Question

Most reactions that can be done under acidic conditions can also be done under basic conditions – for example, conversion of a ketone into a ketal. That said, the reason Fischer esterifications require acid catalysis is that protonating the acid makes the electrophile (the carbonyl carbon) a better electrophile. We could presumably make the nucleophile (an alcohol) a better nucleophile, which is usually done through deprotonation.

Explain why reacting an alkoxide with a carboxylic acid will not result in a Fischer esterification. What happens instead and why?

Answer 1

When a carboxylic acid is treated with an alcohol and an acid catalyst, an ester is formed (along with water). This reaction is called the Fischer esterification.

The reaction is actually an equilibrium. The alcohol is generally used as solvent so is present in large excess. Many different acids can be used; it’s common to see just “H+”, although H2SO4 (sulfuric acid) and TsOH (tosic acid) are also often used.

The byproduct of each of these reactions is water. Note that the fourth example is an intramolecular reaction that forms a cyclic ester. Cyclic esters are also called lactones.

Mechanism:

For such a seemingly simple reaction (replacement of OH by OR) there are actually a lot of steps. Protonation of the carbonyl oxygen by acid (Step 1, arrows A and B) makes the carbonyl carbon a much better electrophile. It undergoes 1,2-addition by the alcohol (Step 2, arrows C and D) whereupon the proton from the alcohol is transferred to one of the OH groups (Step 3, arrows E and F). Subsequent 1,2-elimination of water (Step 4, arrows G and H) leads to the protonated ester, and the ester is then deprotonated (Step 5, arrows I and J).

Notes:

• All of these steps are in equilibrium
• The mechanism for proton transfer can alternatively be drawn as follows

Note that the acid is a catalyst here (regenerated at the end) and serves two purposes. First, it makes the carbonyl carbon a better electrophile (Setting up step 2) and also allows for the loss of H2O as a leaving group (much better leaving group than HO–)