When sodium benzoate labeled with carbon-14 at the carboxylate carbon () undergoes decarboxylation u — Hydrocarbons Chemistry Question

💡 Solution & Explanation

In the decarboxylation reaction of sodium benzoate using soda-lime ($NaOH + CaO$ and heat), the entire carboxylate group ($-COONa$) is removed from the benzene ring and combines with $NaO^-$ to form sodium carbonate ($Na_2CO_3$). Thus, the $^{14}C$ label leaves the aromatic ring and ends up exclusively in the sodium carbonate byproduct, leaving the benzene ring unlabeled.