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关于一氧化二氮产生羟基自由基的热力学考量——绝非儿戏。

Thermodynamic considerations on the generation of hydroxyl radicals from nitrous oxide--no laughing matter.

作者信息

Koppenol W H

机构信息

Department of Chemistry, Louisiana State University, Baton Rouge 70803-1800.

出版信息

Free Radic Biol Med. 1991;10(1):85-7. doi: 10.1016/0891-5849(91)90025-x.

DOI:10.1016/0891-5849(91)90025-x
PMID:1646753
Abstract

The one-electron reduction of nitrous oxide is a possible pathway to the hydroxyl radical. The one- and two-electron reduction potentials EO' (N2O/OH,N2) and EO' (N2O/H2O, N2) are calculated to be 0.32 V and 1.32 V at pH 7, respectively, for all species dissolved in water. Although nitrous oxide is thermodynamically capable of oxidising a variety of biomolecules, it is kinetically rather inert. The reason that nitrous oxide does not produce hydroxyl radicals readily might be that the one-electron reduction proceeds through an N2O- intermediate which is energetically very unfavourable: EO (N2O/N2O-) = -1.1 V.

摘要

一氧化二氮的单电子还原是生成羟基自由基的一条可能途径。对于溶解于水中的所有物种,在pH值为7时,单电子和双电子还原电位EO'(N2O/OH,N2)和EO'(N2O/H2O,N2)经计算分别为0.32 V和1.32 V。尽管一氧化二氮在热力学上能够氧化多种生物分子,但在动力学上它相当惰性。一氧化二氮不易产生羟基自由基的原因可能是单电子还原通过一个能量上非常不利的N2O-中间体进行:EO(N2O/N2O-) = -1.1 V。

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