Write reactions for the preparation of Methane from [1+1] i) CH3MgBr ii) CH3COONa

i) Preparation of Methane from $\mathrm{CH}{}_3\mathrm{MgBr}$

Methane can be prepared from methylmagnesium bromide ($\mathrm{CH}{}_3\mathrm{MgBr}$) by reacting it with water. The reaction proceeds as follows:

$$\mathrm{CH}{}_3\mathrm{MgBr}+\mathrm{H}{}_2\mathrm{O}\to \mathrm{CH}{}_4+\mathrm{Mg}(\mathrm{OH})\mathrm{Br}$$

Explanation:

• $\mathrm{CH}{}_3\mathrm{MgBr}$ is a Grignard reagent.
• When treated with water (a weak acid), the Grignard reagent undergoes hydrolysis, replacing the $\mathrm{MgBr}$ group with a hydrogen atom, forming methane ($\mathrm{CH}{}_4$).

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ii) Preparation of Methane from $\mathrm{CH}{}_3\mathrm{COONa}$

Methane can be prepared from sodium acetate ($\mathrm{CH}{}_3\mathrm{COONa}$) by heating it with soda lime (a mixture of $\mathrm{NaOH}$ and $\mathrm{CaO}$). The reaction is as follows:

$$\mathrm{CH}{}_3\mathrm{COONa}+\mathrm{NaOH}\stackrel{\mathrm{\Delta }}{\to }\mathrm{CH}{}_4+\mathrm{Na}{}_2\mathrm{CO}{}_3$$

Explanation:

• Sodium acetate decomposes upon heating with soda lime.
• The $\mathrm{CH}{}_3$ group is converted to methane ($\mathrm{CH}{}_4$), while the acetate ion ($\mathrm{CH}{}_3\mathrm{COO}{}^{-}$) decomposes to form sodium carbonate ($\mathrm{Na}{}_2\mathrm{CO}{}_3$).