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Mansoura University Pharmacognosy Department Carbohydrates
Carbohydrates make up the bulk of organic substances on earth and perform numerous roles in living things Structurally
A carbohydrate is an organic compound with the general formula Cm(H2O)n, that is, consists only of carbon, hydrogen and oxygen, with the last two in the 2:1 atom ra o. So The name carbohydrate means hydrate of carbon and derives from the formula Cm(H2O)n. Carbohydrates can be also described as polyhydroxy aldehydes, polyhydroxy ketones or compounds that can be hydrolysed to them.
Examples:
Glucose (blood sugar): C6H12O6, or alternatively C6 (H2O) 6 Sucrose (table sugar): C12H22O11, or alternatively C12 (H2O) 11
Exceptions:
1‐carbohydrates, Not obey the hydrate role: Deoxy sugars: Rhamnose C6H12O5, Digitoxose C6H12O4, Cymarose C7H17O4
2‐Compounds obey the hydrate role, but not carbohydrates:
Formaldehyde HCHO (CH₂O), Acetic acid C₂H₄O₂
Monosaccharides
A. Classification
I. According to function group
Aldoses: Monosaccharides with aldehyde group. e.g: glucose & xylose ketoses : those containing a ketone group. e.g: Fructose
II. According to number of oxygen‐ bearing carbon atoms
Bioses, trioses, pentoses, hexoses and heptoses containing 2, 3, 5,6, 7 oxygen carbon atoms, Pentoses and hexoses are the most common. Pentoses: e.g: xylose (Wood sugar), e.g Branched pentose: Apiose Hexoses: e.g: glucose (Dextrose, Grape sugar) Derivatives of hexoses: e.g: Deoxy sugar Deoxy sugar: Present in cardiac glycosides in which 1 oxygen atom is removed from carbon no. 2 or no. 6 to obtain 2‐deoxy sugar or 6‐deoxy sugar or 2,6 deoxy sugar e.g: 2,6‐deoxy sugar: Digitoxose C₆H₁₂O₄ e.g: 6‐deoxy sugar= Methyl pentose: rhamnose C₆H₁₂O₅ Disaccharides
Classified into: A. Reducing: (e.g: maltose, lactose). B. Non‐reducing: (e.g: sucrose).The linkage between 2 saccharides is through their anomeric carbons, so none of the reducing groups are free. Sucrose: 1‐2 linkage (aldose + ketose)
Qualitative identification of carbohydrates
General tests for carbohydrates
The following 2 reac ons are posi ve with all carbohydrates.
1‐ Molisch's test:
Notes:
1‐CHO: aqueous sol. If not sol, boil some solid with dil. H₂SO₄. 2‐Avoid shaking tube.
2‐ Sulfuric acid test:
Heating sugars with conc. H2SO4 results in dehydration and charring with decomposition of sugars. 3‐ Reduction of Fehling's solution
* Positive for monosaccharide and reducing disaccharides.
* It is composed of 2 solu ons mixed together just before use:‐
Fehling A: CuSO₄
Fehling B: Rochell salt (potassium sodium tartarate) + KOH * Needs hot alkaline medium.
* Red precipitate is due to Cu2O ppt.
4‐ Reduction of Barfoed's solution
* Barfoed's solution is composed of 5% cupric acetate/acetic acid.
* Positive for monosaccharides only, ‐ ve with reducing disaccharides.
* Heat on W.B. for 3 min., the result is red ppt. on the wall of tube.
5‐ Osazone formation:
Notes
1‐Before heating, you should have clear solution. 2‐Avoid sudden cooling as it may break the crystal, examine it without cover. 3‐Monosaccharide gives osazone on hot a er 15 min. of hea ng, while disaccharides crystallize of osazones is not so easy and take > 30 min. 4‐Osazone formation needs  hydroxy aldehyde or ketone, so it is positive with reducing sugar having free OH in alpha position. 5‐Glucose and fructose gives the same osazone; this is because osazone formation involves the reaction with C₁ and C₂ and the asymmetry in both carbon atoms is destroyed and the remaining configura ons of the last 4 groups are identical in both. 6‐Sucose does not give the osazone because the aldehydic group of glucose and ketonic group of fructose are united together and mask each other. 7‐Polysacchrides don't give osazone. Mechanism
During osazone formation, the hydroxyl group adjacent to the carbonyl group is oxidized to keto group which is then attacked by phenyl hydrazine to from osazone. Special tests
I. Tests for pentoses and methyl pentoses
1. Bial's test.
Bial's reagent: Freshly prepared Orcinol in Conc. HCl + traces of FeCl₃
Xylose: blue color (typical test: positive) Rhamnose: green color
N.B. Bial’s test can be used to distinguish pentoses from hexoses. In the presence of concentrated HCl, pentoses react to give furfural, whereas hexoses give 5‐hydroxymethyfurfural, as we have just seen. Furfural reacts with orcinol and ferric ions, which are present in Bial’s reagent, along with conc. HCl, to give a blue‐green color. Again, don’t worry about the exact reac on. Hexoses, which give 5‐hydroxyfurfural on dehydration, react with Bial’s reagent to give a brownish color. Di‐ and polysaccharides give the same results but at a much slower rate. 2. Aniline acetate paper test (Furfural test).
Mechanism:
Action of acids: The sugar molecule is very stable towards dilute acids, but with concentrated acids it is distorted, Conc. HCl reacts with aldoses giving different products:‐ i) With pentoses and methyl pentoses, it yields furfural and methyl furfural, which are steam volatile, therefore can be used for the estimation of pentoses. ii) With hexoses the reac on proceeds in 2 steps first hydroxymethyl‐
furfuraldehyde is obtained but this rapidly decomposes giving laevulinic acid which is not steam volatile, this affords a means for the estimation of pentoses in the presence of hexoses. 3. Phloroglucin / HCl test (For Pentoses). 4. Acetone test.
Special tests
II. Tests for Ketoses
1‐ Cobalt nitrate test:
or/ Violet color Violet or purple color * Positive for ketoses: fructose & sucrose
* Glucose: ‐ve ?
(2) Resorcinol test (Slwanoff's test):
* Positive for ketoses: fructose & sucrose, however glucose may give this test but after prolonged heating.
* Select the white crystals for better results
N.B. One can distinguish aldoses from ketoses based on their ability to form furfurals. Since ketoses form furfurals more rapidly than aldoses, ketoses immediately react with resorcinol, which is present in Seliwanov’s reagent along with HCl, to give a colored complex. Don’t worry about the exact reaction though. Aldoses generally exist in solution as pyranoses, whereas ketoses generally exist as furanoses, hence the ability of ketoses to rapidly dehydrate to yield furfurals: Special tests for some sugars
(1) Glucose = (Dextrose) = Grape sugar
1. Reduc on of Amm. AgNO₃ +ve with: glucose, maltose & lactose.
2. Ac on of NaOH.
3. Lead subacetate test.
4. Moisch 5. Fehling 6. Barfoed 7. Osazone
Liquid glucose
Liquid glucose: is the Product of the incomplete hydrolysis of starch.
It consists of glucose; maltose and dextrin.
(2) Fructose = Levulose = Fruit sugar
Used as a food for diabetics especially those with acute keto‐acidosis, as it is sweeter than glucose and produces less urinary secretion than it. 1. Molisch 2. Fehling 3. Barfoed 4. Osazone 5. Resorcinol test 6. Cobalt nitrate test
(3) Xylose = Wood sugar (Pentose)
1. Molisch 2. Fehling 3. Barfoed 4. Bial's test 5. Aniline acetate paper (Furfural test) 6. Acetone test (4) Rhamnose (Methyl pentose)
1. Molisch 2. Fehling 3. Barfoed 4. Bial's test 5. Aniline acetate paper (Fursural test) 6. Acetone test
7. Phloroglucin / HCl test
(5) Sucrose = Cane sugar = Beet sugar = Saccharose = sugar
1. Molisch 2. Fehling test before hydrolysis 3. Fehling test a er hydrolysis: Boil 3ml sugar sol. + 1 ml dil HCl, neutralize with sod. Carbonate powder till no eff., then boil the resulting sol. with equal vol. of Fehling A & B solutions, a red ppt is produced. 4. Resorcinol test 5. Cobalt nitrate test
6. Arsenic acid test: A red color is produced by adding drops of arsenic acid to a sucrose solution. (6) Lactose = Milk sugar
1. Molisch 2. Fehling 3. Osazone (cluster or tu s of needles) (7)Maltose = Malt sugar
1. Molisch 2. Fehling 3. Osazone (rose es of plates)
NOTES
I. Polysaccharides Homo‐polysaccharides (Starch and Dextrin)
A. Starch = Amylum in E.P.
Starch is used for energy storage in plants. It is found in all plant seeds and tubers and is the form in which glucose is stored for later use. Starch can be separated into two principal polysaccharides: amylose and amylopectin. Complete hydrolysis of both amylose and amylopectin yields only D‐glucose. Amylase
Cons tute ~ 25% of starch molecule.
Amylopectin
Cons tute ~ 75% of starch molecule. Partial water soluble
Water insoluble
Composed of straight chains of It is a branched molecule consisting glucose units joined by α‐1,4‐ of glucose units also joined by α‐
glycosidic bonds. 1,4‐glycosidic bonds (at the straight chain), and few by α‐1,6‐
glycosidic bonds (at the branching points). Blue color with I₂
violet/ purple color with I₂ Hydrolysis with maltase enzyme Hydrolysis with maltase enzyme yield maltose. yield maltose + dextrin.
N.B. This enzyme hydrolyze only α‐ 1,4‐linked glucose
Tests for identity
I. Microscopic examination: to detect the source of starch: maize, Corn or wheat starch. II. Iodine test
III. Test B. Dextrin
It results from incomplete hydrolysis of starch by enzymes (partial hydrolysis of starch). I. Classes Amylodextrin
+++ very high molecular weight + I₂
Blue
++ high molecular weight
+ I₂
Red
+ low molecular. weight
+ I₂
Colorless
Erythrodextrin
Achrodextrin
II. Reactions 0.5 gm dextrin + water, boil, cool, a gel is formed, dilute with water then apply the reactions: I2, Ethanol, Fehling, lead acetate tests as before with starch. II. Hetero‐polysaccharides
They can be divided into two distinct groups:
(1) Heteropolysaccharides without acid group, e.g.Neutral hemicellulose.
(2) Heteropolysaccharides contain acid group, e.g. Gums, Mucilages, Pectins, Acidic hemicellulose. The acidic heteropolysaccharides contain one or more uronic acid units in their molecular structure, they are sometimes called derived carbohydrates. A. Gums Item
source
Gum acacia (gum Arabic)
Natural
Gum tragacanth
Artificial gum Natural
Artificial
Composition
1‐ Arabin: Ca, K, Mg salts of arabic acid.
2‐ Oxidase enzyme: turns benzidine or tr. Guaiacum into blue. 1‐ Tragacanthin: water Dextrin
soluble part.
Incomplete starch hydrolysis.
2‐ Bassorin: poly methoxylated acid, Its complete hydrolysis H₂O insoluble part.
gives glucose. 3‐ Galacto‐uronic acid.
4‐ Pentosans, mucilage and starch. Solubility
Aqueous solution
Soluble (no gel formation)
levorotatory
Insoluble (gel) : Bassorin
levorotatory
Insoluble adhesive gel.
dextrorotatorty
Test
Pentose test
Benzid./H₂O₂ Gum acacia (gum Arabic)
Gum tragacanth
+ve +ve Blue/ greenish blue
‐ve Artificial gum ‐ve ‐ve I₂ test
Yellowish brown
Reddish brown
Pb(Ac)₂ No white precipitate
Minute scatered due to blue points
starch
white precipitate
Pb subacetate white precipitate
No white precipitate
‐ve Borax
precipitate
‐ve ‐ve white precipitate
Oxidase enzyme:
Aqueous solution + hydrogen peroxide + few drops of benzidine ; shake ; warm at 37°C.
If produce blue color or greenish blue, Oxidase enzyme is present,,, So it is gum Arabic (gum acacia).
B. Mucilage
1. Agar‐Agar
* It consists of Ca salt of sulfated polysaccharides.
* It can be resolved into 2 parts:
Agarose: It is the principle fraction responsible for the strength of agar.
Agaropectin: It is sulfated galactose. * Insoluble in cold water, Soluble in hot water.
Tests for identification
Boil 1 gm agar power in water to form gell, Then divide the sol. into different fraction to apply the followings reaction: 1. Cooling
Stiff gel.
2. I₂ Reddish Particles.
3. Hcl + BaCl₂
White precipitate.
4. HCl/heat; neutralize + Fehling test
Reduction (red precipitate) due to galactose. 5. Ruthenium test:
Red colored particles (due to its mucilage content).
Honey
It consists of dextrose, fructose, mannitol, sucrose/ in water.
Honey may be adulterated with:
1. Excessive sucrose or liquid glucose.
2. Admixture of commercial glucose & inverted sugar.
3. Too much water
Determination of adulteration:
1‐ Quantitative determination of reducing sugars in the sample before and after hydrolysis (% of reducing Sugar increases with sucrose adulteration).
2‐Test for Commercial and liquid glucose: It contains dextrin which reacts with I₂, to develop a reddish‐brown color. 3‐Test for invert sugar or artificial honey. Invert sugar: it is obtained from sucrose by acid hydrolysis with citric or tartaric acid hea ng to ~115°C. This hydrolysis results in the produc on of furfural derivatives (oxymethyl furfural) which is a decomposition product of fructose and can be tested as follows: Detection of invert sugar or artificial honey (Reactions of oxymethyl furfural) 1. Aniline acetate paper gives bright red color.
2. Resorcinol test gives pink color, a er 20 min. cherry red.
Quantitative estimation of carbohydrates
sugars
Estimated by
Titermetric
Colorimetric
Gravimetric
methods
methods
methods
I. Cupper reduction method This method can be applied for all reducing sugars:  All monosaccharide
e.g. Glucose; Fructose; galactose, mannose,….
 All reducing disaccharides
e.g. Lactose & Maltose
 Principle:
• This method is based on the reduction of known volume of standardized fehling's solution by titrating it directly against the reducing sugar solution. • Such sugars are capable of reducing copper sulfate in hot alkaline medium to cuprous oxide, which forms a red precipitate and the blue color of the solution disappears.  Procedure:
1‐ Add 5 ml of each Fehling A and B (Bulb pipe e) into 250 ml conical flask. 2‐ Add 40 ml water 3‐ Boil on direct on direct flame. 4‐ Place the sugar solution in a burette. 5‐ Titrate the boiling Fehling solution adding 1 ml of the sugar solu on at a time bringing the solution to boiling after each addition. Continue till the blue colour of the mixture disappears. 6‐ Add 4 drops of methylene blue indicator 7‐ Continue titration drop‐wise until the blue colour of the indicator disappears. Read the number of mls of the sugar solution used in this titration. 8‐ Carry out a second titration adding this time all the number of mls of the sugar solu on used in the first tra on, but 1 ml less, and boil the solution. 9‐ Add 4 drops of the indicator and con nue tra on drop‐wise until the accurate end point is reached. 10- Repeat the above procedure until two readings are the same.
 Calculation: ‐ 10 ml fehling sol. contains 0.11 gm cupric oxide which is able to
oxidize 0.05 gm of a monosaccharide or 0.08 gm of a disaccharide  10 ml Fehling ======= 0.05 g monosaccharide
 10 ml Fehling ======= 0.08 g disaccharide
‐ Volume of the sugar consumed by methylene blue to be reduced is
0.2 ml
For monosaccharide (E.P – 0.2) ml 0.05 g mono 100 ml ??? g % glucose = .
. .
.
gm% For disaccharide (E.P – 0.2) ml 0.08 g di 100 ml ??? g % lactose = .
. .
.
gm%  Notes
• Time of reac on not exceeds 3 min. • Methylene Blue (M.B) is used as internal indicator and the E.P is reached when the color of indicator disappear. • Methylene blue has a blue color in the oxidized form, it is reduced by the sugar to be colorless • Cupper reduction is a type of reversed titration as burette contains sample instead of titrant & flak contains titrant instead of sample. II. Iodimetric method This method can be applied for aldoses ; to determine Aldoses in presence of ketoses. Estimation of glucose‐fructose mixture  Principle:
The iodometric method of assay of aldoses is based on the fact that iodine can oxidize aldoses in alkaline medium, whereas it has no effect on ketoses. I2 oxidises glucose in alkaline medium into gluconic acid C6H12O6 + 3KOH + I2
C6H11O6OK + 2KI + 2H2O
N.B: Role of acid & alkali  Procedure
1- Add 25 ml sugar using pippete bulb in a stoppered flask
2- Add 20 ml 0.1 N I2
3- Add 5 ml Na2CO3
4- Incubate at dark for 20 minutes
5- Add 7 ml H2SO4
6-
Titrate the
excess unused I2 against Na2S2O3 using starch as indicator
7- Do blank experiment replacing sugar solution with water
I.
25 ml sugar sol.
(exx)
+ 20 ml 0.1N I2 sol.
+ 5 ml Na2CO3 + 7 ml 2 N H2SO4 II.
Blank
Do blank experiment, but replace the sugar by water?
 Calculation:  Equivalence factor 1 molecule I2 = 180 gm glucose I = 90 gm glucose 1 ml 0.1N Iodine = 0.009 gm glucose (monosacchide) 1 ml 0.1N Iodine = 0.017 gm maltose (disaccharide) %
25
. .
100
% Estimation of glucose‐sucrose mixture  Principle: Glucose is a reducing sugar, so it can be determined by copper reduction method in presence of other non reducing sugars. Sucrose can be determined after acid hydrolysis. Sucrose + H2O glucose + fructose (Inverted sugar)  Procedure: 1) Run the copper reduction method for the mixture solution to get (E.P1) which is equivalent to the glucose content of the mixture. 2) In 100 ml measuring flask, add 20 ml sugar mixture and 2 ml Conc. HCL. Heat on boiling water bath for 15 min. Cool and add gradually sodium carbonate powder till effervescence ceases to get neutralized solution. Complete to the mark with water. Mix well, then run the copper reduction method for this neutralized solution to get (E.P2) which is equivalent to the total sugar after hydrolysis.  Calculation: A (% of glucose) = .
.
.
g glucose B (% of total reducing sugar) = .
.
.
X dilution factor g as monosaccharide Sucrose + H2O glucose + fructose M.wt = 342 180 + 180 342 g sucrose 360 g mono X g sucrose 1 g mono X = 0.95 gm sucrose % sucrose = (B‐A) x 0.95 g sucrose Estimation of glucose‐Maltose mixture  Principle: Both sugars are reducing, so two experiments using copper reduction method have to be run. Then by certain calculation, both sugars can be determined.  Calculation: A (% of reducing sugars before hydrolysis) =
B(% of reducing sugars after hydrolysis) =
as mono  Before hydrolysis 10 ml fehling solu on = 0.05 g mono or 0.08 g di 0.08 g di 0.05 g mono 1 g di X g mono X= 5/8 1 g di = 5/8 = 0.625 g mon*  After hydrolysis Maltose 2 glucose 342 2 x 180 342 g di 360 g mono 1 g di X g mono 1 g di = 360/342 = 1.05 g mono* .
.
.
.
.
.
g as mono x dilution factorg So, a er hydrolysis the reducing power of each 1 g maltose increases by 1.05 ‐ 0.62 = 0.43 g mono but, the increase in reduction power due to hydrolysis of all Maltose present in the mixture= B‐A 1g maltose increases the red power by‐‐‐‐‐‐ 0.43 g mono ? g maltose increases the red power by‐‐‐‐‐‐ B‐A g mono % of maltose= .
g maltose (C) % of glucose = A – (C x 0.62) = g glucose