
Molisch test is answered by:
(A) All carbohydrates
(B) Sucrose
(C) Fructose
(D) Glucose
Answer
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Hint: The answer of this question is the essential part of our diet. It is a macronutrient and it is important for our body. It can also be defined chemically as neutral compounds of carbon, hydrogen and oxygen.
Complete step by step answer:
The question is asking us to identify that compound whose presence is tested by the Molisch test. So, first we should know about the Molisch test. Molisch’s test is a chemical test which is used to check for the presence of carbohydrates in a given substance. So, from this we can now say that the Molisch test is used to check for carbohydrates in a given compound. It is very important for us to know about the process of testing carbohydrates in any substance by the Molisch test. So, detailed explanation about Molisch test is as follows:
> So, till now we know that the Molisch test is used to identify carbohydrates in a given substance. This test is named after Czech-Austrian botanist Hans Molisch. So, this test is named after him. So, in Molisch test we add Molisch’s reagent (a solution of ∝-naphthol in ethanol) to the give substance and the subsequent addition of a few drops of concentrated ${{H}_{2}}S{{O}_{4}}$ (sulphuric acid) to the mixture. So, one question will arise in our mind, which is why we add ${{H}_{2}}S{{O}_{4}}$ to the solution? So, answer to this question is as follows:
> We add ${{H}_{2}}S{{O}_{4}}$ to this reaction because this reaction is based on the dehydration of the carbohydrate by sulphuric acid to produce an aldehyde, which condenses with two molecules of a phenol. It condenses with α-naphthol or other phenols such as resorcinol and thymol. This results in formation of coloured violet rings. The formation of a violet ring at the point of contact between the ${{H}_{2}}S{{O}_{4}}$ and the substance and Molisch’s reagent mixture confirms the presence of carbohydrates in the substance.
So, from the above the discussion now we know our answer. The answer to this question is option A which is carbohydrates. We use the Molisch Test to determine the presence of carbohydrates in compounds.
Note:We should know about the correct procedure to perform the Molisch Test. To begin the testing of carbohydrates in a given substance we add 2-3 drops of Molisch’s reagent to the given substance in a small amount in a test tube and mix it well. Now, we should add a few drops of concentrated sulfuric acid. We should add it drop-wise along the walls of the test tube to help the formation of a layer and avoid mixing. The development of a purple ring at the layer formed by the concentrated acid is a positive indicator for Molisch’s test. If no purple or reddish-purple colour arises, the given substance does not contain any carbohydrate.
Complete step by step answer:
The question is asking us to identify that compound whose presence is tested by the Molisch test. So, first we should know about the Molisch test. Molisch’s test is a chemical test which is used to check for the presence of carbohydrates in a given substance. So, from this we can now say that the Molisch test is used to check for carbohydrates in a given compound. It is very important for us to know about the process of testing carbohydrates in any substance by the Molisch test. So, detailed explanation about Molisch test is as follows:
> So, till now we know that the Molisch test is used to identify carbohydrates in a given substance. This test is named after Czech-Austrian botanist Hans Molisch. So, this test is named after him. So, in Molisch test we add Molisch’s reagent (a solution of ∝-naphthol in ethanol) to the give substance and the subsequent addition of a few drops of concentrated ${{H}_{2}}S{{O}_{4}}$ (sulphuric acid) to the mixture. So, one question will arise in our mind, which is why we add ${{H}_{2}}S{{O}_{4}}$ to the solution? So, answer to this question is as follows:
> We add ${{H}_{2}}S{{O}_{4}}$ to this reaction because this reaction is based on the dehydration of the carbohydrate by sulphuric acid to produce an aldehyde, which condenses with two molecules of a phenol. It condenses with α-naphthol or other phenols such as resorcinol and thymol. This results in formation of coloured violet rings. The formation of a violet ring at the point of contact between the ${{H}_{2}}S{{O}_{4}}$ and the substance and Molisch’s reagent mixture confirms the presence of carbohydrates in the substance.
So, from the above the discussion now we know our answer. The answer to this question is option A which is carbohydrates. We use the Molisch Test to determine the presence of carbohydrates in compounds.
Note:We should know about the correct procedure to perform the Molisch Test. To begin the testing of carbohydrates in a given substance we add 2-3 drops of Molisch’s reagent to the given substance in a small amount in a test tube and mix it well. Now, we should add a few drops of concentrated sulfuric acid. We should add it drop-wise along the walls of the test tube to help the formation of a layer and avoid mixing. The development of a purple ring at the layer formed by the concentrated acid is a positive indicator for Molisch’s test. If no purple or reddish-purple colour arises, the given substance does not contain any carbohydrate.
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