Consider the reaction: $\mathrm{CH}{}_3\mathrm{CH}{}_2\mathrm{COONa}\underset{\mathrm{\Delta }}{\overset{\mathrm{NaOH}+?}{\to }}\mathrm{CH}{}_3\mathrm{CH}{}_3+\mathrm{Na}{}_2\mathrm{CO}{}_3$ Identify the missing reagent.

Solution & Explanation

\textbf{Answer:} (A) $\mathrm{CaO}$. \textbf{Solution:} Heating the sodium salt of a carboxylic acid with soda-lime (a mixture of $\mathrm{NaOH}$ and $\mathrm{CaO}$) brings about decarboxylation, replacing the $-\mathrm{COONa}$ group with $-\mathrm{H}$. Sodium propanoate thus loses $\mathrm{CO}{}_2$ (captured as $\mathrm{Na}{}_2\mathrm{CO}{}_3$) to give ethane. The missing component of soda-lime is $\mathrm{CaO}$.