Infrared (IR) Spectroscopy for Structural Analysis Ridwan Islam

Infrared Spectroscopy  Infrared spectroscopy is a type of absorption spectroscopy that deals with the infrared region of the electromagnetic spectrum.  Infrared radiation which extends from 0.8 to 200 μ affects only the vibrational and rotational energy levels.

 The infrared region is subdivided into- 1. Near infrared region (NIR): extends from 0.8 to 2.5 μ. It is concerned with low energy electronic transitions as well as vibrational and rotational changes. 2. Infrared region (IR): extends from 2.5 to 25 μ (frequency 4000 to 400cm -1 ). It affects mainly the vibrational energy level. 3. Far infrared region (FIR): extends from 25 to 200 μ. It affects only the rotational energy level of the molecule.

 Atoms are represented by balls and the bonds connecting them as springs. If the model is suspended in space and struck, the balls appear to undergo random chaotic motions. The springs will be found to stretch and contract or bend.

 These motions are known as stretching and bending vibrations. The frequency with which these vibrations occur depends on:  - the nature of the atoms joined by the bond  - arrangement of atoms within the molecule (i.e. environment)  - mass of the atom  - strength of the bonds.

 Again, if a spring connecting two balls is struck, it vibrates and this vibration, in turn, influences the rest of the system.  In a similar way, if a chemical bond vibrating at a certain frequency is struck by Infrared radiation of that same frequency, the vibration of that bond will increase by absorption of radiation, i.e. the gain in energy will create a greater amplitude of vibration.

 If a substance is irradiated by IR radiation, the wavelength of which is constantly changing, different bond will absorb energy at different frequencies.  As a result, an Infrared absorption spectrum is obtained in which percentage of intensity of the transmitted IR light is plotted against the wavelength (or wave number) of that light.

Use of IR Spectra  As the frequency of vibration of a bond in a molecule depends on the nature of the atoms joined by the bond, arrangement of atoms within the molecule (i.e. environment), mass of the atom and strength of the bonds, no two molecule of different structure have exactly the same infrared spectrum.  For this reason, the Infrared spectrum can be used for molecules much as a fingerprint can be used for humans.  For example,  -OH group of alcohols absorbs strongly at cm -1  -CO- group of ketone absorbs at 1710 cm -1  -CN group at 2250 cm -1  -CH3 group at cm -1, etc.

Factors affecting vibration and absorption:

1. Bond Strength

2. Mass of the atom:

3. Bending and Stretching vibrations:

4. Hybridization:

5. Resonance:

Calculate the stretching frequencies for C=C bond

Calculate the stretching frequencies for C-H bond

Calculate the stretching frequencies for C-D bond

What to look for while examining infrared data:  peak position  Shape and intensity of the peaks.  Index of hydrogen deficiency (degree of unsaturation)

Index of hydrogen deficiency:

How to approach the analysis of a spectrum: (What you can tell at a glance)

1. Alkanes

2. Alkenes

3. Alkynes:

D I S C U S S I O N: C-H Stretch Region

C-H Bending Vibrations for Methyl and Methylene:

4. Alcohols: Phenols give a C-O absorption at about 1220 cm − 1 because of conjugation of the oxygen with the ring, which shifts the band to higher energy (more double-bond character).

Problem- 1:  Determine the structure or structures possible for a compound with formula C 5 H 12 O. IR spectroscopic analysis revealed a strong and broad peak at about 3350 cm -1. A band appeared near 1050 cm -1 and a pair of peaks were also seen at 1380 and 1370 cm -1.  Hint:

5. Aldehyde:

6. Ketones:

Problem-2:  Determine the structure or structures possible for a compound with formula C 6 H 10 O.The IR spectrum revealed the presence of absorption bands at 1719 cm -1, 1640 cm -1, 3100 cm -1, and absence of peaks near cm -1.  Hint:

7. Aromatic ring:

 Ortho-Disubstituted Rings (1,2-Disubstituted Rings). One strong band near 750 cm − 1 is obtained.

 Meta-Disubstituted Rings (1,3-Disubstituted Rings). This substitution pattern gives the 690-cm − 1 band plus one near 780 cm − 1. A third band of medium intensity is often found near 880 cm − 1.

 para-Disubstituted Rings (1,4-Disubstituted Rings). One strong band appears in the region from 800 to 850 cm − 1.

Problem-3:  Determine the structure or structures possible for a compound with formula C 7 H 9 N. IR analysis revealed the presence of bands at 1600 and 1450 cm − 1, a set of three peaks at about 700, 800, and 900 cm − 1, and two-peak pattern was found between 3300 and 3500 cm − 1.  Hint:

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