How to Model a Copolymer: Structures, Sequences and Examples

Imagine drawing styrene-butadiene rubber as though every section of its chain came from styrene. You would have missed a major part of its chemical structure.

The solution is to model the two different monomer-derived units within the same polymer chain. This is called a copolymer. You must also distinguish the arrangement of those units: they might alternate, appear randomly, form long blocks, or branch from a main chain.

The important point is that a copolymer is not simply a homopolymer with two monomers written beside it. Its structure must show how the different monomers contribute to the polymer.

01The key distinction: homopolymer or copolymer?

Start by counting the different monomer species, not the number of monomer molecules.

• A homopolymer is made from one type of monomer. Polyethene, for example, is made from ethene.
• A copolymer contains units derived from two or more different monomer species. Styrene-butadiene rubber (SBR), for example, contains units derived from styrene and buta-1,3-diene.
• A terpolymer contains units derived from three different monomer species. ABS plastic is a familiar example, containing acrylonitrile-, butadiene-, and styrene-derived units.

A useful comparison is:

Notice that using 500 molecules of ethene does not make polyethene a copolymer. The classification depends on the number of different monomer species, not the total number of molecules.

02How to represent different monomers in one chain

Consider the difference between writing a polymer’s structure as a single repeating unit and showing a chain containing several different units.

For a homopolymer such as polyethene, the repeating structure is straightforward:

\[ [-\mathrm{CH_2-CH_2}-]_n \]

The subscript \(n\) indicates that the structural unit is repeated many times.

For a copolymer, however, you need to retain the identity of each monomer-derived unit. There is no universal rule that says the units must alternate.

Four common arrangements are worth recognising.

Random copolymer

The different monomer-derived units appear in no fixed repeating sequence.

A simplified chain might look like:

\[ -\mathrm{A-A-B-A-B-B-A-B}- \]

Here, A and B represent two different monomer-derived structural units.

This is a schematic sequence, not a claim that every random copolymer has exactly this arrangement.

Alternating copolymer

The two types of units alternate along the chain:

\[ -\mathrm{A-B-A-B-A-B}- \]

If the structure is genuinely alternating, it can be represented using a combined repeat unit:

\[ [-\mathrm{A-B}-]_n \]

Block copolymer

Long sections of one type of unit are joined to sections of another:

\[ -\mathrm{A-A-A-A-B-B-B-B}- \]

The defining feature is that the different units occur in blocks rather than being distributed randomly or alternating one by one.

Graft copolymer

One type of polymer chain forms the main backbone, while branches made from another type of polymer extend from it.

These arrangements are not interchangeable. Writing an alternating structure when the sequence is random makes a claim about the polymer that may not be true.

The modelling rule: represent the chemical identity of each monomer-derived unit, then describe the sequence only to the extent that the information allows.

03Worked structural application: styrene-butadiene rubber (SBR)

SBR is a useful example because its two monomers produce noticeably different structures.

The monomers are:

• Styrene: \(\ce{C6H5-CH=CH2}\)
• Buta-1,3-diene: \(\ce{CH2=CH-CH=CH2}\)

First, work out what each monomer contributes to the polymer chain.

Contribution from styrene

The carbon-carbon double bond in styrene opens during addition polymerisation. The benzene ring remains attached to the carbon backbone as a side group.

The resulting structural unit is:

\[ \ce{C6H5-CH=CH2 -> [-CH2-CH(C6H5)-]} \]

Call this unit S.

If you need to revise why the benzene ring remains intact, see How to Draw Polystyrene from Styrene Without Breaking the Ring.

Contribution from buta-1,3-diene

Buta-1,3-diene has two carbon-carbon double bonds. For this simplified model, assume it undergoes 1,4-addition.

The resulting structural unit is:

\[ \ce{CH2=CH-CH=CH2 -> [-CH2-CH=CH-CH2-]} \]

Call this unit B.

Notice something important: the central carbon-carbon double bond remains in the polymer structure. The monomer does not simply become a four-carbon section containing only single bonds.

Real polybutadiene-derived structures can also contain 1,2-addition units, so the 1,4-addition model is a simplification.

Putting the units together

Suppose a simplified SBR chain contains the sequence:

\[ \mathrm{S-B-B-S} \]

Replace each letter with its corresponding chemical structure:

\[ \begin{aligned} \cdots-&\ce{CH2-CH(C6H5)}\\ &-\ce{CH2-CH=CH-CH2}\\ &-\ce{CH2-CH=CH-CH2}\\ &-\ce{CH2-CH(C6H5)}-\cdots \end{aligned} \]

You now have one continuous polymer chain containing two different monomer-derived units.

This is a copolymer model.

The sequence shown is illustrative. It does not claim that every SBR chain follows this exact arrangement.

The key difference is that the styrene-derived and butadiene-derived sections are chemically connected along the same chain.

04The misconception: two polymers mixed together are not automatically a copolymer

Imagine mixing polystyrene and polybutadiene.

Your container now contains material derived from both styrene and butadiene. Does that make it SBR?

No.

The distinction is whether the different structural units are covalently connected within the same polymer chains.

Consider the two situations:

Situation A: A physical polymer blend

• Polystyrene chains contain styrene-derived units.
• Polybutadiene chains contain butadiene-derived units.
• The two types of chains are physically mixed.
• There is no requirement for styrene-derived and butadiene-derived units to occur in the same chain.

Situation B: A copolymer

• Individual polymer chains contain both styrene-derived and butadiene-derived units.
• The different units are connected through covalent bonds.
• The chain’s composition includes contributions from both monomers.

This distinction matters when interpreting experimental information.

A chemical analysis might show that a sample contains carbon and hydrogen in proportions consistent with both styrene and butadiene. That tells you about its overall composition. It does not, by itself, prove that both monomers occur in the same polymer chains.

You need structural evidence of the chain arrangement to make that conclusion.

Another named example: nitrile rubber (NBR)

Nitrile rubber provides a useful contrast because its monomers have different functional groups.

It is made from:

• Buta-1,3-diene: \(\ce{CH2=CH-CH=CH2}\)
• Acrylonitrile: \(\ce{CH2=CH-CN}\)

The acrylonitrile-derived unit is:

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

The \(\ce{CN}\) group is a nitrile side group. It remains attached to the carbon backbone during addition polymerisation.

The butadiene-derived unit, under the same simplified 1,4-addition assumption, is:

\[ \ce{[-CH2-CH=CH-CH2-]} \]

A possible NBR chain therefore contains both types of structure:

\[ \begin{aligned} \cdots-&\ce{CH2-CH(CN)}\\ &-\ce{CH2-CH=CH-CH2}\\ &-\ce{CH2-CH(CN)}-\cdots \end{aligned} \]

The same modelling method works for both SBR and NBR. Identify each monomer, determine its contribution to the backbone, preserve its side groups, and then assemble the different units into one chain.

ABS is another useful name to recognise. It contains acrylonitrile-, butadiene-, and styrene-derived components, so it is classified as a terpolymer rather than a simple two-monomer copolymer.

05A reliable method for modelling a copolymer

When an exam question gives you unfamiliar monomers or a polymer structure, follow this sequence.

The final step is particularly important.

When can you write \([A-B]_n\)?

This notation represents a repeating sequence of A followed by B:

\[ [-\mathrm{A-B}-]_n \]

It is appropriate for an idealised alternating copolymer.

It is not a safe general representation of every copolymer.

For a random copolymer, a more honest representation is an illustrative chain segment:

\[ -\mathrm{A-A-B-A-B-B-A}-\cdots \]

For a block copolymer, you might show:

\[ -\mathrm{A-A-A-A-B-B-B-B}-\cdots \]

The distinction is not cosmetic. Each representation makes a different claim about the arrangement of the polymer chain.

What can the available evidence actually tell you?

Remember: composition tells you how much of each type of unit is present. Sequence tells you how those units are arranged. Connectivity tells you whether they belong to the same chain.

These are three different pieces of structural information.

06Worked examples

07Questions and solutions

Question 1

A manufacturer physically mixes pre-made polystyrene and polyethene. No chemical reaction joins the polymer chains.

A student describes the resulting material as a copolymer because it contains units derived from two different monomers.

Is the student correct? Classify the material and explain your reasoning.

Solution 1

The material is a physical blend of two homopolymers, not a copolymer.

Polystyrene contains styrene-derived units:

\[ \ce{[-CH2-CH(C6H5)-]_{n}} \]

Polyethene contains ethene-derived units:

\[ \ce{[-CH2-CH2-]_{n}} \]

The two types of polymer chains are mixed together, but they are not covalently connected to each other.

The student’s mistake is counting the different monomers present in the material rather than examining the composition of individual polymer chains.

A copolymer requires different monomer-derived units within the same polymer chain.

Question 2

An unknown addition copolymer contains the following chain segment:

\[ \begin{aligned} \cdots-&\ce{CH2-CH(CN)}\\ &-\ce{CH2-CH=CH-CH2}\\ &-\ce{CH2-CH(CN)}-\cdots \end{aligned} \]

Assume that the buta-1,3-diene-derived units form exclusively through 1,4-addition.

Determine:

(a) The two original monomers.

(b) The name of the resulting copolymer.

(c) Why the \(\ce{CN}\) group is not part of the main carbon backbone.

Solution 2

The monomers are acrylonitrile and buta-1,3-diene. The resulting polymer is nitrile rubber (NBR), a copolymer.

Step 1: Identify the acrylonitrile-derived unit.

The supplied structure contains:

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

Restoring the original carbon-carbon double bond gives:

\[ \ce{CH2=CH-CN} \]

This is acrylonitrile.

Step 2: Identify the butadiene-derived unit.

The second section is:

\[ \ce{-CH2-CH=CH-CH2-} \]

This is consistent with 1,4-addition of:

\[ \ce{CH2=CH-CH=CH2} \]

The monomer is buta-1,3-diene.

Step 3: Classify the polymer.

Both monomers contribute different structural units to the same continuous chain.

Therefore, the material is a copolymer.

Step 4: Explain the position of the nitrile group.

The \(\ce{CN}\) group is attached to one of the two carbon atoms that originally formed the acrylonitrile double bond.

It remains a side group during addition polymerisation. It does not become part of the main carbon backbone.

The important rule is to distinguish the atoms forming the backbone from atoms attached to those backbone atoms.

Question 3

An illustrative copolymer sample contains:

• \(0.250\ \mathrm{mol}\) of styrene-derived units.
• \(0.750\ \mathrm{mol}\) of butadiene-derived units.

Use the following molar masses:

\[ M(\mathrm{S})=104.15\ \mathrm{g\,mol^{-1}} \]

\[ M(\mathrm{B})=54.09\ \mathrm{g\,mol^{-1}} \]

Calculate:

(a) The mole percentage of styrene-derived units.

(b) The mass percentage of styrene-derived units.

Solution 3

The sample contains 25.0 mol% styrene-derived units and approximately 39.1% styrene-derived units by mass.

Step 1: Calculate the total amount of structural units.

\[ n_{\mathrm{total}}=0.250+0.750 \]

\[ n_{\mathrm{total}}=1.000\ \mathrm{mol} \]

Step 2: Calculate the mole percentage of styrene-derived units.

\[ \begin{aligned} \%\mathrm{S} &=\frac{0.250}{1.000}\times100\\ &=25.0\% \end{aligned} \]

Therefore, styrene-derived units make up 25.0 mol% of the sample.

Step 3: Calculate the mass of styrene-derived units.

Rearrange the molar mass equation:

\[ m=nM \]

\[ m(\mathrm{S})=0.250\times104.15 \]

\[ m(\mathrm{S})=26.04\ \mathrm{g} \]

Step 4: Calculate the mass of butadiene-derived units.

\[ m(\mathrm{B})=0.750\times54.09 \]

\[ m(\mathrm{B})=40.57\ \mathrm{g} \]

Step 5: Calculate the total mass and mass percentage.

\[ m_{\mathrm{total}}=26.04+40.57 \]

\[ m_{\mathrm{total}}=66.61\ \mathrm{g} \]

Therefore:

\[ \begin{aligned} \%\mathrm{S} &=\frac{26.04}{66.61}\times100\\ &=39.1\% \end{aligned} \]

The styrene-derived units represent 25.0% of the total number of units but 39.1% of the total mass.

This is because each styrene-derived unit has a greater molar mass than each butadiene-derived unit.

Question 4

An SBR sample has an overall composition of 35.0 mol% styrene-derived units and 65.0 mol% butadiene-derived units.

Two students propose different models:

• Student A represents the polymer as an alternating copolymer, using \([\mathrm{S-B}]_n\).
• Student B argues that because the composition is not 1:1, the polymer must be random.

Evaluate both explanations.

Solution 4

Student A’s alternating model is inconsistent with the supplied composition. However, Student B has not provided enough evidence to establish that the polymer is random.

Step 1: Test the alternating model.

A strictly alternating copolymer contains one A unit for every B unit:

\[ -\mathrm{A-B-A-B-A-B}-\cdots \]

For a sufficiently long chain, this gives approximately:

\[ 50.0\ \mathrm{mol\% \ A} \]

and

\[ 50.0\ \mathrm{mol\% \ B} \]

The supplied sample instead contains 35.0 mol% styrene-derived units.

Therefore, a strictly alternating model does not match the composition.

Step 2: Consider other possible arrangements.

The polymer could have a random arrangement, such as:

\[ -\mathrm{B-S-B-B-S-B-S-B}-\cdots \]

Alternatively, it could contain longer sections of each type of unit:

\[ -\mathrm{B-B-B-B-S-S-B-B-S}-\cdots \]

Both arrangements can have an overall composition different from 1:1.

Step 3: Evaluate what the composition actually establishes.

The 35.0 mol% figure tells us the relative amounts of the two types of units.

It does not tell us whether they are randomly distributed or arranged in blocks.

Student A has assumed that every copolymer must alternate. Student B has assumed that every non-alternating copolymer must be random.

Both assumptions are incorrect.

You need information about the actual sequence of units to distinguish these arrangements.

Question 5

Extension challenge

Two polymer materials, P and Q, have identical overall compositions.

Both contain:

• 27.8% by mass of styrene-derived units.
• 72.2% by mass of butadiene-derived units.

Further structural analysis provides the following information:

A student makes three claims:

1. P and Q must be the same type of polymer because their overall compositions are identical.
2. P must be an alternating copolymer because it contains two different monomer-derived units.
3. P can be represented as \([\mathrm{S-B-B-B-B-B}]_n\) because its overall composition corresponds to approximately one styrene-derived unit for every five butadiene-derived units.

Evaluate all three claims.

Your answer must:

(a) Calculate the mole percentage of styrene-derived units.

(b) Classify P and Q.

(c) Determine which structural conclusions can actually be drawn about P.

(d) Explain what further evidence would be needed to identify its sequence arrangement.

Solution 5

Material P is a copolymer, while material Q is a physical blend of two homopolymers. Their identical overall compositions do not make them structurally equivalent.

Furthermore, P cannot automatically be described as alternating or as having a repeating sequence of one styrene unit followed by five butadiene units.

Step 1: Calculate the mole percentage of styrene-derived units.

Assume a 100.0 g sample.

The masses are:

\[ m(\mathrm{S})=27.8\ \mathrm{g} \]

\[ m(\mathrm{B})=72.2\ \mathrm{g} \]

Calculate the amount of each unit:

\[ n(\mathrm{S})=\frac{27.8}{104.15} =0.2669\ \mathrm{mol} \]

\[ n(\mathrm{B})=\frac{72.2}{54.09} =1.334\ \mathrm{mol} \]

The total amount is:

\[ n_{\mathrm{total}}=0.2669+1.334 \]

\[ n_{\mathrm{total}}=1.601\ \mathrm{mol} \]

Therefore:

\[ \begin{aligned} \%\mathrm{S} &=\frac{0.2669}{1.601}\times100\\ &=16.7\% \end{aligned} \]

The sample contains approximately:

• 16.7 mol% styrene-derived units.
• 83.3 mol% butadiene-derived units.

This corresponds to approximately one styrene-derived unit for every five butadiene-derived units.

Step 2: Classify material P.

P is a copolymer.

The decisive information is that every analysed chain contains both types of structural unit.

The two monomer-derived units occur within the same polymer chains, satisfying the definition of a copolymer.

Step 3: Classify material Q.

Q is a physical blend of two homopolymers.

One population contains only styrene-derived units, while the other contains only butadiene-derived units.

Although the combined sample has the same overall composition as P, its individual chains have different structures.

This is the crucial distinction: bulk composition does not establish chain connectivity.

Step 4: Evaluate the alternating model.

P cannot be strictly alternating.

A strictly alternating copolymer requires approximately equal numbers of the two types of units.

Its composition would be close to 50:50, not 16.7:83.3.

Therefore, claim 2 is incorrect.

Step 5: Evaluate the proposed \([\mathrm{S-B-B-B-B-B}]_n\) representation.

This is a possible model, but it is not established by the supplied evidence.

The measured 1:5 ratio gives the relative quantities of the units. It does not establish their order.

For example, the following arrangements could all have the same overall composition:

• A random distribution of S and B units.
• Long blocks containing different proportions of S and B.
• A regular repeating arrangement containing one S unit followed by five B units.
• A graft arrangement, if the structural evidence supports branching.

The proposed notation claims a specific repeating sequence. The composition alone cannot justify that claim.

Step 6: Identify the additional evidence required.

A sequence-sensitive structural analysis would be needed to establish which units are adjacent to one another and whether the chains contain blocks or branches.

For example, suitable NMR evidence can provide information about neighbouring monomer-derived units.

Another bulk composition measurement would not solve the problem. The missing information concerns molecular connectivity and sequence, not the total amounts of each element or monomer-derived unit.

The final judgement is:

This is the level of precision you should aim for: distinguish what the evidence proves from what it merely allows.

07Where to go next

Once you have identified the correct structural units, the next step is to use them in calculations.

For example, calculating the approximate number of units in a polymer chain requires the molar mass of the correct repeating structure. In a copolymer, using the molar mass of just one monomer-derived unit can give the wrong result.

You can build on this method with How to Calculate Repeat Units in a Polymer Chain.

For HSC-style structural questions, keep one final rule in mind:

Never draw a more specific polymer structure than the information supports.

If the question gives two monomers, identify both contributions. If it gives a sequence, reproduce that sequence. If it gives only an overall composition, do not invent an arrangement.

That is how you model a copolymer accurately without accidentally turning it into a homopolymer, a polymer blend, or an alternating structure that the evidence never established.

Extension note: Random, alternating, block, and graft architectures provide useful structural modelling practice. Their inclusion here does not imply that every architecture or named polymer is prescribed HSC Chemistry content.