Basic Hydrolysis of Esters (Saponification)
Basic Hydrolysis of Esters (Saponification): An ester reacts with a strong base such as NaOH or KOH and is hydrolyzed into a carboxylate salt plus an…
What happens
An ester reacts with a strong base such as NaOH or KOH and is hydrolyzed into a carboxylate salt plus an alcohol; after an acidic workup the carboxylate becomes a carboxylic acid. It is a nucleophilic acyl substitution and the carbonyl carbon is retained. It is also how soap is made: fats contain ester bonds, and when fats react with a strong base the esters become carboxylate salts, and long-chain carboxylate salts are soaps. That is where the name saponification comes from. To predict the product, split R–COOR′ into R–COO⁻ + R′OH, and if there is an acidic workup (H₃O⁺) change COO⁻ to COOH.
Reaction summary
Ester + NaOH/H₂O (then H₃O⁺) → carboxylic acid + alcohol (COOR → COOH)
Reagents and conditions
- Starting material: Ethyl acetate (ester)
- Reagents: Sodium hydroxide (NaOH); H₂O; H₃O⁺ (acidic workup)
- Conditions: 1. NaOH, H₂O, Δ; 2. H₃O⁺
- Products: Acetic acid (carboxylic acid); Ethanol (alcohol)
Mechanism
- OH⁻ Attacks: OH⁻ acts as the nucleophile and attacks the carbonyl carbon of the ester. The C=O π bond opens onto oxygen and a tetrahedral intermediate forms.
- Carbonyl Reforms: The oxygen reforms the C=O and the –OR′ group leaves. This is nucleophilic acyl substitution.
- Carboxylate Forms: Under the basic conditions the carboxylic acid does not remain as R–COOH: it is deprotonated to R–COO⁻. So before workup, the actual product is a carboxylate plus an alcohol.
- Acidic Workup: Add H₃O⁺. The carboxylate is protonated, R–COO⁻ → R–COOH, and you have the final carboxylic acid.
Memory trick
Base breaks the ester. COOR → COO⁻. Then acid workup → COOH. Ultimate shortcut: Ester + NaOH = saponification.