Why the Double Bond Disappears in Addition Polymers

Look at ethene, \(\ce{CH2=CH2}\), and the repeating unit of polyethene, \(\left[-\mathrm{CH_2-CH_2-}\right]_n\). The monomer has a carbon-carbon double bond, but the polymer has a carbon-carbon single bond. Where did the second bond go?

The double bond doesn’t disappear without a trace. Its π bond is broken, while the original σ bond remains. New carbon-carbon σ bonds form between neighbouring monomers, allowing them to join into a long chain.

This is the key change in addition polymerisation: the C=C bond becomes a C-C single bond, and new C-C bonds connect the monomer units.

01What actually happens to the double bond?

A carbon-carbon double bond consists of two different types of covalent bond:

• One sigma (σ) bond, formed by direct overlap of atomic orbitals along the line joining the two carbon nuclei.
• One pi (π) bond, formed by sideways overlap of p orbitals, creating electron density above and below the line joining the carbon nuclei.

When an alkene undergoes addition polymerisation, the π bond is broken. The σ bond between the original two carbon atoms remains.

The carbon atoms can then form additional single bonds with carbon atoms from neighbouring monomers.

Consider ethene as an example.

Before polymerisation, each carbon atom has:

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

The total bond order around each carbon is four:

\[ 2(1)+2=4 \]

After polymerisation, each internal carbon atom in the polyethene chain has:

• Two single bonds to hydrogen atoms.
• One single bond to the other carbon from its original ethene molecule.
• One single bond to a carbon from a neighbouring monomer.

Its total bond order is still four:

\[ 1+1+1+1=4 \]

Carbon has not exceeded its usual valency of four. The bonding has simply been rearranged.

Notice something important: the original C-C bond is not completely broken. It changes from a double bond to a single bond.

The new bonds between monomers are separate bonds formed between different carbon atoms.

If you need to refresh the difference between sigma and pi bonding, see Carbon-Carbon Bond Shapes: Single, Double and Triple Bonds.

02How can an experiment check whether the double bond has disappeared?

The structural formula gives us a model of what happens. But how could we check whether the alkene double bond is actually being consumed?

One useful technique is infrared (IR) spectroscopy.

IR spectroscopy measures how molecules absorb infrared radiation. Different bonds absorb radiation at characteristic wavenumbers, allowing chemists to investigate which bonding features are present.

Consider the conversion of styrene into polystyrene.

The apparatus for a simple Fourier-transform infrared (FTIR) measurement follows this path:

1. An infrared source produces radiation.
2. An interferometer processes the radiation.
3. The radiation passes through, or interacts with, the prepared sample.
4. A detector measures the transmitted signal.
5. The instrument produces a spectrum showing absorbance against wavenumber.

What should the experiment show?

Styrene contains two distinct structural features:

• A vinyl C=C bond, which participates in addition polymerisation.
• An aromatic phenyl ring, which remains attached as a side group.

If the vinyl double bond is consumed during polymerisation, its characteristic IR absorption should disappear or become greatly reduced.

However, the aromatic-ring signals should remain.

The following table shows an illustrative qualitative comparison, not actual measured experimental data.

This supports the proposed structural change: the alkene double bond is used during chain formation, while the aromatic ring remains.

There is an experimental limitation.

A weaker absorption peak does not automatically mean a bond has disappeared. Different sample thicknesses can change peak intensities.

A more reliable comparison uses controlled sample preparation and compares the vinyl absorption against aromatic features expected to remain. This helps distinguish an actual structural change from differences in how the samples were measured.

IR spectroscopy also cannot, by itself, establish the complete arrangement of every atom in the polymer chain. It provides evidence about bonding features, rather than a complete structural drawing.

A small calculation: interpreting an IR wavenumber

Suppose an alkene C=C absorption is assigned an illustrative wavenumber of \(1630\ \mathrm{cm^{-1}}\).

The relationship between wavelength and wavenumber is:

\[ \lambda=\frac{1}{\tilde{\nu}} \]

where:

• \(\lambda\) is wavelength, measured in metres.
• \(\tilde{\nu}\) is wavenumber, measured in reciprocal centimetres when using \(\mathrm{cm^{-1}}\).

Substituting:

\[ \begin{aligned} \lambda &=\frac{1}{1630\ \mathrm{cm^{-1}}}\\ &=6.13\times10^{-4}\ \mathrm{cm}\\ &=6.13\times10^{-6}\ \mathrm{m} \end{aligned} \]

Therefore, the radiation has a wavelength of approximately \(6.13\ \mathrm{\mu m}\).

This is an example of how a bond-related absorption can be investigated experimentally. The exact absorption position depends on the molecular environment, so the value supplied here is illustrative rather than a universal fixed value.

03Why do polymer diagrams have open bonds?

Now that we understand the chemical change, we need to understand how it is represented.

A polyethene repeating unit is written as:

\[ \left[-\mathrm{CH_2-CH_2-}\right]_n \]

There are three important features in this notation.

1. The square brackets

The brackets identify the repeating section of the polymer chain.

The section inside the brackets is the smallest structural pattern being repeated in the representation.

2. The open bonds

The horizontal bonds extending through the left and right bracket boundaries show that the chain continues.

They represent connections to neighbouring repeat units.

They are not:

• Unpaired electrons.
• Unfinished chemical bonds.
• Missing hydrogen atoms.
• Radical dots.

The polymer chain is connected through these positions.

3. The subscript \(n\)

The \(n\) indicates that the repeating unit occurs many times.

It does not represent an additional atom or a chemical group. It represents the number of repeat units in the chain.

Why open bonds matter when checking carbon valency

Look at one carbon in the middle of a polyethene chain.

It has two bonds to hydrogen atoms and two bonds to neighbouring carbon atoms.

That gives four single bonds.

If you examine only the atoms drawn inside one pair of brackets, you might mistakenly think each carbon has only three bonds or has an incomplete valency.

The missing connection is represented by the open bond at the bracket boundary.

This is why you must interpret the repeat unit as part of a larger chain, rather than as an isolated molecule.

The chain ends are also usually omitted from the repeating-unit shorthand. In an actual finite polymer molecule, the terminal atoms have appropriate end groups.

04How to reverse a repeating unit into its original monomer

This is where the chemistry becomes useful in HSC questions.

You might be given a polymer repeating unit and asked to identify the original alkene monomer.

For a conventional addition polymer formed from an alkene, work backwards by reversing the bonding change.

Use this decision process:

The central rule is simple:

Restore the double bond between the two carbon atoms that originally formed the alkene.

Do not move the side groups or randomly add hydrogen atoms to make the structure look complete.

05Addition polymerisation versus condensation polymerisation

The reverse-structure method above applies to the conventional addition polymerisation of alkene monomers.

It is not a universal method for identifying every polymer’s monomer.

For example, polyethene forms by addition polymerisation of ethene. Its carbon atoms and hydrogen atoms are retained in the polymer.

A polyester formed by condensation polymerisation follows a different structural process. You cannot simply restore a C=C bond in its repeating unit to recover its monomers.

For a broader comparison of common addition polymers, see Addition polymers explained: PE, PVC, PS and PTFE for HSC Chemistry.

06Questions and solutions

Question 1

An addition polymer has the repeating unit:

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

Identify the original monomer and explain what happens to its carbon-carbon double bond during polymerisation.

Solution 1

The original monomer is chloroethene, \(\ce{CH2=CHCl}\), and its C=C double bond becomes a C-C single bond in the polymer backbone.

The repeating unit contains two backbone carbon atoms: \(\ce{CH2}\) and \(\ce{CH(Cl)}\).

Reconstructing the monomer:

Change the internal C-C single bond into a double bond and remove the continuing bonds at the bracket boundaries.

\[ \ce{CH2-CH(Cl) -> CH2=CHCl} \]

The chlorine atom remains attached to the second carbon because polymerisation does not remove or relocate this side group.

Checking valency:

The first carbon has two C-H single bonds and a C=C double bond, giving a total bond order of four.

The second carbon has one C-H bond, one C-Cl bond, and a C=C double bond, also giving a total bond order of four.

The reconstructed structure is therefore valid.

Question 2

An addition polymer has the following repeating unit:

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

Two students suggest different original monomers:

• Student A: \(\ce{CH2=C(CH3)CN}\)
• Student B: \(\ce{CH2=CH-CH2CN}\)

Determine which student is correct.

Explain why the other proposed monomer would produce a different repeating unit. Include a carbon-valency check.

Solution 2

Student A is correct. The original monomer is \(\ce{CH2=C(CH3)CN}\), because both the methyl group and the cyano group are attached directly to the same alkene carbon.

Start with the repeating unit:

\[ \mathrm{-CH_2-C(CH_3)(CN)-} \]

The two backbone carbons are the \(\ce{CH2}\) carbon and the substituted carbon.

Restoring a double bond between them gives:

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

Now check the alternative proposed by Student B.

\[ \ce{CH2=CH-CH2CN} \]

If this monomer underwent addition polymerisation, its repeat unit would be:

\[ \left[-\mathrm{CH_2-CH(CH_2CN)-}\right]_n \]

Notice the difference.

The cyano group would be attached through an additional \(\ce{CH2}\) group, rather than directly to the second backbone carbon.

The two structures are not equivalent.

Valency check for Student A’s monomer:

The first alkene carbon has two hydrogen atoms and a double bond to the second carbon:

\[ 2+2=4 \]

The second alkene carbon has:

• One double bond to the first carbon.
• One single bond to the methyl group.
• One single bond to the cyano carbon.

Therefore:

\[ 2+1+1=4 \]

Both alkene carbons satisfy the usual carbon valency of four.

The position of the side groups is essential when reconstructing an unsymmetrical alkene.

Question 3

An infrared comparison is performed on styrene monomer and polystyrene.

The following illustrative data are obtained:

Assume the listed features have been correctly assigned.

(a) Calculate the wavelength corresponding to the vinyl C=C absorption at \(1630\ \mathrm{cm^{-1}}\).

(b) Explain what the comparison suggests about the bonding changes during polymerisation.

(c) A student claims that the remaining aromatic absorptions prove that the polymerisation was incomplete. Explain why this conclusion is incorrect.

Solution 3

(a) The wavelength is approximately \(6.13\ \mathrm{\mu m}\). (b) The comparison supports consumption of the vinyl C=C bond during polymerisation. (c) The remaining aromatic absorptions are expected because the phenyl rings are not consumed during the reaction.

(a) Calculate the wavelength.

Use:

\[ \lambda=\frac{1}{\tilde{\nu}} \]

Substitute the given wavenumber:

\[ \begin{aligned} \lambda &=\frac{1}{1630}\\ &=6.13\times10^{-4}\ \mathrm{cm}\\ &=6.13\times10^{-6}\ \mathrm{m} \end{aligned} \]

Therefore:

\[ \boxed{\lambda=6.13\ \mathrm{\mu m}} \]

(b) Interpret the spectra.

The vinyl C=C feature is present in styrene but absent in polystyrene.

This supports the conclusion that the alkene double bond has been consumed during addition polymerisation.

The corresponding carbon atoms remain connected through a single bond in the polymer backbone.

(c) Evaluate the student’s claim.

The student’s conclusion is incorrect because the aromatic ring is a separate structural feature.

Styrene contains a vinyl C=C bond and an aromatic phenyl ring.

Only the vinyl double bond participates in forming the polymer backbone. The aromatic ring remains attached as a side group.

Therefore, the persistence of aromatic absorption features is consistent with successful addition polymerisation.

The important distinction is between the double bond that reacts and other bonding features that remain unchanged.

Question 4

A poly(propene) sample has a number-average molar mass of:

\[ M_n=12\,624\ \mathrm{g\,mol^{-1}} \]

The repeating unit has the molecular formula \(\ce{C3H6}\), with a molar mass of:

\[ M_r=42.08\ \mathrm{g\,mol^{-1}} \]

Assume an ideal linear polymer, ignore end-group masses, and use the number-average molar mass to estimate the average degree of polymerisation.

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

(b) Estimate the number of original C=C double bonds consumed per average chain.

(c) Determine the number of new C-C bonds connecting neighbouring repeat units.

Solution 4

The average degree of polymerisation is 300. Therefore, approximately 300 original alkene double bonds are consumed, and 299 new C-C bonds connect neighbouring repeat units in an average linear chain.

(a) Calculate the degree of polymerisation.

The degree of polymerisation is the number of repeat units in a polymer chain.

Use:

\[ n=\frac{M_n}{M_r} \]

Substitute:

\[ \begin{aligned} n&=\frac{12\,624}{42.08}\\ &=300 \end{aligned} \]

The average chain contains approximately 300 repeat units.

(b) Count the original double bonds.

Each propene monomer contains one C=C double bond.

Each monomer contributes one repeat unit to the polymer.

Therefore:

\[ N_{\mathrm{C=C}}=300 \]

Approximately 300 original double bonds have been converted into single bonds.

(c) Count the inter-unit bonds.

This is where a common counting mistake occurs.

A linear chain of 300 repeat units does not need 300 bonds connecting neighbouring units.

There are 299 spaces between 300 consecutive units.

Therefore:

\[ N_{\mathrm{inter-unit}}=n-1 \]

\[ =300-1=299 \]

There are 299 new inter-unit C-C bonds.

The original C-C sigma bond within each propene-derived unit remains as a single bond.

This distinction becomes particularly important when questions ask you to count bonds rather than simply identify the monomer.

Question 5

An ideal linear polymer chain is formed from exactly four styrene molecules:

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

Assume that:

• Each monomer undergoes conventional addition polymerisation through its vinyl C=C bond.
• No cross-linking or side reactions occur.
• The chain is linear rather than cyclic.
• Terminal groups are omitted from the repeating-unit representation.

A student makes the following claim:

“Four monomers each lose a double bond, so four new bonds must connect the monomers. Also, because this is addition polymerisation, the resulting polymer contains no π bonds.”

Evaluate the claim.

Determine:

(a) The number of original vinyl π bonds consumed.

(b) The number of original C-C sigma bonds within the vinyl groups that remain as single bonds.

(c) The number of new C-C bonds connecting different monomer units.

(d) Whether any π bonding remains in the polymer.

(e) How the answer to part (c) would change if the four units formed a closed cyclic tetramer instead of a linear chain.

Solution 5

The student’s claim is incorrect in two ways. The linear chain consumes four vinyl π bonds but forms only three inter-unit C-C bonds. All four original vinyl sigma bonds remain as single bonds, and the four aromatic phenyl rings retain their π bonding.

This problem requires us to distinguish three different things:

• The π bonds that are consumed.
• The original sigma bonds that remain.
• The new sigma bonds that connect separate monomer units.

(a) Count the consumed π bonds.

Each styrene molecule contains one vinyl C=C bond.

There are four styrene molecules.

Therefore:

\[ N_{\pi,\mathrm{consumed}}=4 \]

Four vinyl π bonds are consumed during polymerisation.

The aromatic π systems are not included in this count because they do not participate in the chain-forming reaction.

(b) Count the original sigma bonds that remain.

Each vinyl C=C bond contains one sigma bond and one pi bond.

The pi component is broken, but the original sigma bond remains between the same two carbon atoms.

There are four original styrene molecules, so:

\[ N_{\sigma,\mathrm{retained}}=4 \]

These four bonds become single bonds within the polymer backbone.

(c) Count the new inter-unit bonds.

There are four monomer units arranged in a linear chain.

To connect four separate units into one continuous chain, only three connections are required:

\[ \mathrm{Unit\ 1-Unit\ 2-Unit\ 3-Unit\ 4} \]

Therefore:

\[ N_{\mathrm{inter-unit}}=4-1=3 \]

The student incorrectly assumed that four monomers require four inter-unit connections.

That would leave an additional connection beyond what is needed for a linear chain.

The distinction is that each monomer contributes an original C-C sigma bond within its own two-carbon segment. The inter-unit bonds are additional connections between those segments.

The total number of backbone C-C single bonds is therefore:

\[ \begin{aligned} N_{\mathrm{backbone}} &=N_{\mathrm{internal}}+N_{\mathrm{inter-unit}}\\ &=4+3\\ &=7 \end{aligned} \]

This also makes sense structurally: a linear backbone containing eight carbon atoms has seven C-C connections.

(d) Determine whether π bonding remains.

Yes. The four phenyl rings remain aromatic.

Styrene contains a vinyl double bond and an aromatic ring. Addition polymerisation uses the vinyl double bond to form the polymer backbone.

The phenyl rings are side groups. Their delocalised π bonding is retained.

Therefore, the resulting polystyrene contains four aromatic π systems.

The statement that addition polymerisation removes all π bonding is incorrect.

(e) Consider a cyclic tetramer.

A cyclic structure has one additional connection: the final monomer unit connects back to the first.

Therefore:

\[ N_{\mathrm{inter-unit}}=4 \]

The cyclic structure would have four inter-unit bonds rather than three.

Its total number of backbone C-C single bonds would be:

\[ 4+4=8 \]

This is the key difference between a linear chain and a closed ring.

The original vinyl double bonds are consumed in both cases. What changes is the number of connections needed to join the monomer units into the final structure.

The broader lesson is that reconstructing an addition-polymer monomer requires more than changing a single bond into a double bond. You must also understand which bonds belong within a monomer, which connect neighbouring monomers, and which parts of the original molecule remain unchanged.

Once you can distinguish those three features, you can move on to more challenging polymer questions involving unsymmetrical monomers, side groups, and identifying a true repeating unit from a longer section of polymer chain.