Question

The laboratory preparation of methane from sodium acetate is ________
A.$C{H_3}COONa + NaOH\xrightarrow{{CaO}}C{H_3}{O_2} + N{a_2}C{O_3}$
B.$C{H_3}COONa + NaOH\xrightarrow{{CaO}}C{H_3}OH + N{a_2}C{O_3}$
C.$C{H_3}COONa + NaOH\xrightarrow{{CaO}}C{H_4} + N{a_2}C{O_3}$
D.None of these

Answer 1

Hint: We have to know the laboratory preparation of alkanes from salts of carboxylic acids is decarboxylation reaction. In the process of decarboxylation, carbon dioxide is removed from the molecules containing $ - COOH$ group. We can get alkane when saturated monocarboxylic acid salt of sodium/potassium undergoes dry distillation with soda lime.

Complete step by step answer:We can prepare alkane in the laboratory using sodium/potassium salts of carboxylic acids. The method of preparing alkane using sodium/potassium salts of carboxylic acid with soda lime is called decarboxylation. In this process, a molecule of carbon dioxide is removed from the compound containing $ - COOH$.
The formed alkane will have one carbon atom less than the parent compound. For lower members, the yield is good whereas for higher membranes, the yield is poor.
We can prepare soda lime by soaking it with quick lime in caustic soda and the products are dried. Quick lime is calcium oxide and caustic soda is sodium hydroxide.
We can write the chemical formula of $NaOH + CaO$.
The reactivity of soda lime is milder than sodium hydroxide (caustic soda). A violent reaction may occur if caustic soda is also used. We also use quick lime along with NaOH to keep sodium hydroxide dry for fusion.
We can write general reaction of decarboxylation as,
$RCOONa + NaOH\xrightarrow{{CaO}}R - H + N{a_2}C{O_3}$
Here, R is the alkyl group.
In the decarboxylation of carboxylic acids, sodium carbonate is also formed as one of the products.
We can prepare methane from sodium acetate using soda lime. We can write the chemical reaction is,
$C{H_3}COONa + NaOH\xrightarrow{{CaO}}C{H_4} + N{a_2}C{O_3}$

$\therefore $Option C is correct.


Note:
We also prepare alkanes from carboxylic acids using Kolbe’s electrolytic methods. In this method, we have to take concentrated solution of sodium salt of carboxylic acid (or) potassium salt of carboxylic acid undergoes electrolysis to give alkane. The formed alkane contains an even number of carbon atoms.
The general reaction of Kolbe’s electrolysis is written as,
$2RCOONa + 2{H_2}O\xrightarrow{{}}R - R + 2NaOH + {H_2} + C{O_2}$
For example, when we electrolyze a concentrated aqueous solution of sodium acetate, we can observe the liberation of ethane gas at the anode.
$2C{H_3}COONa + 2{H_2}O\xrightarrow{{}}C{H_3} - C{H_3} + 2NaOH + {H_2} + C{O_2}$