How to Draw the Repeating Unit of Nylon-6,6

You are given two monomers and asked to draw the repeating unit of nylon-6,6. You recognise the amine and carboxylic acid groups, but how do you join them without losing the wrong atoms?

The answer is:

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

This is the repeating unit of nylon-6,6, a synthetic polyamide.

The key is to preserve the carbon skeletons of both monomers, convert the reacting amine and carboxylic acid groups into amide linkages, and identify the section that repeats along the polymer chain.

01Identify the two monomers

Before drawing the polymer, identify which molecules form it.

Nylon-6,6 is produced from two bifunctional monomers:

Notice that both molecules have a reactive group at each end.

This is important because each monomer can connect to two neighbouring sections of the growing polymer chain.

The name nylon-6,6 also gives you a useful check:

• The first 6 represents the six carbon atoms in hexane-1,6-diamine.
• The second 6 represents the six carbon atoms in hexanedioic acid.

Be careful with the acid. Its structure contains four \(\ce{CH2}\) groups and two additional carbon atoms in its carboxylic acid groups.

That gives six carbon atoms altogether.

02Understand how the amide linkage forms

Now consider what happens when an amine group reacts with a carboxylic acid group.

The important transformation is:

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

The hydroxyl group (\(\ce{-OH}\)) from the carboxylic acid and one hydrogen atom from the amine combine to form water.

The remaining atoms form a new carbon-nitrogen bond.

This produces an amide linkage:

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

This is the connection that holds the sections of nylon-6,6 together.

Why does the reaction continue?

After one amine group reacts with one carboxylic acid group, each original monomer still has another reactive group available.

For example:

• The diamine still has another amine group at its opposite end.
• The dicarboxylic acid still has another carboxylic acid group at its opposite end.

These remaining groups can react with further monomers.

The process therefore produces an alternating sequence of diamine-derived and diacid-derived sections.

This repeated formation of amide linkages builds the polymer chain.

03How to draw the nylon-6,6 repeating unit

You can now construct the repeating unit systematically.

Step 1: Write the two monomers

Start with their complete condensed structural formulas:

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

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

Do not remove any atoms yet.

Step 2: Identify the parts that remain in the polymer

The diamine contributes the six-carbon section:

\[ \ce{-NH-(CH2)6-NH-} \]

The dicarboxylic acid contributes the four-methylene section and its two carbonyl carbons:

\[ \ce{-C(=O)-(CH2)4-C(=O)-} \]

Notice that the carbonyl carbon from each carboxylic acid group stays in the polymer.

It is not removed when water forms.

Step 3: Join the sections using amide linkages

Connect the two monomer-derived sections through their reactive groups.

The resulting structure is:

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

The nitrogen and carbonyl carbon are directly bonded.

That is what makes the connection an amide linkage rather than an ester linkage.

Step 4: Add brackets and the subscript \(n\)

The final repeating unit is:

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

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

The dashes at either end indicate that the chain continues into neighbouring repeating units. They are not free ends of an isolated molecule.

There is also another correct way to write the same repeating unit:

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

The difference is simply where you choose to start drawing the repeating pattern.

Important: A repeating unit is a section of a continuous chain, not a separate molecule with two capped ends.

Think of it as a repeating tile in a long pattern. You can choose where to cut out the tile, but the pattern must still match when the tiles are joined.

Step 5: Check the number of carbon atoms

A quick carbon count helps catch structural mistakes.

Therefore:

\[ 6+4+1+1=12 \]

Each idealised nylon-6,6 repeating unit contains twelve carbon atoms.

This is consistent with the two six-carbon monomers.

04What does the nylon rope experiment actually show?

The formation of nylon-6,6 can also be demonstrated using an experiment known as the nylon rope trick.

It provides a useful way to understand how two bifunctional monomers can produce a continuous polymer fibre.

There is one important difference between this demonstration and the usual carboxylic acid-diamine reaction used to represent nylon-6,6 in structural questions.

The experiment uses an acid chloride instead of a carboxylic acid.

The experimental setup

Imagine a beaker containing two separate liquid layers.

• The bottom layer is an aqueous solution of hexane-1,6-diamine.
• The upper layer is a solution of hexanedioyl dichloride in cyclohexane.

Cyclohexane is less dense than water, so it forms the upper layer.

The two liquids do not mix readily, leaving a distinct boundary between them.

What should happen?

The prediction is that polymerisation will occur where the two solutions meet.

The diamine molecules and acid chloride molecules react at the interface, forming amide linkages.

The reaction produces hydrogen chloride rather than water.

An idealised repeating-unit equation is:

\[ \begin{aligned} &n\,\ce{ClC(=O)-(CH2)4-C(=O)Cl}\\ &+n\,\ce{H2N-(CH2)6-NH2}\\ &\longrightarrow \left[ \ce{-C(=O)-(CH2)4-C(=O)-NH-(CH2)6-NH-} \right]_n\\ &+2n\,\ce{HCl} \end{aligned} \]

This equation uses the long-chain approximation and neglects terminal-group corrections.

The polymer has the same nylon-6,6 backbone as the product formed from adipic acid and hexane-1,6-diamine.

What is observed?

A thin polymer film forms at the boundary between the two liquids.

The film can be drawn upwards using tweezers, producing a continuous nylon strand. As the film is removed, fresh liquid surfaces meet at the interface, allowing further polymer to form.

The reaction is therefore localised at the boundary rather than occurring uniformly throughout either liquid.

A useful design feature is keeping the two solutions layered instead of mixing them thoroughly. Mixing would spread the reaction through the liquid and make it harder to draw out a continuous film.

What does the experiment prove?

The observation supports the idea that the two reactive monomers can form a long polymer chain through repeated amide-link formation.

However, seeing a nylon strand does not directly reveal the precise arrangement of its atoms or measure its repeating unit.

That structure is determined by examining the starting monomers and tracking which atoms remain after each reaction.

There is also an important distinction to remember:

Both routes produce the same ideal nylon-6,6 repeating structure.

This is a useful example of why you must identify the actual reactants before deciding which small molecule is eliminated.

Chemguide provides a further explanation of the nylon rope demonstration and polyamide reactions.

05Why the amide linkage gives nylon useful properties

Drawing the repeating unit is not just about getting the atoms in the right order.

The structure also explains some of nylon’s useful physical properties.

Look again at the amide linkage:

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

It contains two particularly important features:

• The nitrogen-hydrogen bond provides a hydrogen-bond donor.
• The carbonyl oxygen provides a hydrogen-bond acceptor.

This means neighbouring nylon chains can form hydrogen bonds.

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

These are intermolecular attractions. They occur between separate polymer chains, not between the atoms along a single chain.

These attractions help hold neighbouring chains together.

Along with dipole-dipole attractions and London dispersion forces, hydrogen bonding contributes to nylon’s:

• High tensile strength.
• Resistance to abrasion.
• Durability as a fibre.
• Ability to form strong textile materials.

The actual mechanical properties also depend on polymer chain length, crystallinity, molecular orientation, and processing.

A useful contrast: Nylon versus polyester

A common mistake is assuming all condensation polymers have the same functional linkage.

They do not.

Nylon-6,6 is a polyamide, while PET is a polyester.

PET has ester linkages rather than amide linkages.

Although its carbonyl oxygen can accept hydrogen bonds from suitable external donors, its repeating unit does not contain the N-H donor needed for the same repeated amide-to-amide hydrogen-bonding pattern found in nylon.

This distinction is useful when explaining why changing a polymer’s chemical structure can change its intermolecular forces and physical properties.

For a broader comparison of condensation polymerisation, nylon, and polyester formation, see Condensation Polymerisation: Nylon and Polyester Explained for HSC Chemistry.

08Common mistakes when drawing nylon-6,6

Before attempting exam questions, check that you can distinguish the correct structure from these common errors.

09Questions and solutions

Question 1

Hexane-1,6-diamine reacts with hexanedioic acid to produce nylon-6,6.

Draw the ideal repeating unit of the polymer and identify the functional group responsible for connecting the monomer-derived sections.

Solution 1

The repeating unit is:

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

The connecting functional group is the amide linkage, \(\ce{-CONH-}\).

Hexane-1,6-diamine contributes the six-carbon chain, while hexanedioic acid contributes four methylene groups and two carbonyl carbons.

Each amine group reacts with a carboxylic acid group, eliminating water and forming a carbon-nitrogen bond.

The brackets indicate that this structure repeats along the polymer backbone.

Question 2

A student draws the following repeating unit and identifies it as nylon-6,6:

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

Identify the structural error, draw the corrected repeating unit, and explain why the student’s structure does not represent the expected polyamide.

Solution 2

The student’s structure contains an incorrect oxygen atom in the polymer backbone.

The section \(\ce{-(CH2)4-O-CO-}\) introduces an ester-type connection rather than the required amide connection.

Nylon-6,6 forms from a diamine and a dicarboxylic acid. Both carboxylic acid groups react with amine groups, so nitrogen must connect directly to each carbonyl carbon.

The correct repeating unit is:

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

The difference is chemically important.

An amide linkage contains a carbon-nitrogen connection, whereas an ester linkage contains a carbon-oxygen connection.

Replacing nitrogen with oxygen changes the functional group and therefore changes the polymer’s chemical structure.

Question 3

A nylon rope demonstration uses hexanedioyl dichloride and hexane-1,6-diamine.

The two reactants are initially held in separate liquid layers. A polymer film forms at the boundary between the liquids and can be pulled into a continuous strand.

a) Explain why the polymer forms at the interface.

b) Identify the small molecule released during this reaction.

c) Explain why the repeating unit is still nylon-6,6, even though a different starting acid derivative is used.

Solution 3

a) The polymer forms at the interface because this is where the two reactive monomers come into contact.

Hexanedioyl dichloride is present in the organic layer, while hexane-1,6-diamine is present in the aqueous layer.

At their boundary, the acid chloride groups react with the amine groups, forming amide linkages.

Because both monomers have two reactive functional groups, repeated reactions produce long polymer chains.

b) The small molecule released is hydrogen chloride, \(\ce{HCl}\).

An acid chloride reacts with an amine by eliminating hydrogen chloride when an amide linkage forms.

Each ideal repeating unit accounts for two such reactions.

c) The repeating unit remains the same because both reaction routes produce the same arrangement of atoms along the polymer backbone.

In the conventional condensation reaction, hexanedioic acid reacts with hexane-1,6-diamine, eliminating water.

In the rope demonstration, hexanedioyl dichloride replaces hexanedioic acid, and hydrogen chloride is eliminated instead.

The starting functional groups and eliminated molecules differ, but the resulting amide-linked carbon backbone is the same.

This demonstrates why the repeating unit must be determined from the atoms retained in the polymer, rather than simply copying the byproduct from a different reaction route.

Question 4

An engineer compares nylon-6,6 and PET polyester fibres.

Both polymers contain carbonyl groups, but nylon-6,6 has stronger opportunities for hydrogen bonding between neighbouring polymer chains.

Explain this difference using the repeating-unit structures.

Also explain why the presence of hydrogen bonding alone does not determine the tensile strength of a finished fibre.

Solution 4

Nylon-6,6 can form repeated hydrogen bonds between neighbouring chains because its amide linkages contain both N-H bonds and carbonyl oxygen atoms.

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

This allows an attraction of the form:

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

to form between adjacent polymer chains.

PET contains ester linkages rather than amide linkages.

Although PET contains carbonyl oxygen atoms that can accept hydrogen bonds from suitable donors, its ideal repeating unit does not contain an N-H bond. Therefore, it cannot form the same repeated amide-to-amide hydrogen-bonding arrangement.

These additional attractions help hold nylon chains together and contribute to its mechanical properties.

However, hydrogen bonding does not determine tensile strength by itself.

The finished fibre’s strength also depends on molecular orientation, crystallinity, polymer chain length, and manufacturing conditions.

Therefore, it would be incorrect to conclude that every nylon fibre must be stronger than every PET fibre simply because nylon has amide linkages.

Question 5

A polymer chemist prepares nylon-6,6 using:

• 100 molecules of hexane-1,6-diamine.
• 100 molecules of hexanedioic acid.

Assume that all 200 monomer molecules are incorporated into exactly five separate, linear polymer chains.

Each chain is unbranched, non-cyclic, and contains an alternating sequence of diamine-derived and diacid-derived sections.

No monomer remains unreacted.

a) Calculate the total number of amide linkages formed.

b) Determine the exact number of water molecules released.

c) Determine the total number of terminal amine groups and terminal carboxylic acid groups remaining across the five chains.

d) A student argues that 200 water molecules must be produced because every ideal repeating unit accounts for two amide linkages. Another argues that only 100 water molecules are produced because there is one complete amide linkage inside the written repeating-unit brackets.

Explain why both arguments are incorrect for this finite sample.

Solution 5

a) Exactly 195 amide linkages form.

Each polymer chain is linear and contains no rings or branches.

A chain containing \(k\) monomer molecules requires \(k-1\) connections.

For five separate chains containing 200 monomer molecules altogether:

\[ \begin{aligned} N_{\text{links}}&=N_{\text{monomers}}-N_{\text{chains}}\\ &=200-5\\ &=195 \end{aligned} \]

Therefore:

\[ \boxed{195\text{ amide linkages}} \]

The important detail is that the monomers form five separate chains, not one continuous chain.

b) Exactly 195 water molecules are released.

Each amide linkage formed through the carboxylic acid-amine condensation reaction releases one water molecule.

Therefore:

\[ N(\ce{H2O})=195 \]

\[ \boxed{195\text{ water molecules}} \]

This follows directly from the number of connections formed.

c) Five terminal amine groups and five terminal carboxylic acid groups remain.

There are five linear chains, each with two ends.

Therefore, the total number of terminal groups is:

\[ 5\times2=10 \]

Each chain can have:

• Two amine-derived ends.
• Two carboxylic-acid-derived ends.
• One of each type.

Because the total number of diamine and diacid molecules is equal, the overall number of amine-derived and acid-derived ends must also be equal.

Therefore:

\[ \boxed{5\text{ terminal } \ce{-NH2}\text{ groups}} \]

and

\[ \boxed{5\text{ terminal } \ce{-COOH}\text{ groups}} \]

d) Both students have confused the ideal repeating unit with a finite polymer chain.

The student predicting 100 water molecules has counted only the amide linkage that appears explicitly inside the written brackets.

However, the second amide linkage per ideal repeating unit occurs across the bracket boundary.

The terminal carbonyl section of one repeating unit connects to the nitrogen-containing section of the next repeating unit.

Therefore, the ideal repeating unit accounts for two amide linkages, not one.

The student predicting 200 water molecules has correctly recognised the two linkages per ideal repeating unit but has ignored the five separate chain ends.

An ideal infinitely repeating structure has no terminal ends. A real finite linear chain does.

Every separate chain is missing one connection compared with a fully closed repeating pattern.

With five chains, five potential connections are absent.

Therefore:

\[ \begin{aligned} N(\ce{H2O})&=200-5\\ &=195 \end{aligned} \]

The exact result is:

\[ \boxed{195\text{ water molecules}} \]

The deeper lesson is that a repeating unit describes the regular internal structure of a polymer. It does not automatically describe the exact number of bonds or terminal groups in a finite sample.

That distinction becomes especially important when questions move from drawing an ideal repeating unit to drawing an actual short polymer molecule.

10Where to go next

Once you can confidently draw the nylon-6,6 repeating unit, the next useful skill is drawing the actual ends of a finite polyamide chain.

An ideal repeating unit has no terminal groups shown, whereas a short polymer molecule must retain any unreacted amine or carboxylic acid groups at its ends.

Practise that distinction in How to Draw the End Groups of a Short Polyamide Chain.