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The Solvation of Ca(2+) with Gas Phase Clusters of Alcohol Molecules
A comprehensive examination of how the identity of an alcohol molecule can change the behavior of a solvated, alkaline earth dication has been undertaken. The metal dication of Ca(2+) has been clustered with a range of different alcohols to form [Ca(ROH)(n)](2+) complexes, where n lies in the range...
Autores principales: | , |
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Formato: | Online Artículo Texto |
Lenguaje: | English |
Publicado: |
Springer US
2019
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Materias: | |
Acceso en línea: | https://www.ncbi.nlm.nih.gov/pmc/articles/PMC6695372/ https://www.ncbi.nlm.nih.gov/pubmed/31286446 http://dx.doi.org/10.1007/s13361-019-02263-x |
Sumario: | A comprehensive examination of how the identity of an alcohol molecule can change the behavior of a solvated, alkaline earth dication has been undertaken. The metal dication of Ca(2+) has been clustered with a range of different alcohols to form [Ca(ROH)(n)](2+) complexes, where n lies in the range 2–20. Following collisional activation via electron capture from nitrogen gas, complexes for n in the range 2–6 exhibit a switch in reaction product as a function of n. For low values, solvated CaOH(+) is the dominant fragment, but as n increases beyond 4, this is displaced by the appearance of solvated CaOR(+). A separate study of unimolecular metastable decay by [Ca(ROH)(n)](2+) complexes found evidence of charge separation to form CaOH(+)(ROH)(n−1) + R(+). For two isomers of butanol, the n = 3 complexes were found to follow parallel, but different metastable pathways: one leading to the appearance of CaOH(+) and another that resulted in proton abstraction to form ROH(2)(+). These differences have been attributed to the precursor complexes adopting geometries where one ROH molecule occupies a secondary solvation shell. Comparisons were made with a previous study of magnesium complexes; [Mg(ROH)(n)](2+) show that the difference in second ionization energy Mg(+) (15.09 eV) as opposed to Ca(+) (11.88 eV) influences behavior. A complex between Ca(2+) and 1-chloroethanol is shown to favor the formation of CaCl(+) as opposed to CaOH(+) as a unimolecular charge separation product, which is attributed to differences in bond energy in the precursor molecule. [Image: see text] ELECTRONIC SUPPLEMENTARY MATERIAL: The online version of this article (10.1007/s13361-019-02263-x) contains supplementary material, which is available to authorized users. |
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