Question

On electrolysis of aqueous solution of sodium butanoate gives a hydrocarbon. The number of carbon atoms present in the hydrocarbon are:

• A. \(6\)
• B. \(4\)
• C. \(8\)
• D. \(3\)

Hint:

In Kolbe’s electrolysis, two alkyl radicals combine after loss of \(CO_2\). Therefore, the hydrocarbon formed usually has double the number of carbon atoms present in the alkyl radical.

Answer 1

The Correct Option is A

Step 1: Recall Kolbe's electrolysis and the decarboxylation mechanism.
Kolbe's electrolysis involves electrolysis of an aqueous solution of a carboxylate salt. At the anode, carboxylate ions are oxidized: \[ RCOO^- \rightarrow RCOO^\bullet + e^- \] followed by: \[ RCOO^\bullet \rightarrow R^\bullet + CO_2 \]
Step 2: Identify the substrate - sodium butanoate.
Sodium butanoate is CH₃CH₂CH₂COONa. The carboxylate ion is butanoate (CH₃CH₂CH₂COO^-), which has 4 carbons. After decarboxylation: $R^\bullet$ = propyl radical (CH₃CH₂CH₂$^\bullet$), which has 3 carbons.
Step 3: Write the coupling step.
Two propyl radicals combine at the anode: \[ 2CH_3CH_2CH_2^\bullet \rightarrow CH_3CH_2CH_2CH_2CH_2CH_3 \] Product: hexane (C₆H₁₄), a 6-carbon alkane.
Step 4: Count the carbons in the product.
Each butanoate gives a C₃ radical after losing CO₂. Two C₃ radicals couple to give C₆. Number of carbons = 6.
Step 5: General rule for Kolbe's electrolysis.
For RCOO^- with $n$ carbons (including the carboxyl carbon): product = $(n-1) + (n-1) = 2(n-1)$ carbons. For butanoate ($n = 4$): $2(4-1) = 2 \times 3 = 6$ carbons.
Step 6: State the answer.
Electrolysis of sodium butanoate by Kolbe's method gives hexane (C₆H₁₄), which has 6 carbon atoms.
\[ \boxed{6 \text{ carbons (hexane) - option 1}} \]