Why Nylon-6 and Nylon-6,6 Have Different Names

A label on a nylon fibre says **Nylon-6**. Another says **Nylon-6,6**. Both materials are polyamides, both contain amide links, and both can be made into strong fibres. So why does one have a single number while the other has two?

The numbers tell you how many carbon atoms are present in the starting monomer or monomers. Nylon-6 is made from a single six-carbon monomer. Nylon-6,6 is made from two different monomers, each containing six carbon atoms.

The comma is important. It separates the carbon counts of the two starting monomers.

## Why the names are different

The naming system for many nylons is based on the carbon atoms in their starting materials.

- **Nylon-6:** one monomer containing six carbon atoms.
- **Nylon-6,6:** two monomers, with six carbon atoms in each.

The difference becomes clearer when you look at the actual chemicals.

| Polymer | Starting monomer(s) | Carbon count |
|---|---|---|
| Nylon-6 | ε-caprolactam | 6 |
| Nylon-6,6 | Hexane-1,6-diamine | 6 |
| Nylon-6,6 | Hexanedioic acid (adipic acid) | 6 |

Nylon-6 has one number because its conventional starting material is a single six-carbon molecule. Nylon-6,6 has two numbers because its polymer chain is formed from two different six-carbon molecules.

[IMAGE: Annotated comparison of the molecular structures of epsilon-caprolactam, hexane-1,6-diamine and hexanedioic acid. Highlight every carbon atom, identify the two functional groups on each Nylon-6,6 monomer, and show the six-carbon count for each starting material. Clearly distinguish the single cyclic Nylon-6 monomer from the two separate Nylon-6,6 monomers.]

## Nylon-6: one monomer with six carbon atoms

Nylon-6 is commonly manufactured using a molecule called ε-caprolactam (epsilon-caprolactam).

Caprolactam is a cyclic molecule, meaning its atoms form a ring. It contains:

- Five carbon atoms in a row as methylene groups, \(\mathrm{CH_2}\).
- One additional carbon atom in a carbonyl group, \(\mathrm{C=O}\).
- One nitrogen atom.

That gives six carbon atoms altogether.

[IMAGE: Structural formula of epsilon-caprolactam shown as a clearly recognisable seven-membered lactam ring, with five methylene carbon atoms, one carbonyl carbon and one nitrogen atom individually identified. Add a numbered count of the six carbon atoms.]

During polymer formation, the caprolactam ring opens and the molecules join to form a long polyamide chain. This process is called ring-opening polymerisation.

The resulting Nylon-6 repeating unit is:

\[
\boxed{[-\mathrm{NH-(CH_2)_5-CO-}]_n}
\]

Count the carbon atoms in this repeating unit:

- Five carbon atoms from \((\mathrm{CH_2})_5\).
- One carbon atom from the carbonyl group, \(\mathrm{CO}\).

Total: six carbon atoms.

There is another useful way to represent Nylon-6. Its repeating unit can be derived from the open-chain molecule 6-aminohexanoic acid:

\[
\mathrm{H_2N-(CH_2)_5-COOH}
\]

This molecule also contains six carbon atoms. Its amino group and carboxylic acid group allow it to form amide links through condensation reactions.

For HSC structural questions, this open-chain representation is useful when identifying amide linkages and reconstructing monomers. However, remember that the common industrial route to Nylon-6 uses ring-opening polymerisation of caprolactam. Do not automatically classify every Nylon-6 manufacturing route as condensation polymerisation.

### Worked example: Counting the carbons in Nylon-6

A polymer has the repeating unit:

\[
[-\mathrm{NH-(CH_2)_5-CO-}]_n
\]

Explain why it is called Nylon-6 rather than Nylon-5.

**Step 1**: Identify the carbon-containing sections.

There are five carbon atoms in the \((\mathrm{CH_2})_5\) section.

**Step 2**: Include the carbonyl carbon.

The \(\mathrm{CO}\) section contains another carbon atom. It is easy to overlook because it is written alongside oxygen.

**Step 3**: Calculate the total.

\[
5+1=6\text{ carbon atoms}
\]

The polymer is Nylon-6 because its six-carbon monomer gives a repeating unit containing six carbon atoms.

The nitrogen and oxygen atoms do not contribute to the number in the name.

## Nylon-6,6: two monomers with six carbon atoms each

Nylon-6,6 is made from two different monomers:

1. Hexane-1,6-diamine, which contains six carbon atoms.
2. Hexanedioic acid (adipic acid), which also contains six carbon atoms.

The diamine has an amino group at each end:

\[
\mathrm{H_2N-(CH_2)_6-NH_2}
\]

The dicarboxylic acid has a carboxylic acid group at each end:

\[
\mathrm{HOOC-(CH_2)_4-COOH}
\]

Notice that the acid contains only four methylene groups, not six. Its two carbonyl carbons must also be counted.

[IMAGE: Side-by-side annotated structural formulas of hexane-1,6-diamine and hexanedioic acid. Number all six carbon atoms in each molecule. Highlight the two terminal NH2 groups on the diamine and the two terminal COOH groups on the diacid. Make the four internal CH2 groups and two carbonyl carbons in the acid visually distinct.]

When these monomers react, an amino group combines with a carboxylic acid group to form an amide linkage, releasing water.

The repeating unit is:

\[
[-\mathrm{NH-(CH_2)_6-NH-CO-(CH_2)_4-CO-}]_n
\]

There are two different six-carbon sections in this structure:

- The diamine-derived section contains six carbon atoms: \((\mathrm{CH_2})_6\).
- The diacid-derived section contains six carbon atoms: four from \((\mathrm{CH_2})_4\), plus two carbonyl carbons.

Therefore, the name is Nylon-6,6.

[IMAGE: Annotated Nylon-6,6 repeating unit with the diamine-derived six-carbon section and diacid-derived six-carbon section highlighted separately. Label both amide linkages, show the repeating-unit brackets, and distinguish the two carbonyl carbons from the four methylene carbons.]

### Worked example: Identifying the two numbers in Nylon-6,6

A student is given the following monomers:

\[
\mathrm{H_2N-(CH_2)_6-NH_2}
\]

and

\[
\mathrm{HOOC-(CH_2)_4-COOH}
\]

Explain how these structures determine the name of the polymer.

**Step 1**: Count the carbon atoms in the diamine.

The diamine contains six methylene groups:

\[
6\times 1=6\text{ carbon atoms}
\]

**Step 2**: Count the carbon atoms in the dicarboxylic acid.

There are four methylene carbon atoms and two carbonyl carbon atoms:

\[
4+2=6\text{ carbon atoms}
\]

**Step 3**: Write the two counts in the correct order.

The diamine contributes the first number, and the dicarboxylic acid contributes the second.

\[
\boxed{\text{Nylon-6,6}}
\]

The polymer has twelve carbon atoms in its complete repeating unit, but its name is not Nylon-12. The two numbers identify the separate monomer carbon counts, not the combined total.

## What the different structures mean for their properties

Both Nylon-6 and Nylon-6,6 belong to the polyamide family. A polyamide is a polymer containing repeated amide linkages:

\[
\mathrm{-CO-NH-}
\]

These linkages are responsible for many of nylon's useful properties.

Look at the nitrogen and oxygen atoms in an amide group. The N-H bond can donate a hydrogen bond, while the oxygen in the carbonyl group can accept one.

This allows neighbouring polymer chains to form hydrogen bonds with each other.

[IMAGE: Two adjacent nylon polymer chains shown as realistic zigzag molecular backbones. Include the amide groups, with dashed hydrogen bonds between the N-H group of one chain and the carbonyl oxygen of the neighbouring chain. Clearly distinguish covalent bonds within each chain from intermolecular hydrogen bonds between chains.]

These intermolecular hydrogen bonds help hold neighbouring chains together. They make nylon fibres strong, resistant to abrasion, and suitable for applications such as clothing, ropes, fishing lines, and industrial fabrics.

However, Nylon-6 and Nylon-6,6 do not have identical physical properties.

### Comparing their melting points

Typical melting point values for the two polymers are:

| Property | Nylon-6 | Nylon-6,6 |
|---|---:|---:|
| Number of carbon atoms in repeating unit | 6 | 12 |
| Amide linkages per repeating unit | 1 | 2 |
| Typical melting point | 225°C | 265°C |

These melting temperatures are representative values reported for Toray nylon resins, rather than universal values for every grade. Additives, crystallinity, moisture, and processing can affect the measured properties. [Toray technical information](https://www.plastics.toray/technical/amilan/tec_007.html).

Why does Nylon-6,6 generally have a higher melting point?

The answer involves the arrangement of the polymer chains.

Nylon-6 has five methylene groups between successive amide linkages. Nylon-6,6 has alternating sections containing six and four methylene groups.

This difference changes the geometry and packing of the polymer chains in their crystalline regions. The arrangement of the amide groups and the hydrogen bonds between neighbouring chains influence the thermal stability of these regions.

[IMAGE: Side-by-side molecular comparison of crystalline Nylon-6 and Nylon-6,6 chains. Show Nylon-6 with five methylene groups between amide groups and Nylon-6,6 with alternating six-methylene and four-methylene sections. Include adjacent chains and their hydrogen bonds, illustrating their different chain arrangements without suggesting that either polymer lacks hydrogen bonding.]

There is an important detail here.

Nylon-6,6 does not have twice as many amide linkages per carbon atom simply because its repeating unit contains two amide linkages.

Nylon-6 has one amide linkage per six carbon atoms. Nylon-6,6 has two amide linkages per twelve carbon atoms.

The ratio is the same.

The difference in their properties comes from how those groups are arranged along the polymer backbone and how the chains pack together, not simply from having more amide groups.

## How to identify the correct nylon name

When you encounter an unfamiliar polyamide structure, use this approach.

**Step 1: Identify the monomer arrangement.**

Is the polymer formed from one monomer containing both an amino group and a carboxylic acid group, or from two different monomers?

**Step 2: Count every carbon atom.**

Include:
- Carbon atoms in methylene groups, \(\mathrm{CH_2}\).
- Carbon atoms in carbonyl groups, \(\mathrm{C=O}\).

Do not count nitrogen or oxygen atoms.

**Step 3: Write the numbers in the correct order.**

For a diamine and dicarboxylic acid combination:

- First number = carbon atoms in the diamine.
- Second number = carbon atoms in the dicarboxylic acid.

**Step 4: Check the monomers, not just the total.**

A repeating unit containing twelve carbon atoms does not automatically represent Nylon-12 or Nylon-6,6. You need to identify how those carbon atoms are distributed between the starting monomers.

If you want more practice with the actual counting technique, see [How to Count Carbon Atoms in Nylon Monomers](https://tutorgum.com/learn/resources/hsc-chemistry-how-to-count-carbon-atoms-in-nylon-monomers/).

### Common mistakes to avoid

| Incorrect assumption | Correct understanding |
|---|---|
| Nylon-6 contains six \(\mathrm{CH_2}\) groups. | It contains five \(\mathrm{CH_2}\) groups and one carbonyl carbon in its repeating unit. |
| Nylon-6,6 contains 66 carbon atoms. | It has two starting monomers, each containing six carbon atoms. |
| Nylon-6,6 has the name Nylon-12. | The two numbers identify the separate monomer carbon counts. |
| The second number counts nitrogen atoms. | Both numbers count carbon atoms. |
| All nylons form by the same polymerisation mechanism. | Nylon-6 commonly forms by ring-opening polymerisation, while Nylon-6,6 forms by condensation polymerisation. |

For a broader explanation of how amide linkages form and how condensation polymerisation works, read [Condensation Polymerisation: Nylon and Polyester Explained for HSC Chemistry](https://tutorgum.com/learn/resources/hsc-chemistry-condensation-polymers-nylon-and-polyesters/).

## Questions and solutions

### Question 1

A diamine with the structure

\[
\mathrm{H_2N-(CH_2)_5-NH_2}
\]

reacts with a dicarboxylic acid with the structure

\[
\mathrm{HOOC-(CH_2)_4-COOH}
\]

to form a linear polyamide.

Determine the name of the resulting nylon. Explain how you obtained both numbers.

[IMAGE: The two supplied monomer structures drawn separately, with all carbon atoms shown as individual bonded atoms. Leave carbon numbering and the final polymer name unlabelled so students must perform the counting themselves.]

### Solution 1

The polymer is **Nylon-5,6**.

The first number comes from the diamine.

\[
\mathrm{(CH_2)_5}=5\text{ carbon atoms}
\]

The second number comes from the dicarboxylic acid.

There are four carbon atoms in the methylene groups and two carbon atoms in the carboxylic acid groups.

\[
4+2=6\text{ carbon atoms}
\]

The diamine contributes five carbons, and the dicarboxylic acid contributes six.

Therefore, the correct name is Nylon-5,6.

The key is to count the carbonyl carbons in the carboxylic acid groups. Missing these two atoms would give an incorrect second number.

### Question 2

A student is given the following polyamide repeating unit:

\[
[-\mathrm{NH-(CH_2)_4-NH-CO-(CH_2)_7-CO-}]_n
\]

The student calls the polymer Nylon-13 because there are thirteen carbon atoms in the repeating unit.

(a) Determine the correct name of the polymer.

(b) Calculate the number of carbon atoms contributed by each original monomer.

(c) Explain why the student's suggested name is incorrect.

[IMAGE: A complete Nylon-like repeating unit matching the supplied formula. Visually separate the diamine-derived section from the dicarboxylic-acid-derived section. Show all carbon atoms, including the two carbonyl carbons, but do not supply their counts or the polymer name.]

### Solution 2

The correct name is **Nylon-4,9**, not Nylon-13.

**Part (a): Identify the polymer name.**

The repeating unit contains a diamine-derived section and a dicarboxylic-acid-derived section.

The first section is:

\[
\mathrm{NH-(CH_2)_4-NH}
\]

It contains four carbon atoms.

The second section is:

\[
\mathrm{CO-(CH_2)_7-CO}
\]

It contains seven methylene carbons and two carbonyl carbons.

**Part (b): Calculate the carbon counts.**

For the diamine:

\[
4\text{ carbon atoms}
\]

For the dicarboxylic acid:

\[
7+2=9\text{ carbon atoms}
\]

Therefore, the polymer is Nylon-4,9.

**Part (c): Explain the incorrect naming.**

The student has correctly counted thirteen carbon atoms in the entire repeating unit:

\[
4+9=13
\]

However, the nylon naming system identifies the carbon count of each starting monomer separately.

It does not simply combine the two numbers into one total.

This is why Nylon-4,9 is the correct name, even though its repeating unit contains thirteen carbon atoms.

### Question 3

Consider the following repeating units:

Polymer A:

\[
[-\mathrm{NH-(CH_2)_5-CO-}]_n
\]

Polymer B:

\[
[-\mathrm{NH-(CH_2)_6-NH-CO-(CH_2)_4-CO-}]_n
\]

A student claims that Polymer B must have twice the amide-linkage density of Polymer A because its repeating unit contains two amide linkages instead of one.

Evaluate this claim by calculating the ratio of amide linkages to carbon atoms in each repeating unit.

Explain why the two polymers can still have different melting points.

[IMAGE: Side-by-side molecular structures of Polymer A and Polymer B using the exact supplied repeating units. Highlight each amide linkage and show the carbon backbone. Do not display calculated ratios or melting points.]

### Solution 3

The student's claim is incorrect. Both polymers have the same ratio of amide linkages to carbon atoms, even though their repeating units have different sizes.

**Step 1: Calculate the ratio for Polymer A.**

Polymer A contains one amide linkage and six carbon atoms.

\[
\text{Ratio}=\frac{1}{6}
\]

**Step 2: Calculate the ratio for Polymer B.**

Polymer B contains two amide linkages and twelve carbon atoms.

\[
\text{Ratio}=\frac{2}{12}
=\frac{1}{6}
\]

**Step 3: Compare the results.**

Both ratios are identical:

\[
\frac{1}{6}=\frac{2}{12}
\]

Therefore, Polymer B does not have twice the amide-linkage density.

**Step 4: Explain the difference in melting points.**

Both polymers form hydrogen bonds between neighbouring chains. However, their amide groups are arranged differently along the polymer backbone.

Polymer A has five methylene groups between successive amide linkages.

Polymer B has alternating sections containing six and four methylene groups.

These different arrangements influence how the chains align and pack in crystalline regions. Consequently, the polymers can have different melting points despite having the same overall amide-linkage-to-carbon ratio.

This is a useful distinction: counting functional groups alone does not tell you everything about a polymer's physical properties.

### Question 4

Nylon-6 can be represented using either of the following starting materials:

**Route A:**

\[
\varepsilon\text{-caprolactam}
\]

This molecule undergoes ring-opening polymerisation.

**Route B:**

\[
\mathrm{H_2N-(CH_2)_5-COOH}
\]

This molecule undergoes condensation polymerisation.

Both routes produce the same Nylon-6 repeating unit:

\[
[-\mathrm{NH-(CH_2)_5-CO-}]_n
\]

Explain:

(a) Why both routes produce the same nylon name.

(b) Why the two routes are classified as different types of polymerisation.

(c) Whether the presence of one amino group and one carboxylic acid group in Route B means the polymer should be called Nylon-6,6.

[IMAGE: Two distinct Nylon-6 formation pathways. At the top, show the cyclic epsilon-caprolactam molecule opening into a linear structure. Below, show the open-chain 6-aminohexanoic acid molecule with its terminal NH2 and COOH groups. Both pathways should lead visually towards the same polyamide repeating structure. Keep the two starting routes distinct.]

### Solution 4

Both routes produce **Nylon-6**, but they use different polymerisation mechanisms.

**Part (a): Explain the common name.**

Both starting materials contain six carbon atoms.

For caprolactam, there are five methylene carbons and one carbonyl carbon.

For 6-aminohexanoic acid, there are also five methylene carbons and one carboxyl carbon.

Both produce the same repeating unit:

\[
[-\mathrm{NH-(CH_2)_5-CO-}]_n
\]

Therefore, both products are Nylon-6.

The name is determined by the carbon count of the starting monomer, not by the particular polymerisation mechanism.

**Part (b): Explain the different mechanisms.**

Route A involves ring-opening polymerisation.

The cyclic caprolactam molecules open and join to form the polymer chain. The ring-opening process is not the same as repeatedly eliminating water during condensation polymerisation.

Route B involves condensation polymerisation.

The amino group of one 6-aminohexanoic acid molecule reacts with the carboxylic acid group of another. An amide linkage forms, and water is eliminated.

The two routes therefore involve different reaction processes, even though their final polymer repeating units are identical.

**Part (c): Explain why the name is not Nylon-6,6.**

No. Nylon-6,6 requires two different starting monomers: a diamine and a dicarboxylic acid.

Route B uses one type of monomer containing both an amino group and a carboxylic acid group.

Having two different functional groups does not mean that there are two different monomer types.

This is the important distinction between the two naming systems.

### Question 5

Two unlabelled polyamide samples are supplied to a laboratory.

Elemental analysis shows that both have the same empirical composition:

\[
\mathrm{C_6H_{11}NO}
\]

Their polymer chains have the following structures.

**Sample A:**

\[
\ldots-\mathrm{NH-(CH_2)_5-CO-NH-(CH_2)_5-CO}-\ldots
\]

**Sample B:**

\[
\ldots-\mathrm{NH-(CH_2)_6-NH-CO-(CH_2)_4-CO}-\ldots
\]

A laboratory technician makes the following argument:

"Both polymers have the same empirical composition, so they must be the same nylon. Also, if I draw two consecutive repeating units of Sample A inside one pair of brackets, I get twelve carbon atoms and two amide linkages. That means Sample A can be renamed Nylon-6,6."

Evaluate the technician's argument.

In your response:

(a) Identify the correct nylon name for each sample.

(b) Predict the products obtained when each polymer is completely hydrolysed.

(c) Explain why placing two repeating units inside one set of brackets does not change the polymer's name.

(d) Explain why identical empirical compositions do not necessarily mean identical polymer structures or physical properties.

[IMAGE: A challenging side-by-side comparison of the two supplied polymer chains. Show Sample A with two consecutive six-carbon polyamide segments and Sample B with its diamine-derived and diacid-derived sections. Include the alternative larger bracket around two consecutive Sample A segments to illustrate the technician's proposed interpretation. Do not label either sample with its correct nylon name.]

### Solution 5

The technician's argument is incorrect. Sample A is **Nylon-6**, while Sample B is **Nylon-6,6**. Having the same empirical composition does not make two polymers structurally identical.

**Part (a): Identify each polymer.**

For Sample A:

\[
[-\mathrm{NH-(CH_2)_5-CO-}]_n
\]

There are five methylene carbon atoms and one carbonyl carbon.

Its repeating unit contains six carbon atoms and comes from one six-carbon monomer.

Therefore, Sample A is Nylon-6.

For Sample B:

\[
[-\mathrm{NH-(CH_2)_6-NH-CO-(CH_2)_4-CO-}]_n
\]

The diamine-derived section contains six carbon atoms.

The dicarboxylic-acid-derived section contains four methylene carbon atoms and two carbonyl carbon atoms, giving another six.

Therefore, Sample B is Nylon-6,6.

**Part (b): Predict the hydrolysis products.**

Complete hydrolysis breaks the amide linkages and restores the functional groups of the starting monomers.

For Sample A, the original monomer is 6-aminohexanoic acid:

\[
\mathrm{H_2N-(CH_2)_5-COOH}
\]

For Sample B, the original monomers are:

\[
\mathrm{H_2N-(CH_2)_6-NH_2}
\]

and

\[
\mathrm{HOOC-(CH_2)_4-COOH}
\]

These are hexane-1,6-diamine and hexanedioic acid, respectively.

The hydrolysis products therefore provide a way to distinguish the original monomer structures, even when the polymers have the same empirical composition.

**Part (c): Explain why changing the bracket size does not change the name.**

This is the most important part of the problem.

A repeating unit is the smallest section of a polymer chain that repeats to produce the complete structure.

For Sample A, one repeating unit contains six carbon atoms.

If the technician places two consecutive repeating units inside the brackets, the new drawing contains twelve carbon atoms and two amide linkages.

However, this is simply two Nylon-6 repeating units written together.

The original monomer is still 6-aminohexanoic acid. Drawing a larger section of the chain does not change the chemical identity of the monomer.

The technician has confused the size of the chosen bracketed section with the number of carbon atoms in the starting monomer.

**Part (d): Explain the identical empirical compositions.**

Sample A has the repeating-unit formula:

\[
\mathrm{C_6H_{11}NO}
\]

Sample B has the repeating-unit formula:

\[
\mathrm{C_{12}H_{22}N_2O_2}
\]

Dividing the second formula by two gives:

\[
\mathrm{C_6H_{11}NO}
\]

Both therefore have the same empirical composition.

However, their atoms are connected in different arrangements.

Sample A has one amide linkage per six-carbon repeating unit. Sample B has two amide linkages per twelve-carbon repeating unit.

Their polymer backbones are different, their starting monomers are different, and their crystalline arrangements can also differ.

For example, typical Nylon-6 resin melts at approximately 225°C, while typical Nylon-6,6 resin melts at approximately 265°C.

The empirical formula alone cannot predict this difference. The arrangement of atoms and the way neighbouring polymer chains pack together also matter.

The key conclusion is that **a polymer's empirical formula does not uniquely identify its structure or its monomers**.

You must examine the connectivity of the polymer chain.

## Where to go next

The naming of Nylon-6 and Nylon-6,6 becomes much easier once you can move confidently between three representations:

1. The original monomer structures.
2. The repeating unit of the polymer.
3. The monomers recovered by breaking the amide linkages.

These are three ways of describing the same underlying chemistry.

The next useful skill is learning to reverse the process: given a polyamide repeating unit, identify the original diamine and dicarboxylic acid, including the correct end groups.

That is the foundation of [How to Draw the Repeating Unit of Nylon-6,6](https://tutorgum.com/learn/resources/hsc-chemistry-how-to-draw-the-repeating-unit-of-nylon-6-6/).

### Further scientific reading

- [IARC: Nylon resins and their monomer structures](https://www.ncbi.nlm.nih.gov/books/NBK424093/pdf/Bookshelf_NBK424093.pdf) - explains the carbon-counting convention used in nylon names.
- [American Chemical Society: Caprolactam](https://www.acs.org/molecule-of-the-week/archive/c/caprolactam.html) - describes the formation of Nylon-6 through ring-opening polymerisation.
- [IUPAC educational resource: Nylon-6 and Nylon-6,6](https://media.iupac.org/didac/Didac%20Eng/Didac04/Content/POL14.htm) - explores hydrogen bonding and the different arrangements of polyamide chains.
- [Toray: Nylon resin technical information](https://www.plastics.toray/technical/amilan/tec_007.html) - provides representative melting temperatures and other material properties.