Conversion of carboxylic acids to esters using acid and alcohols (Fischer Esterification)

by James

Description: 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.

Notes: 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.

Examples:

Notes: The byproduct of each of these reactions is water. Note that the third 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–)

Video Walkthrough

 

 

{ 1 comment… read it below or add one }

Jason Mathias March 6, 2014 at 3:25 pm

I came in expecting an expulsion of the OH similar to the Grignard addition to an Ester – the Electrons rise up the Carbonyl to make C-O(-), then slam back down to make the carbonyl again, and then the O-R leaves because the C-O bond breaks.

Why isn’t this the case here? Is it due to the acidic conditions? Is there a basic version of this reaction that would, in fact, do this addition-elimination reaction?

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