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 carbons as in ethane. The compound 'A' reacts with ethanol to form a sweet-smelling compound 'C' in presence of conc. $H_{2} S O_{4}$ acid. Then the name of compound A and the gas evolved is :

ethanoic acid;

C O_{2}

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methanoic acid;

C O_{2}

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ethanoic acid;

H_{2}

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methanoic acid;

H_{2}

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Solution

The correct option is A ethanoic acid; $C O_{2}$
A contains the same number of carbon as ethane. So the number of carbon atoms is 2.

A produces brisk effervescence when sodium bicarbonate is added to it. Therefore carbon dioxide is evolved. Hence it's an organic acid with two carbons, ethanoic acid- $C H_{3} C O O H$ .

$C H_{3} C O O H + N a H C O_{3} \to C H_{3} C O O N a + C O_{2} + H_{2} O$

We know that esters are sweet smelling substances that are used in perfumes. These are formed when a carboxylic acid reacts with an alcohol in presence of conc. $H_{2} S O_{4}$ .

Since, new compound A (i.e., ethanoic acid) reacts with ethanol (an alcohol) in presence of conc. $H_{2} S O_{4}$ to form a sweet smelling substance which is used in perfumes, therefore C must be an ester, i.e., ethyl ethanoate.

$C H_{3} C O O H + C_{2} H_{5} O H C o n c . H_{2} S O_{4} - ------- \to C H_{3} C O O C H_{2} C H_{3} + H_{2} O$

Option A is correct.

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