Question

Which of the following can be prepared by Kolbe’s electrolysis process?
(i) Methane
(ii) Ethyne
(iii) Ethane
(iv) None of these

Answer 1

Hint :Kolbe’s electrolysis is an electrolysis process of a sodium or potassium salt of a carboxylic acid, producing an alkane dimer product. The product will have more than two carbon atoms, and only single bonds at all times as it is an alkane dimer.

Complete Step By Step Answer:
In Kolbe’s electrolysis, a sodium or potassium salt of a carboxylic acid in aqueous medium is electrolysed to give symmetrical hydrocarbons. During the process, the oxidative decarboxylation of the carboxylic acid salt occurs, which gives radicals and these radicals give symmetrical dimers.
The reaction is as follows:
 $ 2RCO{O^ - }N{a^ + }\, + \,2{H_2}O\xrightarrow{{electrolysis}}R - R\, + \,2C{O_2}\, + \,{H_2}\, + \,2NaOH \\
{\text{where}}\,R = \,C{H_3} - \,,{C_2}{H_5} - ,\,{C_3}{H_7} - ,{\text{et}}\,{\text{cetera}} \\ $
Here, two molecules of the carboxylic acid salt in aqueous medium undergo electrolysis to give symmetrical alkane as a product.
Since the product obtained is always an alkane and a dimer, we can say that the method is useful for preparing ethane and higher members. We must remember that the alkane product will always have more than two carbon at atoms. Hence, we can rule out (i) Methane $ ({\text{C}}{{\text{H}}_4}) $ as an answer. (ii) Ethyne $ ({C_2}{H_4}) $ is an alkyne and has a triple bond between two $ CH $ groups, meaning it cannot be the answer as well. Therefore, the required answer is (iii) Ethane $ \left( {C{H_3} - C{H_3}} \right) $ .

Note :
Sodium or potassium salt of carboxylic acid is used instead of using the acid as such in Kolbe’s electrolysis as the acid alone is a weak electrolyte and hence has difficulty in dissociating into its respective ions. If a mixture of two carboxylic acids is used in this process, the product we obtain is unsymmetrical dimers.