How to Draw Polymer Brackets and Continuing Bonds Correctly

An ethene molecule contains a carbon-carbon double bond, \(\ce{CH2=CH2}\). When ethene forms poly(ethene), those carbon atoms become part of a long chain. But if you draw one section as \(\ce{CH2-CH2}\) and add hydrogen atoms to finish the ends, you’ve drawn ethane instead of a polymer.

The correct method is to draw one repeating section, leave a bond extending from each end of the backbone, draw square brackets with both continuing bonds passing through them, and place a subscript \(n\) outside the bottom-right bracket. For poly(ethene), the compact formula is \(\ce{[-CH2-CH2-]_{n}}\).

Before going further, predict what would happen if you kept the original C=C double bond and added continuing bonds at both ends.

Each carbon would need five bonds: two to hydrogen, two counted in the double bond, and one to the neighbouring repeating unit. Ordinary neutral carbon cannot do that. The double bond must become a single bond as the addition polymer forms.

That gives us the first rule: the brackets cannot rescue an incorrect molecular structure. Every atom still has to satisfy its normal bonding requirements.

01Why polymer structures need continuing bonds

Consider a simplified polymerisation process.

Ethene enters a reaction vessel through a feed line. Under suitable polymerisation conditions, ethene molecules join to form poly(ethene), which is collected as the product.

The material entering and leaving the vessel is different, but the apparatus does not directly show individual carbon-carbon bonds. Chemists use their understanding of addition polymerisation, together with structural evidence, to represent the resulting molecules.

Suppose you initially draw the product as \(\ce{CH3-CH3}\).

Every carbon has four bonds, so the structure is chemically valid. The problem is that it’s a complete ethane molecule with only two carbon atoms. It doesn’t represent the long product.

Instead, draw several connected sections:

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

All six carbon atoms belong to one continuous backbone.

You can now select two adjacent \(\ce{CH2}\) groups as the conventional ethene-derived repeating unit. The bonds to the sections on either side must remain visible.

Think of identifying a repeating pattern on a strip of fabric. Marking one section doesn’t cut the fabric into separate pieces. Similarly, polymer brackets mark a repeating section without breaking the covalent chain.

The analogy has a limit: polymer chains are actual molecules with shared electrons in chemical bonds. The brackets are only a way of describing their structure.

02The four checks for a correct polymer bracket

1. Enclose a complete repeating pattern

For an addition polymer made from an alkene such as propene, \(\ce{CH2=CHCH3}\), the conventional monomer-derived repeating unit is:

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

The two carbon atoms from the original double bond form the polymer backbone. The \(\ce{CH3}\) group remains attached to one of those carbons.

All parts of that group belong inside the brackets because they recur with the backbone.

2. Extend the backbone bonds through the brackets

A continuing bond must extend from the last backbone atom through the bracket line, showing where the adjacent repeat connects.

In a fully displayed structural diagram:

• The left continuing bond passes through the left square bracket.
• The right continuing bond passes through the right square bracket.
• Neither bond stops at a hydrogen atom or an isolated free end.
• The subscript \(n\) sits outside the bottom-right bracket.

The compact formulas used throughout this article show these connections using dashes at the start and end of the bracketed formula. In a displayed drawing, those dashes should pass through the brackets.

3. Check every atom’s valence

In the ordinary neutral organic structures considered here, carbon forms four bonds, oxygen forms two, nitrogen usually forms three, and hydrogen and halogens each form one.

Count the continuing bonds when checking valence. They are real covalent bonds to atoms in neighbouring units.

Here, \(R\) represents a side group attached to the backbone.

4. Expand the repeating unit twice

This is the most reliable check when bracket placement is confusing.

Remove the brackets, write the unit twice, and connect the right continuing bond of the first copy to the left continuing bond of the second.

Does the result reproduce the intended chain?

A drawing can satisfy every carbon’s valence and still repeat the wrong arrangement of atoms. The two-unit expansion catches that mistake.

04Condensation polymers: the continuing bonds can involve oxygen or nitrogen

So far, the backbone has contained only carbon atoms. That changes when the polymer is formed by condensation.

Suppose a diol and a dicarboxylic acid react to form a polyester. Their functional groups join through ester linkages.

An ester linkage contains:

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

The oxygen and carbonyl carbon belong to the polymer backbone. They aren’t side groups that can be ignored when choosing the repeating unit.

Unlike the C=C bond in alkene addition polymerisation, the C=O double bond in the ester group remains present.

05How to tell addition and condensation polymer drawings apart

The bracket rule stays the same, but the chemistry that creates the chain is different.

The easiest trap is applying the addition-polymer method to every polymer.

For a polyester, restoring a C=C bond between two adjacent carbon atoms is not the normal way to identify its condensation monomers. The important connections are the ester linkages.

06Brackets show repetition, not the history of every bond

There’s one final distinction that becomes important in difficult questions.

The boundary of a repeating unit doesn’t necessarily mark the bond that formed during polymerisation.

You can shift the brackets along a genuinely repeating chain and still describe the same polymer, provided that repeating the chosen section reconstructs exactly the same sequence of atoms and bonds.

Similarly, a bracketed formula does not, by itself, establish the actual end groups of a polymer molecule or the complete sequence of reactions used to make it.

A useful final drawing procedure is therefore:

1. Trace the continuous backbone, including any oxygen or nitrogen atoms.
2. Select a section that reproduces the correct atom-and-bond pattern.
3. Preserve all side groups, double bonds, and functional groups in that section.
4. Draw the continuing bonds through the brackets and place \(n\) at the lower right.
5. Expand the unit twice and check both connectivity and valence.

If you’re working backwards from a polymer towards its monomer or monomers, use the backbone and functional groups rather than the mere position of the brackets.

07Questions and solutions

Question 1

Fluoroethene, \(\ce{CH2=CHF}\), undergoes addition polymerisation to produce a linear polymer.

(a) Draw its repeating unit using correct brackets, continuing bonds, and the subscript \(n\).

(b) Explain how the carbon atom carrying fluorine satisfies carbon’s normal valence in the polymer.

Solution 1

The correct repeating unit is \(\ce{[-CH2-CH(F)-]_{n}}\), with a continuing single bond passing through each square bracket.

During addition polymerisation, the carbon-carbon double bond becomes a single bond, while the fluorine remains attached to the same carbon.

The fluorine-bearing carbon has:

• One C-F bond.
• One C-H bond.
• One C-C bond to the other carbon in the repeating unit.
• One C-C bond to the neighbouring repeating unit.

That gives four bonds, as required.

Leaving the original C=C double bond while adding a continuing bond would give the fluorine-bearing carbon five bonds in total. The correct diagram must preserve both valence and chain continuity.

Question 2

A regular addition polymer contains this continuous backbone sequence:

\[
\ce{…-CH2-C(CH3)2-CH2-C(CH3)2-CH2-C(CH3)2-…}
\]

Three students propose the following repeating units:

Student A:

\[
\ce{[-CH2-C(CH3)2-]_{p}}
\]

Student B:

\[
\ce{[-C(CH3)2-CH2-]_{q}}
\]

Student C:

\[
\ce{[-CH2-C(CH3)2-CH2-]_{r}}
\]

Assume all continuing bonds are single bonds and each proposed unit repeats regularly.

(a) Which drawings represent the stated polymer backbone?

(b) Which drawing produces a different polymer sequence, even though each carbon can satisfy valence four?

(c) Identify the alkene monomer that produces the stated regular backbone.

Solution 2

Students A and B have drawn the same polymer with different bracket starting points. Student C has drawn a different repeating sequence. The alkene monomer for the stated chain is 2-methylpropene, \(\ce{CH2=C(CH3)2}\).

For A, expanding two repeating units gives:

\[
\ce{-CH2-C(CH3)2-CH2-C(CH3)2-}
\]

This matches the specified backbone.

For B, expanding two units gives:

\[
\ce{-C(CH3)2-CH2-C(CH3)2-CH2-}
\]

This is the same alternating sequence, just beginning at the other type of backbone carbon.

But expanding C twice gives:

\[
\ce{-CH2-C(CH3)2-CH2-CH2-C(CH3)2-CH2-}
\]

The two \(\ce{CH2}\) groups in the middle become adjacent at the boundary between repeats.

The stated polymer alternates \(\ce{CH2}\) and \(\ce{C(CH3)2}\) groups. Student C’s diagram does not preserve that pattern.

Crucially, Student C’s atoms can still satisfy their normal valences. This is why a valence check alone cannot prove that the repeating unit is correct.

To find the alkene monomer, take the valid two-carbon backbone section, remove its external chain connections, and restore the C=C bond:

\[
\ce{CH2=C(CH3)2}
\]

The two methyl groups stay on the same carbon.

Question 3

A linear polyamide is formed by condensation of:

• Hexane-1,6-diamine, \(\ce{H2N-(CH2)6-NH2}\).
• Butanedioic acid, \(\ce{HOOC-(CH2)2-COOH}\).

A particular finite chain contains exactly three diamine molecules and two dicarboxylic acid molecules. All five molecules are incorporated into one unbranched chain. Both ends retain an unreacted \(\ce{-NH2}\) group.

Assume there are no cyclic products or side reactions, and one water molecule is released whenever a new amide linkage forms.

(a) Draw a valid repeating unit for the corresponding long-chain polymer, including continuing bonds.

(b) Calculate the number of amide linkages and water molecules formed in the finite five-molecule chain.

(c) Explain why \(\ce{-NH2-}\) is incorrect inside the backbone of the repeating unit but appropriate at the actual chain ends.

Solution 3

A correct repeating unit is \(\ce{[-NH-(CH2)6-NH-C(=O)-(CH2)2-C(=O)-]_{n}}\). The finite chain contains four newly formed amide linkages and releases four water molecules.

For part (a), the polyamide backbone contains the diamine-derived segment:

\[
\ce{-NH-(CH2)6-NH-}
\]

and the dicarboxylic-acid-derived segment:

\[
\ce{-C(=O)-(CH2)2-C(=O)-}
\]

The continuing bonds connect carbonyl carbons to nitrogen atoms in adjacent repeats, creating amide linkages of the form \(\ce{-C(=O)-NH-}\).

For part (b), the five molecules must form one continuous alternating chain. Since there are three diamines and two diacids, the arrangement is:

\[
\text{diamine-diacid-diamine-diacid-diamine}
\]

There are four connections between the five original molecules.

\[
\begin{aligned}
N_{\mathrm{links}}&=5-1\\
&=4
\end{aligned}
\]

Here, \(N_{\mathrm{links}}\) is the number of new amide linkages.

Because every amide-forming condensation releases one water molecule, the finite product releases four \(\ce{H2O}\) molecules.

For part (c), an internal amide nitrogen is bonded to two non-hydrogen atoms in the backbone. It also carries one hydrogen, giving three covalent bonds.

A neutral internal \(\ce{-NH2-}\) group would instead have two backbone bonds and two N-H bonds, giving nitrogen four bonds. That is not the ordinary neutral amide structure.

At the actual ends, however, an \(\ce{-NH2}\) group has only one bond to the polymer backbone and two N-H bonds. Its valence is correct.

The tempting mistake is to preserve both hydrogens from every starting amine group. Condensation removes hydrogen from the amine group when the amide bond forms. Only unreacted terminal amine groups retain two hydrogens.

Question 4

A perfectly linear, saturated hydrocarbon chain has the exact structure:

\[
\ce{CH3-(CH2)_{240}-CH3}
\]

Assume all its carbon atoms came from ethene by addition polymerisation, with only two hydrogen atoms added during termination. No carbon atoms are gained or lost in the process.

Two students give different contracted representations for this same chain.

Student A:

\[
\ce{CH3-(CH2-CH2)_{120}-CH3}
\]

Student B:

\[
\ce{H-(CH2-CH2)_{121}-H}
\]

(a) Are both contracted representations valid for the stated product?

(b) How many ethene molecules were incorporated into one chain?

(c) Why can both drawings show different numbers of two-carbon groupings without representing different molecules?

Solution 4

Both representations are valid, and exactly 121 ethene molecules are incorporated into the chain. The different group counts occur because Student A treats the two terminal carbon atoms separately, while Student B includes them in the repeated two-carbon groupings.

First, count every carbon atom in the final molecule.

There are 240 interior \(\ce{CH2}\) groups and two terminal \(\ce{CH3}\) groups:

\[
\begin{aligned}
N_{\mathrm{C}}&=240+2\\
&=242
\end{aligned}
\]

Each ethene molecule contributes two carbon atoms. Since all carbon atoms came from ethene:

\[
\begin{aligned}
N_{\mathrm{ethene}}&=\frac{242}{2}\\
&=121
\end{aligned}
\]

These are counts of atoms and molecules, not molar quantities.

We can independently check the hydrogen count.

The final molecule contains:

\[
\begin{aligned}
N_{\mathrm{H}}&=240(2)+2(3)\\
&=486
\end{aligned}
\]

Meanwhile, 121 ethene molecules contribute \(121(4)=484\) hydrogen atoms. Adding two hydrogen atoms during termination gives 486, exactly matching the final structure.

Now compare the bracket groupings.

Student A places the two terminal \(\ce{CH3}\) carbons outside the repeated internal section. The 120 two-carbon groupings account for only the 240 interior carbon atoms.

Student B groups all 242 carbons into 121 two-carbon sections, with one additional hydrogen shown at each end.

Both reconstruct the same carbon skeleton and the same terminal groups.

The tempting but incorrect conclusion is that Student A’s 120 bracketed pairs must correspond to only 120 ethene molecules. That ignores the two carbon atoms explicitly drawn outside the repeated section.

In fact, for a perfectly regular poly(ethene) chain, even a single \(\ce{CH2}\) group is a repeating structural pattern. HSC drawings usually choose the two-carbon ethene-derived grouping because it makes the relationship to the starting monomer clear.

A bracket is a way to group the atoms in a chain. It does not automatically count every starting molecule unless the grouping and end groups have been accounted for.

Question 5

Two chemists propose different routes to make the same finite linear polyester.

The target chain contains exactly four ethan-1,2-diol-derived segments and four butanedioic-acid-derived segments, alternating along the backbone. Its left end is a free alcohol group, \(\ce{-OH}\), and its right end is a free carboxylic acid group, \(\ce{-COOH}\).

Route A: Combine four ethan-1,2-diol molecules with four butanedioic acid molecules.

The starting structures are:

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

and

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

Route B: Instead, combine four molecules of a preformed monoester that already contains one diol-derived segment joined to one diacid-derived segment:

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

In Route B, every starting monoester has one free alcohol group, one free carboxylic acid group, and one ester linkage already present.

Assume all starting molecules in each route are incorporated into one chain, every new ester linkage forms by condensation with the elimination of one water molecule, and there are no cycles or side reactions.

(a) Can both routes make the same final chain? Write a valid repeating-unit formula for its internal structure.

(b) Calculate how many water molecules are released during the chain-building stage of Route A and Route B.

(c) If each preformed monoester in Route B was itself prepared by reacting one diol molecule with one dicarboxylic acid molecule, how many water molecules are released over the whole synthesis of Route B? Compare the total with Route A.

(d) Explain why the final polymer’s bracketed repeating unit cannot, on its own, tell you how many water molecules were released during a particular chain-building stage.

Solution 5

Yes. Both routes can produce the same polyester chain. Route A releases seven water molecules during chain assembly, while Route B releases three during chain assembly. If Route B’s four monoesters are first prepared from the original diols and diacids, both complete syntheses release seven water molecules in total.

The key is distinguishing bonds already present in starting molecules from bonds formed during a particular stage.

For part (a), both routes produce the same repeating backbone pattern:

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

This unit contains the diol-derived and dicarboxylic-acid-derived sections in the correct alternating order.

The brackets enclose the recurring internal structure. In the actual finite product, the left terminal oxygen is bonded to hydrogen, and the right terminal carbonyl carbon is bonded to an \(\ce{OH}\) group. Those terminal groups are not inserted between internal repeats.

For part (b), Route A begins with eight separate molecules: four diols and four diacids.

To connect eight molecules into one unbranched chain, seven new intermolecular connections are needed.

Each connection is an ester linkage formed by condensation, so:

\[
\begin{aligned}
N_{\mathrm{H_2O,A}}&=8-1\\
&=7
\end{aligned}
\]

Route A therefore releases seven water molecules during chain assembly.

Route B starts with four separate monoester molecules. Each already contains one ester bond.

Joining the four starting molecules into one linear chain requires only three new connections:

\[
\begin{aligned}
N_{\mathrm{H_2O,B}}&=4-1\\
&=3
\end{aligned}
\]

Route B therefore releases three water molecules during its chain-building stage.

The product does not have fewer ester bonds than Route A’s product. Four ester bonds were already present in Route B’s starting molecules, and three additional ones formed:

\[
4+3=7
\]

That is the same total of seven ester linkages in the target finite chain.

For part (c), preparing one Route B monoester from one diol and one diacid creates one ester linkage and eliminates one water molecule.

Four monoesters therefore require four initial esterification reactions, releasing four water molecules.

The subsequent chain assembly releases another three:

\[
\begin{aligned}
N_{\mathrm{total,B}}&=4+3\\
&=7
\end{aligned}
\]

Route A also releases seven water molecules overall, assuming the stated reactions occur completely.

This is consistent with the total number of ester bonds created from the original diol and diacid molecules, and with conservation of atoms.

For part (d), the final repeating-unit diagram records the atom connectivity in the polymer. It does not record when each ester bond was formed.

The tempting argument is that a polymer with four diol-diacid repeating sections must have undergone a fixed number of condensation reactions during its final chain-building stage.

But Route B begins with some of those ester bonds already formed. Route A does not.

The same final structure can therefore be reached through different sequences of bond-forming reactions, with different amounts of water released at an individual stage.

To determine the by-product amount, you must know the starting molecules, their functional groups, the actual chain ends, and which bonds are being formed in the stage under consideration.

08From a correct polymer drawing to its possible monomers

Once a repeating-unit drawing passes the valence check and reconstructs the correct chain when expanded, the next question is what starting materials could produce that structure.

For a straightforward alkene-derived addition polymer, restoring the C=C bond can reveal the monomer. For a polyester or polyamide, you instead identify the functional-group linkages and determine which monomer groups could have formed them.

The important qualification from Question 5 remains: a correct polymer structure can suggest possible starting materials, but it doesn’t necessarily prove a unique synthesis route.

To practise that reverse process in more detail, use How to Find Monomers from a Polymer’s Repeating Unit. The same two habits carry across both tasks: follow the actual bonds, and make every atom account for its valence.