Addition repeats: deduce the monomer and retain side groups
Convert a supplied alkene into its repeat unit and reverse the process, distinguishing atom-retaining addition from water-producing condensation.
Extended only (Supplement) · Chemistry 0620, 2026–2028.
Before you begin
- Draw poly(ethene)'s repeat and interpret continuation bonds.
- Identify the two C atoms forming a supplied monomer's C=C.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
Which monomer carbons become the two backbone carbons of the stated addition repeat?
What you will learn
- Deduce an addition repeat from a given alkene and its monomer from the repeat.
- Identify the backbone and retained side groups and contrast addition with condensation polymerisation.
Keep the two double-bond carbons in the backbone
To deduce the stated addition repeat, locate the alkene's two C=C carbons, change that bond to single, and add one backbone continuation at each end. Retain every substituent attached to those carbons. Propene CH₂=CHCH₃ gives [–CH₂–CH(CH₃)–]ₙ: its methyl group remains a side group on the second backbone carbon. It does not become a third consecutive carbon in the two-carbon backbone repeat.
To reverse the process, take those two backbone carbons, remove the two continuation connections and restore their C=C double bond while retaining the attached atoms/groups. [–CH₂–CHCl–]ₙ therefore corresponds to CH₂=CHCl, and [–CF₂–CF₂–]ₙ to CF₂=CF₂. These are deductions from supplied structures; the method does not require assuming that every polymer has the same substituent or properties.
Compare polymerisation by atom accounting
In these addition examples, one alkene monomer type joins through the double bond and all its atoms remain in the repeated polymer pattern. No water is eliminated as a second product. A condensation polymer instead forms links between suitable functional groups and releases small molecules, such as water for the acid-plus-diamine or acid-plus-diol examples taught next.
A repeat's atom total must be compared with the stated monomer and its side groups, not with the entire unknown chain. For propene, both the monomer and repeat contribution are C₃H₆. The drawn methyl side group's carbon makes four bonds with three H and its attachment; both backbone carbons make four when their continuation bonds are included. Complete finite-chain end groups remain a separate matter.
Pause and explain
Which repeat retains propene's side group correctly?
Worked example
Deduce propene's addition repeat and explain why moving its methyl group into the backbone would be wrong.
Show the worked solution
- The C=C is between the CH₂ and CH carbons of CH₂=CHCH₃.
- Make their bond single, add backbone continuations and retain CH₃ on the second C.
- The repeat is [–CH₂–CH(CH₃)–]ₙ, contributing C₃H₆ without eliminating water.
Answer [–CH₂–CH(CH₃)–]ₙ; the CH₃ remains a side group and the repeat retains C₃H₆.
Try a topic test
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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.
Draw the displayed propene addition repeat with all H and the methyl side group.
Give me a hint
Start with the two C=C carbons and keep the third C attached as a branch.
Compare my reasoning
- Change the CH₂–CH backbone pair's bond to single and add two continuations.
- Retain a CH₃ group on the second carbon and show its three H individually.
- Show the first backbone C's two H and the second's one H; the repeat contributes C₃H₆.
Displayed [–CH₂–CH(CH₃)–]ₙ with the methyl attached to the second backbone C, all six H shown and both continuations.
Look for these in your work
- Backbone and side group are distinguished.
- Atom totals and all carbon bond counts agree.
Deduce the monomers for [–CH₂–CHCl–]ₙ and [–CF₂–CF₂–]ₙ.
Give me a hint
Replace the two external backbone continuations with a double bond between the paired carbons.
Compare my reasoning
- For the chlorine-bearing repeat, retain H₂ on the first C and H/Cl on the second.
- Restore C=C to give CH₂=CHCl.
- For the fluorine-bearing repeat, retain two F on each C and restore C=C to give CF₂=CF₂.
CH₂=CHCl and CF₂=CF₂ respectively.
Look for these in your work
- Every supplied substituent is retained.
- Both reconstructed carbon–carbon bonds are double.
An answer to propene addition removes the methyl group and includes water as a product. Correct it and contrast the stated condensation examples.
Give me a hint
Check retained atoms and the origin of any small molecule.
Compare my reasoning
- Propene's methyl group remains on the second backbone C in [–CH₂–CH(CH₃)–]ₙ.
- The repeat retains C₃H₆, with no eliminated water in this addition model.
- Acid-plus-diamine/diol condensation instead joins functional groups and releases water at the new links.
Retain the methyl side group and all C₃H₆ atoms; addition eliminates no water, whereas the stated acid-based condensation links release water.
Look for these in your work
- Both proposed addition errors are corrected.
- Condensation is distinguished by functional-group linking and small-molecule loss.
Common mistakes
- Deleting a methyl or halogen substituent during polymerisation.
- Putting propene's methyl carbon into a three-carbon consecutive backbone repeat.
- Keeping C=C while also adding the two continuation bonds.
- Eliminating water from the supplied atom-retaining addition examples.
What can you explain now?
Which monomer corresponds to [–CH₂–CHCl–]ₙ?
Compare with the explanation
CH₂=CHCl.
Restore the double bond between the two backbone carbons and retain the chlorine substituent.
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Make it stick
Deduce an addition repeat from a fresh alkene, then reconstruct its monomer. Check retained substituents, bond order and the absence of an eliminated small molecule.
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.