Post on 12-Jul-2015
Introduction and Principle of IR Spectrophotometry
Dr.P.K.Manipabitramani@gmail.comBCKV, West Bengal, India, 09477465968
Origin of IR spectra
Atoms or atomic groups in a molecules are in continuous motion with respect to one another. IR spectra originate from the difference modes of vibration and rotation of a molecule, whereas the UV-visible absorption bands are primarily due to electronic transition.
In order to absorb IR radiation, a molecule must undergo a net change in dipole moment as a consequence of its vibrational or rotational motion. The dipole moment is determined by the magnitude of the charge difference and the distance between the two centers of charge.
The change in bond length or angle due to vibrational or rotational motion must cause a net change in the dipole moment of the molecule.
No net change in dipole moment occurs during the vibration or rotation of homonuclear species such as O2, N2, or Cl2 ; consequently, such compounds cannot absorb in the IR.
Vibrational modes which do not involve a change in dipole moment are said to be IR-inactive. With exception of a few compounds of this type, all molecular species exhibit IR-active.
IR ABSORPTION BY MOLECULES
Molecules with covalent bonds may absorb IR radiation
Molecules absorb radiation when a bond in the molecule vibrates at the same frequency as the incident radiant energy
Absorption is quantized
Molecules move to a higher energy state Molecules vibrate at higher amplitude after absorption
IR radiation is sufficient enough to cause rotation and vibration
Radiation between 1 and 100 µm will cause excitation to higher vibrational states
Radiation higher than 100 µm will cause excitation to higher rotational states
Ball and stick figure of an ethanol molecule. But exactly what is the ball, and for that matter, what is the stick? An atom doesn’t really look like a ball, nor does a chemical bond look like a stick, right?
In a covalently bonded diatomic molecule, vibratory motion of the atoms causes compression and extension of the bond like a mechanical spring obeying Hook’s law
Principle of IR spectroscopy• Molecules are made up of atoms linked by chemical
bonds. The movement of atoms and the chemical bonds like spring and balls (vibration)
• This characteristic vibration are called Natural frequency of vibration.
IR SPECTROSCOPY
The IR region is divided into
Near-IR (NIR): 750 nm – 2500 nm
Mid-IR: 2500 nm – 20000 nm
Far-IR: 20000 nm – 400000 nm
DIPOLE MOMENT (µ)
- The repetitive changes in µ makes it possible for polar molecules to absorb IR radiation
- Symmetrical molecules do not absorb IR radiation since they do not have dipole moment (O2, F2, H2, Cl2)
- Diatomic molecules with dipole moment are IR-active(HCl, HF, CO, HI)
- Molecules with more than two atoms may or may not be IR active depending on whether they have permanent
net dipole moment
µ = Q x r
Q = charge and r = distance between charges
- Asymmetrical distribution of electrons in a bond renders the bond polar
Molecular vibrationsThere are 2 types of vibrations:
1.Stretching vibrations
2.Bending vibrations
1.Stretching vibrations: Vibration or oscillation along the line of bond
• Change in bond length resulting from change in interatomic distance (r)
Occurs at higher energy: 4000-1250 cm-1
2 types:
a)Symmetrical stretching
b)Asymmetrical stretching
a) Symmetrical stretching:2 bonds increase or decrease in length simultaneously.
H
H
C
Symmetrical stretching is IR-inactive (no change in µ)
b) Asymmetrical stretching• in this, one bond length is increased and other is
decreased.
H
H
C
PRINCIPAL MODES OF VIBRATION
Bending
- Change in bond angle or change in the position of a group of atoms with respect to the rest of the molecule
Bending Modes- Scissoring and Rocking- In-plane bending modes (atoms remain in the same plane)
- Wagging and TwistingOut-of-plane (oop) bending modes (atoms move out of plane)
Bending vibrations
• Vibration or oscillation not along the line of bond• These are also called as deformations• In this, bond angle is altered• Occurs at low energy: 1400-666 cm-1
2 types:
a) In plane bending: scissoring, rocking
b)Out plane bending: wagging, twisting
a) In plane bendingi. Scissoring:• This is an in plane blending• 2 atoms approach each other• Bond angles are decrease
H
H
CC
ii. Rocking:• Movement of atoms take place in the same
direction.
H
H
CC
b) Out plane bending
i. Wagging:• 2 atoms move to one side of the plane. They move
up and down the plane.
ii. Twisting:• One atom moves above the plane and another
atom moves below the plane.
H
H
CC
H
H
CC
Molecular vibrations • How many vibrations are possible (=fundamental vibrations)?
A molecule has as many degrees of freedom as the total degree of freedom of its individual atoms. Each atom has three degrees of freedom (corresponding to the Cartesian coordinates), thus in an N-atom molecule there will be 3N degree of freedom.
In molecules, movements of the atoms are constrained by interactions through chemical bonds.
Translation - the movement of the entire molecule while the positions of the atoms relative to each other remain fixed: 3 degrees of translational freedom.
Rotational transitions – interatomic distances remain constant but the entire molecule rotates with respect to three mutually perpendicular axes : 3 rotational freedom (nonlinear), 2 rotational freedom (linear).
Vibrations – relative positions of the atoms change while the average position and orientation of the molecule remain fixed.
PRINCIPAL MODES OF VIBRATION 3N-6 possible normal modes of vibration
N = number of atoms in a molecule; Degrees of freedom = 3N
H2O for example- 3 atoms; - Degrees of freedom = 3 x 3 = 9- Normal modes of vibration = 9-6 = 3
• Vibrations which do not change the dipole moment are Infrared Inactive (homonuclear diatomics).
Selection Rules
The energy associated with a quantum of light may be transferred to the molecule if work can be performed on the molecule in the form of displacement of charge.
Selection rule:
A molecule will absorb infrared radiation if the change in vibrational states is associated with a change in the dipole moment (µ) of the molecule.
µ = qr q: electrical charge, r: directed distance of that charge from some defined origin of coordinates from the molecule.Dipole moment is greater when electronegativity difference between the atoms in a bond is greater. Some electronegativity values are:H 2.2; C 2.55; N 3.04; O 3.44; F 3.98; P 2.19; S 2.58; Cl 3.16
ω = 12π
(k/µ)1/2 ω = oscillation frequency
µ = reduced massk = force constant of the bond
ϖ = 12πc
(k/µ)
µ =mx my
mx + my
Oscillation frequency of a vibrating diatomic molecule (XY) following simple harmonic motion
Mass of atom X and Y respectively. To convert frequency to wave number we must divide W by the velocity of light
Hz
cm-11/2
Eν = 2(ν + ) hω 1
Vibrational energies (Ev) like all other molecular energies are quantized and the allowed vibrational energies for any particular system may be calculated from the Schrodinger eqn
ν= vibrational quantum number
ω = 12π
(k/µ)1/2
Molecule k / aJÅ-2
F2 (F−F) 4.45
O2 (O=O) 11.41
N2 (N≡N) 22.41
Several Force Constants
Note: IR spectra are typically presented in units called wavenumber, or more correctly, reciprocal centimetres (cm-1). Increasing wavenumber corresponds to increasing frequency.
The IR spectrum is basically a plot of transmitted (or absorbed) frequencies vs. intensity of the transmission (or absorption). Frequencies appear in thex-axis in units of inverse centimeters (wavenumbers), and intensities are plotted on the y-axis in percentage units.
IR SPECTRUM IN ABSORPTION MODE
IR SPECTRUM IN TRANSMISSION MODE
The graph shows a spectrum in transmission mode.This is the most commonly used representation and the one found in most chemistry and spectroscopy books. Therefore we will use this representation.
IR bands can be classified as strong (s), medium (m), or weak (w),depending on their relative intensities in the infrared spectrum. A strong band covers most of the y-axis. A medium band falls to about half of they-axis, and a weak band falls to about one third or less of the y-axis.
CLASSIFICATION OF IR BANDS
Group frequenciesWith certain functional or structural groups, it has been found that their vibrational frequencies are nearly independent of the rest of the molecule – group frequencies.
Carbonyl group 1650 to 1740 cm-1 various aldehydes and ketones
For many groups involving only two atoms, the approximate frequency of the fundamental vibration can be calculated from a simple harmonic oscillator model.
Calculations show that for most groups of interest, characteristic frequencies of stretching vibrations should lie in the region 4000 to 1000 cm-1. In practical, the region from 4000 to 1300 cm-1 is often called the group frequency region.
The presence of various group vibrations in the IR spectrum is of great assistance in identifying the absorbing molecule.
Note that the blue colored sections above the dashed line refer to stretching vibrations, and the green colored band below the line encompasses bending vibrations.
The complexity of infrared spectra in the 1450 to 600 cm-1 region makes it difficult to assign all the absorption bands, and because of the unique patterns found there, it is often called the fingerprint region. Absorption bands in the 4000 to 1450 cm-1 region are usually due to stretching vibrations of diatomic units, and this is sometimes called the group frequency region.
Fingerprint regionIn the region from ≈ 1300 to 400 cm-1, vibrational frequencies are affected by the entire molecule, as the broader ranges for group absorptions in the figure below – fingerprint region.
Absorption in this fingerprint region is characteristic of the molecule as a whole. This region finds widespread use for identification purpose by comparison with library spectra.
How to approach the analysis of an IR spectrum
1. Is a carbonyl group present ? C=O 1820~1660 cm–1 (strong absorption)
2. If C=O is present, check the following types. (If absent, go to 3)
Acids -- is OH also present ? -- OH 3400~2400 cm–1 (broad absorption)
Amides-- is NH also present ?-- NH 3500 cm–1 (medium absorption)
Esters -- is C-O also present ? -- C-O 1300~1000 cm–1 (strong absorption)
Anhydrides -- have two C=O absorptions near 1810 and 1760 cm–1.
Aldehydes -- is aldehyde CH present ? -- Two weak absorptions near 2850 and 1760 cm–1 .
Ketones -- The above 5 choices have been eliminated.
3. If C=O is absent
Alcohols / Phenols-- check for OH-- OH 3600~3300 cm–1 (broad absorption) C-O 1300~1000 cm–1 .
Amines -- check for NH -- NH 3500 cm–1 . (medium absorption)
Ethers -- Check for C-O (and absence of OH) -- 1300~1000 cm–1 .
OH stretch
CH stretch
CH bend
CO stretch
OH bend CCstretch
The IR Spectrum of Ethanol
Bands representing common groups don’t always appear in exactly the same position, as the local environment has an effect on the force constant. Eg. Conjugation has the effect of lowering the observed frequency.
How to approach the analysis of an IR spectrum
4. Double bonds and / or aromatic rings
C=C 1650 cm–1 (weak absorption) aromatic ring 1650~1450 cm–1
aromatic and vinyl CH 3000 cm–1
5. Triple bonds C≡N 2250cm–1 (sharp absorption) C ≡ C 2150cm–1 (sharp absorption)
acetylenic CH 3300 cm–1
6. Nitro group --two strong absorptions at 1600~1500 cm–1 and 1390~1300 cm–1
7. Hydrocarbons -- none of the above are found, CH 3000 cm–1 (major absorption)
Wavenumber / cm-1 Strength Vibrational mode
900 w C-C stretch
1080 s C-O stretch
1260 m O-H bend
1400 m C-H bend
2800-3000 s C-H stretch
3650 m O-H stretch
The IR Spectrum of Ethanol;
Tabulating IR data
IR Spectroscopy Identifying Functional Groups
C H stretch
ALKANES
~2850-2950 cm-1
C H bend ~1350-1450 cm-1
CH2 rock ~720 cm-1
C H stretch
ALKENES
~3000-3100 cm-1
C H bend ~800-1000 cm-1
C stretch ~1600-1700 cm-1C
C H stretch
ALKYNES
~3250-3350 cm-1
C H bend ~630 cm-1
C stretch ~2100 cm-1C
O H stretch
ALCOHOLS
~3200-3650 cm-1
O H bend ~1330-1420 cm-1
C stretch ~1000-1260 cm-1O
C C stretch
AROMATICS
~1600 & 1400-1500 cm-1
C H strecth ~3000 cm-1
Identifying Functional Groups
O C sym. stretch
ETHERS
~1050 cm-1
O C asym. stretch ~1250 cm-1
MOLECULES WITH CARBONYL GROUPS (C=O)
N H stretch
AMINES
~3250-3450 cm-1
N H bend ~1600-1650 cm-1
C stretch ~1000-1250 cm-1N
C
C
C
O
R G
Functional group
Ketone
Aldehyde
Carboxylic Acid
Acid Chloride
Acid Fluoride
-G
-H
-OH
-Cl
-F
-R
~1720-1740
~1750-1770
~1775-1815
~1870
~1680-1720
cm-1 Functional group
Ester/Lactone
Amide/Lactam
-G
-NR
-OR
~1650-1700
~1735-1750
cm-1
O
OR
O
R
~1750 & 1815
Note: Conjugation in ANY of these systems results in a lowering of the carbonyl stretching frequency!
Understanding & Identifying Molecular Structure
IR Spectroscopy The origin of broad -OH and -NH bands.
gas
liquid
Hydrogen bonding results in lower electron density at each oxygen, thus lowering the force constant, k, thus lowering (& broadening) the frequency for the mode.
H3C
CH2
O
H
H
O
CH2
H3C
H
O
CH2
H3C
H
O
CH2
H3C
H
O
CH2
H3C
H
O
CH2
H3C
Understanding & Identifying Molecular Structure
Q. The two IR spectra on the right correspond to two different molecules sharing the same molecular formula; C3H6O.
a)Identify which is an alcohol and which is a ketone. b)Propose molecular structures for these two molecules!
C
O
CH3H3C
HC
CH2
H2C
OH
Understanding & Identifying Molecular Structure
Q. The three IR spectra on the right correspond to three different molecules all with a C3 carbon chain but different degrees of unsaturation.a)Identify which of these is propane, propene and propyne.
b) Label each peak with the relevant vibrational mode.
Satisfy yourself that some features unambiguously identify some kinds of functional groups
20003000 1000cm-1
4000
Absorption RegionsIrtutord.exe
IR spectrum of n-butanal (n-butyraldehyde).
THE FINGERPRINT REGION
Although the entire IR spectrum can be used as a fingerprint for the purposes of comparing molecules, the 600 - 1400 cm-1 range is called the fingerprint region.This is normally a complex area showing many bands, frequently overlapping each other. This complexity limits its use to that of a fingerprint, and should be ignored by beginners when analyzing the spectrum. As a student, you should focus your analysis on the rest of the spectrum, that is the region to the left of 1400 cm-1.
y axis is %T or A
x axis is wavenumber (or wavelength)
Io → sample → I
T = I/Io %T = 100 I/Io
T transmission / transmittance
A = -log T
A absorbance (no units)
(Note A (but not T) ∝ concentration)
IR spectrum
IR SPECTRUM OF ALKANES
Alkanes have no functional groups. Their IR spectrum displays only C-C and C-H bond vibrations. Of these the most useful are the C-H bands, which appear around 3000 cm-1. Since most organic molecules have such bonds, most organicmolecules will display those bands in their spectrum.
Besides the presence of C-H bonds, alkenes also show sharp, medium bands corresponding to the C=C bond stretching vibration at about 1600-1700 cm-1.Some alkenes might also show a band for the =C-H bond stretch, appearingaround 3080 cm-1
IR SPECTRUM OF ALKENES
The most prominent band in alkynes corresponds to the carbon-carbon triple bond. It shows as a sharp, weak band at about 2100 cm-1. The reason it’s weak is because the triple bond is not very polar. In some cases, such as in highly symmetrical alkynes, it may not show at all due to the low polarity of the triple bond associated with those alkynes.Terminal alkynes, that is to say those where the triple bond is at the end of a carbon chain, have C-H bonds involving the sp carbon (the carbon that forms part of the triple bond). Therefore they may also show a sharp, weak band at about 3300 cm-1
corresponding to the C-H stretch.Internal alkynes, that is those where the triple bond is in the middle of a carbon chain, do not have C-H bonds to the sp carbon and therefore lack the aforementioned band.
unsymmetricalterminal alkyne (1-octyne)
symmetrical internal alkyne (4-octyne).
Carbonyl compounds are those that contain the C=O functional group. In aldehydes, this group is at the end of a carbon chain, whereas in ketones it’s in the middle of the chain. As a result, the carbon in the C=O bond of aldehydes is also bonded to another carbon and a hydrogen, whereas the same carbon in a ketone is bonded to two other carbons.Aldehydes and ketones show a strong, prominent, stake-shaped band around1710 - 1720 cm-1 (right in the middle of the spectrum). This band is due to thehighly polar C=O bond. Because of its position, shape, and size, it is hard tomiss.Because aldehydes also contain a C-H bond to the sp2 carbon of the C=O bond,they also show a pair of medium strength bands positioned about 2700 and 2800cm-1. These bands are missing in the spectrum of a ketone because the sp2carbon of the ketone lacks the C-H bond.
IR SPECTRUM OF AN ALCOHOL
The most prominent band in alcohols is due to the O-H bond, and it appears as astrong, broad band covering the range of about 3000 - 3700 cm-1. The sheer sizeand broad shape of the band dominate the IR spectrum and make it hard to miss.
IR SPECTRUM OF A CARBOXYLIC ACID
because it has both the O-H bond and the C=O bond.
Therefore carboxylic acids show a very strong and broad band covering a wide range between 2800 and 3500 cm-1 for the O-H stretch. At the same time they also show the stake-shaped band in the middle of the spectrum around 1710 cm-1 corresponding to the C=O stretch.
IR SPECTRUM OF A NITRILE
In a manner very similar to alkynes, nitriles show a prominent band around 2250cm-1 caused by the CN triple bond. This band has a sharp, pointed shape justlike the alkyne C-C triple bond, but because the CN triple bond is more polar, thisband is stronger than in alkynes.