Why Nylon Is Strong but Still Flexible: HSC Chemistry

A nylon cord can carry a heavy load while curving around a reel without snapping. If strong materials had to be rigid, that combination would seem impossible.

Nylon is strong because its amide groups form numerous hydrogen bonds between neighbouring polymer chains. These attractions help hold the chains together and resist them being pulled apart. It can still bend because sections of its carbon backbone can rotate, and less ordered regions of the polymer allow the chains to move and rearrange.

The important distinction is that strength and flexibility are not opposites. They describe different ways a material responds to an applied force.

01Strength and flexibility describe different properties

Before looking at the molecular structure, consider two ways of testing a nylon fishing line.

First, pull it from both ends. You need a substantial force to break it.

Now bend it around a smooth reel. The line curves without snapping.

These tests measure different properties.

Tensile strength is not simply the force needed to break an object. It accounts for the object’s cross-sectional area:

\[ \text{Tensile strength}= \frac{\text{Breaking force}}{\text{Cross-sectional area}} \]

A thick nylon cord can withstand a greater breaking force than a thin one simply because it contains more material.

Stiffness is different. A stiff material resists changing its shape, whereas a flexible material can change shape relatively easily.

Predict this: If nylon contains strong attractions between neighbouring chains, shouldn’t those chains be locked together so tightly that the material cannot bend?

It sounds reasonable, but it assumes that the attractions are permanent chemical cross-links and that every part of the polymer chain is rigid. Neither assumption is correct.

To understand why, we need to look at the structure of nylon itself.

01Start with the chemistry of nylon

Nylon is a synthetic polymer belonging to a group called polyamides.

A useful example for HSC Chemistry is nylon-6,6, which forms through condensation polymerisation between two monomers:

• Hexane-1,6-diamine, containing two amine groups.
• Hexanedioic acid (adipic acid), containing two carboxylic acid groups.

Their condensed structural formulas are:

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

and

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

When an amine group reacts with a carboxylic acid group, an amide link forms and water is released.

The characteristic amide linkage is:

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

The resulting nylon-6,6 repeating unit can be represented as:

\[ \left[ \ce{-NH-(CH2)6-NH-C(=O)-(CH2)4-C(=O)-} \right]_n \]

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

Each repeating unit contains two amide links. These are part of the main covalent polymer backbone.

If you need to revise how the monomers join or how to draw this repeating unit, see Condensation Polymerisation: Nylon and Polyester Explained for HSC Chemistry and How to Draw the Repeating Unit of Nylon-6,6.

Notice something important about the repeating unit.

It contains two different structural features:

1. Amide groups, which create strong attractions between neighbouring polymer chains.
2. Methylene sections, \(\ce{-(CH2)-}\), which contain carbon-carbon single bonds that can rotate.

These two features help explain why nylon can be both strong and flexible.

02Why hydrogen bonding makes nylon strong

Look closely at an amide linkage:

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

It contains two particularly important parts:

• The N-H bond, which provides a hydrogen-bond donor.
• The carbonyl group, whose oxygen provides a hydrogen-bond acceptor.

The reason these groups attract each other comes down to electron distribution.

The nitrogen-hydrogen bond is polar, leaving the hydrogen partially positive. The carbonyl oxygen is partially negative because it attracts electrons strongly.

Consequently, the hydrogen attached to nitrogen in one nylon chain can form a hydrogen bond with the carbonyl oxygen of a neighbouring chain.

This interaction is represented as:

\[ \ce{N-H\bond{...}O=C} \]

These are intermolecular attractions because they occur between separate polymer chains.

Why does this increase strength?

Imagine several nylon chains arranged alongside one another.

Without significant attractions between them, the chains would have a greater tendency to separate or slide past each other when a force was applied.

Hydrogen bonding helps prevent this.

There are many amide groups along each chain, so numerous hydrogen bonds can form between neighbouring chains. Although an individual hydrogen bond is weaker than a covalent bond, the combined effect of many interactions is substantial.

This contributes to:

• High tensile strength.
• Resistance to chains sliding apart.
• Durability when used as a fibre.
• Resistance to abrasion.

There are also dipole-dipole attractions and London dispersion forces between the chains, but hydrogen bonding is particularly important in nylon.

One distinction is worth remembering for HSC Chemistry:

The covalent bonds hold each individual polymer chain together. Hydrogen bonds help hold neighbouring polymer chains together.

Hydrogen bonds do not turn nylon into a permanently cross-linked covalent network. They provide strong intermolecular attraction without preventing all molecular movement.

That last point is essential to understanding flexibility.

03How scientists investigated the arrangement of nylon chains

Knowing that hydrogen bonding is possible does not automatically prove how the chains are arranged inside a solid nylon fibre.

How could scientists investigate a structure far too small to inspect directly?

An important technique was X-ray diffraction.

The problem: finding the arrangement of atoms

In 1947, Charles Bunn and E. V. Garner published an important investigation of the crystal structures of nylon-6,6 and nylon-6,10.

They interpreted X-ray diffraction patterns obtained from nylon fibres and sheets to develop structural models of the crystalline regions.

Their work helped establish how neighbouring polyamide chains pack together and how their amide groups can form hydrogen bonds.

The original paper is available in the Proceedings of the Royal Society A.

How does the apparatus work?

Consider the arrangement of an X-ray diffraction experiment.

1. An X-ray source produces a beam of X-rays.
2. A collimator narrows the beam so it travels in a controlled direction.
3. A nylon fibre is positioned in the path of the beam.
4. The X-rays interact with the electrons in the nylon.
5. Some X-rays are scattered in different directions.
6. A detector or photographic film behind the sample records the resulting diffraction pattern.

What should the experiment show?

Before looking at the result, consider two possible structures.

If the polymer chains are arranged in regularly repeating crystalline regions, the scattered X-rays should reinforce one another at particular angles. This produces distinct diffraction features.

If the chains are arranged randomly, there is less long-range order, producing more diffuse scattering.

What would you expect from nylon?

Nylon is semicrystalline. This means it contains both:

• Crystalline regions, where chains are arranged in an ordered structure.
• Amorphous regions, where chains are arranged less regularly.

Therefore, its diffraction pattern contains evidence of both ordered and disordered regions.

What did the observations establish?

The characteristic diffraction patterns provided evidence that nylon contained regularly arranged crystalline regions.

The structural models developed from these patterns showed how neighbouring polyamide chains could form hydrogen-bonded arrangements.

This supported the explanation that hydrogen bonding contributes to the strength and stability of nylon fibres.

However, there is an important limitation.

X-ray diffraction reveals information about the average arrangement of atoms and the repeating distances within ordered regions. It does not directly show how individual polymer chains move when a fibre bends.

To investigate flexibility, we need to combine structural evidence with mechanical testing.

A complementary investigation would use nylon samples of controlled dimensions and moisture content.

One sample could be pulled lengthways in a tensile-testing machine, while another is placed across two supports and loaded at its midpoint in a bending test.

The tensile test measures the stress needed to break the material. The bending test measures how much it deflects under an applied load.

Combining these measurements with X-ray diffraction would help connect the arrangement of polymer chains to their mechanical behaviour.

04Why nylon can still bend

We now know why neighbouring nylon chains attract one another.

But if these attractions are strong, what allows the material to change shape?

The answer involves two features of the polymer structure.

Flexible sections within the polymer backbone

Look again at the nylon-6,6 repeating unit.

Between the amide groups are sections containing several \(\ce{-CH2-}\) groups.

These methylene sections contain carbon-carbon single bonds.

Unlike a rigid double bond, a carbon-carbon single bond can rotate, allowing the polymer chain to adopt different conformations.

This means that the backbone is not completely rigid.

There is one important qualification: the amide group itself is relatively planar because the carbon-nitrogen bond has partial double-bond character due to electron delocalisation.

It does not rotate freely in the same way as an ordinary carbon-carbon single bond.

Much of the flexibility therefore comes from the methylene sections between the amide groups.

Nylon is semicrystalline, not completely crystalline

There is another important reason nylon can bend.

A nylon fibre is not one enormous, perfectly ordered crystal.

It contains a mixture of crystalline and amorphous regions.

The crystalline regions help the material resist stretching and breaking.

The amorphous regions allow sections of the polymer chains to move and rearrange when a force is applied.

Some polymer chains also connect different crystalline regions. These are often called tie molecules, and they help maintain the overall structure.

What happens when you bend nylon?

Consider a nylon fibre being bent into a curve.

The outside of the bend experiences tension, while the inside experiences compression.

The polymer accommodates this change through several molecular processes:

1. Carbon-carbon single bonds rotate within the methylene sections.
2. The amorphous regions allow chain segments to change their arrangements.
3. Neighbouring chains can undergo limited relative movement.
4. Some hydrogen bonds may become distorted or rearranged as the chains respond to the applied force.

The important point is that the covalent polymer backbone does not need to break for the material to bend.

The hydrogen-bonded crystalline regions continue to provide structural support, while the more mobile sections accommodate changes in shape.

This gives us the central explanation:

• Hydrogen bonding helps nylon resist being pulled apart.
• Flexible carbon-chain sections allow changes in molecular conformation.
• Amorphous regions allow molecular rearrangement.
• Crystalline regions help the material retain strength.

Nylon can therefore have high tensile strength without being completely rigid.

Does more crystallinity always mean more flexibility?

No.

Increasing crystallinity generally increases the amount of ordered material and restricts the movement of polymer segments.

Drawing a nylon fibre can also align its polymer chains along the fibre axis. This improves the ability of the chains to share an applied tensile load.

As a result, a more highly drawn nylon fibre will often have greater tensile strength and stiffness.

However, the material can still bend because some molecular movement remains possible, particularly in its amorphous regions.

The final properties depend on the balance between:

• Hydrogen bonding.
• Chain flexibility.
• Crystallinity.
• Molecular orientation.
• Moisture content.
• Temperature and processing conditions.

This is why two samples of nylon with the same repeating unit can behave quite differently.

05A useful contrast: nylon versus PET polyester

Comparing nylon with polyester helps expose another common misconception.

Both nylon and PET are condensation polymers, but their repeating units contain different functional groups.

Nylon contains both a hydrogen-bond donor and acceptor within its repeating amide linkages.

PET contains ester groups instead. Its repeating unit does not contain the N-H donor needed to form the same repeated amide-to-amide hydrogen-bonding arrangement.

However, this does not automatically mean PET must be more flexible.

Why not?

PET contains a benzene ring in its polymer backbone. This relatively rigid structural section restricts some molecular movements.

Its final properties also depend on chain orientation, crystallinity, molecular mass, and processing.

The lesson is that you cannot predict a polymer’s flexibility simply by identifying its strongest intermolecular force.

You must consider both the attractions between chains and the flexibility of the chains themselves.

08Questions and solutions

Question 1

Nylon-6,6 contains repeating amide linkages, \(\ce{-C(=O)-NH-}\).

Explain how these linkages contribute to the high tensile strength of nylon.

Distinguish between the bonds within an individual polymer chain and the attractions between neighbouring chains.

Solution 1

The covalent bonds hold each individual nylon chain together, while hydrogen bonds between neighbouring chains help give nylon its high tensile strength.

The amide linkage contains a polar N-H bond and a carbonyl group.

The hydrogen attached to nitrogen can form a hydrogen bond with the carbonyl oxygen of an adjacent polymer chain.

These interactions occur repeatedly along neighbouring chains.

Although an individual hydrogen bond is weaker than a covalent bond, the combined effect of many hydrogen bonds helps resist separation and sliding between polymer chains.

The covalent bonds maintain the continuous polymer backbone, while the intermolecular hydrogen bonds contribute to the cohesion between different chains.

This distinction explains why nylon can be strong without having covalent cross-links joining every neighbouring chain.

Question 2

A drawn nylon-6,6 fibre is examined using X-ray diffraction.

Its diffraction pattern contains sharp arcs superimposed on a diffuse background.

The same polymer is then examined as a less-oriented film. The diffraction pattern contains ring-like reflections at similar positions.

Explain:

a) What the sharp arcs indicate about the drawn fibre.

b) What the diffuse background indicates.

c) Why the less-oriented film produces ring-like reflections.

d) Why neither diffraction pattern is sufficient to determine how easily the nylon bends.

Solution 2

The sharp arcs indicate ordered crystalline regions with a preferred molecular orientation, while the diffuse background indicates less ordered amorphous regions.

a) Sharp arcs

The sharp diffraction features arise from regularly arranged atoms within crystalline regions.

Their arc-like appearance indicates that the crystalline regions have a preferred orientation.

In a drawn fibre, many polymer chains tend to align along the fibre axis, producing an oriented structure.

b) Diffuse background

The diffuse scattering comes from regions without the same long-range repeating order.

This is evidence of amorphous material within the semicrystalline polymer.

The nylon therefore contains both ordered and disordered regions.

c) Ring-like reflections

The less-oriented film still contains crystalline regions, but their orientations are distributed more randomly.

Consequently, the diffraction features appear as rings rather than strongly oriented arcs.

The similar positions of the reflections indicate similar repeating structural spacings, even though the overall orientation differs.

d) Limitation of the experiment

X-ray diffraction provides information about molecular arrangement and crystallinity, but it does not directly measure how easily the polymer chains move during bending.

A material may contain substantial crystalline order and still have mobile amorphous regions.

To determine its bending behaviour, mechanical testing is required.

The diffraction evidence must therefore be combined with information about chain flexibility and the response of the polymer to applied forces.

Question 3

Two circular nylon cords are tested to determine their tensile strengths.

The following results are obtained:

Both cords are made from the same grade of nylon and are tested under identical conditions.

a) Calculate the cross-sectional area of each cord.

b) Calculate the tensile strength of each cord in MPa.

c) Identify which cord has the greater tensile strength.

d) Explain why comparing breaking forces alone would produce a misleading conclusion.

Solution 3

Cord A has the greater tensile strength, approximately \(159\ \mathrm{MPa}\), compared with approximately \(133\ \mathrm{MPa}\) for Cord B.

Although Cord B withstands a greater breaking force, it also has a larger cross-sectional area.

a) Calculate the cross-sectional areas.

For a circular cord:

\[ A=\pi r^2 \]

where \(r\) is the radius.

For Cord A:

\[ r_A=\frac{0.80}{2}=0.40\ \mathrm{mm} \]

\[ A_A=\pi(0.40)^2 \]

\[ \boxed{A_A=0.503\ \mathrm{mm^2}} \]

For Cord B:

\[ r_B=\frac{1.20}{2}=0.60\ \mathrm{mm} \]

\[ A_B=\pi(0.60)^2 \]

\[ \boxed{A_B=1.131\ \mathrm{mm^2}} \]

b) Calculate the tensile strengths.

Use:

\[ \sigma=\frac{F}{A} \]

For Cord A:

\[ \sigma_A=\frac{80}{0.503} \]

\[ \sigma_A=159\ \mathrm{N/mm^2} \]

Since:

\[ 1\ \mathrm{N/mm^2}=1\ \mathrm{MPa} \]

\[ \boxed{\sigma_A=159\ \mathrm{MPa}} \]

For Cord B:

\[ \sigma_B=\frac{150}{1.131} \]

\[ \sigma_B=132.6\ \mathrm{N/mm^2} \]

\[ \boxed{\sigma_B=133\ \mathrm{MPa}} \]

c) Compare the results.

Cord A has the higher tensile strength because it withstands a greater stress before breaking.

d) Explain why breaking force alone is misleading.

Cord B breaks under a greater force, but it also has more material across its cross-section.

A thicker cord can withstand a greater total force without necessarily having a higher tensile strength.

Tensile strength accounts for the area over which the force is distributed.

This is why mechanical properties must be compared using appropriate measurements rather than simply comparing the forces recorded by a testing machine.

Question 4

PET polyester has the repeating unit:

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

A student makes the following claim:

“PET does not contain N-H bonds, so it cannot form the same repeated hydrogen bonds as nylon. Therefore, PET must be more flexible than nylon.”

Evaluate this claim using the molecular structures of both polymers.

Solution 4

The student’s conclusion is incorrect because the absence of repeated amide hydrogen bonding does not automatically make PET more flexible than nylon.

It is correct that nylon and PET have different intermolecular interactions.

Nylon contains amide groups:

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

These provide both hydrogen-bond donors and acceptors, allowing repeated hydrogen bonding between neighbouring polymer chains.

PET contains ester groups:

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

Its repeating unit does not contain an N-H hydrogen-bond donor, so it cannot form the same repeated amide-to-amide hydrogen-bonding arrangement.

However, this is only one part of the explanation.

PET contains benzene rings within its polymer backbone. These relatively rigid sections restrict some rotational movement along the chain.

Nylon, in contrast, contains flexible methylene sections between its amide groups.

Both polymers also experience other intermolecular attractions, and their properties depend on crystallinity, molecular orientation, chain length, and processing.

Therefore, the presence or absence of hydrogen bonding alone is insufficient to predict which polymer will bend more easily.

A complete comparison must consider both the intermolecular forces and the flexibility of the polymer backbone.

Question 5

Two nylon-6,6 strips are manufactured using different processing conditions.

They have identical dimensions, molecular mass characteristics, moisture content, and temperature during testing.

Their internal structures are different:

• Strip A has a crystallinity of 55%, but relatively low molecular orientation along its length.
• Strip B has a crystallinity of 40%, but a high degree of molecular orientation along its length.

Both strips have a cross-sectional area of \(2.0\ \mathrm{mm^2}\).

The following illustrative mechanical test results are recorded:

The bending tests use identical support spans and loading arrangements.

A student concludes:

“Strip B must be less flexible because it has the greater tensile strength. Also, Strip A must have the greater tensile strength because it has more crystalline material.”

Evaluate both statements.

Your response should:

a) Calculate the tensile strength of both strips.

b) Identify which strip is more resistant to bending under the stated test conditions.

c) Explain how Strip B can have greater tensile strength despite having lower crystallinity.

d) Explain why the results do not establish that crystallinity alone determines bending stiffness.

e) Propose a further investigation that would help distinguish the effects of molecular orientation and crystallinity.

Solution 5

Strip B has the greater tensile strength, while Strip A is more resistant to bending under the specified test conditions.

The result demonstrates that molecular orientation and crystallinity influence different aspects of a polymer’s mechanical behaviour. It also shows why tensile strength cannot be used as a direct measure of flexibility.

a) Calculate the tensile strengths.

Use:

\[ \sigma=\frac{F}{A} \]

For Strip A:

\[ \sigma_A=\frac{160}{2.0} \]

\[ \boxed{\sigma_A=80\ \mathrm{MPa}} \]

For Strip B:

\[ \sigma_B=\frac{220}{2.0} \]

\[ \boxed{\sigma_B=110\ \mathrm{MPa}} \]

Strip B has the greater tensile strength.

b) Identify which strip is more resistant to bending.

Both strips experience the same central load and have identical dimensions and support arrangements.

However, their measured deflections are different:

• Strip A deflects by 4.0 mm.
• Strip B deflects by 7.0 mm.

Strip A therefore has greater bending stiffness under these particular test conditions.

Strip B is more compliant in bending because it undergoes a greater deflection under the same applied load.

c) Explain why Strip B has greater tensile strength despite lower crystallinity.

The important factor is molecular orientation.

In Strip B, a greater proportion of the polymer chains are aligned along the direction in which the tensile force is applied.

This alignment helps the chains share the applied load more effectively.

The hydrogen-bonded regions also help hold neighbouring chains together, allowing the material to resist separation.

Although Strip B has a lower overall crystalline fraction, its greater molecular orientation improves its ability to withstand tensile stress along its length.

This demonstrates that crystallinity is not the only factor determining tensile strength.

d) Explain the bending results and their limitations.

Strip A has a higher crystalline fraction.

Its greater proportion of ordered regions can restrict molecular movement and contribute to greater resistance to bending.

Strip B has a larger proportion of amorphous material, which can allow greater molecular rearrangement under an applied load.

This provides a possible explanation for why Strip B undergoes a greater bending deflection.

However, there is an important limitation.

Molecular orientation also changes between the two samples. Both crystallinity and orientation have changed simultaneously.

Consequently, we cannot conclude that crystallinity alone caused the difference in bending behaviour.

The measured deflections establish which strip is more resistant to bending in this particular experiment, but they do not isolate the contribution of each structural factor.

This is the subtle distinction: a polymer’s mechanical properties result from several interacting structural features, not one property acting independently.

e) Propose a further investigation.

A more controlled investigation would prepare two additional sets of nylon-6,6 samples.

In the first set, keep molecular orientation approximately constant while varying crystallinity.

In the second set, keep crystallinity approximately constant while varying molecular orientation.

For every sample, control:

• Cross-sectional dimensions.
• Moisture content.
• Temperature.
• Molecular mass characteristics.
• Tensile and bending test conditions.

Use X-ray diffraction to investigate the crystalline structure and molecular orientation.

Then perform tensile and bending tests to measure the mechanical response.

Comparing the results would help determine how much each structural feature contributes to tensile strength and bending stiffness.

The key conclusion is that a polymer can have high tensile strength without having equally high resistance to bending. Molecular orientation helps determine how effectively the chains carry a load, while crystallinity and the mobility of amorphous regions also influence how readily the material changes shape.

09The key idea to carry into your next polymer question

When explaining the properties of nylon, separate three levels of structure:

1. Chemical linkage: Amide groups allow strong hydrogen bonding between neighbouring chains.
2. Polymer backbone: Methylene sections allow rotation and changes in chain conformation.
3. Overall arrangement: Crystallinity and molecular orientation influence strength, stiffness, and flexibility.

A strong HSC Chemistry explanation connects all three.

If a question asks why a polymer is strong, identify the relevant intermolecular forces. If it asks why the polymer can bend, examine the flexibility of its backbone and the arrangement of its crystalline and amorphous regions.

That same structure-property approach is useful when moving on to polyesters, where changing the functional linkage and backbone structure changes the intermolecular interactions and mechanical behaviour.