Question

The alkane that cannot be prepared in pure state by Kolbe’s electrolysis is

• A. ethane
• B. 2,3-dimethyl butane
• C. propane
• D. n-butane
• E. n-hexane

Hint:

Kolbe's electrolysis is like the Wurtz reaction; both are excellent for symmetric alkanes but poor for unsymmetric ones due to the formation of multiple side products.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
Kolbe’s electrolysis involves the electrolysis of an aqueous solution of sodium or potassium salts of fatty acids to produce alkanes at the anode.
Step 2: Key Formula or Approach:
The reaction involves the coupling of two alkyl radicals: \( R\text{-COO}^- \rightarrow R^\cdot + \text{CO}_2 \); then \( R^\cdot + R^\cdot \rightarrow R\text{-}R \).
This method produces symmetrical alkanes with an even number of carbon atoms.
Step 3: Detailed Explanation:
1. Ethane (C\textsubscript{2}H\textsubscript{6}): From sodium acetate (\( CH_3^\cdot + CH_3^\cdot \)). Symmetrical.
2. n-Butane (C\textsubscript{4}H\textsubscript{10}): From sodium propionate (\( C_2H_5^\cdot + C_2H_5^\cdot \)). Symmetrical.
3. Propane (C\textsubscript{3}H\textsubscript{8}): This is an asymmetrical alkane with an odd number of carbons. To prepare it, we would need a mixture of sodium acetate and sodium propionate. This would produce a mixture of ethane, propane, and butane, making it impossible to obtain pure propane.
Step 4: Final Answer:
Propane cannot be prepared in a pure state by this method.