CBSE Class 12 Chemistry 2023: question 24(b)
Set in the CBSE Class 12 board exam of 2023, paper 56/5/1, section B, question 24(b), for 2 marks. Try it first; the hints and our step-by-step solution are below.
Q24(b) · 2 marks · Very short answerAmines
(i) Complete the reaction with the main product formed :
Benzenediazonium chloride ()
(ii) Convert Bromoethane to Propanamine.
Figure in the paper: Benzene ring with N2+ Cl- substituent, followed by reaction arrow with CH3CH2OH above it. See it in the official paper.
Source: official question paper from cbse.gov.in (PDF). See it in the full 2023 board paper, set 56/5/1.
Hints
Open one hint at a time and try again before the next.
Hint 1
Ethanol acts as a mild reducing agent towards diazonium salts, replacing –N₂⁺ by H.
Hint 2
Propanamine has one carbon more than bromoethane: which nucleophile adds one carbon and can later be reduced to –CH₂NH₂?
Step-by-step solution
Each step shows the marks it earns, as an examiner would award them. Checked by an independent re-solve from the question paper.
- 1 mark
(i) Ethanol reduces the diazonium salt to benzene:
Main product: benzene. - 1 mark
(ii) Step-up by one carbon via the nitrile:
(bromoethane → propanenitrile → propan-1-amine)
Answer: (i) Benzene (with N₂, CH₃CHO, HCl); (ii) CH₃CH₂Br → (KCN) CH₃CH₂CN → (LiAlH₄) CH₃CH₂CH₂NH₂
Where the marks are
- Correct product benzene in reaction with ethanol (1 mark)
- Conversion via KCN then reduction to propanamine (1 mark)
Revise the chapter
- Chapter: Amines (notes, formulas, where marks are lost and every past question)
- Topic: Preparation of Amines
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