UNIT 2 CHEMISTRY · PEARSON AAQ · E1.7.7
Formation of Esters
FREE CHEMISTRY REVISION · TOPIC 62
Formation of Esters
Pearson BTEC Level 3 AAQ Applied Science
Unit 2: Principles and Applications of Chemistry · E1.7.7
1. What you need to know
An alcohol reacts with a carboxylic acid in a condensation reaction to form an ester and water. For E1.7.7, predict, name and draw the ester, write a balanced equation and know the acid-catalysed reaction conditions. Reversible esterification is the focus; mechanisms and practical techniques are not required.
2. Condensation and the ester linkage
Carboxylic acid + alcohol ⇌ ester + water
R–C(=O)–OH + H–O–R′ ⇌ R–C(=O)–O–R′ + H₂O
A condensation reaction joins molecules while producing a small molecule, here water. The ester linkage is –C(=O)–O–. The carbonyl carbon has a double bond to one oxygen and a single bond to the other oxygen. That single-bonded oxygen joins to the alcohol-derived carbon group R′.
The displayed bookkeeping removes OH from the acid group and H from the alcohol OH to account for water in the overall equation. It is not a reaction mechanism. R and R′ mark the two carbon groups; they may differ. Both reactants’ carbon skeletons are retained in the ester.
3. Conditions and reversibility
Warm or heat the alcohol and carboxylic acid with an acid catalyst, commonly concentrated sulfuric acid. Heating under reflux is a suitable condition for sustained reaction; detailed apparatus or purification methods are not needed. Sulfuric acid is the catalyst, not the carboxylic acid reactant that gives the ester its alkanoate part.
Esterification by this route is reversible, so use ⇌. In a closed mixture the forward and reverse reactions can reach dynamic equilibrium; conversion is not automatically complete. The reverse reaction is hydrolysis: ester + water reforms the alcohol and carboxylic acid under acidic conditions.
The catalyst speeds both directions and helps equilibrium be reached sooner. It does not change the equilibrium constant or, solely by catalytic action, increase the equilibrium proportion of ester. Excess of one reactant or removal of water can favour ester formation; these are equilibrium ideas, not required practical procedures.
4. Name and draw an ester
- Identify the alcohol: its carbon group gives the first word, the alkyl name, such as methyl or ethyl.
- Identify the carboxylic acid: its name becomes the second word ending in -oate, such as ethanoate or propanoate.
- Draw the acid-derived C(=O)–O group, then attach the alcohol-derived alkyl group to the single-bonded O.
- Check the total carbon count equals the sum in the acid and alcohol. Add water and use a reversible arrow.
| Reactant | Part of ester name | Where it appears |
|---|---|---|
| Methanol / ethanol / propan-1-ol | Methyl / ethyl / propyl | Carbon group bonded through O |
| Ethanoic / propanoic / butanoic acid | Ethanoate / propanoate / butanoate | Part containing C=O, including its carbon |
In these examples, propyl means the straight-chain group from propan-1-ol, –CH₂CH₂CH₃. The name order is alcohol-derived alkyl first, acid-derived alkanoate second; it does not depend on which reactant you write first in the equation.
5. Worked esterification examples
Worked example 1 — ethanol + ethanoic acid
CH₃COOH + CH₃CH₂OH ⇌ CH₃COOCH₂CH₃ + H₂O
Ethanol supplies ethyl; ethanoic acid supplies ethanoate. Product: ethyl ethanoate, CH₃C(=O)OCH₂CH₃. Warm with an acid catalyst such as concentrated H₂SO₄. Atom check: left C₄H₁₀O₃; right ester C₄H₈O₂ plus H₂O gives C₄H₁₀O₃.
Worked example 2 — methanol + propanoic acid
CH₃CH₂COOH + CH₃OH ⇌ CH₃CH₂COOCH₃ + H₂O
The ester is methyl propanoate. The acid contributes three carbons including C=O, and methanol contributes one. Its structure is CH₃CH₂C(=O)OCH₃. Do not call it propyl methanoate: that would reverse the carbon-group assignments.
Worked example 3 — propan-1-ol + ethanoic acid
CH₃COOH + CH₃CH₂CH₂OH ⇌ CH₃COOCH₂CH₂CH₃ + H₂O
The ester is propyl ethanoate, with a two-carbon acid part and three-carbon alcohol part: five carbons overall. The O joins to CH₂, not directly to a new carbonyl group. Use acid catalysis and heating.
Worked example 4 — ethanol + butanoic acid
CH₃CH₂CH₂COOH + CH₃CH₂OH ⇌ CH₃CH₂CH₂COOCH₂CH₃ + H₂O
The product is ethyl butanoate. Its acid part has four carbons; the ethyl group adds two, making C₆H₁₂O₂. The reactants together are C₆H₁₄O₃ and the water completes the atom balance.
Worked example 5 — butan-1-ol + propanoic acid
CH₃CH₂COOH + CH₃CH₂CH₂CH₂OH ⇌ CH₃CH₂COOCH₂CH₂CH₂CH₃ + H₂O
The product is butyl propanoate. Follow the O–butyl side to count four carbons, then count three on the propanoate side, including C=O: seven altogether. The same linkage and naming rule apply to longer groups.
6. Work backwards from an ester
Worked example 6 — choose the reactants
Target: ethyl propanoate, CH₃CH₂C(=O)OCH₂CH₃. The ethyl name identifies ethanol; propanoate identifies propanoic acid. Reverse the overall joining: replace the O–ethyl connection with acid OH on the carbonyl side, and restore OH on the ethyl side. React CH₃CH₂COOH with CH₃CH₂OH, warm with an acid catalyst.
Ethyl ethanoate and methyl propanoate both have molecular formula C₄H₈O₂, but their atoms are connected differently. A molecular formula alone cannot identify an ester. Always inspect which carbon group is attached to the single-bonded oxygen.
Common misconceptions
- The alcohol gives the -oate part: it gives the first alkyl word.
- An ester has C–O–O: its linkage is C(=O)–O.
- Water is a reactant in forward esterification: it is a product.
- Concentrated sulfuric acid supplies the ethanoate carbon: the carboxylic acid supplies that part; sulfuric acid catalyses.
- A catalyst forces complete conversion: the reaction is reversible and equilibrium still matters.
- An ester name identifies only the total number of carbons: it also identifies how they are divided across the linkage.
Examination tips
- Write the ester as R–C(=O)–O–R′ to check the linkage.
- Name the alcohol-derived alkyl group first.
- Count the carbonyl carbon as part of the acid-derived group.
- Include H₂O and a reversible arrow in the equation.
- State warming/heating with an acid catalyst such as concentrated H₂SO₄.
- Check C, H and O totals; use structures to distinguish isomers.
Check your understanding
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