MDB 201(202) CARBOHYDRATE.pptx

CHEMISTRY OF CARBOHYDRATES  Carbohydrateare substances containing carbon, Hydrogen and Oxygen having general formula Cn(H2O)n.  They are simply hydrates of carbon. They are the main source of energy in the body.  CHO are referred to as polyhydroxy aldehyde OR  Ketone derivatives  OR...

CHEMISTRY OF CARBOHYDRATES  Carbohydrateare substances containing carbon, Hydrogen and Oxygen having general formula Cn(H2O)n.  They are simply hydrates of carbon. They are the main source of energy in the body.  CHO are referred to as polyhydroxy aldehyde OR  Ketone derivatives  OR  Compounds that yield these derivatives on hydrolysis  NATURE OF CARBOHYDRATES (CHO)  NATURE 1: ENERGY PRODUCTION SITE  Change the basic fuel to a refined fuel that the machine is designed to use  Carry this refined fuel to the places that needs it  Burn this refined fuel in the special equipment set up at theses places  EXAMPLE IN CHO  NATURE OF CARBOHYDRATES CONT’D  Digestion of CHO (Basic fuel) and change to glucose  Absorption and transportation of the refined fuel (Blood circulation) to cells that need glucose,  Glucose is burned in these cells and energy produced through the process of metabolism.  NATURE 2:  CHO are group of carbonyl compounds (aldehyde or ketones) that also contains several hydroxyl groups.  NATURE 3: CHO are composed of C, H and O Cn(H2O)n when n = 5 ; C5H10O5 ETC. NATURE 4: Glucose is the most important carbohydrate. Most dietary CHO is absorbed into the bloodstream as glucose and other sugars are converted into glucose in the liver. NATURE 5: Dietary importance  CHO are widely available and easily grown  Low cost in comparison with other food items  Easily stored and they can be kept for longer periods without spoilage  Mostof the total calories in our diet come from carbohydrates.  TYPES& CLASSIFICATION OF CHO  Carbohydrate can be classified into four groups: monosaccharide, disaccharides, oligosaccharides and polysaccharides.  1. Monosaccharide Monosaccharide are the simplest carbohydrates which contain free aldehyde (-CHO) and ketone (>C=O) groups that have two or more hydroxyl (-OH) groups. The general formula of monosaccharide is Cn(H2O)n. Monosaccharides are sugars that can not be further hydrolysed into simple carbohydrates. They can be classified on the basis of number of carbon atoms for example triose, tetrose, pentose, hexose, heptoses etc. and on the basis of functional group they possess for example, aldoses (those having aldehyde groups) or ketoses (those having ketone groups). NAME OF SUGAR ALDOSES KETOSES Trioses (C3H6O3) Glyceraldehyde Dihydroxyacetone Tetroses (C4H8O4) Erythrose Erythrulose Pentoses (C5H10O5) Ribose Ribulose Hexoses (C6H12O6) Glucose Fructose Heptoses (C7H14O7) Glucoheptose Sedoheptulose  Both the classifications i.e. number of carbon atoms and nature of functional groups may be combined to classify the sugar. For example glycerose (glyceraldehyde) is an aldotriose, ribose is an aldopentose and fructose is a ketohexose.  The smallest monosaccharide are glyceraldehyde and dihydroxyacetone for which n=3 (where n= number of carbon atoms)  Structure of Glyceraldehyde Structure of dihydroxyacetone  Properties of monosaccharide  Monosaccharide exists in both as straight chain structure and cyclic structure. Sugars with five membered rings and with six membered rings are most stable. Cyclic structures are the result of hemiacetal formation by intermolecular reaction between carbonyl group and a hydroxyl group.  α - D-Glucose β–D- Glucose  1. Chiral center All monosaccharide except dihydroxy acetone contain one or more asymmetric (chiral) carbon atoms thus are optically active isomers (enantiomers). A molecule with n chiral centers can have 2n stereoisomers. Glyceraldehyde with one chiral centre has 21 =2 and glucose with four chiral centers, have 24=16 stereoisomers.  2. D and L isomerism One of the two enantiomers of glyceraldehyde is designated the D isomers and the other L isomers. The orientation of the –OH group that is most distant from the carbonyl carbon determines whether the sugar belongs the D or L sugars. When the –OH group on this carbon is on the right the sugar is D isomers, when is on the left the sugar is L isomers. Most of sugars present in biological system are D sugars.  3. Anomers In aqueous solution all monosaccharide with five or more carbon atoms in the backbone occur as cyclic forms. Formation of cyclic structure is result of a reaction between alcohols and aldehydes or ketones to form derivatives called hemiacetal or hemiketals. The ring structure of monosaccharide are either similar to pyran (a six membered ring) or furan (a five membered ring). In linear form of monosaccharide, which is in equilibrium with the cyclic forms, the anomeric carbon is easily oxidised, making the sugar a reducing sugar. D-glucose exists in solution as an intramolecular hemiacetal in which the free –OH at C-5 has reacted with aldehyde C-1 producing two anomers called α and β. D-fructose also forms hemiketal in which –OH at C-5 has reacted with keto at C-2 producing two anomers called α and β  ASSIGNMENT  Isomeric forms of monosaccharide that differ only in their configuration about the hemiacetal or hemiketal carbon atom are called anomers, and the carbonyl carbon atom is called the anomeric carbon. The interconversion of α and β anomers in aqueous solution is called mutarotation, in which one ring form opens briefly into the linear form, then closes again to produce β anomers. Thus, a solution of β-D- glucose and solution of α-D- glucose eventually form identical equilibrium mixtures having identical properties. This mixture consists of about one third of α-isomers and two third β-D-glucose.  Epimers Isomers having different configuration of –OH only at one carbon atoms are known as epimers. The most important epimers of glucose are mannose (epimers at C-2) and galactose (epimers at C-4) 2. Disaccharides  Two monosaccharides can join covalently when –OH group of one sugar reacts with the anomeric carbon of other to form a disaccharide. The bond formed between the monosaccharide is called O- glycosidic bond.  The reaction represents the formation of hemiacetal. Glycosidic bond can be hydrolysed by acid but not cleave by the base. Thus disaccharides can be hydrolysed to yield their constituent monosaccharides by boiling with dilute acid.  Hydrolysis of sucrose yields a mixture of glucose and fructose called ‘invert sugar’ because fructose is strongly levorotatory and change (converts) the weaker dextrorotatory action of sucrose. Some common Disaccharides  When the anomeric carbon is involved is a glycosidic bond, the easy interconversion of linear and a cyclic form is prevented. Because the carbonyl carbon can be oxidised only when the sugar is in its linear form, formation of a glycosidic bond makes the sugar non-reducing.  The end of chain in disaccharide and polysaccharide with free anomeric carbon is called the reducing end. The disaccharides maltose contains two D-glucose residues joined by glycosidic linkage and lactose is made up of D-galactose and D-glucose residues. In both the disaccharides C-1 (anomeric carbon) of one sugar makes glycosidic bond with C-4 of another.  Since the anomeric carbon of one sugar residue is available for oxidation, maltose and lactose are reducing disaccharides. Whereas in sucrose, the two monosaccharides glucose and fructose are linked via their anomeric carbon (C1 of glucose and C2 of fructose form glycosidic bond), no anomeric carbon atom is free, therefore sucrose is a non-reducing sugar The stability of sucrose makes it ideal molecule for storage and transport of energy in plants. Trehalose, like sucrose is a non-reducing sugar is major constituent of circulating fluid of insects, serving as energy storage compound. 3. OLIGOSACCHARIDES  Oligosaccharides consist of carbohydrates with 3 to 10 molecules of monosaccharides. RAFFINOSE α-Glucose + β-Galactose + β-Fructose STACHYOSE 2 Galactose + Glucose + Fructose VERBASCOSE 3 Galactose + Glucose Fructose 4. Polysaccharides & TYPES  Polysaccharides are polymers composed of ten or more monosaccharide units. These monosaccharide units are joined together by glycosidic linkages.  Polysaccharides made up of a single type of monosaccharide units are called as homopolysaccharides, whereas polysaccharides composed of two or more types of monosaccharides are called heteropolysaccharides.  Some polysaccharides are made up of sugar derivatives like, for example glucosamine, a glucose derivative, is the repeating monosaccharide in chitin. Polysaccharides differ from each other in the type of repeating monosaccharide unit, in the degree of branching, and in the type of glycosidic linkage between the monomeric units. Homopolysaccharides  Homopolysaccarides yield a single type of monosaccharide on hydrolysis. They serve as both storage (e.g., starch, glycogen, dextran, inulin) and structural (e.g., cellulose, chitin, xylan, pectin) polymers.  Storage polysaccharides serve as storage form of monosaccharide that is used as fuels. Starch is an example of storage polysaccharide in plants and glycogen is the storage polysaccharide in animals.  Structural polysaccharides such as cellulose and chitin serve as structural elements in plant cell wall and animal exoskeleton, respectively.  Examples of Homopolysaccharides are given below  Assignment  Draw the structure of Starch and Amylopectin indicating the glycosidic bond involved as well as the reducing and non reducing end.  Write short note on the following Homopolysaccharides  Starch  Glycogen  Cellulose  Inulin  Dextran  Xylan  Chitin  Pectin Heteropolysaccharides  Heteropolysaccarides yield a mixture of monosaccharide on hydrolysis. They are present in extracellular matrix of plants, animals and bacteria.  Heteropolysaccharides, unlike the homopolysaccharides, provide extracellular support for organisms. Heteropolysaccharides in the extracellular space of animal tissues form a matrix that holds individual cells together and provides shape, support and protection to the cells and tissues.  Examples include Heteropolysaccharides  Agar Agar is gelatinous polysaccharide produces in cell wall of marine red algae such as species of Gelidium, Gracilaria, Gigartina etc. It is a mixture of sulphated heteropolysaccharides made up of D-galactose and L- galactose derivatives either-linked between C3 and C6. Agarose is the agar component with fewest charged groups (sulfates, pyruvates). It has molecular weight of 150000. If agar and agarose are dissolved in hot water they form sol which upon cooling sets to a gel.  Agarose gels are used as inert support for the electrophoretic separation of nucleic acids. Agar is used to form a surface for the growth of bacterial and plant tissue cultures.  Peptidoglycan Peptidoglycan constitutes the rigid component of bacterial cell wall. It is heteropolysaccharide of alternating β(14) linked N-acetyl-D-glucosamine and N-acetyl muramic acid residues. The linear polysaccharide chains are cross linked by short peptides attached to N-acetyl muramic acid. Cross linking by peptide turns the polysaccharide chains into a strong sheath that envelops the entire cell and prevent osmotic rupture of the cell.  Lysozyme, which is an enzyme present in human tears kills bacteria by hydrolyzing the β(14) glycosidic linkage of peptidoglycan. Some bacteriophages also produce lysozyme for releasing themselves from the host.  An antibiotic penicillin kill bacteria by preventing the cross linking process and therefore making the cell wall weak to resist osmotic lysis. PHYSICAL PROPERTIES OF CARBOHYDRATES  Monosaccharides, Disaccharides and Polysaccharides are Odorless  CHO like monosaccharide are colorless crystalline solids  CHO comprising monosaccharides, oligosaccharides and polysaccharides are solid at room temperature  CHO like monosaccharide are soluble in water but insoluble on organic solvents e.g chloroform. However polysaccharides such as cellulose are insoluble in water but will dissolve in ammoniacal solution of cupric salts CHEMICAL PROPERTIES OF CARBOHYDRATES  Osazone formation: Osazone are carbohydrate derivatives formed when sugars are reacted with an excess of phenyl hydrazine e.g Glucosazone  Oxidation: Sugars readily undergo oxidation to produce carboxylic acids and hence termed reducing sugars. Aldehydes are easier to oxidize because they have an open C= O bond. However, ketones can be oxidized only if they tautomerize to form an aldose first.  Reduction : The C= O groups in open chain forms of CHO can be reduced to alcohols by sodium borohydride NaBH4 or catalytic hydrogenation. The products are known as alditols.  Hydrolysis: CHO undergo hydrolysis to produce α and β isomers. The –OR bond at the anomeric carbon hydrolyses to form a –OH bond.  Glycoside formation: CHO form glycosides when the anomeric hydroxyl group undergoes condensation with the hydroxyl group of another CHO molecule, eliminating a water molecule.  Benedict’s test: Reducing sugars when heated in the presence of an alkali gets converted to powerful reducing specie known as enediols. When Benedict’s reagent solution and reducing sugar are heated together, the solution changes its color to orange-red or brick red FUNCTIONS OF CARBOHYDRATES  Most abundant dietary source of Energy { 4KCal/gm of CHO oxidized).  They are the storage form of energy in the form of glycogen {Glycogen is stored in the liver and Muscle}  Non-digestible CHO e.g cellulose, agar, gum and pectin serve as dietary fibers  They are precursors of organic compounds e.g ribose, deoxyribose (DNA, RNA), glycolipids,glycoproteins and proteoglycans.  They participate in the structure of cell membrane; The main constituents of cell membrane is glycolipids and glycoproteins.  They play a role in lubrication, cellular intercommunication and immunity e.g Hyaluronic acid is importance in lubrication of joints  They serve as structural components e.g glycosaminoglycans in human, cellulose in plants, chitin in insects.  Involved in detoxification e.g gluconic acid is involve in detoxification process. Quantitative analysis and test for carbohydates  Molisch’s Test  Molisch’s test makes use of Molisch’s solution (contains α-naphthol in 95% alcohol) and concentrated H2SO4.  Principle: Molisch’s test detects the carbohydrate presence, which principle is based upon the dehydration reaction. The carbohydrates in the sample get dehydrated into aldehyde by the addition of concentrated H2SO4.  The aldehyde formed is either furfural (produced by the dehydration of pentoses or pentosans) or hydroxymethylfurfural (produced by the dehydration of hexoses or hexosans). Then, the α-naphthol in the molisch reagent reacts with the aldehyde and develops a Procedure: 1.Take a sample and add water to make 2 ml of solution. 2.Add 2 drops of Molisch’s reagent. 3.Then, pour 5 ml of concentrated H2SO4 from the side of the inclined test tube. 4.Observe the colour change in the tube. Result interpretation: Positive result: Violet ring appears at the junction of two layers, i.e. between the sugar and acid. Negative result: Green or brown colour appears.  Benedict’s Test  It is also used for the analysis of carbohydrates as reducing sugars. Reducing sugar consists of a free aldehyde or ketone group. Benedict’s test makes use of Benedict’s solution as a chemical reagent, which contains copper sulphate, sodium citrate, and sodium carbonate with a pH of 10.5.  Principle: The reducing sugar will reduce into the enediols by reacting with an alkaline reagent, i.e. Benedict’s solution. The reducing sugar gives a green to brick-red colour precipitate depending upon the sugar concentration. The colour change is due to the reduction reaction of copper (II) to copper (I) in the solution that develops a red-coloured precipitate. Procedure: 1.Prepare 1 ml of a solution by adding a test sample and water. 2.Add 2 ml of Benedict’s solution to the test tube. 3.Heat the solution in the water bath for 3 minutes. 4.Observe the test tube for any colour changes. Result interpretation: Positive result: Green to brick-red precipitate forms. Negative result: Colour remains unchanged.  Fehling’s Test  This method is also useful for the analysis of reducing sugars. It makes use of Fehling’s solutions A and B as chemical reagents. Fehling’s solution A contains copper sulphate pentahydrate in 1 L of distilled water. Fehling’s solution B contains potassium sodium tartrate and sodium hydroxide in 1 L of distilled water.  Principle: In Fehling’s test, a reduction reaction occurs between the aldehyde or keto groups of the reducing sugar and the alkaline cupric hydroxide that later reduces into cuprous oxide. This cuprous oxide gives a brick-red coloured precipitate in the solution. Procedure: 1.Add 2 ml of the test sample to a test tube. 2.Then, add Fehling’s A and B solutions in equal proportion to the above sample. 3.After that, place the test tube in a hot water bath for 3 minutes. 4.At last, observe the colour change in a test tube. Result Interpretation Positive result: Brick-red precipitate forms. Negative result: Colour remains unchanged.  Barfoed’s Test  This method is also used for the analysis of monosaccharide- reducing sugars. Barfoed’s test makes use of Barfoed’s solution, which contains copper acetate in the dilute acetic acid with a pH of 4.6.  Principle: In Barfoed’s test, the reducing monosaccharide is oxidized by the copper ion in the solution to form a carboxylic acid and copper (I) oxide, which results in the formation of a red-coloured precipitate.  Procedure: 1. Add 1 ml of the test sample to a test tube. 2. Then add 2 ml of Barfoed’s solution to the test tube. 3. Place the tube in a water bath for boiling for up to 1 min. 4. At last, observe the colour change in a test tube. Result interpretation: Positive result: Gives red-coloured precipitate. Negative result: Colour remains unchanged.  Bial’s Test  It is most commonly used for the detection of pentose sugars or pentoses. By this method, we can also differentiate pentoses from hexoses by means of colour change. Bial’s test makes use of Bial’s solution, which contains orcinol, hydrochloric acid and ferric chloride.  Principle: Here, the pentoses react with the Bial’s reagent and convert it into furfural derivatives due to dehydration by HCl. Then, orcinol and furfural condense in ferric ion presence and develop a green-coloured compound. Procedure: 1.Take 2 ml of the test solution in the test tube. 2.Then, add 5 ml of Bial’s reagent. 3.Heat the solution in a water bath for 1 min. 4.Cool and observe the solution for any colour change. Result interpretation: Positive result: Gives green-coloured precipitate. Negative result: The solution remains unchanged.  Seliwanoff’s Test  It is most commonly used for the detection of Keto- sugars or Ketoses. By this method, we can also differentiate Ketoses from Aldoses by means of colour change. Seliwanoff’s test uses Seliwanoff’s reagent that contains 0.5 g of resorcinol per litre of 10% HCl.  Principle: In Seliwanoff’s test, ketoses react with the HCl of Seliwanoff’s reagent and yield furfural derivatives due to dehydration. Then, resorcinol and furfural react to give a deep red colour to the solution. Procedure: 1.Take 1 ml of test solution in the test tube. 2.Then, add 3 ml of Seliwanoff’s reagent. 3.Heat the solution in a water bath for 1 min. 4.Cool and observe the solution for the colour change. Result interpretation: Positive result: Gives a deep red colour to the solution. Negative result: Colour remains unchanged.  Iodine Test  It is widely used for the detection of starch in the solution. In an iodine test, iodine acts as an “Indicator”.  Principle: In the Iodine test, an iodine solution reacts with the starch (contains α-amylose and amylopectin polymers). This reaction between starch and iodine results in a starch-iodine complex, which gives a blue- black colour to the solution.  Procedure: 1. Take 2 ml of the test solution in the test tube. 2. Then, add 2 drops of iodine solution. 3. Observe the solution for the colour change. Result interpretation: Positive result: Gives blue-black colour to the solution. Negative result: Gives a brown-yellow colour to the solution. Colorimetric assays of carbohydrate The Anthrone Method  The Anthrone method is an example of a colorimetric method of determining the concentration of the total sugars in a sample.  Sugars react with the anthrone reagent under acidic conditions to yield a blue-green color. The sample is mixed with sulfuric acid and the anthrone reagent and then boiled until the reaction is completed. The solution is then allowed to cool and its absorbance is measured at 620 nm.  There is a linear relationship between the absorbance and the amount of sugar that was present in the original sample. This method determines both reducing and non-reducing sugars because of the presence of the strongly oxidizing sulfuric acid. This method is non-stoichiometric and so it is necessary to prepare a calibration curve using a series of standards of known carbohydrate concentration. The Phenol - Sulfuric Acid method  The Phenol - Sulfuric Acid method is an example of a colorimetric method that is widely used to determine the total concentration of carbohydrates present in foods.  A clear aqueous solution of the carbohydrates to be analyzed is placed in a test-tube, then phenol and sulfuric acid are added. The solution turns a yellow-orange color as a result of the interaction between the carbohydrates and the phenol.  The absorbance at 420 nm is proportional to the carbohydrate concentration initially in the sample. The sulfuric acid causes all non-reducing sugars to be converted to reducing sugars, so that this method determines the total sugars present. This method is non-stoichemetric and so it is necessary to prepare a calibration curve using a series of standards of known carbohydrate concentration.