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Question: A compound \(X\) is obtained by the reaction of alkaline \(KMn{O_4}\), with another compound \(Y\) f...

A compound XX is obtained by the reaction of alkaline KMnO4KMn{O_4}, with another compound YY followed by acidification . Compound XX also reacts with compound YY in presence of a few drops of H2SO4{H_2}S{O_4} to form a sweet smelling compound ZZ. The compound X,YX,Y and ZZ are respectively.
A. Ethanol, Ethene, Ethanoic acid
B. Ethanoic acid, Ethanol, Ethyl ethanoate
C. Ethanoic acid, Ethanal, Ethene
C. Ethanol, Ethanoic acid, Sodium Ethanoate

Explanation

Solution

We know that alkaline KMnO4KMn{O_4} is a very good oxidising agent which is used in organic chemistry. H2SO4{H_2}S{O_4} is a dehydrating agent. It removes water molecules. It is given in the question that ZZ is a sweet-smelling compound, that means it is an organic compound whose functional group is ether.

Complete step by step solution:
In the question we are given that compound XX is obtained by the reaction of alkaline KMnO4KMn{O_4}, with another compound YY followed by acidification. This means XXwill be having an acidic group. We know that when we will react with ethanol with alkaline KMnO4KMn{O_4}, we will obtain an organic compound which will have an acidic functional group, that is ethanoic acid. The chemical equation can be written as CH3−CH2−OH→acidificationalkKMnO4CH3−COOHC{H_3} - C{H_2} - OH\xrightarrow[{acidification}]{{alkKMn{O_4}}}C{H_3} - COOH.
From this equation we have got XX is ethanoic acid and YYis ethanol. Now according to the question we will have to react with ethanol and ethanoic acid in the presence of H2SO4{H_2}S{O_4} to produce ZZ. ZZ is a sweet smelling compound. The reaction between ethanol and ethanoic acid can be written as CH3−COOH+CH3−CH2−OH→H2SO4CH3−COOCH2−CH3C{H_3} - COOH + C{H_3} - C{H_2} - OH\xrightarrow{{{H_2}S{O_4}}}C{H_3} - COOC{H_2} - C{H_3}. From this reaction we have got ethyl ethanoate as the product which is our compound ZZ. It is sweet smelling in nature.
From the above explanation it is clear to us that XX is ethanoic acid, YY is ethanol and ZZ is ethyl ethanoate.

So. the correct answer of the given question is option: B.

Note: Always remember that when ethanol reacts with alkaline KMnO4KMn{O_4} ethanoic acid is formed. Ethanol and ethanoic acid react in the presence of H2SO4{H_2}S{O_4} to form ethyl ethanoate. Remember the reactions CH3−CH2−OH→acidificationalkKMnO4CH3−COOHC{H_3} - C{H_2} - OH\xrightarrow[{acidification}]{{alkKMn{O_4}}}C{H_3} - COOH and CH3−COOH+CH3−CH2−OH→H2SO4CH3−COOCH2−CH3C{H_3} - COOH + C{H_3} - C{H_2} - OH\xrightarrow{{{H_2}S{O_4}}}C{H_3} - COOC{H_2} - C{H_3}.