How to Draw the End Groups of a Short Polyamide Chain

Suppose an HSC Chemistry question gives you two molecules of hexane-1,6-diamine and two molecules of hexanedioic acid, then asks you to draw the short polyamide chain they form. You know the polymer contains amide links, but what should the two ends of the chain look like?

This is where copying a repeating unit can lead you astray. A repeating unit shows the pattern inside a polymer. It doesn’t necessarily show the actual ends of a finite molecule.

For a linear, uncapped nylon-6,6 chain made from equal numbers of diamine and dicarboxylic acid molecules, the two ends are:

• One terminal amine group, \(\ce{-NH2}\).
• One terminal carboxylic acid group, \(\ce{-COOH}\).

The key is to keep track of which functional groups have reacted and which are still available at the ends.

Predict something before we start: if four monomer molecules join into one straight chain, how many amide links will form? Three, not four. Four connected units need only three connections between them. That simple counting rule makes drawing short polyamides much easier.

01Start with the two monomers

We’ll use nylon-6,6 as our example. It forms through a condensation reaction between two bifunctional monomers.

A bifunctional monomer has two reactive functional groups, allowing it to connect to neighbouring molecules at both ends.

Notice that each monomer has a reactive group at either end.

What happens when the monomers connect?

An amine group reacts with a carboxylic acid group to form an amide link and release water.

The general reaction is:

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

Two things happen:

1. The \(\ce{-OH}\) from the carboxylic acid and one \(\ce{-H}\) from the amine combine to form water.
2. The remaining carbonyl carbon connects to the nitrogen, forming an amide linkage.

The amide linkage is:

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

You may also see it written as \(\ce{-CONH-}\).

Be careful not to confuse this with an ester linkage, \(\ce{-COO-}\). An ester contains an oxygen between the carbonyl carbon and the next section of the chain. An amide contains nitrogen.

02How to draw the ends of a short polyamide chain

There are four steps to follow.

Step 1: Write out the monomers.

Start with the full structural formulas of the diamine and dicarboxylic acid.

Don’t remove any atoms yet.

Step 2: Arrange the monomers in an alternating sequence.

For nylon-6,6, the arrangement is:

\[ A-B-A-B-A-B \]

Here:

• \(A\) represents a diamine molecule.
• \(B\) represents a dicarboxylic acid molecule.

Each neighbouring pair must connect through an amine and a carboxylic acid group.

Step 3: Form the amide links.

At every connection:

• Change the reacting \(\ce{-NH2}\) group into \(\ce{-NH-}\).
• Change the reacting \(\ce{-COOH}\) group into the carbonyl section of an amide.
• Remove one water molecule per connection.

Step 4: Leave the unused groups at the ends.

This is the important step.

The two outer functional groups have not reacted with another monomer. Therefore, they remain as complete functional groups.

For an equimolar chain, one end remains \(\ce{-NH2}\), and the other remains \(\ce{-COOH}\).

Think of connecting train carriages. Four carriages need three couplings, leaving one exposed end at either side. The analogy helps with counting, although real polymer connections form through chemical reactions between atoms, not detachable mechanical couplings.

03What happens when there are unequal numbers of monomers?

Here’s where students often make an incorrect assumption.

You might think that every nylon-6,6 chain must have one amine end and one carboxylic acid end. That is only true when the chain contains equal numbers of diamine and dicarboxylic acid molecules.

If the monomer ratio changes, the terminal groups can change too.

A quick guide to terminal groups

Assuming one linear, alternating chain with no branching, rings, or end-capping:

If the difference is greater than one, you cannot incorporate all those monomers into one simple alternating linear chain. Some monomers must remain unreacted or form other chains.

A useful contrast: Polyester end groups

Don’t assume that every condensation polymer has amine and carboxylic acid ends.

Consider PET, a polyester formed from ethane-1,2-diol and terephthalic acid.

The monomers contain:

• Hydroxyl groups, \(\ce{-OH}\).
• Carboxylic acid groups, \(\ce{-COOH}\).

They form ester linkages:

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

For a finite linear chain containing equal numbers of these monomers, the terminal groups are one \(\ce{-OH}\) and one \(\ce{-COOH}\).

There is no terminal amine because the monomers do not contain amine groups.

The general rule is simple: identify the functional groups in the original monomers, then work out which groups have reacted.

For a broader comparison of these polymerisation reactions, see Condensation Polymerisation: Nylon and Polyester Explained for HSC Chemistry.

04Why do the end groups matter?

The terminal groups are important for drawing a short polyamide correctly, but they also have chemical consequences.

A terminal amine group can accept a proton, while a terminal carboxylic acid group can donate a proton.

These groups therefore influence the acid-base behaviour and potential reactivity of a short polyamide molecule.

However, the properties of a long nylon chain depend much more heavily on its repeating structure and interactions between neighbouring chains.

Amide links and hydrogen bonding

Look at the amide linkage again:

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

It contains two particularly important features:

• The N-H bond can act as a hydrogen-bond donor.
• The carbonyl oxygen can act as a hydrogen-bond acceptor.

This allows neighbouring polyamide chains to form hydrogen bonds.

These attractions help hold the chains together and restrict their relative movement.

This contributes to nylon’s strength and toughness, making it useful for fibres, ropes, and other applications requiring durable materials.

There is an important distinction here: the amide links are covalent bonds within each polymer chain. The hydrogen bonds are attractions between neighbouring chains.

The terminal groups become proportionally less significant as a chain gets longer.

For a chain containing two pairs of monomers, there are two terminal groups and three amide links. The ends are a substantial part of the structure.

For a chain containing 1,000 pairs, there are still only two terminal groups, but there are 1,999 amide links.

This is why end groups deserve particular attention when drawing short oligomers, even though the repeating structure dominates the behaviour of long polymer chains.

For more on the relationship between amide linkages, hydrogen bonding, and nylon’s properties, see Polyamides: How Nylon Forms, Its Structure and Properties.

05How could a chemist test which end groups are present?

So far, we’ve used the monomer structures to predict the terminal groups. But could a chemist test that prediction experimentally?

One possible approach is end-group analysis using acid-base titration.

Consider an illustrative experiment involving a purified, soluble short polyamide sample.

The apparatus works as follows:

1. A known amount of the soluble polyamide sample is placed in a stirred flask.
2. A burette containing a standard base solution is positioned above the flask.
3. The base is added gradually while the sample is stirred.
4. A suitable indicator or electrode detects the neutralisation endpoint.
5. The volume of base required is recorded.

The carboxylic acid end groups react with hydroxide ions:

\[ \ce{R-COOH + OH^- -> R-COO^- + H2O} \]

Each mole of carboxylic acid groups reacts with one mole of hydroxide ions.

What does this experiment actually prove?

There is an important limitation.

The titration measures the amount of carboxylic acid groups. It does not directly prove that every chain also has one amine end.

For example, unreacted adipic acid could interfere with the measurement if it had not been removed from the sample.

A better experimental design would first purify the polymer to remove unreacted monomers. The chemist could then use a separate suitable analysis to measure the terminal amine groups.

Even if the two measurements give equal totals, they do not necessarily prove that every individual chain has one group of each type. A mixture of chains with different terminal groups could produce the same overall totals.

This distinction is important: a chemical measurement tells you what is present in the sample, while the structural model explains how those groups are arranged.

The titration is an illustrative analytical extension, not a requirement for drawing the chain. A real analysis would also need a solvent and measurement method suitable for the particular polyamide.

06Questions and solutions

The following questions move from straightforward chain drawing towards more demanding problems involving monomer ratios, experimental evidence, and multiple polymer chains.

Question 1

A single linear nylon-6,6 chain contains three molecules of hexane-1,6-diamine and three molecules of hexanedioic acid.

Determine:

a) The number of amide linkages.

b) The terminal functional groups.

c) The number of water molecules produced during formation of the chain.

Solution 1

The chain contains five amide linkages, one terminal amine group, one terminal carboxylic acid group, and produces five water molecules.

Part a: Number of amide linkages

There are six monomer molecules in one linear chain.

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

\[ N_{\text{links}}=6-1=5 \]

Therefore, five amide linkages form.

Part b: Terminal groups

Initially, there are:

\[ 3\times2=6\text{ amine groups} \]

and

\[ 3\times2=6\text{ carboxylic acid groups} \]

Five amide linkages consume five groups of each type.

This leaves:

\[ 6-5=1 \]

unused amine group and one unused carboxylic acid group.

Therefore, the terminal groups are \(\ce{-NH2}\) and \(\ce{-COOH}\).

Part c: Water molecules

Each amide linkage produces one water molecule.

\[ n(\ce{H2O})=5 \]

The chain therefore produces five water molecules.

The key is that equal numbers of the two bifunctional monomers leave one unused functional group of each type in a single linear chain.

Question 2

A chemist prepares a short linear nylon-6,6 chain using two molecules of hexane-1,6-diamine and three molecules of hexanedioic acid.

A student claims that the chain must have one \(\ce{-NH2}\) end and one \(\ce{-COOH}\) end because nylon-6,6 always has these terminal groups.

Evaluate the student’s claim.

Determine the actual terminal groups, the number of amide linkages, and the number of water molecules produced.

Solution 2

The student’s claim is incorrect. This chain has two terminal \(\ce{-COOH}\) groups, four amide linkages, and produces four water molecules.

Step 1: Count the monomer molecules.

There are five monomer molecules:

\[ N_{\text{monomers}}=2+3=5 \]

Therefore:

\[ N_{\text{links}}=5-1=4 \]

Step 2: Count the original functional groups.

The two diamine molecules provide:

\[ 2\times2=4\text{ amine groups} \]

The three dicarboxylic acid molecules provide:

\[ 3\times2=6\text{ carboxylic acid groups} \]

Step 3: Determine the remaining groups.

Four amide linkages consume four amine groups and four carboxylic acid groups.

Therefore:

\[ N_{\text{amine remaining}}=4-4=0 \]

\[ N_{\text{acid remaining}}=6-4=2 \]

Both remaining groups are carboxylic acid groups.

Step 4: Calculate water production.

\[ n(\ce{H2O})=4 \]

The chain has two carboxylic acid ends because there was one more dicarboxylic acid molecule than diamine molecule.

The tempting mistake is to memorise the terminal groups of an equimolar nylon-6,6 chain and apply them to every possible chain. Always count the original functional groups before drawing the ends.

Question 3

A student is asked to draw a short linear polyester formed from two molecules of ethane-1,2-diol and two molecules of terephthalic acid.

They draw one terminal \(\ce{-NH2}\) group and one terminal \(\ce{-COOH}\) group.

Explain why this structure is incorrect.

Identify the correct linkage and terminal groups.

Solution 3

The student’s structure is incorrect because this polymer is a polyester, not a polyamide. Its terminal groups are one \(\ce{-OH}\) group and one \(\ce{-COOH}\) group.

Step 1: Identify the monomers.

Ethane-1,2-diol contains two hydroxyl groups:

\[ \ce{HO-CH2-CH2-OH} \]

Terephthalic acid contains two carboxylic acid groups:

\[ \ce{HOOC-C6H4-COOH} \]

Neither monomer contains an amine group.

Step 2: Identify the linkage.

A hydroxyl group reacts with a carboxylic acid group to form an ester linkage:

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

This is different from the amide linkage:

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

Step 3: Determine the terminal groups.

There are four monomer molecules, so the chain contains three ester connections.

Each connection consumes one hydroxyl group and one carboxylic acid group.

The two unused outer groups are therefore:

• One \(\ce{-OH}\) group.
• One \(\ce{-COOH}\) group.

The terminal groups depend on the functional groups present in the original monomers. You cannot apply the nylon-6,6 end-group rule to a polyester.

Question 4

A chemist analyses a purified, soluble short polyamide sample using an acid-base titration.

The sample requires 12.0 mL of \(0.0200\ \mathrm{mol\,L^{-1}}\) sodium hydroxide to neutralise its terminal carboxylic acid groups.

Assume each polymer chain contains exactly one terminal \(\ce{-COOH}\) group.

a) Calculate the amount, in moles, of carboxylic acid groups in the sample.

b) Calculate the amount, in moles, of polymer chains.

c) A student claims that the result proves every chain contains exactly one terminal \(\ce{-NH2}\) group as well. Evaluate this conclusion.

Solution 4

The sample contains \(2.40\times10^{-4}\) mol of carboxylic acid groups and, under the stated assumption, the same amount of polymer chains. However, the titration alone does not prove that every chain has an amine end group.

Part a: Calculate the amount of carboxylic acid groups.

Convert the volume into litres:

\[ V=12.0\times10^{-3}=0.0120\ \mathrm{L} \]

Use:

\[ n=cV \]

Substitute:

\[ \begin{aligned} n(\mathrm{OH}^{-})&=0.0200\times0.0120\\ &=2.40\times10^{-4}\ \mathrm{mol} \end{aligned} \]

The neutralisation reaction has a 1:1 mole ratio.

Therefore:

\[ n(\ce{-COOH})=2.40\times10^{-4}\ \mathrm{mol} \]

Part b: Calculate the amount of polymer chains.

The question states that each chain contains exactly one carboxylic acid end group.

Therefore:

\[ n(\text{chains})=2.40\times10^{-4}\ \mathrm{mol} \]

Part c: Evaluate the student’s conclusion.

The conclusion is not established by this experiment.

The titration measures the total number of carboxylic acid groups. It does not directly measure the number of terminal amine groups.

For example, some chains could have capped ends or different terminal structures.

The chemist would need a separate analysis to measure the amine groups.

Even equal amounts of amine and carboxylic acid groups would not, by themselves, prove that the two groups occur together on every individual chain.

The distinction is between measuring the number of functional groups in a sample and determining how those groups are distributed across individual molecules.

Question 5

A chemist produces two separate, linear, unbranched nylon-6,6 chains.

Across the two chains, the chemist incorporates:

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

Assume:

• All nine monomer molecules are incorporated into the two chains.
• Each chain contains alternating diamine and dicarboxylic acid residues.
• No cyclic structures or end-capping reactions occur.
• Each amide linkage forms through condensation between one amine group and one carboxylic acid group.

Determine:

a) The total number of amide linkages across both chains.

b) The total number of terminal \(\ce{-NH2}\) groups.

c) The total number of terminal \(\ce{-COOH}\) groups.

d) One possible distribution of monomer molecules between the two chains that is consistent with your answer.

Explain why treating the entire sample as one continuous chain would give an incorrect result.

Solution 5

Across the two chains, there are seven amide linkages, three terminal \(\ce{-NH2}\) groups, and one terminal \(\ce{-COOH}\) group.

The crucial detail is that the sample contains two separate chains, not one. Each chain needs two ends.

Part a: Calculate the number of amide linkages.

There are nine monomer molecules in total.

For one linear chain, the number of connections is one fewer than the number of monomer molecules.

For two separate chains, we subtract two connections from the total number of monomer molecules:

\[ N_{\text{links}}=9-2 \]

\[ \boxed{N_{\text{links}}=7} \]

Seven amide linkages form.

Part b: Calculate the terminal amine groups.

Initially, five diamine molecules provide:

\[ 5\times2=10\text{ amine groups} \]

Each of the seven amide linkages consumes one amine group.

Therefore:

\[ N_{\ce{-NH2}}=10-7=3 \]

There are three terminal amine groups.

Part c: Calculate the terminal carboxylic acid groups.

Initially, four dicarboxylic acid molecules provide:

\[ 4\times2=8\text{ carboxylic acid groups} \]

Seven amide linkages consume seven of these groups.

Therefore:

\[ N_{\ce{-COOH}}=8-7=1 \]

There is one terminal carboxylic acid group.

Check the total number of ends:

\[ 3+1=4 \]

Two linear chains must have four terminal groups altogether. Our result satisfies this condition.

Part d: Construct one possible arrangement.

One valid arrangement is:

• Chain 1 contains three diamine molecules and two dicarboxylic acid molecules.
• Chain 2 contains two diamine molecules and two dicarboxylic acid molecules.

Chain 1 has two terminal amine groups because it contains one extra diamine molecule.

Chain 2 has one terminal amine group and one terminal carboxylic acid group because its monomer quantities are equal.

Across both chains, this gives:

\[ 3\text{ amine ends}+1\text{ acid end} \]

The total number of connections is also consistent:

\[ 4+3=7 \]

Why the tempting answer is wrong

A student might see five diamine molecules and four dicarboxylic acid molecules and conclude that one extra diamine must produce exactly two amine ends.

That answer assumes there is only one chain.

If all nine monomers formed one continuous chain, there would be eight connections and two amine ends. But the question specifies two separate chains.

The second chain introduces another pair of terminal groups. Consequently, the sample has four ends rather than two.

This is the important distinction: monomer quantities alone do not determine the terminal groups across a sample. You must also know how many separate chains have formed.

07The drawing rule to remember

When drawing a short polyamide chain, work from the monomers rather than copying a repeating unit and guessing the ends.

1. Identify the two monomers and their functional groups.
2. Arrange their residues in alternating order.
3. Form an amide linkage at each connection.
4. Count the connections and account for the water molecules.
5. Leave the unused functional groups at the outer ends of each chain.

For an ordinary linear nylon-6,6 chain with equal numbers of diamine and dicarboxylic acid molecules, the result is one \(\ce{-NH2}\) end and one \(\ce{-COOH}\) end.

If the monomer ratio changes, the terminal groups can change too. If multiple chains are present, count the ends of each chain rather than treating the sample as one continuous molecule.

Once you can draw these structures correctly, the next useful step is connecting the number of amide linkages and the arrangement of the polymer chains to their hydrogen bonding, molecular mass, and physical properties.