What Is Removed When an Amide Link Forms in Polyamides?

A nylon fibre contains thousands of repeated amide links, each written as \(\ce{-CO-NH-}\). But the monomers that form this link originally contain a carboxylic acid group, \(\ce{-COOH}\), and an amine group, \(\ce{-NH2}\).

Something has to leave before the new bond can form. What disappears? Does the entire \(\ce{-OH}\) group leave? Does the nitrogen lose both its hydrogen atoms?

The answer is water, \(\ce{H2O}\). When a carboxylic acid reacts with a primary amine to form an amide link, the \(\ce{-OH}\) from the carboxylic acid and one hydrogen from the amine combine to form one water molecule.

The carbonyl oxygen stays in the polymer, and the nitrogen retains one hydrogen. Getting these details right is the key to writing correct polyamide polymerisation equations and counting the water produced.

01What is removed when an amide link forms?

Consider a carboxylic acid reacting with a primary amine.

The carboxylic acid contains the functional group \(\ce{-COOH}\), while the amine contains \(\ce{-NH2}\). During condensation, these groups combine to form an amide link.

The overall reaction is:

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

Here, \(R\) and \(R'\) represent the remaining parts of the organic molecules.

Let’s track exactly what happens.

The two removed components combine:

\[ \ce{OH + H -> H2O} \]

This produces one water molecule for every amide link formed.

Two mistakes to avoid

Mistake 1: Removing both hydrogens from the amine.

The nitrogen loses only one hydrogen. The other remains attached, giving the \(\ce{-NH-}\) portion of the amide link.

Mistake 2: Removing the carbonyl oxygen.

The oxygen double-bonded to carbon stays in the polymer. It is the hydroxyl part of the carboxylic acid that is removed.

The resulting amide functional group is:

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

You can revise the structures and reactions of these functional groups in HSC Chemistry: Carboxylic Acids, Amines and Amides Explained.

One important qualification: the equation represents the overall condensation reaction, not necessarily the individual reaction steps. Carboxylic acids and amines can initially form ammonium carboxylate salts. Heating and removing water allow the polycondensation process to produce amide links.

02How does this work in polyamide formation?

A single amide link forms when one carboxylic acid group reacts with one amine group.

To make a polymer, however, we need monomers with multiple reactive groups.

For a typical polyamide such as nylon 6,6, we use:

• A dicarboxylic acid containing two \(\ce{-COOH}\) groups.
• A diamine containing two \(\ce{-NH2}\) groups.

Each end can react with another monomer, allowing the molecules to join into a long chain.

Example: Making nylon 6,6

Nylon 6,6 is made from two monomers:

1. Hexanedioic acid (adipic acid)

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

It contains two carboxylic acid groups.

2. Hexane-1,6-diamine

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

It contains two primary amine groups.

The resulting polymer has the repeating structure:

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

Notice the two different parts of the chain:

• The \(\ce{-CO-NH-}\) sections are the amide links.
• The carbon-containing sections between them come from the original monomers.

Why are two water molecules produced per repeat unit?

This is an important distinction.

A repeat unit of nylon 6,6 contains two amide links when the connections to neighbouring repeat units are included.

Each amide link produces one water molecule.

Therefore:

\[ \boxed{\text{2 amide links} \rightarrow \text{2 water molecules}} \]

The idealised polymerisation equation is:

\[ \begin{aligned} &n\ce{HOOC-(CH2)4-COOH}\\ &+n\ce{H2N-(CH2)6-NH2}\\ &\longrightarrow \ce{[-CO-(CH2)4-CO-NH-(CH2)6-NH-]_{n}}\\ &+2n\ce{H2O} \end{aligned} \]

This equation uses ideal repeat-unit accounting, with chain-end effects neglected.

For a finite linear chain containing \(n\) molecules of each monomer, there are actually \(2n-1\) amide links, rather than \(2n\). The two ends of the chain remain as terminal functional groups.

We will return to this distinction in the final practice question.

For a broader explanation of how the monomers join and how nylon differs from polyester, see Condensation Polymerisation: Nylon and Polyester Explained for HSC Chemistry.

03Why does the amide link matter for nylon’s properties?

Removing water explains how the polymer forms. But the resulting amide link also helps explain why nylon has useful physical properties.

Look at the structure:

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

The amide group contains two particularly important features:

• The \(\ce{N-H}\) bond can act as a hydrogen bond donor.
• The carbonyl oxygen, \(\ce{C=O}\), can act as a hydrogen bond acceptor.

This allows neighbouring polyamide chains to form hydrogen bonds with one another.

These hydrogen bonds increase the attraction between neighbouring chains.

As a result, nylon can have:

• High tensile strength.
• Good resistance to mechanical deformation.
• Relatively high melting temperatures compared with many polymers lacking similarly strong interchain attractions.

The hydrogen bonds are not part of the covalent polymer backbone. They act between separate chains.

This distinction matters when explaining polymer properties in HSC Chemistry: the covalent bonds hold each chain together, while intermolecular attractions help hold neighbouring chains together.

04A useful contrast: Amide links versus ester links

You might have seen polyester formation alongside nylon in condensation polymerisation.

Both reactions can release water, but the functional groups involved are different.

The important difference is the group that remains in the polymer.

For an amide link:

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

For an ester link:

\[ \ce{-COOH + HO- -> -COO- + H2O} \]

Both release water, but one produces an amide and the other produces an ester.

This also affects their intermolecular attractions. Polyamides contain N-H groups that can donate hydrogen bonds. The polyester backbone does not have these same N-H donor groups.

Does every amide-forming reaction release water?

Not necessarily.

The water-producing reaction described above uses a carboxylic acid and an amine.

An alternative reaction uses an acyl chloride and an amine:

\[ \ce{RCOCl + H2N-R' -> RCONH-R' + HCl} \]

In this case, hydrogen combines with chlorine to produce hydrogen chloride, \(\ce{HCl}\), rather than water.

This is an alternative condensation route and an extension beyond the central dicarboxylic acid-diamine model used here.

The useful rule is to examine the reacting functional groups before identifying the small molecule removed. Do not assume that every reaction forming an amide must release water.

05Worked examples

06Questions and solutions

Question 1

A polymerisation reaction forms 0.0400 mol of amide links.

Calculate the amount, in moles, and the mass, in grams, of water produced.

Assume water is the only small-molecule by-product.

Solution 1

The reaction produces 0.0400 mol of water, with a mass of 0.720 g.

Each amide link produces one water molecule.

Therefore:

\[ n(\ce{H2O})=0.0400\text{ mol} \]

Using \(m=nM\):

\[ \begin{aligned} m(\ce{H2O}) &=0.0400\times18.0\\ &=0.720\text{ g} \end{aligned} \]

The important point is that we count amide links, not the number of polymer chains. A single polymer chain can contain many thousands of amide links.

Question 2

A new polyamide is produced using:

• Butanedioic acid: \(\ce{HOOC-(CH2)2-COOH}\).
• Pentane-1,5-diamine: \(\ce{H2N-(CH2)5-NH2}\).

a) Write the repeating unit of the resulting polyamide.

b) Identify the functional group connecting the monomers.

c) Calculate the number of water molecules produced per ideal repeat unit.

d) Explain why a student who predicts one water molecule per pair of monomers is incorrect.

Solution 2

The repeating unit contains two amide links and produces two water molecules per ideal repeat unit.

a) Repeating unit

The dicarboxylic acid contributes:

\[ \ce{-CO-(CH2)2-CO-} \]

The diamine contributes:

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

Combining these gives:

\[ \boxed{\ce{[-CO-(CH2)2-CO-NH-(CH2)5-NH-]_{n}}} \]

b) Functional group

The monomers are connected by amide links:

\[ \ce{-CONH-} \]

These form when the carboxylic acid groups react with the amine groups.

c) Water produced

Each monomer has two reactive ends.

The two carboxylic acid groups can react with two amine groups, producing two amide links per ideal repeat unit.

Since each link produces one water molecule:

\[ \boxed{2\ce{H2O}\text{ per ideal repeat unit}} \]

d) Why one water molecule is incorrect

The tempting prediction is that one pair of monomers produces one water molecule because there is one acid molecule and one diamine molecule.

However, this ignores the fact that both molecules are bifunctional.

Each contains two reactive groups, allowing two amide links to form per ideal repeat unit.

Therefore, two water molecules are produced, not one.

Question 3

A chemist reacts:

• 0.180 mol of a dicarboxylic acid.
• 0.150 mol of a diamine.

Assume each dicarboxylic acid contains two \(\ce{-COOH}\) groups and each diamine contains two \(\ce{-NH2}\) groups.

The reaction proceeds until all amine groups have reacted.

Calculate:

a) The initial amount of carboxylic acid groups.

b) The initial amount of amine groups.

c) The maximum amount of amide links formed.

d) The amount and mass of water produced.

e) The amount of carboxylic acid groups remaining unreacted.

Solution 3

The amine groups are the limiting functional groups, allowing 0.300 mol of amide links and 0.300 mol of water to form.

a) Initial amount of carboxylic acid groups

Each acid molecule contains two carboxylic acid groups.

\[ \begin{aligned} n(\ce{-COOH}) &=2\times0.180\\ &=0.360\text{ mol} \end{aligned} \]

b) Initial amount of amine groups

Each diamine molecule contains two amine groups.

\[ \begin{aligned} n(\ce{-NH2}) &=2\times0.150\\ &=0.300\text{ mol} \end{aligned} \]

c) Maximum amount of amide links

Each amide link requires one carboxylic acid group and one amine group.

There are 0.360 mol of acid groups and 0.300 mol of amine groups.

Therefore, the amine groups are limiting.

\[ \boxed{n(\text{amide links})=0.300\text{ mol}} \]

d) Amount and mass of water

One amide link produces one water molecule.

\[ n(\ce{H2O})=0.300\text{ mol} \]

The mass is:

\[ \begin{aligned} m(\ce{H2O}) &=0.300\times18.0\\ &=5.40\text{ g} \end{aligned} \]

e) Remaining carboxylic acid groups

Initially, there were 0.360 mol of carboxylic acid groups.

Each of the 0.300 mol of amide links consumes one carboxylic acid group.

Therefore:

\[ \begin{aligned} n(\ce{-COOH})_{\text{remaining}} &=0.360-0.300\\ &=0.060\text{ mol} \end{aligned} \]

The excess carboxylic acid groups remain available as unreacted functional groups, including at polymer chain ends.

The key lesson is that you must count functional groups, not just monomer molecules, when calculating the amount of water produced.

Question 4

Two polymers are being compared.

Polymer A is a polyamide containing repeating \(\ce{-CO-NH-}\) groups.

Polymer B is a polyester containing repeating \(\ce{-CO-O-}\) groups.

Assume their chain lengths and degrees of crystallinity are similar.

a) Which polymer can form regular hydrogen bonds between neighbouring chains through N-H donor groups?

b) Explain how the functional groups allow these attractions to form.

c) Explain how these attractions affect the physical properties of the polymer.

d) Does the presence of amide links alone prove that Polymer A must have a higher melting point than Polymer B? Explain.

Solution 4

Polymer A, the polyamide, can form regular interchain hydrogen bonds through its N-H groups and neighbouring carbonyl oxygen atoms.

a) Identify the polymer

Polymer A contains the amide functional group:

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

Polymer B contains the ester functional group:

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

Only Polymer A has the repeating N-H groups required to donate hydrogen bonds in the manner described.

b) Explain the hydrogen bonding

The N-H bond in an amide is polar.

The hydrogen attached to nitrogen can be attracted to the lone pair on the oxygen of a neighbouring carbonyl group.

This produces an interchain hydrogen bond:

\[ \ce{N-H\bond{...}O=C} \]

The attractions form between separate polymer chains, rather than along the covalent backbone.

The polyester has carbonyl oxygen atoms that can accept hydrogen bonds, but its repeating ester backbone does not contain the corresponding N-H donor groups.

c) Effect on physical properties

The hydrogen bonds increase the attraction between neighbouring polyamide chains.

This makes it harder to separate the chains, helping the material resist deformation and contributing to its tensile strength and thermal properties.

d) Can we guarantee a higher melting point?

No. Hydrogen bonding alone does not determine the melting point.

Polymer melting behaviour also depends on factors such as:

• Molecular mass and chain length.
• Crystallinity.
• How regularly the chains pack together.
• Other intermolecular attractions.

Polymer A has an additional source of strong interchain attraction, but we cannot establish an exact melting-point ranking using the amide functional group alone.

The distinction is between identifying a contributing factor and claiming that it determines the entire property.

Question 5

A chemist produces a polyamide from 100.0 mol of hexanedioic acid and 100.0 mol of hexane-1,6-diamine.

The following conditions apply:

• Every monomer molecule becomes part of a polymer chain.
• All chains are linear, unbranched, and non-cyclic.
• Each chain contains equal numbers of acid and diamine monomers.
• Every chain therefore has one terminal \(\ce{-COOH}\) group and one terminal \(\ce{-NH2}\) group.
• No monomer molecules remain unreacted.
• Each amide link produces one water molecule.

After the reaction, the chemist collects 198.0 mol of water.

An excellent student argues that the reaction must be incomplete because 100.0 mol of each monomer should produce 200.0 mol of water.

Is the student correct?

Calculate:

a) The total amount of monomer molecules initially present.

b) The amount of amide links formed.

c) The amount of polymer chains produced, in moles.

d) The average number of monomer molecules in each polymer chain.

e) Explain precisely why the chemist collects 198.0 mol rather than 200.0 mol of water.

Solution 5

The student is incorrect. The reaction can incorporate every monomer molecule into a polymer chain while producing only 198.0 mol of water because each finite linear chain has two terminal functional groups that have not formed an amide link.

This is the difference between an ideal repeat-unit equation and an exact finite-chain calculation.

a) Total amount of monomer molecules

There are two monomers, each present in an amount of 100.0 mol.

Therefore:

\[ \begin{aligned} n(\text{monomers}) &=100.0+100.0\\ &=200.0\text{ mol} \end{aligned} \]

There are 200.0 mol of individual monomer molecules in total.

b) Amount of amide links formed

One amide link produces one water molecule.

Therefore:

\[ \boxed{n(\text{amide links})=198.0\text{ mol}} \]

c) Amount of polymer chains

Initially, there are:

\[ 100.0\times2=200.0\text{ mol} \]

of carboxylic acid groups.

There are also 200.0 mol of amine groups.

The 198.0 mol of amide links consume 198.0 mol of each type of functional group.

The remaining amounts are:

\[ \begin{aligned} n(\ce{-COOH})_{\text{remaining}} &=200.0-198.0\\ &=2.0\text{ mol} \end{aligned} \]

Similarly:

\[ n(\ce{-NH2})_{\text{remaining}}=2.0\text{ mol} \]

Every polymer chain has one terminal carboxylic acid group and one terminal amine group.

Therefore, the number of moles of polymer chains is:

\[ \boxed{n(\text{chains})=2.0\text{ mol}} \]

d) Average number of monomer molecules per chain

There are 200.0 mol of monomer molecules distributed across 2.0 mol of polymer chains.

\[ \begin{aligned} \text{Average monomers per chain} &=\frac{200.0}{2.0}\\ &=100 \end{aligned} \]

Each chain therefore contains, on average:

• 50 acid molecules.
• 50 diamine molecules.
• 100 monomer molecules in total.

e) Why is only 198.0 mol of water produced?

This is the important conceptual point.

A linear chain containing 100 monomer molecules has 99 connections between consecutive monomer molecules.

It does not have 100 connections.

The two terminal groups sit at opposite ends of the chain. They do not connect to each other, so one potential amide link is missing per chain.

Therefore, each chain produces 99 water molecules rather than 100.

Across 2.0 mol of chains:

\[ \begin{aligned} n(\ce{H2O}) &=2.0\times99\\ &=198.0\text{ mol} \end{aligned} \]

This agrees with the measured amount.

The tempting calculation of 200.0 mol of water comes from treating the polymer as an ideal repeating structure and ignoring chain ends.

For the actual finite chains, one water molecule is missing per chain relative to that idealised count.

The remaining functional groups are not evidence of unreacted monomer. They are the natural terminal groups of the completed linear polymer chains.

07What to remember

For the standard condensation reaction between a dicarboxylic acid and a diamine:

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

Remember these four points:

1. The hydroxyl group from the carboxylic acid combines with one hydrogen from the amine to form water.
2. The carbonyl oxygen remains in the polymer.
3. The nitrogen remains in the polymer, retaining one hydrogen in the \(\ce{-NH-}\) link.
4. Each amide link produces one water molecule.

For nylon 6,6, there are two amide links per ideal repeat unit, so the idealised equation produces two water molecules per repeat unit.

The next useful skill is distinguishing a repeating unit from a finite polymer chain. Once you can draw the terminal groups correctly, you can count amide links, water molecules, and polymer mass without accidentally counting a connection that does not exist.