Why Some Polyesters Can Form from One Monomer Type

A flask containing ethanol and ethanoic acid can produce an ester and water. Yet it cannot produce a long polyester chain using this reaction. Now imagine a flask containing only lactic acid molecules. Could these molecules join together repeatedly without a second monomer type?

Yes. The key is that each lactic acid molecule contains two different reactive functional groups: an alcohol group (\(\ce{-OH}\)) and a carboxylic acid group (\(\ce{-COOH}\)). One molecule can therefore react with another, and the resulting molecule still has functional groups available for further reactions.

This is why some condensation polymers can form from a single monomer type.

01Why one monomer type can be enough

Start with lactic acid, also called 2-hydroxypropanoic acid.

Its condensed structural formula is:

\[ \ce{HO-CH(CH3)-COOH} \]

Look carefully at its structure. It contains:

• An alcohol group, \(\ce{-OH}\), attached to the central carbon.
• A carboxylic acid group, \(\ce{-COOH}\), at the other end.

These groups can react with each other when they belong to different lactic acid molecules.

Before looking at the equation, predict what happens when these two functional groups react.

The alcohol group reacts with the carboxylic acid group. One water molecule is eliminated, and a new ester linkage forms between the two molecules.

The general reaction is:

\[ \ce{R-COOH + HO-R' <=> R-C(=O)-O-R' + H2O} \]

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

Notice the new connection:

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

This is an ester linkage.

In the usual HSC structural model, the carboxylic acid group loses its \(\ce{-OH}\), while the alcohol group loses its hydrogen. These combine to form water. The oxygen from the alcohol remains in the new ester linkage.

Why the reaction can continue

Consider what happens when two lactic acid molecules form a dimer:

\[ \begin{aligned} \ce{2HO-CH(CH3)-COOH} &\ce{<=> HO-CH(CH3)-C(=O)-O-} \\ &\ce{-CH(CH3)-COOH + H2O} \end{aligned} \]

The product still has:

• One unreacted alcohol group at one end.
• One unreacted carboxylic acid group at the other end.

These groups can react with additional lactic acid molecules.

The process repeats, creating a longer chain. Every new ester linkage forms through another condensation reaction, releasing another water molecule.

Think of each monomer as a two-ended connector. One end contains an alcohol group, while the other contains a carboxylic acid group. Each connector can join to its neighbours, allowing a chain to grow.

The analogy has one important limitation: these aren’t identical mechanical connectors. The two ends have different chemical structures, and joining them requires an actual chemical reaction.

The important idea is that each monomer already supplies both types of functional groups needed for chain growth.

A molecule containing two reactive functional groups is described as bifunctional.

A single monomer type is therefore sufficient when each monomer molecule contains both complementary functional groups needed to build the polymer.

It does not mean that only one molecule is involved. Many copies of the same molecular species join together.

02Polyester is the family; PLA is one example

We should distinguish between the polymer family and the individual polymer.

A polyester is a family of polymers containing repeated ester linkages in their main molecular chains.

Poly(lactic acid), commonly abbreviated to PLA, is one particular polyester.

Its idealised repeating unit is:

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

The subscript \(n\) represents the number of repeating units in the polymer chain.

There are two structural details worth noticing.

First, the ester linkage is part of the main polymer backbone. It is not simply an ester group attached to the side of an unrelated carbon chain.

Second, the \(\ce{CH3}\) groups are side groups. They are attached to the polymer backbone but do not form the ester linkages.

This matters when interpreting unfamiliar polymer structures. The presence of a methyl group does not mean that another monomer type was used.

The polymer is still formed from repeated lactic acid molecules.

03How this differs from PET

Not all polyesters form from one monomer type.

A familiar contrasting example is polyethylene terephthalate (PET), used in many drink bottles and textile fibres.

PET is also a polyester, but its formation involves two different monomer types.

The diol supplies the alcohol groups, while the dicarboxylic acid supplies the carboxylic acid groups.

These molecules alternate as ester linkages form between them.

The idealised PET repeating unit is:

\[ \ce{[-O-CH2-CH2-O-C(=O)-C6H4-C(=O)-]_{n}} \]

The difference between PLA and PET is not the type of polymer linkage. Both are polyesters.

The difference is how the starting molecules supply the functional groups.

For a closer look at the two-monomer pathway, see How to Draw PET from Ethylene Glycol and Terephthalic Acid.

A useful contrasting case: why ethanol and ethanoic acid cannot make a polyester

Return to the reaction mentioned at the beginning.

Ethanol has one alcohol group:

\[ \ce{CH3CH2OH} \]

Ethanoic acid has one carboxylic acid group:

\[ \ce{CH3COOH} \]

They can react to form ethyl ethanoate and water:

\[ \ce{CH3COOH + CH3CH2OH} \]

\[ \ce{<=> CH3COOCH2CH3 + H2O} \]

However, each reactant has only one functional group available for this esterification.

Once the ester forms, neither molecule has another available alcohol or carboxylic acid group to continue the chain.

The reaction produces a small ester molecule, not a polyester.

This exposes an important misconception:

Having the correct two types of functional groups somewhere in a reaction mixture is not enough. The individual monomers must have the functionality needed for repeated chain formation.

04How an experiment could test the explanation

A chemist investigating this reaction needs to answer two separate questions:

1. Are ester linkages forming?
2. Are enough ester linkages forming to produce long polymer chains?

These are not the same question.

Consider how an investigation could be designed.

The original design: demonstrate esterification

An initial experiment could heat ethanol with ethanoic acid.

The predicted result is an ester and water. The observation of both products would support the proposed esterification reaction.

However, this experiment cannot demonstrate condensation polymerisation.

The problem is the structure of the starting molecules. Each has only one functional group available for ester formation, so the reaction cannot continue into an extended chain.

The experiment demonstrates ester formation, but it does not answer the polymerisation question.

The revised design: use a bifunctional monomer

Replace the ethanol and ethanoic acid with lactic acid.

Now each starting molecule contains both an alcohol and a carboxylic acid group. Repeated esterification is possible.

A simplified investigation could use the following apparatus:

Follow the path through the apparatus.

1. Lactic acid is placed in a dry reaction flask.
2. The flask is heated and stirred to allow esterification to occur.
3. Water vapour leaves the reaction mixture through the outlet.
4. The vapour enters the condenser, where it cools and becomes liquid.
5. The condensed liquid enters a pre-weighed receiver.
6. The polymer-forming material remains in the reaction flask.

The receiver can be weighed before and after the reaction to estimate the mass of water collected.

What should the experiment show?

If condensation occurs, water should be produced, and ester linkages should form in the reaction mixture.

There is another important detail: esterification is reversible.

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

Removing water helps shift the equilibrium towards ester formation.

If too much water remains in the reaction mixture, the reverse reaction can hydrolyse ester linkages and limit chain growth.

This is why continuous water removal is useful in a direct condensation process.

What the measurement cannot prove

Suppose the receiver collects water.

That observation supports the occurrence of condensation, but it does not prove that long polymer chains have formed.

The reaction might have produced:

• Short ester-linked dimers.
• Small oligomers containing several repeating units.
• Longer polyester chains.

All can produce water through esterification.

A stronger investigation would combine several measurements:

• Measure the collected water to estimate the extent of condensation.
• Analyse the product to identify ester functional groups.
• Measure the molecular mass to investigate whether substantial chain growth has occurred.
• Check for unreacted functional groups at the ends of the chains.

The distinction is important: detecting an ester bond is not the same as demonstrating a high-molecular-mass polyester.

An important manufacturing qualification

The direct condensation of lactic acid is a chemically valid way to explain how one monomer type can produce a polyester.

However, it is not the main industrial route for producing high-molecular-mass PLA.

Direct condensation has practical difficulties because removing water becomes increasingly difficult as the reaction mixture becomes viscous. The equilibrium reaction also limits the achievable chain length.

Commercial high-molecular-mass PLA is commonly manufactured through a different route:

1. Lactic acid is converted into lactide, a cyclic dimer.
2. The lactide undergoes ring-opening polymerisation.
3. The opened rings connect to form PLA chains.

The final ring-opening step does not produce water through condensation.

This is a useful distinction between the structural explanation of how a polymer can form and the manufacturing process actually used to produce it.

A peer-reviewed review of PLA synthesis discusses the differences between direct polycondensation and lactide ring-opening polymerisation: Poly(lactic Acid): A Versatile Biobased Polymer for the Future.

05Worked examples

06Questions and solutions

Question 1

A monomer has the following structural formula:

\[ \ce{HO-CH2-COOH} \]

Explain why this molecule can form a polyester using only one monomer type.

Identify the two functional groups involved and state the type of linkage formed.

Solution 1

Yes. This molecule can form a polyester from one monomer type because it contains both an alcohol group and a carboxylic acid group.

The alcohol group is \(\ce{-OH}\), while the carboxylic acid group is \(\ce{-COOH}\).

The alcohol group on one molecule reacts with the carboxylic acid group on another molecule. This eliminates water and produces an ester linkage:

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

The resulting molecule still has an unreacted alcohol group and an unreacted carboxylic acid group at its opposite ends.

These groups can react with additional monomer molecules, allowing the chain to grow.

Therefore, a second monomer type is unnecessary. Every molecule supplies both complementary functional groups required for condensation polymerisation.

Question 2

An idealised polyester chain segment has the following structure:

\[ \begin{aligned} \ce{HO-CH(CH3)-C(=O)-O-}\\ \ce{CH(CH3)-C(=O)-O-}\\ \ce{CH(CH3)-COOH} \end{aligned} \]

Answer the following:

a) Identify the polymer family.

b) Identify the monomer from which this polymer could form through direct condensation.

c) Explain why the presence of the \(\ce{CH3}\) groups does not indicate that a second monomer type was used.

Solution 2

a) The polymer belongs to the polyester family because its main chain contains repeated ester linkages.

The characteristic linkage is:

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

b) The monomer is 2-hydroxypropanoic acid, also known as lactic acid:

\[ \ce{HO-CH(CH3)-COOH} \]

Each monomer contains both an alcohol group and a carboxylic acid group, allowing self-condensation.

c) The \(\ce{CH3}\) groups are side groups attached to the polymer backbone. They do not form the ester linkages or connect neighbouring monomer units.

Each repeating unit still comes from the same lactic acid molecule.

The tempting mistake is to interpret every different-looking group in a polymer structure as evidence of another monomer. Instead, follow the actual covalent connections.

The methyl groups are part of the lactic acid structure, not additional monomers.

Question 3

An idealised reaction begins with \(0.0180\text{ mol}\) of lactic acid monomer units.

All monomer units form open, linear polyester chains containing six monomer units per chain.

Calculate:

a) The amount of polymer chains formed.

b) The amount of ester linkages formed.

c) The mass of water produced.

Assume no cyclic chains or side reactions occur.

Solution 3

The reaction produces \(0.270\text{ g}\) of water.

Step 1: Calculate the amount of polymer chains.

Each chain contains six monomer units.

\[ n(\text{chains})=\frac{0.0180}{6} \]

\[ =0.00300\text{ mol} \]

Step 2: Calculate the amount of ester linkages.

Each open chain containing six monomer units has five ester linkages.

\[ n(\text{links})=0.00300\times5 \]

\[ =0.0150\text{ mol} \]

Step 3: Calculate the amount of water.

One mole of water is produced for every mole of ester linkages formed.

\[ n(\ce{H2O})=0.0150\text{ mol} \]

Step 4: Calculate the mass of water.

\[ \begin{aligned} m(\ce{H2O})&=nM\\ &=0.0150\times18.0\\ &=0.270\text{ g} \end{aligned} \]

The answers are:

• Polymer chains: \(0.00300\text{ mol}\).
• Ester linkages: \(0.0150\text{ mol}\).
• Water produced: \(0.270\text{ g}\).

The result makes sense because an open chain containing six monomer units requires only five ester linkages. One alcohol group and one carboxylic acid group remain at the chain ends.

Question 4

A chemist investigates two possible routes to ester-containing products.

Two identical experimental setups are used.

• Flask A contains \(0.0200\text{ mol}\) of lactic acid.
• Flask B contains \(0.0200\text{ mol}\) of ethanol and \(0.0200\text{ mol}\) of ethanoic acid.

Both experiments produce \(0.126\text{ g}\) of water. Analysis also indicates that ester groups are present in both products.

A student concludes that both experiments have successfully produced polyesters.

Evaluate this conclusion. Explain what the evidence establishes, why the two reactions have different polymer-forming abilities, and what further measurement would help confirm polymer formation in Flask A.

Solution 4

The student’s conclusion is incorrect. Both reactions can produce ester linkages and water, but only Flask A contains a monomer capable of continued chain growth through direct self-condensation.

First, consider the experimental evidence.

The amount of water collected is:

\[ \begin{aligned} n(\ce{H2O})&=\frac{m}{M}\\ &=\frac{0.126}{18.0}\\ &=0.00700\text{ mol} \end{aligned} \]

Both experiments have produced \(0.00700\text{ mol}\) of water.

This supports the occurrence of esterification in both flasks, assuming the collected water comes entirely from the reaction.

However, equal water production does not mean equal polymer formation.

Flask A: lactic acid

Lactic acid contains one alcohol group and one carboxylic acid group.

After forming an ester linkage, a lactic acid molecule can still be connected to another molecule through its remaining functional group.

This allows repeated condensation and the formation of a polyester chain.

Flask B: ethanol and ethanoic acid

Ethanol has one alcohol group, while ethanoic acid has one carboxylic acid group.

Once they react, they form ethyl ethanoate:

\[ \ce{CH3COOCH2CH3} \]

Neither reactant has another available functional group for continuing chain growth through this reaction.

The product is a small ester molecule, not a polyester.

What further measurement is needed?

For Flask A, identifying ester groups is useful, but it does not prove that long polymer chains have formed.

A molecular-mass measurement would help establish whether the reaction produced larger polymer molecules rather than only dimers or short oligomers.

The important experimental lesson is that water production demonstrates condensation, not necessarily polymerisation.

Question 5

A chemist begins with \(0.0200\text{ mol}\) of lactic acid monomer units.

After a reaction, the purified product contains only open linear chains and cyclic oligomers. There is no unreacted monomer, and no side reactions have occurred.

The receiver contains \(0.270\text{ g}\) of water. Assume all water produced by condensation has been collected.

A student argues:

1. The reaction produced \(0.0150\text{ mol}\) of water.
2. Therefore, \(0.0150\text{ mol}\) of ester linkages formed.
3. The amount of polymer chains must be \(0.0200-0.0150=0.00500\text{ mol}\).
4. Therefore, the number-average degree of polymerisation must be 4.00.

Is the student’s conclusion necessarily correct?

Construct two different possible product distributions that satisfy the experimental data but have different number-average degrees of polymerisation.

Solution 5

No. The student’s answer of 4.00 is only correct if every polymer molecule is an open, linear chain.

The overlooked possibility is that some molecules form cyclic oligomers.

A cyclic oligomer has an additional ester linkage compared with an open chain containing the same number of monomer units. Closing the chain into a ring consumes the two remaining terminal functional groups and releases another water molecule.

This means that water production alone cannot establish the total number of polymer molecules when both structures are possible.

Step 1: Calculate the amount of water and ester linkages.

The amount of water is:

\[ \begin{aligned} n(\ce{H2O})&=\frac{0.270}{18.0}\\ &=0.0150\text{ mol} \end{aligned} \]

Therefore:

\[ n(\text{ester links})=0.0150\text{ mol} \]

There are \(0.0200\text{ mol}\) of original lactic acid units.

Step 2: Consider the student’s model: all molecules are open tetramers.

An open tetramer contains four monomer units and three ester linkages.

The amount of polymer chains is:

\[ n(\text{chains})=\frac{0.0200}{4} \]

\[ =0.00500\text{ mol} \]

The amount of ester linkages is:

\[ 0.00500\times3=0.0150\text{ mol} \]

This agrees with the measured water production.

The number-average degree of polymerisation, \(DP_n\), is the average number of monomer units per polymer molecule.

For this model:

\[ DP_n=\frac{0.0200}{0.00500}=4.00 \]

The student’s answer is therefore possible.

But is it the only possibility?

Step 3: Construct an alternative model containing cyclic oligomers.

Consider a mixture containing:

• \(0.00500\text{ mol}\) of open dimers.
• \(0.00200\text{ mol}\) of cyclic pentamers.

Calculate the total monomer units.

Open dimers contribute:

\[ 0.00500\times2=0.0100\text{ mol} \]

Cyclic pentamers contribute:

\[ 0.00200\times5=0.0100\text{ mol} \]

Total:

\[ 0.0100+0.0100=0.0200\text{ mol} \]

The original amount of lactic acid units is accounted for.

Step 4: Check the ester linkages and water production.

Each open dimer contains one ester linkage:

\[ 0.00500\times1=0.00500\text{ mol} \]

Each cyclic pentamer contains five ester linkages:

\[ 0.00200\times5=0.0100\text{ mol} \]

Total ester linkages:

\[ 0.00500+0.0100=0.0150\text{ mol} \]

This produces exactly:

\[ n(\ce{H2O})=0.0150\text{ mol} \]

The predicted water mass is:

\[ m=0.0150\times18.0 \]

\[ =0.270\text{ g} \]

The alternative model therefore agrees with every experimental measurement.

Step 5: Calculate the number-average degree of polymerisation for the alternative model.

There are two types of polymer molecules:

\[ \begin{aligned} n(\text{total molecules}) &=0.00500+0.00200\\ &=0.00700\text{ mol} \end{aligned} \]

Therefore:

\[ DP_n=\frac{0.0200}{0.00700} \]

\[ DP_n=2.86 \]

The two possible models give different answers:

Both produce the same amount of water from the same initial amount of lactic acid.

The reason is that an open chain containing \(N\) monomer units has \(N-1\) ester linkages, whereas a cyclic chain containing \(N\) monomer units has \(N\) ester linkages.

Closing a chain into a ring creates an additional ester linkage without adding another monomer.

An additional structural measurement, such as mass spectrometry of separated oligomers, could help distinguish cyclic and open structures. For the same number of lactic acid units, a cyclic oligomer has lost one additional water molecule compared with its open-chain counterpart.

The final lesson is that the measured amount of water tells you how many ester-forming condensation events occurred. It does not always tell you how many separate polymer molecules formed.

07Where to go next

When you encounter an unfamiliar polyester structure, start by locating the repeated ester linkages in its backbone.

Then work backwards to identify the functional groups that supplied each side of those linkages.

If one monomer contains both an alcohol and a carboxylic acid group, a single monomer type may be sufficient. If the alcohol and carboxylic acid groups belong to separate molecules, two monomer types are needed.

For the next step, practise finding the monomers from a polyester repeating unit.

The key decision is always the same: follow the actual bonds, identify the reactive groups, and determine whether each monomer can connect to the next.