AP EAMCET · Chemistry · Hydrocarbons
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 \(\mathrm{C}_3 \mathrm{H}_8, \mathrm{H}_2\)
- C \(\mathrm{C}_4 \mathrm{H}_{10}, \mathrm{H}_2\)
- D \(\mathrm{H}_2, \mathrm{C}_4 \mathrm{H}_{10}\)
Answer & Solution
Correct Answer
(C) \(\mathrm{C}_4 \mathrm{H}_{10}, \mathrm{H}_2\)
Step-by-step Solution
Detailed explanation
Kolbe's electrolysis method : (iii) \(2 \mathrm{CH}_3-\stackrel{\bullet}{\mathrm{C}} \mathrm{H}_2 \rightarrow \mathrm{CH}_3-\mathrm{CH}_2-\mathrm{CH}_2-\mathrm{CH}_3\)
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