The Molisch Test Shows positive test for: All carbohydrates. Monosaccharides give a rapid positive test. Disaccharides and polysaccharides react slower. Reactions: The test reagent dehydrates pentoses to form furfural (top reaction) and dehydrates hexoses to form 5-hydroxymethyl furfural (bottom reaction). The furfurals further react with -naphthol present in the test reagent to produce a purple product (reaction not shown).
How to perform the test: Two ml of a sample solution is placed in a test tube. Two drops of the Molisch reagent (a solution of -napthol in 95% ethanol) is added. The solution is then poured slowly into a tube containing two ml of concentrated sulfuric acid so that two layers form. A positive test is indicated by: The formation of a purple product at the interface of the two layers.
a negative test (left) and a positive test (right)
Carbohydrates Carbohydrates are a major source of energy for our bodies. Foods high in carbohydrates are potatoes, bread, pasta, and rice. One gram of carbohydrate provides 4 kcal of energy. In general, carbohydrates are polyhydroxyl compounds that contain a carbonyl group. What does polyhydroxyl mean? It means that the compound has many hydroxyl (OH) groups. What are carbonyl groups? A carbonyl group is a carbon double bonded to an oxygen (C=O). If the carbon is attached to two other carbons, in addition to the oxygen, then the carbonyl group is called a ketone (Figure A). If the carbon is bonded to a hydrogen and another carbon, then the carbonyl group is called an aldehyde (Figure B).
Carbohydrates are grouped into three categories: monosaccharides, disaccharides, and polysaccharides. Monosaccharides A monosaccharide is a single "sugar" unit. The most common monosaccharides are made up of six carbon atoms and are called hexoses. Glucose is an example of a hexose (Figure C). Saccharides can exist in two interchangeable forms: an open-chain form and a closed-chain form. Glucose exists almost entirely in the closedchain form.
Monosaccharide
Sources
Glucose
Fruit juices, honey, corn syrup
Fructose
Fruit juices, honey
Disaccharides Disaccharide carbohydrates are made up of two monosaccharide units linked together. When two glucose molecules are linked together they form the disaccharide, maltose (Figure D).
Disaccharide
Sources
Monosaccharides units
Maltose
Germinating grains
glucose + glucose
Lactose
Milk, yogurt, ice cream
glucose + galactose
Sucrose
Sugar cane, sugar beets
glucose + fructose
Polysaccharides When many monosaccharide units are linked together, the resulting carbohydrate is called a polysaccharide. Three important polysaccharides are starch, cellulose, and glycogen which are all made up of glucose molecules linked in different arrangements.
Polysaccharide Starch
Sources Germinating grains
Glycogen
Muscle, liver
Cellulose
Plants, wood, paper, cotton
Reducing Sugars Carbohydrates that can undergo oxidation are called reducing sugars. All monosaccharides are reducing sugars and many disaccharides are also reducing sugars. Lactose and maltose are both reducing sugars, but sucrose is not. You will test for the presence of reducing sugars using Benedict's solution.
Laboratory Activities A. Benedict's Test for Reducing Sugars Fill a 250-mL beaker about half full with water and begin heating it over a Bunsen burner. (Note: You will be using this boiling water bath for more than one series of tests, so you may need to add water occasionally as the water boils away.) Place 10 drops of glucose, fructose, lactose, sucrose, starch, and an unknown in separate labeled test tubes. Add 2 mL of Benedict's reagent to each sample. Place all of the test tubes in the boiling water bath for 5 minutes. The formation of a greenish to reddish-orange color indicates the presence of a reducing sugar. If the solution is the same blue-green color as the Benedict's reagent, there has been no oxidation and the sample is not a reducing sugar. Record your results. Determine if each sample as a reducing or non-reducing sugar.
B. Seliwanoff's Test for Ketone Hexoses Seliwanoff's test is used to distinguish between aldehyde and ketone hexoses (carbohydrates containing 6 carbon atoms). A ketone hexose, also called a ketohexose, will form a deep red color when reacted with Seliwanoff's reagent. An aldehyde hexose, also called a aldohexose, will show a light pink color that takes a longer time to develop when reacted with Seliwanoff's reagent. Place 10 drops of glucose, fructose, lactose, sucrose, starch, and an unknown in separate labeled test tubes. Add 2 mL of Seliwanoff's reagent to each sample. Place all of the test tubes in the boiling water and record the time. After 1 min, observe the colors of the solutions in each test tube. The rapid formation of a deep red color indicates the presence of a ketohexose. Record your observations as either fast color change, slow color change, or no color change. Determine if each sample is a ketohexose or not. C. Iodine Test for Polysaccharides Starch is a polysaccharide that is easily detected by the iodine test. The many glucose units linked together to form starch molecules traps the iodine molecules and produces a dark blue-black complex. Other polysaccharides react to a lesser extent with iodine to form a red-brown or reddish-purple color. Mono- and disaccharides are too small to trap the iodine molecules, so they do not show a color change in the presence of iodine. Place 5 drops of glucose, fructose, lactose, sucrose, starch, and an unknown in separate labeled test tubes. Add 1 drop of iodine solution to each sample. Record your observations. A dark blue-black color is a positive test for starch. D. Identifying your Unknown Carbohydrate Using the results you obtained in the tests above, identify your unknown as either glucose, fructose, lactose, sucrose, starch or none of these. It may be helpful to combine your results from the individual tests into one table similar to the one shown below: Sample Glucose Fructose Lactose Sucrose
Benedict's Test
Seliwanoff's Test
Iodine Test
Starch Unknown E. Testing Foods for Carbohydrates Obtain one sugar sample, one syrup sample, and one food product. Perform the Benedict's test, Seliwanoff's test and iodine test on each sample. Record your results and describe what kinds of carbohydrates are in each sample. For example, if I test a food sample and record the following results: Benedict's test: a reddish-orange color Seliwanoff's test: rapid formation of a deep red color Iodine test: turned dark blue Then I would conclude that my food sample contained a reducing sugar (Benedict's test), a ketohexose (Seliwanoff's test), and starch (Iodine test).
hemicals:
glucose fructose sucrose potassium sodium tartrate sodium hydroxide CuSO4 · 5 H20
Fehling I consists of 7 g of hydrated copper(II) sulfate dissolved in 100 mL of dist. water. Fehling II is made by dissolving 35 g of potassium sodium tartrate and 10 g of sodium hydroxide in 100 mL of dist. water. Fehling's reagent: Equal volumes of Fehling I and Fehling II are mixed to form a deep blue solution.
Apparatus and glass wares:
hot plate thermometer 3 conical measures, graduated, 500 mL 3 glass stirring rods 3 beakers 40 mL
3 beakers 150 mL beaker 1800 mL measuring cylinder 100 mL Hazards and safety precautions: Sodium hydroxide is very corrosive. Causes severe burns. May cause serious permanent eye damage. Very harmful by ingestion. Harmful by skin contact or by inhalation of dust. Copper(II) sulfate is harmful when swallowed. Safety glasses, protective gloves, good ventilation.
Experimental procedure: 5 g of glucose, fructose and sucrose, respectively, are dissolved in 100 mL of dist. water. Three 500 ml conical measures are each filled with the sugar solutions. Each solution is made up to the 500 mL mark with dist. water warmed up to 60 °C. Afterwards 8 mL of Fehling reagent are poured into each of the sugar solutions while stirring.
Results: In two conical measures a yellow-red precipitate is formed. At first in the blue solution the precipitate may appear greenish. Finally the muddy green suspension yields brick red precipitate. Fructose reacts faster than glucose. In the third glass no change can be observed. conical measure 1
glucose solution
brick red precipitate
conical measure 2
fructose solution
brick red precipitate
conical measure 3
sucrose solution
no change
Video clip (Download RealPlayer .rm file) Discussion: Fehling's tests for aldehydes are used extensively in carbohydrate chemistry. A positive result is indicated by the formation of a brick red precipitate. Like other aldehydes, aldoses are easily oxidized to yield carboxylic acids. Cupric ion complexed with tartrate ion is reduced to cuprous oxide.
Fig. 1: Redox reaction • The sucrose does not react with Fehling's reagent. Sucrose is a disaccharide of glucose and fructose. Most disaccharides are reducing sugars, sucrose is a notable exception, for it is a non-reducing sugar. The anomeric carbon of glucose is involved in the glucosefructose bond and hence is not free to form the aldehyde in solution.
Fig. 2: Sucrose • On the other hand, glucose, a reducing sugar, reacts with Fehling's reagent to form an orange to red precipitate. Fehling's reagent is commonly used for reducing sugars but is known to be not specific for aldehydes. For example, fructose gives a positive test with Fehling's solution too, because fructose is converted to glucose and mannose under alkaline conditions. The conversion can be explained by the keto-enol tautomerism.
Fig. 3: Conversion of fructose to glucose and mannose The reduction of Fehling solution using fructose is not only to be attributed to the fact that the ketose is isomerized into an aldose. The treatment of fructose with alkali - e.g. Fehling solution - causes even decompostion of the carbon chain. More products with reducing capability are formed.
Fig. 4: Decomposition of fructose Note: Fehling's test takes advantage of the ready reactivity of aldehydes by using the weak oxidizing agent cupric ion (Cu2+) in alkaline solution. In addition to the copper ion, Fehling's solution contains tartrate ion as a complexing agent to keep the copper ion in solution. Without the tartrate ions, cupric hydroxide would precipitate from the basic solution. The tartrate ion is unable to complex cuprous ion Cu+, so the reduction of Cu2+ to Cu+ by reducing sugars results in the formation of an orange to red precipitate of Cu2O. The structure of copper(II) D-tartrate has been determined to be:
Fig. 5: Copper-tartrate-complex
Bial's Test Shows positive test for: Pentoses Reactions: The test reagent dehydrates pentoses to form furfural. Furfural further reacts with orcinol and the iron ion present in the test reagent to produce a bluish product (reaction not shown).
How to perform the test: Two ml of a sample solution is placed in a test tube. Two ml of Bial's reagent (a solution of orcinol, HCl and ferric chloride) is added. The solution is then heated gently in a Bunsen Burner or hot water bath. If the color is not obvious, more water can be added to the tube. A positive test is indicated by: The formation of a bluish product. All other colors indicate a negative result for pentoses. Note that hexoses generally react to form green, red, or brown products.
two negative tests (left, middle) and a positive test (right)
