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Biomolecular NMR Spectroscopy

Biomolecular NMR Spectroscopy. Part 1 (introduction): Konstantin Ivanov Part 2: Robert Kaptein. Outline for Part 1. I Introduction II Basic Principles III NMR parameters and interactions IV Pulsed Fourier Transform NMR V Spin Relaxation VI Relaxation measurements

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Biomolecular NMR Spectroscopy

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  1. Biomolecular NMR Spectroscopy Part 1 (introduction): Konstantin Ivanov Part 2: Robert Kaptein

  2. Outline for Part 1 I Introduction II Basic Principles IIINMR parametersandinteractions IV Pulsed Fourier Transform NMR V Spin Relaxation VI Relaxation measurements VIINuclearOverhauserEffect (NOE) VIII2D NMR

  3. NMR: Introduction NMR=Nuclear Magnetic Resonance

  4. Spectroscopy “Spectroscopy is the study of the interaction between electromagnetic waves and matter.” Absorption Emission Fluorescence Refraction Diffraction X-ray UV VIS IR RF Solids Liquids Solutions Dispersions Emulsions

  5. Spectroscopy Interaction of electromagnetic waves (e.g. absorption of electromagnetic radiation) with matter rotational electronic Mössbauer vibrational NMR 1016 1014 1012 1010 1022 1020 1018 106 108  in Hz UV VIS -rays X-rays Microwave IR RF

  6. rotational electronic Mössbauer vibrational NMR 1016 1014 1012 1010 1022 1020 1018 106 108  in Hz UV VIS -rays X-rays Microwave IR RF NMR Spectroscopy • Energy differences between energy states are extremely small (very insensitive method) • Lines are narrow (resolution can be quite high) • Spectra can depend on interaction between observed nuclei (structural information)

  7. NMR is a Widely Used Analytical Tool • First observed in 1946, quickly commercially available was widely used since then. • Covers the study of structure, dynamics and function of the complete range of chemical entities. • Preferred technique for rapid structure elucidation of most organic compounds.

  8. Nobel Prices in NMR Richard Ernst (1991) Kurt Wüthrich (2002) Bloch and Purcell (1952)

  9. Typical Applications of NMR • Structure elucidation (small molecules) • Study of dynamic processes • Structural studies on biomacromolecules • Drug Design • MagneticResonance Imaging (MRI)

  10. NMR: Basic Principles

  11. NMR NMR=Nuclear Magnetic Resonance Nuclear: clear Magnetic - ? Resonance -?

  12. Electrons and many nuclei have magnetic moments An analogy from classical electrodynamics When an electrically charged particle rotates  electric current is formed  particle acquires magnetic moment One may think that electrons and nuclei behave similarly: they are charged and spin magnetic moments are identified with spins Spin (angular momentum of particles) is the origin of nuclear magnetism Problem 1: Nobody has ever seen fast rotation of elementary particles Problem 2: Spin is a quantum notion, which has no classical analogue; it disappears when the Plank constant is set equal to zero (but still classical considerations may help)

  13. How does the quantum nature of spin reveal itself? Spin and its projection of the field axis cannot be just any number! Spin, I, can be integer or half integer when it is measured in units of ħ(Plank constant): 12C and 16O have spin 0=no nuclear magnetism Electron, proton, 13C, 15N, 19F, 31P have spin ½ Deuterium, 14N have spin 1 Projection of spin, mI, is also quantized and can have only certain values: it changes from –I to I and spaced by 1 Example: for I=1/2 we have mI={–½;½} for I=1 we have mI={–1;0;1}

  14. How does the spin interacts with external magnetic field?  =  I isthegyromagneticratio: itgivestheratiooftherotationproperty (spin) andmagneticproperty (magneticmoment) Somepeoplesaythatitshouldbecalledmagneto-gyric, sinceitisµ/I From spin to magnetic moment  (likeI) is a vector As Iisquantized, accordingly, also themagneticmoment isquantized: z= ħmI Next question: whatisthisgammaequalto?

  15. Definition of gyromagnetic ratio: g-factor and magneton (i) Onecandefinemagneticmoment, M, foranymovingcharge, q, whichhas angular momentum, G; (ii) Obviousresult: M=γG and γ=q/2mc; (iii) Quantum case: µ=γħI but γ is defined differently (iv) We can still define the magneton as e/2mc but a new dimensionless coefficient appears, which is named g-factor: γeħ= geβe (for electrons) γNħ= –gNβN (for nuclei) In theclassicalcasegisalwaysequalto 1! In thequantumcase forfreeelectronge=2.0023 (comesfromthe Dirac equation) fornucleigN≠1 (andusually positive) but iscanonlybemeasured (thereisnotheoryforgN!) In classical electrodynamics

  16. Table for gN and γN As we know now: γNħ= –gNβN

  17. NMR NMR=Nuclear Magnetic Resonance Nuclear: clear Magnetic: also clear, it comes from nuclear spins Resonance -?

  18. Energy of spin in external field Energy (Hamiltonian) of spin in external magnetic field • EN= –(B,µ)= –gNβN(B,I) When B is parallel to the Z-axis • EN= –gNβNBmI Spin levels are split by the field Spin ½ (many popular nuclei) Consequence: levels are split; splitting is given only by B and γ Zeeman interaction N-spins E=gbB/2 nb Absence of MF Populations E=–gbB/2 na Presence of MF

  19. Now, (Nuclear Magnetic) resonance E=gbB/2 We can apply oscillating magnetic field Only magnetic component is important nb Absence of MF Populations E=–gbB/2 na Presence of MF We can induce transitions between the levels

  20. Now, (Nuclear Magnetic) resonance E=gbB/2 We can apply oscillating magnetic field For nuclear spins splitting between the levels falls in the RF range For instance: for protons B=7 Tesla corresponds to splitting of 300 MHz (in frequency units) Quantum mechanics tells us that: (i) Oscillating field should be perpendicular to B0||z; • (ii) Its frequency should be equal to the splitting  resonance. Then the oscillating field will lead to transitions of population between levels such that ΔmI=±1 nb Absence of MF Populations E=–gbB/2 na Presence of MF We can induce transitions between the levels

  21. Simple (classical) explanation for resonance • Magnetic field creates a torque acting on magnetic moments  spins precess about the Z axis = Larmor precession of spin • The precession frequency is again γB • Precession is an effect, which is also present in the quantum case! Remark: for spin ½ (2-level system) the vector model is even mathematically correct In classical electrodynamics magnetic moments have an important property: they precess about the field

  22. Magnetization components at equilibrium What is the magnetization value prior to NMR experiment? We always start with thermally polarized spins; so the state populations are given by the Boltzmann distribution nb/na = e (–ΔE/kT) ≈ 1 – ΔE/kT At a fieldof 14 T (600 MHz protonfreq.), the relative excessofa-spinsisonly 1 per 104! Thisis 10000 b-spinsand 10001 a-spins Low polarization at equilibrium is one of the reasons why NMR is so insensitive: signal is proportional to (nα – nβ) Nevertheless, the sample carries equilibrium magnetization: A real NMR sample contains ~1017to ~1020atoms  0.01 % of them give rise to the observable signal Spin magnetization along Z-axis is Mz=na(–gNβN*1/2)+ nb(–gNβN*(–1/2))≈N*(gNβN)2B/4kT All other magnetization components are zero for spin system at equilibrium

  23. What happens to Mz when RF-field is applied? Y Y´ X RF X´ Let us switch on the oscillating field 2B1cosωt in the XY-plane in the laboratory frame. Meaning of the coefficient ‘2’ will be explained soon! For clarity, let the B1-vector be parallel to the X-axis, then it takes the form The RHS can be formally presented as a sum of two terms – two vectors rotating with the same speed in opposite directions Graphic representation in both frames: It is convenient to use the vector model again LF

  24. What happens to Mz when RF-field is applied? Z Beff B0–w/g X´ B1 Let us go in the rotating frame From B1 we have only one component parallel to the X-axis, which is constant In the Z direction we have: (i) B0 field; (ii) Field in the opposite direction (because the RF is not an inertial frame of reference) of the strength ω/γ; The effective field vector in the rotating frame is We get precession about the effective field; When ω=ω0 (resonance condition) the Mz vector rotates about X  energy of the RF field is absorbed by the spins Width of the resonance is |ω–ω0|≈ ω1  B1 can flip the spins It is convenient to use the vector model again

  25. NMR NMR=Nuclear Magnetic Resonance Nuclear: clear Magnetic: also clear, it comes from nuclear spins Resonance: now clear, it is resonant absorption of the RF energy

  26. NMR NMR=Nuclear Magnetic Resonance Nuclear: clear Magnetic: also clear, it comes from nuclear spins Resonance: now clear, it is resonant absorption of the RF energy Resonance is observed when (i) RF frequency matches the Larmor precession frequency; (ii) B1 is perpendicular to B0. We should apply B1 with proper frequency and get a line

  27. NMR Parameters and Interactions So far, we discussed only why we can observe resonant effects using nuclear spins; This is, of course, not enough for obtaining any interesting information about molecules; So, we should make a transition from NMR of a single spins to NMR in molecules and introduce NMR-relevant interactions, which contain information about systems under study; We should understand how NMR spectra look like and how to read them.

  28. Important interactions in NMR (i) Zeeman interaction of spins with the field It is different for free spin and spin in the molecule Chemical shifts can be determined from NMR (ii) Spins can interact with each other Scalar couplings give additional NMR lines Dipolar couplings do not give more lines in liquid-state NMR, but result in spin relaxation (to be introduced) (iii) Quadrupolar interactions (quadrupolar moment of nuclei with electric field gradients coming from electrons) Absent for spin ½ For I>1/2 QI result in very fast relaxation: it is almost impossible to observe their NMR in liquids (iv) Relaxation parameters and NOE will be discussed later

  29. Chemical shift… • Modification of the Zeeman interaction • Changes are caused by shielding of the nucleus by electronic currents in the molecule • Shieldingisproportional toexternalfield: • Bloc= -sBz • Beff = Bz + Bloc = Bz (1 – s) • weget a new 'local' Larmourfrequency: • wi = gBeff= g ( 1 – si ) Bz electrons (shielding) Bz Bloc nucleus

  30. Chemical shift… • Modification of the Zeeman interaction • Changes are caused by shielding of the nucleus by electronic currents in the molecule • Chemical shift is dimensionless and is measured in ppm • The value of σdepends on the solvent, temperature, etc. That is why s is measured relative to inert reference substance (usually Si(CH3)4 – TMS) with single narrow NMR line • The δ-scaleof chemical shifts (protons) For H+δ=30,94 For all organic substances chemical shifts fall into the range [–3;15] ppm: 7,27 for benzene; 3,47 for CH3-group in methanol; 0.89 for ethane For other nuclei (for instance , 13C, 19F) the range of possible chemical shifts is much broader (100-1000 ppm) Conclusion Different groups in the molecule have different chemical shifts; Usually chemical shifts are reliably assigned to particular groups  by looking at NMR spectra we can say to what nuclei the lines belong

  31. H1 vicinal geminal C H3 C H2 …and spin-spin interactions • In liquids strong dipole-dipole interactions are averaged out by fast molecular motion • Scalar spin-spin couplings (J-coupling) become important, which are not averaged • Origin: (1) nuclear spins interact with orbital momentum of the electrons; (2) electrons become partly ‘unpaired’ and (3) start interacting with other nuclei by creating small magnetic fields. Nuclear spins interact indirectly via the electrons! J-couplings characterize structure (chemical bonds) • J-couplings are measured in Hz • Protons are never the neighboring nuclei (except for H2), there are chemical bonds in between. • For protons the largest coupling is usually the geminal one (15-20 Hz, negative), vicinal coupling is smaller (7 Hz, positive) Understanding NMR spectrum (chemical shifts+J) is very helpful for understanding the structure

  32. How can we read the spectra now? After that: resonance in molecules Interactions and stick spectra (lineshape does not matter now) Chemical shifts and J-couplings; simple rules for NMR spectra

  33. How does the spectrum look like? (1) Different nuclei give signals at completely different frequencies because of large differences in γ in experiment we can detect signals only from protons, or carbons, or nitrogens; (2) Different nuclei of the same kind (e.g., protons) give NMR signals at different frequencies due to chemical shifts; (3) Lines corresponding to certain nuclei are split due to the presence of other spins ½.

  34. How does the spectrum look like? (1) Different nuclei give signals at completely different frequencies because of large differences in γ in experiment we can detect signals only from protons, or carbons, or nitrogens; (2) Different nuclei of the same kind (e.g., protons) give NMR signals at different frequencies due to chemical shifts; (3) Lines corresponding to certain nuclei are split due to the presence of other spins ½. Having all this we can start drawing spectra

  35. Simplifications (1) Spins will be considered weakly coupled: for each pair of non-equivalents spins difference in Zeeman interactions, ωi–ωj, is much larger than the corresponding coupling Jij; (2) Quantum mechanics (perturbation theory) tells us that we can leave only JijIizIjz terms; (3) Equivalent spins do not interact with each other (in fact, they do, but these couplings cannot be detected). What do we call equivalent spins: Same chemical shifts; Identical couplings to all other nuclei.

  36. To the spectra Spin 1 Spin 2 Spin 3 Different spins give lines at different frequencies, which are given by chemical shifts Frequency for each spin is determined by γi(1+δi)=ωrf signal at δi (in δ-scale) Couplings split the lines, but how should it look like?

  37. To the spectra Spin 1 Spin 2 Spin 3 Different spins give lines at different frequencies, which are given by chemical shifts Frequency for each spin is determined by γi(1+δi)=ωrf signal at δi (in δ-scale) Couplings split the lines, but how should it look like? Let us do it for 2 spins and then generalize

  38. To the spectra Coupling term is JijIizIjz; For equal projections of spins is adds Jij/4 to the energy; For opposite projections of spins is adds –Jij/4 to the energy; Levels are only shifted; but the states remain the same  selection rules for NMR transitions are the same: ΔIz=±1

  39. NMR Two spins bAbB nA/2+nB/2 4 bAaB 3 nA/2–nB/2 dn=n0(sA–sB) –nA/2+nB/2 aAbB 2 Without interaction Each spin gives signals at its characteristic frequency Altogether there are 4 transitions determined by selection rules ΔIz=±1 and two lines in the spectrum: transitions overlap 1 –nA/2–nB/2 aAaB

  40. Two spins bAbB nA/2+nB/2+J/4 4 bAaB 3 nA/2–nB/2–J/4 dn=n0(sA–sB) –nA/2+nB/2–J/4 aAbB 2 With interaction Each spin gives signals at its characteristic frequency Altogether there are 4 transitions determined by selection rules ΔIz=±1 and four lines in the spectrum: transitions do not overlap Reason: each spin sees additional field of J*Mz 1 –nA/2–nB/2+J/4 aAaB NMR

  41. What to do for more spins? No other spins 1 spin ½ 2 spins ½ To be continued… J J Each spin ½ splits NMR line in two lines; their intensities are 2 times lower Reason: each spin sees additional field of J*Mz Important case: groups of equivalent spins J′

  42. Groups of equivalents spins Spin we are looking at sees N nuclei with equal couplings How do they split the NMR line? Relative line intensities in multiplets change as This is called Pascal’ triangle 0 1 2 3 number of interactions 1 1 1 1 2 1 1 3 3 1

  43. An example: AnXm-system (CH3OH) Interactions have only zz-components; Equivalent spins do not interact Intensity of all lines in the multiplet is proportional to the number of spins in the group Splitting is the same for both spins (same J is operative) Distribution is given by Pascal’s triangle rules 12 12 3 3 1 1 NMR spectrum COH CH3

  44. Schematic Spectrum of CH3CH2OH with coupling OH CH2 CH3 The integrals of the signals are still proportional to the number of protons they represent

  45. NMR Resonance is observed when (i) RF frequency matches the Larmor precession frequency; (ii) B1 is perpendicular to B0. We should apply B1 with proper frequency and get a line

  46. NMR Resonance is observed when (i) RF frequency matches the Larmor precession frequency; (ii) B1 is perpendicular to B0. We should apply B1 with proper frequency and get a line This is not really true! One can, in principle, do so but Nowadays nobody is taking spectra using this method

  47. NMR: Pulsed Fourier Spectroscopy Reasons for using pulses: Better signal-to-noise ratio; Many nice tricks can be done: with cw-excitation one can only get a spectrum; with pulsing one can do much more.

  48. RF pulses are used = B1tp B1:strengthof RF-pulse tp: durationof RF-pulse In the rotating frame spin magnetization rotates around B1 when the resonance condition is fulfilled; It we apply a pulse of certain duration we will rotate spins by certain angle around the B1 field; flip angle angular frequency1=  B1: 'speed' offlippingofmagnetization vector due toRF-pulse

  49. z M Mz y My x Effect of RF-Pulses (quantitative) My = Mz sinb b The flip angle b defines the amount of observable transversal magnetization created by an rf-pulse

  50. Pulses • In NMR nobody is doing cw-experiments; • In cw-methods the oscillating field is rather weak and is applied over long period of time to observe the spectra. In pulsed methods it is higher and is applied during a short period of time. This pulse changes the state of the system. After that response of the system to the pulse is studied. • In NMR/EPR pulse applied during time τp, it flips magnetization in the rotating frame by an angle =ω1τp • Pulses are characterized by strength of B1, duration and phase. Strength gives rotation frequency and width of the resonance; Duration gives the flip angle; Phase determines the rotation axis: B1 along x or y. • The most common pulses are 90-degree (Mz goes in the xy-plane) and 180-degree (inverting pulse).

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