Copper-Catalyzed Intermolecular Amidation and Imidation of Unactivated Alkanes

We
report a set of rare copper-catalyzed reactions of alkanes with
simple amides, sulfonamides, and imides (i.e., benzamides, tosylamides,
carbamates, and phthalimide) to form the corresponding N-alkyl products. The reactions lead to functionalization at secondary
C–H bonds over tertiary C–H bonds and even occur at
primary C–H bonds. [(phen)­Cu­(phth)] (1-phth) and
[(phen)­Cu­(phth)2] (1-phth2), which are potential intermediates in the reaction, have
been isolated and fully characterized. The stoichiometric reactions
of 1-phth and 1-phth2 with alkanes, alkyl radicals, and radical probes were investigated
to elucidate the mechanism of the amidation. The catalytic and stoichiometric
reactions require both copper and tBuOOtBu for the generation of N-alkyl product. Neither 1-phth nor 1-phth2 reacted
with excess cyclohexane at 100 °C without tBuOOtBu. However, the reactions of 1-phth and 1-phth2 with tBuOOtBu afforded N-cyclohexylphthalimide (Cy-phth), N-methylphthalimide, and tert-butoxycyclohexane
(Cy-OtBu) in approximate ratios of 70:20:30, respectively.
Reactions with radical traps support the intermediacy of a tert-butoxy radical, which forms an alkyl radical intermediate.
The intermediacy of an alkyl radical was evidenced by the catalytic
reaction of cyclohexane with benzamide in the presence of CBr4, which formed exclusively bromocyclohexane. Furthermore,
stoichiometric reactions of [(phen)­Cu­(phth)2] with tBuOOtBu and (Ph­(Me)2CO)2 at 100 °C without cyclohexane afforded N-methylphthalimide (Me-phth) from β-Me scission of the alkoxy
radicals to form a methyl radical. Separate reactions of cyclohexane
and d12-cyclohexane with benzamide showed
that the turnover-limiting step in the catalytic reaction is the C–H
cleavage of cyclohexane by a tert-butoxy radical.
These mechanistic data imply that the tert-butoxy
radical reacts with the C–H bonds of alkanes, and the subsequent
alkyl radical combines with 1-phth2 to form the corresponding N-alkyl imide product.

我们报道了一例罕见的铜催化烷烃与简单酰胺、磺酰胺及酰亚胺（即苯甲酰胺、对甲苯磺酰胺、氨基甲酸酯与邻苯二甲酰亚胺（phth））发生反应生成对应N-烷基化产物的体系。该反应优先在二级碳氢键（而非三级碳氢键）上实现官能团化，甚至可发生于一级碳氢键。本反应的潜在中间体[(1,10-菲啰啉（phen）)Cu(phth)]（记为1-phth）与[(phen)Cu(phth)₂]（记为1-phth₂）已被分离并完成全表征。我们通过1-phth、1-phth₂分别与烷烃、烷基自由基及自由基探针的计量反应，对该酰胺化反应的机理进行了阐释。催化与计量反应均需要铜与二叔丁基过氧化物（tBuOOtBu）协同作用以生成N-烷基化产物。在无tBuOOtBu的条件下，1-phth与1-phth₂均无法与过量环己烷在100 ℃下发生反应。然而，当1-phth或1-phth₂与tBuOOtBu反应时，可生成N-环己基邻苯二甲酰亚胺（Cy-phth）、N-甲基邻苯二甲酰亚胺（Me-phth）与叔丁氧基环己烷（Cy-OtBu），其摩尔比约为70:20:30。自由基捕获剂实验结果证实，反应过程中生成了叔丁氧基自由基，该自由基可引发烷基自由基中间体的形成。在体系中加入四溴化碳时，环己烷与苯甲酰胺的催化反应仅生成溴环己烷，这一结果进一步证明了烷基自由基中间体的存在。进一步的计量反应实验表明，在100 ℃且无环己烷的条件下，[(phen)Cu(phth)₂]与tBuOOtBu及(Ph(Me)₂CO)₂反应时，可生成N-甲基邻苯二甲酰亚胺（Me-phth），该产物源自烷氧基自由基的β-甲基断裂并生成甲基自由基。分别以环己烷与氘代十二环己烷（d12-cyclohexane）为底物与苯甲酰胺进行反应，结果表明催化反应的决速步为叔丁氧基自由基对环己烷的碳氢键断裂过程。上述机理实验数据表明，叔丁氧基自由基可与烷烃的碳氢键发生反应，生成的烷基自由基随后与1-phth₂结合，最终得到对应N-烷基酰亚胺产物。