What is the mechanism of N-alkylation by reacting an alkyl halide with aniline using tetrabutylammonium iodide (TBAI) as a catalyst and diisopropylethylamine (DIPEA) as a base?
To my opinion, here is the first stage is an exchange of halogen of alkyl halide with iodine-anion of TBAI, thus producing more reactive alkyl iodide. Second stage is a reaction of alkyl halide with aniline giving N-alkyl aniline and againe iodine-anion. DIPEA acts here as a non-nucleophilic base to remove the formed hydrogen halogenide.
Dear Min-Wook,
thank you for sharing this very interesting chemical question wth the RG community. I think that Alexey already provided a very good explnanation of the reaction mechanism. I would lust like to add that DIPEA (= N,N-diisopropylethylamine) is also often called "Hünig's base", named after the German chemist Siegfried Hünig. More more information about this please have a look at the very instructive Wikipedia entry on "N,N-Diisopropylethylamine". Mentioned here is among others the alkylation of amines by alkyl halides as outlined in your question.
Good luck with your work!
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