experiment 5 guide

Experiment 5 Guide

Bulang, Del Gallego, Estrella, Guillermo, Jurilla, Lumaban

3BIO-6

Slide # 4Topic: Carbohydrates

• They can be polyhydroxy aldehydes or polyhydroxy ketones because they have either an aldehyde or a ketone group, along with –OH substituted carbons in a chain

• Polyhydroxy aldehydes are called aldoses. Polyhydroxy ketones are called ketoses. The suffix “ose” means sugar

Slide # 5Topic: Monosaccharide

• Pentose– any of the class of simple

sugars whose molecules contain five carbon atoms

– such as xylose

• Cyclic forms of monosaccharides result from the ability of their carbonyl group to react intramolecularly with hydroxyl group

• Hexose– any of the class of simple

sugars whose molecules contain six carbon atoms

• D-galactose and D-glucose – differ only in the configuration

of the –OH group and –H group on carbon 4

– Epimers (diastereomers that differ only in the configuration at one chiral center

• D-fructose – also known as levulose and fruit sugar

Slide #6Topic: Disaccharide

• A monosaccharide that has cyclic forms (hemiacetal forms) can react with an alcohol to form a glycoside (acetal) = produce a disaccharide

• In disaccharide, one of the monosaccharide reactants function as a hemiacetal, and the other functions as an alcohol

• Acetal - an organic compound formed by the condensation of two alcohol molecules with an aldehyde molecule

• Glycosidic linkage – bond in a disaccharide resulting from the reaction between hemiacetal carbon atom – OH group of one monosaccharide and and –OH group on the other monosaccharide

• Maltose – malt sugar; produced whenever polysaccharide starch breaks down; 2 D-glucose units, one of which must be a D-glucose

Slide #6Topic: Disaccharide

• The glycosidic linkage between two glucose units is an a(1-4) linkage

• The two –OH groups that form the linkage are attached, respectively to carbon 1 of the first glucose unit and to a carbon 4 of the second

• Maltose – reducing sugar because the glucose unit on the right has a hemiacetal carbon atom. Thus this glucose unit can open and close: it is in an equilibrium with its open-chain aldehyde form. This means there are three forms of the maltose molecule: a-molecule, β-molecule, and the open chain form

• In solid state, the β-form is dominant• Hydrolysis of maltose produces two

molecules of D-glucose• Acidic conditions or the enzyme

maltase is needed for the hydrolysis to occur

• Lactose – made up of a β—D-galactose unit and a D-glucose unit joined by a β(1-4) glycosidic linkage

• The glucose hemiacetal center is unaffected when galactosebonds to glucose in the formation of lactose, so lactose is a reducing sugar (the glucose ring can open to give an aldehyde)

Slide # 6Topic: Disaccharide

• Lactose – major sugar found in milk; milk sugar

• Epimerization of glucose glucose yields galactose, and then the β(1-4) linkage forms between a galactose and a glucose unit

• Souring of milk is caused by the conversion of lactose to lactic acid by bacteria in milk

• Epimerization – a chemical process where an epimer is tranformed into its chiral counter part

• Epimer - each of two isomers with different configurations of atoms around one of several asymmetric carbon atoms present

• Sucrose – table sugar; most abundant of all disaccharides; a-D-glucose and β-D-fructose

• The glycosidic linkage is not a (1-4) linkage, instead it is an a. β(1-2) glycosidic linkage

• The –OH group on carbon 2 of D-fructose (the hemiacetal carbon) reacts with the –OH group on carbon 1 of D-glucose (the hemiacetal carbon)

• Sucrose, unlike maltose and lactose is a non-reducing sugar

• No hemiacetal is present in the molecule, because the glycosidic linkage involves the reducing ends of both monosaccharides

Slide # 6Topic: Disaccharide

• Sucrose exist in only one form – there are no a and β isomers, and an open chain form is not possible

• Sucrase – enzyme n9eeded to break the a. β(1-2) linkage in sucrose

• Sucrose hydrolysis (digestions) produces an equimolar mixture of glucose and fructose called invert sugar

Slide # 7Topic: Polysaccharide• Monomer - a molecule that can be

bonded to other identical molecules to form a polymer

• Polysaccharides are also called glycans

• How to distinguish polysaccharide– Based on repeating units

– Homopolysaccharides – starch, glycogen, cellulose, and chitin

– Heteropolysaccharide– Length of the polymer chain– Type of glycosidic linkage– Degree of branching of the

polymer chain

• Starch – glucose monosaccharide units

• If excess glucose enters a plant cell, it is converted to starch; when cell cannot get enough glucose, it hydrolyzes starch to release glucose

• Two different polyglucose polysaccharides can be isolated from most starches: amylose and amylopectin

• Amylose – straight-chain glucose polymer (15%-20%)

• Amylopectin – branched glucose polymer (80%-85%)

Slide # 7Topic: Polysaccharide

• Amylose’s non-branched structure, the glucose units are connected by a(1-4) glycosidic linkages

• Amylopectin – high degree of branching in its polyglucose structure; the branch points involve a(1-6) linkages

• Because of the branching, amylopectin has a larger average molecular mass than the linear amylose

• All of the glycosidic linkages in starch (both amylose and amylopectin) are of the a type

• In amylose, they are all (1-4): in amylopectin, both (1-4) and (1-6) linkages are present

• Because both types of linkages can be broken through hydrolysis within the human digestive tract, starch has nutritional value for humans

• Iodine is often used to test for the presence of starch in solution

• Starch-containing solutions turn a dark blue-black when iodine is added

• As starch is broken down through acid or enzymatic hydrolysis to glucose monomers, the blue-black color disappears


• Glycogen – animal starch; structure similar to amylopectin: all glycosidic linkages are of the a type, and both (1-4) and (1-6) linkages are present

• Glycogen and amylopectin differ in the number of glucose units between branches and in the total number of glucose units present in molecule

• Glycogenesis – formation of glycogen

• Glycogenolysis - decomposition of glycogen

• Structural poysaccharide – serves as a structural element in plant cell walls and animal exoskeleton

• Cellulose – unbranched glucose polymer like amylose but differ in in that the glucose residues present in cellulose have a beta-configuration whereas the glucose residues in amylose have an alpha-configuration.

• the glycosidic linkages in cellulose are therefeore β(1-4) linkages rather than a(1-4) linkages


• The difference in glycosidic linkages type causes cellulose and amylose to have different shapes

• Amylose tend to have a spiral-like structures whereas cellulose molecules tend to have linear structures

• The linear (straight-chain) cellulose molecules, when aligned side by side, become water-insoluble fibers because of interchain hydrogen bonding involving the numerous hydroxyl groups present

• Cellulose is not a source of nutriotion for human beings because humans lack enzymes capable of catalyzing the hydrolysis of β(1-4) linkages in cellulose

• Even animals lack the enzymes for cellulose digestion

• Intestinal tracts of animals such as horses have bacteria that produce cellulase, enzyme that can hydrolyze cellulose β(1-4) linkages and produce free glucose from cellulose

Slide # 8Topic: Reducing / non-reducing sugars

• A reducing sugar is a carbohydrate that gives a positive test with Benedict’s solutions

• Aldoses act as reducing agents in such reactions, they are called reducing sugars

• Under the basic conditions associated with Benedict’s solutions, ketoses are also reducing sugars

• In this situation, ketose undegoes a structural arrangement that produces an aldose and the aldose then reacts

• Thus all monosaccharides, both aldoses and ketoses, are reducing sugars

SLIDE #Topic: Molisch Test

• Amylose– A linear or non-branches polymer of glucose– The glucose units are joined by a a-1-4

glucosidic linkages– Exists in coiled form and each coil contains six

glucose residues– Amylose (unbranched) + amylopectin

(branched) = starch

• Cellulose– Composed of glucose units joined together in

the form of the repeating units of the disaccharide cellobiose w/ numerous cross linkages

– Insoulble structural homoglycan of plant fiber– Main constituent of the cell wall of plant cells– Made of long, unbranched linear chains of

300-3000 Beta-glucose units, joined by Beta-1,4-glycosidic bonds

– Is non-reducing and us nit digested by enzymes of the human digestive system

• Glycogen– Storage form of carbohydrates in animals– Stored in liver and muscles– Made up of many unbranched linear chains,

each having 10-15 glucose units linked by alpha-1,6-glycosidic linkages to form highly branched glycogen molecule

– Forms a colloidal solution in water and is non-reducing

– Gives a red-violet color w/ iodine– MORE BRANCHED than starch


• To detect the presence of carbohydrates• Positive: purple/violet ring interphase

b/w the acid and test layers• Carbohydrates are dehydrated (loss of

water) by conc. H2SO4 to form hydroxymethylfurfural (an aldehyde) will react with (condenses with) two molecules of alpha-naphthol (Molisch reagent) to yield a purple condensation product

• Molisch reagent– 10 % a-naphthol (C10H8OH) dissolved in

ethanol (C2H5OH)– To test all types of carbohydrates

• Reason for obtaining a positive result: alpha-naphthol is not a selective compound and it would react to every sugar

• Positive for: Monosaccharides give a rapid positive test. Disaccharides and polysaccharides react slower.

• This purple color is produced by almost all carbohydrates, with the exception of sugar alcohols (alditols) and 2-amino-2-deoxy sugars.

• The conc. H2SO4 hydrolyzes the glycosidic bonds that are present in oligosaccharides and polysaccharides. The resulting monosaccharides are then dehydrated to produce the furfural derivatives.

• The furfural compounds condense with alpha-naphthol, creating the purple color.

• Pentoses and hexoses form five member oxygen containing rings (furfural) on dehydration

• Positive = presence of carbonyl grp• Furfural derived from the dehydration

of pentoses• 5-hydroxymethyl furfural produces

from hexoses and hexosans


• A green ring alone is seen even in the absence of carbohydrate and it indicates an excess of alpha-naphthol

• Molisch test universal test for carbihydrates and their derivatives

• Conc. H2SO4 not only catalyses the dehydrations of sugars but also brings about the hydrolysis of glycosidic bonds of oligosaccharides and polysaccharides

SLIDE #Topic: Anthrone Test

• For the determination of non-reducing sugars

• Reagent: 2% Anthrone reagent (0.3 g of Anthrone to 100mL conc Sulfuric acid)

• Anthrone– Tricyclic aromatic ketone

• Concentrated sulphuric acid hydrolyses the glycosidic bonds of carbs and produces monsaccharides which are dehydrated by the concentrated acid and form furfural (liquid aldehyde C5H4O2) and their derivatives which react with anthrone to form a blue-green complex

• Carbohydrates + concentrated sulphuric acid -> furfurals + anthrone -> blue-green complex

•

SLIDE #Topic: Iodine Test

• To test for the presence of starch

• Reagent: 0.01 M iodine solution (180 mg of KI in 100 mL water. 130 mg of iodine)

• Starch – Polysaccharide used for energy

storage– Composed of amylose (straight chain

– forms helices where iodine can assemble) and amylopectin (branched, shorter helices – iodine cannot assemble), polymers of alpha-D-glucose

Reaction of Iodine with starch• Deep blue/violet color: reaction of

starch (amylose helices) and iodine -> formation of adsorption complex (adsorbent [starch] + adsorbate [iodine]), formation of poly iodide chains

• During heating: clear colorless solution - breaking of the adsorption complex with heat

• Cooling: deep blue/violet color - reformation of adsorption complex; remained colorless - absence of reformation; mono and disaccharides are too small to trap the iodine molecules; do not show color change in presence of iodine

• Color depends on length and organization of chains

SLIDE #Topic: Iodine Test

• The blue color arises when the electrons of the entrapped iodine molecules interact with the electrons of the starch molecule and the resulting complex absorbs visible light (appears dark)

• Amylose a linear chain component of starch gives a deepblue color

• Amylopectin a branched component of starch gives a purple color

• Glycogen gives a reddish brown color

• Cellulose gives no color w/ iodine

•

SLIDE #Topic: Mucic Acid Test

• Procedure• Fumes generated while boiling – acid

fumes from nitric acid

• Principle: Monosaccharides, upon treating with strong oxidizing agents such as nitric acid, yield saccharic acids. The saccharic acid obtained after the oxidation of galactose is insoluble and separates as gritty crystals. The acid derivative is known as mucic acid.

• Mucic Acid Test: to identify galactose– Galactose is classified as a monosaccharide, an

aldose, a hexose, and is a reducing sugar.• Galactose is a monosaccharide. When combined

with glucose, through a dehydration reaction, the result is the disaccharide lactose

• The position of the -OH group on the carbon (#4) is the only distinction between glucose and galactose

• Glucose is defined as the -OH on C # 4 in a horizontal projection in the chair form, (down in the Haworth structure)

• Galactose is defined as the -OH on C # 4 in a upward projection in the chair form

• Reagent: HNO3 (nitric acid)– Oxidizes primary functional groups to

carboxylic groups


• Oxidation of terminal groups of aldoses– aldehyde primary alcohol groups are

converted to carboxylic groups

• Oxidation of most monosaccharides by nitric acid yields soluble dicarboxylic acids. However, oxidation of galactose yields an insoluble mucic acid.

• Heating of HNO3 with an aldose sugar gives a dicarboxylic acid– Formation of mucic acid/galactaric acid

• Mucic Acid– Melts at 213 °C– Insoluble in cold aqueous solution and

forms crystals– Saccharic acid– THREE principal derivatives of Aldoses:

• Aldonic Acids( alcohol Acids)-the aldehyde is oxidized

• Uronic Acid-those in which the primary alcohol group is oxidized

• Saccharic Acids-both the aldehyde and primary alcohol groups are oxidized

• Results: crystals formed in galactose and lactose– Lactose will also yield a mucic acid, due

to hydrolysis of the glycosidic linkage between its glucose and galactose subunits


SLIDE #Topic: Benedict’s Test

• Procedure: After heating, cool the contents in a test tube rack. Do not hasten cooling by immersion in cold water. WHY?!?

• Benedict’s test: to detect reducing sugars– reducing sugars: meaning that a molecule of

this sugar can react with other molecules by giving electrons to them – reduction

• The sugar sucrose is a non-reducing sugar. It is formed in a condensation reaction making a glycosidic bond between a glucose molecule and a fructose molecule.– Difference in the glycosidic bond formed in

sucrose prevents the sucrose from reacting with the reagent

• Brick red ppt = copper (I) oxide• Benedict’s Reagent

– A copper compound that will oxidize only aldehyde groups (aldoses) and not alcohols

– If you consider cyclic forms of carbihydrates, hemiacetals give positive tests while acetals give negative tests. The reason for this is that the cyclic form interconverts (is in equilibrium) with the linear form that contains an aldehyde

Benedict’s reagent contains:• Copper sulphate – dissociate to give

sufficient cupric ions (as cupric hydroxide) for reduction reactions to occur; furnishes curpic ions (Cu++) in solution

• Sodium citrate – keeps hydroxide in solution w/o getting precipitated; prevents the precipitation of cupric ions as cupric hydroxide by forming a loosely bound cupric-sodium citrate complex w/c on disassociation gives a continuous supply of cupric ions

• Sodium carbonate (Na2CO3) – make the pH of the medium alkaline


• Carbohydrates with a free aldehyde (RCOH) or keto (RCOR) group have the ability to reduce various metallic ions

• Cupric ions are reduced to cuprous ions by the enediols from the sugars in the alkaline medium of the reagent– Benedict’s reagent provides an

alkaline/basic medium– Enediol: powerful reducing agents;

reduce blue cupric hydroxide to insoluble yellow to red cuprous oxide

• Principle: Carbohydrates with free aldehyde or ketone groups have reducing properties. The mild alkali, sodium carbonate, converts glucose into enediol.This enediol reduces copper sulphate to cuprous hydroxide that is unstable and decomposes on boiling to cuprous oxide.


• Sensitive up to 0.1-0.15 gm% of sugar solution– gm% - grams percent or grams per

deciliter

• Color of precipitate depends on the concentration– Blue – absence of reducing sugar– Green – up to 0.5 gm%– Yellow – >0.5 to 1.0 gm%– Orange – 1.0 to 2.0 gm%– Brick Red – ≥2 gm%

• Semi-quantitative test– Because it gives an idea about the sugar

solution’s concentration

• All monosaccharides (glucose, fructose, galactose, mannose) will give a positive reaction

• Disaccharides such as lactose and malose will give a positive reaction

• For negative results, the non-reducing sugar test can be done– Sucrose is broken down into

monosaccharides– Boil the sample with hydrochloric acid –

this hydrolyses any sucrose present, splitting sucrose molecules to give glucose and fructose

– Cool the solution and neutralise it by adding sodium carbonate solution (an alkali solution)

– Carry out the reducing sugar test (Benedict’s test) again

• If you test a sample for both reducing and non-reducing sugars and:– the colours produced are the same: there

are no non-reducing sugars– If the colour in the non-reducing sugar

test is more towards the red end of the spectrum: non-reducing sugars are present


SLIDE #Topic: Barfoed’s Test

Is there a difference between Benedict’s test and Barfoed’s Test?• Yes, yes there is. Barfoed’s test happens in a

more acidic medium while Benedict’s test happens in an alkaline medium. In an acidic medium monosaccharides enolize much more readily than disaccharides.

If monosaccharides reacts faster, does this mean disaccharides also react but in a slower pace?• Yes. Disaccharides react too but

monosacchrides and disaccharides have different time frames since disaccharides are more complex sugars, also, disaccharides are weaker reducing agents (element or compound in a reduction-oxidation (redox) reaction that donates an electron to another species) compared to monosaccharides. In about 1-3 minutes the monosaccharides will react forming the brick red ppt while disaccharides react at around 10 minutes.

What’s the principle?• This test is based on the

reducing nature of sugars due to their having an aldehyde or ketone group which reduce the cupric hydroxide formed in acidic medium to red colored cuprous oxide. This test which is specific for monosaccharides is based on copper reduction, but it differs from Benedict’s reactions in term of the reaction medium. Barfoed’s test is conducted in acidic conditions, whereas the other test is in alkaline medium. As disaccharides are weak reducing agents, they do not reduce cupric ions in Barfoed’s solution under acidic conditions, whereas monosaccharides, being strong reducing agents, are capable of reducing copper ions in acidic conditions. However, on prolonged heating, acetic acid present in the reagent hydrolyzes disaccharides to monosaccharides, which respond to the test.

SLIDE #Topic: Barfoed’s Test

• Barfoed’s reagent– Contains cupric acetate in acetic

acid• Monosaccharides are stronger

reducing agents in acidic medium as compared to disaccharides

SLIDE #Topic: Bial’s Orcinol Test

• What happens?

SLIDE #Topic: Bial’s Orcinol Test① Strong acidity and temperature conditions

permits breaking of glycosidic bonds (a type of covalent bond that joins a carbohydrate (sugar) molecule to another group, which may or may not be another carbohydrate) between saccharide units, hence, the latter undergo their penta/hexatomic ring opening

② The opened ring loses water molecules and convert themselves to furfural or its derivatives

③ Saccharides that contain 5 Carbon atoms or pentoses cannot form polyatomic rings greater than 5 atoms so pentoses generate furfural and give a positive result

④ Furfural acts as an electrophilic agent towards orcinol, it is a better electrophilic agent than its derivatives

⑤ Among furfural derivatives, you may often meet 5-hydroxymethylfurfural (derivatives depends on what sugar sample is present). It is produced by saccharides having more than 5 carbon atoms but still containing pentatomic rings (i.e. Fructose)

IN SHORT: Pentoses, upon dehydration, forms a product called furfural, which is condensed by orcinol to form a blue-green colored solution.

• Furanose (a collective term for sugars that have a chemical structure that has a 5 member ring system [4 Carbon atoms and 1 Oxygen atom]) saccharides can either be a pentose or non-pentose.

• Pentose furanose (xylose) reacts faster and gives off the blue-green solution, thus, a positive result. NOTE: Fructose is a furanose non-pentose. Glucose is a pyranose.

What is a furfural? (See figure 1 for structure)• It is a heterocyclic aldehyde (meaning it

is a cyclic compound that has atoms of at least two different elements as members of its ring). Upon air exposure it turns yellow.

SLIDE #Topic: Bial’s Orcinol Test

• Sensitive test for the detection of pentoses

• Used to differentiate pentose sugars from hexose sugars

• Bial’s reagent– 0.2% orcinol in conc. HCl acid

Slide # 73Topic: Seliwanoff’s Test

• Distinguish ketose and aldose via ketone/aldehyde functionality

• Ketose – if a sugar contains ketone

• Aldose – if a sugar contains aldehyde

• When heated, ketoses are more rapidly dehydrated than aldoses.

• Acid hydrolysis of polysaccharides and oligosaccharides yields simpler sugars followed by furfural

• The dehydrated ketose then reacts with the resorcinol to produce a deep cherry red color. Aldoses may react slightly to produce a faint pink color

• Fructose and sucrose are two common sugars which give a positive test.

• Sucrose gives a positive test as it is a disaccharide consisting of fructose and glucose.

Slide # 73Topic: Seliwanoff’s Test

• Ketose reacts more readily with seliwanoff's test because ketones have 3 carbon atoms whereas aldehyde has 2 hydrogen atoms and one carbon atoms.

• Sugar loses water on being heated with strong acids and forms furfural derivatives. Thus furfurals form colored complexes with phenolic compounds such as resorcinol

experiment 5 guide

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