CBSE Class 12 Chemistry 2023: question 29(b)
Set in the CBSE Class 12 board exam of 2023, paper 56/1/1, section C, question 29(b), for 3 marks. Try it first; the hints and our step-by-step solution are below.
Q29(b) · 3 marks · Short answerAlcohols, Phenols and Ethers
An organic compound 'A', having the molecular formula on treatment with Cu at 573 K, gives 'B'. 'B' does not reduce Fehling's solution but gives a yellow precipitate of the compound 'C' with /NaOH. Deduce the structures of A, B and C.
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
Cu at 573 K removes H₂ from an alcohol; which type of alcohol gives a ketone?
Hint 2
Not reducing Fehling's solution rules out an aldehyde.
Hint 3
A yellow precipitate with I₂/NaOH means a CH₃CO– group is present.
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
Cu at 573 K dehydrogenates alcohols, so A () is an alcohol. B does not reduce Fehling's solution, so B is a ketone, not an aldehyde. A must therefore be a secondary alcohol: **A = (propan-2-ol).**
- 1 mark
. **B = (propanone).** It is a methyl ketone, which explains the positive iodoform test.
- 1 mark
. **C = (iodoform),** the yellow precipitate.
Answer: A = CH₃CH(OH)CH₃ (propan-2-ol); B = CH₃COCH₃ (propanone); C = CHI₃ (iodoform)
Where the marks are
- A = propan-2-ol (1 mark)
- B = propanone (1 mark)
- C = iodoform CHI₃ (1 mark)
Revise the chapter
- Chapter: Alcohols, Phenols and Ethers (notes, formulas, where marks are lost and every past question)
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