pulse radiolysis. These authors obtained a first-order decay of the OH adduct radicals from imidazole and 2-methylimidazole at alkaline conditions (pH ≥ 11) and described this decay ( k first, max = 10 4 s –1 ) as a water elimination reaction (Scheme , blue pathway).…”

“…These authors obtained a first-order decay of the OH adduct radicals from imidazole and 2-methylimidazole at alkaline conditions (pH ≥ 11) and described this decay ( k first, max = 10 4 s –1 ) as a water elimination reaction (Scheme , blue pathway). It is supposed that the formed ·OH adducts might undergo a base-catalyzed water elimination, which includes the OH group and the hydrogen atom at the N1 position. − Only 1-methylimidazole did not show a first-order decay, which leads to the assumption that the water elimination cannot occur for substituted 1-imidazoles . This can be explained by the replacement of the H atom at the N1 position with the substituent.…”

“…Product studies are necessary to identify the oxidation products as well as the reaction mechanism. Several studies predict that the OH radical add to a double bond of the heteroaromatic compound leading to an alkyl radical, − which can further undergo different reactions such as a base-catalyzed water elimination, , formation of peroxyl radicals via the reaction with oxygen, recombination reactions, or ring opening reactions. Nevertheless, the understanding of the degradation pathway of imidazoles is still lacking.…”

Histidinyl Radical Formation in the Self-peroxidation Reaction of Bovine Copper-Zinc Superoxide Dismutase

“…It has been predicted that addition of HO ⅐ to C-5 of the imidazole ring would result in a radical with significant electron density at C-2 (21,22), which could subsequently react with the spin trap to provide the detected adduct. Alternately, the product formed by addition of HO ⅐ to C-5 of the imidazole ring could dehydrate to produce a free radical with significant electron density at C-2 of the imidazole ring (25). It is unlikely, however, that a free radical resulting from the addition of hydroxyl radical at C-2 of the imidazole ring was trapped.…”

Synthesis of unsaturated phosphatidylinositol 4-phosphates and the effects of substrate unsaturation on SopB phosphatase activity

“…Furthermore, assuming that the proton trap/catalyst participated in the racemisation, we selected species with a pK a as low as possible. 2,6-Dichlorobenzoyl chloride (DcbCl) was added portion-wise to a mixture of inositol diol rac-5 and acetylmandelic acid (8a) with N-methylimidazole (NMI, pK a 7.00-7.06 34 ) acting as both base and catalyst, in dichloromethane. 35 The proportion of racemisation fell substantially under these conditions.…”