Identify condensation monomers from a polymer section
Split ester or amide links to recover the monomer residues represented in a section of a condensation polymer.
Section 35 · Polymerisation
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
When reversing a polyester link, which functional groups are restored?
What you will learn
- Locate ester and amide linkages
- Recover corresponding monomer functional groups
- State when the polymer section does not reveal the exact precursor form
Split ester links at the joining bond
In a polyester, locate –CO–O–. Separate the acyl side from the oxygen side and restore –COOH and –OH groups to reconstruct a dicarboxylic acid plus diol, or inspect whether both groups belong to one hydroxycarboxylic-acid residue.
In a given section, identify the repeating residues that are actually shown. A short section may not show every monomer type present in the full chain.
Use nitrogen to recover amide monomers
In a polyamide, locate –CO–NH–. Reverse that link to restore –COOH and –NH₂ groups for a dicarboxylic acid/diamine route, or identify a single aminocarboxylic-acid residue when both groups lie in one monomer unit.
A polymer fragment with a –CO–O– link does not by itself reveal whether the carbonyl precursor was a carboxylic acid or an acyl chloride; both can lead to the same polyester link. Use other information if the exact precursor is required.
Pause and explain
Can a polyester link alone always show whether a carboxylic acid or acyl chloride was the original monomer?
Worked example
Identify the monomers represented in [–OCH₂CH₂OCOC₆H₄CO–]ₙ.
Show the worked solution
- Split the ester links between O and carbonyl C.
- Recover HOCH₂CH₂OH from the diol residue.
- Recover HOOC–C₆H₄–COOH from the diacid residue.
Answer The section represents ethane-1,2-diol and benzene-1,4-dicarboxylic acid (terephthalic acid); the fragment alone does not prove whether the acid or its acyl chloride was used.
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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.
Identify the monomer residues in [–NH(CH₂)₆NHCO(CH₂)₄CO–]ₙ.
Give me a hint
Split –CO–NH– links to restore amino and carboxyl ends.
Compare my reasoning
- Recover H₂N(CH₂)₆NH₂ from the diamine portion.
- Recover HOOC(CH₂)₄COOH from the diacid portion.
- Name the two bifunctional monomers.
Hexane-1,6-diamine and hexanedioic acid are the monomers represented.
Look for these in your work
- Diamine recovered
- Diacid recovered
- Names correct
A section contains repeating –OCH₂CO– residues. What single monomer type could form this polyester?
Give me a hint
Check whether one residue contains both –OH and –COOH after reversing an ester link.
Compare my reasoning
- Restore a hydroxyl group at the oxygen end.
- Restore a carboxyl group at the carbonyl end.
- Keep the CH₂ skeleton between them.
A hydroxycarboxylic acid such as HOCH₂COOH could form the –OCH₂CO– repeat by self-condensation.
Look for these in your work
- Both groups restored
- Skeleton retained
- Single monomer type identified
A polymer fragment shows an ester link but no chlorine. Can you conclude that the polymer was made from a dicarboxylic acid rather than a dioyl chloride?
Give me a hint
Distinguish the final repeat-unit structure from eliminated by-products.
Compare my reasoning
- Both precursors can form an ester link with a diol.
- Chlorine leaves as HCl in the acyl-chloride route.
- The final chain need not retain chlorine.
No. The fragment shows the ester-linked polymer but does not distinguish a dicarboxylic acid from a dioyl chloride precursor; both can yield the same repeat connectivity.
Look for these in your work
- Alternative routes considered
- Leaving group tracked
- Conclusion limited
Common mistakes
- Splitting the wrong bond in –COO–
- Reconstructing an ester monomer instead of the alcohol and acid
- Ignoring whether one monomer contains both groups
- Claiming the fragment distinguishes acid from acyl chloride
What can you explain now?
When reversing a polyester link, which functional groups are restored?
Compare with the explanation
Hydroxyl and carboxyl groups
The ester link can be split to recover alcohol and acid ends.
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Make it stick
How do you reverse an ester or amide link, and what precursor detail may remain unknown?
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.