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

• A

  $\mathrm{C}_2 \mathrm{H}_6, \mathrm{H}_2$
• B

  $\mathrm{C}_3 \mathrm{H}_8, \mathrm{H}_2$
• Correct answer
  C

  $\mathrm{C}_4 \mathrm{H}_{10}, \mathrm{H}_2$
• D

  $\mathrm{H}_2, \mathrm{C}_4 \mathrm{H}_{10}$

Solution

Kolbe's electrolysis method :



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

Hence, the correct answer is option: 'C'.