CBSE Class 12 Chemistry 2023: question 22
Set in the CBSE Class 12 board exam of 2023, paper 56/1/1, section B, question 22, for 2 marks. Try it first; the hints and our step-by-step solution are below.
Q22 · 2 marks · Very short answerHaloalkanes and Haloarenes
(a)
(b)
Identify 'A' and 'B' in the above reactions.
Source: official question paper from cbse.gov.in (PDF). See it in the full 2023 board paper, set 56/1/1.
Hints
Open one hint at a time and try again before the next.
Hint 1
PCl₅ converts an alcohol into an alkyl chloride.
Hint 2
AgCN is covalent, so think about which atom of the cyanide group bonds to carbon.
Hint 3
Alcoholic KOH eliminates HX; then apply Markovnikov's rule for HBr addition.
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.
- ½ mark
(a) replaces –OH by –Cl. **A = (2-chloropropane).**
- ½ mark
AgCN is mainly covalent, so it attacks through the nitrogen atom and gives an isocyanide. **B = (2-propyl isocyanide / isopropyl isocyanide).**
- ½ mark
(b) Alcoholic KOH causes dehydrohalogenation. **A = (propene).**
- ½ mark
HBr adds by Markovnikov's rule, so Br goes to the carbon with fewer H atoms. **B = (2-bromopropane).**
Answer: (a) A = 2-chloropropane, B = 2-propyl isocyanide (CH₃CH(NC)CH₃); (b) A = propene, B = 2-bromopropane
Where the marks are
- (a) A = 2-chloropropane (½ mark)
- (a) B = isopropyl isocyanide (½ mark)
- (b) A = propene (½ mark)
- (b) B = 2-bromopropane (½ mark)
Revise the chapter
- Chapter: Haloalkanes and Haloarenes (notes, formulas, where marks are lost and every past question)
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