The step-down decarboxylation of sodium propanoate () using soda-lime produces which hydrocarbon? — Hydrocarbons Chemistry Question

💡 Solution & Explanation

Decarboxylation with soda-lime ($NaOH + CaO$) removes the entire $-COONa$ group as $Na_2CO_3$, yielding an alkane with one less carbon than the parent carboxylate. Sodium propanoate yields ethane.