How to Find Polyester Monomers from a Repeating Unit

A polyester repeating unit is shown in an HSC Chemistry question, but the monomers are missing. Where should you break the structure to get them back?

You might try cutting the polymer at the square brackets. That seems reasonable: the brackets mark one repeating unit. But the brackets don’t necessarily mark where two monomers originally joined.

The reliable method is to find the ester linkages, break the bond between each carbonyl carbon and its neighbouring single-bonded oxygen, and restore the missing \(\ce{-OH}\) and \(\ce{-H}\) groups.

For a typical polyester, this reveals two monomers: a diol and a dicarboxylic acid. Other polyesters can come from a single hydroxycarboxylic acid, so you must follow the bonds rather than assume there are always two monomers.

01Find the ester linkage before cutting anything

A polyester is a family of polymers containing ester linkages in its backbone. Polyethylene terephthalate (PET), the material used in many drink bottles and textile fibres, is one example of this family.

The important structural feature is:

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

Look carefully at the carbonyl carbon, which is the carbon atom double-bonded to oxygen.

It has two different bonds to oxygen:

• A double bond, \(\ce{C=O}\), which forms part of the carbonyl group.
• A single bond, \(\ce{C-O}\), connecting the carbonyl carbon to another oxygen atom.

When reversing an ordinary ester-forming condensation reaction, you break the second bond: the single bond between the carbonyl carbon and the neighbouring oxygen.

You leave the carbonyl double bond untouched.

How to reverse an ester linkage

Consider a section of a polymer:

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

Here, \(R\) and \(R'\) represent the parts of the molecule attached on either side.

To reconstruct the parent acid and alcohol:

1. Break the single bond between the carbonyl carbon and the oxygen immediately beside it.
2. Attach \(\ce{-OH}\) to the carbonyl carbon, restoring a carboxylic acid group, \(\ce{-COOH}\).
3. Attach \(\ce{H}\) to the separated oxygen, restoring an alcohol group, \(\ce{-OH}\).

The resulting fragments have the structures:

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

This is the reverse structural change of ester formation. In hydrolysis, water provides the elements needed to restore these groups.

A useful test is to imagine joining your proposed monomers again. If removing water and reconnecting their functional groups reproduces the original polymer backbone, you’ve made a consistent deduction.

But if you accidentally cut a carbon-carbon bond, or split an ether linkage that has no adjacent carbonyl group, your proposed monomers won’t reconstruct the given structure.

02Decide whether the polyester comes from one monomer or two

The number and arrangement of functional groups matter more than the number of atoms in the repeating unit.

Case A: A diol and a dicarboxylic acid

A common polyester repeating unit has this pattern:

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

The right-hand end joins the oxygen at the beginning of the next repeating unit.

When you break both ester linkages surrounding the two backbone sections, you obtain:

Diol:

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

Dicarboxylic acid:

\[ \ce{HOOC-R'-COOH} \]

The diol has two alcohol functional groups. The dicarboxylic acid has two carboxylic acid groups.

Each molecule can therefore form ester linkages at both ends, allowing a long chain to develop.

Case B: A hydroxycarboxylic acid

Now consider this different repeating pattern:

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

It has one ester linkage per smallest repeating unit.

What happens if you reverse that linkage?

You obtain:

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

This is one molecule containing both an alcohol group and a carboxylic acid group.

It can react with another molecule of the same type, joining the molecules together through ester linkages.

You don’t need to invent a separate diol and dicarboxylic acid.

The decision table

These patterns cover the main cases. A repeating unit containing two different hydroxycarboxylic acid residues can also have two ester linkages, so the total number of ester bonds alone isn’t enough to identify the monomer types.

03Recognise the PET repeating unit

PET provides a particularly useful example because its repeating unit contains both an aliphatic section and an aromatic ring.

Its structure can be written as:

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

The two carbonyl groups are attached at positions 1 and 4 of the benzene ring.

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

Before identifying the monomers, look at the two different sections.

The \(\ce{-O-CH2-CH2-O-}\) section comes from a diol. Restoring hydrogen atoms to its outer oxygens produces:

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

This is ethane-1,2-diol, commonly called ethylene glycol.

The aromatic section contains a benzene ring attached to two carbonyl carbons. Restoring the carboxylic acid groups produces:

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

With the carboxylic acid groups at positions 1 and 4, this is benzene-1,4-dicarboxylic acid, commonly called terephthalic acid.

Therefore, the conventional structural parent monomers of PET are ethane-1,2-diol and terephthalic acid.

Notice something subtle: the molecular formula \(\ce{C6H4}\) doesn’t tell you which positions on the benzene ring have the carbonyl groups attached.

An otherwise similar polyester using benzene-1,3-dicarboxylic acid would have a different connectivity. You cannot call every polyester containing a benzene ring PET.

This distinction is useful when interpreting abbreviated structural formulae. You need enough structural information to establish the positions of the groups, not just the numbers of atoms.

For the broader explanation of how these reactions build polymer chains, see Condensation Polymerisation: Nylon and Polyester Explained for HSC Chemistry.

04How could an experiment test the predicted monomers?

Finding monomers from a repeating unit is a structural deduction. But can a chemical reaction provide evidence that the deduction is correct?

Suppose an investigator has an unidentified polyester film and proposes that it is PET.

The unanswered question is whether breaking the polymer’s ester bonds produces the expected ethane-1,2-diol and terephthalate-derived compounds.

The proposed apparatus and first prediction

One possible laboratory investigation uses alkaline hydrolysis.

Imagine a round-bottom reaction flask containing small pieces of the polymer and aqueous sodium hydroxide. The flask is heated, and a vertical reflux condenser is attached above it.

Cooling water enters the lower connection of the condenser and leaves through the upper connection. Vapour rising from the heated mixture condenses inside the condenser and returns to the flask.

This allows the mixture to be heated while limiting solvent loss.

What should the products be if the polymer really is PET?

The first prediction might be ethane-1,2-diol and terephthalic acid.

But there’s a complication: the reaction conditions matter.

In aqueous sodium hydroxide, hydrolysis of the ester linkages produces the alcohol and carboxylate salts rather than leaving the dicarboxylic acid in its protonated form.

For complete alkaline hydrolysis of PET, the expected organic products are:

• Ethane-1,2-diol
• Disodium benzene-1,4-dicarboxylate, also called disodium terephthalate

Each ester bond hydrolysed consumes one mole of sodium hydroxide in this idealised reaction.

What the first design cannot establish

Suppose the investigator looks only for free terephthalic acid in the alkaline reaction mixture.

Failing to find it would not prove the original monomer prediction was wrong. The acid has been converted to its carboxylate salt.

A better approach is to acidify a separate portion of the hydrolysis mixture after the reaction.

The carboxylate groups gain hydrogen ions, allowing terephthalic acid to be recovered.

The investigator would then need to identify the recovered compound, rather than relying only on the presence of a precipitate.

A white precipitate alone cannot establish whether the acid is benzene-1,4-dicarboxylic acid or a different aromatic dicarboxylic acid.

Likewise, detecting ethane-1,2-diol would support the proposed diol-derived section but would not, by itself, establish the exact original polymer structure.

This is why the structural formula and the chemical evidence complement each other.

A mass check on the PET interpretation

The expected composition also provides a numerical check.

One mole of ethane-1,2-diol has a molar mass of approximately \(62.1\ \mathrm{g\,mol^{-1}}\).

One mole of terephthalic acid has a molar mass of approximately \(166.1\ \mathrm{g\,mol^{-1}}\).

In a long, ideal PET chain, each effective pair of these monomer residues corresponds to the loss of approximately two water molecules during condensation.

Using \(M\) for molar mass:

\[ \begin{aligned} M_{\text{repeat}} &=62.1+166.1-2(18.0)\\ &=192.2\ \mathrm{g\,mol^{-1}} \end{aligned} \]

This agrees with the formula \(\ce{C10H8O4}\) for the PET repeating unit.

For an illustrative \(1.92\ \mathrm{g}\) sample, the amount of repeating-unit equivalents is:

\[ \begin{aligned} n_{\text{repeat}} &=\frac{1.92}{192.2}\\ &=0.00999\ \mathrm{mol}\\ &\approx0.0100\ \mathrm{mol} \end{aligned} \]

Here, \(n_{\text{repeat}}\) represents the amount in moles of repeating-unit equivalents, not the number of separate polymer molecules.

With approximately two ester bonds per repeat, complete alkaline hydrolysis would require:

\[ \begin{aligned} n_{\ce{NaOH}} &\approx2(0.0100)\\ &=0.0200\ \mathrm{mol} \end{aligned} \]

This calculation assumes very long linear chains, complete hydrolysis, and negligible effects from chain ends.

It is a theoretical amount, not a measured result or an experimentally established yield.

The experimental limitation is also important: hydrolysis products tell you which structural fragments were present, but they don’t necessarily prove which exact chemicals were used to manufacture the polymer. Some polyesters can be produced using alternative acid derivatives or other polymerisation pathways.

05Common mistakes when reversing polyester structures

A final distinction: an ester linkage is \(\ce{-C(=O)-O-}\), whereas an amide linkage contains \(\ce{-C(=O)-NH-}\) or another nitrogen-containing amide group.

If you find an amide linkage in the backbone, don’t automatically apply the polyester method. You may be looking at a polyamide, such as nylon.

06Questions and solutions

Question 1

An ideal linear polyester has the repeating unit:

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

Identify its two structural parent monomers. Write their condensed structural formulae and systematic names.

Solution 1

The monomers are propane-1,3-diol, \(\ce{HO-CH2-CH2-CH2-OH}\), and propanedioic acid, \(\ce{HOOC-CH2-COOH}\).

The diol-derived part contains three carbon atoms between two ester oxygens. Restoring hydrogen atoms produces an alcohol group at each end:

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

This is propane-1,3-diol.

The acid-derived part contains one methylene group between two carbonyl carbons. Restoring hydroxyl groups at those carbonyl carbons gives:

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

There are three carbon atoms in total, so this is propanedioic acid.

A common mistake is to count only the methylene carbon in the second fragment, forgetting that both carboxyl carbon atoms belong to the parent acid.

Reconnecting these monomers through ester bonds gives the original polyester structure.

Question 2

A polyester has the repeating unit:

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

An investigator claims that the three oxygen atoms in the first section must produce three separate alcohol-containing fragments during hydrolysis.

Assume that complete alkaline hydrolysis cleaves ester linkages but leaves ether linkages intact.

(a) Identify the correct structural parent monomers.

(b) Calculate the amount of sodium hydroxide required to completely hydrolyse \(0.0200\ \mathrm{mol}\) of repeating-unit equivalents, neglecting chain ends.

(c) Explain why the investigator’s proposed three-fragment interpretation is incorrect.

Solution 2

The correct parent monomers are 2,2′-oxydiethanol (diethylene glycol) and propanedioic acid. The repeating-unit equivalents require approximately \(0.0400\ \mathrm{mol}\) of sodium hydroxide for complete alkaline hydrolysis.

(a) Identify the monomers

The alcohol-derived section is:

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

Only the two outer oxygens belong to the ester linkages.

The middle oxygen connects two carbon atoms and is part of an ether linkage.

Restoring the two terminal alcohol groups produces:

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

This is 2,2′-oxydiethanol, commonly called diethylene glycol.

The acid-derived section gives:

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

This is propanedioic acid.

In alkaline solution, the fully hydrolysed acid-derived product exists as its propanedioate salt. Acidification restores the neutral dicarboxylic acid.

(b) Calculate sodium hydroxide consumption

There are two ester linkages per effective repeating unit.

Under the stated conditions, one mole of sodium hydroxide is consumed per mole of ester bonds hydrolysed.

\[ \begin{aligned} n_{\ce{NaOH}} &=2n_{\text{repeat}}\\ &=2(0.0200)\\ &=0.0400\ \mathrm{mol} \end{aligned} \]

The theoretical requirement is \(0.0400\ \mathrm{mol}\) of sodium hydroxide.

(c) Evaluate the interpretation

The investigator has confused the presence of oxygen with the presence of an ester group.

An ester contains an oxygen directly attached to a carbonyl carbon. The middle oxygen in the diol-derived section has carbon atoms on both sides and is not attached to a carbonyl carbon.

Because the problem specifies that ether linkages remain intact, the middle oxygen stays within the diethylene glycol molecule.

Breaking it would change the predicted product identities and would require a different reaction.

This is why a carbonyl-based identification rule is more reliable than counting oxygen atoms.

Question 3

Two aromatic polyesters, P and Q, have the same abbreviated repeating-unit formula:

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

However, the ring connections differ:

• In P, the carbonyl-containing substituents attach to benzene positions 1 and 4.
• In Q, they attach to benzene positions 1 and 3.

An investigator completely hydrolyses \(1.92\ \mathrm{g}\) of each polymer and acidifies the products. Approximately \(1.66\ \mathrm{g}\) of aromatic dicarboxylic acid is recovered from each sample.

Assume quantitative recovery, negligible chain-end effects, and a polyester repeating-unit molar mass of \(192.2\ \mathrm{g\,mol^{-1}}\).

(a) Identify the dicarboxylic acid obtained from P and from Q.

(b) Show why the recovered masses could be the same.

(c) Another investigator concludes that both samples must be PET. Assess this conclusion and state what additional structural evidence would distinguish them.

Solution 3

Only P is PET. P contains a benzene-1,4-dicarboxylic acid residue, while Q contains a benzene-1,3-dicarboxylic acid residue. Both can give the same calculated mass of acid because these acids are structural isomers with identical molecular formulae.

(a) Identify the acids

Hydrolysis preserves the carbon skeleton and substituent positions of the aromatic acid-derived section.

For P, the acid is benzene-1,4-dicarboxylic acid, or terephthalic acid.

For Q, the acid is benzene-1,3-dicarboxylic acid, or isophthalic acid.

Both acids have the formula:

\[ \ce{C8H6O4} \]

(b) Calculate the recovered masses

The amount of repeating-unit equivalents in each sample is:

\[ \begin{aligned} n_{\text{repeat}} &=\frac{1.92}{192.2}\\ &=0.00999\ \mathrm{mol}\\ &\approx0.0100\ \mathrm{mol} \end{aligned} \]

One repeat-equivalent contains one aromatic dicarboxylic acid residue.

Both acids have a molar mass of approximately \(166.1\ \mathrm{g\,mol^{-1}}\).

Therefore:

\[ \begin{aligned} m_{\text{acid}} &=nM\\ &=0.0100(166.1)\\ &=1.66\ \mathrm{g} \end{aligned} \]

The expected mass is \(1.66\ \mathrm{g}\) in both cases.

(c) Assess the PET conclusion

The conclusion is incorrect because molecular formula and molar mass don’t establish the positions of groups on a benzene ring.

PET specifically contains the terephthalate residue with its ring substituents in the 1,4 arrangement.

Polymer Q has a different backbone connectivity, even though its elemental composition and theoretical hydrolysis-product masses are the same.

Evidence identifying the substitution pattern of the isolated aromatic acid, such as suitable structural spectroscopy or comparison with authenticated reference compounds, could distinguish them.

The important constraint is connectivity. Two molecules can have the same numbers of carbon, hydrogen, and oxygen atoms but different arrangements of those atoms.

Question 4

A \(9.61\ \mathrm{g}\) sample of high-molar-mass PET undergoes incomplete alkaline hydrolysis.

The reaction consumes \(0.0750\ \mathrm{mol}\) of sodium hydroxide, and all of this consumption is due to cleavage of ester bonds.

Assume:

• The polymer has very long linear chains, so chain-end effects can be neglected.
• Each ester bond broken consumes one mole of sodium hydroxide.
• Each terephthalate residue initially has two ester bonds connecting it to the chain.
• The remaining intact ester bonds may be distributed anywhere along the chains.

Use a PET repeating-unit molar mass of \(192.2\ \mathrm{g\,mol^{-1}}\).

(a) Calculate the amount of ester bonds initially present and the amount remaining intact.

(b) Determine the minimum and maximum possible amounts, in moles, of fully detached terephthalate residues.

(c) Calculate the corresponding range of terephthalic acid masses obtainable from these fully detached residues after acidification and ideal separation. Use \(M=166.1\ \mathrm{g\,mol^{-1}}\).

(d) Explain why the fraction of ester bonds hydrolysed does not uniquely determine the fraction of terephthalate residues fully released.

Solution 4

Initially, approximately \(0.100\ \mathrm{mol}\) of ester bonds were present, and \(0.0250\ \mathrm{mol}\) remain intact. Between \(0.0250\) and \(0.0375\ \mathrm{mol}\) of terephthalate residues could be fully detached, depending on where the bonds were broken.

The surprising feature is that knowing exactly how many ester bonds reacted does not tell us exactly how many complete acid-derived units have been released.

(a) Find the number of ester bonds

The amount of PET repeating-unit equivalents is:

\[ \begin{aligned} n_{\text{repeat}} &=\frac{9.61}{192.2}\\ &=0.0500\ \mathrm{mol} \end{aligned} \]

Each repeat-equivalent contains approximately two ester bonds.

\[ \begin{aligned} n_{\text{bonds, initial}} &=2(0.0500)\\ &=0.100\ \mathrm{mol} \end{aligned} \]

Since \(0.0750\ \mathrm{mol}\) of ester bonds have been cleaved:

\[ \begin{aligned} n_{\text{bonds, intact}} &=0.100-0.0750\\ &=0.0250\ \mathrm{mol} \end{aligned} \]

Thus, 75% of the ester bonds have been broken, leaving 25% intact.

(b) Determine the possible number of fully detached residues

There are \(0.0500\ \mathrm{mol}\) of terephthalate residues.

For a terephthalate residue to become completely detached, both ester bonds connecting it to the rest of the chain must be broken.

If even one of its two ester bonds remains intact, the residue is still attached to another part of the polymer or an oligomer.

First consider the minimum possible number released.

To keep as many terephthalate residues attached as possible, distribute the \(0.0250\ \mathrm{mol}\) of intact bonds so that each remaining bond is attached to a different terephthalate residue.

Then \(0.0250\ \mathrm{mol}\) of terephthalate residues remain attached.

The minimum amount fully detached is:

\[ \begin{aligned} n_{\min} &=0.0500-0.0250\\ &=0.0250\ \mathrm{mol} \end{aligned} \]

Now consider the maximum possible number released.

To keep as few terephthalate residues attached as possible, concentrate the intact bonds in pairs, leaving both ester bonds intact on the same residue wherever possible.

Each such attached residue accounts for two intact bonds.

\[ \begin{aligned} n_{\text{attached,min}} &=\frac{0.0250}{2}\\ &=0.0125\ \mathrm{mol} \end{aligned} \]

Therefore, the maximum amount fully detached is:

\[ \begin{aligned} n_{\max} &=0.0500-0.0125\\ &=0.0375\ \mathrm{mol} \end{aligned} \]

So the range is:

\[ 0.0250\leq n\leq0.0375 \]

where \(n\) is the amount, in moles, of fully detached terephthalate residues.

(c) Calculate the possible acid masses

After acidification, each fully detached terephthalate residue can produce one terephthalic acid molecule.

Minimum mass:

\[ \begin{aligned} m_{\min} &=0.0250(166.1)\\ &=4.15\ \mathrm{g} \end{aligned} \]

Maximum mass:

\[ \begin{aligned} m_{\max} &=0.0375(166.1)\\ &=6.23\ \mathrm{g} \end{aligned} \]

The theoretical range is therefore \(4.15\) to \(6.23\ \mathrm{g}\), assuming the acid derived from fully detached residues is recovered completely and separately from other products.

(d) Explain the limitation

A count of hydrolysed bonds measures how many ester connections have been broken.

It does not specify which connections were broken.

Two samples can consume the same amount of sodium hydroxide but have different distributions of chain breaks, different oligomer lengths, and different amounts of completely liberated monomer-derived fragments.

The tempting approach is to say that 75% of the ester bonds were broken, so exactly 75% of the acid residues must have been recovered. But releasing one acid residue requires breaking both bonds attached to it.

Those two conditions are not equivalent.

Question 5

An ideal polyester has the perfectly alternating repeating unit:

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

Two chemists disagree about the structural parent monomers.

Chemist A proposes:

• Ethane-1,2-diol, \(\ce{HO-CH2-CH2-OH}\)
• 2-Methylpropanedioic acid, \(\ce{HOOC-CH(CH3)-COOH}\)

Chemist B proposes:

• 3-Hydroxypropanoic acid, \(\ce{HO-CH2-CH2-COOH}\)
• 2-Hydroxypropanoic acid, \(\ce{HO-CH(CH3)-COOH}\)

Each chemist points out that their proposed pair has a combined molecular formula of \(\ce{C6H12O6}\).

Both also calculate that eliminating two water molecules would give a repeating-unit formula of \(\ce{C6H8O4}\).

An investigation then completely hydrolyses a sample of the polyester under alkaline conditions, followed by acidification.

Structural analysis identifies two products:

• 3-Hydroxypropanoic acid
• 2-Hydroxypropanoic acid

They are present in equal mole amounts.

Assume ordinary ester formation or hydrolysis does not rearrange the carbon skeleton, and that other functional groups remain unchanged.

(a) Which chemist’s proposed monomer set matches the specified polymer structure? Explain why the other set fails despite satisfying the molecular formula and mass balance.

(b) Does the hydrolysis result alone prove that a separately obtained, structurally unknown polyester has the perfectly alternating sequence shown above?

(c) If equal amounts of the two hydroxycarboxylic acids are mixed and allowed to undergo ordinary condensation polymerisation without sequence control, is the perfectly alternating polymer guaranteed? Explain.

(d) Can the structural and hydrolysis evidence prove that the original polymer was manufactured directly from those two hydroxycarboxylic acids?

Solution 5

Chemist B’s structural parent monomers match the specified polymer. Chemist A’s proposed pair has the correct overall atomic composition but would produce a different molecular connectivity. The hydrolysis products support Chemist B’s structural assignment, but they cannot independently prove the original chain sequence or manufacturing route.

(a) Identify the correct monomers

The central issue is where each carbonyl carbon connects to the neighbouring atoms.

Consider the given repeat:

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

followed by:

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

Each of these sections begins with an oxygen and ends with a carbonyl carbon.

Breaking the ester linkages between successive sections restores an alcohol group at one end and a carboxylic acid at the other.

The first resulting molecule is:

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

This is 3-hydroxypropanoic acid.

The second is:

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

This is 2-hydroxypropanoic acid.

Both match Chemist B’s proposed molecules and the identified hydrolysis products.

Now test Chemist A’s proposal.

Ethane-1,2-diol and 2-methylpropanedioic acid would produce the following sequence through normal condensation:

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

followed by:

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

This is different from the specified structure.

In Chemist A’s polymer, the \(\ce{CH2-CH2}\) section is connected to an oxygen at each end. In the specified polymer, the \(\ce{CH2-CH2}\) section has an oxygen at one end and a carbonyl carbon at the other.

No amount of rearranging the repeating-unit brackets can change that chemical connectivity.

Yet the atom count works for both proposals.

Chemist A’s monomers have formulae:

\[ \ce{C2H6O2} \quad\text{and}\quad \ce{C4H6O4} \]

Chemist B’s monomers each have the formula:

\[ \ce{C3H6O3} \]

Both pairs sum to \(\ce{C6H12O6}\).

Subtracting the elements of two water molecules gives \(\ce{C6H8O4}\) in either case.

This demonstrates why an empirical or molecular formula is not enough to reconstruct a polymer.

It preserves the numbers of atoms, but it does not preserve their arrangement.

(b) Can hydrolysis prove perfect alternation?

No. Complete hydrolysis destroys the sequence information.

Imagine two chains containing the same two hydroxycarboxylic acid residues.

One chain is arranged in a perfectly alternating sequence:

\[ \mathrm{A-B-A-B-A-B} \]

Another contains a different arrangement:

\[ \mathrm{A-A-B-A-B-B} \]

Here, A and B represent the two different monomer-derived residues.

Both chains contain three A residues and three B residues.

If every ester linkage is completely hydrolysed, both can produce the same two hydroxycarboxylic acids in equal mole amounts.

The hydrolysis products establish the identities and overall amounts of the fragments, but not their original order along the chain.

In the question, the perfectly alternating repeat is supplied as a structural fact. The hydrolysis results agree with it, but hydrolysis alone would not establish that alternation for a separate unknown sample.

(c) Does equal monomer composition guarantee alternation?

No. Equal amounts of two hydroxycarboxylic acids do not force them to join in alternating order.

Each type contains both an alcohol group and a carboxylic acid group.

Consequently, under ordinary condensation conditions without sequence control, the following connections are all chemically possible:

• A joining to another A
• A joining to B
• B joining to A
• B joining to another B

A reaction mixture with equal initial amounts of A and B can therefore form chains containing different sequences.

An equal monomer ratio describes composition, not sequence.

To guarantee the particular alternating structure in the question, some form of sequence control or a suitably organised precursor would be required.

(d) Can we identify the original manufacturing route?

No. The evidence identifies structural parent monomers and is consistent with those fragments being incorporated into the polymer.

But the same connectivity could potentially be constructed using suitably activated derivatives or preassembled structures, rather than directly condensing the two free hydroxycarboxylic acids.

The specific polymerisation route cannot be recovered from the repeating unit and hydrolysis products alone.

This is the deepest limitation of reversing a polymer structure:

The final connectivity can identify chemically consistent parent fragments without uniquely identifying the synthetic history or, after complete hydrolysis, the original sequence of those fragments.

07Where this method leads next

Once you’ve identified the monomers, the next useful step is to predict what happens when their functional groups react. A diol and a dicarboxylic acid can build a polyester through repeated ester formation, while a diamine and a dicarboxylic acid form a polyamide.

The distinction also helps explain why different polymers behave differently. An ester linkage contains oxygen atoms but no nitrogen-hydrogen bond. An amide linkage can contain an \(\ce{N-H}\) group, which changes the hydrogen-bonding interactions available between neighbouring polymer chains.

If you’re ready to connect the structural identification method to polymer properties, read How to Compare Hydrogen Bonding in Polyamides and Polyesters.

For any new repeating-unit question, begin with the same decision: locate the actual ester bonds, preserve the carbon skeleton, and only then reconstruct the functional groups of the parent molecules.