Ester hydrolysis (saponification): practice and mechanism
Hydroxide cleaves the ester at the C(=O)–O bond: the carboxylic acid (after acidification) and the alcohol form.
Example
Hydroxide attacks the carbonyl carbon and the alkoxide leaves. Carboxylate and alcohol form; under basic conditions the reaction is irreversible. Acidification gives the carboxylic acid.
Mechanism step by step
- 1
OH⁻ attacks the ester carbonyl carbon: tetrahedral intermediate.
- 2
C=O re-forms and the alkoxide (R'O⁻) leaves; the carboxylic acid forms.
- 3
The acid immediately gives a proton to the alkoxide: carboxylate + alcohol (irreversible). H₃O⁺ protonates to the acid.
Practice tasks
Which product forms, which reagent is needed? After each answer you see the explanation.
Which main product forms?
Typical mistakes
- Two alcohols form when reducing with LiAlH₄. NaOH is not a reducing agent; the acid group remains.
- An aldehyde would only come from reduction (e.g. DIBAL-H). Saponification keeps the oxidation level: carboxylic acid.
- It is the bond between the carbonyl carbon and oxygen that breaks, not the O–alkyl bond. The alcohol part is released as R–OH, not as a diol.
Often confused with: Reduction with lithium aluminium hydride · Fischer esterification
Frequently asked questions
Why is it called saponification?
That is how soap is made: fats (glycerol esters) are split with NaOH into glycerol and the sodium salts of fatty acids.
Why is basic cleavage irreversible but acidic not?
The carboxylate anion is no longer an electrophile and does not react back with the alcohol.