How to Identify Addition and Condensation Polymers from Structure

Two polymer repeat units can both contain an ester group, yet one can be an addition polymer and the other a condensation polymer. If you see an ester group and immediately write “condensation”, you can get the answer wrong.

Consider these two illustrative polymer structures:

\[ P:\quad \ce{[-CH2-CH(C(=O)OCH3)-]_{n}} \]

\[ Q:\quad \ce{[-O-CH2-CH2-O-C(=O)-(CH2)4-C(=O)-]_{n}} \]

Both contain an ester group, \(\ce{-C(=O)-O-}\).

However, P is an addition polymer, while Q is a condensation polymer (a polyester).

The difference is where the ester group sits. In P, it is attached to the carbon chain as a side group. In Q, the ester linkages form part of the main polymer chain.

The most reliable way to distinguish addition and condensation polymers from structure alone is to trace the polymer backbone and identify the bonds connecting successive monomer residues.

01The decisive clue: trace the polymer backbone

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

Side groups are attached to this chain but do not form part of its continuous path.

This distinction matters because the atoms present in a polymer do not automatically tell you how it was made. Their positions and connections matter much more.

Addition polymers: follow the carbon chain

In the standard alkene-based addition polymerisation examples, the monomers contain a carbon-carbon double bond.

During polymerisation:

• The carbon-carbon double bond becomes a single bond.
• The two carbon atoms form additional bonds to neighbouring monomer molecules.
• The carbon atoms form a continuous backbone.
• Any other groups attached to those carbon atoms remain as side groups.
• No small molecule, such as water, is eliminated.

For example, ethene forms poly(ethene):

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

The original double bond is no longer present. The carbon atoms are connected through single bonds in the polymer backbone.

The key structural clue is a saturated carbon-carbon backbone with substituents attached to it.

For example, consider poly(chloroethene), commonly called PVC:

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

The chlorine atoms might initially look like evidence of a different polymerisation process.

They aren’t.

Each chlorine atom is attached to a backbone carbon. It does not connect one repeating unit to the next.

The backbone is still made entirely of carbon-carbon single bonds, so PVC is an addition polymer.

Condensation polymers: look for functional groups connecting the chain

Condensation polymers have a different structural pattern.

Instead of simply opening an alkene double bond, functional groups on different monomers react to form new links between them.

Two important examples are:

Polyesters

Look for the ester linkage:

\[ \ce{-C(=O)-O-} \]

Polyamides

Look for the amide linkage:

\[ \ce{-C(=O)-NH-} \]

These linkages form part of the polymer backbone.

For the standard HSC examples, they are evidence of condensation polymerisation because they form when functional groups on different monomers react, usually eliminating water.

For example, a carboxylic acid reacts with an alcohol to form an ester:

\[ \ce{R-COOH + HO-R' -> R-COO-R' + H2O} \]

Here, \(R\) and \(R'\) represent the remaining parts of the molecules.

Notice that the ester oxygen remains in the polymer. It becomes part of the connection between the two monomer-derived sections.

Quick comparison table

There is one particularly important trap here.

The absence of a C=C bond in the finished polymer does not prove that it is a condensation polymer.

Addition polymerisation consumes the original double bond. Therefore, both addition and condensation polymers can have backbones containing only single bonds.

You need to examine which atoms are connected, not simply whether double bonds are present.

02Read the brackets and open bonds correctly

Before classifying an unfamiliar polymer, make sure you understand what the structural formula represents.

Consider:

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

There are three important features.

• Square brackets identify the repeating unit.
• The subscript \(n\) indicates that the unit repeats many times.
• The short bonds extending beyond the brackets show where the chain continues.

Those outer bonds are not unfinished bonds waiting for hydrogen atoms.

They connect to neighbouring repeating units.

For example, expanding the structure gives:

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

The methyl groups remain attached to the backbone, but the continuous chain passes through the alternating carbon atoms.

Why you should expand more than one repeat unit

Sometimes a condensation linkage is partly hidden at a bracket boundary.

Consider this alternative way of writing the polyester structure:

\[ \ce{[-C(=O)-(CH2)4-C(=O)-O-CH2-CH2-O-]_{n}} \]

Inside the brackets, you can see one complete ester linkage.

But what happens when the repeating units join?

The right-hand oxygen of one unit connects to the left-hand carbonyl carbon of the next unit.

This creates another ester linkage:

\[ \ce{-O-C(=O)-} \]

The connection is still present even though it crosses the bracket boundary.

This is why examining only the atoms inside one set of brackets can lead to an incorrect classification.

A useful rule is to expand at least two repeating units whenever you are unsure where the backbone connections occur.

03A model-kit investigation: test the chain path

Suppose you are given two unfamiliar polymer structures and asked to identify their polymerisation types.

You notice that both contain ester groups.

Your first prediction might be that both are condensation polymers.

How could you test that interpretation?

A molecular model provides a useful way to investigate the connectivity.

Apparatus

• A molecular model kit.
• Two printed polymer repeating-unit structures.
• At least three copies of each repeating unit.

Set-up

Build three repeating units of each polymer and connect the open bonds at their boundaries.

Trace the continuous path from one end of the resulting chain to the other.

Prediction

If an ester group connects successive monomer residues, the backbone path should pass through its carbonyl carbon and ester oxygen.

If the ester group is a side group, the backbone should continue without passing through that ester linkage.

Observation

In poly(methyl acrylate), the backbone continues through carbon-carbon bonds. The ester groups branch away from it.

In the polyester example, the backbone passes through the ester linkages. The oxygen atoms are part of the continuous chain.

There is a limitation to this investigation.

A model kit checks whether your interpretation of the structural formula is consistent with the bonding and connectivity. It does not establish the actual reaction mechanism or prove how the polymer was manufactured.

There is also a problem with using only one repeating unit: it can hide connections at the boundaries.

Building several connected units solves that problem.

The conclusion is that you should classify the polymer by the connections along its backbone, not by the functional groups you can spot anywhere in the structure.

04Use valence to check your interpretation

Once you have identified the likely polymerisation type, check that the structural formula is chemically reasonable.

For the ordinary neutral organic structures considered here, remember:

A double bond counts as two bonds when checking carbon valence.

Checking an ester linkage

Consider:

\[ \ce{R-C(=O)-O-R'} \]

The carbonyl carbon has:

• One double bond to oxygen: 2.
• One single bond to \(R\): 1.
• One single bond to the ester oxygen: 1.

Therefore:

\[ 2+1+1=4 \]

The carbonyl carbon satisfies its usual valence.

The ester oxygen also has two single bonds: one to the carbonyl carbon and one to the next carbon-containing section.

Checking an addition polymer

For poly(ethene):

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

Each backbone carbon has:

• Two bonds to hydrogen.
• One bond to the carbon on its left.
• One bond to the carbon on its right.

Total:

\[ 2+1+1=4 \]

The bonds connecting neighbouring repeating units must be included in this count.

A common mistake is to retain the original C=C double bond while also adding the continuing backbone bonds. That would give an ordinary carbon five bonds.

The brackets cannot fix an incorrect valence.

05A reliable decision process

When an exam gives you an unfamiliar polymer structure, use this sequence.

Step 1: Identify the repeating unit.

Locate the square brackets and the subscript \(n\).

Step 2: Identify the open bonds.

Remember that the bonds extending beyond the brackets connect to neighbouring repeating units.

Step 3: Trace the backbone.

Follow the continuous path from the left-hand open bond to the right-hand open bond.

Step 4: Inspect the backbone linkages.

• If the standard alkene-derived structure has a saturated carbon-carbon backbone and side groups, identify it as an addition polymer.
• If ester or amide linkages connect the monomer-derived sections along the backbone, identify it as a condensation polymer.

Step 5: Check the side groups.

Do not mistake chlorine, nitrogen-containing groups, or ester groups attached to the backbone for links between repeating units.

Step 6: Check valence.

Confirm that the carbon, oxygen, and nitrogen atoms have chemically reasonable bonding.

The most common classification mistakes

06Worked examples

07What can structure alone actually tell you?

There is an important distinction between identifying a polymer’s structure and proving its complete reaction history.

For standard HSC-style comparisons, the following are reliable structural clues:

• A saturated carbon-carbon backbone with side groups is characteristic of an alkene-derived addition polymer.
• Ester linkages in the backbone indicate a polyester structure.
• Amide linkages in the backbone indicate a polyamide structure.
• Side groups do not determine the polymerisation type simply because they contain oxygen, nitrogen, or other atoms.

However, a structural formula records the final molecular connectivity. It does not record every step of the synthesis.

In more advanced polymer chemistry, some polymers containing ester linkages can also be produced by routes such as ring-opening polymerisation. Therefore, an ester linkage alone cannot prove the exact manufacturing mechanism in every possible case.

For an ordinary course-level classification question, use the expected addition-versus-condensation model. If a question supplies additional information about the monomers or reaction conditions, use that evidence as well.

08Questions and solutions

Question 1

A polymer has the repeating unit:

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

Identify whether this is an addition or condensation polymer. Explain your answer using its structure.

Solution 1

It is an addition polymer.

The backbone consists of carbon-carbon single bonds, with no ester or amide linkages connecting the repeating units.

The structure is consistent with ethene undergoing addition polymerisation. Its C=C bond opens, allowing successive monomer molecules to connect through new carbon-carbon single bonds.

The original monomer is:

\[ \ce{CH2=CH2} \]

No small molecule is eliminated during this process.

Question 2

Two polymers have the following repeating units:

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

\[ Q:\quad \ce{[-NH-(CH2)6-NH-C(=O)-(CH2)4-C(=O)-]_{n}} \]

A student argues that neither polymer can be an addition polymer because neither contains a carbon-carbon double bond.

Evaluate this statement and classify both polymers.

Solution 2

The student is incorrect. P is an addition polymer, while Q is a condensation polymer.

The absence of a C=C bond in the finished polymer does not distinguish the two reaction types.

Polymer P

The structure is:

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

The continuous backbone consists of carbon-carbon single bonds.

The chlorine atoms are attached to the backbone as side groups. They do not connect successive repeating units.

This is poly(chloroethene), or PVC, an addition polymer formed from chloroethene:

\[ \ce{CH2=CHCl} \]

The original double bond opens during polymerisation.

Polymer Q

The structure contains the linkage:

\[ \ce{-C(=O)-NH-} \]

This is an amide linkage.

Importantly, the nitrogen and carbonyl carbon are part of the continuous backbone. They connect sections derived from different monomers.

The structure is that of nylon 6,6, a condensation polyamide formed from hexane-1,6-diamine and hexanedioic acid.

Water is eliminated when the amine and carboxylic acid functional groups react.

The exact mistake in the student’s reasoning is assuming that a polymer must retain a C=C bond to be an addition polymer. In reality, the double bond is consumed during addition polymerisation.

Question 3

An unfamiliar polymer has the following repeating unit:

\[ \ce{[-CH2-CH(C(=O)OCH3)-]_{n}} \]

A student identifies it as a condensation polymer because it contains an ester group.

Is the student correct?

Determine the likely polymerisation type, explain the position of the ester group, and identify a possible monomer.

Solution 3

The student is incorrect. This is an addition polymer because the ester group is attached to the backbone rather than connecting successive repeating units.

Start by tracing the backbone:

\[ \ce{...-CH2-CH-CH2-CH-...} \]

Each \(\ce{CH}\) carbon has an ester-containing side group:

\[ \ce{-C(=O)OCH3} \]

The continuous chain does not pass through the ester oxygen.

Instead, it continues directly through carbon-carbon bonds.

The likely monomer is methyl acrylate:

\[ \ce{CH2=CH-C(=O)-OCH3} \]

During polymerisation, the C=C bond opens, while the ester-containing group remains attached to the carbon chain.

This example exposes an important misconception: having an ester group is not the same as having an ester linkage in the polymer backbone.

Question 4

An unknown polymer is represented by:

\[ \ce{[-C(=O)-(CH2)4-C(=O)-O-CH2-CH2-O-]_{n}} \]

A student examines only the bonds inside the square brackets and claims that the structure contains just one ester linkage per repeating unit.

Determine the polymerisation type and explain why examining the bracket boundary is important.

Your answer should include an expanded section containing two consecutive repeating units.

Solution 4

The polymer is a condensation polyester, and its repeating pattern contains two ester linkages when the connections between neighbouring units are included.

The student’s observation is incomplete because one ester linkage is formed across the bracket boundary.

Start with the supplied structure:

\[ \ce{[-C(=O)-(CH2)4-C(=O)-O-CH2-CH2-O-]_{n}} \]

The complete expanded chain includes:

\[ \ce{...-O-C(=O)-(CH2)4-C(=O)-O-CH2-CH2-O-} \]

The right-hand oxygen connects to the carbonyl carbon at the beginning of the next repeating unit.

This creates the additional ester linkage:

\[ \ce{-O-C(=O)-} \]

The other ester linkage is already visible inside the repeating unit:

\[ \ce{-C(=O)-O-} \]

Both form part of the continuous polymer backbone.

The structure is consistent with condensation polymerisation between a diol and a dicarboxylic acid.

The important lesson is that square brackets indicate repetition, not the complete boundary of the chemical connections. A bond can connect two units even when it is not drawn entirely inside either set of brackets.

Question 5

Two polymers have the following repeating units:

\[ P:\quad \ce{[-CH2-CH(C(=O)OCH3)-]_{n}} \]

\[ Q:\quad \ce{[-C(=O)-(CH2)4-C(=O)-O-CH2-CH2-O-]_{n}} \]

A student makes two claims:

1. Both polymers must be condensation polymers because both contain ester groups.
2. Four repeating pairs of polymer Q must produce eight water molecules because each repeating pair contains two ester linkages.

Investigate both claims.

For polymer Q, assume that:

• The polymer is formed from four molecules of ethane-1,2-diol and four molecules of hexanedioic acid.
• The eight monomer molecules form one alternating, linear, unbranched, acyclic polymer chain.
• Each ester linkage forms through a reaction between one carboxylic acid group and one alcohol group.
• Each such condensation reaction produces one water molecule.

Determine:

(a) The polymerisation type of P and Q.

(b) The number of ester linkages in the finite chain of Q.

(c) The number of water molecules produced.

(d) The functional groups remaining at the two chain ends.

(e) Whether the answer would change if the chain were cyclic rather than linear.

Solution 5

P is an addition polymer, Q is a condensation polyester, and the specified linear chain produces seven water molecules, not eight.

The two claims fail for different reasons. The first ignores the position of the ester group. The second ignores the difference between a repeating pattern and the actual connections in a finite linear chain.

Part (a): Classify both polymers.

For P:

\[ \ce{[-CH2-CH(C(=O)OCH3)-]_{n}} \]

The backbone consists of carbon-carbon single bonds.

The ester group is a side group, so it does not connect neighbouring repeating units.

Therefore, P is an addition polymer.

For Q:

\[ \ce{[-C(=O)-(CH2)4-C(=O)-O-CH2-CH2-O-]_{n}} \]

The backbone passes through ester linkages.

The oxygen at the right-hand boundary of one repeating unit connects to the carbonyl carbon at the beginning of the next.

Therefore, Q is a condensation polyester.

The presence of an ester group in both structures does not make their polymerisation types the same.

Part (b): Count the ester linkages in the finite chain.

There are four diol molecules and four dicarboxylic acid molecules.

That gives:

\[ 4+4=8\text{ monomer molecules} \]

In one linear chain, eight connected molecules require seven connections between neighbouring molecules.

Therefore:

\[ N_{\text{links}}=8-1=7 \]

Each connection is an ester linkage.

The finite polymer chain contains seven ester linkages.

Part (c): Calculate the number of water molecules.

Each ester linkage forms through the reaction of one carboxylic acid group with one alcohol group.

Each reaction releases one water molecule.

Therefore:

\[ N(\ce{H2O})=N_{\text{ester links}} \]

\[ N(\ce{H2O})=7 \]

Seven water molecules are produced.

Part (d): Identify the remaining end groups.

There are initially:

• Eight hydroxyl groups from the four diol molecules.
• Eight carboxylic acid groups from the four dicarboxylic acid molecules.

Seven ester linkages consume seven hydroxyl groups and seven carboxylic acid groups.

This leaves:

• One unreacted hydroxyl group, \(\ce{-OH}\).
• One unreacted carboxylic acid group, \(\ce{-COOH}\).

These are the two terminal groups of the linear chain.

Part (e): What if the chain were cyclic?

A cyclic chain would produce eight water molecules.

The final monomer at one end could connect to the monomer at the other end, forming an additional ester linkage and closing the ring.

The cyclic structure would therefore have eight connections between its eight monomer molecules.

This is the crucial distinction:

• A linear chain containing eight monomer molecules has seven connections.
• A cyclic chain containing eight monomer molecules has eight connections.

The student’s calculation of eight water molecules assumes that the final connection exists. But the question explicitly states that the chain is linear and acyclic.

The repeating-unit representation describes the repeating pattern without showing the actual terminal groups. It cannot be used to assume that the two ends of a finite chain are connected.

This is why the number of repeat units does not always equal the number of condensation reactions multiplied by the number of linkages in an ideal repeating pattern.

09Where this method leads next

Once you can distinguish addition and condensation polymers from their structures, the next step is working backwards to identify their monomers.

For an addition polymer, you reconstruct the original alkene by restoring the C=C bond between the appropriate backbone carbons.

For a condensation polymer, you reverse the ester or amide linkage and restore the original functional groups.

These are two related but different reconstruction problems:

• Addition polymers explained: PE, PVC, PS and PTFE for HSC Chemistry
• How to Identify Which Atoms Leave During Condensation Polymerisation

Keep one decision in mind whenever you are given an unfamiliar polymer structure:

Follow the backbone first. Identify the linkages second. Classify the polymer last.