
The compound most likely to decolorize a solution of potassium permanganate is
A.
B.
C.
D.
Answer
232.8k+ views
Hint: Potassium permanganate is an inorganic compound having the chemical formula . It is a purple solution.
This solution is decolorized in the presence of double or triple bonds.
Complete Step by Step Solution:
Potassium permanganate is an inorganic compound possessing the chemical formula KMnO4.
It is a purplish-black crystalline salt and undergoes dissolution in water as and as a result, its solution is pink or purple.
Dilute solutions of transforming alkenes into diols (glycols).
This reaction is also utilised as a qualitative test for the presence of double or triple bonds in a molecule.
Hence, called Baeyer's reagent.
The decolorization occurs as originally purple permanganate solution after reaction generates a brown precipitate of Manganese dioxide.
The oxidation state of manganese is +7 and it gets reduced to +4 in. It undergoes reduction and oxidises the double and triple bonds leading to the formation of single bonds.
We have to find out which of the given options will decolorize potassium permanganate.
A.
This compound only contains single bonds. Hence, it will not decolorize the solution of potassium permanganate.
So, A is incorrect.
B.
This is the structure of Naphthalene. This will not decolorize potassium permanganate as it is an aromatic compound and the double bond here is delocalized.
So, B is incorrect.
C.
This compound contains a double bond which can be attacked by potassium permanganate and eventually will get oxidised.
So, this will decolorize the solution of potassium permanganate.
So, C is correct.
D.
This compound only contains single bonds. Hence, it will not decolorize the solution of potassium permanganate.
So, D is incorrect.
So, option C is correct.
Note: Still it must be remarked that bromine acts as a better reagent in estimating the presence of unsaturation i.e., double or triple bonds quantitatively. This is because , being an extremely strong oxidising agent, it can react with groups other than compounds having double and triple bonds.
This solution is decolorized in the presence of double or triple bonds.
Complete Step by Step Solution:
Potassium permanganate is an inorganic compound possessing the chemical formula KMnO4.
It is a purplish-black crystalline salt and undergoes dissolution in water as and as a result, its solution is pink or purple.
Dilute solutions of transforming alkenes into diols (glycols).
This reaction is also utilised as a qualitative test for the presence of double or triple bonds in a molecule.
Hence, called Baeyer's reagent.
The decolorization occurs as originally purple permanganate solution after reaction generates a brown precipitate of Manganese dioxide.
The oxidation state of manganese is +7 and it gets reduced to +4 in. It undergoes reduction and oxidises the double and triple bonds leading to the formation of single bonds.
We have to find out which of the given options will decolorize potassium permanganate.
A.
This compound only contains single bonds. Hence, it will not decolorize the solution of potassium permanganate.
So, A is incorrect.
B.
This is the structure of Naphthalene. This will not decolorize potassium permanganate as it is an aromatic compound and the double bond here is delocalized.
So, B is incorrect.
C.
This compound contains a double bond which can be attacked by potassium permanganate and eventually will get oxidised.
So, this will decolorize the solution of potassium permanganate.
So, C is correct.
D.
This compound only contains single bonds. Hence, it will not decolorize the solution of potassium permanganate.
So, D is incorrect.
So, option C is correct.
Note: Still it must be remarked that bromine acts as a better reagent in estimating the presence of unsaturation i.e., double or triple bonds quantitatively. This is because , being an extremely strong oxidising agent, it can react with groups other than compounds having double and triple bonds.
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