Sohag University Faculty of Science Chemistry Department

EXPERIMENTAL ORGANIC
CHEMISTRY

For
B.Sc. Students

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2
 Introduction
Volumetric analysis

 Volumetric analysis is one of the most useful techniques in

analytical chemistry for quantative analysis.
 This technique is rapid and very good accuracy can be obtained.
 Since the concentration is known and since the reaction between
analyte and titrant is known, so the amount of analyte can be
calculated.
 Volumetric analysis depends on measuring the volume of titrant
required to just completely react with the analyte.

Titration process

 In this titration, the test substance (analyte) reacts with a reagent added
(standard solution).
 The titration process need to burette and conical flask.

Standard solution (titrant)

That solution has a known concentration and added from the burette.

Analyte

That solution has an unknown concentration and placed in conical flask.

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The requirement of titration

l- Reaction should be stoichiometric.

2- Reaction should be quantitative.
3- Reaction is rapid and specific.
4- No side reaction occurs during the titration.
 If the interference substance occurs, these must be removed before
start titration.
5- It must be a change in the color of the solution.
 A color change usually brought by addition of the indicator ,
 The color is dependent on the properties of the solution e.g. pH.

Equivalence point

Is the point at which an equivalent or stoichiometric amount of titrant is

added , It can be determined using an indicator .

CLASSIFICATION OF VOLUMETRIC METHOD

The reactions employed in volumetric analysis classified into two main
classes:

A-Those in which no change in valence occurs , these are dependent upon

the combination of ions .

 This type of reaction divided into three main classes.

I-Neutralization reactions:

 These include the titration of free bases or those formed from salts of
weak acids by hydrolysis with a standard acid (acidimetry).
 And the titration of free acid or those formed by the hydrolysis of
salts of weak bases, with a standard base (alkalimetry) .

These reactions involve the combination of hydrogen and hydroxide

ions to form water.

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 H+ + OH- → H2O

HCl + NaOH → NaCl + H2O

II-Complex formation reactions:

 These depend upon the combination of ions, other than hydrogen or

hydroxide ions, to form a soluble slightly dissociated ion or compound
as in the titration solution of a cyanide with silver nitrate or of chloride
ion with mercuric nitrate solution
 Also ethylene diamine tetra acetic acid (EDTA) is very important
reagent for complex formation titrations.
Ag+ + 2CN- → [Ag(CN)2]-

III- Precipitations reactions:

 These depend upon the combination of ions, other than hydrogen or

hydroxide ions, to form a simple precipitate as in the titration of silver
ion with a solution of chloride.

Ag+ + Cl- → AgCl ↓

White p.p.t

B- Oxidation reduction reaction

 These involve a change of valence or otherwise expressed, transfer of

electrons for example :-

I2 + 2Na2S2O3 → 2 NaI + Na2S4O6

Types of standard substance: :

 It's classified into :-

1- Primary standard material.
2- Secondary standard material.

 Properties of primary standard substances :

1-It must have high purity.

2-It must have a high molecular weight.
3-It must be stable in concentration during the time.

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 4-It must change the properties during the titration.
Examples:
Na2CO3 K2Cr2O7 C2O4H2.2H2O

Sodium carbonate pot.dicromate oxalic acid dehydrate

 Properties of secondary standard substances :

There are one or more conditions which must be found in primary

standard substance, in a mistake.

For example if the material is not enough pure, a solution is prepared to give
approximately the desired concentration and it standardized by using primary
standard solution.

Examples:-
HCI NaOH KMnO4
Hydrochloric acid sodium hydroxide pot.permanganate

* Standardization of sodium hydroxide by using standard oxalic acid.

** Standardization of hydrochloric acid by using standard sodium carbonate.
Preparation of standard solution:

We can prepare several standard solutions in different types of

concentrations by using volumetric flasks.

Volumetric flasks funnel Beaker

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Concentration of solutions

1- Molarity:
Number of moles of solute which dissolve in one litre of solvent.

g=

2- Normality:

Number of equivalent weight of solute which dissolve in one litre of solvent.

g=

3- g/ L (gram per litre) :-

Number of grams of solute which dissolve in one litre of solvent.

4- % (percentage):

Number of grams of solute which dissolve in 100 ml of solvent.

Molar solution (1M):

It's the solution in which one mole (gram molecular weight) of the solute is
contained per litre (1000 ml) of the solvent.

Normal solution (IN):

It's the solution in which one gram equivalent of the solute is contained per
litre (1000 ml) of the solvent.

Molecular weight (M.wt):

It's summation of atomic weight for substance.

Equivalent weight (Eq.wt) :-

Eq.wt of acid =

Eq.wt of base =

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 Eq.wt of salt =

Eq.wt of oxidizing agent =

Eq.wt of reducing agent =

(N×V)analyte = (N\ V\ )titrant

[ ] [ ]

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***Write the M.wt and Eq.wt for the following compounds :
1 -Oxalic acid dehydrate = ………………………………………………………………

2-0xalic acid anhydrous = ………………………………………………………………

3-Sodium hydroxide = ………………………………………………………………

4-Sodium carbonate = ………………………………………………………………

5-Hydrochloric acid = ………………………………………………………………

6-Tartaric acid = ………………………………………………………………

7-Succinic acid = ………………………………………………………………

8-citric acid anhydrous = ………………………………………………………………

9-citric acid monohydrate = ………………………………………………………………

10-Acetic acid = ………………………………………………………………

11-Formic acid = ………………………………………………………………

12-Malonic acid = ………………………………………………………………

13-Benzoic acid = ………………………………………………………………

14-Phthalic acid = ………………………………………………………………

15-Cinnamic acid = ………………………………………………………………

16-Salicylic acid = ………………………………………………………………

17-Aniline hydrochloride = ………………………………………………………………

18-Glycine = ………………………………………………………………

19-Formaldehyde = ………………………………………………………………

20-Phenol = ………………………………………………………………

21-Catechol = ………………………………………………………………

22-Pyrogallol = ………………………………………………………………

23-Aniline = ………………………………………………………………

24- O-phenylenediamine = ………………………………………………………………

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Write on titration curves for acid-base titration and its types
of indicator used.

1- Estimation of an aliphatic acid

Theoretical bases
 The formula of an aliphatic acid is R(COOH)n , where n is the number
of basicity.
 We can calculate the eq.wt of an aliphatic acid by direct titration in
aqueous solution.
 The most direct method for the determination of water-soluble
carboxylic acids (aliphatic acids) is titration with standard alkali
( sodium hydroxide) using phenolphthalein as indicator.
 Standard sodium hydroxide solution may be employed and is prepared
from the solid of analytical reagent purity and standardize using
standard reagent.

R(COOH)n + n NaOH → R(COONa)n + n H2O

strong base

 The pH at the stoichiometric end point or the equivalence point of the

neutralization reaction will naturally depend upon the ionization
constant of the carboxylic acid.
 Neutralization produces the salt of the weak acid and a strong base,
consequently the resulting solution will be slightly alkaline, having a pH
greater than 7.
 Most carboxylic acids are neutralized at a pH of about 8.4 ; rendering
phenolphthalein a satisfactory indicator (pH interval 8.3-10.1 ;
colorless to red ).

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  In general, the choice of an indicator for a titration is dictated by the
acid to be determined, i.e., the colour change interval of the indicator
must include the equivalence point.

strong base = Aliphatic acid

SO,
(N.V)aliphatic acid = (N\.V\)NaOH
OR …

[ ] [ ]

SO,

The neutralization equivalent is the number of grams of acid required to

neutralize one litre of normal alkali .
Procedure
1- Prepare a solution of 0.1N sodium hydroxide in 250 ml Volumetric flask.

2- Prepare a solution of 0.1N oxalic acid in 100 ml volumetric flask.

2- Standardize sodium hydroxide by using oxalic acid in presence of

phenolphthalein as indicator.
3- Weight out accurately 1.5 : 2 gm of sample (theoretical wt) and dissolve it
in 100 ml or 250 ml volumetric flask.
4- Transfer 10 ml or 25 ml of sample to a 250ml conical flask, and then titrate
with standard sodium hydroxide in presence of phenolphthalein as
indicator.
5- Repeat step 5 about 3-5 times and calculate mean volume (M .V.).
6- Calculate eq.wt of an aliphatic acid from this expression :-

[ ] [ ]

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Where , Factor =

NOTE

Also, we can calculate all the following :-

1- Calculate M.wt of an aliphatic acid from this expression :
M.wt = Eq.wt n , n= number of basicity
2- Calculate number of basicity of an aliphatic acid from this
expression :-
n=

3- Calculatethe the purity of sample, when sample is known

from these expressions :

[ ] [ ]

Calculation
1- Preparation of …… N sodium hydroxide in ….ml volumetric
flask

2- Preparation of …… N oxalic acid in ….ml volumetric flask

3- Standardization of sodium hydroxide using ….N oxalic acid

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No. S.P. E.P. M.V. indicator

Normality of sodium hydroxide = ……………….. N

UnKnown no. (………)

calculate the eq.wt of an aliphatic acid
 Wt. of sample = ….. gm , dissolved in ….. ml volumetric flask
 ….. ml of sample + ph.ph ….. N sodium hydroxide .
No. S.P. E.P. M.V. indicator

Eq.wt of sample = ……………………

UnKnown no. (………)

calculate the M.wt of an aliphatic acid
 Wt. of sample = ….. gm , dissolved in ….. ml volumetric flask
 ….. ml of sample + ph.ph ….. N sodium hydroxide .

No. S.P. E.P. M.V. indicator

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 M.wt of sample = ……………………

UnKnown no. (………)

calculate the number of basicity of an aliphatic acid
 Wt. of sample = ….. gm , dissolved in ….. ml volumetric flask
 ….. ml of sample + ph.ph ….. N sodium hydroxide .
No. S.P. E.P. M.V. indicator

Number of basicity of sample = ……………………

UnKnown no. (………)

calculate the purity of …………… acid
 Wt. (theoretical) = ….. gm , dissolved in ….. ml volumetric flask
 M.wt of ………acid = ………. , n= ………
 Eq.wt of ……...acid = ………..
 ….. ml of sample + ph.ph ….. N sodium hydroxide .

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 No. S.P. E.P. M.V. indicator

Wt. practical of sample = ……. gm

% purity = ………………………..
2- Estimation of an aromatic acid
Theoretical base
 The formula of an aromatic acid is Ar(COOH)n , where n: is the
number of basicity.
 We can calculate the eq.wt of an aromatic acid by back titration
(indirect method) in aqueous solution, why?.
The indirect method for the determination of water insoluble carboxylic
acids (aromatic acid) is considered by dissolving it in excess of aqueous
alkali (sodium hydroxide); the excess of alkali is back titrated with
standard hydrochloric acid .

Ar(COOH)n + NaOH Ar(COONa)n + n H2O + NaOH HCl

Strong base excess

Strong acid (HCI) ≡ Sodium hydroxide (excess)

So, volume of sodium hydroxide (excess) V\ can be calculated

from this expression :
(N.V)HCl = (N\.V\)NaOH excess

so, volume of sodium hydroxide V, that equivalent to aromatic

acid calculated as :
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 VNaOH = [ Vtotal – V\excess] ≡ Aromatic acid
so, the eq.wt of an aromatic acid gives by this expression :

[ ] [ ]

Another method
Also acids which insoluble in water may be dissolved in
aqueous or absolute methanol or ethanol and titrated with
standard alkali (sodium hydroxide) using phenolphthalein as
indicator .
These types of titrations are recommended in which a blank
titration be carried out using the same amount of solvent and
indicator;
 the difference between the two titrations gives the volume
of alkali consumed by the organic acid

Ar(COOH)n + n NaOH Ar(COONa)n + n H2O

Strong base

Procedure
1- Prepare a solution of 0.5N sodium hydroxide in 250 ml volumetric flask.
2- Prepare a solution of 0.5N oxalic acid in 100 ml volumetric flask.
2- Prepare 0.25N hydrochloric acid in 250 ml volumetric flask.
3- Standardize sodium hydroxide by using oxalic acid (0.5N) in presence
of phenolphthalein as indicator.
4- Standardize hydrochloric acid by using sodium hydroxide in presence of
phenolphthalein as indicator.
5- Weight out accurately 0.2-0.5 gm of sample (theoretical wt), place it in a
250 ml conical flask
 and add 20 ml from the prepared sodium hydroxide solution with
shaking until all the sample dissolve,
 Then titrate the mixture by using hydrochloric acid or oxalic acid
(0.25N) in presence of phenolphthalein as indicator.

7-Calculate V\ and V of sodium hydroxide from these expressions :

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 (N.V)HCl = (N\.V\)NaOH excess
VNaOH = [ Vtotal – V\excess] ≡ Aromatic acid

8- Calculate the eq.wt of an aromatic acid from this expression :

[ ] [ ]

NOTE
Also, we can calculate all the following :
1- Calculate the M.wt of an aromatic acid from this expression :

M.wt = Eq.wt × n ; n= number of basicity

2- CalcuIate the number of basicity of an aromatic acid from this expression

n=

3-Calculate purity of sample, when sample is known from these expressions :

[ ] [ ]

Calculation
1-Prepare ……N sodium hydroxide in ….. ml volumetric flask
2-Prepare …...N oxalic acid in ……….. ml volumetric flask
3-Prepare …...N hydrochloric acid in ……….. ml volumetric flask

4-standerdization of sodium hydroxide using ……N oxalic acid

 10 ml of …N oxalic acid + ph.ph ….N sodium hydroxide

No. S.P. E.P. M.V. indicator

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 SO, Normality of sodium hydroxide = …………….N

5-standerdization of hydrochloric acid using ……N sodium

hydroxide

 10 ml of ……N hydrochloric acid + ph.ph ….N sodium

hydroxide
No. S.P. E.P. M.V. indicator

Normality of hydrochloric acid = …………..N

UnKnown no. (………)

calculate the eq.wt of an aromatic acid
 Wt. of sample = ….. gm , dissolved in V(total)NaOH= ….. ml
+ ph.ph ….. N hydrochloric acid .
S.P. E.P. V Indicator

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 Eq.wt of sample = ……………………

UnKnown no. (………)

calculate the M.wt of an aromatic acid
 Wt. of sample = ….. gm , dissolved in V(total)NaOH= ….. ml
+ ph.ph ….. N hydrochloric acid .
S.P. E.P. V Indicator

M.wt of sample = …………………× n

UnKnown no. (………)

calculate the number of basicity of an aromatic acid
 Wt. of sample = ….. gm , dissolved in V(total)NaOH= ….. ml
+ ph.ph ….. N hydrochloric acid .
S.P. E.P. V Indicator

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 Number of basicity of sample =

UnKnown no. (………)

calculate the purity of ……………acid
 Wt. (theoretical) of sample = ….. gm , dissolved in
V(total)NaOH= ….. ml + ph.ph ….. N hydrochloric acid .

S.P. E.P. V Indicator

M.wt of ……. acid = ………… , n = …….

Eq.wt of …… acid = …………

Wt(practical) of sample = ……..gm ; %purity = ………..

3-Estimation of purity of aniline hydrochloride
Theoritical base
The amine salt (e.g., aniline hydrochloride) Is a salt
formed from a very weak base (aniline) and a strong acid
(hydrochloric acid).

C6H5NH2 + HCl → C6H5NH2 . HCl

Aniline hydrochloride is strongly acidic, so the titration may

be carried out directly with standard sodium hydroxide solution
in an aqueous medium.

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 This is because considerable hydrolysis occurs in aqueous
solution and progressive neutralization of the free acid results
ultimately in complete hydrolysis to the amine and strong acid.

C6H5NH2 . HCl C6H5NH2 + HCl NaOH

Strong base
SO ,
Strong base ≡ Hydrochloric acid ≡ Aniline hydrochloride

[ ] [ ]

 When eq.wt of aniline hydrochloride =………….. ;

SO , (practical wt ) of sample can be calculated.
 The percentage purity of the amine salt is calculated from the
formula :

Procedure
1- Prepare a solution of 0.IN sodium hydroxide in 250 ml
volumetric flask.
2- Prepare a solution of 0.1 N oxalic acid in 100 ml volumetric
flask .
3- Standardize sodium hydroxide by using oxalic acid in
presence of phenolphthalein as indicator.
4- Weight out accurately (1.5-2.0) gm of sample (theoretical
wt.) and dissolve it in 100 ml or 250 ml volumetric flask.
5- Transfer 10 ml or 25 ml of sample to a 250 ml conical flask,
and then titrate with standard sodium hydroxide solution in
presence of phenolphthalein as indicator.
6- Repeat step 5 about ( 3-5 ) times and calculate mean volume
(M.V.).
7- Calculate (practical wt.) of sample from this expression :
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 [ ] [ ] × Factor

Where ; Factor =

8- Calculate the percentage purity of sample from the

following
expression :

NOTE THAT :
[ Eq.wt = M.wt ]aniline hydrochloride

Calculation
1- Preparation of ……. N sodium hydroxide in …. ml
volumetric flask .

2- Preparation of ……. N oxalic acid in …. ml volumetric

flask .

3- Standerization of sodium hydroxide using …. N oxalic

acid
 10 ml of ….. N oxalic acid + ph.ph ….. N sodium hydroxide
No. S.P. E.P. M.V. indicator

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 Normality of sodium hydroxide = ………….. N

UnKnown no. (………)

 Wt. (theoretical) = ….. gm , dissolved in ….. ml volumetric
flask
 ….. ml of sample + ph.ph ….. N sodium hydroxide .
No. S.P. E.P. M.V. indicator

% Purity = ……………………%

UnKnown no. (………)

 Wt. (theoretical) = ….. gm , dissolved in ….. ml volumetric

flask

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 ….. ml of sample + ph.ph ….. N sodium hydroxide .
No. S.P. E.P. M.V. indicator

% Purity = ……………………%

UnKnown no. (………)

 Wt. (theoretical) = ….. gm , dissolved in ….. ml volumetric
flask
 ….. ml of sample + ph.ph ….. N sodium hydroxide .
No. S.P. E.P. M.V. indicator

% Purity = ……………………%

UnKnown no. (………)

 Wt. (theoretical) = ….. gm , dissolved in ….. ml volumetric

flask
 ….. ml of sample + ph.ph ….. N sodium hydroxide .

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No. S.P. E.P. M.V. indicator

% Purity = ……………………%

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 Chromatography

 Definition

Chromatography is the modern technique used for qualitative

and quantitative analyses of organic and inorganic compounds.

The most important uses of chromatography are :

purification and separation of mixtures

In this technique the components of a mixture can be separated

by allowing the sample to be distributed between two phases, one
of them is stationary (stationary phase), while the other moves
(mobile phase).

The stationary phase may be solid or liquid supported on a solid

 Types of chromatographic techniques in this study :

l- Column chromatography

2- Paper chromatography.

3- Thin layer chromatography (TLC technique}

I- Column Chromatography
A column is a long narrow tube with dimensions depend on the
weight of the sample, e.g. the column with 20-30 cm. long and 1-3 cm

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 diameter packed with 50-100 gm of stationary phase supported on a
plug of cotton or glass wool, may retain several grams of the sample.

Column used in different chromatographic techniques, one of them

named adsorption chromatography.

Adsorption chromatography :

 This technique of chromatography is based upon the selective

adsorption from solution to the active surface.
 When related substances exhibit different degree of adsorption, the
separations take place.

Column chromatography technique depend on spread the components of

a mixture through a column of two phases: solid Stationary phase and
mobile phase. So this technique is named ―Liquid-Solid chromatography

Liquid – Solid Chromatography Technique

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For Example ,

 A solution of leaf pigments in an organic solvent is passed slowly

through a tube packed with a white adsorbent, such as alumina, the
individual pigments tend to be retained in different coloured
zones, the most strongly adsorbed components being retained at
the top, and those with less affinity for the adsorbent at lower
levels.
 The initial separation of the different coloured layers is usually not
very distinct, but by passing a fresh portion of the original solvent
or of another solvent through the tube, the zones become more
sharply defined and may spread over the length of the whole
column.
 The resolution into chlorophyll-b (yellowish-green), chlorophyll-a
(bluish-green), xanthophyll (yellow), and carotene (yellow) is
clearly visible.

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Choice of stationary phase ( Adsorbents ) :

The separation process for organic mixture is successfully occur ,

when solid stationary phase has these conditions :

1- Chemically inert.
2- Must be adsorbent material.

The most widely used adsorbent is activated aluminium oxide

―alumina ― and silica gel .

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 Other adsorbents include magnesium oxide , magnesium carbonate ,
magnesium trisilicate , calcium carbonate , barium carbonate , calcium
hydroxide , calcium sulphate , glucose , lactose , starch and cellulose .

They are all employed in the form of uniform white powders , the inorganic
compounds may usually be improved by heating at 200-230 c

Choice of mobile phase :

Mobile phase is the moving phase " Eluent ", which separate the mixture to
its components.

The choice of solvent will naturally depend in the first place upon the
solubility relations of the substance.

The solvents generally employed possess oiling points etween 0 and 100
C

The most widely used medium is light petroleum ( .p. not a ove 0 ); others
are cyclohexane, carbon disulphide , benzene, chloroform, carbon tetrachloride,
methylene chloride, ethyl acetate, ethyl alcohol, acetone, ether, pyridine and
acetic acid.

Solvents have a triple role:

1- They serve to introduce the mixture to the column.

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2- They effect the process of development by which the zones of the
chromatogram are separated to their fullest extent.

When used for this purpose, the solvents are termed developers.

The developer is generally a solvent in which the components of the

mixture are not too soluble and is usually a solvent of low molecular weight.

3- They remove the required content of each zone from the mechanically
separated parts of the column.

Solvents utilized for removing the various components of a mixture as

separated on a column are called eluents.

Reference has already been made to the choice of solvent for introducing
the mixture to the column. Adsorption takes place most readily from non-polar
solvents , such as petroleum ether or benzene, to highly polar solvents such as
alcohols, esters and pyridine.

The choice of an eluent is governed by a few simple and obvious rules.

 It should be a liquid which is a good solvent for the components to be

eluted.
 The eluent may be well adsorbed itself, so that its solvent action is
assisted by its displacing action at the interface; sometimes some
strongly adsorbed substance may be added to the eluent to promote this
displacement.
 The eluent should be easily removable from the desorbed component ;
low- boiling eluents may be used to elute high-boiling substances; basic
or acidic solvents may be employed to elute stable neutral copmpounds ;
and neutral solvents to elute acidic or basic subtances .

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32
 Paper Chromatography

Definition:
Paper chromatography is the cellulose filter paper, that has capillary
action, let it be known direction.

Uses :

This is a very simple form of chromatography that is used widely for

qualitative and quantitative analyses.

Theory :

A sample mixture is spotted onto a strip of filter paper with a

micropipette, and the chromatogram is "developed" by placing the

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bottom of the paper (but not the sample spot) in a suitable solvent.

The solvent" mobile phase- eluent" is drawn up the paper " stationary
phase" by capillary action and the sample components move up the paper
at different rates depending on :

1- Their solubility .
2- Their degree of retention by the paper.

Following development, the individual solute spots are noted or are

made visible by treatment with a reagent that forms a colored derivative.
The spots will generally move at a certain fraction of the rate at which
the solvent moves and they are characterized by the Rf value
(Retention factor = Retardation factor) :
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 Rf = 1

So,

(Rf ) A = , (Rf ) B = , (Rf ) C =

The distance which the solvent moves is measured from the starting
line to the maximum solvent front, whereas the distance the solute moves
is measured from the starting line to the center of the solute spot.

The Rf value , then, is characteristic for a given paper and solvent

combination .

Notes :

1-The mechanism of separation :

 The cellulose filter paper used is very hydrophilic and will normally
have a thin coat of water adsorbed from the air.
 So the mechanism of separation is a liquid-liquid chromatography
(partition chromatography) in which the sample distributes between the
stationary water phase and the developing solvent.
 The developing solvent is usually a mixture of an organic solvent with
water and sometimes, a water-immiscible solvent is used to develop the
chromatogram.

2-Two-dimensional paper chromatography :

 A principal advantage of this technique is that greater separating power

can be achieved by using two-dimensional paper chromatography;
 A large square piece of paper is used, and the sample is spotted at a
bottom corner of the paper.
 After development with a given solvent system, the paper is turned (90
deg ) and further development is obtained with a second solvent system.

35
 Thus if two or more solutes are not completely resolved with the first
solvent, it may be possible to resolve them with a second solvent .

Two-dimensional paper chromatography is required for complex

mixtures, such as protein hydrolyses. Almost any mixture of organic
components can be separated.

Separation of chlorophyll mixture by using paper chromatography

Extraction of chlorophyll components from green plants :

There are two methods :

1- By grinding green plants with sand in mortar, using chloroform as

solvent
2- By refluxing green plants in methanol or chloroform in round-
bottom Flask .

mortar

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Choice of solvent :

The most choice of eluent for large separation by using paper

chromatography is :

chloroform : petroleum ether (40-60 C) as 2 : 8 ratio

procedure

1- A pencil line is drawn across the paper a few centimeters from the
bottom and the sample is spotted on it.

The spot must be made as small as possible for maximum separation

and minimum tailing .

2- It is best done dropwise with a blower to evaporate the solvent

after each drop .
3- The paper is placed in chamber (Jar) with its end dipping in the
developing solvent, a Closed chamber must be used to saturate the
atmosphere with the solvent and prevent it from evaporating .
4- The developing may take time, but it requires no operator time.

The amount of development will depend on the complexity of the mixture

being separated. Several samples and standards can be spotted along the
bottom and developed simultaneous .

5- Dry the paper

6- 1f the solutes fluorescence , they can be detected by an ultraviolet
light
7- Apencil line is drawn around the spots for permanent
identification, color developing reagent are often used;
for example, amino acids and amines are detected by spraying the
paper with a solution of ninhydrin, which is converted to a blue or
purple color.

37
 8- Afier the spots identified, they may be cut out and the solutes
eluted to determined quantitatively by a micro method, and then
determined qualitatively by Rf values.

Calculation:

 Num er of components = …………

 Distance solvent front move "E" =……… Paper

chromatography

1- Distance solute moves ―a‖ = …………. cm

( Rf ) a = ……………..

2- Distance solute moves ― ‖ = …………. cm

( Rf ) b = ……………..

3- Distance solute moves ―c‖ = …………. cm

( Rf ) c = ……………..

4- Distance solute moves ―d‖ = …………. cm

( Rf ) d = ……………..

5- Distance solute moves ―e‖ = …………. cm

( Rf ) e = ……………..

38
 Thin-layer Chromatography (TLC)

Definition

Thin-layer chromatography (TLC) is very similar to paper

chromatography except that the stationary phase is a thin layer of
finely divided adsorbent supported on glass or aluminum plate,
plastic strip, etc.

Uses :

This technique is used for separation of mixtures or purification of

different chemical compounds.

Separations include vitamins, cholesterol, fatty acids, lipids in

serum, amino acids, dyes, glycerides, pesticides, and sugars

Preparation of TLC plates :

 The most commonly used adsorbents, include alumina, silica gel, and
cellulose.
 A slurry of the powdered substance is formed and spread on a
plate with an adapter to assure uniform thickness.
Often a binder such as plaster of paris (chloroform or distilled water) is
added to improve the adhesive properties of the material
 After being dried, it is activated by heating in an oven at 110 C for several
hours.
 Commercially prepared plates and strips on plastic are available.

Theory :

 The technique of development is the same as for paper

chromatography.
 The detection of the spots is sometimes easier than with paper
chromatography because more universal techniques can be used.

39
 A sample mixture is spotted onto a plate of TLC with a micropipette,
and the chromatogram is "developed" by placing the bottom of the plate
(but not the sample spot) in a suitable solvent.
 The solvent "mobile phase – eluent " is drawn up the plate of TLC
"silica gel – stationary phase " by capillary action and the sample
components move up the plate at different rates so the mechanism of
separation is liquid—solid chromatography.

Qualitative analysis :

The spots will generally move at different rate and they are
characterized by Rf value (Retention factor = Retardation factor) :

Rf = 1

So,

(Rf ) A = , (Rf ) B = , (Rf ) C =

The common technique for organic compounds is spraying the

plate with sulfuric acid and then heating it to char the Compounds
and develop black spots .

40
Quantitative analysis

The spots can be scraped off the plate and the solutes eluted for
quantitative determination.
The TLC plates must be have less thickness and must be
powder silica gel used to give good accuracy.

Separation of chlorophyl components using TLC -

chromatography

Extraction of chlorophyl components from green plants:

There are two method :

I- By grinding green plants with sand in mortar, chloroform

as solvent.
II- By refluxing green plants in methanol or chloroform in round
bottom flask.

mortar

Choice of solvent :

The most choice of eluent for large separation by using TLC

chromatography is :

chloroform : methanol
as 10 : 0.5 ratio

Procedure

1-Preparation of silica-gel TLC plate :

A slurry of the powdered silica gel with distilled water (1:2 ratio), is
formed and spread on the plate; and then dried, activated by heating in an
oven at 110 C for several hours.

41
 2-By using a pencil, TLC plate marked a few centimeters from the
bottom in any side (right or left) of plate without make a mistake isurface
of silica gel.

 And the sample is spotted on this for future reference in Rf

measurements. The spot must be made as small as possible for
maximum separation and minimum tailing.

3-It is best done dropwise with a blower to evaporate the solvent each
drop.

4-The TLC plate is placed in a chamber (Jar) with its end dipping in the
developing solvent, a closed chamber must be used to saturate the
atmosphere with the solvent and prevent it from evaporating from the
surface of the silica gel as it moves up.

5-The developing may take time, but it requires no operator time

6-Several samples and standards can be spotted along the bottom and
developed simultaneously .

7-Dry the TLC plate horizontal .

8-The spots for identification components were detected.

9-After the spots identified, they may be cut out and the solutes eluted
and determined quantitatively by a micro method, and then determined
qualitatively by Rf values.

Calculation:

 Number of components = …………

 Distance solvent front move "E" =………
1- Distance solute moves ―a‖ = …………. cm

( Rf ) a = ……………..

2- Distance solute moves ― ‖ = …………. cm

42
 ( Rf ) = ……………..

3- Distance solute moves ―c‖ = …………. cm

( Rf ) c = ……………..

4- Distance solute moves ―d‖ = …………. cm

( Rf ) d = ……………..

5- Distance solute moves ―e‖ = …………. cm

( Rf ) e = ……………..

43
 Synthesis of aspirine

In this experiment you will synthesize aspirin (acetylsalicylic acid, C 9H8O4 ),

purify it, and determine the percent yield. The purity of the product will be
confirmed by qualitative analysis and by measuring its melting point range.

Procedure
Weigh out 2.00 g (+ 0.01 g) of salicylic acid (C7H6O3), 8 ml acetic anhydride in
a small conical flask, carefully add 1 drop of concentrated sulfuric acid (18 M
H2SO4), a catalyst, to the flask and rotate a mixture in water bath as shown in
Figure 1.

Figure 1.

After the water begins to boil, heat for an additional 15 min. After heating, turn
the bunsen burner off and CAREFULLY remove the flask from the water bath
(remember it is hot!) and allow the flask and contents to cool on the lab bench
for about 3 min. After the flask has cooled for about 3
min., CAUTIOUSLY add 15 mL of ice water to the flask to facilitate the
decomposition of the excess acetic anhydride. Swirl the flask to mix the
contents. If crystals do not appear, you can scratch the walls of the flask with a

44
stirring rod to induce crystallization. Filter , dry and weight ppt then measure
M.P and calculate the purity.

45
 Synthesis of chalcone

Synthesis

Procedure

Mix 0.05 mol of any arom.ketone compound with 10 ml ethanol and 10 ml

sod.hydroxide in aconical flask. A solution of 0.05 mol aro.aldhyde in 5 ml
ethanol was added, stir the reaction mixture at 15oC for 15 min. Cool the
reaction mixture in an cold ethanol then filter the ppt and wash with ice
cold alc. Calculate the yield and measure the melting point.

46
 Dyes
What is the dye?
The dye is a substance that is used as a carrier of colors on textiles
,papers, leather, etc.so that those colors fixed on these products
protect it against many factors such as heat, light ,washing ….etc.

Dyes differ from pigments in that the second is inorganic

compounds (devoid of carbon)that are insoluble in most solvents
,where they are found in suspended form in some liquids such as
inks,paints and others or mixed with other substances ,unlike dyes
that consist of organic materials (containing carbon )and usually
soluble in most solvents .currently dyes market has more than 9000
colors for more than 50000 brands around the world .

 Is every colored material a dye?

No

The reason is that certain conditions must be met for the colored
material in order to be adye ,which are:

 According to its physical properties :

1- It should have a suitable color.
2- The property of dye fastness on the fabric
3- Resistant to light and washing.
4- Not affected by water or acidic or basic solutions whether they are
concentrated or not.

47
 According to chemical properties:
1- Chromogen
A colorless chemical compound that can be converted by chemical
reaction into a compound that can be described as color or it can be said
that it is the structure of the dye such as the aromatic benzene ring and
naphthalene.

3- Chromophore

It is the main compound of the dye color that works to transfer the
compound from the ultraviolet region to visible area ,thus the color
appears ,like NO2 and SO3 groups, and it is also a functional group linked
to a benzene ring and the presence of color ln lts constituent compounds
is attributed to it, and it has been observed that some compounds such as
quinines ,azo compounds,nitro compounds have clear colors.

48
3- Auxochrome

It is know as the colered group and it Is responsible for the dye fixing on the
fabric like OH group.

 Classification of dyes:

In terms of chemical composition or chromophore group :

1- Nitro dyes: they are dyes in which chromophore group is_NO2

2- Azo dyes : they are dyes in which chromophore group is -N=N-
3- Anthraquinone dyes :they are dyes in which chromophore group is
anthraquinone

In term of application to the fabric:

1- Direct dyes :It’s a class of dyes that are applied directly to the fa ric
in dyeing bath whether the medium is neural or alkaline .It produces
full shades on cotton and linen whithout underlaying and it can also
be applied on rayon matt silk and wool.

2- Acidic dye: Is a dye that applied to a low PH textile dye asnd it is

manily used to dye wool not cotton fabric .some acidic dyes are used
as food colorants and some can be used to dye organelles in the
medical field.

49
3- Basic dyes: basic dyes are cationic dyes that are soluble in water and
are applid manly to acrylic fibers ,but it finds some uses for wool and silk
usually acitic acid is added to the dyeing bath to help the dye absorbed
into the fabric (increase the ability between the dye and fabric ) it is used
in papers coloring.

There are other kinds of dyeas like vat dyes and reactive dyes …etc.
but our studies will be limited on basic dyes and acidic dyes.

***Textiles***
Textiles refer to materials made of fibers ,thin threads ,natural
,manufactured or mixed threads.
Textile are created by interwining these threads in specific patterns which
leads to the formation of the o called cloth .

** Textiles can be classified as either natural or synthetic .natural fibers

can also be classified according to the origion as animal,plant, or
inorganic.

1- Natural fibers:
They are harvested from plants or by cutting animal fur .the most
common ones you should know are wool,silk (from animals) cotton and
linen (from plants).

2- Manufactured fiber:
Man –made fibers like polyester fibers ,acrylic fibers and nylon fibers.

50
  Dyeing bath
It is the solution that contains the dye,the raw material and the
chemicals.

 Exhaustion
It expresses the extent of the dye transfer from the dyeing bath to
the raw material and it expressed as a percentage .

 Leavness
It is the property of the dye spreading at the same depth over all
parts of the fabric .
 Shade
It is the amount of dye required to complete the dyeing process and
it expressed as a percentage .

 Liquor ratio
The ratio between the weight of the raw material to the voiume of
the dyeing bath.

51
 Dyeing of wool by acid dye

Wool: they are protein compounds consist of many amino acids to

form the wool fabric .

** Mechanism of dyeing wool by acidic dye:

Procedure:
1- In a 1:50 dyeing bath apply the required dye according to the
required shade then add 10%from global salt .
2- Stirr and add one drop from concentrated sulfuric acid .
3- Put the raw material insaide the dyeing bath on a cold for two
minutes then put the beaker on direct flame for half an hour .
4- Take out the raw material ,wash and iron it well.

Dyeing of cotton by basic dye

Basic dye:
It is used in dyeing animal fiber such as wool and it is also called
cationic dye because the positive part is responsible for the dyeing
process .It is also used in dyeing plant fibers such as cotton .In
order to the dyeing proces be complated a catalyst must be added
to facilitate the ability of the dye to the fabric.

52
 *** The dyeing proces takes place in three stage :
1- Treatment the raw material in the dyeing bath that contains
tannic acid which act a an intermediary to increase the ability
between the dye and the fabric by makung a complex.
2- Treatment the raw material in tartamestic [Na,K tartarate] which
works to fix the dye on the fabric.
3- The dyeing process itself.

Procedure:
1- In a 1:80 dyeing bath treat the sample with 3%tannic acid
solution at room temperature for five minutes then remove the
sample without squeezing it .
2- In a 1:80 dyeing bath treat the sample with a 3% tartamestic for
ten minutes.
3- In a 1:80 dyeing bath apply the required dye according to the
required shade then put tow drops from acetic acid and let the
dye bath at the room temperature for ten minutes whit stirring ,
4- Put the beaker on the dirct flame for 30 minutes .
5- Wash and let it to dry.
**** **** ****

Exercises

1- When 2 gm. of citric acid monohydrate was dissolved and

completed into 100 ml by distilled water. Titration of 10 ml of this
solution, required 20 ml sodium hydroxid (0.1 N). Calculate the
purity of sample.

2- Calculate the M.wt of aliphatic acid , when 3 gm of the sample

(purity = 82 %) dissolved and completed into 100 ml by distilled

53
 water. The titration of 13 ml of this solution, required 21.3 ml of
sodium hydroxide (0.2 N) .
( number of basicity = 2 )

3- Calculate the purity of' aniline hydrochloride , when 2 gm of the

sample dissolved in 100 ml distilled water , and titration of 10 ml
of sample solution ,required 3.3 ml of sodium hydroxide (0.45 N) .
( M.wt of aniline hydrochloride = 129.59 )

4- Titration of 30 ml of an aliphatic acid solution ( 0.5 gm of the acid

in 250 ml distilled water) , required 7.6 ml of sodium hydroxide
(0.1 N).
Calculate number of basicity of the acid .
( M.wt of acid = 158 )

5- Calculate the eq .wt for aliphatic and aromatic acids in a mixture

( the sample wt. = 4 gm )The mixture was dissolved in 100 ml
distilled water, and filtered off the solution mixture . 10 ml of the
filtrate was titrated with 15 ml of 0.5 N sodium hydroxide
solution. The precipitate (Wt. = 1.5 gm) Was dissolved in 30 ml
0.5 N sodium hydroxide solution . the excess sodium hydroxide
was titrated with 12 ml 0.32 N hydrochloric acid .

6- Calculate the percentage of formaldehyde in sample of formalin

solution, when 2 ml of sample was diluted into 500 ml by distilled
water ; and 25 ml of this solution was treated with ((50 ml I2 , 0.1
N) , 5 ml NaOH , 2N ) , and then acidified with HCl 2 N , the
liberated iodine titrated with sodium thiosulphate solution in
presence of starch as indicator, the volume consumed was 32 ml.
(10 ml iodine react 9.4ml sodium thiosulphate)

7- For that reaction :

R-(NH2)2 + (CH3CO)2O …………….

54
Answer the following

i- complete the equation.

ii- give reason for important use of pyridine.
iii- give reason for important use of blank for determination of
eq.wt of amine .
iv- when 2 gm of sample are used for that reaction, and the
excess of acetic anhydride converted into acetic acid, and
the reaction
mixture is completed into 250 ml by dist. water.
The titration of 25 ml of this solution required 30 ml of 0.1 N
sodium hydroxide.
Calculate M.wt, Eq.wt and percentage of amino groups in
sample .
( 25 ml of the blank required 67 ml of 0.1 N NaOH). (Exa. 2002)
8- Titration of 12 ml of an aliphatic acid solution (0.5 gm of the acid
in 100 ml dist. water) required 7.6 ml of NaOH (0.1 N). Calculate
the M.wt of the acid
(basicity of acid 2) (Exa. 2002)

9- Calculate the eq.wt of amino acid, when 3 gm of the sample

dissolved and completed into 100 ml by dist. water.
Titration of 13ml of this solution after neutralization and addition
of 10 ml of neutralized formaldehyde solution, required 16.6 ml of
NaOH (0.2N).
(purity of sample = 76 %). (Exa. 2003)

10- For pure wt. gm of hydroquinone, the percentage of hydroxyl

groups = ………. (Exa. 2004)

11- Calculate the percentage of vinegar, when 15 ml of sample

diluted to 100 ml. Titration of 10 ml of this solution, required 13.5
ml of sodium hydroxide (0.1 N).
(density of sample = 0.9 gm/cm3),(Exa. 2004)

55
 12- When 2 gm of crude citric acid have purity 76 %, was dissolved
and completed into 100 ml by dist. water, and titration of 10 ml of
this solution, required 20 ml of NaOH (0.1 N). Calculate the
number of hydrated water in sample.
(Exa. 2005)

13- Calculate the following :

1- purity of ß-naphthylamine in sample .
2- percentage of amino groups in crude ß-naphthylamine sample.
3- percentage of amino groups in pure ß-naphthylamine sample.
When 3 gm of crude ß-naphthylamine dissolved in 20 ml pyridine
and reacted with 4 ml acetic anhydride, and the excess of acetic
anhydride hydrolyzed by water into acetic acid . Then the reaction
mixture was completed into 250 ml by distilled water. Titration of
30 ml of this solution required 51 ml of 0.12 N NaOH.
(Titration of 30 ml of the blank prepared at the same condition,
required 39 ml of 0.2 N NaOH) , (Exa. 2006)

14- A mixture (3.5 g) of aliphatic acid and aniline hydrochloride was

dissolved in 100 ml distilled water. Titration of 13 ml of this solution
requires 15.2 ml NaOH (0.428N) using phenolphthalein as indicator,
while titration of 12 ml of this solution requires 13.9 ml NaOH (0. 1N)
using methyl orange as indicator.

Calculate the Number of basicity of this aliphatic acid .

M.wt of aniline hydrochloride =129.59
M.wt of an aliphatic acid = 104

56