How to Identify an Amide Link in a Polymer Chain

A polymer chain can contain carbon, hydrogen, oxygen, and nitrogen without necessarily being a polyamide. The key is to find one particular connection: a carbonyl carbon directly bonded to a nitrogen atom.

That connection is called an amide link, written as \(\mathrm{-CO-NH-}\) or, showing the carbonyl bond explicitly, \(\mathrm{-C(=O)-NH-}\). If this linkage repeats along the polymer backbone, you are looking at a polyamide.

The trick is to identify the bond itself, not simply look for nitrogen and oxygen somewhere in the structure. Let’s work through a reliable method using nylon as our main example.

01How to recognise an amide link

Start by looking for a carbon atom double-bonded to oxygen and single-bonded to nitrogen:

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

This is the characteristic amide linkage.

Read the structure from left to right:

• \(\mathrm{C=O}\) is the carbonyl group.
• The carbonyl carbon is directly attached to nitrogen.
• The nitrogen is attached to the next section of the polymer chain.

That final connection is what makes the group an amide rather than a carboxylic acid, an amine, or an ester.

The identification test

When you are given an unfamiliar polymer structure, use this sequence:

1. Find every oxygen atom that is double-bonded to carbon.
2. Identify the carbon attached to that oxygen.
3. Check what is directly bonded to that carbon.
4. If it is bonded to nitrogen, you have an amide link.
5. Check whether this connection forms part of the repeating polymer backbone.

The last step matters. An amide group attached to a side chain does not automatically mean that the main polymer backbone is a polyamide.

Don’t confuse an amide with an ester or an amine

This is where identification questions become less straightforward. Several organic functional groups contain oxygen, nitrogen, or both.

The difference is which atom is directly attached to the carbonyl carbon.

Consider the difference between these two structures:

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

The first is an amide link. The second is an ester link.

Both contain a carbonyl group, but only the first has nitrogen directly attached to the carbonyl carbon.

A useful way to remember this is that the presence of nitrogen somewhere in a polymer is not enough. You must identify the actual carbon-nitrogen connection.

For a more detailed explanation of the underlying functional groups, see HSC Chemistry: Carboxylic Acids, Amines and Amides Explained.

An important distinction: backbone or side chain?

There is one more detail that can catch you out.

An amide group can appear in a polymer without connecting the repeating units together.

Consider polyacrylamide, which has the repeating unit:

\[ \left[ -\mathrm{CH_2-CH(C(=O)NH_2)-} \right]_n \]

This structure contains an amide functional group, \(\mathrm{-CONH_2}\).

However, follow the main polymer backbone. It runs through the carbon atoms:

\[ \mathrm{-CH_2-CH-CH_2-CH-} \]

The amide groups branch away from this backbone. They are called pendant or side-chain groups.

Compare this with nylon 6,6, where the amide links form part of the main chain itself.

For the backbone-linkage classification used in this guide:

• Nylon 6,6 is a polyamide because amide links connect sections of its main chain.
• Polyacrylamide has amide functional groups attached to a carbon-based backbone, rather than amide links joining its repeating units.

This distinction becomes particularly useful when an exam question provides a complicated polymer structure rather than naming the material.

02Nylon 6,6: tracing the amide link back to its monomers

Nylon 6,6 is a useful example because its repeating structure makes the amide linkage easy to recognise.

It forms from two monomers:

Monomer 1: Hexane-1,6-diamine

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

This molecule has an amine group at each end.

Monomer 2: Hexanedioic acid (adipic acid)

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

This molecule has a carboxylic acid group at each end.

Notice that both monomers have two reactive functional groups. This is important because the reaction can continue at both ends, allowing a long polymer chain to form.

How does the amide link form?

A primary amine reacts with a carboxylic acid through a condensation reaction.

The general reaction is:

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

The OH from the carboxylic acid and one H from the amine are removed together as water.

The remaining carbonyl carbon forms a bond directly with nitrogen.

That new \(\mathrm{-C(=O)-NH-}\) connection is the amide link.

Because both nylon 6,6 monomers have two reactive ends, the process repeats and produces a long chain.

The repeating unit of nylon 6,6

The repeating unit is:

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

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

You can identify the amide linkages by locating the carbonyl carbons directly bonded to nitrogen.

There are two amide linkages per idealised repeating unit when the connections across the repeating-unit boundaries are included.

This is an important detail. The brackets show the repeating pattern, not a separate molecule with two free ends. The final carbonyl at the right connects to the nitrogen at the beginning of the next repeating unit.

What do the numbers in nylon 6,6 mean?

The two numbers refer to the number of carbon atoms in each starting monomer.

Be careful with the acid. Its structure contains four methylene groups, but it also has two carbonyl carbon atoms.

The acid therefore contains six carbon atoms, not four.

The name nylon 6,6 reflects the two monomers, not the total carbon count of the repeating unit.

Does every nylon form from two different monomers?

No. Nylon 6 provides an important contrast.

Nylon 6 is commonly manufactured through ring-opening polymerisation of caprolactam, a cyclic molecule.

Its repeating unit is:

\[ \left[ -\mathrm{NH-(CH_2)_5-C(=O)-} \right]_n \]

It still contains amide links, even though its common manufacturing route does not require a separate diamine and dicarboxylic acid.

Nylon 6 can also be produced through condensation polymerisation of 6-aminohexanoic acid.

The key point is that you identify an amide link from the polymer’s structure, not from assumptions about how it was manufactured.

03Why amide links affect polymer properties

Identifying an amide link tells you more than the polymer’s chemical family. It also helps explain why materials such as nylon have useful mechanical properties.

Look closely at the amide group:

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

The carbonyl bond is polar because oxygen attracts the shared electrons more strongly than carbon.

The N-H bond is also polar. This creates sites that can participate in hydrogen bonding.

In nylon 6,6:

• The carbonyl oxygen can accept a hydrogen bond.
• The N-H hydrogen can donate a hydrogen bond.
• Neighbouring polymer chains can form many hydrogen bonds with one another.

There are two different types of bonding to keep separate:

Within each chain: Covalent bonds, including the amide links, hold the atoms together.

Between neighbouring chains: Hydrogen bonds create additional attractions that help hold the chains together.

This distinction is important. Hydrogen bonds do not create the polymer backbone. They help neighbouring backbones stay associated.

In nylon fibres, chains can become aligned alongside one another. This allows many hydrogen bonds to form between neighbouring chains.

The combined effect contributes to:

• High tensile strength.
• Good durability.
• Relatively high melting temperatures compared with many polymers without these interactions.

However, amide links alone do not determine every property. Chain alignment, crystallinity, molecular mass, and moisture content also influence the behaviour of the material.

04How could an experiment identify an amide-containing polymer?

So far, we have identified amide links by reading a structural formula. But what if a chemist receives an unknown polymer sample without being given its molecular structure?

One useful experimental technique is infrared spectroscopy, particularly attenuated total reflectance Fourier-transform infrared spectroscopy (ATR-FTIR).

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

The apparatus and prediction

A small polymer sample is placed against a special crystal in the ATR-FTIR instrument.

Infrared light travels through the crystal and interacts with the sample near its surface. A detector measures the radiation that emerges.

The instrument produces a spectrum showing absorption against wavenumber, measured in \(\mathrm{cm^{-1}}\).

Before examining the results, make a prediction.

If the unknown polymer contains amide groups, which bonds should produce characteristic absorption bands?

The important features are:

These are representative ranges for common polyamides. Exact positions vary with the polymer, hydrogen bonding, and measurement conditions. Published measurements of polyamide materials show these characteristic absorption regions (example ATR-FTIR data).

What would the results tell us?

A strong amide I band, an amide II band, and an N-H stretching band together provide evidence for amide groups.

A polyester, by comparison, has ester linkages rather than amide linkages. It can show a strong carbonyl absorption, but it does not have the amide N-H bond.

There is an important limitation, though.

Infrared spectroscopy can help identify the functional groups present, but it cannot always establish whether an amide group lies in the main polymer backbone or hangs from a side chain.

That distinction requires structural information in addition to the spectrum.

For an HSC structural-identification question, you will usually be expected to recognise the linkage directly from a supplied molecular structure. The experimental technique is an extension that shows how chemists can investigate an unknown material.

05Worked examples

06Questions and solutions

Question 1

A polymer chain contains the following structural section:

\[ \cdots-\mathrm{CH_2-C(=O)-NH-CH_2} \]

\[ \mathrm{-C(=O)-NH-CH_2}-\cdots \]

Identify the functional linkage and state the polymer family. Explain which atoms establish your answer.

Solution 1

The polymer contains amide links in its backbone, so it belongs to the polyamide family.

There are two carbonyl groups in the displayed section. In each case, the carbonyl carbon is directly bonded to nitrogen.

Both connections have the characteristic structure:

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

The nitrogen is part of the continuous chain, rather than an unrelated atom elsewhere in the molecule.

This is the defining structural feature of an amide link. Simply identifying oxygen and nitrogen would not be sufficient.

Question 2

A student is given the repeating unit of nylon 6,6:

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

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

The student claims that there is only one amide link per repeating unit because only one \(\mathrm{-NH-C(=O)-}\) connection appears inside the brackets.

(a) Explain why this interpretation is incorrect.

(b) How many amide links are associated with each idealised repeating unit?

(c) For an idealised polymerisation equation, how many water molecules are released per pair of monomer molecules?

(d) If a finite linear chain contains \(n\) molecules of each monomer, how many amide-forming condensation events occur?

Solution 2

There are two amide links per idealised repeating unit, not one. The student has overlooked the connection across the repeating-unit boundary.

(a) The brackets indicate a repeating pattern, not the physical ends of a separate molecule.

The final carbonyl group in one repeating unit bonds to the nitrogen at the beginning of the next unit.

Therefore, a complete repeating pattern includes the linkage crossing the boundary as well as the linkage visible within the brackets.

(b) There are two amide links per idealised repeating unit.

One appears between the second nitrogen and the first carbonyl group. The other forms between the final carbonyl group and the nitrogen of the next repeating unit.

(c) Two water molecules are released per idealised pair of monomers.

Each amide-forming condensation removes one water molecule:

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

Two amide bonds therefore correspond to two water molecules in the idealised repeating-unit equation.

(d) A finite linear chain containing \(n\) diamine molecules and \(n\) dicarboxylic acid molecules has \(2n\) monomer molecules.

Connecting \(2n\) separate molecules into one linear chain requires:

\[ 2n-1 \]

connections.

Therefore, exactly \(2n-1\) water molecules are produced if every connection forms through condensation and there are no other reactions.

The familiar \(2n\) water coefficient is the idealised repeating-unit representation, which neglects the small end-group correction for a finite linear chain.

This is why the repeating-unit picture and an exact finite-chain equation are not always identical.

Question 3

An unfamiliar polymer has the following repeating structure:

\[ \left[ -\mathrm{NH-(CH_2)_5-C(=O)-} \right]_n \]

A student identifies it as nylon 6 but argues that it must have been produced from a diamine and a dicarboxylic acid, just like nylon 6,6.

(a) Confirm whether the structure contains an amide link.

(b) Determine the number of carbon atoms in its repeating unit.

(c) Explain why the student’s proposed manufacturing route is not necessarily correct. Identify a common monomer used to produce this polymer.

Solution 3

The polymer is nylon 6, and its structure contains one amide linkage per idealised repeating unit. However, it does not have to be manufactured from a diamine and a dicarboxylic acid.

(a) The structure contains:

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

The carbonyl carbon is directly bonded to nitrogen, confirming an amide link.

(b) Count the carbon atoms.

There are five carbon atoms in the five \(\mathrm{CH_2}\) groups and one carbonyl carbon.

Therefore:

\[ N_{\mathrm C}=5+1=6 \]

The repeating unit contains six carbon atoms, consistent with nylon 6.

(c) The student’s assumption is incorrect because polyamides can be produced through different polymerisation routes.

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

The cyclic monomer opens, allowing the molecules to join into a chain containing the repeating amide structure.

Nylon 6 can also be formed through condensation polymerisation of 6-aminohexanoic acid, which contains both an amine group and a carboxylic acid group.

The important distinction is that the polymer’s structure establishes the presence of amide links, but it does not uniquely establish the route used to manufacture it.

Question 4

Two unknown polymer samples are analysed using ATR-FTIR spectroscopy.

The following results are illustrative data, not measurements from actual samples.

(a) Which sample is more consistent with a polyamide?

(b) Which polymer family is Sample B more consistent with?

(c) Explain why a carbonyl absorption by itself is insufficient to identify an amide link.

(d) Can these results establish whether the amide groups in Sample A are in the polymer backbone rather than in side chains? Explain.

Solution 4

Sample A is more consistent with a polyamide, while Sample B is more consistent with a polyester. However, the spectrum alone cannot establish the precise position of Sample A’s amide groups within its molecular structure.

(a) Sample A has three characteristic features:

• An N-H stretching absorption near \(3300\ \mathrm{cm^{-1}}\).
• A strong amide I absorption near \(1635\ \mathrm{cm^{-1}}\).
• An amide II absorption near \(1540\ \mathrm{cm^{-1}}\).

Together, these support the presence of amide groups.

(b) Sample B is more consistent with a polyester.

Its strong carbonyl absorption and C-O absorption are consistent with ester groups.

Unlike Sample A, it does not show the characteristic N-H stretching absorption associated with common nylon amide links.

(c) A carbonyl group occurs in several different functional groups.

For example:

• Amides contain \(\mathrm{-C(=O)-N-}\).
• Esters contain \(\mathrm{-C(=O)-O-}\).
• Carboxylic acids contain \(\mathrm{-C(=O)-OH}\).

Therefore, detecting a carbonyl absorption does not establish which functional group is present.

The surrounding atoms and other characteristic absorption bands must also be considered.

(d) No. The spectrum supports the presence of amide groups, but it does not establish whether they form the main polymer backbone or appear as pendant side groups.

Both arrangements can produce characteristic amide absorption bands.

Additional structural information would be needed to distinguish them. This is a limitation of identifying a polymer from infrared spectroscopy alone.

Question 5

Two polymers have the following repeating units:

Polymer A:

\[ \left[ -\mathrm{CH_2-CH(C(=O)NH_2)-} \right]_n \]

This is polyacrylamide.

Polymer B:

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

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

This is nylon 6,6.

A student makes three claims:

1. Both polymers must be classified as backbone polyamides because both contain amide groups.
2. Only Polymer B can form hydrogen bonds between neighbouring chains because its amide groups are part of the backbone.
3. Polymer B must have greater tensile strength because it has two amide links per repeating unit, while Polymer A has only one amide group per repeating unit.

Evaluate all three claims.

Also explain whether an FTIR spectrum showing amide I and amide II bands would be sufficient to distinguish the two polymers.

Assume both samples contain dry, neutral polymer chains and no other functional groups.

Solution 5

All three claims are incorrect or unjustified.

Only Polymer B has amide links connecting the repeating sections of its main backbone. However, both polymers can form hydrogen bonds between neighbouring chains, and their tensile strengths cannot be ranked using the number of amide groups alone.

Claim 1: Both polymers are backbone polyamides.

This is incorrect.

Polymer A contains the amide functional group:

\[ \mathrm{-C(=O)-NH_2} \]

However, its main backbone runs through carbon atoms:

\[ \mathrm{-CH_2-CH-CH_2-CH-} \]

The amide groups are attached to this backbone as side chains. They do not connect the repeating units together through amide linkages.

Polymer B is different.

Its main chain contains:

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

The amide links form part of the continuous polymer backbone.

Therefore, Polymer B is the backbone polyamide in this comparison.

Claim 2: Only Polymer B can form hydrogen bonds.

This is also incorrect.

Hydrogen bonding depends on the presence and accessibility of suitable donor and acceptor sites, not simply on whether the amide group is in the backbone.

Both polymers contain primary or secondary amide groups with:

• N-H bonds that can donate hydrogen bonds.
• Carbonyl oxygen atoms that can accept hydrogen bonds.

In Polymer A, neighbouring chains can form hydrogen bonds through their pendant amide groups.

In Polymer B, neighbouring chains can form hydrogen bonds through the amide groups within their backbones.

The positions of the groups differ, but both polymers have the necessary chemical features for hydrogen bonding.

Claim 3: Polymer B must have greater tensile strength.

This conclusion cannot be established from the supplied structures alone.

Polymer B has two amide links per idealised repeating unit, whereas Polymer A has one pendant amide group per repeating unit.

That difference is relevant, but counting functional groups is not enough to predict tensile strength.

The final mechanical properties also depend on:

• How closely the polymer chains can pack.
• The degree of crystallinity.
• The alignment of chains within the material.
• The molecular mass and chain length.
• The extent and arrangement of intermolecular attractions.

A polymer with more hydrogen-bonding sites does not automatically have greater tensile strength. Their arrangement and the overall structure of the material matter.

Can FTIR distinguish the two polymers?

Not conclusively from the amide I and amide II bands alone.

Both polymers contain amide groups, so both can produce characteristic amide absorption bands.

The spectrum can establish evidence for amide functionality, but it cannot independently establish whether those groups are in the backbone or attached as side chains.

You would need additional structural evidence to make that distinction.

This is the central lesson of the question: identifying an amide functional group and identifying an amide link in the polymer backbone are related, but they are not identical tasks.

Once you can identify the amide link reliably, the next step is to reconstruct the original monomers and determine how they joined together. That leads directly into condensation polymerisation, including the formation of nylon 6,6 and the differences between polyamides and polyesters. For that broader connection, continue with Polyamides: How Nylon Forms, Its Structure and Properties.