Which alkane is produced when sodium salt of butanoic acid is heated with soda lime?

C H_{3} C H_{3}

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C H_{3} C H_{2} C H_{2} C H_{3}

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C H_{4}

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C H_{3} C H_{2} C H_{3}

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Solution

The correct option is D $C H_{3} C H_{2} C H_{3}$
When sodium salt of butanoic acid is heated with soda lime, it readily undergoes decarboxylation. Decarboxylation is removal of $C O_{2}$ from a molecule. Removal of $C O_{2}$ from $C H_{3} C H_{2} C H_{2} C O O^{-}$ produces propane.
${C H_{3} C H_{2} C H_{2} C O O}^{-} {N a}^{+} + N a O H C a O - - \to {N a}_{2} C O_{3} + C H_{3} C H_{2} C H_{3} P r o p a n e$

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