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Release of Formic Acid from Copper Formate: Hydride Proton‐Coupled Electron and Hydrogen Atom Transfer All Play their Role

机译:铜甲酸酯中甲酸的释放:氢化物质子耦合电子和氢原子转移均发挥作用

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摘要

Although the mechanism for the transformation of carbon dioxide to formate with copper hydride is well understood, it is not clear how formic acid is ultimately released. Herein, we show how formic acid is formed in the decomposition of the copper formate clusters Cu(II)(HCOO)3 and Cu(II)2(HCOO)5 . Infrared irradiation resonant with the antisymmetric C−O stretching mode activates the cluster, resulting in the release of formic acid and carbon dioxide. For the binary cluster, electronic structure calculations indicate that CO2 is eliminated first, through hydride transfer from formate to copper. Formic acid is released via proton‐coupled electron transfer (PCET) to a second formate ligand, evidenced by close to zero partial charge and spin density at the hydrogen atom in the transition state. Concomitantly, the two copper centers are reduced from Cu(II) to Cu(I). Depending on the detailed situation, either PCET or hydrogen atom transfer (HAT) takes place.
机译:尽管人们很清楚二氧化碳与氢化铜转化为甲酸的机理,但尚不清楚最终如何释放甲酸。在这里,我们展示了甲酸在甲酸铜簇Cu(II)(HCOO)3 -和Cu(II)2(HCOO)5 -。以反对称CO拉伸模式共振的红外辐射激活了团簇,导致释放了甲酸和二氧化碳。对于二元簇,电子结构计算表明,通过从甲酸盐到铜的氢化物转移,首先消除了CO2。甲酸通过质子偶联电子转移(PCET)释放到第二个甲酸根配体上,这在过渡态时氢原子的部分电荷和自旋密度接近于零即可证明。相应地,两个铜中心从Cu(II)还原为Cu(I)。根据具体情况,发生PCET或氢原子转移(HAT)。

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