How to Recover Diamine and Dicarboxylic Acid from Polyamides

A nylon fibre contains the chemical building blocks needed to make it in the first place. The challenge is that those building blocks have been joined together by amide bonds, so you cannot simply pull the polymer apart and collect the original molecules.

To recover the original diamine and dicarboxylic acid, you need to hydrolyse the polyamide. Water breaks the amide bonds, reversing the condensation reaction that formed the polymer. Under acidic conditions, the dicarboxylic acid is regenerated, while the diamine initially forms a protonated diammonium ion. You can then adjust the conditions to obtain the free diamine.

The key is to work backwards from the amide linkage. Once you recognise which part of the bond came from the carboxylic acid and which part came from the amine, recovering the monomers becomes a systematic process.

01Start with the amide bond

Consider the functional group that joins two sections of a polyamide chain:

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

This is an amide linkage. It forms when a carboxylic acid group reacts with an amine group, releasing water.

During condensation:

• The carboxylic acid loses its hydroxyl group, \(\mathrm{-OH}\).
• The amine loses a hydrogen atom, \(\mathrm{-H}\).
• The remaining carbonyl carbon and nitrogen become connected.

During hydrolysis, the process runs in reverse. Water supplies the groups needed to restore the original functional groups.

\[ \mathrm{R-CO-NH-R' + H_2O} \]

\[ \mathrm{\longrightarrow R-COOH + H_2N-R'} \]

This is the simplified overall representation of amide hydrolysis. The actual form of the amine product depends on the pH of the reaction mixture.

Notice something important: hydrolysis does not remove the carbon skeletons or rearrange the carbon atoms. It breaks the bond between the carbonyl carbon and nitrogen, restoring the functional groups that existed in the original monomers.

That gives us our first useful rule:

To identify the monomers of a polyamide, break each amide linkage between the carbonyl carbon and nitrogen, then restore an -OH to the carbonyl carbon and an H to the nitrogen.

02Recovering a diamine and a dicarboxylic acid

A typical synthetic polyamide is formed from two bifunctional monomers:

• A diamine, containing two primary amine groups, \(\mathrm{-NH_2}\).
• A dicarboxylic acid, containing two carboxylic acid groups, \(\mathrm{-COOH}\).

Both monomers have two reactive ends. This allows them to join repeatedly and form a long polymer chain.

For example, nylon 6,6 is produced from:

• Hexane-1,6-diamine: \(\mathrm{H_2N(CH_2)_6NH_2}\)
• Hexanedioic acid (adipic acid): \(\mathrm{HOOC(CH_2)_4COOH}\)

Each diamine molecule reacts with carboxylic acid groups at its two ends. Each dicarboxylic acid molecule reacts with amine groups at its two ends.

When the polymer is hydrolysed completely, both types of monomer are regenerated.

\[ \begin{aligned} &\mathrm{[-NH(CH_2)_6NHCO(CH_2)_4CO-]_n}\\ &\quad \xrightarrow{\text{hydrolysis}}\\ &\mathrm{H_2N(CH_2)_6NH_2}\\ &\qquad +\mathrm{HOOC(CH_2)_4COOH} \end{aligned} \]

The equation represents the idealised repeat-unit relationship. The value of \(n\) represents the number of repeat units, and the small contribution of the polymer’s end groups is ignored.

The recovered products are hexane-1,6-diamine and hexanedioic acid.

A useful way to read the polymer backwards

Do not try to recognise the whole polymer as one enormous molecule. Instead, locate the repeating amide linkage and work through it systematically.

The carbonyl carbon belongs to the dicarboxylic acid. The nitrogen belongs to the diamine.

This is the most important structural distinction. Students sometimes assign the carbonyl group to the amine because it appears directly beside nitrogen in the polymer. That is incorrect: the carbonyl carbon came from the carboxylic acid group.

03Why the reaction conditions matter

There is one complication with the simplified hydrolysis equation: it shows a neutral amine as the product, but an acidic reaction mixture does not generally contain the diamine in that form.

In acidic solution, the amine groups accept hydrogen ions, forming ammonium groups.

\[ \mathrm{-NH_2 + H^+ \rightleftharpoons -NH_3^+} \]

Because the original molecule is a diamine, both ends can become protonated.

\[ \mathrm{H_2N-R-NH_2 + 2H^+ \rightleftharpoons H_3N^+-R-NH_3^+} \]

The product is a diammonium ion with an overall charge of \(+2\). If hydrochloric acid is used, chloride ions act as counterions.

The important distinction is:

• Acid hydrolysis regenerates the carboxylic acid and produces a protonated diamine.
• Subsequent basification converts the protonated diamine into the free diamine.

To release the free diamine, hydroxide ions remove the additional hydrogen ions:

\[ \mathrm{H_3N^+-R-NH_3^+ + 2OH^-} \]

\[ \mathrm{\longrightarrow H_2N-R-NH_2 + 2H_2O} \]

This is a separate acid-base reaction, not another polymer-breaking reaction.

The amide bonds have already been hydrolysed. Adding a base changes the protonation state of the diamine; it does not reconstruct the polymer or break the carbon skeleton apart.

Acid hydrolysis: recover the dicarboxylic acid and diamine salt

For acid hydrolysis, the polyamide is heated with aqueous acid. The acid promotes cleavage of the amide linkages, while water supplies the atoms needed to restore the carboxylic acid and amine groups.

For nylon 6,6, the idealised reaction is:

\[ \begin{aligned} &\ce{[-NH(CH2)6NHCO(CH2)4CO-]_{n}}\\ &\ce{+ 2nH2O + 2nHCl}\\ &\ce{-> nHOOC(CH2)4COOH}\\ &\ce{+ n[H3N(CH2)6NH3]Cl2} \end{aligned} \]

The products are:

1. Hexanedioic acid (adipic acid), \(\mathrm{HOOC(CH_2)_4COOH}\).
2. Hexane-1,6-diammonium chloride, \(\mathrm{[H_3N(CH_2)_6NH_3]Cl_2}\).

The two chloride ions balance the \(+2\) charge on each diammonium ion.

To obtain the free diamine, the diammonium salt must be treated with a base:

\[ \ce{[H3N(CH2)6NH3]^2+ + 2OH^-} \]

\[ \ce{-> H2N(CH2)6NH2 + 2H2O} \]

There is an important separation issue here. If you add sodium hydroxide to the entire hydrolysis mixture before separating the products, you will also neutralise the dicarboxylic acid.

You could end up with sodium adipate rather than the adipic acid you intended to recover.

A sensible recovery strategy is therefore:

1. Complete the acid hydrolysis.
2. Separate the dicarboxylic acid from the diammonium-containing solution, using an appropriate separation method based on the products’ solubilities.
3. Treat the separated diammonium salt solution with a base to release the free diamine.

The exact isolation method depends on the particular acid and diamine. The reaction equation identifies the products; their solubilities determine how easily they can be separated.

Alkaline hydrolysis: why the products are different

Now consider what happens if the polyamide is hydrolysed under alkaline conditions using aqueous sodium hydroxide.

The amide linkage still breaks, but the carboxylic acid product does not remain as \(\mathrm{-COOH}\). In the presence of excess hydroxide, it forms a carboxylate ion, \(\mathrm{-COO^-}\).

The nitrogen-containing product is the neutral amine.

\[ \ce{R-CO-NH-R' + OH^- -> R-COO^- + R'-NH2} \]

For nylon 6,6:

\[ \begin{aligned} &\ce{[-NH(CH2)6NHCO(CH2)4CO-]_{n}}\\ &\ce{+ 2nNaOH}\\ &\ce{-> nH2N(CH2)6NH2}\\ &\ce{+ nNaOOC(CH2)4COONa} \end{aligned} \]

The products are:

• Hexane-1,6-diamine.
• Sodium hexanedioate (sodium adipate).

Notice that the alkaline reaction gives you the free diamine directly, but the dicarboxylic acid is present as a salt.

To recover the dicarboxylic acid, first separate the diamine from the sodium dicarboxylate. Then acidify the separated salt:

\[ \ce{NaOOC(CH2)4COONa + 2HCl} \]

\[ \ce{-> HOOC(CH2)4COOH + 2NaCl} \]

The two sodium ions are replaced by hydrogen ions, restoring both carboxylic acid groups.

Acid versus alkaline hydrolysis

This comparison is worth learning because reaction conditions determine the form of the recovered functional groups.

A useful exam decision rule is:

• If the question specifies acidic conditions, protonate the amine groups.
• If it specifies alkaline conditions, show the carboxylate salt and the neutral amine.
• If it asks for the original neutral monomers, consider whether an additional acid-base reaction is required.

For a refresher on why amines accept protons and carboxylic acids donate them, see Organic Acids and Bases: HSC Chemistry Structures and Reactions.

04What happens during an actual recovery experiment?

The reaction equation gives us the expected products. An experiment must establish whether the polymer has actually been broken down completely and whether the products have been separated successfully.

Imagine a chemical recycling experiment using nylon 6,6.

Small pieces of nylon are placed in a suitable corrosion-resistant reaction vessel containing aqueous acid. The vessel is stirred and heated under controlled conditions. The reaction conditions must be chosen for the particular polymer.

This is not necessarily a simple open-flask experiment. Nylon can be resistant to hydrolysis, and complete depolymerisation may require substantially more vigorous conditions than a school laboratory heating setup. Some research-scale processes use specialised high-temperature reactors.

What should the experiment produce?

If hydrolysis is complete, the polymer’s amide linkages have been cleaved and the original monomer-derived molecules are recovered.

If hydrolysis is incomplete, some shorter polymer fragments remain. These are called oligomers. They still contain one or more amide linkages, so they are not the same as fully recovered monomers.

This distinction matters when interpreting experimental evidence. A nylon fibre becoming weaker or breaking apart does not, by itself, prove that all its amide bonds have been hydrolysed.

A successful recovery experiment needs evidence that the expected products have actually formed. Depending on the method, this could include identifying the isolated compounds, measuring their masses, or analysing the functional groups present.

There is also a useful connection to polymer properties. Amide linkages are covalent bonds within the polymer backbone. Hydrogen bonds between neighbouring chains help give nylon its strength. Hydrolysis breaks the covalent backbone itself, so drying the material cannot simply restore the original long chains.

For the broader explanation of how condensation produces nylon, and how its structure affects its properties, see Polyamides: How Nylon Forms, Its Structure and Properties.

05A reliable method for recovering the monomers from a structure

When an exam question gives you an unfamiliar polyamide, follow this sequence.

Step 1: Locate every amide linkage.

Look for:

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

Do not mistake an amide linkage for an ester linkage, which contains \(\ce{-CO-O-}\).

Step 2: Identify the bond that hydrolysis breaks.

Break the bond between the carbonyl carbon and the nitrogen.

Do not break the carbon-carbon bonds in the backbone.

Step 3: Restore the original functional groups.

On the carbonyl side, restore the carboxylic acid group by adding \(\mathrm{-OH}\).

On the nitrogen side, restore the amine group by adding hydrogen.

Step 4: Check the reaction conditions.

For acidic hydrolysis, show \(\mathrm{-NH_3^+}\) rather than \(\mathrm{-NH_2}\).

For alkaline hydrolysis, show \(\mathrm{-COO^-}\) rather than \(\mathrm{-COOH}\).

Step 5: Reconstruct and name the monomers.

Follow each carbon chain without changing its length. Check that the diamine has two amine groups and the dicarboxylic acid has two carboxylic acid groups.

Step 6: Check whether hydrolysis is complete.

If an amide linkage remains, you have not recovered all the original monomers.

There is one more detail to watch: a repeating unit is an idealised representation of the internal polymer structure. The actual ends of a finite polymer chain can contain unreacted functional groups. Those end groups affect the exact number of amide bonds and water molecules involved.

06Worked examples

A contrasting case: not every polyamide gives two monomers

There is one important limitation to the method.

Not all polyamides are formed from a separate diamine and dicarboxylic acid. Nylon 6 is a useful example.

Its repeating unit is:

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

Nylon 6 is commonly produced by ring-opening polymerisation of a single monomer, \(\varepsilon\)-caprolactam.

Complete hydrolysis of its amide linkages regenerates 6-aminohexanoic acid:

\[ \ce{H2N(CH2)5COOH} \]

This molecule contains both an amine group and a carboxylic acid group.

It is not a separate diamine and dicarboxylic acid.

The lesson is simple: identify the original monomers from the actual polymer structure rather than assuming every polyamide follows the nylon 6,6 pattern.

Extension note for NSW HSC Chemistry

The detailed acid and alkaline hydrolysis pathways and monomer-isolation methods in this article are extension applications of polyamide structure, condensation chemistry, and acid-base reactions.

They provide useful practice in interpreting organic structures, but should not be mistaken for a claim that complete nylon depolymerisation is a prescribed HSC practical.

07Questions and solutions

Question 1

A synthetic polyamide has the following repeating unit:

\[ \left[ -\mathrm{NH}-(\mathrm{CH_2})_3 -\mathrm{NH}-\mathrm{CO} -(\mathrm{CH_2})_2-\mathrm{CO}- \right]_n \]

Determine the two original monomers formed by complete hydrolysis.

Give the condensed structural formula and IUPAC name of each monomer.

Solution 1

The two original monomers are propane-1,3-diamine and butanedioic acid.

Step 1: Identify the amide linkages.

The repeating unit contains two \(\mathrm{-CO-NH-}\) linkages.

Break each bond between the carbonyl carbon and nitrogen.

Step 2: Recover the diamine.

The nitrogen atoms are connected through three methylene groups.

Adding hydrogen to each nitrogen produces:

\[ \mathrm{H_2N-(CH_2)_3-NH_2} \]

This is propane-1,3-diamine.

Step 3: Recover the dicarboxylic acid.

The carbonyl-containing section has two carbonyl carbons and two methylene groups.

Restoring an \(\mathrm{-OH}\) group to each carbonyl carbon gives:

\[ \mathrm{HOOC-(CH_2)_2-COOH} \]

This is butanedioic acid.

Step 4: Check the functional groups.

The diamine has two primary amine groups, while the acid has two carboxylic acid groups.

Both structures can therefore form the original alternating polyamide chain.

Question 2

An idealised sample of nylon 6,6 undergoes complete acid hydrolysis.

After hydrolysis, the adipic acid has been separated, and excess hydrochloric acid has been removed. The remaining solution contains \(0.0200\ \mathrm{mol}\) of hexane-1,6-diammonium ions.

A student adds \(0.0300\ \mathrm{mol}\) of hydroxide ions to the solution.

Calculate:

(a) The amount of neutral hexane-1,6-diamine produced.

(b) The amount of diammonium ions remaining.

Explain why two moles of hydroxide ions are required to liberate one mole of diamine.

Solution 2

The reaction produces \(0.0150\ \mathrm{mol}\) of neutral diamine, leaving \(0.00500\ \mathrm{mol}\) of diammonium ions.

Step 1: Write the neutralisation equation.

Each diammonium ion contains two protonated amine groups.

\[ \ce{H3N+-R-NH3+ + 2OH^-} \]

\[ \ce{-> H2N-R-NH2 + 2H2O} \]

The mole ratio is:

\[ 2\ \mathrm{mol\ OH^-}:1\ \mathrm{mol\ diamine} \]

Step 2: Calculate the amount of diammonium ions neutralised.

\[ \begin{aligned} n(\text{diammonium neutralised}) &=\frac{0.0300}{2}\\ &=0.0150\ \mathrm{mol} \end{aligned} \]

Each neutralised diammonium ion produces one neutral diamine molecule.

Therefore:

\[ \boxed{n(\text{diamine})=0.0150\ \mathrm{mol}} \]

Step 3: Calculate the amount remaining.

Initially, there were \(0.0200\ \mathrm{mol}\) of diammonium ions.

\[ \begin{aligned} n(\text{remaining}) &=0.0200-0.0150\\ &=0.00500\ \mathrm{mol} \end{aligned} \]

Therefore:

\[ \boxed{n(\text{diammonium remaining})=0.00500\ \mathrm{mol}} \]

The important point is that each diamine molecule has two amine groups. Both must lose their additional hydrogen ions to produce the neutral diamine.

A common mistake is to use a 1:1 ratio between hydroxide and diammonium ions. That would only neutralise one of the two protonated nitrogen groups.

Question 3

A polyamide has the following idealised repeating unit:

\[ \left[ -\mathrm{NH}-(\mathrm{CH_2})_4 -\mathrm{NH}-\mathrm{CO} -(\mathrm{CH_2})_3-\mathrm{CO}- \right]_n \]

A sample containing \(0.0125\ \mathrm{mol}\) of repeat units undergoes complete alkaline hydrolysis using sodium hydroxide.

(a) Identify both organic products, giving their condensed structural formulas and names.

(b) Calculate the amount of sodium hydroxide consumed.

(c) Calculate the amount of hydrochloric acid required to convert all the sodium dicarboxylate into the original dicarboxylic acid.

(d) Explain why the diamine should be separated before acidifying the dicarboxylate solution if the aim is to recover both neutral monomers.

Solution 3

The products are butane-1,4-diamine and sodium pentanedioate. The reaction consumes \(0.0250\ \mathrm{mol}\) of sodium hydroxide, and \(0.0250\ \mathrm{mol}\) of hydrochloric acid is required to regenerate the dicarboxylic acid.

Step 1: Reconstruct the diamine.

The nitrogen atoms are separated by four methylene groups.

\[ \mathrm{H_2N-(CH_2)_4-NH_2} \]

This is butane-1,4-diamine.

Step 2: Reconstruct the dicarboxylate.

The acid-derived section contains three methylene groups between two carbonyl carbons.

Under alkaline conditions, both carboxylic acid groups form carboxylate ions:

\[ \mathrm{NaOOC-(CH_2)_3-COONa} \]

This is sodium pentanedioate.

The complete organic products are therefore:

• Butane-1,4-diamine.
• Sodium pentanedioate.

Step 3: Calculate the amount of sodium hydroxide consumed.

Each idealised repeat unit contains two amide linkages.

Each linkage requires one hydroxide ion for alkaline hydrolysis.

Therefore:

\[ \begin{aligned} n(\ce{NaOH}) &=2\times n(\text{repeat units})\\ &=2\times0.0125\\ &=0.0250\ \mathrm{mol} \end{aligned} \]

Step 4: Calculate the amount of hydrochloric acid required.

Each sodium pentanedioate molecule contains two carboxylate ions.

Both require a hydrogen ion to regenerate the dicarboxylic acid.

\[ \ce{NaOOC(CH2)3COONa + 2HCl} \]

\[ \ce{-> HOOC(CH2)3COOH + 2NaCl} \]

The mole ratio is 1:2 between sodium pentanedioate and hydrochloric acid.

Since the amount of sodium pentanedioate is \(0.0125\ \mathrm{mol}\):

\[ \begin{aligned} n(\ce{HCl}) &=2\times0.0125\\ &=0.0250\ \mathrm{mol} \end{aligned} \]

Step 5: Explain the separation requirement.

The diamine is a base. If hydrochloric acid is added to a mixture containing both the diamine and sodium pentanedioate, the acid can also protonate the diamine.

This produces the diammonium salt rather than leaving the diamine in its neutral form.

Separating the diamine first allows the carboxylate fraction to be acidified without changing the recovered diamine’s protonation state.

The key is to consider the acid-base behaviour of both products, not just the reaction you intend to perform.

Question 4

A single uncapped, linear polyamide chain contains four diamine molecules and four dicarboxylic acid molecules.

The monomers are arranged alternately along the chain.

(a) Determine the initial number of amide linkages.

(b) Exactly six amide linkages undergo hydrolysis. How many original monomer molecules have been completely separated from the chain, and how many remain joined together?

(c) Calculate the amount of water consumed per mole of original polymer chains.

Explain why counting two amide linkages per monomer pair gives an incorrect answer for the initial finite chain.

Solution 4

The original chain contains seven amide linkages. After six are hydrolysed, six monomer molecules are free, while one diamine and one dicarboxylic acid remain joined by a single amide linkage. Six moles of water are consumed per mole of original chains.

Step 1: Count the original monomer molecules.

There are:

• Four diamine molecules.
• Four dicarboxylic acid molecules.

Therefore, there are eight monomer molecules in total.

Step 2: Count the connections.

A linear chain containing eight connected molecules requires seven connections.

This follows from a simple counting principle: the first molecule needs no connection to a previous molecule, while each additional molecule requires one new connection.

Therefore:

\[ N(\text{amide linkages})=8-1=7 \]

The initial chain has seven amide linkages, not eight.

Step 3: Determine the products after six hydrolysis events.

Six of the seven linkages have been broken.

Only one amide linkage remains.

The eight original monomer molecules are now distributed as:

• Six individual, completely separated monomer molecules.
• One two-monomer fragment containing a diamine and a dicarboxylic acid still joined by an amide bond.

The six free monomers consist of three diamines and three dicarboxylic acids.

The remaining fragment contains one of each monomer type.

Step 4: Calculate the water consumed.

Each hydrolysed amide linkage consumes one water molecule.

Therefore:

\[ n(\ce{H2O})=6\ \mathrm{mol} \]

per mole of original polymer chains.

Step 5: Explain the end-group effect.

The idealised repeating unit describes the internal structure of a long polymer chain. It does not account for the two terminal functional groups of a finite chain.

In this case, the chain has one terminal amine group and one terminal carboxylic acid group.

These groups have not formed an amide linkage with another monomer.

Consequently, the chain contains one fewer amide linkage than the total obtained by simply multiplying two linkages by four monomer pairs.

This is why end groups must be considered when working with short polymer chains.

Question 5

A chemist analyses a nylon 6,6 sample containing:

• \(0.100\ \mathrm{mol}\) of diamine-derived monomer units.
• \(0.100\ \mathrm{mol}\) of dicarboxylic-acid-derived monomer units.

The sample consists entirely of one of two possible structures:

Model A: Uncapped, linear chains. Each chain contains equal numbers of diamine and dicarboxylic acid molecules, with one terminal amine group and one terminal carboxylic acid group.

Model B: Cyclic chains. Each chain contains equal numbers of both monomers, but every functional group has reacted, leaving no terminal groups.

The chemist performs complete hydrolysis and measures a total consumption of \(0.195\ \mathrm{mol}\) of water.

Assume:

• Every amide linkage consumes exactly one mole of water during hydrolysis.
• Hydrolysis is complete.
• No other reaction consumes water.
• The sample contains only one of the two proposed structures, with no capped chains or unreacted monomers.

Determine:

(a) Which structural model is consistent with the measured water consumption?

(b) The amount of polymer chains present.

(c) The average number of diamine-dicarboxylic acid monomer pairs per chain.

(d) Whether measuring the 1:1 mole ratio of recovered diamine to dicarboxylic acid would be sufficient to distinguish the two models.

Solution 5

The sample must consist of linear chains (Model A). There are \(0.00500\ \mathrm{mol}\) of polymer chains, with an average of 20 diamine-dicarboxylic acid pairs per chain.

The key is that linear and cyclic chains with identical monomer compositions have different numbers of amide linkages.

Step 1: Calculate the maximum number of amide linkages if all chains are cyclic.

The sample contains \(0.100\ \mathrm{mol}\) of diamine-derived units and \(0.100\ \mathrm{mol}\) of dicarboxylic-acid-derived units.

Each pair can form two amide linkages.

Therefore:

\[ \begin{aligned} n(\text{amide linkages}) &=2\times0.100\\ &=0.200\ \mathrm{mol} \end{aligned} \]

For cyclic chains, all functional groups have reacted.

There are no terminal groups, so all \(0.200\ \mathrm{mol}\) of potential linkages are present.

Complete hydrolysis would consume:

\[ n(\ce{H2O})=0.200\ \mathrm{mol} \]

This does not match the measured \(0.195\ \mathrm{mol}\).

Therefore, Model B is inconsistent with the evidence.

Step 2: Determine the relationship for linear chains.

A linear chain containing \(k\) diamine molecules and \(k\) dicarboxylic acid molecules has:

\[ 2k\ \text{monomer molecules} \]

Connecting these molecules in a single open chain requires:

\[ 2k-1\ \text{amide linkages} \]

One amine group and one carboxylic acid group remain at the ends.

For the entire sample, this means:

\[ n(\text{amide linkages}) = 2n(\text{monomer pairs}) - n(\text{chains}) \]

Each mole of amide linkages consumes one mole of water.

Therefore:

\[ n(\ce{H2O}) = 2n(\text{monomer pairs}) - n(\text{chains}) \]

Step 3: Calculate the amount of polymer chains.

Substitute the measured values:

\[ 0.195=2(0.100)-n(\text{chains}) \]

Rearrange:

\[ \begin{aligned} n(\text{chains}) &=0.200-0.195\\ &=0.00500\ \mathrm{mol} \end{aligned} \]

Therefore:

\[ \boxed{n(\text{chains})=0.00500\ \mathrm{mol}} \]

Step 4: Calculate the average number of monomer pairs per chain.

The total amount of monomer pairs is \(0.100\ \mathrm{mol}\).

Divide by the amount of polymer chains:

\[ \begin{aligned} \text{Average pairs per chain} &=\frac{0.100}{0.00500}\\ &=20 \end{aligned} \]

Each average chain therefore contains:

• 20 diamine molecules.
• 20 dicarboxylic acid molecules.
• 40 original monomer molecules in total.
• 39 amide linkages.

Check the water consumption:

\[ 0.00500\times39=0.195\ \mathrm{mol} \]

This agrees exactly with the measured value.

Step 5: Decide whether the recovered monomer ratio distinguishes the models.

No. Both linear and cyclic chains contain equal numbers of diamine and dicarboxylic acid monomer units.

Complete hydrolysis therefore produces:

\[ \text{diamine}:\text{dicarboxylic acid}=1:1 \]

regardless of whether the original chains were linear or cyclic.

The difference lies in the number of amide linkages, not the identity or ratio of the monomers.

A student who assumes every repeat pair must always contain two amide bonds would predict \(0.200\ \mathrm{mol}\) of water consumption and incorrectly reject the experimental measurement.

The missing \(0.00500\ \mathrm{mol}\) of water consumption is not evidence of incomplete hydrolysis. It reflects the one missing amide linkage per linear chain.

This is the crucial distinction: the repeating unit describes the internal polymer pattern, while the actual chain structure determines how many linkages exist.

08The key idea to carry forward

Recovering monomers from a polyamide is a matter of reversing the amide linkages without changing the carbon skeletons.

For a polyamide made from a diamine and a dicarboxylic acid:

• Break the carbonyl carbon-nitrogen bonds.
• Restore \(\mathrm{-COOH}\) and \(\mathrm{-NH_2}\).
• Account for protonation under acidic or alkaline conditions.
• Check whether the hydrolysis is complete.
• Consider terminal groups when calculating the exact number of bonds in a finite chain.

The next useful step is learning to draw the actual end groups of short polyamide chains. Once you can distinguish an idealised repeating unit from a finite polymer molecule, both monomer recovery and polymer stoichiometry become much more reliable.