Learnosha CBSE 12
Chemistry Alcohols, Phenols and Ethers 5 Marks
Appeared in: 2026 (3 sets)
(A) An organic compound 'A', with molecular formula $\text{C}_2\text{H}_6\text{O}$ reacts with active metals such as sodium to give compound 'B' and hydrogen gas. 'A' on treatment with iodine and sodium hydroxide gives 'C' and in presence of $\text{H}_2\text{SO}_4$ at $413\text{ K}$ gives 'D' ($\text{C}_4\text{H}_{10}\text{O}$). 'D' on reaction with excess of HI gives 'E'. Identify 'A', 'B', 'C', 'D' and 'E' and write all the reactions involved. OR (B) (a) Write the reagents which are used in the given conversions : (i) Phenol to $2, 4, 6 - \text{tribromophenol}$ (ii) Propene to propan-$1-\text{ol}$ (iii) Butan-$2-\text{one}$ to butan-$2-\text{ol}$ (b) Explain the mechanism of acid catalyzed hydration of alkene to form corresponding alcohol.
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Solution (A)

Step 1 (Identification of Compounds): [1 Mark]

  • 'A' is Ethanol ($\text{C}_2\text{H}_5\text{OH}$)
  • 'B' is Sodium ethoxide ($\text{C}_2\text{H}_5\text{ONa}$)
  • 'C' is Iodoform ($\text{CHI}_3$)
  • 'D' is Diethyl ether ($\text{C}_2\text{H}_5\text{OC}_2\text{H}_5$)
  • 'E' is Ethyl iodide ($\text{C}_2\text{H}_5\text{I}$)

Step 2 (Chemical Reactions): [3 Marks]

1. Reaction with Na: $2\text{C}_2\text{H}_5\text{OH} + 2\text{Na} \rightarrow 2\text{C}_2\text{H}_5\text{ONa} + \text{H}_2(g)$

2. Iodoform test: $\text{CH}_3\text{CH}_2\text{OH} + 4\text{I}_2 + 6\text{NaOH} \rightarrow \text{CHI}_3(s) + \text{HCOONa} + 5\text{NaI} + 5\text{H}_2\text{O}$

3. Dehydration at 413 K: $2\text{C}_2\text{H}_5\text{OH} \xrightarrow{\text{H}_2\text{SO}_4, 413\text{K}} \text{C}_2\text{H}_5\text{OC}_2\text{H}_5 + \text{H}_2\text{O}$

4. Reaction with excess HI: $\text{C}_2\text{H}_5\text{OC}_2\text{H}_5 + 2\text{HI} \rightarrow 2\text{C}_2\text{H}_5\text{I} + \text{H}_2\text{O}$

Step 3 (Final Conclusion): [1 Mark]

The sequence confirms 'A' is ethanol, which undergoes nucleophilic substitution, oxidation (iodoform test), and intermolecular dehydration to form the ether 'D', which finally cleaves to form the alkyl halide 'E'.

Solution (B)

(a) Reagents for Conversions: [1.5 Marks]

  • (i) $\text{Br}_2$ water
  • (ii) Hydroboration-oxidation: 1. $\text{B}_2\text{H}_6$ / $\text{THF}$, 2. $\text{H}_2\text{O}_2 / \text{OH}^-$
  • (iii) $\text{NaBH}_4$ or $\text{LiAlH}_4$ or $\text{H}_2/\text{Ni}$

(b) Mechanism of Acid Catalyzed Hydration: [3.5 Marks]

Step 1 (Protonation): The alkene is protonated by the acid catalyst to form a carbocation.

$$\text{CH}_3-\text{CH}=\text{CH}_2 + \text{H}_3\text{O}^+ \rightleftharpoons \text{CH}_3-\text{CH}^+-\text{CH}_3 + \text{H}_2\text{O}$$

Step 2 (Nucleophilic Attack):

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