
A gas formed by the action of alcoholic KOH on ethyl iodide decolourises alkaline \[{\rm{KMn}}{{\rm{O}}_{\rm{4}}}\] . The gas is:
A. \[{{\rm{C}}_{\rm{2}}}{{\rm{H}}_{\rm{6}}}\]
B. \[{\rm{C}}{{\rm{H}}_{\rm{4}}}\]
C. \[{{\rm{C}}_{\rm{2}}}{{\rm{H}}_{\rm{2}}}\]
D. \[{{\rm{C}}_{\rm{2}}}{{\rm{H}}_{\rm{4}}}\]
Answer
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Hint: The dehydrohalogenation reaction is the reaction of an alkyl halide in which the removal of a hydrogen atom and a halogen atom takes place in presence of the action of alcoholic KOH. The obtained product in the reaction is an alkene.
Complete Step by Step Answer:
Let’s understand the dehydrohalogenation reaction of ethyl iodide by the action of potassium hydroxide which is alcoholic in nature. In this reaction, a molecule of hydrogen iodide is removed from the compound, and an alkene forms. The reaction is as follows:
\[{{\rm{C}}_{\rm{2}}}{{\rm{H}}_{\rm{5}}} - {\rm{I}} + \,{\rm{Alc}}{\rm{.}}\,{\rm{KOH}} \to {{\rm{C}}_{\rm{2}}}{{\rm{H}}_{\rm{4}}}\]
So, the formation of ethene takes place in the above reaction.
Now, we will understand the decolorization of \[{\rm{KMn}}{{\rm{O}}_{\rm{4}}}\]. We know that pink colour of potassium permanganate (\[{\rm{KMn}}{{\rm{O}}_{\rm{4}}}\]) disappears when it undergoes a reaction with unsaturated compounds (alkene and alkyne). This reaction is termed Baeyer’s reaction. And Baeyer’s reagent is cold, acidified, and dilute potassium permanganate. And in this reaction, the formation of 1,2-diol occurs. The chemical reaction is shown below:

Image: Baeyer’s test
So, the gas produced in the dehydrohalogenation reaction of ethyl iodide, that is, ethene (\[{{\rm{C}}_{\rm{2}}}{{\rm{H}}_{\rm{4}}}\]) can decolorize the potassium permanganate.
Therefore, option D 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.
Complete Step by Step Answer:
Let’s understand the dehydrohalogenation reaction of ethyl iodide by the action of potassium hydroxide which is alcoholic in nature. In this reaction, a molecule of hydrogen iodide is removed from the compound, and an alkene forms. The reaction is as follows:
\[{{\rm{C}}_{\rm{2}}}{{\rm{H}}_{\rm{5}}} - {\rm{I}} + \,{\rm{Alc}}{\rm{.}}\,{\rm{KOH}} \to {{\rm{C}}_{\rm{2}}}{{\rm{H}}_{\rm{4}}}\]
So, the formation of ethene takes place in the above reaction.
Now, we will understand the decolorization of \[{\rm{KMn}}{{\rm{O}}_{\rm{4}}}\]. We know that pink colour of potassium permanganate (\[{\rm{KMn}}{{\rm{O}}_{\rm{4}}}\]) disappears when it undergoes a reaction with unsaturated compounds (alkene and alkyne). This reaction is termed Baeyer’s reaction. And Baeyer’s reagent is cold, acidified, and dilute potassium permanganate. And in this reaction, the formation of 1,2-diol occurs. The chemical reaction is shown below:

Image: Baeyer’s test
So, the gas produced in the dehydrohalogenation reaction of ethyl iodide, that is, ethene (\[{{\rm{C}}_{\rm{2}}}{{\rm{H}}_{\rm{4}}}\]) can decolorize the potassium permanganate.
Therefore, option D 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.
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