Grignard reagents and carbon chain extensionEdexcel International A Level Chemistry: Subtopic test
10 questions, 27 marks
Edexcel International A Level Chemistry
Grignard reagents and carbon chain extension
Total 27 marks
Name
Class
Date
- 1A research student prepares ethylmagnesium bromide, C₂H₅MgBr, from bromoethane and magnesium. The first attempt is carried out in apparatus that has been washed and left to drain, and the student is told that this is a poor choice for this reaction.(a)Which conditions are used to form a Grignard reagent from a haloalkane?[1 mark]
- AMagnesium ribbon in aqueous ethanol, warmed gently
- BMagnesium turnings in dry ethoxyethane (ether)
- CMagnesium powder in water at room temperature
- DMagnesium turnings in dilute hydrochloric acid
(b)What happens to the magnesium when bromoethane is converted into ethylmagnesium bromide?[1 mark]- AIt is reduced from 0 to +2
- BIt is oxidised from 0 to +1
- CIt is oxidised from 0 to +2
- DIts oxidation state stays at 0
(c)Explain why the apparatus and the ether must be completely dry for this preparation.[2 marks]Total for question 1: 4 marks
- 2A student plans to convert 1-bromopropane into butanoic acid in two stages. In stage 1 the haloalkane is turned into a Grignard reagent. In stage 2 the Grignard reagent is added to an excess of crushed solid carbon dioxide, and the mixture is then treated with dilute hydrochloric acid.(a)What type of reaction is the attack of the Grignard reagent on carbon dioxide?[1 mark]
- AElectrophilic addition
- BFree-radical substitution
- CNucleophilic substitution
- DNucleophilic addition
(b)How does the number of carbon atoms in the carboxylic acid compare with the number in 1-bromopropane?[1 mark]- AOne more carbon atom
- BThe same number of carbon atoms
- COne fewer carbon atom
- DTwo more carbon atoms
(c)Explain why the Grignard reagent attacks the carbon atom of carbon dioxide.[2 marks]Total for question 2: 4 marks
- 3A chemist has a bottle of methylmagnesium bromide, CH₃MgBr, in dry ethoxyethane. She wants to make three different alcohols by reacting separate portions of the reagent with three different carbonyl compounds, each followed by dilute acid: methanal, ethanal and propanone.(a)Deduce the name of the alcohol formed from each carbonyl compound, and state whether it is a primary, secondary or tertiary alcohol.[3 marks](b)Describe the reaction between methylmagnesium bromide and ethanal, and explain why dilute acid is added in a second step.[4 marks]
Total for question 3: 7 marks
- 4A pharmaceutical chemist needs the tertiary alcohol 2-methylbutan-2-ol, (CH₃)₂C(OH)CH₂CH₃. The only organic starting materials available are bromoethane and propanone, and a single carbon–carbon bond-forming step must be used. In the first attempt the chemist obtained mostly ethane gas and only a very low yield of the alcohol.(a)Describe how the chemist could make 2-methylbutan-2-ol from these two starting materials, giving the reagents, conditions and a safety precaution.[6 marks](b)Explain why the first attempt gave mostly ethane and a low yield of alcohol, and suggest changes that would improve the yield, justifying your answer.[6 marks]
Total for question 4: 12 marks
End of questions
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).