Study Material-Organic Chemistry - Some Basic Principles and Techniques Class 11 PDF

Certainly! "Some Basic Principles and Techniques" is a crucial topic in the field of organic chemistry, especially for Class 11 students. This topic covers fundamental principles and techniques that are essential for understanding the behavior of organic compounds.

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Chapter – 33


The determination of structure of an organic compound involves the following steps
(i) Purification of the compounds
(ii) Qualitative analysis for determining the elements present
(iii) Quantitative analysis of elements detected under (ii)
(iv) Determination of molecular mass and
(v) Determination of structural formula by physicochemical and spectroscopic methods.



In order to obtain the compound in its purest form, the followings techniques can be used
(i) Filtration
(ii) Crystallization
(iii) Sublimation
(iv) Distillation
(v) Differential extraction
(vi) Chromatography
Once the compound has been purified, its purity is checked by determining its melting or boiling
point. Most of the pure compounds have sharp melting and boiling points.
1.1 FILTRATION
The process of filtration is used to separate insoluble solid component of a mixture from the
soluble component of a mixture in a given solvent. For example, a mixture of naphthalene and urea can be
separated using water as solvent. Urea dissolves in water while naphthalene remains insoluble. Upon
filtration, naphthalene remains on the filter paper while urea is recovered from the filtrate by evaporating
water.
If the water soluble component of a mixture is appreciably soluble in hot water but only sparingly
soluble in cold water, then to separate the mixture, filtration of a hot solution is required. If the solution to
be filtered is sufficiently large, then filtration is done through a ‘hot water funnel’ to avoid formation of
crystals in the funnel and its stem. The jacket of the hot water funnel keeps the solution hot in the glass
funnel placed in it.
For example, a mixture of anthracene and benzoic acid is separated by dissolving the mixture in
hot water and filtering the hot solution. Benzoic acid dissolves in hot water but anthracene does not. Upon
filtration, anthracene remains as a residue on the filter paper while benzoic acid crystallizes from the
filtrate on cooling.
PURIFICATION

1

PRACTICAL ORGANIC CHEMISTRY

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Hot water funnel
Filtering the solution of the substance using hot water
funnel to prevent crystallization during filteration

Sometimes, the filtration is very slow and takes a long time. In such cases, filtration is carried out
under reduced pressure using a Buchner funnel and water suction pump are shown in figure.
Crystals
Filter paper
Perforations
Buchner
funnel
Suction
pump
Filtrate
To sink
Filtration flask
Quick filtration process using a
Buchner funnel and a suction pump
1.2 CRYSTALLISATION
The process by which an impure compound is converted into its crystals is known as
crystallisation. This is one of the most commonly used techniques for purification of solid organic
compounds. It is based on the difference in the solubilities of the compound and the impurities in a suitable
solvent. The impure compound is dissolved in a suitable solvent in which it is sparingly soluble at room
temperature but appreciably soluble at higher temperature. The solution is concentrated to get nearly a
saturated solution. When this saturated solution is cooled, crystals of pure substance will separate out
which are removed by filtration. The filtrate, i.e., mother liquor contains the impurities alongwith small
quantity of the compound. If the compound is highly soluble in one solvent and too little soluble in another
solvent, then crystallization can be carried out in a mixture of these solvents taken in a suitable ratio.
A suitable solvent which fulfills the following conditions is selected for crystallisation:
(a) It should not react chemically with the impure substance.
(b) It should dissolve more of the substance upon heating than at room temperature so that the
excess of the substance is thrown out upon cooling.
(c) Either the impurities should not dissolve at all in the solvent, or if they dissolve, they should be
soluble to such an extent that they remain in the solution, i.e., in the mother liquor upon crystallisation.
The various solvents which are commonly employed for crystallisation are water, alcohol, ether,
chloroform, carbon tetrachloride, benzene, acetone, ethyl acetate, petroleum ether etc.
Coloured impurities if present are removed first by boiling the solution of the impure substance
with activated charcoal for about 15-20 minutes. The charcoal is filtered out and the filterate is allowed to
cool when crystals of pure substance will separate out.
If the compound and the impurities have comparable solubilities, then repeated crystallizations
may be necessary to purify the substance. For example,
(i) Crystallisation of sugar

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Suppose we have a sample of sugar containing an impurity of common salt (sodium chloride). This
can be purified by shaking the impure solid with hot ethanol at 348 K. The sugar will dissolve whereas the
common salt remains insoluble. The hot solution is filtered, concentrated and then allowed to cool when
crystals of sugar will separate out. In this case, had water been used as a solvent, the purification of sugar
would not have been possible since both sugar and common salt are readily soluble in water.
(ii) Crystallisation of benzoic acid
Suppose we have a mixture of benzoic acid and naphthalene. This mixture can be purified by
treating the impure solid with hot water. Benzoic acid will dissolve while naphthalene remains insoluble.
The hot solution is filtered and then allowed to cool when crystals of benzoic acid separate out. The
crystals are separated by filtration and dried. Had benzene been used as a solvent, instead of water in this
case, the purification of benzoic acid would not have been possible since both benzoic acid and
naphthalene are quite soluble in benzene.
1.3 SUBLIMATION
It involves the direct conversion of a solid into the gaseous state on heating without passing
through the intervening liquid state and vice versa on cooling.
Only those substances whose vapour pressures become equal to the atmospheric pressure much
before their respective melting points are capable of undergoing sublimation. Such substances are called
sublimable. Since the number of such compounds is expected to be small, therefore, the process of
sublimation is not of general application.
The process of sublimation is very useful in the purification of such solids which sublime on
heating and are associated with non-volatile impurities.
The impure substance is taken in a china dish covered with a perforated filter paper over which an
inverted funnel is placed. The stem of the funnel is plugged with a little cotton as shown in the figure. On
heating the dish on a sand bath, vapours of the volatile solid rise up, pass through the holes in the filter
paper and condense on the cooler walls of the funnel leaving behind the non-volatile impurities in the dish.
Cotton plug
Sublimate
Perforated filter paper
Crude organic substance

Camphor, naphthalene, anthracene, benzoic acid, iodine etc. are purified by this process.
In case of organic compounds which are decomposed by heat, sublimation is done under reduced
pressure.
1.4 DISTILLATION
1.4.1 Simple Distillation
Liquids are generally purified by simple distillation.
Distillation involves conversion of a liquid into vapours by heating followed by condensation of
the vapours thus produced by cooling.
The method is commonly used for those liquids which are sufficiently stable at their boiling points
and which contain non-volatile impurities. For example, simple organic liquids such as benzene, ethanol,
acetone, chloroform, carbon tetrachloride, toluene, xylenes, etc. can be purified by simple distillation.
Procedure : The apparatus used for simple distillation is shown in figure. When the flask is
heated, the temperature rises gradually and the liquid starts boiling when its vapour pressure becomes equal
to the atmospheric pressure. These vapours as they pass through the condenser are condensed.

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Impure
liquid
To sink
Water
Condenser
Adapter
Pure liquid
From
tap
Apparatus for Simple Distillation

Only the liquid which distills at a constant temperature is collected in a receiver. This gives us the
pure liquid.
Purification of a mixture of liquids: Simple distillation can also be used for the separation and
purification of a mixture of two or more miscible organic liquids provided their boiling points differ by 30–
50 K. The separation is based upon the fact that at the b.p. of the more volatile liquid (low boiling) of the
mixture, the vapours almost exclusively consist of the more volatile liquid. Likewise, at the b.p. of the less
volatile liquid (high boiling), vapours almost entirely consist of the less volatile liquid since the more
volatile liquid has already distilled over. Thus, the separation of the liquid mixture into individual
components can be achieved at their respective boiling points; the more volatile component distills over
first while the less volatile component distills over afterwards. The non volatile impurities and impurities
of liquids having boiling points much higher than those of the two liquids separated above are, however,
left in the distillation flask. This method can be used to separate :
(a) a mixture of ether (b.p. 308 K) and toluene (b.p. 384 K),
(b) a mixture of hexane (b.p. 342 K) and toluene (b.p. 384 K),
(c) a mixture of benzene (b.p. 353 K) or chloroform (b.p. 334 K) and aniline (b.p. 457 K) and so
on.
1.4.2 Fractional distillation
If the boiling points of the two liquids of the mixture are very close to one another i.e. differ by 10
K or so, the separation cannot be achieved by the simple distillation method as described above. This is due
to the reason that at the b.p. of the more volatile liquid of the mixture there will be sufficient vapours of the
less volatile liquid as well. As a result, both the liquids of the mixture will distill together and the
separation is denied.
The separation of such a liquid mixture into individual components can, however, be achieved by
fractional distillation, which involves repeated distillations and condensations. Fractional distillation is
carried out using a fractionating column. It usually consists of a long glass tube with a wide bore either
packed with glass beads, small stones, porcelain rings or coke, or blown into a number of spherical or pear-
shaped bulbs. The actual purpose of the fractionating column is to increase the cooling surface area and to
provide hurdles or obstructions to the ascending vapours and descending liquid.
Procedure: The apparatus used for fractional distillation is shown in figure.

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To sink
Water
Condenser
Adapter
Cold water
from tap
Apparatus for Fractional Distillation using a Fractional column
Distillate
Mixture of
liquids
More volatile component

Less volatile

component

Receiver

Suppose we have a mixture of two liquids ‘A’ and ‘B’ of which ‘A’ is more volatile than ‘B’.
When such a liquid mixture is heated, the temperature rises slowly and mixture starts boiling. The vapours
formed mainly consist of the more volatile liquid ‘A’ with little of the less volatile liquid ‘B’. As these
vapours travel up the fractionating column, the vapours of the less volatile liquid ‘B’ condense more
readily than those of the more volatile liquid ‘A’. Therefore, the vapours rising up become richer in ‘A’
and the liquid flowing down becomes richer in ‘B’. This process of distillations and condensations is
repeated at every point in the fractionating column. As a result of series of successive distillations, by the
time the vapours reach the top of the column and escape into the condenser, they consist mainly of the
more volatile component ‘A’.
Similarly, after a series of successive distillations, the remaining liquid in the distillation flask gets
enriched in higher boiling component. Each successive condensation and vaporization is called theoretical
plate. Commercially, columns with hundreds of theoretical plates are available.
Application
(a) One of the technological applications of fractional distillation is to separate crude oil in
petroleum industry into various useful fractions such as gasoline, kerosene oil, diesel oil,
lubricating oil etc.
(b) Fractional distillation has also been used to separate acetone (b.p. 329 K) and methyl alcohol
(b.p. 338 K) from pyroligneous acid obtained by destructive distillation of wood.
1.4.3 Distillation under reduced pressure or Vacuum distillation
This method is used for the purification of high boiling liquids and liquids which decompose at or
below their boiling points.
Principle: A liquid boils when its vapour pressure becomes equal to the external pressure.
Obviously, the same liquid would boil at a lower temperature if the pressure acting on it is reduced. Since
the liquid now boils at a lower temperature, its decomposition does not occur.
With the commonly used water pumps in the laboratory, a pressure of 10–20 mm Hg can be
obtained. Under these conditions, the boiling points are reduced by about 100 degrees. However, with
vacuum pumps, pressure of the order of 0.1 mm Hg can be easily obtained.
Apparatus : The apparatus used for vacuum distillation is shown in figure.

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Given below are some of the examples where vacuum distillation has been used to purify liquids:
(i) Glycerol which decomposes at its boiling point (563 K) can be distilled without decomposition
at 453 K under 12 mm Hg pressure.
(ii) Concentration of sugarcane in sugar industry.
1.4.4 Steam distillation
This is a convenient method for the separation and purification or organic compounds (solid or
liquids) from non-volatile organic or inorganic impurities. This method is applicable to only those
compounds which are volatile in steam, insoluble in water, possess a high vapour pressure (10–15 mm Hg)
at 373 K and contain non-volatile impurities.
Steam distillation is particularly valuable when the substance to be purified boils above 373K at
760mm and decomposes at or below its boiling point. This is due to the fact that steam distillation makes
the high-boiling substances to distil at low temperature and hence avoids their decomposition. In this
respect, steam distillation is comparable to vacuum distillation even though there is no reduction in the
total pressure acting on the solution during the process of distillation.
Principle : In this method, a mixture of two immiscible liquids, i.e., water and an organic liquid is
heated. Each would exert its own vapour pressure independently of the other and the mixture will begin to
boil at a temperature when the sum of the vapour pressures of the organic liquid (p1) and that of water (p2)
becomes equal to the atmospheric pressure (p).
p = p1 + p2
Unless the vapor pressure of water or that of the organic liquid is zero, the temperature at which
the mixture boils must be lower than the normal boiling point of both the organic liquids and the water. In
other words, the organic liquids boils at a temperature lower than its normal boiling point and hence the
decomposition is avoided.
Procedure : The apparatus used for steam distillation is shown in figure.

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The impure organic compound mixed with water is taken in a round bottomed flask and
steam is passed. The mixture starts boiling when the combined vapour pressure becomes equal to
the atmospheric pressure. At this temperature, steam mixed with vapours of the compound passes
over to the condenser where they are condensed and collected in the receiver. The distillate
contains the desired substance and water which can easily be separated with the help of aseparating
funnel.
Some of the compounds which can be purified by this process are o-nitrophenol, bromo-benzene, aniline,
nitrobenzene, essential oils, turpentine oil etc.
1.5 DIFFERENTIAL EXTRACTION
This method is used to recover organic compounds (solids or liquids) from their aqueous solutions.
The process essentially involves the shaking of the aqueous solution of the organic compound in a
separating funnel with a suitable solvent which is immiscible with water but in which the organic
compound is very highly soluble. Ether, benzene, chloroform, carbon tetrachloride are some of the solvents
which are generally employed for extraction.
Procedure : The aqueous solution is mixed with a small quantity of the organic solvent in a
separating funnel. The funnel is stoppered and the contents are shaken thoroughly for sometimes when the
organic solvent dissolves the organic compound present in the aqueous solution. The separating funnel is
now allowed to stand for some time when the organic solvent and water form two separate layers. The
lower aqueous layer (when the organic solvent used is benzene or ether) is run out by opening the tap of
the funnel and the organic layer separated. The aqueous solution is again poured into the funnel, mixed
again with a small quantity of the organic solvent and the process is repeated several times till the entire
amount of the organic compound is extracted.

The organic layers from all the steps are taken in a distillation flask. The organic solvent is distilled
off leaving the organic compound in the distillation flask.
The efficiency of the process of extraction depends upon the number of times the extraction is
repeated. It has been found that with a given amount of the solvent, larger the number of extractions,
greater is the amount of the material extracted.

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This method is normally applicable to non-volatile compounds. For example, benzoic acid can be
extracted from its water solution using benzene.
If the organic compound is less soluble in the organic solvent, a very large quantity of the solvent
would be required to extract even a very small quantity of the compound. In such cases, the technique of
continuous extraction is employed. In this process, the same solvent is repeatedly used for extraction of the
compound. The most commonly used apparatus for this process is called soxhlet extractor.
1.6 CHROMATOGRAPHY
Chromatography is the most modern and versatile method used for the separation, purification and
testing the purity of organic compounds. This method was first discovered by Tswett, a Russian botanist, in
1906 for the separation of coloured substances (plant pigments) into individual components. Now this
method is widely used for separation, purification, identification and characterisation of the components of
a mixture, whether coloured or colourless.
Chromatography is essentially a physical method of separation. It is defined as follows :
The technique of separating the components of a mixture in which separation is achieved by the
differential movement of individual components through a stationary phase under the influence of a mobile
phase.
1.6.1 Types of chromatography
The stationary phase can be either a solid or tightly bound liquid on a solid support while the
mobile phase can be either a liquid or a gas. Depending upon the nature of the stationary and the mobile
phases, the different types of chromatographic techniques commonly used are given in table below.
Table : Some common types of chromatography
S. No. Type of Chromatography Mobile/
Stationary
Phase
Uses
1. Adsorption or Column
chromatography
Liquid/Solid Large scale separations
2. Thin layer chromatography
(TLC)
Liquid/Solid Qualitative analysis (identification and
characterisation of organic compounds).
3. High performance liquid
chromatography (HPLC)
Liquid/Solid Qualitative and quantitative analysis
4. Gas liquid chromatography
(GLC)
Gas/Liquid Qualitative and quantitative analysis
5. Paper or partition
chromatography
Liquid/Liquid Qualitative and quantitative analysis of polar
organic compounds (sugars, -amino acids)
and inorganic compounds.
Depending upon the principle involved, chromatography can be divided into the following two
categories :
(a) Adsorption chromatography
(b) Partition chromatography
(a) Adsorption chromatography
Principle: This method is based upon the differential adsorption of the various components of a
mixture on a suitable adsorbent such as silica gel or alumina. Since some compounds are more strongly
adsorbed than the other, they will travel through the column at different rates and thus get separated.
Types of adsorption chromatography
Adsorption chromatography is of the following two types :
(I) Column Chromatography

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(II) Thin layer chromatography
(I) Column chromatography:
Column chromatography is the simplest of all the chromatographic techniques and is widely used.
The while process is carried out in a long glass column provided with a stopcock at the bottom. The
various steps involves in this process are:
(i) Preparation of the Column: A plug of cotton or glass wool is placed at the bottom of a clean
and dry glass column. Above this, a thin layer of acid-washed sand is placed to support the adsorbent. A
suitable adsorbent such as alumina (Al2O3), silica gel, magnesium oxide, starch, charcoal etc. is made into
slurry with a suitable solvent (preferably non-polar) such as hexane or petroleum ether. The slurry is then
carefully packed in the column by gentle tapping so that no air bubble is entrapped in the column. This
constitutes the stationary phase.
Band
formation
begins
Solvent
(Mobile Phase)
Sand + Mixture of
compounds
(A+B+C)
Adsorbent
(Stationary
Phase)
Glass wool
Sand
A
B + C
Sand
A
B
C
Sand
A
B
Sand
A
Sand
(i) (ii) (iii) (iv) (v) (vi)
Band
formation
completed
Completed
‘C’ eluted
Compound
‘B’ eluted
Compound
‘A’ eluted
Column Chromatography – stages (i), (ii), (iii), (iv), (v) and (vi) represents the progressive separation of the
mixture into three individual components.


(ii) Adsorption: The mixture to be separated (or the impure compound to be purified) is dissolved
in a minimum volume of a suitable highly polar solvent and applied on the top of the column of the
adsorbent with the help of a dropper or a microsyringe. As the solution travels down, the mixture is
adsorbed in a narrow band. A thin layer of acid-washed sand is again placed at the top of the column

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followed by a loose plug of cotton or wool. The sand layer prevents the column from being disturbed
during the addition of solvent from time to time.
After the application of the sample, a little amount of the solvent is placed over the sand layer and
the column is allowed to stand for about 15–20 minutes as shown in stage I. During this period, the various
components of the mixture (say A, B and C) are adsorbed to different extents depending upon their polarity
(say A > B > C) within a narrow band. Thus, within the narrow band, component A is strongly adsorbed,
component B is moderately adsorbed while component C is weakly adsorbed.
(iii) Elution: It is the process of extraction of the adsorbed components from the adsorbent with the
help of solvents of increasing polarity. The solvents usually employed in the increasing order of polarity
are petroleum ether, carbontetrachloride, benzene, chloroform, diethyl ether, ethyl acetate, acetone, alcohol
etc.
A solvent or a mixture of solvents which is used to extract the column constitutes the mobile phase
and is usually called an eluent. As the eluent passes down the column, it dissolves the different compounds.
The least strongly adsorbed component of the mixture i.e. component C is eluted first by the least polar
solvent followed by moderately strongly adsorbed component B by solvent of intermediate polarity, while
the most strongly adsorbed component A is eluted last of all by the solvents of higher polarity.
In this way, the various components of the mixture can be separated into different fractions.
Distillation or evaporation of the solvent from different fractions gives the various components of the
mixture in pure form.
This technique is being widely used throughout the world for the purification of different
substances and the separation of mixtures. For example, a mixture of naphthalene (hydrocarbon) and
benzophenone (ketone) can be separated over a column of alumina and by using petroleum ether
containing benzene as eluent. Naphthalene being less polar is weakly adsorbed while benzophenone being
more polar is strongly adsorbed over the column. Elution of the column will first elute naphthalene and
then benzophenone.
(II) Thin layer chromatography (TLC)
It is another type of adsorption chromatography in which separation of the components of a
mixture is achieved over a thin layer of an adsorbent. A thin layer (0.2 mm thick) of an adsorbent such as
silica gel (or alumina) is spread over a plastic or glass plate of suitable size.
A suitable TLC plate is taken and two pencil lines are drawn across the width of the plate about 1
cm from each end. The lower pencil line is called the starting line while the upper line is called the finish
line or solvent front.
A solution of the mixture to be separated is applied as a small spot with the help of a capillary on
the starting line. The plate is then placed in a closed jar containing a suitable solvent.

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As the solvent moves up, the components of the mixture also move up along the plate to different
distances depending upon their degree or extent of adsorption. When the solvent front reaches the finish
line, the plate is removed and then dried in air.
The spots of coloured components are visible on TLC plate due to their original colour. The spots
of the colourless components which are invisible to the eye can be observed using the following
visualization methods.
(i) Ultraviolet light: Organic compounds which fluoresce can be detected by placing the plate
under UV lamp having light of 254 nm. Since all organic compounds do not produce fluorescence under
UV light, this method is not of general applicability.
(ii) Iodine vapours: This is the most commonly used detection reagent. The developed TLC plate
is placed in a covered jar containing a few crystals of iodine. Spots of compounds which adsorb iodine will
show up as brown spots.
(iii) Chemical methods: Sometimes a suitable chemical reagent may be sprayed on the plate. For
example, amino acids can be detected by spraying the plate with ninhydrin solution. Similarly,
aldehydes/ketones can be detected by spraying the plate with the solution of
2,4-dinitrophenylhydrazine.
The various components on the developed TLC plate are identified through their retention factor,
i.e., Rf values. It is defined as : (Y) front solvent theby travelled Distance
(X) compound theby travelled Distance
R
f
.
Since the solvent front always moves faster on the TLC plate than the compounds,
Rf values are usually expressed as a decimal fraction.
(b) Partition chromatography:
We have discussed above the column chromatography or TLC is a liquid/solid chromatography
i.e., the mobile phase is a liquid while the stationary phase is a solid. In contrast, partition chromatography
is a liquid/liquid chromatography in which both the mobile phase and the stationary phase are liquids.
Paper chromatography is a type of partition chromatography. In paper chromatography, a special
quality paper called chromatographic paper is used. Although paper consists mainly of cellulose, the
stationary phase in paper chromatography is not the cellulose but the water which is adsorbed or
chemically bound to it. The mobile phase is another liquid which is usually a mixture of two or three
solvents with water as one of the components.
Principle : Paper chromatography works on the principle of partition i.e., it is based upon
continuous differential partitioning (or distribution) of the various components of the mixture between the
stationary and the mobile phases.
Process: A suitable chromatographic paper is selected and a starting line is drawn across the
width of the paper at about 1 or 2 cm from the bottom. A spot of the mixture of components to be separated
is applied on the starting line with the help of a fine capillary or syringe. The chromatographic paper is
then suspended in a suitable solvent mixture.

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The solvent rises up the paper by capillary action and flows over the spot. The different
components of the mixture travel through different distances depending upon their solubility in or
partitioning between the stationary and the mobile phases. When the solvent reaches the top end of the
paper, the paper is taken out and allowed to dry. The paper strip so developed is called the chromatogram.
The spots of the separated coloured compounds are visible at different heights from the starting line and
are identified by their Rf values as discussed under TLC. The spots of the colourless compounds may,
however, be observed either under ultraviolet light or by the use of an appropriate spray reagent.
The type of chromatography discussed above is called ascending paper chromatography. Alternatively, the
paper can be folded into a cyclinder and the two ends dipped together. This is also sometimes called
circular chromatography.

Illustration

Question: Name the methods by which we can separate the following:
(i) o–nitrophenol and p–nitrophenol
(ii) Benzoic acid and NaCl
Solution: (i) Steam distillation can be used to separate the o–nitrophenol from p–nitrophenol since the
former has got intramolecular hydrogen bonding and is steam distillable.
(ii) Benzoic acid can be sublimed while NaCl cannot be. Hence, sublimation can be used to
separate the two.

Illustration

Question: If a liquid ‘X’ has a vapour pressure of 0.02 atm and steam has a vapour pressure of p atm.
What is the external pressure at which ‘X’ is steam distilled.
Solution: External pressure = 0.02 + p atm.


Qualitative analysis of an organic compound involves following operations.
(i) Preliminary examination:
(a) Physical state (b) Colour (c) Odour (d) Ignition test (e) Solubility test
(ii) Detection of elements
(iii) Detection of functional groups
(iv) Determination of melting or boiling point and
(v) Preparation of derivatives.
2.1 DETECTION OF ELEMENTS
The elements generally present in organic compounds are carbon, hydrogen and oxygen. The
detection of these elements in the organic compounds is not essential. Organic compounds also contain
nitrogen, halogens, phosphorus and sulphur, in addition to these elements.
Thus, detection of elements basically refers to nitrogen, halogens, phosphorous & sulphur.
2.1.1 Detection of Carbon and Hydrogen
The organic compound is mixed with an equal amount of dry copper oxide and strongly heated.
The carbon present in the compound is oxidised to carbon dioxide (turns limewater milky) and hydrogen to
water (turns anhydrous copper sulphate blue).
C + 2CuO 
 CO2 + 2Cu
2H + CuO 
 H2O + Cu
CO2 + Ca(OH)2  CaCO3 + H2O
limewater white

QUALITATIVE ANALYSIS

2

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CuSO4H2O + 4H2O  CuSO45H2O
white blue
2.1.2 Lassaigne’s test
Nitrogen, sulphur and halogens in an organic compound are detected by Lassaigne’s test through
Lassaigne’s extract. Organic compounds contain these elements mostly linked through covalent bonds. In
order to make them free, we convert them into ionic form by treating the compound with sodium. The ionic
compound formed are watersoluble and releases the element in ionic form, which is easy to detect.
(a) Preparation of Lassaigne’s Extract
A small piece of dry sodium metal is heated gently in a fusion tube till it melts to a shining
globule. Then, a small amount of organic substance is added and the tube is heated strongly till it
becomes red hot. The red hot tube is plunged into distilled water contained in a china dish. The
fusion tube should break into pieces on adding to water. If it is not broken, it is broken using a
glass rod. Then the solution is boiled for some time. It is then cooled and filtered. The filtrate is
known as sodium extract or Lassaigne’s extract.
Sodium reacts with elements of the organic compound to give following reactions.
C + N + Na  NaCN
S + 2Na  Na2S
X + Na  NaX (where X = Cl, Br or I)
When nitrogen and sulphur both are present in the organic compound, then sodium thiocynate is
formed.
Na + C + N + S  NaSCN
All the sodium salts being soluble in water, can be easily detected.
2.1.3 Detection of Nitrogen
If the organic compound contains nitrogen, the sodium fusion extract will contain sodium cyanide
(NaCN). A small portion of the extract is boiled with ferrous sulphate solution and acidified with
dilute sulphuric acid. A blue precipitate of Prussian blue (ferric ferrocyanide) confirms the
presence of nitrogen.
2NaCN + FeSO4  Fe(CN)2 + Na2SO4
Fe(CN)2 + 4NaCN  Na4[Fe(CN)6]
(excess)
FeSO4 42SOH
byOxidised
  Fe2(SO4)3
3Na4[Fe(CN)6] + 2Fe2(SO4)3  Fe4[Fe(CN)6]3 + 6Na2SO4
Prussian blue
If the organic compound contains sulphur along with nitrogen and the sodium, present is in
insufficient amount then sodium thiocynate is formed, which gives blood red colouration by
reaction with ferric ions.
3NaSCN + Fe
3+
 [Fe(SCN)3] + 3Na
+

Blood red colour
However, the absence of blood red colouration does not necessarily mean that sulphur is absent in
the organic compound because NaCNS combines with Na to give NaCN and Na2S.
NaSCN + HCl  HSCN + NaCl
4HSCN + Co(NO3)2  H2[Co(SCN)4] + 2HNO3
Blue colour
2.1.4 Detection of Sulphur

15. All right copy reserved. No part of the material can be produced without prior permission




If the organic compound contains sulphur, the sodium fusion extract will contain sodium sulphide.
It is divided into two portions and following tests are performed.
Sodium Nitroprusside Test:
Add freshly prepared sodium nitroprusside solution to one portion. A deep violet colour confirms
the presence of sulphur.
colourViolet
]NOS)CN(Fe[Na]NO)CN(Fe[NaSNa
54522 

Lead Acetate Test:
Acidify the second portion with acetic acid and then add lead acetate solution. A black precipitate
of lead sulphide confirms the presence of sulphur.
COONaCH2PbSPb)COOCH(SNa
3232
ppt.black
acetate lead

2.1.5 Detection of Halogens
If the organic compound contains halogen, the sodium fusion extract will contain sodium halide.
The sodium fusion extract is boiled with dilute nitric acid to decompose sodium cyanide or sodium
sulphide, (if present) otherwise a white precipitate of silver cyanide or silver sulphide will be
formed even in the absence of halogen. The solution is then cooled and silver nitrate solution is
added. The characteristic precipitate confirms the presence of a halide.
33 NaNOAgXAgNONaX 
(a) White precipitate, soluble in aqueous ammonia indicates chlorine.
(b) Light yellow precipitate, sparingly soluble in aqueous ammonia indicates bromine.
(c) Pale yellow precipitate, insoluble in aqueous ammonia indicates iodine.
2.1.6 Detection of Phosphorus
The organic compound containing phosphorus is fused with sodium peroxide and extracted with water.
The aqueous solution is boiled with concentrated nitric acid and ammonium molybdate is added. A
canary yellow precipitate indicates the presence of phosphorus.
2.2 DETECTION OF FUNCTIONAL GROUPS
Detection of functional groups is an important step in the analysis of organic compounds.
It entirely depends on the correct determination of the constituent elements.
2.2.1 Tests for Carboxylic Acid Group
(i) Litmus test: Place a small quantity of the organic compound or its aqueous solution on a
piece of moist blue litmus paper. If the litmus paper turns red, carboxylic acid group may be
present.
(ii) Sodium bicarbonate test: Add a small quantity of the organic compound to sodium
bicarbonate solution taken in a testtube. Compound dissolves with brisk effervescences.
RCOOH + NaHCO 3  RCOONa + H2O + CO2 
(iii) Take a small quantity of the organic compound and add some alcohol in a testtube. Add a
few drops of conc. sulphuric acid and warm on a water bath, a fruity smell indicates the
formation of an ester. OHRRCOOOHRRCOOH
2
Ester
SOH
42


2.2.2 Tests for Aldehyde Group
(i) Schiff’s test: Take small amount of Schiff’s reagent in a testtube and shake with organic
compound (do not heat). A pink colour appears instantaneously.

16. All right copy reserved. No part of the material can be produced without prior permission




H2N

Cl

C
NH2
NH2
H2SO3
or SO2 gas, 2H2O
H2N

Cl

C
RCHO
HO3S
NHSOH
O
pRosaniline hydrochloride Schiff’s reagent (colourless)
H2N

Cl

C
NH2
NH2
(Pink colour)
+ RCHO.H2SO3
NHSOH
O

(ii) Fehling’s test: Mix equal volumes of Fehling’s (A) and Fehling’s (B) solution in a testtube.
Add a small amount of the organic compound and boil for some time. A red precipitate of Cu2O is
obtained. 
.pptdRe
OCuRCOOHCuO2RCHO
2

(iii) Tollen’s test: Add few drops of Tollen’s reagent ppt. to a small quantity of the organic
compound and heat on a water bath. A shining silver mirror is formed on the inner walls of
testtube.
reagents'Tollen mirrorSilver
Ag2OHNH2NHRCOONHOHOH])NH(Ag[2RCHO
434223 
(iv) Benedict’s test: Take small amount of Benedict’s reagent in a testtube. Add a few ml of the
organic compound (or a solution prepared in alcohol or water). Heat the mixture to boiling.
Formation of red precipitate indicates the presence of aldehydic group. OH3NaRCOOOCuNaOH)OH(Cu2RCHO
222 


2.2.3 Tests for Ketone Group
(i) Ketones, unlike aldehydes, do not restore the pink colour of Schiff’s reagent nor do they
reduce Fehling’s solution or ammoniacal silver nitrate solution. However, they form yellow or
red crystalline precipitate with 2,4dinitro phenylhydrazine and also with sodium bisulphite
reagent. This test should be conduct only after the confirmation of absence of aldehydes.

17. All right copy reserved. No part of the material can be produced without prior permission




C=O + H2NHN
R
R
NO2  C=NHN
R
R
NO2
NO2
(2,4 DNP)
NO2

Ketones add on sodium hydrogen sulphite to form crystalline bisulphite compounds.
(Bisulphite addition
compound)
C=O + NaHSO3 
R
R
C
R
R
OH
SO3Na

(ii) Sodium nitroprusside test: Treat some organic compound with few drops of
freshly prepared solution of sodium nitroprusside followed by excess of NaOH solution. A
winered colour is obtained.
2.2.4 Tests for Phenol Group
(i) Litmus test: Place a trace of the organic compound on the moist blue litmus paper.
It turns red.
(ii) Neutral FeCl3 test: When phenols are treated with neutral ferric chloride solution, they form
coloured complexes. The colour of the complex may be violet, red, blue
or green. This is a characteristic reaction of compounds having enolic group (=COH). All
stable enols respond to this test.
With relatively milder oxidizing agents such as ferric chloride (Fe
3+
), a resonance stabilized
phenoxyl radical is formed. This radical soon undergoes dimerization, which may take place
by orthoortho, orthopara, parapara coupling.
+ Fe
3+

OH

O O
H

O

H
O
H
O O
H H
+
O
H
O
H
+
H
O
H
O
+ Fe
+2
+ H
+

The phenoxyl radicals are highly coloured and therefore, products are also coloured.
(iii) Aniline Dye test: Dissolve a little aniline in dilute hydrochloric acid, cool in
icecold water and add aqueous solution of sodium nitrite dropwise. To this cold
solution, add organic compound already dissolved in sodium hydroxide solution.
A red or orange precipitate is obtained. NaClHNOHClNaNO
22 

18. All right copy reserved. No part of the material can be produced without prior permission





Cl

NH2 + HNO2 + HCl
Benzenediazonium
chloride
0°C
NN + 2H2O

Cl

NN + OH
NaOH
NaCl
H2O
N=N OH
Phenol phydroxy azobenzene (dye)

This test is applicable for the detection of small quantities of phenols with the para position
free.
(iv) Phthalein test (Fluorescein test) : Heat some organic compound and phthalic anhydride with
a few drops of concentrated sulphuric acid in a testtube. Pour the heated mixture into water
taken in a beaker and then add sodium hydroxide solution, till it became alkaline. A beautiful
pink or purple colour appears.
C=O H2SO4
O C O
+
OH
OH
C
OH
OH
2NaOH
C
O=CO

Na
+
O

Na
+
O
(Pink or purple)
H2O
O C O

(v) Bromine water test: Take aqueous or alcoholic solution of the organic compound in a
testtube. Add excess of bromine water. A yellowish white precipitate is obtained.
+ 3Br2 
OH
Phenol
OH
(2,4,6tribromophenol)
Br
Br Br
H2O

(vi) Ceric ammonium nitrate test: Add few drops of the ceric ammonium nitrate to
hot aqueous solution of the organic compound. A green or brown precipitate
is obtained. .pptbrownorGreenPhenol
3425643566324 NONH2])OHHC()NO(Ce[OHHC2])NO(Ce[)NH( 

19. All right copy reserved. No part of the material can be produced without prior permission




HO
NaNO2 + H2SO4
HO N=O
Tautomerize
O NOH
(Green colour)
Phenol
H2SO4
O N
Indophenol (red)
OH
NaOH
O N O

Na
+
+
Indophenol salt (blue)

2.2.5 Tests for Alcohol Group
Alcohols may be considered as neutral compounds. They are soluble in water or dioxane.
(i) Sodium test: Take some organic compound in a testtube. Add a thin slice of sodium metal
to it. A brisk effervescence due to the evolution of hydrogen takes place. (Organic compound
should be free from water). 

2HNaRO2Na2ROH2

(ii) Ester test: Warm the given liquid with acetic acid in presence of sulphuric acid.
A fragrant fruity smell indicates the formation of ester. OHRCOCHCOOHCHROH
223
SOH
3
42



Alcohols give crystalline esters with 3,5dinitrobenzoyl chloride. These esters have sharp
melting points and can be used for testing a given alcohol.
ROH + ClC
NO2
O NO2
ROC
O
+ HCl
NO2
NO2

(iii) Oxidation test: Take some organic compound in a test tube, add to it K2Cr2O7 solution and
concentrated H2SO4. Heat the testtube after fitting a delivery tube. The distillate is passed
into another testtube containing water. The solution becomes green coloured. Test the
aqueous solution of the distillate for carboxylic acid group. A positive test indicates the
presence of primary alcoholic group in the given compound. OHHRCOOHRCH
22
]O[
2 

(iv) Ceric ammonium nitrate test: Shake a few drops of ceric ammonium nitrate with
the organic compound. A red colour is produced. This test is useful only when the compound
contains less than 10 carbon atoms per molecule. 342436324 NONH2])ROH()NO(Ce[])NO(Ce[)NH(ROH2 

Distinction between primary, secondary and tertiary alcohols
Lucas test: Add some amount of alcohol to 2 ml of Lucas reagent (anhydrous ZnCl2 +
concentrated HCl) in a testtube. Shake the mixture and allow it to stand at room temperature. If a
cloudy precipitate of alkyl chloride is obtained immediately then it is a tertiary alcohol, if obtained
within 5 minutes then it is a secondary alcohol, while
no cloudy precipitate even after long standing means it is a primary alcohol. This test is
applicable for alcohols with less than 6 carbon atoms.

20. All right copy reserved. No part of the material can be produced without prior permission




ROH + ZnCl2
Cl

RO
…..
ZnCl2 
+ 
R

[Zn(OH)Cl2]

+
a
H
 RCl
Cloudy ppt.


2.2.6 Tests for Primary Amines (NH2)
(i) Carbylamine test: Heat the organic compound with alcoholic KOH and chloroform in a
testtube. A highly offensive smell is evolved due to the formation of isocyanides.
RNH2 + CHCl3 + 3KOH RN C + 3KCl + 3H2O

This test is applicable for aliphatic and aromatic primary amines but not for 2° and
3° amines.
(ii) Dye test: This test is applicable only for aromatic primary amines. Dissolve a small amount
of the organic compound in dilute HCl. Cool it in icecold water bath. Add cold solution of
sodium nitrite with constant stirring and then alkaline naphthol solution. An orangered
dye is obtained.
NaNO2 + HCl  HNO2 + NaCl


NH2 + HNO2 + HCl
naphthol
NN + 2H2O
Cl


NN +
Cl
OH
N=N
+ HCl
Orangered dye
OH
0°–5°C
0°–5°C

2.2.7 Tests for Secondary Amines (NH)
Liebermann’s nitroso test:
Dissolve some organic compound concentrated HCl and then add small amount of
water. Cool the solution in icecold water bath and then add cold dilute NaNO2 solution.
A yellow oily emulsion is produced. Take above emulsion in a test tube and add phenol and
concentrated H2SO4 to it. A green colour appears. Addition of water changes
green colour to red, which changes to deep blue on adding NaOH solution.
2.2.8 Tests for Nitro Group
(i) Reduction test: Take a small amount of the organic compound in a testtube. Add a few ml
of concentrated hydrochloric acid and one or two pieces of granulated zinc or tin. Heat the
mixture gently for few minutes. Decant the clear liquid into a testtube. Cool it in icecold
water bath and then add cold dilute solution of sodium nitrite. Pour this cold solution into an
alkaline solution of naphthol. An orangered dye is obtained. This test is applicable only
to aromatic nitro compounds.
]H[2ZnClHCl2Zn
2
OH2NHHC]H[6NOHC
2256256 
NaClHNOHClNaNO
22 
OH2ClNHCHClHNONHHC
22562256 


21. All right copy reserved. No part of the material can be produced without prior permission




naphthol


NN +
Cl
OH
N=N
OH
+ HCl
Orange red dye

(ii) Mulliken’s test: Take a small amount of the organic compound in a testtube. Add a few ml
of water or alcohol, 1 ml of calcium chloride or ammonium chloride solution and then a pinch
of zinc dust. Boil the mixture for few minutes. Cool and then filter. Treat this filtrate with
ammoniacal silver nitrate solution. A grey or black precipitate is obtained.
Zn + 2H2O  Zn(OH)2 + 2[H]
C6H5NO2 + 4[H]  C6H5NHOH + H2O
C6H5NHOH + Ag2O  C6H5NO + H2O + 2Ag 
Grey or black ppt.
This test is applicable for aliphatic as well as aromatic nitro compounds.



After determining the elements present in the organic compound, the subsequent step is to find out
the percentages of elements in the compound. The methods used for determining the percentage of
elements involve combustion, oxidation and precipitation etc. of an organic compound such that they are
converted into inorganic compounds, which can be easily estimated by gravimetric or volumetric
techniques.
3.1 ESTIMATION OF CARBON AND HYDROGEN
3.1.1 Liebig’s Combustion Method
A known mass of the organic compound is heated in a current of dry oxygen
(free from CO2) in the presence of cupric oxide till all the carbon is oxidised to carbon dioxide and
all the oxygen is oxidised to water.
CxHy + 






4
y
x O2  xCO2 + y/2 H2O

Water is absorbed in a previously weighed Utube containing anhydrous calcium chloride or anhydrous
magnesium perchlorate. Carbon dioxide is absorbed in a previously weighed Utube containing a strong
solution of potassium hydroxide or ascarite (NaOH + CaO). The weights of carbon dioxide and water thus
formed are determined and the amounts of carbon and hydrogen in the organic compound can be calculated
as
QUANTITATIVE ANALYSIS

3

22. All right copy reserved. No part of the material can be produced without prior permission




Moles of CO2 formed = 44
formedCOofMass
2 = Moles of C in CO2
= Moles of C in organic compound
 Mass of carbon in organic compound = 12
44
formedCOofMass
2

 Percentage of carbon in organic compound =100
compoundorganicofMass
formedCOofMass
44
12
2


Moles of H2O formed = 18
formedOHofMass
2
Moles of H in H2O = 2
18
formedOHofMass
2
 = Moles of H in the organic compound.
Mass of hydrogen in the organic compound = 18
12formedOHofMass
2 
 Percentage of hydrogen in organic compound =
100
compoundorganicofMass
formedOHofMass
18
2
2


3.2 ESTIMATION OF NITROGEN
Nitrogen can be estimated either by Dumas’ method or Kjeldahl’s method.
3.2.1 Duma’s Method
Known mass of an organic compound is heated with cupric oxide in an atmosphere of carbon
dioxide. The carbon and the hydrogen get oxidised to carbon dioxide and water while the nitrogen is set
free. Traces of oxides of nitrogen, which may be formed in some cases are reduced to elemental nitrogen
by passing over heated copper spiral. The gaseous mixture is collected over an aqueous solution of
potassium hydroxide. CO2 is absorbed by the caustic potash solution while H2O condenses and nitrogen is
collected by downward displacement of KOH solution.

The volume of nitrogen is measured and this is equivalent to the nitrogen present in the compound.
The volume of nitrogen is reduced to STP, its weight calculated and from this the percentage of nitrogen
present in the organic compound is estimated.
The following reactions take place during Duma’s process.
C + 2CuO  CO2 + 2Cu ; 2H + CuO  H2O + Cu
2N + CuO  N2 + Oxides of nitrogen
Oxides of nitrogen + Cu  CuO + N2

23. All right copy reserved. No part of the material can be produced without prior permission





Let the mass of organic compound = w g,
Volume of nitrogen collected = V cm
3
,
Pressure of moist N2 = P mm of Hg,
Room temperature = t°C,
Aqueous tension at t°C = a mm of Hg,
 Pressure of dry nitrogen = (P  a) mm of Hg,
Let us first convert the volume of nitrogen at given temperature & pressure to the volume of N2 at
STP.
22
NNVP = RTn
2N
)t273(0821.0760
10V)aP(
n
3
N
2




 Volume of N2 at STP = )t273(0821.0760
4.2210V)aP(
3


 = x L (say)
Now, 22.4 L of N2 at STP weigh = 28 g
 x L of N2 at STP will weigh =g
4.22
28






x
Percentage of nitrogen = 100
compoundorganicofMass
nitrogenofMass
 = w22.4
10028

x
where x is the volume of N2 (in litre) at STP.
3.2.2 Kjeldahl’s Method
This method is simpler and more convenient than Duma’s method. This method is largely used for
the estimation of nitrogen in food stuffs, drugs, fertilizers and many other organic compounds. However,
this method cannot be used for
(i) organic compounds containing nitrogen in the ring such as pyridine, quinoline etc.
(ii) organic compounds containing nitro (NO2) and diazo (N=N) groups.
Principle: A known weight of the organic compound is heated with concentrated H2SO4 so that
nitrogen is quantitatively converted into ammonium sulphate. The solution is then heated with excess of
sodium hydroxide. The ammonia gas evolved is passed into a known but excess volume of standard acid
(HCl or H2SO4). The acid left unused is estimated by titrating the solution with standard alkali. From the

24. All right copy reserved. No part of the material can be produced without prior permission




amount of acid left unused, the amount of acid used for neutralization of ammonia can be calculated. From
this, percentage of nitrogen can be calculated. The chemical reactions involved are
C, H, S  
4SO2H.Conc CO2 + H2O + SO2
(From organic compound)
N  
4SO2H.Conc (NH4)2SO4
(From organic compound) Ammonium sulphate
(NH4)2SO4 + 2NaOH  Na2SO4 + 2NH3 + 2H2O
2NH3 + H2SO4  (NH4)2SO4
(n = 1)
H2SO4 + 2NaOH  Na2SO4 + 2H2O
(excess)


Let the mass of organic compound = w g,
Total volume of standard acid taken = V1 cm
3
,
Normality of acid = N1
Now, the excess acid after dilution is titrated against standard alkali.
Volume of standard alkali used for neutralization of unused acid = V2 cm
3
,
Normality of standard alkali = N2,

Equivalents of unused acid = Equivalents of alkali = N2V2  10
3

Initial equivalents of acid = N1V1  10
3

Equivalents of acid consumed by NH3 = (N1V1  N2V2)10
3
= Equivalents of NH3 reacted.
Moles of NH3 reacted = (N1V1 N2V2)  10
3

Moles of NH3 liberated = (N1V1  N2V2)  10
3
= Moles of nitrogen in NH3
= Moles of nitrogen in organic compound.
Mass of nitrogen in the organic compound = (N1V1  N2V2)  10
3
 14
Percentage of nitrogen = 100
compoundorganicofMass
nitrogenofMass

= 100
w
1410)VNVN(
3
2211


 = w
)VNV(N1.4
2211
where V1 and V2 are the volume of standard acid and alkali respectively, in cm
3
.


3.3 ESTIMATION OF HALOGENS
3.3.1 Carius Method
In this method, a known mass of the organic substance is heated with fuming nitric acid in the
presence of silver nitrate in a special sealed tube known as Carius tube. Carbon, hydrogen or
sulphur present in the compound will be oxidised to CO2, H2O and H2SO4 respectively. The halogen
forms a precipitate of silver halide (AgX). The precipitate is filtered, washed, dried and weighed.
C + 2O 
3HNO CO2

25. All right copy reserved. No part of the material can be produced without prior permission




2H + O 
3HNO H2O
S + H2O + 3O 
3HNO H2SO4
X + AgNO3  AgX 
(halogen)

The Carius tube is a long narrow tube closed at one end and (during the experiment) sealed at the
other end. A weighed quantity of the organic substance is taken in the tube with fuming nitric acid
and silver nitrate, then the open end is sealed and the tube is heated in a furnace. The organic
compound decomposes and silver halide is formed. At the end of the experiment, the tube is cooled,
the sealed end broken and the contents are transferred to a filter. The precipitated silver halide is
filtered, washed, dried and weighed.
Moles of AgX =AgXofmassMolar
AgXofWeight = Moles of halogen in AgX
= Moles of halogen in organic compound
Mass of halogen in organic compound = AgXofmassMolar
enloghaofmassatomicAgXofWeight 
100
compoundorganicofwt.halogen)ofmassatomic(108
AgXofwt.halogenofmassAtomic



enloghaofPercentage




3.4 ESTIMATION OF SULPHUR
Sulphur is estimated by Carius method. The organic compound containing sulphur is heated with
fuming nitric acid. The sulphur in the compound is oxidised to sulphuric acid, which is then precipitated as
barium sulphate by adding excess of barium chloride solution. The precipitate is filtered, washed and dried.
From the weight of barium sulphate formed, the percentage of sulphur in the organic compound can be
estimated.
Moles of BaSO4 = 4
4
BaSOofmassMolar
BaSOofWeight = Moles of S in BaSO4
= Moles of S in organic compound.
Mass of sulphur in organic compound = 32
BaSOofmassMolar
BaSOofWeight
4
4

 100
compoundorganicofweightBaSOofmassMolar
BaSOofweight32
sulphurofPercentage
4
4




100
compoundorganicofweight233
BaSOofweight32
4






3.5 ESTIMATION OF PHOSPHORUS
A known mass of the organic compound is heated with fuming nitric acid, when the phosphorus
present in the organic compound is oxidised to phosphoric acid. The phosphoric acid obtained is treated

26. All right copy reserved. No part of the material can be produced without prior permission




with magnesia mixture (a mixture of MgCl2, NH4CI and NH4OH) to give a precipitate of Mg(NH4)PO4,
which on ignition gives Mg2P2O7.

Moles of Mg2P2O7 = 722
722
OPMgofmassMolar
OPMgofWeight

Moles P in Mg2P2O7 = 722
722
OPMgofmassMolar
OPMgofweight2 = Moles of P in organic compound
 Mass of phosphorous in organic compound = 31
OPMgofmassMolar
OPMgofweight2
722
722

 100
compoundorganicofweightOPMgofmassMolar
OPMgofweight312
phosphorusofPercentage
722
722





100
compoundorganicofweight222
OPMgofweight62
722




3.6 ESTIMATION OF OXYGEN
There is no direct method for the estimation of oxygen in a given organic compound. It is estimated by
subtracting the sum of the percentages of all other elements in the compound from 100.
Percentage of oxygen = 100  (% of all other elements)



From the percentage composition of the elements present in an organic compound, we can
calculate the empirical formula. To know the molecular formula, it is necessary to fix the molecular
weight. The following are some of the methods for determining the molecular weights of organic
compounds.
4.1 SILVER SALT METHOD
This method is used for determining the molecular mass of organic acids. Most of the organic
acids form insoluble silver salts, which upon ignition decompose to give residue of metallic silver.
RCO2H 
3NH RCO2NH4 34
3
NONH
AgNO

 RCOOAg 
Heat Ag
Silver salt Silver residue
A small amount of organic acid is dissolved in water, then a slight excess of NH3 solution is added.
The solution is then boiled to expel excess NH3 and then excess of AgNO3 is added, whereby white
precipitate of insoluble silver salt is obtained. The precipitate is filtered, washed, dried and weighed.
This weighed silver salt is taken in a crucible and ignited till decomposition is complete.
The crucible is cooled and then weighed. Heating, cooling and weighing the crucible is continued till we
get constant weight.
Let the weight of silver salt = w g and the weight of silver residue obtained = x g.
Since, one atom of hydrogen (H) is equivalent to one atom of silver (Ag), therefore, one equivalent
of silver salt will contain one atom of silver in place of hydrogen. Thus, if E be the equivalent mass of acid,
then equivalent mass of silver salt will be
(E  H + Ag) or (E  1 + 108) or (E + 107).
Equivalent of silver salt = Equivalent of silver
silverofweightEquivalent
silverofWeight
saltsilverofmassEquivalent
saltsilverofWeight

107E
w
 = 108
x
DETERMINATION OF MOLECULAR MASS

4

27. All right copy reserved. No part of the material can be produced without prior permission




Equivalent mass of Acid = E = 107108
w







x
Molecular mass of the acid = Equivalent mass  Basicity
Molecular mass of the acid = n107108
w














x (where n is the basicity of acid)
4.2 PLATINICHLORIDE METHOD
This method is used for determining the molecular masses of the bases. This method is based on
the fact that organic bases i.e. amines combine with chloroplatinic acid, H2PtCl6 to form insoluble double
salts known as chloroplatinates or platinichlorides. These salts when ignited leave a residue of metallic
platinum.
The base is dissolved in HCl and a solution of platinic chloride, PtCl4 is added, when a yellow
crystalline salt is precipitated. The salt is then filtered, washed, dried and weighed.
This weighed amount of salt is ignited in a crucible, which on decomposition gives metallic platinum that
is weighed. Heating, cooling and weighing the crucible is continued till we get constant weight.
Let the weight of chloroplatinate = w g and the weight of platinum obtained = x g.
Let E be the equivalent mass of the monoacidic base, then the molecular formula of the
chloroplatinate will be B2H2PtCl6.
 Molecular mass of chloroplatinate = 2E + 2 + 195 + (35.5  6) = 2E + 410
2RNH2 + H2PtCl6  (RNH3)2PtCl6
or 2B + H2PtCl6  B2H2PtCl6
One molecule of chloroplatinate on heating gives one atom of platinum.
B2H2PtCl6 
Heat Pt
salttheofmassMolar
salttheofWeight = platinumofmassAtomic
platinumofWeight
410E2
w
 = 195
x
2E = 











 410195
w
x
Molecular mass of base = 
2
n 











 410195
w
x (where n is the acidity of base)

4.3 RELATIVE LOWERING OF VAPOUR PRESSURE METHOD
The molecular weight (M1) of a nonvolatile organic compound can be determined by measuring
the relative lowering of vapour pressure of a solvent caused by dissolving a known mass of the solute
(nonvolatile organic compound) in a given mass of the solvent.
Relative lowering of vapour pressure = 2
2
1
1S
W
M
M
w
P
PP



 (for very dilute solutions)
where 
P = Vapour pressure of solvent in pure state,
PS = Vapour pressure of solution,
w1 = Weight of nonvolatile organic compound,
M1 = Molecular weight of nonvolatile organic compound,
W2 = Weight of solvent
and M2 = Molecular weight of solvent.

28. All right copy reserved. No part of the material can be produced without prior permission




4.4 ELEVATION OF BOILING POINT METHOD
The molecular weight (M1) of a nonvolatile organic compound can be determined by measuring
the elevation in boiling point of a solvent caused by dissolving a known mass of the solute (nonvolatile
organic compound) in a given mass of the solvent.
Elevation in boiling point = Tb = Tsoln  Tsolvent = 21
1
b
WM
1000w
K



where Kb = Molal boiling point elevation constant or ebullioscopic constant,
w1 = Weight of nonvolatile organic compound,
M1 = Molecular weight of nonvolatile organic compound and
W2 = Weight of solvent.
4.5 FREEZING POINT DEPRESSION METHOD
The molecular weight (M1) of a nonvolatile organic compound can be determined by measuring
the depression in freezing point of a solvent caused by dissolving a known mass of the solute (nonvolatile
organic compound) in a given mass of the solvent.
Depression in freezing point = Tf = Tsolvent  Tsoln = 21
1
f
WM
1000w
K



where Kf = Molal freezing point depression constant or cryoscopic constant,
w1 = Weight of nonvolatile organic compound,
M1 = Molecular weight of nonvolatile organic compound and
W2 = Weight of solvent.


The empirical formula of a compound is defined as the simplest formula that gives the ratio of
atoms of the various elements in a molecule. This is calculated from the percentage composition of the
elements.
The molecular formula of the compound gives the actual number of atoms of each element present
in a molecule of the compound.
Molecular formula = Empirical formula  n (where n is an integer)
n = massformula Empirical
massformula Molecular

Illustration

Question: 0.2 g of an organic monobasic acid gave 0.509 g of CO2 and 0.08856 g of water on complete
combustion. 0.244 g of the acid required for neutralization 20 ml of N/10 alkali. Determine the
molecular formula of the acid.
Solution: Weight of carbon in the organic acid = 0.509  44
12 = 0.1388 g
Percentage of carbon in the organic acid = 2.0
1388.0  100 = 69.4
Weight of hydrogen in the organic acid = 0.08856  18
2 = 0.00984 g
Percentage of hydrogen in the organic acid = 2.0
00984.0  100 = 4.92
Percentage of oxygen = 100 – (69.4 + 4.92) = 25.68
DETERMINATION OF EMPIRICAL AND MOLECULAR FORMULA

5

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Equivalents of acid = Equivalents of alkali = 3
10
10
1
20


Moles of organic acid = 3
10
10
1
20

 = 0.002 (monobasic acid n = 1)
Molecular weight of acid = 0.002
0.244 = 122
Element Percentage weight % weight/Atomic
weight
Whole number ratio
Carbon 69.4 12
4.69
= 5.78 605.1
78.5 = 3.6
Hydrogen 4.92 1
92.4
= 4.92 605.1
92.4 = 3.06
Oxygen 25.68 16
68.25
= 1.605 605.1
605.1 = 1
 Empirical formula is C3.6H3.06O1 or C7H6O2
Molecular formula = (C7H6O2)n
(C7H6O2)n = 122
[(12  7) + (1  6) + (16  2)] n = 122
n = 1
Hence, Molecular formula = n  emperical formula = C7H6O2



When it is required to identify the constituents of a mixture of organic substances, it is
necessary to separate each component from the mixture and to isolate it in a pure state before
proceeding with the examination. To identify the constituents of a mixture without separating
them is an extremely difficult feat.
Owing to the great number of possibilities, no definite rule for procedure can be laid down and
advantage will have to be taken of any facts that emerge in the preliminary examination.
The preliminary examination of the mixture should include the physical state of the sample, its behaviour
on distillation, determination of the elements present, behaviour on ignition, its solubility in water and
whether an acid or an alkaline reaction is given.
If the mixture appear to contain some volatile liquid, it may be heated on the waterbath
in a distilling flask attached to a condenser until no more of the liquid passes over. The residue
in the flask or the original mixture (if it is solid) is treated with an excess of dry ether, any
insoluble portion being filtered off and washed with the same solvent, the washings being added
to the filtrate. By this means the majority of salts, carbohydrates and other polyhydroxylic
compounds, sulphonic acids and similar substances insoluble in ether, may be separated from
the main portion. Such a residue is to be examined independently, extraction with cold methyl
alcohol being carried out as a preliminary step towards further separation.
If the mixture have been found to contain nitrogen, the ethereal solution is shaken in a separating
funnel with dilute sulphuric acid. In the absence of nitrogen, this operation can be omitted. By this means
basic substances are removed from the mixture on separating the aqueous and ethereal layers. The bases
may be recovered by rendering the aqueous solution alkaline and again extracting with ether.
The ethereal solution after this treatment should be washed with a small quantity of water, the
washings being discarded-and shaken with dilute caustic soda solution. This has the effect of removing all
compounds of an acidic character. The treatment of the aqueous portion will be discussed below.
SEPARATION OF MIXTURES OF ORGANIC COMPOUNDS

6

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The ether now contains only neutral substances. From these any aldehydic and most of the ketonic
compounds can be removed by shaking with a concentrated solution of sodium bisulphite. The aldehydes
and ketones can be recovered from the resulting precipitate or aqueous solution by acidification with dilute
sulphuric acid followed by distillation, extraction or filtration.
The alkaline solution containing the acidic substances should be saturated with carbon dioxide and
extracted with ether. By this procedure all phenolic compounds, which contain no carboxyl or nitro
groups, oximes and similar weak acids are, liberated and pass into the ether layer. On adding dilute
sulphuric acid until evolution of carbon dioxide ceases, carboxylic acids and nitrophenols are liberated and
can be isolated by extraction, filtration or distillation.
At this stage all ethereal solutions should be evaporated, condensed and thus recovered. All
residues should be tested afresh for constituent elements. The results of the above operations are briefly
summarized in the scheme shown below.
The various fractions denominated (I), (II), etc., may contain the following types of compound.
(I) Hydrocarbons, ethers, alcohols, ketones, esters, aliphatic halogen compounds and conceivably
aldehydes, acetals, nitriles, aliphatic amines, alkyl nitrates and nitrites whose boiling points lie
below 100

C.
(II) Salts of organic bases with mineral acids, carbohydrates and other polyhydroxylic compounds,
amino acids, sulphonic acids of all types.
(III) Aliphatic and aromatic primary, secondary, and tertiary amino compounds, possibly some amides
also.
(IV) Aldehydic and ketonic compounds containing no groups, which would have placed them in another
fraction.
(V) Hydrocarbons, ethers, alcohols, higher ketones, esters and aliphatic or aromatic halogen
compounds, not included in other fractions.
(VI) Simple and substituted phenols, ketoenolic substances.
(VII) Carboxylic acid, nitrophenols (in which the nitro groups are present in either ortho or para
position).
Much information as to the nature of the constituents may be obtained by the mere performance of
the above separations.
For further separation of the individual substances contained in the different fractions no definite
scheme can be drawn up, but can be detected by some characteristic reactions and characteristic derivative
formations.






Illustration

Question: Draw a flow diagram for the separation and recovery in almost quantitative yield of a mixture
of the waterinsoluble compounds PhCHO, PhNMe2, PhCl, pMeC6H4OH and PhCOOH.
Solution:

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Ether solution
PhCHO, PhNMe 2 PhCl, pMeC6H4OH, and PhCOOH.

Extract with
dil aq. HCl

Lower aqueous HCl layer

Upper ether layer

PhNMe2H
+
Cl


Add aq. NaOH, ether

Extract with
aq. NaOH

PhNMe2 in ether
pMeC6H4O

Na
+
, PhCOO

Na
+

Add dry ice (CO2), ether

Lower NaOH layer

Upper ether layer

Add aq. NaHSO3

Lower aq. NaHCO3 layer

Upper ether layer

PhCOO

Na
+

Add dil HCl, ether

PhCOOH in ether

pMeC6H4OH in ether

PhCH(OH)SO
3 Na
+
Add Na2CO3, Ether

PhCHO in ether

PhCl
Upper ether layer

Lower aq. layer

in ether






























SOLVED SUBJECTIVE EXAMPLES

Example 1:

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0.21 g of an organic compound gave on combustion 0.462 g of carbon dioxide and 0.1213 g of water.
Calculate the percentage of carbon and hydrogen in it.
Solution:
% of carbon in the organic compound = 100
21.044
12462.0


 = 60%
and % of hydrogen in the organic compound = 100
21.018
21213.0


 = 6.41%

Example 2:
0.9 g of an organic compound containing only carbon, hydrogen and nitrogen on combustion, gave 2.2
g of carbon dioxide and 0.6 g of water. What is the empirical formula of the compound?
Solution:
Weight of carbon = 44
12 x 2.2 = 0.6 g
Weight of hydrogen = 18
2 x 0.6 = 0.0667 g
Weight of carbon + hydrogen = 0.6667 g
Weight of nitrogen = 0.9 – 0.6667 = 0.2333 g
Element Weight Weight/Atomic
Weight
Whole number ratio
Carbon 0.6 12
6.0
= 0.05 0167.0
05.0 = 3
Hydrogen 0.0667 1
0667.0
= 0.0667 0167.0
0667.0 = 4
Nitrogen 0.2333 14
2333.0
= 0.0167 0167.0
0167.0 = 1
 Empirical formula = C3H4N

Example 3:
0.45 g of an organic compound gave on combustion 0.792 g of CO2 and 0.324 g of water. 0.24 g of the
same substance was Kjeldahlised and the ammonia formed was absorbed in 100 ml of N/8 H2SO4. The
excess acid required 154 ml of N/20 NaOH for complete neutralization. Calculate the empirical formula of
the compound.
Solution:
Percentage of carbon:
44 g of CO2 contains = 12 g of C
 0.792 g of CO2 contain = 44
792.012 = 0.216 g of C
 percentage of C in the organic compound = 45.0
216.0 x 100 = 48
Percentage of hydrogen:
18 g of H2O contains = 2 g of H
 0.324 g of H2O contain = 18
324.02 = 0.036 g of H
 percentage of H in the organic compound = 45.0
036.0 x 100 = 8
Percentage of nitrogen:

33. All right copy reserved. No part of the material can be produced without prior permission




Initial equivalents of acid taken = 3
10
8
1
100


Equivalents of excess of acid = Equivalents of NaOH = 3
10
20
1
154


Equivalents of acid used with ammonia = 


















 33
10
20
1
15410
8
1
100
= 4.8  10
-3

Equivalents of ammonia reacted = 4.8  10
3
= Moles of NH3 reacted
= Moles of NH3 produced.
Mass of nitrogen present in organic compound = 4.8  10
3
 14 = 0.0672 g
 percentage of nitrogen in organic compound = 24.0
0672.0  100 = 28
Percentage of oxygen = 100 – (48 + 8 + 28) = 16
To determine empirical formula:
Element Percentage weight % weight/amount Whole number ratio
Carbon 48 12
48
= 4 1
4 = 4
Hydrogen 8 1
8
= 8 1
8 = 8
Nitrogen 28 14
28
= 2 1
2 = 2
Oxygen 16 16
16
= 1 1
1 = 1

 Empirical formula = C4H8N2O


Example 4:
0.492 g of an organic compound gave 0.396 g of carbon dioxide and 0.2028 g of water on
combustion. 0.37 g of the compound gave 0.639 g of silver bromide. What is the molecular formula of the
compound if its vapour density is 54.5?
Solution:
Weight of carbon in the organic compound = 44
12396.0  = 0.108 g
Percentage of carbon in the organic compound = 492.0
108.0  100 = 21.95%
Weight of hydrogen in the organic compound = 18
22028.0  = 0.02253 g
Percentage of hydrogen in the organic compound = 492.0
02253.0  100 = 4.58%
Percentage of bromine in the organic compound = 188
80  37.0
639.0  100 = 73.47%

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Element Percentage weight % weight/atomic weight Whole number ratio
Carbon 21.95 12
95.21
= 1.83 9183.0
83.1 = 2
Hydrogen 4.58 1
58.4
= 4.58 9183.0
58.4 = 5
Bromine 73.47 80
47.73
= 0.9183 9183.0
9183.0 = 1
 Empirical formula = C2H5Br
Molecular formula = (C2H5Br)n
Molecular weight = 2  vapour density = 2  54.5 = 109
(C2H5Br)n = 109
[(2  12) + (5  1) + 80] n = 109
109 n = 109
n = 1
 Molecular formula is C2H5Br.

Example 5:
A mixture of pamino benzoic acid and phydroxy benzoic acid is taken in diethyl ether.
How will you separate them (in not more than three steps)? Write the reagents and all the required
conditions to separate them. Give confirmatory test for the presence of each functional group.
Solution:
Step 1. Extract with aq. dil. HCl
pamino benzoic acid, phydroxy benzoic acid
(in ether)
Ether layer Aqueous layer
pHOC6H4CO2H pCO2HC6H4NH3
+
Cl


Separate the two layers using separating funnel.
Step 2. Basification of aqueous layer with NaOH.
Test for NH2 group: When compound containing NH2 group is heated with CHCl3 and KOH, a very
pungent smell of isocyanide is obtained.
RNH2 + CHCl3 + 3KOH  RNC + 3KCl + 3H2O.
Test for COOH group: The compound containing COOH group gives brisk effervescence on treatment
with NaHCO3 solution due to evolution of CO2.
Test for OH group: A blue or green colouration is obtained when the compound is treated with neutral
FeCl3 solution.

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MIND MAP
Mixture
Distilled on water-bath
NonVolatile
Treated with ether
Volatile
(I)
Insoluble in ether
(II)
Soluble in ether
Treated with dilute H2SO4
Neutral or acidic
Treated with dilute NaOH
Basic
(III)
Neutral
Treated with NaHSO3
Acidic
Treated with CO2
Aldehydes
and
Ketones
(IV)
Neutral
(V)
Acids
and
Nitrophenols
(VII)

Phenols
(VI)


The various fractions denominated (I), (II), etc., may contain the following types of compound.

(I) Hydrocarbons, ethers, alcohols, ketones, esters, aliphatic halogen compounds and conceivably
aldehydes, acetals, nitriles, aliphatic amines, alkyl nitrates and nitrites whose boiling points lie
below 100

C.
(II) Salts of organic bases with mineral acids, carbohydrates and other polyhydroxylic compounds,
amino acids, sulphonic acids of all types.
(III) Aliphatic and aromatic primary, secondary, and tertiary amino compounds, possibly some amides
also.
(IV) Aldehydic and ketonic compounds containing no groups, which would have placed them in another
fraction.
(V) Hydrocarbons, ethers, alcohols, higher ketones, esters and aliphatic or aromatic halogen
compounds, not included in other fractions.
(VI) Simple and substituted phenols, ketoenolic substances.
(VII) Carboxylic acid, nitrophenols (in which the nitro groups are present in either ortho or para
position).

36. All right copy reserved. No part of the material can be produced without prior permission







EXERCISE – I

CBSE PROBLEMS

1. Name two solvents which are commonly used to dissolve organic solids.

2. Suggest a method to separate o-nitrophenol from p-nitrophenol.

3. Name the adsorbent commonly used in adsorption chromatography.

4. Which elements are normally not detected in an organic compound?

5. For which substances Kjeldahl’s method is not useful?

6. What is a standard solution?

7. What is the principle of fractional distillation? Discuss the function of fractionating column.

8. Discuss the principle of steam distillation.

9. Write the chemical equations involved in the detection of nitrogen by Lassaigne’s test.

10. An organic compound contains 69% carbon and 4.8% hydrogen, the remainder being oxygen.
Calculate the masses of carbon dioxide and water produced when 0.20 g of this compound is
subjected to complete combustion.

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EXERCISE – II

JEE & NEET-SINGLE CHOICE CORRECT

1. In Lassaigne’s test the compound containing sulphur gives deep violet colour with sodium
nitroprusside. The violet colour is due to the complex
(a) Na4[Fe2(CN)5NOS] (b) Na3[Fe(CN)5NOS]
(c) Na4[Fe(CN)5NOS] (d) Na[Fe(CN)NOS]

2. The complex Fe4[Fe(CN)6]3 has ______________ colour.
(a) Black (b) Violet
(c) Orange (d) Blue

3. 2, 4–DNP is used to detect carbonyl compounds. What is the full form of 2, 4–DNP?
(a) 2, 4–Dinitrophenol. (b) 2,4–Dinitrophenyl.
(c) 2,4–Dinitrophenyl hydrazine. (d) 2,4–Dinitrophorphorous.

4. Alcohols and phenols can be distinguished by
(a) acidic ferric chloride solution. (b) neutral ferric chloride solution.
(c) basic ferric chloride solution. (d) crystals of ferric chloride.

5. Phenolphthalein turns pink in alkali. The colour develops due to the formation of
(a) Benzenoid structure. (b) Quinonoid structure.
(c) Roseaniline. (d) None of these

6. In sodium bicarbonate test for carboxylic acid group the CO2 evolved comes from
(a) RCOOH (b) NaHCO3
(c) RCOONa (d) None of these

7. 59 g of an amide obtained from a carboxylic acid, RCOOH on hydrolysis gave 17 g ammonia. The
acid is
(a) formic acid (b) acetic acid
(c) propionic acid (d) butyric acid

8. A compound contains 38.8% carbon, 16% hydrogen and 45.2% nitrogen. The formula of the
compound would be
(a) CH3NH2 (b) CH3CN
(c) C2H5CN (d) CH2(NH2)2

9. In Kjeldahl’s method of estimating nitrogen, potassium sulphate in added to the sample being
analysed. Potassium sulphate
(a) lowers the boiling point of H2SO4. (b) maintains regular temperature.
(c) raises the boiling point of H2SO4. (d) reacts with the sample.

10. Detection of chlorine is possible without preparing sodium extract in
(a)
CH2Cl

(a) CH2 = CH – CH2Cl
(c) Both (a) and (b) (d)
Cl

11. Kjeldahl’s test for nitrogen in based on
(a) Redox titration (b) Complexometric titration
(c) Simple titration (d) Back titration

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12. Platinum salt of an organic base contains 32.5% platinum. Hence, equivalent weight of the base is
(a) 95 (b) 195
(c) 300 (d) 600

13. In Lassaigne’s test of halogens, the sodium extract is first boiled with HNO3 as to
(a) decompose NaCN and Na2S. (b) make silver halides isoluble.
(c) increase the solubility of AgNO3. (d) dissolve AgCN.

14. In Kjeldahl’s method nitrogen present in the sample is converted to
(a) nitrogen dioxide. (b) aluminium sulphate.
(c) ammonium sulphate. (d) ammonium phosphate.

15. Mulliken’s test for nitro group is valid for
(a) aliphatic nitro compounds. (b) aromatic nitro compounds.
(c) both (a) and (b). (d) none of them.

16. Absolute alcohol is prepared from rectified spirit by
(a) fractional distillation. (b) steam distillation.
(c) azeotropic distillation. (d) vacuum distillation.

17. Platinichloride combines with amines to form
(a) single salts. (b) complexes.
(c) double salts. (d) none of these.

18. An organic compound weighing 20 g when subjected to combustion with oxygen gives 10 g of
water. The percentage of hydrogen in the organic compound is
(a) 67.2% (b) 33.33%
(c) 2.85% (d) 5.55%

19. Magnesia mixture is used to detect phosphate ions present in a salt. It consists of a
mixture of
(a) NaCl, CaCl2, NH4OH (b) MgCl2, NH4Cl, NH4OH
(c) MgCl2, NaCl, NH4OH (d) MgCl2, KCl, NH4OH

20. Carbylamine test gives an offensive smell. The smell is due to the
(a) formation of cyanide. (b) formation of amines.
(c) formation of ether. (d) formation of isocyanides.

21. Dye test is performed by adding diazotized solution to
(a) -naphthol (b) -naphthol
(c) phenol (d) alcohol

22. Duma’s method for estimation of nitrogen is performed by heating organic compound with
(a) CuO (b) Cu2O
(c) CuCl2 (d) CuSO4

23. Fehling solution is a mixture of
(a) Cu(OH)2 + Na–citrate (b) Cu(OH)2 + Na–tartarate
(c) Cu(OH)2 + Na–lactate (d) Cu(OH)2 + NaK–formate

24. Which of the following compound does not give Lassaigne’s test for nitrogen?
(a) Hydrazine hydrochloride (b) Hydroxylamine
(c) Hydrazoic acid (d) All of them

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25. Copper wire test of halogens is known as
(a) Lassaigne test (b) Fusion test
(c) Leibig’s test (d) Beilstein test
EXERCISE – III



IIT-JEE- SINGLE CHOICE CORRECT

1. The percentage of sulphur in the organic compound, when 0.2595 g of a sulphur containing organic
compound in a quantitative analysis by Carius method yielded 0.35 g of barium sulphate, is
(a) 14.52% (b) 16.52%
(c) 18.52% (d) 19.52%

2. A compound containing 80% carbon and 20% hydrogen is likely to be
(a) C6H6 (b) C2H5OH
(c) C2H6 (d) CHI3

3. If 0.228 g of silver salt of dibasic acid gave a residue of 0.162 g of silver on ignition, then molecular
weight of the acid is
(a) 70 (b) 80
(c) 90 (d) 100

4. Which element does not give borax bead test?
(a) Pb (b) Cu
(c) Cr (d) Fe

5. In Lassaigne’s test sodium metal is used because
(a) It is very reactive
(b) Its melting point is low
(c) Its compounds are soluble in water
(d) All of these.

6. In which of the following compounds, nitrogen cannot be tested by Lassaigne’s test?
(a) C6H5NH2 (b) N2H4
(c) CH3CONH2 (d) C6H5NO2

7. To determine the weight of a halogen in organic compound, the compound is heated with fuming
HNO3 in presence of
(a) Ag (b) AgNO3
(c) Ag2CO3 (d) Ag2SO4

8. Liquid benzene burns in oxygen according to, 2C6H6 + 15O2  12CO2(g) + 6H2O(g). How many
litre of O2 at STP are needed to complete the combustion of 39 g of liquid benzene?
(a) 11.2 litre (b) 22.4 litre
(c) 84 litre (d) 74 litre

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9. Which of the organic compound will give red colour with sodium nitroprusside in Lassaigne’s test?
(a) C6H5CN (b)
CO2H H2N

(c)
NH2 –C–NH2
O

(d)
SO3Na H2N

10. Organic compound (A)  
33AgNO/HNO

white ppt. The organic compound (A) could be
(a) NH4Cl (b)
Cl

(c)
Cl

(d)
CH2Cl

11. Detection of the chlorine is possible without preparing sodium extract in

(a)

Cl O2N
CH3
CH3
(b) CHCl3

(c)
CH2CH2Cl

(d) CH2 = CHCH2Cl

12. 6 g of the organic compound on heating with NaOH gave NH3, which is neutralized by 200 ml of 1
N HCl. Percentage of nitrogen in the organic compound would be
(a) 12% (b) 60%
(c) 46.67% (d) 26.67%
13. An organic compound (A) contains 20% C, 46.66% N and 6.66% H. It gave NH3 gas on heating with
NaOH. The organic compound (A) could be
(a) CH3CONH2 (b) C6H5CONH2
(c) NH2CONH2 (d) CH3NHCONH2
14. The sulphur content of cystine is 26.7 percent. Given that cystine contains two sulphur atoms,
molecular weight of cystine is approximately
(a) 120 (b) 240
(c) 100 (d) 60
15. A colourless salt gives violet colour in Bunsen flame it may be
(a) Na2CO3 (b) Na2CrO4
(c) K2CO3 (d) BaCO3
16. Fractional crystallization is carried out to separate
(a) organic solids mixed with inorganic solids.
(b) organic solids highly soluble in water.
(c) organic solids having small difference in their solubilities in suitable solvent.
(d) organic solids having great difference in their solubilities in a suitable solvent.
17. Two substances when separated out on the basis of their extent of adsorption by one material, the
phenomenon is
(a) chromatography (b) paper chromatography
(c) sublimation (d) steam distillation
18. An organic compound X contains y and z impurities. Their solubility differs slightly. They may be
separated by
(a) simple crystalization (b) fractional crystallization
(c) sublimation (d) fractional distillation
19. Fuels from crude oil are separated from one another by
(a) fractional distillation (b) crystallization

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(c) steam distillation (d) selective adsorption
20. Sodium extract of an organic substance gives a blood red colour with FeCl3. It contains the elements
(a) N (b) S
(c) N and S both (d) N or S
21. 0.099 g of an organic compound when heated with fuming nitric acid in the presence of silver nitrate
in a carius tube gave 0.287 g silver chloride. The percentage of chlorine in the compound is about
(a) 29.6 (b) 71.7
(c) 35.4 (d) 64.2
22. A mixture of camphor and NaCl can be separated by
(a) sublimation (b) evaporation
(c) filtration (d) decantation
23. Glycerol is decomposed below its boiling point. It may be purified by
(a) steam distillation. (b) simple distillation.
(c) fractional distillation. (d) distillation under reduced pressure.
24. Indophenol is
(a) red (b) blue
(c) green (d) colourless
25. 0.22 g of organic compound CxHyO which occupied 112 ml at NTP and on combustion gave 0.44 g
CO2. The ratio of X to Y in the compound is
(a) 1 : 1 (b) 1 : 2
(c) 1 : 3 (d) 1 : 4

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EXERCISE – IV


ONE OR MORE THAN ONE CHOICE CORRECT

1. Which of the following compounds are not violet in colour?
(a) Na4[Fe(CN)5NSO3]3 (b) Na3[Fe(CN)6NSO]2
(c) Na2[Fe(CN)5NSO]3 (d) Na4[Fe(CN)5NOS]

2. The weights of carbon, hydrogen and oxygen in an organic compound are in the ratio
6 : 1 : 8 respectively. The molecular formula of the compound may be :
(a) CH2O (b) C2H4O2
(c) CH2O2 (d) C3H6O3

3. Which of the organic compounds will give white precipitate with AgNO3?
(a) 
ClNHHC
356 (b) NaCl
(c) C6H5Cl (d) 2,4,6-Trinitrochlorobenzene

4. Which of the following compound does not give Lassaigne’s test for nitrogen?
(a) Hydrazine hydrochloride (b) Hydroxylamine
(c) Hydrazoic acid (d) Methyl amine

5. A solution (miscible) of benzene + CHCl3 can not be separated by
(a) sublimation (b) filtration
(c) distillation (d) crystallisation

6. HCOOH and CH3COOH can be distinguished by
(a) Tollen’s reagent (b) Fehling’s solution
(c) KMnO4 (d) NaHCO3

7. Which of the following compound(s) is/are not expected to give Lassaigne’s test of nitrogen?
(a) NH2CONHNH2.HCl (b) NH2NH2.2HCl
(c) NH2CONH2 (d) C6H5–N=N–C6H5

8. Which of the following reagent(s) can be used to separate a mixture of aniline and phenol?
(a) Water (b) NaOH

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(c) NaHCO3 (d) HCl

9. Which of the following compound(s) will respond to the Beilstein test positively?
(a) m-chlorobenzoic acid (b) Chlorobenzene
(c) Urea (d) Sodium sulphide

10. If a sample of Lassaigne’s extract is warmed with FeSO4 solution and then treated with
a few drops of a ferric chloride solution, a blood-red colour is obtained. This indicates the presence
of
(a) N (b) S
(c) halogen (d) none of these

11. Which of the following compounds may give blood red colouration while performing Lassaigne’s
test for nitrogen?
(a) (NH2)2CO (b) (NH2)2C=S
(c) p-NH2C6H4SO3H (d) C6H5SO3H

12. The desiccants used for absorbing water during Liebig’s method for estimation of carbon and
hydrogen are :
(a) Anhydrous CaCl2 (b) Anhydrous Na2SO4
(c) MgSO4 .7H2O (d) Mg(ClO4)2

13. The empirical formula of a compound is CH2. To which of the hydrocarbon series does it belong?
(a) alkanes (b) alkenes
(c) alkynes (d) Cycloalkanes

14. An organic compound contains about 52% carbon. It could be :
(a) ethanol (b) dimethyl ether
(c) acetic acid (d) phenol

15. Presence of halogen in a compound is tested by :
(a) Iodoform test (b) Millon’s test
(c) Silver nitrate test (d) Beilstein test

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EXERCISE – V


MATCH THE FOLLOWING
Note: Each statement in column I has one or more than one match in column II.

1.

Column I (Compounds)
Column II (Properties)
I. KOH (A) Deliquescent
II. MgSO4.7H2O (B) Hygroscopic
III. NaOH (C) Efflorescent
IV. NH4NO3 (D) Alkali

2.

Column I
(Estimation / detection of elements)
Column II
(Methods)
I. Estimation of halogens (A) Kjeldahl’s method
II. Estimation of carbon and hydrogen (B) Dumas method
III. Estimation of nitrogen (C) Carius method
IV. Estimation of Phosphorus (D) Leibig’s method

Note: Each statement in column I has only one match in column II.

3.

Column I
(Impure compounds/mixtures)
Column II
(Methods of purification)
I. Benzene and toluene (A) Steam distillation
II. Aniline (B) Fractional distillation

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III. Hydrogen peroxide (C) Fractional crystallization
IV. Na2Cr2O7 and Na2SO4 (D) Distillation under reduced pressure

REASONING TYPE

Directions: Read the following questions and choose
(A) If both the statements are true and statement-2 is the correct explanation of statement-1.
(B) If both the statements are true but statement-2 is not the correct explanation of statement-1.
(C) If statement-1 is True and statement-2 is False.
(D) If statement-1 is False and statement-2 is True.

1. Statement-1: Lassaigne’s extract is boiled with dilute HNO3 before testing for halogens.
Statement-2: Na2S and NaCN are decomposed by HNO3.
(a) (A) (b) (B) (c) (C) (d) (D)



2. Statement-1: To estimate sulphur in the organic compound, it’s first heated with fuming HNO3
and then by adding BaCl2.
Statement-2: BaCl2 is insoluble and gets precipitated.
(a) (A) (b) (B) (c) (C) (d) (D)

3. Statement-1: Fehling’s test cannot be used to detect the presence of ketonic group but can be used
for aldehydic group.
Statement-2: Ketone is stronger reducing agent than aldehyde.
(a) (A) (b) (B) (c) (C) (d) (D)

4. Statement-1: Libermann’s nitroso test cannot be used to detect the presence of tertiary amine.
Statement-2: Tertiary amines cannot form oily emulsion of nitroso phenol.
(a) (A) (b) (B) (c) (C) (d) (D)

5. Statement-1: An organic compound on diazotization followed by reaction with alkaline solution
of -naphthol gives orange dye.
Statement-2: The organic compound is an aromatic amino compound which forms diazonium salt
and undergoes coupling reaction to form azo dye.
(a) (A) (b) (B) (c) (C) (d) (D)

LINKED COMPREHENSION TYPE

PASSAGE 1

The chemical bonds of organic compounds are generally covalent. Therefore, like inorganic
compounds, no direct method is available for detection of elements. In sodium-fusion method covalent
bonds of hetro atoms are broken by heating of organic compounds with sodium metal. This results in the
formation of inorganic ions involving these elements. These ions can in turn be readily identified by
inorganic qualitative methods.

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1. In the detection of nitrogen by Lassaigne’s test, blue/green colour is obtained. The colour is due to
(a) NaFe
III
[Fe
II
(CN)6] (b) Na2Fe
II
[Fe
II
(CN)6]
(c) Na4[Fe(CN)6] (d) Na3[Fe(CN)6]

2. In the Lassaigne’s test, sodium can be replaced by one of the following in the detection of essential
elements in organic compounds:
(a) NaHCO3 (b) Na2CO3/Zn (c) NaHCO3/Zn (d) CaCO3

3. Which of the organic compounds will give red colour in Lassaigne’s test?
(a) NaCNS (b)
NH2CNH2
S (c) (d) CHCl3







PASSAGE 2

An organic compound (A) gives positive Libermann’s reaction and on treatment with CHCl3/KOH
followed by hydrolysis gives (B) and (C). Compound (B) gives colour with Schiff’s reagent but not (C),
which is steam volatile. (C) on treatment with LiAlH4 gives (D), C7H8O2 which on oxidation gives (E).
Compound (E) reacts with (CH3CO)2O/CH3COOH to give a pain reliever (F).

1. Liebermann reaction uses reagent :
(a) NaOH (b) NaNO2/H2SO4
(c) NaNO2/NaOH (d) NaNO2/H2SO4 then NaOH


2. Other possible test for the compound (A) can be :
(a) using FeCl3 (neutral) (b) phthalein test
(c) azo-dye test (d) all of these

3. (B) and (C) are isomeric but (C) does not give colour with Schiff’s reagent. It is due to the fact that :
(a) (B) contains aldehyde group and (C) contains keto group in addition to phenolic group.
(b) (B) contains keto group and (C) contains aldehyde group in addition to phenolic group.
(c) carbonyl reactivity in (C) is masked by intramolecular H-bonding.
(d) carbonyl group is lost in hydrolysis step.









NH2CNH2
O

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EXERCISE – VI


SUBJECTIVE PROBLEMS

1. Write the chemical equations involved in the detection of nitrogen, sulphur, halogens and
phosphorus present in an organic compound, using Lassaigne’s extract.

2. 0.25 g of an organic compound (X) containing carbon, hydrogen and oxygen was analysed by the
combustion method. The increase in mass of calcium chloride tube and potash bulbs at the end of the
experiment was found to be 0.15 g and 0.1837 g respectively. Calculate the percentage composition
of the compound.

3. 0.2325 g of an organic compound was analysed for nitrogen by Duma’s method. 31.7 ml of moist
nitrogen was collected at 25°C and 755.8 mm pressure. Calculate the percentage of nitrogen in the
sample (aqueous tension of water at 25°C and 23.8 mm)

4. 0.2585 g of an organic compound containing iodine was heated with excess of strong nitric acid and
silver nitrate in a Carius tube. The precipitate of silver iodide was filtered, washed and dried. Its
weight was found to be 0.3894 g. Calculate the percentage of iodine in the compound.

5. 0.4 g of an organic compound was Kjeldahlised and the ammonia evolved was absorbed into 50 ml
of semi normal sulphuric acid solution. The residual acid solution was diluted with distilled water
and the volume was made upto 150 ml. 20 ml of this diluted solution required 31 ml of N/20 NaOH
solution for complete neutralization. Calculate the percentage of nitrogen in the compound.

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6. 1.75 g of a dibasic acid were dissolved in water and the volume was made upto 250 ml. In a titration,
20 ml of 0.1 N KOH consumed 18.45 ml of the acid solution for complete neutralization. Calculate
the molecular mass of the acid.

7. 0.6387 g of the platinichloride of a monoacid base on ignition gave 0.209 g of platinum. Calculate
the molecular mass of acid.

8. An organic compound contains C = 48.66% and H = 8.11%. If the vapour density of the compound
is 37, write all possible formulae of the compound.

9. A monoacidic organic base gave following results.
(i) 0.10 g gave 0.2882 g of CO2 and 0.0756 g of H2O.
(ii) 0.2 g gave 21.8 ml of nitrogen at 15°C and 760 mm.
(iii) 0.40 g of the platinichloride quantitatively gave 0.125 g platinum. Calculate the molecular
formula of the compound.

10. An aromatic compound contains 69.4% carbon and 5.8% hydrogen. A sample of 0.303 g of this
compound was analysed for nitrogen by Kjeldahl’s method. The ammonia evolved was absorbed in
50 ml of 0.05 M H2SO4. The excess of the acid required 25 ml of 0.1 M sodium hydroxide for
neutralization. Determine the molecular formula of the compound if its molecular mass is 121. Draw
the possible structures for this compound.

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ANSWERS

EXERCISE – I


CBSE PROBLEMS

1. CS2, C6H6

2. Steam distillation

3. Alumina or Al2O4

4. C and H

5. NO2, –N=N– groups and heterocyclic ‘N’ containing compounds.

6. Solution of known concentration.

7. Different boiling points.

8. Impurities must be of lower boiling point than water.

10. CO2 = 0.506 g ; H2O = 0.0864 g.

EXERCISE – II

JEE & NEET-SINGLE CHOICE CORRECT

1. (c) 2. (d) 3. (c) 4. (b) 5. (b)
6. (b) 7. (b) 8. (a) 9. (c) 10. (c)
11. (d) 12. (a) 13. (a) 14. (c) 15. (c)
16. (c) 17. (c) 18. (d) 19. (b) 20. (d)
21. (b) 22. (a) 23. (b) 24. (d) 25. (d)

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EXERCISE – III

IIT-JEE-SINGLE CHOICE CORRECT

1. (c) 2. (c) 3. (c) 4. (a) 5. (d)
6. (b) 7. (b) 8. (c) 9. (d) 10. (d)
11. (d) 12. (c) 13. (c) 14. (b) 15. (c)
16. (a) 17. (a) 18. (b) 19. (a) 20. (c)
21. (b) 22. (a) 23. (d) 24. (a) 25. (b)


EXERCISE – IV

ONE OR MORE THAN ONE CHOICE CORRECT

1. (a, b, c) 2. (a, b, d) 3. (a, d) 4. (a, b, c) 5. (a, b, d)
6. (a, b, c) 7. (b, d) 8. (b, d) 9. (a, b, c) 10. (a, b)
11. (b, c) 12. (a, d) 13. (b, d) 14. (a, b) 15. (c, d)

EXERCISE – V

MATCH THE FOLLOWING

1. I  (A), (D) ; II  (C) ; III  (A), (D) ; IV  (B)
2. I  (C) ; II  (D) ; III  (A), (B) ; IV  (B)
3. I  (B) ; II  (A) ; III  (D) ; IV  (C)

REASONING TYPE


1. (a) 2. (c) 3. (c) 4. (a) 5. (a)


LINKED COMPREHENSION TYPE

PASSAGE1

1. (a) 2. (b) 3. (b)

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PASSAGE2

1. (d) 2. (d) 3. (c)


EXERCISE – VI


SUBJECTIVE PROBLEMS

2. C = 20.04%, H = 6.66%, O = 73.30%

3. 15.03%

4. 81.4%

5. 46.81%

6. 129.15

7. 92.96

8. C2H5COOH, HCOOC 2H5, CH3COOCH3

9. C7H9N
10.
C7H7NO, C6H5–C–NH2, C6H5–CH=N–OH and
CHO
NH2
O
(cis & trans)