Why Polyamide Monomers Need Two Reactive Functional Groups

An amine reacts with a carboxylic acid to form an amide. Suppose you want to turn that reaction into a long nylon fibre. You mix hundreds of molecules of methanamine and ethanoic acid, hoping they’ll keep joining together.

What stops them from forming one enormous molecule?

Each molecule has only one suitable reactive functional group. Once that group forms an amide bond, there’s no second connection point available for the chain to keep growing.

That’s why making a linear polyamide by condensation polymerisation requires monomers with at least two suitable reactive groups. Each building block must be able to connect to another molecule without using up every possible connection.

The important distinction is between making an amide bond and making a chain containing many amide bonds.

01Why the second reactive group matters

Think of joining sections of an extension lead. A section with a connection at only one end can attach to something, but it cannot continue the line. A section with connections at both ends can sit between two neighbours.

For condensation polymerisation, those chemical connection points are functional groups.

The analogy has a limit: amines and carboxylic acids are not interchangeable plugs and sockets. They undergo chemical reactions that change bonds and remove atoms. The groups must also be chemically compatible.

One reactive group makes an amide, not a long chain

Consider these two molecules:

• Ethanoic acid: \(\ce{CH3COOH}\)
• Methanamine: \(\ce{CH3NH2}\)

Ethanoic acid contains one carboxylic acid group, \(\ce{-COOH}\). Methanamine contains one primary amine group, \(\ce{-NH2}\).

Under conditions that promote amide formation, the carboxylic acid group reacts with the amine group. An amide linkage, \(\ce{-C(=O)-NH-}\), forms, and water is eliminated.

The product is N-methylethanamide:

\[ \ce{CH3C(=O)NHCH3} \]

Now inspect the product. It contains an amide group, but no remaining free carboxylic acid or amine group capable of extending the chain by the same condensation reaction.

Adding another molecule of methanamine or ethanoic acid won’t make this amide grow into a nylon chain.

An important experimental detail: simply mixing a carboxylic acid and an amine at room temperature usually favours formation of an ammonium carboxylate salt. Converting that mixture into an amide requires suitable reaction conditions. The structural argument above assumes the amide-forming reaction has occurred.

Two reactive groups allow the chain to continue

Now replace those molecules with the monomers used to make nylon-6,6.

Hexane-1,6-diamine contains two amine groups:

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

Hexanedioic acid, also called adipic acid, contains two carboxylic acid groups:

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

When one amine group reacts with one carboxylic acid group, an amide bond forms.

But each original molecule has another reactive group at its opposite end.

That means the product can react again with another monomer or another growing chain. Repeating this process produces a long chain with many amide linkages along its backbone.

Chemists call a monomer with two suitable reactive groups bifunctional. In this example, the diamine and dicarboxylic acid are both bifunctional.

The growing nylon-6,6 backbone contains alternating sections derived from the diamine and the dicarboxylic acid. The two types of section are joined by amide bonds.

Which monomers can keep growing?

Notice the last row.

A monomer does not need two amine groups or two carboxylic acid groups specifically. It needs two compatible connection points.

6-Aminohexanoic acid contains one of each. Its amine group can react with the carboxylic acid group of another molecule of the same substance.

This gives a possible condensation route to nylon-6.

Industrially, nylon-6 is commonly manufactured by ring-opening polymerisation of caprolactam, a cyclic amide. That is a different reaction pathway, so the two-free-functional-groups rule should not be applied to every method of making polyamides.

The rule we’re studying concerns linear polyamides formed by step-growth condensation between available amine and carboxylic acid groups, or their suitable reactive derivatives.

02How the nylon rope experiment makes chain growth visible

The historical challenge in developing synthetic polyamide fibres was not simply discovering how to make an amide bond. Chemists needed to produce molecules long enough to form useful fibres.

Nylon-6,6, developed from hexane-1,6-diamine and hexanedioic acid in the 1930s, demonstrated that suitably chosen bifunctional monomers could form fibre-producing polymers.

A later, widely used teaching demonstration called the nylon rope trick makes the chain-forming reaction easier to observe.

It uses a different reactive form of the acid monomer, so we need to distinguish the demonstration from the ordinary carboxylic acid condensation equation.

The apparatus: two liquids with a reaction boundary

In a nylon-6,6 version of the demonstration, a beaker contains two liquid layers.

The lower layer is an aqueous solution containing hexane-1,6-diamine.

The upper layer is an organic solution containing hexanedioyl dichloride, also called adipoyl chloride:

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

This molecule has two acid chloride groups, \(\ce{-COCl}\), rather than two \(\ce{-COOH}\) groups.

The liquids do not mix freely, so they form a boundary called an interface.

At that boundary, amine groups encounter acid chloride groups and react to form amide bonds.

A thin nylon film forms where the two reactants meet.

The film can be gripped with tweezers and drawn upwards into a continuous thread. As it is lifted away, fresh reactants meet at the interface and more material can form.

Why does a thread form rather than just a collection of unrelated small molecules?

Both reactants have two reactive groups. Each can become connected to neighbouring molecular sections at both ends, allowing long covalently bonded chains to develop.

The ability to draw a continuous filament is consistent with polymer formation.

It does not, by itself, measure the average molecular mass of the nylon or prove that every molecule in the film is a long chain.

A small but important chemistry difference: HCl, not water

When a carboxylic acid and an amine form an amide through condensation, water is the small-molecule product.

When an acid chloride reacts with an amine, hydrogen chloride is produced instead:

\[ \ce{R-COCl + H2N-R' ->} \]

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

Here, \(R\) and \(R'\) represent the carbon-containing parts of the reactants.

In the demonstration, the hydrogen chloride can react further with amine or other basic species present in the aqueous solution.

You should not use a water-production equation to describe the acid chloride version of the experiment.

The experiment illustrates polyamide formation through repeated amide bond formation, but its by-product differs from that of the corresponding dicarboxylic acid reaction.

The reagents and solvents used in this demonstration present significant hazards. It is a supervised laboratory demonstration, not a home practical.

What the original demonstration cannot establish

The nylon rope trick shows that the selected bifunctional reagents can produce a polymer film and thread.

But does that observation prove the second reactive group is necessary?

Not by itself.

A useful experimental comparison would replace the bifunctional acid chloride with a molecule containing only one acid chloride group, while keeping the diamine as the other reactant.

For example, hexanoyl chloride has the structure:

\[ \ce{CH3-(CH2)4-COCl} \]

It has one reactive acid chloride group.

Predict what happens if hexanoyl chloride reacts with both amine ends of one hexane-1,6-diamine molecule.

Two hexanoyl chloride molecules can attach to the diamine, one at each end.

The resulting molecule contains two amide bonds, but no remaining amine or acid chloride group through which it can extend into a long chain.

The monofunctional acid chloride has acted as a chain terminator.

This control is a proposed comparison, not a claim that an additional experiment was performed or that a particular solid was observed.

It also has a limitation.

Changing the reagent can change solubility, reaction rate, and the appearance of the products. If a solid appeared in the control, that alone would not prove polymer formation. Small organic compounds can also be insoluble.

A stronger investigation would combine observations of fibre formation with evidence about molecular size and remaining end groups.

The chemistry predicts why the structures behave differently. The physical observations then test whether the predicted long chains actually form.

03Two reactive groups are necessary, but they aren’t enough

A bifunctional monomer has the capacity to become part of a long chain. That does not guarantee that a long chain will form.

Three additional conditions matter.

First, the reactive groups must be compatible. A mixture containing only diamines cannot form the intended nylon-6,6 amide backbone without a suitable carbonyl-containing partner.

Second, the proportions of the functional groups matter. A large excess of one type can leave many chains with unreacted ends of the same type, limiting further growth.

Third, enough successful reactions must occur. In step-growth polymerisation, long chains generally require a very high proportion of the reactive groups to form bonds.

Side reactions and the formation of cyclic molecules can also compete with useful linear chain growth.

A monofunctional reactant can deliberately stop a chain

This leads to a useful reversal of the original question.

If having two reactive groups allows chain growth, what happens when a chemist deliberately adds some molecules with only one suitable reactive group?

Those molecules can cap reactive chain ends.

A monocarboxylic acid can react with an available terminal amine, creating an amide bond. Since the monocarboxylic acid has no second carboxylic acid group, that end cannot continue reacting through the same pathway.

This can limit further polymer growth.

However, the number and length of chains also depend on the original ratio of diamine to diacid. Adding a chain stopper does not automatically shorten every chain that has already formed.

04Why one amide bond doesn’t give a material all the properties of nylon

A single small molecule can contain an amide group. Nylon contains many amide groups joined along long molecular backbones.

These are chemically related, but they don’t necessarily behave as materials in the same way.

In nylon-6,6, the carbonyl oxygen of an amide group can accept a hydrogen bond from an N-H group on a nearby polymer chain.

These attractions help neighbouring chains associate.

But intermolecular hydrogen bonds are not the same as the covalent amide bonds that build each chain.

Hydrogen bonding between separate short molecules does not magically join them into one long covalent backbone.

Long chains can also become entangled and align during fibre processing. Chain length, molecular packing, crystallinity, and processing all influence the material’s strength and flexibility.

So two different questions must be kept separate:

• Can the molecules react repeatedly to create long covalent chains?
• Once those chains exist, how strongly do they interact, and how are they arranged?

The number of reactive groups primarily helps answer the first question. The polymer’s structure and intermolecular forces help answer the second.

For the wider comparison between nylon, polyesters, and different condensation reactions, see Condensation Polymerisation: Nylon and Polyester Explained for HSC Chemistry.

05Questions and solutions

Question 1

6-Aminohexanoic acid has the structure:

\[ \ce{H2N-(CH2)5-COOH} \]

Explain why this single type of monomer can undergo self-condensation to form a linear polyamide, even though it has only one amine group.

For two molecules joining together, state the number of amide bonds and water molecules formed, and identify the functional groups remaining at the ends of the product.

Solution 1

6-Aminohexanoic acid can form a linear polyamide because every molecule has two compatible reactive groups: one amine and one carboxylic acid. Two molecules can form one amide bond and one water molecule, while leaving one free amine and one free carboxylic acid group at the opposite ends of the product.

The carboxylic acid group of one molecule reacts with the amine group of another.

The reaction removes an \(\ce{-OH}\) from the acid and a hydrogen from the amine, producing \(\ce{H2O}\) and joining the molecules through \(\ce{-C(=O)-NH-}\).

The resulting two-unit molecule can be written as:

\[ \ce{H2N-(CH2)5-C(=O)-NH-} \]

\[ \ce{(CH2)5-COOH} \]

These two lines represent one continuous structure.

The free \(\ce{-NH2}\) group remains at one end, and the free \(\ce{-COOH}\) group remains at the other.

Further molecules can therefore attach.

The tempting mistake is to think a monomer needs two groups of the same type. It does not. One amine and one carboxylic acid group provide two suitable connection points for this self-condensation pathway.

Question 2

A chemist considers two possible versions of an interfacial nylon experiment.

Both use an aqueous solution of hexane-1,6-diamine as the lower liquid layer.

The upper organic layer contains one of the following reagents:

• Experiment A: hexanedioyl dichloride, \(\ce{ClCO-(CH2)4-COCl}\)
• Experiment B: hexanoyl chloride, \(\ce{CH3(CH2)4COCl}\)

Assume each acid chloride group reacts with an available amine group, and ignore hydrolysis and other competing reactions.

Predict which system can produce a long linear polyamide.

For experiment B, determine the maximum number of amide bonds one diamine molecule can form with hexanoyl chloride molecules. Explain why this product cannot grow further through the same reaction.

Finally, suppose an insoluble solid appears in both experiments. Would that observation alone establish that both produced long-chain polymers? Explain.

Solution 2

Experiment A can produce a long linear polyamide, whereas experiment B can produce amide-containing molecules but cannot build a long backbone using only these two reactants. In experiment B, one diamine molecule can form at most two amide bonds with two molecules of hexanoyl chloride.

In experiment A, the acid chloride has two \(\ce{-COCl}\) groups.

The diamine also has two amine groups.

Each molecule can therefore connect to two neighbouring sections, allowing continued step-growth polymerisation.

In experiment B, hexanoyl chloride has only one \(\ce{-COCl}\) group.

Each of the two amine ends of a diamine molecule can react with a separate hexanoyl chloride molecule.

The resulting diamide has the general structure:

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

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

where \(R = \ce{CH3(CH2)4}\). The two lines form one continuous molecule.

Both amine groups have become amide groups, and neither hexanoyl-derived end has another acid chloride group.

Two amide bonds have formed, releasing two molecules of HCl before any further acid-base reactions.

There is no suitable unused functional group for the structure to continue growing through this pathway.

An insoluble solid would not prove that both experiments produced long-chain polymers.

A short diamide can have limited solubility and form a solid. Observing a precipitate is evidence that a physical change occurred, not direct proof of high molecular mass.

The better evidence would concern molecular size, chain-end chemistry, and the ability of the material to form a sustained filament.

The trap is equating amide formation or solid formation with polymer formation. The second reactive group is what allows repeated backbone extension.

Question 3

An idealised condensation reaction begins with:

• Six molecules of hexane-1,6-diamine
• Six molecules of hexanedioic acid

The reaction produces exactly ten molecules of water through amide bond formation.

Assume that each amide bond releases one water molecule, no rings or branched structures form, and all starting molecules remain accounted for.

A student argues:

“Ten water molecules means ten amide bonds, so the product must be one long polymer chain.”

Evaluate the student’s claim.

Determine the number of separate molecular components remaining, calculate the numbers of unreacted amine and carboxylic acid groups, and describe two different product arrangements consistent with all the evidence.

Would measuring the total number of remaining amine and acid groups distinguish your two arrangements?

Solution 3

The student’s claim is incorrect. Exactly two separate molecular components remain, with two unreacted amine groups and two unreacted carboxylic acid groups in total. The evidence does not establish the lengths of the individual components.

First, count the initial reactive groups.

Six diamine molecules provide:

\[ 6(2)=12 \]

amine groups.

Six diacid molecules provide twelve carboxylic acid groups.

Ten water molecules mean ten amide-forming condensation reactions occurred. Each consumes one group of each type.

The remaining groups are therefore:

\[ \begin{aligned} \text{Free amines}&=12-10=2\\ \text{Free acids}&=12-10=2 \end{aligned} \]

Now count the separate molecules.

Initially there are twelve molecular components. Each intermolecular amide-forming reaction joins two components into one.

With ten intermolecular connections and no rings:

\[ 12-10=2 \]

separate molecular components remain.

Two different arrangements are possible.

Arrangement A: Two chains of six monomer units each

Each chain can have an alternating sequence of three diamine-derived and three diacid-derived units.

Each six-unit chain has five amide bonds.

Together, the two chains contain ten amide bonds, two terminal amine groups, and two terminal carboxylic acid groups.

Arrangement B: One chain of eleven units and one unreacted diacid molecule

The eleven-unit chain can contain six diamine-derived sections and five diacid-derived sections.

It has ten amide bonds and two free amine ends.

The remaining hexanedioic acid molecule has two free carboxylic acid groups.

This arrangement also gives ten amide bonds, two free amine groups, and two free carboxylic acid groups.

Both satisfy the supplied evidence.

Measuring only the total remaining functional groups would not distinguish them because the totals are identical.

Evidence of molecular size or the presence of unreacted diacid would help distinguish the products.

The tempting reasoning is that ten amide bonds must mean one long chain. But forming a bond does not tell us which two molecular components joined. The same number of bonds can be distributed among different chains.

Question 4

Two different diamines are considered for condensation with hexanedioic acid.

Diamine X is hexane-1,6-diamine:

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

Diamine Y is N,N’-dimethylhexane-1,6-diamine:

\[ \ce{CH3-NH-(CH2)6-NH-CH3} \]

Assume suitable reaction conditions, comparable functional-group conversion, no competing reactions, and the formation of linear chains.

A student argues:

“Diamine Y cannot make a polyamide because each nitrogen has only one hydrogen attached.”

Assess this claim.

Then compare the ability of the resulting polymers to form hydrogen bonds between neighbouring polyamide chains. Can you conclude that the polymer from X must have greater tensile strength than the polymer from Y?

Solution 4

Both diamines can form linear polyamides under the stated assumptions, but the products have different hydrogen-bonding capabilities. The polymer made from X retains amide N-H groups, whereas the polymer made from Y does not. This difference alone is insufficient to guarantee which finished fibre will have greater tensile strength.

Diamine X has two primary amine groups.

Each primary amine can react with one carboxylic acid group to form an amide linkage. Its nitrogen retains one hydrogen after that reaction.

As a result, the polymer backbone contains secondary amide groups with N-H bonds.

Diamine Y has two secondary amine groups.

Each secondary amine contains one N-H bond and can react with a carboxylic acid group to form an amide.

After the reaction, that amide nitrogen is bonded to the carbonyl carbon, the backbone carbon group, and a methyl group.

There is no N-H bond left at that amide nitrogen.

The product contains tertiary amide linkages, but it remains a polyamide because the defining carbonyl carbon-nitrogen connection is present repeatedly along the backbone.

The student has confused the number of hydrogen atoms on each nitrogen with the number of reactive amine groups on each monomer.

Each molecule of Y still has two distinct amine functional groups, so it can connect at both ends.

The physical-property difference comes next.

In the polymer from X, N-H groups can donate hydrogen bonds to carbonyl oxygen atoms on neighbouring chains.

In the polymer from Y, the tertiary amide groups cannot donate those N-H hydrogen bonds because they have no N-H bonds. Their carbonyl oxygen atoms can still act as hydrogen-bond acceptors, and other intermolecular attractions remain possible.

Therefore, under otherwise comparable arrangements, the polymer from X has more opportunity for interchain N-H···O=C hydrogen bonding.

But tensile strength also depends on molecular mass, chain alignment, packing, crystallinity, and processing.

The correct distinction is:

• Two amine functional groups allow both monomers to participate in chain growth.
• The hydrogen-bond donor groups remaining after polymerisation influence the interactions between the finished chains.

The trap is treating the chemistry that forms the covalent backbone and the forces between completed backbones as if they were the same requirement.

Question 5

Extension challenge – cyclic products and the limits of bond counting

An idealised reaction map starts with fifteen identical molecules of 6-aminohexanoic acid:

\[ \ce{H2N-(CH2)5-COOH} \]

Each molecule has one amine group and one carboxylic acid group.

After condensation, all fifteen monomer-derived units are recovered in exactly two separate amide-containing products.

The following information is supplied:

Assume that all bonds formed through the stated amide condensation reaction, each amide bond releases one water molecule, no branching occurs, and there are no other reactions or products.

A student argues:

“Fifteen starting molecules and fourteen condensation bonds means one linear chain of fifteen units. The water result proves it.”

Determine the molecular arrangement of products P and Q.

Calculate the number of amide bonds in each product, explain why the total water result is consistent with two separate products, and assess whether water production close to one molecule per starting monomer proves that a useful long-chain linear polyamide has formed.

Solution 5

Product P must be a cyclic four-unit polyamide containing four amide bonds, while product Q must be a linear eleven-unit chain containing ten amide bonds. Together, they account for all fourteen water molecules. The student’s water-counting argument overlooks the possibility of ring formation.

Start with product Q.

It contains eleven monomer-derived units, one free amine, and one free carboxylic acid group.

For eleven units to form one unbranched, open chain, ten connections are needed.

Therefore:

\[ N_{\mathrm{Q}}=11-1=10 \]

where \(N_{\mathrm{Q}}\) is the number of amide bonds in product Q.

Ten amide-forming reactions produce ten water molecules.

Its two terminal functional groups remain available because an open chain has not connected its ends together.

Now examine product P.

It contains four monomer-derived units but has no free amine or carboxylic acid groups.

If those four units formed an ordinary open chain, they would require only three amide bonds.

That would leave one free amine and one free carboxylic acid group at the chain ends.

But the end-group results show neither.

Under the stated assumptions, the two ends must have reacted with one another, forming an additional amide bond within the same molecule.

The result is a cyclic polyamide containing four amide bonds.

Therefore:

\[ N_{\mathrm{P}}=4 \]

The total number of amide bonds is:

\[ \begin{aligned} N_{\mathrm{total}}&=N_{\mathrm{P}}+N_{\mathrm{Q}}\\ &=4+10\\ &=14 \end{aligned} \]

That accounts for all fourteen water molecules.

But why does the familiar rule of one fewer bond than starting molecules appear to fail?

It doesn’t fail. Its assumption has changed.

For a single open linear chain containing \(n\) connected building blocks, the number of connections is \(n-1\).

A ring containing \(n\) units has \(n\) connections, because the last connection closes the chain back onto itself.

That last reaction consumes another pair of functional groups and produces another water molecule without joining two previously separate molecular components.

This is the conceptual hinge.

The count of water molecules tells us how many amide bonds formed. It does not necessarily tell us how many separate molecules remain, because an intramolecular reaction can make a bond without reducing the number of molecules.

A single linear fifteen-unit chain would also contain fourteen amide bonds and produce fourteen water molecules.

So the water result alone cannot distinguish that chain from the actual mixture.

The independent evidence that two products exist, together with their sizes and end groups, rules out the student’s interpretation.

Finally, fourteen condensation reactions from fifteen starting amino-acid molecules represent a high fraction of reacted functional groups, but this does not guarantee one long, useful linear polymer.

The reaction has produced a short cyclic molecule as well as a longer open chain.

Cyclic oligomers can contain amide bonds and have all their reactive functional groups consumed without being long polymer chains.

Long fibre-forming chains require suitable monomer functionality, sufficient chain extension, control of competing reactions, and appropriate physical structure.

The most important lesson is that high conversion of reactive groups and high molecular mass are related, but they are not identical ideas. Ring formation breaks the simple inference from bond count to linear chain length.

06The next useful question: what remains at the chain ends?

When you’re given unfamiliar polyamide monomers, start by counting suitable reactive groups rather than looking for the word amide.

One amine and one carboxylic acid group on each monomer can support self-condensation. Two amines on one monomer and two carboxylic acid groups on another can support alternating chain growth. A monofunctional molecule can make an amide bond but cannot extend through a second site.

Then check what happens after those groups react. A chain may still have available ends, be permanently capped, or have closed into a ring.

Those possibilities explain why two reactions can produce the same number of amide bonds but very different molecular products.

The natural next step is to learn how to represent those remaining groups correctly in a finite polymer structure. How to Draw the End Groups of a Short Polyamide Chain develops that skill using the actual structures of the monomers, rather than relying only on a repeating-unit formula.