Polyamides: How Nylon Forms, Its Structure and Properties

A nylon fishing line can be surprisingly thin while still carrying a substantial load. What stops it from stretching apart or snapping? The answer is not simply that its individual polymer chains are strong. The chains also attract one another.

Polyamides are polymers containing repeating amide links, \(-\mathrm{C(=O)-NH}-\), in their main chain. These links allow neighbouring polymer chains to form hydrogen bonds, producing strong, durable materials. Nylon is the best-known synthetic example.

Most importantly, polyamides can be formed through condensation polymerisation. Nylon 6,6, for example, is made by reacting a diamine with a dicarboxylic acid, eliminating water as the polymer forms. Nylon 6 provides an interesting contrast because it is commonly manufactured through ring-opening polymerisation instead.

01What makes a polymer a polyamide?

The easiest way to recognise a polyamide is to examine its repeating unit.

An amide functional group contains a carbonyl group directly bonded to a nitrogen atom:

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

The carbonyl group contains a carbon atom double-bonded to oxygen, while the nitrogen is bonded to hydrogen and another part of the carbon chain.

When this amide linkage repeats throughout a polymer’s backbone, the material is classified as a polyamide.

The name is quite literal: poly means many, and amide refers to the functional group connecting the repeating sections.

Nylon 6,6: the classic example

Nylon 6,6 is formed from two monomers:

• Hexamethylenediamine, also called hexane-1,6-diamine: \(\ce{H2N-(CH2)6-NH2}\)
• Adipic acid, also called hexanedioic acid: \(\ce{HOOC-(CH2)4-COOH}\)

Notice something important about both molecules.

Hexamethylenediamine contains two amine groups, one at each end. Adipic acid contains two carboxylic acid groups, also one at each end.

This makes both monomers bifunctional, meaning each molecule has two reactive functional groups.

Because both ends can react, the monomers can connect repeatedly to form long polymer chains.

The repeating unit of nylon 6,6 is:

\[ \boxed{ \left[ -\mathrm{NH}-(\mathrm{CH_2})_6- \mathrm{NH}-\mathrm{C(=O)}- (\mathrm{CH_2})_4-\mathrm{C(=O)}- \right]_n } \]

Here, \(n\) represents the number of repeating units in the polymer chain.

The name nylon 6,6 also tells you something about its monomers. Each starting molecule contains six carbon atoms.

The first six refers to the six carbon atoms in hexamethylenediamine. The second six refers to the six carbon atoms in adipic acid, including its two carboxyl carbon atoms.

This naming convention becomes useful when comparing different nylons.

02How are polyamides formed? The development of nylon

The development of nylon provides a particularly useful example of how chemists connect molecular structure with the properties of a new material.

In the late 1920s, American chemist Wallace Hume Carothers began investigating whether long polymer molecules could be constructed by joining smaller organic molecules together.

At the time, the structure of polymers was still being debated. Some scientists questioned whether polymers were genuinely enormous molecules held together by covalent bonds, or aggregates of smaller molecules.

Carothers approached the problem by constructing polymers from molecules with reactive groups at both ends.

His work at DuPont eventually led to the development of nylon.

The original problem: producing a useful synthetic fibre

Silk was an important natural fibre, but chemists wanted to develop synthetic alternatives with useful mechanical properties.

Carothers initially investigated reactions between dicarboxylic acids and diols.

A dicarboxylic acid contains two \(-\mathrm{COOH}\) groups, while a diol contains two \(-\mathrm{OH}\) groups.

When these molecules react, they form ester linkages and release water. Repeating the reaction produces a polyester.

The difficulty was achieving sufficiently long polymer chains.

Water was produced during the reaction. If too much remained in the reaction mixture, it limited further condensation and restricted the molecular mass of the polymer.

What would happen if the water could be continuously removed?

Removing a reaction product favours the forward condensation reaction. This helps the monomers continue joining together rather than allowing the reverse reaction to limit chain growth.

The apparatus and the first successful approach

Carothers’ team used a laboratory device called a molecular still to remove water during its early polyester experiments.

The arrangement can be understood by following the substances through the apparatus:

1. A heated reaction vessel contained the dicarboxylic acid and diol.
2. As the reaction proceeded, water was produced.
3. The volatile water was removed from the reaction mixture through the distillation system.
4. A receiver collected the removed material, while the increasingly viscous polymer remained in the heated vessel.
5. Reduced pressure assisted the removal of volatile products.

In April 1930, Carothers’ associate Julian Hill produced a higher-molecular-mass polyester using the molecular still.

Hill observed that the hot polymer could be pulled into a thin filament. After cooling, the filament could be stretched further, producing a stronger and more elastic fibre.

This additional stretching process became known as cold drawing.

However, the polyester fibres still had significant limitations. Their melting points were too low, and they were sensitive to many organic solvents.

The experiment had demonstrated that synthetic polymer fibres were possible, but the materials were not yet suitable for the intended applications.

The revised approach: making polyamides

Carothers’ team began investigating reactions between dicarboxylic acids and diamines.

Instead of forming ester linkages, these reactions formed amide linkages.

On 28 February 1935, Gerard Berchet, working within Carothers’ research group, prepared nylon 6,6 using hexamethylenediamine and adipic acid.

The reported preparation involved heating the reactants in meta-cresol. Water distilled off during the reaction, and the temperature was subsequently raised to approximately \(255\text{-}260\,^\circ\mathrm{C}\) to remove the cresol under vacuum.

The resulting polymer could be formed into fibres.

After cold drawing, these fibres were strong and elastic, with a high melting point and resistance to many solvents.

DuPont selected the material for full-scale production in July 1935.

The American Chemical Society’s historical account documents the development of the molecular still, the early polyester experiments, and the eventual selection of nylon 6,6. A complementary account of the 1935 synthesis is provided by the University of Birmingham.

What did the experiments actually establish?

It is useful to separate the historical observations from the modern chemical explanation.

The experiments provided evidence for the formation of very large molecules connected by chemical bonds.

Modern chemistry explains the particularly useful properties of nylon through the polar amide groups and the hydrogen bonding between neighbouring chains.

Importantly, the original fibre experiments did not directly demonstrate individual hydrogen bonds. That is the modern structural explanation of the observed properties.

03The chemistry of condensation polymerisation

Let’s now examine exactly how the nylon 6,6 chain forms.

An amide linkage forms when a carboxylic acid reacts with an amine.

The general reaction is:

\[ \ce{R-COOH + H2N-R' -> R-CONH-R' + H2O} \]

The OH group from the carboxylic acid and one hydrogen from the amine combine to form water.

The remaining atoms form the amide linkage.

This process can happen repeatedly because the nylon monomers each have two reactive ends.

For nylon 6,6:

• Hexamethylenediamine supplies the two amine groups.
• Adipic acid supplies the two carboxylic acid groups.
• Amide links connect the monomers.
• Water is released as a by-product.

The resulting polymer contains alternating sections derived from the two monomers.

How much water is produced?

Each individual amide linkage forms one water molecule.

An idealised nylon 6,6 repeat unit contains two amide linkages, so its repeat-unit calculation accounts for the removal of two water molecules.

There is a small but useful distinction when considering a finite polymer chain.

If one linear chain contains \(n\) diamine molecules and \(n\) dicarboxylic acid molecules, it contains \(2n-1\) connections between monomer molecules.

Therefore, it eliminates \(2n-1\) water molecules.

The chain retains two terminal functional groups, one at each end.

For very long chains, the effect of these terminal groups becomes negligible. This is why calculations using the idealised repeat-unit mass are usually appropriate.

Why does each monomer need two functional groups?

Consider what would happen if one monomer only had one reactive functional group.

It could connect to another molecule, but it could not continue extending the chain from both ends.

It would effectively terminate chain growth.

This is why bifunctional monomers are essential for producing the long, linear nylon 6,6 chains used in fibres.

It also explains why an imbalance in the quantities of the two monomers can restrict the final polymer chain length.

04Why do polyamides have useful properties?

The amide linkage does more than connect monomers.

It also determines how neighbouring polymer chains interact.

This is the key to understanding why nylon is useful as a fibre and as an engineering material.

Hydrogen bonding between polymer chains

Look again at the amide group:

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

The N-H bond is polar, and the carbonyl oxygen has a partial negative charge.

This allows hydrogen bonding between neighbouring polymer chains.

The hydrogen attached to nitrogen in one chain can be attracted to the carbonyl oxygen in another chain:

\[ \mathrm{N-H\cdots O=C} \]

The N-H group acts as the hydrogen-bond donor, while the carbonyl oxygen acts as the hydrogen-bond acceptor.

Think of two long ribbons with many small magnetic attractions distributed along their lengths.

Pulling one ribbon away from the other requires overcoming many attractions, rather than breaking just one.

The analogy is not exact: hydrogen bonds are directional electrostatic attractions, not actual magnets, and they can break and reform.

However, it captures an important idea. Many individual attractions between chains can collectively produce substantial cohesion.

There are two different types of bonding to distinguish:

Do not confuse these two.

Hydrogen bonds help hold the chains together, but they do not replace the covalent bonds forming the polymer backbone.

Cold drawing and molecular orientation

Hydrogen bonding is not the only reason nylon fibres are strong.

The arrangement of the polymer chains also matters.

During fibre production, molten nylon can be forced through tiny holes in a spinneret to form continuous filaments.

After cooling, the filaments are stretched.

This stretching aligns the polymer chains more closely with the direction of the fibre.

Why does this improve strength?

When the chains are more aligned:

• More polymer chains contribute to carrying a load along the fibre.
• The chains can pack more effectively in ordered regions.
• Interchain attractions help resist the movement of neighbouring chains.
• The material becomes stronger along the direction of drawing.

Nylon is therefore a semi-crystalline polymer. It contains both ordered crystalline regions and less ordered amorphous regions.

The crystalline regions provide structural order and strength, while the amorphous regions contribute to flexibility and toughness.

This combination helps explain why nylon can be strong without behaving like a completely rigid solid.

Important properties and their uses

The relationship between molecular structure and macroscopic properties is particularly important in HSC Chemistry.

The relationship between structure and properties is not identical for every polyamide. Different monomers, chain lengths, and processing conditions change the final material.

For example, a nylon fibre and a moulded nylon gear may have the same basic chemical family but different molecular orientation and physical properties.

Does nylon absorb water?

Yes. This is an important detail that is sometimes overlooked.

The polar amide groups attract water molecules, allowing nylon to absorb moisture from its surroundings.

Water can form hydrogen bonds with the amide groups and enter the less ordered amorphous regions of the polymer.

This can increase chain mobility, making the material more flexible and reducing its stiffness.

It can also cause swelling and dimensional changes.

The amount of water absorbed depends on the nylon type, temperature, humidity, shape, and processing conditions. Toray’s technical information on nylon describes this relationship between water absorption and dimensional change.

This means that nylon is not completely waterproof, despite its useful resistance to many chemicals and solvents.

05Nylon 6 versus nylon 6,6: an important distinction

There is a common misconception that all polyamides must be formed by reacting a diamine with a dicarboxylic acid.

Nylon 6 demonstrates why this is incorrect.

Both nylon 6 and nylon 6,6 contain amide linkages in their polymer backbones, but they can be manufactured using different polymerisation processes.

Nylon 6 is commonly produced by opening the ring of ε-caprolactam, a cyclic amide.

The opened molecules join to form a long chain containing repeating amide linkages.

The important distinction is that the polymer is classified according to its structure, not simply the reaction used to manufacture it.

Both materials are polyamides because both contain repeated amide linkages in their backbones.

How is a polyamide different from a polyester?

This is another useful comparison because both materials can be formed through condensation polymerisation.

The difference lies in the functional group connecting their monomers.

Both materials can form strong fibres.

However, polyamides have the additional repeated N-H···O=C hydrogen-bonding interactions between neighbouring chains.

This helps explain their characteristic combination of strength, toughness, and relatively high melting temperatures.

The same reasoning can be used to distinguish nylon from polypropylene.

Polypropylene has a predominantly non-polar carbon backbone and does not have the amide groups responsible for hydrogen bonding in nylon.

For a comparison of its formation, structure, and uses, see Polypropylene: Addition Polymerisation, Structure and Properties.

06Worked examples

07Questions and solutions

Question 1

A polymer has the following repeating unit:

\[ \left[ -\mathrm{NH}-(\mathrm{CH_2})_6- \mathrm{NH}-\mathrm{CO}- (\mathrm{CH_2})_4-\mathrm{CO}- \right]_n \]

Identify:

a. The two monomers used to produce this polymer.

b. The functional group connecting the monomers.

c. The small molecule eliminated during polymer formation.

Solution 1

The monomers are hexamethylenediamine and adipic acid. The polymer contains amide linkages, and water is eliminated during condensation polymerisation.

a. Identify the monomers

The section containing six methylene groups and two nitrogen-containing ends comes from hexamethylenediamine:

\[ \ce{H2N-(CH2)6-NH2} \]

The section containing four methylene groups between two carbonyl groups comes from adipic acid:

\[ \ce{HOOC-(CH2)4-COOH} \]

b. Identify the linkage

The repeating functional group is:

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

This is an amide linkage.

c. Identify the eliminated molecule

The reaction between a carboxylic acid and an amine eliminates water.

The OH from the carboxylic acid and an H from the amine form:

\[ \ce{H2O} \]

Therefore, nylon 6,6 is a condensation polyamide.

Question 2

A chemist prepares nylon 6,6 using:

• \(0.100\ \mathrm{mol}\) of hexamethylenediamine.
• \(0.0900\ \mathrm{mol}\) of adipic acid.

Assume that only linear chains form, all carboxylic acid groups react, and no side reactions occur.

a. Calculate the maximum amount, in moles, of amide linkages that can form.

b. Calculate the amount, in moles, of amino groups remaining unreacted.

c. Explain why the imbalance between the monomers limits the average polymer chain length.

Solution 2

A maximum of \(0.180\ \mathrm{mol}\) of amide linkages forms, leaving \(0.0200\ \mathrm{mol}\) of unreacted amino groups. The excess diamine limits chain growth because there are insufficient carboxylic acid groups to connect all the monomers into longer chains.

a. Calculate the maximum number of amide linkages

Each adipic acid molecule contains two carboxylic acid groups.

Therefore, the total amount of carboxylic acid groups is:

\[ n(\ce{COOH})=2(0.0900) \]

\[ =0.180\ \mathrm{mol} \]

Each amide linkage requires one carboxylic acid group and one amino group.

Since the carboxylic acid groups are limiting:

\[ \boxed{n(\text{amide links})=0.180\ \mathrm{mol}} \]

b. Calculate the remaining amino groups

Each hexamethylenediamine molecule contains two amino groups.

The initial amount of amino groups is:

\[ n(\ce{NH2})=2(0.100) \]

\[ =0.200\ \mathrm{mol} \]

Each amide linkage consumes one amino group.

Therefore:

\[ n(\ce{NH2}\text{ remaining}) =0.200-0.180 \]

\[ \boxed{0.0200\ \mathrm{mol}} \]

c. Explain the effect on chain length

The excess diamine leaves additional amino groups at the ends of the polymer chains.

Once all available carboxylic acid groups have reacted, these remaining amino groups cannot form further amide linkages.

This restricts further chain growth.

For high-molecular-mass condensation polymers, the two monomers must be supplied in very nearly equal quantities so that their reactive functional groups remain balanced.

Question 3

Two polymers are prepared using different methods.

Polymer A is made by reacting hexamethylenediamine with adipic acid.

Polymer B is made by opening the rings of ε-caprolactam molecules.

A student claims that Polymer B cannot be a polyamide because it is formed from only one monomer and does not eliminate water during polymerisation.

Evaluate this claim. Include the repeating unit of each polymer in your answer.

Solution 3

The student’s claim is incorrect. Both polymers are polyamides because both contain repeating amide linkages in their backbones, even though their monomers and polymerisation mechanisms differ.

Polymer A: Nylon 6,6

Nylon 6,6 is produced through condensation polymerisation.

Its monomers are:

\[ \ce{H2N-(CH2)6-NH2} \]

and

\[ \ce{HOOC-(CH2)4-COOH} \]

Water is eliminated as amide linkages form.

Its repeating unit is:

\[ \left[ -\mathrm{NH}-(\mathrm{CH_2})_6- \mathrm{NH}-\mathrm{CO}- (\mathrm{CH_2})_4-\mathrm{CO}- \right]_n \]

Polymer B: Nylon 6

Nylon 6 is commonly produced through ring-opening polymerisation of ε-caprolactam.

The cyclic monomer opens, allowing the molecules to connect into a long chain.

Its repeating unit is:

\[ \left[ -\mathrm{NH}-(\mathrm{CH_2})_5- \mathrm{CO}- \right]_n \]

No small molecule is eliminated during this ring-opening polymerisation.

The key distinction is that polymer classification depends on the chemical structure of the repeating unit, not whether one or two monomers were used.

Both polymers contain repeated amide linkages, so both are polyamides.

Question 4

During Carothers’ early research, polyester fibres could be pulled into filaments and cold-drawn to improve their strength and elasticity.

However, these early polyester materials had relatively low melting points and were sensitive to many organic solvents.

The research team subsequently developed nylon 6,6, which had a higher melting point and better resistance to many solvents.

Explain why cold drawing alone could not solve the limitations of the early polyester fibres. In your answer, compare the roles of molecular orientation, covalent bonding, and hydrogen bonding.

Solution 4

Cold drawing improved molecular orientation, but it could not change the chemical functional groups within the polyester backbone. The development of nylon introduced amide linkages, allowing additional hydrogen bonding between neighbouring chains and improving the material’s thermal and mechanical performance.

Step 1: Consider what cold drawing changes.

Cold drawing stretches the polymer filament, causing its chains to become more aligned.

This improves the ability of the chains to carry tensile loads along the fibre.

It can also improve molecular packing and orientation.

However, it does not change the chemical composition of the polymer.

Step 2: Consider the interactions in the early polyester fibres.

Polyester chains contain ester linkages:

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

These groups are polar and can contribute to dipole-dipole interactions.

However, the ester linkage does not contain an N-H hydrogen-bond donor.

Therefore, it cannot produce the same repeated amide-to-amide hydrogen-bonding network found in nylon.

Step 3: Explain the improvement provided by nylon.

Nylon contains amide linkages:

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

The N-H group can form hydrogen bonds with carbonyl oxygen atoms on neighbouring chains.

These additional attractions increase the cohesion between polymer chains.

They help restrict chain movement and contribute to nylon’s relatively high melting point and mechanical strength.

Step 4: Combine the effects.

Nylon’s useful fibre properties result from two complementary structural features:

• Hydrogen bonding increases attraction between neighbouring polymer chains.
• Cold drawing aligns the chains, allowing them to carry loads more effectively.

Therefore, changing the chemical structure and controlling the physical arrangement of the polymer chains both contributed to the development of a more useful synthetic fibre.

Question 5

Two nylon 6,6 samples are prepared from the same monomers and have the same average polymer chain length.

Only their processing conditions differ.

Sample A is strongly drawn and has an estimated crystalline fraction of 65%.

Sample B is less drawn and has an estimated crystalline fraction of 30%.

Both samples are initially dried and then exposed to identical humidity and temperature conditions.

The following illustrative data are obtained:

After drying, both samples return close to their original masses and recover most of their original tensile strength.

A separate portion of Sample B is exposed to prolonged treatment with hot dilute hydrochloric acid. It is then neutralised, washed, and dried.

This treated portion remains significantly weaker than the untreated material and has a lower solution viscosity when measured at the same polymer concentration and temperature.

Answer the following:

a. Calculate the percentage moisture uptake of each sample, using the initial dry mass as the reference.

b. A student argues that Sample A absorbs less water because it contains fewer amide groups. Evaluate this explanation.

c. Explain why absorbing water can temporarily soften nylon and why drying allows much of its original mechanical performance to return.

d. Explain why the hot-acid-treated sample remains weaker after drying. Identify the type of chemical bond affected and explain the significance of its lower solution viscosity.

e. Which sample would be the better starting point for an application requiring dimensional stability in humid conditions? Explain your choice and identify one limitation of your prediction.

Solution 5

Sample A absorbs 2.0% water, while Sample B absorbs 4.5%. The difference is explained by their molecular packing and the accessibility of their amide groups, rather than a difference in the number of amide groups in their chemical structures.

a. Calculate the percentage moisture uptake

Use:

\[ \%\text{ moisture uptake} = \frac{\text{mass gain}}{\text{initial dry mass}} \times100 \]

For Sample A:

\[ \begin{aligned} \%\text{ uptake} &=\frac{10.20-10.00}{10.00}\times100\\ &=\boxed{2.0\%} \end{aligned} \]

For Sample B:

\[ \begin{aligned} \%\text{ uptake} &=\frac{10.45-10.00}{10.00}\times100\\ &=\boxed{4.5\%} \end{aligned} \]

Sample B absorbs more than twice the percentage of water absorbed by Sample A.

b. Explain the difference in water absorption

The student’s explanation is incorrect.

Both samples are made from nylon 6,6 and have the same average chain length.

Therefore, they have approximately the same number of amide groups per unit mass.

The difference is how accessible these groups are to water molecules.

Sample A has a higher crystalline fraction. Its chains are more closely packed and organised, making it harder for water to penetrate these regions.

Sample B has a larger amorphous fraction, where polymer chains are less regularly packed.

Water can enter these regions more readily and interact with the polar amide groups.

This explains why Sample B absorbs more water despite having the same underlying chemical structure.

c. Explain the reversible softening of nylon

Water molecules can form hydrogen bonds with the amide groups in nylon.

They can interact with the carbonyl oxygen and the hydrogen attached to nitrogen.

These water molecules compete with some of the hydrogen bonds that normally attract neighbouring nylon chains.

As a result:

• Some interchain hydrogen-bonding interactions are weakened.
• Polymer chains gain greater mobility.
• The material becomes more flexible.
• Swelling and changes in mechanical properties can occur.

This is called plasticisation.

When the sample dries, water molecules leave the polymer.

Many of the original polymer-to-polymer hydrogen bonds can then reform, allowing much of the original stiffness and strength to return.

The important distinction is that ordinary moisture absorption does not necessarily break the covalent polymer backbone.

d. Explain the irreversible weakening after acid treatment

The hot-acid-treated sample undergoes a different process.

Under sufficiently severe conditions, acid can promote hydrolysis of the amide linkages.

Hydrolysis breaks the covalent bonds connecting sections of the polymer backbone.

A simplified representation is:

\[ \ce{ R-CONH-R' + H2O + H+ -> R-COOH + R'-NH3+ } \]

This reaction produces carboxylic acid and protonated amine end groups.

The original long polymer chains are broken into shorter chains.

Shorter polymer chains have lower average molecular masses and generally produce lower solution viscosities under comparable measurement conditions.

Drying cannot reconnect the broken covalent bonds.

This explains why the acid-treated material remains weaker after drying.

The distinction is fundamental:

• Moisture plasticisation mainly changes interactions between existing polymer chains.
• Hydrolysis chemically breaks the covalent bonds within the polymer chains.

Hydrolysis under severe conditions is an extension of the core HSC structure-property discussion, but it provides a useful explanation of how polyamides can undergo chemical degradation.

e. Choose the better material for dimensional stability

Sample A is the better starting point.

It absorbs less water, so it is expected to undergo less moisture-induced swelling and dimensional change.

Its higher crystalline fraction also restricts the movement of polymer chains.

However, this prediction has a limitation.

Strong drawing can produce anisotropic properties, meaning the material may behave differently along and across the fibre direction. The performance of an actual engineering component would also depend on its dimensions, manufacturing conditions, and operating environment.

Therefore, Sample A appears more suitable based on the supplied data, but further testing would be needed before selecting a commercial material.

08Connecting polyamides to the wider study of polymers

The most important idea to take from polyamides is that a polymer’s properties depend on more than the strength of its individual covalent bonds.

The functional groups in its backbone influence the attractions between neighbouring chains. The arrangement of those chains then determines how effectively the material can resist stretching, heating, and other forms of physical stress.

For nylon 6,6, the key relationships are:

• Diamine and dicarboxylic acid monomers form a polymer through condensation polymerisation.
• Amide linkages connect the monomers and release water.
• Hydrogen bonding between neighbouring chains contributes to strength and thermal resistance.
• Molecular orientation and crystallinity influence the mechanical properties of the finished fibre.
• Polar amide groups also allow moisture absorption, which can affect flexibility and dimensional stability.

For your next comparison, focus on the difference between amide and ester linkages. Both can form condensation polymers, but their different functional groups produce different intermolecular interactions.

That is the connection that allows you to move beyond memorising examples and start predicting the properties of an unfamiliar polymer from its repeating structure.

09Sources and further reading

• NSW Education Standards Authority: Chemistry Stage 6 Syllabus (2017)
• NSW Education Standards Authority: Chemistry 11-12 Syllabus (2025)
• American Chemical Society: Wallace Carothers and the Development of Nylon
• University of Birmingham: Nylon
• The Open University: Introduction to Polymers – Polymer Families
• Toray: Physical Properties of Nylon

NSW syllabus note: The Chemistry Stage 6 Syllabus (2017), which applies to students completing the HSC in 2026 and 2027, includes condensation polymers such as nylon and polyesters in its Module 7 polymers content. The Chemistry 11-12 Syllabus (2025) is scheduled for implementation from 2028, with its first HSC examination in 2029. Students should check the syllabus applicable to their cohort.