HC1_MAP_Intro_UV_FLU_JCR(Merged).pdf

Molecular Analytics Premaster L1 Introduction UV-vis & Fluorescence Jesper C. Ruiter Section BioAnalytical Chemistry Vrije Universiteit Amsterdam [email protected] Friday 8 September 2023 Meet the team Jesper Ruiter, MSc Department Chemistry and Pharmaceutical sciences Section Bioanalytical Chemi...

Molecular Analytics Premaster L1 Introduction UV-vis & Fluorescence Jesper C. Ruiter Section BioAnalytical Chemistry Vrije Universiteit Amsterdam [email protected] Friday 8 September 2023 Meet the team Jesper Ruiter, MSc Department Chemistry and Pharmaceutical sciences Section Bioanalytical Chemistry Prof. dr. Govert Somsen Department Chemistry and Pharmaceutical sciences Section Bioanalytical Chemistry Coordinator 2 Introduction: Approach during my lectures • • Together Active participation☺ • • • • Weekly quizzes Eight in total Count towards your final grade (10%) For you to see if you understand the material Own Exam questions • • Questions Open (just like my door) • • • • Mentimeter Best three will be in the exam☺ Excited! 3 Where are we? L1 L4 L5 L2 T2 T1 L3 4 Contents from book covered in Jesper’s lectures Date 8 Sept Friday Lectures Molecular Analytics Intro + UV-vis + Fluorescence Chapter from Pavia Ch. 10 (10.1 – 10.9 & 10.1410.17) + slides Ch. 2 15 Sept Friday Infrared spectroscopy 22 Sept Friday Mass spectrometry 29 Sept Friday NMR part 1 Ch. 3 (3.3 EI only) From Ch. 4 the material that was covered in the lecture Ch. 5 6 Oct Friday NMR part 2 Ch. 6 (6.1 – 6.5 & 6.10 – 6.13) Book = Pavia, D. L.; Lampman, G. M.; Kriz, G. S.; Vyvyan, J. R. Introduction to Spectroscopy, Fifth edition.; Cengage Learning: Stamford, CT, 2015. 5 Learning objectives of today • Understanding what spectroscopy is and what it is used for • Describe the absorption process in such a manner that your fellow students understand • Knowing what Lambert-Beer’s law is and how to work with it to determine the concentration and molar absorption coefficient • Describe what a chromophore is • Knowing which electronic transitions there are • Understanding what a Jablonski diagram is • Understanding what conjugation is and the effect of it on the λmax • Understanding what vibrational relaxation, internal conversion are • Understanding what fluorescence is • Explaining what Kasha’s rule is 6 From the top: What is spectroscopy? Spectroscopy: study of the absorption and emission of light and other radiation by matter, as related to the dependence of these processes on the wavelength of the radiation 7 Using light-matter interaction in Molecular Analysis Using light to quantitatively measure compounds Examples: Determining the active component in drugs (and quantity) Absorbance of UV light by paracetamol Concentration determination of proteins in a solution by adding Cu(II) Absorbance of VIS light by Cu2+-protein complex O2 saturation of blood (and heart rate Absorbance of light by hemoglobin monitoring) 8 Light (1/3) Electromagnetic Radiation Wave character Light has a wavelength, λ (m) and a frequency ν (Hz of s-) λ= 𝑐 ν ν= 𝑐 λ Speed of light c = 2.998∙108 m/s (in vacuum) 9 Light (2/3) Electromagnetic Radiation Particle character Packages of light (photons) with energy E (J) ℎ𝑐 E = hν = λ Planck’s constant h = 6.626∙10-34 Js Double slit experiment 10 Electromagnetic spectrum Energy UV-light: λ = 10–400 nm Visible light (Vis): λ = 400–800 nm Infrared (IR): λ 3–100 µm 11 Light (3/3) Electromagnetic radiation Colour- or wavelength selection White light is a mix of all colours - Unravel with a prism or a grating - Select using a slit - Grating also works for UV light 12 UV-Vis light • Vacuum UV Middle and near UV Visible (Vis) (10-200 nm) (200-400 nm) (400-800 nm) Very harmful • Colourless • Coloured • Harmful • Harmless Range of a UV-Vis spectrometer (190-800 nm) 13 The electromagnetic spectrum in detail 14 Measuring an absorption spectrum Measuring the light absorbance of a sample lb Cuvette with sample (in solution typically) 15 UV-Vis absorption spectrum Absorption of light (=absorbance) as a function of the wavelength We see peaks at 255 nm and 395 nm. What does that mean? 16 Quantitatively measuring UV-Vis absorbance I measurement Two measurements needed l I0 measurement l Transmittance (T) = the ratio of the amount of light that passes through the solution (I) with respect to the amount of light that passes through the blank (I0) 𝐼 𝑇= 𝐼0 %T = 100T Calculation Imagine I0 = 1000 en I = 500 then T = I/I0 = 500/1000 = 0.5 17 T as function of the concentration Concentration (mg/L) 100 %T 0 100 2 50 4 25 6 12.5 8 6.25 10 3.12 80 60 %T 40 20 0 0 2 4 6 8 10 Concentration concentratie (mg/l) Relation between T and concentration is not linear 18 Defining the unit absorbance Absorbance (A) • Question: If the transmittance is high, is the absorbance then high or low? 𝐼 𝐴 = −𝑙𝑜𝑔𝑇 = −𝑙𝑜𝑔 𝐼0 𝑇 = 10−𝐴 19 Defining the unit absorbance Absorbance (A) • Question: If the transmittance is high, is the absorbance then high or low? 𝐼 𝐴 = −𝑙𝑜𝑔𝑇 = −𝑙𝑜𝑔 𝐼0 𝑇 = 10−𝐴 Calculation Say T = 0.5 what is the value of A? 20 Defining the unit absorbance Absorbance (A) • Question: If the transmittance is high, is the absorbance then high or low? 𝐼 𝐴 = −𝑙𝑜𝑔𝑇 = −𝑙𝑜𝑔 𝐼0 𝑇 = 10−𝐴 Calculation Say T = 0.5 what is the value of A? A = -log(T), so A = -log(0.5) = 0.3 21 Absorbance (A) vs concentration 𝐴 = −𝑙𝑜𝑔𝑇 Concentration (mg/L) %T A 0 100 0.0 2 50 0.3 4 25 0.6 6 12.5 0.9 8 6.25 1.2 10 3.12 1.5 1.5 1.0 A 0.5 0.0 0 2 4 6 8 10 Concentration (mg/l) concentratie Do you notice anything? Relation between A and concentration is linear 22 Lambert-Beer law 23 Lambert-Beer law 𝐴= 𝜀∙𝑐∙𝑙 A = absorbance ε = molar absorption coefficient (L·mol-1·cm-1) bl c = concentration (mol/L or M) l = optical pathlength of cuvette (cm) ε is a constant for a certain sample at a given wavelength An absorption spectrometer measures T and determines A By measuring T (and calculating A) you can determine the c or e of a compound (l = usually 1 cm) 24 Determining ε (1/2) How do we do that?: determining ε of a compound by measuring A of a solution with a known concentration c Question You fill a cuvette (pathlength = 1.00 cm) with a solution of paracetamol with a concentration of 5.00·10-5 mol/L and you measure an absorbance (A) of 0.500 at a wavelength of 240 nm. What is the molar absorption coefficient of paracetamol at 240 nm? 25 Question You fill a cuvette (pathlength = 1.00 cm) with a solution of paracetamol with a concentration of 5.00·10-5 mol/L and you measure an absorbance (A) of 0.500 at a wavelength of 240 nm. What is the molar absorption coefficient of paracetamol at 240 nm? 𝐴= 𝜀∙𝑐∙𝑙 26 Determining ε (2/2) How do we do that?: determining ε of a compound by measuring A of a solution with a known concentration c Question You fill a cuvette (pathlength = 1.00 cm) with a solution of paracetamol with a concentration of 5.00·10-5 mol/L and you measure an absorbance (A) of 0.500 at a wavelength of 240 nm. What is the molar absorption coefficient of paracetamol at 240 nm? 𝐴 = 𝜀 ∙ 𝑐 ∙ 𝑙 → 𝜀240 = 𝐴 0.500 = = 1.00 ∙ 104 𝐿 ∙ 𝑚𝑜𝑙 −1 ∙ 𝑐𝑚−1 −5 𝑐∙𝑙 5.00 ∙ 10 × 1.00 27 Determining an unknown concentration (1/2) How do we do that?: determining the concentration of a compound with known ε in a solution by measuring A of the solution Question You fill a cuvette (pathlength = 1.00 cm) with a solution of paracetamol with an unknown concentration and you measure an absorbance (A) of 0.350 at a wavelength of 240 nm. The ε240 of paracetamol is 1.00x104 L/mol·cm. What is the concentration paracetamol in the solution? 28 Question You fill a cuvette (pathlength = 1.00 cm) with a solution of paracetamol with an unknown concentration and you measure an absorbance (A) of 0.350 at a wavelength of 240 nm. The ε240 of paracetamol is 1.00x104 L/mol·cm. 𝐴= 𝜀∙𝑐∙𝑙 What is the concentration paracetamol in the solution? 29 Determining an unknown concentration (1/2) How do we do that?: determining the concentration of a compound with known ε in a solution by measuring A of the solution Question You fill a cuvette (pathlength = 1.00 cm) with a solution of paracetamol with an unknown concentration and you measure an absorbance (A) of 0.350 at a wavelength of 240 nm. The ε240 of paracetamol is 1.00x104 L/mol·cm. What is the concentration paracetamol in the solution? 𝐴 = 𝜀∙𝑐∙𝑙 →𝑐 = 𝐴 0.350 = = 3.50 × 10−5 𝑚𝑜𝑙 ∙ 𝐿−1 4 𝜀 ∙ 𝑙 1.00 ∙ 10 × 1.00 30 Limits on Absorbance measurements & Beer’s Law: Lambert-Beer law does not ‘’work’’ at very high concentrations: Analyte molecules are no longer isolated from each other l At very high absorption (>2) I is too low to reliably measure At very low absorption (<0,01) it is difficult to distinguish I from I0 Most reliable when A = 0.2-0.8 31 Determining concentrations using spectroscopy The compound needs to absorb either UV or visible light: Reason, look in the literature or measure an absorption spectrum and determine the absorption wavelength Use a solvent that does not absorb light at the absorption wavelength of that compound Determine ε with a reference or construct a calibration curve 32 Determining concentrations using spectroscopy Calibration curve Determine A of a number of solutions with known concentration and construct a calibration curve slope = el Measure A of a unknown solution and determine the concentration using the calibration curve Real samples often show background absorbance Blank measurement: measure absorbance of sample matrix without the analyte Subtract the blank form the measurement value 33 Okay but… bl 34 Tijd voor koffie ☺ 35 Absorption of light by molecules Absorption of UV-Vis photons correlates with an electronic transition in a molecule Energy level diagram • Absorption occurs when the photon energy exactly matches the energy S2 difference (ΔE) between the S1 = (first) excited ground and excited state state with energy E1 S1 • For example, a molecule is irradiated with UV-VIS light and the Energy hν ΔE = E1 – E0 = hν S0 = ground state with energy E0 S0 UV-Vis absorption photon energy matches with the ΔE: • An electron is promoted from an occupied molecular orbital to a higher unoccupied molecular orbital → absorption 36 Electronic transitions -Energy of UV-Vis photons = electronic transitions in molecules -Absorption of photons by our molecule = electron promoted to higher energy state -Change in electron distribution E Electronic transitions in molecules: LUMO Unoccupied Excitation • • • Molecules: molecular orbitals (MO) HOMO: highest occupied molecular orbital LUMO: lowest unoccupied molecular orbital HOMO Valence-electrons Core electrons 37 Ground state and excited states Energy level diagram S1 LUMO Energy E S1 S2 Notice! The direction of HOMO S0 Different options possible for excitation S0 E S2 Excited states have a different electron distribution E • • the arrow stays the same! 38 Chromophore • Chromophore = part of the molecule that is responsible for the absorption of UV-Vis • Wat functional groups absorb in the UV-Vis range? π-bond electrons (double bonds) C=C Absorption C≡C No absorption Hetero-atom with a double bond and free electron pairs C=O C=S C=N C≡N N=O Metal ions (mainly visible light) Fe2+ Zn2+ Mg2+ Cu2+ 39 Type of chemical bonds & absorption • Photochemical reaction: Excitation of electronic transition like σ → σ*, n → σ*, n → ϖ*, ϖ → ϖ* • σ → σ∗ transition • Saturated compounds like alkanes • sigma bonds are very strong, therefore high energy required to transfer electron • n → σ∗ transition • saturated compounds with one hetero atom (oxygen, nitrogen, fluorine, chlorine) • transitions require comparatively less energy than the σ → σ∗ • the energy required for n → σ∗ transition decreases with • • the increase in the size of the halogen atom decrease in electronegativity of the atom • π → π ∗ transition • unsaturated centers like unsaturated hydrocarbons and carbonyl compounds • transitions require comparatively less energy than the n → σ∗ but the n → π∗ transition required the lowest energy (if present) • n → π ∗ transition • unshared pair on a hetero atom is excited to π∗ antibonding orbital • least amount of energy than all types of transition in UV/VIS spectroscopy 40 Molecular orbitals: ethane • Chemical bond: combining atomic orbitals to molecular orbitals 6 x 1s AOs H 2 x 2s AOs C 6 x 2p AOs C lowest unoccupied molecular orbital (LUMO) highest occupied molecular orbital (HOMO) Ethane C-H s-bond (MO) C-C s-bond (MO) σ*C-H LUMO σ*C-C HOMO σC-C σC-H singlet ground state S0 41 Molecular orbitals: excitation to S1 • Electronic excitation: electron excited to a higher, unoccupied molecular Ethane orbital • Absorpbed photon energy matches exactly the energy difference between the S0-S1 state (between the HOMO and LUMO) lowest unoccupied molecular orbital (LUMO) σ*C-H σ*C-C LUMO photon highest occupied molecular orbital (HOMO) HOMO First electronic excited state S1 σC-C σC-H 42 Molecular orbitals: excitation to S2 • Electronic excitation: electron excited to a higher, unoccupied molecular orbital • Absorpbed photon energy matches the energy difference between the S0-Sn Ethane state lowest unoccupied molecular orbital (LUMO) σ*C-H σ*C-C LUMO photon highest occupied molecular orbital (HOMO) HOMO Higher singlet excited state Sn σC-C σC-H 43 ϖ → ϖ* transition: ethene • • • C-H s-bond (MO) C-C s-bond (MO) C-C p-bond (MO) Ethene σ*C-H σ*C-C LUMO π*C-C π → π* HOMO πC-C σC-C σC-H Singlet ground state S0 First singlet electronic excited state S1 44 n → ϖ* transition: formaldehyde Formaldehyde σ*C-H σ*C-O LUMO π*C-O n → π* HOMO nO πC-O σC-O σC-H Singlet ground state S0 First singlet electronic excited state S1 45 Jablonski diagram • • In stead of drawing all the Molecular Orbitals: The total energy of the molecule in the ground state is depicted as one energy level Energy level diagram Absorption spectrum S2 Energy S1 S0 Wavelength (nm) 46 Electronic transitions • • • • • • Absorption of photon: electron promoted to higher energy state Change in electron distribution UV-VIS: change of lone-pair electrons (n-electrons) or double bond electrons (ϖ-electrons) Excitation into anti-bonding orbitals (n → ϖ*, ϖ → ϖ*) Hetero-atoms in molecule (O, N, S, Cl): n → σ* HOMO → LUMO 47 σ* (antibonding) n → σ* π → σ* n → π* σ → π* π → π* π (bonding) σ (bonding) σ → σ* π → π*: 170 nm <λmax< 800 nm Medium-strong n → σ*: λmax< 250 nm Weak n (non-bonding) Occupied in ground state Energy π* (antibonding) Unoccupied in ground state Electronic transitions - absorption n → π*: 240 nm <λmax< 800 nm Weak-medium s → s*: VUV, s → p* deep-UV 48 UV-VIS – experimental observations • Typically we scan from 200-800 nm: • So, we can observe: π–π*, n–π* and n–σ* transitions (why not the σ–σ* ?) Intensity Functional group π–π* medium-strong double bonds n–π* weak-medium hetero atoms with n-electrons and double bonds n–σ* weak hetero atoms with n-electrons • Dominant signatures in UV-Vis spectroscopy: molecules with π electrons (double bonds such as C=C, C=O, etc.) 49 Conjugation (1/3) • Conjugation: alternating C-C en C=C bonds within a molecule • λmax shifts to longer wavelength • Molar absorption coefficient increases λmax 165 nm (no conjugation) 222 nm 180 nm (no conjugation) 256 nm 290 nm 334 nm 364 nm The longer the conjugated system, the higher the absorption wavelength 50 Conjugation (2/3) The energy gap dividing the bonding and antibonding orbitals becomes progressively smaller with increasing conjugation ℎ𝑐 E=h∙ν= λ 51 Conjugation (3/3) • Bla 1,3-butadiene λmax = 222 nm beta-carotene λ max = 450 nm red orange (620-780 nm) (580-620 nm) violet (400-430 nm) yellow (560-580 nm) blue green (430-490 nm) (490-560 nm) 52 color wheel The effect of substituents on λmax of conjugated systems • Effect of substituent on λmax of the ϖ-systeem -alkyl -C=C Red shift λmax ~5 nm ~30 nm Effect of λmax because of donation of n-elektronen to the ϖ-systeem -O-alkyl -S-alkyl -N-(alkyl)2 Red shift λmax ~6 nm ~30 nm ~60 nm 53 The effect of substituents on λmax • Auxochromic group: decrease excitation energy (shift to higher wavelengths) or increase ε • Electron donating groups (EDG) raise the HOMO • Electron withdrawing groups (EWG) lower the LUMO LUMO HOMO 54 Examples zwak p-EDG middel/sterk p-EDG sterk p-EWG sterk p-EDG 55 Jablonski diagram • • In stead of drawing all the Molecular Orbitals: The total energy of the molecule in the ground state is depicted as one energy level Energy level diagram Absorption spectrum S2 Energy S1 S0 Wavelength (nm) 56 Fluorescence • The possible result of absorption 57 What happens after a molecule absorbs light? (1/4) • Where does the absorbed energy go? • after UV-Vis absorption (excitation), most compounds show non-radiative decay to the electronic ground state (S0) VR: Vibrational relaxation = Molecules excited to electronic states VR S1 and S2 rapidly lose any excess vibrational energy and relax to the ground vibrational level of that electronic state. This S2 nonradiational process is termed vibrational relaxation IC VR S1 IC: Internal conversion = describes intermolecular processes that leave the molecule in a lower-energy electronic state without Energy IC IC emission of radiation. Internal conversion is a crossover between two states of the same multiplicity (singlet → singlet or triplet → triplet) VR S0 Vibrational levels 58 Intermezzo: electron spin • Pauli exclusion principle: states that no two electrons in an atom can have the same set of four quantum numbers. • This restriction requires that no more than two electrons occupy an orbital and furthermore the two have opposed spin states (spin up, s = +1/2 spin down, s = -1/2) Equation for multiplicity: 2S + 1 (S = the sum of all electron spins) Singlet: 2*(+1/2 + -1/2) + 1 = 2*(0) + 1 = 1 Triplet: 2*(+1/2 + +1/2) + 1 = 2*(1) + 1 = 3 59 What happens after a molecule absorbs light? (2/4) • Two things can happen Most compounds: non- Some compounds: radiative decay to the Show emission of light electronic ground state Energy level diagram Energy level diagram Energy level diagram S2 S2 S2 S1 S1 S1 hν hν S0 S0 UV-VIS absorbance S0 Non-radiative decay Fluorescence 60 What happens after a molecule absorbs light? (3/4) • Where does the absorbed energy go? • After UV-Vis absorption (excitation), most compounds show non-radiative decay to the electronic ground state (S0) VR Kasha’s rule Excited molecule very quickly decays to lowest vibrational level of S1 state by non-radiative vibrational relaxation (VR) and internal conversion (IC) with a 100% yield S2 IC VR S1 IC VR S0 IC Decay from S1 to S0 state: Non-radiative decay: molecule loses energy via VR and IC Energy is transferred to heat 61 What happens after a molecule absorbs light? (4/4) • Where does the absorbed energy go? • after UV-Vis absorption (excitation), some compounds show emission of light (fluorescence) VR Kasha’s rule Excited molecule very quickly decays to lowest vibrational level of S1 state by non-radiative vibrational relaxation (IVR) and internal conversion (IC) with a 100% yield S2 IC Makes fluorescence selective VR S1 hν S0 hν Decay from S1 to S0 state: Molecule loses energy by emitting a photen Energy is partly transferred to light 62 Fluorescence spectra (emission) - tyrosine S2 S1 S0 63 Fluorescence: excitation and emission • Fluorescence = emission of light after absorption (excitation) Energy level diagram S2 S1 hν Energy of emitted photon cannot be higher than energy of absorbed photon Wavelength of the emission is larger than the wavelength of excitation E = hν = ℎ𝑐 λ λem > λex S0 Fluorescence 64 Fluorescence: excitation and emission • Fluorescence = emission of light after absorption (excitation) Energy level diagram S2 S1 hν Energy of emitted photon cannot be higher than energy of absorbed photon Wavelength of the emission is larger than the wavelength of excitation E = hν = ℎ𝑐 λ λem > λex ΔE of the emission is smaller (or equal) to the ΔE of the excitation S0 From the formula E = Fluorescence ℎ𝑐 follows (since h λ and c are constants) That λ must be larger 65 Absorption and fluorescence Quinine in Tonic UV Flu Quinine Excitation spectrum (absorbance) Emission spectrum (fluorescence) 66 Fluorescent molecules -Electronic states: UV-VIS region of EM spectrum: π-electrons -Only a few compounds can fluoresce -Typical molecules that have a planar and conjugated structure -Examples: tryptophane, pyrene, naproxen, etc. 67 Measuring fluorescence Light of emission goes in all directions Fluorescence is measured at a 90o angle with respect to the excitation light Measure light against a fully dark background: can be performed very sensitively! Iflu: Intensity of the measured fluorescence signal For low concentrations: 𝐼𝑓𝑙𝑢 = 2,3𝑘Φ𝐼0 𝜀𝜆 𝑙𝑐 k: Instrumental constant Φ: Quantum yield Φ= number of photons emitted number of photons absorbed I0: Intensity of the light source Fluorescence signal is directly proportional to the concentration 0<Φ<1 68 Jablonski diagram 69 Jablonski diagram Abs: S0 → S2 of S0 → S1 IC: S2 → S1 Flu: S1 → S0 ISC: S1 → T1 Phos: T1 → S0 Will be covered in the course Biomolecular spectroscopy 70 Take home message You know the most important points of this lecture if you can answer the following questions: • What is spectroscopy and name a few examples of what it can be used for? • What is absorbance? • What is Lambert-Beer’s law? How can you use it to calculate unknown concentrations and molar absorption coefficients? • What is a chromophore? • Which electronic transitions are there? • What is a Jablonski diagram? Can you draw one yourself showing absorbance and fluorescence? • What is conjugation and what happens when a molecule contains a longer conjugated system? • What are vibrational relaxation and internal conversion? • What is fluorescence? • What is Kasha’s rule? 71 Next lecture Friday 15 September at 13:30-15:15 ‘’Infrared spectroscopy’’ NU-4C07 Check the student lecture assignment on Canvas 72 End of today Questions? 73 Molecular Analytics Premaster L2 Infrared spectroscopy Jesper C. Ruiter Section BioAnalytical Chemistry Vrije Universiteit Amsterdam [email protected] Friday 15 September 2023 Previously on MAP • What is spectroscopy and name a few examples of what it can be used for? • What is absorbance? • What is Lambert-Beer’s law? How can you use it to calculate unknown concentrations and molecular absorption coefficients? • What is a chromophore? • Which electronic transitions are there? • What is a Jablonski diagram? Can you draw one yourself showing absorbance and fluorescence? • What is conjugation and what happens when a molecule contains a longer conjugated system? • What are vibrational relaxation and internal conversion? • What is fluorescence? • What is Kasha’s rule? 2 Where are we? L1 L4 L5 L2 T2 T1 L3 3 Check-in 4 Learning objectives of today • Understanding what a molecular vibration is • Determining the number of vibrational modes of a molecule • Understanding when a molecule absorbs IR radiation • Calculating the vibrational frequency • Become familiar with a number of group frequencies and recognising these in IR-spectra for spectral assignment 5 The electromagnetic spectrum in detail 6 The IR part of the EM spectrum IR regions (+/-) Near infrared (NIR): λ = 0.8–2.5 µm Mid infrared (MIR/mid-IR): λ = 2.5–~15 µm Far infrared (FIR/Far-IR): λ = ~15–1000 µm 7 Wavenumbers 8 Wavenumbers • IR spectroscopy: Energy is expressed in wavenumbers (cm-1) • Can be calculated from the reciprocal of the wavelength (λ) in cm (1) • Wavenumber (𝑣) ҧ convert to a frequency (𝑣)? → multiply with the speed of light (c) in cm/sec (2) 𝑣ҧ (𝑐𝑚−1 ) 1 = λ (𝑐𝑚) (1) 𝑐 (𝑐𝑚Τsec) 𝑣 𝐻𝑧 = 𝑣ҧ ∙ 𝑐 = λ (𝑐𝑚) (2) 9 Infrared spectroscopy Energy of mid-IR photons correspond to vibrational transitions Energy level diagram v3 v2 v1 v0 S1 Vibrational levels hν Energy E1 This energy is now in the order of ….. cm-1 (or micron) E = E1 - E0 = h Energy S2 Excited vibrational state with energy E1 E0 IR absorbance S0 v3 v2 v1 v0 Vibrational transition Vibrational levels Vibrational ground state with energy E0 IR-absorbance Molecule vibrates with larger amplitude • • Vibrational frequencies stay the same 10 What is a molecular vibration? Molecular vibration Lengthening and compression of a covalent bond 11 IR-spectrum of 3-methyl-2-butanone with a C=O stretch vibration ~1750 cm-1 Types of vibrations Stretching vibrations symmetric Bending / out-of-plane vibrations asymmetric scissoring rocking Bond length increases/decreases around center of gravity wagging twisting Bond bends around central position; bond length does not change • Choice is huge • Which IR vibrations do we observe for a molecule? 12 Vibrational modes (1/2) • A molecule with N atoms has 3N-6 vibrational modes • A linear molecule with N atoms has 3N-5 vibrational modes Question How many vibrational modes does H2O have? 13 Vibrational modes (1/2) • A molecule with N atoms has 3N-6 vibrational modes • A linear molecule with N atoms has 3N-5 vibrational modes Question How many vibrational modes does H2O have? • H2O is not linear, so we use 3N-6 • H2O has (3 * 3) – 6 = 3 vibrational modes Symmetric stretch 3652 cm-1 Bending 1596 cm-1 Asymmetric stretch 3756 cm-1 14 Vibrational modes (2/2) • A molecule with N atoms has 3N-6 vibrational modes • A linear molecule with N atoms has 3N-5 vibrational modes Question How many vibrational modes does HCl have? 15 Vibrational modes (2/2) • A molecule with N atoms has 3N-6 vibrational modes • A linear molecule with N atoms has 3N-5 vibrational modes Question How many vibrational modes does HCl have? • HCl is a linear molecule, so we use 3N-5 • HCl has (3 * 2) – 5 = 1 vibrational modes stretch 2886 cm-1 H Cl 16 IR-absorbance When do molecules absorb IR radiation? Two requirements 1: When frequency of IR radiation equals the frequency of a vibration 2: When the magnitude and/or direction of the electrical dipole moment changes Dipole moment: µe = qr v3 v2 S0 q: charge r: distance r v1 v0 δ- δ+ The IR absorbance spectrum of a molecule can, in principle, show all vibrations that obey these requirements δ- δ+ δ- δ+ r r µe changes during stretch vibration 17 Vibrations of diatomic molecules HCl shows 1 vibration • Number of vibrations: 3N-5 = (3x2) - 5 = 1 • HCl and N2 both could have potentially 1 stretch vibration δ- δ+ H stretch 2886 cm-1 Cl N2 does not absorb IR radiation r δ- δ+ r δ- δ+ µe changes upon stretching vibration r r µe does not change during N stretch 2744 cm-1 N r stretching vibration r 18 Example: IR-absorbance of CO2 (1/2) Question How many vibrational modes does CO2 have? CO2 is linear, so 3N-5 (3 * 3) – 5 = 4 vibrational modes 19 Example: IR-absorbance of CO2 (2/2) Symmetric stretch (1340 cm-1)  µe =0, does not change δ- δ- 2δ+ δ- δ- 2δ+ δ- 2δ+ δ- 2δ+ δ- Asymmetric stretch (2350 cm-1)  µe changes δ- δ- 2δ+ δ- 2δ+ δ- δ- 2 bending –in and out-of-plane (with the same frequency 666 cm-1)  µe changes δ- δ2δ+ δ- 2δ+ δ20 2δ+ δ- δ- How to describe a molecular vibration • Mechanical model of a stretching vibration • Two masses connected by a spring • Harmonic oscillator Vibrational Frequency • Motion of mass as a function of time (Hooke’s law) 𝐹 = 𝑚𝑎 • Energy can be obtained via integration (potential energy) 1 𝑑𝑉 1 2 𝑉 𝑟 = 𝑘 𝑟 − 𝑟𝑒 2 𝐹= 𝑉 𝑥 = 𝑘𝑥 2 𝑑𝑥 2 • Parabolic function: Potential energy diagram • k (spring constant) → bond strength • Large: strong bond, steep parabola • Small: weaker bond, wider parabola (easy to expand & compress, large x) 21 How to describe a molecular vibration • Mechanical model of a stretching vibration • Two masses connected by a spring • Harmonic oscillator Vibrational Frequency • Motion of mass as a function of time (Hooke’s law) 𝐹 = 𝑚𝑎 • Energy can be obtained via integration (potential energy) 1 𝑑𝑉 1 2 𝑉 𝑟 = 𝑘 𝑟 − 𝑟𝑒 2 𝐹= 𝑉 𝑥 = 𝑘𝑥 2 𝑑𝑥 2 • Parabolic function: Potential energy diagram • k (spring constant) → bond strength • Large: strong bond, steep parabola • Small: weaker bond, wider parabola (easy to expand & compress, large x) 22 Modelling vibrations: Harmonic oscillator • Consider a chemical bond to be a spring which connects two masses (atoms) • Pulling and releasing the spring causes oscillation Frequency  of harmonic oscillator (classical model) 1 = 2 k  m1m2 = m1 + m2 : frequency (Hz) k: force constant (N/m) µ: reduced mass (kg) m1: mass atom 1 (kg) Other frequencies are not allowed m2: mass atom 2 (kg) 23 Frequency/wavenumber of diatomic stretch vibration 1 = 2 k  =  •  =  1 k  = 2c  c Larger bond strength (k) causes higher frequency • • 1 Lighter atoms cause a higher frequency When atom 1 is much larger than atom 2, the frequency is mainly determined by atom 2 m1>>m2 : 1 m1m2 m1m2 = =  = m2 2 m1 + m2 m1 k m2 24 Vibration frequency of NO • The bond strength of NO is 1590 N/m • 1: Calculate the vibrational frequency (ν) of the stretch vibration (in Hz) van NO • 2: Does NO show a peak in its IR-spectrum? (why yes/no?) 1 = 2 k  = m1m2 m1 + m2 • First try yourself (2 min) • Discuss with your neighbours (2 min) • Discussion together ☺  = 1  =  c 1 k  = 2c  25 Vibration frequency of NO - 1 • The bond strength of NO is 1590 N/m • 1: Calculate the vibrational frequency (ν) of the stretch vibration (in Hz) van NO • 2: Does NO show a peak in its IR-spectrum? (why yes/no?) 1 = 2 k  mm = 1 2 m1 + m2  = 1  =  c 1 k  = 2c  𝑚𝑁 = 14.01 × 1.66 ∙ 10−27 = 2.33 ∙ 10−26 𝑘𝑔 𝑚𝑂 = 16.00 × 1.66 ∙ 10−27 = 2.66 ∙ 10−26 𝑘𝑔 (2.33 ∙ 10−26 × 2.66 ∙ 10−26 ) 𝜇= = 1.24 ∙ 10−26 𝑘𝑔 −26 −26 (2.33 ∙ 10 + 2.66 ∙ 10 ) ν= 1 1590 × = 5.70 ∙ 1013 𝐻𝑧 −26 2𝜋 1.24 ∙ 10 26 Vibration frequency of NO - 2 • The bond strength of NO is 1590 N/m • 1: Calculate the vibrational frequency (ν) of the stretch vibration (in Hz) van NO • 2: Does NO show a peak in its IR-spectrum? (why yes/no?) • Answer 2 • Yes, the dipole moment will change upon NO-vibration δ+ δ- N O ν 5.70 ∙ 1013 𝐻𝑧 ν = νത ∙ 𝑐 ⇒ νത = = = 1900 𝑐𝑚−1 10 𝑐 3.00 ∙ 10 27 Vibrational ground state (1/2) • • • At room temperature most molecules are in the vibrational ground state (ν0) Boltzmann distribution 𝑁1 − =𝑒 𝑁0 ∆𝐸 𝑘𝐵 𝑇 N1: number of molecules in v1 Question: What percentage of the molecules at room temperature is located in the first excited vibrational state • N0: number of molecules in v0 ΔE: E1-E0 (J) kB: Boltzmann constant (1.38x10-23 J/K) T: temperature (K) Calculation ∆𝐸 𝑖𝑛 𝑚𝑖𝑑 𝐼𝑅 𝑖𝑠 𝑏𝑖𝑗𝑣. λ = 5.00 𝜇𝑚 = 5.00 ∙ 10−6 𝑚 𝑐 3.00 ∙ 108 𝑚/𝑠 ν= = = 6.00 ∙ 1013 𝐻𝑧 −6 λ 5.00 ∙ 10 𝑚 ∆𝐸 = ℎν = 6.626 ∙ 10−34 × 6.00 ∙ 1013 = 3.98 ∙ 10−20 𝐽 𝑁1 − =𝑒 𝑁0 3.98∙10−20 1.38∙10−23 ×298 = 0.000063 = 0.0063% 28 Vibrational ground state (2/2) • v3 When absorbing mid-IR: • The molecule stays in the electronic ground state (S0) • The molecule transfers from the vibrational ground state v2 S1 v0 (v0) into the first excited vibrational state (v1) • Transition from the v0 to the second excited vibrational state (v2) is forbidden, resulting in a very weak v1 v3 v2 S0 v1 v0 absorbance 29 What about polyatomic molecules? • Non-linear molecules with N atoms have 3N-6 vibrations: • The number of stretch vibrations = the number of bonds • The remaining vibrations involve bending • Example: ethyl acetate (C4H8O2, N = 14) • Number of vibrations = (3 * 14) – 6 = 36, of which 13 stretch vibrations • IR absorption spectrum of a chemical compound often shows large number of absorption 30 IR transmission spectrum • Y-axis: transmission • X-axis: cm-1 • • From high to low Stretch frequencies in entire spectrum • Bending frequencies always <1500 cm-1 • Ethyl acetate Absorbances > 1500 cm-1 are caused by stretch vibrations 31 Group frequencies for general assignment 32 Group frequencies • Vibrational frequency of two atoms in a molecule is mainly determined by their bond, and slightly by other parts • Functional groups absorb in specific region of IR spectrum • These group frequencies are valid for all compounds • For example • C-H stretch: 3300 – 2850 cm-1 • C=O stretch: 1850 – 1550 cm-1 • The exact position of the vibrational frequency depends on the other bonds present For example • C-H (sp3) stretch: 2950 – 2850 cm-1 • C-H (sp2) stretch: 3100 – 3010 cm-1 • C-H (sp) stretch: ~3300 cm-1 33 IR spectrum of ethyl acetate • fingerprint area IR- C-H C-O Ethyl acetate C=O 34 How do we measure IR-spectra? (1/2) • Dispersive IR spectrometer 35 How do we measure IR-spectra? (2/2) • Fourier transform IR (most modern IR spectrometers operate on this principle) 36 FTIR example: styrene IR-spectrum Interferogram Fourier transform 37 Use of IR-spectroscopy • Structure elucidation: spectral interpretation and assignment of functional groups • Comparison with reference spectra→ identification of compounds Possible database matches identified by metabolic screening 38 IR-table 39 IR-table 40 Your turn! 41 Coffee break ☺ 42 Alkanes, Alkenes, and Alkynes – Group 1 Alkanes: - C-H stretch near 3000 cm-1 - CH2 bending absorption around 1465 cm-1 - CH3 bending absorption around 1375 cm-1 - When CH2 ≥ 4, bending motion occurs around 720 cm-1 - C-C many weak peaks Alkenes: - =C-H stretch for sp2 C-H from 3095 – 3010 cm-1 - =C-H out-of-plane bending between 1000 – 650 cm-1 - C=C stretch between 1660 – 1600 cm-1 Alkynes: - ≡C-H stretch for sp C-H near 3300 cm-1 - C≡C stretch near 2150 cm-1 - Disubstituted or symmetrically substituted triple bonds give either no or weak absorption Benzene and Aromatic Rings – Group 2 List here the relevant bonds of the functional group with the following information: - Only sp2 hybridized orbitals. C=C conjugated bonds. A benzene to benzene substitution bond. And sp2 H-C= bonds. - Both bending and stretching of C-H bonds and stretching of C=C bonds - 3000 cm-1 sp2 stretch. 2000-1660 cm-1 , C=C stretch 1500-1400 cm-1 , C-H 750-700 cm-1 oop bending 1,1'-Biphenyl Mono-substitution overtones C-H out-ofplane bending vibrations for substituted benzenoid compounds. No broad sp3 peak =C-H sp2 stretch Aromatic conjugated C=C stretch Benzene mono substitutions, C-H Outof-plane bending 44 Alcohols and phenols – Group 3 - O–H stretch, 3600-3200 cm-1 C–O stretch, 1260-1050 cm-1 Aliphatic alcohol O-H stretch C-O stretch Phenol C-O stretch O-H stretch 45 Ethers – Group 4 - C-O Bond Stretch vibration 1000-1300 cm-1 46 Aldehyde – Group 5 •C-H •C=O •Both bonds show stretching vibration •C-H 2695 – 2830 cm^-1​ •C=O •1720 – 1740 cm^-1 (Aliphatic) •1685 – 1710 cm^-1 (unsaturated) C-H C=O 47 Ketone – Group 6 C=O stretching 1750cm-1 C-H bending 2850cm-1 H stretching (cm-1) 48 Carboxylic acids: Hexanoic Acid – Group 7 Relevant bonds of hexanoic acid Bond Type of vibration Wavenumbers (cm-1) O-H Stretch 2500 – 3300 C=O Stretch 1721 O-H Bend 1419 C-O Stretch 1296 O-H Bend 948 ‘Why did the carboxylic acid go to therapy?’ Because it had too many attachment issues with its hydrogen atoms! 49 Orgchemboulder. IR: carboxylic acids (orgchemboulder.com), accessed on 13-09-2023 Esters (Ethyl acetate) – Group 8 Wave number Functional (cm-1) group Stretch/bend 3000-2840 C-H Stretch 1755-1735 C=O Stretch 1100-1300 C-O Stretch 50 Amides – Group 9 - - Primary amide (-NH2) - C=O stretching 1690 cm-1 - N-H stretching 3500 cm-1 Secondary amide (-NHR) - C=O stretching 1680 cm-1 Tertiary amide (-NRR’) - C=O stretching 1680 cm-1 51 Acid chlorides: Group 10 Acid chlorides are able to make anhydrides and can be used to synthesize most of the carbonyl functional groups. - C=O bond (1815-1785 cm-1, stretch) - C-Cl bond (730-550 cm-1, stretch) 52 Anhydrides – Group 11 List here the relevant bonds of the functional group with the following information: - C=O and C-O - Stretch (symmetric and asymmetric) - Wavenumbers (1830-1800cm-1 and 1775-1740cm-1)(1300-1900cm-1) Insert example IR-spectrum here Indicate which peak corresponds with which bond using arrows 53 Amines – Group 12 - N-H stretch at 3500-3300 cm-1 N-H bend at 1640-1560 cm-1 C-N stretch 1250-1000 cm-1 N-H out-of-plane bending at 800 cm-1 N-H stretch N-H bend C-N stretch N-H Oop bending 54 Source: https://www.chemicalbook.com/SpectrumEN_75-64-9_IR2.htm Nitriles – Group 13 List here the relevant bonds of the functional group with the following information: - Nitrile - Stretch - ~ 2250 cm⁻¹ 55 IR-spectral assignment tips • Finding group frequencies (wavenumbers) / diagnostic signatures • Start in the 4000-3000 cm-1 range • OH stretch and NH stretch very diagnostic (broader, very intense) • Double/triple bond stretch • C=O present? Check 1820-1660 cm-1 depending on where the peak is observed you can get an indication which functional group is present in the molecule 56 Take home message You know the most important points of this lecture if you can answer the following questions: • What is a molecular vibration? • How do you determine the number of vibrational modes of a (non)-linear molecule? • When do molecules absorb IR radiation? • How do you calculate the vibrational frequency of, for example, C=O? • Can you identify the groups discussed in this lecture in IR spectra? 57 Next lecture Friday 22 September at 13:30–15:15 ‘’Mass spectrometry’’ NU-4C07 58 End of today Questions? 59 Hexaan 1 Hexane • Bekijk dit spectrum • Welke kenmerkende pieken zie je? 60 Hexaan 2 61 Hexanol 1 Hexanol • Bekijk dit spectrum • Welke pieken wordt veroorzaakt door de OH groep? 62 Hexanol 2 63 Hexanol 3 64 3-heptanon 1 3-heptanon • Bekijk dit spectrum • Welke piek wordt veroorzaakt door de keton groep? 65 3-heptanon 2 66 Heptanal 1 heptanal • Bekijk dit spectrum • Welke pieken worden veroorzaakt door de aldehyde groep? 67 Heptanal 2 68 Heptanal 3 69 Hex-1-een 1 hex-1-een • Bekijk dit spectrum • Welke pieken worden veroorzaakt door de C=C binding? 70 Hexaan 71 Hex-1-een 2 72 Hex-1-een 3 73 1-Heptyn 1 1-heptyn • Bekijk dit spectrum • Welke pieken worden veroorzaakt door de drievoudige binding? 74 1-Heptyn 2 75 1-Heptyn 3 76 IR-regions 77 IR-regions 78 IR-regions 79 IR-regions 80