Polyesters Explained: PET Monomers, Ester Links and Uses

A plastic drink bottle and a polyester school uniform might seem like completely different materials. Chemically, however, they can be made from the same polymer: polyethylene terephthalate (PET).

The clue is in the name polyester. Both materials contain long molecular chains joined by ester linkages. What changes between different polyesters is the arrangement of atoms within those chains, which affects their properties and uses.

To understand polyesters, you need to recognise three things: the monomers used to build them, the ester bonds connecting those monomers, and the repeating structure that forms the polymer.

01What is a polyester?

A polyester is a polymer containing repeating ester linkages within its main molecular chain.

The word describes a family of polymers, not one particular chemical. Different polyesters can have different monomers, repeating units, chain structures, and physical properties.

One important example is polyethylene terephthalate (PET), commonly used to manufacture:

• Plastic drink bottles and food packaging.
• Polyester clothing and textile fibres.
• Packaging films and other mouldable plastic products.

PET is a specific member of the polyester family. It is not another name for every polyester.

For NSW HSC Chemistry students, this connects to condensation polymerisation and the relationship between molecular structure and material properties. The Chemistry Stage 6 Syllabus (2017) remains the syllabus being taught in 2026 and 2027. The replacement Chemistry 11-12 Syllabus (2025) begins implementation in 2028, with its first HSC examination in 2029.

02What makes an ester linkage?

Before looking at PET, identify the functional group that gives polyesters their name.

An ester contains the arrangement:

\[ \mathrm{R-C(=O)-O-R'} \]

The two R groups represent the remaining sections of the molecule. The important feature is the sequence:

\[ \mathrm{-C(=O)-O-} \]

This is the ester linkage.

Notice that the carbonyl carbon, \(\mathrm{C=O}\), is directly bonded to an oxygen atom, which is then bonded to another carbon-containing section.

In a polyester, these ester linkages occur repeatedly along the main chain. They are not simply attached to the side of an otherwise unrelated polymer.

This gives us a useful identification rule:

• If ester groups occur repeatedly in the polymer backbone, the polymer belongs to the polyester family.
• If a molecule contains just one ester group, that does not make it a polyester.
• If a polymer contains oxygen atoms but only ether linkages, it is not a polyester.

For example, PET contains repeated ester linkages, whereas a polyether contains repeated ether linkages of the form \(\mathrm{-C-O-C-}\).

The distinction is about connectivity, not simply which elements are present.

The formal definition is consistent with the IUPAC definition of a polyester: its ester linkages must be part of the polymer backbone.

03Which monomers are used to make PET?

PET is made from two monomers:

1. Ethane-1,2-diol, commonly called ethylene glycol.
2. Benzene-1,4-dicarboxylic acid, commonly called terephthalic acid.

Look at their structures:

Both monomers have two reactive functional groups. Chemists describe this as being bifunctional.

This is important because each monomer can react at both ends, allowing the molecular chain to continue growing.

A useful way to picture this is to imagine each monomer as a connector with a joining point at either end. The diol and dicarboxylic acid can alternate, linking together to form a long chain.

The analogy has a limit: these aren’t pre-made plastic building blocks. Each connection requires a chemical reaction that forms new covalent bonds.

How does an ester linkage form?

The key reaction occurs between a carboxylic acid group and an alcohol group.

\[ \ce{R-COOH + HO-R' -> R-COO-R' + H2O} \]

Here:

• \(\mathrm{R}\) represents the remaining carbon-containing section of the carboxylic acid.
• \(\mathrm{R'}\) represents the remaining carbon-containing section of the alcohol.
• \(\ce{-COO-}\) is the newly formed ester linkage.
• \(\ce{H2O}\) is water, released as a by-product.

Where does the water come from?

The hydroxyl group from the carboxylic acid contributes an \(\ce{-OH}\), while the alcohol contributes a hydrogen atom from its hydroxyl group. Together, these form one water molecule.

The oxygen from the alcohol remains in the new ester linkage.

This reaction is called esterification. When it happens repeatedly between bifunctional monomers, a polyester chain forms.

How does this produce a repeating unit?

Return to the two PET monomers.

The diol contributes the section:

\[ \ce{-O-CH2-CH2-O-} \]

The terephthalic acid contributes the section:

\[ \ce{-C(=O)-C6H4-C(=O)-} \]

These sections alternate along the polymer chain.

The repeating unit of PET is:

\[ \boxed{ \left[ \ce{-O-CH2-CH2-O-C(=O)-C6H4-C(=O)-} \right]_n } \]

The symbol \(n\) represents the number of repeating units in the polymer chain. A large value of \(n\) represents a long polymer.

Notice that the repeating unit contains both the flexible ethylene section and the benzene ring.

This is more useful than memorising the structure. Once you can recognise the two monomer-derived sections, you can reconstruct the polymer from its starting materials.

For a more detailed drawing method, see Condensation Polymerisation: Nylon and Polyester Explained for HSC Chemistry.

04Why must the monomers have two functional groups?

This is one of the most important ideas when deciding whether a reaction can produce a long polyester.

A monomer needs two reactive ends if it is going to connect to other molecules on both sides.

Consider the difference between these two situations.

Situation A: Two bifunctional monomers

• A dicarboxylic acid has two \(\ce{-COOH}\) groups.
• A diol has two \(\ce{-OH}\) groups.
• Each monomer can form two ester linkages.
• Repeated reactions allow long chains to develop.

Situation B: Monofunctional molecules

Consider ethanoic acid, \(\ce{CH3COOH}\), and ethanol, \(\ce{CH3CH2OH}\).

They can react to produce ethyl ethanoate:

\[ \ce{CH3COOH + CH3CH2OH -> CH3COOCH2CH3 + H2O} \]

An ester has formed, and water has been released.

However, each starting molecule has only one reactive functional group. Once the ester forms, neither molecule has another equivalent reactive end available to continue building a chain.

The product is a small ester molecule, not a long polyester.

This exposes a common misconception: releasing water does not automatically mean a polymer has formed.

The monomers must also have the correct functionality to allow repeated chain growth.

05How is PET actually manufactured?

The simplified reaction between terephthalic acid and ethylene glycol explains the chemistry. Industrial manufacture adds another problem: how do you keep the reaction progressing towards longer polymer chains?

This is where the production process becomes interesting.

Following the reaction through the apparatus

In the direct esterification route, the process follows several stages.

Stage 1: Esterification

Terephthalic acid and ethylene glycol enter a heated reactor.

The carboxylic acid and hydroxyl groups react to form ester linkages. Water is produced as the molecular chains begin to develop.

The prediction is straightforward: if esterification is occurring, water should be produced and can be collected from the reactor’s vapour stream.

Stage 2: Prepolymerisation

The partially reacted material moves into the next stage.

Further reactions increase the length of the molecular chains. Water and other volatile substances are removed from the reaction mixture.

Stage 3: Polycondensation

The material undergoes further chain growth at high temperature and reduced pressure.

Excess ethylene glycol is removed, helping the reaction proceed towards longer PET chains.

The final product is a molten polymer that can be extruded and formed into pellets.

Industrial PET production involves esterification followed by further polycondensation, with removal of volatile products. The detailed process is described in research on PET synthesis and industrial polycondensation.

Why does removing water matter?

Esterification is reversible.

\[ \ce{Carboxylic\ acid + Alcohol <=> Ester + Water} \]

If water remains in the reaction mixture, the reverse reaction can compete with ester formation.

Removing water favours the forward reaction according to Le Chatelier’s principle.

The same general idea applies during later stages of PET manufacture, where removing excess ethylene glycol helps drive further polycondensation.

There is an important distinction here:

• The HSC reaction model focuses on the formation of ester linkages and the release of a small molecule.
• Industrial production uses controlled conditions and additional stages to obtain sufficiently long polymer chains.

Experimental limitation: Collecting water is evidence that esterification has occurred, but it does not prove that a high-molecular-mass polymer has formed. The reaction might have produced many short oligomers instead.

To establish chain growth, chemists also need information about the molecular mass or viscosity of the resulting polymer.

The industrial reactor details are useful extension context, rather than a requirement to memorise manufacturing conditions for the 2017 HSC syllabus.

06What are the properties and uses of polyesters?

The structure of a polyester helps explain why different members of this family are useful for different purposes.

PET is particularly useful because its repeating unit combines several structural features.

An important detail is that PET contains polar ester groups but does not have the \(\ce{N-H}\) or \(\ce{O-H}\) groups needed to donate hydrogen bonds to neighbouring pure PET chains.

Its intermolecular attractions therefore differ from those in polyamides such as nylon.

PET in clothing and plastic bottles

Polyester clothing

PET can be manufactured into fibres for clothing.

During fibre production, the polymer is drawn out, helping align its molecular chains. This alignment contributes to the strength of the resulting fibres.

The aromatic rings also contribute rigidity to the polymer structure.

Plastic drink bottles

PET can also be processed into packaging, including drink bottles.

Its combination of strength, relatively low mass, and suitability for forming makes it useful for this application.

Here is the interesting part: PET used in clothing and PET used in bottles can have the same underlying chemical repeating unit.

Their final properties are influenced by more than just their chemical formula. Molecular mass, chain orientation, crystallinity, and manufacturing conditions also matter.

The word polyester on a clothing label describes the material family commonly represented by PET fibres. It does not mean that every polyester product has identical properties.

07Polyester versus other polymer families

It is easy to confuse polyester with other condensation polymers, particularly nylon.

Both can form through condensation reactions, but the functional groups connecting their molecular chains are different.

The key distinction is the functional group within the backbone.

Nylon contains amide linkages, whereas PET contains ester linkages. Polypropylene has neither of these groups in its backbone.

This means that:

• PET is a polyester and a condensation polymer.
• Nylon-6,6 is a polyamide and a condensation polymer.
• Polypropylene is an addition polymer, not a polyester.

For a broader comparison of polymer formation, see the tutorgum explanation of condensation polymerisation.

08How to identify a polyester in an HSC question

When you are given an unfamiliar polymer structure, use this process.

Step 1: Find the backbone.

Follow the continuous molecular chain rather than looking at isolated side groups.

Step 2: Look for the ester linkage.

Identify the sequence:

\[ \ce{-C(=O)-O-} \]

Step 3: Check whether it repeats along the backbone.

One ester group in a small molecule does not make it a polyester. The ester linkages must form part of the polymer chain.

Step 4: Identify the monomers if required.

For the common PET structure:

• The section containing \(\ce{-O-CH2-CH2-O-}\) comes from ethane-1,2-diol.
• The section containing \(\ce{-C(=O)-C6H4-C(=O)-}\) comes from terephthalic acid.

Step 5: Explain the formation reaction.

State that carboxylic acid and alcohol groups react to form ester linkages, releasing water during the idealised direct esterification process.

This gives you a complete explanation rather than simply naming the polymer.

09Questions and solutions

The following questions move from basic structural recognition to more challenging reasoning about chain growth and experimental evidence. All numerical data are illustrative.

Question 1

A polymer has the following molecular chain segment:

\[ \ce{ ...-CH2-CH2-O-C(=O)-C6H4-C(=O)-O-CH2-CH2-... } \]

Identify the polymer family and explain which structural feature supports your answer.

Solution 1

The polymer belongs to the polyester family because its backbone contains repeating ester linkages.

The characteristic ester linkage is:

\[ \ce{-C(=O)-O-} \]

In the supplied structure, the carbonyl carbon is bonded to an oxygen atom, which connects to another carbon-containing section of the chain.

This arrangement appears repeatedly along the backbone.

The presence of oxygen alone would not be sufficient to classify the polymer as a polyester. The connectivity of the atoms is what matters.

The structure is consistent with PET because it contains both ethylene-derived sections and terephthalate sections.

Question 2

A student attempts to produce a long polyester by reacting ethane-1,2-diol with ethanoic acid.

They argue that both molecules contain suitable functional groups for esterification, and water will be released.

Explain why this reaction does not provide the correct monomer functionality for producing a long, linear polyester. Identify the structural change required to make PET instead.

Solution 2

The proposed monomers can form ester molecules, but they cannot produce a long linear polyester through repeated alternating chain growth.

Ethane-1,2-diol contains two hydroxyl groups:

\[ \ce{HO-CH2-CH2-OH} \]

Ethanoic acid contains only one carboxylic acid group:

\[ \ce{CH3COOH} \]

The ethanoic acid molecule can form one ester linkage, but after that reaction it has no second carboxylic acid group available to connect to another diol molecule.

It therefore acts as a terminating molecule rather than a bridging monomer.

The reaction can produce small ester molecules, including ethylene glycol diacetate if both hydroxyl groups react with separate ethanoic acid molecules.

To produce PET, replace ethanoic acid with terephthalic acid:

\[ \ce{HOOC-C6H4-COOH} \]

Terephthalic acid has two carboxylic acid groups, allowing it to connect to two ethylene glycol molecules or growing chain sections.

Both monomers are now bifunctional, so repeated esterification can produce a long linear polyester.

The important distinction is not simply whether esterification occurs. It is whether the resulting molecules retain reactive groups that allow further chain growth.

Question 3

A different polyester is prepared using:

• Butane-1,4-diol: \(\ce{HO-(CH2)4-OH}\)
• Hexanedioic acid: \(\ce{HOOC-(CH2)4-COOH}\)

(a) Explain why the resulting polymer belongs to the polyester family.

(b) Write its repeating unit.

(c) Compare the expected flexibility of its molecular backbone with PET, assuming comparable chain lengths and broadly similar processing.

Solution 3

The polymer is a polyester because its backbone contains repeating ester linkages, even though it does not contain a benzene ring.

(a) Each molecule contains two reactive functional groups.

The hydroxyl groups of butane-1,4-diol react with the carboxylic acid groups of hexanedioic acid.

Each reaction forms an ester linkage and releases water.

The resulting polymer therefore contains repeating \(\ce{-C(=O)-O-}\) groups in its backbone.

(b) The repeating unit is:

\[ \left[ \ce{-O-(CH2)4-O-C(=O)-(CH2)4-C(=O)-} \right]_n \]

The butane-1,4-diol contributes the four-carbon section between the two oxygen atoms.

The hexanedioic acid contributes the four-carbon section between the two carbonyl groups.

(c) The new polyester is expected to have a more flexible backbone than PET, all else being comparable.

The new polymer contains flexible aliphatic carbon chains in both monomer-derived sections.

PET contains benzene rings, which are relatively rigid and restrict rotation within the backbone.

Replacing these aromatic sections with flexible aliphatic chains generally increases the ability of the polymer backbone to bend and rotate.

This does not mean flexibility is determined by one structural feature alone. Molecular mass, crystallinity, intermolecular forces, and processing also affect the final material.

Question 4

An industrial trial uses direct esterification to produce PET.

A receiver connected to the reactor collects 1.80 g of water.

Assume:

• Water is produced only through esterification.
• All produced water is collected.
• The molar mass of water is 18.0 g mol\(^{-1}\).

(a) Calculate the amount of water collected in moles.

(b) Determine the amount of ester linkages formed.

(c) Explain why this measurement alone cannot establish that high-molecular-mass PET has been produced.

Solution 4

The trial produces 0.100 mol of water, corresponding to 0.100 mol of esterification events. However, this does not establish the average length of the polymer chains.

(a) Calculate the amount of water.

Use:

\[ n=\frac{m}{M} \]

where \(n\) is the amount in moles, \(m\) is the mass in grams, and \(M\) is the molar mass in grams per mole.

Substitute:

\[ \begin{aligned} n(\ce{H2O})&=\frac{1.80}{18.0}\\ &=0.100\text{ mol} \end{aligned} \]

(b) Each esterification event between a carboxylic acid and an alcohol produces one water molecule.

Therefore:

\[ n(\text{ester linkages})=0.100\text{ mol} \]

(c) The amount of water measures the total number of esterification events, not how those linkages are distributed between polymer chains.

For example, the reaction could produce:

• A smaller number of long polymer chains.
• A larger number of short oligomers.

Both situations could produce the same total amount of water if the same number of esterification events occurred.

Further evidence, such as molecular mass measurements or suitable viscosity measurements, is required to assess polymer chain growth.

This distinction is important when interpreting experimental data. A measurement can confirm that a reaction has occurred without proving that the intended final product has been obtained.

Question 5

Two PET production trials use the following illustrative quantities.

Both trials contain 1.00 mol of terephthalic acid and 1.00 mol of ethane-1,2-diol.

• Trial A: Contains only these two bifunctional monomers.
• Trial B: Also contains 0.020 mol of benzoic acid, \(\ce{C6H5COOH}\), which has only one carboxylic acid group.

Both trials produce 1.98 mol of water through esterification.

Assume:

• Every esterification event joins two previously separate molecular fragments.
• No cyclic polymers or branching form.
• All benzoic acid molecules in Trial B become incorporated once as chain-end caps.
• All starting molecules are incorporated into the resulting linear molecular chains.
• Each ester linkage produces one water molecule.

(a) Explain why benzoic acid can act as a chain stopper.

(b) Calculate the amount of polymer chains formed in each trial.

(c) Calculate the average number of starting monomer molecules per chain in each trial.

(d) Determine which trial produces longer average chains, and explain why the identical water production does not mean the two trials produce equivalent polymers.

Solution 5

Trial A produces longer average chains. Although both trials form the same total number of ester linkages, Trial B contains additional monofunctional molecules that increase the number of separate chains.

(a) Benzoic acid acts as a chain stopper because it has only one reactive carboxylic acid group.

It can react with a hydroxyl group on a growing polyester chain to form an ester linkage.

However, once this reaction occurs, the benzoic acid-derived end has no second functional group available to connect to another monomer.

It therefore terminates that end of the chain.

(b) Calculate the amount of polymer chains.

Each ester linkage joins two previously separate molecular fragments.

Therefore:

\[ n(\text{chains})= n(\text{starting molecules}) - n(\text{ester linkages}) \]

For Trial A:

The initial amount of monomer molecules is:

\[ 1.00+1.00=2.00\text{ mol} \]

The amount of ester linkages formed is 1.98 mol.

Therefore:

\[ \begin{aligned} n(\text{chains})&=2.00-1.98\\ &=0.020\text{ mol} \end{aligned} \]

For Trial B:

The initial amount of monomer molecules is:

\[ 1.00+1.00+0.020=2.020\text{ mol} \]

Therefore:

\[ \begin{aligned} n(\text{chains})&=2.020-1.98\\ &=0.040\text{ mol} \end{aligned} \]

(c) Calculate the average number of starting monomer molecules per chain.

Use:

\[ \text{Average molecules per chain} = \frac{\text{Initial amount of molecules}} {\text{Amount of chains}} \]

For Trial A:

\[ \frac{2.00}{0.020}=100 \]

The average chain contains 100 starting monomer molecules.

For Trial B:

\[ \frac{2.020}{0.040}=50.5 \]

The average chain contains 50.5 starting monomer molecules.

(d) Trial A produces longer average chains, with approximately twice as many starting monomer molecules per chain.

The tempting prediction is that both trials should produce polymers of similar length because both release the same amount of water.

However, water production measures the number of esterification events, not the number of polymer chains.

Trial B starts with additional monofunctional molecules. These can attach to growing chains but cannot connect onwards to another monomer.

Consequently, the same number of ester linkages is distributed across more separate molecular chains.

This is why the average chain length is lower in Trial B.

The deeper lesson is that polymer formation depends on both the extent of reaction and the functionality and proportions of the monomers. Counting ester linkages alone is not enough to predict the final molecular structure.

10The key idea to carry into your next topic

A polyester is identified by its repeated ester linkages in the polymer backbone. PET is one specific example, formed in the idealised HSC model from ethane-1,2-diol and terephthalic acid through condensation polymerisation.

When analysing an unfamiliar polyester, focus on three decisions:

1. Can you identify the ester linkage?
2. Can you trace the monomer-derived sections and explain how they connect?
3. Can you relate the molecular structure to the material’s properties and uses?

The next useful comparison is with polyamides. Replacing an ester linkage with an amide linkage changes the intermolecular forces between polymer chains, particularly their ability to form hydrogen bonds. This helps explain why two condensation polymers, such as PET and nylon, can behave quite differently.

Continue with How to Compare Hydrogen Bonding in Polyamides and Polyesters.