IR SPECTROSCOPY

The molecules remain in its ground electronic state during this vibrational transition. The “true IR-region” extends from 2.5 to 15(. The region from 0.8 to 2.5( is called “near infrared”and that from 15( to 200( is called “far IR-region.”

Purpose of IR spectroscopy

The technique can be employed: ----------

To establish the identity of two compounds.

To determine the structure of a new compound.

In revealing the structure of a new compound, it is quite useful to predict the presence of certain fictional groups which absorb at definite frequencies. For example: the hydroxyl group in a compound absorbs at 3600 – 3200cm-1. Carbonyl group of ketone gives a strong band at 1710cm-1.

So IR spectroscopy is a very reliable technique for disclosing the identity of a compound.

Application of IR spectroscopy

IR (Infra-red) spectrum of a compound provides the following more information: ----------

In this technique, almost all groups absorb characteristically within a definite range. The shift in the position of absorption for a particular group may change with the changes in the structure of the molecule.

Impurities in a compound can be detected from the nature of the bands which no longer remain sharp and well defined.

The position or the band not only tells the presence of a particular group but also reveals a good deal about the environments affecting the group.

Presence of conjugation with carbonyl group can be detected a sit shifts vC=O stretching to the lower wave number.

In case of hydrogen bonding, the wave number of absorption is shifted downwards for both the donor as well as the acceptor group. It can also make distinction between intermolecular hydrogen bonding and intra-molecular hydrogen bonding.

Due to the different positions of absorptions, it is also possible to know the axial and the equatorial positions of certain groups in a cyclic structure.

The force Constants responsible for the absorption peaks can be used to calculate bond distances and bond angles in some simple cases.   

Expression of IR absorption

The absorption of Infra-red (IR) radiations can express either in terms of wavelength (() or in wave number ((). Mostly infra-red spectra of organic compounds are plotted as percentage transmittance against wave number. The relationship between wavelength and wave number is as follows: ------

If wavelength (() is 2.5( = 2.5 ( 10-4cm then:

The wavelength 16( corresponds to wave number 625cm-1. Thus in terms of wave number the ordinary infrared region covers 4000 – 625cm-1.

Band intensity is either expressed in terms of absorbance (A) or Transmittance (T).

A = log10 (1/T)

Mode of vibration

When Infra-red light is passed through the sample the vibrational and the rotational energies of the molecules are increased. There are two types of vibration may occur: ------

Fundamental vibration

Non fundamental vibration

Fundamental vibration: Two kinds of fundamental vibrations are: -----

Stretching and

Bending

Stretching:

In this type of vibration, the distance between two atoms increases or decreases but the atoms remain in the same bond axis, i.e. such vibration corresponds to changes in bond lengths. Two types of stretching vibration are: ------

Symmetric stretching: In this type the movement of atoms with respect to a particular atom in a molecule (central atom) is in the same direction.

Asymmetric stretching: In this type of vibration, one atom approaches the central atom while the other departs in direction from it. Required more energy to stretch.  

Bending:

In this type of vibration, the positions of the atoms change with respect to the original bond axis. So, bending vibration corresponds to change in bond angles. Bonding vibrations are of four types: 

Scissoring: In this type two atoms approach each other. (a)

Rocking: In this type, the movement of the atoms takes place in the same direction. (b)

Wagging: Two atoms move up and below the plane with respect to the central atom. (c)

Twisting: In this type, one of the atoms moves up the plane while the other moves down the plane with respect to the central atom. (d)    

More energy is required to stretch a bond than that required to bend it. Therefore, stretching absorptions of a bond appear at higher frequencies (higher energy) as compared to bending absorptions of the same bond. All these fundamental absorptions arise from excitation from the ground state to the lowest energy excited state. But the IR spectrum is complicated due to the presence of some weak absorption bands called non fundamental vibration.  

Non-Fundamental vibration:

Overtone: Overtone corresponds to integral multiples of the frequency of the fundamental. These are relatively weak bands.

Combination band: When two vibrational frequencies in a molecule couple to give rise to a vibration of a new frequency within the molecule and when such a vibration is IR active it is called a combination band.

Difference band: The observed frequency, in this case, results from the difference between two interacting bands.   

If there are two fundamental bands at X and Y, then the additional bands can be expressed as: ------

2X, 2Y (overtone)

X+Y, X+2Y, 2X+Y etc. (combination band)

X-Y, X-2Y, 2X-Y etc (difference band)

Additional bands are usually 10 – 100 times less intense as compared to the fundamental band.

Fermi Resonance

When a fundamental vibration couples with an overtone or combination band, the couple vibration is called Fermi resonance. Fermi resonance is often observed in carbonyl compounds. Fermi resonance can be explained by saying that a molecule transfers its energy from fundamental to overtone and back again. This type of resonance gives to a pair of transitions of equal intensity. An example of Fermi resonance is given by aldehydes in which C – H stretching absorption usually appears as a doublet ((2820cm-1 and (2720cm-1) due to the interaction between C – H stretching (fundamental) and the overtone of C – H deformation (bending).

Calculation of fundamental vibrations

For linear molecule:

In case of 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 of n atoms: -------

                        Total degrees of freedom: Translational + Rotational + Vibrational = 3n

                        Translational degrees of freedom: 3

                        Rotational degrees of freedom: 2

                        Vibrational degrees of freedom: 3n – 3 – 2 = 3n – 5

Example: In the linear molecule of carbon dioxide (CO2), the number of vibrational degrees of freedom can be calculated as follows:                  (linear).

                       

                        No. of atoms (n): 3

                        Total degree of freedom = 3 ( 3 = 9

                        Translational = 3

                        Rotational = 2

                        Vibrational degrees of freedom for CO2 = 9 – 5 = 4.

  

For non linear molecule:

In case of non-linear molecule, there are three degrees of rotation as the rotation about all the three axes (X, Y, Z) will result in a change in the position of the atoms. So, for a non-linear molecule of n atoms: -------

                        Total degrees of freedom: Translational + Rotational + Vibrational = 3n

                        Translational degrees of freedom: 3

                        Rotational degrees of freedom: 3

                        Vibrational degrees of freedom: 3n – 3 – 3 = 3n – 6

Example: In the non-linear molecule of benzene (C6H6), the number of vibrational degrees of freedom can be calculated as follows:

                       

                        No. of atoms (n): 12

                        Total degree of freedom = 3 ( 12 = 36

                        Translational = 3

                        Rotational = 3

                        Vibrational degrees of freedom for C6H6 = 36 – 6 = 30

Infra-red inactive:

Molecules displaying partial or total symmetry give rise to simpler spectra than might be anticipated. This is due to the fact that for particular vibration to absorb infrared energy, a change in the dipole moment of the molecule must result. The double bond in symmetrical alkenes (C=C) and in do not absorb infrared radiation due to lack of changing in dipole moment. Such bands are said to be “Infra-red inactive”.

Finger print region:

The region below 1500cm-1 (900 – 1400cm-1) is IR region is termed as “Finger print region”.

This region is important for comparing the identity of two compounds and also for the detection of certain functional groups like ester, ether etc. this area of spectrum is usually complex since many absorption bands are located here, especially those due to bending vibrations as well as C – C, C – O, C – N stretching vibrations.  As there are more bending than stretching vibrations in a molecule the finger print region contain many absorption bands of varying intensities which makes this region of particular importance in establishing the identity of a compound by comparison with on authentic sample. Two extremely similar molecules often have virtually identical vibrations in the other two regions (600 – 900, 1400 – 4000cm-1), but almost invariably, there are differences though sometimes slight in the finger print region.       

Calculation of vibrattional frequency (Hook’s Law):

The Infra-red light is absorbed when the oscillating dipole moment of the molecule interacts with the oscillating electric vector of an infrared beam. For a hetero-nuclear diatomic molecule Hook’s Law to express the value of stretching vibrational frequency may be represented as: -------

        

The term “(” or reduced mass of a system can be given by: -------

Frequency

Where,

m1 and m2 = Masses of atoms concerned in gms in a particular bond.

K = Force constant of the bond and relates to the strength of the bond.

For single bond K= 5 ( 105 dyne/cm, For double bond, K= 10 ( 105 dyne/cm and For triple bond K= 15 ( 105 dyne/cm.

C = Velocity of the radiation. (3 ( 10-10 cm/sec)

So, the value of vibrational frequency depends upon:

Bond strength

Reduced mass.

If the bond strength increases or the reduced mass decreases, the value of vibrational frequency increases.

Factors influencing stretching vibration

Considering a particular group like C=O, the factors influencing C=O stretching can be stated as follows: 

Conjugation effects:

The introduction of a C=C bond adjacent to a carbonyl group results in delocalization of the (-electrons in the carbonyl and double bonds. This conjugation increases the single bond character of the C=O bond and hence lowers its force constant, resulting in a lowering of the frequency of carbonyl absorption.

Generally, the introduction of an ( - ( double bond in a carbonyl compound results in a shift of about 30cm-1 to lower frequency from the base value. A similar lowering occurs when an adjacent aryl group is introduced. 

But amides do not follow the rule, conjugation does not reduce the C=O frequency. The introduction of (, ( unsaturation  causes an increase in frequency from the base value. Apparently, the introduction of SP2 hybridized carbon atoms removes electron density from the carbonyl group and strengthens the bond instead of interacting by resonance as other C=O example.

Substitution effects:

When the carbon next to the carbonyl is substituted ((-substituted) with a chlorine (or other halogen) atom, the carbonyl band will shift to a higher frequency. The electron withdrawing effect removes electrons from the carbon of the C=O band. This is compensated for by a tightening of the (-band (stretching) which increases the force constant and leads to an increase in the absorption frequency. This effect holds for all carbonyl compounds. 

In ketone, two bands result from the substitution of an adjacent chlorine atom: one is due to the conformation in which the chlorine is rotated next to the carbonyl (equatorial position) and the other from the conformation in which the chlorine is away from the group (axial position). When the chlorine is next to the carbonyl, non-bonded electrons on the oxygen atom are repelled. This results in a stronger bond and higher absorption frequency.   . 

Hydrogen bonding effects:

Hydrogen bonding to a carbonyl compound lengthens the C=O bond and lowers the stretching force constant K. hydrogen bonding is very weak and small energy will be required to stretch such a bond (O - - - H). This results in a lowering of the absorption frequency. Two types of hydrogen bonds are readily distinguished in IR technique. Generally, the inter molecular hydrogen bonds give rise to broad bands where as bands arising from

intra-molecular hydrogen bonds are sharp and well defined, but both lowers the absorption frequency. Examples include:     

The decrease in the C=O frequency of carboxylic acid (intermolecular bonding)

The lowering of ester C=O frequency in methyl salicylate (intra-molecular hydrogen bonding)

An extreme example of intra-molecular hydrogen bonding is observed in enolic (-diketons, which have broad and intense absorptions at values of 1640cm-1 or lower whereas the keto tautomer absorbs at frequencies near where normal ketones absorb.    . 

Ring size effects:

Six-membered rings with carbonyl groups are unstrained and absorb at about the same values. Decreasing the ring increases the frequency of the C=O absorption. Some functional group like amide, acid chloride, anhydride, carboxylic acid, ester etc. which can form rings, give increased frequencies of absorption with increased angle strain. For ketones and esters, there is often a 30cm-1 increase in frequency for each carbon removed from the unstrained six-membered ring values.       

Questions: “Infra-red radiation causes vibrational transitions only” – rationalize the statement. 

Answer:

The true infrared region corresponds to 2.5 to 16(, or 4000 to 625cm-1. If the source of radiation is the far infrared or microwave ragion, the rotational energy of the molecule is affected, whereas when the energy supplied is on the 2.5 to 16( range, changes in the vibrational as well as rotational energies of the molecule result. For all purposes however, the 2.5 to 16( region of the electromagnetic spectrum can be considered to affect the only vibrational energy levels of the irradiated molecule. Rotational transitions are rarely observed when solids or solutions are being examined; they appear at fine structure associated with the vibrational absorption bands. For this reason, the irradiation of a molecule by means of infrared radiation causes only vibrational transitions.       

Questions: “Though Benzene (C6H6) has 30 fundamental bands theoretically, all are seldom obtained” - why?

Answer:

Theoretically Benzene has 30 fundamental bands (since, non linear molecule no. of possible bands:

3n – 6 = 3 ( 12 - 6 = 30). But it has been observed that the theoretical number of fundamental vibrations is seldom obtained because of the following reasons: -----------

Fundamental vibrations that fall outside the region under investigation i.e. 2.5 to 16(.

Fundamental vibrations that are too weak to b observed as bands.

Fundamental vibrations that is so close that they overlap i.e. degenerate vibrations.

Certain vebrational bands do not appear for want of the required change in dipole moment in a molecule.

There many appear some additional bands called combination bands, difference bands and overtones. Thus due to all of these, a large number of bands will be observed as compared to the theoretical number. 

Questions: “Amine, Alcohol both absorbs IR in the same region” – Then why they differ infrequency type?

Answer:

The N–H and O–H region overlap (O-H: 3650 – 3200cm-1 and N-H: 3500 – 3300cm-1) in the IR region. The

N-H absorption usually has two sharp absorption bands of lower intensity (or one double natured band) while

O-H when it is in the N-H region, usually gives a road absorption peak. 

This is due to the fact that the O-H group is a strong absorber because of its large charge and dipole moment., whereas the N-H bond generally absorbs very weakly.

Questions: “Ketone absorb at a lower frequency than aldehydes” – why?

Answer:

Ketone absorb at a lower frequency than aldehydes because of the additional alkyl group present on a ketone. This second alkyl group is electron donating (compared to H) and supplies electrons to the C=O bond. This electron releasing effect weakens the C=O bond in the ketone and lowers the force constant subsequently absorption frequency.  

Questions: “The C=O absorption of frequency is much less in amides as compared to that in esters” – why?

Answer:

Nitrogen atom is less electronegative than oxygen atom. So the electron pair on nitrogen atom in amide is more labile and participates more in conjugation. Due to this greater degree of conjugation, amides get comparatively single bond type nature and to some extent and the C=O absorption frequency is much less in amides as compared to that in esters.      

Questions: “Carboxylic acid in very dilute solution absorbs at about 1760cm-1 whereas in concentrated solution frequency lowers to 1710cm-1 ” – why?

Answer:

A carboxylic acid exists in a monomeric form in very dilute solution, and it absorbs at about 1760cm-1 because of the electron withdrawing effect. The electronegative element of carboxylic acid may tend to draw in the electrons between the carbon and oxygen atoms through its electron withdrawing effect, so that the C=O bond becomes somewhat stronger a high frequency (higher energy) absorption then results in a strong bond in 1760cm-1.  

But in concentrated solution, acids tend to dimerize via hydrogen bonding. Hydrogen bonding weakens the C=O bond and lowers the stretching force constant, K resulting in a lowering of the carboxyl frequency of saturated acids to about 1710cm-1.

Preparation of sample

Infra-red spectra of gases, liquids and solids can be obtained. The sample must be dry because water absorbs infrared radiations near 3700cm-1 and 1630cm-1 and absorption bands in these regions may be erroneously assigned to the substance being analyzed.

Solid: The solid may examined

Neat

In suspension

In solution

Neat solid:

Neat solid sample preparation is occasionally used for that solid compound that melts without decomposition.

Method

A small quantity is melted and placed between two alkali halide plates (which transmit infrared radiation)

They are clamped in a suitable holder

The capillary film which forms in allowed to solidify.

Then the whole in a suitable holder is placed in the sample beam path.

No reference is necessary i.e. air is the reference. 

Solid in Suspension:

The method most commonly used in the examination of the solid suspended in an inert oil or inert solid. In both instances the substance must be reduced to very small particles because the orientation of a crystalline material in the infrared beam affects the intensity of the absorption bands.

Mull Method:

The finely powdered sample (5gm) is dispensed in a drop of a suitable mulling agent. The most popular mulling agent is the mineral oil nujol or a mixture of high molecular weight alkanes:

Fluorinated oil eg. Fluorolube

Hexochlorobutadiene.

The mull is transferred to the surface of a flat alkali halide plate and a second plate is carefully placed on the top.

The two plates are manually pressed together using a rotary motion to ensure an even spread of the mull. By varying the amount of pressure applied films of varying thickness can be obtained.

The plates are held together in a suitable holder which is placed in the sample beam path.

No reference cell is necessary.    

Disadvantages:

The mulling agents themselves absorb infrared radiation in certain regions and thus observe some of the absorption bands of the substance being analyzed:

Nujol absorbs strongly near 3000cm-1 (C-H stretching) and less strongly near 1460 and 1735cm-1 (C-H bending). It is transparent in thin layer over the 1375 – 650cm-1 region.

Fluorolube is transparent in thin layers from 1330 – 5000cm-1. Thus two spectra one in each medium would yield maximum information.

Hexachlorobutadiene has no hydrogen atoms and will be transparent in the regions where nujol absorbs.

Pressed Disk Method:

Pure, dry and finely powdered potassium bromide or chloride is intimately mixed with the sample preferably in a Wig-L-Bug or its equivalent. Generally a concentration of about 1% w/w is suitable i.e. 2mg of the substance is ground with 200 – 300mg of KBr. With big molecular weight substances a somewhat larger concentration is often required

The mixture is compressed in a die under vacuum at room temperature and at high pressure

(40000-50000 lb/inch2). This treatment produces a solid transparent disk.

The transparent disk is mounted in a holder and placed in the sample beam path.

A reference disk of pure KBr can be used but as spectral grade KBr is virtually transparent over the 5000-650cm-1 range, it is the common practice to use no reference cell.

Disadvantages:

During the grinding process, physical and chemical changes can occur and give rise to anomalous spectra.

In solution:

Solids can also be examined in a solution in a suitable solvent. The three most commonly used solvent are:

Carbon tetrachloride (CCl4)

Chloroform (CHCl3)

Carbon disulfide (CS2)

The concentration required depends on the pathlength and on the molecular weight of the substance being examined. Generally, a 5% w/v solution of a substance of molecular weight 100 to 500 will usually give a good spectrum using a cell of 0.1mm pathlength. With 1mm cells a 1.5% w/v solution is often satisfactory. The prepared solution is transferred to a solution cell by means of a hypodermic syringe. Solvent evaporation is avoided, especially in quantitative analysis.  

Disadvantages:

The solvent CCl4, CHCl3 or CS2 absorb strongly in certain regions and thus obscure what may be important absorption bands. In addition many polar substances are not sufficiently soluble in those solvents.