Reactions of carbohydrates

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   De La Salle University-Dasmariñas Dasmariñas, Cavite, Philippines

|  The reactions of given samples of carbohydrates were observed from different test that gives different color of product to identify their specific type and distinguish from their reactions. There are nine carbohydrate samples that were tested with two unknown who were observed to determine their identity from comparing their results to the other given samples. Six different tests were used and a 0.20ml of each carbohydrate sample was placed in different test tube and added with 0.20ml of the test reagent. The identity of the unknown sample 1 was determined to be Fructose while unknown sample 2 was identified as Lactose.  |   Carbohydrates are the main energy source of the human body and they are the single most abundant class of organic molecule that can be found in nature. It is the structural building blocks of the cells and components of numerous metabolic pathways. A broad range of cellular phenomena such as cell recognition and binding depend on carbohydrates.Ñ
Natural carbohydrates are subdivided into monosaccharides, or simple sugars containing three to nine carbon atoms, polysaccharides, or polymers of monosaccharides, and an intermediate category of oligosaccharides, with two to ten monosaccharide units joined. The most important oligosaccharides to humans economically and biologically are the disaccharides.Ñ


!|" !  A 0.20ml of known carbohydrate samples and two unknown samples were place in separate test tubes. They were subjected to different test to observe their reactions. The different test performed were shown on Table 1.1 R
      
    Test

Amount of reagent

Procedure

Molisch test

0.20ml Molisch 0.20ml conc. H2SO4

reagent,

Iodine test

0.20ml iodine reagent

Benedict¶s test

0.20ml benedict reagent

Water bath at 100°C

Barfoed¶s test

0.20ml barfoed reagent

Water bath at 100°C

Seliwanoff¶s test Bial¶s test

0.20ml seliwanoff reagent 0.20ml bial reagent

Water bath at 100°C Water bath at 100°C

Visible color changes were observed on each reaction. The unknown samples were then identified by comparing from the known carbohydrate samples.

|!"    Carbohydrates can be identified or differentiated through different tests. One of these was the use of condensation reagents which react with carbohydrates to produce highly colored products. The tests that were used in the experiment with condensation reagents were Molisch¶s, Bial¶s and Seliwanoff¶s test. In Iodine test, the carbohydrates must possess certain structural features that allow it to form a condensation product. The condensation reaction tests were useful to classify the type of carbohydrates and to identify the specific type of carbohydrates. The other two tests used were for reducing sugars or saccharides were the Benedict¶s and Barfoed¶s test. The determination of positive reactions from the different tests was shown in Table 2 with their corresponding positive color change. Reactions of the carbohydrate samples to each test were shown in Table 3 and the identity of two unknown carbohydrate sample were determined as Fructose and Lactose. R
       
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$  Molisch Deep purple color Iodine Blue-black complex Benedict Rust-colored precipitate Barfoed Red precipitate Seliwanoff Aldoses (blue green); Ketoses (red) Bial Pentoses (blue/green); Hexoses (muddy brown to gray

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lucose alactose Fructose Lactose Sucrose Starch Ribose Unknown 1 Unknown 2 Legend: ++

       
    %
  ++ ++ + ++ ++ + ++ ++ + ++ + ++ + + + ++ ++ ++ + + ++ + + + + fast reaction; + slow reaction; - no reaction


 ++ -

Identity of Unknown 1: Fructose Identity of Unknown 2: Lactose A positive for Molisch¶s test has a presence of carbonyl group which results to a deep purple color product. It uses concentrated sulfuric acid as dehydrating acid to dehydrate carbohydrate. All of the carbohydrate samples got a positive test because they obviously have carbonyl groups. Monosaccharide gives a rapid positive test while disaccharide and starch were slower. The Iodine

test is used to distinguish starch so the starch sample got a positive test while the remaining samples got a negative results or no reaction. An illustration of pertinent reaction for Molisch¶s test was shown in Figure 1.

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        Benedict¶s test which is a mild basic solution reduces carbohydrates and it uses a mixture of copper(II) sulfate, sodium nitrate and sodium carbonate. The presence of red copper(I) oxide precipitate indicates that the saccharide has reduce the copper(II) ions. Only sucrose and starch got a negative test. The Barfoed¶s test is a copper(II) ions in acidic medium that is use for reducing monosaccharide and disaccharide. Samples with a negative test were lactose and sucrose because they were disaccharides, starch because it was a polysaccharide and the unknown sample 2. Illustrations for the reaction of carbohydrate in Benedict¶s and Barfoed¶s test were shown in Figure 2 and 3.

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           The Seliwanoff¶s test used hydrochloric acid as dehydrating acid and this test is for distinguishing aldoses from ketoses. The formation of red product was the basis for a positive test for the ketoses. Fructose got a positive result because it was basically a ketose, while the sucrose was formed from glucose and fructose so it gives also a positive reaction and the rest of the samples got negative result because they were all aldoses while the starch was a polysaccharide. Bial¶s test was the other test that identifies a specific type of carbohydrates because it can distinguish pentoses from hexoses. In the experiment, positive result was for pentoses which will give a blue or green product. Only the ribose got a positive result from all the other sample because it was the only pentose while the other monosaccharide like glucose, galactose were hexose and the two unknown sample got a negative results that signifies they

were not pentoses. Illustrations for the reaction of carbohydrates in Seliwanoff and Bial¶s test were shown in Figure 4 and 5. From the results of the experiment, the two unknown carbohydrate sample were identified to be Fructose for unknown sample 1 and Lactose was the unknown sample 2 because they got same results from the given known sample of fructose and lactose.

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A faster way to identify the unknown carbohydrate is using a general test to a more specific type of test. A good order of test that can be used to identify an unknown sample was shown in Figure 6.


 

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 v $  R  %   &!"      |!&!|! ! (1) Legaspi, .A. 2009. Essentials of Biochemistry Laboratory rd (2) McKee, T. McKee, J.R. 2003. Biochemistry-The Molecular Basis of Life. 3 Edition. Mcraw-Hill Companies, Inc (3) Langley R. Moore J.T. 2008. Biochemistry for Dummies. Wiley Publishing, Inc. Indianapolis, Indiana (4) Seeley, R.R. Stephens, T.D. Tate, P. 2005. Essentials of Anatomy & Physiology. 5th Edition. The Mcraw-Hill Companies, Inc. New York