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One of the reactions that you can do with a carboxylic acid is to react it with ammonia or an amine, which produces the corresponding ammonium carboxylate salt. When this salt is heated, amide is obtained by elimination of water (although it seems that the yield is not that good). $$\ce{RNH2 + RCOOH->RCOO- ^+NH3R->(heat)->RCOONHR + H2O}$$ So what sort of mechanism can we draw for the elimination of water? It seems to me a bit complicated because the amine is already protonated in the salt, so it does not have any lone pair remaining for nucleophilic attack. Does the acid base reaction go backward under high temperatures? Or is it something else?

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  • $\begingroup$ Uhh..dehydration? $\endgroup$ Apr 1, 2019 at 8:17
  • $\begingroup$ @NilayGhosh, but can we draw an electron pushing mechanism? $\endgroup$
    – S R Maiti
    Apr 14, 2019 at 20:21

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The question is about mechanism of amidation not from R-NH2/R'-COOH but from their salt R-NH3+/R'-COO-. The former is well known and the only mechanism of amidation. The salt R-NH3+/R'-COO- is always in equilibrium with R-NH2 + R'-COOH. Equilibrium constant increases with increasing base strength of R-NH2 and increasing acid strength of R'-COOH, and vice versa. Salts with higher constants have higher ionic characters, and their pyrolysis temperatures become higher for amidation, signifying that completely ionic salts cannot be amidated. If they are heated to very high temperatures, they will be not amidated but just degraded.

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In this reaction the lone pair on the ammonia or amine reacts with the carbon attached to the oxygen atoms since it is partially electropositive due to the presence of extremely electronegative oxygen atoms. It does NOT react with the hydrogen on the alcohol group. This will give us an ammonium salt and the electrons involved in the pi-bond between the carbon and oxygen will be taken by the oxygen atom since carbon can't have 5 bonds . On heating the compound produced above an intra-molecular acid-base reaction takes place protonating the alcohol group and making it a good leaving group . The negative charge on the other oxygen atom forms a pi-bond again, forcing the protonated alcohol to leave as water ; leaving us with an Amide . The following image shows the reaction mechanism : enter image description here

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