Question

An organic compound ‘A’ produces a gas with brisk effervescence when sodium bicarbonate is added to it. The compound ‘A’ contains the same number of carbon as in ethane. The compound ‘A’ reacts with ethanol to form a sweet smelling compound ‘C’ in presence of concentrated ${ H }_{ 2 }{ SO }_{ 4 }$ acid. Then the name of the compound and the gas evolved is:
A. Ethanoic acid; ${ CO }_{ 2 }$
B. Methanoic acid; ${ CO }_{ 2 }$
C. Ethanoic acid; ${ H }_{ 2 }$
D. Methanoic acid; ${ H }_{ 2 }$

Answer 1

Hint: Sodium bicarbonate reacts with an acid to produce effervescence. There are 2 carbon atoms in ethane. Ethanol reacts with a carboxylic acid to give a sweet-smelling compound in the presence of concentrated ${ H }_{ 2 }{ SO }_{ 4 }$.

Complete answer:
Let us understand in a step by step process and analyse all the hints given in the question to get the compound A and the gas evolved.
It is given in the question that an organic compound ‘A’ reacts with sodium bicarbonate to produce a gas with brisk effervescence. Sodium bicarbonate reacts with an acid to produce carbon dioxide with brisk effervescence. So, the gas evolved is carbon dioxide (${ CO }_{ 2 }$). Since compound ‘A’ is an organic compound, it must be a carboxylic acid.
It is also given that compound ‘A’ has the same number of carbon as in ethane. Ethane contains 2 carbon atoms. So, compound ‘A’ will also have 2 carbon atoms. A carboxylic acid with 2 carbon atoms is ethanoic acid.
The compound ‘A’ reacts with ethanol to give a sweet-smelling compound ‘C’ in the presence of concentrated ${ H }_{ 2 }{ SO }_{ 4 }$ acid. This reaction is called the esterification reaction and ethanoic acid reacts with ethanol in the presence of concentrated ${ H }_{ 2 }{ SO }_{ 4 }$ acid to give ethyl ethanoate. So, compound ‘C’ is ethyl ethanoate.
The reactions of the entire process are given below.
${ CH }_{ 3 }COOH+Na{ HCO }_{ 3 }\longrightarrow { CH }_{ 3 }COONa+{ H }_{ 2 }O+{ CO }_{ 2 }\uparrow$
${ CH }_{ 3 }COOH+{ { CH }_{ 3 }CH }_{ 2 }-OH\overset { Conc.\quad { { H }_{ 2 }SO }_{ 4 } }{ \longrightarrow } { CH }_{ 3 }COO-O-{ { CH }_{ 2 }CH }_{ 3 }+{ H }_{ 2 }O$
So, the compound ‘A’ is ethanoic acid and the gas evolved is ${ CO }_{ 2 }$.

So, option A is the correct one.

Note: Always remember that sodium bicarbonate reacts with acids to produce carbon dioxide, not hydrogen gas. Don’t confuse between them.