IB DP Chemistry Reactivity 3.3: Electron sharing reactions – 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 – the IB holds copyright in its own papers. Use these alongside the official past papers available through your school or the IB store.

The questions below target Reactivity 3.3, Electron sharing reactions, drawn from the Chemistry guide for the IB Diploma Programme (February 2023 edition, first assessment 2025). Nothing in the sub-topic is skipped: the set spans Reactivity 3.3.1 to 3.3.3, all for SL and HL; Reactivity 3.3 has no additional higher level understandings. Questions 1 to 3 test 3.3.1, questions 4 and 5 test 3.3.2, questions 6 to 9 test 3.3.3, and questions 10 and 11 combine them.

Stuck on a question? Reread the study guide or the revision notes. The IB DP chemistry hub and printable checklist cover the other units, and our free diagnostics point you to what to revise next.

Where a question gives bond enthalpies, use the values supplied.

Questions

1. Define the term radical and explain why radicals are highly reactive. [2]

2. Using valence electron counts, identify which of the following species are radicals: CN, CN⁻, ClO, H₂O₂, CHCl₂. [3]

3. The species H•, NO, C₂H₆•⁺ and CO₂•⁻ all have an unpaired electron.

(a) State whether each one is an atom, a neutral molecule, a cation or an anion. [2] (b) Show by counting valence electrons that C₂H₆⁺ and CO₂⁻ must be radicals. [2] (c) Explain why the chloride ion, Cl⁻, is an ion but not a radical. [1]

4. Chlorine reacts with alkanes in ultraviolet light.

(a) Write an equation for the homolytic fission of chlorine. [1] (b) Describe how single-barbed arrows are drawn to show this process. [2] (c) State the name of this step in the chain reaction and one condition, other than UV light, that can cause it. [1]

5. Radicals can combine as well as form.

(a) State the process that is the reverse of homolytic fission and write an equation for one example of it involving bromine radicals. [2] (b) Chlorine radicals in the stratosphere can break down ozone, O₃, but not oxygen, O₂. Suggest what this indicates about the oxygen–oxygen bonds in the two allotropes, and explain your answer in terms of bonding. [2]

6. Cyclobutane, C₄H₈, reacts with chlorine in UV light to give chlorocyclobutane, C₄H₇Cl.

(a) Write the two propagation steps that form chlorocyclobutane. [2] (b) Write three possible termination steps. [3] (c) State the molecular formula of the hydrocarbon formed in one of your termination steps. [1]

7. Octane reacts with bromine in UV light. One product is 2-bromooctane, CH₃CHBr(CH₂)₅CH₃, formed with hydrogen bromide. Use these average bond enthalpies (kJ mol⁻¹): C–H 414, Br–Br 193, C–Br 285, H–Br 366.

(a) Calculate ΔH for each of the two propagation steps. [2] (b) Calculate ΔH for the overall reaction from your answers to (a), and confirm it using the bonds broken and made in the overall equation. [2] (c) State the sign of ΔH for every termination step and explain why. [1]

8. Alkanes are often described as kinetically stable but thermodynamically unstable.

(a) Explain this description. [2] (b) Give two features of alkane molecules that account for their low reactivity. [2]

9. Propane is mixed with bromine vapour and exposed to UV light. The products include 1-bromopropane, 2-bromopropane, 2,2-dibromopropane and some hydrocarbons of formula C₆H₁₄.

(a) Write two propagation steps that convert 2-bromopropane into 2,2-dibromopropane. [2] (b) Explain why two monobromo products form. [2] (c) Write an equation for the formation of hexane in this mixture. [1]

10. A sample of 6.01 g of 2,4-dimethylpentane, (CH₃)₂CHCH₂CH(CH₃)₂, is reacted with chlorine in UV light. After separation, 3.15 g of 3-chloro-2,4-dimethylpentane, C₇H₁₅Cl, is collected.

(a) Deduce how many monochloro structural isomers 2,4-dimethylpentane can form, and name them. [2] (b) Calculate the percentage yield of 3-chloro-2,4-dimethylpentane. [3] (c) The yield is far below the theoretical maximum. Using the radical mechanism, explain two causes. [2] (d) Suggest how the proportion of monochloro products could be increased. [1]

11. Chlorotrifluoromethane, CClF₃, is a CFC. In the stratosphere it absorbs UV radiation. Use these values (kJ mol⁻¹): C–Cl 324, C–F 492.

(a) Write an equation for the homolytic fission that takes place, and show by counting valence electrons that the carbon-containing product is a radical. [2] (b) Explain why fluorine radicals are not typically released. [2] (c) The chlorine radicals then react in a cycle: Cl• + O₃ → ClO• + O₂, followed by ClO• + O → Cl• + O₂. Classify these steps and explain why a single chlorine radical can destroy many ozone molecules. [3]

Answers

1. A molecular entity with an unpaired electron [1]. The unpaired electron makes it react readily to pair up, by removing an atom from a molecule or combining with another radical [1] [2] Examiner insight: “A species with a single electron” is too vague; the key word is unpaired.

2. CN: 4 + 5 = 9, odd, radical [1]; ClO: 7 + 6 = 13, odd, radical [1]; CHCl₂: 4 + 1 + 14 = 19, odd, radical, while CN⁻ (10) and H₂O₂ (14) are even and not radicals [1] [3] Examiner insight: Show the electron count for each species; a bare list of names shows no reasoning, so a single slip leaves nothing creditworthy.

3. (a) H• atom; NO neutral molecule [1]; C₂H₆•⁺ cation; CO₂•⁻ anion [1] (b) C₂H₆⁺: 2 × 4 + 6 − 1 = 13, odd [1]; CO₂⁻: 4 + 2 × 6 + 1 = 17, odd [1] (c) Cl⁻ has 8 valence electrons, all paired (a full octet), so there is no unpaired electron [1] Examiner insight: Remember to adjust the count for charge: subtract for a positive charge and add for a negative one.

4. (a) Cl₂ → 2Cl• [1] (b) Each arrow has a single barb (half arrowhead) and starts at the centre of the Cl–Cl bond [1]; one arrow goes to each chlorine atom, showing one electron moving to each [1] (c) Initiation; heat (high temperature) [1] Examiner insight: Dots on the radicals are part of the equation; Cl₂ → 2Cl is not the same answer.

5. (a) Bond formation between two radicals, each giving one electron (a termination step) [1]; Br• + Br• → Br₂ [1] (b) The oxygen–oxygen bonds in O₃ are weaker than the bond in O₂ [1]; O₃ has resonance, so each O–O bond has a bond order of about 1.5, while O₂ has a full double bond [1] Examiner insight: For (b), “ozone is less stable” does not answer the question; compare bond strength and give the bonding reason.

6. (a) Cl• + C₄H₈ → C₄H₇• + HCl [1]; C₄H₇• + Cl₂ → C₄H₇Cl + Cl• [1] (b) Cl• + Cl• → Cl₂ [1]; C₄H₇• + Cl• → C₄H₇Cl [1]; C₄H₇• + C₄H₇• → C₈H₁₄ [1] (c) C₈H₁₄ (two four-membered rings joined by a C–C bond) [1] Examiner insight: Each propagation step must have one radical on each side; count the dots before moving on.

7. (a) Step 1, Br• + C₈H₁₈ → C₈H₁₇• + HBr: +414 − 366 = +48 kJ mol⁻¹ [1]. Step 2, C₈H₁₇• + Br₂ → C₈H₁₇Br + Br•: +193 − 285 = −92 kJ mol⁻¹ [1] (b) +48 + (−92) = −44 kJ mol⁻¹ [1]; broken 414 + 193 = 607, made 285 + 366 = 651, so 607 − 651 = −44 kJ mol⁻¹, which agrees [1] (c) Negative (exothermic): a bond forms and no bond breaks [1] Examiner insight: Write “bonds broken minus bonds made” and keep the sign on every value; a dropped sign turns the endothermic first step into an exothermic one.

8. (a) The products of combustion are much lower in energy, so reaction with oxygen is very exothermic (thermodynamically unstable) [1]; but the activation energy is high, so at room temperature the reaction does not happen without a spark or flame (kinetically stable) [1] (b) C–C and C–H bonds are strong [1]; the bonds are essentially non-polar because C and H have similar electronegativities, so nucleophiles and electrophiles are not attracted [1] Examiner insight: “Kinetically stable” refers to rate, not to energy content; link it to activation energy, not to ΔH.

9. (a) Br• + CH₃CHBrCH₃ → CH₃C•BrCH₃ + HBr [1]; CH₃C•BrCH₃ + Br₂ → CH₃CBr₂CH₃ + Br• [1] (b) Propane has hydrogens in two environments, on the end CH₃ groups and on the central CH₂ [1]; Br• can remove either, giving 1-bromopropane or 2-bromopropane [1] (c) Two propyl radicals combine in a termination step: CH₃CH₂CH₂• + CH₃CH₂CH₂• → CH₃(CH₂)₄CH₃ [1] Examiner insight: In (a) the second bromine goes on the carbon that carries the dot, so show the radical’s structure, not just C₃H₆Br•.

10. (a) Three hydrogen environments (the four CH₃ groups, the two CH carbons, the central CH₂), so three isomers [1]: 1-chloro-2,4-dimethylpentane, 2-chloro-2,4-dimethylpentane and 3-chloro-2,4-dimethylpentane [1] (b) M(C₇H₁₆) = 100.20 g mol⁻¹, n = 6.01 ÷ 100.20 = 0.05998 mol [1]; M(C₇H₁₅Cl) = 134.64 g mol⁻¹, so maximum mass = 0.05998 × 134.64 = 8.08 g [1]; yield = 3.15 ÷ 8.08 × 100 = 39.0 % [1] (c) Chlorine can replace hydrogens in the other two environments, giving the other isomers [1]; further substitution gives dichloro and more highly substituted products, and termination gives other hydrocarbons [1] (d) Use an excess of the alkane so that Cl• is more likely to meet alkane than chloroalkane [1] Examiner insight: Carry unrounded values for the moles and the maximum mass, then round only the percentage at the end.

11. (a) CClF₃ → CF₃• + Cl• [1]; CF₃: 4 + 3 × 7 = 25, odd, so it is a radical [1] (b) The C–F bond (492) is 168 kJ mol⁻¹ stronger than C–Cl (324) [1], so the UV radiation reaching the stratosphere has enough energy to break C–Cl but not C–F [1] (c) Both are propagation steps, because each uses one radical and produces one radical [1]; the Cl• used in the first step is regenerated in the second [1]; so the cycle repeats many times until a termination step removes the radical [1] Examiner insight: Say that the chlorine radical is regenerated; that is the point the explanation of the chain needs.

Where marks are usually lost

• Leaving the dot off a radical, or putting it on the wrong atom.
• Giving a halogen radical a negative charge.
• Showing homolytic fission with a full curly arrow or with one arrow only.
• Writing a propagation step that produces H• instead of HX.
• Listing only Br• + Br• → Br₂ when several termination steps are asked for.
• Forgetting further substitution or position isomers when explaining a mixture of products.
• Mixing up kinetic and thermodynamic stability.
• Rounding moles early in yield calculations.
• Comparing O₃ and O₂ without mentioning bond order or bond strength.

Next steps

• Reactivity 3.3 revision notes for a final read-through
• Reactivity 3.3 study guide for full explanations
• More mechanism questions: Reactivity 3 mechanisms practice
• The IB DP chemistry course hub
• Tick off this unit on the printable checklist
• Take all free 10-minute diagnostics
• Book a free trial class

Official syllabus

These questions practise Reactivity 3.3 (Electron sharing reactions), understandings 3.3.1 to 3.3.3, SL and HL. Source: Chemistry guide, IB Diploma Programme, International Baccalaureate, February 2023 (first assessment 2025).