Organic chemistry IEdexcel A-Level Chemistry: Topic test

1. 1
   In an oil refinery, dodecane, C₁₂H₂₆, from the kerosene fraction is cracked. In one reaction, one molecule of dodecane forms one molecule of octane, C₈H₁₈, and one molecule of an alkene.

   (a)

   What is the molecular formula of the alkene?

   [1 mark]

   • AC₄H₆
   • BC₄H₈
   • CC₄H₁₀
   • DC₂H₄

   (b)

   Which statement about reforming is correct?

   [1 mark]

   • AIt breaks long-chain alkanes into shorter alkanes and alkenes.
   • BIt separates crude oil into fractions with different boiling points.
   • CIt removes sulfur compounds from fuels so that less sulfur dioxide is formed.
   • DIt converts straight-chain alkanes into branched and cyclic hydrocarbons, which burn more efficiently in engines.

   (c)

   Suggest why oil refineries crack long-chain hydrocarbons such as dodecane.

   [2 marks]

   Total for question 1: 4 marks

2. 2
   1,2-Dichloroethene, CHCl=CHCl, exists as two stereoisomers. One boils at 48 °C and the other at 60 °C.

   (a)

   Which statement describes the Z isomer of 1,2-dichloroethene?

   [1 mark]

   • AThe two chlorine atoms are on the same side of the C=C bond.
   • BThe two chlorine atoms are on opposite sides of the C=C bond.
   • CBoth chlorine atoms are bonded to the same carbon atom.
   • DThe two carbon atoms are joined by a single bond that rotates freely.

   (b)

   Which compound also shows E/Z isomerism?

   [1 mark]

   • ACH₂=CHCH₂CH₂CH₃
   • B(CH₃)₂C=CHCH₃
   • CCH₃CH=CHCH₂CH₃
   • DCH₃CH₂CH₂CH₂CH₃

   (c)

   Explain why 1,2-dichloroethene shows E/Z isomerism.

   [2 marks]

   Total for question 2: 4 marks

3. 3
   Butane reacts with chlorine in ultraviolet light by radical substitution to form chlorobutanes.

   (a)

   Write equations, showing the radicals, for the initiation step and the two propagation steps in the formation of a chlorobutane, C₄H₉Cl.

   [3 marks]

   (b)

   A small amount of octane, C₈H₁₈, is also found in the product. Write an equation for its formation and explain why this step is a termination step. State one reason why the product mixture also contains dichlorobutanes.

   [4 marks]

   Total for question 3: 7 marks

4. 4
   Poly(tetrafluoroethene), PTFE, is used to coat non-stick pans. It is made by addition polymerisation of tetrafluoroethene, CF₂=CF₂. PTFE is very unreactive. Mean bond enthalpies: C–F 467 kJ mol⁻¹ and C–Cl 346 kJ mol⁻¹. When PTFE is burned, toxic hydrogen fluoride is released.

   (a)

   Describe how PTFE forms from tetrafluoroethene, and use the bond enthalpies to explain why carbon–fluorine compounds react much more slowly with nucleophiles than carbon–chlorine compounds. State one consequence of this lack of reactivity for the disposal of waste PTFE.

   [6 marks]

   (b)

   Evaluate landfill, incineration and recycling as ways of disposing of waste PTFE, and give a justified conclusion.

   [6 marks]

   Total for question 4: 12 marks

5. 5
   But-1-ene, CH₂=CHCH₂CH₃, reacts with hydrogen chloride at room temperature by electrophilic addition. Two different carbocation intermediates are possible.

   (a)

   What is the major organic product?

   [1 mark]

   • A1-chlorobutane
   • B1,2-dichlorobutane
   • C2-chlorobutane
   • D2-chloro-2-methylpropane

   (b)

   Which statement about the mechanism is correct?

   [1 mark]

   • AThe H–Cl bond breaks heterolytically and the hydrogen atom is attacked by the π electrons, so it acts as the electrophile.
   • BThe H–Cl bond breaks homolytically to form hydrogen and chlorine radicals.
   • CA chloride ion attacks the π bond first, acting as a nucleophile.
   • DThe reaction is nucleophilic addition, because HCl is a polar molecule.

   (c)

   Explain why 2-chlorobutane is formed in greater amounts than 1-chlorobutane.

   [2 marks]

   Total for question 5: 4 marks

6. 6
   Bromoethane, CH₃CH₂Br, is heated with an excess of concentrated ammonia solution in a sealed tube. Ethylamine, CH₃CH₂NH₂, is formed.

   (a)

   What is the role of ammonia in this reaction?

   [1 mark]

   • AAn electrophile, because it accepts a pair of electrons
   • BA free radical, because it has an unpaired electron
   • CAn oxidising agent, because it removes electrons from bromoethane
   • DA nucleophile, because it donates a lone pair of electrons

   (b)

   Which reagent and conditions would convert bromoethane into ethene?

   [1 mark]

   • AAqueous potassium hydroxide, warmed
   • BEthanolic potassium hydroxide, heated
   • CPotassium cyanide in ethanol, heated
   • DConcentrated ammonia in a sealed tube

   (c)

   Explain why the carbon atom attached to the bromine atom is attacked by ammonia.

   [2 marks]

   Total for question 6: 4 marks

7. 7
   A student has samples of butan-2-ol and butan-1-ol, and a solution of acidified potassium dichromate(VI).

   (a)

   Butan-2-ol is warmed with the acidified potassium dichromate(VI). State the colour change, name the organic product and write an equation using [O] for the oxidising agent.

   [3 marks]

   (b)

   Butan-1-ol can be oxidised to butanal or to butanoic acid. Explain how the choice of apparatus and conditions decides which is formed, and describe a test that distinguishes butanal from butanone.

   [4 marks]

   Total for question 7: 7 marks

8. 8
   But-2-ene, CH₃CH=CHCH₃, is converted into butanone in three steps. Step 1: but-2-ene reacts with hydrogen bromide to give 2-bromobutane. Step 2: 2-bromobutane is warmed with aqueous potassium hydroxide to give butan-2-ol. Step 3: butan-2-ol is heated with acidified potassium dichromate(VI) to give butanone.

   (a)

   Explain why but-2-ene shows E/Z isomerism, and describe the mechanism of step 1.

   [6 marks]

   (b)

   Explain the chemistry of steps 2 and 3, and explain how the product of step 2 would differ if ethanolic potassium hydroxide were used instead.

   [6 marks]

   Total for question 8: 12 marks

End of questions