Spectroscopy Notes

Spectroscopy

Spectroscopy is the branch of science dealing with the interaction of electromagnetic radiations with matter.
In spectroscopy, light energy interacts with matter.
Absorption spectra arise when matter absorbs light energy and transitions from a lower to a higher energy state.
Emission spectra result when matter returns from an excited state to the ground state, emitting energy in the form of light radiations.

Electromagnetic Spectrum

The electromagnetic spectrum is the arrangement of electromagnetic radiations based on increasing or decreasing wavelength, energy, or frequency.

Beer-Lambert Law

States that the absorbance of a sample is directly proportional to the concentration and path length of the solution when monochromatic radiations pass through it.
Formula: $A = \epsilon \cdot c \cdot l$
- $A$ = absorbance
- $c$ = concentration of solution
- $l$ = path length
- $\epsilon$ = molar absorptivity
Molar absorptivity ( $\epsilon$ ) is also known as the intensity of absorption or absorptivity coefficient.
Units of molar absorptivity:
- $L \cdot mol^{-1} \cdot cm^{-1}$ (when concentration is in mol/L)
- $L \cdot g^{-1} \cdot cm^{-1}$ (when concentration is in g/L)
Incident radiation: $I_0$
Emitted radiation: $I$

Absorbance and Transmittance

Absorbance (A) is the logarithm of the reciprocal of transmittance (T).
$A = log{10} (1/T) = log{10} (I_0/I)$
Transmittance (T) is the ratio of the intensity of light transmitted by the sample (I) to the intensity of light incident on the sample ( $I_0$ ).
$T = I/I_0$
$\%T = (I/I_0) \times 100$
Thus, $A = \epsilon cl = log(I_0/I)$
The equation shows a linear relation between absorbance and the concentration of absorbing molecules.

Derivation of Beer-Lambert Law

The decrease in the intensity of monochromatic radiations is directly proportional to the intensity of incident radiation (I), path length (dl), and concentration (c).
$dI \propto -I \cdot c \cdot dl$
$dI = -K \cdot I \cdot c \cdot dl$
- where k is the proportionality coefficient.
Integrating the equation over 0 to l:
$\int{I0}^{I} \frac{dI}{I} = -k \cdot c \int_{0}^{l} dl$
$ln \frac{I}{I_0} = -kcl$
Converting to $log{10}$ : $log{10} \frac{I}{I_0} = \frac{-kcl}{2.303}$
Replacing $\frac{k}{2.303}$ with $\epsilon$ : $log{10} \frac{I}{I0} = -\epsilon cl$
$log{10} \frac{I0}{I} = \epsilon cl$
$A = \epsilon cl$

Numerical Example

A compound with a concentration of $10^{-3}$ g/L results in an absorbance value of 0.20 at a wavelength of 510nm using a 1cm cell. Calculate its absorptivity values (Molecular weight of the compound = 400).
Given:
- $c = 10^{-3} g/L$
- $l = 1 cm$
- $A = 0.20$
Using $A = \epsilon cl$
$\epsilon = A/cl = 0.20/(10^{-3} \times 1) = 200 L \cdot g^{-1} \cdot cm^{-1}$
Molar concentration $c_m = \frac{10^{-3}}{400} mol/L = 0.25 \times 10^{-5} mol/L$
Molar absorptivity $\epsilonm = A/(cm \cdot l) = \frac{0.20}{(0.25 \times 10^{-5} \times 1)} = 8 \times 10^{4} mol^{-1} \cdot cm^{-1}$
When UV light is passed through a solution, the radiant power is reduced to 50%. Calculate the absorbance.
Let the radiant power of UV light be $P_0$
The radiant power of emitted light $P = 0.5 P_0$
$A = log(\frac{P0}{P}) = log(\frac{P0}{0.5 P_0}) = log(2) = 0.3010$

Ultraviolet & Visible Spectroscopy

Deals with the interaction of UV-visible radiations with matter.
Radiation range:
- Ultraviolet: 200-400 nm
- Visible: 400-850 nm
UV region divisions:
- Near UV: 400nm - 250nm
- Far UV:
- Vacuum UV region: <190nm
Ordinary UV-visible measurements are carried out from 200-800 nm.
The region below 200 nm is called vacuum UV; measurements are done under vacuum to avoid absorbance signals from air components like $CO_2$
UV-visible spectroscopy is also known as electronic spectroscopy.
Causes electronic transitions in a molecule.
Electrons gain energy and are promoted from HOMO (Highest Occupied Molecular Orbital) to LUMO (Lowest Unoccupied Molecular Orbital).
The HOMO-LUMO gap decreases as conjugation increases.
The absorption of UV-visible rays depends on the type & nature of electrons in a molecule.

Types of Electronic Transitions

Electrons are promoted from bonding orbitals to anti-bonding orbitals.
- $\sigma \to \sigma^*$ transitions:
  - Requires very high energy.
  - Electrons are strongly held.
  - Lies in the vacuum UV region, so it cannot be observed.
  - Example: Alkanes
- $n \to \sigma^*$ transitions:
  - Occur in saturated compounds containing atoms with lone pairs (O, N, S, halogens).
  - Require less energy than $\sigma \to \sigma^*$ transitions.
  - Example: alkyl halides, aldehydes, alcohols.
- $\pi \to \pi^*$ transitions:
  - Occur in unsaturated molecules like alkenes, alkynes, carbonyl compounds, aromatic compounds.
  - Require less energy than $n \to \sigma^*$ transitions.
- $n \to \pi^*$ transitions:
  - An electron from a non-bonding orbital is promoted to an anti-bonding $\pi^*$ orbital.
  - Occur in compounds containing unsaturation and heteroatoms (C=O, C=N, -N=O).
  - Require minimum energy and show absorption at longer wavelengths.
Sequence of energy required for electronic transitions:
- $\sigma \to \sigma^* > n \to \sigma^* > \pi \to \pi^* > n \to \pi^*$
Energy order: $\sigma \to \sigma^* > n \to \sigma^* > \pi \to \pi^* > n \to \pi^*$

Chromophore

Functional groups containing multiple bonds capable of absorbing radiations in the UV-visible range.
They absorb due to $n \to \pi^$ and $\pi \to \pi^$ transitions.
Examples: $NO_2$ , $N=O$ , $C=O$ , $C\equiv N$ , $C=N$ , $C=C$ , $C=S$ .
Compounds containing chromophores are known as chromogens.
Chromophores may or may not impart color to the compound.

Auxochrome

A saturated group that does not act as a chromophore itself, but when attached to a chromophore, shifts absorption maxima towards longer wavelengths and increases the intensity of absorption.
It is known as a color-enhancing group.
Auxochromes contain non-bonding electrons, which help to extend the conjugation of the chromophore.
Examples: $NH_2$ , $OH$ , $OR$ , $COOH$ .

Absorption and Intensity Shift

Addition or removal of auxochromes results in an increase/decrease in the absorption maxima as well as intensity of absorption.

Absorption Shifts

$\lambda_{max}$ changes.
- Bathochromic Shift (Red Shift):
  - When absorption maxima ( $\lambda_{max}$ ) shifts to a longer wavelength.
  - Due to the presence of an auxochrome or change of solvent.
  - Example: The addition of an auxochrome group -NH2 to benzene shifts the absorption ( $\lambda_{max}$ ) from 255nm to 280nm.
  - Example: P-nitrophenol shows bathochromic shift in alkaline medium, because negatively charged oxygen delocalizes more effectively than the unshared pair of electrons.
- Hypsochromic Shift (Blue Shift):
  - When absorption maxima ( $\lambda_{max}$ ) shifts to a shorter wavelength.
  - Due to the removal of an auxochrome (or presence of any group which causes removal of conjugation) or by change of solvent.
  - Example: Aniline shows blue shift in acidic medium as it loses conjugation

Intensity Shifts

Intensity of absorption ( $\epsilon$ ) changes.
- Hyperchromic Shift:
  - When absorption intensity ( $\epsilon$ ) of a compound increases.
  - If an auxochrome is introduced to the compound, the intensity of absorption increases.
  - Example: When -CH3 group is introduced to pyridine, the value of increases from 2750 to 3560, which is due to the induction effect of -CH3 group.
  - -CH3 group results in increase in the electron density which results in increase in value of $\epsilon$ .
- Hypochromic Shift:
  - When absorption intensity ( $\epsilon$ ) of a compound decreases.
  - Caused by the introduction of a group which distorts the chromophore by forcing the rings out of coplanarity resulting in loss of conjugation.

Factors Causing Shifts in UV-Vis Spectroscopy

Conjugation:
- Increase in conjugation decreases the energy differences between HOMO & LUMO.
- Higher the extent of conjugation, the more significant the bathochromic shift.
- Due to increase in conjugation, the electronic transition becomes possible even at lower energy.
- $E = h\nu = h \cdot c/\lambda$
Auxochrome:
- Auxochromes extend conjugation through resonance, resulting in a bathochromic shift.
Effect of solvent:
- Polar Solvent:
  - In polar solvents, $\pi \to \pi^*$ transitions shift towards longer wavelengths (Redshift).
  - Dipole interactions with polar solvent molecules lower the energy of the excited state ( $\pi^*$ ) more than that of the ground state ( $\pi$ ).
  - $\pi^*$ orbitals are stabilized by hydrogen bonding with polar solvent.
- Non-polar solvent
  - In polar solvent, $n \to \pi^*$ transitions shift (Blue Shift)
  - molecules lower the energy of the ground state (n) more than that of the excited state ( $\pi^*$ ). In orbitals are
  - $n \to \pi^*$ orbitals are stabilized by hydrogen bonding with polar solvent).

Instrumentation of UV-Visible Spectroscopy

Light Source:
- Hydrogen and deuterium lamps, Xenon arc lamp, Tungsten halogen lamp.
Wavelength Selector:
- Monochromators are used as wavelength selectors.
- Components: Slit, mirrors, lens, grating/prism.
Sample containers or cells or cuvettes:
- Made of material transparent to UV-visible rays (e.g., quartz).
- Common size: 1 cm (can vary from 0.1 to 10cm).
Detectors:
- Photodetectors, photographic films, human eye.
UV-visible spectrum:
- A graph that shows absorption at different wavelengths.
- Relative maxima are known as $\lambda_{max}$ .

Applications of UV-Visible Spectroscopy

Qualitative and quantitative analysis.
Characterizing aromatic compounds and conjugated olefins.
Finding out molar concentration of the solute under study.
Detecting impurities in organic compounds.
Detection/differentiation of isomers.
Determining the molecular weight of a compound (using Beer-Lambert law).

Numerical Problems to Solve

Identify the chromophoric groups present in cyclopentene, toluene, butanone & methanethiol in U.V. spectroscopy.
- Cyclopentene: $C=C$
- Toluene: Benzene ring
- Butanone: $C=O$
- Methanethiol: $C-S$
Why the $\lambda_{max}$ for the diene (I) is observed at a larger nm than diene (II)?
- Structure II has two exocyclic double bonds.
- Base value of $\lambda_{max}$ for (I) = 214nm.+4x5 = 234nm
- Base value of $\lambda_{max}$ for (II) =217non + 2x5 nm + 2 = 227nm+20=247nm
Can U.V. visible spectral data be useful to distinguish between the following compounds?
1. Ethyl benzene & styrene
2. $CH2=CH-CH2-CH=CH2$ and $CH2=CH-CH=CH-CH_3$
- styrene has extended conjugation compared to ethyl benzene, so it shows absorption at a higher wavelength in UV spectra.
- $CH2=CH-CH=CH-CH2$ has a conjugated system whereas $CH2=CH-CH2-CH=CH_2$ is a non-conjugated di-ene.

Woodward-Fieser Rule

Calculation of $\lambda_{max}$ (nm) in conjugated dienes.
- Base Values :
  - Acyclic or Heteroannular dienes : 214
  - Homoannulas dienes : 253
- Increments :
  - Double bond extending conjugation : 30
  - R alkyl substituent or ring residue : 5
  - Exocyclic double bond : 5
  - Polar Groupings:
    -OCOCH3 : 0
- OR-6
  -cl,-Br. 5
NR₂ 60
Homoannular diene
Heteroannular diene
Exocyclic double bond
Double bond extending conjugation.
Examples:
1. Basic value = 253nm, alkyl/ring residue = 4x5 = 20nm, Exocyclic double bond = 5nm. Calculated = 278nm
2. Basic value = 253nm, Ring residue = 5x5 = 25nm, Exocyclic douver bond =3x5=15nm, Double bond extending cony. =2x30-60nm. Calculated = 353 nm

InfraRed or Vibrational Spectroscopy

When a molecule absorbs electromagnetic radiations in the IR region, it undergoes vibrational and rotational transitions.
Regions- 4000-400 $cm^{-1}$

Criteria

Correct wavelength of radiation, means the natural frequency of vibrations.
Change in the net dipole moment of the molecule should occur.
Homo nuclear diatomic molecules are IR inactive. All the hetero nuclear diatomic molecules are IR active due to presence of permanent dipole moment.

Principle of IR spectroscopy

IR spectroscopy is based on Hooke's law, which says- that suppose two atoms or masses are connected through spring (bond), then the frequency of vibration can be represented by.
$\nu = \frac{1}{2\pi} \sqrt{\frac{K}{\mu}}$
- $\nu$ = frequency
- $K$ = force constant of the bond
- $\mu$ = reduced mass
$\overline{\nu} = \frac{\nu}{c} = \frac{1}{2\pi c} \sqrt{\frac{K}{\mu}}$
- $\frac{K}{\mu} *$ \overline{\nu} $= wave number ($ cm^{-1})

Reduced Mass

Reduced mass is given by:
\mu = \frac{m1 \times m2}{m1 + m2}
- where m1 & m2 are the masses of the atoms.

Numericals based on Hooke's Law

If one of the fundamental vibrational modes of H2O $occurs at 3652$ cm^{-1} $, what would be the new wording vibration frequency of the corresponding mode in$ D2O?
Given wave number for H_2O $= 3652$ cm^{-1}
we'll consider that force constant of H2O & D2O $are same then$ K{H2O} = K{D2O}
{\nu{H2O}} ={\frac{1}{2\pi c} \sqrt {\frac {K{H2O}}{\mu{H2O}}}}
also,{\nu{D2O}} ={\frac{1}{2\pi c} \sqrt {\frac {K{D2O}}{\mu{D2O}}}}
\frac{{\nu{D2O}}}{{\nu{H2O}}} = \sqrt {\frac {\mu{H2O}}{\mu{D2O}}}
Reduced mass for H2O $:$ \mu{H_2O} = \frac{1 \times 16}{1+16} = 0.94
Reduced mass for D2O $:$ \mu{D_2O} = \frac{2 \times 16}{2+16} = 1.78
{\nu{D2O}} ={{\nu{H2O}}} \sqrt {\frac {\mu{H2O}}{\mu{D2O}}} = 3652 \sqrt{\frac{0.94}{1.78}} = 2654 cm^{-1}

Fundamental Vibrations

Arise when a molecule undergoes transition from ground state to first excited state
Linear: 3n-5
Non-Linear Molecules: 3n-6
- n = no. of atoms present in the molecule.
Example: Benzene (Non linear molecule)
- 3 \times 12 - 6 = 30
=3 \times 3 - 5 = 4

Types of Fundamental Vibrations

There are two basic fundamental vibrations observed in the XY2 bent molecule.
- Stretching Vibrations
  - Changes distance between two atoms but the bond angle remains the same
  - Occur at high frequency (4000-1250 cm-1).
- Bending Vibrations
  - Distance between atoms remains constant, but bond angle changes.
  - occur at lower frequency (1400-666 cm-1).

Stretching

Symmetric:
- Both atoms are stretched or compressed.
Asymmetric:
- One atom undergoes stretching, while the other one undergoes compression.

Bending Vibrations

In-plane vibrations:
- The plane of vibrations does not change.
  - Scissoring
  - Rocking
Out-of-plane vibrations:
- Here the plane of vibrations changes (Deformations)
  - Wagging
  - Twisting

Vibrational Modes of CO_2

Symmetric stretch:
- doesn't change dipole moment, so it is IR inactive.
Asymmetric stretch:
- changes dipole moment, so it is IR active.
Bending
changes dipole moment, so it is also IR active
So, CO_2 has 3 IR active vibrational modes.

Interpretation of IR Spectra

Functional Group Region (4000-1500 cm-1):
- Associated with stretching vibrations shown by different functional groups.
- Useful for identification of functional groups in compounds.
Finger Print Region (1500-400 cm-1):
- The number of bending vibrations are more than the number of stretching vibrations.
- Molecules that contain the same functional groups show similar peaks above 1500 cm-1 but different peaks in the finger print region.
- Each functional group has its unique absorption pattern in this region

IR Correlation Chart

lists the types of vibrations and the corresponding frequency range.

Factors Affecting Vibrational Frequency

Conjugation:
- As conjugation increases, stretching frequency decreases because force constant decreases due to conjugation.
Bond Order:
- Higher the bond order, larger is the band frequency.
  - Absorption frequency increases with increasing bandwidth
Hydrogen bonding:
- Intermolecular H-bonding : Decreases position & increase bands frequency or
Ring strain:
- As the ring size decreases, vibrational frequency of C=O increases.
Inductive effect and Resonance effect:
Lengthening of bond, as a result vibrational frequency decreases

Instrumentation of IR Spectroscopy

Radiation Source:
- Nernst glower, Tungsten lamps, Mercury arc.
Sample cells and sampling of substances:
- Solid:
  - Solid films, mull technique is used for preparing solid samples.
- Liquid:
  - Samples can be held as a liquid Sample cell made of alkali halides.
- Gas:
  - Sampling of gas is similar to sampling of liquids.
Monochromators (Wavelength Selectors).
Detectors:
- Thermocouples, Bolometers and pyro electric etc.
Recorders: are used to record IR spectrum.

Applications of IR Spectroscopy

Identification of different functional groups
Distinguish between inter and intramolecular H-bonding:
Identification of purity of the compound:
Study of chemical reaction
Identification Geometric Isomers

Numerical Problems to Solve

Two isomers X and Y having molecular formula C3H6O give IR band near 3550 cm-1 and 1717 cm-1 respectively. Assign structural formula to X and Y consistent with their IR absorption band.
- The peak at 3550cm corresponds to alcoholic group (-OH).
- CH2 = CH-CH2OH
- The peak at 1717 cm^{-1} corresponds to -C=O group stretching frequency
  CH3COCH3
Determine the structure of the compound Cro which shows the following absorption bands in IR spectrum: (1) 2950 cm- and (4) 1720 cm-1. The compound gives negative best with tollen's reagent.
- As the compound (C3H6O $) shows absorption band at 2950$ cm^{-1} $and 1720$ cm^{-1} in IR spectrum. It means, this compound has carbonyl functional group.

Further, the comp gives negative test with tollen's reagent, so it cannot be an aldehyde. Thus, the compound is ketone. It
must be acetone.

A compound having a molecular formula C2H4O2 $while studied for it's IR analysis resulted the following peak in its spectrum: 2900-2950$ cm^{-1} $, 1710$ cm^{-1} $and 3500-3650$ cm^{-1} $. The compound also gave effervescence with$ Na2CO_3. Suggest structure of the compound

C-H stretching of alkyl grip say -CH grip ,-C=0 stretching of c=o grip -
(carboxylic grip),O-|| stretches of carbonylic acids.CH-COOH (Ethanoic acid)-
EtHanoic acid + NaCo3 - followed by +120+CO2$$

How will you distinguish between the following pairs of compounds on the basis of IR spectroscopy?

-Cthy wood shows strong absorption at 2500
3'000S/$ because of O-H bond while
5 does nor now.(ii) G2H12CH shows strong absorption band at
*3