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Physical Chemistry II CHEM 3340 Molecular Spectroscopy: Magnetic Resonance Magnetic Resonance Transition between energy levels in the presence of magnetic eld Small energies are involved: radio (mostly) and microwaves Related to the interaction of magnetic eld with spin angular momentum fi One of th...
Physical Chemistry II CHEM 3340 Molecular Spectroscopy: Magnetic Resonance Magnetic Resonance Transition between energy levels in the presence of magnetic eld Small energies are involved: radio (mostly) and microwaves Related to the interaction of magnetic eld with spin angular momentum fi One of the most commonly used spectroscopic technique fi Spin Inherent property of subatomic particles Spin quantum number, I Electron, neutron, proton: I = 1/2 Magnitude of angular momentum: [I(I+1)]0.5 ħ A component of angular momentum on a speci ed axis (’z’ axis): mI ħ, mI=I, I-1,...., -I 2I+1 different orientations fi I=1/2 Nuclear Spin Quantum Number I cannot be negative! I =0: 16O, 32S, 12C I=1/2: 1H, 13C, 19F, 31P I=1: 14N, 2H should be able to tell if the atom is goin to show up on NMR or not and calculate interger For magnetic resonance, I> 0 Nuclei in magnetic elds The energy of this interaction depends: ◊ the extent of magnetic eld: B0 ◊ magnetic quantum number, mI ◊ type of nucleus E = - γ ħ mI B0 γ: nuclear magnetogyric ratio of a nucleus: should be experimentally determined Magnetic momentum around z axis: μz E = - μz B0 μz = γ ħ mI fi fi Particles with nonzero spin interact with magnetic eld fi Magnetogyric Ratio γ is usually expressed in terms of nuclear magneton Nuclear magneton: μN = eħ/(2mp) = 5.051×10-27 J/T where mp is the mass of proton γħ is on the order of the nuclear magneton: γħ =gI μN where gI is the g-factor for the nucleus Energy with the g factor E = - gI μN mI B0 Nuclear Spin Properties g is usually > 0: magnetic moment and spin are parallel in rare cases g< 0 It is assumed in further treatment: g>0 Electron in magnetic eld Bohr magneton: μB = (eħ/2me) = 9.274×10-24 J/T The g value of electron: ge = 2.0023 (comes from Dirac’s relativistic theory and an additional correction) Energy of interaction between electron and magnetic eld E = ge μB mI B0 Note: there is no - sign because of the negative charge of electron Because μB >>> μN, the energies related to spin nuclei are a lot smaller then the energies related to electron spin (because the electron is so light) fi It is better to replace the mass of proton with the mass of electron in the nuclear magneton fi Magnetic Resonance Spectroscopy Example: electron, I= 1/2 mI = ms = +1/2, -1/2 Energies: E = ge μB ms B0 Energy difference: ΔE = E(↑) - E(↓) = ge μB B0 (0.5 - -0.5) = ge μB B0 Corresponding electromagnetic radiation: hν = ge μB B0 Electron spin resonance: ESR with B0 =0.3 T, ν = 10 GHz, λ = 3 cm, -> Microwave Nuclear Magnetic Resonance Example: 1H, I=1/2 mI=+1/2, -1/2 Energy levels E = - γ ħ mI B0 Energy separation: ΔE = E(↓) - E(↑) = γ ħ B0 Absorbed electromagnetic radiation: hν = γ ħ B0 Larmor frequency: ν = γ ħ B0/h With typical elds: B0 = 12 T fi ν = 500 MHz -> radiowave Intensity of NMR lines Intensity is proportional to I ∝ (Nα - Nβ)B0 Nα: number of molecules in the low energy state Nβ: number of molecules in the high energy state B0: strength of magnetic eld Nα and Nβ can be calculated for a population of N particles (N = Nα +Nβ) from the Boltzmann distribution knowing the energy difference between the two states ΔE = γ ħ B0, and temperature (T) N⇥ =e N N N⇥ N N = =1 E =1 kT B0 kT N N⇥ N (1 N⇥ /N ) 1 (1 = = N + N⇥ N (1 + N⇥ /N ) 1 + (1 N⇥ N E/(kT ) B0 = 2kT N fi B0 /kT ) B0 /kT ) N B0 N⇥ = 2kT Intensity of NMR lines N N B0 N⇥ = 2kT For typical values, the two populations have similar numbers For electron, B0 = 1T, T=300K: (Nα - Nβ)/N ≈ 0.0022 The population sizes are similar, the intensity of spectral lines is very weak Solution: I ∝ (Nα - Nβ)B0 , Nα - Nβ ∝ B0 I ∝B02 fi fi ◊ The intensity is proportional to B02 ◊ High magnetic eld can increase line intensities ◊ They use superconducting magnet to produce intense, uniform magnetic eld (10 T or more) The shielding constant Not all nuclei (e.g., 1H) have the same ∆E (and thus ν) in the spectrum Each proton feels local magnetic eld (Bloc), which is different from the applied B0 Bloc = B0 - σB0=B0 (1-σ) Shielding constant, σ: usually positive, but can be negative The local magnetic eld is affected by the induced motion of electrons in the molecule; this motion generates circulating currents and thus magnetic eld that typically weakens the applied eld Nuclei in different chemical groups have different shielding constant Resonance frequency fi ν = Bloc γ/(2π) = (1-σ) B0 γ/(2π) fi fi fi ν = γ ħ B0/h= γ B0/(2π) Chemical Shift NMR spectroscopy does not report the absorption frequencies ν and the corresponding B0 Frequencies are very close to each other -> the change is important from a reference material TMS (Si(CH3)4): ν0 TMS is strongly shielded -> ν-ν0 is usually positive To eliminate B0 dependency, this difference is divided by ν0 (this values is really small, ν-ν0 is usually about 100 Hz, ν0 is about 500 MHz ) Chemical Shift, δ = (ν-ν0) / ν0 x 106 Typical chemical shifts: proton: 0-14 The Chemical Shifts of Proton When H is connected to electronegative atom: the H is deshielded δ is large Different groups have different chemical shifts NMR spectrum of ethanol δ increases from right to left There are three sets of lines> three different sets of H atoms The intensity of the lines correspond to the number of H atoms: 1: 2: 3 The positions of the lines correspond to the different groups Contributions to Shielding Constant σ = σ(local) + σ(neighbor) + σ(solvent) Three contributions: ◊ Local: contributions of the electrons of the atom that contains the nucleus ◊ Neighbor: contributions from groups of atoms from rest of the molecule ◊ Solvent: from solvent molecules Local contributions to shielding constant Approximately proportional to the electron density of the atom containing the nucleus Electronegativity effect: the electron density can decrease due to the presence of a neighboring element -> decreased shielding -> increased chemical shift Special effect: paramagnetic contribution The applied eld can force electrons to effectively circulate on ‘empty’ molecular orbitals fi The electronegativity trend holds for neighboring H, but fails for H in CH3 Neighboring group contributions Due to currents induced in nearby atoms Ring current: in aromatic substances magnetic eld causes circulation of electrons ◊ Atoms in the plane: deshielded (δ is large) ◊ Atoms above or below the plane: shielded (δ is small) fi Solvent contribution Molecule interacts with solvent: hydrogen bond formation, steric interactions, etc. Example: benzene ring of a solvent can affect the shielding/deshielding These interactions can help even further differentiate among protons in the molecules: proper choice of solution is essential Fine Structure fi Why do we have so many lines in the spectrum? Fine structure: magnetic nucleus may contribute to the local eld experienced by the other nuclei Spin-spin coupling: there is an interaction energy between the spins of nucleus A and X E = hJ mA mX mA, mX: magnetic spin quantum number of the two nuclei J: coupling constant [Hz], does NOT depend on magnetic eld fi The Fine Structure of an A-X spin system Assumption: the chemical shifts of A and X nucleus are very different E = - γ ħ mI B0(1-σX) A A X X E = - γ ħ mI B0(1-σA) βA E = 0.5 γ ħ B0(1-σA) E = 0.5 γ ħ B0(1-σX) βX ∆E = γ ħ B0(1-σX) ∆E = γ ħ B0(1-σA) αA E = -0.5 γ ħ B0(1-σA) αX E = -0.5 γ ħ B0(1-σX) A-X spin system without spin-spin coupling mA mX -0.5 -0.5 -0.5 0.5 0.5 -0.5 0.5 0.5 A X E = - γ ħ mA B0(1-σA) - γ ħ mX B0(1-σX) E = 0.5 γ ħB0(1-σA) + 0.5 γ ħ B0(1-σX) E = 0.5 γ ħB0(1-σA) - 0.5 γ ħ B0(1-σX) Transitions for nucleus A ∆E = γ ħ B0(1-σA) E = - 0.5 γ ħB0(1-σA) + 0.5 γ ħ B0(1-σX) ∆E = γ ħ B0(1-σA) E = - 0.5 γ ħB0(1-σA) - 0.5 γ ħ B0(1-σX) Without spin-spin coupling the resonances do not change A-X spin system with spin-spin coupling A X E = - γ ħ mA B0(1-σA) + - γ ħ mX B0(1-σX) + mA, mB -0.5, -0.5 hJmA mX Transitions for nucleus A ∆E = γ ħ B0(1-σA)+0.5 hJ -0.5, 0.5 0.5, -0.5 ∆E = γ ħ B0(1-σA) - 0.5 hJ 0.5, 0.5 With spin-spin coupling: two lines instead of one with difference in energy: hJ Corresponding frequency: hJ/h = J A-X spin system: chemical shifts of nucleus A There are two lines instead of one One line: J/2 Hz lower frequencies Second line: J/2 Hz higher frequency J value is independent of the magnetic eld fi A-X spin system with spin-spin coupling Transitions for nucleus X A X Similar argument Energy difference without spin-spin coupling ∆E = γ ħ B0(1-σX) Energy difference with spin-spin coupling ∆E = γ ħ B0(1-σX) + 0.5 hJ ∆E = γ ħ B0(1-σX) - 0.5 hJ With spin-spin coupling: two lines instead of one with frequency difference J Fine structure of AX2 spin system Resonance frequency of X is easier to discuss If the two X are chemically (and magnetically) equivalent, they do not couple (JX-X =0) CH3 - CH2 -OH X2 A group of equivalent nuclei resonates like a single nucleus AX2 Doublet-> two lines Intensity doubled because of two nucleus Fine structure of AX2 spin system Resonant frequency of A The line splits twice by frequency amount J Intensity ratio: 1:2:1 Fine Structure of AX3 spin system Splitting of the lines of A Intensity ratio: 1:3:3:1 Distance between lines: J A2X3 would be the same but with doubled intensity Fine Structure of AXN spin system Pascal Triangle Intensities can be obtained from Pascal triangle AX4 Example ◊ CH3: the H couples with two H in CH2 : triplet with 1:2:1 intensity (very high resolution NMR could show further splitting of each line into doublets due to H in OH ◊ CH2: the H couples with 3 H in CH3: four lines with intensities: 1:3:3:1 (high resolution NMR can split each of these lines into doublets) ◊ OH: Single line because the H in OH jumps from one molecule to another due to H bonding and the coupling effects are average out Coupling Constant, J Notation: NJAX Coupling between nuclei A and X over N bonds Varies with the angle between the bonds Karplus Equation J = A + B cos ϕ + C cos 2ϕ Important J: 3JHH A = 7Hz, B=-1Hz, C = 5 Hz It can be used to determine bond angles Coupling over many bonds (>4 has very weak effects (negligible) on the spectrum! Magnetic Resonance Imaging When a gradient of magnetic eld is applied, resonance is only possible at a certain value of magnetic eld that corresponds to a plane fi fi Protons are detected only in a certain plane By changing the plane, three dimensional reconstruction of proton concentration is possible 3D mapping of H ‘concentrations’ Reaction in a tubular reactor BrO3 -+ MA+ H+ -> MA-Br + CO2 + H2O MRI: Medical Applications Light shading: large proton concentration -> imaging of soft tissue It is possible to get 3D image of the proton concentrations in the brain Electron Spin Resonance (ESR or EPR) Similar to NMR, but the spin of the electron is used Can be used only when the molecule has unpaired electron (O2, radicals, d metal complexes) The shielding constants do not change much -> good technique for detecting radicals, but less information about structure Frequencies fall in microwave range Usually many peaks (hyper ne structure) due to interaction with nuclei fi ESR spectrum Peculiarity: usually the derivative of absorption is shown Benzene radical ion: C6H6- Importance: medical applications for effects of radicals on cancer etc.
