Fischer esterification: practice and mechanism
Acid + alcohol → ester + water (acid-catalysed, equilibrium): the ethyl group then sits on the oxygen.
Example
Under acid catalysis the alcohol attacks the carbonyl carbon of the acid; after loss of water an ester forms. The reaction is an equilibrium; excess alcohol or removing water drives it towards the ester.
Mechanism step by step
- 1
The acid protonates the carbonyl oxygen, making the carbonyl carbon more electrophilic.
- 2
Ethanol attacks with its oxygen; after proton transfer a tetrahedral intermediate is present.
- 3
A protonated OH leaves as water, C=O re-forms, and loss of H⁺ gives the ester.
Practice tasks
Which product forms, which reagent is needed? After each answer you see the explanation.
Which main product forms?
Typical mistakes
- Here the alcohol part sits on the carbon (ketone). But the alcohol attacks with its oxygen; the C(=O)–O–Et group forms.
- That is the tetrahedral intermediate. It loses water and the C=O group re-forms: the ester.
Often confused with: Ester hydrolysis (saponification) · Reduction with lithium aluminium hydride
Frequently asked questions
How do you increase the yield?
Use excess alcohol or remove water (Dean–Stark trap); both shift the equilibrium.
Where does the oxygen in the ester come from?
The bridging O comes from the alcohol; the water released contains the oxygen of the acid OH (¹⁸O labelling experiments).
How do you shift the esterification equilibrium?
Use excess alcohol or remove the water formed, for example with a Dean–Stark trap.
Where does the ester oxygen come from?
The ether oxygen of the ester comes from the alcohol; the water lost carries the OH oxygen of the acid (¹⁸O labelling).