How to Compare Hydrogen Bonding in Polyamides and Polyesters

Put a nylon fishing line beside a PET drink bottle. Both polymers contain polar carbonyl groups, so you might predict that both sets of polymer chains can form similar hydrogen-bond networks. That prediction is wrong.

For the standard HSC comparison, nylon-6,6 (a polyamide) contains N-H hydrogen-bond donors and carbonyl oxygen acceptors. This allows neighbouring nylon chains to form hydrogen bonds. PET (a polyester) contains ester oxygen atoms that can accept hydrogen bonds, but its repeating unit has no N-H or O-H donor. Consequently, dry PET chains cannot form the same hydrogen-bond network between their repeating units.

The important distinction is that having an oxygen atom does not automatically allow a molecule or polymer to form hydrogen bonds with itself. It needs both a hydrogen-bond donor and an acceptor.

01Start with the polymer family, then identify the example

Before comparing intermolecular forces, make sure you know which polymers you are comparing.

A polyamide is a polymer containing repeating amide links. A polyester is a polymer containing repeating ester links.

These are polymer families, not individual polymer names.

Two common examples are:

• Nylon-6,6 is a polyamide.
• Polyethylene terephthalate (PET) is a polyester.

Their repeating links are different:

The name of the family tells you which functional group appears repeatedly along the polymer backbone. The named example tells you the particular structure you should inspect.

This distinction matters because not every polymer within a family has identical end groups or exactly the same number of hydrogen-bonding sites.

If you need to revise how these polymers form through condensation polymerisation, see Condensation Polymerisation: Nylon and Polyester Explained for HSC Chemistry.

For this comparison, we are interested in what happens after the polymer chains have formed.

02The key test: find a hydrogen-bond donor and an acceptor

Hydrogen bonding requires two different roles.

1. Hydrogen-bond donor

A donor contains a hydrogen atom covalently bonded directly to a highly electronegative atom, typically nitrogen, oxygen, or fluorine.

For example:

• \(\ce{O-H}\)
• \(\ce{N-H}\)

Because the electronegative atom attracts the shared electrons, the hydrogen becomes partially positive.

2. Hydrogen-bond acceptor

An acceptor contains an electronegative atom with an available lone pair of electrons that can interact with the partially positive hydrogen.

Oxygen atoms, particularly carbonyl oxygen atoms, commonly act as hydrogen-bond acceptors.

A hydrogen bond can therefore form between an N-H group on one polymer chain and a carbonyl oxygen on another:

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

The N-H group is the donor. The carbonyl oxygen is the acceptor.

Notice that the hydrogen bond forms between different parts of neighbouring molecules or chains. It is not the same as the covalent bonds holding the polymer backbone together.

Polyamides: why nylon can form hydrogen bonds

Consider the repeating unit of nylon-6,6:

\[ \ce{[-NH-(CH2)6-NH-CO-(CH2)4-CO-]_{n}} \]

The polymer contains repeating amide links:

\[ \ce{-CO-NH-} \]

Look at the two important parts of this link.

• The N-H bond contains a partially positive hydrogen, making it a hydrogen-bond donor.
• The carbonyl group, \(\ce{C=O}\), contains an oxygen atom with lone pairs, making it a hydrogen-bond acceptor.

When two nylon chains lie alongside one another, the N-H group of one chain can interact with the carbonyl oxygen of another.

This produces interchain hydrogen bonding.

The attractions between neighbouring chains increase the cohesion of the polymer. They help hold the chains together and contribute to the strength and toughness associated with many nylon materials.

An important detail: the nitrogen atom in an ordinary amide link is not usually treated as the hydrogen-bond acceptor. Its lone pair is delocalised towards the carbonyl group. The carbonyl oxygen is the relevant acceptor.

Polyesters: why PET behaves differently

Now consider PET, or polyethylene terephthalate.

Its repeating unit is:

\[ \ce{[-O-CH2-CH2-O-CO-C6H4-CO-]_{n}} \]

PET contains repeating ester links:

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

An ester group has the following general connectivity:

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

The carbonyl carbon is double-bonded to one oxygen and single-bonded to another oxygen. That second oxygen is connected to another carbon-containing group.

Both oxygen atoms have lone pairs, although the carbonyl oxygen is generally the more important hydrogen-bond acceptor.

Here is the critical difference: neither oxygen has a hydrogen attached to it.

The PET repeating unit therefore contains hydrogen-bond acceptors but no conventional hydrogen-bond donors.

The hydrogen atoms in its \(\ce{-CH2-}\) groups do not count as conventional hydrogen-bond donors. They are bonded to carbon, not directly to oxygen or nitrogen.

This also explains a common source of confusion.

An ester is not the same as a carboxylic acid.

Compare:

• Ester: \(\ce{-COO-R}\)
• Carboxylic acid: \(\ce{-COOH}\)

The carboxylic acid contains an O-H bond and can donate hydrogen bonds. The ester has an oxygen attached to carbon rather than hydrogen, so it does not provide that donor.

For PET, this means neighbouring repeating units cannot form the same conventional hydrogen-bond network as nylon-6,6.

Direct comparison: nylon-6,6 versus PET

Assuming dry polymer samples and focusing on their repeating units rather than their chain ends:

The main conclusion is that nylon-6,6 can form extensive hydrogen bonding between neighbouring chains, whereas PET cannot form the same network through its repeating units.

However, don’t conclude that PET has no intermolecular forces.

Its ester groups are polar, so PET experiences dipole-dipole attractions. It also experiences London dispersion forces.

These attractions, together with the covalent polymer backbone and the way its chains pack, contribute to PET’s physical properties.

03A useful complication: PET can still accept hydrogen bonds

There is a subtle distinction between saying:

• PET cannot hydrogen-bond to itself through its repeating units.
• PET cannot participate in hydrogen bonding at all.

The first statement is correct for the standard dry PET repeat-unit model. The second is incorrect.

Imagine placing a PET fibre in contact with water.

Water contains O-H bonds, so its hydrogen atoms can act as hydrogen-bond donors. The carbonyl oxygen atoms in PET can act as acceptors.

A hydrogen bond can therefore form between water and PET:

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

The water molecule supplies the donor hydrogen. PET supplies the acceptor oxygen.

This is hydrogen bonding between different substances, not between two PET repeating units.

There is one further qualification.

Real polymer chains have ends. Depending on how a polyester is manufactured, some chain ends may contain hydroxyl (\(\ce{-OH}\)) or carboxylic acid (\(\ce{-COOH}\)) groups.

These groups can donate hydrogen bonds.

However, chain-end groups are relatively sparse compared with the repeating ester groups along a long polymer chain. They do not change the basic comparison between the repeating structures of nylon-6,6 and PET.

Moisture, additives, and other substances can also introduce additional hydrogen-bond donors.

This is why it is more accurate to describe PET as lacking hydrogen-bond donors in its repeating unit than to claim that every PET sample is completely incapable of hydrogen bonding.

04What does this mean for polymer properties?

The difference in hydrogen bonding helps explain why nylon and PET can behave differently as materials.

In nylon, hydrogen bonding increases the attraction between polymer chains. This contributes to chain cohesion and helps nylon fibres resist mechanical stress.

PET does not have the same repeating-unit hydrogen-bond network. However, its polar ester groups, London dispersion forces, and ability to form well-packed polymer structures still contribute to useful mechanical properties.

Both materials can be manufactured into strong fibres.

The important point is that hydrogen bonding is one factor affecting polymer properties, not the only factor.

Other important factors include:

• Length of the polymer chains.
• Degree of crystallinity.
• Alignment and orientation of the chains.
• Flexibility of the polymer backbone.
• Manufacturing and processing conditions.

For example, drawing a polyester fibre can align its chains and significantly affect its mechanical properties.

Therefore, avoid writing that nylon must always be stronger than PET simply because it can form hydrogen bonds. The actual properties depend on the complete polymer structure and how the material is manufactured.

05A repeatable method for HSC questions

When an exam asks you to compare hydrogen bonding in a polyamide and a polyester, follow this process.

Step 1: Identify the polymer family.

Look for the repeating functional group.

• \(\ce{-CO-NH-}\) indicates an amide link.
• \(\ce{-CO-O-}\) indicates an ester link.

Step 2: Find the hydrogen-bond donors.

Look for hydrogen atoms directly bonded to nitrogen or oxygen.

Do not count hydrogen atoms bonded to carbon.

Step 3: Find the hydrogen-bond acceptors.

Look for electronegative atoms with available lone pairs, particularly carbonyl oxygen atoms.

Step 4: Determine whether neighbouring chains can interact.

A donor on one chain must be able to interact with an acceptor on another chain.

Having acceptors on both chains is not enough.

Step 5: Connect the structure to the properties.

Explain how hydrogen bonding affects the attraction between chains. Then acknowledge other intermolecular forces and structural factors where relevant.

A strong HSC response might read:

Nylon-6,6 is a polyamide containing N-H hydrogen-bond donors and carbonyl oxygen acceptors. These groups allow hydrogen bonds to form between neighbouring polymer chains, increasing intermolecular attraction and contributing to the strength and toughness of the material. PET is a polyester containing ester oxygen atoms that can accept hydrogen bonds, but its repeating unit contains no N-H or O-H donor groups. Consequently, PET cannot form the same hydrogen-bond network between its repeating units. It still experiences dipole-dipole and London dispersion forces.

06How could an experiment test this prediction?

The structural comparison gives us a prediction. But how could a chemist obtain experimental evidence that the two polymer structures behave differently?

One useful technique is infrared spectroscopy, particularly Fourier transform infrared spectroscopy (FTIR).

This technique measures how a material absorbs infrared radiation at different wavenumbers.

Following the signal through an FTIR instrument

Consider an attenuated total reflectance (ATR-FTIR) instrument.

A thin polymer film is pressed against a crystal, commonly made from diamond or another suitable infrared-transmitting material.

The infrared beam enters the crystal and reflects internally. At the surface where the crystal contacts the polymer, an electromagnetic field interacts with the sample.

The polymer absorbs particular infrared frequencies when those frequencies match the vibrational energies of its chemical bonds.

A detector measures the returning infrared signal. The instrument then produces a spectrum showing absorbance against wavenumber.

What should the spectra show?

If the structural model is correct, we expect different characteristic absorption bands.

These are approximate reference regions, not fixed values for every sample.

Reference spectra of nylon-6,6 and PET demonstrate these characteristic functional-group patterns.

There is another interesting observation.

When an N-H group forms a hydrogen bond, the N-H covalent bond is slightly weakened and lengthened. Its stretching vibration generally shifts towards a lower wavenumber and becomes broader.

This means the N-H stretching region can provide evidence of hydrogen bonding in nylon.

However, an infrared spectrum does not directly photograph a hydrogen bond. It measures molecular vibrations. The interpretation comes from comparing the observed absorption bands with the expected chemical structure.

A flaw in the original experimental design

Suppose the PET sample has not been dried properly.

PET can interact with water through hydrogen bonding. Absorbed water may therefore produce broad O-H absorption in the spectrum.

A student might mistakenly interpret this as evidence that the PET repeating unit contains an O-H donor.

The experiment needs better control.

A revised investigation would:

1. Dry both polymer samples under suitable, controlled conditions.
2. Record their spectra using the same measurement method.
3. Expose a separate PET sample to controlled humidity.
4. Record its spectrum again.
5. Compare the dry and humidified spectra.

If a broad O-H absorption appears after humidification, alongside a change in the ester carbonyl absorption, this supports an interaction between water and the PET ester groups.

It does not demonstrate that PET repeating units have suddenly become hydrogen-bond donors.

There are also limitations. Polymer crystallinity, chain orientation, temperature, and moisture content can affect infrared absorption bands. A single shifted peak is not enough to calculate the total strength of a polymer’s intermolecular attractions.

The FTIR investigation is an extension of the structural comparison. For the core HSC reasoning, identifying the donor and acceptor groups is the essential method.

09Questions and solutions

Question 1

A polymer has the following repeating unit:

\[ \ce{[-NH-(CH2)5-CO-]_{n}} \]

Identify its polymer family and explain how two neighbouring chains can form a hydrogen bond.

Solution 1

The polymer is a polyamide, and neighbouring chains can form hydrogen bonds through N-H donors and carbonyl oxygen acceptors.

The repeating unit contains an amide link, \(\ce{-CO-NH-}\).

The N-H bond supplies a partially positive hydrogen atom, making it a hydrogen-bond donor. The carbonyl oxygen contains lone pairs, making it an acceptor.

An N-H group on one chain can therefore interact with the carbonyl oxygen on another:

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

This creates an intermolecular attraction between the chains, rather than another covalent bond within the polymer backbone.

Question 2

A PET fabric is stored in a humid room. After several hours, the fabric has absorbed a small amount of water.

A student claims:

“PET cannot form hydrogen bonds because it has no O-H groups. Therefore, the absorbed water must have chemically reacted with the polymer.”

Evaluate this claim.

Assume that the PET backbone remains chemically unchanged and no hydrolysis has occurred.

Solution 2

The student’s claim is incorrect. PET can form hydrogen bonds with water because its ester oxygen atoms can accept hydrogen bonds donated by water molecules.

Water contains O-H bonds. These bonds are polar, so their hydrogen atoms can act as hydrogen-bond donors.

The carbonyl oxygen in PET acts as an acceptor.

Therefore, water can interact with PET through:

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

This interaction is intermolecular. It does not require PET to contain an O-H group itself.

The absorbed water can remain associated with the polymer through these attractions without forming new covalent bonds.

The distinction is:

• PET cannot form the same hydrogen-bond network between its own repeating units because it lacks donors.
• PET can accept hydrogen bonds from water because water supplies the donor hydrogen.

Drying the fabric may remove the absorbed water and reduce these interactions without chemically changing the PET backbone.

Question 3

A chemist obtains the following illustrative ATR-FTIR data for two dry polymer samples.

The chemist then exposes Sample B to controlled humidity.

A broad absorption band appears between \(3400\) and \(3600\ \mathrm{cm^{-1}}\), while its ester carbonyl band shifts from \(1718\) to \(1712\ \mathrm{cm^{-1}}\).

Identify both polymers and explain the significance of the changes after humidification. Does the new spectrum demonstrate hydrogen bonding between PET repeating units?

Solution 3

Sample A is a polyamide, most consistent with nylon, while Sample B is PET, a polyester. The changes after humidification support hydrogen bonding between water and PET, not between PET repeating units.

The reasoning follows from the characteristic functional groups.

Sample A: polyamide

The broad absorption near \(3290\ \mathrm{cm^{-1}}\) is consistent with N-H stretching.

The strong amide I and amide II bands provide further evidence of amide linkages.

This supports the presence of a polyamide structure containing N-H hydrogen-bond donors and carbonyl oxygen acceptors.

Sample B: polyester

The strong carbonyl absorption near \(1718\ \mathrm{cm^{-1}}\), together with the C-O bands, is consistent with ester groups.

The absence of an N-H band supports the expected PET structure, which has no N-H donor in its repeating unit.

Effect of humidification

The new broad absorption between \(3400\) and \(3600\ \mathrm{cm^{-1}}\) is consistent with O-H stretching from absorbed water.

The carbonyl band also shifts to a lower wavenumber:

\[ 1712-1718=-6\ \mathrm{cm^{-1}} \]

This change is consistent with water interacting with PET’s carbonyl oxygen through hydrogen bonding.

The water supplies the donor hydrogen, while PET supplies the acceptor oxygen.

Conclusion: The humidified spectrum supports water-PET hydrogen bonding. It does not demonstrate that PET repeating units can donate hydrogen bonds to neighbouring PET chains.

The distinction is essential: the appearance of an O-H absorption band does not automatically mean that the polymer backbone contains O-H groups.

Question 4

Consider two idealised polymer chains.

• A PET chain contains 200 repeating units and has one hydroxyl group (\(\ce{-OH}\)) at each end.
• A nylon-6,6 chain also contains 200 repeating units. Each repeating unit contains two N-H hydrogen-bond donor sites.

Assume all donor groups are accessible, and each donor can donate one hydrogen bond.

1. Calculate the maximum number of hydrogen-bond donor sites on each chain.
2. Calculate the ratio of donor sites in the nylon chain to those in the PET chain.
3. Explain whether the PET chain can form any hydrogen bonds with another PET chain.

Solution 4

The nylon-6,6 chain has 400 donor sites, while the PET chain has only 2 donor sites. Nylon therefore has 200 times as many donor sites per chain under the stated assumptions.

Part 1: Calculate the donor sites

For PET, there are two terminal hydroxyl groups:

\[ N_{\mathrm{PET}}=2 \]

For nylon-6,6, there are two N-H donor sites per repeating unit:

\[ N_{\mathrm{nylon}}=200\times2 \]

\[ N_{\mathrm{nylon}}=400 \]

Part 2: Calculate the ratio

\[ \text{Ratio}=\frac{400}{2} \]

\[ \boxed{200:1} \]

The nylon chain has 200 times as many donor sites as the PET chain in this idealised comparison.

Part 3: Interpret the result

PET can still form hydrogen bonds with another PET chain through its terminal hydroxyl groups.

An O-H group at one end of a PET chain can donate a hydrogen bond to a carbonyl oxygen on a neighbouring PET chain.

However, these donor sites are restricted to the chain ends. Most of the PET backbone still consists of ester repeating units without hydrogen-bond donors.

In nylon-6,6, N-H donor sites occur repeatedly along the backbone, allowing much more extensive opportunities for interchain hydrogen bonding.

The important limitation is that the donor-site count does not tell us the exact number of hydrogen bonds that will form. Actual bonding depends on chain arrangement, accessibility, and molecular orientation.

Question 5

A materials laboratory prepares a dry physical blend of nylon-6,6 and PET.

The samples are processed under identical conditions, and their infrared spectra are collected after controlled drying.

The following illustrative measurements are obtained:

No broad O-H absorption is detected in the blend, and no new covalent-linkage bands are identified.

The blend also demonstrates greater toughness than initially predicted from the properties of the two separate polymers.

A student makes the following argument:

“PET cannot hydrogen-bond with nylon because PET has no hydrogen-bond donors. The shifted spectral bands must therefore indicate that a new covalent copolymer has formed.”

Evaluate this argument.

In your response:

• Identify the most plausible interaction between the two polymer families.
• Explain the significance of both spectral shifts.
• Determine whether the evidence proves that a covalent copolymer has formed.
• Explain how the interaction could affect the toughness of the blend.

Solution 5

The most plausible explanation is interchain hydrogen bonding between nylon N-H donors and PET ester carbonyl oxygen acceptors. The spectral changes support this interaction, but they do not prove that a covalent copolymer has formed.

This problem tests an important distinction: a substance does not need to donate hydrogen bonds to participate in hydrogen bonding.

Step 1: Identify the available donors and acceptors.

Nylon-6,6 contains:

• N-H hydrogen-bond donors.
• Carbonyl oxygen hydrogen-bond acceptors.

PET contains:

• Ester carbonyl oxygen acceptors.
• No N-H or O-H donors in its repeating unit.

Therefore, PET cannot donate a conventional hydrogen bond through its repeating unit, but it can accept one from nylon.

Step 2: Consider the possible interaction between the chains.

An N-H group on a nylon chain can interact with a carbonyl oxygen on a PET chain:

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

This is a hydrogen bond between two different polymer chains.

Nylon can also retain its own nylon-to-nylon hydrogen bonds.

PET does not need to supply a donor for this cross-polymer interaction to occur.

Step 3: Interpret the N-H stretching shift.

The nylon N-H stretching band moves from:

\[ 3305\ \mathrm{cm^{-1}} \]

to:

\[ 3275\ \mathrm{cm^{-1}} \]

The change is:

\[ 3275-3305=-30\ \mathrm{cm^{-1}} \]

The shift towards a lower wavenumber is consistent with a change in the N-H hydrogen-bonding environment.

The nylon N-H groups may now be interacting with carbonyl oxygen atoms on PET chains, rather than interacting exclusively with other nylon chains.

Step 4: Interpret the PET carbonyl shift.

The PET carbonyl band also shifts:

\[ 1710-1718=-8\ \mathrm{cm^{-1}} \]

This is consistent with the PET carbonyl oxygen accepting hydrogen bonds from nylon N-H groups.

The two shifts therefore support the same proposed interaction from the perspectives of both polymer components.

However, spectral shifts can also be influenced by chain packing, crystallinity, and molecular orientation. The controlled drying and identical processing make the hydrogen-bonding explanation more convincing, but the measurements do not establish the exact proportion of cross-chain hydrogen bonds.

Step 5: Evaluate the copolymer claim.

The student is confusing intermolecular interactions with covalent chemical bonding.

A hydrogen bond is an attraction between different molecular groups. It does not join two polymer backbones through a new covalent bond.

A physical blend contains separate polymer chains mixed together. A copolymer contains different monomer-derived units covalently incorporated into the same polymer chain.

The observed spectral shifts do not establish that such covalent chain formation has occurred.

Step 6: Explain the increased toughness.

Hydrogen bonding at the interface between nylon and PET can improve the attraction between the two polymer components.

This can improve interfacial adhesion and help transfer mechanical stress between the different polymer chains.

Consequently, the blend may show improved toughness without requiring the formation of a new covalent polymer.

Other factors, including crystallinity and the distribution of the two polymers, can also affect the measured properties.

Final conclusion: The most defensible interpretation is that nylon-6,6 and PET form some interfacial hydrogen bonds through nylon N-H donors and PET carbonyl oxygen acceptors. This helps explain the spectral shifts and potentially improved toughness. It does not prove covalent copolymerisation.

The key insight is that PET cannot donate hydrogen bonds through its repeating units, but it can still participate in hydrogen bonding as an acceptor.

10Syllabus context and what to study next

As at November 2026, NSW students are still studying the Chemistry Stage 6 Syllabus (2017). NESA schedules implementation of the Chemistry 11-12 Syllabus (2025) from 2028, with the first HSC examination under the new syllabus in 2029. This resource is framed around the current 2017 course. The FTIR discussion is an extension of the structural reasoning, rather than a claim that the polymer-specific spectral values are prescribed content. See NESA’s Chemistry syllabus information.

The most useful skill to take from this comparison is the ability to identify hydrogen-bond donors and acceptors directly from a chemical structure.

Apply the same reasoning when comparing alcohols, amides, carboxylic acids, and other organic compounds. Once you can identify which groups donate and which accept, you can make much more accurate predictions about intermolecular attractions, solubility, and physical properties.

For another application of this method, see Intermolecular Bonding in Alcohols: HSC Chemistry Explained.