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Topic 3.3.3: Halogenoalkanes - Exam Practice

AQA A-Level Chemistry • Organic Chemistry • Exam-Style Questions

Exam Practice

Topic 3.3.3: Halogenoalkanes Exam Practice

Test your understanding of nucleophilic substitution mechanisms, elimination pathways, and CFC-mediated ozone breakdown with exam-style questions.

Halogenoalkanes › Exam Practice

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📋 Structured Questions

Complete each question on paper, then check your answers against the mark scheme.

Question 1: Rates of Hydrolysis

4 marks

(a) Outline an experiment to compare the rates of hydrolysis of 1-chlorobutane, 1-bromobutane, and 1-iodobutane. State the observations you would expect. [3]

(b) Explain the order of reactivity observed in part (a) in terms of bond strength. [1]

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(a)

  • Add equal volumes of each halogenoalkane to separate test tubes, then add ethanol (as a mutual solvent), water, and aqueous silver nitrate solution (AgNO3). [1]
  • Place test tubes in a water bath at 50-60 degrees C and time how long it takes for a precipitate to appear in each tube. [1]
  • Observations: Yellow precipitate (AgI) forms fastest with 1-iodobutane. Cream precipitate (AgBr) forms next with 1-bromobutane. White precipitate (AgCl) forms slowest/last with 1-chlorobutane. [1]

(b)

  • The bond strength/enthalpy of the C-X bond decreases down Group 7 (C-Cl > C-Br > C-I). The weaker C-I bond requires less energy to break, allowing 1-iodobutane to undergo hydrolysis fastest. [1]
Examiner tip: Do not use bond polarity to explain rates of hydrolysis! Although the C-Cl bond is the most polar and should attract nucleophiles fastest, the bond enthalpy (strength) is the dominant factor. The weaker the bond, the faster the reaction.

Question 2: Mechanisms and Conditions

6 marks

(a) Write the mechanism for the reaction of 1-bromopropane with excess ammonia to form propylamine. Include all relevant curly arrows, dipole charges, and lone pairs. [4]

(b) Under different conditions, the reaction of 2-bromopropane with potassium hydroxide can yield an alkene. State the reagent and conditions required for this reaction, and name the mechanism. [2]

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(a)

  • Nucleophilic attack: Curly arrow starts from the lone pair of the NH3 nitrogen to the C delta+ atom of the C-Br bond. [1]
  • Leaving group: Dipoles shown as C delta+ and Br delta-. Curly arrow from the C-Br bond to the Br atom. [1]
  • Intermediate: Correct drawing of the CH3CH2CH2NH3+ ion with a positive charge on the nitrogen. [1]
  • Deprotonation: A second ammonia molecule uses its lone pair to attack one of the hydrogens on the NH3+ group, with a curly arrow going from the N-H bond to the positive nitrogen. Forms CH3CH2CH2NH2 + NH4+ + Br-. [1]

(b)

  • Reagent/Conditions: Potassium hydroxide (KOH) dissolved in ethanol (ethanolic), heated under reflux. [1]
  • Mechanism Name: Elimination. [1]
Examiner tip: For the reaction with ammonia, remember it is a two-step mechanism. The positive charge must be on the nitrogen atom in the intermediate, not on carbon. Two molecules of ammonia are required overall: one as a nucleophile and one as a base.

Question 3: CFCs and Ozone Depletion

3 marks

(a) Write two equations to show how chlorine radicals act as a catalyst in the decomposition of ozone (O3) in the upper atmosphere. [2]

(b) Explain why trichlorofluoromethane (CF3Cl) is capable of causing ozone depletion, whereas 1,1,1,2-tetrafluoroethane (CF3CH2F) is not. [1]

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(a)

  • Equation 1: Cl. + O3 -> .ClO + O2 [1]
  • Equation 2: .ClO + O3 -> Cl. + 2O2 (or .ClO + O -> Cl. + O2) [1]

(b)

  • CF3Cl contains a C-Cl bond, which has a relatively low bond enthalpy and can be broken by UV light to form chlorine radicals (Cl.). CF3CH2F contains only C-H and C-F bonds. The C-F bond is extremely strong and cannot be broken by UV light in the stratosphere, so no fluorine radicals are released to catalyze ozone breakdown. [1]
Examiner tip: Ensure the radical dots are clearly visible next to the Cl and ClO species. In part (b), make sure to emphasize the difference in bond strength between C-Cl and C-F bonds.

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