Topic 7 of 12
Condensation and hydrolysis
The functional group and reaction medium determine the products.
A-Level 8873, revised syllabus (2026-2027)
Make and break ester or amide links
The functional group and reaction medium determine the products.
Condensation joins molecules with elimination of a small molecule, often water. Hydrolysis uses water to break a bond; aqueous acid or alkali and heat can facilitate it. Ester and amide links connect carbonyl-containing fragments to O or N respectively. Find that connecting bond before trying to name the products.
| Link formed | Reagents and conditions | Organic product |
|---|---|---|
| Ester, -C(=O)-O- | Ethanoic acid + ethanol; concentrated H2SO4 catalyst, heat. | CH3COOH + HOCH2CH3 ⇌ CH3COOCH2CH3 + H2O |
| Amide, -CONH- | Ethanoic acid + ethylamine in the presence of DCC (dicyclohexylcarbodiimide). | CH3CONHCH2CH3, N-ethylethanamide. The acid-derived fragment remains attached to the carbonyl carbon; the ethyl group remains on nitrogen. |
The ester link -C(=O)-O- is neutral: the final dash is a covalent bond continuing to a carbon group. In a carboxylate ion, -COO-, the superscript minus is a charge. These are different structures. In particular, ester links in a polyester backbone do not give the unchanged polymer a negative charge.
Keep the two fragments attached to the right atoms
Structural fragments show the carbonyl carbon joined to oxygen in an ester and to nitrogen in an amide. Hydrolysis cleaves the acyl C-O or acyl C-N bond, restoring an acid-derived fragment and an alcohol- or amine-derived fragment. The diagram is a connectivity guide, not a reaction mechanism.
| Conditions | Equation | Why this form appears |
|---|---|---|
| Aqueous acid and heat | CH3COOCH2CH3 + H2O ⇌ CH3COOH + CH3CH2OH | The acidic mixture contains the carboxylic acid. Acid catalyses the reaction. |
| Aqueous NaOH and heat | CH3COOCH2CH3 + NaOH → CH3COONa + CH3CH2OH | Any acid formed is deprotonated by the alkali. The isolated acid-derived product in this medium is the carboxylate salt. |
| Conditions | Net ionic equation | Product check |
|---|---|---|
| Aqueous acid and heat | CH3CONH2 + H2O + H+ → CH3COOH + NH4+ | The nitrogen product is protonated in acid; write ammonium, not free ammonia as the final major form. |
| Aqueous alkali and heat | CH3CONH2 + OH- → CH3COO- + NH3 | The acid fragment is a carboxylate; nitrogen leaves as ammonia. With NaOH, the salt is sodium ethanoate. |
Worked example
Transfer hydrolysis to an unfamiliar ester
Predict the organic products when methyl propanoate, CH3CH2COOCH3, is heated with aqueous NaOH.
- Identify the acid-derived fragment CH3CH2CO- and the O-attached methyl fragment.
- Breaking the ester link gives a propanoate ion and methanol in the alkaline mixture.
- Include Na+ as counter-ion: CH3CH2COOCH3 + NaOH → CH3CH2COONa + CH3OH. All four carbons are accounted for.
Sodium propanoate and methanol form. Propanoic acid would require subsequent acidification.
Product isolation is a reasoning task rather than a memorised apparatus list. After making an ester, ask which reagents remain: excess acid, acid catalyst, alcohol and water can all accompany the product. An aqueous wash can remove water-soluble impurities; an appropriate carbonate wash can neutralise acid. If two liquid layers form, separation can isolate the ester-containing layer; drying then removes traces of water. Use supplied solubility and boiling-point data to choose further separation, and never identify a layer solely from whether it is on top.