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Phenolic Hydrogen Transfer by Molecular Oxygen and Hydroperoxyl Radicals. Insights into the Mechanism of the Anthraquinone Process

[Image: see text] Hydrogen atom transfer (HAT) by (3)O(2) and HO(2)(•) from arenols (ArOH), aryloxyls (ArO(•)), their tautomers (ArH), and auxiliary compounds has been investigated by means of CBS-QB3 computations. With (3)O(2), excellent linear correlations have been found between the activation en...

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Detalles Bibliográficos
Autores principales: Korth, Hans-Gert, Mulder, Peter
Formato: Online Artículo Texto
Lenguaje:English
Publicado: American Chemical Society 2020
Acceso en línea:https://www.ncbi.nlm.nih.gov/pmc/articles/PMC7040920/
https://www.ncbi.nlm.nih.gov/pubmed/31922747
http://dx.doi.org/10.1021/acs.joc.9b03286
Descripción
Sumario:[Image: see text] Hydrogen atom transfer (HAT) by (3)O(2) and HO(2)(•) from arenols (ArOH), aryloxyls (ArO(•)), their tautomers (ArH), and auxiliary compounds has been investigated by means of CBS-QB3 computations. With (3)O(2), excellent linear correlations have been found between the activation enthalpy and the overall reaction enthalpy. Different pathways have been discerned for HATs involving OH or CH moieties. The results for ArOH + HO(2)(•) → ArO(•) + H(2)O(2) neither afford a linear correlation nor agree with the experiment. The precise mechanism for the liquid-phase autoxidation of anthrahydroquinone (AnH(2)Q) appears to be not fully understood. A kinetic analysis shows that the HAT by chain-carrying HO(2)(•) occurs with a high rate constant of ≥6 × 10(8) M(–1) s(–1) (toluene). The second propagation step pertains to a diffusion-controlled HAT by (3)O(2) from the 10-OH-9-anthroxyl radical. Oxanthrone (AnOH) is a more stable tautomer of AnH(2)Q with a ratio of 13 (298 K) in non-hydrogen-bonding (HB) solvents, but the reactivity toward (3)O(2)/HO(2)(•) is much lower. Combination of the computed free energies and Abrahams’ HB donating (α(2)(H)) and accepting (β(2)(H)) parameters has afforded an α(2)(H)(HO(2)(•)) of 0.86 and an α(2)(H)(H(2)O(2)) of 0.50.