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).
• $A = log<em>{10} (1/T) = log</em>{10} (I_0/I)$
• 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:
• $\int<em>{I</em>0}^{I} \frac{dI}{I} = -k \cdot c \int_{0}^{l} dl$
• $ln \frac{I}{I_0} = -kcl$
• Converting to $log<em>{10}$: $log</em>{10} \frac{I}{I_0} = \frac{-kcl}{2.303}$
• Replacing $\frac{k}{2.303}$ with $\epsilon$: $log<em>{10} \frac{I}{I</em>0} = -\epsilon cl$
• $log<em>{10} \frac{I</em>0}{I} = \epsilon cl$
• $A = \epsilon cl$

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 $\epsilon<em>m = A/(c</em>m \cdot l) = \frac{0.20}{(0.25 \times 10^{-5} \times 1)} = 8 \times 10^{4} mol^{-1} \cdot cm^{-1}$

• 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 $P = 0.5 P_0$

• $A = log(\frac{P<em>0}{P}) = log(\frac{P</em>0}{0.5 P_0}) = log(2) = 0.3010$

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 $n \to \pi^<em>$ and $\pi \to \pi^</em>$ 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

1. Identify the chromophoric groups present in cyclopentene, toluene, butanone & methanethiol in U.V. spectroscopy.
   • Cyclopentene: $C=C$
   • Toluene: Benzene ring
   • Butanone: $C=O$
   • Methanethiol: $C-S$
2. 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
3. Can U.V. visible spectral data be useful to distinguish between the following compounds?
   1. Ethyl benzene & styrene
   2. $CH<em>2=CH-CH</em>2-CH=CH<em>2$ and $CH</em>2=CH-CH=CH-CH_3$
   • styrene has extended conjugation compared to ethyl benzene, so it shows absorption at a higher wavelength in UV spectra.
   • $CH<em>2=CH-CH=CH-CH</em>2$ has a conjugated system whereas $CH<em>2=CH-CH</em>2-CH=CH_2$ 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}$)</li></ul></li> </ul> <h5 id="reducedmass">Reduced Mass</h5> <ul> <li>Reduced mass is given by:</li> <li>$\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$?</li> <li>Given wave number for$H_2O$= 3652$cm^{-1}$</li> <li>we'll consider that force constant of$H2O & D2O$are same then$K{H2O} = K{D2O}$</li> <li>${\nu{H2O}} ={\frac{1}{2\pi c} \sqrt {\frac {K{H2O}}{\mu{H2O}}}}$</li> <li>also,${\nu{D2O}} ={\frac{1}{2\pi c} \sqrt {\frac {K{D2O}}{\mu{D2O}}}}$</li> <li>$\frac{{\nu{D2O}}}{{\nu{H2O}}} = \sqrt {\frac {\mu{H2O}}{\mu{D2O}}}$</li> <li>Reduced mass for$H2O$:$\mu{H_2O} = \frac{1 \times 16}{1+16} = 0.94$</li> <li>Reduced mass for$D2O$:$\mu{D_2O} = \frac{2 \times 16}{2+16} = 1.78$</li> <li>${\nu{D2O}} ={{\nu{H2O}}} \sqrt {\frac {\mu{H2O}}{\mu{D2O}}} = 3652 \sqrt{\frac{0.94}{1.78}} = 2654 cm^{-1}$</li> </ul> <h4 id="fundamentalvibrations">Fundamental Vibrations</h4> <ul> <li>Arise when a molecule undergoes transition from ground state to first excited state</li> <li>Linear:$3n-5$</li> <li>Non-Linear Molecules:$3n-6$<ul> <li>n = no. of atoms present in the molecule.</li></ul></li> <li>Example: Benzene (Non linear molecule)<ul> <li>$3 \times 12 - 6 = 30$</li></ul></li> <li>=$3 \times 3 - 5 = 4$</li> </ul> <h4 id="typesoffundamentalvibrations">Types of Fundamental Vibrations</h4> <ul> <li>There are two basic fundamental vibrations observed in the XY2 bent molecule.<ul> <li>Stretching Vibrations<ul> <li>Changes distance between two atoms but the bond angle remains the same</li> <li>Occur at high frequency (4000-1250 cm-1).</li></ul></li> <li>Bending Vibrations<ul> <li>Distance between atoms remains constant, but bond angle changes.</li> <li>occur at lower frequency (1400-666 cm-1).</li></ul></li></ul></li> </ul> <h5 id="stretching">Stretching</h5> <ul> <li>Symmetric:<ul> <li>Both atoms are stretched or compressed.</li></ul></li> <li>Asymmetric:<ul> <li>One atom undergoes stretching, while the other one undergoes compression.</li></ul></li> </ul> <h5 id="bendingvibrations">Bending Vibrations</h5> <ul> <li>In-plane vibrations:<ul> <li>The plane of vibrations does not change.<ul> <li>Scissoring</li> <li>Rocking</li></ul></li></ul></li> <li>Out-of-plane vibrations:<ul> <li>Here the plane of vibrations changes (Deformations)<ul> <li>Wagging</li> <li>Twisting</li></ul></li></ul></li> </ul> <h4 id="vibrationalmodesofddco_2dd">Vibrational Modes of$CO_2$</h4> <ul> <li>Symmetric stretch:<ul> <li>doesn't change dipole moment, so it is IR inactive.</li></ul></li> <li>Asymmetric stretch:<ul> <li>changes dipole moment, so it is IR active.</li></ul></li> <li>Bending<br /> changes dipole moment, so it is also IR active</li> <li>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

  1. 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$</li> <li>The peak at 1717$cm^{-1}$corresponds to -C=O group stretching frequency<br />$ CH3COCH3 $</li></ul></li> <li><strong>Determine the structure of the compound Cro which shows the following absorption bands in IR spectrum: (1) 2950 cm- and (4) 1720 cm-1. The compound gives negative best with tollen's reagent.</strong><ul> <li>As the compound ($C3H6O$) shows absorption band at 2950$cm^{-1}$and 1720$cm^{-1}$in IR spectrum. It means, this compound has carbonyl functional group.</li></ul></li> </ol> <ul> <li>Further, the comp gives negative test with tollen's reagent, so it cannot be an aldehyde. Thus, the compound is ketone. It<br /> must be acetone.</li> </ul> <ol> <li><strong>A compound having a molecular formula$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$. Suggest structure of the compound</strong></li> </ol> <ul> <li>C-H stretching of alkyl grip say -CH grip ,-C=0 stretching of c=o grip -<br /> (carboxylic grip),O-|| stretches of carbonylic acids.CH-COOH (Ethanoic acid)-</li> <li>EtHanoic acid +$NaCo3$- followed by$+120+CO2$$
     1. 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