Question

Name the organic compound prepared in the following reaction:

${\mathrm{CH}}_3\mathrm{COONa}+\hspace{0.17em}\mathrm{NaOH}\stackrel{\mathrm{CaO}}{\to }$

Answer 1

Decarboxylation:

Decarboxylation is the elimination of Carbon dioxide ($C{O}_2$) from a Carboxylic acid ($C_n{H}_{2n+1}COOH$).

When Sodium salts of Acetic acid ($C{H}_{3}COOH$) are heated with soda lime ($NaOH+CaO$) alkanes are formed.

$\underset{\underset{\mathrm{acetic}\mathrm{acid}}{\mathrm{Sodium}\mathrm{salts}\mathrm{of}}}{{\mathrm{CH}}_3\mathrm{COONa}}+\hspace{0.17em}\underset{\underset{\mathrm{hydroxide}}{\mathrm{Sodium}}}{\mathrm{NaOH}}\stackrel{\mathrm{CaO}}{\to }\underset{\mathrm{Methane}}{{\mathrm{CH}}_4}+\underset{\mathrm{Sodium}\mathrm{carbonate}}{{\mathrm{Na}}_2{\mathrm{CO}}_3}$

In the above reaction Methane ($C{H}_4$) is going to be formed as a major product.