DISCUSSIONMolisch’s testBased on the result that we obtained in the experiment, it was observed that the appearance of purple ring at the junction of 2 layers. The reaction is based on the fact that concentrated H2SO4 catalyses the dehydration of sugars to form furfural (from pentoses) or hydroxymethyl furfural (from hexoses).These furfurals then condense with sulfonated alpha-naphthol to give a purple coloured product (furfuryl-diphenyl-methane-dyes). Polysaccharides and glycoproteins also give a positive reaction. In the event of the carbohydrate being a poly- or disaccharide, the acid first hydrolyses it into component monosaccharides, which then get dehydrated to form furfural or its derivatives.
So the hyphothesis was archieved in this Molisch’s test.Researcher generally use the Molisch test to determine the presence of carbohydrate in variety of substances, from elastin protein to unripened guavas (Misra & Seshadri, 1968? Stein &Miller, 1938).Iodine testBased on the result that obtained in this experiment, it was observed that the appearance of brown wine color compound. The iodine test is used to test for the presence of starch. Iodine forms a coordinate complex between the helically coiled polysaccharide chain and iodine centrally located within the helix due to absorption. The color obtained depends upon the length of the unbrached or linear chain available for complex formation. Test tube changed color to brown this test tube was the positive control as it was a starch solution (Choiet al., 2013).
So the hyphothesis was archieved.Barfoed’s testBarfoed’s test is used to detect presence of reducing sugars. Appearance of a red precipitate as a thin film at the bottom of the test tube within 3-5 min was indicative of reducing monosaccharide. If the precipitate formation takes more time, then it is a reducing disaccharide. Based on the result of this Barfoed’s test, hypothesis of this experiment was not achieved because there was no formation of brick-red precipitation was observed at the bottom of test tube within 2-3 minutes.
There are some reasons on why there was no formation of brick-red precipitation. Firstly, adding less amount of Barfoed reagent as compared to solution. Moreover, do not note time when put test tube in boiling water bath may cause over heating that can affect the result. Next, do not remove the test tube regularly and check if precipitates are formed or not. So, there was not formation of brick-red precipitation.
Benedict’s testThe Benedict’s test identifies reducing sugars (monosaccharide’s and some disaccharides), which have free ketone or aldehyde functional groups. Some sugars such as glucose are called reducing sugars because they are capable of transferring hydrogens (electrons) to other compounds, a process called reduction. When reducing sugars are mixed with Benedicts reagent and heated, a reduction reaction causes the Benedicts reagent to change color. When Benedict’s solution and simple carbohydrates are heated, the solution changes to orange red/ brick red. This reaction is caused by the reducing property of simple carbohydrates. The copper (II) ions in the Benedict’s solution are reduced to Copper (I) ions, which causes the color change.The red copper(I) oxide formed is insoluble in water and is precipitated out of solution. This accounts for the precipitate formed.
As the concentration of reducing sugar increases, the nearer the final colour is to brick-red and the greater the precipitate formed. Based on the result, the hypothesis was not archieved. There were reasons why the hypothesis cannot be archieved because heated the sample in a boiling water bath for more than 3 minutes. Next, the sample do not allow to cool under running tap water.Seliwanoff’s testSeliwanoff’s test is a chemical test which distinguishes between aldose and ketose sugars.
Ketoses are distinguished from aldoses via their ketone/aldehyde functionality. If the sugar contains a ketone group, it is a ketose. If a sugar contains an aldehyde group, it is an aldose.
This test relies on the principle that, when heated, ketoses are more rapidly dehydrated than aldoses. When reacted with Seliwanoff reagent, ketoses react within 2 minutes forming a cherry red condensation product. Aldopentoses react slowly, forming the coloured condensation product. Based on the result obtained in this experiment,it was observed that present of fructose(ketose) that form cherry red shading. This is because if the ketose is presence,then the Seliwanoff’s reagent would dehydrate the ketose and form hydroxymethyl furfural. The hydroxymethyl furfural will further react with resorcinol giving off a red condensation product.Therefore, the presence of a ketose molecule is indicated by the formation of a red colored solution.Bial’s testBial’ Reagent is used to test the presence of pentose sugar namely Ribose sugar and deoxyribose sugar.
Bial’s Reagent contains Orcinol, Hydrochloric acid and Friec chloride. When a sugar Is added to Bial’s reagent, it will dehydrate to form furfural which will then react with Orcinol to produce a colored product. When pentose is present it will produce blue or green precipitate. Whereas when a hexose sugar is added, it will change into muddy brown precipitate. Theoretically, the bial’s test for sucrose, it will remain as blue-green solution since it is a pentose sugar.
For fructose will change its colour to muddy brown solution since fructose are hexoses sugar. Based on the result obtained, the hypothesis was not archieved since there no change colour of solution. CONCLUSIONREFERENCESCowart, S. L., Stachura, M. E.
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