In Kolbe's electrolysis of sodium propanoate, products formed at anode and cathode are respectively

1. $\mathrm{C}_2 \mathrm{H}_6, \mathrm{H}_2$
2. $\mathrm{C}_3 \mathrm{H}_8, \mathrm{H}_2$
3. $\mathrm{C}_4 \mathrm{H}_{10}, \mathrm{H}_2$
4. $\mathrm{H}_2, \mathrm{C}_4 \mathrm{H}_{10}$

Solution

Kolbe's electrolysis method :



(iii) $2 \mathrm{CH}_3-\stackrel{\bullet}{\mathrm{C}} \mathrm{H}_2 \rightarrow \mathrm{CH}_3-\mathrm{CH}_2-\mathrm{CH}_2-\mathrm{CH}_3$

Asked in: AP EAMCET 2024 (22 May Shift 1)

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