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1. Aim
To identify the functional groups present in an organic compound and
To characterize unknown organic compound.
2. The need for organic qualitative analysis
Qualitative organic chemistry has been in use since long before the advent of modern
spectroscopy. Modern spectroscopic techniques have assisted the chemist by providing
spectra that can be interpreted to give more detail of the interaction between atoms and
functional groups. Some of you have difficulty identifying the structures using
exclusively nuclear magnetic resonance (NMR) spectra, infrared (IR) spectra, and mass
spectra (MS). The information obtained through chemical tests allows you to narrow
down the possible functional groups.
3. General note
Only distilled water should be used throughout the qualitative analysis procedures.
Impurities in tap water could lead to “false positive” results.
When using a medicine dropper, you can assume 20 drops is approximately
equivalent to 1 ml.
The ignition test will give you an indication as to whether or not your unknown is
aromatic.
If a reaction occurs when a solvent is added to a compound, is evidenced by a color
change for instance, the compound is consider to soluble in that solvent.
If any amount of the compound appears to dissolve in a particular solvent, consider it
to be soluble in that solvent.
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4. Systematic Identification
You will work in groups on this project. Each group will be given one unknown (solid or
liquid) compound that the group will identify using various chemical classification tests,
spectral data and melting points of prepared derivatives.
All necessary reagents are provided for you in the laboratory.
Precautionary note! The result of all the tests (especially negative results) should be
regarded as strongly suggestive but not absolutely reliable.
When you have identified to which functional group class of compounds your
unknown belongs, you will prepare one or two derivatives, as necessary, to
accurately determine the actual identify of your unknown. All unknown
compounds contain at least one derivatizable functional group.
Each student in a group must hand in his/her own discussions and interpretation of the
results for each unknown; because you will be graded on your presentation and
answering questions from the examiner.
You will not be able to rely on your laboratory instructor for help every step of the
way. The laboratory technician assistant will not be able to answer questions such
as “what test should I do next” and “Is this a positive on negative result?” both
because s/he could in advertently steer you in the wrong direction and because you
must learn to trust your own judgment and move your own decisions in a problem
solving process such as this.
Once you predict the structure of your unknown based on the chemical analysis and the
derivative that you synthesized you are expected to report it.
After preparing the appropriate derivative for your unknown, you will be given MS,
NMR, and IR of the unknown.
The process of identifying unknowns is analogous to solving a puzzle. You must be
systematic don’t jump to conclusion.
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Five basic of experimental inquiry are useful for identifying an unknown compound.
These five areas of inquiry are:-
1. Preliminary examination
2. Physical constants
3. Classification by solubility
4. Classification test for functional groups
5. Synthesis of solid derivatives
6. Spectroscopic analysis
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Flow Chart for Identification of Organic Compounds
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4.1Preliminary examination
If it is carried out intelligently, the preliminary examination may provide more
information with less effort than any other part of the procedure. It includes physical
state (solid or liquid), color, odor, and ignition/flame test.
The ignition test involves a procedure in which a drop or two of a liquid or about 50 mg
of a solid is heated gently on a small spatula or crucible. A yellow, sooty flame is
indicative of aromatic or a highly unsaturated aliphatic compound; a yellow but non-
sooty flame is characteristics of aliphatic hydrocarbons. The oxygen content of a
substance makes its flame more colorless or blue; high oxygen content lowers or prevents
flammability.
4.2Physical constants
If the unknown is a solid, its melting point is measured by the capillary tube method. For
an unknown that is a liquid, the boiling point is determined by the micro boiling-point
technique.
Other physical constants may be of use for liquids are the refractive index and the
density.
4.3Classification by solubility
Solubility classification groups compounds on the basis of their solubilities in water,
ether, 5% NaHCO3, 5%NaOH, 5%HCl and conc. H2SO4. The results of the solubility
tests can help classify molecules in to categories of functional group.
The following flow chart and procedure will help you to determine the solubility of your
unknown compounds.
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Figure 1: Solubility Flow Chart
Producer: Place about 1 ml of the solvent in small test tube. Add one drop of your
unknown liquid or a few crystals of an unknown solid directly into the
solvent. Gently tap the test tube with your finger to ensure mixing, and then
observe whether any mixing lines appear in the solution. The disappearance
of solid or the appearance mixing line indicates solubility. If the compound is
soluble add another drop or a few more crystals to ensure solubility. You may
heat the solution gently in a worm water both to assist in dissolution.
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4.4Functional group Classification tests
Solubility results should narrow down the types of functional groups present in your
molecule. To verify the presence of these functional groups, and learn more about
structural and substitution patterns in your unknown, a series of functional group tests
have been developed and provided.
You need not to do every test for a functional group, but you should read through the
different tests to determine if that test would give you additional insights.
Plan ahead! Write out the procedures for the functional group tests, and expected
out comes, before you come to lab.
4.4.1 Test for unsaturation
Organic compounds containing >C=C< and/or —C≡C– bonds are called unsaturated
compounds. These compounds undergo addition reaction with bromine water or the
solution of bromine in carbon tetrachloride, chloroform or glacial acetic acid. Addition of
bromine to an alkene results in the formation of vicinal dibromide. The reddish orange
color of the solution of bromine in carbon tetrachloride disappears on reaction with an
alkene. The reaction is as follows :
Alkenes decolorize the neutral/alkaline KMnO4 solution and vicinal glycols are formed
(Bayer’s test). Reaction takes place as follows:
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A. Bromine water test
Procedure: Dissolve 0.1 g or 5 drops of organic compound in 2 ml of carbon
tetrachloride in a test tube and add 2% solution of bromine in carbon
tetrachloride or bromine water drop by drop with continuous shaking.
Decolorization of bromine solution indicates the presence of unsaturation in
organic compound.
B. Bayer’s test
Procedure: Dissolve 25-30 mg of organic compound in 2 ml of water or acetone (free of
alcohol) and add 1% potassium permanganate solution containing equal
volume of 1% sodium carbonate solution. The discharge of the color of more
than one drop of potassium permanganate indicates the presence of
unsaturation in the organic compound. Carrying out the reaction under
alkaline conditions removes the possibility of confusion due to substitution
in aromatic compounds.
4.4.2 Test for alcoholic (R–OH) group
Alcoholic compounds on reaction with ceric ammonium nitrate give a red colouration
due to the formation of a complex.
(NH4)2 [Ce(NO3)6] + 3ROH [Ce(NO3)4(ROH)3] + 2NH4NO3
ceric ammonium nitrate Red complex
Distinction between primary, secondary and tertiary alcohols can be done on the basis of
iodoform test and Lucas test. Ethanol and secondary alcohols which contain CH 3
—CH(OH)R group (iodoform reaction) gives positive iodoform test. To carry out
reaction, potassium iodide and sodium hypochlorite solution are added to the compound
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in the presence of sodium hydroxide solution. Probably sodium hypochlorite first oxides
potassium iodide into potassium hypoiodite, which oxidizes CH3—CH(OH)R group to
CH3COR group and then iodinates it in the alkaline medium of the reaction mixture by
replacing the α-hydrogen attached to the carbon atom adjacent to carbonyl group by
iodine. Iodoform is formed after cleavage of C—C bond.
CH3CH2OH potassium CH3CHO potassium CI3CHO NaOH CHI3 + HCOONa hypoiodite hypoiodite
Lucas reagent contains zinc chloride and concentrated hydrochloric acid. This reagent
reacts with primary, secondary and tertiary alcohols at different rates. Tertiary alcohols
react almost instantaneously, secondary alcohols react in about 1-5 min and primary
alcohols react very slowly. The reaction may take 10 min to several days.
RCH2OH + HCl ZnCl2 No reaction/Slow reaction
R2CHOH + HCl ZnCl2 R2CHCl + H2O
R3COH + HCl ZnCl2 R3CCl + H2O
Alcohols are soluble in Lucas reagent but the formed alkyl halides are not soluble.
Therefore, formation of two layers in the reaction medium indicates the occurrence of the
reaction.
Primary alcohols – Layers do not separate
Secondary alcohols – Layers separate within 1-5 min
Tertiary alcohols – Layers separate immediately
A. Ceric ammonium nitrate test
Procedure: Take 1 mL solution of organic compound dissolved in a suitable solvent.
Add a few drops of ceric ammonium nitrate solution. Appearance of red
color shows the presence of alcoholic –OH group.
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Note: The red color disappears after keeping the reaction mixture for some time. The
color also disappears if excess of ceric ammonium nitrate solution is added. Therefore,
use of excess of ceric ammonium nitrate solution should be avoided.
B. Iodoform test
First method
Procedure: Take 0.2 ml of the compound in a test tube; add 10 ml of 10% aqueous KI
solution and 10 ml of freshly prepared NaOCl solution. Warm gently;
yellow crystals of iodoform separate.
Second method
Procedure: Dissolve 0.1 g or 4 to 5 drops of compound in 2 ml of water. If it does not
dissolve, add dioxane drop by drop to get a homogeneous solution. Add 2
ml of 5% sodium hydroxide solution followed by potassium iodide-iodine
reagent [Potassium iodide- iodine reagent is prepared by dissolving 20 g of
potassiumiodide and 10 g of iodine in 100 ml of water] dropwise with
continuous shaking till a definite dark color of iodine persists. Allow the
reactants to remain at room temperature for 2-3 minutes. If no iodoform
separates, warm the reaction mixture in a water bath at 60°C. Add more
drops of potassium iodide–iodine reagent. If color of iodine disappears
continue addition of reagent till the color of iodine persists even after two
minutes of heating at 60°C. Remove excess iodine by adding a few drops of
sodium hydroxide solution with shaking. Dilute the mixture with equal
volume of water and keep it at room temperature for 10-15 min. A yellow
precipitate of iodoform is obtained if test is positive.
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C. Lucas test
Procedure: Take 1 ml of compound in a test tube. Add 10 ml of Lucas reagent. Shake
well and note the time for the separation of two distinct layers.
Note: Lucas test is applicable to only those alcohols which are soluble in the reagent
because the test is based on separation of alkyl halides as separate layer.
4.4.3 Test for phenolic (AR-OH) group
The –OH group attached directly to the ring carbon of an aromatic ring is called phenolic
–OH group. Phenols are weakly acidic, therefore they are soluble in NaOH solution but at
the same time they are not sufficiently acidic to be soluble in sodium hydrogencarbonate
solution. Phenols give colored complex with neutral ferric chloride solution. For
example, phenol gives a
complex of violet color as follows :
6C6H5OH + FeCl3 [Fe(C6H5O)6]3– + 3HCl + 3H+
Colored complex
Resorcinol, o–, m– and p–cresol give violet or blue coloration, catechol gives green color
which rapidly darkens. 1 and 2–Naphthol do not give characteristics colors. Phenols
condense with phthalic anhydride in the presence of concentrated H2SO4, Phenol
condenses to give phenolphthalein which gives a dark pink color with NaOH solution.
This is called phthalein dye test.
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A. Ferric chloride test
Procedure: Take 2 ml of aqueous or alcoholic solution of the organic compound in a test
tube, add neutral ferric chloride solution dropwise and note the color change.
Appearance of a blue, green, violet or red color indicates the presence of
phenolic –OH group.
B. Phthalein dye test
Procedure: Take 0.1 g of organic compound and 0.1 g of phthalic anhydride in a clean
dry test tube and add 1-2 drops of conc. H2SO4. Heat the test tube for about 1
min in an oil bath/water bath. Cool and pour the reaction mixture carefully
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into a beaker containing 15 ml of dilute sodium hydroxide solution.
Appearance of pink, blue, green, red etc. colors indicates the presence of
phenolic –OH group in the compound. However, the color disappears on
addition of large excess of sodium hydroxide solution.
4.4.4 Test for carbonyl (—CO—) group
Both aldehydes and ketones contain carbonyl group (>C = O) and are commonly known
as carbonyl compounds. Identification of aldehydes and ketones is done by two important
reactions of carbonyl group i.e.
(i) Addition reaction on double bond of >C = O group and
(ii) Oxidation of carbonyl group.
Addition reactions of derivatives of ammonia are important from the point of view of
identification of carbonyl compounds. Addition is generally followed by elimination
resulting in the formation of unsaturated compound.
These reactions are catalysed by an acid or a base and do not occur under strongly acidic
or basic conditions. Each reaction requires an optimum pH for its occurrence. Therefore,
maintenance of pH is very important while carrying out these reactions.
As far as oxidation is concerned, aldehydes are easily oxidized to carboxylic acids while
ketones require relatively stronger oxidizing agents. Distinction can be made between
these two types of carbonyl compounds on the basis of difference in their reactivity.
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Following tests are performed for the identification of aldehydic and ketonic groups: On
reaction with 2,4-dinitrophenylhydrazine (2,4-DNP), they form the respective 2,4–
dinitrophenyl hydrazones.
These two carbonyl compounds (aldehydes and ketones) are distinguished on the basis of
tests using mild oxidizing reagents, like Tollen’s reagent and Fehling’s reagent or
Benedict’s reagent. Tollen’s reagent is an alkaline solution of silver cation complexed
with ammonia, and Fehling’s and Benedict’s reagents are alkaline solutions containing
cupric ions complexed with tartarate and citrate ions respectively. Fehling’s reagent is
freshly prepared by mixing equal amounts of Fehling’s solution A and Fehling’s solution
B. Fehlings reagent deteriorates on keeping while Fehling’s solutions A and B are quite
stable. Fehling’s solution A is an aqueous copper sulphate solution while Fehling’s
solution B is an alkaline solution of sodium potassium tartarate (Rochelle’s salt). The
reagent contains Cu2+ ion complexed with tartarate ions. The structure of the complex is
given below:
Copper tartarate complex
Benedict modified the original Fehling’s test by using a single solution which is more
convenient for the test. Benedict’s solution is more stable than Fehling’s reagent and can
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be stored for a long time. It is an alkaline solution containing a mixture of copper
sulphate and sodium citrate (2Na3C6H5O7.11H2O).
Complex formation decreases the cupric ion concentration below that necessary for
precipitation of cupric hydroxide. These two reagents oxidize aldehydes while ketones
remain unaffected. The chemistry of these tests is as follows:
RCHO + 2[Ag (NH3)2]+ + 2OH– → 2Ag + 3NH3 + H2O + RCOONH4
From Tollen’s reagent
RCHO + 2Cu2+ (complexed) + 5OH– → RCOO– + Cu2O + 3H2O
Fehling’s solution
However, aromatic aldehydes do not give positive Fehling’s test. In Benedict test also,
Cu2+ ions are reduced to Cu+ ions in the same manner as in the case of Fehling’s reagent.
Aldehydes also give pink color with Schiff’s reagent (the reagent is prepared by
decolorizing aqueous solution of p–rosaniline hydrochloride dye by adding sodium
sulphite or by passing SO2 gas). Ketones do not respond to this test.
A. Test given by both aldehydes and ketones: (2,4-Dinitrophenylhydrazine test)
Procedure: Take 2-3 drops of the liquid compound in a test tube or in case of solid
compound, dissolve a few crystals of it in 2-3 ml alcohol. Add a few drops
of an alcoholic solution of 2,4-dinitrophenylhydrazine. Appearance of
yellow, orange or orange-red precipitate confirms the presence of carbonyl
group. If precipitate does not appear at room temperature, warm the mixture
in a water bath for a few minutes and cool.
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B. Tests given by aldehydes only
Following tests namely Schiff’s test, Fehling’s test and Tollen’s test are given by
aldehydes only.
Schiff’s test
Procedure: Take 3-4 drops of the liquid compound or dissolve a few crystals of organic
compound in alcohol and add 2-3 drops of the Schiff’s reagent. Appearance of pink
colour indicates the presence of an aldehyde.
Fehling’s test
Procedure: Take nearly 1 ml of Fehling’s solution A and 1 ml of Fehling’s solution B in
a clean dry test tube. To this add 2-3 drops of the liquid compound or about
2 ml of the solution of the solid compound in water or alcohol. Heat the
content of the test tube for about 2 minutes in a water bath. Formation of
brick red precipitate of copper (I) oxide indicates the presence of an
aldehyde. This test is not given by aromatic aldehydes.
Benedicts test
Procedure: Add 5 drops of the liquid compound or the solution of the solid organic
compound in water or alcohol to 2 ml Benedict’s reagent. Place the test tube
in boiling water bath for 5 minutes. An orange-red precipitate indicates the
presence of an aldehyde.
Tollen’s test
Procedure: (i) Take 1 ml of freshly prepared (~ 2 %) silver nitrate solution in a test tube.
Add 1-2 drops of sodium hydroxide solution to it and shake, a dark
brown precipitate of silver oxide appears. Dissolve the precipitate by
adding ammonium hydroxide solution drop-wise.
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(ii) To the above solution, add an aqueous or an alcoholic solution of the
organic compound.
(iii) Heat the reaction mixture of step (ii) in a water bath for about 5 min.
Formation of a layer of silver metal on the inner surface of the test tube
which shines like a mirror, indicates the presence of an aldehyde.
4.4.5 Test for carboxylic (—COOH) group
Organic compounds containing carboxyl functional groups are called carboxylic acids.
The term carboxyl derives its name from the combination of words carbonyl and
hydroxyl because carboxylic functional group contains both of these groups. These acids
turn blue litmus red and react with sodium hydrogencarbonate solution to produce
effervescence due to the formation of carbon dioxide. This is a test that distinguishes
carboxylic acids from phenols.
RCOOH + NaHCO3 RCOONa + H2O + CO2
They also react with alcohols in the acidic medium to produce esters.
RCOOH + R'OH Conc.H2SO4 RCOOR' + H2O
A. Litmus test
Procedure: Put a drop of the liquid compound or a drop of the solution of the compound
with the help of a glass rod on a moist blue litmus paper. If the blue color of
the litmus paper changes to red, the presence of either a carboxylic group or
a phenolic group is indicated.
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B. Sodium hydrogencarbonate test
Procedure: Take 2 mL of saturated aqueous solution of sodium hydrogencarbonate in a
clean test tube. Add a few drops of the liquid compound or a few crystals of
solid compound to it. The evolution of brisk effervescence of CO2 indicates
the presence of carboxyl group.
C. Ester test
Procedure: Take about 0.1 g compound in a test tube, add 1 ml ethanol or methanol and
2-3 drops of concentrated sulphuric acid. Heat the reaction mixture for 10-
15 minutes in a hot water bath at about 50°C. Pour the reaction mixture in a
beaker containing aqueous sodium carbonate solution to neutralize excess
sulphuric acid and excess carboxylic acid. Sweet smell of the substance
formed indicates the presence of carboxyl function in the compound.
4.4.6 Test for amino (—NH2, —NHR, —NR2) group
Organic compounds containing amino group are basic in nature. Thus they easily react
with acids to form salts, which are soluble in water. Both, aliphatic and aromatic amines
are classified into three classes namely–primary (–NH2), secondary (-NH-) and tertiary
(-N<), depending upon the number of hydrogen atoms attached to the nitrogen atom.
Primary amine has two hydrogen atoms, secondary has one while tertiary amine has no
hydrogen atom attached to nitrogen.
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A. Solubility test
Procedure: Take 1 ml of given organic compound in a test tube and add a few drops of
dilute HCl to it. Shake the contents of the test tube well. If the organic
compound dissolves, it shows the presence of an amine.
C6H5NH2 + HCl C6H5NH3+ Cl–
(Anilinium chloride soluble in water)
C. Carbylamine test
Procedure: Take 2-3 drops of the compound in a test tube and add 2-3 drops of
chloroform followed by addition of an equal volume of 0.5 M alcoholic
potassium hydroxide solution. Heat the contents gently. An
obnoxious/unpleasant smell of carbylamine confirms the presence of
primary amino group in the compound.
Caution! Do not inhale the vapours. Destroy the product immediately by adding
concentrated hydrochloric acid and flush it into the sink.
D. Azo dye test
Procedure: (i) Dissolve nearly 0.2 g of the compound in 2 ml of dilute hydrochloric acid
in a test tube. Cool the content of the test tube in ice.
(ii) To the ice cooled solution adds 2 ml of 2.5% cold aqueous sodium nitrite
solution.
(iii) In another test tube, dissolve 0.2 g of β-naphthol in dilute sodium
hydroxide solution.
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(iv) Add diazonium chloride solution prepared in step (ii) into the cold β-
naphthol solution slowly with shaking. The formation of a scarlet red dye
confirms the presence of aromatic primary amine.
5. Synthesis of solid derivatives
The preliminary examination and group classification tests indicate the particular class
(functional group) to which an unknown organic compound may belong. Further
characterization and identification depends on the selection and preparation of a suitable
solid derivative and accurate determination of its melting point.
Plan ahead! Write out the procedures for the derivative test before you come to
laboratory.
The last step in the identification of an unknown compound, the experiment any
remaining doubt, is the synthesis of a solid derivatives whose melting point can be
compared with the melting points of known compounds.
Derivatization procedures have somewhat diminished in importance with the advent of
spectroscopy. However, this procedure still provides both physical data and an in insight
into the chemistry of the unknown, especially when the possibilities for the identity of the
unknown have similar boiling and melting points and similar spectra.
Derivatization is the process of conversion of one common organic compound into
another. Many derivatives are really synthesis.
Useful solid Derivatives
Alcohol
3,5-Dinitrobenzoate
1-Naphthylurethanes
Phenylurethanes
4-Nitrobenzoate
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Phenols
1-Naphthylurethanes
3,5-Dinitrobenzates
Phenylurethanes
4-Nitrobenzoates
Acetates
Aldehydes and Ketones
2-4-dinitrophenylhydrazones
Semicarbazones
4-Nitrophenylhydrazones
Phenylhydrazones
Oximes
Oxidation to acid (for ardehyde)
Carboxylic acids
Amides
Anilides
4-Toluidides
Neutralization equivalent
4- Nitrobenzyl esters
Amines- 1O and 2O
Acetamides
benzamides
4-toluenesulfoamides
Benzenesulfonamides
Phenylthioureas
Amine hydrochlorides
Amines- 3O
Methyl 4–toluenesulfonate
Methiodides
21
Picrate
Amine hydrochlorides
5.1Preparation of derivatives for alcohols
The most general derivatives of alcohols are urethanes and benzoate esters; the former are
best for 1O and 2O alcohols, whereas the latter are useful for all types of alcohols.
Urethane derivatives are prepared when the alcohol is treated with either phenyl
isocyanate ar naphthyl isocyanate.
Benzoate esters are easily prepared from the reaction of the alcohol with 4-nitrobenzoly
chloride or 3,5-dinitrobenzoyl chloride.
22
The reaction of alcohols with 3-nitrophthalic anhydride produces hydrogen 3-
nitrophthalate derivatives.
A. Synthesis of Benzoate
Procedure: To 1 ml of benzoyl chloride in a test tube and 2 ml (1 g) of alcohol and 2
ml of strong NaOH solution. Warm the tube gently and pour the contents in
25 ml water. Filter the solid formed, wash with water, dry and determine the
melting point.
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B. Synthesis of acetate
Procedure 1: To 1 gm substance in a dry test tube adds 1 ml acetyl chloride. Warm and
then pour in water after cooling. Filter the derivative formed; wash with
water: dry and determine the melting point.
Procedure 2: Add 4 ml of acetic anhydride and 3 drops of conc. H2SO4 to a test tube with
1gm of substance. Warm in a water bath for 15 min. Cool and pour in a
beaker with 10 ml of water. Filter the derivative formed, wash with water,
crystallize from water, dry and determine the melting point.
C. Synthesis of phenyl urethanes
Procedure: Place 1 ml of alcohol and 3 gm phenyl isocyanate in a dry test tube and keep
the tube in boiling water for 10 min. Cool and allow to stand until crystals of
phenyl urethane derivative appears. Filter, wash with water, crystallize from
alcohol, dry and determine the melting point.
D. Synthesis of 3,5-dinitrobenzoates
Procedure: Place 0.5 g of 3,5-dinitrobenzoic acid and 1g of PCl5 in a dish. To this
semisolid add 1 ml of alcohol and stir. Warm on a water both and allow
standing for about 2 min. Filter the solid formed, wash with cold water,
crystallize from alcohol on petroleum ether, dry and determine the melting
point.
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5.2 Preparation of derivatives for Phenols
Benzoate, acetates, and bromo compounds are useful derivatives of phenols.
A. Synthesis of Benzoates
Procedure: To the phenol (0.5 g) is added 5% sodium hydroxide (10 ml) in a well-
corked boiling tube or a small conical flask. Benzoyl chloride (2 ml) is
added in small quantities at a time, and the mixture shaken vigorously with
occasional cooling under the tap or in ice-water. After 15 min the solid
benzoate separates out: the solution should be alkaline at the end of the
reaction; If not alkaline, or if the product is oily, add a solid pellet of sodium
hydroxide and shake again. Collect the benzoate, wash thoroughly with cold
water, and recrystallise from alcohol or light petroleum.
B. Synthesis of Acetates
Procedure: Acetates of many simple phenols are liquids; however, this is a suitable
derivative for polyhydric and substituted phenols. The phenol (0.5 g) is
dissolved in 10% sodium hydroxide solution and an equal quantity of
crushed ice is added, followed by acetic anhydride (2 ml). The mixture is
vigorously shaken in a stoppered test tube until the acetate separates. The
product is filtered and recrystallised from alcohol.
C. Synthesis of Bromo derivatives
Procedure: The phenol (0.3 g) is suspended in dilute hydrochloric (10 ml) and bromine
water added drop wise until no more decolorization occurs. The bromo
25
derivative which precipitates out is filtered off and recrystallised from
alcohol.
5.3 Preparation of derivatives for aldehydes and ketones
For low-molecular-weight, water-soluble aldehydes and ketones, it is often advantageous
to prepare the semicarbazones. This is usually synthesized from the reaction of the
aldehyde or ketone with semicarbazide hydrochloride.
The most useful derivatives of carbonyl compounds are hydrazones, 2,4-dinitrophenyl
hydrazones, Phenylhydrazones, 4-nitrophenylhydrazones. These devivatives are prepared
from the reaction of the aldehyde or ketone with hydrazine, 2,4-dinitrophenyl hydrazine,
phenyl hydrazine a 4-nitrophenyl hydrazine respectively.
26
A. Synthesis of Semicarbazones
Procedure: Dissolve 1 ml on 1 g of the aldehyde or ketone, 1 g of semicarbazide
hydrochloride and 1.5g of sodium acetate in 10 ml of water in a test tube.
Shake the mixture vigorously, and place the test tube in a beaker or boiling
water for 5 min. Remove the test tube from the beaker, allow it to cool, and
place it in a beaker of ice. Scratch the sides of the tube with a glass rod.
Isolate the crystals of the semicarbazone by filtration and recrystilize from
water or 25 to 50% ethanol.
B. Synthesis of 2,4- Dinitrophenylhydrazones
Procedure: dissolve 100 mg or 3 drops of the unknown carbonyl compound in 2ml of
ethanol and add 3 ml of the 2,4–dinitrophenylhydrazine reagent. A large
quantity of crystals usually forms immediately; however, heating the
reaction mixture in water both at 50OC for 2 min may be necessary to
produce a derivative. Recrystallize the 2,4-dinitrophenylhydrazone from 30
ml of 95% ethanol by heating on a stream both. If the derivative dissolves
immediately, add water slowly until the cloud point is reached or until a
maximum of 5 ml water is used. If the derivative does not dissolve, add ethyl
acetate to the hot mixture until the derivative dissolves. Filter the hot solution
and allow stand at room temperature until crystallization is complete. Isolate
the crystals by suction filtration.
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Preparation 2,4-dinitrophenylhydrazine (2,4-DNPH) reagent: In a small flask, add
0.4 g of 2,4-dinitrophenylhydrazine to 2 ml of conc. H2SO4. Add 3 ml of water drop wise,
with stirring, until the 2,4-DNPH is dissolves. Add 10 ml of 95% ethanol.
C. Synthesis of oximes
Procedure: dissolve 0.5 g of the hydroxylamine hydrochloride in 3 ml of water. Add
2 ml of 10% NaOH and 0.2 g of the aldehyde or ketone. [If the carbonyl
compound is water insoluble, add a minimal amount of ethanol (EtOH) to
the mixture to give a clear solution]. Warm the mixture on steam bath for
10 min and cool in an ice bath. Scratch the sides of the flask with a glass rod
to hasten crystallization.
5.4 Preparation of derivatives for Carboxylic acids
Three satisfactory derivatives of carboxylic acids are amides, anilides and p-toluidides.
These derivatives are prepared by treating the corresponding acid chlorides with
ammonia, aniline on p-toluidine.
The acid chlorides are most conveniently prepared from the acid and thionyl chloride
/Phosphorous pentachloride.
ROCl + 2NH3 RCONH2 + NH4Cl
acid chloride amide
ROCl + 2H2NC6H5 RCONHC6H5 + C6H5NH3+
Cl-
acid chloride aniline anilide
ROCl + 2H2NAr RCONHAr + ArNH3+
Cl-
acid chloride p-toluidine p-toluidide
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RCO2H + SOCl2 RCOCl + SO2 + HCl
RCO2H + PCl2 RCOCl + POCl3 + HCl
A. Synthesis of Amides
Procedure 1: perform this reaction in the hood. Mix 1 g of the acid with 5 ml of SOCl2.
Add one drop of N,N–dimethylformamide (DMF). Heat under reflux for
15-30 min. Pour the mixture cautiously in to 15 ml of ice-cold, conc.
aqueous ammonia. Extract the solution with three 10 ml portions of
methylene chloride. Isolate the methylene chloride (CH2Cl2) layer and
evaporate to dryness. Recrystallize the crude amide from ethanol.
Procedure 2: Or mix 1 g of the acid with 1 g PCl5 in a dry dish and heat for a minute. To
half the crude acid chloride add 10 ml conc. NH4OH. Cool and filter, wash
the residue with water. Recrystallize the product from alcohol or water, dry
and determine the melting point.
B. Synthesis of anilides and 4-toluidides
Procedure: perform the experiment in the hood. Mix 1 g of the acid with 2 ml of SOCl2.
Heat the mixture at reflux for 3 min. Cool the mixture, and add a solution of
1-2 g of aniline or 4-toluidine in 30 ml of benzene, and heat the mixture in a
hot-water bath for 2 min. Pour the benzene solution in to a separatory
funnel and wash successively with 2 ml of water, 5 ml of 5% HCl, 5 ml of
5% NaOH recover the benzene solution into a separation funnel and wash
successively with 2 ml of water, 5 ml of 5% HCl, 5 ml of 5% NaOH
solution and 2 ml of water. In the hood recover the benzene by distillation,
using a steam bath, and recrystallize the resulting amide from water or
ethanol.
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5.5 Preparation of derivatives for amines
Suitable derivatives of primary and secondary amines are acetamides, benzamides, and
benzenesulphonamides.
These transformations are satisfactory for derivatizing most primary and secondary
amines, but tertiary amines do not undergo such reactions. However, such amines form
salts that constitute solid derivatives. Thus a useful crystalline salt is formed upon
reaction with methyl iodide to afford a methiodide.
Picrate is also a useful derivative of tertiary amine. It is formed when picric acid is
reacted with tertiary amine.
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A. Synthesis of Acetamides
Procedure: Reflux gently in a small dry flask under a dry condenser the amine (1 g) with
acetic anhydride (3 ml) for 15 min. Cool the reaction mixture and pour into
20 ml cold water. Boil to decompose the excess acetic anhydride. Cool and
filter by suction the insoluble derivative. Recrystallise from ethanol.
B. Synthesis of Benzamides
Procedure: Suspend 1 g of the amine in 20 ml of 5% aqueous sodium hydroxide in a
well-corked flask, and add 2 ml benzoyl chloride (fume hood!), about 0.5 ml
at a time, with constant shaking. Shake vigorously for 5 - 10 min until the
odor of the benzoyl chloride has disappeared. Ensure that the mixture
remains alkaline. Filter off the solid derivative, wash with a little cold water
and recrystallise from ethanol.
C. Synthesis of Benzenesulphonamides
Procedure: To 1 g of the amine in 20 ml of 5% sodium hydroxide solution in a well-
corked flask adds 1 ml benzenesulphonyl chloride (fume hood!). Shake the
mixture until the odor of the sulphonyl chloride disappears. Check that the
solution is alkaline. Acidify if necessary to obtain the precipitated
derivative. Concentrated hydrochloric acid added dropwise should be used.
Filter the product, wash with a little cold water and suck dry. Recrystallise
from ethanol.
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D. Synthesis of Methiodides
Procedure: Mix 0.3 g of the amine with 0.3 ml methyl iodide in a test tube, warm the
mixture for several min. and then cool it in an ice bath-water bath. If
necessary scratch at the air-liquid interface with glass rod to induce
crystallization. Isolate the product by vacuum filtration and purity it by
recrystallization from absolute ethanol or methanol or from ethyl acetate.
E. Synthesis of Picrate
Procedure: Add 0.3 – 0.5 g of your unknown to 10 ml of 95% ethanol. If the sample
does not dissolve, shake the mixture until a saturated solution results and
then filter. Add the filtrate to 10 ml of a saturated solution of picric acid in
95% ethanol, and heat the solution to boiling. Allow the solution to cool
slowly, and remove the yellow crystals of picrate by filtration.
6. Spectroscopic analysis
IR and NMR spectroscopy are crucial to organic structural determination. IR analysis is
an excellent functional group probe, which can be used in conjunction with the functional
group tests. Use of both IR and chemical tests may lead to structural diagnosis. NMR also
aids in the structural determination. NMR is essentially a method of determining the
relative positions and numbers of spin-active nuclei. Both 1H NMR and 13C NMR spectra
can yield useful information concerning the types of protons or carbon present, such as
aromatic or aliphatic; the number of adjacent protons (for 1H NMR); and the number of
protons attached to a particular carbon. Once some preliminary structures are chosen, MS
can be used to narrow down the choices by utilization of fragmentation patterns and
molecular weight.
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Discussion Questions
(i) What is Bayer’s reagent?
(ii) Why do alkenes and alkynes decolorize bromine water and alkaline KMnO4?
(iii) Explain why for the confirmation of unsaturation in a compound both the tests
namely test with bromine water and test with Bayer’s reagent should be performed.
(iv) Why does phenol decolorize bromine water?
(v) How will you distinguish between phenol and benzoic acid?
(vi) Why does benzene not decolorize bromine water although it is highly unsaturated?
(vii) Why does formic acid give a positive test with Tollen’s reagent?
(viii) Outline the principle of testing glucose in a sample of urine in a pathological
laboratory?
(ix) Why is Benedict’s reagent more stable than Fehling’s reagent?
(x) How would you distinguish an aldehyde from a ketone by chemical tests?
(xi) How would you separate a mixture of phenol and benzoic acid in the laboratory by
using chemical method of separation?
(xii) Write the chemistry of diazotization and coupling reactions.
(xiii) How can you distinguish between hexylamine (C6H13NH2) and aniline (C6H5NH2)?
(xiv) How can you distinguish between ethylamine and diethylamine?
(xv) How can CH3OH and C2H5OH be distinguished chemically?
(xvi) Why is solution of iodine prepared in potassium iodide and not in water?
(xvii) What is haloform reaction? What types of compounds generally give this reaction?
(xviii) How can you distinguish the compounds 2-pentanone and 3-pentanone by simple chemical test?
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