How to Reconstruct an Alkene Monomer from an Addition Polymer

A polymer is represented by the repeating unit \(\ce{[-CH2-CH(CH3)-]_{n}}\). You are asked to identify the original alkene monomer.

The answer is propene, \(\ce{CH2=CHCH3}\).

The trick is to reverse what happened during addition polymerisation. Find the two carbon atoms in the polymer backbone that originally formed the alkene double bond, remove their connections to neighbouring repeat units, and restore the double bond between them.

The methyl group stays exactly where it is. You do not remove it, move it, or turn it into part of the double bond.

01Why reversing the structure works

An alkene contains a carbon-carbon double bond, \(\ce{C=C}\). During addition polymerisation, this double bond opens, allowing the alkene molecules to connect and form a long carbon chain.

The original double bond becomes a single bond in the polymer backbone. Each of the two carbon atoms also forms a new bond with a neighbouring monomer.

For example, propene undergoes addition polymerisation to form poly(propene):

\[ \ce{n CH2=CHCH3 -> [-CH2-CH(CH3)-]_{n}} \]

Here, \(n\) represents the number of repeating units in the polymer chain.

Notice three things:

• The two carbon atoms originally joined by a double bond are still next to each other.
• Their double bond has become a single bond.
• The methyl group, \(\ce{CH3}\), remains attached to the same carbon atom.

No atoms are removed during this process. This is why the original monomer can be reconstructed by reversing the changes to the carbon-carbon bonds.

Think of each alkene molecule as a pair of train carriages connected together. During polymerisation, the connection between the two carriages becomes a single bond, while each carriage connects to neighbouring pairs. Any side groups attached to a carriage stay attached.

The analogy helps you track connectivity, but real chemical bonds involve shared electrons, not mechanical connections. Carbon valence must still be satisfied.

02Understanding the brackets and open bonds

Before reconstructing the monomer, you need to understand what the polymer brackets actually represent.

Consider:

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

This represents a repeating section of a much longer chain:

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

The brackets identify the repeating unit. The subscript \(n\) indicates that this unit is repeated many times.

The dashes at the left and right edges of the brackets are called continuing bonds or open bonds. They show where the repeating unit connects to neighbouring units.

They do not represent additional hydrogen atoms.

03The five-step method for reconstructing an alkene

Use this method whenever you are given the structural formula of a conventional addition polymer and asked to identify its original alkene monomer.

Step 1: Identify the carbon backbone.

Find the carbon atoms that form the main chain of the polymer.

For a simple addition polymer made from one alkene, the monomer contributes two carbon atoms to the backbone.

Step 2: Identify the original alkene carbon atoms.

These are the two carbon atoms that were originally connected by the double bond.

In a simple two-carbon repeating unit, they are the two backbone carbons shown inside the brackets.

Step 3: Keep all side groups attached.

Look for branches such as:

• \(\ce{CH3}\), a methyl group
• \(\ce{Cl}\), a chlorine atom
• \(\ce{C6H5}\), a phenyl group

These groups were already attached to the alkene before polymerisation.

They remain attached to the same carbon when you reconstruct the monomer.

Step 4: Restore the carbon-carbon double bond.

Remove the two connections to neighbouring polymer units.

Then change the single bond between the original alkene carbon atoms back into a double bond.

Step 5: Check carbon valence.

Every carbon atom must have four covalent bonds.

Remember that a double bond counts as two bonds.

If a carbon has too many or too few bonds, you have probably reconstructed the structure incorrectly.

04Why carbon valence confirms your answer

Carbon valence is your final accuracy check.

Return to the poly(propene) repeating unit:

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

Consider the two backbone carbon atoms separately.

The first carbon, \(\ce{CH2}\):

In the polymer, it has:

• Two bonds to hydrogen atoms.
• One bond to the second backbone carbon.
• One continuing bond to the previous repeat unit.

That gives four bonds.

When reconstructing the monomer, the continuing bond is removed. The bond between the two backbone carbons becomes a double bond.

The carbon now has two C-H bonds and a C=C double bond. Its valence is still four.

The second carbon, \(\ce{CH(CH3)}\):

In the polymer, it has:

• One bond to hydrogen.
• One bond to the methyl group.
• One bond to the first backbone carbon.
• One continuing bond to the next repeat unit.

Again, that gives four bonds.

After reconstruction, it has one C-H bond, one C-CH3 bond, and a C=C double bond.

Its valence is also four.

This check is particularly useful when the monomer contains several different substituents.

For more practice checking carbon’s four-bond requirement, see How to Check Carbon Valence in Polymer Structural Formulae.

07Common HSC addition polymers and their monomers

The following table is a useful reference when practising reverse structure identification.

One final limitation: this reverse method applies to conventional addition polymers made from alkene monomers. Do not automatically apply it to condensation polymers. Condensation polymerisation involves joining functional groups and eliminating small molecules, so reconstructing the original monomers requires a different approach.

For a broader explanation of how these polymers form, see Addition polymers explained: PE, PVC, PS and PTFE for HSC Chemistry.

08Questions and solutions

The following questions move from straightforward monomer identification to more demanding structural reasoning. Pay particular attention to the side groups, carbon valence, and the meaning of the repeating-unit brackets.

Question 1

A polymer has the repeating unit:

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

Identify the original alkene monomer and explain why the single bond between the two backbone carbon atoms must be restored to a double bond.

Solution 1

The original monomer is ethene, \(\ce{CH2=CH2}\).

The repeating unit contains two backbone carbon atoms, each originally part of the alkene double bond.

During addition polymerisation, the double bond becomes a single bond as the carbon atoms form connections to neighbouring monomers.

To reverse the process:

1. Remove the continuing bond on the left of the first carbon.
2. Remove the continuing bond on the right of the second carbon.
3. Restore the double bond between the two carbon atoms.

The reconstructed structure is:

\[ \ce{CH2=CH2} \]

Each carbon now has two hydrogen bonds and a double bond to the other carbon, giving four bonds in total.

The reaction used to form the polymer was addition polymerisation.

Question 2

Two different addition polymers have the following repeating units:

Polymer A

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

Polymer B

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

Reconstruct the original alkene monomer for each polymer.

Give the systematic name of each monomer and explain why the two polymers cannot have been formed from the same alkene.

Solution 2

Polymer A comes from chloroethene, while polymer B comes from 2-chloroprop-1-ene. They have different monomers because their backbone carbon atoms carry different substituents.

Polymer A

Start with:

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

The chlorine atom stays attached to the second carbon.

Restoring the double bond gives:

\[ \ce{CH2=CHCl} \]

The monomer is chloroethene, commonly called vinyl chloride.

Polymer B

Start with:

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

The second carbon has two substituents: chlorine and a methyl group.

Neither group is removed. Restoring the double bond gives:

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

The monomer is 2-chloroprop-1-ene.

Why are they different?

In polymer A, the second backbone carbon has one hydrogen and one chlorine substituent.

In polymer B, it has one chlorine and one methyl substituent, with no hydrogen.

This difference must be preserved when reconstructing the monomers.

A useful final check is carbon valence. In polymer B, the second monomer carbon has a double bond to \(\ce{CH2}\), a single bond to chlorine, and a single bond to the methyl group. That gives four bonds.

You cannot simply identify the chlorine and assume both polymers came from chloroethene. The complete connectivity matters.

Question 3

Polytetrafluoroethene (PTFE) has the repeating unit:

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

A student argues that this polymer cannot have been formed by addition polymerisation because its repeating unit contains no hydrogen atoms.

Is the student correct?

Reconstruct the original monomer, identify its type, and use carbon valence to justify your answer.

Solution 3

The student is incorrect. PTFE is formed by addition polymerisation of tetrafluoroethene, \(\ce{CF2=CF2}\).

The absence of hydrogen does not prevent addition polymerisation. The essential structural feature is a carbon-carbon double bond, not the presence of hydrogen.

Start with:

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

Step 1: Identify the backbone carbons.

There are two carbon atoms, each bonded to two fluorine atoms.

Step 2: Remove the continuing bonds.

Each carbon loses one connection to a neighbouring repeat unit.

Step 3: Restore the double bond.

The single bond between the two backbone carbons becomes a double bond:

\[ \ce{CF2=CF2} \]

Step 4: Check carbon valence.

Each carbon has:

• Two single bonds to fluorine atoms.
• One double bond to the other carbon.

That gives four bonds per carbon.

The monomer is tetrafluoroethene, a fluorinated alkene, and the polymer forms through addition polymerisation.

The tempting mistake is to assume that every alkene-derived monomer must contain hydrogen. What matters here is the C=C bond and the ability of its carbon atoms to form additional bonds.

Question 4

An unknown addition polymer has the repeating unit:

\[ \ce{[-CH2-C(CH3)2-]_{n}} \]

Its repeating-unit molecular formula is \(\ce{C4H8}\).

A student suggests that the original monomer must have been but-1-ene because but-1-ene also has the molecular formula \(\ce{C4H8}\).

Identify the actual monomer.

Explain why molecular formula alone is insufficient to identify the original alkene. Compare the given polymer structure with the structures that would form from but-1-ene and but-2-ene.

Solution 4

The actual monomer is 2-methylpropene, \(\ce{CH2=C(CH3)2}\).

Although the three candidate alkenes have the same molecular formula, they have different structural arrangements. The repeating-unit connectivity identifies which isomer was used.

Step 1: Reconstruct the given polymer.

The repeating unit is:

\[ \ce{[-CH2-C(CH3)2-]_{n}} \]

There are two backbone carbon atoms.

The second carbon has two methyl groups attached to it.

Remove the continuing bonds and restore the double bond:

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

This is 2-methylpropene.

Step 2: Compare with but-1-ene.

But-1-ene has the structure:

\[ \ce{CH2=CHCH2CH3} \]

Its polymer repeating unit would be:

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

Notice that the ethyl group, \(\ce{CH2CH3}\), remains as one branch.

This is not the structure supplied in the question.

Step 3: Compare with but-2-ene.

But-2-ene has the structure:

\[ \ce{CH3CH=CHCH3} \]

Its polymer repeating unit would be:

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

Here, each backbone carbon has one methyl group attached.

Again, this differs from the given structure.

Step 4: Understand why the molecular formula is insufficient.

All three monomers have the molecular formula \(\ce{C4H8}\).

Their repeating units also have the same molecular formula because addition polymerisation does not eliminate atoms.

However, their atoms are connected differently.

The key distinction is between molecular formula and structural formula.

A molecular formula tells you how many atoms are present. A structural formula tells you how those atoms are connected.

When reconstructing a monomer, connectivity is what allows you to distinguish structural isomers.

Question 5

An idealised alternating addition copolymer has the following repeating unit:

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

Assume that:

• The polymer was formed from exactly two different monofunctional alkene monomers.
• Each monomer contributes exactly two adjacent carbon atoms to the polymer backbone.
• The monomers alternate regularly throughout the chain.
• No atoms are lost or rearranged during polymerisation.

A student argues:

“The repeating unit contains four backbone carbon atoms, so it must have come from one alkene with four carbon atoms. I only need to restore one double bond.”

Determine the two original monomers.

Explain why the student’s reasoning is incorrect. Your explanation must identify the possible monomer-derived carbon pairs, show that both give the same two monomers, and explain why the bracket boundaries do not necessarily identify the original monomer boundaries.

Solution 5

The two original monomers are ethene, \(\ce{CH2=CH2}\), and propene, \(\ce{CH2=CHCH3}\).

The student’s mistake is assuming that one bracketed repeating unit must contain exactly one monomer residue.

That is not always true. In an alternating copolymer, one repeating unit can contain contributions from two different monomers.

Step 1: Label the four backbone carbon atoms.

Represent the repeating unit as:

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

where:

• A is \(\ce{CH2}\).
• B is \(\ce{CH2}\).
• C is \(\ce{CH2}\).
• D is \(\ce{CH(CH3)}\).

The chain continues as:

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

The prime symbols indicate carbon atoms in the next repeating section.

Step 2: Identify the first possible pairing.

One valid way to group the carbon atoms is:

\[ \underbrace{\ce{A-B}}_{\text{ethene}} - \underbrace{\ce{C-D}}_{\text{propene}} \]

The first pair contains two \(\ce{CH2}\) groups.

Restoring their double bond gives:

\[ \ce{CH2=CH2} \]

This is ethene.

The second pair contains \(\ce{CH2}\) and \(\ce{CH(CH3)}\).

Restoring their double bond gives:

\[ \ce{CH2=CHCH3} \]

This is propene.

Therefore, the first pairing identifies ethene and propene.

Step 3: Check the alternative pairing.

Now shift the grouping by one carbon:

\[ \underbrace{\ce{B-C}}_{\text{ethene}} - \underbrace{\ce{D-A'}}_{\text{propene}} \]

The B-C pair again gives ethene:

\[ \ce{CH2=CH2} \]

The D-A’ pair gives propene:

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

This is the same propene structure written in the opposite direction.

The alternative pairing therefore produces exactly the same two monomers.

Step 4: Explain why the student’s single-monomer assumption fails.

The repeating unit contains four backbone carbon atoms because it represents one ethene-derived section and one propene-derived section.

It does not represent one four-carbon alkene molecule.

For a monofunctional alkene, the two carbons originally involved in the C=C bond form the two-carbon section of the polymer backbone. The remaining atoms, if any, stay attached as substituents.

Here, the four backbone atoms form two separate two-carbon monomer-derived sections.

Restoring only one double bond would reconstruct only one of those sections. The other would still be an addition-polymer-derived pair with an unresolved bond pattern.

Step 5: Understand the role of the brackets.

The brackets identify a convenient repeating section of the polymer chain. They do not label the original boundaries of every monomer molecule.

Depending on where you begin reading the chain, the monomer-derived pairs can be grouped in different ways.

In this example, both possible groupings give the same two monomers: ethene and propene.

The important conclusion is that a repeating unit and a monomer are not always the same-sized structural section.

This is why you must examine the full connectivity and repeating pattern rather than automatically assuming that one bracket contains one monomer.

09What to remember when reconstructing an alkene monomer

The reverse method is straightforward once you understand what changed during addition polymerisation:

1. Identify the two carbon atoms that originally formed the alkene double bond.
2. Remove the continuing bonds connecting the polymer to neighbouring repeat units.
3. Restore the single bond between those two carbons to a double bond.
4. Keep every side group attached to its original carbon.
5. Check that every carbon has four bonds.
6. Check whether the bracket contains one monomer-derived section or multiple sections, particularly in a copolymer.

The most common errors are removing a side group, adding hydrogen atoms where continuing bonds used to be, or restoring the wrong double bond.

Once you can reverse the structure reliably, the natural next step is learning to move in the opposite direction: given an unfamiliar alkene, predict its addition polymer, draw the repeating unit, and place the brackets and continuing bonds correctly.