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IB Chemistry · Reactivity 3: What are the mechanisms of chemical change?

Electron sharing reactions (radicals) practice questions.

36 exam-style questions on Electron sharing reactions (radicals) — topic R3.3 of Reactivity 3: What are the mechanisms of chemical change? in the IB Chemistry syllabus — with answers and explanations. Samples below; the full set is free with an account.

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About this topic

Syllabus code
R3.3
Subject
IB Chemistry
Concept
R3 · Reactivity 3: What are the mechanisms of chemical change?
Questions
36 with worked answers
Question types
14 multiple-choice questions · 8 fill-in-the-gap questions · 8 true/false questions · 5 flashcards · 1 matching task
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Sample questions on Electron sharing reactions (radicals)

State the term for the step in a radical chain mechanism in which two radicals combine to form a molecule with no unpaired electrons.

  • Initiation
  • Propagation
  • Termination
  • Substitution

Answer: Termination

Termination removes radicals from the system by combining two of them, giving a species with all electrons paired; this reduces the total number of radicals.

In the chlorination of methane, which equation represents the initiation step?

  • Cl₂ → 2Cl•
  • CH₄ + Cl• → CH₃• + HCl
  • CH₃• + Cl₂ → CH₃Cl + Cl•
  • CH₃• + Cl• → CH₃Cl

Answer: Cl₂ → 2Cl•

Initiation is the homolytic fission of the Cl–Cl bond by UV light to form two chlorine radicals; the C–H bond is much stronger and does not break first.

Identify the type of bond fission that produces radicals.

  • Heterolytic fission
  • Homolytic fission
  • Nucleophilic fission
  • Ionic fission

Answer: Homolytic fission

In homolytic fission each atom takes one electron from the shared pair, producing two radicals; heterolytic fission gives ions because one atom takes both electrons.

The monochlorination of methane requires ultraviolet light. State the role of the UV light.

  • It oxidises the methane
  • It provides energy to break the Cl–Cl bond homolytically
  • It acts as a chemical catalyst that is regenerated at the end
  • It ionises the chlorine molecule into Cl⁺ and Cl⁻

Answer: It provides energy to break the Cl–Cl bond homolytically

UV photons supply the energy for homolytic fission of Cl₂ into two radicals in the initiation step; it does not ionise the chlorine.

Determine the two propagation steps for the reaction of methane with chlorine.

  • Cl₂ → 2Cl• ; CH₄ → CH₃• + H•
  • CH₄ + Cl• → CH₃• + HCl ; CH₃• + Cl₂ → CH₃Cl + Cl•
  • CH₃• + Cl• → CH₃Cl ; Cl• + Cl• → Cl₂
  • CH₄ + Cl₂ → CH₃Cl + HCl ; CH₃Cl + Cl₂ → CH₂Cl₂ + HCl

Answer: CH₄ + Cl• → CH₃• + HCl ; CH₃• + Cl₂ → CH₃Cl + Cl•

In propagation a radical reacts to form a product plus a new radical, so the chain continues; each step regenerates a radical (Cl• then CH₃•).

Suggest why the radical substitution of methane with chlorine produces a mixture of products such as CH₃Cl, CH₂Cl₂, CHCl₃ and CCl₄.

  • Chlorine radicals can further substitute the hydrogens remaining on already-chlorinated products
  • The whole reaction reaches a dynamic equilibrium in which every possible chlorinated product is equally favoured
  • Methane contains impurities
  • Chlorine is a poor oxidising agent

Answer: Chlorine radicals can further substitute the hydrogens remaining on already-chlorinated products

As long as radicals and C–H bonds are present, further substitution occurs, so chloromethane is chlorinated again, giving a mixture of poly-substituted products.

That was 6 of 36 Electron sharing reactions (radicals) questions.

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