CHAPTER III
GENERAL TESTS
First Test for Flavonoids
1. A small amount of the collected filtrate or corn
extract was put in a test tube
2. 5 gtts. of 5% NaOH was added to the extract
followed by addition of 2mL 10% HCl.
3. A yellow solution that turns colorless upon
addition of HCl indicates the presence of
flavonoids.
The flavonoids being tested in the corn extract contain polyphenolic molecules
which are weakly acidic in nature. In the first general test for flavonoids, the extract was
first treated with a small amount of a weak concentration of NaOH (5%). The addition of
5% NaOH to the extract didn’t give very evident reactions because at first, it only gives
a colorless solution which contains sodium phenoxide. Also, only 5 gtts of a weak
concentration of NaOH was added. (G.A. Ayoola, September 2008)
NaOH is completely ionic, containing sodium cations and hydroxide anions. It is a
sufficiently strong base which deprotonates phenol entirely. The purpose of 5% NaOH in
this test was to deprotonate the polyphenolic molecules cointained in flavonoids. NaOH
turns phenols into phenoxides which are much more soluble in water than phenols. The
alkaline condition may have been necessary to disrupt bonding between the flavonoids
and other components of the brain. (G.A. Ayoola, September 2008)
Flavonoids give the yellow coloration to the extract. The objective of this
experiment was really to decolorize the yellow color of the corn extract containing
flavonoids. This was done through the addition of HCl to the mixture of the corn extract
and 5% NaOH. Only a minimal amount of NaOH was put in the extract to avoid the
tendency of HCl to neutralize with the mixture. The addition of sufficient amounts of
hydrochloric acid prevents the development of colored complexes of many phenols and
all enols. This is the reason why upon addition of HCl to the mixture, the solution slowly
decolorizes. (G.A. Ayoola, September 2008)
The experimental result of the first general test for flavonoids was consistent to
its theoretical positive result which is the decolorization of the extract. This indicates the
presence of polyphenolic flavonoids in the corn extract. (G.A. Ayoola, September 2008)
Third Test for Flavonoids
1. A portion of the extract was heated with 10 ml
of ethyl acetate over a steam bath for 3
minutes.
2. The mixture was filtered and 4 ml of the filtrate
was shaken with 1 mL of dilute ammonia
solution.
3. A yellow coloration indicates the presence of
flavonoids.
This general test for flavonoids involves the addition of ethyl acetate to a portion
of the corn extract. The mixture was subjected to heat for 3 minutes in a water bath and
then it was filtered. The corn extract is suspected to contain polyphenolic flavonoids.
The ethyl acetate which is an ester reacted with these flavonoids by decolorizing it upon
subjection to heat. With the addition of dilute ammonia solution, an amine, to the phenyl
ester, the yellow coloration of the extract was recovered indicating the presence of
flavonoids. (G.A. Ayoola, September 2008)
Fourth Test for Flavonoids
1. 4mL of aqueous NaOH was added to 2ml of each of
ethanol extract.
2. Formation of a yellow precipitate indicates the presence of
flavonoids in the extract.
Flavonoids are soluble in NaOH solution since the former contains weakly acidic
polyphenols and the latter is a strong alkaline medium. When phenol is reacted with
aqueous NaOH, it gives a colorless solution which contains sodium phenoxide. The
hydroxyl part of phenol loses H+ forming phenoxide ion and H2O. Though the negative
charge of phenoxide ion is now delocalized around the ring, electrons are drawn
towards the oxygen because of its electronegativity. (T.I. Olabiyi, 2008)
Since aqueous solution of sodium hydroxide is strongly alkaline due to its
complete dissociation into Na+ and OH-, sodium cation is attracted by the oxygen of
phenoxide ion. That’s why at first, formation of colorless globule-like sodium phenoxide
in the sample is seen. But after some time, yellow precipitate accumulated. This is due
to the fact that phenol is fairly insoluble in H2O. While some phenols solubilized with
NaOH, some also reacted with the H2O produced by the first reaction which is
characterized by the late formation of yellow precipitate. (T.I. Olabiyi, 2008)
SPECIFIC TESTS
Chromic Acid Test or Jone’s Test
1. Dissolve 1 gtt. of the extract or a small amount of the
solid sample in 1 mL of acetone in a small vial.
2. Add 2 gtts. of 10% aqueous K2Cr2O7 solution and 5
gtts 6M H2SO4 .
3. The appearance of the green Cr3+ ion or a blue-green
solution indicates oxidation of the compound or the
presence of primary or secondary alcohols in the
sample.
Chromic acid is capable of oxidizing many kinds of organic compounds. In
organic chemistry, dilute solutions of hexavalent chromium can be used to oxidize
primary or secondary alcohols to corresponding aldehydes and ketones while tertiary
alcohols are unaffected. Oxidation is signaled by a color change from orange to a bluegreen coloration.
The principle involved in the Jone’s test is oxidation. Oxidation is a qualitative
analytical test for the presence of primary or secondary alcohols in a sample. Oxidation
is readily detected by the appearance of the green Cr3+ ion or a blue-green solution.
Primary and secondary alcohols are rapidly oxidized by chromium trioxide in acidic,
aqueous acetone, whereas tertiary alcohols are stable to oxidation.
Chromic acid in aqueous H2SO4 and acetone is known as the Jone’s reagent
which oxidizes primary and secondary alcohols to carboxylic acids and ketones
respectively, while rarely affecting unsaturated bonds.
Primary and secondary alcohols and aldehydes react with chromic acid or Jone’s
reagent and give a blue-green precipitate in this test. Aldehydes are oxidized to
carboxylic acids, secondary alcohols are oxidized to ketones and primary alcohols are
oxidized first to aldehydes and then to carboxylic acids. Ketones and tertiary alcohols
are unreactive under the test conditions.
The polyphenolic flavonoids in the extract contain secondary alcohols in its
chemical structure which caused the rapid oxidation of the compound by chromic acid
and the formation of a blue-green solution. The experimental result of Jone’s test to the
corn extract is consistent with the theoretical positive result which proves that the
prepared corn extract contains oxidizable secondary alcohols.
Lucas’ Test
Lucas’ Reagent: Dissolve 16 grams of anhydrous zinc chloride in 10 mL of conc. HCl
with cooling.
1. Add about 50 mg or 2-3 gtts of the extract to 1 mL of the
reagent in a small vial. Cap the vial and shake vigorously
for a few seconds.
2. Allow to stand at room temperature.
Turbidity indicates the presence of alcohol compounds in the
sample. Tertiary alcohols react immediately with Lucas’ reagent to
produce turbidity while secondary alcohols do so in five minutes.
Primary alcohols do not react appreciably with Lucas reagent at room temperature.
Coordination of the zinc chloride with the hydroxyl results in the formation of a
sufficiently good leaving group. The carbon-oxygen cleavage can occur when
reasonably stable carbocation is produced.
Alkyl Chloride
(insoluble-cloudiness)
Lucas’ test for alcohol is a test to differentiate primary, secondary and tertiary
alcohols. It is based on the difference in reactivity of the three classes of alcohols with
hydrogen halides.
The principle behind this is the SN1 or substitution nucleophilic by first order
reaction which caused the turbidity (halide compounds insoluble in water.)
The experimental result is the formation of a turbid solution in a few minutes
which is consistent with the theoretical result of Lucas’ test for secondary alcohols. This
indicates the presence of secondary alcohols in the corn extract.
Bromine Test
The bromine test is a qualitative test for the
presence of unsaturated C–C bonds and phenols.
1. The sample is treated with a small amount of
elemental bromine — either as an aqueous
solution, or as a solution in dichloromethane or
carbon tetrachloride.
2. A positive test for the presence of unsaturation
and/or phenol is indicated by the disappearance of
the deep brown coloration of bromine, which
happens because the bromine has been
consumed by reaction with the unknown sample.
3. The formation of a white precipitate indicates the
formation of a brominated phenol.
The decolorization of bromine water is often used as a test for the presence f a
carbon-carbon double bond. The addition of bromine to cyclohexene gives a racemic
trans product. Since the product’s color differs from the original deep brown solution, the
bromine is rapidly decolorized when added to an alkene. When bromine water is
decolorized, there is a presence of unsaturated fatty acid.
Alkenes and alkynes will readily add bromine across the multiple bonds unless
there are electron withdrawing groups on the multiple bonds. One observes the rapid
disappearance of the red-brown bromine color. Aromatic compounds can react with
bromine more slowly to give bromine substitution and the formation of HBr, which can
sometimes be observed by placing a piece of wet litmus paper over the mouth of the
test tube.
The principle behind this reaction is the addition of a halogen into the double
bond. These results to an alkene with a halogen attached. The experimental result is
the same with the theoretical result where there is the formation of white precipitate due
to the discharging of the bromine color without the evolution of the hydrogen bromide
gas.
Baeyer’s Test
The Baeyer’s test for unsaturation is for determining the
presence of carbon-carbon double bonded compounds, called
alkenes or carbon-carbon triple bonded compounds, called
alkyne bond.
1. Place 5 gtts. of sample in a dry test tube.
2. Add 2 gtts. of 2% KMnO4 solution. Shake the test tube
vigorously and observe the rate and extent by which the
reagent is decolorized.
3. Note the formation of a brown suspension. Decolorization
of the reagent is immediate if it occurs within 1 minute.
The reaction is important because it doesn’t work on alkanes (compounds with
carbon-carbon single bonds) or aromatic compounds. A negative reaction for this test
indicates that the sample is either an alkane or an aromatic compound.
Potassium permanganate (KMnO4) is the oxidizing agent. Permanganate
converts cyclohexene into a diol. Since a syn-hydroxylation takes place, the reaction is
thought to involve the formation of an intermediate cyclic manganate ester, which is
readily hydrolyzed under the reaction conditions to yield the glycol. In the course of the
reaction, purple permanganate is reduced to brown manganese dioxide (MnO 2). Other
easily oxidized compounds such as aldehydes, some alcohols, phenols, and aromatic
amines should be accounted for analysis.
This reaction is used as a qualitative test for the presence of an alkene
(compounds with carbon-carbon double bonds) or an alkyne (compounds with carboncarbon triple bonds).
The experimental result of this test was consistent with the positive result of
Baeyer’s test which is the formation of a brown suspension or the decolorization of the
reagent. This indicates the presence of alkenes in flavonoids contained in the corn
extract. In the case of flavonoids specifically quercetin, though it is an aromatic
compound, it still contains alkenes separated from the aromatic part which caused the
positive reaction for this test.
Test for Aromaticity: Nitration
1. Place 2 mL of conc. HNO3 in an Erlenmeyer flask.
Immerse the flask in a water bath and gradually add 2
ml conc. H2SO4. Cool the resulting mixture to room
temp. This will serve as the nitrating mixture.
2. Place 5 gtts. of the sample in a dry test tube. Add 8
gtts. of the nitrating mixture and shake the test tube to
ensure complete mixing. Note the formation of a
yellow oily layer or droplet. Dilute with 20 gtts of water.
The presence of a yellow oily layer or oil droplets indicates aromaticity of the
sample. In this case, there is the occurence of the positive result which proves that the
constituent extracted is aromatic in its structure.
Benzene undergoes electrophilic aromatic substitution or SE in which an
electrophile substitutes for one of the hydrogens attached to the benzene ring. It has a
2- step mechanism: first, benzene reacts with an electrophile (nitronium ion) forming a
carbocation intermediate, second, a base (H2O) in reaction mixture pulls off a proton
from the carbocation intermediate forming the more stable nitrobenzene. The first step
is slow that it consumes energy while energy is released in the second fast step.
Nitration Chemical Reaction
Bial’s Orcinol Test
Bial’s reagent: a solution of orcinol, HCl and ferric
chloride
1. Two ml of a sample solution is placed in a test
tube.
2. Two ml of Bial's reagent was added to the
extract.
3. The solution was then heated gently in a hot
water bath for 2 minutes. If the color is not
obvious, more water can be added to the tube.
4. The formation of a blue-green solution indicates the presence of pentoses or
nucleotides containing pentoses while the formation of a yellow-green precipitate
indicates the presence of hexoses and disaccharides.
Bial’s orcinol test uses concentrated hydrochloric acid (HCl) as the dehydrating
acid and orcinol with a trace of ferric chloride (FeCl3) as the condensation reagent.
Bial’s orcinol test is used to distinguish between pentoses and hexoses. Pentose sugar
forms furfural derivatives (dehydration forming furfural) and hexose sugar forms 5hydroxymethylfurfural (dehydration forming 5-hydroxymethylfurfural) when it is heated
with concentrated HCl. It is then reacted with orcinol (condensation with orcinol) in the
presence of ferric ion (complexation with Fe+3) which gives a characteristic bluegreen/moss green solution for pentose sugar and yellow-green precipitate/muddy
brown-gray condensation product for hexose sugar.
The experimental result of this test, a yellow-green precipitate, proves that
the specific constituent found in the plant has a hexose sugar in its structure.
For pentose:
pentose → dehydration product: → blue-green/moss green solution
furfural
For hexose:
hexose → dehydration product: → yellow-green precipitate/
5-hydroxymethylfurfural muddy brown-gray condensation
product