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DR.DIGAMBAR D. GAIKWAD

DEPT. OF CHEMISTRY,

GOVT. COLLEGE OF ARTS & SCIENCE,

AURANGABAD-431001

E-mail:-gaikwad_dd@yahoo.com

Mobile No. 9422552449

------------------------------------------------------------------------------------------------------------

CHAPTER 3. ELECTRONIC SPECTROSCOPY The spectroscopy deals with study of electromagnetic radiation with matter. Thus

it gives the information about the transition between rotational and vibrational energy

levels with the addition of electronic transition. In this technique the matters come in

contact with different types of radiations such as such as UV, Visible, Micro-wave,

Radio-waves produce different kinds of excitations of the molecules and this excitation

provides important information about the structure of molecules and atoms. This

information given by the molecule are recorded with the help of a device

Spectrophotometer, in the form of a graph also known as Spectrum. A molecule exposed

to radiation absorbs part of it and gets excited to higher energy level. The type of

wavelength of radiation absorbed by the molecule in order to reach the state depends

upon the structural features of the molecule. By studying the spectrum technique

obtained, it is possible to throw light on the chemical constitution of the molecules. The

spectroscopy can be studded by following two basic points

i) Atomic spectroscopy:- This spectroscopy is concerned with the

interaction of electromagnetic radiation with atoms which are commonly

in their lowest energy states

ii) Molecular spectroscopy:-This spectroscopy deals with interaction of

electromagnetic radiation with molecules.

Fig. 3.1 Propagation of electromagnetic radiation

Where

X →Axis of propagation of radiation

Y → Axis represents direction of magnetic field

Z → Axis represents direction of electric field.

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Thus results in transition between rotational and vibrational energy levels in

addition to electronic transitions. Spectroscopy is, therefore, the study of molecular

responses when it is exposed to certain kind of radiations.

The absorption of different types of radiations such as UV, Visible, Micro-wave,

Radio-waves produce different kinds of excitations of the molecules and this excitation

provides important information about the structure of molecules. Since in these methods

the nature of radiation absorbed is studied, the Spectroscopic methods designated as

absorption Spectroscopy (The Ultra-Violet region, which extends from 200-400 nm and

the visible ranging from 400-800 nm)

In absorption Spectroscopy, the molecule has different kinds of radiations.

Therefore it is necessary to study the properties of radiations. Electromagnetic radiations

has dual nature i.e. particle and wave form. In studying Spectroscopy more importance is

given on wave form of the radiation. Radiations when propagate in wave form produce

electric and magnetic fields. The directions of their propagation are mutually

perpendicular to each other as shown in the above fig. (1).

The electromagnetic wave are characterized by following parameters

1. Amplitude (a)

2. Energy (E)

3. Wave length (λ)

4. Frequency (γ)

5. Wave number (υ)

The units commonly used for the following parameters are as fallows

Wave length (λ):-It is the distance between two successive maxima on an

electromagnetic waves

1 A° (Angstrom) 10-8

cm = 10-10

m

1 nm (Nanometer) 10-7

cm = 10-9

m

1 µ (Micron) 10-4

cm = 10-6

m

Frequency (γ):- The number of wavelength units passing through a given points

in unit time is called as the frequency

1Hz (Hertz) = 1 Cycle per second (cps)

1 KHz (Kilo Hertz) = 103 Hertz

1MHz = 106 Hertz

Wave number (ν):- The frequency as the wave number which is defined as the

number of waves per centimeter in vacuum. This quantity is generally denoted by ν

The measured in term of number of waves/cm = cm-1

or Kaysers (K)

1 kayser = 1 cm-1

The spectrum, of electromagnetic radiation is shown in bellow with increasing

wavelengths or decreasing frequencies.

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Fig. 3. 2 Electromagnetic spectrum.

The visible portion is shown on an expanded scale

Wavelength (λ)

Types of radiations according to wavelength Table No. 1

The above table shows that, cosmic rays have the shortest wavelengths and

highest energy similarly radio waves have the longest wavelengths and lowest energy. As

we move from left hand side to right hand side in the above table the wavelength goes on

increasing at the same time energy and frequency go on decreasing. The visible region

lies between Infra- red and Ultra -violet region. Although all types of radiations travel as

waves with same velocity but they differ from one another in certain properties for

example, X - rays can pass through glass and muscle tissues. Radio waves pass through

Cosmic

Y -

ray

X-

ray

UV

Visible

IR

Micro

wave

Radar

TV

Radio

nm 10-5

10-3

0.1 100 400 800 103

High energy

Energy(E)

107 10

9 10

10 10

11

Low E

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air. Visible and Infra-fed radiation can be sent by reflection or diffraction in a prism, it is

observed that when radiation of certain frequency and energy (ΔE=hυ) are passed

through an organic compound, the electrons of the component atoms are excited. In

addition the vibrational and rotational energies of the molecules as a whole are quantized.

The wavelength are measured and recorded in the form of a spectrum with the help of a

device called Spectrophotometer. If we plot the changes in absorption against

wavelength, we get absorption bands which are highly characteristic of a compound and

the technique provides an excellent tool to elucidate the molecular structure of an

unknown compound. Interaction of radiation with Matter when radiation strikes the

matter, the molecule absorbs part of it. The wavelength or frequency of radiation

absorbed depends on the structural features of molecules. As a result of absorption of

energy, molecule undergoes excitation higher energy state. The type of excitation

produced depends upon the energy of the radiation used.

i. Rotational excitations:- These excitations occur if microwaves are used (λ= 105

to 107 nm)

ii. Vibrational excitations:- These excitations occur if IR radiations are used. Since

the energy of IR radiation is higher than microwave (800 – 10-5

) along with

vibrational excitations, rotational excitation also takes place.

iii. Electronic excitations:- These excitations occur if the radiations from Visible

and UV region are used; they bring about electronic excitations from bonding

and anti-bonding levels. As energy of these radiations is higher than IR

radiations, along with electronic excitations, vibrational and rotational excitations

also take place.

Microwaves = Rotational excitations

Infrared = Vibrational + Rotational excitations

UV and Visible = Electronic + Vibrational + Rotational excitations

In the study of spectroscopy, the molecule is exposed to the various types of

radiations. The wavelength of the radiation slowly changed from minimum to maximum

in the given region and the absorption at every wave length is recorded. A graph

wavelength vs absorbance is then plotted.

Fig. 3.3 Graph of wavelength vs Absorbance Above graph shows that the absorption is maximum at a particular wavelength.

The wavelength at which there is maximum absorption observed is called as wavelength

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maximum. This is a characteristic property of the molecule and helps in determining its

structure.

Ultraviolet and Visible Spectroscopy

These are given following regions

Vacuum Ultraviolet 1-180nm

Ultraviolet 180-400nm

Visible 400-750nm

The Ultra violet region, which extends from 180-400 nm and the visible region

from 400-75nm are more useful to organic chemists. They bring about electronic

excitations in the molecule from bonding to anti-bonding levels. Information about the

structure of the molecule containing double bond or triple bond or conjugated bonds is

given by UV-Visible spectroscopy. It also helps in differenting conjugated and isolated

dienes, carbonyl compounds and α, β unsaturated carbonyl compounds and cis and trans

isomers. Since the energy level of a molecule is quantized, the energy required to bring

about the excitation is a fixed quantity. Thus, the electromagnetic radiation with only a

particular value of frequency will be able to cause excitation. Allowed to fall on sample

of the molecule, energy is absorbed and electrons will be promoted to the higher energy

levels.

Beer - Lambert law The Beer-Lambert law is the linear relationship between absorbance and

concentration of an absorbing species. This law states that, "The fraction of the incident

light absorbed is proportional to the number of molecules and the path of light that is

absorbed by the solution which is proportional to its concentration. The absorbance of

light is directly proportional to the thickness of the media through which the light is being

transmitted multiplied by the concentration of absorbing chromospheres.

This law states that

Log Io = ξ x C x L

It

Where

Io=Intensity of incident light

It = Intensity of transmitted light

A = Absorbance

ξ = Extension coefficient

C = Concentration of solution (ml)

L = Length of the cell (cm)

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The UV spectra are usually recorded as absorption (A) vs wavelength (λ). The

intensity of peak is found by plotting ξ or log ξ vs wavelength. The intensity of

absorption depends upon following two things

i) Size of the molecule

ii) Change in dipole moment

Spectrophotometer:-

The device which detects the percentage transmittance of light radiation when

light of certain intensity and frequency range is passed through the sample. The modem

UV-Visible spectrophotometer consists of following components

i. Radiation source:- A deuterium or hydrogen lamps discharge of the range

180-400 nm or tungsten filament lamp of wavelength greater than 375 nm

are used as radiation source.

ii. Sample container:- It is a quartz or fused silica transparent cell of 1 cm

path length. (Glass cell cannot be used as glass absorbs light strongly below

300 nm).

iii. Monochromator:- The incident radiation is dispersed with the help of a

rotating prism and wavelengths thus separated are passed through a

specially devised slit to a monochromatic beam of desired wavelength. It is

then divided into two beams of equal intensity. Thus light from the first

dispersion is passed through a slit called Monochromator and then sent to

second dispersion, light passes through the exit slit.

iv. Detector:- In Spectrometer the reference beam subtracts the absorption of

the solvent from the absorption of the solution. The signed for the intensity

of absorbance Vs corresponding wavelength is automatically recorded on the

graph with the help of detector.

v. Amplifier: The amplifier is coupled to a small servomotor, which drives an

optical wedge into the reference beam until the Photoelectric cell receives

light of equal intensities from the sample as well as reference beams.

vi. Recorder:- Amplifier is also coupled with a small servomotor which in turn

is coupled to a pen recorder. It records the absorption bands automatically.

vii. Sample and Solvent:- It is essential that a spectrum should be recorded in

dilute solutions and solvent must be transparent within the range of

wavelength being examined. Only 0.1 mg of pure and dry sample is dissolved

in about 100 ml non absorbing solvent like cyclohexane, 1, 4-dioxane, water

or 95% ethyl alcohol. (Absolute alcohol cannot be used as it contains traces of

benzene which shows peak at 225 nm). Sometimes a sample in pure gaseous

state may be used.

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Fig. 3.4 Ultraviolet Spectrophotometer

Working One of the beams of selected monochromatic light is passed through the sample

solution and the other beam of equal intensity is passed through the reference solvent.

The intensities of the respective transmitted beams are then compared over the whole

wavelength range of the instrument. The spectrophotometer electronically subtracts the

absorption of the solvent in the reference beam from the absorption of the solution. The

signal for the intensity of absorbance versus corresponding wavelength is automatically

recorded on the graph. The spectrum is usually plotted as absorbance (A) against

wavelength (λ).

Types of Electronic Transitions

UV - Visible radiations are more energetic. Absorption of these radiations by an organic

compound brings about electronic excitations. The process of electronic excitation is

accompanied by a large number of vibrational and still larger numbers of rotational

changes.

E = E electronic + E vibrational + E rotational

Fig. 3.5 Energy levels diagram

There are three kinds of electrons present in organic molecule viz.

i) σ- electrons present in Sigma bonds

ii) π - electrons present in pi bonds

iii) n - nonbonding electrons present as unshared electrons

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Fig. 3.6 Electronic transitions

The amount of ultraviolet radiation absorbed depends upon the structure of the

compounds and their wavelength of radiation. The following types of electronic

transitions.

1. σ →σ* Transition:

The electron is excited from bonding to anti-bonding orbital that is σ

→σ* Transition. High energy is required for this transition as sigma (σ ) bonds

are very strong. This transition accurse at shorter wavelength that is 165nm such

as saturated hydrocarbon like CH4, CH3CH3

C C C C6 6*

2. π → π * Transition:

The excitation of electron from bonding (π ) to an anti-bonding (π *)

orbital. This transition occurs in compounds containing double or triple bonds.

e.g. Alkenes, Alkynes, Carbonyl and Cyanides compounds etc. Such transitions

requires comparatively less energy than that required for n →σ* transitions and

conjugation of double bonds lowers the energy required for the transition and

absorption occurs at longer wavelengths.

C C C C

3. n →σ* Transition:

When one electron from nonbonding lone pair to anti-bonding orbital

that is n →σ* Transition. This type of transition occurs with compounds

containing single or double bonds containing heteroatom like O, N and S atoms.

This transition contain lower energy than σ →σ* Transition.

C X C X

e.g. Alcoholes, Aldehyde, Ketones, Eters and Amines etc.

4. n → π * Transition:

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It represents the Transition of one electron of a lone pair to an anti-

bonding orbital. Thus transition require lower energy than compared to π→ π *

transition and hence absorbance occur at longer wavelength. Compound

containing hetroatom gives this type of transition such as CN, CO, CS, NO etc.

C O C O

The decreasing order of energy required for these transitions can be represented as

σ →σ* > n →σ* > π → π * > n → π *

Transition probability:

It is not necessary that when a compound absorbs UV light, there will be

promotion of an electron from bonding or lone pair to an ant-bonding or non-bonding

orbital. The probability of a particular electronic transition is found to depend on the

value of molar extinction coefficient.

The extinction coefficient

ξmax = 0.87 x 1020

p.a.

Where

p = Transition probability with values from 0 to 1

a = Target area of the absorbing system

It is observed that the value of ξmax is found to be around 105 when the

chromophore has a wavelength of the order 10A0 or 10

-7cm. The chromophore with low

transition probability will be having ξmax < 1000. Hence there is a direct relationship

between the area of the chromophore and the absorption intensity ξmax. In addition there

are some other important factors which give their impact on transition probability.

Depending upon the symmetry and the value of ξmax, the transition may be classified as

allowed and forbidden.

i. Allowed transition: The transition with the values of Extinction coefficient (ξ

max) more than 104 are usually called allowed transitions.

e.g.

π → π * transitions in 1, 3 butadiene (217 mm ξ max 22,000).

ii. Forbidden transitions: The forbidden transition is a result of the excitation of

one electron from the loan pair present on the hetero atom to an ant-bonding π *

orbital. The values of ξmax for forbidden transitions are generally below 104.

e.g.

n → π *transition in carbonyl compounds (ξmax. 10-100)

In order to decide whether the transition is allowed or forbidden for symmetrical and

totally unsymmetrical molecules, it is important to consider following factors:

a. Geometry of the molecular orbital in the ground state

b. Geometry of the molecular orbital in the excited state

c. Orientation of the electric dipole of the incident light that might induce transition

Terms used in UV spectra

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Following four terms are used

1. Chromophore:-

Chromophore was considered as any system which is responsible for imparting

colour to the compound. Nitro group is the chromophore which imparts yellow colour.

The term chromophore may be defined as any isolated covalently bonded group that

shows a characteristic absorption in the UV or Visible region.

The origin of this word comes from the Greek that is Chroma means colour and

Phoros means bearing. So chromophore means colour bearing unit of the molecule e.g.

Ethylenic, Acetylenic, Carbonyls, Acids, Esters group etc. Carbonyls, group is an

important chromosphere although the absorption of light by an isolated group does not

produce any colour in UV spectroscopy. There are two types of chromophore 1) The

chromophore which contain π electrons and they under go n → π * e.g. Ethylene,

Acetylene 2) The chromophore which contain both n and π electrons such chromophore

undergo two types of transition n → π * and π → π * e.g, Carbonyls, Nitriles, Azo

compound, Nitro compounds etc.

2. Auxochrome:-

An auxochrome can be defined as any group which does not itself acts as a

chromophore but whose presence brings about a shift of the absorption band towards the

red end of the spectrum. Asoption at longer wavelength is due to combination of

chromophore an as auxochrome gives rise to another chromophore. An auxochromic

group is called as colour enhancing group e.g. -OH, -OR, -NH2, -NHR, -NR2, -SH etc.

Aniline λ max = 280nm

(ξmax = 1430)

The effect of auxochromic group is due to its ability to extend the conjugation of

a chromophore by the sharing of non bonding electrons. Thus a new chromophore results

which has different value of absorption maximum as well as extinction coefficient.

3. Bathochromic Shift (Red Shift) Absorption due to auxochrome or by the change of solvent towards longer

wavelength is known as Bathochromic shift or red shift. The n → π * transition for

carbonyl compound gives bathochromic shift when the polarity of solvent is decrease.

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N

OO

OH

N

OO

O

I II

Alkaline OH-

H+

λ max = 255nm λ max = 265nm

(ξmax = 900) (ξmax = 150000)

P-Nitrophenol

The negatively charge present on oxygen atom delocalized more due to red shift.

4. Hypsochromic shift (Blue shift)

It is an effect in which the absorption is shifted towards shorter wavelength. It

may be caused by the removal of conjugation and also by changing the polarity of the

solvent. e.g. Aniline shows blue shift in acidic medium and an unshared pair on nitrogen

of aniline is not available for delocalization in cation.

NH2 NH3Cl

HCl

λ max = 230nm λ max = 262nm

5. Hyperchromic shift

Absorption of electromagnetic radiations having greater intensity is called

hyperchromic shift .

e.g.. Pyridine and 2-methyl pyridine

N N

HCl

CH3 λ max = 257nm λ max = 262nm

(ξmax = 2750) (ξmax = 3560)

The introduction of an auxochrome usually increase intensity of absorption.

6. Hypochromic shift

Absorption of electromagnetic radiation having lesser intensity is called

hypochromic shift. The presence of extra group which destroy the geometry of the

molecule cause the hypochromic effect.

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CH3

λ max = 250nm λ max = 237nm

(ξmax = 19000) (ξmax = 10250)

Fig. 3.7: Absorption and intensity shifts

Effect of solvent on UV spectra: The polarity of solvent also affects the various types of band in compounds

a. K* Bands:- This type of bands occurred due to π → π * transition takes

place compounds containing conjugated system of bonds such as dienes,

polynes and enones etc. The intensity of this bands more than 104.

Butadiene π → π * transition 226nm (λmax)

Acetophenone π → π * transition 224nm (λmax)

b. R*Bands:-

This bands also called as fprbiden bands.This type of bands

occurred due to n→ π * transition takes place compounds containing

single chromophoric groups and at least one lone pair of electron the

heteroatom such as N, O and S. These are less intensity with ξmax values

below of these bands more than 104.

Aldehyde n→ π * transition 293nm (λmax)

Ketone n→ π * transition 270nm (λmax)

c. B-Bands:-

This type of band is formed due to π → π * transition in

aromatic compound chromophoric group or hetero - aromatic compound.

The observed wavelength and intense less than K - band.

Acetophenone π → π * transition 278nm (λmax)

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Benzaldehyde π → π * transition 280nm (λmax)

d. E-Bands:-

This band is formed due to electronic transition in the benzenoid

system of three ethylenic bonds which are related to cylic conjugation.

This band given as two types E1-bands & E2-bands.

Compounds E1-bands (λmax) E2-bands (λmax)

Benzene 184nm 204nm

Napthalene 221nm 286nm

Anthracene 256nm 375nm

Quinoline 288nm 270nm.

UV Spectra of Aromatic Compound:- As the conjugation in the compound increases, the λmax and ξ max value also

get shifted to higher value i.e. Bathochromic and Hyperchromic effect are observed.

While studying the series of aromatic compounds like Benzene, Naphthalene, Anthacene,

Phenanthrene, etc the increase in aromaticity or conjugation increases the Bathochromic

and Hyperchromic effect. Benzene absorbs at 184nm; 60,000 ξ max, 204nm; ξ max

7400 with allowed transition and shows B-band at 254 nm with ξ max 204 i.e. Forbidden

transition. Benzene also shows intensity band between 230 and 270nm.The B-band at

254nm given by benzene is fineness structure in hexane solvents while in alcohol it is

completely destroyed. It is noted that absorption maxima for Poly-nuclear aromatic

hydrocarbon moves to longer wavelength. While comparing the UV spectra of benzene

with other aromatic compounds shows increase in the value of λmax and ξmax.

Absorption of Aromatic Compound:

S.N.

Compound

Name

λmax)nm)

ξmax Transition.

1

Benzene 184 60,000

Allowed

204

7400

Allowed

254

204

Forbidden

2

Naphthalene

480

11,000

3

Pentacene

580

12600

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Woodward - Fischer Rules:-

In 1945 Robert Burns woodward gave certain rules for correlating (λmax) with

molecular structure. In 1959 Louis Frederic Fieser modified these rules and the modified

rule is known as Woodward-Feiser rules. It is used to calculate the position and λmax) for

a given structure by relating the position and degree of substitution of chromophore. Each

type of diene or triene system is having a certain fixed value at which absorption takes

place, this constitute the base value or parent value. This rule state that it is used for

finding absorption maxima (λmax) in an ultraviolet–visible spectrum of a given

compound. Longer the conjugated system grater is the wavelength of absorption

maximum. The intensity of absorption also increases with the increase in the lenth of the

chromophore. This is used in the calculation are the type of chromophores present, the

substituents on the chromophores and shifts due to the solvent. Examples are conjugated

carbonyl compounds, conjugated dienes and polyenes.

Double bond conjugation diene correlation described as fallows

a. Alicyclic or Open chain diene:-In chemistry, an open-chain compound

or acyclic compound is a compound with a linear structure, rather than a

cyclic one. An open-chain compound having no side chains is called a

straight-chain compound. HC

CH2

H2C

CH

b. Homo-anular diene:-Cyclic diene having conjugated double bond in the

same ring.

c. Hetro-annular diene:- Cyclic diene having conjugated double bond in

the different ring.

d. Endo-cyclic double bonds:- Double bond present in the ring

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e. Exo-cyclic double bonds:-Double bond in which one of the doubly

bonded atoms is a part of ring system. Here first ring has one exocyclic

and endocyclic double bond. Second ring has only one endocylic double

bond.

Parent value and increments for different groups and substituents:

Parent Value a) Butadiene or cyclic conjugated butadiene 217 nm

b) Acyclic triene 245 nm

c) Homoannular conjugated diene 253 nm

d) Heteroannular conjugated diene 215nm

Increments for substitution a) Alkyl substituents 5 nm

b) Ring residue 5 nm

c) Exocyclic double bond 5 nm

d) Double bond extending conjugation 30nm

e) Bicyclic or strain correction 15 nm

Auxochrome -Cl,-Br +5nm

-OR +6nm

-SR +30 nm

-NR2 +60 nm

-OCOCH3 0 nm

If the cyclic diene or open chain conjugated diene is substituted by -Cl or -Br

then 17nm is added in basic value.

The difference between calculated value and observed value of (λmax) should be

lower than 5 nm.

Woodward - Fischer Rules for calculating (λmax) in α, β carbonyl compounds

Woodward Fischer suggested some empirical rule for the calculation of λmax) in α, β

unsaturated compound, which was modified by Scott, which are as follows

In α, β unsaturated ketone is taken as 215mμ.

In a cyclic ketone, if α, β unsaturated carbonyl group is a part of six member

cyclic ring then basic value is taken as a 215 nm, but if α, β unsaturated carbonyl

group is a part of five member ring then base value is taken as 202 nm. The (λmax)

for such compounds are generally 104

Structural increments for calculation of (λmax) for given α, β unsaturated carbonyl

compound

1) Exo-cyclic double bond +5 nm

2) For each double bond extending conjugation +30 nm

3) For a homo-anular conjugated diene +39 nm

Page 16 of 38

4) For each double bond endocyclic. +5nm

5) Increment of various auxochrome:

Chromophore

α β γ δ

Ring residue +10 +12 +18 +18

-R +10 +12 +18 +18

-OR +35 +30 +17 +31

-OH +35 +30 - +50

-OCH3 +6 +6 +6 +6

-Cl +15 +12 - -

-Br +25 +35 - -

-SR - +85 - -

-NR2 - +95 - -

Application of Woodward - Fischer Rules to calculate (λmax) of the following

compounds

1. Calculate (λmax) for the given structure

H3C

C

H3C

CH C

O

CH3

Ans:

The above given compound is α, β-unsaturated ketone

Basic value = 215 mµ

2 alkyl β-substituent (2x12) = 24 mµ

---------

Calculated value = 239 mµ

2. Calculate (λmax) for the given structure

H2C C C

O

CH3

CH3

Ans:

The above given compound is α, β-unsaturated ketone

Basic value = 215 mµ

1 alkyl substituent = 10 mµ

------------

Page 17 of 38

Calculated value = 225 mµ

3. Calculate (λmax) for the given structure

CH3

CH3

H3C

Ans:

The above given compound is homoannular conjugated diene

Basic value = 253mµ

3 Ring residues (3x5) = 15mµ

1 exo-cyclic double bond = 05mµ

----------

Calculated value = 273 mµ

4. Calculate (λmax) for the given compound

CH3

CH3

CH3

Ans:

The above given compound is heteroannular conjugated diene

Basic value = 215mµ

3 Ring residues (3x5) = 15mµ

1 exo-cyclic double bond = 05mµ

----------

Calculated value 235 mµ

5. Calculate (λmax) for the given compound

Ans:

The above given compound is heteroannular conjugated diene

Basic value = 215mµ

4 Ring residues (4x5) = 20mµ

----------

Calculated value = 235 mµ

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6. Calculate (λmax) for the given compound

CH3

Ans:

The above given compound is heteroannular conjugated diene

Basic value = 215mµ

3 Ring residues (3x5) = 15mµ

1 exo-cyclic double bond = 05mµ

----------

Calculated value = 235mµ

7. Calculate (λmax) for the given compound

O

H3C

CH3

Ans:

The above given compound is homoannular α, β-unsaturated ketone

Basic value = 215mµ

1 α-ring residues = 10mµ

1 δ-ring residues = 18mµ

1 exo-cyclic double bond = 05mµ

1 double bond in conjugation = 30mµ

1 homoanular conjugated diene = 39mµ

----------

Calculated value = 317mµ

8. Calculate (λmax) for the given compound

O

Ans:

The above given compound is α, β-unsaturated cyclic ketone

Basic value = 215mµ

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1 α-ring residues = 10mµ

2 β-ring residues(2x12) = 24mµ

double bond exocyclic to two ring (2x5) = 10mµ

----------

Calculated value = 259mµ

9. Calculate (λmax) for the given compound

CH3

H3C CH3

Ans:

The above given compound is homoannular diene

Basic value = 253mµ

4 alkyl residues (4x5) = 20mµ

----------

Calculated value = 273mµ

10. Calculate (λmax) for the given compound

CH3

CH3 Ans:

The above given compound is homoannular diene

Basic value = 253mµ

3 alkyl residues (3x5) = 15mµ

----------

Calculated value = 268mµ

Page 20 of 38

B. INFRA-RED SPECTROSC0PY

Infrared spectroscopy is one of the important spectroscopic techniques for the

detection of functional groups in pure unknown organic compounds. The infrared

electromagnetic spectrum can be divided into following type

Near IR region 0.8 μm - 2.5 μm

IR region 2.5 μm - 15 μm

Far IR region 15 μm - 200 μrn

The absorption of IR radiation are expressed in terms of wavelength (λ).

Different functional groups absorb characteristic frequencies of IR radiation. Thus, IR

spectroscopy is an important and popular tool for structural elucidation and compound

identification. In this electromagnetic spectrum, that is light with a longer wavelength and

lower frequency than visible light. As with all spectroscopic techniques, it can be used to

identify and study chemicals. A common laboratory instrument that uses this technique is

a Fourier transform infrared (FTIR) Spectrometer. The higher-energy near-IR,

approximately 14000–4000 cm−1

(0.8–2.5 μm wavelength) can excite overtone or

harmonic vibrations. The mid-infrared, approximately 4000–400 cm−1

(2.5–25 μm) may

be used to study the fundamental vibrations and associated rotational-vibrational

structure. The far-infrared, approximately 400–10 cm−1

(25–1000 μm), lying adjacent to

the microwave region, has low energy and may be used for rotational spectroscopy. The

names and classifications of these sub regions are conventions, and are only loosely

based on the relative molecular or electromagnetic properties

IR spectrum gives detailed information about the functional groups present in

the given organic compound and spectrum obtained within 3 to 7 minutes and requires 4

to 5 mg of the sample which can be recovered.

Principle of IR spectroscopy The absorption of IR radiations causes an excitation of molecule from lower

vibrational level to higher vibrational level. But as the vibrational level is associated with

a number of closely spaced rotational levels, the IR spectra is considered as vibrational -

rotational spectra. The necessary condition for the molecule to absorb in IR region is to

change in dipole moment. In general, vibrational transitions which will lead to a change

in the dipole moment of the molecule are called IR active transitions. Otherwise they are

said to be IR inactive and will show no absorption. For ex. Vibrational transition in H2,

N2, Cl2 do not results in a change in dipole moment of a molecule. Hence these molecules

are not absorbed in IR and are called as IR inactive. But Vibrational transition in I-CI,

CO, CHCl3 results in a change in dipole moment of a molecule. Hence these molecules

are absorbed in IR and are called as IR active. Electromagnetic radiation of IR region is

absorbed to various extents by all substances. The absorption process involves excitation

of the molecule to higher vibrational states and therefore is quantized. Even at 0° K the

atomic nuclei are vibrating about the bond which binds them together.

Presentation of the IR spectra The IR radiation does not have sufficient energy to cause the excitation of

electrons. However it causes atoms and groups of atoms of organic compounds to vibrate

faster about the covalent bonds which connect them. The vibrations are quantized and as

Page 21 of 38

they occur, the compound absorbs IR energy in particular regions of the spectrum. The

position of an absorption band can be specified by following units,

1. Frequency (γ) - nm or cm-1

or Hz.

2. Wavelength (λ) -µm (1 μm = 10-4

cm).

3. Wave numbe(υ ) -cm-1

v = 1/λ, (with λ in cm)

v = 10,000/λ (with λ in μm)

An IR spectrum is the graph of, % transmittance vs increasing wavelength or decreasing

frequency. Each spectrum called a band or peak represents absorption of IR radiation at

that frequency by the sample. A 100% transmittance means 0% absorption and if all the

radiation is absorbed, the transmittance is 0%.

Fig.3.13. Regions in the IR spectrum

Radiation source: The various popular sources of IR radiations are:

1. Incandescent lamp: In the Near IR instruments an ordinary incandescent lamp is

generally used. However, this fails in the Far IR because it is glass enclosed and has a

low spectral emissivity.

2. Nernst Glower: It consists of a hollow rod which is about 2 mm in diameter and 30

mm in length. The glower is composed of rare earth oxides such as Zirconium, Yttrium

and Thorium. Nernst Glower is non-conducting at room temperature and must be heated

by external means to bring it to a conducting state. Glower is generally heated to a

temperature between 1000-1800 °C. It provides maximum radiation at about 7100 cm-1

(1.4 μm).

The main disadvantage of Nernst Glower is that it emits IR radiation over a wide

wavelength range; the intensity of radiation remains steady and constant over long

periods of time. One main disadvantage of Nernst Glower is its frequent mechanical

failure. Another disadvantage is that its energy is also concentrated in the visible and

Near IR regions of the spectrum.

90

3. Globar source: It is a rod of centered silicon carbide which is about 50 mm in length

and 4 mm in diameter. When it is heated to temperature between 1300 and 1700 °C, it

strongly emits radiation in the IR region. It emits maximum radiation at 5200

cm"1.Unlike the Nernst Glower, it is self-starting. As its temperature coefficient is

positive, it can be conveniently controlled with a variable transformer. It also works at

wavelengths longer than 650 cm-1

.The main disadvantage is that it is a less intense source

than the Nernst Glower.

4. Mercury Arc: In the Far IR instruments, high pressure Mercury Arc is generally

employed. Beckmann devised the quartz mercury lamps for the same region in a unique

Page 22 of 38

manner. At the shorter wavelengths, the heated quartz envelope emits the radiation

whereas at the longer wavelengths the mercury plasma provides radiation through the

quartz.

Types of vibrational modes The atoms in a molecule are not rigid but elastic. Covalent bond between atoms

show like a spring. When Infrared radiation is passed through organic compounds

vibrational and rotational energies of the molecule get increased. A nonlinear molecule

undergoes two kinds of fundamental vibration.

a) Stretching vibrations: The change in the internuclear distance between two atoms

without changing bond axis is called stretching vibration. In this distance between two

atoms increases or decreases. These vibration are further divided into two types

i) Symmetrical stretching: In this type the movement of atoms is in the same

direction with respect to particular reference atom.

ii) Asymmetric stretching: In this type of vibration, the movement of a atom

is towards to the centre while that of another atom is opposite to centre

with respect to particular reference atom.

b) Bending vibrations: The change in angle between two covalent bonds, due to

change in the position of atoms with respect to the original bond axis is called as bending

vibrations. There are four types of bending vibrations,

a) In plane bending vibration

i) Scissoring:

In this type, two atoms approach each other with respect to central atom.

Page 23 of 38

ii) Rocking:

In this type, the movement of atoms takes place in the same direction

with respect to central atom,

b) Out of plane bending

i) Wagging:

In this type, two atoms move "up and down" the plane with respect to

central atom,

(+)

(-)

ii) Twisting:

In this type, one of the atom moves up the plane while the other moves

down the plane with respect to central atom. (+)

(+)

Bending vibration is more as compared to stretching vibration and requires less

energy. Hence bending vibration occurs at higher wavelengths than stretching.

Page 24 of 38

Hooks Law:-

This law first discovered by Robert Hooke the 17th century physicist who

discovered it in 1660 Hooke’s Law. This law gives the relationship between the forces

applied to a spring and its elasticity. The value of stretching vibrational frequency of a

bond can be calculated by the application of Hooks law which is expressed as fallows

m1 m2

k

1/2π

µ = m1m2

m1+m2

Where

k -Force constant

γ -Frequency

μ -Reduced mass

m1 -mass of first atom

m2 -mass of second atom

Absorption peaks caused by stretching vibration are usually the most intense

peaks in the IR spectrum. If the bond strength increase and the masses of the atom

decrease, the value of vibrational frequency increases. On the basis of Hooks law. We

can predict that O-H bond requires more energy for stretching vibration than C-C bond.

Hence in IR spectrum O-H stretching bond will appear in higher wave number region

9high energy region 3600cm-1

) and C-C stretching bond will appear in lower number

region(low energy region 1200cm-1

).

Calculation of reduced mass

1. For O-H stretching vibration

µ = m1m2

m1+m2

µ = 16x1

16+1

µ = 16

17

µ = 0.9411

2. For C-C stretching vibration

µ = m1m2

m1+m2

Page 25 of 38

µ = 12x12

16+12

µ = 144

24

µ = 6

Fundamental modes of vibration The fundamental modes of vibrations are determined by stretching and bending

modes. At certain quantized frequencies stretching and bending vibrations of a bond can

occur. Radiations are absorbed at a particular vibaration only if it causes change in the

dipole moment of the molecule. The vibrational frequency of a bond increases with

increases in bond strength. There is very important requirement for a molecule to show

and infra read spectrum its states that dipole moment of the molecule must change during

the vibration. Homonuclear diatomic molecules such as H2O2, N2 etc are IR inactive

where as hetronuclear diatomic molecule such as CO, NO, CN, HCl are IR active

The number of fundamental modes of vibration for a molecule can be calculated

from the total number of atoms present in the molecule. The molecule containing 'n'

number of atoms will have '3n' degrees of freedom. The fundamental modes of vibration

are depends on the geometry of molecule and can be calculated as follows:

1) For linear molecule:

There are only two degrees of rotation. It is due to the fact that the

rotation of such a molecule about its axis of linearity does not bring about any

change in the position of the atoms while rotation about the other two axes

changes the position of the atoms. Thus, for a linear molecule containing n

atoms,

Total degrees of freedom =3n

Translational degrees of freedom =3

Rotational degrees of freedom =2

Hence, Vibrational degrees of freedom = 3n-(3+2)

= 3n-5

Hence for linear molecule, containing n atoms, there are (3n-5)

possible fundamental modes of vibrations.

e.g.

Nitrous oxide (NO)

Total atoms (n) =2

Applicable formula (3n-5)

Fundamental modes of vibration = 3x2-5

= 6-5

= 1

2) For non-linear molecule:

Page 26 of 38

There are three degrees of rotation as the rotation about all the three axes

(X, Y, and Z) will result in a change in the position of the atoms. Thus, for a non-

linear molecule containing n atoms,

Total degrees of freedom = 3n

Translational degrees of freedom = 3

Rotational degrees of freedom = 3

Hence, Vibrational degrees of freedom = 3n-(3+3)

= 3n-6

Hence for linear molecule, containing n atoms, there are 3n-6 possible

fundamental modes of vibrations.

e.g.

Methane Molecule (CH4)

Total atoms (n) =5

Applicable formula (3n-6)

Fundamental modes of vibration = 3x5-6

= 15-6

= 9

Following table no. 1.

Shows fundamental modes of vibrations for different molecules.

Sr.

No.

Molecule

Total no. of

atoms (n)

Geometry of

the molecule

Applicable

formula

Fundamental

modes of

vibration

1

NO

2

Linear

3n-5

1

2

CO2

3

Linear

3n-5

4

3 C2H2 4 Linear 3n-5 7

3

H2O

3

Non-linear

3n-6

3

4

NH3

4

Non-linear

3n-6

6

5

CH4

5

Non-linear

3n-6

9

6

C6H6

12

Non-linear

3n-6

30

Spectral range IR region 2.5 to 15 um (4000cm"

1 to 667 cm"

1) is subdivided into three sub regions,

a. Functional group region 2.5 to 7.7 μm (4000-1300 cm-1

)

b. Finger print region 7.7 to 11 μrn (1300 to 909 cm-1

)

c. Aromatic region 11 to 15μm (909-667 cm-1

)

Page 27 of 38

a. Functional group region:

This region is very useful for organic chemists. Bands due to stretching

vibrations of functional groups such as O-H, N-H, C-H, S-H in the range 2500- 3700 cm-1

are shown in this part of IR spectrum.

The absorption band at 2100-2500 cm'1 shows the presence of CN, CH=CH bond

and is known as triple bond region. The IR absorption band at 1500-1600 cm-1

indicates

the presence of aromatic C=C stretching frequency, similarly at 1620 - 1680 cm-1 is due

to alkenes, C=C stretching frequency. The important IR absorption band at 1700 cm-1,

clearly indicates the presence of carbonyl group (C=O) may be aldehydes or ketone. Thus

functional groups in the molecule can be identified from absorption bands in IR.

a. Finger print region:

The region between 7.7-11 μrn (1300 to 909 cm-1) is known as finger print

region because the pattern of absorptions in this region are unique to any particular

compound, just as a person's fingerprints are unique. This is the most complex part of IR

spectrum and contains number of absorption bands. These absorption bands appear due to

stretching and bending vibrations. This region is not of much useful because of its

complexity but helpful for a sample comparison. Some molecules containing the same

functional group show similar absorptions above 1500 cm-1

but their spectra differ in

finger print region If the finger print region of two samples are identical, then these

samples are also identical.

b. Aromatic region:

The region between 11-15μm (909-667 cm-1

) is known as aromatic region. This

part helps to detects aromaticity of the compound. The lack of absorption in this region

shows presence of non-aromatic compound. The bands occur in this region is due to

bending vibrations. This region also useful in determination of substitution patterns on

aromatic compounds such as ortho, meta, para substitution. Though most of the covalent

bonds have characteristic and invariable wavelength of absorption, but sometime due to

environmental conditions like conjugation, hydrogen bonding, electron donating or

withdrawing group, IR spectra are slightly shifted. The exact position of absorption

depends on force constant, masses of the atoms, environment of the bond. The table

shows characteristic IR absorption frequencies of some functional groups

Page 28 of 38

Table No. 2. Infrared absorption of some functional groups

Frequency range (cm-1

)

Bond

Type of Compound

[A] Bonds to Hydrogen atom:

3600-3650

-O-H (free)

Alcohols, Phenols

3200-3500

-O-H(bonded)

Hydrogen bonded alcohol

& phenol

3450-3600

sharp

Inter molecular 'H' bond

3200-3550

broad

Intra molecular 'H' bond

2500-3200

Very broad

Cheleted 'H' bond

3300-3350

-N-H

Amines, Amides

3200-3310

=C-H

Acetylenic

3000-3100

=C-H

Ethylenic & aromatic

2850-2950

-C-H

Saturated alkanes

2700-2900

-CHO

Aldehydes

2550-2900

-S-H

Thiols

2500-3600

-COOH

Carboxylic acids (very broad)

[B] Triple Bond Region:

2240-2260

-ON

Nitriles (Non conjugated)

2215-2240

-C=N

Nitriles (Conjugated)

2100-2140

-OC-H

Terminal alkynes

2200-2260

-OC-H

Non terminal alkynes

[C] Double Bond Region:

1740-1850

R O R

O O

Anhydrides

Page 29 of 38

1770-1815

R X

O

Acid halides

1730-1750

R O

O

R

Esters

1730-1740

R H

O

Aldehydes

1705-1720

R R

O

Ketones

1700-1725

R OH

O

Acids (Carboxylic)

1680-1700

R NH2

O

Amides

1620-1680

HC CH

Alkenes

1590-1690

-C=N-

Imines, Oximes

1500-1600

C C

Aromatic

1370-1540

-NO2

Nitro group

[D] Single Bond Region

1000-1300

C O

Alcohol, Phenol, Ester, Ether etc.

1100-1120

C C

Saturated alkanes

500-750

C X

Halo-alkanes

[E] Aromatic Region (=C-H Bending)

Page 30 of 38

690-710

730-770

CH

Mono-substituted benzene

735-770

CH(O)

Ortho-substituted benzene

750-810 CH(m)

Meta-substituted benzene

790-850

CH(p)

Para-substituted benzene

Structure of Organic Compounds

1. Structure of H2O molecule Total atoms (n) =3

Applicable formula (3n-6)

Fundamental modes of vibration = 3x3-6

= 9-6

= 3

There for water is a triatomic nonlinear molecule. In this molecule two stretching

frequency such as 3650cm-1

, 3760cm-1

and one bending vibration appears at 1600cm-1

.

All the three vibrations gives change in dipole moment and hence water molecule are IR

inactive

2. Structure of CO2 molecule

Total atoms (n) =3

Applicable formula (3n-5)

Fundamental modes of vibration = 3x3-5

= 9-5

= 4

The symmetrical stretching mode produces no change in the dipole moment of

the molecule and therefore it is IR inactive.

In the asymmetrical stretching mode, the two C=O bonds stretch out of phase;

one C=O bond stretches as the other contracts. The asymmetrical stretching mode, since

it produces a change in the dipole moment, is IR active; the absorption (2350 cm"1) is at a

higher frequency (shorter wavelength) than observed for a carbonyl group in aliphatic

ketones. This difference in carbonyl absorption frequencies displayed by the carbon

Page 31 of 38

dioxide molecule results from strong mechanical coupling or interaction. In contrast, two

ketonic carbonyl groups separated by one or more carbon atoms show normal carbonyl

absorption near 1715 cm"1 because appreciable coupling is prevented by the intervening

carbon atom(s). The bending vibrations in 3 and 4 above are equivalent and there

frequencies must be identical because turning the plane of vibrations cannot alter the

frequencies.

3. Structure of Acetone molecule

Acetone is a saturated aliphatic ketones i.e. acetone shows a strong C=O

(carbonyl) stretching absorption band at 1715cm-1

.

It's relatively constant position, high intensity and relatively freedom from

interfering bands makes this one of the easiest band to recognize in IR spectra. Changes

in the environment of the carbonyl can either lower or raise the absorption frequency

from this normal value. The absorption frequency observed for a neat sample is increased

when absorption is observed in nonpolar solvents. Polar solvents reduce the frequency of

absorption. The overall range of solvent effects does not exceed 25 cm-1

. The band at

3000 cm-1

suggests a methyl stretching vibration whereas the bands at 1370 and 1430 cm-

1 can be assigned to symmetrical and asymmetrical methyl bending vibrations. The

presence of the combination band for C-C-C near 1218 cm-1

further confirmed the

presence of carbonyl group in acetone.

4. Structure of Alkane(Ethane) molecule

Ethane gives two absorption strong absorption bands in high frequency region.

One band at 2940cm-1

due to-C-H asymmetric stretching and another band at 2880cm-1

due to symmetric stretching vibration. Asymmetric C-H bending vibration appears at

1465cm-1

while symmetric C-H bending vibration appears at 1380cm-1

5. Structure of 1-propanol molecule

Page 32 of 38

IR spectra of 1-propanol give two characteristic O-H and C-O stretching

bands. These vibrations are sensitive to hydrogen bonding due to O-H group of

alcohol gives a sharp band at 3610cm-1

. A strong broad band appears in3600cm-1

region due to O-H stretching involved in hydrogen bonding. Primary alcohol show

characteristic strong broad band at 1050cm-1

.

6. Structure of Benzene molecule

IR spectrum of Benzene molecule gives C-H stretching band accurs in

3100cm-1

, 3000cm-1

region. Absorption bands due to C-C stretching accurs at

1600cm-1

, 1580cm-1

, 1500cm-1

and 1450cm-1

. In plane C-H bending absorption

bands appear in 1300cm-1

-1000cm-1

region and out of plane C-H bending

absorption bands appear in 900cm-1

-675cm-1

region.

7. Structure of Benzmide molecule

Page 33 of 38

IR spectrum of benzmide molecule gives

N-H stretching - 3370cm-1

due to asymmetric stretching

N-H stretching -3170cm-1 due to symmetric stretching

C=O -1660cm-1 doe to conjugation with one

pair of electron on nitrogen atom

N-H -1631cm-1 bending in amide group

G. PURIFICATION OF ORGANIC COMPOUNDS

1) Sublimation Purification of solid organic compounds is done by the sublimation. Its used for

those compounds which on heatng pass readily and directly from solid to vapour state,

with a sbsquint ready reversal of this process when the vapour is cooled in a small

evaporating basin the dry crude material is placed which is then kept on a wire gauze. It

is then covered with a filter paper which is pierced by a number of small holes made in

an upward direction. A glass funnel is then inverted over the paper as shown in the figure.

The basin is then gently heated by a small Bunsen flame, which should be carefully

protected from side draughts by screens, so that the material in basin receives a steady

uniform supply of heat. The material vaporizes and the vapor passes up through the holes

into the cold funnel. Here it cools and condenses as fine crystals on the upper surface of

the paper and on the walls of glass funnel. Heating is stopped when almost the whole of

the material in evaporating basin has vaporized, and then the pure sublimed material

collected. In using such an apparatus, it is clearly necessary to adjust the supply of heat so

that the crude material in basin is being steadily vaporized, while the funnel does not

became more than luke-warm.-Cotton

Page 34 of 38

Fig. 3.20 Sublimation

2) Crystallization or Recrystalllization:- The important and delicate method in which the purification of solid organic

compounds takes place is the crystallization process. It consists of dissolving the

substance in minimum amount of a solvent in which the substance is more soluble in hot

than in cold. By filter in with hot solution the impurities are removed then the filtrate is

allowed to cool when crystals of pure substance are obtained. These are filtered ad if

necessary, again dissolved for crystallization. It should be noted that on cooling of the

filtrate yields tiny but purer crystals whereas on slow cooling yields bigger but impure

crystals. Mother liquor is the liquid left after separation of crystallized product. It is

concentrated by evaporation and cooled to obtain a fresh crop of crystals of pure

substance which are further subjected to crystallization. Maximum amount of pure

substance is recovered by repeating the process several times. The following suggestions

are helpful in increasing the efficiency of crystallization.

Fig. 3.20 Crystallization

a) Selection of suitable solvent: The suitable solvent is selected which is of great

importance in purification of substance by crystallization. The solvent is selected on the

'hit and trial' basis and it is necessary to try a number of solvents before arriving at a

conclusion. If the solvent dissolves the substance in cold or large excess of hot solvent is

require to dissolve the substance it is obviously unsuitable for crystallization. Only that

solvent is to be selected which does not dissolve or dissolves a very small amount of

substance in cold but dissolves its appreciable amount on boiling or heating. If none of

the solvent appears good enough for crystallization, a combination of two or more

Page 35 of 38

solvents may be employed. Solubility of nonpolar substances is generally in nonpolar

solvents, where as solubility of polar substances is in polar solvents. Some of the

common solvents in order of increasing polarity are: petroleum ether, carbon

tetrachloride, benzene, chloroform, diethyl ether, acetone, ethyl acetate, ethyl alcohol and

water.

b) Drying of crystallized substance: the safe and reliable method for drying, through

much time consuming, is the use of desiccators with vacuum arrangement known as

vacuum desiccators. On a watch glass the crystallized substance is spread over and kept

in the vacuum desiccators and vacuum is applied. The substance is allowed to stay under

these conditions for several hours and then the vacuum is released very slowly when the

dry crystallized substance is obtained.

c) Decolourisation of undesirable colours: The impurity of colouring matters is

removed by boiling the substance in solution with a sufficient quantity of finely

powdered animal charcoal for about 15 to 30 minutes and then filtering the hot solution.

The animal charcoal adsorbed the coloured impurities and the pure decolorized substance

crystallizes out from the filtrate on cooling. As the animal charcoal is insoluble in all the

common solvents it can be employed in the solvent used for crystallization.

3) Paper Chromatography

It was the invention of Cambridge school of workers, A.J.P. Martin and his

coworkers. Mixture of organic substances such as dyes and amino acids was separated by

the use of this technique. Now a day cations and anions of inorganic substances can also

be seprated by the use of this technique.

Principal In this type of chromatography the substances is distributed between two phases.

i. Stationary phase

ii. Moving phase

On the paper components of the mixture are separated and they migrate at

different rates and appear as spots at different points.

Working In this technique, a drop of test solution is applied as a small spot on a filter paper

and the spot is dried. In a closed chamber paper is kept and the edge of the filter paper is

dipped into a solvent called developing solvent. As soon as the filter paper gets the liquid

through its capillary axis and when it reaches the spot of the test solution (a mixture of

two or more substances), the various substances are moved by solvent system at various

speeds. When the solvent has moved these cations to a suitable height (15 - 18 cm) the

Page 36 of 38

paper is dried and the various spots are visualized by suitable reagents called visualizing

reagents. The Rf values that is migration parameters gives the movement of substances

relative to the solvent.

Migration parameters On the chromatograms the position of migrated spots is indicated by a term

called as RF. RF: The R is related to the migration of the solute front relative to the

solvent. Distance travelled by the solute from the origin line. Distance travelled by the

solvent from the origin line.

The RF value of substance depends on the following factors:

The nature of the compound

The size of the vessel in which the operation is carried out

The temperature

The solvent employed

The medium used for separation, i.e., the quality of paper.

Fig. 3.21 paper Chromatography

Questions:- 1. Discuss briefly the principal of IR spectroscopy.

2. Write note on fundamental vibration of nonlinear molecule.

3. Write a short note on finger print region.

4. Define Electronic spectroscopy. What is its absorption range?

5. Calculate the energy associated with the radiations having wave number 3 x 10

per cm.

6. Write a short note on Electromagnetic spectrum.

7. Define the following terms:

i) Bathochromic shift ii) Hypsochromic shift iii) A Chromophore

iv) Hyperchromic

8. Define the term chromophore. How will you detect the presence of carbonyl

group in aldehydes and ketones?

9. Discuss the various types of electronic transitions and explain the effect of the

polarity of the solvent on the each type of transition.

10. "A conjugated diene absorbs at a higher wavelength with with higher value of

extinction coefficient as compare to a diene in which double bonds are isolated."

Comment on this statement with examples giving the chemistry involved.

Page 37 of 38

11. What types of transitions are observed in a, p-unsaturated carbonyl compounds?

How absorption maximum and intensity are shifted when carbonyl group is not

conjugated? Discuss the effect of solvent polarity on R-band.

12. Aniline absorbs at 280 nm (ξmax 8600), however in acidic solution, the main

absorption band is seen at 203 nm. Explain.

10. The position of absorption of acetone shift in different solvent 279 nm (hexane),

272 nm (ethanol), and 264.5 nm (water).explain.

11. (a) The UV spectrum of acetone shows two peaks at

λmax = 280 nm ξ max 15and λmax = 1900 nm ξ max 100

i) Identify the electronic transition for each

ii) Which one of these is more intense?

(b) Why is methanol a good solvent for UV but not for IR determination?

12. "Increase in polarity of the solvent shift π → π *band to longer wavelength but

n→ π *and n - σ* bands to shorter wavelength". Comment on the statement.

13. a) Write notes on K band, B band and R band.

b) Which solvents are generally used, in UV spectroscopy and why.

14. Identify the types of transitions in each of the following compounds

i) Ethanol =290nm ξ max 17

ii) Acetic anhydride = 227 nm ξ max 44

iii) Pentadiene l, 3 = 224nm ξ max 22650

iv) Crotonaldehyde = 214 nm ξ max 15850

v) Styrene = 282 nm ξ max 450

15. What type of transitions will be expected to occur in each of the following?

i) CH3CHO ii) (CH3)2NCH=CH2

16. What are different types of vibrations? How fundamental modes of vibrations are

calculated for linear and non linear molecules.

17. Calculate fundamental modes of vibrations for the following molecules:

Benzene, Hydrogen peroxide, Carbon disulphide, Ammonia, Water, and

Methane.

18. What are essential requirements for a substance to absorb in the IR region?

19. How will you distinguish between the following pairs of compounds by using IR

spectroscopy?

a) Ethanol & Diethyl ether

b) m-cresol & anisole

c) Acetone & Acetic acid

20. Explain the following:

a) Acetaldehyde absorbs at 1725 cm-1

, acetone absorbs at 1715 cm-1

While benzaldehyde absorbs at 1710 cm-1

.

b) Benzaldehyde absorbs at 1710 cm-1

while salicyladehyde absorbs at

1698 cm-1.

21. An organic compound C8H8O shows absorption frequencies at 1680, 1600, 1500,

1430, 1360, 690 cm-1

. Assign the structure.

22. Explain the following

a) Sublimation

b) Crystallization

Page 38 of 38

c) Paper chromatography.