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One of my labs had the following reaction:

Reaction

I understand the Pictet Spengler mechanism. However, I am confused about the decarboxylation. I cannot come up with a way in which this decarboxylation would be acid catalyzed. Here is what I have for the Pictet mechanism:

Mechanism

I have been trying to figure out the acid catalyzed decarboxylation for several days now but I cannot. This is the closest I can come up with, but I'm not sure that this is correct:

Proposed Mechanism

Thanks!

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    $\begingroup$ If you protonate the beta position of the indole, the resultant immonium salt is isoelectronic with a a beta-keto acid, which readily decarboxylate. $\endgroup$
    – user55119
    Apr 22, 2021 at 0:39

1 Answer 1

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Your suggested mechanism should be acceptable (need two more steps, protonation and deprotonation, see last two steps of my scheme below), if you have isolated the carboxylic acid product (compound $\bf{1}$ in my scheme). If compound $\bf{1}$ is not isolable, I suggest following path would be taken to decarboxylation:

Protonation and decarboxylation

Your reaction is an analog to Pictet Spengler reaction (Wikipedia). When the intermediate $\bf{I}$ is formed during the reaction as depicted in the mechanism, 1,2-hydride transfer is possible to give more stable carbocation $\bf{II}$, which is resonance stabilized $(\bf{\ce{II <-> III}})$. The intermediate $\bf{III}$ can easily undergo decarboxylation to give some what stable enamine product, $\bf{2}$. In acidic medium, this enamine $\bf{2}$ would undergo protonation to give the intermediate $\bf{IV}$, which would then deprotonated to gain aromaticity and give thermodynamically stable indole product, $\bf{3}$ (the intended product).

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