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3– Chemical exchange

and the McConnell Equations

NMR is a technique which is well suited to study

dynamic processes, such as the rates of chemical

reactions. The time window which can be

investigated in NMR is in the range of nanoseconds

to seconds. It is sampled using different experimental

observables, such as relaxation rates, lineshape

analysis, etc.

In this chapter, we will describe how we can use

NMR to study chemical exchange processes of the

type:

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A2A1k

k

local field: B BResonance frequency:

01 02

Ωω01

1 Ωω02

2

(lab frame)(rot. frame)

Examples include internal rotations about a bond,

ring puckering, proton exchange, and bond

rearrangements.

To describe chemical exchange, we can continue with

our semi-classical description and use a modified

Bloch equations (also known as McConnell equations,

after H.M. McConnell who derived them in 1958).

3.1 McConnell Equations

Let’s take a spin which can be in state A1 or A2

(which is equivalent to two different environments)

as shown above. If there is no exchange, we can

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write Bloch equations for both states, namely:

d ~M1

dt= γ ~M1 × ~B1 − R1( ~M1 − ~M10)

d ~M2

dt= γ ~M2 × ~B2 − R2( ~M2 − ~M20)

(3.1)

Beware: R2 here is NOT 1T2

. Similarly for R1. R1

and R2 represent the full relaxation matrix for our

spin in state A1 and A2, respectively.

If we assume now that chemical exchange occurs

instantaneously, i.e. that we do not give the system

the time to gradually change from A1 to A2 and vice

versa as this would affect the parameters in equation

3.1 (think of relaxation, for instance). In this case,

then we can add first order kinetics terms to the

Bloch equations above, such that the overall change

in magnetization over time is given by

d ~M1

dt= γ ~M1 × ~B1 − R1( ~M1 − ~M10) + k( ~M2 − ~M1)

d ~M2

dt= γ ~M2 × ~B2 − R2( ~M2 − ~M20) + k( ~M1 − ~M2)

(3.2)

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These are the McConnell equations.

3.2 Lineshape Analysis

Keeping on with the example above, let us now

apply a β= (π/2)−x pulse, so that the initial

conditions for the transverse components in the

rotating frame are given by

Mr1y(0) = M r

2y(0) = M0/2

Mr1x(0) = M r

2x(0) = 0

Mr1y(0) + M r

2y(0) = M0 (3.3)

Using equation 2.22 from the previous chapter, we

can write the modified Bloch equations in the

rotating frame as

dMr1x

dt= −Ω1M

r1y −

Mr1x

T(1)2

+ k(M r2x − Mr

1x)

dMr1y

dt= Ω1M

r1x −

Mr1y

T(1)2

+ k(M r2y − Mr

1y) (3.4)

and similarly for the M r2x and Mr

2y components.

To make life simpler (!), we can use the definition for

the signal which we had in equation 2.28 to write the

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overall magnetization of 1 and 2 as

M1 = Mr1y − iMr

1x

M2 = Mr2y − iMr

2x (3.5)

such that equation 3.4 can be rewritten as

dM1

dt= iΩ1M1 −

M1

T(1)2

+ k(M2 − M1)

dM2

dt= iΩ2M2 −

M2

T(2)2

+ k(M1 − M2) (3.6)

Often we can assume that relaxation is negligible so

that equation 3.6 can be written in matrix form as

d

dt

M1

M2

=

iΩ1 − k k

k iΩ2 − k

M1

M2

(3.7)

The solution to this equation (you may be familiar

with it already if you have done kinetics) is:

M1(t) = c11eΛ1t + c12e

Λ2t

M2(t) = c21eΛ1t + c22e

Λ2t

where Λ1 and Λ2 are the two eigenvalues of the

kinetic matrix and can be obtained by solving the

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equation

(iΩ1 − k − Λ1)(iΩ2 − k − Λ2) − k2 = 0 (3.8)

so that

Λ1,2 =i

2(Ω1 + Ω2) − k ±

√

k2 −1

4(Ω1 − Ω2)2 (3.9)

with the subscript 1 corresponding to the sum and

the subscript 2 corresponding to the difference.

Keeping in mind that we are applying a β= (π/2)−x

pulse, we can apply our initial conditions and solve

for the coefficients cij (i = 1, 2; j = 1, 2) so that

c11 =1

4

M0[i2 (Ω1 − Ω2) + k + W ]

W

c12 =1

4

M0[−i2 (Ω1 − Ω2) − k + W ]

W

c21 =1

4

M0[−i2 (Ω1 − Ω2) + k + W ]

W

c22 =1

4

M0[i2 (Ω1 − Ω2) − k + W ]

W(3.10)

with

W =

√

k2 −1

4(Ω1 − Ω2)2 (3.11)

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Finally the signal we measure experimentally is the

sum of the signal we get from our spin in state A1

and in state A2, or mathematically

s(t) = M1(t) + M2(t) (3.12)

with the two time-dependent magnetization

components given by equation 3.8, i.e. depending on

cij and Λ1,2.

To try to make sense of these equations and to

understand the physical implications, let us now

consider two cases:

Case 1: slow exchange

In the case of slow exchange, the following condition

applies:

k << |Ω1 − Ω2| (3.13)

i.e. the NMR spectrum will consist of two lines, one

corresponding to state A1 at Ω1 and one for state A2

at position Ω2.

In this case,

Λ1 = iΩ1 − k

Λ2 = iΩ2 − k

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c11 = c22 = M0/2

c12 = c21 = 0 (3.14)

since

W ≈

√

−1

4(Ω1 − Ω2)2 =

i

2(Ω1 − Ω2) (3.15)

Therefore

s(t) =1

2M0[e

(iΩ1−k)t + e(iΩ2−k)t] (3.16)

Fourier transforming this gives the frequency domain

signal:

S(ω) =1

2M0[

1

k − i(Ω1 − ω)+

1

k − i(Ω2 − ω)] (3.17)

with a real part

ReS(ω) =1

2M0[

k

k2 + (Ω1 − ω)2+

k

k2 + (Ω2 − ω)2]

(3.18)

In other words, the absorptive peaks measured in the

NMR spectrum will consist of two Lorentzian lines

with a half width at half height of k.

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Ω2Ω1

2k 2k

Case 2: fast exchange

In the fast exchange limit,

k >> |Ω1 − Ω2| (3.19)

i.e. only an average of the two species/lines will be

seen, at an average position of (Ω1 + Ω2)/2.

In this case,

Λ1,2 =i

2(Ω1 + Ω2)− k ±

√

k2 −1

4(Ω1 − Ω2)2 (3.20)

can be simplified to

Λ1,2 ≈i

2(Ω1 +Ω2)−k±k[1−

1

8k2(Ω1 −Ω2)

2] (3.21)

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and

c11 = c21 = M0/2

c12 = c22 = 0 (3.22)

As a result, the signal is given by

s(t) = M0exp[iΩ1 + Ω2

2−

(Ω1 − Ω2)2

8k]t (3.23)

which when Fourier transformed gives

ReS(ω) = M0[∆ 1

2

(Ω − ω)2 + (∆ 1

2

)2] (3.24)

with

Ω =Ω1 + Ω2

2

∆ 1

2

=(Ω1 − Ω2)

2

8k(3.25)

or graphically

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+

2

2∆1/2

Ω1 Ω 2

One can obtain a general equation for the real part

of the frequency domain signal from

ReS(ω) =M0k(Ω1 − Ω2)

2

2[(ω − Ω1)2(ω − Ω2)2 + 4k2(ω − Ω1+Ω2

2 )2](3.26)

Therefore with increasing rate constants (or

increasing temperature), the lineshapes vary in the

following manner:

•In the slow exchange limit, we have two lines with

the full width in half given by 2π∆FWHH = 2∆ 1

2

=

2k or k = π∆FWHH where ∆FWHH is in Hz and k is

in s−1 (BEWARE OF THE UNITS and the factors

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of 2π!!!).

•As the temperature increases the lines broaden,

until we reach the coalescence temperature - at this

temperature, the two lines merge into one broad

“flat-topped” line. The rate at which coalescence

occurs is given by kc = π|Ω1−Ω2|√2

, where kc is in s−1

and |Ω1 − Ω2| is in Hz.

•Past this temperature, we have one line, with a

linewidth which is inversely proportional to the rate

of exchange, i.e. k= π(Ω1−Ω2)2

2∆F W HHwhere again k is in

s−1 and ∆FWHH and |Ω1 − Ω2| are in Hz. This is

equivalent to the second equation in 3.25.

For example:

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Given experimentally determined exchange rate

constants as a function of temperature, we can

determine the activation energy Ea or the activation

enthalpy ∆H 6= and entropy ∆S 6= from the Arrhenius

equation

k(T ) = Aexp(−Ea/RT ) (3.27)

or the Eyring equation

k(T ) =kBT

he

∆S6=

R e−∆H

6=

R , (3.28)

respectively.

3.3 Exchange from 2D NMR

The chemical exchange problem is a nice way to

introduce the concept of higher dimensional NMR, so

in this section, we will show how 2D experiments can

be created and applied to the chemical exchange

problem.

Let us start again from our exchange process:

A2A1k

k

local field: B BResonance frequency:

01 02

Ωω01

1 Ωω02

2

(lab frame)(rot. frame)

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which gives rise to two lines when the exchange is

slow. We will now perturb our system in such a way

as to invert the magnetization for the A1 state

selectively (see box 1), so that the initial condition at

time 0 is

M1z(0) =−M0

2

M2z(0) =M0

2(3.29)

In this case, the exchange equation can be written as:

d

dt

M1z(t)

M2z(t)

=

−k k

k −k

M1z(t)

M2z(t)

(3.30)

or

d[M1z − M2z]

dt= −2k(M1z − M2z)

d[M1z + M2z]

dt= 0. (3.31)

The latter equation describes the conservation of the

total magnetization of the system over time.

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Box 1. Selective Pulses As we saw in the last chapter,

the magnitude of B1 has an effect on the frequency do-

main profile of the rf pulse applied. To make a pulse

selective for one region of the spectrum, we can there-

fore use a low power pulse (small B1). Alternatively,

we can use rf pulses which have a special pulse pro-

file (in the frequency domain) - called shaped pulses.

Examples include Gaussian pulses:

and others (e.g. BURP, EBURP, REBURP) which

have a range of inversion/excitation profiles.

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Solving these equations gives

[M1z − M2z](τm) = [M1z − M2z](0)e−2kτm (3.32)

[M1z + M2z](τm) = [M1z + M2z](0) (3.33)

or

M1z(τm) =1

2[M1z + M2z](0)

+1

2[M1z − M2z](0)e−2kτm

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M2z(τm) =1

2[M1z + M2z](0)

−1

2[M1z − M2z](0)e−2kτm

(3.34)

Therefore measuring the magnetization using the

sequence:

π/2τm

as a function of τm, one can determine k, as long as

relaxation effects can be neglected. This experiment

can also be applied to more complex systems (more

than two state exchange) but requires that a series of

experiments, each with the appropriate selective

pulse, be run.

In the example above, we have used a selective pulse

to distinguish the spins in the A1 state from those in

the A2 state. Is there another way to make this

distinction? We know that if we apply a single

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(π/2)y pulse and let the system evolve, we can

“label” our magnetization in terms of Ω (recall

equation 2.26).

So if we perform the following experiment,

τm

π/2( )yπ/2( )y π/2( )y

t 1

A B C D E

F

then for the spins in the A1 state, we have

~M1A =

1

2M0~ez

~Mr,B1 =

1

2M0~ex

~Mr,C1 =

1

2M0[~excos(Ω1t1) + ~eysin(Ω1t1)]

M1zD = −

1

2M0cos(Ω1t1)

(3.35)

and similarly for the spins in the A2 state.

Therefore at point D we have a similar initial

condition as we had in equation 3.29, but now it is

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modulated (or “labelled”) by a cos(Ωt1) term, so

that for the exchange process, the initial condition is

M1z(t1, 0) = −1

2M0cos(Ω1t1)

M2z(t1, 0) = −1

2M0cos(Ω2t1)

(3.36)

The components of the magnetization after exchange

M1z(t1, τm) and M2z(t1, τm) will be measured in the

transverse using the last (π/2)y pulse such that

~Mr,F1 = M1z(t1, τm)~ex

~Mr,F2 = M2z(t1, τm)~ex

(3.37)

The observed x-component of the magnetization is

thus

Mrx(t1, τm, t2) = M1z(t1, τm)cos(Ω1t2)

+ M2z(t1, τm)cos(Ω2t2)

(3.38)

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where the right hand side can be rewritten as

−1

4M0 [ (1 + e−2kτm)cos(Ω1t1)cos(Ω1t2)

+ (1 + e−2kτm)cos(Ω2t1)cos(Ω2t2)

+ (1 − e−2kτm)cos(Ω1t1)cos(Ω2t2)

+ (1 − e−2kτm)cos(Ω2t1)cos(Ω1t2)]

(3.39)

In other words, the first two terms oscillate with the

same frequency in both t1 and t2 and the two last

terms represent a change in frequency and therefore

represent the chemical exchange phenomenon. If one

Fourier transforms the data in t2 and t1, then the

following spectrum is obtained

ω 1ω 2

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The diagonal peaks, which have a relative amplitude

of 12 (1 + e−2kτm), represent the non-exchanged

magnetization, whereas the cross-peaks, with relative

intensity of 12 (1 − e−2kτm), represent the exchanging

magnetization from the chemical exchange process.

Plotting the diagonal and cross-peak intensity as a

function of the mixing time (τm), we get

0

0.2

0.4

0.6

0.8

1

0 1 2 3 4 5 6

(0.5*(1+exp(-2*1*x)))(0.5*(1-exp(-2*1*x)))

τm

diagonal

cross−peak

If one takes into account relaxation effects, then this

is changed to

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0

0.2

0.4

0.6

0.8

1

0 1 2 3 4 5 6

(0.5*(exp(-x/10))*(1+exp(-2*1*x)))(0.5*(exp(-x/10))*(1-exp(-2*1*x)))

diagonal

cross−peak

τm

where the diagonal peak intensities are12 (1 + e−2kτm)e−t/T1 and the cross-peak intensities

are 12 (1 − e−2kτm)e−t/T1

Questions:

Why is T1 only included here?

Does T2 contribute?

In what way?