Question

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

Answer 1

Hint: We know that alkaline $KMn{O_4}$ is a very good oxidising agent which is used in organic chemistry. ${H_2}S{O_4}$ is a dehydrating agent. It removes water molecules. It is given in the question that $Z$ 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 $X$ is obtained by the reaction of alkaline $KMn{O_4}$, with another compound $Y$ followed by acidification. This means $X$will be having an acidic group. We know that when we will react with ethanol with alkaline $KMn{O_4}$, we will obtain an organic compound which will have an acidic functional group, that is ethanoic acid. The chemical equation can be written as $C{H_3} - C{H_2} - OH\xrightarrow[{acidification}]{{alkKMn{O_4}}}C{H_3} - COOH$.
From this equation we have got $X$ is ethanoic acid and $Y$is ethanol. Now according to the question we will have to react with ethanol and ethanoic acid in the presence of ${H_2}S{O_4}$ to produce $Z$. $Z$ is a sweet smelling compound. The reaction between ethanol and ethanoic acid can be written as $C{H_3} - COOH + C{H_3} - C{H_2} - OH\xrightarrow{{{H_2}S{O_4}}}C{H_3} - COOC{H_2} - C{H_3}$. From this reaction we have got ethyl ethanoate as the product which is our compound $Z$. It is sweet smelling in nature.
From the above explanation it is clear to us that $X$ is ethanoic acid, $Y$ is ethanol and $Z$ is ethyl ethanoate.

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

Note: Always remember that when ethanol reacts with alkaline $KMn{O_4}$ ethanoic acid is formed. Ethanol and ethanoic acid react in the presence of ${H_2}S{O_4}$ to form ethyl ethanoate. Remember the reactions $C{H_3} - C{H_2} - OH\xrightarrow[{acidification}]{{alkKMn{O_4}}}C{H_3} - COOH$ and $C{H_3} - COOH + C{H_3} - C{H_2} - OH\xrightarrow{{{H_2}S{O_4}}}C{H_3} - COOC{H_2} - C{H_3}$.