How to Draw PET from Ethylene Glycol and Terephthalic Acid

A student draws PET with an ethylene glycol unit joined directly to a benzene ring. The carbon skeleton looks reasonable, but something is missing: the ester bonds that actually hold the polymer together.

So where should the oxygen atoms go?

To draw PET from ethylene glycol and terephthalic acid, connect the oxygen of each alcohol group to the carbonyl carbon of a carboxylic acid group, forming ester linkages. The repeating unit is:

\[
\ce{[-O-CH2CH2-O-CO-C6H4-CO-]_{n}}
\]

Here, \(\ce{C6H4}\) represents a benzene ring with its two carbonyl groups attached at opposite positions (1 and 4). Each \(\ce{-CO-}\) represents a carbonyl group, \(\ce{-C(=O)-}\), so the important linkage is \(\ce{-C(=O)-O-}\).

PET, or polyethylene terephthalate, is one particular polyester. Polyester is the family of polymers containing repeated ester links; PET is the named example made from these two particular monomers.

01Start with the two monomers

Before drawing any polymer, write out both starting molecules and find their reactive functional groups.

Ethylene glycol: the molecule with two alcohol groups

Ethylene glycol is also called ethane-1,2-diol.

Its structure is:

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

Notice that it has an \(\ce{-OH}\) group at each end.

These are alcohol functional groups, not carboxylic acid groups.

The two carbon atoms in the middle remain connected by a single bond throughout the polymerisation reaction.

Terephthalic acid: the molecule with two carboxylic acid groups

Terephthalic acid is also called benzene-1,4-dicarboxylic acid.

Its condensed structure is:

\[
\ce{HOOC-C6H4-COOH}
\]

The benzene ring has six carbon atoms. Its two \(\ce{-COOH}\) groups attach at positions 1 and 4, directly opposite each other.

Each carboxylic acid group contains a carbonyl, \(\ce{C=O}\), and a hydroxyl group, \(\ce{-OH}\).

The useful pattern is simple: both molecules have two reactive ends.

That allows them to join into a chain rather than forming just one ester molecule.

Think of each monomer as a piece with a connection point at either end. One piece can connect to the next, which can connect to another. The important limitation is that chemical groups must actually react to form those connections, and the molecules can sometimes form rings rather than long straight chains.

02Understand the ester bond before drawing the polymer

An alcohol and a carboxylic acid can react to form an ester and water.

The general reaction is:

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

Here, \(\ce{R}\) and \(\ce{R’}\) represent carbon-containing parts of the reactants.

Now make a prediction. When the new ester bond forms, does the oxygen from the alcohol disappear into the water?

No. In the usual HSC bond-bookkeeping model:

• The carboxylic acid loses its \(\ce{-OH}\) group.
• The alcohol loses the hydrogen from its \(\ce{-OH}\) group.
• Those removed parts form \(\ce{H2O}\).
• The remaining alcohol oxygen bonds to the carbonyl carbon of the acid.

The new connectivity is:

\[
\ce{R-C(=O)-O-R’}
\]

This tells you exactly where the ester link belongs in PET.

One side comes from terephthalic acid’s carbonyl carbon. The other comes from ethylene glycol’s alcohol oxygen.

The description above tracks the net structural change. It isn’t a complete acid-catalysed mechanism or a claim about the movement of every individual atom during an actual reaction.

Why condensation polymerisation can produce a chain

Suppose ethylene glycol reacts with one carboxylic acid group on terephthalic acid.

An ester bond forms, but the joined molecule still has an alcohol group at one end and a carboxylic acid group at the other.

Those remaining groups can react with other molecules.

Repeated reactions produce a polyester chain, with a small molecule eliminated at each ester-forming condensation step.

That is condensation polymerisation.

There is no carbon-carbon double bond in ethylene glycol to open as there would be in the addition polymerisation of ethene. Despite the word “ethylene” in PET’s name, PET is not polyethylene.

03Draw the PET repeating unit in four steps

Step 1: Keep the ethylene glycol backbone

Start with:

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

The two carbon atoms stay together. Each oxygen can eventually form an ester connection to a terephthalic acid unit.

Step 2: Keep the terephthalic acid backbone

Write:

\[
\ce{-CO-C6H4-CO-}
\]

Each \(\ce{-CO-}\) means \(\ce{-C(=O)-}\). Preserve both carbonyl groups and the 1,4-substitution pattern on the benzene ring.

Do not accidentally place the two carbonyl groups next to each other on the ring.

Step 3: Join the two fragments through ester bonds

Connect the glycol oxygen to a carbonyl carbon:

\[
\ce{-O-CH2CH2-O-CO-C6H4-CO-}
\]

This is the correct sequence of atoms.

One ester connection appears inside the written fragment, between the glycol oxygen and the following carbonyl carbon.

The second occurs where this fragment joins the next copy of itself.

Step 4: Put brackets around one complete repeat

The result is:

\[
\ce{[-O-CH2CH2-O-CO-C6H4-CO-]_{n}}
\]

The brackets identify the repeating section. The subscript \(n\) represents the number of repeats in the idealised chain.

The open bonds at the ends of the bracket connect to neighbouring repeating units. They are not free ends of a complete polymer molecule.

A quick check using the molecular formula

Each PET repeat contains:

Therefore, the repeat-unit formula is:

\[
\ce{C10H8O4}
\]

The formula is a useful check, but it cannot prove the connectivity. Two differently connected molecules can contain exactly the same numbers of each atom.

The formal repeat-unit balance is:

\[
\ce{C2H6O2 + C8H6O4}
\]

\[
\ce{-> C10H8O4 + 2H2O}
\]

This describes the composition of a repeat in an ideal long chain, where the terminal groups are ignored.

It does not mean that a finite chain made from \(n\) molecules of each monomer must release exactly \(2n\) water molecules. Its ends matter, as we’ll calculate shortly.

04What would the reaction actually tell an investigator?

A structural drawing is a prediction about which bonds form. How could a chemist test whether condensation is happening?

First consider a simpler reaction.

Ethanoic acid reacts with ethanol to form ethyl ethanoate:

\[
\ce{CH3COOH + CH3CH2OH}
\]

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

This produces an ester, but not a polymer. Each molecule has only one functional group available for this particular reaction. Once the two molecules join, there isn’t another acid-alcohol pair at their ends to continue chain growth.

That first design demonstrates ester formation. It cannot demonstrate polyester formation.

The logical revision is to replace the single-ended reactants with ethylene glycol and terephthalic acid, which each have two reactive groups.

What a PET production apparatus would reveal

Consider a simplified direct-esterification process.

Terephthalic acid and ethylene glycol enter a heated, stirred reaction vessel. As esterification takes place, volatile material leaves through an outlet above the reaction mixture.

The vapour travels to a condenser, where some of it becomes liquid and enters a collection vessel. The growing ester-containing material remains in the reaction vessel.

What should the collected material tell us?

If acid and alcohol groups react through condensation, water should be generated.

In the direct-esterification route used to manufacture PET, water is indeed produced during the initial esterification stage. Later chain growth can also occur through transesterification reactions, which release ethylene glycol rather than water.

There is an important limitation.

Collecting water shows that reactions capable of generating water have occurred. It does not, by itself, establish that the product is high-molecular-mass PET.

Short oligomers can form the same ester links. Another polyester can also produce water.

The condensate may contain other volatile substances, so its total mass is not automatically the mass of pure water.

A better investigation would combine several measurements:

• Identify the starting monomers, especially the 1,4 arrangement on terephthalic acid.
• Confirm that the product contains ester functional groups.
• Measure the molecular-mass distribution or remaining end groups to investigate how far polymerisation has progressed.
• Analyse the condensate composition before using its mass in a stoichiometric calculation.

This is why a correctly drawn ester linkage is evidence of the intended chemical structure, but water production alone is not proof of a long PET chain.

05Distinguish PET from other possible products

A common mistake is to identify a polymer from the presence of an ester group alone.

That only identifies the polymer family.

Both terephthalic acid and isophthalic acid have the molecular formula \(\ce{C8H6O4}\).

Both can react with ethylene glycol to produce polyesters.

But their carbonyl groups occupy different positions around the benzene ring, so the resulting polymer backbones are different.

A correct molecular formula is necessary, but it isn’t sufficient to identify PET.

The same reasoning distinguishes polyester from polyamide. A polyester contains repeated ester links, \(\ce{-C(=O)-O-}\), whereas a polyamide contains repeated amide links, \(\ce{-C(=O)-NH-}\) in the common case of a secondary amide.

The distinction also affects how neighbouring polymer chains interact, which is explored in how to compare hydrogen bonding in polyamides and polyesters.

06Worked examples

07Questions and solutions

Question 1

Ethylene glycol and terephthalic acid react through condensation polymerisation.

(a) Write the structural formula of the PET repeating unit.

(b) Name the functional-group linkage that connects the monomers.

(c) Explain which atoms are removed when this linkage forms in the standard condensation model.

Solution 1

The repeating unit is \(\ce{[-O-CH2CH2-O-CO-C6H4-CO-]_{n}}\), with the two carbonyl substituents at positions 1 and 4 on the benzene ring.

(a) The two-carbon glycol section joins to the terephthalic acid section through oxygen-containing links. The repeat brackets identify the section that continues along the chain.

(b) The functional-group linkage is an ester:

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

This identifies PET as a polyester.

(c) In the standard condensation model, a carboxylic acid group loses \(\ce{-OH}\), and an alcohol group loses \(\ce{H}\). They combine to form \(\ce{H2O}\).

The remaining alcohol oxygen forms a bond to the carbonyl carbon.

The common mistake is to remove the entire alcohol \(\ce{-OH}\) group. That would incorrectly eliminate the oxygen needed for the ester linkage.

Question 2

A manufacturer substitutes benzene-1,3-dicarboxylic acid (isophthalic acid) for terephthalic acid while keeping ethylene glycol as the second monomer.

A student notices that the resulting polyester has the same repeat-unit molecular formula as PET and that water is produced during direct condensation.

They conclude that the product must still be PET.

(a) Explain whether the student is correct, using the repeating structure.

(b) Explain why the molecular formula and water production cannot settle the question.

(c) Identify one additional piece of chemical evidence that would distinguish the products.

Solution 2

The student is incorrect. The product is a polyester, but it is not PET because the benzene ring has a different substitution pattern.

(a) With isophthalic acid, the chain still contains the sequence:

\[
\ce{-O-CH2CH2-O-CO-}
\]

followed by a benzene ring and another carbonyl group.

However, the two carbonyl groups attach at positions 1 and 3 of the aromatic ring, not positions 1 and 4.

The product is poly(ethylene isophthalate), whereas PET is poly(ethylene terephthalate).

This change alters how the aromatic units connect along the backbone. It cannot be corrected simply by rotating the drawing or moving the repeating-unit brackets.

(b) Both acids have molecular formula \(\ce{C8H6O4}\). Both react with ethylene glycol through esterification and can produce repeat units with formula \(\ce{C10H8O4}\).

Both reactions also eliminate water.

Therefore, the same elemental composition and condensation by-product are consistent with two different structures.

The tempting conclusion ignores structural isomerism: the arrangement of the atoms matters, not just their numbers.

(c) Identifying the dicarboxylic acid starting material, including the positions of its two carboxyl groups, would distinguish them.

A suitable structural analysis of the monomer or polymer could also establish whether the aromatic substitution pattern is 1,3 or 1,4.

Water collection alone cannot make that distinction.

Question 3

Two idealised molecules are made from exactly two ethylene glycol molecules and two terephthalic acid molecules each.

• Molecule L is a linear alternating oligomer with one free alcohol group and one free carboxylic acid group.
• Molecule R is a closed cyclic oligomer with no free alcohol or carboxylic acid groups.

Both contain only ester connections between the monomer fragments.

Each molecule is then subjected to controlled hydrolysis in which exactly one ester bond is broken, with no other reactions.

(a) State how many separate molecules result from each hydrolysis.

(b) Determine the total number of free alcohol and carboxylic acid groups in the products from each starting molecule.

(c) Explain why the same hydrolysis reaction can produce different numbers of molecules.

Solution 3

One hydrolysis of L produces two separate molecules, while one hydrolysis of R produces one open-chain molecule. The difference comes from the starting connectivity, not from a different chemical reaction.

(a) In a linear chain, breaking any one connection separates the chain into two pieces.

Therefore, hydrolysing one ester bond in L produces two molecules.

A cyclic molecule has two routes connecting neighbouring parts of its ring. Breaking one bond opens the ring, but the remaining bonds still connect all the fragments.

Therefore, hydrolysing one ester bond in R produces one linear molecule.

(b) Acidic hydrolysis of an ester produces a carboxylic acid group and an alcohol group:

\[
\ce{R-COO-R’ + H2O}
\]

\[
\ce{-> R-COOH + HO-R’}
\]

For L, there was already one free \(\ce{-OH}\) and one free \(\ce{-COOH}\).

Hydrolysis creates one more of each, giving:

• Two free alcohol groups.
• Two free carboxylic acid groups.
• Two separate product molecules.

For R, there were initially no free functional-group ends.

Hydrolysis creates one of each, giving:

• One free alcohol group.
• One free carboxylic acid group.
• One open-chain product molecule.

(c) The tempting assumption is that breaking one covalent bond must split a molecule into two molecules.

That is true for an internal bond in a linear, unbranched chain, but not for a bond in a simple closed ring.

A ring stays connected after one bond is broken because another path remains between the atoms.

This is a structural consequence of connectivity, even though exactly one water molecule is consumed in each hydrolysis.

Question 4

A chemist studies two idealised direct-condensation mixtures.

Mixture A contains three ethylene glycol molecules and three terephthalic acid molecules. They form one linear oligomer with a free alcohol end and a free carboxylic acid end.

Mixture B contains the same six molecules plus one methanol molecule, \(\ce{CH3OH}\). All seven molecules become part of one linear product. Every carboxylic acid group reacts, and exactly one glycol alcohol group remains unreacted.

Assume that each ester-forming reaction eliminates one water molecule, with no cycles or side reactions.

(a) Calculate the number of ester links and water molecules formed in each mixture.

(b) Identify the two ends of each product.

(c) A student claims that mixture B must have formed a better-growing PET chain because more water was produced. Evaluate this claim.

Solution 4

Mixture A produces five ester links and five water molecules. Mixture B produces six ester links and six water molecules, but its extra link involves methanol, which caps one end of the product.

(a) Mixture A begins with six separate molecules.

Joining six molecules into one linear chain requires:

\[
6-1=5\text{ links}
\]

Therefore:

\[
N(\ce{H2O})=5
\]

Here, \(N(\ce{H2O})\) is the number of water molecules produced in the idealised molecular assembly.

Mixture B begins with seven separate molecules.

Joining seven molecules into one linear chain requires:

\[
7-1=6\text{ links}
\]

Therefore:

\[
N(\ce{H2O})=6
\]

This also agrees with the functional-group count. The three terephthalic acid molecules supply six carboxylic acid groups, and all six react.

(b) Mixture A has one free \(\ce{-OH}\) end and one free \(\ce{-COOH}\) end.

In mixture B, methanol reacts with a carboxylic acid end to give a methyl ester:

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

That end is no longer a free carboxylic acid group.

The other end remains a free glycol alcohol group, \(\ce{-OH}\).

(c) The student’s conclusion is incorrect.

Mixture B produces more water because an additional ester bond forms with methanol. It has not incorporated more ethylene glycol or terephthalic acid units.

Methanol has only one alcohol functional group. Once it forms the methyl ester, it has no second reactive alcohol group available to continue the same condensation process.

It therefore acts as a chain stopper at that end.

The trap is treating water yield as a direct measure of polymer-chain growth. A small molecule can consume a reactive end, release water, and prevent further extension at that end.

Question 5

Extension challenge – beyond a typical HSC repeat-unit question

An idealised molecular batch begins with exactly six ethylene glycol molecules and six terephthalic acid molecules.

Only direct ester-forming condensation reactions occur. Every new ester bond produces one water molecule. No starting molecules are lost, and the resulting products contain all twelve starting monomer fragments.

An analysis gives the following results:

All fragments are connected only through ester links. Products may be linear or cyclic, but there is no branching.

A student argues:

“Ten water molecules and four remaining functional-group ends prove that the product consists of exactly two linear chains. There cannot be any rings.”

Is this conclusion necessarily correct?

Construct two different product arrangements that satisfy every measurement, using complete alternating sequences of ethylene glycol and terephthalic acid fragments.

Explain why even the water measurement and end-group measurement together cannot distinguish the arrangements. State what additional kind of evidence could resolve the ambiguity.

Solution 5

The student’s conclusion is not necessarily correct. Two linear chains are consistent with the results, but a mixture containing two linear chains and one cyclic oligomer can produce exactly the same water count and the same number of free end groups.

The key constraint is that cyclic molecules have ester links but no free ends.

First arrangement: two linear chains

Let \(G\) represent one ethylene glycol fragment and \(T\) represent one terephthalic acid fragment.

Consider two identical linear chains:

\[
G-T-G-T-G-T
\]

Each chain contains three \(G\) fragments and three \(T\) fragments.

Each chain contains six initially separate fragments joined by five ester links.

Therefore, across two chains:

\[
\begin{aligned}
N(\text{ester links})&=2(5)\\
&=10
\end{aligned}
\]

Under the stated assumptions, this produces ten water molecules.

Each alternating six-fragment chain has one free alcohol end and one free carboxylic acid end.

Across both chains, there are two free alcohol groups and two free carboxylic acid groups.

All twelve monomer fragments are accounted for.

So the student’s proposed arrangement is possible.

Second arrangement: two shorter linear chains and one cyclic oligomer

Now consider two linear chains, each with four fragments:

\[
G-T-G-T
\]

Each chain contains three ester links.

Together, the two linear chains contain six ester links, and they leave two free alcohol groups and two free carboxylic acid groups.

These chains consume four \(G\) fragments and four \(T\) fragments altogether.

That leaves two \(G\) fragments and two \(T\) fragments.

Arrange those four remaining fragments in an alternating closed ring. The ring has four ester links, including the bond that closes it.

It has no free ends.

The total number of ester links is now:

\[
\begin{aligned}
N(\text{ester links})
&=3+3+4\\
&=10
\end{aligned}
\]

Again, ten water molecules form.

The two linear chains supply exactly two free alcohol groups and two free carboxylic acid groups.

The cyclic molecule supplies none.

All six \(G\) fragments and all six \(T\) fragments have been used.

Therefore, this second arrangement satisfies every measurement as well.

Why the measurements fail to distinguish them

Count the functional groups initially.

Six ethylene glycol molecules provide twelve alcohol groups.

Six terephthalic acid molecules provide twelve carboxylic acid groups.

Every ester-forming condensation consumes one alcohol group and one carboxylic acid group.

Ten ester-forming reactions therefore leave:

\[
\begin{aligned}
N(\text{free OH})&=12-10=2\\
N(\text{free COOH})&=12-10=2
\end{aligned}
\]

Those counts are fixed by the number of ester links formed. They do not tell us how the links are distributed among the product molecules.

Even the overall elemental composition would be identical because both arrangements begin with the same atoms and eliminate the same ten water molecules.

The decisive difference is connectivity.

Arrangement A contains two molecules, both linear.

Arrangement B contains three molecules, one of which is cyclic.

To distinguish them, the investigation would need evidence about the molecular-mass distribution or the presence of cyclic species, rather than only total water production and remaining functional groups.

This is the deeper lesson behind PET’s simple repeating-unit formula: knowing which bonds exist does not automatically tell you how many chains exist, how long they are, or whether some molecules form rings.

08What this means when you draw PET in an exam

For a standard NSW HSC Chemistry question, your job is normally to establish the correct repeat-unit connectivity, not to analyse every possible chain length or cyclic oligomer.

Keep these decisions in order:

1. Identify ethylene glycol’s two alcohol groups and terephthalic acid’s two carboxylic acid groups.
2. Preserve the two-carbon glycol backbone and the benzene ring’s 1,4-substitution pattern.
3. Form ester connections between glycol oxygen atoms and acid carbonyl carbons.
4. Draw one complete repeating unit with brackets and \(n\).
5. Check that the repeated chain contains two ester links per repeat and that the structure has the correct repeat-unit composition.

Remember that the ideal repeating structure describes PET’s characteristic backbone. It leaves out the exact end groups, chain lengths, and possible minor structural variations of a real sample.

Once the repeat unit is correct, the next useful question is what those ester groups and aromatic rings do to the material’s properties. In particular, identifying the ester oxygen atoms prepares you to compare PET with polyamides, where the presence of amide N-H groups changes the possibilities for hydrogen bonding between polymer chains.