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
{ 42 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?
This page will show you the reaction including the H2SO4 showing you why it is used as a catalyst there are probably many more catalysts that work but sulphuric acid is mass produced and very cheao
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
sir why does the protonation not occur at the hydroxyl group of alcohol?
It does, to some extent, but that reaction is reversible and it doesn’t help the reaction progress forward.
1. Why is the carboxylic acid and alcohol not reacted in equimolar amounts
2. Why is it necessary that the aqueous solution is basic?
1. The reaction is an equilibrium. Using an excess of alcohol drives it to the right (ester) side.
2. The aqueous solution is ACIDIC in order to 1) speed up addition of the nucleophile to the C=O group, and 2) convert OH into OH2+, making it a better leaving group.
Would this reaction work using a base as a catalyst?
No, it would not. Base just forms the conjugate base (carboxylate) and the reaction stops there.
How about in the case of mevalonate forming a cyclic ester to become mevalolactone? Mevalonate is in its carboxylate form…I am trying to understand the proton transfers that occur in this cyclization reaction/equilibrium under physiological conditions.
in certain reaction we use thionyl chloride instead of acid catalyst,why??
In those cases you are converting a carboxylic acid to an acid chloride. It’s not the Fischer esterification.
What is the purpose of water during this step. In the condensation flask?
Water is formed during this reaction as a byproduct.
Hi,
I was wondering if the reaction could still take place if the CO2H was attached to a benzene ring with a Nitro group in the para position? Would this react with a Phenol?
Thanks in advance
Yep, you’d make an ester.
Why is the ester hydrolysis not observed in this reaction? (acid is in excess)
Ester hydrolysis is not observed because the ALCOHOL is in excess of water (we use the alcohol as solvent). To get the reaction to run the other direction we use water as solvent.
thank you :)
How does the catalyst generally work in this reaction? What reacts with the sulfuric acid and does it (H2SO4) need to be concentrated or diluted and why?
Thank you :-)
Dilute H2SO4 or TsOH is fine. It’s catalytic.
Ok, what could reasons be for a hydrolysis reaction that is expected to be a first order reactions turns a zero order reaction ??
Why hydrogen is removed from alcohol and hydroxyl group from carboxyl group during esterification why not reverse of it? And how can we determine this fact?
Choice of solvent. To get the ester, use the alcohol as solvent. To do the reverse, use water as solvent.
hi, can i dilute the acetic acid directly in the ethanol? and what is the amount of ethanol and acetic acid (in L or molarity) be in order to make the esterification irreversible? thanks.
One drop of concentrated H2SO4 usually does the trick.
Hey James,
Suppose I wanted to create an ester linkage by grafting a aromatic carboxylic acid on to a long chain polyol with a medium density of hydroxyl groups. Would using an acid catalyst work and give me a high yield? or would I need to use a different type of esterfication rxn/catalyst? right now I am using nmp as a solvent
Any suggestions?
If there any possible carboxylic acid can react with secondary alcohol
Yes, isopropanol for example. The major restriction is the fact that the alcohol must be used as solvent.
Can we use hydrochloric acid for the reaction with a Dean-Stark trap to remove water formed. Would the water in hydrochloric also distil over.
HCl can be used. As for volatility, the HCl will react with solvent or water forming a conjugate acid (e.g. H3O+ Cl-) . This is a charged species and is not volatile, therefore it will not boil over.
HCl is slightly less preferred because sometimes at elevated temperature the Cl- counterion can act as a nucleophile, leading to undesired substitution byproducts. This is not a problem with the HSO4- ion.