Question

\({\rm{C}}{{\rm{H}}_{\rm{2}}} = {\rm{C}}{{\rm{H}}_{\rm{2}}} \overset{Alkaline KMnO_{4},KOH/H_{2}O}{\rightarrow} X\). Product X in the above reaction is [RPMT 2003]
A) Ethylene glycol
B) Glucose
C) Ethanol
D) All of these

Answer 1

Hint: We know that potassium permanganate is an oxidising agent (strong). It is used in a set of chemical reactions. Here, we have to identify the product obtained on the reaction ethene with potassium permanganate in the basic medium.

Complete Step by Step Answer:
We know that the strong oxidising nature of potassium permanganate gives oxygen in an alkaline or neutral medium. The oxygen obtained has a role in the oxidation of alkene to form 1,2-diol. Therefore, the reaction of ethene with the potassium permanganate in the alkaline medium gives the product of ethylene glycol or 1,2-diol.

The oxidation reaction is as follows:

Image: Reaction of ethene with potassium permanganate in alkaline medium

Let's understand some facts regarding the above reaction. The reaction is also termed hydroxylation reaction because of the attachment of the hydroxyl groups across the double bond of the ethene. In this reaction, the potassium permanganate loses its pink colour and a precipitate of manganese dioxide is obtained. This test is termed Baeyer's test. Therefore, product X in the reaction is ethylene glycol.
Hence, option A is right.

Note: The cold, acidified and potassium permanganate(dilute) is termed Baeyer's reagent. It is useful in determining alkenes and alkynes. And in this test, if the decolourisation of potassium permanganate happens, confirms the unsaturated nature of the compounds. An alkene when undergoes a reaction with cold, acidified and dilute potassium permanganate gives vicinal diols.