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The Kolbe’s electrolysis of potassium succinate gives $C{O_2}$ and _____?
A) ethene
B) ethane
C) methane
D) methanol

Answer
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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.

Complete answer:
1) Kolbe’s electrolysis reaction can be defined as a type of reaction which involves formation of symmetrical hydrocarbons by the coupling of the radicals generated from the carboxylic acid at an anode through electrolysis process.
2) The electrolysis reaction of the potassium succinate will be as follows:
${\left( {C{H_2}COOK} \right)_2} + {H_2}O\xrightarrow{{Electrolysis}}C{H_2} = C{H_2} + 2C{O_2} + {H_2} + 2KOH$
3) The $C{O_2}$ molecules produced in the above reaction come as the decarboxylative product from the carboxylic groups of the potassium succinate.
4) The radical intermediates produced in the reaction during the decarboxylation combine to give ethene as a product.

Therefore, option A will be the correct answer which is ethene.

Additional information:
The intermediates formed in this reaction are free radicals and it has been studied that radicals were used for the initiation of polymerization reaction for the making of low molecular weight polymers and polystyrene. The radicals are formed at the anode through the electrolysis reaction. This electrolysis reaction was modified to happen under the condition using liquid ammonia. This reaction is mostly useful for the preparation of hydrocarbons.

Note:
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.