What Happens to a Polyester Chain During Hydrolysis?

Leave a PET drink bottle in water and it stays a bottle. But the polymer chains inside it contain chemical links that water can break. So what actually happens when those links undergo hydrolysis?

Hydrolysis breaks ester bonds in the polyester backbone, splitting long polymer molecules into shorter chains. With sufficient hydrolysis, those chains can eventually break down into their original building blocks. In acidic conditions, the products contain carboxylic acid and alcohol groups. In alkaline conditions, carboxylate ions form instead of free carboxylic acids.

Here’s the important prediction: if polyester contains bonds that can be hydrolysed, should a PET bottle left in water quickly turn into its monomers?

No. PET hydrolyses very slowly under ordinary conditions. Breaking an ester bond is chemically possible, but that doesn’t mean the reaction happens quickly. Temperature, pH, reaction time, and how easily water or hydroxide reaches the bonds all affect what happens.

01Which bond breaks when a polyester undergoes hydrolysis?

A polyester is a family of polymers containing ester linkages in its backbone.

An ester linkage has the structure:

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

Here:

• \(\ce{R}\) represents the carbon-containing section attached to the carbonyl carbon.
• \(\ce{R'}\) represents the section attached to the single-bonded oxygen.
• \(\ce{C=O}\) is the carbonyl group.
• The \(\ce{C(=O)-O}\) connection is the part of the ester structure involved in the usual hydrolysis reaction.

In the standard hydrolysis of an ester, the single bond between the carbonyl carbon and the neighbouring oxygen is broken.

Under acidic conditions, the overall reaction is:

\[ \ce{RCOOR' + H2O -> RCOOH + R'OH} \]

Water is consumed. The two products contain:

1. A carboxylic acid group, \(\ce{-COOH}\).
2. An alcohol group, \(\ce{-OH}\).

One way to picture this is a chain made from connected metal links. If you open a connection in the middle, one chain becomes two shorter chains.

The analogy explains why breaking a backbone bond separates a polymer molecule. But it misses the chemistry: hydrolysis doesn’t simply pull two pieces apart. It changes the chemical groups at the newly created ends.

One broken ester bond does not mean complete breakdown

Imagine one long, unbranched polyester molecule containing hundreds of ester links.

If a single internal ester bond hydrolyses, the chain becomes two shorter molecules. Both may still contain many ester groups and behave like polymeric material.

If more ester bonds hydrolyse, further fragments form. Some become short molecules called oligomers. Complete hydrolysis of a simple polyester made from a diol and a dicarboxylic acid can recover the corresponding small molecular building blocks.

The progression is:

Long polymer chain → shorter chains → oligomers → monomer-derived products.

These stages are possible, not a guaranteed sequence of isolated substances. Different ester bonds may react at different times, and a partly hydrolysed sample can contain several chain lengths at once.

02What happens specifically to a PET chain?

PET stands for poly(ethylene terephthalate), also called polyethylene terephthalate.

PET is one particular polyester. The word polyester names the wider family, just as polyamide names the family containing nylon.

PET is built from two starting compounds:

In terephthalic acid, the two carboxylic acid groups are attached at opposite positions on the benzene ring.

During PET formation, the alcohol and carboxylic acid groups react to create ester bonds.

The chain therefore contains alternating glycol-derived and terephthalate-derived sections, connected through ester linkages.

Two recognisable parts of its structure are:

The glycol-derived section:

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

The terephthalate-derived section:

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

The connection between these sections includes the ester group, \(\ce{-C(=O)-O-}\).

For the full reaction that joins these monomers, see Condensation Polymerisation: Nylon and Polyester Explained for HSC Chemistry.

Follow one PET bond as it hydrolyses

Suppose an ester connection joins a terephthalate-derived section to an ethylene glycol-derived section.

Under acidic hydrolysis:

• The bond between the ester carbonyl carbon and the connecting oxygen breaks.
• The terephthalate-derived fragment gains a carboxylic acid group.
• The ethylene glycol-derived fragment gains an alcohol group.

That can happen at an internal connection, creating two shorter chains. Hydrolysis of the remaining ester connections can eventually produce terephthalic acid and ethylene glycol.

03How could an experiment show that polyester chains are breaking?

A piece of PET becoming smaller isn’t enough evidence of hydrolysis.

It might have fractured into tiny pieces, or material might have been lost during handling. Neither observation alone establishes that an ester bond has been broken.

The better question is: what chemical evidence would distinguish hydrolysis from a physical change?

The initial experimental design

Consider a proposed, teacher-supervised investigation comparing PET in hot water with PET in hot aqueous sodium hydroxide.

The apparatus uses heated reaction flasks connected to vertical reflux condensers.

Starting at the bottom of each apparatus:

1. A heated flask contains the liquid and a measured sample of PET.
2. Vapour rises from the heated solution into the vertical condenser.
3. Cooling water circulates around the condenser’s inner tube, entering at the lower connection and leaving at the upper connection.
4. Condensed vapour flows back down into the reaction flask, limiting solvent loss during heating.

A control flask contains PET and water without added sodium hydroxide. Another contains polyethene and aqueous sodium hydroxide.

Polyethene is useful as a comparison because its carbon-carbon backbone has no ester linkages.

What should the first comparison show?

If hydrolysis occurs to a substantial extent in the alkaline PET sample, some polymer material may be converted into smaller, soluble products.

The PET in water is expected to react much more slowly under otherwise comparable conditions.

The polyethene control should not undergo the corresponding ester hydrolysis because it has no ester bonds.

But there is a problem with this design.

Suppose a sample loses some of its original mass after filtration and drying. Does that prove covalent bonds were hydrolysed?

No. Material might have been lost during transfer, particles might have passed through the filter, or substances might have dissolved without chemical bond cleavage.

The experiment needs a way to identify chemical products.

The revised design: examine the liquid after treatment

A stronger investigation would use both the remaining solid and the filtered solution.

For alkaline PET hydrolysis, the expected dissolved products include ethylene glycol and terephthalate ions.

After separating the remaining polymer, the liquid can be acidified. Terephthalate ions gain protons and form terephthalic acid, which has low solubility in water and can precipitate.

The collected solid would then need to be identified, rather than merely described as a white precipitate. For example, its infrared spectrum could be compared with a suitable terephthalic acid reference.

These are expected chemical outcomes, not a claim that the proposed experiment has been performed or that particular yields were measured.

The measurements now have different jobs:

The revision answers the original weakness: it looks for chemical evidence instead of relying only on a change in appearance or recovered mass.

A practical limitation remains. PET hydrolysis can be slow, and its rate depends strongly on reaction conditions and surface accessibility. Hot corrosive solutions also require professional laboratory controls. This is a model of experimental reasoning, not a home experiment.

04Acidic and alkaline hydrolysis: same ester connection, different products

It’s easy to remember that hydrolysis reverses the formation of an ester.

That is a useful starting point, but it isn’t the full story.

The product containing the carboxyl group depends on the reaction conditions.

Hydrolysis under acidic conditions

The net reaction is:

\[ \ce{RCOOR' + H2O -> RCOOH + R'OH} \]

The products are a carboxylic acid and an alcohol.

An acid can catalyse the reaction. Ester hydrolysis under acidic conditions is reversible, so reaction conditions influence how far it proceeds.

For complete acidic hydrolysis of PET, the characteristic small-molecule products are:

• Terephthalic acid.
• Ethylene glycol.

Hydrolysis under alkaline conditions

In aqueous sodium hydroxide, hydroxide ions react with ester groups.

The net ionic reaction is:

\[ \ce{RCOOR' + OH- -> RCOO- + R'OH} \]

The products are a carboxylate ion and an alcohol.

The carboxylate ion remains paired with counterions in the solution. Sodium hydroxide therefore produces sodium carboxylate salts.

For complete alkaline hydrolysis of PET, the corresponding products are:

• Disodium terephthalate.
• Ethylene glycol.

If acid is added afterwards, terephthalate ions are protonated to give terephthalic acid.

A quick comparison

A common mistake is to describe the immediate alkaline products as terephthalic acid and ethylene glycol.

The glycol identification is correct, but the acid identification misses the pH.

In strongly alkaline solution, the carboxylic acid groups exist predominantly as carboxylate ions.

Acidification is a separate step.

05Does hydrolysis always shorten a polymer chain?

Here’s a useful contrasting case.

Suppose two different polymers each contain ester groups.

One has those ester groups joining the main chain. The other has ester groups hanging off a carbon-carbon backbone as side groups.

If both undergo ester hydrolysis, must both main chains break?

No. The location of the ester group determines the structural consequence.

Compare PET, poly(vinyl acetate), and polyethene

Poly(vinyl acetate) provides a particularly useful counterexample.

Its repeating backbone contains a carbon-carbon chain carrying acetate ester side groups.

Under alkaline hydrolysis, those side groups can be converted to alcohol groups on the polymer, with acetate ions released into solution.

But the carbon-carbon backbone can remain connected.

This gives you a useful decision rule for interpreting unfamiliar polymer structures:

1. Locate the ester functional group.
2. Trace the continuous covalent path along the main polymer backbone.
3. Decide whether the ester bond lies on that path or is attached as a side group.
4. If it lies on the backbone, its hydrolysis can cause chain scission.
5. If it is only in a side group, ester hydrolysis alone need not shorten the backbone.

The presence of oxygen atoms or ester groups somewhere in a polymer is not enough to predict depolymerisation.

The connectivity matters.

06Questions and solutions

Question 1

A single long, unbranched polyester molecule undergoes acidic hydrolysis at exactly one ester bond in the middle of its backbone.

No other chemical reactions occur.

(a) How many polymer-derived molecules result?

(b) Identify the two new functional groups formed at the broken connection.

(c) How many water molecules are consumed?

Solution 1

The hydrolysis produces two shorter molecules, creates one carboxylic acid group and one alcohol group, and consumes one water molecule.

(a) The original molecule contains one continuous backbone. Breaking one internal ester connection divides it into two separate molecular fragments.

(b) The net acidic hydrolysis reaction is:

\[ \ce{RCOOR' + H2O -> RCOOH + R'OH} \]

The fragment on the carbonyl side gains a carboxylic acid group, \(\ce{-COOH}\).

The fragment on the alcohol-derived side gains an alcohol group, \(\ce{-OH}\).

(c) Exactly one ester bond is hydrolysed, so one water molecule is consumed.

The important distinction is that hydrolysis produces two molecules from the original chain, but those molecules are not necessarily monomers. Most of the ester connections may still be intact.

Question 2

A teacher proposes an investigation comparing the hydrolysis behaviour of PET and polyethene.

Three samples, each initially weighing 0.500 g, are treated for the same period at the same controlled temperature.

The following results are illustrative data, not actual experimental measurements.

Assume the acid product in sample B has been reliably identified and all three samples were handled using equivalent filtration and drying methods.

(a) Which sample gives the strongest evidence that ester hydrolysis occurred?

(b) Explain why the difference in recovered mass is insufficient by itself to establish hydrolysis.

(c) Can the experiment establish that every ester bond in sample B has hydrolysed?

(d) Explain the purpose of sample C.

Solution 2

Sample B provides the strongest evidence of PET hydrolysis, but the data do not demonstrate complete depolymerisation.

(a) Sample B shows substantial loss of recovered solid and, more importantly, a liquid-phase product identified as terephthalic acid after acidification.

PET contains terephthalate units joined through ester bonds. Under alkaline hydrolysis, those connections can break and produce terephthalate ions, which are converted to terephthalic acid when acid is added.

Identifying that chemical product is much stronger evidence than simply noticing that less plastic remains.

(b) Sample B loses:

\[ \begin{aligned} \Delta m&=0.500-0.310\\ &=0.190\ \mathrm{g} \end{aligned} \]

As a percentage of the starting solid mass:

\[ \begin{aligned} \%\text{ loss} &=\frac{0.190}{0.500}\times100\\ &=38.0\% \end{aligned} \]

This is a 38.0% loss of recovered solid mass, not proof that 38.0% of the ester bonds were hydrolysed.

Some material could have been lost during physical handling. Conversely, ester bonds could have broken in the polymer remaining as solid without that material entering the solution.

Mass recovery and extent of bond cleavage are different measurements.

(c) Complete hydrolysis cannot be established.

The remaining 0.310 g might contain unreacted PET or partially hydrolysed polyester fragments. Even material already in the solution could include oligomers rather than only monomer-derived products.

To establish the extent of hydrolysis, the investigator would need additional information about the remaining ester bonds and the identity and amount of all relevant products.

(d) Sample C is a structural control.

Polyethene contains carbon-carbon backbone bonds but no ester groups.

Its lack of detected terephthalic acid under the same alkaline treatment supports the interpretation that the product identified in sample B originates from PET’s ester-containing structure.

The tempting shortcut is to treat solid mass loss as the entire result. The chemical identity of the products is what makes the interpretation much stronger.

Question 3

A 0.961 g sample of pure, high-molecular-mass PET is treated with 0.0120 mol of sodium hydroxide.

Assume all PET ester bonds undergo complete alkaline hydrolysis, the sodium hydroxide reacts only through ester hydrolysis, and the polymer’s original end groups make a negligible contribution.

The reaction mixture is then treated with hydrochloric acid.

Use:

• PET repeating-unit molar mass: \(192.2\ \mathrm{g\,mol^{-1}}\).
• Terephthalic acid molar mass: \(166.1\ \mathrm{g\,mol^{-1}}\).
• Hydrochloric acid concentration: \(0.1000\ \mathrm{mol\,L^{-1}}\).

(a) Calculate the amount of sodium hydroxide left after complete hydrolysis.

(b) Calculate the theoretical mass of terephthalic acid obtainable after acidification.

(c) Calculate the minimum volume of hydrochloric acid required to neutralise the excess hydroxide and protonate both carboxylate groups on every terephthalate ion.

For part (c), assume stoichiometric proton transfer is complete, ignore acid-base equilibria, and do not include any additional acid excess needed to maintain a low final pH.

(d) Explain why calculating the hydrochloric acid requirement from the leftover sodium hydroxide alone gives the wrong result.

Solution 3

The reaction leaves 0.00200 mol of sodium hydroxide, can theoretically produce 0.831 g of terephthalic acid, and requires at least 120 mL of 0.1000 mol L\(^{-1}\) hydrochloric acid under the stated assumptions.

(a) First calculate the amount of PET repeating units.

\[ \begin{aligned} n_{\mathrm{rep}} &=\frac{0.961}{192.2}\\ &=0.00500\ \mathrm{mol} \end{aligned} \]

The numerator is in grams and the molar mass in grams per mole.

Each repeating unit contributes two ester bonds in the high-molecular-mass approximation.

Therefore:

\[ \begin{aligned} n(\ce{NaOH})_{\mathrm{used}} &=2(0.00500)\\ &=0.0100\ \mathrm{mol} \end{aligned} \]

Subtract this from the initial amount:

\[ \begin{aligned} n(\ce{NaOH})_{\mathrm{left}} &=0.0120-0.0100\\ &=0.00200\ \mathrm{mol} \end{aligned} \]

(b) One mole of PET repeating units provides one mole of terephthalate-derived product.

After acidification:

\[ n_{\mathrm{acid}}=0.00500\ \mathrm{mol} \]

Using \(m=nM\):

\[ \begin{aligned} m_{\mathrm{acid}} &=(0.00500)(166.1)\\ &=0.831\ \mathrm{g} \end{aligned} \]

This is the theoretical maximum, assuming complete product recovery.

(c) Hydrochloric acid has two separate jobs.

First, it neutralises the remaining sodium hydroxide.

This requires 0.00200 mol of hydrochloric acid.

Second, it converts the terephthalate ions into terephthalic acid.

Each terephthalate ion has two negatively charged carboxylate groups, so it needs two protons to form the neutral dicarboxylic acid.

Therefore:

\[ \begin{aligned} n(\ce{HCl})_{\mathrm{salt}} &=2(0.00500)\\ &=0.0100\ \mathrm{mol} \end{aligned} \]

The total amount of hydrochloric acid required is:

\[ \begin{aligned} n(\ce{HCl})_{\mathrm{total}} &=0.00200+0.0100\\ &=0.0120\ \mathrm{mol} \end{aligned} \]

Using \(c=n/V\), where \(c\) is concentration in moles per litre and \(V\) is volume in litres:

\[ \begin{aligned} V&=\frac{n}{c}\\ &=\frac{0.0120}{0.1000}\\ &=0.120\ \mathrm{L}\\ &=120\ \mathrm{mL} \end{aligned} \]

(d) Neutralising the leftover hydroxide is not the same as completing the acidification of the reaction products.

The solution also contains disodium terephthalate. Its carboxylate groups must accept protons to produce terephthalic acid.

A calculation based only on the unused sodium hydroxide would overlook this second acid requirement.

The trap comes from treating all the acid added after hydrolysis as though its only purpose were to neutralise excess alkali.

Question 4

Two different polymers each begin as one long, unbranched molecule.

• Polymer A is a polyester containing ester connections in its main backbone.
• Polymer B is poly(vinyl acetate), with a carbon-carbon backbone and acetate ester side groups.

Under controlled alkaline conditions, exactly three ester groups are hydrolysed in each original molecule.

Assume no other reactions occur. Each hydrolysed ester in polymer A is an internal backbone connection, and polymer B’s carbon-carbon backbone remains intact.

All reaction mixtures are then acidified.

(a) Determine the number of polymer-backbone fragments produced from A and B.

(b) Count all organic product molecules formed from each original polymer molecule, including small molecules released from side groups.

(c) Explain how the two reactions can produce the same total number of organic molecules but have completely different effects on the polymer backbone.

(d) Would a decrease in the intensity of the ester carbonyl absorption in an infrared spectrum, by itself, prove that either backbone has become shorter?

Solution 4

Polymer A forms four separate backbone fragments, while polymer B retains one continuous polymer backbone. Counting all organic products after acidification, however, both reactions produce four molecules.

(a) Polymer A is initially one linear molecule.

Each of its three internal ester cleavages divides one existing backbone fragment into two.

Therefore:

\[ N_{\mathrm{fragments}}=1+3=4 \]

Polymer B has acetate ester groups attached to the carbon-carbon main chain.

Hydrolysing those groups changes the side-chain chemistry but does not break the carbon-carbon backbone.

It therefore remains one polymer-backbone molecule.

(b) Polymer A gives four backbone-derived molecules.

For polymer B, the three acetate ester groups are hydrolysed.

Each releases one acetate ion into alkaline solution, which becomes acetic acid after acidification.

The organic products are:

• One modified polymer molecule, with alcohol groups replacing the three acetate ester side groups.
• Three acetic acid molecules.

That is four organic molecules altogether.

(c) The total molecule counts are equal, but the connectivity is different.

In polymer A, the three cuts each divide the main chain. There are four separate pieces of the original backbone.

In polymer B, the original main chain is still one continuous molecule. Three small chemical groups have been detached from it.

This exposes a misconception: producing additional molecules during hydrolysis does not automatically mean that the polymer backbone has depolymerised.

You must identify exactly which bonds are broken and whether those bonds lie along the backbone.

(d) No. Loss of ester carbonyl absorption would provide evidence that ester functional groups were consumed, but it would not independently establish backbone shortening.

Both polymers could lose ester groups while only A experiences the specified backbone scission.

To distinguish the effects, the investigator could compare polymer chain-length or molar-mass distributions before and after treatment, using an appropriate polymer analysis method.

The decisive issue is not simply whether ester groups disappear. It is whether their disappearance disconnects the main chain.

Question 5

Extension challenge – beyond a typical HSC examination question

A chemist studies a batch containing only short, linear PET-derived oligomers made from ethylene glycol and terephthalic acid.

Every starting chain is unbranched and contains alternating glycol-derived and terephthalate-derived sections.

There are exactly two possible kinds of chain:

• Type G: both ends are ethylene glycol-derived units with free alcohol groups.
• Type T: both ends are terephthalic acid-derived units with free carboxylic acid groups.

There are no cyclic molecules, mixed-ended chains, free monomers, or side reactions.

Before hydrolysis, an independent analysis identifies 0.200 mmol of free carboxylic acid end groups.

The entire batch then undergoes complete alkaline hydrolysis. After the appropriate separation and acidification steps, product analysis establishes the following total amounts:

The sodium hydroxide measurement includes both ester hydrolysis and any neutralisation of pre-existing carboxylic acid groups.

(a) Determine the amount, in mmol, of starting Type G chains and Type T chains.

(b) Determine how many mmol of ester bonds were hydrolysed.

(c) Explain why the full 2.00 mmol of sodium hydroxide consumption cannot be attributed to ester-bond cleavage.

(d) A student claims that the recovered products prove the starting material was contaminated, because PET must contain equal numbers of ethylene glycol and terephthalic acid units. Evaluate the claim.

(e) Suppose the initial free carboxylic acid measurement had not been available. Could the total sodium hydroxide consumption replace it to determine the numbers of Type G and Type T chains? Explain.

Finally, state whether the supplied measurements uniquely determine the lengths of the individual starting chains.

Solution 5

The batch originally contained 0.500 mmol of Type G chains and 0.100 mmol of Type T chains. Exactly 1.80 mmol of ester bonds were hydrolysed; the remaining 0.200 mmol of sodium hydroxide consumption came from neutralising the original carboxylic acid end groups. The data determine the amounts of the two chain types, but not their individual lengths.

Part (a): Use the original acid end groups to count Type T chains.

Every Type T molecule has two free carboxylic acid ends.

Every Type G molecule has none.

The amount of original free carboxylic acid groups is 0.200 mmol.

Therefore:

\[ \begin{aligned} n_{\mathrm T} &=\frac{0.200}{2}\\ &=0.100\ \mathrm{mmol} \end{aligned} \]

Here, \(n_{\mathrm T}\) is the amount of Type T chains.

Now consider the monomers recovered after complete hydrolysis.

Every Type G chain contains exactly one more glycol unit than terephthalate unit.

For example, its shortest possible alternating sequence is:

\[ G-T-G \]

Here, \(G\) represents one glycol-derived section, and \(T\) represents one terephthalate-derived section.

Every Type T chain contains exactly one more terephthalate unit than glycol unit.

Its shortest sequence is:

\[ T-G-T \]

The measured excess of glycol over terephthalic acid is:

\[ \begin{aligned} \Delta n&=1.40-1.00\\ &=0.400\ \mathrm{mmol} \end{aligned} \]

Let \(n_{\mathrm G}\) be the amount of Type G chains.

The glycol excess equals the contribution from Type G chains minus the opposite contribution from Type T chains:

\[ n_{\mathrm G}-n_{\mathrm T}=0.400 \]

Substitute \(n_{\mathrm T}=0.100\ \mathrm{mmol}\):

\[ \begin{aligned} n_{\mathrm G} &=0.400+0.100\\ &=0.500\ \mathrm{mmol} \end{aligned} \]

So there were:

• 0.500 mmol of Type G chains.
• 0.100 mmol of Type T chains.

The total original amount of polymer molecules was:

\[ \begin{aligned} n_{\mathrm{chains}} &=0.500+0.100\\ &=0.600\ \mathrm{mmol} \end{aligned} \]

Part (b): Count the ester connections that originally joined the molecular sections.

After complete hydrolysis, the total number of monomer-derived sections is represented by:

\[ \begin{aligned} n_{\mathrm{sections}} &=1.40+1.00\\ &=2.40\ \mathrm{mmol} \end{aligned} \]

Each starting molecule was a linear sequence with no cycles.

For one linear molecule made by connecting \(k\) originally separate sections, the number of connections is \(k-1\).

Across the whole batch, the number of ester connections is therefore the total number of sections minus the total number of starting chains:

\[ \begin{aligned} n_{\mathrm{ester}} &=2.40-0.600\\ &=1.80\ \mathrm{mmol} \end{aligned} \]

Complete hydrolysis breaks all these ester bonds.

So 1.80 mmol of ester bonds were hydrolysed.

Part (c): Account for the hydroxide consumed without breaking a bond.

The original Type T chains already contained free carboxylic acid groups.

When aqueous sodium hydroxide was added, these groups were neutralised:

\[ \ce{RCOOH + OH- -> RCOO- + H2O} \]

This reaction consumes hydroxide but does not break an ester linkage.

The original acid end groups consumed 0.200 mmol of hydroxide.

Ester hydrolysis consumed another 1.80 mmol.

Together:

\[ \begin{aligned} n(\ce{OH-})_{\mathrm{total}} &=1.80+0.200\\ &=2.00\ \mathrm{mmol} \end{aligned} \]

This agrees with the measured reagent consumption.

The tempting mistake is to assume that every hydroxide ion consumed represents one broken ester bond.

That assumption fails when the starting polymer already contains acidic groups.

Part (d): Explain the unequal monomer amounts.

The student’s contamination claim is incorrect.

For a long PET chain, using a 1:1 ratio of ethylene glycol-derived and terephthalate-derived units is often a useful approximation.

But a short chain has end groups, and those ends affect the exact monomer counts.

A Type G chain contains one extra glycol unit. A Type T chain contains one extra terephthalate unit.

The measured 0.400 mmol excess of glycol follows directly from the difference between the amounts of the two types of chain.

There is no need to invoke contamination.

Part (e): Decide whether hydroxide consumption could replace the acid end-group measurement.

It could not.

This is the key constraint that makes the problem more difficult than an ordinary hydrolysis calculation.

Every terephthalate-derived section contains two carboxyl-derived positions.

Before hydrolysis, each position is either part of an ester connection or already a free carboxylic acid group.

During complete alkaline treatment:

• Each ester connection consumes one hydroxide ion through hydrolysis.
• Each original free carboxylic acid group consumes one hydroxide ion through neutralisation.

Therefore, the total hydroxide consumption is always two equivalents per terephthalate-derived section under the stated assumptions.

Using the measured terephthalic acid amount:

\[ \begin{aligned} n(\ce{OH-})_{\mathrm{total}} &=2n_{\mathrm{terephthalate}}\\ &=2(1.00)\\ &=2.00\ \mathrm{mmol} \end{aligned} \]

This result is true regardless of how those two carboxyl-derived positions were divided between original acid end groups and ester bonds.

The sodium hydroxide measurement therefore does not provide an independent way to determine how many starting chains had acid ends.

Without the initial free acid end-group measurement, different combinations of Type G and Type T chains could satisfy the measured monomer totals and the same hydroxide consumption.

Can we determine the lengths of all the original chains?

No.

Even with the initial acid end-group result, we know the total amounts of the two chain types, not how the monomer-derived sections were distributed among individual chains.

For example, the following two arrangements both satisfy every measurement.

Arrangement A:

• 0.500 mmol of chains with sequence \(G-T-G\).
• 0.100 mmol of chains with sequence \(T-G-T-G-T-G-T-G-T\).

The glycol total is:

\[ 2(0.500)+4(0.100) =1.40\ \mathrm{mmol} \]

The terephthalate total is:

\[ 1(0.500)+5(0.100) =1.00\ \mathrm{mmol} \]

Arrangement B:

• 0.300 mmol of chains with sequence \(G-T-G\).
• 0.200 mmol of chains with sequence \(G-T-G-T-G\).
• 0.100 mmol of chains with sequence \(T-G-T-G-T\).

These give:

\[ \begin{aligned} n(G) &=0.300(2)+0.200(3)\\ &\quad+0.100(2)\\ &=1.40\ \mathrm{mmol} \end{aligned} \]

And:

\[ \begin{aligned} n(T) &=0.300(1)+0.200(2)\\ &\quad+0.100(3)\\ &=1.00\ \mathrm{mmol} \end{aligned} \]

Both arrangements have 0.500 mmol of glycol-terminated chains, 0.100 mmol of acid-terminated chains, 1.80 mmol of ester bonds, and 0.200 mmol of original acid end groups.

They therefore give the same total sodium hydroxide consumption of 2.00 mmol.

But their chain lengths are different.

This reveals a limitation of complete hydrolysis as an analytical method: destroying every ester connection can tell us which building blocks were present and how much of each was present, while erasing information about their original sequence and distribution among chains.

To resolve the original chain-length distribution, an investigator would need additional structural or molar-mass evidence from the material before hydrolysis.

06Using hydrolysis products to reason backwards

When an unfamiliar polyester hydrolysis question appears, the most reliable approach is to work from connectivity rather than memorising one product equation.

First identify whether the hydrolysed ester groups are part of the polymer backbone. Then determine whether the conditions are acidic or alkaline.

Next, trace the fragments on both sides of each broken ester connection. Work out whether those fragments remain oligomers or become complete monomer-derived products.

For quantitative problems, count the actual hydrolysed ester bonds whenever possible. Don’t automatically equate the number of repeating units, starting monomers, or hydroxide ions consumed with the number of chain scissions.

The next useful skill is to reverse this process: take the structures of a diol and a dicarboxylic acid, identify the two sites that can form ester bonds, and reconstruct the polymer. You can practise that with How to Draw PET from Ethylene Glycol and Terephthalic Acid.

The same bond that joins the monomers during condensation is the connection to examine when explaining how hydrolysis breaks the resulting polyester chain.