Question

The disappearance of the characteristic purple colour of \[{\rm{KMn}}{{\rm{O}}_{\rm{4}}}\] in its reaction with an alkene is the test for unsaturation. It is known as
A) Markovnikov’s test
B) Baeyer’s test
C) Wurtz’s test
D) Grignard test

Answer 1

Hint: Potassium permanganate, \[{\rm{KMn}}{{\rm{O}}_{\rm{4}}}\] , is of the purple colour. Its decolourisation is a test for unsaturation. And this test helps to determine the presence of a triple or double bond in organic compounds.

Complete Step by Step Answer:
Let's discuss Baeyer's test in detail. The cold, acidified, and dilute potassium permanganate is termed Baeyer's reagent. It is used in the determination of 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. This test is called Baeyer's test. The reaction is,

The permanganate possesses the strong nature of the oxidising agent. Therefore, they oxidise alkene to alkane 1,2-diol. But, in the acidic condition, the permanganate(VII) ions undergo reduction to give manganese (II) ions. The reaction is,
\[{\rm{C}}{{\rm{H}}_{\rm{2}}} = {\rm{C}}{{\rm{H}}_{\rm{2}}} + 2{{\rm{H}}_{\rm{2}}}{\rm{O}} + {\rm{2Mn}}{{\rm{O}}_{\rm{4}}}^ - + 6{{\rm{H}}^ + } \to 5{\rm{C}}{{\rm{H}}_{\rm{2}}}{\rm{OH}} - {\rm{C}}{{\rm{H}}_{\rm{2}}}{\rm{OH}} + {\rm{2M}}{{\rm{n}}^{2 + }}\]

Therefore, the Bayer test is the test for unsaturation of alkene which is confirmed by the decolorization of the purple coloured \[{\rm{KMn}}{{\rm{O}}_{\rm{4}}}\] .
 Hence, option B is right.

Note: It is to be noted that only the acidic nature of potassium permanganate gives the test of unsaturation. If the medium is alkaline, the reaction will change. In such conditions, the purple coloured permanganate (VII) changes to green coloured manganate (VII) ions first and then changes to manganese (IV) oxide which is of dark brown colour.