Question

Ethene gives an acidic \[KMn{O_4}\]solution.
(a) Ethylene glycol
(b) Ethylene oxide
(c) Formaldehyde
(d) Acetaldehyde

Answer 1

Hint: The potassium permanganate (\[KMn{O_4}\]) can lead to the formation of various products in cold, acidic, and alkaline reaction mediums. The cold or alkaline \[KMn{O_4}\]can oxidise an alkene into 1,2-diol or 1,2-glycols.

Complete Step by Step Answer:
Because potassium permanganate (\[KMn{O_4}\]) acts as an oxidising agent therefore it has the ability to provide oxygen in neutral as well as alkaline reaction medium and this oxygen can fuse with an alkene. When ethene reacts with alkaline potassium permanganate formation of ethane-1,2-diol or ethylene glycol occurs.
The reaction of ethene with alkaline potassium permanganate can be represented as:

Image: Formation of Ethene-1,2-diol from an alkene.
Due to the formation of hydroxyl groups this reaction is known as hydroxylation reaction.
During the oxidation reaction of alkene by alkaline potassium permanganate formation of a cyclic diester of manganese intermediate is observed, which at the end of the reaction causes the formation of syn-diol or we can say cis-diol.

Image: Formation of the cyclic diester of manganese during the oxidation of alkene by potassium permanganate.

During the reaction of oxidation of an alkene, the purple colour of potassium permanganate disappears and a brown colour precipitate of manganese dioxide is formed. Therefore, we can say this reaction was used for the test of unsaturation and this unsaturated test is named Baeyer’s test.

Therefore, from the above discussion, it is quite clear that potassium permanganate (\[KMn{O_4}\]) is the best candidate for the formation of syn-diol. Hence option (a) will be the correct answer.

Note: The compounds which possess carbon-carbon double and triple bonds are considered unsaturated compounds. And for the detection of these bonds the bromine and Baeyer’s test are used to conduct. Potassium permanganate (\[KMn{O_4}\]) is also called Beyer’s reagent.