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I am a first-year chemistry graduate student and I am currently learning how to synthesize proteins. I have reached a stage in the process where I need to deprotect a carboxylic acid moiety protected by t-butyl using trifluoroacetic acid (TFA). I went online trying to find the mechanism for this step, but in terms of step-wise mechanism with electron movement arrows, the only thing I was able to find was this: Deprotection of t-butyl carboxylic acid using TFA

(Taken from here, no references therein)

While this looks legit, I am wondering about the t-butyl cation and its fate. My reaction is a bit simpler, and is presented below: Deprotection of t-butyl with TFA

And my question is: is this truly where the deprotection ends? Does the t-butyl cation remains in solution as pictured?

(I know that a t-butyl cation is considered to be relatively stable on account of it being a tertiary carbocation, but still I was not sure that it can remain in solution like that.)

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    $\begingroup$ Probably gives isobutene, depending on what solvent you're using $\endgroup$
    – Waylander
    Nov 19, 2017 at 16:31
  • $\begingroup$ The solvent is dichloromethane. What is its role? $\endgroup$
    – Don_S
    Nov 19, 2017 at 18:22
  • $\begingroup$ In which case DCM plays no part as it will not react with tBu+, other solvents may react with it e.g. toluene $\endgroup$
    – Waylander
    Nov 19, 2017 at 18:42
  • $\begingroup$ According to a post-doc colleague, step no. 4 IS the final state of the molecules in the solution (i.e. t-butyl can exist in the solution as a cation), but if water will be added to the solution, then the t-butyl cation will take an OH group and the TFA will become protonated. In case anyone was wondering... $\endgroup$
    – Don_S
    Nov 21, 2017 at 9:49
  • $\begingroup$ It is worth remembering that when you cleave a tBu ester, tBu+ can hang around and potentially react with e.g. electron-rich aromatics $\endgroup$
    – Waylander
    Nov 21, 2017 at 12:22

2 Answers 2

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I realize this question is old but the mechanism is nicely described here (steps 1-3): http://www.umich.edu/~chemh215/W06HTML/SSG2/ssg6/main_files/Page346.htm

Briefly, the released t-butyl carbocation is subsequently deprotonated by the anionic TFA, resulting in the formation of 2-methyl-propene. This regenerates the protonated TFA that is needed to continue the reaction, and thus the TFA is catalytic (i.e. not consumed in the reaction).

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The reaction is, in principle, catalytic in TFA. If t-butyltrifluoroacetate is the by-product, then the catalyst is consumed. If only isobutylene is formed, catalytic is OK. Using excess TFA covers all contingencies. In the initial step, protonation is more likely on the C=O oxygen. Higher electron density.

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