How to Identify Which Atoms Leave During Condensation Polymerisation

A polymer repeat unit contains the link \ce{-C(=O)-O-}. Which oxygen was removed when that ester link formed?

It is tempting to point at the oxygen between the carbonyl carbon and the next carbon chain. But that oxygen is still in the polymer. For the standard reaction between a carboxylic acid and an alcohol, the atoms removed are the carboxylic acid’s -OH group and one hydrogen from the alcohol’s -OH group. Together, they form water.

The same idea applies to polyamides such as nylon. The carboxylic acid’s -OH leaves with one hydrogen from an amine, while the nitrogen stays in the polymer’s \ce{-C(=O)-NH-} linkage.

The key rule: find the new bond in the polymer, then work backwards to restore the functional groups. Don’t simply remove every oxygen or hydrogen you can see.

01Start by identifying the polymer linkage

Before deciding which atoms have left, identify the type of chemical bond connecting the monomer residues.

Two important linkages appear in the standard HSC condensation polymer examples.

Polyester: look for an ester linkage

An ester linkage has the structure:

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

It forms when a carboxylic acid reacts with an alcohol.

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

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

Notice what happens to the atoms:

• The carboxylic acid loses its -OH group.
• The alcohol loses one hydrogen atom.
• These atoms form \(\ce{H2O}\).
• The alcohol’s oxygen remains in the polymer, connecting the two carbon-containing sections.
• The carbonyl oxygen also remains.

This last point is particularly important. Neither oxygen in the finished ester linkage is the oxygen removed as water.

Polyamide: look for an amide linkage

An amide linkage has the structure:

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

It forms when a carboxylic acid reacts with an amine.

\[ \ce{R-COOH + H2N-R' -> R-CONH-R' + H2O} \]

The changes are:

• The carboxylic acid loses its -OH group.
• The primary amine loses one hydrogen.
• These atoms form \(\ce{H2O}\).
• The nitrogen remains in the polymer.
• The carbonyl oxygen remains.

The nitrogen originally belonged to the amine functional group. It becomes part of the amide linkage.

Quick comparison

For both types, the oxygen in the water comes from the carboxylic acid’s hydroxyl group. One hydrogen comes from that same group, and the other comes from the alcohol or amine.

This is the standard structural atom-bookkeeping model for these reactions.

02Reverse the structure to find the missing atoms

Once you’ve recognised the linkage, you can work backwards.

Use this four-step method whenever an exam question gives you a polymer structure and asks which atoms were eliminated.

Step 1: Find the functional group in the polymer.

Look for either:

• -C(=O)-O- for a polyester
• -C(=O)-NH- for a polyamide

Step 2: Identify the bond formed during polymerisation.

For a polyester, it is the single bond between the carbonyl carbon and the ester oxygen.

For a polyamide, it is the single bond between the carbonyl carbon and the nitrogen.

Do not break the carbonyl C=O double bond.

Step 3: Reverse the bond formation.

Imagine splitting the bond between the carbonyl carbon and the oxygen or nitrogen.

Then restore the original functional groups:

• Add -OH to the carbonyl carbon to reconstruct the carboxylic acid.
• Add H to the oxygen to reconstruct an alcohol, or H to the nitrogen to reconstruct a primary amine.

Step 4: Check that the atoms make sense.

Check that:

• Every carbon has four bonds.
• A neutral oxygen has two bonds.
• A neutral nitrogen has three bonds.
• The reconstructed monomers contain the correct functional groups.

If your reconstructed structure leaves an oxygen with only one bond or a nitrogen with an incomplete valence, you may have missed a hydrogen.

03Understand brackets and open bonds before reversing anything

There is another detail that catches students out: the brackets around a polymer’s repeating unit.

Consider:

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

The square brackets identify the repeating unit. The subscript \(n\) indicates that this section is repeated many times.

But what do the dashes at the left and right edges mean?

They indicate that the chain continues beyond the repeating unit.

They are not dangling bonds waiting for hydrogen atoms. They are connections to neighbouring repeating units.

This distinction changes how you reconstruct monomers.

Suppose you see an open bond at the left edge of a repeating unit. You should not automatically attach a hydrogen there. That bond normally connects to the preceding unit.

Similarly, an open bond at the right edge connects to the following unit.

Only add terminal hydrogens or hydroxyl groups when you are actually reconstructing the ends of a finite polymer chain.

There is another consequence worth remembering:

The number of repeating units is not necessarily the number of water molecules produced.

Each newly formed condensation linkage produces one water molecule in the reactions discussed here. The exact number of linkages depends on how the monomers connect and whether the polymer is linear or cyclic.

We’ll return to this in the final question.

04Worked examples: identify the atoms that leave

05Use valence to check your answer

Valence is a useful way to catch incorrect reconstructions.

For the structures in this article, remember three basic rules:

Consider an ester linkage:

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

The carbonyl carbon has:

• A double bond to oxygen.
• A single bond to \(R\).
• A single bond to the ester oxygen.

That gives four bonds in total.

When you reverse the condensation, the carbonyl carbon replaces its bond to the ester oxygen with a bond to OH. Its valence is still four.

The ester oxygen originally has two single bonds. After the C-O bond is broken, it has one bond to \(R'\). Adding H gives it the second bond it needs.

The same reasoning applies to an amide:

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

The nitrogen has three bonds: one to the carbonyl carbon, one to \(R'\), and one to hydrogen.

After separating the C-N bond, adding another H restores a neutral primary amine.

A good exam answer should satisfy both the functional-group pattern and the valence rules.

06Don’t confuse condensation polymerisation with addition polymerisation

Not every polymer loses a small molecule during its formation.

This is an important distinction because students sometimes see a repeating unit and immediately assume that water must have been eliminated.

Consider polyvinyl chloride (PVC), which has the repeating unit:

\[ \ce{[-CH2-CHCl-]_{n}} \]

Its monomer is chloroethene:

\[ \ce{CH2=CHCl} \]

During addition polymerisation, the carbon-carbon double bond opens and forms single bonds connecting the monomers.

The chlorine remains attached to the carbon chain.

No water or other small molecule is eliminated.

Compare the two reaction types:

For the standard carboxylic acid reactions in this article, the eliminated molecule is water. Other condensation reactions can produce different small molecules, so always inspect the actual starting functional groups.

The useful decision is simple: don’t look at the fact that a substance is a polymer. Look at the bond that joins its repeating sections.

07Questions and solutions

Question 1

A polyester contains the following linkage:

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

It was formed from a carboxylic acid and an alcohol.

Identify exactly which atoms are eliminated to form one water molecule. Explain which two oxygen atoms remain in the polymer.

Solution 1

The eliminated atoms are the carboxylic acid’s -OH group and one hydrogen from the alcohol’s -OH group. The carbonyl oxygen and the ester oxygen both remain in the polymer.

The carboxylic acid contributes one oxygen and one hydrogen from its hydroxyl group.

The alcohol contributes one hydrogen from its hydroxyl group.

Together, these form:

\[ \ce{H2O} \]

The oxygen from the alcohol remains attached to the carbonyl carbon, creating the ester linkage.

The other oxygen remains double-bonded to carbon as the carbonyl oxygen.

The common mistake is to remove the visible ester oxygen because it appears to be part of the new connection. In fact, that oxygen is retained and is essential to the ester linkage.

Question 2

A polymer contains the following structural section:

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

Assume it was formed through condensation between a carboxylic acid and a primary amine.

a) Identify the bond that must be broken to reconstruct the original monomers.

b) State which functional group must be restored on each side.

c) Identify the atoms that formed water.

Solution 2

The bond to break is the single bond between the carbonyl carbon and nitrogen. The original carboxylic acid and primary amine are restored by adding -OH to the carbonyl carbon and one H to nitrogen.

Part a

The relevant bond is:

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

Do not break the carbonyl C=O double bond or the bond between nitrogen and the remaining carbon chain.

Part b

The carbonyl side regains -OH, producing a carboxylic acid:

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

The nitrogen side gains one H, producing a primary amine:

\[ \ce{H2N-R'} \]

Part c

The eliminated atoms are:

• The oxygen and hydrogen from the carboxylic acid’s -OH.
• One hydrogen from the primary amine.

These form \(\ce{H2O}\).

The nitrogen itself remains in the polymer. Removing it would destroy the amide linkage and would not reconstruct the original amine correctly.

Question 3

A polymer has the repeating unit:

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

A student claims that one water molecule must have been eliminated for every repeating unit because the polymer was formed through a polymerisation reaction.

Identify the monomer and reaction type. Explain why the student’s prediction is incorrect.

Solution 3

The monomer is propene, and the polymer forms through addition polymerisation. No water molecules are eliminated because all atoms in the monomer are retained in the polymer.

The monomer is:

\[ \ce{CH2=CH-CH3} \]

During polymerisation, the carbon-carbon double bond opens to form the single bonds of the polymer backbone.

The methyl group remains attached to the second carbon.

The resulting repeating unit is:

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

There is no ester or amide linkage, and no small molecule is released.

The student has confused polymerisation in general with condensation polymerisation specifically.

The bracket edges also represent continuation of the carbon chain, not missing atoms that must be replaced with hydrogen.

Question 4

An unknown polyester has the following repeating unit:

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

Assume it was formed from one diol and one dicarboxylic acid.

a) Reconstruct both monomers.

b) Identify the oxygen atom in the section -CH2-O-CH2-.

c) A student suggests that this central oxygen must also be eliminated as water. Explain why this is incorrect, using valence and functional-group identification.

Solution 4

The monomers are ethane-1,2-diol, \ce{HO-CH2-CH2-OH}, and the dicarboxylic acid \ce{HOOC-CH2-O-CH2-COOH}. The central oxygen is part of an ether linkage, not an ester linkage, so it remains in the polymer.

Part a: Reconstruct the monomers

First, identify the ester linkages:

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

Break the single bond between the carbonyl carbon and ester oxygen.

Restore -OH to the carbonyl carbon and H to the ester oxygen.

The diol-derived section becomes:

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

This is ethane-1,2-diol.

The acid-derived section becomes:

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

This is a dicarboxylic acid containing an ether oxygen between its two carbon-containing sections.

Part b: Identify the central oxygen

The central oxygen is located here:

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

It connects two carbon atoms through single bonds.

This is an ether linkage.

Part c: Explain why it remains

An ether oxygen already has two single bonds, one to each neighbouring carbon.

It satisfies its normal valence of two.

More importantly, it is not directly attached to a carbonyl carbon. Therefore, it is not part of the ester linkage that formed during condensation.

The student has assumed that every oxygen atom must be removed because water contains oxygen.

That is incorrect. Only the oxygen from the carboxylic acid’s hydroxyl group is eliminated in the standard reaction considered here.

The central ether oxygen stays in the dicarboxylic acid residue and therefore remains in the polymer.

This question demonstrates why identifying the functional group is more reliable than simply counting oxygen atoms.

Question 5

Five molecules of hydroxyethanoic acid, with the structure

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

undergo condensation to form one oligomer.

The product is confirmed to be a single linear chain, rather than a ring. Structural analysis also identifies one terminal alcohol group and one terminal carboxylic acid group.

a) How many ester linkages have formed?

b) How many water molecules have been eliminated?

c) Identify the atoms that form each water molecule.

d) Another student argues that the repeating unit

\[ \ce{[-O-CH2-C(=O)-]_{5}} \]

must produce five water molecules because there are five repeating units. Explain the error.

e) If the same five monomer molecules instead formed a cyclic oligomer, with all their functional groups reacting, how many water molecules would form? What would happen to the terminal functional groups?

Solution 5

The linear chain has four ester linkages and produces four water molecules. If the five monomers instead form a fully condensed cyclic oligomer, five ester linkages and five water molecules are produced.

This question tests an important distinction: the number of monomers does not automatically equal the number of bonds joining them.

Part a: Count the ester linkages

Five monomer molecules are arranged in one linear chain.

To connect five separate molecules into one continuous chain, four connections are needed.

Therefore:

\[ N_{\text{links}}=5-1=4 \]

There are four ester linkages.

Part b: Calculate the water molecules

Each ester linkage forms through one condensation event.

Each event eliminates one water molecule.

Therefore:

\[ N_{\ce{H2O}}=4 \]

Four water molecules are produced.

Part c: Identify the atoms in each water molecule

At each ester linkage:

• The carboxylic acid contributes its -OH.
• The alcohol contributes one H from its -OH.

These combine to form:

\[ \ce{H2O} \]

Across the four condensation events, four carboxylic acid hydroxyl groups and four alcohol hydrogens are eliminated.

Part d: Explain the error in the student’s reasoning

The student has confused the number of repeating units with the number of condensation events.

A linear chain of five monomer residues has only four connections between neighbouring residues.

One alcohol group and one carboxylic acid group remain unreacted at the two ends of the chain.

The actual structure can be represented as:

\[ \begin{aligned} &\ce{HO-CH2-C(=O)-O-CH2-C(=O)-}\\ &\ce{O-CH2-C(=O)-O-CH2-C(=O)-}\\ &\ce{O-CH2-C(=O)-OH} \end{aligned} \]

There are five hydroxyethanoic acid residues, but only four ester connections.

The bracket notation describes the repeating structure. It does not automatically specify the terminal groups or the exact number of condensation events.

Part e: Consider the cyclic structure

In a cyclic oligomer, the final monomer connects back to the first monomer.

This creates one additional ester linkage.

The total number of connections is now:

\[ N_{\text{links}}=5 \]

Therefore:

\[ N_{\ce{H2O}}=5 \]

All five alcohol groups and all five carboxylic acid groups have reacted.

There are no remaining terminal alcohol or carboxylic acid groups because the chain has closed into a ring.

This is the subtle point: the same five monomer molecules can produce different numbers of water molecules depending on how the polymer structure closes.

For a linear chain of \(m\) monomers, there are \(m-1\) connections. For a cyclic structure in which every monomer connects to its neighbours, there are \(m\) connections.

The number of water molecules follows the number of condensation linkages, not simply the number of repeating units.

Once you can identify the linkage, restore the correct functional groups, and account for the chain ends, you can reliably determine which atoms leave during condensation polymerisation.

The next useful step is reconstructing complete monomer structures from more complicated repeating units. See How to Find Monomers from a Polymer’s Repeating Unit for further practice, or Condensation Polymerisation: Nylon and Polyester Explained for HSC Chemistry for the broader polymer chemistry.