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Questions on elimination from halogenoalkanes, and CFCs and the ozone layer

Elimination of HX from a halogenoalkane with ethanolic potassium hydroxide, the mechanism with hydroxide acting as a base, the competition between substitution and elimination, the alkenes formed from 2-bromobutane, the ozone layer and how it absorbs ultraviolet light, CFCs and the chlorine radical cycle, chlorine reservoirs and the Antarctic spring, the research that identified CFCs as the cause, the Montreal Protocol and HFCs as replacements.

19 original questions · 64 marks · the elimination from halogenoalkanes, and CFCs and the ozone layer notes · Alkanes and halogenoalkanes · Flashcards for this unit

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• 6 marks · Explain · EH-Q21 AQA · OCR A · CAIE

  Explain how CFCs released at ground level lead to the breakdown of ozone in the stratosphere, and why a small amount of CFC breaks down a very large amount of ozone.

  Mark scheme

  Indicative content

  1. CFCs are unreactive, so they are not broken down in the lower atmosphere and drift up into the stratosphere over years
  2. there, ultraviolet light breaks the weaker C–Cl bond homolytically, releasing chlorine radicals, while the stronger C–F bond survives
  3. a chlorine radical takes an oxygen atom from ozone: Cl• + O₃ → ClO• + O₂
  4. the ClO• radical reacts with a second ozone molecule, ClO• + O₃ → 2O₂ + Cl•, so overall 2O₃ → 3O₂
  5. the chlorine radical is regenerated in the second step, so it is a catalyst and is not used up by the cycle
  6. each radical goes round the cycle many thousands of times before termination, such as ClO• + NO₂ → ClONO₂, removes it

  Levels

  • Level 3 (5–6 marks) the whole chain is explained: the inertness that lets CFCs reach the stratosphere, homolysis of the C–Cl bond rather than C–F, both propagation equations giving the overall 2O₃ → 3O₂, and the regeneration of the chlorine radical as the reason a little CFC destroys a great deal of ozone, which makes chlorine a catalyst
  • Level 2 (3–4 marks) ultraviolet light forms chlorine radicals and both ozone equations are given; how the CFC survives to reach the stratosphere, the regeneration of the radical, or the overall equation is not there
  • Level 1 (1–2 marks) CFCs are said to destroy ozone directly, or the C–F bond is said to break; one equation or none, and nothing explains why a small amount does so much damage

• 4 marks · State · EH-Q1 AQA · Edexcel · CAIE

  State the reagent and conditions for the elimination of hydrogen bromide from 2-bromopropane, and describe a test that shows the organic product is an alkene and not an alcohol.

  Mark scheme

  1. B1 potassium hydroxide dissolved in ethanol; reject aqueous KOH
  2. B1 the mixture is heated under reflux
  3. B1 bromine water is added to the product, or the gas is bubbled through it
  4. B1 an alkene turns the orange bromine water colourless; an alcohol leaves it orange; reject 'clear' for colourless

• 4 marks · Draw · EH-Q2 AQA · Edexcel · CAIE

  Draw the mechanism for the elimination reaction of 2-bromopropane with hydroxide ions.

  Mark scheme

  1. B1 the dipole: δ+ on the carbon bonded to bromine and δ− on the bromine
  2. B1 a curly arrow from a lone pair on the oxygen of OH⁻ to an H atom on a carbon next to the C–Br carbon; reject an arrow that ends on the carbon
  3. B1 a curly arrow from that C–H bond to the C–C bond, forming the C=C
  4. B1 a curly arrow from the C–Br bond to the bromine, giving propene, water and Br⁻

• 4 marks · Deduce · EH-Q4 AQA · Edexcel · CAIE

  Deduce the structures and names of every alkene formed when 2-bromopentane, CH₃CHBrCH₂CH₂CH₃, is heated with ethanolic potassium hydroxide.

  Mark scheme

  1. B1 removing H from carbon 1 gives pent-1-ene, CH₂=CHCH₂CH₂CH₃
  2. B1 removing H from carbon 3 gives pent-2-ene, CH₃CH=CHCH₂CH₃
  3. B1 pent-2-ene has two different groups on each carbon of the C=C and restricted rotation, so it exists as E and Z isomers
  4. B1 three alkenes in all: pent-1-ene, E-pent-2-ene and Z-pent-2-ene; allow cis and trans for Z and E

• 4 marks · Compare · EH-Q7 AQA · Edexcel · CAIE

  Compare the conditions and products of the reactions of a halogenoalkane with hydroxide ions that favour substitution and that favour elimination.

  Mark scheme

  1. B1 substitution is favoured by an aqueous solvent; elimination by ethanol as the solvent
  2. B1 substitution is favoured by warm conditions; elimination by heating strongly under reflux
  3. B1 primary halogenoalkanes favour substitution; tertiary ones favour elimination
  4. B1 substitution gives an alcohol; elimination gives an alkene, with water and a halide ion

• 4 marks · Explain · EH-Q12 AQA · OCR A · CAIE

  Explain why CFCs reach the stratosphere and why it is the C–Cl bond rather than the C–F bond that breaks there, and explain why HFCs such as CH₂FCF₃ do not deplete the ozone layer.

  Mark scheme

  1. B1 CFCs are very unreactive, so they are not broken down in the lower atmosphere and slowly drift upwards
  2. B1 the C–Cl bond (346 kJ mol⁻¹) is weaker than the C–F bond (467 kJ mol⁻¹), so ultraviolet light breaks C–Cl homolytically and leaves C–F intact
  3. B1 HFCs contain no chlorine, so no chlorine radicals can be released
  4. B1 their C–F bonds are too strong to be broken by the ultraviolet light in the stratosphere, so no radicals form

• 4 marks · Evaluate · EH-Q14 AQA · OCR A · CAIE

  Evaluate the use of HFCs as replacements for CFCs in refrigerators and air conditioning.

  Mark scheme

  1. B1 HFCs have the properties that made CFCs useful: they are volatile, non-toxic and non-flammable
  2. B1 they do not deplete ozone, because they contain no C–Cl bond to release chlorine radicals
  3. B1 but they are powerful greenhouse gases, absorbing infrared radiation and contributing to global warming
  4. B1 so they solve the ozone problem but not the climate one, which is why their use is now being phased down in favour of hydrocarbons and other alternatives

• 4 marks · Outline · EH-Q20 AQA · OCR A · CAIE

  Outline the evidence from different research groups that showed CFCs to be the cause of ozone depletion, including what has been measured since the Montreal Protocol.

  Mark scheme

  1. B1 laboratory data were used to predict that CFCs reach the stratosphere and that chlorine radicals from them break down ozone
  2. B1 measurements over Antarctica, later confirmed by satellites, showed a large springtime fall in ozone
  3. B1 aircraft measurements found high concentrations of ClO• exactly where the ozone was depleted
  4. B1 since CFCs were phased out, stratospheric chlorine has fallen and the ozone layer has begun slowly to recover, as the explanation predicts

• 3 marks · Calculate · EH-Q8 AQA · Edexcel · CAIE

  Calculate the percentage yield of 2-methylpropene from the elimination of HBr from 2-bromo-2-methylpropane, using the data in the table. The gas volume is measured at room temperature and pressure, where the molar gas volume is 24.0 dm³ mol⁻¹.

  Quantity	Value
  Mass of 2-bromo-2-methylpropane used	6.85 g
  Mr of 2-bromo-2-methylpropane	136.9
  Volume of 2-methylpropene collected	0.912 dm³

  Mark scheme

  1. M1 amount of 2-bromo-2-methylpropane = 6.85 ÷ 136.9 = 0.05004 mol
  2. M1 theoretical volume of 2-methylpropene = 0.05004 × 24.0 = 1.201 dm³
  3. A1 percentage yield = 0.912 ÷ 1.201 × 100 = 75.9%, to three significant figures; error carried forward from M1

• 3 marks · Explain · EH-Q9 AQA · OCR A · CAIE

  Explain how ozone is formed in the stratosphere and why the ozone layer is beneficial.

  Mark scheme

  1. B1 ultraviolet light splits oxygen molecules into atoms: O₂ → 2O
  2. B1 an oxygen atom combines with an oxygen molecule: O + O₂ → O₃
  3. B1 ozone absorbs harmful ultraviolet radiation (UV-B), which would otherwise cause skin cancer and cataracts

• 3 marks · Give · EH-Q10 AQA · OCR A · CAIE

  Give equations to show how chlorine radicals catalyse the decomposition of ozone, and the overall equation.

  Mark scheme

  1. B1 Cl• + O₃ → ClO• + O₂
  2. B1 ClO• + O₃ → 2O₂ + Cl•
  3. B1 overall: 2O₃ → 3O₂

• 3 marks · Explain · EH-Q11 AQA · OCR A · CAIE

  Explain why a chlorine radical is described as a catalyst for the decomposition of ozone, and why a small amount of CFC can break down a large amount of ozone.

  Mark scheme

  1. B1 the chlorine radical is used in the first step and regenerated in the second
  2. B1 so it speeds up the decomposition without being used up
  3. B1 each radical goes round the cycle many thousands of times before a termination step removes it

• 3 marks · Calculate · EH-Q15 AQA · OCR A · CAIE

  Calculate the mass of ozone broken down by the chlorine from 5.0 g of CCl₂F₂, using the data in the table.

  Quantity	Value
  Mr of CCl₂F₂	121.0
  Chlorine radicals released per CCl₂F₂ molecule	1
  Ozone molecules broken down per radical	2.0 × 10⁴
  Mr of O₃	48.0

  Mark scheme

  1. M1 amount of CCl₂F₂, and of Cl•, = 5.0 ÷ 121.0 = 4.13 × 10⁻² mol
  2. M1 amount of O₃ = 4.13 × 10⁻² × 2.0 × 10⁴ = 826 mol
  3. A1 mass of O₃ = 826 × 48.0 = 4.0 × 10⁴ g, or 40 kg, to two significant figures; error carried forward from M1

• 3 marks · Suggest · EH-Q18 AQA · Edexcel · CAIE

  Suggest every alkene formed when 3-bromohexane, CH₃CH₂CHBrCH₂CH₂CH₃, is heated with ethanolic potassium hydroxide.

  Mark scheme

  1. B1 removing H from carbon 2 gives hex-2-ene, CH₃CH=CHCH₂CH₂CH₃
  2. B1 removing H from carbon 4 gives hex-3-ene, CH₃CH₂CH=CHCH₂CH₃
  3. B1 each has two different groups on both carbons of its C=C, so there are four alkenes: E- and Z-hex-2-ene, E- and Z-hex-3-ene

• 3 marks · Explain · EH-Q19 AQA · OCR A · CAIE

  Explain why the loss of ozone is greatest over Antarctica in the southern spring.

  Mark scheme

  1. B1 most stratospheric chlorine is normally held in reservoir compounds, HCl and ClONO₂, which do not attack ozone
  2. B1 in the polar winter, ice clouds form and on their surfaces the reservoir compounds react to release Cl₂
  3. B1 when sunlight returns in spring, ultraviolet light splits the Cl₂ into chlorine radicals all at once, which destroy ozone rapidly

• 3 marks · Justify · EH-Q22 AQA · Edexcel · CAIE

  Justify the choice of potassium hydroxide dissolved in ethanol, heated under reflux, when propene is to be made from 2-bromopropane, rather than aqueous potassium hydroxide warmed gently.

  Mark scheme

  1. B1 with little water present the hydroxide ion acts as a base, removing an H⁺ from a carbon next to the C–Br carbon, so elimination gives propene; in water it acts mainly as a nucleophile and substitution to propan-2-ol competes
  2. B1 elimination has the higher activation energy of the two reactions, so the hot conditions of reflux favour it, whereas gentle warming favours substitution
  3. B1 heating under reflux lets the volatile ethanol and 2-bromopropane be heated for long enough without loss, while the propene, a gas, leaves through the condenser to be collected

• 2 marks · Explain · EH-Q3 AQA · Edexcel · CAIE

  Explain the difference in the role of the hydroxide ion in elimination and in nucleophilic substitution.

  Mark scheme

  1. B1 in elimination the hydroxide ion acts as a base, accepting an H⁺ from the carbon next to the C–X carbon
  2. B1 in substitution it acts as a nucleophile, donating its lone pair to the δ+ carbon

• 2 marks · Suggest · EH-Q6 AQA · Edexcel · CAIE

  Suggest why 2-bromo-2-methylpropane gives only one alkene on elimination, and name it.

  Mark scheme

  1. B1 the three carbons next to the C–Br carbon are identical CH₃ groups, so removing H from any of them gives the same product
  2. B1 the alkene is 2-methylpropene, (CH₃)₂C=CH₂

• 2 marks · Suggest · EH-Q16 OCR A

  Suggest two equations to show how nitrogen monoxide radicals, NO•, from aircraft engines can catalyse the decomposition of ozone in the same way as chlorine radicals.

  Mark scheme

  1. B1 NO• + O₃ → NO₂• + O₂
  2. B1 NO₂• + O → NO• + O₂, regenerating NO•; allow NO₂• + O₃ → 2O₂ + NO• by analogy with the chlorine cycle

Practise elimination from halogenoalkanes, and CFCs and the ozone layer one question at a time