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$\ce{CaO}$ being a basic oxide should not react with $\ce{NaOH}$. Then how come decarboxylation takes place using soda lime? I am confused.

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closed as unclear what you're asking by Mithoron, andselisk, Jan, Todd Minehardt, Nilay Ghosh Dec 26 '17 at 17:49

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Sodalime when used in decarboxylation, the reaction does not take place between NaOH and CaO. In Oakwood reaction, Carboxyl group is the one which gets reduced by NaOH being base.

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The decarboxylation using soda lime does not involve any reaction between $\ce{NaOH}$ and $\ce{CaO}$.

It is in fact the $\ce{NaOH}$ which reacts with the carboxylic acid to form the alkane, alongwith the formation of $\ce{Na_2CO_3}$.

$\ce{CaO}$ is simply used to make the reagent easier to handle. $\ce{NaOH}$ is highly hygroscopic and easily forms a concentrated sodium hydroxide solution when exposed to air. Sodalime does not absorb moisture that easily.

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