Question

$\begin{array}{c} \text{CH}_3 \\ \text{I} \\ \text{C} \\ \text{III} \\ \text{CH} \end{array} \xrightarrow[\text{H}_2\text{SO}_4]{\text{KMnO}_4} \text{A} \xrightarrow[\text{CaO}]{\text{NaOH}} \text{B}$; Compound B is

• A. C$_2$H$_6$
• B. CH$_4$
• C. CH$_3$COOH
• D. CH$_3$COONa

Answer 1

Answer: (B)

CH$_4$

Answer 2

The reaction proceeds in two steps:

1. Oxidation of Propyne (CH₃-C≡CH) with KMnO₄/H₂SO₄: Strong oxidizing agents like hot acidic KMnO₄ cause oxidative cleavage of alkynes. For terminal alkynes, the triple bond breaks, and the terminal ≡CH group is oxidized to CO₂. The other part of the alkyne (CH₃-C≡) is oxidized to a carboxylic acid (CH₃COOH).

   CH₃-C≡CH + [O] --(KMnO₄, H₂SO₄)--> CH₃COOH + CO₂

   So, compound A is Acetic Acid (CH₃COOH).

2. Decarboxylation of Acetic Acid (A) with NaOH/CaO: This is a soda lime decarboxylation reaction. Carboxylic acids (or their sodium salts, which would be formed in situ with NaOH) undergo decarboxylation when heated with soda lime (a mixture of NaOH and CaO), losing a carbon atom as CO₂ (which forms Na₂CO₃).

   CH₃COOH + NaOH --(CaO, Δ)--> CH₄ + Na₂CO₃

   (More accurately, CH₃COONa + NaOH --(CaO, Δ)--> CH₄ + Na₂CO₃)

   So, compound B is Methane (CH₄).

Compound B is CH₄.