Answer the following. (a) Write the mechanism of the following reaction : CH_3CH_2OH [443 K]H^+ CH_2=CH_2 (b) Write the main product in each of the following reactions : (i) CH_3-CH_2-CH=CH_2 [b) 3H_2O_2/OH^-]a) B_2H_6 (ii) phenol [b) CO_2, H^+]a) aq. NaOH

ChemistryCBSE Class 12 · Board exam 2024, set 56/1/1 · 3 marks · Short answer

CBSE Class 12 Chemistry 2024: question 26

Set in the CBSE Class 12 board exam of 2024, paper 56/1/1, section C, question 26, for 3 marks. Try it first; the hints and our step-by-step solution are below.

Q26 · 3 marks · Short answerAlcohols, Phenols and Ethers

Answer the following.

(a)
Write the mechanism of the following reaction :
$C H_{3} C H_{2} OH H^{+} 443 K C H_{2} = C H_{2}$
1 mark
(b)
Write the main product in each of the following reactions :
(i) $C H_{3} - C H_{2} - CH = C H_{2} a) B_{2} H_{6} b) 3 H_{2} O_{2} /O H^{-}$
(ii) phenol $a) aq. NaOH b) C O_{2}, H^{+}$
2 marks

Figure in the paper: Part (b)(ii) shows phenol (benzene ring with OH). See it in the official paper.

Source: official question paper from cbse.gov.in (PDF). See it in the full 2024 board paper, set 56/1/1.

Hints

Open one hint at a time and try again before the next.

Hint 1

Acid first converts the poor leaving group –OH into a good leaving group.

Hint 2

B₂H₆/H₂O₂ gives the opposite of Markovnikov; phenoxide + CO₂ is a named reaction putting –COOH ortho to –OH.

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.

(a) Step 1 – protonation: $C H_{3} C H_{2} - \ddot{O} H + H^{+} ⇌ C H_{3} C H_{2} - O + H_{2}$ . Step 2 – loss of water (slow): $C H_{3} C H_{2} O + H_{2} \to C H_{3} C + H_{2} + H_{2} O$ . Step 3 – loss of proton: $H - C H_{2} - C + H_{2} \to C H_{2} = C H_{2} + H^{+}$ .
1 mark
(b)(i) Hydroboration–oxidation gives anti-Markovnikov addition of water: $C H_{3} C H_{2} C H_{2} C H_{2} OH$ (butan-1-ol).
1 mark
(b)(ii) Kolbe's reaction: phenol → sodium phenoxide → on CO₂ and acidification gives salicylic acid (2-hydroxybenzoic acid, o-HO–C₆H₄–COOH).
1 mark

Answer: (a) Protonation → carbocation (loss of H₂O) → loss of H⁺ gives ethene; (b)(i) butan-1-ol; (ii) salicylic acid

Where the marks are

Three-step mechanism: protonation, carbocation formation, deprotonation (1 mark)
Butan-1-ol (1 mark)
Salicylic acid (1 mark)

Revise the chapter

Chapter: Alcohols, Phenols and Ethers (notes, formulas, where marks are lost and every past question)

Practise this chapterBoard-style questions from Alcohols, Phenols and Ethers, one at a time, marked against the answer key and marking points.Back to the full paper2024 board paper, set 56/1/1: every question, by section.

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CBSE Class 12 Chemistry 2024: question 26

Hints

Step-by-step solution

Revise the chapter

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