SPECTROSCOPIC METHODS OF ORGANIC MOLECULES. Basics in IR... · consequence of changing stretching...

SPECTROSCOPIC METHODS OF ORGANIC MOLECULES

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Professor Ioannis P. GerothanassisDepartment of Chemistry, University of Ioannina

FEBRUARY 2018

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SPECTROSCOPIC METHODS OF ORGANIC MOLECULES

Aspirin (acetylsalicylic acid)

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5001000150020002500300035004000

wavenumber/ cm-1

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Wavenumber = 1/ wavelength

Professor Ioannis P. GerothanassisDepartment of Chemistry, University of Ioannina

INFRARED SPECTROSCOPY



• Purpose of each Electromagnetic Radiation

Wade, Jr. L.G. Organic Chemistry 5th ed. Pearson Education In., 2003

What is Infrared? (Cont.)Humans, at normal body temperature, radiate most

strongly in the infrared, at a wavelength of about 10microns (A micron is the term commonly used inastronomy for a micrometer or one millionth of ameter). In the image to the left, the red areas are thewarmest, followed by yellow, green and blue(coolest).

The image to the right shows a cat in the infrared. The yellow-white areas are the warmest and the purple areas are the coldest. This image gives us a different view of a familiar animal as well as information that we could not get from a visible light picture. Notice the cold nose and the heat from the cat's eyes, mouth and ears. 4

Fourier Transform Infrared (FT-IR) Spectroscopy, Prof. V. KrishnakumarProfessor and Head, Department of Physics, Periyar University, Salem – 636 011, India

IR: Masses, Atoms and Springs

F = force, restoring back to equilibrium

position

k = characteristic stretching constant

Δr = displacement from the equilibrium

position

A Model: Picture the atoms of a diatomic molecule as point masses connected by springs (bonds).

As a first approximation use Hooke’s Law

F = -k Δr

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http://www.columbia.edu/itc/chemistry/chem-c1403/lectures/Fall2005/%20Lecture22A_IR_112805.pdf

ν =1

2πkmr

mr =m1m2

m1 + m2

IR Stretching Frequencies of two bonded atoms:

ν = frequencyk = spring strength (bond stiffness)mr = reduced mass (~ mass of largest atom)

What Does the Frequency, ν, Depend On?

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Let’s examine values as absolute numbers. Which is the vibrational frequency of the C – H bond?

Why ?

where mc and mH are the atomic weights of carbon and hydrogen, respectively, and 6.02x1023 is the Avogadro number.Taking into consideration that

K = 5 X 105 dynes cm-1

C – H = 9 x 1013 Hz or cycles s-1 8

Therefore, the C – H bond vibrates 9 x 1013 times s-1

To avoid such extreme numbers the term wavenumber is introduced

(cm-1)

where C is the velocity of light = 3 x 1010 cm s-1

Therefore, for the C – H bond

≈ 3000 cm-1

9

The absorption intensity depends on how efficiently the energy of an electromagnetic wave of frequency ν can be

transferred to the atoms involved in the vibration

The greater the change in dipole moment during a vibration, the higher the intensity of absorption of a photon

What does the absorption intensity depend on?

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What is the intensity of an IR signal of:O2 or N2 or H2?

Ans: In order to absorb infrared radiation, a molecular vibration must cause a change in the

dipole moment of the molecule

O2, N2 and H2 DO NOT ABSORB IR LIGHT!

They are not “Greenhouse” gases 11



• In the IR region of the electromagnetic spectrum, the absorption of radiation by a sample is due to changes in the vibrational energy states of a molecule.

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http://social.ocr.org.uk/files/ocr/An Introduction to Infrared and UV-Visible Spectroscopy.ppt

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• Only vibrations that cause a change in ‘polarity’ give rise to bands in IR spectra – which of the vibrations for CO2 are infrared active?

Symmetric stretch

Asymmetric stretch

Bending (doubly degenerate)

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Symmetrical stretching:2 bonds increase or decrease in length simultaneously.

H

H

C

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Introduction and Principle of IR Spectrophotometry, Rajaram Kshetri B.pharm, IOM

Asymmetrical stretchingone bond length is increased and other is decreased.

H

H

C

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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, rockingb) Out plane bending: wagging, twisting

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In plane bendingi. Scissoring:• This is an in plane blending• 2 atoms approach each other• Bond angles are decrease

H

H

CC18


ii. Rocking:• Movement of atoms take place in the same direction.

H

H

CC

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Out plane bendingi. 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

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• How much movement occurs in the vibration of a C-C bond?

0.1 Å

1.54 Åstretching vibration For a C-C bond with a bond length of 1.54 Å, the variation is about 0.1 Å.

bending vibration4o 0.1 Å

For C-C-C bond angle a change of 4o

is typical. This moves a carbon atom about 0.1 Å.

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Simple harmonic oscillator

0

3

12

54

ω

( )212 eqE K r r= −

12osc

kωπ µ

=

Energy

Oscillation

12 oscE hν ν ω = +

12 osc

E hhc

ννε ν ω = = +

Permitted energy levels ofthe harmonic oscillator

ν - Vibrational quantum number

If ν = 0, 12 osc

E hhc

ννε ω = =

Zero-point energy

Selection rule for the harmonic oscillator undergoing vibrational changes ∆ν = ± 1

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• Bond – not a perfect Elastic nature

• Does not obey exactly simple harmonic motion

• Dissociates

• Not a ideal parabola

( ){ } 21 expeq eqE D a r r = − −

A – constant; Deq – Dissociation energy

Morse curve

Anharmonic Oscillator

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IR Stretching Frequencies: What Do they Depend On?

Directly on the strength of the bonding between the two atoms (ν ~ k)

Inversely on the reduced mass of the two atoms (ν ~ 1/m)

Expect: ν will increase with increasing bond strength (bond order) and decreasing mass 24


Examples of stretching frequencies and correlations with bond strengths (bond order)

C-C

C=C

C C

Bond strength*

350

600

840

Bond order

1

2

3

ν

2200 cm-1

1600 cm-1

1000 cm-1

*In kJ/mol For same reduced mass! 25


Structural unit Frequency, cm-1

Stretching vibrations (single bonds)

O—H (alcohols) 3200-3600

O—H (carboxylic acids) 3000-3100

N—H 3350-3500

Infrared Absorption Frequencies: functional groups

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The presence of the hydrogen atom in X—H bonds results in high values (effect of reduced mass).


Structural unit Frequency, cm-1

Stretching vibrations (single bonds)

sp C—H 3310-3320

sp2 C—H 3000-3100

sp3 C—H 2850-2950

sp2 C—O 1200

sp3 C—O 1025-1200

Infrared Absorption Frequencies

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4000-3000cm-1

3000-2000cm-1

2000-1500cm-1

1500-1000cm-1

O-H N-HC-H

C≡CC≡N

C=CC=O

C-OC-FC-Cl

deformations

Increasing energy

Increasing frequency 28


The spectrophotometer give the spectra of certain wave by indicating the quality of wave (i.e. its wavenumber) and the quantity of wave that how much wave is absorbed by the molecule (Transmittance).

Parameters

Transmittance is the fraction of incident light electromagnetic radiation at a specified wavelength that passes through a sample.(Lambert-Beer Law) It is a ratio so it has no unit. 29

Infrared Spectroscopy, S. YOUSUF RAZA ZAIDI

INTERPRETATION OF INFRARED SPECTRA

The most useful regions are as follows:• 3200-3700 cm-1: various types of O-H and N-H stretching

vibrations.• 2700-3100 cm-1: different types of C-H stretching vibrations. • 1680-1750 cm-1:C=O stretches feature very strongly in IR

spectra and the type of carbonyl group can be determined from the exact position of the peak.

Too many bonds absorb in the region of 600-1600 cm-1 to allow confident assignment of individual bands. However, this region is useful as a fingerprint of a molecule, i.e. if the spectrum is almost identical to an authentic reference spectrum then the structure can be assigned with some confidence.

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Infrared SpectroscopyFor isopropyl alcohol, CH(CH3)2OH, the infrared absorption bandsidentify the various functional groups of the molecule.

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The IR SpectrumThere are two type of IR Spectra from which we can obtained the information about the quality of molecule.

1. The Functional Group region: Identifies the functional group with the consequence of changing stretching vibrations. Ranges from 4000 to 1600 cm-1.

2. The Fingerprint region: Identifies the exact molecule with the consequence of changing bending vibrations. Ranges from 1600 to 625cm-1.

Focus your analysis on this region. This is where most stretching frequencies appear.

Fingerprint region: complex and difficult to interpret reliably.

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IR SPECTRUM OF ALKANESAlkanes 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 organic molecules will display those bands in their spectrum.

Graphics source: Wade, Jr., L.G. Organic Chemistry, 5th ed. Pearson Education Inc., 2003

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IR SPECTRUM OF ALKENESBesides 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, appearing around 3080 cm-1 as shown below. However, this band could be obscured by the broader bands appearing around 3000 cm-1 (see next slide)

Graphics source: Wade, Jr., L.G. Organic Chemistry, 5th ed. Pearson Education Inc., 2003

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IR SPECTRUM OF ALKENESThis spectrum shows that the band appearing around 3080 cm-1 can be obscured by the broader bands appearing around 3000 cm-1.

Graphics source: Wade, Jr., L.G. Organic Chemistry, 6th ed. Pearson Prentice Hall Inc., 2006

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IR SPECTRUM OF ALKYNESThe 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.

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IR SPECTRUM OF ALKYNES


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IR SPECTRUM OF A NITRILEIn a manner very similar to alkynes, nitriles show a prominent band around 2250 cm-1 caused by the CN triple bond. This band has a sharp, pointed shape just like the alkyne C-C triple bond, but because the CN triple bond is more polar, this band is stronger than in alkynes.


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IR SPECTRUM OF AN ALCOHOLThe most prominent band in alcohols is due to the O-H bond, and it appears as a strong, broad band covering the range of about 3000 - 3700 cm-1. The sheer size and broad shape of the band dominate the IR spectrum and make it hard to miss.


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IR SPECTRUM OF ALDEHYDES AND KETONES

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 around 1710 - 1720 cm-1 (right in the middle of the spectrum). This band is due to the highly polar C=O bond. Because of its position, shape, and size, it is hard to miss.

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 2800 cm-1. These bands are missing in the spectrum of a ketone because the sp2

carbon of the ketone lacks the C-H bond.

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IR SPECTRUM OF ALDEHYDES AND KETONES


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IR SPECTRUM OF A CARBOXYLIC ACIDA carboxylic acid functional group combines the features of alcohols and ketones 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.


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IR SPECTRA OF AMINESThe most characteristic band in amines is due to the N-H bond stretch, and it appears as a weak to medium, somewhat broad band (but not as broad as the O-H band of alcohols). This band is positioned at the left end of the spectrum, in the range of about 3200 - 3600 cm-1.

Primary amines have two N-H bonds, therefore they typically show two spikes that make this band resemble a molar tooth. Secondary amines have only one N-H bond, which makes them show only one spike, resembling a canine tooth. Finally, tertiary amines have no N-H bonds, and therefore this band is absent from the IR spectrum altogether. The spectrum below shows a secondary amine.


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IR SPECTRUM OF AMIDESThe amide functional group combines the features of amines and ketones because it has both the N-H bond and the C=O bond. Therefore amides show a very strong, somewhat broad band at the left end of the spectrum, in the range between 3100 and 3500 cm-1 for the N-H stretch. At the same time they also show the stake-shaped band in the middle of the spectrum around 1710 cm-1 for the C=O stretch. As with amines, primary amides show two spikes, whereas secondary amides show only one spike.


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Infrared SpectroscopyA molecule can be characterized (identified) by its molecularvibrations, based on the absorption and intensity of specific infraredwavelengths, however, IR depends upon the state of the matter (solid-liquid-solution-gas phase)

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Human Breath

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OH H

OH H

O C O

O C O

O C O

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