How to Draw Polystyrene from Styrene Without Breaking the Ring

Styrene looks like a trap when you’re drawing polymers. Its structure, \(\ce{C6H5CH=CH2}\), contains a carbon-carbon double bond in the side chain and an aromatic benzene ring. If you treat every double bond in the molecule as an ordinary alkene bond, you might accidentally destroy the benzene ring in your polymer drawing.

So, which part actually reacts?

In ordinary addition polymerisation of styrene, the side-chain C=C bond opens, while the benzene ring stays intact. The ring becomes a phenyl side group attached to the polymer backbone.

The correct repeat unit of polystyrene is:

\[
\boxed{\left[-\ce{CH2-CH(C6H5)-}\right]_n}
\]

The important detail is that the two carbon atoms from the original alkene form the polymer backbone. The \(\ce{C6H5}\) group remains attached to one of those carbons.

01Identify which double bond reacts

Start with the structure of styrene:

\[
\ce{C6H5-CH=CH2}
\]

There are two features worth recognising:

• The \(\ce{CH=CH2}\) section is the vinyl group. It contains an ordinary carbon-carbon alkene double bond.
• The \(\ce{C6H5}\) section is a phenyl group. It consists of a benzene ring with one hydrogen replaced by a bond to the vinyl group.

These two features have different roles during polymerisation.

Why doesn’t the benzene ring break?

Benzene is aromatic. Its six pi electrons are delocalised across the entire ring, giving it a particularly stable electronic structure.

When benzene is drawn using alternating single and double bonds, those double bonds are a representation of its delocalised bonding. They should not be treated as three separate alkene sites.

Under ordinary addition polymerisation conditions, the reactive site in styrene is the separate alkene double bond in the vinyl group.

The benzene ring retains its aromatic structure.

This distinction matters because breaking the aromatic ring would produce a completely different structure from polystyrene.

What happens to the vinyl group?

Think of each styrene molecule as a carriage with two connectors and a nameplate on its side.

The two alkene carbons become the connections that join neighbouring monomers. The phenyl group is like the nameplate: it stays attached to its original carbon as the carriages join into a longer chain.

The analogy helps identify the backbone and side group, although the real process involves the formation of covalent bonds rather than mechanical connectors.

The structural change is:

\[
\ce{C6H5-CH=CH2}
\]

becomes part of a chain containing:

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

The original double bond becomes a single bond, and new single bonds connect the monomer to neighbouring units.

Importantly, no atoms need to be removed from the styrene molecule.

The difference between the backbone and the side group

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

In polystyrene, the backbone consists of alternating \(\ce{CH2}\) and \(\ce{CH(C6H5)}\) groups.

The phenyl groups are attached to the backbone but do not form part of it. These are called pendant groups or side groups.

There is also an important naming distinction here. The attached group is phenyl, \(\ce{C6H5-}\), not benzene, \(\ce{C6H6}\). One hydrogen has been replaced by the bond connecting the ring to the polymer backbone.

02How to draw the polystyrene repeat unit

Use the following method whenever you are asked to draw polystyrene from styrene.

Step 1: Write the monomer structure.

\[
\ce{C6H5-CH=CH2}
\]

Step 2: Identify the reactive double bond.

Find the C=C bond in the vinyl group.

Do not change the aromatic benzene ring.

Step 3: Convert the alkene double bond into a single bond.

The two carbon atoms remain connected, but the double bond becomes a single bond.

Step 4: Add bonds for chain continuation.

Each of the two original alkene carbons needs a connection to a neighbouring repeat unit.

This produces a continuous carbon backbone.

Step 5: Keep the phenyl group attached.

The \(\ce{C6H5}\) group stays bonded to the carbon that originally carried it.

Step 6: Draw the repeat unit inside square brackets.

The completed structure is:

\[
\boxed{\left[-\ce{CH2-CH(C6H5)-}\right]_n}
\]

The \(n\) outside the brackets indicates that this repeating structural unit occurs many times in the polymer chain.

Three common drawing mistakes

A useful final check is to count the atoms.

Styrene has the molecular formula \(\ce{C8H8}\). The polystyrene repeat unit also has the formula \(\ce{C8H8}\).

\[
\ce{C6H5 + CH2 + CH = C8H8}
\]

No atoms have disappeared, and no small molecule is released during this addition polymerisation.

03A useful contrast: Poly(propene) and benzene

Comparing styrene with another monomer makes the drawing rule easier to recognise.

Consider propene:

\[
\ce{CH2=CH-CH3}
\]

Its polymer is poly(propene), with the repeat unit:

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

Notice what happens to the methyl group, \(\ce{CH3}\). It remains attached to the backbone.

Styrene follows exactly the same structural pattern, except its methyl-like substituent is replaced by a phenyl group.

The important pattern is that the groups attached to the alkene carbons are retained when the alkene becomes part of the polymer backbone.

Benzene itself is different from styrene. A benzene molecule, \(\ce{C6H6}\), does not have a separate vinyl C=C bond. Its aromatic ring is not treated as an ordinary alkene during this type of addition polymerisation.

This does not mean benzene can never react. It means that aromatic substitution and other ring reactions are different processes from the addition polymerisation being considered here.

04How could an experiment support this structure?

There is a useful experimental question behind the structural model:

How could we check that the vinyl double bond has reacted while the benzene ring remains intact?

Simply observing that liquid styrene becomes solid polystyrene would not answer this. A change in physical state does not tell us which bonds have changed.

Infrared spectroscopy provides a way to investigate the molecular structure.

The apparatus and prediction

An infrared spectrometer sends infrared radiation from a source towards a sample. The radiation passes through a sample holder containing styrene or a thin polystyrene film before reaching a detector.

The detector measures how much radiation is transmitted at different wavenumbers. The resulting spectrum contains absorption features associated with different types of molecular vibrations.

Before looking at the results, make a prediction.

If the vinyl group reacts but the aromatic ring remains, what should happen to the characteristic absorption features?

We would expect:

• Absorption features associated with the vinyl C=C and terminal \(\ce{=CH2}\) group to disappear or become strongly reduced.
• Absorption features associated with the aromatic ring to remain.

What would the observations show?

The following is an illustrative qualitative comparison, not data from a claimed experimental measurement.

These observations are consistent with the proposed polystyrene structure.

The vinyl group has been consumed during polymerisation, while the aromatic ring remains in the product.

Why one absorption peak is not enough

There is a potential problem with the original experimental approach.

A student might look at an absorption feature in the region associated with C=C stretching and conclude that the original alkene double bond must still be present.

That conclusion is not justified.

Aromatic rings also produce infrared absorption features associated with their delocalised bonding. Some of these occur in regions that overlap with alkene absorptions.

Therefore, a remaining absorption in the relevant region does not automatically mean the vinyl C=C bond survived.

A better experimental design compares the full spectra of both samples under consistent conditions. It examines several features, including the terminal vinyl C-H vibrations and the aromatic ring absorptions, rather than relying on one peak.

What can we conclude?

The experimental observations support the structural model in which styrene’s vinyl group reacts and its phenyl group remains intact.

However, an infrared spectrum is not a direct picture of molecular connectivity. It provides evidence about the types of bonds and groups present, but does not necessarily establish the exact arrangement of every atom.

Knowing the original monomer and the reaction taking place is therefore important when assigning the polymer’s structure.

05Questions and solutions

Question 1

Styrene has the condensed structural formula \(\ce{C6H5CH=CH2}\).

Draw the repeat unit formed during ordinary addition polymerisation. Use square brackets and \(n\), and identify the group that remains attached to the polymer backbone.

Solution 1

The correct repeat unit is:

\[
\boxed{\left[-\ce{CH2-CH(C6H5)-}\right]_n}
\]

The phenyl group, \(\ce{C6H5}\), remains attached to the polymer backbone.

The reasoning is that ordinary addition polymerisation occurs across the vinyl C=C bond. This double bond becomes a single bond, and the two alkene carbons form the continuous polymer backbone.

The benzene ring is not broken or incorporated into the backbone. It remains an intact pendant group.

Question 2

A mixture contains styrene and benzene. The reaction conditions are suitable for ordinary free-radical addition polymerisation of styrene.

A student argues that both molecules should become incorporated into the polymer because both molecular structures can be drawn with alternating double bonds.

Assume benzene is acting as an inert solvent and no separate aromatic substitution reaction occurs.

Evaluate the student’s claim. In your answer:

• Identify which molecule supplies the polymer repeat units.
• Explain why the double bonds in benzene should not be treated as an ordinary alkene.
• Draw the resulting polymer repeat unit.
• Explain why your conclusion applies to the specified reaction conditions rather than every possible reaction involving benzene.

Solution 2

The student’s claim is incorrect. Styrene supplies the repeat units, while benzene does not become incorporated into the polymer under the specified conditions.

The key difference is the type of bonding in each molecule.

Styrene contains a separate alkene double bond:

\[
\ce{C6H5-CH=CH2}
\]

This is the reactive site for ordinary free-radical addition polymerisation.

Benzene, however, has an aromatic ring with delocalised pi electrons. Its alternating double bonds are a representation of this delocalised bonding, not three independent alkene sites.

Under the specified conditions, the vinyl group in styrene reacts to form the polymer backbone.

The resulting repeat unit is:

\[
\boxed{\left[-\ce{CH2-CH(C6H5)-}\right]_n}
\]

The phenyl group remains attached to the backbone carbon.

It would be incorrect to conclude that benzene can never undergo a chemical reaction. Benzene can participate in other reactions, including substitution reactions, under suitable conditions.

The important distinction is that those reactions are not the ordinary addition polymerisation taking place here.

Question 3

A chemist begins with \(0.780\ \mathrm{g}\) of styrene and achieves a polymerisation conversion of \(75.0\%\).

The unreacted styrene is removed, leaving polystyrene with a number-average molar mass of \(20.8\ \mathrm{kg\,mol^{-1}}\).

Calculate:

(a) The mass of polystyrene produced.

(b) The amount, in moles, of polymer chains produced.

(c) The average number of styrene repeat units per polymer chain.

Assume:

• The repeat unit has a molar mass of \(104\ \mathrm{g\,mol^{-1}}\).
• All converted styrene becomes part of the polymer.
• No small molecule is released during polymerisation.
• The mass of the polymer chain ends is negligible.

Solution 3

The mass of polystyrene produced is \(0.585\ \mathrm{g}\), the amount of polymer chains is \(2.81\times10^{-5}\ \mathrm{mol}\), and the average degree of polymerisation is 200.

(a) Calculate the mass of polystyrene.

First, calculate the fraction of styrene that has reacted:

\[
\text{Conversion}=0.750
\]

Since no small molecule is released, all the mass of the converted styrene is retained in the polymer.

\[
\begin{aligned}
m_{\text{polymer}}
&=0.750\times0.780\\
&=0.585\ \mathrm{g}
\end{aligned}
\]

Therefore:

\[
\boxed{m_{\text{polymer}}=0.585\ \mathrm{g}}
\]

(b) Calculate the amount of polymer chains.

Convert the number-average molar mass into grams per mole:

\[
20.8\ \mathrm{kg\,mol^{-1}}
=20\,800\ \mathrm{g\,mol^{-1}}
\]

Use:

\[
n=\frac{m}{M}
\]

Substituting:

\[
\begin{aligned}
n_{\text{chains}}
&=\frac{0.585}{20\,800}\\
&=2.81\times10^{-5}\ \mathrm{mol}
\end{aligned}
\]

Therefore:

\[
\boxed{n_{\text{chains}}=2.81\times10^{-5}\ \mathrm{mol}}
\]

(c) Calculate the average number of repeat units per chain.

First, calculate the initial amount of styrene:

\[
\begin{aligned}
n_{\text{styrene}}
&=\frac{0.780}{104}\\
&=7.50\times10^{-3}\ \mathrm{mol}
\end{aligned}
\]

At \(75.0\%\) conversion, the amount incorporated into the polymer is:

\[
\begin{aligned}
n_{\text{repeat units}}
&=0.750\times7.50\times10^{-3}\\
&=5.625\times10^{-3}\ \mathrm{mol}
\end{aligned}
\]

Every polymer chain contains a number of repeat units, so:

\[
DP_n=
\frac{n_{\text{repeat units}}}
{n_{\text{chains}}}
\]

Substituting:

\[
\begin{aligned}
DP_n
&=\frac{5.625\times10^{-3}}
{2.8125\times10^{-5}}\\
&=200
\end{aligned}
\]

Therefore:

\[
\boxed{DP_n=200}
\]

The average polymer chain contains 200 styrene repeat units.

Notice that the amount of repeat units and the amount of polymer chains are different quantities. Confusing these two is an easy way to lose marks in a multi-step calculation.

Question 4

A chemist compares pure styrene with a polystyrene film using transmission infrared spectroscopy.

The following qualitative observations are recorded. These are illustrative results.

One absorption feature remains in a region commonly associated with C=C stretching.

A student concludes that the original alkene double bond must still be present in the polymer.

(a) Evaluate the student’s conclusion.

(b) Explain how the observations support the proposed structure of polystyrene.

(c) Suggest one improvement to the experimental design that would make the interpretation more reliable.

Solution 4

The student’s conclusion is not justified. The remaining absorption could be associated with the aromatic benzene rings rather than the original vinyl C=C bond.

(a) Evaluate the conclusion.

Aromatic rings produce their own infrared absorption features associated with their delocalised bonding.

Some aromatic and alkene absorptions occur in overlapping regions.

Therefore, observing an absorption in a region associated with C=C stretching does not prove that the original alkene bond remains.

The student has assumed that every absorption in this region must come from an ordinary alkene.

(b) Explain the observations.

The terminal vinyl \(\ce{=CH2}\) feature disappears after polymerisation.

This supports the idea that the vinyl group has reacted.

The vinyl C=C feature also disappears or becomes strongly reduced, consistent with the conversion of the alkene double bond into single bonds in the polymer backbone.

Meanwhile, the aromatic ring absorptions remain.

This supports the proposed structure in which the phenyl groups retain their aromatic rings.

Together, these observations are consistent with:

\[
\left[-\ce{CH2-CH(C6H5)-}\right]_n
\]

(c) Improve the experimental design.

Rather than measuring only one region of the spectrum, compare the full infrared spectra of styrene and polystyrene.

In particular, examine the terminal vinyl C-H vibrations alongside the aromatic ring features.

Using comparable sample thicknesses and consistent measurement conditions would also improve the reliability of the comparison.

This makes it easier to distinguish the disappearance of the vinyl group from the continued presence of aromatic rings.

The experiment supports the structural model, but infrared spectroscopy alone does not necessarily establish the exact connectivity of every atom.

Question 5

A laboratory receives an unknown polymer sample.

The following information is available:

• Its repeat-unit formula is \(\ce{C8H8}\).
• Its number-average molar mass is \(20.8\ \mathrm{kg\,mol^{-1}}\).
• Its infrared spectrum shows aromatic absorptions.
• Its infrared spectrum shows no terminal alkene absorptions.

Two possible structural models are proposed:

Model A: Polystyrene

\[
\left[-\ce{CH2-CH(C6H5)-}\right]_n
\]

Model B: A hypothetical alternative polymer

\[
\left[-\ce{C6H4-CH2-CH2-}\right]_n
\]

In Model B, the \(\ce{C6H4}\) rings are para-disubstituted, meaning that the aromatic rings form part of the polymer backbone rather than remaining as pendant groups.

Both models have the repeat-unit formula \(\ce{C8H8}\). Model B is a hypothetical alternative connectivity, not an assertion that it is a particular commercial polymer.

Assume the infrared report is qualitative and does not provide detailed information about aromatic substitution patterns or backbone connectivity.

(a) Calculate the average degree of polymerisation for either model.

(b) Determine whether the supplied evidence uniquely identifies the unknown polymer as polystyrene.

(c) Explain what additional information would allow the laboratory to justify Model A rather than Model B.

Solution 5

The average degree of polymerisation is 200, but the supplied evidence does not uniquely establish that the unknown polymer is polystyrene.

This question tests an important distinction: knowing which atoms are present is not necessarily the same as knowing how those atoms are connected.

(a) Calculate the average degree of polymerisation.

Both proposed repeat units have the molecular formula:

\[
\ce{C8H8}
\]

Using the supplied atomic molar masses:

\[
\begin{aligned}
M_r&=8(12.0)+8(1.0)\\
&=104\ \mathrm{g\,mol^{-1}}
\end{aligned}
\]

Convert the polymer’s number-average molar mass:

\[
20.8\ \mathrm{kg\,mol^{-1}}
=20\,800\ \mathrm{g\,mol^{-1}}
\]

The average degree of polymerisation is:

\[
\begin{aligned}
DP_n&=\frac{M_n}{M_r}\\
&=\frac{20\,800}{104}\\
&=200
\end{aligned}
\]

Therefore:

\[
\boxed{DP_n=200}
\]

Both models give the same result because their repeat units have the same molar mass.

(b) Does the evidence uniquely identify polystyrene?

No. Both models are consistent with the supplied information.

Consider what each measurement actually tells us.

First, the repeat-unit formula establishes the numbers of carbon and hydrogen atoms. It does not establish which carbon atoms are bonded to one another.

Second, the infrared spectrum indicates that aromatic structures remain present and that terminal alkene features are absent.

However, both proposed models contain aromatic rings and lack terminal alkene groups.

Third, the molar mass determines the average number of repeat units, but it does not reveal whether the phenyl rings are pendant groups or part of the polymer backbone.

The tempting approach is to see \(\ce{C8H8}\), recognise that this is the formula of styrene, and immediately identify the sample as polystyrene.

The problem is that the alternative model also has the formula \(\ce{C8H8}\).

The evidence establishes composition and some bonding features, but it does not uniquely establish connectivity.

(c) What additional information would justify Model A?

The clearest additional information would be confirmation that the original monomer was styrene and that the reaction was ordinary addition polymerisation across its vinyl C=C bond.

Starting with:

\[
\ce{C6H5-CH=CH2}
\]

and adding monomers through the vinyl group necessarily gives:

\[
\boxed{\left[-\ce{CH2-CH(C6H5)-}\right]_n}
\]

The phenyl group remains attached to the backbone carbon because the reaction occurs at the separate alkene double bond.

Alternatively, a more detailed structural investigation using a connectivity-sensitive technique, such as nuclear magnetic resonance spectroscopy, could help distinguish the two models.

The important lesson is that molecular formula, molar mass, and evidence of functional groups do not always establish a unique molecular structure. You also need information about how the atoms are connected.

When drawing other addition polymers, use the same backbone-first approach: identify the reactive alkene, retain its attached substituents, and only then construct the repeating unit. That is the reliable way to move from a monomer’s structure to the correct polymer without accidentally changing the molecule you were supposed to preserve.