How to Find the Monomer of a Fluorinated Addition Polymer

Given the polymer repeat unit \(\ce{[-CF2-CF2-]_{n}}\), how would you find the original monomer?

The answer is tetrafluoroethene, \(\ce{CF2=CF2}\). You can work this out directly from the polymer structure, without memorising the monomer’s name.

The key is to reverse one structural change: restore a carbon-carbon double bond between the two backbone carbon atoms, while keeping every other atom attached to its original carbon.

Now predict the monomer for a different fluorinated polymer:

\[
\ce{[-CH2-CF2-]_{n}}
\]

Would you still write \(\ce{CF2=CF2}\), because the polymer contains fluorine? No. That would lose two hydrogen atoms. The correct monomer is \(\ce{CH2=CF2}\), known as vinylidene fluoride or 1,1-difluoroethene.

This small difference is the key to identifying fluorinated addition polymer monomers accurately.

01The reverse method: from polymer to monomer

First, remember what happened during addition polymerisation.

An alkene monomer contains a carbon-carbon double bond. During addition polymerisation, that double bond opens, allowing the carbon atoms to connect with neighbouring monomer molecules.

The result is a long carbon chain containing carbon-carbon single bonds.

To reverse this process, you need to identify the two carbon atoms that originally formed the double bond.

You can review the forward process in Addition polymers explained: PE, PVC, PS and PTFE for HSC Chemistry.

Three steps to find the monomer

Step 1: Identify the carbon backbone.

Find the two adjacent carbon atoms that form the repeating section of the polymer.

These are the atoms that originally formed the double bond in the alkene monomer.

Step 2: Keep all attached atoms in place.

This is particularly important for fluorinated polymers.

Fluorine atoms do not disappear during addition polymerisation. Neither do hydrogen or chlorine atoms.

If a carbon has two fluorine atoms attached in the polymer, it must still have those two fluorine atoms when you reconstruct the monomer.

Step 3: Change the C-C single bond into a C=C double bond.

Restore the double bond between the two backbone carbon atoms.

You now have the original alkene monomer.

The general idea is:

\[
\ce{C-C -> C=C}
\]

Only the bond order changes. You do not add or remove atoms.

An important detail: carbon valency

Carbon forms four covalent bonds.

When you restore the double bond, check that each carbon still has four bonds in total.

For example, in vinylidene fluoride:

\[
\ce{CH2=CF2}
\]

The first carbon has two C-H single bonds and one C=C double bond.

The second carbon has two C-F single bonds and one C=C double bond.

Both carbons satisfy their valency of four.

This is a useful final check if you’re unsure whether you’ve reconstructed the structure correctly.

02Recognising common fluorinated polymers

Several fluorinated addition polymers follow exactly the same method, but their monomers have different arrangements of hydrogen and fluorine atoms.

Here are four useful structural examples.

A few naming distinctions are worth learning:

• PTFE is polytetrafluoroethylene, made from tetrafluoroethene.
• PVDF is poly(vinylidene fluoride), made from vinylidene fluoride.
• PVF is poly(vinyl fluoride), made from vinyl fluoride.
• PCTFE is poly(chlorotrifluoroethylene), made from chlorotrifluoroethene.

Notice that PVDF and PVF are different polymers. Their monomers contain different numbers of fluorine atoms.

Also, PTFE is not the only possible answer when a question mentions a fluorinated polymer. You must use the actual structure supplied.

03Worked examples

04When one monomer is not enough: fluorinated copolymers

So far, we’ve considered polymers made from one type of monomer. These are called homopolymers.

However, some fluorinated addition polymers are copolymers. Their chains contain units derived from two different monomers.

This changes the reconstruction process slightly.

Instead of looking for one original alkene, you must identify each distinct monomer-derived section of the polymer.

Example: FEP

FEP, or fluorinated ethylene-propylene copolymer, is made using two monomers:

• Tetrafluoroethene: \(\ce{CF2=CF2}\)
• Hexafluoropropene: \(\ce{CF2=CF-CF3}\)

The two monomers produce different sections of the polymer chain.

The second structure introduces an important detail.

The \(\ce{CF3}\) group is attached to one of the backbone carbon atoms. It is a side group, not an additional carbon atom in the main polymer backbone.

When reconstructing the monomer, restore the double bond between the two backbone carbons and leave the \(\ce{CF3}\) group attached.

This is an extension of the basic HSC reconstruction method. The European Environment Agency’s technical report on fluorinated polymers also identifies FEP as a copolymer of tetrafluoroethene and hexafluoropropene.

The distinction matters because applying the single-monomer method to an entire copolymer can give an incorrect result.

05Common mistakes to avoid

Before attempting exam questions, check these common errors.

The last mistake is particularly important in unfamiliar questions. Two polymers can have the same elemental composition but different structures and different monomers.

06Questions and solutions

Question 1

An addition polymer has the repeating unit:

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

Determine:

(a) The structural formula of its monomer.

(b) The common and systematic names of the monomer.

(c) The molar mass of the monomer, using the following relative atomic masses:

\[
A_r(\ce{C})=12.0,\quad
A_r(\ce{H})=1.0,\quad
A_r(\ce{F})=19.0
\]

Solution 1

The monomer is \(\ce{CH2=CHF}\), fluoroethene, commonly called vinyl fluoride. Its molar mass is \(46.0\text{ g mol}^{-1}\).

Step 1: Restore the double bond between the two backbone carbon atoms.

The original repeat unit is:

\[
\ce{-CH2-CHF-}
\]

Changing the C-C single bond into a C=C double bond gives:

\[
\ce{CH2=CHF}
\]

The fluorine atom remains attached to the second carbon.

Step 2: Identify the monomer’s name.

The systematic name is fluoroethene.

Its common name is vinyl fluoride.

Step 3: Determine the molecular formula.

Counting the atoms gives:

\[
\ce{C2H3F}
\]

Step 4: Calculate the molar mass.

Use the relative atomic masses supplied:

\[
\begin{aligned}
M &= 2(12.0)+3(1.0)+19.0\\
&=24.0+3.0+19.0\\
&=46.0\text{ g mol}^{-1}
\end{aligned}
\]

The monomer has a molar mass of \(46.0\text{ g mol}^{-1}\).

The important check is that the monomer and repeat unit contain exactly the same atoms. Addition polymerisation does not remove a small molecule from the monomer.

Question 2

A student is given the following polymer repeat unit:

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

They identify the original monomer as tetrafluoroethene:

\[
\ce{CF2=CF2}
\]

Explain why this identification is incorrect.

Determine the correct monomer’s structural formula and name.

Solution 2

The correct monomer is \(\ce{CF2=CFCl}\), chlorotrifluoroethene, also called chlorotrifluoroethylene.

The student’s answer is incorrect because it replaces the chlorine atom with another fluorine atom.

Step 1: Count the attached atoms.

The supplied polymer repeat unit contains:

• Two carbon atoms.
• Three fluorine atoms.
• One chlorine atom.

All these atoms must be present in the original monomer.

Step 2: Restore the double bond.

The two backbone carbon atoms are connected by a single bond.

Changing this into a double bond gives:

\[
\ce{CF2=CFCl}
\]

Step 3: Identify the monomer’s name.

The monomer is chlorotrifluoroethene.

Its more systematic name is 1-chloro-1,2,2-trifluoroethene.

Step 4: Explain the student’s mistake.

Tetrafluoroethene has the structure:

\[
\ce{CF2=CF2}
\]

It contains four fluorine atoms and no chlorine atoms.

The supplied polymer contains chlorine, so tetrafluoroethene cannot be its monomer under the stated addition-polymerisation model.

The key constraint is conservation of atoms. Changing the bond order does not change the identity or number of atoms.

Question 3

An idealised 1:1 alternating copolymer has the following repeating sequence:

\[
\ce{[-CH2-CH2-CF2-CF2-]_{n}}
\]

Determine:

(a) The two original monomers.

(b) The name of each monomer.

(c) Why the four-carbon repeating sequence does not represent a single four-carbon monomer.

Assume the polymer contains alternating units derived from the two monomers.

Solution 3

The two monomers are ethene, \(\ce{CH2=CH2}\), and tetrafluoroethene, \(\ce{CF2=CF2}\).

The four-carbon repeating sequence represents one unit derived from each monomer, not one four-carbon monomer.

Step 1: Identify the two distinct carbon pairs.

The repeating sequence is:

\[
\ce{-CH2-CH2-CF2-CF2-}
\]

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

The second pair contains two \(\ce{CF2}\) groups.

Step 2: Reconstruct the first monomer.

The two \(\ce{CH2}\) carbon atoms are connected by a single bond.

Restoring the double bond gives:

\[
\ce{CH2=CH2}
\]

This is ethene.

Step 3: Reconstruct the second monomer.

The two \(\ce{CF2}\) carbon atoms are also connected by a single bond.

Restoring the double bond gives:

\[
\ce{CF2=CF2}
\]

This is tetrafluoroethene.

Step 4: Explain the four-carbon repeating sequence.

The polymer is a copolymer, so two different monomers contribute to its repeating structure.

One ethene molecule contributes two \(\ce{CH2}\) groups, while one tetrafluoroethene molecule contributes two \(\ce{CF2}\) groups.

The bond connecting the ethene-derived section to the tetrafluoroethene-derived section remains a single bond.

The tempting mistake is to assume that every bracketed polymer section must come from one monomer. That is not true for copolymers.

Question 4

FEP is a fluorinated copolymer containing two distinct polymer sections:

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

and

\[
\ce{-CF2-CF(CF3)-}
\]

Determine both original monomers.

Explain why the \(\ce{CF3}\) group in the second section must remain attached to a backbone carbon rather than becoming part of the main carbon chain.

Solution 4

The two monomers are tetrafluoroethene, \(\ce{CF2=CF2}\), and hexafluoropropene, \(\ce{CF2=CF-CF3}\).

The \(\ce{CF3}\) group remains a side group attached to the second backbone carbon.

Step 1: Identify the first monomer.

The first polymer section is:

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

Restoring the double bond gives:

\[
\ce{CF2=CF2}
\]

This is tetrafluoroethene.

Step 2: Examine the second polymer section.

The second section is:

\[
\ce{-CF2-CF(CF3)-}
\]

There are two carbon atoms in the main backbone section.

The third carbon belongs to the attached \(\ce{CF3}\) group.

Step 3: Restore the double bond between the backbone carbons.

The resulting monomer is:

\[
\ce{CF2=CF-CF3}
\]

This is hexafluoropropene.

Step 4: Explain why the \(\ce{CF3}\) group stays attached.

The monomer originally contained three carbon atoms, but only two formed the alkene double bond.

The third carbon was already part of a side group attached to one of those alkene carbons.

During addition polymerisation, the double bond opens and the two alkene carbons become part of the polymer backbone. The \(\ce{CF3}\) group remains attached to the same carbon.

The important distinction is between the carbon atoms that form the polymer backbone and those that form side groups.

Question 5

Extension

A laboratory analyses an unknown fluorinated addition polymer.

The results show:

• A saturated carbon-carbon polymer backbone.
• An empirical composition of \(\ce{C2H2F2}\).
• No additional elements in the polymer.

A student concludes that the original monomer must have been vinylidene fluoride:

\[
\ce{CH2=CF2}
\]

The student’s reasoning is that PVDF has the repeat unit \(\ce{[-CH2-CF2-]}\), which has exactly the reported composition.

Evaluate this conclusion.

Construct two chemically distinct polymer structures consistent with the supplied information:

(a) One homopolymer.

(b) One idealised 1:1 alternating copolymer.

Identify the monomer or monomers for each structure, and explain what additional evidence would be needed to determine which polymer was actually analysed.

Solution 5

The student’s conclusion is not justified. The empirical composition \(\ce{C2H2F2}\) does not uniquely identify the monomer because it gives the ratio of atoms, not how those atoms are connected.

Two different polymer structures satisfy the supplied information.

Step 1: Construct the first possibility: PVDF.

Vinylidene fluoride has the monomer structure:

\[
\ce{CH2=CF2}
\]

Its addition polymerisation produces:

\[
\ce{n CH2=CF2 -> [-CH2-CF2-]_{n}}
\]

Each repeat unit contains:

• Two carbon atoms.
• Two hydrogen atoms.
• Two fluorine atoms.

Its repeat-unit formula is therefore:

\[
\ce{C2H2F2}
\]

This is a valid homopolymer model.

Step 2: Construct the second possibility: an idealised ETFE copolymer.

ETFE is made from two different monomers:

• Ethene: \(\ce{CH2=CH2}\)
• Tetrafluoroethene: \(\ce{CF2=CF2}\)

For an idealised 1:1 alternating sequence, the polymer structure can be represented as:

\[
\ce{[-CH2-CH2-CF2-CF2-]_{n}}
\]

Each four-carbon sequence contains:

• Four carbon atoms.
• Four hydrogen atoms.
• Four fluorine atoms.

Its composition is:

\[
\ce{C4H4F4}
\]

Dividing the subscripts by two gives the empirical formula:

\[
\ce{C2H2F2}
\]

This is also consistent with the laboratory’s elemental analysis.

Step 3: Compare the two structures.

The two possibilities have different carbon connectivity.

In PVDF, the backbone alternates between:

\[
\ce{-CH2-CF2-}
\]

In the idealised ETFE structure, the chain contains:

\[
\ce{-CH2-CH2-CF2-CF2-}
\]

The elemental composition is the same, but the arrangement of atoms is different.

Step 4: Identify the additional evidence required.

Elemental analysis alone cannot distinguish these structures.

A structure-sensitive technique, such as carbon-13 nuclear magnetic resonance spectroscopy (\(^{13}\mathrm{C}\) NMR), could provide information about the different carbon environments and their connectivity.

That evidence would help determine whether the sample contains the alternating \(\ce{CH2-CF2}\) structure of PVDF or the separate ethene-derived and tetrafluoroethene-derived sections of ETFE.

The deeper lesson: Finding a monomer requires structural information, not simply an elemental formula. The reverse-bond method works when you know which carbon atoms are connected and which groups are attached to them. Without that information, more than one polymer structure may fit the same chemical composition.

07Where to go next

Once you can reconstruct a monomer from its polymer structure, two useful skills follow.

First, practise moving in the opposite direction: given an unfamiliar alkene, predict the repeating unit it will produce. This is the foundation of Addition polymers explained: PE, PVC, PS and PTFE for HSC Chemistry.

Second, learn how to use a polymer’s molar mass and repeat-unit mass to estimate the number of repeating units in a chain. See How to Calculate Repeat Units in a Polymer Chain.

Syllabus note (2026): The current Year 12 course remains the Chemistry Stage 6 Syllabus (2017) through 2027. The Chemistry 11-12 Syllabus (2025) begins implementation for Year 11 in 2028, with the first HSC examination under the new syllabus in 2029. PTFE is used here as the familiar HSC anchor. The additional polymer structures provide transfer practice, while the copolymer and empirical-composition problems are extension material rather than claims about prescribed syllabus examples. See NESA’s syllabus transition information.