Write the method of preparation for hydrocarbons by the decarboxylation of carboxylic acids.

(N/A) Decarboxylation by sodalime: Carboxylic acids lose carbon dioxide to form hydrocarbons when their sodium salts are heated with sodalime ($NaOH$ and $CaO$ in the ratio of $3:1$). The reaction is known as decarboxylation.
$\underset{\text{Sodium carboxylate}}{RCOONa}$ $\xrightarrow[\text{Sodalime, Heat}]{NaOH \text{ and } CaO} \underset{\text{Hydrocarbon}}{RH} + Na_{2}CO_{3}$
In this decarboxylation reaction,due to the removal of $CO_{2}$,the number of carbon atoms decreases by one compared to the original acid salt.
$(b)$ Decarboxylation by Kolbe electrolysis: Alkali metal salts of carboxylic acids also undergo decarboxylation on electrolysis of their aqueous solutions and form hydrocarbons having twice the number of carbon atoms present in the alkyl group of the acid. The reaction is known as Kolbe electrolysis.
$2RCOO^{-}Na^{+} + 2H_{2}O \xrightarrow{\text{Electrolysis}} R-R + 2CO_{2} + 2NaOH + H_{2} \uparrow$
Examples of sodalime decarboxylation:
$(i)$ $CH_{3}COONa + NaOH \xrightarrow{\text{Sodalime}, \Delta} CH_{4} + Na_{2}CO_{3}$
$(ii)$ $\text{Sodium benzoate} \xrightarrow{\text{Sodalime}, \Delta, -Na_{2}CO_{3}} \text{Benzene}$
$(iii)$ $\text{Sodium phthalate} \xrightarrow{\text{Sodalime}, \Delta, -Na_{2}CO_{3}} \text{Benzene}$
$(iv)$ $\text{Sodium salicylate} \xrightarrow{\text{Sodalime}, \Delta, -Na_{2}CO_{3}} \text{Phenol}$