How to Find Monomers from a Polymer’s Repeating Unit

A polymer is shown with the repeating unit \(\ce{[-CH2-CH(CH3)-]_{n}}\). You can see that the carbon atoms are joined by single bonds, but the original monomer might have contained a double bond. Where did it go?

The monomer is propene, \(\ce{CH2=CHCH3}\). To find it, remove the bonds connecting neighbouring repeating units and restore a carbon-carbon double bond between the two backbone carbon atoms. Keep the methyl group attached to the same carbon.

This is the essential reverse-identification method for addition polymers. The challenge is recognising which bonds connect the repeating units, which atoms belong to the polymer backbone, and which groups must stay exactly where they are.

01Understand what the repeating unit represents

Before trying to reconstruct a monomer, you need to read the polymer structure correctly.

Consider:

\[ \ce{[-CH2-CH(CH3)-]_{n}} \]

There are three important features.

The square brackets identify the repeating unit. This is the smallest structural section used to represent the repeating pattern in the polymer.

The subscript \(n\) indicates that this unit is repeated many times. It does not represent the number of carbon atoms or the number of monomers in a single repeating unit.

The dashes at either end represent bonds connecting the repeating unit to neighbouring units in the polymer chain.

These are particularly important when finding the original monomer.

A polymer chain containing several polypropylene units can be represented as:

\[ \ce{...-CH2-CH(CH3)-CH2-CH(CH3)-CH2-CH(CH3)-...} \]

Each section contains two backbone carbon atoms and one methyl side group.

Think of the repeating units as sections of a train. The connections at either end join one section to the next. The brackets identify one section, while the open bonds show where the chain continues.

The analogy has a limit: polymer chains are not separate objects clipped together. They are continuous molecules with covalent bonds connecting their atoms.

What are the backbone and side groups?

The polymer backbone is the continuous chain of atoms running through the polymer.

A side group is attached to the backbone but does not form part of that continuous chain.

For polypropylene:

• The backbone contains alternating \(\ce{CH2}\) and \(\ce{CH}\) carbon atoms.
• Each \(\ce{CH}\) carbon has a \(\ce{CH3}\) side group.
• The side groups remain attached during the reverse reconstruction.

This distinction becomes essential when the repeating unit contains larger or more complicated side groups.

02The four-step method for finding an addition polymer’s monomer

For the common HSC examples of addition polymers formed from alkenes, use this method.

Step 1: Identify the carbon backbone

Find the two adjacent carbon atoms that form the repeating unit’s main chain section.

For a simple alkene-derived polymer, each monomer contributes two carbon atoms to the backbone.

For example:

\[ \ce{[-CH2-CH(Cl)-]_{n}} \]

The two backbone carbons are:

• \(\ce{CH2}\)
• \(\ce{CH(Cl)}\)

The chlorine atom is a side group. It is not part of the carbon backbone.

Step 2: Remove the chain-continuation bonds

The bonds extending from either end of the repeating unit connect it to neighbouring units.

When reversing the polymerisation, remove these external connections.

Do not remove the bonds holding side groups to the backbone.

Step 3: Restore the carbon-carbon double bond

The original alkene monomer contained a C=C double bond.

During addition polymerisation, this double bond becomes a single bond as the monomer molecules connect to form a chain.

To reverse the process, restore the double bond between the two backbone carbon atoms.

Step 4: Check the substituents and valence

Keep every side group attached to its original carbon.

Then check that each carbon has four covalent bonds in total, counting a double bond as two bonds.

If your reconstructed structure gives a carbon five bonds, something has gone wrong.

03Use carbon valence to catch mistakes

A common mistake is to add hydrogen atoms to the ends of the repeating unit instead of restoring the original double bond.

Consider poly(chloroethene), commonly called PVC:

\[ \ce{[-CH2-CH(Cl)-]_{n}} \]

What monomer would you predict?

A tempting answer is chloroethane, \(\ce{CH3CH2Cl}\). After all, the repeating unit contains two carbon atoms joined by a single bond, with chlorine attached to one of them.

But this reconstruction is incorrect.

The problem is that the external dashes are not missing hydrogen atoms. They represent bonds connecting neighbouring repeating units.

The correct reconstruction restores the double bond:

\[ \ce{CH2=CHCl} \]

The monomer is chloroethene, also known as vinyl chloride.

Why does the correct structure satisfy carbon valence?

Carbon forms four covalent bonds.

For chloroethene, \(\ce{CH2=CHCl}\):

Both carbon atoms satisfy their usual valence of four.

This is a useful final check whenever you reconstruct a monomer from a condensed structural formula.

Common addition polymers and their monomers

The same reverse method works for several familiar polymers.

Notice that the side groups change, but the basic two-carbon backbone pattern remains.

The key is to restore the double bond without changing the arrangement of the other atoms.

04What if the polymer is a condensation polymer?

There is an important limitation to the method above.

Not every polymer is made by opening an alkene double bond.

Some polymers form through condensation polymerisation. In this process, monomers with suitable functional groups join together while eliminating a small molecule, commonly water.

Two important examples are:

• Polyesters, which contain ester linkages.
• Polyamides, which contain amide linkages.

Their repeating units have different structural features from those of simple alkene-derived addition polymers.

Recognising the important functional groups

An ester linkage contains:

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

An amide linkage contains:

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

These groups form part of the polymer backbone.

This is different from PVC, where chlorine is attached as a side group and does not form part of the backbone.

How do you reverse a condensation polymer?

Instead of restoring a C=C double bond, identify the functional group that connects the monomers.

For a polyester:

1. Find the ester linkage, \(\ce{-C(=O)-O-}\).
2. Break the bond between the carbonyl carbon and the ester oxygen.
3. Restore an \(\ce{-OH}\) group to the carbonyl side.
4. Restore an \(\ce{H}\) atom to the oxygen side.

This reconstructs the original carboxylic acid and alcohol functional groups.

For a polyamide, the equivalent process involves breaking the carbon-nitrogen bond in the amide linkage, restoring \(\ce{-OH}\) to the carbonyl side and \(\ce{H}\) to the nitrogen side.

The forward reaction eliminates water when these functional groups combine.

A simplified ester-forming reaction is:

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

Here, \(R\) and \(R'\) represent organic groups.

The important distinction is that you are reconstructing the original functional groups, not creating an alkene.

05A quick decision guide

When you are given an unfamiliar repeating unit, first decide which type of polymerisation could have produced it.

A few common mistakes are worth avoiding:

• Treating the external bonds as missing hydrogen atoms.
• Moving a side group to a different carbon.
• Assuming every atom inside the brackets belongs to the backbone.
• Restoring a double bond in a polymer that was formed by condensation.
• Forgetting to check carbon valence.
• Assuming that every repeating unit uniquely identifies the original manufacturing process.

That final point becomes important in the harder question below.

06Questions and solutions

Question 1

A polymer has the repeating unit:

\[ \ce{[-CF2-CF2-]_{n}} \]

Identify the original monomer, give its systematic name, and state the type of polymerisation involved.

Solution 1

The original monomer is tetrafluoroethene, \(\ce{CF2=CF2}\), and the polymer forms through addition polymerisation.

Step 1: Identify the backbone.

The repeating unit contains two carbon atoms joined by a single bond.

Each carbon is also attached to two fluorine atoms.

Step 2: Remove the external chain connections.

The bonds extending from either end connect the unit to neighbouring units.

Remove these connections without changing the carbon-fluorine bonds.

Step 3: Restore the double bond.

Place a C=C double bond between the two backbone carbon atoms.

The resulting monomer is:

\[ \ce{CF2=CF2} \]

Step 4: Check the valence.

Each carbon has two C-F single bonds and one C=C double bond.

Its total bond order is therefore four.

The structure is valid.

The polymer is polytetrafluoroethene (PTFE), formed by addition polymerisation of tetrafluoroethene.

Question 2

A polymer has the repeating unit:

\[ \ce{[-CH2-C(CH3)(CH2CH3)-]_{n}} \]

Identify the original monomer and give its systematic name.

Explain why the central carbon atom does not require an additional hydrogen atom.

Solution 2

The monomer is \(\ce{CH2=C(CH3)CH2CH3}\), named 2-methylbut-1-ene.

The central carbon does not require a hydrogen because its four available bonds are already accounted for.

Step 1: Identify the backbone.

The two backbone carbon atoms are the left-hand \(\ce{CH2}\) and the central carbon.

The central carbon has two side groups:

• A methyl group, \(\ce{CH3}\).
• An ethyl group, \(\ce{CH2CH3}\).

Both groups must remain attached to the central carbon.

Step 2: Remove the chain-continuation bonds.

Remove the external bonds connecting the repeating unit to neighbouring units.

Do not remove either side group.

Step 3: Restore the double bond.

Place a C=C double bond between the two backbone carbons.

The reconstructed structure is:

\[ \ce{CH2=C(CH3)CH2CH3} \]

Step 4: Check the carbon valence.

The left carbon has two hydrogen atoms and a double bond to the central carbon.

Its total bond order is four.

The central carbon has:

• A double bond to the left carbon: bond order 2.
• A single bond to the methyl group: bond order 1.
• A single bond to the ethyl group: bond order 1.

Therefore:

\[ 2+1+1=4 \]

The central carbon already has four bonds. Adding a hydrogen would give it five, which is not valid for an ordinary neutral carbon atom.

The correct monomer is 2-methylbut-1-ene.

Question 3

A polyamide has the repeating unit:

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

Identify the two original monomers and classify the polymerisation reaction.

Explain how the amide linkages help you identify the monomers.

Solution 3

The two monomers are 1,6-diaminohexane and hexanedioic acid (adipic acid). The polymer forms through condensation polymerisation.

Step 1: Identify the functional group.

The polymer contains amide linkages:

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

These form when an amine group reacts with a carboxylic acid group, eliminating water.

This tells us that the polymer is a polyamide rather than an alkene-derived addition polymer.

Step 2: Reconstruct the diamine.

The nitrogen-containing section comes from a molecule with two amino groups.

Restore hydrogen atoms to the nitrogen atoms to obtain:

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

This is 1,6-diaminohexane.

It contains six carbon atoms and two amino groups.

Step 3: Reconstruct the dicarboxylic acid.

The carbonyl-containing section comes from a molecule with two carboxylic acid groups.

Restore hydroxyl groups to the two carbonyl carbons:

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

This is hexanedioic acid, also known as adipic acid.

It contains six carbon atoms in total: four methylene carbons and two carbonyl carbons.

Step 4: Identify the reaction type.

The amino groups react with the carboxylic acid groups to form amide linkages.

Water is eliminated during the formation of these links.

The polymer is nylon 6,6, a condensation polymer.

The important reasoning is that the repeating unit contains sections derived from two different monomers. Treating the entire bracketed structure as one alkene monomer would ignore the nitrogen and carbonyl groups that define the polymer’s structure.

Question 4

Two polymers are represented by the following structural blocks.

Polymer A:

\[ \ce{[-CH2-CH(CH3)-CH2-CH(CH3)-]_{n}} \]

The bracket deliberately contains two consecutive identical two-carbon motifs.

Polymer B:

\[ \ce{[-CH2-CH(CH2CH3)-]_{n}} \]

Assume both polymers are homopolymers formed by addition polymerisation from a single alkene monomer.

Identify the original monomer for each polymer and explain why simply counting all the carbon atoms inside the brackets can produce an incorrect answer.

Solution 4

Polymer A is formed from propene, \(\ce{CH2=CHCH3}\), while Polymer B is formed from but-1-ene, \(\ce{CH2=CHCH2CH3}\).

The difference comes from how the carbon atoms are arranged between the backbone and side groups.

Step 1: Analyse Polymer A.

Its bracketed structure contains four backbone carbon atoms:

\[ \ce{-CH2-CH(CH3)-CH2-CH(CH3)-} \]

However, the structure contains two identical two-carbon motifs:

\[ \ce{-CH2-CH(CH3)-} \]

Each motif is derived from one propene molecule.

The bracket has simply been drawn around two consecutive propene-derived sections.

To reverse the polymerisation, separate these two sections by breaking the bond connecting them, remove the external chain connections, and restore a C=C double bond within each two-carbon section.

Each reconstructed monomer is:

\[ \ce{CH2=CHCH3} \]

Therefore, Polymer A is polypropylene.

Step 2: Analyse Polymer B.

The repeating unit contains only two backbone carbon atoms:

\[ \ce{-CH2-CH(CH2CH3)-} \]

The ethyl group is a side group attached to the second backbone carbon.

Remove the external chain connections and restore the double bond between the two backbone carbons.

This gives:

\[ \ce{CH2=CHCH2CH3} \]

The monomer is but-1-ene.

Step 3: Compare the two structures.

The tempting approach is to count every carbon inside the brackets and assume that all of them belonged to one original monomer.

That is incorrect because carbon atoms can occupy different positions in the structure.

For Polymer A, the bracket contains two propene-derived units, each contributing two backbone carbons and one methyl group.

For Polymer B, one but-1-ene molecule contributes two backbone carbons and an ethyl side group.

The location of the side groups and the number of backbone atoms are more useful than the total number of carbon atoms inside the brackets.

Question 5

Extension: Testing the limits of reverse identification

A polymer has the repeating unit:

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

A student concludes that the original monomer must have been 6-aminohexanoic acid:

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

They also conclude that the polymer must have formed through condensation polymerisation.

However, the same polymer structure can also be produced through ring-opening polymerisation of caprolactam, a cyclic amide containing the sequence \(\ce{-NH-(CH2)5-C(=O)-}\) within its ring.

Evaluate the student’s conclusion.

What can the repeating unit establish about the polymer, and what information is needed to identify the actual starting material and reaction pathway?

Solution 5

The student’s conclusion is not justified. The repeating unit establishes the polymer’s repeating molecular structure, but it does not uniquely identify the starting monomer or the reaction pathway.

Both routes can produce the same repeating structure.

Step 1: Examine the proposed condensation route.

The proposed monomer is:

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

This molecule contains:

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

The amino group of one molecule can react with the carboxylic acid group of another.

An amide linkage forms, and water is eliminated.

Repeated reactions produce the polymer:

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

The student’s proposed monomer and reaction type are therefore chemically possible.

Step 2: Consider the alternative ring-opening route.

Caprolactam is a cyclic amide.

Its nitrogen and carbonyl carbon are already connected within a ring.

During ring-opening polymerisation, the cyclic structure opens and the monomer units connect to form the same repeating backbone.

The resulting polymer has the same repeating unit:

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

In the idealised ring-opening representation, the ring opens without the characteristic elimination of water between separate amino acid molecules.

Step 3: Compare what the two routes tell us.

Both routes produce the same repeating unit, even though the starting molecules and reaction pathways differ.

Step 4: Identify the limitation of the repeating unit.

The repeating unit records the final arrangement of atoms in the polymer backbone.

It does not preserve every detail of how the polymer was manufactured.

The brackets also omit the exact chain ends, which can sometimes provide additional information about the starting materials and reaction conditions.

To establish the actual route, we would need additional information, such as the starting material used, the manufacturing process, or evidence from the reaction products and polymer end groups.

The tempting approach is to reverse every amide bond, restore an amino group and a carboxylic acid group, and declare that the original monomer has been identified.

That reconstruction is valid for the condensation route, but it is not proof that this was the route actually used.

The deeper lesson is that a polymer’s repeating unit identifies its repeating structure, not necessarily its complete synthetic history.

This ring-opening comparison is an extension beyond the standard HSC reverse-identification examples. It tests the limits of what can be inferred from a structural formula rather than introducing another monomer to memorise.

07What to do with the monomer once you have identified it

Finding the monomer is only the first part of understanding a polymer.

Once you know the original structure, you can investigate how its functional groups and side groups influence the polymer’s properties.

For example, the methyl side groups in polypropylene affect how closely neighbouring polymer chains can pack together. The ester and amide linkages in condensation polymers also influence polarity and intermolecular attractions.

This connects structural identification to the broader question of why different polymers have different properties and uses.

For that next step, see Monomers vs Polymers: Health and Environmental Impacts.