Form a polyamide from an aminocarboxylic acid
An aminocarboxylic acid carries both –NH₂ and –COOH, allowing self-condensation through amide links.
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 functional groups on an aminocarboxylic acid form the polyamide link?
What you will learn
- Recognise amino and carboxyl groups on one monomer
- Describe self-condensation into an amide chain
- Deduce the monomer residue in a repeat
One molecule has both linking groups
An aminocarboxylic acid contains an amino group and a carboxyl group in the same molecule. The –NH₂ group of one molecule reacts with the –COOH group of another to form an amide link, –CONH–.
Each monomer therefore has two reactive ends and can join two neighbours in a chain. Water is eliminated as the amide links form.
Keep the carbon chain between N and COOH
For H₂N–R–COOH, the polyamide repeat can be written [–NH–R–CO–]ₙ. The carbon skeleton R stays between nitrogen and carbonyl carbon.
Do not split an aminocarboxylic acid into separate diamine and diacid molecules. Its two different groups on one molecule are enough for chain-forming condensation.
Pause and explain
What type of link connects aminocarboxylic-acid residues?
Worked example
Write the repeat unit from 6-aminohexanoic acid, H₂N(CH₂)₅COOH.
Show the worked solution
- The monomer has –NH₂ and –COOH.
- Join N from one residue to the carbonyl C of the next.
- Keep all five CH₂ groups in the residue.
Answer [–NH(CH₂)₅CO–]ₙ, with water 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 the repeating residue from H₂N(CH₂)₄COOH.
Give me a hint
Connect the amino end of one molecule to the carboxyl end of another.
Compare my reasoning
- Keep four methylene groups between N and carbonyl C.
- Form –NH–CO– links between residues.
- Use bracket notation for the repeating polyamide.
[–NH(CH₂)₄CO–]ₙ; water is eliminated as amide bonds form.
Look for these in your work
- Carbon count retained
- Amide link correct
- By-product stated
Explain how this monomer forms a polymer without a separate diamine.
Give me a hint
Count its different reactive groups.
Compare my reasoning
- It has one –NH₂ group.
- It also has one –COOH group.
- Those groups link successive molecules at both ends.
Each aminocarboxylic-acid molecule is bifunctional, with an amino group and a carboxyl group, so repeated condensation makes a chain.
Look for these in your work
- Both groups identified
- Bifunctionality explained
- Chain formation described
A proposed repeat is [–NH(CH₂)₃CO–]ₙ for H₂N(CH₂)₅COOH. Find the missing atoms.
Give me a hint
Compare the count of methylene groups in monomer and repeat.
Compare my reasoning
- The monomer has five CH₂ groups.
- Those carbon atoms remain in the repeat residue.
- The proposed repeat is short by two CH₂ groups.
The repeat should be [–NH(CH₂)₅CO–]ₙ; condensation removes atoms from the linking groups as water, not two carbon atoms from R.
Look for these in your work
- Monomer count checked
- Residue traced
- Correction justified
Common mistakes
- Drawing a polyester link
- Removing a CH₂ group from the monomer residue
- Assuming a second monomer is required
- Omitting water from the condensation description
What can you explain now?
Which functional groups on an aminocarboxylic acid form the polyamide link?
Compare with the explanation
–NH₂ and –COOH
The amino and carboxyl groups form –CONH–.
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Make it stick
Why can an aminocarboxylic acid form a polyamide by itself?
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