Preparing amines and diazonium dyesEdexcel International A Level Chemistry: Flashcards
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Reagents and conditions to make a primary amine from a halogenoalkane?
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- Reagents and conditions to make a primary amine from a halogenoalkane?
- Excess concentrated ammonia in ethanol, heated in a sealed tube.
- Why use excess ammonia when making a primary amine?
- To reduce further substitution that gives secondary and tertiary amines.
- How can a halogenoalkane be converted to an amine with one extra carbon?
- KCN in ethanol under reflux to form the nitrile, then reduce with LiAlH₄ in dry ether (or H₂ with Ni).
- Equation for reducing CH₃CH₂CN to an amine?
- CH₃CH₂CN + 4[H] → CH₃CH₂CH₂NH₂
- How is nitrobenzene reduced to phenylamine?
- Heat under reflux with tin and concentrated HCl, then add excess NaOH.
- Why is NaOH added after the tin and HCl reflux?
- To convert phenylammonium ions into free phenylamine.
- Equation for nitrobenzene to phenylamine using [H]?
- C₆H₅NO₂ + 6[H] → C₆H₅NH₂ + 2H₂O
- How is nitrous acid made for diazotisation?
- In situ from sodium nitrite and hydrochloric acid.
- What temperature is needed to make a diazonium salt, and why?
- Below 10 °C, because the ion decomposes above this to phenol and nitrogen.
- Formula of the benzenediazonium ion?
- C₆H₅N₂⁺
- What is needed to couple a diazonium ion with phenol?
- Phenol dissolved in aqueous NaOH, kept cold.
- Name the dye from benzenediazonium chloride and phenol.
- 4-hydroxyazobenzene, a yellow-orange azo dye.
- Why is an azo compound coloured?
- The N=N group extends the delocalised π system over both rings, so visible light is absorbed.
Exam questions on Preparing amines and diazonium dyes
- A chemist wishes to make butylamine, CH₃CH₂CH₂CH₂NH₂, starting from 1-bromopropane, CH₃CH₂CH₂Br. Two different routes to primary amines are being considered.Write an equation for the reaction of 1-bromopropane with excess ammonia, using structural formulae, and state the conditions used.2 marks
- Phenylamine is prepared in a school laboratory from nitrobenzene, C₆H₅NO₂ (Mr = 123). In one preparation, 12.3 g of nitrobenzene is reduced to phenylamine, C₆H₅NH₂ (Mr = 93), and the percentage yield is 60%.Write an equation for the reduction of nitrobenzene to phenylamine, using [H] to represent the reducing agent, and explain why the reaction mixture is finally made alkaline.2 marks
- A student prepares benzenediazonium chloride from phenylamine and then uses the solution straight away to make an azo dye.Describe how benzenediazonium chloride is prepared from phenylamine. Give the reagents, the temperature and an equation.3 marks
Written by the Exaim team, led by Shaun Daswani (Head of Upper Secondary, Improve ME Institute; MSc Financial Mathematics, Imperial College London; BSc, UCL) and Jason Daswani (operational lead, Improve ME Institute; LSE).