Form a polyamide from a diamine and a dioyl chloride
A diamine reacts at both ends with a dioyl chloride to form amide links and hydrogen chloride.
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 by-product distinguishes diamine/dioyl chloride polyamide formation from diamine/diacid formation?
What you will learn
- Recognise both bifunctional monomers
- Form amide links from –NH₂ and –COCl
- Distinguish HCl from water as the by-product
Amino groups attack acyl chloride ends
A dioyl chloride has two –COCl groups; a diamine has two –NH₂ groups. Their reaction forms amide links, –CONH–, connecting the monomer residues into a polyamide.
In this route, chlorine from –COCl and hydrogen from the amino group are removed as HCl. The carbonyl carbon and the nitrogen stay in the polymer link.
Compare with the diacid route
A dicarboxylic acid and diamine form amide links with water eliminated. A dioyl chloride and diamine form the same amide-link pattern but eliminate HCl.
When the question asks for the polymer, show the linked residues. When it asks for by-products, match them to the monomer functional group rather than naming water automatically.
Pause and explain
Which atoms stay together in the new amide link?
Worked example
Describe the reaction between a diamine and a dioyl chloride.
Show the worked solution
- Pair –NH₂ with –COCl.
- Form –CO–NH– and retain the carbonyl carbon and nitrogen.
- Repeat at both ends; account for HCl.
Answer A diamine and dioyl chloride form an alternating polyamide through –CONH– links, eliminating HCl.
Try a topic test
Choose Easy, Normal or Difficult. Save your answers and open worked review after finishing.
Test this topic →From guided practice to a new situation
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.
Represent the repeat residues for H₂N–R–NH₂ and ClCO–R′–COCl.
Give me a hint
Join nitrogen to each acyl carbon and show the repeat connectivity.
Compare my reasoning
- Keep –NH–R–NH– from the diamine.
- Keep –CO–R′–CO– from the dioyl chloride.
- Join them with –CO–NH– links and remove Cl as HCl.
An equivalent repeat is [–NH–R–NH–CO–R′–CO–]ₙ; HCl is eliminated at the amide-forming links.
Look for these in your work
- Residues retained
- Amide links shown
- HCl identified
A learner writes water for every polyamide condensation. What reagent clue corrects the answer here?
Give me a hint
Look for the leaving group on the carbonyl monomer.
Compare my reasoning
- The monomer is a dioyl chloride, not a dicarboxylic acid.
- Each –COCl group supplies chlorine.
- The amino hydrogen and chlorine leave as HCl.
This dioyl chloride route eliminates HCl. Water is the by-product when –COOH groups condense with –NH₂ groups.
Look for these in your work
- Monomer class checked
- Leaving group tracked
- Routes contrasted
Explain why the polymer is still a polyamide when the by-product changes from water to HCl.
Give me a hint
Classify it by the link in the polymer, not by the by-product.
Compare my reasoning
- The carbonyl carbon bonds to nitrogen.
- The repeated link is –CONH–.
- That link defines a polyamide in either route.
It remains a polyamide because the repeating link is –CONH–; the starting carbonyl group determines whether water or HCl is eliminated.
Look for these in your work
- Bond formed identified
- Polymer class justified
- By-product distinguished
Common mistakes
- Drawing –COO– instead of –CONH–
- Writing water as the only by-product
- Removing the carbonyl carbon
- Leaving chlorine in the polymer link
What can you explain now?
Which by-product distinguishes diamine/dioyl chloride polyamide formation from diamine/diacid formation?
Compare with the explanation
HCl
The acyl chloride route eliminates HCl.
After trying it yourself, choose your next review. This is your self-assessment.
Your review choice appears on Today. Sign in to sync it across devices.
Make it stick
How do the monomer functional groups determine the amide link and HCl by-product?
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.