由扭曲的酸 (cobalt phthalocyanine) 催化的二氧化物的质子合电子转移减少
Tatsuhiko Honda1, Takahiko Kojima, Shunichi Fukuzumi
1Department of Material and Life Science, Graduate School of Engineering, Osaka University, ALCA, Japan Science and Technology Agency (JST), Suita, Osaka 565-0871, Japan.
Journal of the American Chemical Society
|February 4, 2012
概括
这项研究研究了使用铁衍生物和甲酸催化剂的二氧化物 (O2) 到过氧化 (H2O2) 的催化降解. 这项研究揭示了催化剂活性和反应机制是如何根据还原剂强度和质子化变化的.
科学领域:
- 电化学和催化剂的应用
- 无机化学 无机化学 有机化学
- 氧化降解反应 氧化降解反应
背景情况:
- 对二氧化物 (O2) 到过氧化 (H2O2) 的选择性两电子还原对于各种化学过程至关重要.
- 类酸是已知的催化剂,但它们在酸性条件下减少O2的活性和机制需要详细的研究.
- 铁衍生物具有可调节的还原电位,因此非常适合研究电子转移机制.
研究的目的:
- 通过使用铁衍生物和扭曲的甲酸催化剂 (Co(II) (((Ph(8) Pc)) 来研究与质子合的电子转移减少O2到H2O2 .
- 阐明反应机制,包括催化剂质子化的作用和活性物种的身份.
- 为了确定不同铁素还原剂的催化循环中的速度决定性步骤.
主要方法:
- 使用铁衍生物,特别是二甲基铁 (Me(2) Fc) 和十甲基铁 (Me(10) Fc),作为减少剂.
- 在酸性条件下使用扭曲 (α-octaphenylphthalocyaninato) (Co) (II) (Ph) (Pc)) 作为催化剂.
- 在催化循环中对整体催化反应和单个氧化还原步骤进行了详细的动力学研究.
- 为了进行比较分析,获得并确定了一种单质子复合物的晶体结构,[Co ((III) Cl (((2) ((Ph ((8) PcH)).
主要成果:
- 通过使用Me(2) Fc和Me(10) Fc作为减电剂,实现O2的选择性两电子降解为H2O2,产生相应的铁离子.
- 证明了与O2反应的活性物种根据减少剂的强度在Co(II)(Ph(8)Pc) 和质子化Co(I)(Ph(8)PcH) 之间发生转变.
- 确定了确定速率的步骤,即O2的质子合电子转移减少Co(II) ((Ph(8) Pc) (与Me(2) Fc) 或[Co(II) ((Ph(8) PcH) ] ((+) 的电子转移减少Me(10) Fc (与Me(10) Fc).
结论:
- 在酸性条件下,O2到H2O2的催化还原是有效地通过Co (II) (Ph) (8) (Pc) 与铁衍生物进行的.
- 催化剂质子化显著影响反应路径和活性催化物种的身份.
- 了解电子转移,质子转移和催化剂结构之间的相互作用是优化O2降解催化剂的关键.
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