Question

in decarboxylation of sodium salt of propanoic acid,the alkane x is formed and its kolbes electrolysis of its aqueous solution another alkane y is liberated.x,y are​

Answer 1

Answer:

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Answer 2

Alkane x is ETHANE.

Alkane y is BUTANE.

Explanation:

Decarboxylation of $CH_{3}CH_{2}COO^{-}Na^{+}$:

• In the decarboxylation reaction, when a sodium salt of acid is heated it gives sodium carbonate and an alkane.
• Decarboxylation reaction is done in presence of soda-lime (NaOH/CaO).
• The alkane formed contains a carbon atom less than that from the reactant.
• So, sodium salt of propanoic acid on heating will give ethane and sodium carbonate.
• The reaction involved is:

         $CH_{3}CH_{2}COO^{-}Na^{+}\rightarrow CH_{3}-CH_{3}+Na_{2}CO_{3}$

• Alkane x is "ethane".

Kolbe's electrolysis of $CH_{3}CH_{2}COO^{-}Na^{+}$:

• In kolbe's electrolysis, a dimer of the alkane is formed and carbon dioxide and sodium metal is also formed.
• So, 1 molecule of sodium salt of propanoic acid will give 1 molecule of $C_{2}H_{5}$ will will combine with other molecule of ethyl and form butane, along with the liberation of carbon dioxide and Na.
• The reaction involved is:

        $CH_{3}CH_{2}COO^{-}Na^{+}\rightarrow C_{4}H_{10}+Na+CO_{2}$

• Alkane y is "butane".