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使用含NAD⁺模型配体的Ru(II)配合物产生氢化物供体的机制:脉冲和稳态辐解研究

Mechanism of hydride donor generation using a Ru(II) complex containing an NAD+ model ligand: pulse and steady-state radiolysis studies.

作者信息

Polyansky Dmitry E, Cabelli Diane, Muckerman James T, Fukushima Takashi, Tanaka Koji, Fujita Etsuko

机构信息

Chemistry Department, Brookhaven National Laboratory, Upton, New York 11973-5000, USA.

出版信息

Inorg Chem. 2008 May 19;47(10):3958-68. doi: 10.1021/ic702139n. Epub 2008 Mar 18.

Abstract

The mechanistic pathways of formation of the NADH-like [Ru(bpy) 2(pbnHH)] (2+) species from Ru(bpy)2(pbn) were studied in an aqueous medium. Formation of the one-electron-reduced species as a result of reduction by a solvated electron (k=3.0 x 10(10) M(-1) s(-1)) or CO2(*-) (k=4.6 x 10(9) M(-1) s(-1)) or reductive quenching of an MLCT excited state by 1,4-diazabicyclo[2.2.2]octane (k=1.1 x 10(9) M(-1) s(-1)) is followed by protonation of the reduced species (p K a = 11). Dimerization (k7a=2.2 x 10(8) M(-1) s(-1)) of the singly reduced protonated species, Ru(bpy) 2(pbnH(*)), followed by disproportionation of the dimer as well as the cross reaction between the singly reduced protonated and nonprotonated species (k8= 1.2 x 10(8) M(-1) s(-1)) results in the formation of the final Ru(bpy)2(pbnHH) product together with an equal amount of the starting complex, Ru(bpy)2(pbn). At 0.2 degrees C, a dimeric intermediate, most likely a pi-stacking dimer, was observed that decomposes thermally to form an equimolar mixture of Ru(bpy)2(pbnHH) and Ru(bpy)2(pbn) (pH<9). The absence of a significant kinetic isotope effect in the disproportionation reaction of Ru(bpy)2(pbnH(*)) and its conjugate base (pH>9) indicates that disproportionation occurs by a stepwise pathway of electron transfer followed by proton transfer.

摘要

在水介质中研究了由[Ru(bpy)₂(pbn)]²⁺形成类NADH的[Ru(bpy)₂(pbnHH)]²⁺物种的机理途径。通过溶剂化电子还原(k = 3.0×10¹⁰ M⁻¹ s⁻¹)或CO₂⁻(k = 4.6×10⁹ M⁻¹ s⁻¹),或通过1,4 - 二氮杂双环[2.2.2]辛烷对MLCT激发态进行还原猝灭(k = 1.1×10⁹ M⁻¹ s⁻¹)形成单电子还原物种,随后还原物种发生质子化(pKa = 11)。单还原质子化物种[Ru(bpy)₂(pbnH*)]²⁺的二聚化(k7a = 2.2×10⁸ M⁻¹ s⁻¹),随后二聚体的歧化以及单还原质子化和非质子化物种之间的交叉反应(k8 = 1.2×10⁸ M⁻¹ s⁻¹)导致最终的[Ru(bpy)₂(pbnHH)]²⁺产物与等量的起始配合物[Ru(bpy)₂(pbn)]²⁺一起形成。在0.2℃时,观察到一种二聚中间体,最可能是π - 堆积二聚体,其热分解形成[Ru(bpy)₂(pbnHH)]²⁺和[Ru(bpy)₂(pbn)]²⁺的等摩尔混合物(pH < 9)。[Ru(bpy)₂(pbnH*)]²⁺及其共轭碱的歧化反应中不存在显著的动力学同位素效应(pH > 9),这表明歧化是通过电子转移后质子转移的分步途径发生的。

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