electronic transitions (uv-vis)

Electronic transitions (UV-vis)

Bond vibrations (IR) Nuclear spin (NMR)

UV-Vis

When UV-Vis light is absorbed by a molecule a pi electron is moved to higher energy levels.

*h

*

This electronic transition is known as a pi to pi star (*) transition.

UV-Vis

The highest energy molecular orbital containing electrons is called the Highest Occupied Molecular Orbital - HOMO for short.

The lowest energy * orbital containing no electrons is called the Lowest Unoccupied Molecular Orbital - LUMO for short.

UV-Vis

A * transition can therefore also be described as a HOMO LUMO transition.

The wavelength of UV-Vis light absorbed by a compound is related to the energy difference between the HOMO and LUMO.

Conjugation reduces the energy difference between the HOMO and LUMO, therefore more conjugated systems absorb light of longer wavelengths.

UV-Vis spectroscopy

0

0.2

0.4

0.6

0.8

1

200 400 600

Wavelength (nm)

Ab

so

rba

nc

e AB

max is the wavelength at which a compound has maximun absorbance of light.

max max

UV-Vis spectroscopy

The value of max can be estimated using the Woodward-Fieser Rules

217 nm 256 nm 232 nm

To estimate max values:

• Identify the base structure to give a starting value.

• Each conjugated C=C adds 30 - 40 to base value.

• Each additional alkyl group adds 5 - 10.

Estimate the max values for the following:

Generally you will find that max is found by simply scanning the UV-Vis spectrum.

The main use of UV-Vis spectroscopy is in quantitative analysis.

Beer-Lambert Law: A = lc

Where A is absorption and is equal to the log(Io/I) where I is the intensity of light.

= extinction coefficient or molar absorptivity

c = concentration

l = path length through the cell in cm (usually 1 cm)

For a given compound being measured in the same instrument, l is constant. Thus, the relationship between A and c is linear.

UV-Vis spectroscopy: Beer-Lambert law

Quantitative UV-Vis analysis is done at max of a compound to obtain maximum sensitivity, stability and reproducibility. The value of max is determined using a scanning UV-Vis spectrophotometer.

Once max is known an adsorption vs concentration curve is generated.

The Beer-Lambert Law: A = lc tell us that the relationship between adsorption and concentration is linear.

The concentration of the compound in a sample is determined by measuring the adsorption of the sample at max and finding the concentration from the adsorption vs concentration graph. The concentration can also be determined using linear regression analysis.


What is the concentration of A in a sample the has an adsorption of 0.32 @ 485 mn?


Concentration of A (M)

Adsorption @ 485 mn

0.0 0.0

0.01 0.09

0.02 0.21

0.04 0.39

0.08 0.82

Bonds between atoms act like springs – bond lengths are really average distances (e.g. C-C single bond = 1.48 Å; C-C bond in aromatic ring = 1.397 Å).

How easily atoms move toward or away from each other (bond stiffness) is related to the strength of the bond – stronger bonds vibrate at a higher frequency and require higher energy infrared light to change the vibration.

Infrared (IR) Spectroscopy

Infrared (IR) Spectroscopy

Masses of bonded atoms also affect vibrational frequencies. Bigger atoms (more mass or higher atomic weight) in a bond lower the frequency of the vibration.

In order to interact with IR light, a bond has to undergo a change in dipole moment as a result of the vibration.

This means the bond has to have a dipole moment to begin with. The bigger the dipole moment, the more strongly IR light will be absorbed.

Polar bonds = IR active bonds

Nonpolar, symmetric bonds = IR inactive bonds

O

H

HH H

H

H

HH CH3 CH3

H

H H

H

H

H CH3

H H

CH3 CH3

H

CC

CC C CC C

C C C C

There are different ways in which bonds can vibrate.

More complicated molecules have more fundamental vibrational modes and more complicated spectra.

symmetric stretching antisymmetric stretching in-plane bending (scissoring)

Number of fundamental vibrational modes = 3n – 6 where n = the number of atoms in the molecule.

Water 3 atoms 3(3) – 6 = 3 fundamental vibrational modes

C6H12 18 atoms 3(18) – 6 = 48 fundamental vibrational modes

Other vibrational modes are possible (out-of-plane bending) and one part of a molecule can affect the vibrating bonds in another part. Secondary interactions increase the complexity of the IR spectrum of a molecule.

Number of molecular vibrations

Double beam IR instrument:

A modern Fourier-Transform IR spectrometer reads all wavelengths at once, requiring a lower power IR source and completing the spectrum in a matter of seconds. The interferogram produced by the FT-IR is converted by a computer program to the same kind of spectrum produced from an instrument that reads one wavelength at a time.

Example IR spectrum:

Wavenumber (cm1)

4000 3000

2000

1500

1000

500

Tra

nsm

itta

nce

(%

T)

50

100

0

Fingerprint region – 600 cm1 to 1400 cm1 – region of the spectrum where most of the complex rocking and out-of-plane vibrations are found. Each compound (except enantiomers) has a unique fingerprint.

Each dip in %T toward 0 represents light that is being absorbed by vibrating polar bonds. The dips are called peaks. Peaks can be strong, weak, or medium. Peaks can be broad or sharp.

IR spectra have two primary uses:

• Certain peaks and peak patterns are associated with specific structural features. Therefore IR is used to identify important structural features (i.e. functional groups) present in the compound.

IR spectra have two primary uses:

• Since almost all compounds have unique fingerprint regions, IR spectra can confirm or deny the identity of a compound. If the fingerprint region of an unknown compound is compared to a standard reference sample and the fingerprint regions match, the compounds should be the same. If fingerprint regions do not match, the compounds must be different.

1 - two spikes 2 - one spike

ester C-0 often just above 1200 cm1

sp3 C-H stretch

sp2 C-H stretch

sp C-H stretch

aldehydes also show two C-H stretching peaks for the CHO group at about 2700 cm1 and 2800 cm1

Additional things to keep in mind:

• Sometimes the absence of a peak is just as helpful as the presence of a peak in determining what kind of compound you have. (e.g. No strong peak near 1700 cm1 eliminates ketones, aldehydes, esters, and carboxylic acids as possibilities for your compound. No strong, broad peak near 1640 cm1 eliminates amide.)

Example 1:

sp3 C-H stretch

no O-H

no N-H

no C=O stretch

no C=C stretch

no sp or sp2 C-H stretch

no C-O stretch

Conclusion: This compound is a simple alkane.

C-H bending

Example 2:

O-H stretch of acid

C=O stretch

aromatic C=C stretch

Conclusion: Aromatic carboxylic acid

Example 3:

O-H stretchsp2 C-H stretch

sp3 C-H stretch

aromatic C=C stretch

harmonics associated with monosubstituted ring

peaks in this area can tell you if a ring is monosubstituted, o-, p-, or m-disubstituted etc.

Conclusion: aromatic alcohol with alkyl portion

C-O stretch

Example 4:

sp2 C-H stretch

C=O stretch above 1750 cm1 indicates ester aromatic

C=C stretch

C-O stretch of ester

Conclusion: This is an aromatic ester

Example 5:

1 N-H stretch

sp3 C-H stretch

C-H bending

Shows up with 1 amines (not C=C)

Conclusion: aliphatic 1 amine

Example 6:

2 N-H stretch

sp3 C-H stretch

no C=O

no C=C

Conclusion: aliphatic 2 amine

Example 7:

1 N-H stretch

sp2 C-H stretch

C=O of amide – also conjugated aromatic C=C

Conclusion: aromatic 1 amide (conjugated)

Example 8:

sp C-H stretchsp3 C-H stretch

CΞC stretch

Conclusion: terminal alkyne

Example 9:

CΞN stretch

sp3 C-H stretch

too high for N-H

Conclusion: aliphatic nitrile

Example 10:

1 N-H stretch

sp3 C-H stretch

C=O stretch of amide

Conclusion: aliphatic 1 amide

Example 11:

C=O stretch of ketone, aldehyde, or acid

sp3 C-H stretch

no O-H stretch

no aldehyde C-H stretch

Conclusion: aliphatic ketone

Example 12:

C=O stretch of ketone, aldehyde, or acid

aldehyde C-H stretch

aromatic C=C

Conclusion: aromatic aldehyde

electronic transitions (uv-vis)

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