How to Tell if a Ring Is in a Polymer Backbone or Side Group

Two polymer structures can both contain a benzene ring, yet the ring plays a completely different role in each. In one polymer, the ring is part of the main chain. In the other, it hangs off the chain as a side group.

The quickest way to tell the difference is to trace the covalent bonds connecting one repeating unit to the next. If the polymer chain passes through the ring, the ring belongs to the backbone. If the chain continues without passing through the ring, the ring is a side group.

For example, the benzene ring in polyethylene terephthalate (PET) belongs to the backbone. The phenyl ring in polystyrene is a side group.

The position of the ring on the page can be misleading. What matters is how its atoms are connected.

01Start by tracing the polymer backbone

The backbone is the continuous covalently bonded framework that connects the repeating units of a polymer.

A side group, also called a pendant group, is an atom or group of atoms attached to the backbone but not forming part of that connecting framework.

Think of the backbone as a main road connecting two suburbs. A side group is like a building connected to that road. The building might be large, but you do not need to travel through it to get from one suburb to the other.

The analogy is useful, but the actual decision comes down to chemical bonds, not the physical position or size of a structure.

The practical test

When you are given a polymer structure, follow this process.

1. Find the repeating unit. Look for the square brackets and the subscript \(n\), or identify the repeating pattern if the brackets are missing.

2. Trace the connecting chain. Follow the covalent bonds from one end of the repeating unit to the other.

3. Check the ring’s connections. Does the chain enter the ring and leave through another ring atom, or is the ring attached to the chain as a branch?

4. Check whether the chain needs the ring. If you can trace the continuous backbone without passing through the ring, the ring is not part of that backbone.

There are three useful possibilities:

The third case is important. Two connections do not automatically make a ring part of the backbone. You still need to trace the actual chain.

02Why the benzene ring in PET belongs to the backbone

Let’s apply the method to PET, or polyethylene terephthalate.

PET is a member of the polyester family. A polyester is a polymer containing repeated ester links in its backbone.

The ester functional group has the general structure:

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

Here:

• \(\ce{R}\) and \(\ce{R'}\) represent carbon-containing groups.
• The carbonyl carbon is double-bonded to one oxygen.
• That carbonyl carbon is also single-bonded to another oxygen.
• The single-bonded oxygen connects to the next carbon-containing group.

The important connectivity is:

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

Do not confuse this with a carboxylic acid group, \(\ce{-COOH}\). An ester has its oxygen connected to another carbon-containing group rather than to hydrogen.

The repeating unit of PET

The repeating unit of PET is:

\[ \ce{[-O-CH2-CH2-O-C(=O)-C6H4-C(=O)-]_{n}} \]

The \(\ce{C6H4}\) represents a benzene ring with two substituents at positions 1 and 4. This arrangement comes from terephthalic acid.

Now trace the polymer chain:

\[ \begin{aligned} \ce{...-O-CH2-CH2-O-C(=O)-} \\ \ce{-C6H4-C(=O)-O-CH2-CH2-O-...} \end{aligned} \]

The chain enters the terephthalate section through one carbonyl group, passes through the benzene ring, and exits through the other carbonyl group.

The ring has two connections to the continuing polymer chain, at different positions on the ring.

That makes it part of the backbone.

The benzene ring in PET is not a pendant side group. It is incorporated into the main polymer chain.

Polyester is the family; PET is the example

There is another distinction worth getting right in HSC Chemistry questions.

A polyester is a polymer family. PET is one specific member of that family.

Not every polyester contains a benzene ring. The defining feature of the polyester family is the repeated ester linkage, not the presence of an aromatic ring.

Similarly, the presence of a benzene ring does not automatically make a polymer a polyester.

03The contrasting case: polystyrene

Now consider polystyrene.

Its repeating unit is:

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

The \(\ce{C6H5}\) represents a phenyl group: a benzene ring attached to the rest of the structure.

Follow the backbone:

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

Each \(\ce{C6H5}\) group is attached to one of the backbone carbon atoms.

The main chain continues through the carbon-carbon bonds. It never needs to pass through the benzene ring.

Therefore:

• The continuous carbon-carbon chain is the backbone.
• The phenyl rings are pendant side groups.
• Polystyrene is an addition polymer, not a polyester.

Notice something important: PET and polystyrene both contain benzene rings, but the rings have different structural roles.

The difference is not the chemical identity of the ring. It is the connectivity.

04A useful trap: an ester group can also be a side group

Here is a slightly more subtle example.

Consider a polymer formed from vinyl benzoate, with the repeating unit:

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

This structure contains:

• A continuous carbon-carbon backbone.
• An ester group, \(\ce{-O-C(=O)-}\).
• A phenyl ring, \(\ce{C6H5}\).

A student might see the ester group and immediately classify the material as a polyester.

That conclusion is incorrect.

The polymer backbone is:

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

The ester-containing group is attached to the backbone through the oxygen bonded to a backbone carbon.

The polymer chain does not pass through the ester group or the benzene ring.

Consequently:

• The benzene ring is part of a pendant side group.
• The ester group is also pendant.
• The polymer is an addition polymer with pendant ester groups, not a polyester with ester links in its backbone.

This gives you a useful rule:

Finding an ester group does not automatically identify a polyester. Check whether the ester linkage is actually part of the polymer backbone.

The same principle applies to other functional groups. Their presence somewhere in a polymer does not necessarily mean they define its polymer family.

05How could an experiment distinguish the two arrangements?

So far, we have identified the ring’s position by reading a structural formula.

But imagine a chemist has an unknown polymer sample. They know it contains aromatic rings and ester groups, but they do not know whether those ester groups are in the backbone or in side groups.

How could they investigate?

An infrared spectrum might provide evidence for aromatic structures and ester carbonyl groups. However, simply identifying those functional groups does not establish how they are connected.

A more informative investigation would examine what happens when the ester bonds are hydrolysed.

The experimental design

Consider two candidate polymers:

• PET, with ester linkages in its backbone.
• Poly(vinyl benzoate), with ester groups in its side groups.

A simplified investigation uses two separate reaction vessels, each containing a sample of one polymer and aqueous sodium hydroxide.

A vertical reflux condenser is attached above each vessel. The reaction mixture is heated under conditions suitable for complete ester hydrolysis.

As the mixture heats, vapour travels upwards into the condenser. It cools, condenses, and flows back into the reaction vessel. This prevents substantial solvent loss during heating.

After the reaction, the products are cooled, separated where appropriate, and analysed. The investigator examines both the small molecules produced and whether a high-molecular-mass polymer chain remains.

What should happen?

The two polymers behave differently because their ester groups occupy different positions.

PET: ester bonds in the backbone

Each ideal PET repeating unit contains two ester linkages.

Complete alkaline hydrolysis breaks these linkages, separating the polymer into ethylene glycol and terephthalate ions.

The idealised reaction per repeat unit is:

\[ \begin{aligned} &\ce{[-O-CH2-CH2-O-C(=O)-C6H4-C(=O)-]}\\ &\ce{+ 2OH^-}\\ &\ce{-> HO-CH2-CH2-OH}\\ &\ce{+ ^-OOC-C6H4-COO^-} \end{aligned} \]

The polymer backbone has been broken into smaller molecules.

The benzene ring remains attached to two carboxylate groups in the terephthalate product.

Poly(vinyl benzoate): ester groups in side groups

In this polymer, the ester group is attached to the carbon-carbon backbone rather than forming part of it.

Hydrolysis removes the benzoate side group, but the carbon-carbon backbone remains intact:

\[ \begin{aligned} &\ce{[-CH2-CH(O-C(=O)-C6H5)-]_{n}}\\ &\ce{+ nOH^-}\\ &\ce{-> [-CH2-CH(OH)-]_{n}}\\ &\ce{+ nC6H5COO^-} \end{aligned} \]

The aromatic benzoate fragment is released, but the polymeric carbon-carbon chain survives.

The predicted outcomes can be compared:

These are theoretical predictions for idealised structures, not reported experimental measurements.

Why the first design is not enough

Suppose the investigator only used infrared spectroscopy to identify ester and aromatic groups.

Both polymers would provide evidence for those functional groups. That observation would not distinguish their connectivity.

The revised investigation examines the products of hydrolysis and whether the original polymer chain survives.

There is an important limitation: incomplete hydrolysis can leave partially broken chains and unreacted ester groups. Real polymer samples may also contain chain ends, additives, or crosslinks. The interpretation above assumes complete hydrolysis of the relevant ester bonds and no competing reactions.

The key experimental lesson is that identifying a functional group is not the same as establishing its position in a polymer.

06Worked examples: applying the structural method

07Questions and solutions

Question 1

An aromatic polymer contains the following structural segment:

\[ \ce{...-NH-C(=O)-C6H4-C(=O)-NH-...} \]

The two carbonyl groups are attached to different carbon atoms of the benzene ring.

Identify the polymer family and explain whether the benzene ring belongs to the backbone or is a side group.

Solution 1

The benzene ring belongs to the backbone, and the polymer belongs to the polyamide family.

The repeated amide linkages contain the functional group:

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

This identifies the polymer family as a polyamide.

Now trace the chain through the supplied structure. It enters the benzene-containing section through one carbonyl group and leaves through the other.

The ring therefore connects two sections of the polymer chain.

It is not a pendant side group.

The tempting mistake is to assume that an aromatic ring must be a side group because it looks like a separate feature attached to a chain. The actual connections show otherwise.

Question 2

Consider two polymers with the following repeating units.

Polymer A:

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

Polymer B:

\[ \ce{[-CH2-C6H4-CH2-]_{n}} \]

For Polymer B, the two \(\ce{-CH2-}\) groups connect to the benzene ring at positions 1 and 4.

A student claims that both polymers must have pendant benzene rings because both contain phenyl-type aromatic structures.

Evaluate this claim. Explain the different roles of the rings.

Solution 2

The student’s claim is incorrect. Polymer A has pendant benzene rings, whereas Polymer B incorporates its benzene rings into the backbone.

For Polymer A, the structure is polystyrene.

Its backbone follows:

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

Each phenyl group attaches to a backbone carbon at one point. The chain remains continuous without passing through the ring.

Therefore, the rings are side groups.

For Polymer B, the chain follows a different route:

\[ \ce{...-CH2-C6H4-CH2-C6H4-...} \]

The methylene groups connect to different positions on each benzene ring.

The polymer chain passes through the aromatic rings to reach the next repeating section.

Therefore, the rings are part of the backbone.

The tempting mistake is to classify both structures based on the presence of a benzene ring. However, the same chemical ring can be a side group in one polymer and a backbone component in another.

Connectivity, not chemical identity, determines its role.

Question 3

Ethylene glycol can react with either terephthalic acid or isophthalic acid to form a polyester.

Both acids contain a benzene ring with two carboxylic acid groups.

• Terephthalic acid has its carboxylic acid groups at positions 1 and 4.
• Isophthalic acid has its carboxylic acid groups at positions 1 and 3.

A student argues that the benzene ring in the isophthalic-acid-derived polymer must become a side group because its two attachment positions are not opposite each other.

Is this reasoning correct? Explain what changes between the two polymers and what remains the same.

Solution 3

The student’s reasoning is incorrect. The benzene ring remains part of the backbone in both polyesters.

The important feature is that both carboxylic acid groups react to form ester linkages.

In both cases, the polymer chain connects to the ring-containing section through two separate carbonyl groups.

The chain therefore passes through the benzene ring, regardless of whether the attachment positions are 1,4 or 1,3.

What changes is the geometry of the polymer backbone.

With terephthalic acid, the substituents are opposite each other. This produces a more extended arrangement through the aromatic section.

With isophthalic acid, the substituents are at positions 1 and 3. The chain connections are angled differently, changing the geometry of the repeating structure.

However, the ring still connects two sections of the polymer chain.

The misconception is assuming that a ring must have opposite attachment positions to belong to the backbone. That is not necessary. The relevant condition is that the ring lies along the chain’s connecting route.

Question 4

Consider a polymer with an otherwise continuous carbon-carbon backbone:

\[ \ce{...-A-B-C-D-...} \]

Here, A, B, C, and D represent carbon atoms in the backbone.

A benzene ring is also attached to A and D through two separate carbon-carbon bonds.

The original backbone route from A through B and C to D remains intact.

A student argues that the benzene ring must be part of the backbone because it has two connections to backbone atoms.

Is this conclusion correct? Classify the ring’s structural role and explain the significance of the existing A-to-D route.

Solution 4

The conclusion is incorrect. The benzene ring forms an additional bridge between two backbone positions rather than being a necessary part of the main-chain route.

The key information is that the original backbone remains continuous:

\[ \ce{A-B-C-D} \]

You can trace the polymer chain from A to D without entering the benzene ring.

The ring creates an additional covalent connection between A and D, but it does not replace the existing backbone route.

This is different from PET, where the chain passes through the benzene-containing section to connect successive repeating units.

The ring in this question is best described as an additional bridge or intrachain crosslink. If it connected two different polymer chains, it would act as an interchain crosslink.

The tempting shortcut is to count the number of connections:

• One connection suggests a side group.
• Two connections suggest a backbone component.

But this shortcut is incomplete.

Two connections can also create an additional loop or bridge. You must check whether the ring is actually part of the chain route or simply provides another connection between points already joined by the backbone.

Question 5

Two unknown aromatic polymers are being investigated. Each contains ester groups and benzene rings.

Two candidate structural models are proposed.

Model P: PET

\[ \ce{[-O-CH2-CH2-O-C(=O)-C6H4-C(=O)-]_{n}} \]

Model Q: Poly(vinyl benzoate)

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

Two samples are prepared, each containing 0.0100 mol of ideal repeating units.

Assume complete alkaline hydrolysis, negligible chain-end effects, and no cleavage of carbon-carbon backbone bonds.

The following illustrative results are obtained:

A student makes the following argument:

“Sample X must have had its benzene ring as a side group because terephthalate is recovered as a separate molecule after hydrolysis. If the ring had been part of the backbone, it would have remained connected to the polymer.”

Evaluate the argument.

Identify which structural model matches each sample, determine the position of each benzene ring, and explain why the student’s interpretation of the hydrolysis products is incorrect.

Solution 5

The student’s argument is incorrect. Sample X matches PET, whose benzene ring is part of the backbone. Sample Y matches poly(vinyl benzoate), whose benzene ring is a pendant side group.

The critical distinction is that hydrolysis can break bonds in the polymer backbone itself. Recovering a small molecule after hydrolysis does not prove that the molecule was originally a side group.

Step 1: Identify Sample X.

Sample X consumes 0.0200 mol of hydroxide for 0.0100 mol of repeat units.

The ratio is:

\[ \frac{0.0200}{0.0100}=2 \]

Therefore, two hydroxide ions are consumed per ideal repeat unit.

This matches Model P, PET, which contains two ester linkages per repeat unit.

Step 2: Interpret the products from Sample X.

Sample X produces:

• 0.0100 mol of ethylene glycol.
• 0.0100 mol of terephthalate.

These are the products expected when PET undergoes complete alkaline hydrolysis.

The two ester linkages in the backbone are broken, separating the ethylene glycol-derived and terephthalate-derived sections.

The benzene ring is still present in the terephthalate product.

However, that does not mean it was originally a side group. It was part of the polymer backbone before the ester bonds were hydrolysed.

The absence of a remaining high-molar-mass polymer is consistent with complete cleavage of the backbone ester linkages.

Step 3: Identify Sample Y.

Sample Y consumes 0.0100 mol of hydroxide for 0.0100 mol of repeat units.

The ratio is:

\[ \frac{0.0100}{0.0100}=1 \]

This matches Model Q, poly(vinyl benzoate), which contains one pendant ester group per repeat unit.

Step 4: Interpret the products from Sample Y.

Sample Y produces 0.0100 mol of benzoate ions.

Its high-molar-mass polymer chain remains because the ester groups were attached to the backbone rather than forming part of it.

The hydrolysis removes the benzoate side groups, leaving a polymeric carbon-carbon backbone containing hydroxyl substituents:

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

The benzene ring was part of the released benzoate side group.

Step 5: Identify the exact flaw in the student’s argument.

The student assumes that a ring recovered as a separate molecule must originally have been a side group.

That assumption ignores the possibility of breaking the backbone itself.

In PET, the ester bonds connect the repeating units. Breaking those bonds releases the terephthalate-containing section as a small molecule.

In poly(vinyl benzoate), the ester bonds connect pendant groups to an otherwise continuous carbon-carbon backbone. Breaking those bonds releases the benzoate group without destroying the main polymer chain.

The difference is not whether the aromatic fragment is eventually released. It is which bonds must be broken to release it.

Final conclusion:

• Sample X is PET. Its benzene ring belongs to the backbone.
• Sample Y is poly(vinyl benzoate). Its benzene ring belongs to a pendant side group.

The strongest interpretation combines the structural formula, the number of ester linkages, the hydrolysis products, and whether the polymer backbone survives.

07The method to remember

When an HSC Chemistry question asks whether a ring belongs to the polymer backbone or is a side group, use these checks:

Keep these three examples in mind:

• PET: polyester with benzene rings in the backbone.
• Polystyrene: addition polymer with pendant phenyl groups.
• Poly(vinyl benzoate): addition polymer with pendant ester-containing aromatic groups.

The next useful step is to connect polymer structure to material properties. Once you can distinguish the backbone from its side groups, you can start explaining how chain rigidity, molecular packing, and intermolecular forces influence the behaviour of the polymer.

For more practice constructing PET from its monomers, see How to Draw PET from Ethylene Glycol and Terephthalic Acid.

If you are moving on to intermolecular forces, How to Compare Hydrogen Bonding in Polyamides and Polyesters develops the next structural decision: identifying which parts of a polymer can form hydrogen bonds between neighbouring chains.