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.


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.


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).


  • 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



{ 7 comments… read them 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?


Cornelius October 4, 2014 at 7:43 pm

It’s because -OR is a bad leaving group. If to say that it was -OCl or -OTs, then your thought process would most likely be valid. As is, it would be hard to generate an alkoxide (-OR) in acidic conditions too. I am not sure this reaction would work in base (well not with these starting materials) as it wont go very far from a simple acid-base reaction at the first step.


Mastaneh Nikravesh September 4, 2014 at 3:13 am

I actually have a question, what would happen if excess watr were added to the ester product and the solution was refluxed under acidic conditions?
I believe that the rxn would occur in the reverse because the rxn is an equilibrium rxn and follows le chatalier’s principle.
Thank you in advance!


James October 7, 2014 at 3:13 pm

Yes, that’s correct. That’s acidic hydrolysis of an ester.


Collin November 3, 2014 at 11:15 pm

Would we use methanol for a methyl ester or diazomethane?


James November 6, 2014 at 11:26 am

Hi Collin –

Diazomethane is great on small scale, and if you want to make the methyl ester. Probably one of my favorite reactions to run because it’s so darn easy.

However the higher analogs of diazomethane (e.g. diazoethane, diazopropane) are not stable enough for routine use. So it’s limited to methyl esters for the most part.

The Fischer esterification is one of those robust, old school, tried and true reactions that can be used to make a huge variety of esters.

Hope that answers your question! James


Hamada Abulkhair November 14, 2014 at 11:37 am

why protonation in the first step occures at the carbonyl oxygen not at the hydroxyl oxygen?


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