Carboxylic Acid Derivatives

By James Ashenhurst


Last updated: December 29th, 2022 |

Transesterification of Esters

  • Conversion of one ester into another ester via exchange of -OR groups is called transesterification
  • Transesterification can be performed under basic or acidic conditions
  • Under basic conditions, transesterification occurs via a two-step addition-elimination mechanism
  • Intramolecular transesterification results in the formation of cyclic esters (lactones)
  • Under acidic conditions, transesterification follows the PADPED pattern (protonation-addition-deprotonation-protonation-elimination-deprotonation)
  • Watch out for transesterification as a side-reaction in the course of other reactions involving alkoxide bases (e.g. the Claisen condensation)

summary of transesterification under basic and acidic conditions - alkoxides

Table of Contents

    1. Transesterification
    2. Transesterification Under Basic Conditions
    3. Intramolecular Transesterification
    4. Transesterification Under Acidic Conditions
    5. Mechanism of Transesterification Under Acidic Conditions
    6. Summary
    7. Notes
    8. Quiz Yourself!
    9. (Advanced) References and Further Reading

1. Transesterification

What is transesterification? It’s the conversion of one ester into another through exchange of -OR groups.

-general scheme for a tranesterification reaction could occur under acidic or basic conditions

There are two sets of conditions for transesterification. It can be done under basic conditions or under acidic conditions.

2. Transesterification Under Basic Conditions

Transesterification under basic conditions can be achieved through adding an alkoxide to an ester.

In this example, adding sodium ethoxide (NaOCH2CH3) to this methyl ester results in formation of the ethyl ester, giving off sodium methoxide as the leaving group.

Examples of transesterification under basic conditions using alkoxides on esters

The usual choice of solvent for transesterification is the conjugate acid of the alkoxide (i.e. ethanol in this case). Note 1

The mechanism for transesterification under basic conditions is fairly straightforward if you’ve covered nucleophilic acyl substitution; it proceeds through a two step  addition-elimination mechanism. (See post: Nucleophilic Acyl Substitution)

In the first step, a nucleophilic alkoxide ion performs nucleophilic addition upon the ester.  This results in a tetrahedral intermediate, which then undergoes elimination of the RO(-) group of the starting ester to give the final ester.

Mechanism of transesterification under basic conditions to give a new ester - ethyl ester from methyl ester

For many, this explanation will be sufficient. However, if you’re wondering how to get this reaction to favor one direction versus the other, see Note 1

3. Intramolecular Transesterification

One interesting application of base-catalyzed transesterification is in the formation of lactones from ester-alcohols.

This usually requires a base, as the neutral alcohol is generally not nucleophilic enough by itself to promote this reaction.

scheme for formation of lactones via transesterification under basic conditions

For these purposes, it’s often useful to employ a base that won’t react with the carbonyl group of our final product (and in so doing cause another transesterification).

An excellent choice of base here is hydride ion (H-), in the form of sodium hydride (NaH) or potassium hydride (KH).

Hydride ion deprotonates OH to give the alkoxide and hydrogen gas (H2), which bubbles away and can no longer participate in any reactions. The alkoxide then acts as a nucleophile to form the lactone through the two-step addition-elimination mechanism.

mechanism for intramolecular transesterification - lactone formation - under basic conditions using sodium hydride

Generally speaking the formation of five- and six-membered rings is favored relative to opening of the ring, for reasons related to entropy. [Note 2]

4. Transesterification Under Acidic Conditions

It’s also possible to perform transesterification under acidic conditions.

example of transesterification of esters under acidic conditions to interconvert esters

Here are two examples.

example of transesterification of esters under acidic conditions including lactone formation

Note that this reaction can also be used in an intramolecular fashion to give cyclic esters (lactones).

In order to get the reaction to proceed in a desired direction, it helps to use the alcohol nucleophile as the solvent.

  • For example, to make the ethyl ester from a methyl ester, use ethanol as solvent.
  • To achieve the reverse transformation, use methanol.

5. Acidic Transesterification – Mechanism

So how does this reaction work?

Good news – it’s essentially the same mechanism as the Fischer esterification reaction, except our leaving group is ROH instead of H2O.

As you may recall, the mechanism for the Fischer esterification is PADPED. So is the mechanism for a great deal of other reactions (See post: Making Music With Mechanisms)

If you’re unfamiliar, PADPED stands for

  • Protonation (of the ester carbonyl oxygen)
  • Addition (of alcohol to the carbonyl)
  • Deprotonation (of the alcohol)
  • Protonation (of the alcohol that is going to leave)
  • Elimination (of the alcohol leaving group)
  • Deprotonation (of the carbonyl).

Here is the mechanism:

mechanism of acid-catalyzed esterification reaction - padped mechanism - all steps in equilibrium

The reaction begins with protonation of the carbonyl by the acid catalyst  [Note 3] (Step 1, form O-H) which makes the carbonyl carbon into a better electrophile (See post:  Nucleophilic Addition to Carbonyls).

In the next step, the alcohol attacks the carbonyl to give a tetrahedral intermediate (Step 2, form C-O, break C-O (pi)) which is then deprotonated (Step 3, break O-H) and then protonated on the other oxygen (Step 4, form O-H).

Since the conjugate acid is a better leaving group, this group then undergoes elimination  (Step 5, form C-O (pi) break C-O) to give the protonated (new) ester, which then undergoes a final deprotonation (Step 6, break O-H) to give the new ester.

All of these steps are in equilibrium.

Since most aliphatic alcohols are of comparable acidity, there is no strong driving force for this reaction unless one uses a large excess of alcohol (i.e. if we use it as the solvent). [Note 4].

6. Summary

One of the important reasons to learn about transesterification is just to be aware that it can happen in the context of other reactions.

For example, in the Claisen condensation (See post: Claisen and Dieckmann Condensations) if we choose our base unwisely, we may end up peforming nucleophilic acyl substitution in addition to doing our Claisen!

To summarize:

  • Transesterification is the conversion of one ester to another via exchange of the alkoxy (OR) groups.
  • It can be performed under both basic and acidic conditions.
  • The mechanism under basic conditions is a two-step addition-elimination sequence.
  • Under acidic conditions, the mechanism is PADPED (Protonation-Addition-Deprotonation-Protonation-Elimination-Deprotonation).
  • Be on the lookout for intramolecular transesterifications (lactone formation)
  • Also, be alert to the possibility for transesterifications under basic conditions in the context of other reactions (e.g. enolate formation with RO(-) ).


Note 1. One detail of this reaction may be unsatisfying to some, so let’s use this footnote to go into more detail.

Nucleophilic acyl substitution reactions with basic nucleophiles tend to be favored in the direction whereby the leaving group is a weaker base than the nucleophile. (See post: How to Use a pKa Table)

However, if the two alkoxides (the nucleophile and leaving group) are of comparable basicity, then there won’t be a significant driving force for the reaction based on basicity alone.

For example, the conjugate bases of ethanol (pKa 15.9)  and methanol (pKa 15.5) have extremely similar basicities, meaning they have similar leaving group abilities. (See post: What Makes a Good Leaving Group?)

get transesterification to occur under basic conditions by using a large excess of nucleophile

The question becomes:  how do you get the reaction to favor one direction or the other?

There are two ways to ensure that the mechanism favors the desired product.

One way is to use the conjugate acid of the alkoxide as the solvent.

Let’s say we want to substitute CH3CH2O– for CH3O–. We should use CH3CH2OH as our solvent.

This means there will be a large excess of CH3CH2OH relative to the alkoxide leaving group CH3O(-).

Acid-base reactions are fast. (See post: Acid-Base Reactions Are Fast). So as soon as that CH3O(-) is formed, it undergoes an acid-base reaction with the solvent CH3CH2OH . Even though they are of comparable acidity, equilibrium will greatly favor formation of CH3CH2O(-) since the concentration of CH3CH2OH is several orders of magnitude higher than the concentration of CH3OH.

using alcohol as solvent in transesterification reaction under basic conditions drives equilibrium toward desired product

The other way is to use a large excess (e.g. 5-10 equivalents at least) of the alkoxide nucleophile.

This will push the equilibrium towards the desired product; in this case it’s advisable to avoid using an alcoholic solvent, since it could result in the solvent alkoxide acting as the nucleophile and resulting in mixtures of esters.

Note 2. There is a slight entropic preference for ring formation.

Note 3. When HCl is added to an alcohol, it will immediately protonate the alcohol to give the conjugate acid of the alcohol  (e.g. CH3OH2(+) Cl(-) ) but HCl is drawn here for simplicity.

Note 4. Aromatic alcohols (e.g. phenol) are usually not very good candidates as nucleophiles for transesterification reactions. Besides the fact that phenol is a solid (and can’t be used as solvent) strong acid will protonate the aromatic ring.

Quiz Yourself!

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(Advanced) References and Further Reading

The interchange of alkoxy groups in esters can be carried out with acidic or basic catalysis, and follows a similar mechanism as hydrolysis. This is an equilibrium process which is driven to completion by using a large excess of the alcohol (usually as the solvent, in the case of low-boiling alcohols such as methanol or ethanol).

    1. Catalysis in ester cleavage. I. Nucleophilic catalysis by acetate ion in the methanolysis of p-nitrophenyl acetate
      Richard L. Schowen and Cherie Gass Behn
      Journal of the American Chemical Society 1968, 90 (21), 5839-5844
      This paper uses radioactively labeled (with tritium) acetate ion to study the basic transesterification mechanism in methanol.
    2. Catalysis in ester cleavage. II. Isotope exchange and solvolysis in the basic methanolysis of aryl esters. Molecular interpretation of free energies, enthalpies, and entropies of activation
      Carl G. Mitton, Richard L. Schowen, Michael Gresser, and John Shapley
      Journal of the American Chemical Society 1969, 91 (8), 2036-2044
      DOI: 10.1021/ja01036a029
    3. Catalysis in ester cleavage. III. Solvent isotope effects and transition-state solvation in the basic methanolysis of esters
      Carl G. Mitton, Michael Gresser, and Richard L. Schowen
      Journal of the American Chemical Society 1969, 91 (8), 2045-2047
      : 10.1021/ja01036a030
    4. Production of Biodiesel through Base-Catalyzed Transesterification of Safflower Oil Using an Optimized Protocol
      Umer Rashid and Farooq Anwar
      Energy & Fuels 2008, 22 (2), 1306-1312
      This is a popular application of the transesterification reaction, and is commonly used in undergraduate chemistry laboratories. Biodiesel, or fatty acid methyl esters (‘FAMEs’) can be produced through the methanolysis of naturally occurring triglycerides.


Comment section

5 thoughts on “Transesterification

  1. Hi there,
    If anyone can answer this question, I would be eternally grateful. This article was very helpful but I am stuck on one thing. I created biodiesel in the lab, and I used methoxide (MeO-) as the nucleophile for the transesterification reaction by combining NaOH with MeOH. I also used a large excess of just MeOH as the solvent. However, I don’t understand how the glycerol is formed in the mechanism. Is the pre-glycerol alkoxide intermediate protonated by the water that is created in the reaction between NaOH and MeOH, or is it protonated by the excess MeOH solvent that I added? Initially, I thought it must be the water since that would regenerate the -OH (catalyst from NaOH). However, Note 1 in this article makes me think it’s the excess MeOH solvent that protonates it. But this leaves me confused about the catalyst part… Can anyone explain this mechanism to me? Oleic Acid + Methanol = Methyl Esters + Glycerol using a base catalyst (NaOH + MeOH = H2O + NaOMe) and excess methanol solvent.

    1. Hi, the acidity of methanol, water and glycerol are all very similar – within an order of magnitude of each other which in acidity terms is a very, very small difference. Any of these species can act as acids here. I don’t know your exact conditions but if you were using MeOH as solvent then its concentration will be >> the concentration of any other possible acid in solution.
      If you drew MeOH as the acid you would largely be correct; it doesn’t really matter since all protons are identical anyway and there will be reverse acid-base equilibria occurring in solution between all the hydroxyl groups.

      I should note that even in neutral water, the protons on H2O undergo very fast exchange with other H2O molecules. Protons are very dynamic. If you ever take an NMR spectrum of an alcohol or a carboxylic acid you will note that these peaks tend to be quite broad, which is due to their relatively fast motion compared to more static protons such as C-H that do not undergo these kinds of exchange processes.

      Hope that’s helpful – James

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