What is Z in the following reaction?CH_3-CH_2 | vedclass

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(C) The reaction given is a decarboxylation reaction,also known as the soda-lime decarboxylation.When a sodium salt of a carboxylic acid is heated wit

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ethane

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(C) The reaction given is a decarboxylation reaction,also known as the soda-lime decarboxylation.When a sodium salt of a carboxylic acid is heated with soda lime $(NaOH + CaO)$,it loses a molecule of $CO_2$ to form an alkane with one carbon atom less than the parent carboxylic acid.$CH_3-CH_2-CO_2^{\ominus} Na^{\oplus} + NaOH \xrightarrow{CaO, \Delta} CH_3-CH_3 + Na_2CO_3$Here,sodium propanoate ($3$ carbons) reacts to form ethane ($2$ carbons).Therefore,$Z$ is ethane.

What is $Z$ in the following reaction?
$CH_3-CH_2-CO_2^{\ominus} Na^{\oplus} \stackrel{NaOH / CaO}{\longrightarrow} Z$

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What is $Z$ in the following reaction?$CH_3-CH_2-CO_2^{\ominus} Na^{\oplus} \stackrel{NaOH / CaO}{\longrightarrow} Z$