Question:medium
Which of the following statement is INCORRECT with Kolbe's electrolytic process?
Which of the following statement is INCORRECT with Kolbe's electrolytic process?
Show Hint
In Kolbe electrolysis, two identical alkyl radicals combine, so the alkane formed usually has an even number of carbon atoms.
Updated On: May 14, 2026
- Ethane can be prepared by this method.
- Presence of alkyl groups in \(\alpha\)-position decrease the yield of alkanes.
- The reaction proceeds via methyl free radical.
- At anode alkane and \(CO_2\) gas is formed.
- An alkane obtained at anode contains odd number of carbon atoms.
Show Solution
The Correct Option is
Solution and Explanation
Step 1: Understanding the Concept:
Kolbe's electrolytic process is a method for preparing alkanes (and other products) by the electrolysis of an aqueous solution of a sodium or potassium salt of a carboxylic acid. We need to evaluate each statement based on the mechanism and products of this reaction.
Step 2: The Mechanism of Kolbe's Electrolysis:
Let's consider the electrolysis of sodium acetate (CH\(_3\)COONa) as an example.
\[ 2\text{CH}_3\text{COONa} + 2\text{H}_2\text{O} \xrightarrow{\text{Electrolysis}} \text{CH}_3\text{-CH}_3 + 2\text{CO}_2 + \text{H}_2 + 2\text{NaOH} \] At the Anode (Oxidation):
1. Acetate ions are oxidized: \( 2\text{CH}_3\text{COO}^- \rightarrow 2\text{CH}_3\text{COO}^\cdot + 2e^- \)
2. The acetate free radical is unstable and loses CO\(_2\): \( 2\text{CH}_3\text{COO}^\cdot \rightarrow 2\text{CH}_3^\cdot + 2\text{CO}_2 \)
3. Two methyl free radicals combine (dimerize) to form ethane: \( \text{CH}_3^\cdot + \text{CH}_3^\cdot \rightarrow \text{CH}_3\text{-CH}_3 \)
At the Cathode (Reduction):
Water is reduced: \( 2\text{H}_2\text{O} + 2e^- \rightarrow \text{H}_2 + 2\text{OH}^- \)
Step 3: Evaluating the Statements:
(A) Ethane can be prepared: As shown in the example above, electrolyzing sodium acetate yields ethane at the anode. This statement is correct.
(B) Effect of \( \alpha \)-substitution: The reaction proceeds via free radicals. The stability of free radicals is 3\(^\circ\)>2\(^\circ\)>1\(^\circ\)>methyl. Highly substituted (branched) alkyl free radicals are more stable, but they are also more prone to side reactions like disproportionation, which leads to the formation of alkenes and alkanes with fewer carbon atoms, thus decreasing the yield of the main dimerized alkane. This statement is correct.
(C) Methyl free radical intermediate: The statement is about a "methyl free radical". This is true if the starting material is an acetate salt. More generally, the reaction proceeds via alkyl free radicals (R\(^\cdot\)). If we start with sodium propanoate, an ethyl radical is formed. The statement as written is specific to the preparation of ethane, but it points to the correct mechanism type. This statement is considered correct in the context of the reaction mechanism.
(D) Products at the anode: As seen in the mechanism, the alkane (R-R) and carbon dioxide (CO\(_2\)) are both formed at the anode from the oxidation of the carboxylate ion. This statement is correct.
(E) Number of carbon atoms in the alkane: The final alkane is formed by the dimerization of two alkyl free radicals (R\(^\cdot\) + R\(^\cdot\) \( \rightarrow \) R-R). If the alkyl group R has 'n' carbon atoms, the resulting alkane R-R will have '2n' carbon atoms, which is always an even number. Therefore, an alkane obtained at the anode in Kolbe's electrolysis always contains an even number of carbon atoms. The statement that it contains an odd number of carbon atoms is incorrect.
Step 4: Final Answer:
The incorrect statement is (E) because the alkane product from Kolbe's electrolysis is a dimer of the alkyl group from the carboxylic acid salt, and thus always has an even number of carbon atoms.
Kolbe's electrolytic process is a method for preparing alkanes (and other products) by the electrolysis of an aqueous solution of a sodium or potassium salt of a carboxylic acid. We need to evaluate each statement based on the mechanism and products of this reaction.
Step 2: The Mechanism of Kolbe's Electrolysis:
Let's consider the electrolysis of sodium acetate (CH\(_3\)COONa) as an example.
\[ 2\text{CH}_3\text{COONa} + 2\text{H}_2\text{O} \xrightarrow{\text{Electrolysis}} \text{CH}_3\text{-CH}_3 + 2\text{CO}_2 + \text{H}_2 + 2\text{NaOH} \] At the Anode (Oxidation):
1. Acetate ions are oxidized: \( 2\text{CH}_3\text{COO}^- \rightarrow 2\text{CH}_3\text{COO}^\cdot + 2e^- \)
2. The acetate free radical is unstable and loses CO\(_2\): \( 2\text{CH}_3\text{COO}^\cdot \rightarrow 2\text{CH}_3^\cdot + 2\text{CO}_2 \)
3. Two methyl free radicals combine (dimerize) to form ethane: \( \text{CH}_3^\cdot + \text{CH}_3^\cdot \rightarrow \text{CH}_3\text{-CH}_3 \)
At the Cathode (Reduction):
Water is reduced: \( 2\text{H}_2\text{O} + 2e^- \rightarrow \text{H}_2 + 2\text{OH}^- \)
Step 3: Evaluating the Statements:
(A) Ethane can be prepared: As shown in the example above, electrolyzing sodium acetate yields ethane at the anode. This statement is correct.
(B) Effect of \( \alpha \)-substitution: The reaction proceeds via free radicals. The stability of free radicals is 3\(^\circ\)>2\(^\circ\)>1\(^\circ\)>methyl. Highly substituted (branched) alkyl free radicals are more stable, but they are also more prone to side reactions like disproportionation, which leads to the formation of alkenes and alkanes with fewer carbon atoms, thus decreasing the yield of the main dimerized alkane. This statement is correct.
(C) Methyl free radical intermediate: The statement is about a "methyl free radical". This is true if the starting material is an acetate salt. More generally, the reaction proceeds via alkyl free radicals (R\(^\cdot\)). If we start with sodium propanoate, an ethyl radical is formed. The statement as written is specific to the preparation of ethane, but it points to the correct mechanism type. This statement is considered correct in the context of the reaction mechanism.
(D) Products at the anode: As seen in the mechanism, the alkane (R-R) and carbon dioxide (CO\(_2\)) are both formed at the anode from the oxidation of the carboxylate ion. This statement is correct.
(E) Number of carbon atoms in the alkane: The final alkane is formed by the dimerization of two alkyl free radicals (R\(^\cdot\) + R\(^\cdot\) \( \rightarrow \) R-R). If the alkyl group R has 'n' carbon atoms, the resulting alkane R-R will have '2n' carbon atoms, which is always an even number. Therefore, an alkane obtained at the anode in Kolbe's electrolysis always contains an even number of carbon atoms. The statement that it contains an odd number of carbon atoms is incorrect.
Step 4: Final Answer:
The incorrect statement is (E) because the alkane product from Kolbe's electrolysis is a dimer of the alkyl group from the carboxylic acid salt, and thus always has an even number of carbon atoms.
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