When an organic compound X is warmed with alkaline solution of ${KMnO}_{4}$ a new compound Y is formed which has vinegar like smell. On heating compounds X and Y with few drops of concentrated sulphuric acid another compound Z with fruity smell is formed. Name the
(i) Compound X and Y
(ii) process of conversion of X and Y
(iii) compound Z
(iv) process of formation of Z

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Solution

Step 1
(i) Alkaline Potassium permanganate $({KMnO}_{4})$ solution gives an oxidized product.
Compound Y gives a vinegar-like smell so it must be Ethanoic acid $({CH}_{3} COOH)$ .
So, X must be an Ethanol $(C_{2} H_{5} OH)$ .
Step 2
(ii) The conversion of X (Ethanol) and Y (Ethanoic acid) in the presence of alkaline Potassium permanganate solution:
The chemical reaction is:
${CH}_{3} {CH}_{2} OH \overset{alk . {KMnO}_{4}}{\to} {CH}_{3} COOH + H_{2} O (Ethanol) (Ethanoic acid) (Water) (X) (Y)$
Step 3
(iii) The reaction of Ethanoic acid and Ethanol in the presence of concentrated Sulphuric acid $(H_{2} {SO}_{4})$ forms the ester that is Ethyl ethanoate $({CH}_{3} {COOC}_{2} H_{5})$ and water.
Ethyl ethanoate smells like a fruity smell. This reaction is called the Esterification reaction.
So, compound Z is Ethyl ethanoate.
Step 4
(iv) The chemical equation of Ethanoic acid and Ethanol in the presence of concentrated Sulphuric acid is written as:
${CH}_{3} COOH + C_{2} H_{5} OH \overset{conc H_{2} {SO}_{4}}{\to} {CH}_{3} {COOC}_{2} H_{5} Ethanoic aicd Ethanol Ethylethanoate (X) (Y) (Z)$

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