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 ${\mathrm{H}}_2{\mathrm{SO}}_4$ acid. Then the name of the compound and the gas evolved is:

Sodium bicarbonate and an organic component "A" combine to form a gas with brisk effervescence. When sodium bicarbonate and an acid interact, carbon dioxide is produced with brisk effervescence. Carbon dioxide has been the gas that evolved.

Thus,  Compound A must be a carboxylic acid as it is an organic compound. The amount of carbon in compound "A" is equal to that of ethane. Ethanoic acid is a carboxylic acid that contains two carbon atoms.

In the presence of concentrated ${\mathrm{H}}_2{\mathrm{SO}}_4$ acid, chemical "A" combines with ethanol to produce the sweet-smelling compound "C." The esterification process occurs when ethanoic acid and ethanol interact with concentrated ${\mathrm{H}}_2{\mathrm{SO}}_4$ acid to produce ethyl ethanoate. Ethyl ethanoate is therefore compound "C." The overall reaction can be written as:

${\mathrm{CH}}_3\mathrm{COOH}+{\mathrm{NaHCO}}_3⟶{\mathrm{CH}}_3\mathrm{COONa}+{\mathrm{H}}_2\mathrm{O}+{\mathrm{CO}}_2\uparrow$

$\begin{array}{rcl} & & {\mathrm{CH}}_3\mathrm{COOH}+{\mathrm{CH}}_3{\mathrm{CH}}_2-\mathrm{OH}\stackrel{\mathrm{Conc}.{\mathrm{H}}_2{\mathrm{SO}}_4}{\to }\\ & & {\mathrm{CH}}_3\mathrm{COO}-\mathrm{O}-{\mathrm{CH}}_2{\mathrm{CH}}_3+{\mathrm{H}}_2\mathrm{O}\end{array}$

Hence, the name of the compound is ethanoic acid and gas evolved is carbon dioxide (${\mathrm{CO}}_2$).

Thus, the correct option is (A).