An Experimental−Theoretical Study of the Factors That Affect the Switch between Ruthenium-Catalyzed Dehydrogenative Amide Formation versus Amine Alkylation
作者:Ainara Nova、David Balcells、Nathan D. Schley、Graham E. Dobereiner、Robert H. Crabtree、Odile Eisenstein
DOI:10.1021/om101015u
日期:2010.12.13
either in the coordination sphere of the metal (path I) or after dissociation from the metal (path II). Path I yields the Ru-bound zwitterionic form of the hemiaminal protonated at nitrogen, which eliminates H2, forming the amide product. In path II, the free hemiaminal dehydrates, giving an imine, which yields the amine product by hydrogenation with the reduced form of the catalyst generated in the initial
钌(II)二胺配合物可通过两种途径催化氨基醇H 2 N(CH 2)n OH的分子内环化:(i)一种通过氧化还原中性氢借入途径产生环状仲胺而失水; (ii)第二种通过涉及H 2损失的净氧化反应生成相应的环状酰胺。在产物具有六元环的情况下,该反应是最有效的。酰胺和胺的途径密切相关:DFT计算表明,胺和酰胺的形成均始于氨基醇5-氨基-1-戊醇氧化为相应的氨基醛,同时还原了催化剂。氨基醛的分子内缩合发生在金属的配位域中(路径I)或与金属解离后(路径II)。路径I产生了在氮气中质子化的Ru结合的两性离子形式的两性离子形式,从而消除了H 2。,形成酰胺产物。在途径Ⅱ中,游离的血醛缩醛脱水,得到亚胺,该亚胺通过用在初始氨基醇氧化中生成的还原形式的催化剂进行氢化而得到胺产物。为了形成酰胺,半缩醛必须在关键中间体中保持与金属结合,并且必须从同一中间体中消除H 2才能提供一个用于β消除的空位。H 2的消除受关