Why Polyester Formation Needs a Diol in Direct Condensation

A reaction mixture can produce water and ester bonds without producing a polyester. Ethanol, for example, can react with a carboxylic acid to form an ester. But once its single \(\ce{-OH}\) group has reacted, it cannot connect to another acid molecule.

This is why the common condensation route to polyesters uses a diol rather than an ordinary monohydric alcohol. A diol has two alcohol groups, one at each end. After one forms an ester bond, the other remains available to form another bond, allowing the molecular chain to grow.

The important difference is not how reactive the alcohol is. It is how many opportunities each molecule has to connect to other molecules.

01First, distinguish polyester from PET

Polyester is a family of polymers. Its defining feature is the presence of repeated ester linkages within the main polymer chain.

An ester linkage has the structure:

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

PET, or poly(ethylene terephthalate), is one particular member of this family.

It is formed from two specific monomers:

Both molecules have two reactive ends.

When these monomers join, they produce PET, whose repeating structure can be represented as:

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

Here, \(\ce{C6H4}\) represents the para-substituted benzene ring from terephthalic acid, and \(n\) represents the number of repeating units.

Notice that the ester linkages occur along the main chain. The polymer is not simply a molecule containing an ester group somewhere on its structure.

The key question is why ethane-1,2-diol works for chain formation while ethanol does not.

02Why two OH groups make a difference

What happens when an alcohol forms an ester?

An alcohol reacts with a carboxylic acid to produce an ester and water.

The general reaction is:

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

Here, \(R\) and \(R’\) represent the remaining carbon-containing parts of the molecules.

During the usual HSC condensation bookkeeping:

• The carboxylic acid loses its \(\ce{-OH}\) group.
• The alcohol loses the hydrogen from its \(\ce{-OH}\) group.
• These removed atoms form one water molecule.
• The oxygen remaining from the alcohol forms the ester connection to the carbonyl carbon.

The important point is that the alcohol oxygen stays in the ester linkage.

Now consider what happens when one end of ethane-1,2-diol reacts with a carboxylic acid.

\[
\ce{R-COOH + HO-CH2-CH2-OH}
\]

After one esterification:

\[
\ce{R-COO-CH2-CH2-OH + H2O}
\]

Look at the product.

One alcohol group has reacted, but the other \(\ce{-OH}\) group is still present.

That remaining group can react with another carboxylic acid molecule. The process can repeat, extending the chain.

What happens with an ordinary single-OH alcohol?

Now replace ethane-1,2-diol with ethanol:

\[
\ce{CH3CH2OH}
\]

Ethanol has only one alcohol group.

It can still react with a carboxylic acid:

\[
\ce{R-COOH + HO-CH2CH3}
\]

The product is:

\[
\ce{R-COO-CH2CH3 + H2O}
\]

The ethanol has formed an ester, but its only \(\ce{-OH}\) group has been consumed.

There is no second alcohol group available to connect to another carboxylic acid molecule.

This is where the two molecules behave differently:

A useful way to picture this is to imagine connecting pieces of a chain.

A diol has two connection points. It can attach to one molecule at one end and another molecule at the other end.

A monohydric alcohol has only one connection point. Once attached, it becomes an end cap rather than a bridge.

The analogy has a limit: actual chemical connections require suitable functional groups to react. They are not simply physical clips. But it captures the essential difference in molecular connectivity.

Why the acid must also have two functional groups

There is another detail worth noticing.

Having a diol does not automatically guarantee polyester formation.

The other monomer must also be capable of continuing the chain.

In PET, terephthalic acid has two \(\ce{-COOH}\) groups. Both can form ester linkages with alcohol groups on neighbouring molecules.

For a straightforward linear polyester formed by direct condensation, both monomers therefore have two reactive groups.

This property is called functionality. It describes the number of reactive groups on an individual molecule that can participate in the polymer-forming reaction.

The distinction becomes clearer when we compare different functionalities:

For the standard linear polyester model:

• A monofunctional alcohol tends to terminate chain growth.
• A bifunctional alcohol allows chain growth.
• A trifunctional alcohol can introduce branching rather than simply extending a linear chain.

This explains why the number of reactive groups per molecule matters more than the total number of OH groups in a reaction mixture.

04A useful exception: Not every polyester requires a separate diol

The explanation so far applies to the common condensation route involving a diol and a dicarboxylic acid.

It is not a universal requirement for every polyester.

For example, a hydroxycarboxylic acid contains both an alcohol group and a carboxylic acid group in the same molecule:

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

Such a molecule can react at both ends and participate in polyester formation without requiring a separate diol.

Some polyesters can also be produced by ring-opening polymerisation of cyclic esters, which follows a different reaction pathway.

The deeper principle is therefore not that every polyester must contain a diol-derived section.

It is that the molecules involved in chain formation need suitable functionality to connect repeatedly.

06How could an experiment test the difference?

Suppose a chemist wants to determine whether a reaction has produced a polyester rather than simply a collection of small ester molecules.

Measuring the amount of water released seems like a sensible starting point.

But is it enough?

Consider an illustrative investigation comparing two reaction vessels.

The original investigation

Each reaction vessel contains:

• A heated, stirred mixture of terephthalic acid and an alcohol.
• A vapour outlet connected to a condenser.
• A separate collection vessel beneath the condenser.

As the reaction proceeds, volatile material travels out of the heated vessel and into the condenser.

The condenser cools the vapour, allowing liquid to collect in the receiver. The less volatile reaction products remain in the original vessel.

The chemist measures the collected water.

What should the investigation show?

If esterification occurs, water should be produced in both mixtures.

This is the important observation to predict.

Even the reaction involving ethanol can produce water because esterification does not require the alcohol to have two OH groups.

Therefore, finding water in the collection vessel demonstrates that esterification has occurred, but it does not establish that a long polymer chain has formed.

The two reaction mixtures can produce the same amount of water while producing very different molecular structures.

What was missing from the original design?

The original investigation measured the extent of condensation, not the molecular size of the product.

A small ester molecule and a long polyester chain can both contain ester linkages and both can be produced through reactions that release water.

The water measurement cannot distinguish them.

There is also an experimental limitation: the collected condensate may contain unreacted alcohol or other volatile substances. Its total mass cannot automatically be treated as the mass of pure water.

How could the investigation be improved?

A more complete investigation would examine the material remaining in the reaction vessel.

The chemist could:

• Confirm that ester functional groups are present.
• Determine the molecular-mass distribution of the product.
• Investigate the remaining alcohol and carboxylic acid end groups.
• Identify the collected condensate before using its mass in a water calculation.

A suitable molecular-size measurement would help distinguish small ester molecules from high-molecular-mass polymer chains.

The structural prediction also matters. Knowing whether the starting monomers are mono-, bi-, or trifunctional helps explain which molecular architectures are possible.

This is the key experimental limitation: evidence that esterification occurred is not automatically evidence that polymerisation produced a long chain.

07How to recognise the correct answer in an HSC question

When a question asks why a diol is used instead of a monohydric alcohol, focus on the number of reactive groups per molecule.

A reliable way to reason through an unfamiliar reaction is:

1. Identify the alcohol and carboxylic acid functional groups on each monomer.
2. Count how many of those groups each molecule contains.
3. Imagine the first ester linkage forming.
4. Check whether each molecule still has another reactive group available.
5. Decide whether the molecules can continue connecting into a chain.

Remember these distinctions:

A monohydric alcohol is not incapable of esterification. It simply lacks the second reactive OH group needed to bridge another molecular section.

It can even be used as a chain-stopping molecule in a polyester-forming reaction.

The other important qualification is that having two reactive groups makes linear chain formation possible, not automatic. Functional-group balance and a high extent of reaction are also important for producing long chains.

For a detailed structural treatment, see How to Draw PET from Ethylene Glycol and Terephthalic Acid.

Once you understand why the monomers can connect, the next useful step is counting how many ester linkages actually form in a finite chain. This is particularly important when calculating the amount of water released, as explained in How to Count Water Molecules in a Short Polyester Chain.

08Questions and solutions

Question 1

Propan-1-ol, \(\ce{CH3CH2CH2OH}\), is reacted with hexanedioic acid, \(\ce{HOOC-(CH2)4-COOH}\).

Explain whether these two monomers can form a long-chain polyester through direct condensation.

Solution 1

No. Propan-1-ol is monofunctional, so it cannot connect two different acid-derived sections into a growing polyester chain.

Hexanedioic acid contains two carboxylic acid groups, meaning it can potentially react at both ends.

However, propan-1-ol contains only one alcohol group.

When propan-1-ol reacts with one carboxylic acid group, its only OH group is consumed, producing an ester.

The other carboxylic acid group on the hexanedioic acid molecule could react with another propan-1-ol molecule. This would produce a small diester rather than a polymer.

The tempting mistake is to assume that because both acid groups can react, the chain must grow. But the alcohol molecules cannot bridge between successive acid molecules.

For linear polyester formation by this route, both monomers need two suitable reactive groups.

Question 2

Ethane-1,2-diol is reacted with ethanoic acid.

A student argues that a polyester must form because ethane-1,2-diol has two OH groups and can therefore connect to two different molecules.

Evaluate this argument.

Include the expected product if both OH groups of the diol react.

Solution 2

The student’s argument is incorrect. Ethane-1,2-diol is bifunctional, but ethanoic acid is monofunctional in this esterification reaction.

Ethanoic acid has only one \(\ce{-COOH}\) group.

Each ethanoic acid molecule can therefore form only one ester linkage. It cannot connect onward to another alcohol molecule through a second carboxylic acid group.

If both OH groups of ethane-1,2-diol react, the product is a small diester:

\[
\ce{HO-CH2-CH2-OH + 2CH3COOH}
\]

\[
\ce{-> CH3COO-CH2-CH2-OOCCH3 + 2H2O}
\]

The diol has been connected to two acid-derived sections, but neither acid-derived section has another reactive carboxylic acid group available to extend the structure.

The important trap is focusing only on the alcohol.

A diol provides two reactive ends, but linear polyester formation by this route requires the other monomer to have two suitable reactive ends as well.

Question 3

A reaction mixture contains:

• \(0.150\ \text{mol}\) of ethane-1,2-diol.
• \(0.100\ \text{mol}\) of terephthalic acid.

Assume direct esterification occurs, all carboxylic acid groups react, and there are no side reactions.

Calculate:

(a) The initial amount of alcohol groups.

(b) The initial amount of carboxylic acid groups.

(c) The maximum amount of ester linkages and water that can form.

(d) The amount of alcohol groups remaining after all carboxylic acid groups have reacted.

Explain how the unequal amounts of functional groups affect the potential chain length.

Solution 3

The mixture contains \(0.300\ \text{mol}\) of alcohol groups and \(0.200\ \text{mol}\) of carboxylic acid groups. The maximum amount of ester linkages and water is \(0.200\ \text{mol}\), leaving \(0.100\ \text{mol}\) of alcohol groups unreacted.

(a) Calculate the initial amount of alcohol groups.

Each ethane-1,2-diol molecule contains two OH groups.

\[
\begin{aligned}
n(\ce{-OH})
&=2(0.150)\\
&=0.300\ \text{mol}
\end{aligned}
\]

(b) Calculate the initial amount of carboxylic acid groups.

Each terephthalic acid molecule contains two \(\ce{-COOH}\) groups.

\[
\begin{aligned}
n(\ce{-COOH})
&=2(0.100)\\
&=0.200\ \text{mol}
\end{aligned}
\]

(c) Calculate the maximum ester linkages and water produced.

Each ester linkage requires one alcohol group and one carboxylic acid group.

The carboxylic acid groups are present in the smaller amount, so they are the limiting functional groups.

Therefore:

\[
n(\text{ester links})=0.200\ \text{mol}
\]

Each ester linkage produces one water molecule under the stated direct-condensation model.

\[
n(\ce{H2O})=0.200\ \text{mol}
\]

The corresponding mass of water is:

\[
\begin{aligned}
m(\ce{H2O})
&=nM\\
&=0.200(18.0)\\
&=3.60\ \text{g}
\end{aligned}
\]

(d) Calculate the remaining alcohol groups.

Initially, there were \(0.300\ \text{mol}\) of OH groups.

The \(0.200\ \text{mol}\) of esterification reactions consume \(0.200\ \text{mol}\) of these groups.

\[
\begin{aligned}
n(\ce{-OH})_{\text{remaining}}
&=0.300-0.200\\
&=0.100\ \text{mol}
\end{aligned}
\]

The reaction mixture contains excess diol.

This imbalance limits the potential average chain length because all available carboxylic acid groups can be consumed while alcohol groups remain unreacted.

The excess alcohol groups cannot create additional ester connections without more carboxylic acid groups.

A common mistake is to assume that having more diol must produce a longer polymer. In step-growth polymerisation, the balance between the two types of functional groups is also important.

The exact number and length of polymer chains cannot be determined from these quantities alone.

Question 4

A chemist compares two reaction vessels.

Both initially contain \(0.100\ \text{mol}\) of terephthalic acid.

• Vessel A also contains \(0.100\ \text{mol}\) of ethane-1,2-diol.
• Vessel B also contains \(0.200\ \text{mol}\) of ethanol.

Both reactions produce \(0.180\ \text{mol}\) of water.

(a) Calculate the maximum possible amount of ester linkages in each vessel.

(b) Calculate the percentage of the maximum possible esterification that has occurred.

(c) Does the water measurement prove that both reactions produced polyesters? Explain.

(d) Suggest one additional measurement that would help distinguish the products.

Solution 4

Both vessels have a maximum of \(0.200\ \text{mol}\) of possible ester linkages, and both have reached 90% of this theoretical maximum. However, only vessel A contains the bifunctional alcohol needed for linear polyester chain growth.

(a) Calculate the maximum ester linkages.

Each terephthalic acid molecule contains two carboxylic acid groups.

\[
\begin{aligned}
n(\ce{-COOH})
&=2(0.100)\\
&=0.200\ \text{mol}
\end{aligned}
\]

For vessel A:

\[
n(\ce{-OH})=2(0.100)
=0.200\ \text{mol}
\]

For vessel B:

\[
n(\ce{-OH})=1(0.200)
=0.200\ \text{mol}
\]

Both vessels contain equal amounts of alcohol and carboxylic acid functional groups.

Therefore, the maximum possible amount of ester linkages is:

\[
0.200\ \text{mol}
\]

(b) Calculate the percentage of maximum esterification.

The actual amount of ester linkages is \(0.180\ \text{mol}\), as indicated by the amount of water produced.

\[
\begin{aligned}
\text{Percentage}
&=\frac{0.180}{0.200}\times100\\
&=90\%
\end{aligned}
\]

Both reactions have reached 90% of their theoretical maximum esterification.

(c) Evaluate whether both reactions produced polyesters.

No. Equal water production does not establish equal polymer formation.

Vessel A contains ethane-1,2-diol. Each molecule has two OH groups, allowing it to connect to two acid-derived sections and support linear polyester formation.

Vessel B contains ethanol. Each molecule has only one OH group and can therefore form only one ester connection.

Ethanol can cap the carboxylic acid groups, producing small ester molecules rather than an extended polyester chain.

The tempting conclusion is that equal water production means equal polymerisation. However, water measures the number of esterification reactions, not how those reactions connect the monomers.

(d) Suggest an additional measurement.

A suitable measurement of the product’s molecular-mass distribution would help determine whether high-molecular-mass polymer chains have formed.

Analysis of the remaining functional groups would also provide useful information about the molecular ends.

These measurements address the limitation of collecting water alone.

Question 5

Extension challenge: Monomer functionality and polymer architecture.

Three reaction batches each contain \(0.500\ \text{mol}\) of terephthalic acid.

The alcohol component differs between the batches:

Each batch therefore starts with:

• \(1.00\ \text{mol}\) of carboxylic acid groups.
• \(1.00\ \text{mol}\) of alcohol groups.

After reaction, each batch has formed exactly \(0.700\ \text{mol}\) of ester linkages and released \(0.700\ \text{mol}\) of water.

Assume:

• Only direct esterification occurs.
• Each ester linkage consumes one OH group and one carboxylic acid group.
• No monomers are lost.
• No intramolecular ring closure occurs.

A student argues:

“All three batches have the same number of reactive groups and form the same amount of water. Therefore, they must produce the same type of polyester.”

Evaluate this claim.

In your answer:

(a) Explain the possible polymer architecture in batch A.

(b) Explain why batch B cannot form an extended polyester chain.

(c) Determine whether batch C must contain a branching point. Justify your conclusion quantitatively.

(d) Explain why the water measurements alone cannot establish that the three products have the same molecular structure.

Solution 5

The student’s claim is incorrect. All three batches produce the same amount of water, but their alcohol molecules have different functionalities, allowing very different molecular architectures.

The key is to distinguish the total number of OH groups from the number of OH groups on each individual molecule.

(a) Explain the architecture of batch A.

Batch A contains ethane-1,2-diol, which has two OH groups per molecule.

Terephthalic acid also has two reactive groups per molecule.

Both monomers are therefore bifunctional.

Each diol molecule can connect to two acid-derived sections. Each acid molecule can connect to two alcohol-derived sections.

This allows repeated connections in both directions, producing linear polyester chains.

Because neither monomer has more than two reactive groups, branching points cannot form through the stated reactions.

The exact number and length of the chains cannot be determined from the amount of water alone.

(b) Explain why batch B cannot form an extended polyester chain.

Batch B contains ethanol, which has only one OH group per molecule.

Once an ethanol molecule forms an ester linkage, it cannot connect to another carboxylic acid group.

Even though the total amount of OH groups is equal to the amount in batch A, those groups are distributed across twice as many alcohol molecules.

Ethanol therefore acts as a chain cap.

The reaction can produce ester-containing molecules, but it cannot produce an extended polyester chain through this route.

(c) Determine whether batch C must contain a branching point.

Yes. Under the stated assumptions, at least one glycerol molecule must have all three of its OH groups reacted.

Glycerol contains three OH groups per molecule.

Batch C contains:

\[
\frac{1}{3}\ \text{mol glycerol}
\]

Therefore, the total amount of OH groups is:

\[
n(\ce{-OH})=3\left(\frac{1}{3}\right)
=1.00\ \text{mol}
\]

We know that \(0.700\ \text{mol}\) of ester linkages have formed.

Each linkage consumes one OH group, so:

\[
n(\ce{-OH})_{\text{reacted}}
=0.700\ \text{mol}
\]

Now consider the alternative.

Suppose no glycerol molecule had all three OH groups reacted.

Each glycerol molecule could then contribute a maximum of two ester linkages.

The maximum possible amount of ester linkages without any three-connected glycerol molecules would be:

\[
\begin{aligned}
n(\text{ester links})_{\max}
&=2\left(\frac{1}{3}\right)\\
&=\frac{2}{3}\\
&=0.667\ \text{mol}
\end{aligned}
\]

But the actual amount is:

\[
0.700\ \text{mol}
\]

Since:

\[
0.700>0.667
\]

the measured reaction extent cannot be achieved unless at least one glycerol molecule has all three OH groups reacted.

That glycerol molecule connects to three different acid-derived sections, creating a branching point under the assumption that no intramolecular rings form.

This is the important distinction between a diol and a triol.

A diol can connect in two directions and support a linear chain. A triol can connect in three directions and introduce branching.

At sufficiently extensive reaction, such branching can lead to a cross-linked polymer network.

(d) Explain why equal water measurements do not establish equal polymer structures.

Each batch produces \(0.700\ \text{mol}\) of water because each forms \(0.700\ \text{mol}\) of ester linkages.

Water production tells us how many esterification reactions have occurred under the stated assumptions.

It does not tell us how those ester linkages are distributed between molecules.

The three batches demonstrate three different possibilities:

• Batch A can form linear polyester chains because both monomers are bifunctional.
• Batch B produces capped ester molecules because ethanol is monofunctional.
• Batch C produces branched polyester structures because glycerol can connect through three OH groups.

The tempting approach is to compare only the total number of OH and COOH groups. But this ignores the number of reactive groups attached to each individual molecule.

That distribution determines whether molecules can extend a chain, terminate a chain, or create a branching point.

The deeper lesson is that polymer structure depends on molecular connectivity, not simply the number of ester bonds formed.