(i) Give mechanism of the following reaction : CH_3CH_2OH [413\ K]H_2SO_4 CH_3CH_2 – O – CH_2CH_3 + H_2O (ii) Illustrate hydroboration – oxidation reaction with an example.

ChemistryCBSE Class 12 · Board exam 2023, set 56/5/1 · 3 marks · Short answer

CBSE Class 12 Chemistry 2023: question 30(b)

Set in the CBSE Class 12 board exam of 2023, paper 56/5/1, section C, question 30(b), for 3 marks. Try it first; the hints and our step-by-step solution are below.

Q30(b) · 3 marks · Short answerAlcohols, Phenols and Ethers

(i) Give mechanism of the following reaction :
$C H_{3} C H_{2} OH H_{2} S O_{4} 413 K C H_{3} C H_{2} -O-C H_{2} C H_{3} + H_{2} O$
(ii) Illustrate hydroboration – oxidation reaction with an example.

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

The acid first converts –OH into a better leaving group.

Hint 2

A second alcohol molecule acts as the nucleophile; finish by removing H⁺.

Hint 3

Hydroboration uses B₂H₆ and then alkaline H₂O₂; the –OH ends up on the less substituted carbon.

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.

(i) Step 1 – Protonation of ethanol:
$C H_{3} C H_{2} - \ddot{O} H + H^{+} \to C H_{3} C H_{2} - O + H_{2}$
½ mark
Step 2 – Nucleophilic attack ( $S_{N} 2$ ) by a second ethanol molecule on the protonated alcohol, with loss of water:
$C H_{3} C H_{2} - \ddot{O} H + C H_{3} C H_{2} - O + H_{2} \to C H_{3} C H_{2} - O + (H) - C H_{2} C H_{3} + H_{2} O$
1 mark
Step 3 – Deprotonation:
$C H_{3} C H_{2} - O + (H) - C H_{2} C H_{3} \to C H_{3} C H_{2} - O - C H_{2} C H_{3} + H^{+}$
½ mark
(ii) Hydroboration–oxidation: diborane adds to an alkene, and the trialkylborane is oxidised by alkaline $H_{2} O_{2}$ to an alcohol (anti-Markovnikov product):
$3C H_{3} CH = C H_{2} + B H_{3} \to (C H_{3} C H_{2} C H_{2})_{3} B 3 H_{2} O_{2} /O H^{-} 3C H_{3} C H_{2} C H_{2} OH + B (OH)_{3}$
(propene → propan-1-ol)
1 mark

Answer: (i) protonation → SN2 attack by ethanol → deprotonation gives ethoxyethane; (ii) propene + B₂H₆, then H₂O₂/OH⁻ → propan-1-ol

Where the marks are

Protonation of ethanol (½ mark)
Nucleophilic attack by second ethanol with loss of water (1 mark)
Deprotonation to give ether (½ mark)
Correct hydroboration–oxidation equation (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 paper2023 board paper, set 56/5/1: every question, by section.

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CBSE Class 12 Chemistry 2023: question 30(b)

Hints

Step-by-step solution

Revise the chapter

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