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In this mechanism, I got a pretty different mechanism from the solution. From the model answer, there is an intramolecular proton transfer just before the formation of the carboxylate ion. However, in the second intermediate, I think the electrostatic repulsion between nitrogen and oxygen is large, increase separation between nitrogen and hydrogen, hence reducing chance of intramolecular transfer.

My purposed mechanism rely on the H2O given in reaction mixture to undergo proton transfer, forming OH- (matching basic condition) and then pull the proton from R-COOH to form the carboxylate salt (product).

Is my mechanism plausible? What mistakes I have made if the model answer is the only correct mechanism?

Also, please suggest ways that I can improve this type of questions, thank you very much.

234ff
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    Both answers are correct. The intramolecular reaction is fast, but you're in water as a solvent. In reality, you probably have some mix of both mechanisms taking place. If I were grading, I would accept either answer. – Zhe Jul 07 '20 at 15:29
  • @Zhe Thanks a lot. One question - you said the intramolecular reaction is 'fast', can this be explained by "https://chemistry.stackexchange.com/questions/16592/why-is-proton-transfer-so-fast" such that even there is a huge separation (O <-> N electronic repulsion), it occurs very fast? – 234ff Jul 07 '20 at 15:37
  • I don't understand your statement about the huge separation. There may be some charges present, but it's largely going to be attenuated by the solvent. And the intramolecular proton transfer is on the time scale of the conformational change. – Zhe Jul 07 '20 at 17:59
  • I quite don't get your issue with mechanisms. Things like what donates proton are largely irrelevant for mechanistic analysis of reaction - labile protons are transferred all the time. – Mithoron Jul 07 '20 at 21:02
  • @Mithoron OK so as a beginner, when there are any protons on any of a molecule, proton transfer can happen provided that it suits the reaction condition? (e.g. cannot generate H3O+ in this basic condition) – 234ff Jul 08 '20 at 01:12
  • You comment isn't particularly comprehensible... If you don't get what labile means, then it's "easily exchangeable" - quite not "any". – Mithoron Jul 08 '20 at 17:07

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