Esters

Esters are one of the most practically important families of organic compounds. They are responsible for the pleasant smells of many fruits and flowers, and they are used industrially as solvents, plasticisers, and in the production of biodiesel. At A-Level, you need to understand how esters are formed, how they are named, their hydrolysis reactions, and their varied applications.

Formation of Esters

Esters are formed by a condensation reaction between a carboxylic acid and an alcohol, with the loss of a water molecule:

Carboxylic acid + Alcohol <=> Ester + Water

For example:

CH3COOH + CH3OH <=> CH3COOCH3 + H2O (ethanoic acid + methanol <=> methyl ethanoate + water)

This reaction requires:

A concentrated sulfuric acid catalyst (H2SO4)
Heating under reflux to maintain the temperature without losing volatile components

The reaction is reversible and reaches an equilibrium, so yields are typically moderate unless steps are taken to remove the water or use excess of one reactant.

Alternative Routes to Esters

Fischer esterification is not the only way to make esters. Two other routes are faster and give better yields:

Route	Reagents	Advantages	Disadvantages
Fischer esterification	RCOOH + R'OH, conc. H2SO4, reflux	Simple, cheap reagents	Slow, reversible, moderate yield
Acyl chloride + alcohol	RCOCl + R'OH	Fast, irreversible, room temperature	Acyl chloride must be prepared; HCl by-product
Acid anhydride + alcohol	(RCO)2O + R'OH	Cheaper than acyl chloride, safer	Slower than acyl chloride; RCOOH by-product

In the exam, if asked for the best way to prepare an ester efficiently, the acyl chloride route is usually the answer.

Naming Esters

Esters are named from the alcohol part first, then the acid part:

The part from the alcohol gives the first name (using the alkyl prefix) -- e.g., methyl, ethyl, propyl
The part from the acid gives the second name (replacing -oic acid with -oate) -- e.g., methanoate, ethanoate, propanoate

Alcohol	Acid	Ester Name	Molecular Formula
Methanol	Ethanoic acid	Methyl ethanoate	CH3COOCH3
Ethanol	Propanoic acid	Ethyl propanoate	CH3CH2COOC2H5
Propan-1-ol	Butanoic acid	Propyl butanoate	CH3(CH2)2COOC3H7
Ethanol	Methanoic acid	Ethyl methanoate	HCOOC2H5

A useful way to remember: the ester bond is -COO-. Everything on the C=O side comes from the acid; everything on the single-bonded O side comes from the alcohol.

Worked Example: Naming from Structure

Given the structure CH3CH2COOCH2CH2CH3:

Identify the ester linkage: -COO-
On the C=O side: CH3CH2CO- comes from propanoic acid
On the O side: -OCH2CH2CH3 comes from propan-1-ol
Name: propyl propanoate

Hydrolysis of Esters

Hydrolysis is the reverse of ester formation -- the ester is broken down by reaction with water. There are two types:

Acid Hydrolysis

The ester is heated under reflux with dilute acid (e.g., dilute H2SO4 or dilute HCl):

CH3COOCH2CH3 + H2O <=> CH3COOH + CH3CH2OH

This is reversible -- it is simply the reverse of esterification. The products are the original carboxylic acid and alcohol. Because it is an equilibrium process, the yield is limited.

Base (Alkaline) Hydrolysis

The ester is heated under reflux with aqueous NaOH or KOH:

CH3COOCH2CH3 + NaOH --> CH3COONa + CH3CH2OH

This is irreversible because the carboxylic acid produced immediately reacts with the base to form a carboxylate salt (sodium ethanoate in this case). Since the acid is removed from the equilibrium as a salt, the reaction goes to completion. This makes base hydrolysis much more efficient than acid hydrolysis.

Base hydrolysis of esters is also called saponification when applied to fats and oils.

Comparison of Hydrolysis Methods

Feature	Acid Hydrolysis	Base Hydrolysis
Reagent	Dilute H2SO4 or HCl	Aqueous NaOH or KOH
Reversible?	Yes (equilibrium)	No (goes to completion)
Organic product	Carboxylic acid	Carboxylate salt
Other product	Alcohol	Alcohol
Yield	Moderate (limited by equilibrium)	High (reaction driven to completion)
Why irreversible?	N/A	Acid immediately neutralised to salt

Uses of Esters

Esters have numerous practical applications:

Solvents

Many esters are excellent solvents because they dissolve both polar and non-polar substances. Ethyl ethanoate (commonly called ethyl acetate) is widely used as a solvent in nail polish remover, adhesives, and in the laboratory.

Flavourings and Fragrances

Many esters have characteristic fruity or floral smells:

Ester	Smell	Alcohol + Acid
Ethyl butanoate	Pineapple	Ethanol + Butanoic acid
Pentyl ethanoate	Banana/pear	Pentan-1-ol + Ethanoic acid
Octyl ethanoate	Orange	Octan-1-ol + Ethanoic acid
Methyl butanoate	Apple	Methanol + Butanoic acid
Ethyl methanoate	Rum	Ethanol + Methanoic acid
Benzyl ethanoate	Jasmine	Benzyl alcohol + Ethanoic acid

These esters are used in the food industry as artificial flavourings and in the perfume industry.

Plasticisers

Esters such as dioctyl phthalate are added to PVC (polyvinyl chloride) to make it flexible. Without plasticisers, PVC would be rigid and brittle. The ester molecules sit between the polymer chains and reduce the intermolecular forces, allowing the chains to slide past each other more easily.

Biodiesel

Biodiesel is produced by reacting vegetable oils (which are esters of glycerol with long-chain fatty acids) with methanol in the presence of a catalyst. This transesterification reaction replaces glycerol with methanol, producing fatty acid methyl esters (FAME) -- biodiesel -- and glycerol as a by-product.

flowchart LR
    A[Triglyceride<br>Glycerol + 3 Fatty acids] -->|3 CH3OH<br>NaOH catalyst| B[3 FAME molecules<br>Biodiesel]
    A -->|3 CH3OH<br>NaOH catalyst| C[Glycerol<br>By-product]

Fats and Oils -- Naturally Occurring Esters

Fats and oils are triesters (triglycerides) formed from glycerol (propane-1,2,3-triol) and three fatty acid molecules (long-chain carboxylic acids). Each of the three -OH groups on glycerol forms an ester bond with a fatty acid.

Fats are solid at room temperature and tend to contain saturated fatty acids (no C=C double bonds), leading to straight chains that pack closely together with strong London forces
Oils are liquid at room temperature and tend to contain unsaturated fatty acids (one or more C=C double bonds), which create kinks in the chains that prevent close packing and weaken London forces

Property	Saturated Fats	Unsaturated Oils
Chain shape	Straight	Kinked at each C=C
Packing	Close -- strong London forces	Poor -- weak London forces
State at room temp	Solid	Liquid
Melting point	Higher	Lower
Example	Butter, lard	Olive oil, sunflower oil

Hydrogenation of Oils

Unsaturated oils can be converted to saturated fats by reacting with hydrogen gas in the presence of a nickel catalyst at 200 degrees C. This is how margarine is produced from vegetable oils. The C=C double bonds are reduced to C-C single bonds, straightening the chains and raising the melting point.

Saponification -- Making Soap

When a fat or oil is heated with concentrated aqueous NaOH, the ester bonds are hydrolysed in an irreversible reaction:

Triglyceride + 3NaOH --> Glycerol + 3 Sodium carboxylate (soap)

The sodium carboxylates produced are soaps. Each soap molecule has:

A hydrophilic (water-loving) head -- the carboxylate ion (COO-Na+)
A hydrophobic (water-hating) tail -- the long hydrocarbon chain

This dual nature allows soap molecules to form micelles around grease droplets. The hydrophobic tails dissolve in the grease while the hydrophilic heads face outward into the water, allowing the grease to be washed away.

Common Exam Mistakes

Naming esters in the wrong order. The alcohol part comes first in the name, then the acid part. Methyl ethanoate comes from methanol and ethanoic acid, not the other way round.
Forgetting that acid hydrolysis is reversible. This is why the yield is low. Base hydrolysis goes to completion because the acid is immediately neutralised.
Confusing transesterification with hydrolysis. In transesterification, one alcohol replaces another (glycerol is replaced by methanol). In hydrolysis, the ester is broken down completely.
Not knowing the reagents for saponification. Concentrated aqueous NaOH (or KOH) heated under reflux. Not dilute acid.

Summary

Esters bridge the gap between laboratory chemistry and real-world applications. Their formation from acids and alcohols (or more efficiently from acyl chlorides), their hydrolysis back to the starting materials, and their roles as solvents, flavourings, plasticisers, biodiesel, and soaps make them one of the most practically relevant functional groups in organic chemistry.

A-Level Deep Dive: Esters

Spec mapping

Edexcel 9CH0 specification, Topic 17 — Carboxylic acids and their derivatives, sub-strands 17.5–17.7 covers the formation of esters from carboxylic acids and alcohols (acid-catalysed equilibrium reaction with conc. H2SO4 catalyst), the formation of esters from acyl chlorides + alcohols (vigorous, irreversible) and acid anhydrides + alcohols (milder, irreversible), acid hydrolysis of esters (reversible — gives back the carboxylic acid + alcohol) and base hydrolysis (irreversible — gives the carboxylate ion + alcohol; the soap-making/saponification reaction), and naming of esters as alkyl alkanoates (refer to the official specification document for exact wording). It also covers IR identification of esters via C=O at ≈ 1735 cm⁻¹ and C–O at ≈ 1050–1300 cm⁻¹. Esters are examined directly on Paper 2 with synoptic appearances on Paper 3 and connect explicitly to CP15 (preparation of an ester) and to industrial chemistry (biodiesel, polyester fibres, flavourings, plasticisers).

Worked example with full mark scheme

Question (8 marks):

(a) Write the equation for the reaction between methanol (CH3OH) and ethanoic acid (CH3COOH) in the presence of concentrated sulfuric acid catalyst. Name the organic product. (2)

(b) Predict and explain the effect on the equilibrium yield of the ester of (i) using a large excess of methanol, and (ii) continuously distilling off the ester as it forms. (4)

Solution with mark scheme:

(a) Step 1 — write the equation.

CH3OH + CH3COOH ⇌ CH3COOCH3 + H2O

M1 — correct equation including reversible arrow.

A1 — product correctly named methyl ethanoate.

(b) Step 1 — apply Le Chatelier's principle to (i).

Adding excess methanol increases [methanol] on the left-hand side; the system responds by shifting the equilibrium to the right to remove the added methanol, increasing the yield of methyl ethanoate.

M1 — Le Chatelier reasoning + correct direction.

A1 — explicit statement that yield increases.

Step 2 — apply Le Chatelier to (ii).

Removing methyl ethanoate as it forms decreases [ester] on the right-hand side; the system shifts to the right to replace the removed ester. Methyl ethanoate (b.p. 57 °C) is the lowest-boiling species in the mixture, so it can be selectively distilled off.

M1 — Le Chatelier reasoning.

A1 — explicit yield increase + identification that the ester is the lowest-boiling component.

Concentrated H2SO4 is hygroscopic and acts as a dehydrating agent, removing water (the other product of esterification) and shifting the equilibrium to the right. Dilute H2SO4 contains too much water to act as an effective dehydrating agent, so the equilibrium would lie further to the left.

M1 — concentrated acid removes water/dehydrates.

A1 — link to equilibrium shift.

Total: 8 marks (M5 A3).

Specimen question modelled on the Edexcel 9CH0 Paper 2 format

Question (7 marks): An ester X has the molecular formula C5H10O2 and a fruity smell. Mild acid hydrolysis of X gives an alcohol Y and a carboxylic acid Z. Y and Z each contain three carbon atoms.

(a) Identify Y, Z and X. (3) (b) Write equations for the acid hydrolysis and base hydrolysis (using NaOH(aq)) of X. (2) (c) State two observations that distinguish acid from base hydrolysis. (2)

Mark scheme decomposition by AO: