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 electrolytic method involves the electrolysis of an aqueous solution of sodium or potassium salts of carboxylic acids. The process results in the coupling of alkyl radicals at the anode to form alkanes.
Key Formula or Approach:
The general reaction at the anode is:
\[ 2RCOO^- \rightarrow R-R + 2CO_2 + 2e^- \]
This means the method is primarily designed to produce symmetric alkanes ($R-R$) with an even number of carbon atoms.

Step 2: Detailed Explanation:
1. Ethane: Formed by coupling two methyl radicals ($CH_3-CH_3$). It is symmetric and can be prepared pure.
2. 2,3-dimethyl butane: Formed by coupling two isopropyl radicals. It is symmetric and can be prepared pure.
3. Propane: Propane ($CH_3-CH_2-CH_3$) is an unsymmetric alkane with an odd number of carbons. To prepare it, one would need a mixture of sodium acetate and sodium propionate. This would produce a mixture of ethane ($CH_3-CH_3$), propane ($CH_3-CH_2-CH_3$), and butane ($CH_3CH_2-CH_2CH_3$). Because these alkanes have similar boiling points, obtaining pure propane is difficult.

Step 3: Final Answer:
Propane cannot be prepared in a pure state by Kolbe's electrolysis.