How to Check Carbon Valence in Polymer Structural Formulae

A polymer is drawn with the repeating unit \(\ce{[-CH2-CH(CH2CH3)-]_{n}}\). The second backbone carbon seems to have only three bonds: one to hydrogen, one to an ethyl group, and one to the carbon on its left. Is the structural formula wrong?

No. Its fourth bond continues through the right-hand bracket to the next repeating unit.

To check carbon valence in a polymer structural formula, count the total bond order around each carbon, including bonds that continue through the brackets. A single bond counts as one, a double bond counts as two, and a neutral carbon in these ordinary organic structures should have a total of four.

The important distinction is between a genuinely missing bond and a bond that simply continues outside the part of the molecule being shown.

01Why polymer brackets can make carbon look underbonded

Consider a simpler polymer:

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

Each carbon belongs to a \(\ce{CH2}\) group. That means it has two single bonds to hydrogen atoms.

Within the brackets, the two carbon atoms are joined by a single bond.

If you count only what appears between the brackets, each carbon seems to have three bonds:

• Two bonds to hydrogen.
• One bond to the other carbon.

Where is the fourth?

Look at the short lines extending towards the brackets. They represent bonds to the neighbouring repeating units.

An expanded section of the polymer is:

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

Every internal carbon now has two carbon-hydrogen bonds and two carbon-carbon bonds.

Total bond order: four.

The brackets haven’t removed any chemical bonds. They’ve shortened the drawing.

The subscript \(n\) means that the enclosed structural sequence repeats many times. It doesn’t mean that carbon has \(n\) bonds, and it doesn’t specify the exact groups at the two ends of a real polymer molecule.

A molecular-model test: where do the open bonds go?

You can test this interpretation with a molecular model kit.

Start with two carbon atoms joined by a single bond. Attach two hydrogen atoms to each carbon, representing one \(\ce{-CH2-CH2-}\) repeating unit.

Leave one connection available on the outside of each carbon.

At this stage, the model is incomplete as a whole molecule. You could be tempted to attach another hydrogen atom at each end. That would produce ethane, \(\ce{CH3CH3}\), not a section of a continuing polymer chain.

The first design has a problem: a single isolated repeating unit cannot show what lies beyond its boundaries.

Now revise the model.

Place an identical unit on each side. Connect the right-hand carbon of the left unit to the left-hand carbon of the central unit, then connect the right-hand carbon of the central unit to the left-hand carbon of the right unit.

The two central carbon atoms now each have four bonds without any additional hydrogen atoms.

This is a test of the structural representation, not an experiment that establishes how the polymer was manufactured. Molecular models can help check connectivity and valence, but they cannot establish a reaction mechanism or reveal the actual length of a polymer chain.

02The bond-counting method

For ordinary neutral organic structures, begin with these familiar bonding patterns.

These are the usual neutral bonding patterns, not rules for every possible ion, radical, or unusual chemical species. For standard HSC polymer structures, they’re the right starting point.

Carbon’s four bonds are related to its four valence electrons, but four valence electrons and four bonded neighbours are not the same statement. For the electron-configuration background, see spdf notation and valence electrons.

Count bond order, not just neighbouring atoms

Compare these two arrangements:

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

and

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

In the first structure, a carbon in the middle of a saturated chain has four single bonds when its hydrogen atoms and both carbon neighbours are included.

In the second structure, the carbonyl carbon has three neighbouring atoms but a total bond order of four:

• One carbon-carbon single bond: 1.
• One carbon-oxygen double bond: 2.
• One carbon-oxygen single bond: 1.

Therefore:

\[ 1+2+1=4 \]

A carbon doesn’t need four different neighbouring atoms to satisfy its normal valence.

The bracket only supplies bonds at the boundary

This prevents a second mistake.

Suppose someone proposes the following fully specified repeating unit for a neutral, saturated polymer:

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

The middle \(\ce{CH}\) carbon has:

• One bond to hydrogen.
• One bond to the carbon on its left.
• One bond to the carbon on its right.

Its total is three.

The bonds extending beyond the brackets belong to the two outer \(\ce{CH2}\) carbons. They cannot provide an additional bond to the middle carbon.

So the middle carbon is genuinely underbonded for the intended neutral, saturated structure. If there is no other substituent, it should be written as \(\ce{CH2}\), not \(\ce{CH}\).

There’s one qualification: skeletal structural formulae often leave hydrogen atoms implicit. A bare carbon vertex in a skeletal drawing may have an understood hydrogen attached. A group explicitly labelled \(\ce{CH}\), however, specifies one hydrogen.

04What carbon valence tells you about polymerisation

Checking valence verifies whether a proposed structure is chemically plausible. It doesn’t, by itself, identify how that structure was made.

You need a second check: inspect the bonds and functional groups that make up the polymer backbone.

Addition polymerisation: the carbon-carbon double bond changes

In an ordinary alkene-derived addition polymer, the monomer contains a C=C double bond.

During polymerisation, one bonding component of the double bond is replaced by new connections between monomer units. The original carbon-carbon single-bond framework remains.

For ethene, the simplified equation is:

\[ \ce{nCH2=CH2 -> [-CH2-CH2-]_{n}} \]

The monomer contains four hydrogen atoms attached to two carbon atoms. The repeating unit also contains four hydrogen atoms attached to those two carbons.

The formula hasn’t gained extra hydrogens to fill the bonds at the bracket boundaries. Those bonds connect to other carbon atoms in the polymer.

Condensation polymerisation: functional groups form links

A condensation polymer can contain oxygen or nitrogen atoms within its backbone.

For example, polyester formation from a carboxylic acid group and an alcohol group produces an ester linkage:

\[ \ce{R-COOH + R'-OH -> R-C(=O)-O-R' + H2O} \]

Here, \(R\) and \(R'\) represent the remaining organic groups.

The carbonyl double bond, \(\ce{C=O}\), remains in the ester structure. You shouldn’t convert it into a carbon-carbon double bond when reconstructing the monomers.

The new connection is between the carbonyl carbon and the alcohol-derived oxygen.

A similar idea applies to polyamides, where an amide linkage contains:

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

For a fuller reverse-identification procedure, see how to find monomers from a polymer’s repeating unit.

06A reliable decision process for unfamiliar polymer structures

Use this order rather than trying to guess the monomer immediately.

The order matters. First establish that the drawing works. Then consider how the structure could have formed. Finally, test whether the reaction evidence actually supports that explanation.

07Questions and solutions

Question 1

A polymer has the repeating unit:

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

(a) Explain why both backbone carbons satisfy carbon’s usual valence.

(b) Identify a suitable alkene monomer and state the polymerisation type.

Solution 1

Both backbone carbons have a total bond order of four. A suitable monomer is fluoroethene, \(\ce{CH2=CHF}\), which forms the polymer through addition polymerisation.

For the first carbon, \(\ce{CH2}\):

\[ 2+1+1=4 \]

The terms represent two C-H bonds, the bond to the other backbone carbon, and the bond to the preceding repeating unit.

For the second carbon, \(\ce{CH(F)}\):

\[ 1+1+1+1=4 \]

These are the bonds to hydrogen, fluorine, the neighbouring backbone carbon, and the next repeating unit.

A possible monomer is:

\[ \ce{CH2=CHF} \]

In this molecule, each carbon has a double bond to the other carbon and two additional single bonds.

The tempting mistake is to add hydrogen atoms at the bracket boundaries. That would replace the chain connections with terminal C-H bonds, describing a different molecule rather than the interior of the polymer.

Question 2

An unknown polymer has two carbon atoms in its backbone repeating unit.

Three proposed structural formulae are:

Structure A

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

Structure B

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

Structure C

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

Additional structural information establishes that:

• The polymer is neutral and its carbons have their usual valence.
• The repeating unit has molecular composition \(\ce{C3H5Br}\).
• The bromine atom and methyl group are attached to the same backbone carbon.

Determine which structure is consistent with all the information.

Explain why each rejected structure fails, and identify a suitable alkene monomer for the accepted structure.

Solution 2

Structure B is the only structure consistent with all the information. Its two backbone carbons each satisfy valence four, its repeat-unit formula is \(\ce{C3H5Br}\), and the bromine and methyl groups share the same backbone carbon.

Structure A fails the valence test.

Its second backbone carbon has single bonds to:

• The first backbone carbon.
• The next repeating unit.
• Bromine.
• The methyl carbon.
• Hydrogen.

Total bond order:

\[ 1+1+1+1+1=5 \]

That’s invalid for the proposed neutral carbon. Structure A also specifies six hydrogen atoms per repeat, not five.

Structure B has:

• Two hydrogen atoms on the first backbone carbon.
• Three hydrogen atoms on its methyl group.
• No hydrogen directly attached to the second backbone carbon.

Its formula is therefore \(\ce{C3H5Br}\), and the second backbone carbon has four single bonds.

Structure C is more subtle. Both of its backbone carbons also have correct valence, and its repeat-unit formula is also \(\ce{C3H5Br}\).

However, bromine and the methyl group are attached to different backbone carbons, contradicting the structural evidence.

This is why matching a molecular formula isn’t enough. Different structures can have the same numbers of each atom while differing in connectivity.

Reconstructing an addition monomer from Structure B gives:

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

The monomer is 2-bromoprop-1-ene.

Its first double-bonded carbon has bond order \(2+2=4\), and the second has \(2+1+1=4\).

The final decision requires three separate constraints: valid valence, correct atom counts, and correct connections. Structures B and C both pass the first two checks, but only B passes the third.

Question 3

A polyamide has the repeating unit:

\[ \ce{[-NH-(CH2)4-NH-C(=O)-CH2-C(=O)-]_{n}} \]

A student makes two claims:

1. The right-hand carbonyl carbon requires an additional hydrogen atom because it appears to have only three bonds.
2. The polymer must have formed by addition polymerisation because the carbon-carbon bonds shown in the backbone are single bonds.

Evaluate both claims.

Identify suitable condensation monomers, and explain how the valence of the nitrogen atoms and the right-hand carbonyl carbon is satisfied.

Solution 3

Both claims are incorrect. The right-hand carbonyl carbon already has total bond order four, and the amide linkages are consistent with condensation polymerisation rather than alkene addition polymerisation.

The right-hand carbonyl carbon has:

• A C=O double bond: bond order 2.
• A C-C single bond to the adjacent \(\ce{CH2}\): bond order 1.
• A bond to the nitrogen of the next repeating unit: bond order 1.

Therefore:

\[ 2+1+1=4 \]

An extra hydrogen would make the carbon overbonded.

The nitrogen atoms in the repeating unit are each shown as \(\ce{NH}\). Each nitrogen has a bond to hydrogen and two bonds to carbon atoms, giving three single bonds.

For example, the nitrogen at the left bracket is attached to its hydrogen, the neighbouring methylene carbon, and the carbonyl carbon of the preceding repeating unit.

Suitable monomers are:

\[ \ce{H2N-(CH2)4-NH2} \]

butane-1,4-diamine, and

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

propanedioic acid.

Their amine and carboxylic acid groups can form amide linkages, eliminating water.

The student’s second claim confuses the type of bond present in the final carbon skeleton with the reaction that formed the polymer.

A polyamide can contain many C-C single bonds. Those bonds do not mean its backbone was formed by opening an alkene double bond.

The decisive structural feature is the amide linkage, \(\ce{-C(=O)-NH-}\), where carbon-nitrogen bonds connect the monomer-derived sections.

Question 4

A manufacturer begins with vinyl ethanoate, an alkene containing an ester side group:

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

It undergoes addition polymerisation to form a polymer with the repeating unit:

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

The polymer is then treated with water under conditions that hydrolyse some ester side groups.

Each converted side group changes from \(\ce{-O-C(=O)-CH3}\) into \(\ce{-OH}\), releasing one molecule of ethanoic acid, \(\ce{CH3COOH}\).

In an illustrative sample:

• The starting polymer contains \(0.0200\ \mathrm{mol}\) of ester-bearing repeating units.
• Exactly \(0.0150\ \mathrm{mol}\) of ethanoic acid is produced and recovered.
• No other reaction produces ethanoic acid, and no ester-bearing polymer units are lost.

A student writes the final polymer as:

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

They conclude that the original monomer must have been \(\ce{CH2=CHOH}\), and that every repeating unit now contains an \(\ce{-OH}\) side group.

Evaluate the conclusion. Calculate the fraction of ester groups converted, check the valence of the carbon bonded to the side group, and explain what cannot be determined from the given data.

Solution 4

The conclusion is not justified. The original monomer used was vinyl ethanoate, and the data show that only 75.0% of its ester side groups were hydrolysed. The final sample therefore contains both hydroxyl-bearing and ester-bearing units.

First, the original addition polymerisation creates the saturated carbon backbone:

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

The ester is a side group, not an ester linkage connecting successive backbone units.

During the subsequent hydrolysis, the side group changes, but the carbon backbone does not need to be broken.

For a converted repeating unit, the backbone \(\ce{CH}\) carbon has:

• One C-H bond.
• One C-O bond to the side group.
• Two C-C bonds to neighbouring backbone carbons.

Its total bond order is:

\[ 1+1+1+1=4 \]

This is true whether the oxygen belongs to an ester side group or an \(\ce{-OH}\) group. Both proposed local structures satisfy carbon valence.

Next, calculate the extent of hydrolysis.

The amount of ethanoic acid produced equals the amount of ester side groups converted:

\[ n_{\mathrm{converted}}=0.0150\ \mathrm{mol} \]

The conversion percentage is:

\[ \begin{aligned} \%\text{ converted} &=\frac{0.0150}{0.0200}\times100\\ &=75.0\% \end{aligned} \]

The amount of unconverted ester-bearing units is:

\[ \begin{aligned} n_{\mathrm{remaining}} &=0.0200-0.0150\\ &=0.00500\ \mathrm{mol} \end{aligned} \]

So 25.0% of the original ester side groups remain.

The final polymer cannot be represented accurately as a sample containing only:

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

Its backbone contains both types of structural unit:

\[ \ce{-CH2-CH(OH)-} \]

and

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

The molar ratio of hydroxyl-bearing to remaining ester-bearing units is \(3:1\).

However, the data do not establish how these two types are distributed along individual polymer chains. They could be grouped in sections or spread more irregularly. Some chains might contain different proportions from others.

Most importantly, reconstructing the hypothetical monomer \(\ce{CH2=CHOH}\) from the final hydroxyl-bearing unit doesn’t establish that it was the material originally polymerised.

The stated manufacturing history identifies vinyl ethanoate as the actual alkene feedstock. Hydrolysis subsequently changed the side groups.

The tempting shortcut assumes that a final repeating structure must preserve the identity of its original monomer. This example shows why carbon valence and structural evidence must be combined with the reaction history.

Question 5

Extension challenge: Can a repeating unit hide the shape of an entire molecule?

A chemist considers an idealised condensation reaction between two bifunctional monomers:

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

and

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

The first monomer has two amino groups, and the second has two carboxylic acid groups.

Assume that:

• Every product molecule contains exactly four molecules’ worth of each starting monomer.
• Each product is one connected molecule, formed only by amide bonds between these monomer-derived sections.
• Every amide bond formed eliminates exactly one molecule of water.
• No side reactions, branching, or losses of starting material occur.
• All products being compared have the same molecular composition and connectivity pattern apart from whether the chain closes into a ring.

In a hypothetical measurement, \(0.00100\ \mathrm{mol}\) of product molecules forms and \(0.00800\ \mathrm{mol}\) of water is released.

Two students propose different explanations.

Student A claims that the product consists of open, linear molecules containing four repeating pairs of monomers. They count two amide linkages per repeating pair and predict eight water molecules for each product molecule.

Student B suggests that each product forms a closed ring, with the final available amino and carboxylic acid groups reacting with each other.

Determine which proposed structure agrees with the data.

Your explanation must:

(a) Establish how many amide bonds are present in an open chain and in a closed ring.

(b) Calculate the amount of water each model predicts.

(c) Identify the functional groups that must remain at the ends of the proposed linear molecule.

(d) Explain how both proposed structures could satisfy carbon valence even though only one matches the reaction evidence.

(e) State why the usual bracketed repeating-unit formula alone cannot settle the question.

Solution 5

The product must be cyclic under the stated assumptions. A linear molecule formed from the eight starting monomer molecules would produce seven water molecules, whereas the measured water-to-product ratio corresponds to eight. The additional amide link closes the chain into a ring.

The crucial idea is that a complete molecule and a bracketed repeating-unit formula are not quite the same thing.

(a) Count the bonds needed to make each structure

Start with four diamine molecules and four dicarboxylic acid molecules.

There are eight separate molecules altogether.

To connect eight separate objects into one open chain, you need seven links.

You can picture this without chemistry first: putting eight separate beads into one row requires seven connections between them.

Here, each connection is an amide bond, formed by an amino group reacting with a carboxylic acid group.

Therefore, an open chain contains seven newly formed amide links and produces seven water molecules.

Once the eight monomer-derived sections form one connected open chain, two reactive ends remain.

Making one more amide bond between these ends closes the molecule into a ring.

The ring contains eight newly formed amide links.

(b) Compare the water predictions

Let \(N=0.00100\ \mathrm{mol}\) be the amount of product molecules formed.

For the open-chain model:

\[ \begin{aligned} n_{\mathrm{w,open}}&=7N\\ &=0.00700\ \mathrm{mol} \end{aligned} \]

For the ring model:

\[ \begin{aligned} n_{\mathrm{w,ring}}&=8N\\ &=0.00800\ \mathrm{mol} \end{aligned} \]

The hypothetical measurement is:

\[ n_{\mathrm{w,measured}} =0.00800\ \mathrm{mol} \]

This agrees with the cyclic model and not the open-chain model.

The numerical difference may look small, but it’s an entire bond per product molecule. Because the product is constructed from only eight monomer molecules, that difference cannot be ignored.

(c) Determine the linear molecule’s end groups

Each diamine starts with two amino groups:

\[ \ce{-NH2} \]

Each dicarboxylic acid starts with two carboxylic acid groups:

\[ \ce{-COOH} \]

With four molecules of each type, there are eight amino groups and eight carboxylic acid groups available initially.

Seven amide bonds consume seven amino groups and seven carboxylic acid groups.

That leaves:

• One unreacted \(\ce{-NH2}\) group.
• One unreacted \(\ce{-COOH}\) group.

Those are the terminal functional groups of the open-chain product.

The carbonyl carbon in the terminal \(\ce{-COOH}\) group has:

• A double bond to one oxygen: bond order 2.
• A single bond to the hydroxyl oxygen: bond order 1.
• A single bond to the neighbouring carbon: bond order 1.

Its total bond order is four.

In the cyclic product, the remaining amino and carboxylic acid groups react with each other, producing the eighth amide link and an eighth water molecule. Neither of those free terminal groups remains.

(d) Why carbon valence alone cannot choose the structure

Consider the repeating pattern:

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

Every carbonyl carbon has a C=O double bond and two single bonds, giving total bond order four.

Every saturated methylene carbon has two hydrogen bonds and two backbone bonds, again giving four.

These local bonding arrangements can occur in either a long open-chain structure or a closed-ring structure.

The distinction is what happens at the ends.

In an open chain, the final reactive positions must be completed with terminal functional groups. In a ring, the final positions bond to each other.

Both can satisfy ordinary carbon valence.

Valence is therefore necessary for checking the structures, but it doesn’t determine which global arrangement was formed.

(e) Why the repeating unit is insufficient

The bracketed formula records the repeating connectivity of the backbone, but it doesn’t automatically specify whether a finite molecule has open ends or forms a closed ring.

The open-chain model would be tempting if you treated every bracket boundary as an indefinite continuation and ignored the final two ends.

Student A’s specific error was counting two fully formed amide links per repeating pair while also describing a finite open chain. Their count effectively includes the bond that would join the final carbonyl carbon back to the first nitrogen.

That last connection is precisely what the open chain doesn’t have.

Under the problem’s assumptions, the amount of water provides independent evidence of how many amide links formed. It distinguishes the two structures where the local carbon-valence test cannot.

A real polymer sample is usually more complicated, containing a distribution of chain lengths and possibly different end groups. The exact molecule-by-molecule bookkeeping here is an idealised extension, not a general rule that every condensation-polymer sample forms rings.

The broader lesson is that checking carbon valence answers one question: Is this arrangement of bonds chemically valid?

Identifying the reaction pathway, the monomers actually used, and the complete molecular structure may require further information.

Once you’ve established a valid repeating unit, the next useful step is to distinguish the bonds that define the backbone from those formed or changed during the reaction. That is how a structural formula becomes evidence for a possible chemical process, rather than just a drawing that happens to give every carbon four bonds.