Gordon Jonathan, Hildebrandt Sven, Dewese Kendra R, Klare Sven, Gansäuer Andreas, RajanBabu T V, Nugent William A
Department of Chemistry and Biochemistry, The Ohio State University, 100 West 18th Avenue, Columbus, OH 43210, USA.
Kekulé-Institut für Organische Chemie und Biochemie, Universität Bonn, Gerhard-Domagk-Straße 1, 53121 Bonn, Germany.
Organometallics. 2018 Dec 24;37(24):4801-4809. doi: 10.1021/acs.organomet.8b00793. Epub 2018 Nov 30.
The role of CpTi(H)Cl in the reactions of CpTiCl with trisubstituted epoxides has been investigated in a combined experimental and computational study. Although CpTi(H)Cl has generally been regarded as a robust species, its decomposition to CpTiCl and molecular hydrogen was found to be exothermic (ΔG = -11 kcal/mol when the effects of THF solvation are considered). In laboratory studies, CpTi(H)Cl was generated using the reaction of 1,2-epoxy-1-methylcyclohexane with CpTiCl as a model. Rapid evolution of hydrogen gas was demonstrated, indicating that CpTi(H)Cl is indeed a thermally unstable molecule, which undergoes intermolecular reductive elimination of hydrogen under the reaction conditions. The stoichiometry of the reaction (CpTiCl:epoxide = 1:1) and the quantity of hydrogen produced (1 mole per 2 moles of epoxide) is consistent with this assertion. The diminished yield of allylic alcohol from these reactions under the conditions of protic versus aprotic catalysis can be understood in terms of the predominant titanium(III) present in solution. Under the conditions of protic catalysis, CpTiCl complexes with collidine hydrochloride and the titanium(III) center is less available for "cross-disproportionation" with carbon-centered radicals; this leads to by-products from radical capture by hydrogen atom transfer, resulting in a saturated alcohol.
通过实验与计算相结合的研究,对CpTi(H)Cl在CpTiCl与三取代环氧化物反应中的作用进行了考察。尽管CpTi(H)Cl通常被认为是一种稳定的物种,但发现它分解为CpTiCl和分子氢是放热反应(考虑四氢呋喃溶剂化作用时,ΔG = -11千卡/摩尔)。在实验室研究中,以1,2-环氧-1-甲基环己烷与CpTiCl的反应为模型生成了CpTi(H)Cl。实验证明有氢气快速放出,这表明CpTi(H)Cl确实是一个热不稳定分子,在反应条件下会发生分子间氢的还原消除反应。反应的化学计量比(CpTiCl:环氧化物 = 1:1)以及产生的氢气量(每2摩尔环氧化物产生1摩尔氢气)与这一论断相符。在质子催化与非质子催化条件下,这些反应中烯丙醇产率的降低可以从溶液中主要存在的三价钛来理解。在质子催化条件下,CpTiCl与盐酸可力丁形成配合物,三价钛中心较难与碳中心自由基发生“交叉歧化反应”;这会导致通过氢原子转移进行自由基捕获而产生副产物,生成饱和醇。