IR Spectra Interpretation

 IR Spectra Interpretation

IR Spectra Interpretation

In order to understand infrared spectra, absorption bands in the spectrum of an unknown chemical must be compared to known absorption frequencies for various types of bonds.

A molecule's chemical bonds vibrate when it absorbs infrared light. The bonds are elastic and flexible. Infrared spectroscopy is a form of vibrational spectroscopy because of this. Consequently, due to the differences in their structural makeup, different molecules vibrate at various frequencies. This is why infrared spectroscopy allows for the differentiation of molecules.

The first criterion for one chemical compound to absorb infrared light is the presence of a vibration during which there is a non-zero change in the dipole moment with respect to distance. The second prerequisite for infrared absorbance is that there must be an equivalence between the energy of the light intruding on a molecule and the difference in vibrational energy levels within the molecule.

A good spectrum should have a minimum of five characteristics: low noise, little to no baseline offset, a flat baseline, peaks that are on scale, and no spectral distortions. Below is an example of the IR spectrum that has all the characteristics.

In order to interpret the IR spectra, you should know the following factors that affect the vibrational frequency. 

Factors Influencing Vibrational Frequency
1. Coupled Vibration

If a simple C-H bond is present in the compound then its stretching occurs at one frequency if the CH2 group is present, here two stretching’s occur it can be symmetric or asymmetric. This is coupled vibration. Normally stretching shows at a higher frequency than bending. Symmetric stretching causes a lowering of wave number or frequency and asymmetric stretching increases wave number more than normal stretching.

2. Fermi Resonance

when the overtone band and fundamental band overlap, Fermi resonance occurs.

The Overton band is the band that forms when a transition occurs from 0 to 2 transitional levels in the vibrational frequency of molecules.

3. Electronic effects

This effect is produced due to the change of substituents in the neighbor group.

It is of three types

a) Inductive effect (I)

It can be positive and negative (Occur in sigma bond), Position of bonds remains same but pulling of electrons towards more electronegative atoms occurs that create dipoles in the molecules.

+I ………      Decrease wavenumber (alkyl group)

-I  ……      Increase wavenumber (Electronegative atom)

1.

Positive Inductive effect

   

Negative Inductive effect

b)   b) Mesomeric or Resonance effect

Lengthening of bond occur, decrease wavenumber (Occur in pi bond). It has dual effect, if single bond convert to double bond than wavenumber increase and if double bond convert to single bond than wavenumber decrease.

Resonance or Mesomeric effect

c) Field effect

When two atoms interact through space they cause repulsion between them which in turn increases the wave number. Specifically, in those molecules that have ortho group substituents. Cis and trans form interaction produce, in case of cis vibrational frequency or wave number increases and in trans form wave number decreases.

4. Hydrogen Bonding

a) Intramolecular hydrogen bonding

Sharp bands and well-defined bands form a spectrum which in turn increases the wave number.  For example, sharp bands are formed in the case of amines due to the lower electronegativity of Nitrogen as compared to alcohols (oxygen low E. N).

b) Intermolecular hydrogen bonding

Present between the two molecules, forms broad bands in the spectrum, it is concentration dependent, in dilute solutions N-H stretching occurs at 3500 cm-1 but in the condensed phase or after hydrogen bonding it decreases waves number to 3300 cm-1. O-H stretching occurs at 3650 cm-1 in dilute solutions but in concentrated solutions it decreases stretching frequency to 3450 cm-1. So, Hydrogen bonding shifts the wave number to the Lower level.

5. Bond Order

Bond order is directly related to the strength of a bond. As we know bond order is the number of bonds between two atoms. More bonds, the more energy is required to stretch or bend the bond. So, as in triple bond there are three bonds between atoms that are formed by sharing of three pair of electrons between these two atoms its vibrational frequency increase.

In a single bond, there is only one sigma bond.

In a double bond, there is one pi and one sigma bond.

In a triple bond, there are two pi bonds and one sigma bond, so triple bond is strongest bond.

As we know sigma bond is strong than the pi bond but the pi bonds are stiffer and vibrate faster increase the frequency of vibrations so when IR radiation falls on the sample the vibrational frequency for a single bond is lower than the double bond which is lower than the triple bond. 

Bond Angle

The bond angle is also related to bond order. In a single bond, an angle is 109.5^o, more s-character, more strength of the bond, and greater vibrational frequency value. S-character in molecules increases the strength of molecule’s bonds, resulting in shortness of bond. Short bonds are strong bonds that require more force to compress or stretch.

 Now, following in the table that shows the values of ir vibration of specific functional groups, you should have to compare the values by seeing the values from table and seeing the spectra.

Table 1.

Functional Groups   C-H Alkanes (stretch) -CH3 (bend) -CH2- (bend)	Frequency (cm-1)   3000-2850 cm-1 1450 & 1375 cm-1 1465 cm-1
C=C	1680-1600cm1
Aromatic	1600 & 1475 cm-1
C ☰ C	2260-2100 cm-1
C=O                                             Aldehyde	1740-1720 cm-1
Ketones	1725-1705 cm-1
Carboxylic acid	1725-1700 cm-1
Ester	1750-1730 cm-1
Amide	1700-1640 cm-1
Anhydride	1810 and 1760 cm-1
Acid chlorides	1800 cm-1
C-O	1300-1000 cm-1
O-H Free H-Bonded	3650-3600 cm-1 3400-3200 cm-1
N-H	3100-3500 cm-1
C-N amines	1350-1000 cm-1
C=N Imines and oximes	1690-1640 cm-1
Nitriles	2260-2240 cm-1
X=C=Y	2270-1940 cm-1
N=O Nitro	1550 and 1350 cm-1
S-H Mercapto	2550 cm-1
S=O	1375-1300 cm-1
C-X
Fluorides	1400-1000 cm-1
Chlorides	785-540 cm-1
Bromides	< 667 cm-1
Iodides	< 667 cm-1

Detailed process to understand Spectra 

• First, when you want to interpret the spectrum, not every peak needs to be examined. Rather, IR is excellent for identifying a few distinct functional groups, such as alcohols and carbonyls. In this way, it’s complementary to other techniques (like NMR) which don’t yield this information as quickly.

• Two specific regions of the spectrum—3200-3400 cm-1 and 1650-1800 cm-1—provide 80% of the data that is most pertinent to our needs.

• The instrumental resolution, sampling strategy, and presence or absence of spectral manipulation (such as baseline adjustment, smoothing, and subtraction) should all be known before viewing a spectrum.

• Strongly IR-absorbing atmospheric gases like water vapor and carbon dioxide have ambient concentrations that are high enough for their peaks to show up in mid-IR spectra.

• The group wavenumbers presented in Table I should be noted whether they are present or absent when you read the spectrum from left to right like a sentence in a book. In order to quickly evaluate whether or not certain significant functional groups are present in a sample, you scan the spectrum from left to right and use the peak ranges in Table I. The easiest bands to see are the most powerful, but these peaks are typically the ones that are most helpful for diagnosis. Less intense peaks still matter, but it is advisable to take care of the larger ones first because they are simpler to see and allocate.

• Peaks with lower intensity are known as secondary bands for a specific functional group. There are two reasons why secondary bands need to be found. First off, a lot of functional groups have numerous peaks in the mid-IR, thus finding as many peaks as you can for each peak for a functional group increases the likelihood of a proper interpretation. Second, these bands must be assigned to prevent the mistaken assumption that they are the result of a functional group that might not actually exist.

• Write down the functional groups you believe to be present as the peaks are allocated during your interpretation process. Put the pieces of the functional groups you've discovered together like a puzzle piece to create real chemical structures. Next, check to see if the structure you have sketched is compatible with the spectrum. For instance, the OH may be joined to the methylene or the ring in a sample that includes an OH group, CH2, and benzene ring. Your true choice between these two options can be determined by looking at the spectrum.

• The factors that are discussed above are the main reason for increasing or decreasing the frequency of IR. For example, carbonyl stretching frequency that given in table is different for different groups containing carbonyls like aldehydes, ketones, etc. From the above discussion you will also be able to give specific frequency value to specific functional groups in molecules by considering the above factors.