Copper-catalyzed intermolecular amidation and imidation of unactivated alkanes.
- Author(s): Tran, Ba
- Li, Bijie
- Driess, Matthias
- Hartwig, John
- et al.
Published Web Locationhttps://doi.org/10.1021/ja411912p
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.