How to Tell If Two Polymer Drawings Show the Same Structure

Two drawings of a polymer can look different even when they represent exactly the same chain. One might put a carbon atom at the left edge of the brackets, while another puts it at the right. Does that make them different polymers?

Not necessarily. To compare two polymer structures, ignore the bracket positions temporarily, join several repeating units together, and trace which atoms are bonded to which. If the continuous bonding pattern matches after shifting the starting point or reversing the drawing, the structures are equivalent.

The crucial test is connectivity, not appearance. Two drawings can have the same atoms, molecular formula, and even the same possible monomer, yet represent different polymer structures.

01What are the brackets and open bonds actually telling you?

Consider this repeating unit:

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

This represents a polymer made from repeating sections containing two backbone carbon atoms.

The \(\ce{CH2}\) carbon is attached to two hydrogen atoms. The \(\ce{CH(Br)}\) carbon is attached to one hydrogen atom and one bromine atom.

Both carbon atoms are also connected to neighbouring carbon atoms along the chain.

Three details matter when reading the structure.

• The brackets mark the section chosen to represent the repeating pattern.
• The open bonds extending through the brackets show where this section joins the next one. They are not unpaired electrons or missing hydrogen atoms.
• The subscript \(n\) indicates the number of repeating units in the idealised polymer representation. It does not tell you the number of carbon atoms in one unit.

Joining three copies produces:

\[ \ce{...-CH2-CH(Br)-CH2-CH(Br)-} \]

\[ \ce{CH2-CH(Br)-...} \]

The chain continues through the bonds at the edges of the brackets.

Now imagine selecting a different starting point. You could describe the same pattern using:

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

Nothing about the carbon-carbon bonds or the positions of the bromine atoms has changed.

Think of repeating units like a pattern on wallpaper

Imagine wallpaper with alternating circles and squares.

One person draws a box around a circle followed by a square. Another draws a box around a square followed by a circle.

The boxes look different, but the wallpaper pattern is unchanged.

Polymer brackets work similarly. They identify a convenient repeating section, not a chemically separate piece of the molecule.

The analogy has a limit: wallpaper shapes sit next to one another, while polymer atoms are joined by real covalent bonds. Those bonds, including which atoms they connect, determine whether two molecular structures are equivalent.

02A reliable method for comparing two polymer drawings

A common first attempt is to compare the atoms inside the brackets.

Suppose both drawings contain four carbon atoms and eight hydrogen atoms. A student might conclude that they must be the same polymer.

That test is useful, but incomplete. It measures composition, not bonding.

A better approach is to treat each drawing as a proposed molecular model and test whether both produce the same extended chain.

Step 1: Find the continuous backbone

The backbone is the connected sequence of atoms running through the polymer.

For a typical addition polymer made from an alkene, this backbone consists of carbon atoms joined by single bonds.

Side groups such as \(\ce{-CH3}\), \(\ce{-Cl}\), and \(\ce{-Br}\) remain attached to particular backbone carbons.

Don’t confuse a side group with the next atom along the backbone.

Step 2: Connect at least two repeating units

Imagine making two copies of each displayed structure on separate paper strips.

Starting with the first strip, connect its right-hand open bond to the left-hand open bond of the second strip. Keep every side group in its original position.

This gives a longer section of the actual molecular structure.

Why do this?

Because the bond crossing a bracket boundary is just as important as any bond drawn inside the brackets.

Sometimes a difference only becomes visible when the units are connected.

Step 3: Compare the complete bonding patterns

Ask three questions:

1. Does each backbone atom have the same neighbours?
2. Are the same side groups attached to the corresponding atoms?
3. Does the pattern repeat in the same order?

You may slide the starting point along the chain or reverse the direction in which you read it.

You may not move a side group onto a different carbon, change a single bond into a double bond, or rearrange the order of atoms.

If you find a difference, extend the chain another unit to check whether it is real or simply caused by the bracket position.

Step 4: Check valence

For the ordinary neutral organic structures used in HSC Chemistry:

A double bond counts as two bonds when checking valence.

This catches impossible structures, but it doesn’t prove two valid structures are identical. Different molecules can both satisfy every valence rule.

Step 5: Identify the likely polymerisation reaction

After confirming the bonding pattern, look for evidence of how the polymer might form.

A polymer structure can suggest a likely synthesis, but it does not always uniquely establish the actual manufacturing reaction. Different synthetic routes can sometimes produce the same connectivity.

For the 2026 HSC, this material sits within the Chemistry Stage 6 Syllabus (2017). The replacement Chemistry 11-12 Syllabus (2025) has its first HSC examination scheduled for 2029.

04The trap: identical atoms do not guarantee identical polymers

Consider two carbon-backbone patterns.

Pattern A:

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

Pattern B:

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

In both cases, there is one methyl side group for every two backbone carbon atoms.

Both patterns can be assembled from propene-derived units. Both have valid carbon valences.

Are they equivalent?

No.

In Pattern A, every carbon carrying a methyl group is separated from the next such carbon by a \(\ce{CH2}\) group.

In Pattern B, two methyl-substituted backbone carbons are directly connected.

Extending Pattern B periodically also produces adjacent pairs of \(\ce{CH2}\) carbons across the repeating-unit boundaries.

This is a difference in regiochemistry: the way the unsymmetrical propene-derived units connect along the chain.

Pattern A has the usual head-to-tail arrangement. Pattern B has a periodically repeating head-to-head and tail-to-tail arrangement.

Turning the page or moving brackets cannot turn one arrangement into the other. Doing so would require changing carbon-carbon connections.

The key distinction is:

• Different bracket boundaries can represent the same polymer connectivity.
• Different connections between atoms represent different constitutional structures.

There is another limitation. Two drawings with identical connectivity may still omit information about stereochemistry, chain length, branching elsewhere, or terminal groups. Matching simplified repeating-unit drawings establishes equivalence at the structural level shown, not necessarily identity of every real polymer sample.

06What if the drawings predict different chemical behaviour?

Structural comparison is a test of what the diagrams claim. It is not, by itself, an experimental identification of an unknown material.

A second test becomes useful when two valid candidate structures contain different functional groups.

For example, compare:

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

with:

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

The first contains an ester linkage.

The second contains a ketone carbonyl, bonded to carbon atoms on both sides.

Both contain a carbonyl group, \(\ce{C=O}\), but the carbonyl carbon has different neighbours.

That difference predicts different reactions.

A polyester containing ester bonds can undergo hydrolysis, breaking those ester linkages. A chain containing a ketone and an ether, but no ester linkage, would not undergo the same ester-cleavage reaction.

Suppose an investigator wanted to distinguish two such proposed structures. They could compare the predicted products of selective ester hydrolysis with the products actually identified.

The sequence of reasoning would be:

1. Predict which bonds should be cleaved in each candidate.
2. Use a reaction that selectively cleaves ester bonds under the stated conditions.
3. Identify the resulting organic products.
4. Compare the observation with the predictions.

The important observation would be the identity of the products, not merely whether the samples changed appearance.

This is a hypothetical investigative method, not a claim about a particular experiment. Actual identification would require suitable chemical analysis and controls.

It also exposes the weakness of the original comparison by formula alone: two structures with the same atoms can predict different reaction products.

07Questions and solutions

These questions progress from bracket shifts to comparisons where identical formulas, common monomers, and even apparently matching functional groups can be misleading.

Question 1

Two students draw a fluorine-containing polymer as:

Drawing A:

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

Drawing B:

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

Determine whether the two drawings represent the same polymer connectivity.

Identify the alkene monomer and explain how carbon valence confirms your reconstruction.

Solution 1

The two drawings represent the same polymer connectivity, and the corresponding alkene monomer is fluoroethene, \(\ce{CH2=CHF}\).

Joining the repeating units in Drawing A gives:

\[ \ce{...-CH2-CH(F)-CH2-CH(F)-...} \]

Drawing B gives exactly the same alternating sequence, beginning at the fluorine-substituted carbon.

No bonds or side groups have changed.

To reconstruct the alkene monomer, remove the external continuation bonds from a two-carbon unit and restore the carbon-carbon double bond:

\[ \ce{CH2=CHF} \]

The first carbon has two C-H bonds and a C=C double bond, giving a total bond order of four.

The second carbon has one C-H bond, one C-F bond, and a C=C double bond. Its total bond order is also four.

Both carbons satisfy their normal valence.

The polymer is consistent with addition polymerisation of fluoroethene. Reversing the order in which the repeating unit is written does not produce a different polymer.

Question 2

Two chemists propose different repeating patterns for polymers formed from propene.

Pattern A:

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

Pattern B:

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

Here, \(n\) and \(m\) are the numbers of the respective bracketed units. Assume both structures extend periodically as drawn.

A student argues:

“These must be the same polymer. They contain one methyl group per two backbone carbon atoms, and both can be made from propene.”

Evaluate this claim.

Explain the precise structural difference, verify that both bonding patterns are chemically valid, and identify why changing the bracket positions cannot make the structures identical.

Solution 2

The student’s claim is incorrect. The polymers have the same overall chemical composition per propene-derived unit, but different backbone connectivity.

In Pattern A, the methyl-bearing carbons alternate with \(\ce{CH2}\) carbons:

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

Every methyl-bearing backbone carbon is connected to a \(\ce{CH2}\) carbon on either side.

In Pattern B, connecting two repeating units gives a different pattern. There are adjacent methyl-bearing carbons inside each displayed unit and adjacent \(\ce{CH2}\) carbons across the unit boundaries.

Therefore, Pattern B contains both:

• \(\ce{CH(CH3)-CH(CH3)}\) connections.
• \(\ce{CH2-CH2}\) connections.

Pattern A contains neither type of adjacent pair.

Both structures have valid carbon valence. A \(\ce{CH(CH3)}\) backbone carbon has two bonds to neighbouring backbone carbons, one bond to its methyl group, and one bond to hydrogen. A \(\ce{CH2}\) backbone carbon has two C-C and two C-H bonds.

Both arrangements are consistent with different orientations of propene-derived units during addition polymerisation.

However, their regiochemical sequences differ.

Pattern A is the regular head-to-tail arrangement. Pattern B alternates head-to-head and tail-to-tail connections.

A tempting shortcut is to compare their empirical composition.

Pattern A contributes \(\ce{C3H6}\) per two backbone carbons, while Pattern B contributes \(\ce{C6H12}\) per four backbone carbons. Both reduce to the same elemental ratio.

That establishes matching composition, not matching connectivity.

Moving brackets can change where a repeated sequence starts, but it cannot remove the adjacent methyl-bearing carbons in Pattern B.

The two polymers are therefore constitutionally different at the level represented.

Question 3

A polyamide is represented using two structural segments:

\[ D=\ce{-NH-CH2-CH2-NH-} \]

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

Two diagrams show \([DS]_n\) and \([SD]_n\), respectively.

A student claims that the second drawing must be different because its brackets begin at a carbonyl group rather than a nitrogen atom.

Another student suggests that the polymer could be synthesised from ethanamine, \(\ce{CH3CH2NH2}\), and propanedioic acid, \(\ce{HOOCCH2COOH}\).

Determine whether the two drawings represent the same polymer. Then evaluate the proposed monomer pair and identify a suitable alternative.

Solution 3

The two drawings represent the same polyamide connectivity, but ethanamine and propanedioic acid are not a suitable pair for producing the displayed long-chain structure.

First, connect the repeating units.

The two drawings produce:

\[ \cdots D-S-D-S-D-S\cdots \]

and:

\[ \cdots S-D-S-D-S-D\cdots \]

These are identical alternating backbone patterns with different starting points.

The amide connections are present regardless of whether the brackets place them inside or between repeating units.

An amide linkage contains:

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

The carbonyl carbon is bonded directly to nitrogen. This connection is preserved in both drawings.

Next, examine the proposed monomers.

Ethanamine has only one amine functional group. Once that group forms an amide bond, the molecule cannot connect to another monomer through a second amine group.

It would act as a chain-ending component in the proposed condensation reaction, rather than providing the two nitrogen atoms needed for the displayed \(D\) segment.

A suitable diamine is:

\[ \ce{H2N-CH2-CH2-NH2} \]

This is ethane-1,2-diamine.

The dicarboxylic acid is:

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

This is propanedioic acid.

Each molecule contains two functional groups capable of forming amide linkages.

Repeated reaction between the amine and carboxylic acid groups produces amide bonds, with water eliminated during condensation.

The tempting approach was to recognise an amine and a carboxylic acid and assume that was sufficient.

The missing condition is functionality: each monomer must be able to form the connections required to extend the particular polymer chain.

The structures are equivalent, and a suitable condensation synthesis uses ethane-1,2-diamine with propanedioic acid. The repeating structure is consistent with that route, although the drawing alone does not prove which synthetic route was used.

Question 4

Two addition polymers are proposed.

Polymer A:

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

Polymer B:

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

An analyst calculates that both repeating units have the molecular formula \(\ce{C3H5Cl}\).

Using relative atomic masses C = 12.0, H = 1.0, and Cl = 35.5:

1. Calculate the molar mass of the repeating unit for each polymer.
2. Determine whether the structures are equivalent.
3. Reconstruct a suitable alkene monomer for each polymer.
4. Explain why neither the matching molar masses nor the common polymerisation type establishes structural identity.

Solution 4

The structures are different, although both repeating units have a molar mass of \(76.5\ \mathrm{g\,mol^{-1}}\) and both are consistent with addition polymerisation of chlorinated alkenes.

Part 1: Calculate molar mass.

Each repeating unit has formula \(\ce{C3H5Cl}\).

\[ \begin{aligned} M_{\mathrm{repeat}} &=3(12.0)+5(1.0)+35.5\\ &=76.5\ \mathrm{g\,mol^{-1}} \end{aligned} \]

This applies to both structures.

Part 2: Compare connectivity.

In Polymer A, chlorine and the methyl group are attached to the same backbone carbon.

That carbon also connects to the backbone carbons on its left and right, giving four single bonds. It has no hydrogen atom attached.

In Polymer B, chlorine and the methyl group are attached to different backbone carbons.

Each of those carbons has one hydrogen atom, one side group, and two backbone connections.

Both arrangements satisfy carbon valence, but their side-group positions differ.

A rotation, reflection, or shift of the repeating-unit brackets cannot transfer chlorine from one carbon to another.

Part 3: Reconstruct the monomers.

Polymer A gives:

\[ \ce{CH2=C(Cl)CH3} \]

The monomer is 2-chloroprop-1-ene.

Polymer B gives:

\[ \ce{CH(Cl)=CHCH3} \]

The monomer is 1-chloroprop-1-ene.

These two alkenes have the same molecular formula but different connectivity.

Part 4: Explain the measurement limitation.

The molar mass calculation counts each type of atom. It contains no information about which atom is bonded to which.

Similarly, recognising that both polymers could form through addition polymerisation tells us the broad reaction type, not the exact monomer structure.

The tempting conclusion is that equal formula plus equal reaction type means equal polymer.

The missing constraint is the location of the chlorine and methyl substituents.

These are different constitutional polymers, and the reverse-identification step reveals two different alkene monomers.

Question 5

A chemist is investigating an unfamiliar linear polymer. Three proposed repeating-unit drawings are supplied.

Drawing A:

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

Drawing B:

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

Drawing C:

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

All three repeating units have the same molecular formula, \(\ce{C3H4O2}\), and satisfy ordinary carbon and oxygen valence rules.

An illustrative chemical test is specified: under conditions that selectively cleave ester linkages but do not cleave ordinary ether or carbon-carbon bonds, the polymer is completely broken down into a single type of organic molecule:

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

This molecule is 3-hydroxypropanoic acid.

The chemist proposes that the original linear polymer was made by condensation of 12 molecules of this hydroxy acid, with one hydroxyl end group and one carboxylic acid end group remaining.

Assume one molecule of water is eliminated for every ester bond formed, and use C = 12.0, H = 1.0, and O = 16.0.

Determine:

1. Which proposed drawings represent equivalent continuous polymer structures.
2. Which drawing or drawings are consistent with the hydrolysis result, and why the third candidate behaves differently.
3. The number of ester linkages in the proposed 12-monomer chain.
4. The molar mass of that complete chain.
5. Why multiplying the molar mass of the repeating unit by 12 gives a different result.

Solution 5

Drawings A and B describe the same polyester connectivity. Drawing C describes a different structure containing ether and ketone groups rather than ester linkages. The proposed 12-monomer linear chain contains 11 ester linkages and has a molar mass of \(882\ \mathrm{g\,mol^{-1}}\), including its terminal groups.

The important hinge is that an ester bond can cross a bracket boundary. Looking only for \(\ce{-C(=O)-O-}\) inside the printed brackets gives the wrong answer.

Part 1: Establish structural equivalence.

Expand Drawing A:

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

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

At the boundary between repeating units, the carbonyl carbon connects directly to the oxygen of the next unit.

So the continuous structure contains:

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

This is an ester linkage.

Drawing B displays the same ester linkage inside its brackets:

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

Sliding the starting point to the carbonyl carbon transforms the representation of A into B without breaking or rearranging any bonds.

Therefore, A and B are equivalent.

Part 2: Examine Drawing C and the hydrolysis evidence.

Drawing C gives the repeating backbone sequence:

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

Across its bracket boundary, the final \(\ce{CH2}\) bonds to the next oxygen.

The carbonyl carbon, however, is bonded to a \(\ce{CH2}\) group on each side.

This gives a ketone group:

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

The chain also contains ether connections, \(\ce{-C-O-C-}\), but no ester linkage.

The critical distinction is the immediate neighbour of the carbonyl carbon.

• In A and B, the carbonyl carbon is directly bonded to an oxygen atom by a single bond, producing an ester.
• In C, the carbonyl carbon is bonded to carbon atoms on both sides, producing a ketone.

Drawing C has the same repeating-unit formula as A and B, but it is not constitutionally equivalent.

Under the specified selective hydrolysis conditions, the ester bonds in A or B can be cleaved, giving the original hydroxy acid:

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

Drawing C contains no ester bonds to undergo that reaction. It therefore cannot explain the stated complete ester hydrolysis into 3-hydroxypropanoic acid.

The experimental observation is consistent with A and B, not C.

Part 3: Count ester linkages in the finite chain.

The proposed synthesis begins with 12 molecules of 3-hydroxypropanoic acid.

To join 12 separate molecules into one unbranched linear chain, 11 connections must be formed.

Each connection is an ester bond formed by condensation.

Therefore:

\[ N_{\mathrm{ester}}=12-1=11 \]

Here, \(N_{\mathrm{ester}}\) is the number of ester bonds formed between the original monomers.

Eleven molecules of water are eliminated.

This is a different counting problem from looking at one idealised repeating unit. The finite chain has two terminal groups and does not connect back onto itself.

Part 4: Calculate the molar mass of the actual chain.

The starting monomer is 3-hydroxypropanoic acid, \(\ce{C3H6O3}\).

Its molar mass is:

\[ \begin{aligned} M_{\mathrm{monomer}} &=3(12.0)+6(1.0)+3(16.0)\\ &=90.0\ \mathrm{g\,mol^{-1}} \end{aligned} \]

The molar mass of water is:

\[ M_{\ce{H2O}}=18.0\ \mathrm{g\,mol^{-1}} \]

The 12 monomers have a combined molar mass of:

\[ 12(90.0)=1080\ \mathrm{g\,mol^{-1}} \]

Formation of 11 ester bonds removes 11 water molecules.

Therefore:

\[ \begin{aligned} M_{\mathrm{chain}} &=1080-11(18.0)\\ &=1080-198\\ &=882\ \mathrm{g\,mol^{-1}} \end{aligned} \]

The resulting finite molecule has formula \(\ce{C36H50O25}\), consistent with this calculated mass.

Part 5: Explain the apparent molar-mass contradiction.

The ideal repeating unit has formula \(\ce{C3H4O2}\).

Its molar mass is:

\[ \begin{aligned} M_{\mathrm{repeat}} &=3(12.0)+4(1.0)+2(16.0)\\ &=72.0\ \mathrm{g\,mol^{-1}} \end{aligned} \]

Multiplying by 12 gives:

\[ 12(72.0)=864\ \mathrm{g\,mol^{-1}} \]

That is \(18.0\ \mathrm{g\,mol^{-1}}\) below the calculated mass of the complete molecule.

Why?

The repeating-unit formula describes an interior section of the polymer without explicitly representing its terminal groups.

A finite chain of 12 hydroxy-acid monomers loses only 11 water molecules, not 12.

The final molecule retains an \(\ce{-OH}\) group at one end and a \(\ce{-COOH}\) group at the other.

Compared with 12 idealised repeat formulas, those end groups contribute the elemental equivalent of one additional \(\ce{H2O}\):

\[ \begin{aligned} M_{\mathrm{chain}} &=12M_{\mathrm{repeat}}+M_{\ce{H2O}}\\ &=864+18\\ &=882\ \mathrm{g\,mol^{-1}} \end{aligned} \]

The tempting approach was to assume that every printed repeating unit corresponds to one complete monomer after condensation and that all 12 units each lose one water molecule.

The missing constraint is the finite linear chain: 12 separate molecules require only 11 bonds to become one connected molecule.

The final conclusions are:

• A and B have identical polymer backbone connectivity despite their different bracket boundaries.
• C has the same elemental composition but different functional groups and connectivity.
• Selective ester hydrolysis supports A and B rather than C.
• The number of condensation reactions depends on the number of actual connections formed.
• Idealised repeating-unit formulas do not automatically include the terminal groups of finite polymer molecules.

This is why structural equivalence must be established by following the complete chain before using molecular formulas, reaction types, or molar-mass calculations.

07What to do when two drawings appear different in an HSC question

Before deciding that a second polymer drawing represents a different substance, extend the chain and follow the bonds across the bracket boundaries.

A useful order is:

1. Identify the continuous backbone.
2. Join two or three repeating units.
3. Shift or reverse the starting position if needed.
4. Compare which atoms are directly bonded and where the side groups attach.
5. Check valence and the repeating-unit formula.
6. Use functional groups and possible monomers as supporting evidence, not as substitutes for the connectivity test.

The most revealing question is often: What is attached to this particular carbon atom on either side?

That question catches differences that formula comparisons miss.

Once the two representations have been established as equivalent, you can reconstruct the monomer or monomers and identify a plausible polymerisation reaction. If they are not equivalent, locate the exact bond or substituent position that changes before attempting the reverse reaction.

A bracket is just a way of describing a repeating pattern. The covalent bonds are the chemistry.