Chemistry, volume II organic and physical chemistry

The main methods of purification of organic solids and liquids are as follows : 1 Purification of Organic Solids i S'imple crystallisation This method is used to purify those organic com

1 ORGANIC CHEMISTRY AND PURIFICATION OF ORGANIC COMPOUNDS 1-1 ~ Organic chemistry 1

Requirement and methods of purification 2

Tests of purity 10

Chromatography 12

Exercises 13

2 OUALITATIVE AND OUANTITATIVE ELEMENTAL ANALYSIS

Qualitative detection of elements 16

Quantitative estimation of elements 19

Exercises 32

} EMPIRICAL, MOLECULAR AND STRUCTURAL FORMULA,

MOLECULAR MASSES OF ORGANIC COMPOUNDS

4 TETRAVALENT CHARACTER OF CARBON, FUNCTIONAL GROUPS AND

NOMENCLATURE OF ORGANIC COMPOUNDS n-8}

Tetrahedral concept of carbon atom 53

General methods of preparation of alkanes 85

General properties of alkanes 88

Exercises 92

6 UNSATURATED HYDROCARBONS (ALKENES AND ALKYNES)

Alkenes or olefins 95

Nomenclature of alkenes 95

General methods of preparation of alkenes 96

General properties of alkenes 98

Alkynes or acetylenes 103

Nomenclature of alkynes 103

General methods of preparation of alkynes 104

General properties of alkynes 105

Acidic nature of hydrogen in acetylene 109

Ascent and descent of alkane series and important conversions 110

Exercises 111

7 HALOGEN SUBSTITUTED ALKANES

Monohalogen derivatives of alkanes 115

84-94

9~-114

l1~-no

Ethyl bromide or bromoethane 119

Dihalogen derivatives of alkanes 122

Trihalogen derivatives of alkanes 123

9 PREPARATION, PROPERTIES AND USES OF COMPOUNDS

(Ethanol, Glycol and Glycerol)

General methods of preparation 164

General properties of aldehydes and ketones 166

16 ARENES AND AROMATICITY

Aromatic hydrocarbons or arenes 259

Heat content or enthalpy 278

Heat capacity of system 279

Spontaneous and non-spontaneous processes 280

Second law ofthermodynamics 281

Radioactive decay as first order reaction 298

Binding energy 300

Nuclear reaction equations 301

Nuclear fission (atomic fission) 302

Nuclear fusion (atomic fusion) 303

Exp 1 To prepare N/10 standard solution of ferrous ammonium sulphate and

find out the strength of the supplied ferrous ammonium sulphate using potassium permaganate solution as an intermediate solution 317 Exp.2 To prepare N/10 standard solution of oxalic acid and find out the

strength of the given oxalic acid solution by using potassium permanganate solution as an intermediate solution 318 Exp.3 To prepare N/30 standard solution of ferrous ammonium sulphate and

find out the strength of supplied ferrous ammonium sulphate solution by using K2Cr207 solution as an intermediate solution

Exp.4 To prepare N/30 standard solution of ferrous ammonium sulphate and

find out the strength of the given ferrous ammonium sulphate solution

by using K2Cr207 solution as an intermediate solution (internal indicator)

Exp.5: To prepare N/30 standard solution of potassium dichromate and find out

the strength of given potassium dichromate solution using ferrous

Exp 7 To prepare N/30 K2Cr207 solution and find out the strength of the given

K2Cr207 solution by titrating it against sodium thiosulphate solution

Exp 5 To determine the solubility of potassium nitrate at room temperature

and also to draw its solubility curve 339

1845, gave the final blow to vital force theory Soon after, methane was also synthesised by Berthelot in 1856, therefore, the term organic lost its original significance

It was Lavoisier who showed that carbon was the essential element of organic compounds Accordingly, organic compounds are now defined as 'the compounds of carbon' and organic chemistry as the study of these compounds

[I] Justification for Separate Study

The reason to study compounds of carbon separately, coming under the heading

of organic compounds, is the large number of their typical characteristics

At present, over two million organic compounds are known and each year the number of new organic compounds discovered or synthesised, sometimes exceed the total number of compounds of all the remaining elements (nearly 75,000 are known), This provides sufficient justification for their study as a separate branch

of chemistry

[II] Characteristics of Organic Compounds

(i) All the organic compounds are covalent compounds

(ii) The carbon present in organic compounds is always tetravalent, i.e •

tetracovalent

(iii) Unlike inorganic compounds, most of them are insoluble in water

2 CHEMISTRY-II (BIOTECH.)

(iv) Most of the organic compounds possess low m.p or b.p as compared to inorganic compounds

(v) Perhaps the most important characteristic property is the phenomenon

of 'isomerism', which may be defined as "when two or more compounds possess the same molecular formula but differ in their properties" Such compounds are called 'isomers' and the phenomenon as 'isomerism' The phenomenon of isomerism is exhibited by organic compounds due to their highly directional covalent bonds

(vi) Another characteristic of organic compounds is the presence of 'catenation property', actually a property of carbon, which means the tendency to link together The carbon atoms unite with each other by all possible means including linking through single, double or triple bonds, forming straight or branched chains or cyclic compounds It is this property of carbon which is mainly responsible for such a large number

of organic compounds known

REQUIREMENT AND METHODS OF PURIFICATION

[I] Requirement of Purification

Since most of the organic compounds are isolated from natural sources where they are present along with other organic compounds with identical properties, it

is essential to purifY them before subjecting them to qualitative and quantitative analysis Unlike inorganic compounds, the purification of organic compounds is tedious as large number of them decompose on heating, are sensitive to other reagents and resist the solvent action of water

[II] Methods of Purification

Separation and purification of organic compounds depend mainly on the difference of physical properties of organic compounds The main methods of purification of organic solids and liquids are as follows :

(1) Purification of Organic Solids

(i) S'imple crystallisation

This method is used to purify those organic compounds which are mixed either with insoluble impurities or less soluble impurities

Principle: Each organic compound is more soluble in a particular solvent at higher temperature but less soluble at lower

temperature

The process includes the preparation of a

saturated solution of the solid at a higher

temperature and then separating the solid in pure

form by cooling the solution

The method is also employed in inorganic

chemistry The only difference in the case of organic

compounds is the use of various organic solvents,

e.g., alcohol, benzene, ether, acetone, chloroform

etc., apart from the use of water The success of the

method depends upon the selection of suitable

solvent A good solvent must have low boiling point Fig 1 Hot water funnel

ORGANIC CHEMISTRY AND PURIFICATION OF ORGANIC COMPOUNDS 3

Whatman tilter paper ~~:=~

Trap

Fig 2 Filtration by using filter pump

and easy vaporisation Moreover, the solid to be purified should be more soluble

at high temperature but less at a low temperature so that cooling should lead to crystallisation

Method : The crystallisation of an organic solid involves the following procedure:

A saturated solution ofthe solid to be purified is prepared at high temperature

It is decolorised, if necessary, by heating or refluxing with animal charcoal, and filtered while hot through a hot water funnel (Fig 1) using filter pump (Fig 2) A hot water funnel is·an ordinary funnel surrounded by a double walled copper jacket between which hot water is filled This keeps the funnel hot and thus prevents cooling and consequent crystallization of the solid over the filter paper during filtration

The filtrate is then allowed to cool in a shallow container when pure solid starts depositing in the form of crystals leaving behind impurities in the mother liquor Crystals thus obtained are separated by filtration or by centrifuging Filtration is generally done on a Buchner funnel applying a little

vacuum by using a filter pump (Fig 2) to expedite the

procedure

Normally, after the first filtration, the crystals,

separated on filter paper, are washed once or twice by

the same solvent in order to remove any adhering

impurity Crystals thus obtained may further be

purified by recrystallisation

Crystals of pure compound are kept in a vacuum

desiccator (Fig 3) A vacuum desiccator differs from

an ordinary one in that it has on the top a tube which

is connected to a suction pump to suck out the air from

4 CHEMISTRY-II (BIOTECH.)

(ii) Fractional crystallisation

The method is used to purify those organic compounds which are mixed with impurities soluble in a particular solvent along with the organic compound but have different solubilities

Principle: A saturated solution of a mixture of two compounds with different solubilities in a particular solvent is prepared at a higher temperature On cooling slowly, the compound with tess solubility in the solvent crystallises out first

The method is used for the separation of a mixture of two or more solids which are soluble in the same solvent but at different temperatures

Method: In this method, the saturated solution of the mixture is prepared in hot solvent and then, after decolorisation and filtration, it is subjected to gradual cooling The various fractions obtained at different temperatures are separated from time to time The method can be easily understood by the following example Suppose two solids 'N WId 'B' are present in a mixture Both are soluble in the same solvent but '/I: is more soluble than 'B' A saturated solution of the mixture

is prepared in hot solvent, decolorised and filtered The hot filtrate is then allowed

to cool down slowly, when the less soluble 'B' separates first along with small quantity of 'No This bunch of crystals is separated Further cooling leads to crystallisation of more soluble solid 'N alongwith a small quantity of'B' This bunch

of crystals is also separated Then these two bunches of crystals are furher subjected to fractional crystallisation separately The process is repeated several times till the separation is complete

(iii) Sublimation

Normally, a solid, when heated, first melts into a liquid and on strong heating,

is changed into vapours and vice versa on cooling However, there are certain solids which on heating directly go into vapour state and on cooling vapours form the Bolid This phenomenon is called sublimation

Heat Solid "> Vapour

Cool Principle: Like liquid compounds, solid compounds also have vapour pressure though it is low A solid on heating absorbs energy

equivalent to its latent energy of sublimation This

results in a sudden rise in the kinetic energy of

molecules which directly go to vapour state

without passing through the liquid state

This property of sublimation is found very

useful in purification and separation of those

compounds which possess it Generally, the

method is used to purify a solid which sublimes,

from other compounds which do not sublime A few

examples include naphthalene, camphor,

anthracene, anthraquinone, benzoic acid etc

Method : The procedure involves heating of

impure compound in a porcelain dish covered with

a perforated asbestos sheet on which a funnel is

inverted (Fig 4) The walls of the funnel are kept

ORGANIC CHEMISTRY AND PURIFICATION OF ORGANIC COMPOUNDS 5

cool from outside with the help of a wet cloth The solid, which sublimes, changes into vapours which, on coming in contact with cooled walls of funnel, condense and deposit on it in the form of pure crystals

(iv) Extraction by solvents

This method involves dissolution of organic solid compounds present in plants

or mixed with other solids, in a suitable solvent, filtering it off and removing the solvent finally by evaporation or distillation The method is discussed under following three heads for clear understanding

(a) Separation of organic solids mixed with inorganic solids: Most of the organic solids are insoluble in water but soluble in organic solvents, while inorganic solids are insoluble in organic solvents but soluble in water Therefore,

such a mixture is shaken with suitable organic solvent, e.g., chloroform, ether,

benzene etc It is filtered to remove insoluble inorganic compounds Finally, the solvent is distilled off from the filtrate to get the solid organic compound

The reverse of his process is also

applied sometimes, i.e., the inorganic

compounds are removed by shaking the

mixture with water or dilute acid in which

they are soluble Organic solid is then

obtained by filtration

Soxhlet extraction : As the organic

compound to be separated may not be

completely soluble in a solvent at room

temperature, the mixture has to be heated

with organic solvent This is done in a

special apparatus known as 'Soxhlet

extractor' (Fig 5) The apparatus consists

of a strong wide tube 'C' with a side tube

'T' on the left hand side a siphon tube '8' on

the right hand side and a water condenser

fitted at the top of the wide tube

The solid mass is taken in the wide

tube A suitable solvent is taken in the

flask and heated on wire gauze or sand

plate When the solvent of the flask boils,

its vapours ascend through the side tube

'T' and pass through water condenser

where they get condensed The hot

condensed solvent drops fall on the solid

mass placed in the wide tube and dissolve

out its soluble constituents As the level of

the solution in the wide tube rises above

the height of siphon tube, it comes out

through the siphon tube into the flask

containing solvent The process continues

and more and more of the soluble

constituent passes in solution is collected

in the flask The solid is then obtained, as

6 CHEMISTRY-II (BIOTECH.)

Soxhlet extraction is the best method to extract alkaloids, essential oils from flowers and leaves and other products

A mixture consists of more than one organic solid and is shaken with different organic solvents one by one After each shaking, the solutiOli is filtered The different filterings are collected separately Separate distillation of each provides different organic compounds

(c) Separation of organic compound from its solution or suspension in water: To separate an organic compound from its solution

or suspension in water, organic solvents are used which

dissolve organic compounds but are themselves immiscible

with water, e.g., ether, benzene, chloroform, carbon

tetrachloride etc For such separation, perhaps the best

solvent is ether for two reasons First, it is a very good

solvent for most of the organic compounds and second, it is

inactive in nature

This method uses a separating funnel (Fig 6) The

separating funnel is half filled with the solutioil or

suspension of organic compound in water Now, the organic

solvent selected for the purpose is added After shaking

vigorously, it is allowed to settle down till two layers, one

of water and the other of solvent separate out The solvent

layer containing the organic compDund is collected

separately from the aqueous layer The remaining aqueous

Fig 6 Separating funnel

layer is again subjected to above treatment with a fresh quantity of organic solvent The solvent layer is then kept over anhydrous calcium chloride, which absorbs any water present Finally, it is filtered and the solvent is distilled off when the pure organic compound is obtained as residue

(2) Purification of Organic Liquids

(i) Simple distillation

Principle: Conversion of an organic liquid into vapours, sending vapours to other place and condensing the vapours back to liquid is known as distillation

The method may, therefore, be divided safely into two processes, one of vaporisation and the other of condensation

Distillation = Vaporisation + Condensation The method is suitable to separate or purify an organic liquid mixed with solid impurities or liquid impurities widely differing in boiling points (more than 40°C) The simple distillation apparatus (Fig 7) consists of a distillation flask with a side tube, in which the impure liquid is taken Attached to the side tube is a Liebig's condenser, the other end of which is connected to a receiver to collect the distilled liquid Distillation flask generally carries a thermometer fixed through a cork Sometimes, glass beads are also added to the flask to prevent bumping of liquid Method: The distillation flask is half filled with impure liquid and is heated Vapours formed are passed through the condenser and the liquid obtained is collected in a receiver

In case the liquid boils below 100°C, a water bath is used, otherwise a sand bath is advisable If the boiling point of the liquid to be distilled is above 150·C, air condenser should be used in place of Liebig's condenser

ORGANIC CHEMISTRY AND PURIFICATION OF ORGANIC COMPOUNDS

Thermometer

Air condenser '~================

-_/-Fig 7 Simple distillation

(ii) Fractional distillation

7

If the mixture consists of liquids with a difference of 15·C to 40·C in their boiling points, fractional distillation is the ideal method, provided that none of the liquids present decomposes at its b.p

Principle: (i) The boiling point of a mixture of two miscible liquids is always

in between the boiling points of the two liquids

(ii) The vapours obtained at the boiling point of the mixture have higher content of the vapours of low boiling liquid

(iii) As the vaporisation of mixture increases, the boiling point of the mixture also goes higher

The above principle is based on Raoult's law

The apparatus used for this purpose is similar to that of simple distillation except that a fractionating column containing a side tube is introduced between the ordinary flask and the Liebig's condenser (Fig 8) There are several types of fractionating columns used A fractionating column is a long tube of wide bore blown into a series of bulbs, either spherical or pear shaped Alternately, the tube

is filled with glass beads or broken glass tube (Fig 9) The actual purpose of a fractionating column is to increase the cooling surface and to provide obstruction

to the passage of ascending vapour or descending liquid

Method: To follow the process exactly, let us consider, for example, a mixture

of two liquids 'N and 'B' with boiling points 56°C and 78°C, respectively The mixture is heated in a flask fitted with a fractionating column and a Liebig's

Fractional distillation in industries is carried out using carefully designed

(according to need) big fractionating columns, e.g., separation of alcohol from water

is carried in a special type of fractionating column called "Coeffy still" Separation

of various components of petroleum is also carried out using a special type of fractionating column

(iii) Vacuum distillation or reduced pressure distillation

Many organic liquids decompose at their boiling points Thus, they cannot be purified by distillation under atmospheric pressure Such compounds are, therefore, purified by distillation under reduced pressure A common example is glycerine Glycerine decomposes much below its boiling point of 290·C However,

at 12 mm it boils at 180·C and can be safely distilled

ORGANIC CHEMISTRY AND PURIFICATION OF ORGANIC COMPOUNDS 9

Principle: The boiling point of a liquid is the temperature at which its vapour pressure becomes equal to atmospheric pressure Hence, by reducing the atmospheric pressure we can lower the boiling point of the liquid Thus, the boiling point of a liquid, which decomposes at its boiling point, can be lowered to such a temperature at which it does not decompose

Manometer

Fig 10 Vacuum distillation

in distillation under reduced pressure, a special type of distillation flask, known

as Claisen flask, is used to minimise this effect The Claisen flask (Fig 10) has two necks, one of which carries a thermometer and consists of a side tube The other neck carries a capillary tube partially closed at the top by a pressure tubing screw-clip arrangement, through which air can be regulated (this checks super heating) The side tube of the first neck is connected to a Liebig's condenser and

a receiver Receiver is attached to a vacuum pump to lower the pressure which is registered by a manometer

Reduced pressure distillation is used not only to avoid decomposition but it also serves to economise the fuel in industries A common example is the evaporation of sugarcane juice in vacuum pans in the sugar industry

(iv) Steam distillation

This method is used to purify those organic compounds (solids or liquids) which are (;) immiscible with water, (ii) possess a high molecular weight, and (iii) have

a fairly high vapour pressure around 100°C, e.g.,o-nitrophenol, aniline, nitrobenzene etc Steam distillation is also used in case of compounds which decompose at their boiling points However, the impurities present should be non-volatile

Principle: The boiling point of a liquid is the temperature at which its vapour pressure becomes equal to atmospheric pressure In steam distillation, the vapours

of liquid and water both are present The two together can become equal to atmospheric pressure at a temperature which will be lower than the t~mperature

at which either the liquid or water boils

Method: The apparatus (Fig 11) consists of a round bottom flask, which on one side is attached to a steam generator through a bent tube and on the other hand is connected with Liebig's condenser through another bent tube by using a two hole cork The compound to be purified is taken in the flask, clamped at an angle to prevent the substance from being splashed into the condenser The tube from the steam generator is dipped deep into the impure liquid Now, water is boiled in steam generator and the steam formed is bubbled through the liquid to

be purified Flask containing the impure liquid is also heated on a sand bath to avoid much condensation of steam in it The boiling of the impure liquid soon starts and the vapours of organic compound along with steam pass over and get condensed in the condenser

Safety Tube

Fig 11 Steam distillation

The next step is the separation of organic compound from water lfthe organic compound is a solid, it may be separated by simple filtration In case it is a liquid,

a separating funnel is used for the purpose

TESTS OF PURITY

Since the presence of even traces of moisture may be enough to affect the properties of a compound, a wet organic compound cannot be considered as completely pure

Drying of solid is generally achieved by first pressing it between the folds of filter paper and finally by keeping it in a desiccator, preferably vacuum desiccator containing anhydrous CaCI2, which absorbs moisture Solids with high melting points and sufficient stability are normally dried in an air or steam oven

Organic liquids are generally dried by keeping them over some desiccating agent which does not react chemically with the liquid Common desiccating agents include quicklime, anhydrous calcium chloride, anhydrous potassium carbonate,

ORGANIC CHEMISTRY AND PURIFICATION OF ORGANIC COMPOUNDS 11

fused copper sulphate, solid potassium hydroxide etc Finally, they are distilled over with a small quantity of phosphorus penta oxide or sodium

After drying, it is necessary to test the purity of organic compounds As the physical constants of a compound are always sharp and constant, they are used as criteria of purity Generally, determination of

melting and boiling point is used as the

criterion of purity, as even traces of impurities

change the melting or boiling point of a

ascertained by the specific gravity or refractive Capacity

[I) For Solids

Each pure and dry organic solid possesses a

sharp melting point Presence of even traces of

Furthermore, the melting point no longer

remains sharp if the solid is impure, i.e., it

melts between a range of temperature

For determination of melting point, Thiele's

method is used The apparatus, as shown in

figure 12, is used for this purpose First of all,

a thin capillary tube, sealed at the lower end by

heating, is filled one-third by the finely

powdered substance It is then attached to Fig 12 Melting point determination

thermometer by wetting it from outside with a

little concentrated sulphuric acid The thermometer is then lowered into Thiele's tube containing concentrated sulphuric acid

Thiele's tube is then heated slowly The exact

temperature at which the opaque solid in the

capillary tube becomes transparent, is the

melting point of the solid The solid is purified

again by recrystallisation and melting point

is redetermined No change in the two

melting points confirms the purity of the

compound

(ii) By mixed melting point : The

purified sample of the substance is mixed

with a little of pure sample and the melting

point of the mixture is determined If the

melting point of the mixture is the same as

that of pure substance, it confirms the purity

of the sample

[II] For Liquids

By determination of boiling point:

The purity of liquids is ascertained by their

boiling points Each pure liquid possesses a Fig 13 Boiling point detennination

12 CHEMISTRY-I! (BIOTECH.)

sharp and constant boiling point If the liquid is available in large quantity, it is boiled in a distillation flask fitted with a thennometer The constant temperature

at which the liquid distils over, is its boiling point

If the liquid is available in small quantity, it is detennined by Thiele's method (Fig 13) The liquid is taken in a small fusion tube attached to a thermometer with

a rubber band A small capillary tube, sealed at the upper end, is inverted in the liquid and the thennometer is lowered into Thiele's tube containing concentrated sulphuric acid The Thiele's tube is then heated slowly The tempeature at which

a brisk evolution of bubbles comes out from the open end of the capillary tube is the boiling point of the liquid

Sometimes, a simplified method is used A small amount of liquid is taken in

a boiling tube fitted with a thennometer with the help of a cork, so that the liquid vapours may escape out The tube is heated until the mercury shoots up in the thermometer The temperature at which mercury becomes stationary is the boiling point of the liquid

CHROMATOGRAPHY

It is a modern technique used for the

separation of mixture, isolation of compounds

identification of the constituents of a mixture

and also in the concentration of product that

occurs in nature in very dilute state

Principle: Chromatography is based on the

distribution of a mixture between two phases,

one stationary and the other moving

The mixture is dissolved in the moving phase

which may be a liquid or gas and is passed over

a stationary phase which may be a packed

column or a paper strip Different constituents of

the mixture migrate at different rates through

the stationary phase The different rates of

migration depend upon the solubility of the

constituents in the liquid phase and their

adsorption on the stationary phase Based on the

nature of the stationary and the moving phase,

chromatography may be classified into followiI.g

types

[I] Adsorption Chromato- graphy (Column

Chromato- graphy)

This method is based on the differential

adsorption of the constituents of a mixtue on

different adsorbents Important adsorbents

include alumina, silica gel, magnesium oxide,

sucrose etc

The solvent used to dissolve the mixture

should be non-polar like petroleum ether

Cotton ;.; - wool

Fig 14 Apparatus of column

chro-matography

I

ORGANIC CHEMISTRY AND PURIFICATION OF ORGANIC COMPOUNDS 13

The apparatus (Fig 14), consisting of a long glass tube with a stop cock at the lower end, is packed with a suitable adsorbent For this purpose, adsorbent is made into a slurry in the solvent to be used for dissolving the mixture It is then poured into the glass tube The adsorbent settles down while the excess of solvent is allowed to flow down through the stop cock

Now, a concentrated solution of the mixture to be separated, is prepared and introduced on the top of the adsorbent As the solution runs through the adsorbent, the components of the mixture are left behind at different distances depending upon their solubilities in solvent and their rates of adsorption on the adsorbent The layers of the adsorbent with different components of the mixture are called zones, bands or chromatograms Zones or bands may possess different colours Finally, pure solvent is allowed to run through the adsorbent, which carries along with it the components adsorbed at different distances, one by one, out of the tube This process has been named as elution The solution flows down through the stop cock with each disappearing band being collected separately in different receivers They are then concentrated and crystallised separately

Sometimes, the adsorbent column as a whole is pushed out of the tube and different bands are cut down separately These sections of adsorbent are then dissolved separately in suitable solvents Insoluble adsorbent is then removed by filtration and the pure components are obtained by crystallisation This process is generally used when the components form colourless bands so that they may be treated with some reagent to obtain coloured bands, e.g., amino acids obtained by hydrolysis of proteins are sprayed with ninhydrin when they give blue colour

[III] Thin Layer Chromotography (TLC)

It is a special type of adsorption chromatography in which a glass strip coated with a thin layer of adsorbent is used as the stationary phase instead of a column

EXERCISES [I] Long Answer Type Questions

1 Describe with diagram the steam distillation method of purification of organic compounds

2 A liquid mixture has two components - methyl alcohol (B.P = 65'C) and acetone (B.P

= 56'C) Describe a method for their separation

3 Boiling points of two liquids have a difference of W'C Describe with diagram the method for their separation

4 Describe purification of organic compounds with steam distillation method giving diagram In which condition is this method used?

5 On which principle is steam distillation based? How are organic compounds purified

by this method? Explain with an example Describe the method with a suitable diagram

6 An organic compound is insoluble in water but is steam volatile.l Describe the method

of its purification with a diagram

7 Give a method of purification of organic compounds, which decompose at their boiling points

8 On which principle fractional distillation is based? Describe the method with diagram

9 What is the importance of fractional distillation in the purification of organic compounds? Explain with an example

14 CHEMISTRY-II (BIOTECH.)

10 How is the purity of an organic compound determined ? Explain adsorption chromatography with an example

11 Write a note on 'tests of purity' (Meerut 2007)

12 Desribe fractional crystallisation method for the purification of organic compounds

13 What is fractional crystallisation? Explain with a suitable example

14 What is the importance of steam distillation in purification of organic compounds? Describe with a suitable example and diagram

15 Chlorobenzene is water insoluble but is steam volatile Describe with a diagram the method for its purification

16 Glycerine decomposes at its boiling point Describe with diagram the method for its purification

17 What is fractional distillation? Describe the method with diagram

18 How is an organic compound detected by mixed melting point determination method

23 On which principle is vacuum distillation method based ? Which type of organic compounds are purified by this method? Describe it with labelled diagram

24 Write a note on the following:

(i) Fractional distillation (ii) Sublimation

25 Write a note on column chromatography (Meerut 2006)

[II] Short Answer and Very Short Answer Type Questions

1 How can two immiscible liquids be separated by a simple method?

[Ans Two immiscible liquids may be separated by using a separating funnell

2 Generally, which physical constant of an organic solid and a liquid is used to determine their purity?

[Ans Melting point or mixed melting point in case of solids and boiling point in case

of liquids are used to test the purity of organic compounds!

3 How can naphthalene mixed with common salt be purified?

[Ans Nephthalene may be obtained in pure form by sublimation method]

4 Which method is used to separate aniline from 50% benzene?

[Ans Aniline may be separated from 50% benzene by fractional distillation]

5 Which method should be used to separate methyl alcohol and acetone from a mixture

100'C and B.P offormic acid = 100.6'Cl]

8 Nitrobenzene (B.P = 211 'C) is immiscible with water (B.P = 100'C) A mixture of the two is subjected to steam distillation Will the temperature at which distillation takes place between 100' and 211'C, greater than 211'C or less than 100'C ?

[Ans Less than 100'C]

ORGANIC CHEMISTRY AND PURIFICATION OF ORGANIC COMPOUNDS 15

9 For a substance to be separated by distilllation, it shdould be with water [Ans Steam, insoluble]

10 Give your answer in 'yes' or 'no'

(i) Purity of an organic solid is tested by determinatitm of its M.P

(ii) Vegetable oils are purified by fractional distillation

(iii) A mixture of aniline and toluene may be separated by chemical method

(iv) Purification of glycerol is done by steam distillation

(v) Soxhlet extractor is used for the separation of organic solids from their mixture [Ans (i) Yes, (ii) No, (iii) Yes, (iv) No, (v) Yes]

000

CHAPTER 2

QUALITATIVE AND QUANTITATIVE ELEMENTAL ANALYSIS

For investigation and characterisation of organic compounds obtained in a pure state, it is necessary to subject these to complete molecular diagnosis The various steps involved include:

1 Qualitative Analysis

This method includes the detection of elements commonly present in organic compounds such as carbon, hydrogen, nitrogen, halogen and sulphur Some compounds also contain phosphorus and metals

2 Quantitative Analysis

This method involves the determination of the percentage of various elements present in organic compounds

QUALITATIVE DETECTION OF ELEMENTS

[I] Detection of Carbon and Hydrogen

As we know, organic compounds are the compounds of carbon so there is no need of its detection However, it may be detected alongwith hydrogen as follows: Small quantity of organic compound, in which the detection of carbon and hydrogen is to be done, is taken along with some copper oxide in a hard glass test tube fitted with a one hole cork A bent tube with a bulb in its middle, is inserted into the cork The other end of the tube reaches inside a test tube containing lime water A small quantity of anhydrous copper sulphate is taken in the bulb Now the hard glass tube is heated strongly, resulting in the oxidation of carbon and hydrogen, if present in the compound, to carbon dioxide and water, respectively Carbon dioxide produced, turns the lime water milky, while the water formed, turns the anhydrous copper sulphate blue The oxidation reactions may

Colourless Blue

QUALITATIVE AND QUANTITATIVE ELEMENTAL ANALYSIS 17

If the organic compound is a liquid or gas, the vapours of compound are passed over preheated copper oxide Vapours escaping out are then tested for carbon

[II] Detection of Oxygen

The detection of oxygen in an organic compound cannot be done by qualitative analysis To ascertain its presence, help is taken of quantitative estimation First the percentage of all the other elements is determined and if the total of their percentage is less than 100, it is subtracted from 100 The difference is the percentage of oxygen in the organic compound

[III] Detection of Nitrogen, Sulphur and Halogens

The best and the simplest method for testing these elements is Lassiagne's test

Lassaigne's solution : A small piece of freshly cut sodium is taken in an ignition tube It is heated slightly till the sodium piece shines Now a small quantity of organic compound, in which the detection of above elements is to be done is introduced The amount of organic compound should be just sufficient to cover the sodium piece The ignition tube is heated, first slowly and then strongly, till it becomes red hot The tube should always be heated in slanting position keeping its mouth away from the body with the help of tongs The red hot tube is then plunged into about 10-12 ml of distilled water taken in a china-dish or small beaker The tube.will break itself If it does not, it should be broken with a glass rod The process is repeated with another ignition tube with fresh quantity of organic compound and this tube is also dropped into the same distilled water The distilled water containing both the broken tubes is boiled for 2-3 minutes and filtered hot The filtrate, which should be colourless, is called Lassaigne's solution

or sodium extract

Principle : Since organic compounds are covalent compounds, they do not ionise The elements present in organic compounds have to be converted into inorganic compounds, i.e., ionic compounds so that the ions formed may be tested

Heating with sodium produces ionic compounds of the elements present, e.g.,

nitrogen forms sodium cyanide, sulphur produces sodium sulphide and halogens form sodium halides The reactions may be written as :

(i) Na + C + N ~ NaCN

Nitrogen Sodium cyanide

Sulphur Sodium sulphide

(iii) 2Na + X2 -~ 2NaX

Halogen Sodium halide

(a) 2Na + Cl2 -~ 2NaCl

(iv) 2Na + 2H20 -~ H2 + 2NaOH

Lassaigne's solution (sodium extract) is used to test the elements present by the following methods :

18 CHEMISTRY-II (~IOTECH.)

(1) Test for nitrogen: To 0.5 ml of sodium extract, 0.5 ml of freshly prepared aqueous solution of ferrous sulphate and a few drops of sodium hydroxide are added The green precipitate so formed is boiled and cooled Now the green precipitate so produced is dissolved in minimum quantity of hydrochloric acid [if sulphuric acid is used in place of hydrochloric acid then a few drops of ferric chloride should also be added) A green or prussian blue colour of solution confirms the presence of nitrogen in the compound Reactions involved in this test, are:

Ferrous sulphate Ferrous hydroxide

Sod cyanide Sod ferrocyanide

Ferric chloride Ferric ferrocyanide

(Pruss ian blue)

If the organic compound contains both nitrogen and sulphur, a red colour is formed instead of green or blue owing to the formation of ferric sulphocyanide

Sod sulphocyanide

Ferric sulphocyanide (Red) (2) Test for sulphur:

(a) Take 0.5 ml of sodium extract in a test tube A few drops of freshly prepared sodium nitroprusside are added Formation of a purple colour confirms the presence of sulphur

Sod nitroprusside Sod thionitroprusside

(Purple)

(b) In another test tube, take 0.5 ml of sodium extract and add sufficient acetic acid to make the sodium extract acidic Then a small quantity of lead acetate solution is added to this acidic solution Formation of a black precipitate of lead sulphide confirms the presence of sulphur

Lead acetate Lead sulphide

Now, silver nitrate solution is added:

(a) Formation of white precipitate, soluble in ammonium hydrmcide, confirms the presence of chlorine

QUALITATIVE AND QUANTITATIVE ELEMENTAL ANALYSIS 19 (b) Formation of pale yellow precipitate, soluble only in excess of ammonium hydroxide, confirms the presence of bromine '

(c) Formation of yellow precipitate, insoluble even in excess of ammonium hydroxide, confirms the presence of iodine

Various reactions involved may be written as :

Diamine silver chloride (Soluble)

NaBr + AgN03 ~ NaN03 + AgBr

Silver bromide (Pale yellow ppt.) AgBr + 2NH40H ~ Ag(NH3)2Br + 2H20

Diamine silver bromide (Partially soluble)

AgI + AgN03 ~ NaN03 + AgI

Silver iodide (Yellow ppt.) AgI + NH40H ~ Insoluble

(b) Layer test for bromine and iodine: To 1 ml of sodium extract, few drops

of chloroform or carbon tetrachloride or carbon disulphide are added Being heavy, these solvent drops settle down at the bottom Now chlorine water is added drop

by drop shaking the solution each time If the drops settled at the bottom turn : (a) pale brown, it confirms the presence of bromine

(b) violet, it confirms the presence of iodine

Actually, chlorine displaces bromine or iodine present in the sodium extract as sodium bromide or sodium iodide The displaced bromine or iodine dissolves in the solvent (chloroform etc.) to impart it pale brown or violet colour, respectively

2N aBr + Cl2 ~ 2N aCI + Br2 CHCl3 + Br2 ~ Pale brown 2NaI + Cl2 -~ 2NaCl + 12 CHel3 + 12 -~ Violet

QUANTITATIVE ESTIMATION OF ELEMENTS

After the detection of elements, the percentage of elements present is determined The percentage of carbon and hydrogen is determined in a single method, while the percentage of other elements is determined separately Following methods are used for the quantitative estimation of various elements

[I] Estimation of Carbon and Hydrogen

The estimation of carbon and hydrogen is done by Liebig's combustion method

20 CHEMISTRY-II (BIOTECH.), Principle: The method is based on the fact that when an organic compound [CxHyl is heated with pure and dry cupric oxide in a stream of pure and dry oxygen, then carbon and hydrogen present in the compound oxidise quantitatively into CO2 and H20

in the weight of the two represents respectively the quantities of H20 and CO2 fonned

Calculations for the percentage of carbon and hydrogen

Weight of organic compound taken = m g

Increase in the weight of CaCl2 tube = x g

Increase in the weight of KOH bulb = y g

Therefore,

The weight of H20 fonned from the hydrogen present in the compound = x g

and the weight of CO2 formed from the carbon present in the compound = y g

18 g of water [H20 (1 x 2 + 16)1 contains 2 g of H

x g of water contains 2 1~ x g of H

m g of organic compound contains 2 1~ x g of H

Therefore, the percentage of H in organic compound

2 x

= - x - x 100

18 m

Similarly, 44 g of carbon dioxide [C02 (12 + 16 x 2)1 contains 12 g of C

y g of carbon dioxide contains 1~: y g of C

m g of organic compound contains 124: Y g of C

Percentage of C in organic compound = ~! x ~ x 100

However, if the compound also contains S, N or halogens, the results by this method are never accurate and some modifications have to be done

(i) If S is present: The sulphur present in the compound oxidises to S02 which is also absorbed by KOH So, sulphur is first removed by passing the gases through lead chromate which reacts with S02 to fonn non-volatile PhS04

2PbCr04 -1 2Pb + Cr203 + 5[0]

Pb + S02 + 2[0] -1 PbS04

Non-volatile

QUALITATIVE AND QUANTITATIVE ELEMENTAL ANALYSIS 21 (ii) If N is present: Nitrogen present in a compound oxidises to its oxides which are also absorbed by KOH A shining copper wire gauze is introduced before the bulb of KOH This reduces the nitrogen oxides to nitrogen which is not absorbed by KOH

4Cu + 2N02 -~ 4CuO + N2 2Cu + 2NO -~ 2CuO + N2

(iii) If halogens are present: Halogens present in the compound form volatile copper halides, some of which also decompose into free halogens Both of these are absorbed by KOH To prevent halogens or volatile copper halides from reaching the KOH bulb, a shining silver wire gauze is introduced before the KOH bulb The silver reacts with free halogens or volatile copper halides to form silver halides

Halogens Silver halide

Problem 1 0.2475 g of organic compound on combustion gave 0.4950 g of C02 and 0.2025 g of H20 Calculate the percentage of carbon and hydrogen in the compound

Solution Weight of organic compound = 0.2475 g

Weight of CO2 formed = 0.4950 g Weight of H20 formed = 0.2025 g

Soltuion Weight of organic compound = 0.45 g

Weight of CO2 formed = 0.792 g Weight of H20 formed = 0.324 g

22 CHEMISTRY-I! (BIOTECH.) [II] Estimation of Oxygen

There is no suitable direct method for the estimation of oxygen First the percentages of all other elements present are estimated The total of these percentages is subtracted from 100 [if the sum is less than 100], which would be

• the percentage of oxygen

Problem 3 0.2127 g of an organic compound containing C, Hand 0, on combustion gave 0.4862 g of C02 and 0.1989 g of H20 Calculate the percentage of carbon, hydrogen and oxygen

Solution Weight of organic compound = 0.2127 g

Weight of CO2 formed = 0.4862 g Weight of H20 formed = 0.1989 g Percentage of carbon in the compound

[III] Estimation of Nitrogen

Many methods are used to find out the percentage of nitrogen but the following two methods are cvmmonly used

(i) Dumas' method : This method can be used for the estimation of nitrogen

in all types of organic compounds

(ii) Kjeldahl's method : This method is applicable only for those compounds

in whcih nitrogen is attached directly to either carbon or hydrogen

(i) Duma's Method

This method also involves the combustion of organic compound, as is done in the case of carbon and hydrogen

Principle: A known weight of organic compound is mixed with cupric oxide and the mixture is heated in an atmosphere of CO2, The gases produced on combustion are passed through hot spirals of copper C, H and S present in the compound oxidise to CO2, H20 and S02, respectively, while nitrogen is obtained

as a gas If some oxides of nitrogen are formed during combustion, they are reduced

to nitrogen again by the hot copper spiral

4Cu + 2N02 ~ 4CuO + N2 2Cu + 2NO ~ 2CuO + N2 The mixture of CO2, H20, S02 and N2 is passed into a calibrated nitrometer filled with 40% KOH solution CO2 and S02 are absorbed by KOH, while water vapours condense Only N2 is collected in the nitrometer displacing KOH solution The volume of nitrogen formed at atmospheric pressure and room temperature is noted This volume of nitrogen is then converted at N.T.P

QUALITATIVE AND QUANTITATIVE ELEMENTAL ANALYSIS

Calculation for the percentage of nitrogen

Weight of organic compound = m g

Volume of nitrogen at room temperature = v ml

Problem 4 0.2046 g of an organic compound gave 30.4 ml of nitrogen at 15°C

and 732.7 mm pressure Calculate the percentage of nitrogen [Aqueous tension at

Weight of organic compound = 0.2046 g

Weight of 1 litre of nitrogen = 1.25 g

Weight of 1 ml of nitrogen = ;o~~ = 0.00125 g

Copper sulphate is used as catalyst, while potassium sulphate is used to

elevate the boiling point of H2S04, The method suffers from the drawback that some nitrogen is also evolved alongwith ammonia, hence the estimation results are not 100 percent correct

Calculation for the percentage of nitrogen

Let,

Weight of organic compound = W g

Volume of acid consumed = V ml

Normality of acid consumed = N

V ml of N normal acid = V ml of N normal NH3

: 1000 ml of N normal of ammonia contains

NI2 sulphuric acid Calculate the percentage of nitrogen

Solution (a) First Method

Weight of organic compound = 1.029 g

Volume of acid consumed = 14 ml

Normality of acid consumed = ~

P ercen age t 0 f 't m rogen = 1.4 NY 1.4 x 1 x 14 9 5 W = 1.029 x 2 = •

(b) Second Method

'.' 1000 ml of N normal H2S04 contains = 49 g of pure H2S04

QUALITATIVE AND QUANTITATIVE ELEMENTAL ANALYSIS 25

: 14 ml of~ normal H2S04 contains = 49 x 10010\ 2 g of pure H 2S04

Let V ml ofthis acid be left unused This required 73.7 ml ofN NaOH for back titration

or

Vml ofN H2S04 = 73.7 ml ofN NaOH

V xl = 73.7 x 1

V= 73.7 ml Volume of acid left over = 73.7 ml N H2S04

Volume of acid used to neutralize ammonia

= (100 - 73.7) ml N H2S04

= 26.3 ml N H2S04

According to formula,

P ercen age t 0 fm rogen 't = 1.4 W NV = 1.4 x 1 x 26.3 0.788 = • • 46 72

Problem 7 0.70 g of an organic compound was kjeldahlized and liberated ammonia was absorbed in 100 ml ofNI10 H2S0, After absorption, the left over acid required 10 ml of NI5 NaOH for complete neutralization Calculate the percentage

of nitrogen in the compound

Thus, volume of acid left unused = 20 ml of N /10 H2S04

The acid consumed = (100 - 20) == 80 ml of N 110 H2S04

W = Weight of compound

P ercen age t 0 fN -- 1.4 x 0.2030 x 25.304 5.427 1.325

[IV] Estimation of Halogens

Estimation of halogens is done by the following two methods

(i) Carius method (ii) Piria and Schiff's method

(1) Carius method

This method is suitable for those organic compounds which decompose easily

Principle: In this method, a known amount of an organic compound is heated

to a high temperature (180°-200°C) in a sealed tube alongwith fuming nitric acid containing a few crystals of silver nitrate Halogen present in the compound is converted into insoluble silver halide It is then filtered off, washed, dried and weighed With the help of the amount of silver halide obtained, the percentage of halogen is then calculated

Reactions:

Fuming Halogen acid

HX + AgN03 -~ HN03 + AgX

Silver halide

Calculation for the percentage of halogen

Weight of organic compound = W g

Weight of silver halide = WI g

The molecular weight of AgX includes the atomic weight of X

Weight of halogen in W g of compound

= Atomic weight of halogen x W

QUALITATIVE AND QUANTITATIVE ELEMENTAL ANALYSIS 27

Atomic weight of X x WI x 100 Percentage of halogen = M I o ecu ar welg I ht fAX W

If the compound contains chlorine, then

1 g molecular weight of AgCl (108 + 35.5 = 143.5) contains 35.5 g of chlorine 143.5 g AgCI contains = 35.5 g of chlorine

35.5 x WI WI g of AgCI contains ::= 143.5 g of chlorine

by Carius method Calculate the percentage of chlorine in the organic compound

Solution Weight of organic compound = 0.1890 g

Weight of silver chloride = 0.2870 g

(2) Piria and Schiff's method

Some organic compounds do not decompose completely in the Carius method, hence for such compounds Carius method is not suitable The best method for such compounds is Piria and Schiff's method

Principle: In this method, the finely powdered organic compound [0.1 - 0.3 g] is mixed with an intimate mixture of pure lime and sodium carbonate (4 : 1) This mixture is taken in a platinum crucible The crucible is then completely filled with pure lime and sodium carbonate mixture and is inverted in a bigger crucible More mixture is added to bigger crucible till it completely covers the smaller crucible It is then heated strongly The product is dissolved in excess of nitric acid and the halogen is precipitated with silver nitrate solution Percentage of halogen

is calculated from the weight of silver halide obtained as in Carius method

[V] Estimation of Sulphur

Sulphur is also estimated like halogens by Carius method as follows

Principle: In this method, a known amount of an organic compound is heated with fuming nitric acid in the Carius tube Sulphur present in the compound

28 CHEMISTRY-II (BIOTECH.)

oxidises to sulphuric acid Sulphuric acid thus formed is taken out in a beaker Carius tube is washed with water and the washing is also taken in the same beaker To this solution, sufficient amount of barium chloride solution is added Sulphuric acid is completely converted into insoluble barium sulphate The precipitate of barium sulphate is then filtered off, washed with hot water, dried and weighed From the weight of barium sulphate obtained, the percentage of sulphur is calculated Reactions involved in the method may be written as:

Heat

[C, H, S] + HN03 ~ CO2, H20, H2S04

Organic compound Fuming

H2S04 + BaCl2 ~ 2HCI + BaS04

Ppt

Calculation of percentage of sulphur

1 g molecular weight of BaS04 [137 + 32 + 64 = 233] contains 32 g of sulphur

Weight of organic compound;::: W g

Weight of BaS04 ;::: m g

233 g of BaS04 contains 32 g of sulphur

m g of BaS04 contains ;::: ;323 x m g of sulphur

Solution Weight of organic compound = 0.1254 g

Weight of BaS04 obtained = 0.1292 g

Solution Weight of organic compound = 0.32 g

QUALITATIVE AND QUANTITATIVE ELEMENTAL ANALYSIS

Solution According to the formula,

12 Weight of CO2 Percentage of carbon = 44 x W ht f elg 0 compoun d x 100

12 0.099

= 44 x 0.123 x 100 = 21.96

12 Weight of H20 Percentage of hydrogen = 18 x liT • ht f d x 100

vv elg 0 com poun

of C, Hand Br in the compound

Solution Percentage of carbon

Solution According to formula

12 Weight of CO2 Percentage of carbon = 44 x UT • vvelg ht f 0 compoun d x 100

= 44 x 0.45 x 100 = 48

2 Weight of H20 Percentage of hydrogen = 18 x W ht f d x 100

So, volume of acid left = 15.4 ml of N /2 H2S04

Volume of acid consumed = 25 - 15.4

= 9.6 ml of N /2 H2S04

CHEMISTRY-I! (BIOTECH.)

ercen age 0 m rogen = W = 0.24 X 2 = •

Problem 15 0.225 of an organic compound on combustion gave 0.396 g of C02 and 0.162 g of H20 0.24 g of this compound was Kjeldahlised the ammonia liberated was absorbed in 25 ml of NI2 H2S04 The excess of acid for complete neutralization required 40 ml of NI5 NaOH Calculate the percentage of C, Hand

Volume of acid consumed = (25 - 16) = 9 ml of N /2 H2S04

Problem 16 Following results were obtained on analysis of an organic compound:

(i) 0.369 g of compound gave 0.396 g of C02 and 0.189 g of H20

(ii) 0.246 g of the compound gave 0.379 g of AgBr Calculate the percentage of

C, Hand Br

I

,

QUALITATIVE AND QUANTITATIVE ELEMENTAL ANALYSIS

Solution According to formula,

12 Weight of CO2 Percentage of carbon = 44 x W ht f d x 100

mg 0 compoun

12 0.396

= 44 x 0.369 x 100 = 29.27

2 Weight of H20 Percentage of hydrogen = 18 x W ht f elg d x 100

Problem 18 0.2 gm of organic compound on analysis by Kjeldahl's method produced NH3 which was absorbed in 64 ml of NI2 H2804 Remaining acid was diluted to 500 ml by distilled water 20 ml of this diluted acid required 12.0 ml of NI10 NaOH for complete neutralisation Find out the percentage ofnitrogen in the compound

Solution As 20 ml of remaining acid is neutralized by 12 ml of N/10 NaOH

500 ml of diluted acid requires 12 ;0500 = 300 ml of N/10 N aOH

N

300 ml of N/l0 NaOH == 300 ml N/l0 H2S04 == 60 ml

2 H2S04 Acid used to neutralize NH3 = (64 - 60) = 4 ml

According to formula,

1.4 NV 1.4 x 1 x 4

Percentage of N = W = 0.2 x 2 = 14

32 CHEMISTRY-II (BIOTECH.)

EXERCISES [I] Long Answer Type Questions

1 Mention Carius method to estimate halogens in an organic compound

2 How are carbon and hydrogen detected qualitatively?

3 How will you detect Nand S in an organic compound qualitatively?

4 How will you estimate C and H in an organic compound quantitatively?

5 How is nitrogen estimated in an organic compound quantitatively?

[II] Short Answer and Very Short Answer Type Questions

1 Which elements are tested by Lassaigne's test?

[Ans Nitrogen, sulphur and halogens (chlorine, bromine and iodine) are tested by Lassaigne's testl

2 Why are organic compounds fused with sodium in this test?

[Ans Organic compounds are fused with sodium so as to convert the non-ionic (covalent) organic compounds into ionic sodium saltsl

3 What is sodium extract? Which compounds may be present in it ?

[Ans Organic compounds are fused with sodium in an ignition tube which is then plunged into distilled water This is boiled and filtered The filtrate thus obtained is called sodium extract This extract may contain: NaCN, NaCNS, Na2S, NaCI, NaBr, NaI and NaOHl

4 Why is sodium extract generally alkaline in nature?

[Ans Some of the sodium used for fusion with organic compounds remains unreacted This unreacted sodium then reacts with water to produce alkali, i.e., sodium hydroxide]

5 During the preparation of sodium extract, particularly when the hot ignition tube is plunged into water, it catches fire It is due to the burning of which compound? [Ans The unreacted sodium reacts with water to produce hydrogen gas Since the reaction is exothermic, hydrogen burns with flame

8 During the test for nitrogen, a red colour is sometimes obtained, why?

[Ans When nitrogen and sulphur both are present in the compound, they form sodium sulphocyanide on fusion with sodium This reacts with ferric chloride to give a red coloured ferric sulphocyanide, Fe(CNS)3l

9 During the test for halogens, why is sodium extract first boiled with a few drops of conc HN03 ?

[Ans During the test for halogens, sodium extract is boiled with a few drops of conc HN03 in order to decompose sodium cyanide and sodium sulphide present in it, so that they may not interfere in the test

NaCN + HN03 ~ NaN03 + HCN Na2S + 2HN0 ~ 2NaN0 + H2Sl

: QUALITATIVE AND QUANTITATIVE ELEMENTAL ANALYSIS 33

10 What is the role of chloroform or carbon tetrachloride in the layer test for bromine and iodine?

[Ans Chloroform or carbon tetrachloride dissolves the bromine or iodine displaced by chlorine, hence its layer becomes pale brown or violet)

11 Why is sodium extract made acidic during the lead acetate test for sulphur? The black precipitate obtained is due to the formation of which compound?

[Ans As we know, the sodium extract is alkaline in nature due to the presence of sodium hydroxide, it is mixed with acetic acid to make it acidic When lead acetate is mixed with this acidic sodium extract, a black precipitate of lead sulphide is formed

Na2S + (CH3COOhPb ~ 2CH3COONa + PbS]

(Black ppt.)

12 What is the role of copper sulphate and potassium sulphate in lijeldahl's method ?

[Ans Copper sulphate acts as a catalyst, while potassium sulphate is used to elevate the boiling point of sulphuric acid]

13 Which absorbents are used for carbon dioxide and water evolved in the Leibig's combustion method ?

[Ans CO2 is absorbed by potassium hydroxide solution, while anhydrous CaCl2 or conc H2S0 4 is used to absorb water]

14 Which of the following reagents are used in the Carius method for the estimation of sulphur? What are their functions?

Sodium chloride, silver chloride, barium chloride, sulphuric acid, nitric acid

[Ans Barium chloride and nitric acid Nitric acid to oxidise sulphur of the compound into sulphuric acid and barium chloride to convert sulphuric acid formed into insoluble barium sulphate}

15 Why do we use copper spiral in Dumas' method?

[Ans To reduce oxides of nitrogen produced during combustion back to nitrogen!

16 Which of the following reagents are used in the Carius method for the estimation of halogens? What are their functions?

Sodium chloride, Lunar caustic, Barium chloride, Sulphuric acid and Nitric acid

[Ans Nitric acid and Lunar caustic Nitric acid converts halogen of the compound into corresponding halogen acid, while Lunar caustic changes halogen acid formed into silver halide!

17 Why are oxygen and cupric oxide used in Leibig's combustion method? Write equations also

[Ans Oxygen and cupric oxide are used to oxidise carbon and hydrogen of the compound into CO2 and ~O, respectively

The equations may be written as:

C + 2CuO ~ CO2 + 2Cu 2H + CuO ~ H20 + CuI

18 During the heating of nitrogenous organic compound and conc H2S0 4 in KJeldahl flask, some nitrogen is also formed along with ammonia What will be the effect of nitrogen evolved on the percentage of nitrogen ?

[Ans The percentage of nitrogen will come out to be slightly less)

19 (i) On which principle is the method of estimation of carbon and hydrogen based?

[Ans Organic compound is heated with dry and pure copper oxide in order to oxidise the carbon and hydrogen present in the compound into carbon dioxide and water, respectively Knowing the amount of compound taken, and CO2 and H20 formed, the percentage of carbon and hydrogen is calculated]

(ii) Who gave the method for estimation of carbon and hydrogen?

[Arts Leibig]