AS Chemistry: Alkanes and Alkenes — Practice Questions

These are original questions written for Marlbridge, for revision and practice on this content. They are not reproduced past-paper questions, and they do not replicate the exam’s exact structure, question count or mark tariffs — examination boards hold copyright in their own papers. Use these alongside the official past papers available free from your board.

Related: Alkanes and Alkenes revision notes

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Questions

1. Explain why alkanes are relatively unreactive. [2]

2. Methane reacts with chlorine in the presence of UV light.

(a) Name the mechanism. [1] (b) Write equations for the initiation, both propagation steps and one termination step. [4] (c) Explain why this reaction gives a mixture of products. [2]

3. Explain, in terms of bonding, why alkenes are more reactive than alkanes. [2]

4. But-1-ene reacts with hydrogen bromide.

(a) Name the mechanism. [1] (b) Give the two possible products and identify the major one. [2] (c) Explain, using carbocation stability, why one product predominates. [3]

5. Describe a chemical test to distinguish an alkane from an alkene, stating the reagent and both observations. [3]

6. Ethene undergoes addition polymerisation.

(a) Draw or describe the repeat unit. [2] (b) (Background — “atom economy” is not a 9701 syllabus term, though the chemistry is useful context.) Explain why no atoms are wasted when ethene undergoes addition polymerisation. (c) Explain why poly(ethene) is not biodegradable. [2]

7. Propene is reacted separately with (i) hydrogen and a nickel catalyst, and (ii) cold, dilute, aqueous, acidified potassium manganate(VII). For each reaction, name the type of reaction, state the conditions, and give the organic product. [6]

8. A student has two unlabelled gas jars, one containing ethane and one containing ethene. Describe how cold dilute acidified potassium manganate(VII) could be used to identify which is which, including both observations. [3]

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9. Propene reacts with hydrogen bromide. Name the major organic product and explain why it is the major product. [3]

Answers

1. They contain only strong C–C and C–H sigma bonds [1] which are non-polar, so there is no δ+ centre to attract nucleophiles and no region of high electron density to attract electrophiles [1].

2. (a) Free radical substitution [1]. (b) Initiation: Cl₂ → 2Cl• (UV light) [1]. Propagation: Cl• + CH₄ → •CH₃ + HCl [1] and •CH₃ + Cl₂ → CH₃Cl + Cl• [1]. Termination: any two radicals combining, e.g. •CH₃ + Cl• → CH₃Cl [1]. (c) The chlorine radical can attack the product as well as the starting material [1], giving further substitution to CH₂Cl₂, CHCl₃ and CCl₄ [1].

3. The alkene contains a C=C double bond with a π bond [1], which is a region of high electron density that attracts electrophiles [1].

4. (a) Electrophilic addition [1]. (b) 1-bromobutane and 2-bromobutane [1]; 2-bromobutane is the major product [1]. (c) The mechanism proceeds via a carbocation intermediate [1]. The secondary carbocation is more stable than the primary [1], because the electron-donating alkyl groups on either side spread the positive charge [1].

5. Add bromine water [1]. The alkene decolourises it from orange to colourless [1]; the alkane produces no change [1].

6. (a) Open the double bond to a single bond, draw two backbone carbons with the bonds extending through the brackets, and write n outside [1] [1]. (b) (Background only.) All the atoms of the monomer end up in the single product; nothing is eliminated, since addition polymerisation simply opens double bonds and links monomers together with no by-product. (c) The C–C backbone is unreactive and non-polar [1], with no bonds that enzymes or water can attack [1].

7. (i) Hydrogenation (reduction/addition) [1]. Conditions: nickel catalyst, about 150 °C [1]. Product: propane [1]. (ii) Oxidation, adding two hydroxyl groups across the carbon-carbon double bond [1]. Conditions: cold, dilute, aqueous, acidified KMnO₄, no heating required [1]. Product: propane-1,2-diol [1].

8. Add a few drops of cold dilute acidified potassium manganate(VII) to each gas jar and shake [1]. In the jar containing ethene, the purple colour decolourises as the alkene is oxidised to the diol [1]. In the jar containing ethane, there is no reaction, so the purple colour persists, because the C–C sigma bonds have no region of high electron density for the oxidant to attack [1].

9. 2-bromopropane [1]. The reaction goes through the secondary carbocation, CH₃C⁺HCH₃, which is more stable than the primary carbocation [1], because it has two alkyl groups that donate electrons (positive inductive effect) towards the positive carbon [1].

Examiner insight (June 2025): less precise answers described 2-bromopropane itself as the carbocation or intermediate. The intermediate is the carbocation; the product forms when bromide ions attack it. Try the real question next: Cambridge International AS & A Level Chemistry 9701, June 2025, Paper 22, Question 3.

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Where marks are usually lost

• Writing an initiation step without UV light.
• Forgetting that propagation must regenerate the radical.
• Stating Markovnikov’s rule instead of explaining via carbocation stability.
• Leaving the double bond in a repeat unit, or omitting the extending bonds.
• Confusing the bromine water test (orange-brown to colourless) with the cold dilute acidified KMnO₄ test (purple to colourless) — both detect unsaturation, but they are different reagents and should not be described interchangeably.
• Forgetting that hydrogenation needs a nickel catalyst and heat, not just “hydrogen gas.”
• Naming the diol product of KMnO₄ oxidation incorrectly, or forgetting that two -OH groups are added across the former double bond, one on each carbon.

Comparing the alkene addition reactions

Test/reaction	Reagent	Conditions	What happens
Test for unsaturation	Bromine water	Room temperature	Orange-brown → colourless
Test for unsaturation	Cold dilute acidified KMnO₄	Room temperature	Purple → colourless; diol forms
Hydrogenation	H₂	Ni catalyst, ~150 °C	Saturated alkane forms
Hydration	Steam (H₂O(g))	H₃PO₄ catalyst, ~300 °C, 60 atm	Alcohol forms

Both bromine water and cold dilute acidified KMnO₄ decolourise in the presence of a C=C double bond, which is why either can be used to distinguish an alkene from an alkane — the key is that alkanes give no observable change with either reagent, since their C–C and C–H bonds have no accessible pi electrons for the oxidant or electrophile to attack.

For the full mechanisms and explanations behind these reactions, see the Hydrocarbons: Alkanes and Alkenes study guide; for condensed recall notes covering the same reaction table, see the Alkanes and Alkenes revision notes.