Discussion
Which of the following sodium salt of carboxylic acid is used for the preparation of n-hexane by Kolbe's electrolytic method?
Answer: E
The salt is \(CH_{3}CH_{2}CH_{2}COONa\) (sodium butanoate).
In Kolbe's electrolysis, two alkyl radicals from the carboxylate (after loss of \(CO_{2}\)) join, so the alkane has twice the carbon chain of the alkyl group. n-Hexane (\(C_{6}H_{14}\)) needs two propyl groups, from butanoate:
\[ \begin{aligned} &2CH_{3}CH_{2}CH_{2}COO^{-} \\ &\rightarrow C_{6}H_{14} + 2CO_{2} + 2e^{-} \end{aligned} \]
Sodium acetate gives ethane and sodium propanoate gives butane. Sodium formate gives hydrogen, not an alkane. The five-carbon salt in option D (as printed) would give octane.
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