AminesAQA A-Level Chemistry: Topic test

1. 1
   A dye manufacturer makes 4-methylphenylamine, CH₃C₆H₄NH₂, from 4-nitrotoluene, CH₃C₆H₄NO₂. The 4-nitrotoluene is heated under reflux with tin and concentrated hydrochloric acid. The cooled mixture is then made alkaline with aqueous sodium hydroxide and the amine is separated.

   (a)

   What is the role of the tin in this reaction?

   [1 mark]

   • AA catalyst that is regenerated
   • BAn oxidising agent
   • CA reducing agent
   • DA dehydrating agent

   (b)

   Which nitrogen-containing species is present in the flask at the end of the reflux, before the sodium hydroxide is added?

   [1 mark]

   • ACH₃C₆H₄NH₂
   • BCH₃C₆H₄NH₃⁺
   • CCH₃C₆H₄NO₂
   • DCH₃C₆H₄NH⁻

   (c)

   In one preparation, 8.22 g of 4-nitrotoluene (Mᵣ = 137.0) gave 4.82 g of 4-methylphenylamine (Mᵣ = 107.0). Calculate the percentage yield of 4-methylphenylamine.

   [2 marks]

   Total for question 1: 4 marks

2. 2
   A mixture of phenylamine, C₆H₅NH₂, and hexane, C₆H₁₄, is shaken in a separating funnel with dilute hydrochloric acid. Two layers form. The lower aqueous layer is run off, and aqueous sodium hydroxide is then added to it until the solution is alkaline.

   (a)

   Which species is mainly present in the aqueous layer after shaking with the acid?

   [1 mark]

   • AC₆H₅NH₃⁺ and Cl⁻
   • BC₆H₅NH₂
   • CC₆H₁₄
   • DC₆H₅NO₂

   (b)

   Which statement correctly explains why phenylamine is a weaker base than ammonia?

   [1 mark]

   • AThe nitrogen atom in phenylamine has no lone pair of electrons
   • BThe phenyl group donates electrons to nitrogen, which makes the lone pair less available
   • CPhenylamine is a larger molecule, so it has stronger London forces
   • DThe lone pair on nitrogen is delocalised into the benzene ring, which makes it less available to accept H⁺

   (c)

   Explain why ethylamine is a stronger base than ammonia.

   [2 marks]

   Total for question 2: 4 marks

3. 3
   A chemist heats dimethylamine, (CH₃)₂NH, with excess iodomethane, CH₃I, in ethanol in a sealed tube. The reaction proceeds in stages and ends with an ionic product.

   (a)

   Outline the mechanism for the first stage, in which dimethylamine reacts with iodomethane to form trimethylamine. Describe each step, including the movement of electron pairs.

   [3 marks]

   (b)

   The trimethylamine reacts with more iodomethane to form tetramethylammonium iodide. Write an equation for this reaction and explain why the substitution cannot continue beyond this product.

   [4 marks]

   Total for question 3: 7 marks

4. 4
   A research group needs pure pentylamine, CH₃(CH₂)₄NH₂. Route 1: heat 1-bromopentane in a sealed tube with a large excess of ethanolic ammonia. Route 2: heat 1-bromobutane with potassium cyanide in aqueous ethanolic solution under reflux to form pentanenitrile, then reduce the nitrile with LiAlH₄ in dry ether. The pentylamine will later be compared with ammonia and phenylamine as a base.

   (a)

   Evaluate the two routes and decide which should be used to make pure pentylamine.

   [6 marks]

   (b)

   Explain why pentylamine is a stronger base than ammonia and ammonia is a stronger base than phenylamine. Include an equation for pentylamine acting as a base in water, and describe how the order could be confirmed experimentally.

   [6 marks]

   Total for question 4: 12 marks

5. 5
   Hexane-1,6-diamine, H₂N(CH₂)₆NH₂, is made industrially by reducing hexanedinitrile, NC(CH₂)₄CN, with hydrogen gas at high pressure in the presence of a nickel catalyst. The diamine is one of the monomers used to make nylon-6,6.

   (a)

   How many moles of hydrogen are needed to convert one mole of hexanedinitrile completely into hexane-1,6-diamine?

   [1 mark]

   • A2
   • B3
   • C6
   • D4

   (b)

   Which reagent could be used in the laboratory to reduce a nitrile to a primary amine?

   [1 mark]

   • ALiAlH₄ in dry ether
   • BAcidified potassium dichromate(VI)
   • CAqueous sodium hydroxide
   • DEthanolic potassium cyanide

   (c)

   In one batch, 54.0 kg of hexanedinitrile (Mᵣ = 108.0) gave 48.7 kg of hexane-1,6-diamine (Mᵣ = 116.0). Calculate the percentage yield of the diamine.

   [2 marks]

   Total for question 5: 4 marks

6. 6
   Ethylamine, CH₃CH₂NH₂, is a colourless, water-soluble amine. A pharmaceutical technician reacts it with hydrochloric acid to make a crystalline salt, which is easier to store and handle than the amine.

   (a)

   What is the formula of the salt formed?

   [1 mark]

   • ACH₃CH₂NH₂Cl
   • B[CH₃CH₂NH₃]⁺Cl⁻
   • CCH₃CH₂Cl
   • D[CH₃CH₂NH]⁻H₃O⁺

   (b)

   Ethylamine is a weak base. What does this mean for its reaction with water, CH₃CH₂NH₂ + H₂O ⇌ CH₃CH₂NH₃⁺ + OH⁻?

   [1 mark]

   • AThe equilibrium lies mainly to the right because ethylamine is a strong proton acceptor
   • BThe equilibrium lies mainly to the left because ethylamine has no lone pair
   • CThe equilibrium lies mainly to the left because only a small proportion of the amine molecules accept a proton
   • DNo equilibrium is set up because amines do not react with water

   (c)

   Aqueous sodium hydroxide is added to a solution of the salt and a strong smell of ethylamine is noticed. Write an equation for the reaction and explain what it shows about the relative strengths of hydroxide ions and ethylamine as bases.

   [2 marks]

   Total for question 6: 4 marks

7. 7
   Propan-2-amine, (CH₃)₂CHNH₂, is made by heating 2-bromopropane, (CH₃)₂CHBr, with a large excess of concentrated ammonia in ethanol in a sealed tube. The other product is ammonium bromide, NH₄Br.

   (a)

   Calculate the percentage atom economy for making propan-2-amine in this reaction. Use Mᵣ values: (CH₃)₂CHBr = 122.9, NH₃ = 17.0, (CH₃)₂CHNH₂ = 59.0.

   [3 marks]

   (b)

   Explain why a large excess of ammonia is used, and why this favours the formation of the primary amine.

   [4 marks]

   Total for question 7: 7 marks

8. 8
   A company makes the cationic surfactant dodecyltrimethylammonium bromide, [C₁₂H₂₅N(CH₃)₃]⁺Br⁻, for use in fabric conditioners. Step 1: 1-bromododecane is heated with a large excess of ethanolic ammonia to give dodecylamine, C₁₂H₂₅NH₂. Step 2: dodecylamine is heated with excess bromomethane, CH₃Br.

   (a)

   Explain how the quaternary ammonium salt is formed in Step 2 and why the reaction stops at this product. Explain also how its structure makes it a cationic surfactant.

   [6 marks]

   (b)

   Dodecylamine can act as both a base and a nucleophile, but the quaternary ammonium salt can act as neither. Explain why, with an equation for each of the two reactions of dodecylamine. Compare the base strength of dodecylamine with that of phenylamine.

   [6 marks]

   Total for question 8: 12 marks

End of questions