1. 1
   When a mixture of methane and chlorine is kept in the dark, no reaction occurs. When the mixture is exposed to ultraviolet (UV) light, chloromethane and hydrogen chloride are formed by a radical chain reaction. Bond enthalpies: Cl–Cl 242 kJ mol⁻¹; C–H 414 kJ mol⁻¹.

   (a)

   Which species is a radical?

   [1 mark]

   • ACl⁻
   • B•CH₃
   • CCH₃⁺
   • DCl₂

   (b)

   Which equation represents a propagation step in this reaction?

   [1 mark]

   • ACl• + CH₄ → •CH₃ + HCl
   • BCl₂ → 2Cl•
   • C•CH₃ + Cl• → CH₃Cl
   • DCH₄ → •CH₃ + H•

   (c)

   Explain why UV light is needed for the reaction to take place, including an equation for the step it causes.

   [2 marks]

   Total for question 1: 4 marks

2. 2
   Bond enthalpies, in kJ mol⁻¹: F–F 159, Cl–Cl 242, Br–Br 193, I–I 151, C–H 414, H–Cl 431, H–Br 366.

   (a)

   Which halogen molecule undergoes homolytic fission most readily on heating?

   [1 mark]

   • AF₂
   • BCl₂
   • CBr₂
   • DI₂

   (b)

   Which statement describes the homolytic fission of a Br–Br bond?

   [1 mark]

   • AOne bromine atom takes both electrons, forming Br⁺ and Br⁻.
   • BEach bromine atom gains an extra electron, forming two Br⁻ ions.
   • CEach bromine atom takes one electron from the bond, forming two radicals.
   • DThe bond breaks to form one radical and one bromine molecule.

   (c)

   Use the bond enthalpies to compare the enthalpy changes of the propagation steps Cl• + CH₄ → •CH₃ + HCl and Br• + CH₄ → •CH₃ + HBr.

   [2 marks]

   Total for question 2: 4 marks

3. 3
   Ethane and chlorine were mixed in a sealed flask and irradiated with UV light. Analysis of the products showed hydrogen chloride, chloroethane (the main organic product), 1,1-dichloroethane, 1,2-dichloroethane and a small amount of butane.

   (a)

   Write equations for the initiation step and the two propagation steps that produce chloroethane.

   [3 marks]

   (b)

   Explain, using equations, the formation of butane and of the dichloroethanes in the product mixture.

   [4 marks]

   Total for question 3: 7 marks

4. 4
   A chemist wants to make pure 2-chloropropane, CH₃CHClCH₃, by reacting propane with chlorine in UV light at room temperature. A propane molecule contains six hydrogen atoms on its two end (primary) carbon atoms and two hydrogen atoms on its central (secondary) carbon atom. With a large excess of propane, the monochlorinated products were 45% 1-chloropropane and 55% 2-chloropropane, with only traces of dichloropropanes. With excess chlorine, large amounts of dichloropropanes and more highly chlorinated propanes formed.

   (a)

   Explain the mechanism by which the monochloropropanes are formed, including initiation, propagation and termination steps, and explain why the reaction is described as a chain reaction.

   [6 marks]

   (b)

   Evaluate radical substitution as a method for making pure 2-chloropropane. Use all the data, including a comparison of the observed product ratio with the ratio expected if every hydrogen atom were equally likely to be replaced.

   [6 marks]

   Total for question 4: 12 marks

End of questions