Prepare alkenes by cracking long-chain alkanes
Recognise cracking as a route to alkenes as well as smaller alkane products.
AS Level; Cambridge International Chemistry 9701, 2025–2027 syllabus. Apply the stated reagents and conditions, balance equations, and keep mechanistic claims within the named examples. Written practice is self-checked
Before you begin
- Use displayed and structural formulas, covalent bond polarity, balanced equations, and the organic reaction terms from Topic 13.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
Which product class can cracking form from a long-chain alkane?
What you will learn
- Identify cracking as an alkene preparation
- Balance a product equation
- Distinguish cracking from elimination and hydrogenation
Cracking can form a C=C product
A long-chain alkane can break into smaller molecules, including an alkene. The process involves heating and may use a catalyst such as aluminium oxide.
For example, C₁₀H₂₂ → C₈H₁₈ + C₂H₄ gives a smaller alkane and ethene. This is one balanced example, not the unique product distribution from cracking.
Use the whole equation as evidence
Count each element on both sides to check a proposed cracking equation.
Cracking differs from dehydration or hydrogen halide elimination because it splits a larger hydrocarbon rather than removing a small molecule from one functionalized molecule.
Pause and explain
Which statement about cracking is most accurate?
Worked example
Explain why C₁₀H₂₂ → C₈H₁₈ + C₂H₄ is a valid representative alkene preparation by cracking.
Show the worked solution
- The reactant is a long-chain alkane.
- The products have fewer carbon atoms and include an alkene.
- Check ten carbons and twenty-two hydrogens on each side.
Answer The balanced cracking equation forms the smaller alkane octane and the alkene ethene from a larger alkane.
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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.
Identify which product in C₉H₂₀ → C₆H₁₄ + C₃H₆ is an alkene.
Give me a hint
Use the CₙH₂ₙ pattern and the C=C class.
Compare my reasoning
- C₃H₆ has the open-chain alkene formula pattern.
- C₆H₁₄ is an alkane formula.
- The C₃H₆ product is propene in a suitable structure.
C₃H₆ is the alkene product; the equation is balanced.
Look for these in your work
- Formula pattern used
- Alkane distinguished
- Alkene identified
State why an alkene product from cracking is useful beyond being a fuel.
Give me a hint
Link C=C reactivity to chemical feedstocks.
Compare my reasoning
- An alkene has a reactive C=C bond.
- It can undergo addition or addition polymerisation.
- Those reactions make other chemicals and polymers.
Alkenes are reactive feedstocks for addition reactions and polymer manufacture.
Look for these in your work
- C=C noted
- Reaction pathway named
- Use explained
Check whether C₁₀H₂₂ → C₇H₁₆ + C₃H₆ + H₂ is balanced.
Give me a hint
Count hydrogen atoms after checking carbon.
Compare my reasoning
- Carbon is balanced: 10 = 7 + 3.
- Hydrogen on the right is 16 + 6 + 2 = 24.
- The left has 22, so the proposed products are not balanced.
It is unbalanced because the right side has 24 hydrogen atoms but the left has 22.
Look for these in your work
- Carbon check stated
- Hydrogen check stated
- Conclusion justified
Common mistakes
- Calling cracking a reaction that only makes alkanes
- Assuming the product split is universal
- Ignoring atom conservation
What can you explain now?
Which product class can cracking form from a long-chain alkane?
Compare with the explanation
An alkene
Cracking can produce smaller alkanes and alkenes.
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Make it stick
How does an alkene produced by cracking become a useful chemical feedstock?
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.