Biomolecular NMR Spectroscopy

Biomolecular NMR Spectroscopy

Part 1 (introduction): Konstantin IvanovPart 2: Robert Kaptein

Outline for Part 1I IntroductionII Basic PrinciplesIII NMR parameters and interactionsIV Pulsed Fourier Transform NMRV Spin RelaxationVI Relaxation measurementsVII Nuclear Overhauser Effect (NOE)VIII 2D NMR

NMR: IntroductionNMR=Nuclear Magnetic Resonance

Spectroscopy

“Spectroscopy is the study of the interaction between electromagnetic waves and matter.”

X-rayUVVISIRRF

SolidsLiquidsSolutionsDispersionsEmulsions

AbsorptionEmissionFluorescenceRefractionDiffraction

Spectroscopy

Interaction of electromagnetic waves (e.g. absorption of electromagnetic radiation) with matter

102 2

102 0

101 8

101 6

101 4

101 2

101 0

108

106

in Hz

-rays X-rays UV VIS IR Microwave RF

Mössbauerelectronic

vibrationalrotational

NMR

NMR Spectroscopy

102 2

102 0

101 8

101 6

101 4

101 2

101 0

108

106

in Hz

-rays X-rays UV VIS IR Microwave RF

Mössbauerelectronic

vibrationalrotational

NMR

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)

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.

Nobel Prices in NMR

Bloch and Purcell (1952)

Bloch and Purcell (1952)

Richard Ernst (1991)

Richard Ernst (1991)

Kurt Wüthrich(2002)

Kurt Wüthrich(2002)

Typical Applications of NMR

Structure elucidation (small molecules)

Study of dynamic processes

Structural studies on biomacromolecules

Drug Design

Magnetic Resonance Imaging (MRI)

NMR: Basic Principles

NMRNMR=Nuclear Magnetic Resonance

Nuclear: clearMagnetic - ?Resonance -?

Electrons and many nuclei have magnetic moments

An analogy from classical electrodynamicsWhen 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 magnetismProblem 1: Nobody has ever seen fast rotation of elementary particlesProblem 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)

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 magnetismElectron, proton, 13C, 15N, 19F, 31P have spin ½Deuterium, 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}

= I is the gyromagnetic ratio: it gives the ratio of the rotation property (spin) and magnetic property (magnetic moment)Some people say that it should be called magneto-gyric, since it is µ/I

As I is quantized, accordingly, also the magnetic moment is quantized:z = ħmI Next question: what is this gamma equal to?

From spin to magnetic moment

(like I) is a vector

How does the spin interacts with external magnetic field?

In classical electrodynamics

Definition of gyromagnetic ratio: g-factor and magneton

(i) One can define magnetic moment, M, for any moving charge, q, which has angular momentum, G;(ii) Obvious result: 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 the classical case g is always equal to 1!In the quantum case for free electron ge=2.0023 (comes from the Dirac equation)for nuclei gN≠1 (and usually positive) but is can only be measured (there is no theory for gN!)

As we know now: γNħ= –gNβN

Table for gN and γN

Nucleus Spin Natural abundance % gN gN, rad/s*Gs1H ½ 99,98 5,585 267532H(D) 1 0.015 0.857 410713C ½ 1,108 1,405 672814N 1 99,63 0,403 193415N ½ 0.37 –0,567 –2712

19F ½ 100 5,257 2517931P ½ 100 2,263 10840

Electron

½ 2,0023 1.76107


Nuclear: clearMagnetic: also clear, it comes from nuclear spins

Resonance -?

Energy of spin in external fieldEnergy (Hamiltonian) of spin in external magnetic field

EN= –(B,µ)= –gNβN(B,I)

When B is parallel to the Z-axisEN= –gNβNBmI

Spin levels are split by the fieldSpin ½ (many popular nuclei)

Consequence: levels are split; splitting is given only by B and γ

N-spins

Absence of MF

Presence of MF2/1

2/1E=gbB/2

E=–gbB/2

Populations

nb

na

Zeeman interaction

We can induce transitions between the levelsNow, (Nuclear Magnetic) resonance

We can apply oscillating magnetic field

Only magnetic component isimportant

Absence of MF

Presence of MF

E=gbB/2

E=–gbB/2

Populations

2/1

2/1 nb

na

We can induce transitions between the levelsNow, (Nuclear Magnetic) resonance

We can apply oscillating magnetic fieldFor nuclear spins splitting between the levels falls in the RF rangeFor 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

Absence of MF

Presence of MF

E=gbB/2

E=–gbB/2

Populations

2/1

2/1 nb

na

Simple (classical) explanation for resonance

Magnetic field creates a torque acting on magnetic moments spins precess about the Z axis = Larmor precession of spinThe precession frequency is again γBPrecession is an effect, which is also present in the quantum case!

In classical electrodynamics magnetic moments have an important property: they precess about the field

zyx

zyx

zyx

BBBdt

d

eee

μBμ

Remark: for spin ½ (2-level system) the vector model is even mathematically correct

Magnetization components at equilibriumWhat 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 field of 14 T (600 MHz proton freq.), the relative excess of a-spins is only 1 per 104! This is 10000 b-spins and 10001 a-spinsLow 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 ~1017 to ~1020 atoms 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

What happens to Mz when RF-field is applied?It is convenient to use the vector model againLet 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:

)0,0,(cos2cos2 11 tBt B

)0),sin(),(cos()0,sin,(cos 11 ttBttB

X

Y

LF X´

Y´

RF

What happens to Mz when RF-field is applied?It is convenient to use the vector model againLet us go in the rotating frameFrom B1 we have only one component parallel to the X-axis, which is constantIn 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 spinsWidth of the resonance is |ω–ω0|≈ ω1 B1 can flip the spins

)/)(,0,( 01 BeffB B0–w/g

B1

Beff

X´

Z



Resonance: now clear, it is resonant absorption of the RF energy



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

NMR Parameters and InteractionsSo 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.

Important interactions in NMR

(i) Zeeman interaction of spins with the field It is different for free spin and spin in the moleculeChemical shifts can be determined from NMR

(ii) Spins can interact with each otherScalar couplings give additional NMR linesDipolar 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

Chemical shift…• Modification of the Zeeman interaction

• Changes are caused by shielding of the nucleus by electronic currents in the molecule

zNNzNNNZeem IBgIBgE ˆ)1(ˆ

electrons(shielding)

nucleus

Bz

Bloc

Shielding is proportional to external field:Bloc = -s Bz

Beff = Bz + Bloc = Bz (1 – s)

we get a new 'local' Larmour frequency: wi = g Beff = g ( 1 – si ) Bz

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 σ depends 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-scale of chemical shifts (protons)

For H+ δ=30,94For 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 ethaneFor 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

zNNzNNNZeem IBgIBgE ˆ)1(ˆ

66 1010

B

B

B

BB

res

resTMS

…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

ji

zjziijji

jiij SSJJE )( SS

CC

H1

H2

H3geminalvicinal

After that: resonance in moleculesInteractions and stick spectra

(lineshape does not matter now)Chemical shifts and J-couplings;

simple rules for NMR spectra

How can we read the spectra now?

(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.

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

How does the spectrum look like?

(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: (a) Same chemical shifts;(b) Identical couplings to all other nuclei.

Simplifications

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?

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

To the spectra

Spin 1 Spin 2 Spin 3

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

To the spectra

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

Two spins

bAbB

bAaB

aAbB

aAaB

nA/2+nB/2

nA/2–nB/2

–nA/2+nB/2

–nA/2–nB/2

dn=n0(sA–sB)

NMR

1

2

3

4

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

Two spins

bAbB

bAaB

aAbB

aAaB

nA/2+nB/2+J/4

nA/2–nB/2–J/4

–nA/2+nB/2–J/4

–nA/2–nB/2+J/4

dn=n0(sA–sB)

1

2

3

4

NMR

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

What to do for more spins?

J

No other spins

1 spin ½

2 spins ½To be continued…

J

J′

Spin we are looking at sees N nuclei with equal couplingsHow do they split the NMR line?

Relative line intensities in multiplets change as This is called Pascal’ triangle

Groups of equivalents spins

1

1 1

1 2 1

1 3 3 1

0 1 2 3 number of interactions

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

COH CH3

1 13 3

NMR spectrum

12 12

Schematic Spectrum of CH3CH2OH with coupling

OH CH2 CH3

The integrals of the signals are still proportionalto the number of protons they represent

NMRResonance is observed when (i) RF frequency matches the Larmor precession frequency; (ii) B1 is perpendicular to B0.


NMRResonance is observed when (i) RF frequency matches the Larmor precession frequency; (ii) B1 is perpendicular to B0.


This is not really true!(a) One can, in principle, do so but(b) Nowadays nobody is taking spectra using this method

NMR: Pulsed Fourier Spectroscopy

Reasons for using pulses:(1) Better signal-to-noise ratio;(2) Many nice tricks can be done:

with cw-excitation one can only get a spectrum;with pulsing one can do much more.

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;

RF pulses are used

= B1tp

flip angle

B1: strength of RF-pulsetp: duration of RF-pulse

angular frequency 1 = B1:'speed' of flipping of magnetization vector due to RF-pulse

My = Mz sin b

Effect of RF-Pulses (quantitative)

The flip angle b definesthe amount of observabletransversal magnetizationcreated by an rf-pulse

x

y

z

My

Mz

M

b

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).




=ω1τp



• Initial z-magnetization goes into x (or y-magnetization)Mz cos()Ix+sin()Iz




=ω1τp



Now the question is: how do we observe the spectrum?

(1) Spin magnetization goes from z to y (rotation around x-axis);(2) Spins keep precessing: magnetization rotates around the field axis, which is z-axis;(3) A coil should be introduced for detection: changes of the spin magnetization will change magnetic flux that goes through the coil electric current will be induced, which can be measured

Spin evolution

(1) Spin magnetization goes from z to y (rotation around x-axis);(2) Spins keep precessing: magnetization rotates around the field axis, which is z-axis;(3) A coil should be introduced for detection: changes of the spin magnetization will change magnetic flux that goes through the coil electric current will be induced, which can be measured

Spin evolution

B0

NMR detecting coil

To measuring device

Spin evolution(1) Spin magnetization goes from z to y (rotation around x-axis);(2) Spins keep precessing: magnetization rotates around the field axis, which is z-axis;(3) A coil should be introduced for detection: changes of the spin magnetization will change magnetic flux that goes through the coil electric current will be induced, which can be measured

The measured signal isM0z*sin()*cos(ωt) for x-magnetization

M0z*sin()*sin(ωt) for y-magnetization

Mx and My are shifted with respect to each other by 90-degree

Spin evolution(1) Spin magnetization goes from z to y (rotation around x-axis);(2) Spins keep precessing: magnetization rotates around the field axis, which is z-axis;(3) A coil should be introduced for detection: changes of the spin magnetization will change magnetic flux that goes through the coil electric current will be induced, which can be measured

The measured signal isM0z*sin()*cos(ωt) for x-magnetization


Mz is not changed and not measured

Spin evolutionThe measured signal isM0z*sin()*cos(ωt) for x-magnetization


Fourier Transform of the observed signals gives the spectrum: in our simplest case it has one lineWhen there are more frequencies:(1) We excite all NMR transitions: flip Mz for all molecules;

(2) We detect the signal that has more frequencies;(3) Fourier transform gives the same spectrum as we expect for cw-detection;(4) The spectrum is taken in one shot!(5) The observed signal is called Free Induction Decay (FID)

Spin evolutionThe measured signal isM0z*sin()*cos(ωt) for x-magnetization


Fourier Transform of the observed signals gives the spectrum: in our simplest case it has one lineWhen there are more frequencies:(1) We excite all NMR transitions: flip Mz for all molecules;

(2) We detect the signal that has more frequencies;(3) Fourier transform gives the same spectrum as we expect for cw-detection;(4) The spectrum is taken in one shot!(5) The observed signal is called Free Induction Decay (FID) – why decay?

Spin evolutionWithout relaxation

Continuous rotation induces continuous harmonic oscillation in receiver coil

Spin evolutionReality is different

Free Induction Decay

t0

1

-1

NMR: Relaxation

NMR: RelaxationNew concept appears: spin relaxation

Relaxation brings any system to its equilibrium: no transverse magnetization and Boltzmann z-magnetization

Relaxation also causes the FI Decay

Longitudinal Relaxation

Restoring the equilibrium magnetization Mz

Transitions between the a- and b-states

The time constant connected with this is T1

Longitudinal or spin-lattice relaxation

T1 defines the maximum repetition rate of an NMR experiment

Transverse Relaxation

Does not involve transitions between a- and b-states (no restoring of the equilibrium state)

Instead loss of phase coherence, vanishing of observable magnetization Mx,y

The time constant connected with this is T2

Transverse or spin-spin relaxation

Time after which FID does contain no signal

Practical Aspects of Relaxation

T1: The time we have to wait before we can repeat our experiment (repetition rate).Before we can repeat the experiment, the equilibrium-distribution has to be restored!

T2: The time we have to record an FID. After that time no observable signal does exist anymore. The FI has Decayed!

Couse of relaxation: randomly fluctuating interactions

In fluctuating environment there are time-dependent energy terms

here A-spin components are time-independent operator and f(t) is a random functionWe can always assume that the time average of f(t) is zero, but the auto-correlation function is non-zero

We will assume that noise is stationary. Then G-function is independent of tIt drops down to zero with the correlation time τc, which is defined as

The new function J(ω) is the noise spectral density

0

1)0(,)( GdGc

deGJ i)()(

time

t t

)()()( Gtftf

)(ˆ)(ˆ tfAtV

Random transitions between the states induce both T1 and T2

Transitions can be caused by dipolar couplings

or transverse fluctuating fields

Transverse fields can mix α-states and β-states

Effciency is associated with J(ω)=energy transfer

with environment=spin-lattice relaxation)

Inhomogeneity of Bz gives T2

Distribution of (static) Bz fields:

Different Larmor frequencies on different

locations in the sample

Dephasing of magnetization

Transverse magnetization vanishes

Energy exchange with the lattice is not

required=spin-spin relaxation

A mathematical description is given by the following set of differential equations:

dMz (Mz – Meq)

dt T1

dMx Mx

dt T2

dMy My

dt T2

Longitudinal magnetization goes to Meq with time T1

Transverse magnetization decays with time T2

=

=Mz(t)-Meq = [Mz(0)-Meq] e –t/T1

Mx(t) = [Mx(0)-Meq] e –t/T2

My(t) = [My(0)-Meq] e –t/T2

Relaxation of Spins

=

What happens to FID NMR and spectrum?

FT

t wtime domain f(t) frequency domain g(w)

g(w) = f(t) cos wt dt

0

Our case, x-magnetization:M0z*sin()*cos(ω0t) M0z*sin()*cos(ω0t)*exp(–t/T2)FT gives Lorentzian line:g(ω)~1/[1+((ω–ω0)T2)2] position is ω0; width is 1/T2

Moving Dipoles

The additional field (Bdip),that is felt by m2 dependson the orientation of m2

relative to m1 (the angle ):

)1Θcos3(rμ

B 231dip

z

Random Fluctuating Fields

Both T1 and T2 relaxation is caused by randomly fluctuating magnetic fields (dipole-dipole interaction):

Random tumbling of molecules (reorientation relative to B0; and dipolar fields from neighboring nuclei)

Movement of parts of the molecule relative to each other.

The Tumbling of Molecules

Molecules are moving constantly (Brown's molecular motion)

A measure for the speed of the tumbling of a molecule is tc, the rotational correlation time:

t < tc t tc t >> tc

For t << tc the molecule hardly moved, at t >> tc orientation is random

The Tumbling Of Molecules

Rotational Correlation Time tc

tc is dependent from the size of the molecule V, the viscosity of the solvent and the temperature T :

tc = V

kT

For biomacromolecules in water at room temperature:

Mr

2.4tc 10-12 Mr : molecular mass

J(w) = 2tc

1+w2tc2

The Spectral Density Function

J(w)

w1/tc

T1 and tc

Longitudinal relaxation is fastest, when the spectral density has a maximum at the frequency wo. This is the case for 1/ tc = wo:

T1

tc1/ tc = wo

= 2g2 B 2 J (wo)1T1

T2 and tc

1/T2

tc

For large molecules the rate of transverse relaxation is simply proportional to tc:

1T2

g2 B 2 tc

T2 and tc

More general case: there are two contributions to T2

One is simply proportional to tc;The other one is like T1

1T2

g2 B 2 tc+ 12T1

Reason:Contribution 1 is from dephasingContribution 2 is from transition between the states

Mechanisms of Relaxation (Summary)

Energy transfer with environment (spin-lattice) -a b transitions Motions with Larmor Frequencies Fluctuating fields

Can occur without energy transfer with environment (instead spin-spin interaction)

No -a b transitions are necessary Dephasing of transverse magnetization Fluctuating and static fields

T1:

T2:

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Biomolecular NMR Spectroscopy