Esters: Formation, Uses, and Hydrolysis Revision Notes | A Level AQA Chemistry

1. What Are Esters and Why Do They Matter?

That distinctive smell of nail varnish remover, the artificial strawberry flavour in sweets, the flexibility of a plastic water bottle — all of these involve esters. Esters are a family of organic compounds that appear throughout everyday life, and understanding their formation, properties, and hydrolysis is a core requirement of the AQA A Level specification.

Esters contain the functional group $-\text{COO}-$, sometimes written as $-\text{C}(=\text{O})\text{O}-$. You can think of this group as the bridge between the carbon chain from the acid and the carbon chain from the alcohol. The name of an ester always ends in -anoate (or more generally -yl -anoate), which is a reliable signal in exam questions.

2. Formation of Esters: Esterification

Esters are formed by the reaction of a carboxylic acid with an alcohol. This reaction is called esterification, and it is a condensation reaction — meaning a small molecule (water) is eliminated as the ester forms. The general equation is:

For the reaction to proceed at a useful rate, a catalyst is required. AQA specifies concentrated sulphuric acid as the acid catalyst. The acid catalyst protonates the carbonyl oxygen, making the carbonyl carbon more electrophilic and susceptible to nucleophilic attack by the lone pair on the oxygen of the alcohol. Heating under reflux is also used to increase the rate while preventing loss of volatile reactants or products.

A specific and important example is the formation of ethyl ethanoate from ethanoic acid and ethanol. This ester has a characteristic fruity smell and is widely used as a solvent.

Worked example: Writing the Equation for Ester Formation

Write the equation for the formation of ethyl ethanoate from ethanoic acid and ethanol. Include state symbols and identify the bond that forms and the small molecule eliminated.

1. $\text{CH}_3\text{COOH}_{(l)}$ — ethanoic acid — The carboxylic acid provides the $-\text{COO}-$ part of the ester. The $-\text{OH}$ on the acid is involved in the condensation.
2. $\text{C}_2\text{H}_5\text{OH}_{(l)}$ — ethanol — The alcohol provides the alkyl group that bonds to the oxygen in the ester. The $-\text{OH}$ of the alcohol is also involved.
3. The $-\text{OH}$ from the acid and the $\text{H}-$ from the alcohol's $-\text{OH}$ combine to form $\text{H}_2\text{O}$ — This is the condensation step. One water molecule is released for every ester bond formed.
4. $\text{CH}_3\text{COOH}_{(l)} + \text{C}_2\text{H}_5\text{OH}_{(l)} \rightleftharpoons \text{CH}_3\text{COOC}_2\text{H}_5_{(l)} + \text{H}_2\text{O}_{(l)}$ — The double arrow is essential — esterification is reversible. Concentrated $\text{H}_2\text{SO}_4$ acts as the catalyst; it is written above the arrow, not in the equation itself.
5. $\text{CH}_3\text{COOC}_2\text{H}_5$ = ethyl ethanoate — Naming rule: the alkyl group from the alcohol comes first (ethyl), then the name of the acid with '-ic acid' replaced by '-ate' (ethanoate).

Answer: $\text{CH}_3\text{COOH}_{(l)} + \text{C}_2\text{H}_5\text{OH}_{(l)} \rightleftharpoons \text{CH}_3\text{COOC}_2\text{H}_5_{(l)} + \text{H}_2\text{O}_{(l)}$, catalysed by concentrated $\text{H}_2\text{SO}_4$. The ester formed is ethyl ethanoate.

Because esterification is reversible, the yield of ester at equilibrium is typically around 65% when equimolar quantities of acid and alcohol are used. To shift the equilibrium towards the ester (applying Le Chatelier's principle), you can use an excess of one reactant, remove water as it forms, or remove the ester by distillation.

3. Naming Esters

Naming esters follows a consistent two-part pattern. The first part of the name comes from the alcohol (the alkyl group, e.g. methyl, ethyl, propyl). The second part comes from the carboxylic acid, with the ending changed from -ic acid to -ate (e.g. ethanoic acid becomes ethanoate). So ethanol + propanoic acid gives ethyl propanoate.

4. Uses of Esters

Esters have a wide range of commercial and industrial uses, all of which the AQA specification requires you to know. Their usefulness stems from a combination of properties: many are volatile liquids with pleasant smells, they are generally good solvents for organic compounds, and their physical properties can be tuned by varying the length and structure of the carbon chains on either side of the ester group.

• Solvents: Ethyl ethanoate is widely used as a solvent in glues, nail varnish removers, and paints. It dissolves many organic compounds that water cannot, because it is a polar molecule but lacks the hydrogen-bonding network that makes water a poor solvent for non-polar substances.
• Plasticisers: Long-chain esters (such as dioctyl phthalate) are added to polymers like PVC to make them more flexible. The ester molecules sit between polymer chains, reducing intermolecular forces and allowing the chains to slide past each other more readily.
• Perfumes and cosmetics: Many esters have distinctive, pleasant fruity or floral smells. Isoamyl acetate (3-methylbutyl ethanoate) smells of banana; octyl ethanoate has a citrus note. Their volatility means they evaporate readily from skin, releasing the scent.
• Food flavourings: Synthetic esters are used to create artificial fruit flavours in sweets, drinks, and confectionery. Ethyl butanoate gives a pineapple flavour; methyl butanoate contributes to apple flavouring.

The link between structure and use is worth noting for evaluation questions. Short-chain esters (low molecular mass) are more volatile and are therefore preferred for perfumes and flavourings, where the compound must reach the nose or taste receptors. Longer-chain esters have higher boiling points and lower volatility, making them more suitable as plasticisers where they must remain within the polymer matrix over time.

5. Hydrolysis of Esters

Esterification is reversible, and the reverse reaction — breaking the ester back down using water — is called hydrolysis. At A Level, you need to know two distinct routes: acid hydrolysis and alkaline hydrolysis. They differ in the conditions used, the reversibility of the reaction, and crucially, the organic product obtained from the acid component.

5a. Acid Hydrolysis

In acid hydrolysis, the ester is heated under reflux with a dilute acid — typically dilute sulphuric acid or dilute hydrochloric acid. The acid acts as a catalyst. The products are the original carboxylic acid and the original alcohol. Because this is the reverse of esterification, the reaction is reversible, and an equilibrium is established.

For example, acid hydrolysis of ethyl ethanoate gives ethanoic acid and ethanol. To drive the equilibrium towards complete hydrolysis, an excess of water is used. The acid catalyst speeds up the reaction but does not alter the equilibrium position.

5b. Alkaline Hydrolysis (Saponification)

Alkaline hydrolysis uses a hot aqueous alkali — typically sodium hydroxide solution. The ester is heated under reflux with excess NaOH. This reaction is irreversible, which is the critical distinction from acid hydrolysis. The products are the alcohol and the sodium salt of the carboxylic acid, not the free acid itself.

For ethyl ethanoate, alkaline hydrolysis with NaOH gives ethanol and sodium ethanoate ($\text{CH}_3\text{COONa}$). To obtain the free carboxylic acid from alkaline hydrolysis, you would need to acidify the salt with dilute acid after the reaction — a separate step not part of the hydrolysis itself.

Worked example: Comparing Acid and Alkaline Hydrolysis of Methyl Propanoate

Write balanced equations for (a) acid hydrolysis and (b) alkaline hydrolysis of methyl propanoate ($\text{C}_2\text{H}_5\text{COOCH}_3$). Identify all products and state whether each reaction is reversible.

1. $\text{C}_2\text{H}_5\text{COOCH}_3$ = methyl propanoate Acid component: propanoic acid ($\text{C}_2\text{H}_5\text{COOH}$) Alcohol component: methanol ($\text{CH}_3\text{OH}$) — Read the ester name: methyl (from methanol) + propanoate (from propanoic acid). This tells you exactly which acid and alcohol will be regenerated on hydrolysis.
2. $\text{C}_2\text{H}_5\text{COOCH}_3{}_{(l)} + \text{H}_2\text{O}_{(l)} \rightleftharpoons \text{C}_2\text{H}_5\text{COOH}_{(l)} + \text{CH}_3\text{OH}_{(l)}$ Conditions: dilute $\text{H}_2\text{SO}_4$ (or dilute $\text{HCl}$), heat under reflux — The double arrow is essential — acid hydrolysis is reversible. The products are the free carboxylic acid and the alcohol.
3. $\text{C}_2\text{H}_5\text{COOCH}_3{}_{(l)} + \text{NaOH}_{(aq)} \rightarrow \text{CH}_3\text{OH}_{(aq)} + \text{C}_2\text{H}_5\text{COONa}_{(aq)}$ Conditions: hot aqueous NaOH, heat under reflux — The single arrow indicates the reaction is irreversible. The acid product is the sodium salt (sodium propanoate), not the free acid. One mole of NaOH is consumed per mole of ester.
4. Acid hydrolysis: Left: C$_4$H$_8$O$_2$ + H$_2$O = C$_4$H$_{10}$O$_3$; Right: C$_3$H$_6$O$_2$ + CH$_4$O = C$_4$H$_{10}$O$_3$ ✓ Alkaline hydrolysis: Left: C$_4$H$_8$O$_2$ + NaOH; Right: CH$_4$O + C$_3$H$_5$O$_2$Na — balanced ✓ — Always verify that atom counts balance on both sides. Sodium appears once on each side in the alkaline equation.

Answer: (a) Acid hydrolysis (reversible): $\text{C}_2\text{H}_5\text{COOCH}_3 + \text{H}_2\text{O} \rightleftharpoons \text{C}_2\text{H}_5\text{COOH} + \text{CH}_3\text{OH}$ (dilute acid catalyst, heat). (b) Alkaline hydrolysis (irreversible): $\text{C}_2\text{H}_5\text{COOCH}_3 + \text{NaOH} \rightarrow \text{CH}_3\text{OH} + \text{C}_2\text{H}_5\text{COONa}$ (hot aqueous NaOH).

6. Molecule Structures: Visualising Esters

Seeing the ester functional group in a structural formula helps you identify it quickly in exam questions. The $-\text{COO}-$ group sits between the two carbon chains: the chain from the acid on the left and the chain from the alcohol on the right. Below are the structures of two common esters.

The SMILES string above represents ethyl ethanoate ($\text{CH}_3\text{COOC}_2\text{H}_5$). In the 2D view, locate the $\text{C}(=\text{O})\text{O}$ fragment — that is the ester group. The methyl group ($\text{CH}_3$) to its left comes from ethanoic acid; the ethyl group ($\text{C}_2\text{H}_5$) to its right comes from ethanol. Recognising this split is exactly what you need to identify hydrolysis products in an exam.

7. Exam Technique and Common Mistakes

Ester questions on AQA papers tend to cluster around three areas: writing equations for formation and hydrolysis, naming esters correctly, and evaluating the differences between acid and alkaline hydrolysis. The marks are often lost through imprecise language or incorrect reaction arrows rather than a lack of knowledge.

For 'evaluate' or 'compare' questions, structure your answer using the key differences in the table from Section 5: reversibility, the nature of the acid product (free acid vs salt), and the role of the reagent (catalyst vs reactant). A top-band answer will explicitly link the irreversibility of alkaline hydrolysis to the stability of the carboxylate anion.

Esters are formed by the reversible condensation of a carboxylic acid and an alcohol, catalysed by concentrated sulphuric acid, with water as the by-product. Their volatility, pleasant odours, and solvent properties make them valuable as solvents, plasticisers, perfumes, and food flavourings. Hydrolysis breaks esters back down: acid hydrolysis (reversible, dilute acid catalyst) regenerates the carboxylic acid and alcohol; alkaline hydrolysis (irreversible, NaOH reactant) produces the alcohol and the carboxylate salt.

8. Bringing It Together: Spec-Point Summary

Pause before the next quiz block: esters dissolve many organic molecules because they are polar enough for dipole interactions, but they do not form the extensive hydrogen-bonded network that makes water such a strong solvent for many ionic and very polar substances.