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在酸性水溶液中用氧钒(V)离子氧化羟基脲。

Oxidation of hydroxyurea with oxovanadium(V) ions in acidic aqueous solution.

作者信息

Gabricević Mario, Besić Erim, Birus Mladen, Zahl Achim, Eldik Rudi van

机构信息

Faculty of Pharmacy and Biochemistry, University of Zagreb, A. Kovacića 1, 10000 Zagreb, Croatia.

出版信息

J Inorg Biochem. 2006 Oct;100(10):1606-13. doi: 10.1016/j.jinorgbio.2006.05.008. Epub 2006 May 27.

Abstract

Hydroxyurea (HU) effectively reduces vanadium(V) into vanadium(IV) species (hereafter V(V) and V(IV) species, respectively) in acidic aqueous solution via the formation of a transient complex followed by an electron transfer process that includes the formation and subsequent fading out of a free radical, U* (U* identical with H(2)N-C(=O)N(H)O*). The electron paramagnetic resonance (EPR) spectra of U* in H(2)O/D(2)O solutions suggest that the unpaired electron is located predominantly on the hydroxamate hydroxyl-oxygen atom. Visible and V(IV)-EPR spectroscopic data reveal HU as a two-electron donor, whereas formation of U*, which reduces a second V(V), indicates that electron transfer occurs in two successive one-electron steps. At the molarity ratio [V(V)]/[HU]=2, the studied reaction can be formulated as: 2 V(V)+HU-->2 V(IV)+0.98 CO(2)+0.44 N(2)O+1.1 NH(3)+0.1 NH(2)OH. Lack of evidence for the formation of NO is suggested to be a consequence of the slow oxidation of HNO due to the too low reduction potential of the V(V)/V(IV) couple under the experimental conditions used. The nuclear magnetic resonance ((51)V-NMR) spectral data indicate protonation of (H(2)O)(4)V(V)O(2)(+), and the protonation equilibrium constant was determined to be K=0.7 M(-1). Spectrophotometric titration data for the V(V)-HU system reveal formation of (H(2)O)(2)V(V)O(OH)U(+) and (H(2)O)(3)V(V)OU(2+). Their stability constants were calculated as K(110)=5 M(-1) and K(111)=22 M(-2), where the subscript digits refer to (H(2)O)(4)V(V)O(2)(+), HU and H(+), respectively.

摘要

在酸性水溶液中,羟基脲(HU)通过形成瞬态络合物,随后经过一个电子转移过程,有效地将五价钒(V)还原为四价钒物种(以下分别称为V(V)和V(IV)物种),该电子转移过程包括一个自由基U*(U与H₂N - C(=O)N(H)O相同)的形成及随后的消失。H₂O/D₂O溶液中U的电子顺磁共振(EPR)光谱表明,未成对电子主要位于异羟肟酸酯的羟基氧原子上。可见光谱和V(IV)-EPR光谱数据表明HU是双电子供体,而还原第二个V(V)的U的形成表明电子转移以两个连续的单电子步骤发生。当摩尔比[V(V)]/[HU]=2时,所研究的反应可表示为:2V(V) + HU→2V(IV) + 0.98CO₂ + 0.44N₂O + 1.1NH₃ + 0.1NH₂OH。在所用实验条件下,由于V(V)/V(IV)电对的还原电位过低,导致HNO缓慢氧化,因此没有证据表明形成了NO。核磁共振(⁵¹V-NMR)光谱数据表明(H₂O)₄V(V)O₂⁺发生了质子化,质子化平衡常数测定为K = 0.7 M⁻¹。V(V)-HU体系的分光光度滴定数据表明形成了(H₂O)₂V(V)O(OH)U⁺和(H₂O)₃V(V)OU²⁺。它们的稳定常数计算为K₁₁₀ = 5 M⁻¹和K₁₁₁ = 22 M⁻²,下标数字分别指(H₂O)₄V(V)O₂⁺、HU和H⁺。

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