Ester Formation

To transform a carboxylic acid into an ester, the OH group must be turned into a good leaving group and the acid treated with an alcohol, which is a weakly basic nucleophile.

Protonation of the carboxylic acid at the carbonyl oxygen and its two resonance forms, with the positive charge on the carbon

The reaction is carried out with a catalytic amount of a strong mineral acid (hydrochloric or sulfuric acid) capable of protonating the weakly basic C=O group.

Protonation of the C=O group increases the electrophilicity of its carbon and hence its reactivity, even towards weak nucleophiles such as the oxygen of an alcohol.

Addition of the alcohol (R′OH) to the protonated acid: a tetrahedral intermediate forms with an emergent water molecule as a good leaving group

In the presence of an excess of alcohol, which is usually the solvent as well, the nucleophilic attack takes place.

One of the two lone pairs of the alcohol oxygen attacks the carboxylic acid molecules that have been protonated by the catalytic amount of mineral acid. A tetrahedral intermediate is thus formed with an incipient water molecule as an excellent leaving group.

Elimination of water from the tetrahedral intermediate and loss of the proton, which gives the ester

Loss of water restores the protonated carbonyl group, which is now bonded to the alcohol oxygen.

Deprotonation leads to the ester and gives back the proton for more molecules of starting carboxylic acid to react.

Ester formation takes place in strongly acidic medium, in the presence of an excess of alcohol (the solvent) and in the absence of water.

Conversely, an ester can be turned (hydrolysed) into a carboxylic acid by treatment with the same acid in an excess of water.

The mechanism of both reactions is the same: ester formation is read from left to right and ester hydrolysis from right to left (principle of microscopic reversibility).