Pure methane can be produced by -

The correct option is C

Soda-lime decarboxylation

The correct option is (C):

Explanation for the correct answer:

Soda-lime decarboxylation:

1. Basically in this chemical decarboxylation reaction, the $-\mathrm{COOH}$ or the $-\mathrm{COONa}$ group is just removed and then further is replaced with a Hydrogen atom.
2. As the soda lime used is manufactured by the addition of Sodium hydroxide solution to the solid Calcium oxide which is also known as quicklime.
3. So when the $\mathrm{NaOH}\mathrm{and}\mathrm{the}\mathrm{CaO}$ gets to react with the a compound which consists of $-\mathrm{COOH}$ or the $-\mathrm{COONa}$ group in the presence of heat, there would be removal of the Hydrogen atom or the $\mathrm{Na}$ atom.
4. Reaction of soda-lime decarboxylation for the formation of methane is: $\underset{\mathrm{Sodium}\mathrm{acetate}}{{\mathrm{CH}}_3\mathrm{COONa}\left(\mathrm{s}\right)}+\underset{\mathrm{Sodium}\mathrm{hydroxide}}{\mathrm{NaOH}\left(\mathrm{aq}\right)}\underset{\mathrm{Cao}}{\overset{∆}{\to }}\underset{\mathrm{Methane}}{{\mathrm{CH}}_4\left(\mathrm{g}\right)}+\underset{\mathrm{Sodium}\mathrm{carbonate}}{{\mathrm{Na}}_2{\mathrm{CO}}_3\left(\mathrm{s}\right)}$

Explanation for the incorrect options:

Option (A):

Wurtz reaction:

1. It is also a type of chemical reaction which is actually a coupling reaction.
2. In this reaction the Sodium metal would be there which when reacted with the two alkyl halides and in the presence of dry ether solution.
3. Then there would be the formation of higher alkane and that is along with the compound which consists of Sodium and the halogen.
4. This reaction is not responsible for the formation of Methane (${\mathrm{CH}}_4$) because, it contains only one Carbon.
5. But in the Wurtz reaction, there would be presence of two alkyl halides for the preparation of symmetrical alkane.
6. Example of Wurtz reaction:

Option (B):

Kolbe’s electrolytic method:

1. This is also a chemical method in which there is the usage of Sodium salt of the fatty acid.
2. It is responsible for the formation of alkane as a product.
3. Basically in this reaction, there would be the decarboxylation of the Sodium salt of the fatty acid.
4. As this method is responsible for the formation of the symmetrical alkanes, but Methane has only one Carbon so it will not follow Kolbe electrolytic method.
5. Example of Kolbe's electrolytic reaction:

Option (D):

Reduction with ${\mathrm{H}}_2$:

1. It is also a chemical reaction in which there would be a gain of Hydrogen atom and the loss of the Oxygen atom.
2. There would be the reduction of alkene to alkane using Hydrogen in the presence of the proper catalysts in the heat.
3. But there is the presence of only one Carbon in methane and alkene is formed by using 2 Carbon (minimum) or more.
4. Example of Reduction with ${\mathrm{H}}_2$:

Therefore, the correct answer is: Option C: Soda-lime decarboxylation.