’
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:
6 )O2(H 6, or alternatively C6O12H6Glucose (blood sugar): C
11 )O2(H12 , or alternatively C11O22H12sugar): CSucrose (table
Exceptions:
carbohydrates, Not obey the hydrate role: ‐1
4O17H7, Cymarose C4O12H6Digitoxose C ,5O12H6Deoxy sugars: Rhamnose C
Compounds obey the hydrate role, but not carbohydrates:‐2
Formaldehyde HCHO (CH₂O), Acetic acid C₂H₄O₂
Monosaccharides
A. Classification
I. According to function group
: glucose & xylose e.g .Monosaccharides with aldehyde group Aldoses:
: Fructose. e.gthose containing a ketone group ketoses :
bearing carbon atoms ‐According to number of oxygenII.
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
sDisaccharide
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)
of carbohydrates Qualitative identification
General tests for carbohydrates
The following 2 reac ons are posi ve with all carbohydrates.
h's test:cMolis ‐1
Notes:
1‐CHO: aqueous sol. If not sol, boil some solid with dil. H₂SO₄.
2‐Avoid shaking tube.
Sulfuric acid test: ‐2
Heating sugars with conc. H2SO4 results in dehydration and charring with
decomposition of sugars.
solution Fehling'sReduction of ‐3
* 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.
O ppt.2is due to Cu ed precipitate* R
Barfoed's solutionReduction of ‐4
* 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.
Osazone formation: ‐5
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
for pentoses and methyl pentoses sI. Test
.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.
.. Aniline acetate paper test (Furfural test)2
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.
).For Pentoses( Phloroglucin / HCl test3.
.4. Acetone test
Special tests
for Ketoses sII. Test
Cobalt nitrate test: ‐1
Violet color or/
Violet or purple color
* Positive for ketoses: fructose & sucrose
* Glucose: ‐ve ?
Resorcinol test (Slwanoff's test):) 2(
* 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
Glucose = (Dextrose) = Grape sugar(1)
.lactose maltose & glucose,+ve with: ₃1. Reduc on of Amm. AgNO
.2. Ac on of NaOH
.3. Lead subacetate test
h cMois 4.
Fehling 5.
6. Barfoed
Osazone7.
Liquid glucose
Liquid glucose: is the Product of the incomplete hydrolysis of starch.
It consists of glucose; maltose and dextrin.
Fructose = Levulose = Fruit sugar(2)
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
(Pentose) Xylose = Wood sugar (3)
1. Molisch
2. Fehling
3. Barfoed
4. Bial's test
5. Aniline acetate paper (Furfural test)
6. Acetone test
(Methyl pentose) Rhamnose (4)
1. Molisch
2. Fehling
3. Barfoed
4. Bial's test
5. Aniline acetate paper (Fursural test)
6. Acetone test
7. Phloroglucin / HCl test
Sucrose = Cane sugar = Beet sugar = Saccharose = sugar (5)
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.
Lactose = Milk sugar (6)
1. Molisch
2. Fehling
3. Osazone (cluster or tu s of needles)
Maltose = Malt sugar(7)
1. Molisch
2. Fehling
3. Osazone (rose es of plates)
NOTES
I. Polysaccharides
)Starch and Dextrin( polysaccharides‐Homo
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.
AmylopectinAmylase
Cons tute ~ 75% of starch
molecule.
Cons tute ~ 25% of starch
molecule.
Water insolublePartial water soluble
It is a branched molecule consisting
of glucose units also joined by α‐
1,4‐glycosidic bonds (at the
straight chain), and few by α‐1,6‐
glycosidic bonds (at the branching
points).
Composed of straight chains of
glucose units joined by α‐1,4‐
glycosidic bonds.
violet/ purple color with I₂ Blue color with I₂
Hydrolysis with maltase enzyme
yield maltose + dextrin.
Hydrolysis with maltase enzyme
yield maltose.
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
Blue + I₂+++ very high molecular weight
Amylodextrin
Red + I₂ ++ high molecular weight
Erythrodextrin
Colorless + I₂ + low molecular. weight
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.
polysaccharides‐II. Hetero
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
Artificial gum Gum tragacanthGum acacia (gum Arabic)
Item
Artificial NaturalNatural source
DextrinIncomplete starch hydrolysis.
Its complete hydrolysis gives glucose.
water :Tragacanthin‐1soluble part.
poly Bassorin: ‐2methoxylated acid, H₂O insoluble part.
uronic acid.‐Galacto ‐3
, Pentosans ‐4mucilage and starch.
Ca, K, Mg Arabin: ‐1salts of arabic acid.
Oxidase enzyme: ‐2turns benzidine or tr. Guaiacum into blue.
Composition
Insoluble adhesive gel.Insoluble (gel) : Bassorin
Soluble (no gel formation)
Solubility
dextrorotatortylevorotatorylevorotatoryAqueous solution
Artificial gum Gum tragacanthGum acacia (gum Arabic)Test
‐ve +ve +ve Pentose test
‐ve ‐ve Blue/ greenish blue
Benzid./H₂O₂
Reddish brown
Minute scatered blue points due to
starch
Yellowish brown
I₂ test
white precipitate
white precipitate
No white precipitate
Pb(Ac)₂
‐ve No white precipitate
white precipitate
Pb subacetate
‐ve ‐ve precipitateBorax
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
Agar‐1. 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:
Stiff gel.
1. Cooling
Reddish Particles.
2. I₂
White precipitate.
3. Hcl + BaCl₂
Reduction (red precipitate) due to galactose.
4. HCl/heat; neutralize + Fehling test
Red colored particles (due to its mucilage content).
5. Ruthenium test:
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.
of carbohydrates Quantitative estimation
sugarsEstimated by
Titermetricmethods
Colorimetricmethods
Gravimetricmethods
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
‐ 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.
10 ml Fehling ======= 0.05 g monosaccharide 10 ml Fehling ======= 0.08 g disaccharide
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
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
C6H12O6 + 3KOH + I2 C6H11O6OK + 2KI + 2H2O
I.
II. Blank Do blank experiment, but replace the sugar by water?
25 ml sugar sol.+ 20 ml 0.1N I2 sol. (exx)+ 5 ml Na2CO3
+ 7 ml 2 N H2SO4
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) =.
. .g as mono
B(% of reducing sugars after hydrolysis) =.
. . x dilution factorg
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*
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