Experiment 20
Identification of Some Carbohydrates
Carbohydrates are the direct product of the photosynthetic combination of carbon
dioxide and water. By weight, they are the most common organic compounds on earth.
Since most have the empirical formula CnH2nOn = Cn(H2O)n it was initially believed that
they were hydrates of carbon. Hence the name. In actuality they are
polyhydroxylaldehydes and ketones and exist as cyclic hemi- and full acetals. The cyclic
forms may be five-membered rings (furanose) or six-membered rings (pyranose). They
are classified as mono-, di-, and polysaccharides. The term sugar applies to mono-, di-,
and oligosaccharides, which are all soluble in water and thereby distinguished from
polysaccharides, which are not soluble in water.
The most commonly encountered carbohydrates are starch, glycogen, inulin,
cellulose, sucrose, fructose, arabinose, mannose, glucose, maltose, galactose, and lactose.
The method of analysis used in this experiment is one of elimination. The assumption is
made that the compound is a single carbohydrate and tests are applied sequentially until a
positive result is obtained. The carbohydrate giving the positive test is thus identified.
The sugars that we will identify are shown below.
Figure 20.1 Monosaccharides
Aldohexoses
O
H
C
CH2 OH
O
HO
HO
H C
OH
HO
OH
D-glucose
O
C
H
H C
OH
H C
OH
CH2 OH
O
C
OH
HO
OH
OH
C
H
HO
C
H
H C
OH
CH2 OH
O
Aldopentose
HO
HO
OH
O
C
OH
D-Arabinose
HO
C
H
H
H C
OH
H C
OH
OH
C
H
HO
C
H
H C
OH
H C
OH
D-mannose
OH
HO
D-galactose
HOCH 2 OH
O
HO
HO
HO
CH2 OH
H
H C
CH2 OH
O
H
C
OH
CH2 OH
ketohexose
HOCH2 O
HO
HO
OH
CH2 OH
O C
HO
CH2 OH
D-Fructose
C
H
H C
OH
H C
OH
CH2 OH
Figure 20.2 Disaccharides
CH2 OH
OH
O
HO
HO
CH2 OH
O
HO
HO
O
sucrose
D-glucose + D-fructose with
(1-1) α-glucosidic linkage
OH
CH2 OH
maltose
OH
OH
HO
O
OH
lactose D-galactose + D-glucose with (1-4)
β-glucosidic linkage
HO
OH
O
O
HO
D-glucose + D-glucose with (1-4)
α-glucosidic linkage
O
HO
HO
CH2 OH
HO
CH2 OH
CH2 OH
O
O
HO
HO
Figure 20.3 Polysaccharides
CH2 OH
O
HO
CH2 OH
O
CH2 OH
HO
O
HO
starch
n O
HO
O
HO
O
n
CH2 OH
polymer of D-glucose with (1-4) α-glucosidic linkage
HO
CH2 OH
HO
O
HO
O
O
HO
HO
O O
polymer of D-fructose
with (2-1) B-D-fructose
linkages
CH2 OH
O
CH2 OH
O
O
O
O
HO
Oh
OH
O
O
HO
n
glycogen
polymer of D-glucose with (1-4)
α-glucosidic linkages and branch at α (1−6)
HO
OH
O
O
O
HO
inulin
OH
HO
HO
O O
HO
O
O
HO
n
cellulose
polymer of D-glucose with (1-4)
β-glucosidic linkage
Sugars that have a free aldehyde group (or can from one under the reaction
conditions) are oxidized by cupric ion (Cu2+) to produce the carboxylic acid and cuprous
ion (Cu1+). These are called reducing sugars. All monosaccharides are reducing sugars
since keto-sugars such as fructose can isomerize to aldo-sugars. In our experiment we
will use Fehling’s solution as the source of our copper II ion. This is prepared by mixing
together, at the moment it is to be used, equal volumes of a solution of copper sulfate
(Fehling’s solution A) and an alkaline solution of sodium potassium tartrate (Fehling’s
solution B)
Figure 20.4 Reducing Sugars
O
C
CH2 OH
+
O C
-
H
O
+
CHOH
OH
Cu
2+
C
-
+ 2 OH
red ppt
R
carboxylic acid
aldo-sugar
keto-sugar
Cu+
+
CHOH
R
R
OH
Test 1 Molisch Test
The Molisch test is a general test for carbohydrates. Under strongly acidic
conditions, monosaccharides undergo dehydration to provide furfural (R = H) from
pentoses sugars or 5-hydroxymethylfurfual (R = CH2OH) from hexose sugars as shown
in Figure 20.4. Furfural or its derivative can then react with two moles of α-naphthol to
provide the violet-colored quinoid. These reactions with α-naphthol are examples of
electrophilic aromatic substitution which occurs at the para position of the ring that is
activated by the phenolic hydroxy group. Polysaccharides undergo at least partial
hydrolysis under the conditions of the test, and the resulting monosaccharides, as well as
any other monosaccharides present, are dehydrated to provide furfural or 5hydroxymethyl furfural. Concentrated sulfuric acid is used as the acid catalyst and
dehydrating agent.
Figure 20.5 Molisch Test: Dehydration and Quinoid Formation
R
OH
O
H
R
Pinacol Rearrangement
H
H
H
H
H
CH2 OH
OH
HO
Pentose: R = H
Hexose: R = CH2 OH
CHO
O
OH
HO
furfural (R= H)
OH
H+
H+
O
- H2 O
OH
R
H C
H C
OH
C
H
H
α-naphthol
R=
R
R
CHO
R
H
OH
OH
O
- 2H2 O
O
OH
R' = H, CH2 OH
oxidation
H C
R
OH
Quinoid (violet)
C
R
O
R'
Procedure
Place about 5 mg of any carbohydrate in 1 mL of water in a small test tube and
mix with two drops of a 10% solution of α-naphthol in water. Using a medicine dropper,
add 1.0 mLconcentrated sulfuric acid down the side of the test tube so that it forms a
layer underneath the water layer (tilt the test tube). Be careful not to mix the layers. A
purple color between the two layers indicates a carbohydrate.
Test II Solubility
Add about 50 mg of the carbohydrate to 1.0 mL of cold water. If the substance is
insoluble it is a polysaccharide. The mono- and disaccharides are all soluble in cold
water.
Note 1: Make sure that the substance is given sufficient time to dissolve.
Note 2: Glycogen forms a colloidal suspension so that the mixture is milky. This test
alone is “almost” sufficient to prove that the compound is glycogen.
Test III Identification of the Polysaccharides
(A) Starch: Treat a suspension of the carbohydrate with a drop of 10% Iodine-Potassium
Iodide solution. If a blue-black color results, the substance is starch.
(B) Glycogen: Iodine-Potassium Iodide gives a faint red color with glycogen. However,
if it is not observed the existence of a colloidal suspension in the solubility test should be
taken as confirmation of glycogen.
(C) Inulin: The polysaccharide inulin is a polyfructose. The reagent used causes
hydrolysis. Seliwanoff’s reagent (0.5 g resorcinol in a liter of 10% HCL) is used. When
heated in this reagent hexose sugars undergo dehydration to form hydroxymethyl
furfural. This then condenses with resorincol (1,3-dihydroxybenzene) to give a red
product. The exact structure of the red colored product is not known but it is believed to
be the result of multiple electrophilic aromatic substitutions just as in the Molisch Test.
The reaction is shown in Figure 20.6.
Figure 20.6
O
HOCH2
O
OH
C H
H
H+
O
+
HOCH2
OH
colored products
C
OH OH
OH
resorcinol
To 5.0 mL of the reagent, add a spatula tip of the carbohydrate and boil for 30
seconds (Do not boil longer than 30 seconds). The development of a red color confirms
the inulin.
(D) Cellulose: Cellulose is identified by the fact that it gives a negative response to all of
the above tests.
Test IV Identification of the Water Soluble Carbohydrates
(A) Sucrose: Add about 10 g. of the carbohydrate to a mixture containing 1.0 mL of
Fehling’s A and 1.0 mL of Fehling’s B solutions and heat to boiling. A negative test,
failure to reduce the blue Cu (II) ion to red Cu (I), confirms sucrose.
(B) Fructose: The test for fructose is identical with that for inulin. If a small amount of
the unknown added to 5.0 mL Seliwanoff reagent gives a cherry red color on heating for
30 seconds or less, fructose is confirmed.
(C) Arabinose: Add 0.5 g of arabinose to 5 mL water in a medium test tube and stir to
dissolve. Add 5 drops of Fehling’s solutions A and B and boil the test tube for one
minute. Formation of a colored precipitate is confirmation of arabinose. Go to the Bial’s
test for additional confirmation.
(D) Aniline Acetate Test: Add about 0.3 g of the unknown to 5 mL of 3.0 M HCl in a
small test tube. Boil the test tube in a water bath for about one minute and then insert a
cylindrical roll of freshly prepared aniline acetate paper (prepared by dipping a piece of
filter paper in an equimolar solution of aniline and glacial acetic acid). Continue boiling
for a minute longer. Arabinose will color the paper cherry red (see Figure 20.7).
Figure 20.7 Aniline Acetate Reaction
O
H H OH
HOH2 C
C
C
C
O
C
OH OH H
H+
O
H
CH3 CO2 H
Furfural
Arabinose
NH2
C H
H
O
C N
H
red
(E) Bial’s reaction: An alternate for pentoses like arabinose involves Bial’s reagent (1.5
g of orcinol dissolved in 500 mL of concentrated HCl to which 20 drops of ferric chloride
are added). To 5 mL of the reagent, add about 25 mg of the unknown and heat gently to
the boiling point and then cool see Figure 20.8). If the solution turns green, the test is
positive. In some cases a green flocculent precipitate may form.
Figure 20.8 Bial's Reaction
O
H H OH
HOH2 C
C
C
C
OH OH H
Arabinose
CH 3
C
O
+
H
O
H
Furfural
C H
HO
OH
green condensation products
HCl, FeCl3
Test V Osazone Test
Carbohydrates that have an aldehyde or ketone carbonyl group, either free or in
equilibrium with a hemiacetal, will react with phenylhydrazine to form bright yellow,
crystalline derivatives called osazones. These derivatives may usually be identified
readily both by their melting points and by their crystalline forms. The reactions involved
are shown in Figure 20.9.
Figure 20.9 Osazone Formation
O
C
H C
HO
C
H C
H C
H
HC NNHC 6 H5
C NNHC 6 H5
OH
NHNH2
H
OH
+ 3
H3 O+
OH
CH2 OH
HO
C
H
+
H C
OH
H C
OH
C 6H 5NH2
+ NH3
CH2 OH
D-glucose
osazone
The accepted mechanism for the reaction is shown in figure 20.10. Following the
formation of the phenylhydrazone 20.1, there is an internal oxidation-reduction reaction
that involves the tautomeric migration of two hydrogens from C-2 to the hydrazone
moiety to give the carbonyl product 20.2. The newly formed carbonyl group condenses
with a second equivalent of phenylhydrazine to give 20.3 which undergoes subsequent
tautomerization to 20.4. Following a 1,4-elimination of aniline, which produces 20.5, a
third equivalent of phenylhydrazine condenses with the imine group to give the osazone
20.6 and ammonia. The formation of the intramolecular hydrogen bond in 20.6 prevents
another internal oxidation-reduction sequence between the secondary alcohol at C-3 and
the hydrazone group at C-2 and the introduction of a third phenylhydrazine unit.
Figure 20.10 Mechanism for Osazone Formation
O
C
H C
HO
C
H
NHNH2
OH
+
H3 O+
H
R
HC NNHC 6 H5
H C OH
internal
HO C H
redox rxn
R
HO
CH2 NHNHC 6 H5
CH2 NHNHC 6 H5
C
O
C NNHC 6 H5
C
H
HO
R
20.1
NH3
+
H
R
C
N
OH
H
N
20.6
CHNHNHC 6 H5
HC NH
C NNHC 6 H5
C 6H 5NHNH2
C 6 H5
H
R 20.3
20.2
NHC6 H5
N
C
HO
C
H
R
20.5
C NHNHC 6 H5
HO
C
H
R
20.4
Although the time it takes for osazones to form may be used to make qualitative
distinctions among the carbohydrates, the test will not be used in this way here. It will
rather be used to divide the remaining fine sugars in such a way so that they may be
identified.
The reagent is prepared by dissolving 0.80 g of phenylhydrazine hydrochloride
and 1.2 g of sodium acetate in 8 mL of water. To about 2 mL of this reagent, add about
0.1 g of the unknown in a small test tube. Immerse the test tube in a beaker of boiling
water and note the time of immersion. Shake the test tube occasionally without removing
it from the boiling water.
(A) The immediate formation of a dirty white precipitate confirms mannose. Mannose
under these conditions does not form an osazone. The observed precipitate is the
phenylhydrazone of the sugar. Mannose is the only one of the remaining carbohydrates
that behaves this way.
(B) The formation of an orange precipitate in less than 30 minutes indicates that the
unknown is either glucose or galactose. If no precipitate forms in the allotted time then
the unknown is either maltose or lactose.
(C) Glucose is oxidized by concentrated nitric acid to the water-soluble saccharic acid,
whereas galactose is oxidized by the same reagent to the water insoluble mucic acid.
This difference may be used to distinguish between the two hexoses.
Test VI Mucic Acid Test
This test is specific for galactose. Hot nitric acid oxidizes a sugar to the
carboxylic acid (Figure 20.11). Aldoses are oxidized at both ends of the ring-opened
form to provide dicarboxylic acids. Ketoses oxidize to give a mixture of dicarboxylic
acids resulting from chain fragmentation. Galactaric acid (or mucic acid) from the
oxidation of galactose tends to be much less soluble in the oxidizing medium (and water)
than the saccharic acids obtained from other aldoses. This is partly due to the high
molecular symmetry of galactraric acid (it is a meso compound).
O
C
H C
H
OH
HO
C
H
HO
C
H
H C
Figure 20.11 Mucic Acid Test
O
O
OH
H
C
C
OH
CH2 OH
galactose
H C
HNO 3
OH
H C
O
C
H C
OH
HO
C
OH
OH
H
HO
C
H
HO
C
H
H C
OH
H C
OH
OH
H C
OH
H C
OH
C
H C
HO
C
OH
O
galactaric acid (mucic acid)
water insoluble
C
H
CH2 OH
HNO 3
OH
O
glucaric acid
(saccharic acid)
water soluble
In an evaporating dish, place about 2 mg of your unknown, 6.0 mL of water and
6.0 mL of concentrated nitric acid. Connect a large funnel to the aspirator with a rubber
tube (Or a hood may be used instead). Evaporate over a small free flame stirring to avoid
spattering a burning. When the contents become pasty, cool, dilute with cold water and
filter. Wash with cold water. The presence of a white precipitate on the filter confirms
galactose. The absence confirms glucose.
We must now distinguish between maltose and lactose. Since maltose is a dimer
of glucose and galactose, oxidation of the former with nitric acid gives a soluble
saccharic acid, whereas the latter gives a mixture of saccharic and mucic acids.
Therefore the mucic acid test serves also to distinguish between the two disaccharides.
The melting point of mucic acid may be determined. With slow heating, it is 214 °C,
while with rapid heating it will be found between 223-224 °C.