Form a polyamide from a diamine and a dicarboxylic acid
Join –NH₂ groups from a diamine to –COOH groups from a dicarboxylic acid, forming amide links and water.
A Level; Cambridge International Chemistry 9701, 2028–2030 syllabus, Topic 35. Track the functional groups, repeat-unit connectivity and small-molecule by-products. Written practice is self-checked
Before you begin
- Recognise alkenes, carboxylic acids, acyl chlorides, alcohols, amines, amides and ester linkages.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
Which linkage forms from a diamine and a dicarboxylic acid?
What you will learn
- Identify the two bifunctional monomers
- Form –CONH– links
- Preserve the residues and account for water
Amino and carboxyl groups form an amide
A diamine has two –NH₂ groups and a dicarboxylic acid has two –COOH groups. An amino group and a carboxyl group react to form an amide link, –CONH–, with water eliminated.
The diamine and diacid residues alternate in the chain. With hexane-1,6-diamine and hexanedioic acid, the resulting polyamide is commonly called nylon 6,6.
Retain both carbon frameworks
The repeat unit must include the carbon chain between the two amine groups and the chain between the two carboxyl groups. At each link, the amide nitrogen connects to a carbonyl carbon.
Both monomers have two reactive ends, so repeated condensation can extend the chain. Do not mistake the product for a polyester: nitrogen is part of each –CONH– link.
Pause and explain
What is eliminated when a diamine reacts with a dicarboxylic acid?
Worked example
Give the repeat for hexane-1,6-diamine plus hexanedioic acid.
Show the worked solution
- Keep –NH(CH₂)₆NH– from the diamine.
- Keep –CO(CH₂)₄CO– from the diacid.
- Join through –NH–CO– amide links.
Answer An equivalent repeat is [–NH(CH₂)₆NHCO(CH₂)₄CO–]ₙ; water is eliminated as amide links form.
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Write your answer and reasoning first. Use a hint only if you are stuck. The model solution is for self-checking; your written work is not automatically marked.
Write a repeat unit for H₂N–(CH₂)₄–NH₂ and HOOC–(CH₂)₈–COOH.
Give me a hint
Alternate the diamine and diacid residues through –CONH– links.
Compare my reasoning
- Keep four CH₂ groups between the two N atoms.
- Keep eight CH₂ groups between the two carbonyls.
- Show the nitrogen-to-carbonyl amide connections.
[–NH(CH₂)₄NHCO(CH₂)₈CO–]ₙ; water is eliminated as the amide links form.
Look for these in your work
- Both chain lengths correct
- Amide links correct
- Water stated
Why is nylon 6,6 a polyamide rather than a polyester?
Give me a hint
Inspect the atoms in the repeating link.
Compare my reasoning
- Its link includes nitrogen.
- The characteristic link is –CONH–.
- An ester link would be –COO–.
Nylon 6,6 is a polyamide because its chain contains –CONH– links; it does not have the –COO– link that defines a polyester.
Look for these in your work
- Nitrogen noted
- Amide link named
- Ester contrast made
A suggested repeat for the diamine/diacid pair drops one terminal N atom. Why is that not a valid repeat?
Give me a hint
Trace the two ends of the diamine residue.
Compare my reasoning
- The diamine has an amino group at each end.
- Both ends can form amide links along the chain.
- Both nitrogen atoms therefore appear in the repeated residue.
Both nitrogens from the diamine are part of the repeating polyamide chain; omitting one breaks the residue and changes the monomer ratio/connectivity.
Look for these in your work
- Monomer ends counted
- Nitrogen atoms tracked
- Connectivity explained
Common mistakes
- Drawing an ester link instead of an amide
- Omitting nitrogen from the repeat
- Dropping a carbon from a monomer residue
- Treating the entire chain as one monomer
What can you explain now?
Which linkage forms from a diamine and a dicarboxylic acid?
Compare with the explanation
An amide link, –CONH–
The amino nitrogen bonds to the carboxyl carbon.
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Make it stick
Which groups make a polyamide from a diamine and dicarboxylic acid, and what link results?
If you needed the solution, close it and solve the problem again from a blank page. A correct answer is stronger when you can explain the reason for each step.