Conversion of carboxylic acids to esters using acid and alcohols (Fischer Esterification)
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 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–)
Video Walkthrough
{ 13 comments… read them below or add one }
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?
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.
Hello,
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!
Yes, that’s correct. That’s acidic hydrolysis of an ester.
Would we use methanol for a methyl ester or diazomethane?
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
why protonation in the first step occures at the carbonyl oxygen not at the hydroxyl oxygen?
How many mole of acid catalyst that we need to use in this esterification, and is the amount of acid used will effect the reaction?
For practical purposes about 5-10 mol % of acid is common.
Why is H2SO4 the preferred acid to use as a catalyst over other acids?
Minor typo: it’s the fourth example reaction that’s an intramolecular reaction, not the third. Great job, as always!
Can fischer esterification happen in aqueous solution? In other words, can the protonation in step 1 and deprotonation in step 5 happen via hydrogen ions and hydroxide ions of water, respectively?
Thanks
Hey Ted – the Fischer esterification is in equilibrium with the reverse reaction (acidic hydrolysis of esters to give carboxylic acids).
In order to get the reaction to proceed from acid —> ester, you need to add a large excess of alcohol so that the concentration of alcohol will be much greater than water . This is what drives the reaction toward the ester [think Le Chatelier]
So if you use water as the solvent, nothing will happen to the carboxylic acid. It will just sit there. You need to use an alcohol solvent.
For example if you want to make a methyl ester, add methanol as solvent.
Hope this helps – James