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Chemistry · Alcohols, Ethers and Phenols

JEE Main 2024 — 5 April, Shift 2 — Question 71

CH3CH2OH  →(ii) KMnO4(iii) NaOH, CaO, Δ(i) Jones’ ReagentP\mathrm{CH_3CH_2OH} \;\xrightarrow[\substack{\text{(ii) KMnO}_4 \\ \text{(iii) NaOH, CaO, } \Delta}] {\text{(i) Jones' Reagent}} P

Consider the above reaction sequence and identify the major product P .

  1. Option A:

    Methane

    Correct
  2. Option B:

    Methanal

  3. Option C:

    Methoxymethane

  4. Option D:

    Methanoic acid

Answer: A

Step-by-step solution

  1. Oxidation with Jones’ reagent / KMnO4_4: CH3CH2OH→[O]CH3COOH\mathrm{CH_3CH_2OH \xrightarrow{[O]} CH_3COOH}

Ethanol (primary alcohol) is oxidised completely to ethanoic acid.

  1. Reaction with NaOH: CH3COOH+NaOH→CH3COONa+H2O\mathrm{CH_3COOH + NaOH \rightarrow CH_3COONa + H_2O} Ethanoic acid forms sodium ethanoate.

  2. Heating with soda lime (NaOH + CaO, Δ\Delta): CH3COONa→NaOH/CaO, ΔCH4+Na2CO3\mathrm{CH_3COONa \xrightarrow{NaOH/CaO,\ \Delta} CH_4 + Na_2CO_3} Sodium ethanoate undergoes decarboxylation, giving methane.

Answer key and solution verified before publishing.

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Exam
JEE Main 2024
Paper
5 April, Shift 2
Subject
Chemistry
Chapter
Alcohols, Ethers and Phenols
Topic
Chemical properties of alcohols