NMR Course, LecbJre 6 14.28 Apri114, 1999

Lecture 6

Chapters (pages) in 'Modem NMR Techniques for Chernistry Research' by A. E. Derorne :

.Notbing unfortunately

r-')

r"'-

Other useful books or scientific papers related to the topics in this lecture :

.'Proton and Carbon-13 NMR Spectroscopy. An Integrated Approach' by R. J. Abraham and P.Loftus : Ch. 4

.'Nuclear Magnetic Resonance Spectroscopy. A physicochemical View' by R. K. Harris :Ch. 1.12 -1.18, Ch. 2

.'NMR Spectroscopy' by H. Giinther : Ch. 2.2 -2.3 and Ch. 5

.'A Practical Guide to Pirst-Order Multipiet Analysis in lH-NMR Spectroscopy' by T. R. Hoye,P. R. Hanson and J. R. Vyvyan, J. Org. Chem, 59, p. 4096-4103 (1994)

.Many books on chemistry , spectroscopy and physical chemistry in general cover the basics ofthe analysis of NMR spectra

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Page 1 / 1

NMR Lecture 6 15.00 April 12, 1999

Magnetically equivalent nuclei :

T wo chemically equivalent nuclei do not have to be magnetically equivalentA pair of nuclei can be magnetically equivalent only if theyare chemically equivalent Chemicalequivalence is necessary hut not sufficient

CI Fy

A2X spin system AA 'XXI spin system

Both molecnles above have chemically equivalent protons (and flnorines)The protons (and flnorines) are not magnetically equivalent in the molecule on the right-This is understood by inspecting the conpling constants between HA and F X and between HB

and FX. Those t wo conplings are not identical, one is a meta conpling the other is an orthoconpling, and therefore the two protons can not be magnetically eqnivalent

If all nuclei in a molecule are chemically equivalent (benzene) they must also be magneticallyequivalent since there is no other, different, nucleus which would allow the distinction betweenthe chemical and magnetic equivalence-The individual nuclei in a group of chemically equivalent nuclei must be compared (using 1) to

another nucleus which is not part of that group

HE HF

HA.:§HB Hc

In the system above (left) the t wo labelled carbon atoms (CE and CF) are 13C othelWise therewould be no nucleus for comparison of J.All four protons are chemically equivalent (symmetry). The pairs A,B and C,D are not

magnetically equivalent since they couple differently to E (l3C).Are A and B magnetically equivalent? How do A and B couple to C?By symmetry we see that J AC=JBC and J AD=JBO. A and B are magnetically equivalent (as are Cand D).

The system is therefore of the type A2Ai XX'

In the system on the right, the four protons (A-D) are chemically equivalent.The pairs A,B and C,D are not equivalent since they couple differently to HE.J AC ~ JBC and therefore HA and HB (and Hc and HO) are not magnetically equivalent, only

chemicallyequivalent.This system is then of the type AA ' A" A "'XX'

Equivalence does in general simplify NMR spectra considerably. This is especially true formagnetic equivalence

NMR Lecture 6 15.~ Apri112, 1999

Magnetically equivalent nuclei act as if there are no couplings between them-A system with four nuclei of the type A3X (three magnetically equivalent nuclei) is as simple

as the two-spin system AX (first-order) while four-spin systerns of the type ABCX (noequivalent nuclei) or AA'A"X (three chemically equivalent nuclei) are rather complicated

Analysis of non-trivial spectra (not of first order)

-Quantum mechanics must be used (see section on QM in lecture 5)-The energies of our spin system. here it is examplified by an AB system. is computed using the

Schrödinger equation and the t wo NMR related Hamiltonians

!}{\IJ = E'P Eq. 1-10) -11) A A

.1{ = .1[' + .1[' = -(v AmA + VBmB) + J ABIAIB Eq.2

Where mA and mB are the magnetic quantum numbers for nuclei A and B~

r

Simple example: The AX system

-In case of Av » J AB (flfst-order AX system) then the solution is simply

E = -(v AmA + vBmB) + JABmAmB Eq.3

The energies of the four states given by the four possible cornbinations of the individual spin states

(aa, a~, ~a and ~~) are then :

A 1 1p~ 2(VA +VB) + 4 J

1 1~a 2 (v A -VB) -4 J

1 1a~ -2 (v A -VB) -4 J

1 1aa -2 (v A + VB) + 4 J

f"'

/"'""

States with anti-parallell spins, af} and f}a, are stabilired energetically by the coupling (by ~ )

while the t wo states corresponding to parallell spins, aa and f}f}, are destabilired by the

coupling interaction (by ~ )

The energies of the transitions for nucleus A (spin changes for A only) are :

a f} ~ f} f} : E = v A + t J and a a ~ f} a: E = v A -t J

And the energies of the transitions for nucleus B (spin changes for B only) are :

f} a ~ f} f} : E = v B + t J and a a ~ a f} : E = v B -t J

In our spectrum we will see two doublets where all four signals are of equal intensity .

The transitions come in the following order (starting downfield going toward higher field) :

af} ~ f}f} (Al transition, highest energy), aa ~ f}a (A2 transition), f}a ~ f}f} (Bl transition)

and aa ~ af} (B2 transition, lowest energy)

The AB spin system

-In this case Av z J AB

-For this strongly coupled system, the Hamiltonian has to be solved explicitly-The Schrödinger equation can be written as (using Dirac notation)

I <Pm 19lI'Pn>- E8mnl = O

Where Ömn=l for m=n and is O otherwise

Page 3/11

NMR Lecture 6 15.~ April 12, 1999

-Equation 4 is the secular determinant that needs to be solved to obtain the eigenvalues (energies)and eigenfunctions (w ave functions)

-The elements <'I' m !9fI'l' n> are given from Eq. 2 and the w ave functions 'I' are the basic w ave

functions for this system, ie the states a.a, a.f3 etc

-The secular determinant for this AB system will be a 4 x 4 determinant which is shownschematically below

a.f3 J3a J3J3A B aamT

a

~

a

~

I<aalH laa.>-E: O O O

I I

O: : OI II

O: : OI O O O :

I

a

a

(3

(3

=o

r--

The diagonal elements of this detenninant are computed from Eq. 3. All off-diagonal elementsbetween w ave functions with different mT equal zero.

The determinant has now been broken down to t wo single detenninants (elements) and one 2 x 2

determinant for the a.r3 and r3a. w ave functions

Using the variational method of quantum mechanics we obtain the coefficients for the linear

combination of w ave functions a.r3 and r3a.

Continuing with the variational method and using an angle, e, which is related to the mixingcoefficients we obtain

sin 2e = ~

2e (v A -VB)cos = D

D = V(VB -v A)2 + ]2

The stationary states and the corresponding energies are easily calculated using these new tenns

r-

"

If öv » J (an AX system) then the term D = öv and t = Q. Then follows that e = Q" and the

af} and f}a spin functions are not mixed at all

If ov = O (an A2 system) then the mixing is maximal and 8 = 45"

The same angle can be used to express the relative intensities of the resonance lines-The simple rules involving binornial coefficients (Pascal's triangle) are not useful in strong ly

coupled systerns

Page 4/11

I

O

O

-I

NMR Lectln'e 6 15.00 April 12, 1999

b a

~I ~-I åv/J=16

" VAVB'" "

r'--I I I v' L\v / J = 101'.,1 1, 1 -

, ,

'1,,1 I",r" L\v / J = 5

\ Ir-

~\., I I t'~ A v I J = 2.5

\ I

\ II tt A1'\11(1 L\v/J=l

~ II: A~lIf L\v / J = 0.5

'J: "I~ L\v/J=O

v

r"""'

Simulated NMR spectra with a constant J of 5 Hz and varying v A and VB.

i\v varles between 80 H z and 0 Hz. Note the changes ofintensirles as weIl as the relarlve

posirlons of the resonance lines and v

We can calculate v A and VB as weIl as J AB from the observed lines in our spectrum :

r Fb -FaJ -~

-[ Fd -Fc"

ov = V(Fd -Fa)(Fc -Fb)

(OV Examp1e from one ofthe spectra above T = ]

Fa = 43.96 Hz Fb = 48.96 Hz c = 51.04 Hz

Fb -Fa = 5.0 Hz

Fd-Fc=5.0Hz

OV = V(Fd -Fa)(Fc -Fb) = V(56.04 -43.96)(51.04 -48.96)

Fd = 56.04 H z

J =

= 5.01 H z

The 'experimenta!' spectrum was simulated using the foIlowing settings:

J = 5 H z and v A = 47.5 H z and VB = 52.5 Hz.

The values calculated here agree very weIl with the 'experimental' spectrum

Page 5 111

NMR Lecture 6 15.00 April 12, 1999

Spin systerns with three or more nucleiThese systerns can in principle be calculated as weIl, the general procedure is the same. Cornplicationsdo arise however

"

-We consider a three-spin system of the type ABC

-The basic w ave functions are now <1.<1.<1., <1.<1.13, <1.13<1.,..., 131313

-There are eight of those basic w ave functions (23) leading to an 8 x 8 secular determinant

-The 8 x 8 determinant can be factorized into t wo single elements ( <1.a.a. where mT = + t and

131313 where mT = -~ ) and two 3 x 3 determinants from the states with mT = +

1 12 «1.<1.13, <1.13<1., 13<1.<1.) and mT = -2 «1.1313, 1313<1., 13<1.13)

-The 3 x 3 determinants give cubic equations in energy which can not be solved explicitly.

2 x 2 is the largest determinant we can solve explicitly, but we can solve many of them.U sing iterative computational methods is how the problems with large matrices are solved

-If two of the nuclei are equivalent we obtain an AB2 (or A2B) spin system and this we cansolve explicitly

-Molecules with a two-fold axis of symmetry (and some others too ) orten give AB2 spectrar

Ph

ICH2I

ROOC/CH 'COOR

r

CI

There is no ohservahle coupling hetween the t wo B nuclei and the transition energies and

intensities depend only upon v A, VB and J AB

In general there are 7 or 8 ohserved lines in our spectrum (a ninth should be there hut intensity is

often too low since it is a forhidden transition (app ~ paa))

Using the variational method one ohtains the following energies (in units of frequency) for thenine transitions :

.f'\

Page 6111

NMR Lecture 6 17:17 PM April 12, 1999

1- I ~ 9]2The term C+ in the tablesrefersto:f:2V(vA .VB)2 +J(VA .VB)+ -T

1- I ~ 9]2The term C. in the tables refers to:f:2V(v A. VB)2 .J(V A. VB) + 'T

We can o btain v A and VB as weil as J AB from the lines observed in our spectrum using the

foilowing formula :

F5 + F7VB = 2

JAB = Fl -F4 + F6 -Fg

3v A = F3

Example

I I I I

The following 8 lines are observed in the spectrum (frequencies in Herz)Fl = 32.26 F2 = 28.26 F3 = 25.00 F4 = 21.00F5 = 17.26 F6 = 16.74 F7 = 12.74 F8 = 10.00

32.26- 21.00 + 16.74- 10.00..r = 6.00 H zFl -F4 + F6 -F8

J = 3

VA = F3 = 25.00 H z

F5 + F7 -17.26 + 12.74 = 15.00 H zVB = 2 -2

The ABX Spectrum-Here the Hamiltonian of the ABC is broken down by the X approximation since one nucleus

(X) has a chemical shift very different from the others :

(v X -v A) and (v X -VB) » J AX and JBX

-We can treat m X as a good quantum number (no mixing) separately from the A and B spin

-The 3 x 3 determinant of the three w ave functions with mT = + t ( aap, apa and paa) can

be reduced to a 2 x 2 determinant (can be solved explicitly) and a single element since we candivide the determinant into two depending on m X

Page 7111