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Molisch test is mainly a group test for all carbohydrates, that can be either free or bound to proteins or lipids. It is a very sensitive test and requires precision for the detection of carbohydrates. It is a colorimetric method that is used for the analysis of the presence of carbohydrates in a given analyte. The test was named after an Austrian botanist named Hans Molisch. Molisch’s test is done via using Molisch reagent. A solution of naphthol in 95%ethanol is called a Molisch reagent. It can also be referred to as the purple ring test. Other chemicals like that of concentrated sulfuric acid are also used in the test rather than the Molisch reagent. All carbohydrates give a positive result for the Molisch test. Whereas, tetrose and triose are exceptions. Monosaccharides give a faster positive test in Molisch’s test. Whereas disaccharides and polysaccharides react slowly with Molisch reagent and delays in giving the positive test. Therefore, Molisch’s test is very important for the detection of the presence of carbohydrates in a substance.
Principle of Molisch Test
The reaction is generally based on the fact that the concentrated acid helps in catalysing the dehydration of sugars to form furfural (via pentoses) or hydroxymethylfurfural (via hexoses).
Any one of these aldehydes condenses with two molecules of naphthol in order to form a purple or violet coloured complex at the interface of the acid and the test layer.
If the carbohydrate is a polysaccharide or a disaccharide, a glycoprotein or glycolipid, the acid first hydrolyses it into monosaccharides, which further gets dehydrated to form furfural or its derivatives.
A green ring can be observed if any impurities are present in the reagent as they could interact with the α-naphthol and the acid.
If a concentrated sugar solution is used then a rind ring is observed. This may be due to the charring of the sugar because of the acid.
Objectives of Molisch’s Test
The main objectives of Molisch’s Test are given below:
1. To determine the presence of carbohydrates in a given sample.
2. To differentiate carbohydrates from other biomolecules.
Requirements for Molisch Test
Reagent required
Molisch reagent: Dissolve 3.75 g of α-naphthol in 25 ml of 99% ethanol This reagent should always be prepared fresh.
Concentrated sulphuric acid
Test sample
Materials required
Test tubes
Test tube stand
Pipette
Distilled water
Reaction of Molisch’s Test
The test reagent dehydrates pentoses in order to form furfural (which is represented at top in the image) and also dehydrates hexoses to form 5-hydroxymethyl furfural (which is represented at bottom in the image). These furfurals further react with naphthol that is present in the test reagent to produce a purple product.
Procedure of Molisch Test
Each distilled water and test sugar solution is taken about 2 ml in four test tubes separately.
Then two drops of Molisch reagent were added to each tube.
Then we need to hold the test tube in an inclined position and gently 1 ml of concentrated H2SO4 was added along the side of the test tube. The acid must not be mixed with the solution. A black ring is formed if the concentrated acid is not added slowly, which generates heat from the reaction that can char the carbohydrates.
The test tube is observed for the formation of a purple-coloured ring at a layer between the solution and the acid.
Observation of Molisch Test
Formation of the purple coloured ring can be seen at the interface between the sulphuric acid and the test solution this confirms the presence of carbohydrates, this indicates a positive result.
The sulphuric acid is observed to remain above the test solution as the acid is denser than the test solution.
The absence of purple colour means the test solution lacks carbohydrate molecules and thus, indicates a negative result.
Uses of Molisch Test
It is used to detect the presence of carbohydrates in different samples.
Molisch’s Test can also be used to detect the formation of carbohydrates as a by-product in different reactions and differentiate it from other biomolecules.
Conclusion
Lastly to conclude the topic let us discuss the limitations of the Molisch Test. Trioses and tetroses do not contain the necessary five carbon atoms for furfural formation, thus they do not give a positive result for this reaction. It is not a specific test for carbohydrates. Furfurals or furfural yielding substances, some organic acids such as citric acids, lactic acid, oxalic acid, formic acid, etc. produces a positive result. For example the Molisch test is used to detect D-glucose presence in the sample. The D-glucose is reacted with α- naphthol (i.e. the Molisch reagent) and concentrated sulphuric acid.
As concentrated sulphuric acid is added to the given solution, then D-Glucose loses hydroxyl groups in the form of water. After the loss of three water molecules, 5-(hydroxymethyl) furfural is produced. The furfural now comes in contact with the sulphonated α- naphthol that is present in the Molisch reagent. The furfural again loses another water molecule in the presence of the Molisch reagent, thus giving rise to an intermediate molecule. The intermediate molecule that is formed results in the loss of two electrons and a single proton, thus giving rise to a purple or purplish-red coloured product. The appearance of this purple or purplish-red coloured product confirms the presence of D-glucose in the provided sample.
