Question

# The Kolbe’s electrolysis of potassium succinate gives $C{O_2}$ and _____?A) etheneB) ethaneC) methaneD) methanol

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Hint:
The Kolbe’s electrolysis reaction is basically a decarboxylative dimerization of the two carboxylic acids present in the structure potassium succinate. The decarboxylation will produce a radical intermediate and they get combined to form the product.

${\left( {C{H_2}COOK} \right)_2} + {H_2}O\xrightarrow{{Electrolysis}}C{H_2} = C{H_2} + 2C{O_2} + {H_2} + 2KOH$
3) The $C{O_2}$ molecules produced in the above reaction come as the decarboxylative product from the carboxylic groups of the potassium succinate.
Whenever a dicarboxylic acid is present in a structure then it will always give alkene as a product in Kolbe’s electrolysis reaction. The number of carboxylic acids present in the structure will be the same as the number of $C{O_2}$ molecules liberated in the reaction.