Why Polyamide Chains Form Hydrogen Bonds with Each Other

A nylon thread can be pulled hard without its polymer chains immediately sliding past one another. But those chains aren’t joined to their neighbours by permanent covalent bonds. So what helps hold them together?

Polyamide chains can form hydrogen bonds with each other because their amide linkages, \(\mathrm{-C(=O)-NH-}\), contain two complementary features: an N-H group that can donate a hydrogen bond, and a carbonyl oxygen that can accept one. The hydrogen on one chain is attracted to a carbonyl oxygen on a neighbouring chain, producing an intermolecular \(\mathrm{N-H\cdots O=C}\) interaction.

Individually, these hydrogen bonds are much weaker than covalent bonds. But when many polymer chains lie close together, numerous hydrogen bonds can help resist their movement.

01Why the amide group makes hydrogen bonding possible

Start with an individual amide linkage:

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

There are two important bonds within it.

The first is the carbonyl bond, \(\mathrm{C=O}\). Oxygen attracts the shared electrons more strongly than carbon, so the oxygen carries a partial negative charge, written \(\delta^-\).

The second is the \(\mathrm{N-H}\) bond. Nitrogen attracts the shared electrons more strongly than hydrogen, leaving the hydrogen partially positive, written \(\delta^+\).

Now imagine two polyamide chains lying next to each other.

The partially positive hydrogen on the first chain can approach the partially negative carbonyl oxygen on the second. If their orientation and separation are suitable, a hydrogen bond can form.

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

The dotted part represents the hydrogen bond. The solid bonds within the amide groups remain covalent.

Which atom donates, and which accepts?

For the usual secondary amide linkages found in nylon, use this distinction:

The word donor can be misleading. The N-H group isn’t handing over a proton or transferring an electron. The hydrogen remains covalently attached to nitrogen.

Hydrogen bonding is an intermolecular attraction, not a chemical reaction that creates a new amide linkage.

There’s also a geometric condition. Having donors and acceptors doesn’t guarantee that every possible hydrogen bond forms. The groups must be close enough and correctly oriented.

02Why nylon-6,6 has so many hydrogen-bonding sites

Nylon-6,6 provides a useful example because its repeating structure contains two amide linkages.

It is made from:

• Hexane-1,6-diamine: \(\ce{H2N-(CH2)6-NH2}\)
• Hexanedioic acid (adipic acid): \(\ce{HOOC-(CH2)4-COOH}\)

During condensation polymerisation, the amine and carboxylic acid groups react to create covalent amide bonds, releasing water.

A typical interior repeating unit contains two \(\mathrm{N-H}\) groups and two carbonyl oxygens. That gives two hydrogen-bond donors and two carbonyl acceptor sites per repeat.

The carbon chains between these groups provide flexibility, allowing neighbouring sections to adopt arrangements in which hydrogen bonding is possible.

There are actually two different kinds of bonding involved:

1. Covalent amide bonds join monomer-derived sections into each long nylon chain.
2. Hydrogen bonds can form between the amide groups on neighbouring chains.

The first creates the polymer molecule. The second helps explain how multiple polymer molecules behave as a material.

If you’re still working on how the monomers join, the condensation polymerisation guide covers that reaction in detail. Here, the important question is what happens after the chains have formed.

03How could an experiment reveal hydrogen bonding?

The development of nylon illustrates an important distinction between observing a material’s properties and explaining their molecular origin.

In February 1935, researchers in Wallace Carothers’ group at DuPont prepared the polyamide that became known as nylon-6,6. Fibres could be drawn, or stretched, to improve their useful mechanical properties.

The fibres showed that a synthetic polyamide could form a strong, flexible material.

But strength alone couldn’t establish exactly which intermolecular attractions were responsible. Longer chains, entanglement, molecular alignment, crystallinity, and hydrogen bonding could all contribute.

To investigate the hydrogen-bonding explanation, chemists need evidence that says something about the environment around the amide groups.

One technique is infrared spectroscopy.

The apparatus: following infrared light through a nylon film

In a transmission Fourier-transform infrared (FTIR) experiment, the arrangement can be described in this order:

1. An infrared source produces radiation containing a range of frequencies.
2. The radiation enters an interferometer, where a beam splitter and fixed and moving mirrors produce a varying interference signal.
3. The recombined beam passes through a thin nylon film held in the sample compartment.
4. A detector records the transmitted radiation.
5. A computer processes the signal to produce an infrared spectrum, showing absorption against wavenumber.

A sufficiently thin film is needed because a thick sample may absorb too much infrared radiation for useful transmission measurements.

Infrared radiation can be absorbed when its frequency corresponds to molecular vibrations, such as stretching an \(\mathrm{N-H}\) or \(\mathrm{C=O}\) bond.

These vibrations are affected by hydrogen bonding.

For an N-H group, hydrogen bonding generally shifts its stretching absorption towards a lower wavenumber compared with a similar group that is not hydrogen-bonded.

That gives chemists something measurable.

The first comparison: nylon against polyester

Suppose we compare infrared spectra from dry nylon-6,6 and dry polyethylene terephthalate (PET), a common polyester.

Both contain carbonyl groups, so both have strong infrared absorption associated with \(\mathrm{C=O}\) vibrations.

But the nylon backbone also contains N-H groups, while the idealised PET backbone contains ester links and no N-H groups.

What should we predict?

Nylon should have N-H stretching absorption. PET should not have a corresponding N-H absorption from its repeating backbone.

Real polyamide infrared studies commonly identify hydrogen-bonded N-H stretching near \(3300\ \mathrm{cm^{-1}}\) and amide carbonyl stretching near \(1640\ \mathrm{cm^{-1}}\). The precise positions depend on the polymer’s structure and conditions.

This supports the presence of amide groups in hydrogen-bonding environments.

But the first comparison has a limitation.

The absence of an N-H peak from PET is expected simply because its repeating backbone has no N-H bond. Comparing the two materials doesn’t, by itself, prove that nylon’s hydrogen bonds connect separate polymer chains.

We need a way to investigate changes in the environment of the same amide groups.

The revised experiment: changing temperature

A more informative approach is to measure a dry nylon film at room temperature, warm it under a controlled dry atmosphere, and measure it again.

The prediction is that warming can weaken or disrupt some hydrogen-bonding interactions, shifting the N-H stretching absorption towards higher wavenumbers.

Cooling may allow some hydrogen bonds to reform.

Consider the following hypothetical results.

Illustrative data only. These are constructed values for interpreting an experiment, not measurements reported from a particular nylon sample.

The band moves to a higher wavenumber during warming, then returns close to its original position after cooling.

This is consistent with a reversible change in the hydrogen-bonding environment.

However, temperature also affects molecular vibrations, chain mobility, and potentially crystalline structure. A shift alone cannot prove that a particular number of interchain hydrogen bonds has broken.

The experiment should therefore control moisture, film thickness, and temperature history. Complementary measurements of chain packing, such as X-ray diffraction, can help test the proposed intermolecular arrangement.

The important experimental conclusion is narrower than simply saying ‘heat breaks hydrogen bonds’. Infrared spectroscopy provides evidence about the local chemical environment, while additional structural evidence is needed to establish how the polymer chains are arranged.

04Why hydrogen bonding changes a polyamide’s properties

Imagine two long nylon chains running alongside each other.

If they are positioned so that several N-H groups on one chain can hydrogen-bond with carbonyl oxygens on another, moving the chains apart or sliding them past one another can require disruption and rearrangement of multiple contacts.

A useful analogy is two strips of hook-and-loop tape.

One matching contact is easy to separate. Many matching contacts across a larger area resist separation more strongly.

But the analogy has limits. Hydrogen bonds aren’t physical hooks. They form and break dynamically, and polymer chains also interact through other intermolecular forces, entanglements, and crystalline packing.

Three consequences

Hydrogen bonding can contribute to the following properties of polyamides:

This last consequence is particularly interesting.

If hydrogen bonds help make nylon strong, what happens when water, which also forms hydrogen bonds, enters the material?

You might predict that additional hydrogen bonds always make the material stronger.

That prediction misses an important competition.

05Water competes with hydrogen bonds between nylon chains

A water molecule has hydrogen atoms that can donate hydrogen bonds and an oxygen atom that can accept them.

So water can interact with both complementary features of an amide group.

For example:

• A water hydrogen can hydrogen-bond to a nylon carbonyl oxygen.
• A nylon N-H hydrogen can hydrogen-bond to a water oxygen.

These nylon-water interactions can replace some nylon-nylon interactions.

Water can therefore increase local chain mobility and act as a plasticiser. Depending on the conditions, nylon can become less stiff and easier to deform.

This is a physical change in the hydrogen-bonding environment. It doesn’t automatically mean the covalent amide backbone has been hydrolysed.

At sufficiently harsh conditions, hydrolysis can occur, but ordinary moisture absorption and chemical bond cleavage are different processes.

06Why a polyester provides an important contrast

Consider a polyester containing ester linkages:

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

An ester group contains oxygen atoms that can act as hydrogen-bond acceptors, including its carbonyl oxygen.

But the ester link itself doesn’t contain an N-H or O-H group capable of donating the same kind of strong hydrogen bond found between nylon amide groups.

That means an ordinary PET polyester backbone cannot form the characteristic nylon-type \(\mathrm{N-H\cdots O=C}\) hydrogen bonds with another PET backbone.

Does that mean polyester chains have no intermolecular attractions?

No. PET has London dispersion forces, permanent dipole interactions, and important packing effects. Its properties can also depend strongly on how the chains are oriented and processed.

The difference is the availability of complementary hydrogen-bond donors and acceptors, not whether intermolecular forces exist at all.

Here’s a useful diagnostic:

This comparison concerns the groups shown, not additional functional groups, chain ends, or other possible weak interactions.

One final detail: a polymer that cannot donate hydrogen bonds to itself may still accept hydrogen bonds from a different polymer.

That becomes important when two materials are mixed.

07Questions and solutions

Question 1

Two nylon-6,6 chains lie close together. Each contains the linkage \(\mathrm{-C(=O)-NH-}\).

(a) Identify the hydrogen-bond donor and acceptor.

(b) Describe the hydrogen bond that can form between the chains.

(c) Explain why this interaction does not make the two chains into one covalently bonded molecule.

Solution 1

The N-H group is the hydrogen-bond donor, and the carbonyl oxygen is the acceptor. A hydrogen bond can form between the partially positive N-H hydrogen on one chain and the partially negative carbonyl oxygen on the other.

(a) Nitrogen attracts the bonding electrons in N-H, making the hydrogen partially positive. The carbonyl oxygen has lone pairs and a partial negative charge, allowing it to act as an acceptor.

(b) The interaction is represented by:

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

The N-H and C=O bonds belong to their respective polymer chains. The dotted interaction connects groups on different chains.

(c) No new covalent bond joins the two polymer backbones. Hydrogen bonding is an intermolecular attraction, whereas the amide linkages within each chain are covalent bonds formed during polymerisation.

A common mistake is to say the nitrogen atom forms a new covalent bond with the neighbouring oxygen. That is not what the dotted hydrogen-bond notation means.

Question 2

Two batches of nylon-6,6 fibre have the same chemical composition and comparable polymer chain lengths. Batch A is undrawn. Batch B has been cold-drawn so that its chains are more aligned.

The following results are illustrative.

Assume measurements are taken at the same temperature and moisture content, with comparable test conditions.

(a) Calculate the breaking stress of each fibre in megapascals. Use \(\sigma=F/A\), where \(\sigma\) is stress, \(F\) is force, and \(A\) is cross-sectional area. Recall that \(1\ \mathrm{N/mm^2}=1\ \mathrm{MPa}\).

(b) A student argues that Batch A is stronger because its breaking force is greater. Evaluate that conclusion.

(c) Explain how Batch B can have a larger hydrogen-bonded fraction without forming any new covalent amide groups.

(d) Does the evidence establish that hydrogen bonding alone caused the mechanical difference?

Solution 2

Batch B has the greater breaking stress, even though it breaks under a slightly smaller force. Its chains are more aligned, and a larger fraction of its N-H groups is hydrogen-bonded, but those observations do not prove hydrogen bonding is the only cause of the difference.

(a) Calculate stress using force divided by cross-sectional area.

For Batch A:

\[ \begin{aligned} \sigma_A&=\frac{48}{0.40}\\ &=120\ \mathrm{N/mm^2}\\ &=120\ \mathrm{MPa} \end{aligned} \]

For Batch B:

\[ \begin{aligned} \sigma_B&=\frac{45}{0.25}\\ &=180\ \mathrm{N/mm^2}\\ &=180\ \mathrm{MPa} \end{aligned} \]

Batch B withstands a breaking stress of \(180\ \mathrm{MPa}\), which is \(60\ \mathrm{MPa}\) greater than Batch A.

(b) The student’s conclusion confuses force with stress.

A thicker fibre can carry a larger total force while experiencing a smaller stress. Comparing forces without accounting for cross-sectional area can therefore give the wrong ranking.

(c) Drawing changes the arrangement and orientation of existing polymer chains.

It can place more N-H groups close to suitable carbonyl acceptors. This can increase the fraction involved in hydrogen bonding without changing the number of N-H or amide groups in the polymer.

No additional condensation reaction is needed.

(d) No. The results are consistent with hydrogen bonding contributing to Batch B’s greater breaking stress, but drawing also changes molecular orientation and may change crystallinity.

These factors can affect tensile behaviour independently of, and together with, hydrogen bonding.

The tempting conclusion is that a higher bonded fraction directly explains the entire increase in stress. The missing constraint is that drawing changes several structural features simultaneously.

Question 3

Two batches of nylon-6,6 film undergo different, initially unknown treatments.

Matched specimens from each batch are tested before treatment, while wet, and after thorough redrying. Separate specimens are used for each destructive tensile test.

The following measurements are hypothetical. Assume equal specimen dimensions, no oxidation, and no lasting change in crystallinity from the treatments.

(a) Identify which batch most strongly supports reversible water plasticisation and which supports hydrolysis of the polymer backbone.

(b) Explain the hydrogen-bonding interactions involved in the reversible change.

(c) Explain why Batch Y can retain hydrogen-bonding capability yet have much lower strength after drying.

Solution 3

Batch X is consistent with reversible water plasticisation, while Batch Y provides evidence of covalent amide-bond hydrolysis. The decisive difference is that X nearly recovers its original strength after drying, whereas Y does not and also develops additional chain-end groups.

(a) Batch X falls from \(100\ \mathrm{N}\) to \(70\ \mathrm{N}\) when wet but returns to \(99\ \mathrm{N}\) after drying.

This near-recovery supports a mostly reversible physical change. There is also no detected increase in chemical end groups.

Batch Y falls from \(100\ \mathrm{N}\) to \(72\ \mathrm{N}\) when wet and reaches only \(55\ \mathrm{N}\) after drying.

Its new terminal amine and carboxylic acid groups are consistent with cleavage of amide links. Given the stated assumptions, hydrolysis is the supported explanation.

(b) Water can form hydrogen bonds with nylon carbonyl oxygens and N-H groups.

Some water-polymer interactions replace polymer-polymer interactions. This allows increased chain mobility, which can reduce stiffness or strength under the tested conditions.

Removing the water can allow many interchain hydrogen bonds to reform.

(c) Hydrolysis breaks covalent bonds in the polymer backbone.

A simplified hydrolysis reaction is:

\[ \begin{aligned} &\ce{R-C(=O)-NH-R' + H2O}\\ &\ce{-> R-C(=O)-OH + H2N-R'} \end{aligned} \]

Here, R and R’ represent the remaining carbon-containing sections of the chain. The products are shown in their neutral forms; acidic conditions can protonate the amine.

The cleavage creates shorter polymer chains and new terminal groups.

Those new groups may themselves form hydrogen bonds, and the remaining amide groups can still hydrogen-bond. But hydrogen bonds cannot simply restore the broken covalent backbone.

This is why the presence of hydrogen-bonding groups doesn’t guarantee recovery of the original mechanical strength.

The trap is treating every moisture-related loss of strength as the same phenomenon. Water absorption can be reversible, while hydrolysis changes the polymer’s chemical structure.

Question 4

A chemist prepares two idealised condensation polyamides using hexanedioic acid.

Polyamide A uses hexane-1,6-diamine:

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

Polyamide B instead uses N,N’-dimethylhexane-1,6-diamine:

\[ \ce{CH3NH-(CH2)6-NHCH3} \]

Assume both amine groups of every incorporated diamine have reacted, polymer end groups can be ignored, and no other hydrogen-bond donor groups are present.

(a) State the amide linkage structure in each polymer and identify the key change caused by the methyl groups.

(b) Determine the number of N-H donors and carbonyl oxygen acceptors per interior repeating unit in each polymer.

(c) Predict whether B can form the same characteristic N-H···O=C hydrogen bonds between its own chains as A.

(d) Could an A-B mixture form intermolecular N-H···O=C hydrogen bonds between the two different polymers? Explain.

Solution 4

Polyamide A contains secondary amide groups with N-H donors, while Polyamide B contains tertiary amide groups without N-H donors. B therefore lacks the usual nylon-type hydrogen-bond donation between its own chains, but it can still accept hydrogen bonds from A.

(a) Polyamide A contains the linkage:

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

Polyamide B contains:

\[ \mathrm{-C(=O)-N(CH_3)-} \]

In B, the methyl group occupies the nitrogen position that would otherwise carry hydrogen.

The amide carbonyl remains, but the nitrogen no longer has an N-H bond.

(b) Both repeating units are formed from one diamine-derived section and one dicarboxylic-acid-derived section. Each contains two amide linkages.

The carbonyl acceptor sites are retained when the nitrogen atoms are methyl-substituted.

(c) Polyamide B cannot form the characteristic amide N-H···O=C interactions between its own chains because it has no N-H donor groups.

Its chains still experience other intermolecular forces, including dipole-dipole and dispersion interactions.

Therefore, B should not be described as having no intermolecular attractions.

(d) A-B hydrogen bonding is possible.

An N-H group on A can donate a hydrogen bond to a carbonyl oxygen on B.

\[ \mathrm{N-H_A\cdots O=C_B} \]

The reverse donor arrangement isn’t available because B has no N-H group.

This exposes the important misconception: a polymer doesn’t need to contain both donors and acceptors to participate in hydrogen bonding with a different material. The required complementary groups can come from different molecules.

Question 5

A very thin nylon-6,6 film is pressed against a second polymer film made from tertiary amide linkages of the form \(\mathrm{-C(=O)-N(CH_3)-}\).

The nylon contains N-H donors and carbonyl acceptors. The second polymer contains carbonyl acceptors but no N-H donors.

Consider a representative region of the nylon film containing 100 N-H donor groups.

Before the two films are brought together, a calibrated analysis indicates that 80 of these donors participate in hydrogen bonding. All 80 bonds connect neighbouring nylon chains.

After the films are pressed into contact, three findings are reported:

• The total number of hydrogen-bonded N-H donors in the region remains 80.
• An independent interface-sensitive analysis establishes that 30 of those donors now hydrogen-bond to carbonyl oxygens on the second polymer.
• Adhesion between the two films increases, while the nylon film’s in-plane tensile resistance decreases.

Assume the films remain dry, no covalent reactions occur, and there are no significant changes in chain length, orientation, or crystallinity. Each N-H donor contributes to at most one counted hydrogen bond, and the changes in interchain contacts dominate the mechanical response.

(a) Calculate how many of the original nylon donors now participate in nylon-nylon hydrogen bonds.

(b) Explain how adhesion between the films can increase while cohesion within the nylon layer decreases.

(c) A student argues that the unchanged total number of hydrogen-bonded N-H groups proves that nylon’s original interchain hydrogen-bonding network is unchanged. Evaluate this claim.

(d) Explain why the second polymer can participate in the new interfacial hydrogen bonds despite having no N-H groups.

Solution 5

Only 50 of the nylon N-H donors now form nylon-nylon hydrogen bonds. The other 30 bonded donors interact with the second polymer. The total number of hydrogen-bonded donors stays constant, but the connections have been redistributed.

That redistribution can explain increased adhesion between the films and reduced cohesion within the nylon layer.

(a) Before contact, 80 N-H donors participate in nylon-nylon hydrogen bonding.

After contact, 30 of the 80 hydrogen-bonded donors connect to carbonyl acceptors on the second polymer.

Therefore:

\[ \begin{aligned} N_{\mathrm{nylon-nylon}} &=80-30\\ &=50 \end{aligned} \]

Here, \(N_{\mathrm{nylon-nylon}}\) represents the number of hydrogen bonds donated by nylon N-H groups to neighbouring nylon chains.

The remaining groups are distributed as follows:

No new N-H donors have been created.

(b) The interfacial hydrogen bonds connect nylon chains to the second material.

These contacts can resist separation of the two films, increasing adhesion.

But the 30 donors reassigned to the interface are no longer connecting neighbouring nylon chains.

In the stated thin-film model, the loss of those internal connections allows greater relative movement within the nylon layer and can reduce its in-plane tensile resistance.

The result is not contradictory. Adhesion describes resistance to separation between different materials, while cohesion describes resistance to separation or movement within one material.

The same types of molecular interactions can contribute differently depending on which objects they connect.

(c) The student’s claim is incorrect.

The unchanged total of 80 hydrogen-bonded N-H donors tells us how many donor groups are involved in hydrogen bonding. It doesn’t identify their acceptor partners.

Before contact, all 80 bonded donors connect nylon to nylon.

After contact, only 50 do so. The other 30 connect nylon to the second polymer.

An infrared measurement of N-H hydrogen-bonding environments alone may not distinguish these interaction partners.

The exact missing information is which molecule supplies the acceptor for each hydrogen bond.

That is why the independent interface-sensitive evidence is essential.

(d) The second polymer doesn’t need to donate hydrogen bonds to participate in the interface.

Its carbonyl oxygens can act as acceptors for the N-H donors belonging to nylon.

The interaction still has the general form:

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

But the donor and acceptor now belong to different types of polymer.

The difficult idea here is that counting hydrogen bonds and understanding their effects are different tasks.

Two materials can contain the same number of hydrogen-bonded groups yet have different mechanical behaviour because those bonds connect different molecules.

In a real material, the geometry, distribution, and dynamics of the bonds also matter, so a bond count alone cannot predict a precise strength.

08What hydrogen bonding tells us – and what it doesn’t

The central reason polyamide chains can hydrogen-bond to one another is structural: secondary amide linkages provide N-H donors and carbonyl oxygen acceptors along the polymer backbone.

That leads to a useful reasoning sequence for unfamiliar polymers:

1. Identify the functional groups present in the backbone.
2. Find suitable hydrogen-bond donors and acceptors.
3. Decide whether those groups occur on neighbouring chains and can approach each other.
4. Explain which molecular movements the resulting attractions could resist.
5. Consider competing interactions, especially with water, and other structural factors such as chain orientation.

Infrared spectroscopy can show changes in the hydrogen-bonding environment, but it doesn’t automatically identify which chains are connected. The next experimental step is to combine that chemical evidence with information about molecular packing and orientation.

And when analysing a new polyamide structure, the most useful first move is still to identify exactly where the amide links sit. The guide to identifying amide links in polymer chains develops that structural skill before you apply it to intermolecular forces and material properties.