The step-down decarboxylation of sodium propanoate () using soda-lime produces which hydrocarbon? — Hydrocarbons Chemistry Question
Question
The step-down decarboxylation of sodium propanoate ($CH_3CH_2COONa$) using soda-lime produces which hydrocarbon?
Answer: B
💡 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.
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