How to Draw a Polymer from an Unsymmetrical Alkene

Take propene, \(\ce{CH2=CHCH3}\). When its double bond opens during polymerisation, where does the \(\ce{CH3}\) group go? Does it become part of the main polymer chain, or does it remain attached to the side?

It remains attached to the same carbon. The two carbon atoms that originally formed the double bond become part of the polymer backbone, while the methyl group becomes a side branch.

The resulting polymer has this repeating unit:

\[ n\,\ce{CH2=CHCH3} \longrightarrow \left[-\mathrm{CH_2-CH(CH_3)-}\right]_n \]

This is poly(propene), also called polypropylene.

That is the essential method for drawing a polymer from an unsymmetrical alkene: open the double bond, extend the chain through the two alkene carbons, and keep every side group attached to its original carbon.

01Understand what happens to the double bond

An alkene contains a carbon-carbon double bond, written as C=C.

In an unsymmetrical alkene, the groups attached to the two double-bonded carbon atoms are not identical.

For example, propene has the structure:

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

Look closely at the two carbon atoms in the double bond:

• The first carbon is \(\ce{CH2}\).
• The second carbon is \(\ce{CH}\), which is attached to a \(\ce{CH3}\) group.

These two carbon atoms have different attachments. That is what makes propene an unsymmetrical alkene.

During addition polymerisation, the double bond becomes a single bond. Each of the two alkene carbon atoms then forms an additional carbon-carbon bond to a neighbouring monomer.

The original two carbon atoms remain next to each other in the polymer backbone.

The important detail is that the attached groups do not move.

For propene:

• The \(\ce{CH2}\) carbon becomes one backbone carbon.
• The \(\ce{CH}\) carbon becomes the next backbone carbon.
• The \(\ce{CH3}\) group remains attached to the second backbone carbon.

This gives:

\[ \left[-\mathrm{CH_2-CH(CH_3)-}\right]_n \]

The general pattern

For a common unsymmetrical terminal alkene with the structure:

\[ \mathrm{CH_2=CH-R} \]

where \(R\) represents the group attached to the second alkene carbon, the repeating unit is:

\[ \boxed{ \left[-\mathrm{CH_2-CH(R)-}\right]_n } \]

For example:

Notice that the entire attached group stays together. An ethyl group does not become two separate branches, and it does not automatically become part of the polymer backbone.

If the alkene has more than one substituent on an alkene carbon, keep all of them attached to that carbon. The same principle applies.

02The five-step method for drawing an addition polymer

When an HSC Chemistry question asks you to draw a polymer from an unsymmetrical alkene, follow this sequence.

Step 1: Identify the C=C double bond.

Find the two carbon atoms connected by the double bond. These are the two atoms that will form the repeating backbone segment.

Step 2: Change the double bond into a single bond.

Replace C=C with C-C.

Do not remove either carbon. They remain connected.

Step 3: Keep all attached groups in their original positions.

Any methyl, ethyl, or other group attached to either alkene carbon must stay attached to that same carbon.

This is the step where students commonly make mistakes.

Step 4: Add an open bond at each end.

Each of the two alkene-derived carbon atoms needs an additional bond connecting it to a neighbouring monomer.

These are represented by open bonds extending from the repeating unit.

Step 5: Add brackets and \(n\).

Place square brackets around the repeating unit and write \(n\) outside the lower-right corner.

For propene, the result is:

\[ \left[-\mathrm{CH_2-CH(CH_3)-}\right]_n \]

What do the open bonds actually mean?

Look at the two dashes at the ends of the repeating unit:

\[ \left[ \underbrace{-}_{\text{chain continues}} \mathrm{CH_2-CH(CH_3)} \underbrace{-}_{\text{chain continues}} \right]_n \]

They represent bonds connecting this section to neighbouring repeating units.

They are not extra hydrogen atoms, and they are not unfinished bonds that should be removed.

The brackets indicate that the structure inside repeats along a much longer polymer chain. The \(n\) represents the number of bracketed repeating units.

Check the valency of carbon

Carbon forms four covalent bonds.

A double bond counts as two bonds when checking carbon’s valency. After polymerisation, each alkene carbon has one additional bond connecting it to the polymer chain.

For propene, compare the two carbon atoms before and after polymerisation:

Both carbons have four bonds in the polymer.

This is a useful final check. If a carbon has five bonds, or an attached group has disappeared, revisit your drawing.

03A contrasting case: what if the alkene has two side groups?

Consider 2-methylpropene:

\[ \ce{CH2=C(CH3)2} \]

This molecule looks similar to propene, but there is an important difference.

The second alkene carbon is attached to two methyl groups, not one. It also has no hydrogen attached to it.

When the double bond opens, both methyl groups remain attached to that same carbon.

The correct repeating unit is:

\[ \boxed{ \left[-\mathrm{CH_2-C(CH_3)_2-}\right]_n } \]

Compare the three cases:

But-2-ene is a useful contrasting example because both alkene carbons have a methyl group attached.

The main lesson is that you must inspect the original molecule rather than assume that every substituted alkene carbon has one hydrogen.

The reaction type: addition polymerisation

Why is this reaction called addition polymerisation?

Because the alkene monomers join together by using their carbon-carbon double bonds. The monomers add to the growing chain without eliminating a small molecule.

For a simple alkene addition polymerisation:

• The C=C double bond becomes a C-C single bond.
• New C-C bonds connect neighbouring monomers.
• All atoms from the monomer are retained in the repeating unit.
• No water or other small molecule is eliminated.

This last point is particularly useful when checking your structure.

If your proposed repeating unit contains fewer carbon atoms than the original monomer, you may have accidentally removed a side group.

04How to work backwards from a polymer to its monomer

You can also reverse the process.

Suppose you are given a polymer repeating unit and asked to identify the original alkene monomer.

For a simple repeating unit containing one monomer’s worth of structure, use the reverse of the drawing method.

1. Identify the two backbone carbon atoms that originally formed the double bond.
2. Remove the open chain-continuation bonds at the two ends.
3. Change the single bond between those two carbon atoms back into a double bond.
4. Keep every side group attached to its original carbon.
5. Check that each carbon has four bonds.

For example, consider:

\[ \left[-\mathrm{CH_2-CH(CH_2CH_3)-}\right]_n \]

Remove the two outside chain links and restore the double bond between the backbone carbons.

You obtain:

\[ \ce{CH2=CHCH2CH3} \]

This is but-1-ene.

The ethyl group remains attached to the second alkene carbon throughout the reversal.

One important detail: a polymer can sometimes be represented using a bracket containing two or more of its smallest repeating units. A longer bracket does not automatically mean a different polymer. Check the repeating pattern before reversing the structure.

For more practice with this reverse-identification process, see How to Find Monomers from a Polymer’s Repeating Unit.

05How could an experiment check that the double bond has disappeared?

So far, we have used structural reasoning to predict what happens during polymerisation.

But how could a chemist test whether the C=C bond has actually disappeared?

One possible approach is infrared (IR) spectroscopy.

This is an experimental extension: you do not need to perform this experiment to draw the repeating unit, but it helps explain the chemical change behind the model.

The proposed experiment

An infrared spectrometer directs infrared radiation towards a sample.

The apparatus can be represented as:

1. An infrared source produces radiation.
2. The radiation interacts with a sample of propene or a sample of purified poly(propene).
3. A detector measures the radiation that passes through or is reflected from the sample.
4. The instrument produces a spectrum showing which infrared frequencies have been absorbed.

The experiment compares the monomer with the polymer.

What should the experiment show?

Before polymerisation, propene contains a C=C double bond.

That bond produces an associated infrared absorption feature.

After complete addition polymerisation, the polymer backbone contains C-C single bonds rather than the original alkene double bonds.

Therefore, the expected observation is:

• The propene spectrum contains an alkene C=C absorption feature.
• The purified polymer spectrum does not contain that feature.
• Other features, such as those associated with C-H bonds, remain.

This supports the model that the double bond has been consumed during polymerisation.

Why does the experimental design matter?

Imagine measuring the IR spectrum of the reaction mixture before separating the polymer.

There could still be unreacted propene present. Its C=C bond would produce an absorption feature even if the polymer itself contained no double bonds.

That makes the original design difficult to interpret.

A better approach is to separate and wash the polymer before testing it. Then compare the purified polymer with a pure monomer sample.

This reduces the chance that leftover monomer will interfere with the result.

The experiment supports the loss of the alkene double bond, but IR spectroscopy alone does not establish the complete arrangement of every atom in the polymer. The placement of side groups still requires structural evidence.

For the HSC drawing method, the essential reasoning remains atom conservation, bond formation, and carbon valency.

06What does \(n\) tell you about the polymer?

The \(n\) outside the brackets represents the number of repeating units.

It can also be used to estimate the degree of polymerisation: the average number of repeating units in a polymer chain.

For an addition polymer, the repeating unit has the same elemental composition as the original monomer because no atoms are lost.

Consider poly(propene).

The monomer is:

\[ \ce{CH2=CHCH3} \]

Its molecular formula is \(\ce{C3H6}\).

Using relative atomic masses of C = 12.0 and H = 1.0:

\[ M_r = 3(12.0)+6(1.0)=42.0 \]

The repeating unit therefore has a molar mass of approximately \(42.0\ \mathrm{g\,mol^{-1}}\).

Suppose a polymer sample has an average molar mass of \(84\,000\ \mathrm{g\,mol^{-1}}\).

The approximate degree of polymerisation is:

\[ \begin{aligned} n &\approx \frac{\text{average polymer molar mass}} {\text{repeat unit molar mass}}\\[4pt] &=\frac{84\,000}{42.0}\\ &=2\,000 \end{aligned} \]

The average chain contains approximately 2,000 repeating units.

This estimate assumes that the contribution from the chain-end groups is negligible. It also uses the average molar mass of the polymer, so not every individual chain necessarily contains exactly 2,000 units.

Notice that the side groups count towards the repeating unit’s molar mass, even though they are not part of the main backbone.

07Worked examples

08Questions and solutions

The following questions progress from straightforward polymer drawings to more demanding problems involving atom conservation, reverse identification, and the interpretation of repeating units.

Question 1

Propene, \(\ce{CH2=CHCH3}\), undergoes complete addition polymerisation.

Draw the repeating unit of the polymer. Include the brackets, \(n\), and both open bonds. Explain what happens to the methyl group during the reaction.

Solution 1

The correct repeating unit is:

\[ \boxed{ \left[-\mathrm{CH_2-CH(CH_3)-}\right]_n } \]

The two carbon atoms that originally formed the C=C double bond become the polymer backbone.

The double bond changes into a single bond, and each carbon forms an additional bond connecting it to a neighbouring monomer.

The methyl group remains attached to the second backbone carbon. It becomes a side branch rather than part of the main chain.

The carbon valencies are both satisfied:

• \(\ce{CH2}\) has two C-H bonds, one internal C-C bond, and one chain link.
• \(\ce{CH}\) has one C-H bond, one C-\(\ce{CH3}\) bond, one internal C-C bond, and one chain link.

No atoms are lost, so the repeat unit retains the monomer’s \(\ce{C3H6}\) composition.

Question 2

A student is asked to draw the polymer formed from 2-methylpropene:

\[ \ce{CH2=C(CH3)2} \]

The student submits:

\[ \left[-\mathrm{CH_2-CH(CH_3)-}\right]_n \]

Identify the two structural errors in the student’s answer. Draw the correct repeating unit and explain how carbon valency helps you check it.

Solution 2

The correct repeating unit is:

\[ \boxed{ \left[-\mathrm{CH_2-C(CH_3)_2-}\right]_n } \]

The student’s answer contains two errors.

First, one methyl group has disappeared. The original monomer contains two methyl groups attached to the same alkene carbon. Both must remain in the polymer.

Second, the student has added a hydrogen to the second backbone carbon. The original alkene carbon has no hydrogen attached to it, and no hydrogen is added during this addition polymerisation.

The correct central carbon has:

• One bond to the first backbone carbon.
• Two bonds to its methyl branches.
• One open chain-continuation bond.

That gives four bonds.

The correct repeating unit also contains four carbon atoms and eight hydrogen atoms, matching the original monomer’s \(\ce{C4H8}\) composition.

The mistake is not simply a matter of drawing style. The student’s structure represents a different chemical composition.

Question 3

A polymer has the following repeating unit:

\[ \left[-\mathrm{CH_2-C(CH_3)(CH_2CH_3)-}\right]_n \]

Identify the original alkene monomer and give its systematic name.

Explain how you reconstruct the double bond and check that the resulting monomer satisfies carbon valency.

Solution 3

The original monomer is:

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

Its systematic name is 2-methylbut-1-ene.

To reconstruct it, first identify the two backbone carbons: the \(\ce{CH2}\) carbon and the central carbon.

Remove the two outside chain-continuation bonds. Then change the internal single bond between these carbon atoms into a double bond.

The methyl and ethyl groups stay attached to the central carbon.

The resulting structure is:

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

Check the central carbon:

• The C=C double bond contributes two bonds.
• The C-\(\ce{CH3}\) bond contributes one.
• The C-\(\ce{CH2CH3}\) bond contributes one.

Its total valency is four.

The terminal carbon has two hydrogen bonds and a double bond to the central carbon, also giving four bonds.

The reconstruction is therefore chemically consistent.

Question 4

A sample of poly(2-methylpropene) has an average molar mass of \(42\,000\ \mathrm{g\,mol^{-1}}\).

A student calculates the degree of polymerisation using \(42.0\ \mathrm{g\,mol^{-1}}\) as the molar mass of the repeating unit and obtains \(n=1000\).

Using the relative atomic masses C = 12.0 and H = 1.0:

1. Calculate the correct average degree of polymerisation.
2. Explain why the student’s repeat-unit molar mass is incorrect.

Assume that chain-end contributions are negligible.

Solution 4

The correct average degree of polymerisation is:

\[ \boxed{n\approx750} \]

The error is that the student has failed to include both methyl groups when calculating the repeat-unit molar mass.

The monomer is 2-methylpropene:

\[ \ce{CH2=C(CH3)2} \]

Its molecular formula is:

\[ \ce{C4H8} \]

Because this is addition polymerisation, all four carbon atoms and eight hydrogen atoms are retained in the repeating unit.

Calculate its molar mass:

\[ \begin{aligned} M_{\text{repeat}}&=4(12.0)+8(1.0)\\ &=48.0+8.0\\ &=56.0\ \mathrm{g\,mol^{-1}} \end{aligned} \]

Now calculate the degree of polymerisation:

\[ \begin{aligned} n&\approx\frac{42\,000}{56.0}\\ &=750 \end{aligned} \]

The student’s value of \(42.0\ \mathrm{g\,mol^{-1}}\) corresponds to \(\ce{C3H6}\), meaning one carbon and two hydrogen atoms have been left out of the calculation.

The key point is that the repeating unit includes both backbone atoms and all attached side groups.

A side group may sit outside the main polymer backbone, but its atoms still contribute to the repeat unit’s molar mass.

Question 5

Consider three proposed ways of representing a polymer formed from propene.

Structure A:

\[ \left[-\mathrm{CH_2-CH(CH_3)-}\right]_n \]

Structure B:

\[ \left[-\mathrm{CH_2-CH(CH_3)-CH_2-CH(CH_3)-}\right]_n \]

Structure C:

\[ \left[-\mathrm{CH_2-CH(CH_3)-CH(CH_3)-CH_2-}\right]_n \]

Assume:

• Every monomer is propene, \(\ce{CH2=CHCH3}\).
• Every incorporated C=C bond becomes a C-C single bond.
• No atoms are eliminated.
• Every carbon must have a valency of four.
• The open bonds at the ends of each bracket connect to the next copy of that repeating segment.

Answer the following:

1. Which structures describe the same regular polymer chain?
2. Does Structure C violate atom conservation or carbon valency?
3. If Structure C is valid under these constraints, explain how its carbon-carbon connections differ from those in A and B.
4. If an ordinary HSC question gives only propene as the monomer and asks for its addition polymer, which representation should you normally use? Explain why the monomer’s molecular formula alone cannot establish every detail of the polymer’s structure.

Solution 5

Structures A and B represent the same regular head-to-tail polymer chain, while C represents a different arrangement of the carbon-carbon connections.

However, Structure C is not ruled out by atom conservation or carbon valency alone.

The distinction requires looking beyond the molecular formula.

First, compare A and B.

Structure A is the smallest repeating unit:

\[ \left[-\mathrm{CH_2-CH(CH_3)-}\right]_n \]

Structure B contains two copies of Structure A:

\[ \left[ -\mathrm{CH_2-CH(CH_3)-} -\mathrm{CH_2-CH(CH_3)-} \right]_n \]

The atoms connect in exactly the same repeating pattern.

The only difference is the size of the bracketed segment.

In A, \(n\) counts individual propene repeating units. In B, \(n\) counts pairs of propene repeating units.

Therefore, A and B describe the same polymer, provided the number of bracketed segments is interpreted accordingly.

Next, check the atom count in C.

Structure C contains:

• Four backbone carbon atoms.
• Two methyl branch carbon atoms.
• Twelve hydrogen atoms in total.

Its composition is therefore:

\[ \ce{C6H12} \]

This is exactly the composition of two propene molecules:

\[ 2\ce{C3H6}=\ce{C6H12} \]

No atoms have disappeared.

Now check the valency.

In Structure C:

• Each \(\ce{CH2}\) carbon has two hydrogen bonds, one internal backbone bond, and one open chain link.
• Each \(\ce{CH(CH3)}\) carbon has one hydrogen bond, one methyl branch, and two backbone bonds.

Every carbon has four bonds.

So C satisfies both atom conservation and carbon valency.

What makes C different?

Define the substituted alkene carbon, \(\ce{CH(CH3)}\), as the head of a propene unit, and the terminal \(\ce{CH2}\) carbon as its tail.

In the usual head-to-tail arrangement, the tail of one monomer connects to the head of the next.

That gives the regular sequence:

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

In Structure C, the two substituted carbons connect directly to each other inside the bracket. This is a head-to-head connection.

At the boundary between repeating segments, the two \(\ce{CH2}\) carbons connect to each other. This is a tail-to-tail connection.

The result has the same overall composition as two propene molecules, but a different connectivity.

Finally, which structure should you draw in an HSC question?

Use Structure A:

\[ \boxed{ \left[-\mathrm{CH_2-CH(CH_3)-}\right]_n } \]

It is the standard idealised regular head-to-tail representation of poly(propene).

Structure B is simply a larger way of writing the same repeating pattern. Structure C has different connectivity, even though it satisfies the stated atom and valency constraints.

Determining whether a particular polymer actually contains such alternative arrangements requires evidence about its structure and the conditions under which it formed. The monomer’s molecular formula alone cannot establish that information.

This is the important distinction: having the correct atoms and satisfying carbon valency are necessary conditions for a polymer structure, but they do not always establish its exact connectivity.

Once you can draw the polymer in the forward direction and reconstruct its monomer in reverse, the next useful skill is recognising the carbon skeleton and naming the original alkene correctly. If you need to refresh that foundation, see HSC Chemistry: Alkane, Alkene and Alkyne Formulae.