NMR Spectroscopy

NMR Spectroscopy

CHEM 430Fall 2011

CHEM 430 – NMR

Spectroscopy

Introduction to Spectroscopy

Spectroscopy is the study of the interaction of matter with the electromagnetic (EM) spectrum

1. EM radiation displays the properties of both particles and waves

2. This “packet” of wave and particle properties is called a photon

The term “photon” is implied to mean a small, massless particle that contains a small wave-packet of EM radiation/light

3. The energy E component of a photon is proportional to the frequency n

E = hn

The constant of proportionality is Plank’s constant, h

2CHEM 430 – NMR Spectroscopy


4. Because the speed of light (c ) is constant, the frequency (n) (number of cycles of the wave per second) can complete in the same time, must be inversely proportional to how long the oscillation is, or wavelength (l):

5. Amplitude describes the wave height, or strength of the oscillation

6. Because the atomic particles in matter also exhibit wave and particle properties (though opposite in how much) EM radiation can interact with matter in two ways:• Collision – particle-to-particle – energy is lost as heat

and movement

• Coupling – the wave property of the radiation matches the wave property of the particle and “couple” to the next higher quantum mechanical energy level

3

n = ___

l

cD E = hn =

___

l

hc

CHEM 430 – NMR Spectroscopy


8. Remember atoms and molecules are quantum mechanical particles

9. Where a photon is a wave with some particle character, matter is made of particles with some wave character – wave/particle duality

10. As a result of this, the energy of these particles can only exist at discrete energies – we say these energy levels are quantized

11. It is easy to understand if we visualize the “wave” property of matter as an oscillating string in a box—only certain “energy levels” can exist as the string is bound at both ends:

4

Ener

gyCHEM 430 – NMR

Spectroscopy

The Spectroscopic Process

55

1. Irradiation: Molecule is bombarded with photons of various frequencies over the range desired

hn hn hn

5. Detection: Relaxation as photons are reemitted. Spectrometers differ whether they measure actual emission or absorbance

Ener

gy

2. Absorption: Molecule takes on the quantum energy of a photon that matches the energy of a transition and becomes excited

hn4. Relaxation

rest state rest state

excited state

3. Ex

citati

on


Types of Spectroscopy

66

UVX-rays IRg-rays RadioMicrowave

Energy (kcal/mol) 300-30 300-30 ~10-4> 300 ~10-6

Visib

le

Frequency, n (Hz) ~1015 ~1013 ~1010 ~105~1017

Wavelength, l 10 nm 1000 nm 0.01 cm 100 m~0.01 nm

nuclear excitation (PET)

core electron excitation (X-ray cryst.)

electronic excitation (p to p*)

molecular vibration

molecular rotation

Nuclear Magnetic Resonance NMR & MRI


NMR Spectroscopy

• NMR spectroscopy has emerged as the ultimate spectroscopic method for organic structural analysis

• Currently, the development of novel NMR methods is in its “golden age” with some of the 2-D methods entering their maturation period as routine spectroscopic methods

• A typical NMR sample consists of 1-10 mg of sample, with which a full analysis of 1H, 13C, DEPT, COSY, HMBC, HSQC and NOESY could be done in a few hours on a high-field instrument

• Important spin-offs of NMR spectroscopy include a host of medical and security imaging equipment


Brief History of NMR

• First NMR spectrum of H2O, 1946:

8

Bloch, F.; Hansen, W. W.; Packard, M. Phys. Rev. 1946, 70 474-85.



• First observation of chemical shift1H spectrum of ethanol – 1951 vs. 2011

9

Arnold, J.T., S.S. Dharmatti, and M.E. Packard, J. Chem. Phys., 1951, 19, 507.



• Fourier transform NMR by Ernst - 1966



• 2D NMR – 1975 Jeener and Ernst


Brief History of NMR (MRI)

• First magnetic resonance image – 1973 Lauterbur and

Mansfield

12

2011



• First 3-D spectrum of small protein - 1985 Wüthrich



Nobel Prizes for NMR•1944 Physics Rabi (Columbia)•1952 Physics Bloch (Stanford), Purcell (Harvard)•1991 Chemistry Ernst (ETH)•2002 Chemistry Wüthrich (ETH)•2003 Medicine Lauterbur (University of Illinois in Urbana ) and Mansfield (University of Nottingham)


Basis of NMR Spectroscopy

2.1 Magnetic Properties of Nuclei • The sub-atomic particles within atomic nuclei possess a spin

quantum number just like electrons

• Just as when using Hund’s rules to fill atomic orbitals with electrons, nucleons must each have a unique set of quantum numbers

• The spin quantum number of a nucleus is a physical constant, I

• For each nucleus, the total number of spin states allowed is given by the equation:

2I + 1



• Observe that for atoms with no net nuclear spin, there are zero allowed spin states

• Nuclear Magnetic Resonance can only occur where there are allowed spin states

• Note that two nuclei, prevalent in organic compounds have allowed nuclear spin states – 1H and 13C, while two others do not 12C and 16O

16

Spin Quantum Numbers of Common Nuclei

Element 1H 2H 12C 13C 14N 16O 17O 19F 31P 35ClNuclear Spin Quantum Number

½ 1 0 ½ 1 0 5/2 ½ ½ 3/2

# of spin states 2 3 0 2 3 0 6 2 2 4



2.1 Magnetic Properties of Nuclei • There are three types of nuclei:

• No spin: I = O • 12C, 16O• Cannot be observed by NMR

• Spinning sphere: I = ½ • 1H, 13C, 15N, 19F, 29Si, 31P)• Easiest to observe by NMR

• Spinning ellipsoid I = 1, 3/2, 2…• 2H, 11B, 14N, 17O, 33S, 35Cl• Difficult to observe by NMR



2.1 Magnetic Properties of Nuclei • A nucleus contains protons, which each bear a +1 charge

• If the nucleus has a net nuclear spin, and an odd number of protons, the rotation of the nucleus will generate a magnetic field along the axis of rotation

• A hydrogen atom with its lone proton making up the nucleus, can have two possible spin states—degenerate in energy

18

m

H Hm

I = +½ I = -½



2.1 Magnetic Properties of Nuclei • The magnitude of m varies from atom to atom :

m = għI

• ħ is Planck’s constant divided by 2p

• g is the characteristic gyromagnetic ratio of the nucleus

• The larger g is, the greater the magnetic moment


Nuclear Magnetic Resonance

In the absence of stimulus all nuclear spin

sates are degenerate

When a large magnetic field B0 is applied the two spin

states become non-degenerate

As B0 increases, the larger DE

becomes



2.1 Magnetic Properties of Nuclei • The large external magnetic field is defined as B0 in units of Tesla, T

• The axis of B0 is defined as the z-direction

• Splitting of spins into quantized groups is called the Zeeman effect

21

DE

B0 +z directionCHEM 430 – NMR Spectroscopy


2.1 Magnetic Properties of Nuclei • The force of B0 causes m to move in a circular motion about the z-

direction – precession

• B0 field in z-direction operates on the x component of m to create a force in the y-direction (F = m X B0)

• This occurs with an angular frequency w0 known as the Larmor frequency (rad s-1)



2.1 Magnetic Properties of Nuclei • A quantum mechanical particle can absorb a photon of energy

equal to DE and become promoted to a higher state – spectroscopic process

• As B0 increases so does w0 (B0 w); the constant of proportionality is g:

w0 = g B0

• By equating w0 with Planck’s relationship:DE = hn0 = ħw0 = g B0



2.1 Magnetic Properties of Nuclei • What does this mean for the NMR experiment (observing DE)?

• Solving for the frequency of EM radiation we are observing:• For a bare hydrogen nucleus (H+), g = 267.53 (106 rad/T·s)• In a B0 of 1.41 Tesla:

DE = 60 MHzDE corresponds to the highly weak radio region of the EM spectrum:

l > 5 meters and energies of < 0.02 cal·mol-1

• This causes technical challenges to observing NMR



2.1 Magnetic Properties of Nuclei Boltzmann distribution – more problems with NMR observation•DE at 60 MHz (DE = hn) is 2.39 x 10-5 kJ mol-1 (tiny) – thermal energy at room temperature (298 oK) is sufficient to populate both energy levels

•DE is small, so rapid exchange is occurring between the two populations, but there is always a net excess of protons in the lower energy state

•From the Boltzman’s Law we can calculate the population of each energy state:

Nupper/Nlower = e-DE/kT = e-hn/kT

@ 298 oK the ratio is 1,000,000 / 1,000,009 !

There is an excess population of 9 nuclei in the lower energy state! 25CHEM 430 – NMR Spectroscopy


2.1 Magnetic Properties of Nuclei • As the applied B0 increases, exchange becomes more difficult and the excess

increases:

6. In each case, it is these few nuclei that allow us to observe NMR

7. When radio radiation is applied to a sample both transitions upward and downward are stimulated – if too much radiation is applied both states completely equilibrate – a state called saturation – no observed NMR signal

26

Field, B0

TFrequency

MHzExcess nuclei

1.41 60 91.88 80 127.05 300 489.40 400 6414.1 600 96



2.2 Commonly Studied Nuclides



2.2 Commonly Studied Nuclides

• Remember that the greater DE the easier it is to detect NMR active nuclei and have greater S/N ratios:



2.2 Commonly Studied Nuclides 1. Spin

• In general spin ½ nuclei are the easiest to observe

• Quadrapolar (I > ½) nuclei are more difficult to observe• Unique mechanism for relaxation gives very short relaxation time

• Heisenberg uncertainty principle dictates:DE Dt ~ ħ

• As relaxation times become very short, the uncertainty in energies becomes large and peaks broaden greatly



2.1 Magnetic Properties of Nuclei 2. Natural Abundance

• Remember from Boltzman’s Law we have only a small excess of nuclei we can observe: 60 MHz, 1.41 T: 1,000,000 / 1,000,009 for 1H

•1H is 99.985% of natural hydrogen - only 9 excess nuclei

•Consider the excess population when the nuclei we are observing is 13C – 1.11% of natural carbon•Spin couplings between low-abundant nuclei are also hampered:

• Chance of two 1H-1H on adjacent Cs: 99.985 x 99.985 = 99.97%• Chance of two 13C-13C adjacent to one another: 1.11 x 1.11 = 0.1%



2.2 Commonly Studied Nuclides 3. Natural Sensitivity - g

• Remember - DE = hn0 = ħw0 = g B0

NMR signal a function of only B0 and g

• The larger DE the greater the excess population for observation (Boltzmann distribution)

• 1H is best followed by 19F in routine observations



2.2 Commonly Studied Nuclides 4. Receptivity

• Mathematical product of abundance and g

• Good quick measure of amenable a nuclei is for observation

• Your text uses 13C as a guidepost rather than 1H

• Quick survey:• 1H is 5680 times easier to observe than 13C• 15N is 2.2% as easy to observe as 13C• 19F is 4730 times easier to obersve than 13C; 31P is 2 times easier



2.3 The Chemical Shift• Observation of the NMR phenomenon would be of little use if all

protons resonated at the same frequency

• In organic compounds protons are not bare nuclei, they are surrounded by an s-orbital of containing an e- shared with an e- in a hybridized orbital of another atom to form a covalent bond

• In the presence of an external magnetic field, an induced circulation of electrons opposite to that of a proton is observed since the two are of opposite charges

• This induced circulation generates a magnetic field in opposition to the applied magnetic field – a local diamagnetic current 33CHEM 430 – NMR

Spectroscopy


2.3 The Chemical Shift• Since the magnetic field “felt” by the proton within this electron cloud is

lowered, the resonance condition frequency is also lowered

• This effect of lowering the energy of transition by a cloud of electrons is called diamagnetic shielding or shielding - represented as s

• The opposite effect – if electron density is removed from the vicinity of the proton is called deshielding

• The actual field around the nucleus becomes B0(1 – s); substitution in the energy equation gives:



2.3 The Chemical Shift• Since the magnetic field “felt” by the proton within this electron cloud is

lowered, the resonance condition frequency is also lowered

• This effect of lowering the energy of transition by a cloud of electrons is called diamagnetic shielding or shielding

• The opposite effect – if electron density is removed from the vicinity of the proton is called deshielding

3535

DE

B0 +z directionCHEM 430 – NMR Spectroscopy

NMR Spectroscopy - Introduction 2-6

• In acetic acid, the –CH3 protons are in an e- rich environment relative to the –OH proton.

• Shielding of the electrons opposes B0 and therefore DE is lower than that observed for the –OH proton. Here DE is large as the full effect of B0 is felt

36

DE –CH3

B0 +z direction

DE –OH



37

DE

B0


Basis of NMR Spectroscopy• The effect of electrons on a 1.41 T magnetic field is negligible, but

measurable

• Compare the resonance frequencies for the protons in fluoromethane vs. chloromethane

CH3F CH3Cl

• The stronger inductive w/d of electrons by fluorine reduces the resonance frequency by 72 Hz (not MHz) compared to an operating frequency of the instrument at 60 MHz @ 1.41 T – barely 1 part per million (ppm)

• Using units of 60000072 vs. 60000000 is clunky at best

• There needs to be a reference “proton” by which these “chemical shifts” can be related - the best candidate would be a completely deshielded proton (H+) which does not exist in the solution phase 38CHEM 430 – NMR

Spectroscopy

Basis of NMR Spectroscopy• NMR spectroscopists chose the other end of the spectrum- a proton that

was more shielded than any other known proton (at the time) – those in tetramethylsilane (TMS)

• The 12 chemically identical protons in TMS were used as the standard zero for an NMR spectrum

• The resonance frequency of any proton to be studied (since all were less shielded) would be at parts per million of the operating frequency of the instrument greater than this zero

• This allowed NMR instruments of varying field (and thus operating frequency) strengths to use the same scale

• Here’s how:


Basis of NMR Spectroscopy• In an applied field of 1.41 T, the resonance frequency for a typical proton is

60 MHz, at 2.35 T it is at 100 MHz – a ratio of 5/3

• Thus, for a given proton, the shift in Hz from the TMS standard should be 5/3 greater in the 100 MHz instrument compared to the 60 MHz

• Since these are simple ratios, we can simply factor out the effect of field strength by defining d, or chemical shift to be

d = (shift from TMS in Hz) (spectrometer frequency in MHz)

…or ppm of the instruments operating frequency



90 Mhz spectrum 300 MHz spectrum

4190 Hz 300 HzCHEM 430 – NMR

Spectroscopy

Basis of NMR Spectroscopy• In an applied field of 1.41 T, the resonance frequency for a typical proton is

60 MHz, at 2.35 T it is at 100 MHz – a ratio of 5/3

• Thus, for a given proton, the shift in Hz from the TMS standard should be 5/3 greater in the 100 MHz instrument compared to the 60 MHz

• Since these are simple ratios, we can simply factor out the effect of field strength by defining d, or chemical shift to be

d = (shift from TMS in Hz) (spectrometer frequency in MHz)

…or ppm of the instruments operating frequency


• A detailed study of chemical shifts is the basis of Chapter 3

43


Si

CH3

H3C CH3CH3

0.010downfield d (ppm) upfielddeshielded shieldedhigher DE lower DE

9 8 7 6 5 4 3 2 1

H

C EWG

H

C

C

H

CH

O

15

C

C

H

CO

O

H

SO

O

H

O

Ph

OH

H

C

R

R R

R = H or alkyl



Continuous-Wave (CW) Instrument An NMR spectrometer needs to perform several functions:• Generate a high (>1 Tesla) magnetic field to split the energy levels of the

spin states enough to:– Create an excess nuclei population large enough to observe– Make the radio n that correspond to the transition be

observable

• Ensure that the field is homogeneous (shimming)

• Vary either the applied field or the radiofrequency (RF) to observe different nuclei at their various energies of transition

• Receive the faint signal of the relaxation of the excited nuclei to their ground state

• Process the signal into a usable spectrum vs. a reference 44

Spectrometer Design



Continuous-Wave (CW) Instrument:

45

RF (60 MHz) oscillator

Permanent Magnet

RF Detector

Variable magnetic field – 1.41 T ± few millionths of TCHEM 430 – NMR

Spectroscopy


How it works (CW NMR): • The sample is placed in a 5 mm solution cell or tube (experimental aspects

we will cover shortly) in the center of a large permanent or electromagnet

• A RF oscillator coil at 90° to the sample generates a radio signal at the operating frequency of the instrument (60 MHz for a 1.41 T field)

• The overall magnetic field is varied by a small electromagnet capping the poles of the larger field magnet

• Remember: DE = n = (g/2p) B0, so variations of either magnetic field or frequency will cover the observed spectral width if the other is held constant

• As with older dispersive IR instruments, the sweep of magnetic fields is simultaneous with the movement of the chart paper



How it works (CW NMR): • As a particular proton population comes into resonance, a second receiver

coil at 90° to the transmitter coil will pick up the change in orientation of nuclear spin

• This is recorded by the chart as a voltage response, proportional to the size of the proton population that generated the resonance

• One artifact of CW instruments is that the relaxation of the protons is slower than the movement (sweep) of the chart paper

• This causes the ringing effect – a decreasing oscillation of the signal after the spectrometer has moved past a given resonance

• CW instruments operate by bringing each individual population of protons into resonance individually.



Limitations- CW NMR: • Since the spectrum is collected once, the sample must possess enough

protons to give a suitable excess population that can be observed – need a concentrated sample

• Due to the limitations of the relatively low magnetic field (CW instruments top out at 60-90 MHz) the coupling constants for JHH are relatively large compared to the spectral width – so only simple molecules can be observed and their structures elucidated

• For nuclei of lower magnetogyric ratios, g, or natural abundance (13C most specifically) the ratio of radio noise to signal is high



2.4 Excitation and Relaxation – FT-NMR• Let’s revisit the NMR experiment we considered earlier

• Recall that at even a high applied B0 only a small excess of nuclei are in the lower spin state (+ ½ ) as per the Boltzman distribution

Ex: @ 7.04 T an excess of 50 spins per million

• It is important to also note that 298 K imparts enough energy to the system such that all spins are interchanging rapidly:



2.4 Excitation and Relaxation – FT-NMR• So when we are discussing an NMR sample we have quadrillions of

protons creating a net magnetization M in the z direction (B0)

• The xy components are distributed randomly, and we can think of the rest state as M = Mz

• M precesses at the Larmor frequency. We are using the rotating frame of reference to make the visualization easier.

50

x

y

z

M

z

y

x

Bo



2.4 Excitation and Relaxation – FT-NMR• Important note:

• Each individual spin can only exist in either quantized state (± ½) separated by DE

• The bulk magnetization however can exist at a continuum of states

51

M

z

y

x

Bo



2.4 Excitation and Relaxation – FT-NMR• If a RF (w0 = w) is applied that creates a magnetic oscillation

along the x-axis a torque is applied to M rotating it towards y

• With the application of energy, a small amount of the excess population has flipped spins. Mxy < M0

52

B1

(or off-resonance)

Mo

z

y

B1

z

y

Mxy

x xwo

wo



2.4 Excitation and Relaxation – FT-NMR• If RF energy continues to be applied the excess of + ½ nuclei

disappears – sample at saturation – no NMR signal can be detected

• Natural mechanisms return an excess population to equilibrium

• Any process that returns the z-magnetization to equilibrium with an excess of + ½ spins is called spin-lattice or longitudinal relaxation

• It is usually a first-order process and with a time constant of T1



2.4 Excitation and Relaxation – FT-NMR• Other magnetic nuclei in the NMR sample tumble, generating a

field from its motion

• If the field is at w, excess spin energy can pass to this motional energy as – ½ nuclei become + ½ nuclei

• For this process to be effective the nuclei need to be spatially proximate to the tumbling molecule.

• Most H-atoms are on the outside of molecules and have roughly equal probability to relax by this process - #1Hs ~ signal



2.4 Excitation and Relaxation – FT-NMR• C-atoms differ substantially in their contact with external

molecules – T1 relaxation usually by attached protons

• Therefore T1 relaxation is more efficient for a –13CH3 rather than a quaternary carbon or 13C=O.



2.4 Excitation and Relaxation – FT-NMR• T1 relaxation is responsible for generating the excess of + ½ nuclei

in the first place!

• When the sample is first placed in the B0 field, all spins are degenerate

• Magnetization builds up as spins flip from the effect of interactions with surrounding magnetized nuclei



2.4 Excitation and Relaxation – FT-NMR•Any process that returns xy magnetization to its equilibrium position of zero is called spin-spin or transverse relaxation

•It is a first order process with time constant T2

•By definition T2 < T1

•For T2 relaxation the phases of nuclear spins must become randomized

•The mechanism for this is when spin is transferred between nuclei of opposite spin



2.4 Excitation and Relaxation – FT-NMR•When one spin goes from +1/2 to -1/2 while the other goes from -1/2 to +1/2 there is no net change in z magnetization

•However switches in spin cause dephasing; as the process continues xy magnetization disappears



2.4 Excitation and Relaxation – FT-NMR•Detection – as xy magnetization dephases the vectors begin to rotate about z in the rotating frame of reference

•A second coil 90o to the transmitter coil detects the decay in the xy

59

z

y

Mxy

Receiver coil (x)

x

NMR signal

wo



2.4 Excitation and Relaxation – FT-NMR•The NMR process:



2.4 Excitation and Relaxation – FT-NMR•Introduction to Pulse Sequences

•Shorthand:

61

90x

Pulse

Mixing timeDetection


Pulsed Fourier Transform (FT) Instrument

First, what is a Fourier Transform?

Fourier transforms interconvert mathematical functions in the frequency domain to the time domain:

For purposes of this discussion, we will black box the actual calculations and derivations of these functions, but we need to understand what they do

62

f(n) = ∫ f(t) e-int dt

f(t) = ½ p ∫ f(n) eint dt-

-


Pulsed Fourier Transform (FT) Instrument

If we feed two simple oscillating equations into a FT, here are the results:

63

Spectrometer Design

f (t) = cos (nt)

FT-n +n

f (t) = sin (nt)

FT

-n

+n


NMR Spectroscopy

Pulsed Fourier Transform (FT) Instrument • In the FT instrument, all proton populations are excited simultaneously by a short,

intense burst of RF energy

• Due to a variation of the Heisenberg Uncertainty Principle, even if the RF generator is set at 90 MHz, if the duration of the pulse is short, the radio waves do not have time to establish a solid fundamental frequency

• This can be illustrated by the following cartoon, showing the combination of a short pulse being added to a step function:

64

Spectrometer Design

+ =

tp = pulse duration

off

on

off

on

off off

on


Pulsed Fourier Transform (FT) Instrument 4.If this short pulse is converted into the frequency domain by a FT:

Observe that we have a continuum of frequency content centered at the operating frequency of the instrument

We will talk more about the effects of pulse time and width when we discuss advanced 1-D and 2-D NMR

65

Spectrometer Design

tp

off off

on

FT

n


Pulsed Fourier Transform (FT) Instrument 5.For now, if a sample containing one unique population of hydrogens was excited

over tp by a pulse, it would then “relax” back to its original spin state

6.As each nuclei relaxes it will emit RF radiation of a given frequency; since different nuclei will relax at different rates, the signal decays over time

7.This emission is recorded by the spectrometer as a free-induction decay or FID

66

Spectrometer Design

off off


Pulsed Fourier Transform (FT) Instrument 8.The actual frequency of the FID is the interference signal of the relaxing protons

superimposed with the frequency of the RF source

9.Conversion of this decay signal by FT back into the frequency domain gives us the actual n of resonance for the proton being observed

10.Again to due Heisenburg and other factors, NMR signals are not single lines, but a Lorentzian shaped continuum of lines centered at the n of the signal

67

Spectrometer Design

FT

n

Proton signal Pulse n

time


Pulsed Fourier Transform (FT) Instrument Advantages:Since all nuclei are excited and observed simultaneously, the pulse can be repeated

after each relaxation period (for 1H, about 10 seconds) and the resulting signals added together

Because we are observing weak radiofrequency signals in a sea of RF noise for dilute samples (or those observed once as in CW NMR) noise becomes an issue

If several to hundreds of FIDs are added together, signals will tend to constructively add together and become more pronounced; since noise is random, it will tend to destructively add and become less pronounced

Signal to noise ratio improves as a function of the square root of the scans (FIDs) performed: S/N = f (n)

68

Spectrometer Design



2.4 Excitation and Relaxation – FT-NMR•Back to Pulse Sequences•Standard 1-D observation of 1H or other nuclei

•Delay must be longer than T1 (and T2 < T1)

•Scans are repeated until S/N ratio is high enough

69

90x

Pulse

Detection

n = scans

Delay



2.4 Excitation and Relaxation – FT-NMR•How can you know what T1 is? •We cannot measure it directly – in the z-axis•Pulse sequence is used called inversion recovery:

70

180y (or x) 90y

tD



2.4 Excitation and Relaxation – FT-NMR

71

z

x

y

tD = 0

z

x

y

tD > 0

z

x

y

tD >> 0

z

x

y

z

x

y

z

x

y

FT

FT

FT


NMR Spectroscopy 2-6

Basis of NMR Spectroscopy• If we plot the intensity versus time we get the following curve:

• It is an exponential with a time constant equal to the T1 relaxation time.

• Most 1Hs decay in less than 10 seconds – no need to run routinely

72

time

Inte

nsity

, (I)

I(t) = I * ( 1 - 2 * e - t / T1 )




2.4 Excitation and Relaxation – FT-NMR•How can you know what T2 is? •We cannot measure it directly like T1

•Pulse sequence is used called spin-echo sequence

73

180y (or x)90y

tD tD




2.4 Excitation and Relaxation – FT-NMR•What it does:

74

z

x

y

x

y

x

y

x

y

x

y

tD

180y (or x)

tD

dephasing

refocusing




2.4 Excitation and Relaxation – FT-NMR• If we acquire the FID right after the spin-echo sequence, the intensity of the signal after FT will only be affected by T2 relaxation and not by dephasing due to B0 imperfections.

• Upon repetition for different tD values, we plot the intensity versus 2 * tD and get a graph similar to the one we got for inversion recovery, but in this case the decay rate will be equal to T2.



The Coupling Constant

• Consider the spectrum of ethyl alcohol:• Why does each resonance “split” into smaller peaks?

CHEM 430 – NMR Spectroscopy 76

HOCH2

CH3



• The magnetic effects of nuclei in close proximity to those being observed have an effect on the local magnetic field, and therefore DE

• Two examples of close proximity are geminal and vicinal protons (homonuclear) and protons attached to 13C (heteronuclear):

• On any one of the 108 of these molecules in a typical NMR sample, there is an equal statistical probability each proton is either spin + ½ or – ½



HC

H HC C

H

geminal vicinal

homonuclear

13C1H

heteronuclear


• This creates two different magnetic environments for a proton being observed – one where its neighbor is +½ the other where its neighbor is –½

• If there is more than one proton on an adjacent carbon – all the statistical probabilities exist that each one is either + ½ or – ½ in spin

• The summation of these effects over all of the observed nuclei in the sample is observed as the spin-spin splitting of resonances




• Consider 1-chloro-4-nitrobenzene


NO2

Cl

HA

HX

If we are observing the resonance for the HA proton, in half the molecules in the sample HX would be –½ in the other half the spin would be +½

We would see the resonance of HA “split” into two different resonances (energy states):





NO2

Cl

HA

HX

Likewise, if we are observing the resonance for the HX proton, in half the molecules in the sample HA would be –½ in the other half the spin would be +½

We would see the resonance of HX “split” into two different resonances (energy states):





NO2

Cl

HA

HX

The observed 1H NMR spectrum shows two doublets, one for HA the other for HX

HA and HX are described as a spin-system



• The influence of neighboring spins on peak multiplicity is called spin-spin coupling, indirect coupling or J-coupling

• The difference between the component peaks of a resonance is a measure of how strong the interaction is between adjacent nuclei

• This difference is called the coupling constant, J, measured in Hz


3JH-cis# of bondsseparating nuclei

Nuclei that are coupledand stereochemistry whereappropriate



• Since J is a measure of interaction between two nuclei, it must be the same for both nuclei

• Similarly, J is independent of the applied B0 and will have the same value regardless of the field strength of the NMR


60 MHz – propyl bromide


J-values are =

J-values are =



• Since J is a measure of interaction between two nuclei, it must be the same for both nuclei

• Similarly, J is independent of the applied B0 and will have the same value regardless of the field strength of the NMR




J-values are =

J-values are =



• Recall, we are observing the frequency (E = hn) where a proton goes into resonance


Any change in B0 will cause a change in energy at which the resonance condition will occur for a proton of a given chemical shift



• In solution we are not looking at a single molecule but about 108

• On some molecules the proton being observed may be next to another proton of spin + 1/2 :




• On some molecules the proton being observed may be next to another proton of spin – 1/2 :




• Mechanism for coupling: usually oversimplified in early studies of NMR - If molecules can rotate in solution any spatial effect of one protons magnetic field on another is averaged to zero

• The most common mechanism involves the interaction of electrons along the bonding path between the nuclei.

• Electrons, like protons, act like spinning particles and have a magnetic moment. The X proton ( HX) influences, or polarizes, the spins of its surrounding electrons, making the electron spins favor one Iz state very slightly.

CHEM 430 88



• Thus, a proton of spin +½ polarizes the electron to –½. The electron in turn polarizes the other electron of the bond, and so on, finally reaching the resonating A proton (HA).

• Because J normally represents an interaction through bonds, it is a useful parameter for drawing conclusions about molecular bonding, such as bond order and stereochemistry.

CHEM 430 89



• Observe what effect this has on an isolated ethyl group:

• The two methylene Ha protons have three neighbors, Hb, on the adjacent methyl carbon

• Each one of these hydrogens can be + ½ or – ½ , and since we are not looking at one molecule, but billions, we will observe all combinations


C C

HbHa

R

HbHb

Ha



• The first possibility is that all three Hb protons have a + ½ spin; in this case the three protons combine to generate three small magnetic fields that aid B0 and deshield the protons – pushing part of the resonance for Ha slightly downfield (the magnetic field of a proton is tiny compared to B0)


All 3 Hb protons + ½

C C

HbHa

R

HbHb

Ha

Resonance, n, in absence of coupling

+++



• The second possibility is that two Hb protons have a + ½ spin and the third a - ½ ; in this case the two protons combine to enhance B0 and the other against it, a net deshielding; there are 3 different combinations that generate this state


3 combinations of+½, +½ , -½

C C

HbHa

R

HbHb

Ha


+++

or or

-+++-+++-



• The third possibility is that two Hb protons have a –½ spin and the third +½; here, the two protons combine to reduce B0 and the other enforce it, a net shielding effect; there are 3 different combinations that generate this state


C C

HbHa

R

HbHb

Ha


+++

-+++-+++-

or or

--+-+-+--

3 combinations of+½, -½ , -½



• The last possibility is that all three Hb protons have a – ½ spin; in this case the three protons combine to oppose B0, a net shielding effect; there is one combination that generates this state:


All 3 Hb protons -½

C C

HbHa

R

HbHb

Ha


+++

-+++-+++-

--+-+-+-- ---



• The result is instead of one resonance (peak) for Ha, the peak is “split” into four, a quartet, with the constituent peaks having a ratio of 1:3:3:1 centered at the d (n) for the resonance :


C C

HbHa

R

HbHb

Ha


+++

-+++-+++-

--+-+-+-- ---

-CHa2- is ‘split’ into a quartet by three adj. Hbs



• Similarly, the Hb protons having two protons, on the adjacent carbon each producing a magnetic field, cause the Hb resonance to be split into a triplet


C C

HbHa

R

HbHb

Ha


++-++- --

-CHb3- is ‘split’ into a triplet by two adj. Has



• Rather than having to do this exercise for every situation, it is quickly recognized that a given family of equivalent protons (in the absence of other spin-coupling) will have its resonance split into a multiplet containing n+1 peaks, where n is the number of hydrogens on carbons adjacent to the carbon bearing the proton giving the resonance – this is the n + 1 rule


# of Hs on adj.

C’s

Multiplet

# of peaks

The relative ratios of the peaks are a mathematical progression given by Pascal’s triangle:

0 singlet 1 11 doublet 2 1 12 triplet 3 1 2 13 quartet 4 1 3 3 1 4 quintet 5 1 4 6 4 15 sextet 6 1 5 10 10 5 16 septet 7 1 6 15 20 15 6 1



• Common patterns:



X CH3X

XX

X

tert-butyl - singletmethyl - singlet

iso-propyl – septet - doublet

ethyl – quartet - triplet

n-propyl – triplet - quintet - triplet








• Heteronuclear coupling between 1H and 13C are not apparent in 1H spectra (13C low abundance - 1.1%), In 99 of 100 cases, 1H are attached to nonmagnetic 12C atoms.

• In the 13C spectrum, carbon nuclei are coupled to 1H directly (99% abundant) attached to the carbon.

• Thus, the 13C resonance of a methyl carbon is split into a quartet, that of a methylene carbon CH2 into a triplet, and that of a methine carbon CH into a doublet. A quaternary carbon is not split by one bond coupling.

CHEM 430 101



• Shown here (top) the 13C spectrum of 3- hydroxybutyric acid which contains a carbon resonance with each type of multiplicity.

• From right to left are seen a quartet CH3 , a triplet CH2 , a doublet CH , and a singlet CO2H (C=O)

• Instrumental procedures, called decoupling, are available by which spin– spin split-tings may be removed. These methods, discussed in Section 5- 3, involve irradiating one

CHEM 430 102

HO

O OH


undecoupled

decoupled


• Instrumental procedures, called decoupling, are available by which spin– spin splittings are removed for clarity.

• These methods involve irradiating one nucleus with an additional field B2 while observing another nucleus resonating in the B1 field.

• 13C spectra are usually run decoupled as other spectral techniques are used to establish 1H-13C connectivity more rapidly and with more clarity

CHEM 430 103

HO

O OH


undecoupled

decoupled

NMR Spectroscopy

Spin-spin splitting – 1H NMR

The next level of complexity (which we will cover in detail in Chapter 4) is when protons on adjacent carbons exert different J’s than one another.

Consider the ethylene fragment:

104

The NMR Spectrum - 1H

C

CX

H

H

HThe magnetic influence of the trans- relationship is over the longest distance

The cis-relationship, is over an intermediate distance

The influence of the geminal-relationship is over the shortest distance


NMR Spectroscopy


For ethylene we would then observe three chemically distinct resonances with spin-spin splitting exerted by the other two protons:

J couplings:

105


2Jgem = 0 – 1 Hz

3Jtrans = 11- 18 Hz

3Jcis = 6 - 15 Hz

The observed multiplet for Ha is a “doublet of doublets”

3JAB3JAB

3JACC

CX

Ha

Hc

Hb


NMR Spectroscopy


Similar behavior is observed with aromatic rings; since the ring structure is fairly rigid and electronic effects are conducted over a longer distance, J – couplings are observed across the ring system:

106


In low-field 1H NMR the signal for this proton would be split into a doublet by the proton ortho to it.

On a high field instrument one finds this 3Jortho as well as a 4Jmeta and a 5J para from the effect of the protons meta and para to it

Typically:3Jortho = 7-10 Hz4Jmeta = 1-3 Hz5J para = 0-1 Hz

XH

Hortho

Hmeta

Hpara

H

3Jortho

4Jmeta

5Jpara


NMR Spectroscopy


For our initial treatment of 1H NMR the alkenyl, aromatic and the following J values should be learned:

107


3J = 6-8

3Ja,a = 8-143Ja,e = 0-73Je,e = 0-5

3Jtrans = 4-83Jcis = 6-12


3Jtrans = 11-18

3Jcis = 6-15

3Jallyl = 4-10

3J = 8-11

3J = 5-7

3Jortho = 7-10 Hz4Jmeta = 1-3 Hz5J para = 0-1 Hz

HH

Ha

He

Ha

He

H

H

HH

CH3

H

H

H

H

H

H

H

HH

O

HH

H

H

HHo

Hm

Hp


A typical 1H NMR is recorded from -2 to 15 d (ppm); what is typically reported is the region from 0 to 10 d

Remember, if a proton is shielded (e- circulation reduces “felt” magnetic field) DE for the transition is lowered and the signal is near the high field or upfield region of the spectrum (right)

If the proton is deshielded (e- circulation doesn’t reduce the “felt” magnetic field) DE for the transition is raised and the signal is near the low field or downfield region of the spectrum (left)

108


d or ppm 010

low DEshielded 1H reduces B0upfield

high DEdeshielded 1H sees full B0downfield


The number of signals observed will be equal to the number of unique populations of chemically equivalent protons

To determine if two protons are chemically equivalent, substitute “X” for that each respective hydrogen in the compound and compare the structures

If the two structures are fully superimposible (identical) the two hydrogens are chemically equivalent; if the two structures are different the two hydrogens were not chemically equivalent

A simple example: p-xylene

109


CH3

CH3

H

H

CH3

CH3

H

X

CH3

CH3

X

H

Same structure


The position (v) of each resonance is dependant on the electronic environment around the proton – chemical shift as a result of local diamagnetic shielding

There are three principle effects that contribute to local diamagnetic shielding:1) Electronegativity 2) Hybridization3) Proton acidity/exchange

110



Local Diamagnetic Shielding - Electronegativity

1. Electronegative groups comprise most organic functionalities:

-F -Cl -Br -I -OH -OR -NH2

-NHR -NR2 -NH3+ -C=O -NO2 -NO -SO3H

-PO3H2 -SH -Ph -C=C and most others

In all cases, the inductive w/d of electrons of these groups decreases the electron density in the C-H covalent bond – proton is deshielded – higher DE of transition

111




2. Protons bound to carbons bearing electron withdrawing groups are deshielded based on the magnitude of the withdrawing effect – Pauling electronegativity:

112


CH3F CH3O- CH3Cl CH3Br CH3I CH4 (CH3)4Si

Pauling Electronegativity

4.0 3.5 3.1 2.8 2.5 2.1 1.8

d of H 4.26 3.40 3.05 2.68 2.16 0.23 0.0



3. The magnitude of the withdrawing effect is cumulative:

4. The magnitude of the withdrawing effect is reduced by distance, as the inductive model suggests

113


CH3Cl CH2Cl2 CHCl3d of H 3.05 5.30 7.27

-CH2Br -CH2CH2Br -CH2CH2CH2Brd of H 3.30 1.69 1.25


NMR Spectroscopy

Local Diamagnetic Shielding - Hybridization

1. The hybridization of the carbon the proton is bound exerts a strong electronic effect

2. The greater the s-character, the more tightly bound the electrons are to carbon, raising its effective electronegativity (sp = 50% s, sp2, 33% s and sp3 25% s)

114


Type of H Name of H Chemical Shift, dR-CH3, R2CH2, R3CH alkyl 0.8-1.7

C=C-CH3 allyl 1.6-2.6CC-H Acetylenic 2.0-3.0C=C-H Vinylic 4.6-5.7Ar-H aromatic 6.5-8.5

O=C-H aldehydic 9.5-10.1

Something odd is happening here, as we will discussCHEM 430 – NMR

Spectroscopy

NMR Spectroscopy

Local Diamagnetic Shielding - Proton Acidity/Exchange

1. If an organic molecule that possesses hydrogen atoms of low pKA are dissolved in a deuterated solvent that also has a low pKA, the “visible” protons will exchange with “deuterium” from solvent and become “invisible” to the NMR spectrometer

Such studies are useful, if it is desired to see which H-atoms on an organic are acidic!

115


OH

D2OOD


NMR Spectroscopy

Local Diamagnetic Shielding - Proton Acidity/Exchange

2. Due to H-bonding effects, the resonance for certain functional groups (esp. –OH and –NH2) can change drastically dependent on concentration and the extent of the H-bonding

3. Just as in IR spectroscopy, peaks corresponding to these resonances are broad and often undefined – observing a continuum of bond strengths/electron densities about the observed proton

4. The correlation tables for the position of such protons tend to be broad and unreliable:• Acid –OH 10.5-12.0 d• Phenol –OH 4.0-12.0 d• Alcohol –OH 0.5-5.0 d• Amine –NH2 0.5-5.0 d• Amide –NH2 5.0-8.0 d• Enol CH=CH-OH >15 d

116



NMR Spectroscopy

Some observed 1H resonances can not be fully explained by local diamagnetic shielding effects

Magnetic Anisotropy – literally “magnetic dissimilarity”

For example, by our hybridization model, a proton bound to an sp2 C should be observed at lower d than a proton bound to an sp C

117

Type of H Name of H Chemical Shift, dR-CH3, R2CH2, R3CH alkyl 0.8-1.7

C=C-CH3 allyl 1.6-2.6CC-H Acetylenic 2.0-3.0C=C-H Vinylic 4.6-5.7Ar-H aromatic 6.5-8.5

O=C-H aldehydic 9.5-10.1


NMR Spectroscopy

Magnetic Anisotropy –

1. This effect is primary due to the fact that there is an additional effect of circulating electrons, observed in p-systems

2. In benzene, the 6-p-orbitals overlap to allow full circulation of electrons; as these electrons circulate in the applied magnetic field they oppose the applied magnetic field at the center – just like the circulation of electrons in the 1-s orbital about hydrogen – at the middle!:

118


B0


NMR Spectroscopy


3. On the periphery of the ring, the effect is opposite – the magnetic effect reinforces the applied B0, and DE becomes greater – deshielding effect

119



NMR Spectroscopy


4. This theory can easily be tested by the observation of large aromatic systems that possess protons inside the ring (now a shielding effect):

Or over a ring system: 120


HH

-1.8 d

8.9 d

H2C

CH2 -1.0 d

2.0 d


NMR Spectroscopy


5. In alkynes, a similar situation (to the central protons in large aromatic systems) arises where the terminal proton is in the region of maximum shielding

121



NMR Spectroscopy

General Correlation Chart – 1H NMR

Due to the three effects on local diamagnetic shielding, in conjunction with the effect of magnetic anisotropy 1H NMR chemical shifts are variable

• Avoid using hard and fast rules (tables of numbers)

• Instead, start from the general correlation table and deduce structural features based on the effects just discussed

• After a structural inference has been made, then use the more specific correlation tables to confirm the analysis

122



NMR Spectroscopy

General Correlation Chart – 1H NMR

Here are the general regions for 1H chemical shifts:

123


Si

CH3

H3C CH3CH3

0.010

downfield d (ppm) upfielddeshielded shieldedhigher DE lower DE

9 8 7 6 5 4 3 2 1

H

C EWG

H

C

C

H

CH

O

15

C

C

H

CO

O

H

SO

O

H

O

Ph

OH

H

C

R

R R

R = H or alkyl


NMR Spectroscopy


1. The magnetic effects of nuclei in close proximity to those being observed have an effect on the local magnetic field, and therefore DE

2. Specifically, when proton is close enough to another proton, typically by being on an adjacent carbon (vicinal), it can “feel” the magnetic effects generated by that proton

3. On any one of the 108 of these molecules in a typical NMR sample, there is an equal statistical probability that the adjacent (vicinal) proton is either in the + ½ or – ½ spin state

4. If there is more than one proton on an adjacent carbon – all the statistical probabilities exist that each one is either + ½ or – ½ in spin

5. The summation of these effects over all of the observed nuclei in the sample is observed as the spin-spin splitting of resonances 124



Intensity of Signals—Integration

• The area under an NMR signal is proportional to the number of absorbing protons

• An NMR spectrometer automatically integrates the area under the peaks, and prints out a stepped curve (integral) on the spectrum

• The height of each step is proportional to the area under the peak, which in turn is proportional to the number of absorbing protons

• Modern NMR spectrometers automatically calculate and plot the value of each integral in arbitrary units

• The ratio of integrals to one another gives the ratio of absorbing protons in a spectrum; note that this gives a ratio, and not the absolute number, of absorbing protons


NMR Spectroscopy

Integration – 1H NMR

1. Like instrumental chromatography, in NMR spectroscopy, the area under a peak (or multiplet) is proportional to the number of protons in the sample that generated that particular resonance

2. The NMR spectrometer typically will print this information on the spectrum as an integral line (stepped line on the spectrum below)

3. The height of the integral is proportional to that proton population; by comparing the ratios of the integrals on an NMR spectrum you can determine the number of protons as a least common multiple of these ratios

126



NMR Spectroscopy

Integration – 1H NMR

4. For example observe the integration of the ethanol spectrum below:

127


HOCH2

CH3

-OH

-CH3

1.25 units high

2.5 units high

3.75 units high







When a nuclei of spin +½

encounters a photon where n = E/h, the two

“couple”

The nuclei “tips” its spin state and is now

opposed to B0

The nuclei “relaxes” and

returns to the + ½ spin state



2.1 Excellent Simulators for NMR Phenomenon• 2-D compass needle analogy:

http://www.drcmr.dk/JavaCompass/

• 3-D Bloch equation simulator

http://www.drcmr.dk/BlochSimulator/


http://www.drcmr.dk/JavaCompass/

http://www.drcmr.dk/BlochSimulator/

NMR Spectroscopy

Nuclear Spin States• The sub-atomic particles within atomic nuclei possess a spin quantum number just

like electrons

• As with electrons, the nucleons are organized in energy levels

• Just as when using Hund’s rules to fill atomic orbitals with electrons, nucleons must each have a unique set of quantum numbers

• The total spin quantum number of a nucleus is a physical constant, I

• For each nucleus, the total number of spin states allowed is given by the equation:2I + 1

132

General Theory


NMR Spectroscopy

Nuclear Spin States

6.Observe that for atoms with no net nuclear spin, there are zero allowed spin states

7.All the spin states of a given nucleus are degenerate in energy

133

General Theory

Spin Quantum Numbers of Common Nuclei

Element 1H 2H 12C 13C 14N 16O 17O 19F 31P 35ClNuclear Spin Quantum Number

½ 1 0 ½ 1 0 5/2 ½ ½ 3/2

# of spin states 2 3 0 2 3 0 6 2 2 4


NMR Spectroscopy

Nuclear Magnetic Moments• A nucleus contains protons, which each bear a +1 charge

• If the nucleus has a net nuclear spin, and an odd number of protons, the rotation of the nucleus will generate a magnetic field along the axis of rotation

• Thus, a nucleus has a magnetic moment, m, generated by its charge and spin

• A hydrogen atom with its lone proton making up the nucleus, can have two possible spin states, degenerate in energy

134

General Theory

m

H H

m

+ ½ - ½


Nuclear Spin States5.In the presence of an externally applied magnetic field, these two spin states are no

longer degenerate in energy

6.The spin opposed orientation is slightly higher in energy than the spin aligned orientation

135

General Theory

m

H

Hm

+ ½

- ½

B0 – externally applied magnetic field

DE


Absorption of Energy• The energy difference between the two non-degenerate spin states in the presence

of an applied magnetic field is quantized

• At low B0 is easy to surmise that the potential energy of the spin opposed state would be low, and as B0 grows in strength, so would the potential energy

• Thus, with increasing strength of B0, DE between the two spin states also increases

136

General Theory

B0 – increasing

-1/2 +1/2

DE

-1/2

+1/2


NMR Spectroscopy

Absorption of Energy4.From theory we have already discussed, we say that a quantum mechanical particle

can absorb a photon of energy equal to DE and become promoted to the higher state

5.This energy is proportional to the frequency of the photon absorbed, and in the case of nuclear spin, is a function of the magnetic field applied:

DE = hn = f (B0)

6.Every nucleus has a different ratio of m to angular momentum (each has a different charge and mass) – this is referred to as the magnetogyric ratio, g

DE = hn = f (gB0)

Angular momentum is quantized in units of h/2p, thus:

DE = hn = g (h/2p)B0

137

General Theory


Absorption of Energy7.Solving for the frequency of EM radiation we are observing:

DE = n = (g/2p) B0

8.For a bare hydrogen nucleus (H+), g = 267.53 (106 radians/T·sec)

9.In a field strength of 1 Tesla, DE = 42.5 MHz (for our discussion, at 1.41 T, DE = 60 MHz)

DE = hn = f (B0)

This energy difference corresponds to the highly weak radio frequency region of the EM spectrum – with wavelengths of >5 meters equal to < 0.02 cal·mol-1

138

General Theory

UVX-rays IRg-rays RadioMicrowave


Mechanism of absorption – nuclear magnetic resonance• What we are actually observing for DE is the precessional or Larmor frequency (w)

of the spinning nucleus – this is analogous to a spinning toy top precessing as a result of the influence of the earth’s magnetic field:

139

General Theory

H B0

w


NMR Spectroscopy

Mechanism of absorption – Nuclear Magnetic Resonance2.When a photon of n = 60 MHz encounters this spinning charged system (a bare

proton) the two can couple and change the spin state of the proton

140

General Theory

H

B0

wn = w

HwDEThis state is called

nuclear magnetic resonance, and the nucleus is said to be in resonance with the incoming radio wave


NMR Spectroscopy

Mechanism of absorption – Nuclear Magnetic Resonance3.The energy difference corresponding to 60 MHz (DE = hn) is 2.39 x 10-5 kJ mol-1 (tiny)

– thermal energy at room temperature (298 oK) is sufficient to populate both energy levels

4.The energy difference is small, so rapid exchange is occurring between the two populations, but there is always a net excess of protons in the lower energy state

5.From the Boltzman distribution equation we can calculate the population of each energy state:

Nupper/Nlower = e-DE/kT = e-hn/kT

@ 298 oK the ratio is 1,000,000 / 1,000,009 !

There is an excess population of 9 nuclei in the lower energy state!

141

General Theory


Mechanism of absorption – Nuclear Magnetic Resonance6.As the applied B0 increases, exchange becomes more difficult and the excess

increases:

7.In each case, it is these few nuclei that allow us to observe NMR

8.When radio radiation is applied to a sample both transitions upward and downward are stimulated – if too much radiation is applied both states completely equilibrate – a state called saturation – no NMR signal can be observed 142

General Theory

Frequency (MHz)

Excess nuclei

60 980 12

100 16200 32300 48600 96


NMR Spectroscopy

Chemical Shift 6.Spectroscopic observation of the NMR phenomenon would be of little use if all

protons resonated at the same frequency

7.The protons in organic compounds are not bare nuclei, they are surrounded by an s -orbital of containing an electron shared with an electron in a hybridized orbital of another atom to form a covalent bond

8.In the presence of an external magnetic field, an induced circulation of electrons opposite to that of a proton is observed since the two are of opposite charges

9.This induced circulation generates a magnetic field in opposition to the applied magnetic field – a local diamagnetic current

143

General Theory



2.1 Magnetic Properties of Nuclei • The magnitude of m from atom to atom varies:

m = għI

• ħ is Planck’s constant divided by 2p

• g is the characteristic gyromagnetic ratio of the nucleus

• The larger g is, the greater the magnetic moment



• For the 1H nucleus (proton) this resonance condition occurs at low energy (lots of noise) unless a very large magnetic field is applied

• Early NMR spectrometers used a large permanent magnet with a field of 1.4 Tesla—protons undergo resonance at 60 MHz (1 MHz = 106 Hz)

• Modern instruments use a large superconducting magnet—our NMR operates at 9.4 T where proton resonance occurs at 400 MHz

• In short, higher field gives cleaner spectra and allows longer and more detailed experiments to be performed


Origin of the Chemical Shift

Electrons surrounding the

nucleus are opposite in charge

to the proton, therefore they

generate an opposing b0

DeshiedingFactors which lower e- density allow the

nucleus to “see” more of the B0 being applied – resonance

occurs at higher energy

ShieldingFactors which raise

e- density reduce the amount of B0

the nucleus “sees” – resonance

condition occurs at lower energy


The Proton (1H) NMR Spectrum


The 1H NMR Spectrum

A reference compound is needed—one that is inert and does not interfere with other resonances

Chemists chose a compound with a large number of highly shielded protons—tetramethylsilane (TMS)

No matter what spectrometer is used the resonance for the protons on this compound is set to d 0.00

148

Si

CH3

H3C CH3CH3


The 1H NMR Spectrum

The chemical shift for a given proton is in frequency units (Hz)

This value will change depending on the B0 of the particular spectrometer

By reporting the NMR absorption as a fraction of the NMR operating frequency, we get units, ppm, that are independent of the spectrometer


The 1H NMR Spectrum

• We need to consider four aspects of a 1H spectrum:

a. Number of signalsb. Position of signalsc. Intensity of signals.d. Spin-spin splitting of signals


The Number of Signals

The number of NMR signals equals the number of different types of protons in a compound

Protons in different environments give different NMR signals

Equivalent protons give the same NMR signal



To determine if two protons are chemically equivalent, substitute “X” for that each respective hydrogen in the compound and compare the structures

If the two structures are fully superimposible (identical) the two hydrogens are chemically equivalent; if the two structures are different the two hydrogens were not equivalent

A simple example: p-xylene

152

CH3

CH3

H

H

CH3

CH3

H

Z

CH3

CH3

Z

H

Same Compound



• Examples

153

Important: To determine equivalent protons in cycloalkanes and alkenes, always draw all bonds to show specific stereochemistry:



In comparing two H atoms on a ring or double bond, two protons are equivalent only if they are cis or trans to the same groups.



Proton equivalency in cycloalkanes can be determined similarly:



Enantiotopic Protons – when substitution of two H atoms by Z forms enantiomers:a. The two H atoms are equivalent and give the same NMR signalb. These two atoms are called enantiotopic



Diastereotopic Protons - when substitution of two H atoms by Z forms diastereomersa. The two H atoms are not equivalent and give two NMR signalsb. These two atoms are called diastereotopic


Chemical Shift – Position of Signals

• Remember:

158

Electrons surrounding the

nucleus are opposite in charge

to the proton, therefore they

generate an opposing b0

DeshiedingFactors which lower e- density allow the

nucleus to “see” more of the B0 being applied –

resonance occurs at higher energy

ShieldingFactors which raise

e- density reduce the amount of B0 the nucleus “sees” –

resonance condition occurs at lower

energyCHEM 430 – NMR Spectroscopy


• The less shielded the nucleus becomes, the more of the applied magnetic field (B0) it feels

• This deshielded nucleus experiences a higher magnetic field strength, to it needs a higher frequency to achieve resonance

• Higher frequency is to the left in an NMR spectrum, toward higher chemical shift—so deshielding shifts an absorption downfield

159

Downfield, deshielded Upfield, shielded



• There are three principle effects that contribute to local diamagnetic shielding:

a. Electronegativity b. Hybridizationc. Proton acidity/exchange



Electronegative groups comprise most organic functionalities:

-F -Cl -Br -I -OH -OR -NH2

-NHR -NR2 -NH3+ -C=O -NO2 -NO

-SO3H

-PO3H2 -SH -Ph -C=C and most others

In all cases, the inductive WD of electrons of these groups decreases the electron density in the C-H covalent bond – proton is deshielded – signal more downfield of TMS



Protons bound to carbons bearing electron withdrawing groups are deshielded based on the magnitude of the withdrawing effect – Pauling electronegativity:

162

CH3F CH3O- CH3Cl CH3Br CH3I CH4 (CH3)4Si

Pauling Electronegativity

4.0 3.5 3.1 2.8 2.5 2.1 1.8

d of H 4.26 3.40 3.05 2.68 2.16 0.23 0.0



3. The magnitude of the deshielding effect is cumulative:

As more chlorines are added d becomes larger

4. The magnitude of the deshielding effect is reduced by distance, as the inductive model suggests

163

CH3Cl CH2Cl2 CHCl3d of H 3.05 5.30 7.27

-CH2Br -CH2CH2Br -CH2CH2CH2Brd of H 3.30 1.69 1.25



Hybridization Increasing s-character (sp3 sp2 sp) pulls e- density closer to nucleus

effectively raising electronegativity of the carbon the H atoms are bound to – a deshielding effect

We would assume that H atoms on sp carbons should be well downfield (high d) and those on sp3 carbons should be upfield (low d)



• What we observe is slightly different:

165

Type of H Carbon hybridization

Name of H Chemical Shift, d

R-CH3, R2CH2, R3CH sp3 alkyl 0.8-1.7C=C-CH3 sp3 allyl 1.6-2.6CC-H sp acetylenic 2.0-3.0C=C-H sp2 vinylic 4.6-5.7Ar-H sp2 aromatic 6.5-8.5

O=C-H sp2 aldehydic 9.5-10.1

Chemists refer to this observation as magnetic anisotropy CHEM 430 – NMR

Spectroscopy


Magnetic Anisotropy – Aromatic Protonsa. In a magnetic field, the six p electrons in benzene circulate around

the ring creating a ring current.b. The magnetic field induced by these moving electrons reinforces the

applied magnetic field in the vicinity of the protons.c. The protons thus feel a stronger magnetic field and a higher

frequency is needed for resonance. Thus they are deshielded and absorb downfield.



• Similarly this effect operates in alkenes:



• In alkynes there are two perpendicular sets of p-electrons—the molecule orients with the field lengthwise—opposing B0 shielding the terminal H atom







The area under an NMR signal is proportional to the number of absorbing protons

An NMR spectrometer automatically integrates the area under the peaks, and prints out a stepped curve (integral) on the spectrum

The height of each step is proportional to the area under the peak, which in turn is proportional to the number of absorbing protons

Modern NMR spectrometers automatically calculate and plot the value of each integral in arbitrary units

The ratio of integrals to one another gives the ratio of absorbing protons in a spectrum; note that this gives a ratio, and not the absolute number, of absorbing protons






Spin-Spin Splitting

• Consider the spectrum of ethyl alcohol:• Why does each resonance “split” into smaller peaks?

174

HOCH2

CH3


Spin-Spin Splitting

The magnetic effects of nuclei in close proximity to those being observed have an effect on the local magnetic field, and therefore DE

Specifically, when proton is close enough to another proton, typically by being on an adjacent carbon (vicinal), it can “feel” the magnetic effects generated by that proton

On any one of the 108 of these molecules in a typical NMR sample, there is an equal statistical probability that the adjacent (vicinal) proton is either in the + ½ or – ½ spin state

If there is more than one proton on an adjacent carbon – all the statistical probabilities exist that each one is either + ½ or – ½ in spin

The summation of these effects over all of the observed nuclei in the sample is observed as the spin-spin splitting of resonances


Spin-Spin Splitting

• Recall, we are observing the frequency (E = hn) where a proton goes into resonance

176

Any change in B0 will cause a change in energy at which the resonance condition will occur for a proton of a given chemical shift


In solution we are not looking at a single molecule but about 108

On some molecules the proton being observed may be next to another proton of spin + 1/2 :


Spin-Spin Splitting

• On some molecules the proton being observed may be next to another proton of spin – 1/2 :


Spin-Spin Splitting

Observe what effect this has on an isolated ethyl group:

The two methylene Ha protons have three neighbors, Hb, on the adjacent methyl carbon

Each one of these hydrogens can be + ½ or – ½ , and since we are not looking at one molecule, but billions, we will observe all combinations

179

C C

HbHa

R

HbHb

Ha


Spin-Spin Splitting

The first possibility is that all three Hb protons have a + ½ spin; in this case the three protons combine to generate three small magnetic fields that aid B0 and deshield the protons – pushing the resonance for Ha slightly downfield (the magnetic field of a proton is tiny compared to B0)

180

C C

HbHa

R

HbHb

Ha

All 3 Hb protons + ½

resonance for Ha in absence of spin-spin splittingCHEM 430 – NMR

Spectroscopy

Spin-Spin Splitting

The second possibility is that two Hb protons have a + ½ spin and the third a - ½ ; in this case the two protons combine to enhance B0 and the other against it, a net deshielding; there are 3 different combinations that generate this state

181

C C

HbHa

R

HbHb

Ha

2 Hb protons + ½

or or


Spectroscopy

Spin-Spin Splitting

The third possibility is that two Hb protons have a –½ spin and the third +½; here, the two protons combine to reduce B0 and the other

enforce it, a net shielding effect; there are 3 different combinations that generate this state

182

C C

HbHa

R

HbHb

Ha

2 Hb protons - ½

resonance for Ha in absence of spin-spin splitting

or or


Spin-Spin Splitting

The last possibility is that all three Hb protons have a – ½ spin; in this case the three protons combine to oppose B0, a net shielding effect; there is one combination that generates this state

183

C C

HbHa

R

HbHb

Ha

All 3 Hb protons - ½


Spectroscopy

Spin-Spin Splitting

The result is instead of one resonance (peak) for Ha, the peak is “split” into four, a quartet, with the constituent peaks having a ratio of 1:3:3:1 centered at the d (n) for the resonance

184

C C

HbHa

R

HbHb

Ha


Spectroscopy

Spin-Spin Splitting

Similarly, the Hb protons having two protons, on the adjacent carbon each producing a magnetic field, cause the Hb resonance to be split into a triplet

185resonance for Ha in absence of spin-spin splitting

C C

HbHa

R

HbHb

Ha


Spin-Spin Splitting

Rather than having to do this exercise for every situation, it is quickly recognized that a given family of equivalent protons (in the absence of other spin-coupling) will have its resonance split into a multiplet containing n+1 peaks, where n is the number of hydrogens on carbons adjacent to the carbon bearing the proton giving the resonance – this is the n + 1 rule

186

# of Hs on adj. C’s

Multiplet # of peaks

The relative ratios of the peaks are a mathematical progression given by Pascal’s triangle:

0 singlet 1 11 doublet 2 1 12 triplet 3 1 2 13 quartet 4 1 3 3 1 4 quintet 5 1 4 6 4 15 sextet 6 1 5 10 10 5 16 septet 7 1 6 15 20 15 6 1


1H NMR—Spin-Spin Splitting

• Common patterns:

187

X CH3X

XX

X

tert-butyl - singletmethyl - singlet

iso-propyl – septet - doublet

ethyl – quartet - triplet

n-propyl – triplet - quintet - triplet





Another Example:

189

BrC

CBr

Br

HaHb

Hb

BrBr

Br=



Another Example:



Three general rules describe the splitting patterns commonly seen in the 1H NMR spectra of organic compounds:1.Equivalent protons do not split each other’s signals2.A set of n nonequivalent protons splits the signal of a nearby proton into n + 1 peaks3.Splitting is observed for nonequivalent protons on the same carbon or adjacent carbonsIf Ha and Hb are not equivalent, splitting is observed when:



• Magnetic influence falls off dramatically with distance

• The n + 1 rule only works in the following situations:

192

H H

H HAliphatic compounds that have free rotation about each bond

GHa

HbHc

H

H

Aromatic compounds where each proton is held in position relative to one anotherCHEM 430 – NMR

Spectroscopy


The amount of influence exerted by a proton on an adjacent carbon is observed as the difference (in Hz) between component peaks within the multiplet it generates. This influence is quantified as the coupling constant, J

Two sets of protons that split one another are said to be “coupled”

J for two sets of protons that are coupled are equivalent—therefore on complex spectra we can tell what is next to what

193

This J Is equal to this J

-CH2- -CH3CHEM 430 – NMR

Spectroscopy


The next level of complexity (which at this level, is only introduced) is when protons on adjacent carbons exert different J’s than one another.

Consider the ethylene fragment:

194

C

CX

H

H

HThe magnetic influence of the trans- relationship is over the longest distance

The cis-relationship, is over an intermediate distance

The influence of the geminal-relationship is over the shortest distance



• For this substituted ethylene we see the following spectrum:

195

2Jgem = 0 – 1 Hz

3Jtrans = 11- 18 Hz

3Jcis = 6 - 15 Hz

The observed multiplet for Ha is a “doublet of doublets”

3JAB3JAB

3JACC

CX

Ha

Hc

Hb



In general, when two sets of adjacent protons are different from each other (n protons on one adjacent carbon and m protons on the other), the number of peaks in an NMR signal = (n + 1)(M + 1)

In general the value of J falls off with distance; J values have been tabulated for virtually all alkene, aromatic and aliphatic ring systems



• Some common J-values

197

3J = 6-8

3Ja,a = 8-143Ja,e = 0-73Je,e = 0-5



3Jtrans = 11-18

3Jcis = 6-15

3Jallyl = 4-10

3J = 8-11

3J = 5-7

3Jortho = 7-10 Hz4Jmeta = 1-3 Hz5J para = 0-1 Hz

HH

Ha

He

Ha

He

H

H

HH

CH3

H

H

H

H

H

H

H

HH

O

HH

H

H

HHo

Hm

Hp



• We can now tell stereoisomers apart through 1H NMR:



• A combined example:



• Under usual conditions, an OH proton does not split the NMR signal of adjacent protons

• Protons on electronegative atoms rapidly exchange between molecules in the presence of trace amounts of acid or base (usually with NH and OH protons)


Structure Determination








13C NMR

• The lack of splitting in a 13C spectrum is a consequence of the low natural abundance of 13C

• Recall that splitting occurs when two NMR active nuclei—like two protons—are close to each other. Because of the low natural abundance of 13C nuclei (1.1%), the chance of two 13C nuclei being bonded to each other is very small (0.01%), and so no carbon-carbon splitting is observed

• A 13C NMR signal can also be split by nearby protons. This 1H-13C splitting is usually eliminated from the spectrum by using an instrumental technique that decouples the proton-carbon interactions, so that every peak in a 13C NMR spectrum appears as a singlet

• The two features of a 13C NMR spectrum that provide the most structural information are the number of signals observed and the chemical shifts of those signals


13C NMR


13C NMR

• The number of signals in a 13C spectrum gives the number of different types of carbon atoms in a molecule.

• Because 13C NMR signals are not split, the number of signals equals the number of lines in the 13C spectrum.

• In contrast to the 1H NMR situation, peak intensity is not proportional to the number of absorbing carbons, so 13C NMR signals are not integrated.


13C NMR

• In contrast to the small range of chemical shifts in 1H NMR (1-10 ppm usually), 13C NMR absorptions occur over a much broader range (0-220 ppm).

• The chemical shifts of carbon atoms in 13C NMR depend on the same effects as the chemical shifts of protons in 1H NMR.


13C NMR


13C NMR


Shoolery Tables

• After years of collective observation of 1H and 13C NMR it is possible to predict chemical shift to a fair precision using Shoolery Tables

• These tables use a base value for 1H and 13C chemical shift to which are added adjustment increments for each group on the carbon atom

211

X C ZY

methine

X C YH

methylene

X C HH

methyl

H HH


Shoolery Values for Methylene

X or Y Substituent Constant

X or Y Substituent Constant

-H 0.34 -OC(=O)OR 3.01

-CH3 0.68 -OC(=O)Ph 3.27

-C—C 1.32 -C(=O)R 1.50

-CC- 1.44 -C(=O)Ph 1.90

-Ph 1.83 -C(=O)OR 1.46

-CF2- 1.12 -C(=O)NR2 or H2

1.47

-CF3 1.14 -CN 1.59

-F 3.30 -NR2 or H2 1.57

-Cl 2.53 -NHPh 2.04

-Br 2.33 -NHC(=O)R 2.27

-I 2.19 -N3 1.97

-OH 2.56 -NO2 3.36

-OR 2.36 -SR or H 1.64

-OPh 2.94 -OSO2R 3.13


Shoolery Values for Methine

X ,Y or Z Substituent Constant

X, Y or Z Substituent Constant

-F 1.59 -OC(=O)OR 0.47

-Cl 1.56 -C(=O)R 0.47

-Br 1.53 -C(=O)Ph 1.22

-NO2 1.84 -CN 0.66

-NR2 or H2 0.64 -C(=O)NH2 0.60

-NH3+ 1.34 -SR or H 0.61

-NHC(=O)R

1.80 -OSO2R 0.94

-OH 1.14 -CC- 0.79

-OR 1.14 -C=C 0.46

-C(=O)OR 2.07 -Ph 0.99

-OPh 1.79 213CHEM 430 – NMR Spectroscopy

Shoolery Tables

• For methyl—use methylene formula and table using the –H value

• For methylene—use a base value of 0.23 and add the two substituent constants for X and YIn 92% of cases experimental is within 0.2 ppm

• For methine—use a base value of 2.50 and add the three substituent constants for X, Y and ZError similar to methylene


Shoolery Tables

• Work for aromatics as well (.pdf posted)


Running an NMR Experiment

• Sample sizes for a typical high-field NMR (300-600 MHz):• 1-10 mg for 1H NMR• 10-50 mg for 13C NMR

• Solution phase NMR experiments are much simpler to run; solid-phase NMR requires considerable effort

• Sample is dissolved in ~1 mL of a solvent that has no 1H hydrogens

• Otherwise the spectrum would be 99.5% of solvent, 0.5% sample!



• Deuterated solvents are employed—all 1H atoms replaced with 2H which resonates at a different frequency

• Most common: CDCl3 and D2O

• Employed if necessary: CD2Cl2, DMSO-d6, toluene-d8, benzene-d6, CD3OD, acetone-d6

• Sample is contained in a high-tolerance thin glass tube (5 mm)



• IMPORTANT—no deuterated solvent is 100% deuterated, there is always residual 1H material, and this will show up on the spectrum

• CHCl3 in CDCl3 is a singlet at d 7.27

• HOD in D2O is a broad singlet at d 4.8

• No attempt is made to make solvents for 13C NMR free of 13C, as the resonances are so weak to begin with

• 13C NMR using CDCl3 shows a unique 1:1:1 triplet at d 77.00 (+1, 0, 1 spin states of deuterium coupled with 13C)


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NMR Spectroscopy