Question

Which alkane is produced when sodium salt of butanoic acid is heated with soda lime?
A) $C{H_3}C{H_3}$
B) $C{H_3}C{H_2}C{H_2}C{H_3}$
C) $C{H_4}$
D) $C{H_3}C{H_2}C{H_3}$

Answer 1

The reaction occurs between sodium salt of butanoic acid and soda lime. An alkane is formed the chemical formula of sodium salt of butanoic acid is $C{H_3}C{H_2}C{H_2}COONa$ and soda lime is the mixture of $NaOH$ (Sodium hydroxide) and $Ca{(OH)_2}$ (Calcium Hydroxide), soda lime causes removal of carbon dioxide $C{O_2}$. Thus, this is a decarboxylation reaction.

Complete answer:
The reaction occurs between sodium salt of butanoic acid and soda lime. Butanoic acid is given as $C{H_3}C{H_2}C{H_2}COOH$. In the sodium salt of butanoic acid the $-OH$ group is replaced by $N{a^ + }$. Thus sodium salt of butanoic acid is $C{H_3}C{H_2}C{H_2}COONa$.
Soda lime is a mixture of sodium hydroxide and calcium hydroxide $(NaOH + Ca{(OH)_2})$. It is used in granular form in closed breathing environments to remove $C{O_2}$ (Carbon dioxide) from breathing gases to present $C{O_2}$ retention and carbon dioxide poisoning.
When sodium salt of butanoic acid is heated with soda lime, it readily undergoes decarboxylation. Decarboxylation is the removal of $C{O_2}$ (Carbon dioxide) from a molecule. The removal of $C{O_2}$ from $C{H_3}C{H_2}C{H_2}COONa$ (Sodium salt of butanoic acid) produces propane $(C{H_3}C{H_2}C{H_3})$.
$C{H_3}C{H_2}C{H_2}CO\mathop O^{ - } N{a^ + } + NaOH\xrightarrow{CaO}N{a_2}C{O_3} + C{H_3}C{H_2}C{H_3}$

Thus option (D) is the correct answer.

Note: Decarboxylation is a useful process. It removes carbon dioxide $(C{O_2})$ from a molecule. Soda lime has much less tendency to absorb water the solid sodium salt of a carboxylic acid is mixed with soda lime and heated to obtain an alkane students must note that the alkane formed by this process will always have one carbon less than the reacting sodium salt of carboxylic acid.
An example can be taken of sodium propionate is heated with soda lime, decarboxylation occurs to form ethane. Ethane has two carbons while sodium propionate has three carbons.
$C{H_3}C{H_2}CO\mathop O^{ - } N{a^ + } + NaOH\xrightarrow{CaO}C{H_3}C{H_3}$