Nuclear Magnetic Resonance (NMR)

The most common tool for an organic chemist to determine structure

Any nucleus with either an odd atomic number or odd mass has a “nuclear spin”

A charge species that is spinning creates a current loop

+ + + +

A current loop creates magnetic field lines

proton

Spinning direction

Spinning nucleus creates a current loop

Current loop creates magnetic field lines

I B

This magnetic field is analogous to a bar magnet

Thus a spinning nucleus behaves similar to tiny bar magnets

Nuclear Magnetic Resonance

B

+

B

N S

In solution there are many nuclear spins occurring, hence many tiny bar magnets are present

Each spinning direction is random (there is no energy difference between spinning in any one direction)

and hence the magnetic moments will cancel

Nuclear Magnetic Resonance

+

Consider the bar magnet analogy

While each individual bar magnet will have a random orientation, in the presence of a larger magnetic field the bar magnet will align itself with the field

The same result will occur if we place a spinning nucleus in the field of a large magnetic field

Nuclear Magnetic Resonance

N

S

B0 Field is aligned

In the presence of an external magnetic field the spinning nucleus can align in one of two ways (with or against the field) when S = ½

For any S = ½ nucleus, obtain two quantized spin directions in large external magnetic field (B0)

Most nuclei of interest for organic chemists have a S = ½ state (1H, 13C, 19F)

B0

Nuclear Magnetic Resonance

N S

α-spin state, align with field

β-spin state, align against field

The nuclear spins are therefore no longer random

There will be more spins in the α-state than β-state, the difference in energy is dependent upon the external magnet strength

Nuclear Magnetic Resonance

N S

Orientation when B0 = 0 B0 increasing

ΔE

N S

N S

N S

N S

B0

α-spin state

β-spin state

The difference in energy is dependent upon two terms -external magnetic field and gyromagnetic ratio

Energy difference between α- and β-state spins ΔE = γ (h/2π) B0

γ = gyromagnetic ratio

This term is dependent upon which nucleus is spinning, It is constant for a given nucleus

1H for example has γ = 26,753 sec-1 gauss-1

Nuclear Magnetic Resonance

Resonance

A key remaining point is the nucleus in the α-state can be changed to the β-state

-need the exact amount of energy that separates the two states

We know this energy difference (ΔE)

We can supply this energy by absorption of the appropriate wavelength of light ΔE = hν

With infrared spectroscopy we supplied light in the infrared region to cause a bond vibration

In NMR we supply energy in the radio frequency region to cause the transition of nuclei from the α-state to the β-state

This absorption of energy will cause the spin state to change

In presence of external magnetic field the spin states are quantized with more spins in the α-state than the β-state

With exact combination of external magnetic field (B0) and energy of photons (ΔE, which happens to be in the radio frequency for NMR) the nuclear spins can flip from the

α-spin to β-spin state, thus they are in “resonance”

Called nuclear magnetic resonance (NMR)

Nuclear Magnetic Resonance

N S

N S

N S

N S

B0

α-spin state

β-spin state

N S

N S

N S

B0

α-spin state

β-spin state ΔE = hν

The NMR machine can detect this absorption of energy for a sample

Once the field strength and the radiofrequency are in “resonance” the nucleus will absorb the energy of the radio waves

If this was all a NMR machine did it would be practically useless for an organic chemist -we could detect whether a compound had hydrogens

But that would be the only piece of information

Almost all organic compounds have hydrogens so how does a NMR machine distinguish between different compounds?

Nuclear Magnetic Resonance

Shielding

Need to remember the structure of a compound (consider only an isolated C-H bond)

To reach the nucleus the magnetic field must past through the electron cloud surrounding the nucleus

The electrons surrounding the nucleus are charged species that can rotate in the presence of the external magnetic field

What this means is that the external magnetic field (B0) is effectively reduced by the time it reaches the nucleus (B0 minus the field of the electron cloud)

C HB0

Bnet = B0 - Belectron

This process is called shielding

The electrons are shielding the nucleus from the full external magnetic field -the field at the nucleus is thus always less than the actual B0

The critical point for a NMR experiment is that each hydrogen nucleus will be shielded by different amounts

The amount of shielding is a direct result of the electron density around the nucleus

The greater the electron density, the more shielded the hydrogen nucleus

Therefore a NMR experiment infers the amount of electron density around each hydrogen in the compound!!!

Shielding

Protons in Resonance

The UTD Chemistry department has a 400 MHz NMR to measure samples

With no shielding, what is the magnetic field strength required to cause a proton to be in resonance at this radiofrequency?

ΔE = hν = γ (h/2π) B0

B0 = 2νπ / γ 400 MHz = 400 x 106 sec-1

B0 = 2(4 x 108 sec-1)(3.14) / 26,753 sec-1 gauss-1

B0 = 93,896 gauss for 400 MHz NMR

Shielding

But this is for an unshielded proton The actual proton MUST experience a lower effective magnetic field

If the machine is set at 93,896 gauss the proton will not be in resonance and therefore it will not absorb the energy to change spin states

Must supply a greater external magnetic field to cause the resonance for a shielded proton

The amount of extra magnetic field required is dependent upon the amount of shielding

MORE electron density around the proton, REQUIRES a larger external magnetic field

How can we predict how much the magnetic field must be increased?

By knowing the amount of electron density around a proton

Therefore WHERE ARE THE ELECTRONS

If we know where the electrons are located, we can begin to predict how much the external magnetic field needs to be increased to cause resonance

Shielding

Chemical Shifts

To record the position of an absorption scientists use a chemical shift

A standard is chosen, for NMR the standard chosen was tetramethylsilane (TMS)

Since silane is less electronegative than carbon, the methyl groups (and hence the hydrogens on the methyl groups)

will have more electron density than a typical carbon compound

The chemical shift of TMS is defined as 0.00

The chemical shift for a different hydrogen is defined as follows:

Chemical shift (ppm or δ) = (shift downfield from TMS, Hz) / (total frequency, MHz)

H3CSi CH3

H3C CH3

A NMR spectrum is recorded with increasing magnetic field towards the right

A more shielded hydrogen (more electron density around the hydrogen) would therefore be further to the right (called upfield)

A less shielded hydrogen (less electron density around the hydrogen) would therefore be further to the left (called downfield)

Since the δ is measured by shift downfield, a positive δ number means the hydrogen has less electron density (or less shielding) than TMS

An advantage of δ is the number is the same regardless of which instrument is used

Chemical Shifts

Chemical Shifts

upfield downfield

Predicting chemical shifts is therefore the same as predicting electron density around nucleus

Consider methyl halides

As the electronegativity of the halide increases there is less electron density around the hydrogens

Chemical Shifts

CH3Cl CH3Br CH3I

Chemical Shifts

δ of H’s 3.0 ppm

δ of H’s 2.7 ppm

δ of H’s 2.2 ppm

ClH

HH

BrH

HH

IH

HH

As amount of electron density around hydrogens decrease, the peak is shifted

more downfield in a 1H NMR

The chemical shift is lowered the further removed from electronegative atoms

Saw same trend for pKa effect in acidity

Also notice that more than one peak is obtained for each type of hydrogen

Chemical Shifts

H3CBr

H H

H H

δ3.39

δ1.87

δ1.03

Number of Signals

A NMR supplies more information than merely the amount of electron density around nuclei of interest

(although that alone is a tremendously important piece of information)

In a 1H NMR, each electronically different hydrogen will have its own unique chemical shift

In the example above with 1-bromopropane there are three different types of hydrogens (one α to bromine, one β to bromine and one γ to bromine)

Each different hydrogen has a different chemical shift α = 3.4 ppm, β = 1.9 ppm, γ = 1.0 ppm

By counting number of different peaks we can determine the number of different hydrogens

Number of Signals

The compound for this 1H NMR therefore must have two different types of hydrogens

H3C

O

O CH3

Area of Peaks

In a 1H NMR spectrum, the area of each peak indicates the relative number of hydrogens of that type

O

3H’s δ = 2.1

2H’s δ = 2.3

9H’s δ = 1.0

Integrating the signals will indicate the area under each curve,

which is directly related to number of hydrogens causing signal

Spin-Spin Splitting

We can observe a 1H NMR with additional signals

There are three types of hydrogens in this molecule, but some of the signals display more than one peak

In addition to the electron density surrounding the hydrogen causing shielding, each hydrogen acts like its own magnetic field

If one magnet is close to another, it will feel the effect of that magnetic field

This occurs in a 1H NMR a hydrogen will feel the effect of neighboring hydrogens

In the field of one additional hydrogen therefore we would observe two signals

Spin-Spin Splitting

H H

Each hydrogen is similar to a tiny bar magnet N

S N S

The bar magnets could thus be aligned in same

direction

N S

Or in opposite directions

These two scenarios lead to a different field strength required to get hydrogen in resonance

As the number of neighboring hydrogens increases, the number of signals increases

Can predict the splitting by knowing the number of magnetically equivalent hydrogens causing the splitting

Pascal’s triangle

# of hydrogens causing splitting Peak area

0 1 1 1 1 2 1 2 1 3 4

1 3 3 1 1 4 6 4 1

Spin-Spin Splitting

singlet doublet triplet quartet pentet

Splitting pattern

N+1 Rule

We can use Pascal’s triangle to determine the number of signals expected (and the relative intensity of each signal),

but a quick guide is to remember the observed splitting is equal to N+1

If you have N number of magnetically equivalent hydrogens causing the splitting then you have N+1 peaks in the spectrum

A hydrogen does not cause splitting with itself, but only with neighboring hydrogens

Must be within a short distance to allow the small magnetic field of one hydrogen to affect the magnetic field around another

For alkanes normally observe splitting only for hydrogens attached to adjacent carbons

Knowing these “rules” we can interpret a spectrum

3H’s no adjacent H’s

Therefore singlet

2H’s 3 adjacent H’s

Therefore quartet

3H’s 2 adjacent H’s

Therefore triplet

Nuclear Magnetic Resonance

H3C O CH3

O

Distance of Splitting

In normal organic compounds, splitting is only observed with hydrogens attached to adjacent atoms

Osinglet

triplet

hextet triplet

too far removed

Coupling Constants

The splitting caused by adjacent hydrogens (or any magnetically active species) is called J coupling and it is a constant value

A hydrogen, which splits a neighboring hydrogen by a J value, will also be split itself by the same J value (reciprocal splitting)

The magnitude of this J coupling can also be characteristic of the type of hydrogen

Most alkane hydrogens have a J constant of ~7 Hz

Other characteristic J couplings

Most convenient to distinguish E from Z alkenes

Coupling Constants

H H

H

H

H

H

H

H

H

HH

H

Cis alkene 6-12 Hz

Trans alkene 12-18 Hz

Gem alkene 0-3 Hz

Trans > Cis

Ortho 6-8 Hz

Meta 1-3 Hz

Para 0-1 Hz

Can observe coupling farther

than adjacent carbons with

extended conjugation

Only observe if geminal hydrogens

are different

Complex Splitting

Most of the cases we have observed so far are relatively quite simple In many examples, though, the splitting is caused by more than

one magnetically equivalent adjacent hydrogen

O

HH

Hdoublet

doublet of doublet

doublet of triplet

The β hydrogens of a conjugated carbonyl are more downfield due to resonance

forms placing positive charge at this site

O

H

O

H

Stereochemical Nonequivalence

Remember that all magnetically equivalent hydrogens will have the same chemical shift

If the hydrogens are diastereotopic, however, they will have different chemical shifts

The magnitude of this difference is entirely dependent on the structure (can be quite small and not observable)

If the hydrogens are diastereotopic they will in addition split each other

To determine if hydrogens are diastereotopic, replace hydrogens in question with a different group selectively at each position and consider the stereochemical relationship

H H oxidize HO H H OH These products are identical, hydrogens are called homotopic

H H oxidize HO H H OH These products are enantiomers, hydrogens are called enantiotopic

(identical NMR signals) H H

Br

oxidizeHO H

Br

H OH

Br

These products are diastereomers, hydrogens are called diastereotopic

(NMR signals are different)

Time Dependence

NMR takes a certain amount of time to record

Therefore anything with a lifetime shorter than the recording time will not be observed, in practice need a lifetime of ~1 sec to observe

This is true for any of the alkanes where bond rotation is much quicker than this time scale and hence the observed spectrum is an average of all the bond rotations

If the process is slower than the NMR, however, the NMR can distinguish the different conformations

Sometimes a NMR will be taken at different temperatures to slow down the process to observe by NMR

Some Characteristic Functional Groups

H

We have observed many functional groups (and will see more as course continues), NMR can distinguish these groups due to the characteristic chemical shifts for these functional groups

Aromatic Compounds (e.g. Benzene) View benzene as flat

Remember how the p orbitals are in conjugation In magnetic field, an electron will rotate in a circular motion

The circulating electrons cause a ring current which will generate magnetic field lines, causing deshielding of the

aromatic hydrogens

Due to this deshielding effect, aromatic hydrogens have chemical shifts in the 7-8 ppm range

B0

Alkenes

Similar to benzene, hydrogens on alkenes are attached to sp2 hybridized carbons

The sp2 hybridization causes the electrons to be closer to the carbon, further from the hydrogen

The hydrogens are thus deshielded more than alkanes

Alkenes have chemical shifts in the 5-6 ppm range, Less deshielded than aromatics but more than alkanes

H H

Alkynes

In terminal alkynes, a hydrogen is attached to a sp hybridized carbon

Therefore the electrons are further removed from the hydrogen than either an alkane or alkene due to the greater s character in the bond to carbon

(sp hybridized electrons are closer to carbon than sp2 or sp3)

Would expect hydrogen to be further downfield than an alkene, but need to consider the effect of the induced magnetic field caused by the electrons moving in the orbitals

HH

H HHH

H

H

Bnet = B0 + Bi

Bnet = B0 - Bi

Induced magnetic field adds to applied field, therefore less B0 is

required to get into resonance δ = 5-6 ppm

Induced magnetic field opposes applied field, therefore more B0 is

required to get into resonance δ = low 2 ppm

B0 B0

Bi

Bi

Aldehyde

An aldehyde is similar to an alkene

In addition to the effect of the sp2 hybridized carbon, the aldehyde hydrogen is further deshielded by the electron withdrawing effect of the oxygen of the carbonyl

Due to these effects, an aldehyde hydrogen has a chemical shift of 9-10 ppm

O

H

H

H

H

HO

R

HO

R

H

Hydrogen-Bonded Protons

Protons that can hydrogen bond, e.g. alcohols, amines, carboxylic acids, can exist in many states on the NMR time scale

Each state can have a slightly different chemical shift

Therefore these hydrogens can often appear broader than the traditional sharp peaks for other functional groups

The exact chemical shift can also depend on the degree of hydrogen bonding (which is affected by concentration, temperature, solvent, etc.)

Alcohols 2-6 ppm Amines 2-5 ppm

Carboxylic acids 10-12 ppm

1H NMR Chemical Shift Scale

Bapplied

Bmolecular

Beffective

Applied Magnetic Field: Homogeneous for 1H NMR – therefore spin sample Molecular Magnetic Field: two effects will change the magnetic field experienced by nucleus

1)  Density of electron density around nucleus “shields” nucleus (range ~12 ppm for 1H NMR, ~200 ppm for 13C NMR)

2) Nearby magnetic nuclei (spin-spin splitting) The effective magnetic field at the nucleus is thus Bapplied-Bmolecular

Due to empirically observed effects, chemists can predict the position for functional groups

01234567891011

ppm (!)

TMS (standard)

alkane CH-O, CH-X, CH-N C=C-H Aromatic-H RCHO RCO2H

RO-H

ROCH3

13C NMR

Any nucleus with an odd atomic number or odd mass will have a nuclear magnetic resonance

A proton is the most common example in organic chemistry so by default when someone talks about NMR it usually means a 1H NMR

Carbon-12, the most abundant source of carbon, has an even atomic number and even mass so therefore this nuclei is NMR inactive

The isotope C-13, though, has an odd mass and is therefore NMR active (also has S = ½ like 1H)

Another very useful nucleus to determine organic structures

Carbon-13

The natural abundance of C-13 isotope is ~1%

Therefore only 1% probability that a given carbon atom is NMR active -therefore the signal-to-noise ratio for carbon NMR is less

Put another way – if studying 1-chloroethane only 1 in 100 molecules will be active at C1

In addition the gyromagnetic ratio, γ, for carbon is ~1/4 that of hydrogen (remember that each nucleus has its own unique gyromagnetic ratio)

-this also causes a lower signal-to-noise for carbon relative to hydrogen NMR

13C NMR Spectra

Spectrum shown as the most common “broad-band” decoupled or also called “proton” decoupled

O

H

Unique Features of 13C NMR Relative to 1H NMR

1)  The scale is much larger for 13C than 1H NMR Notice spectrum is 0-200 ppm instead of 0-11 ppm, due to carbon being more shielded than hydrogen

[8 electrons around carbon rather than 2 electrons around hydrogen]

2)  Relative placement of functional groups though is the same in the above spectrum Which functional group was more deshielded in 1H NMR is still more deshielded in 13C NMR

-aldehyde carbon 194 ppm, sp2 carbons 158 and 133 ppm, sp3 carbons below 40 ppm

3)  Each carbon has its own unique chemical shift big advantage with 13C NMR, can instantly tell number of

distinct symmetrically different carbons [the spectrum shown is a proton-decoupled 13C NMR, most common example]

4)  Do not observe 13C-13C splitting Due to low probability of finding two adjacent carbons both 13C isotope with only 1%

probability for each

Splitting

Notice that carbon-carbon splitting is not detected in NMR

Can still detect 13C-1H splitting from hydrogens attached to carbon of interest This is called off-resonance decoupled

Have same N+1 rule

Normally, though, a 13C NMR is proton spin (broad-band) decoupled In this mode a sharp singlet is observed for each carbon

-far easier to interpret spectra

Remember to look for symmetry Only distinguish symmetrically different carbons

Carbon-13 NMR

Peak Areas

Unlike 1H NMR we cannot integrate peaks in 13C NMR to detect relative abundance

In a typical 13C NMR, the peak areas are dependent upon how many hydrogens are attached to carbon, not the relative number of carbons causing the signal

CH3 groups are the biggest followed by CH2, CH and quaternary carbons are the smallest

Position of Functional Groups in 13C NMR

The relative placement is similar to 1H NMR, but the scale is much larger

Observe all carbons (obviously do not “see” carbons that have no hydrogens attached in 1H NMR)

020406080100120140160180200220

ppm (!)

Alkanes

C-X

C-O

Alkynes

Alkenes

Aromatic Carbonyl

Examples

O

Four signals: 1 carbonyl, 2 sp2 hybridized alkene carbons, one sp3 hybridized carbon

Examples

Br

Four signals: all four are sp2 hybridized carbons

DEPT (distortionless enhancement with polarization transfer)

A programmed sequence of radio pulses can be used which allow the determination of number of hydrogens attached to each carbon

In this technique, 4 different 13C spectra are obtained with different pulse sequences: 0˚: a normal broad-band decoupled 13C NMR

45˚: only carbons with hydrogen attached will appear (no quaternary) 90˚: only CH groups will appear

135˚: CH and CH3 groups are positive, CH2 groups are negative

H3C CH3

MRI Imaging

MRI (magnetic resonance imaging) is in theory identical to our discussions MRI = NMR

Consumers are hesitant with anything that has “nuclear” in its descriptor so the name was changed

One main difference is that instead of spinning the sample to make the magnetic field homogeneous throughout the sample in a NMR experiment (thus the chemical shift will be identical for all hydrogens

with the same amount of shielding), in MRI the field is inhomogeneous throughout the sample

This inhomogeneity allows one to measure either placement of protons in a body (where is the water)

or also changes in movement of the protons (i.e. how mobile is water inside a tumor relative to a normal tissue)