Question

Which of the following statement is INCORRECT with Kolbe's electrolytic process?

• A. Ethane can be prepared by this method.
• B. Presence of alkyl groups in \(\alpha\)-position decrease the yield of alkanes.
• C. The reaction proceeds via methyl free radical.
• D. At anode alkane and \(CO_2\) gas is formed.
• E. An alkane obtained at anode contains odd number of carbon atoms.

Hint:

In Kolbe electrolysis, two identical alkyl radicals combine, so the alkane formed usually has an even number of carbon atoms.

Answer 1

Answer: (E)

Step 1: Recall the basis of Kolbe electrolysis.
In Kolbe's electrolytic process, the sodium or potassium salt of a carboxylic acid is electrolysed. At the anode, the carboxylate ion loses an electron and forms a radical after decarboxylation.

Step 2: Write the anode reaction in general form.
At the anode: \[ RCOO^- \rightarrow RCOO^\cdot + e^- \] Then: \[ RCOO^\cdot \rightarrow R^\cdot + CO_2 \] Thus, an alkyl free radical is formed along with carbon dioxide gas.

Step 3: Show how alkane is formed.
Two alkyl radicals combine: \[ R^\cdot + R^\cdot \rightarrow R-R \] So the alkane produced has double the number of carbon atoms present in the alkyl radical \(R\).

Step 4: Check option (1).
If sodium acetate is used, the methyl radical is formed: \[ CH_3COO^- \rightarrow CH_3^\cdot + CO_2 \] Then: \[ CH_3^\cdot + CH_3^\cdot \rightarrow C_2H_6 \] Hence ethane can indeed be prepared by this method, so option \((1)\) is correct.

Step 5: Check options (2), (3), and (4).
The process does proceed through free radicals such as methyl radical in the case of acetate, so option \((3)\) is correct. At the anode, \(CO_2\) is evolved and alkane is formed via radical combination, so option \((4)\) is also correct in the overall description. Also, branching or alkyl substitution at the \(\alpha\)-position can reduce the yield of the desired alkane, so option \((2)\) is acceptable.

Step 6: Check option (5) carefully.
Since the product alkane is formed by coupling two identical radicals, the number of carbon atoms in the alkane is always even: \[ 2n \] Therefore, saying that the alkane obtained at anode contains an odd number of carbon atoms is incorrect.

Step 7: State the final answer.
Hence, the incorrect statement is:
\[ \boxed{\text{An alkane obtained at anode contains odd number of carbon atoms}} \] which matches option \((5)\).