Question

Explain the preparation of ethane from sodium propionate.

Answer 1

It is given that the reactant is sodium propionate. It is also known as sodium propanoate. Thus it is an ester compound. Its chemical formula is ${{C}}{{{H}}_3}{{C}}{{{H}}_2}{{COONa}}$. The chemical formula of ethane is ${{C}}{{{H}}_3} - {{C}}{{{H}}_3}$. In this reaction, $- {{COONa}}$ of sodium propanoate is removed and ${{{C}}_2}{{{H}}_6}$ is formed.

Complete step by step answer:
To get ethane, ${{{C}}_2}{{{H}}_6}$ from sodium propanoate, ${{C}}{{{H}}_3}{{C}}{{{H}}_2}{{COONa}}$, we should first write the chemical equation involved in the reaction.
We know that the sodium propanoate is a sodium salt of an acid, i.e. propanoic acid. While the product obtained is ethane which has two carbon atoms. So let’s assume that a decarboxylation reaction has undergone here.
Thus the reagent, for the decarboxylation to occur, is soda lime. The chemical formula of soda lime is ${{CaHNa}}{{{O}}_2}$. Soda lime is obtained when calcium oxide is mixed with sodium hydroxide. Soda lime is used in respiration. It eliminates carbon dioxide and prevents carbon dioxide poisoning. It is used as a drying agent. It absorbs carbon dioxide.
When sodium propanoate is heated with soda lime, it undergoes decarboxylation. The carboxyl group in sodium benzoate is converted to sodium carbonate. It also forms benzene. The chemical reaction is given below:
${{{C}}_2}{{{H}}_5}{{COONa}} + {{NaOH}}\xrightarrow[\Delta ]{{{{CaO}}}}{{{C}}_2}{{{H}}_6} + {{N}}{{{a}}_2}{{C}}{{{O}}_3}$
Thus the reagent used for converting sodium propanoate to ethane is soda lime.

Note:
During this reaction, a very high amount of temperature is applied so that the carbon dioxide is removed. The temperature is about ${300^ \circ }{{C}}$. Calcium oxide used in this reaction is referred to as a catalyst. Similar to this reaction, sodium benzoate can also be converted to benzene in the presence of soda lime.