The following shows the conversion of an acid to a weinreb amide:

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It was taken from the IChO 2015 preparatory problem set problem 23. In this two-step synthesis, I am slighly puzzled by the first step. The rationale behind first forming the ester is that we would like to synthesise an acid derivative which can react with the amine to form the amide since direct conversion of an acid to an amide is rather difficult. However, I wonder why is the t-butyl ester formed instead of say, a methyl ester, since the former sterically-hinders the attack of the amine in the subsequent step. In fact, conversion to an acid chloride first (e.g. using thionyl chloride), instead of an ester would seem to be more optimal than conversion to an ester first since acid chlorides would be more reactive with the amine.

In summary, I would like to know why ester formation is preferred over acid chloride formation in the first step and also why a rather bulky ester is preferred over a much less bulky one. Regarding the former, I believe it could be that the formation of $\ce {HCl}$ by-product may hinder the 2nd reaction by protonating the amines. Regarding the latter, I believe there may be some electronic factors that need to be considered?


Your question is easily answered by the fact that the conditions of Me3COCl + starting acid give a mixed anhydride not an ester. The tButyl group is chosen as the hindrance it brings makes reaction at the desired C=O carbon more likely. Further reading about mixed anhydride coupling is here

  • $\begingroup$ Oh oops... Yes it is an anhydride... Very careless of me. $\endgroup$ Apr 13 '19 at 10:34
  • $\begingroup$ Is my reasoning regarding why an acid chloride is not generated instead in the first step correct? Reaction of the acid chloride with the amine in the 2nd step would generate HCl which would protonate the amine $\endgroup$ Apr 13 '19 at 10:37
  • $\begingroup$ I think the question has omitted the tertiary base used in both steps. $\endgroup$
    – Waylander
    Apr 13 '19 at 10:43

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