introduction to nmr spectroscopyww2.chemistry.gatech.edu/class/2311/marder/nmr.pdf · introduction...

Introduction to NMR spectroscopy

Nuclei of isotopes which possess an odd number of protons, an odd number of neutrons, or both,have a nuclear spin quantum number, I, such that, I = 1/2n, where n is an integer 0,1,2,3...etc.

Since atoms have charge, a spinning nucleus generates a small electric current which in turn createsa finite but small magnetic field. The magnetic dipole, of the nucleus varies with each element.Thus, any nucleus with a spin can be observed by NMR spectroscopy.

When a nucleus with a spin is placed in a magnetic field the nuclear magnet experiences a torquewhich tends to align it with the external field just like a compass needle in the earth’s magneticfield.

For a nucleus with a spin of 1/2, there are two allowed orientations of the nucleus: parallel to thefield (low energy) and against the field (high energy). Since the parallel orientation is lower inenergy, this state is slightly more populated than the anti-parallel, high energy state.

Turn on mangetic field

Field splits systeminto two allow spin states

spin = –1/2

spin = +1/2

Direction of Applied Field

NMR spectroscopyThe energy difference between the state is proportional to the magnetogyric ratio of the nucleus andthe magnitude of the applied field as shown in the figure and equation below.

Ener

gy

Applied Magnetic Field

ΔE =2π

Happh X νH

Planck's constant9.532 x10-14 Kcal/mol

Magnetogyric ratio(26,753 radian/gauss-sec for protons)

Applied magnetic field (variable)70460 gauss for a routine machine

An applied magnetic field of roughly 7000 gauss corresponds to energy difference of 2.85 x 10-5

Kcal/mol for protons, an extremely small difference

If you have 2,000,000 protons in such a field roughly 999,995 would be spin down aligned againstthe field and 1,000,005 would be spin up with the field.

Since E = hν, the frequency of electromagnetic radiation needed to promote the nuclei from thelower energy to higher energy state, for the condition described in the equation above correspondsto a frequency of = 300 x 106 sec-1 or 300 MHz which is FM radio frequency

In the NMR experiment a radio frequency applied to molecules is applied to the nucleus in themagnetic field.

When he RF radiation energy exactly matches the energy gap between the energy level for the spinaligned with the magnetic field and the spin antiparallel to the magnetic field the nuclei can absorbthe energy and be promoted from the lower energy state to the higher energy state.

spin = –1/2

spin = +1/2Ener

gy o

f Spi

n St

ates

ΔE

apply RF correspondingto exactly ΔE and a spin is promoted to the higher energy state

Chemical Shift

Inte

nsity

Peak corrsponding to absorption of energy

When this condition is met the nuclei are said to be in resonance with the applied field and thisabsorption of radiation leads to a peak in a spectrum.

NMR and chemical structure.

Four types of information from an NMR spectra make it an extremely valuable technique fordetermining chemical structure.

1. The number of sets of peaks: This corresponds to the number of magnetically (chemically)different types of protons.

2. Chemical Shift: The key to NMR spectroscopy is that the exact energy at which this peakcomes (which can be measured extremely accurately) is dependent upon the local environmentof the proton.

3. Splitting: The number of peaks observed for a given proton is dependent upon the number andtype of protons in the vicinity of the one that is being observed.

4. Integration: The area of the peak is proportional to the number of protons that are magneticallyequivalent.

The NMR spectrum of methyl isopropyl ether (CH3-OCH(CH3)2) is shown below:

Note that there are three sets of resonances. One has seven peaks, one has one peak and one hastwo peaks. They show up in different places in the spectrum and they have different integratedareas under each of the resonances.

It is the combination of all of this information that enables us to assign the structure to themolecule.

Chemical Shift

Naively you might expect that all nuclei of a given type would undergo the spin-flip transition atexactly the same applied frequency in a given magnetic field. However, electrons in the moleculehave small magnetic fields associated with them which tend to oppose the applied field.

Electron Cloud

Induced Current

Induced Magnetic fieldat nucleus

Happ

These magnetic fields are said to screen the nuclei from the full strength of the applied field.

The greater the electron density, the greater this 'shielding' will be, hence nuclei which are in electronrich environments will undergo transition at a higher applied field than nuclei in electron poorenvironments.

Thus, the relevant field at the nucleus which determine the energy of the transition now becomesHeff = Happ + Hshielding

Remember that Hshielding is opposite in sign to Happ.

The resulting shift in the NMR signal for a given nuclei is referred to as the chemical shift. For 1Hthe chemical shift is given relative to the absorption of tetramethyl silane ((CH3)4Si)(TMS). The chemical shift in Hz is proportional to the operating frequency and is given by

Chemical shift (Hz) = νH – νTMS.

The delta scale for chemical shifts

Chemical shifts are reported in a manner that is independent of frequency by normalizing both thefrequency of the observed hydrogens and those for TMS hydrogens by the operating frequency ofthe instrument as defined by the equation shown below:

Chemical Shift (δ) = (νH – νTMS (Hz))/operating frequency (Hz) x 106 ppm

The factor of 106 is introduced into the equation to give a simple whole number scale forconvenience.

In this scale TMS is arbitrarily assigned a value of zero. TMS is used because its protons are morehighly shielded than those observed in most common organic molecules and because it ischemically inert. For 1H NMR, the δ scale generally extends from 0-12 ppm and then eachchemically inequivalent nucleus has it own chemical shift that is independent of the operatingfrequency of the NMR.

The ppm scale is plotted from δ = 0 on the right and with increasingly deshielded nuclei (increasingδ) to the left. Peaks to the left are referred to a “downfield” of peak to the right since it wouldrequire a smaller applied field for them to be in resonance with a fixed excitation frequency.• Electronegative groups are "deshielding" and thus protons on carbons in the vicinity of

electronegative atoms are deshielded and appear at higher chemical shifts. The moreelectronegative the group the further down field the proton will be.

• If the electronegative group is directly attached to the carbon the effect is greatest and if it isattached to the α carbon it is somewhat damped and more damped when attached to the βcarbon and so on.

• If multiple electronegative groups are attached then there is an additive (although not strictlyadditive) deshielding effect on the proton.

Protons on oxygen or nitrogen have highly variable chemical shifts which are sensitive toconcentration, solvent, temperature, etc.

Anisotropic Effects

The π-system of alkenes, aromatic compounds and carbonyls strongly deshield attached protonsand move them "downfield" to higher δ values.

This is due to an effect known as anistropic shielding and deshielding.

Happ OH

Happ

H

HHappHapp

H

So now the total magnetic field is given by: Heff = Happ + Hshielding + Hanisotropic shielding

Thus, for aromatic rings, olefins and carbonyl compounds this additional magnetic field adds toHapp and for terminal acetylenes the field substracts from Happ.

Proton NMR Chemical Shifts for Common Functional Groups

Functional Group and NMR shifts

δ

Cyclopropane C3H6 0.22

Ethane CH3-CH3 0.88

Ethylene CH2=CH2 5.84

Acetylene HC≡CH 2.88

Benzene C6H6 7.27

Propene CH2=CH-CH3 1.71

Propyne CH≡C-CH3 1.80

Acetone CH3-CO-CH3 2.17

Cyclohexane C6H1 2 1.44

Methyl chloride CH3Cl 3.10

Methylene chloride CH2Cl2 5.30

Chloroform CHCl3 7.27

Ethanol CH3CH2OH CH3CH2OH CH3CH2OH

1.22 3.70 2.58

Acetic acid CH3-COOH CH3-COOH

2.10 8.63

Acetaldehyde CH3-CHO CH3-CHO

2.20 9.80

Diethyl ether (CH3CH2)2O (CH3CH2)2O

1.16 3.36

Ethyl acetate CH3COOCH2CH3

CH3COOCH2CH3

CH3COOCH2CH3

2.03 1.25 4.12

Trimethylamine N(CH3)3 2.12

Triethylamine N(CH2CH3)3 2.42

Toluene C6H5-CH3 2.32

Benzaldehyde C6H5-CHO 9.96

Chemical Shift Table II. Starting Value of Shift Positions: Methyl, δ = 0.90 methylene, δ =1.20 methine, δ =1.55 Alpharefers to the substituent being on the same carbon as the hydrogen and for β it is one removed.

Functional Group R

Type of Hydrogen

Alpha Shift

Beta Shift

Chlorine

CH3- -CH2- -CH-

2.30 2.30 2.55

0.60 0.55 0.15

Bromine

CH3- -CH2- -CH-

1.80 2.15 2.20

0.80 0.60 0.25

Iodine

CH3- -CH2- -CH-

1.30 1.95 2.70

1.10 0.60 0.35

Aryl

CH3- -CH2- -CH-

1.45 1.45 1.35

0.35 0.55 ----

-HC=O and -RC=O

CH3- -CH2- -CH-

1.25 1.10 0.95

0.25 0.30 ---

-CO2H and -CO2R

CH3- -CH2- -CH-

1.20 1.00 0.95

0.25 0.30 ---

-C=C-

CH3- -CH2- -CH-

0.90 0.75 1.25

0.05 0.10 ---

-O-H and -O-Alkyl

CH3- -CH2- -CH-

2.45 2.30 2.20

0.35 0.15 ----

-O-Aryl

CH3- -CH2- -CH-

2.95 3.00 3.30

0.40 0.45 ----

-OCO-Alkyl

CH3- -CH2- -CH-

2.90 2.95 3.45

0.40 0.45 ----

Amine (R=nitrogen)

CH3- -CH2- -CH-

1.25 1.40 1.35

0.20 0.15 ---

-C≡N

CH3- -CH2- -CH-

1.10 1.10 1.05

0.45 0.40 ---

-NO2

CH3- -CH2- -CH-

3.50 3.15 3.05

0.65 0.85 ---

-C≡C-

CH3- -CH2- -CH-

0.90 0.80 0.35

0.15 0.05 ---

Integration

One further feature of the proton NMR is the fact that the intensity of the absorbance of a givenclass of nuclei (with a certain chemical shift) is proportional to the number of protons giving rise tothe signal; that is, the area under a given peak (the integration) is directly proportional to the numberof that type of proton in the molecule.

Integrations are typically given as simplest whole-number ratios, hence, acetic acid, CH3COOH, willhave two peaks in the proton NMR, one at δ ~ 2, area = 3, and a second at δ ~ 12, area = 1. t-butyl,methylether, (CH3)3COCH3, will also have two peaks in the proton NMR, one at δ ~ 1, area = 3,and a second at δ ~ 3.5, area = 1 (the relative areas or both peaks are the same, but each onerepresents three hydrogens).

Equivalence

Homotopic hydrogens: If you have a methylene group, of the form X- CH2-X and you replace oneof hydrogens of the CH2 with a dummy atom and then you independently replace the otherhydrogen of the CH2 group with a dummy atom, then the two molecules thus created will beidentical to each other, the hydrogens are said to be homotopic. The protons on a methyl group arehomotopic.

Enantiotopic: If you have a methylene group, of the form X- CH2-Y and you replace one of thehydrogens of the CH2 with a dummy atom and then you independently replace the other hydrogenof the CH2 group with a dummy atom, the two molecules thus created will be enantiomers of eachother, the protons are said to be enantiotopic. In a nonchiral environment enantiotopic protons areequivalent. However, in a chiral environment such as a chiral solvent, they can, in principle, havedifferent chemical shifts.

Diastereotopic: If you have a methylene group, for example, in a chiral molecule X- CH2-Y* (wherethe * indicates that Y is a chiral group) and you replace one of the hydrogens of the CH2 with adummy atom and then you independently replace the other hydrogen of the CH2 group with adummy atom the two molecules thus created will be diastereomers to each other thus, in principle,the two hydrogens should have different chemical shifts.

Spin-Coupling/Splitting

Proton NMR

Diethyl ether, exhibits two sets of protons in the NMR spectrum; a CH3 group around δ = 1, and aCH2 group around δ 4 (shifted downfield by its proximity to the electronegative oxygen).

The NMR spectrum of diethyl ether does not have simply two peaks, rather, has two sets of peaks,one with four lines of relative intensity 1:3:3:1 which in total integrate to two protons and one set ofthree lines of relative intensity 1:2:1 which in total integrate to three protons.

This multiplicity is due to the phenomena known as spin coupling and arises because of interactionof among proton magnetic field mediated by bonding electrons.

To understand this better let's first examine a simple case in which one has X2CHa–CHbY2.

Each proton has spin 1/2 and to first approximation there is an equal probability of its spin beingaligned with (or aligned against) the applied magnetic field.

Coupling in X2CHa–CHbY2

So if we observe Ha, for 50% of the molecules in the sample Hb is aligned with the field:

• thus the total field is slightly larger and

• thus for these molecules the resonance will occur at slightly higher frequency

For the other 50% of the molecules in the sample Hb is aligned against the field:

• thus the total field is slightly smaller

• thus for these molecules the resonance will occur at slightly higher frequency

As a result, we observed two peaks for Ha whose total integrated area adds to 1.

Similarly we observed two peaks for Hb whose total integrated area adds to 1.

The energy difference between the two peaks for Ha in Hz is the coupling constant (noted as J): • because there are two peaks Ha would be called a doublet

• when two nuclei are coupled to each other the coupling constants are identical

• thus Hb would also be a doublet

• furthermore, the energy difference in Hz between the two peaks for Hb would equal that of Ha in

Hz since Jab = Jba

Thus, spin–spin coupling is very useful to provide information about the connectivity in organicmolecules since:• protons on adjacent atoms can couple to each other• and if there are multiple occurances of this happening in the spectra one can match up pairs of

coupled protons by checking to see which set have identical coupling constants

As we will see shortly, depending upon the distance and orientation of the protons and thehybridization of the carbons to which they are attached the coupling constants between protons havedistinctive values.

The ethyl group

Now lets examine a slightly more complicated case of an ethyl group as shown below:

HA C

HA

HA

C O

HB

HB

HA C

HA

HA

C O

HB

HB

Resultant net magnetic field felt by HB

β-spin

H

Contribution to magnetic field felt by HA from spectrometer

H

α-spin

Contribution to magnetic field felt by HA from HB's

JBA

Contribution to magnetic field felt by HB from HA's

Contribution to magnetic field felt by HB from spectrometer

H

Resultant net magnetic field felt by HA

H

JAB

JAB

JBA

JBA

2

2

In the spectrum for diethyl ether, the CH3 group is split by the two protons on the adjacent CH2

group into three peaks (a triplet).

This results from the adjacent spin either being +1/2, +1/2 or +1/2, –1/2 or –1/2, +1/2 or –1/2, –1/2.

Notice that for the middle two spin possibilities, the magetic fields associated with each of the twospins exactly cancel, and that there are two possible ways for this two occur:

+1/2, –1/2 or –1/2, +1/2.

Likewise, the absorbance for the CH2 is split by the three protons on the methyl group into (n + 1)= 4 peaks (a quartet). The two ethyl groups in diethyl ether show as a single absorbance since the

molecule has a plane of symmetry, that is, both ethyl groups are chemically and magneticallyequivalent:• typical values for J are typically less than 20 Hz• for simple splitting the number of peaks observed for a proton next to n equivalent protons is n

+ 1• the intensity of the lines for the observed resonance can be shown to be that as the prefactors

from the binomial expansion, (1 + x)n, which follows Pascals triangle, shown below.

n = 0 1 singlet s 1 1 1 doublet d 2 1 2 1 triplet t 3 1 3 3 1 quartet q 4 1 4 6 4 1 pentet p 5 1 5 10 10 5 1 hextet h 6 1 6 15 20 15 6 1 septet sp many mess multiplet m

Splitting Trees

The effects of splitting are often described using a splitting tree which depicts the originalabsorbance being split by a coupling constant J into (n + 1) peaks.

This is illustrated for the example below:

Jab Jab Jab

HC CH2

CH2 CH

note peak heightsnot to absolute scale

CH2 peaks need to be 2X

A proton can be coupled to two inequivalent protons in which case the magnetic field couplingbetween each pair would not necessarily be equal. This is illustrated below:

JabJbc

Jab

JabJab

Jbc Jbc

HaC CHb CHc

CHc CHbCHa

In this case, you can see that for Hb coupling to Ha would yield a doublet with coupling constant Jab.When we consider the effect of coupling to Hc each of these lines are further split into two linesseparated by an amount corresponding to Jbc.

Such a pattern is called a double of doublets.

More complex situations

What happens to a doublet of doublet as Jab– Jbc decreases. This is illustrated below.

The simple coupling patterns we have seen so far assume the difference in chemical shift betweenthe peaks (in Hz) is much larger than the coupling constants between the hydrogens.

When this condition is met the spectrum is said to have first-order coupling. However, when Δδab isnot much larger than Jab then the coupling is said to be non-first-order and the appearance of thespectrum changes. In some cases quite dramatically.

JabJab JabJab JabJab

Δδab >> Jab Δδab > Jab Δδab ~ Jab Δδab = 0

In particular, the inner lines of the pattern gain intensity at the expense of the outer line as shownbelow. The origin of this evolution is in quantum mechanics, which is beyond the scope of thecourse.

These patterns are quite common and being able to recognize them coupled with a knowledge ofwhich hydrogen atoms have similar chemical shift is a useful tool to make assignments, since younow have two pieces of information (i.e. the two hydrogen atoms are likely adjacent and havesimilar environments).

Note that unlike chemical shift coupling constants are due to the internal magnetic field of theprotons and thus coupling constants are independent of spectrometer frequency.

decreasing Jab – Jbc

Jab = Jbc

Things are not always ideal in the real world

A splitting tree such as this is useful in understanding more complex splitting patterns, such asthose that occur in Br-CH2CH2CH2-OD, as shown below:

As shown by the coupling tree, the spectrum is predicted to consist of 9 peaks centered aroundδ 1.53. The actual NMR, as shown above, shows only five peaks in this region.

This is because the central seven peaks are only separated by 2 Hz, and the peaks are not infinitelynarrow. As a result some overlap and appear as single peaks.

Common 1H NMR Splitting Patterns

The examples given above represent only the simplest and most common coupling patterns seen inthe 1H NMR.

A simple ethyl group displays a quartet and a triplet in the ratio 2:3; the chemical shift of the CH2

group is sensitive to the attached substituent and typically varies between δ 4 (for oxygen) to δ 2(for a carbonyl).

An isopropyl group displays a septet (7 peaks) and a doublet in the ratio 1:6; again, the chemicalshift of the CH group is sensitive to the attached substituent and typically varies between δ 4 (foroxygen) to δ 2 (for a carbonyl).

A 1,4-disubstituted aromatic compound displays two doublets in the ratio of 1:1, typically in theregion around δ 7.

Although the chemical shift of a monosubstituted aromatic compound should give rise to three setsof peaks (integration ratio 2:2:1), often only a single peak of integrated intensity 5 is observedaround δ = in the 1H NMR.

This is because the differences in chemical shift are typically small, as are the coupling constants.

Table of coupling constants between Hydrogen (given in Hz)

CH

H

12-20

C CH

H

0-3.5

C CH H

C CH H

C CH

H

O CH

C H

C CH

C CH

C CH

C H

C CC H

H

C C HH

H H H

H

HH

Ha

Ha

HeHe

2-9 6-14 11-18 4-10

10-13 3-7 1-32-3

7-10 2-3 0.1- Jaa = 10-13Jae = 2-5Jee = 2-5

Use of deuterium in NMR

Although deuterium has a nuclear spin, deuterium NMR and proton NMR require differentoperating frequencies.• Deuterium NMR absorptions are not detected under the conditions used for proton NMR.• The coupling constants for proton-deuterium splitting are very small.• Deuterium substitution can be used to simplify NMR spectra and assign resonances.• Solvents used in NMR spectroscopy must either be devoid of protons, or their protons must

not have NMR absorptions that obscure the sample absorptions.

NMR spectra of dynamic systems

• The spectrum of a compound involved in a rapid equilibrium is a single spectrum that is thetime-average of all species involved in the equilibrium.

• The time-averaging effect of NMR is not limited to simple conformational equilibria; the spectraof molecules undergoing any rapid process, such as a chemical reaction, are also averaged byNMR spectroscopy.

• The assignment of the OH proton can be confirmed by adding a drop of D2O to the NMRsample tube shaking and taking the spectra. The OH protons rapidly exchange with the protonsof D2O to form OD groups on the alcohol and thus become invisible to NMR.

NMR of alcohols and amines The chemical shift of the OH proton in an alcohol depends on the degree to which the alcohol isinvolved in hydrogen bonding under the conditions that the spectrum is determined.• The presence of water, acid or base causes collapse of the OH resonance to a single line and

obliterates all coupling associated with this proton.• This effect is caused by a phenomenon called chemical exchange.• This type of behavior is quite general for alcohols, amines, and other compounds with a proton

bonded to an electronegative atom.• Rapidly exchanging protons do not show spin-spin splitting with neighboring protons.• Acid and base catalyze this exchange reaction, accelerating it enough that splitting is no longer

observed. In the absence of acid or base, this exchange is much slower, and splitting of the OHprotons and neighboring protons is observed.

• The assignment of the OH proton can be confirmed by adding a drop of D2O to the NMRsample tube shaking and taking the spectra. The OH protons rapidly exchange with the protonsof D2O to form OD groups on the alcohol and thus become invisible to NMR.

Introduction to 1 3C NMR

Any nucleus with a nuclear spin can be studied by NMR spectroscopy; for a given magnetic fieldstrength, different nuclei absorb energy in different frequency ranges.

The NMR spectroscopy of carbon is called carbon NMR or CMR; the only isotope of carbon thathas a nuclear spin is 1 3C.

• The relative abundance of 1 3C occurs in 1.1% natural abundance.

• The magnetogyric ratio for the 1 3C nucleus is about 0.25 of the 1H nucleus and thesensitivity goes as this ratio to the third power which by itself would lead to a 1 3Cnucleus having 0.159 the sensitivity of the 1H nucleus.

• Because of these two factors, 1 3C spectra are about 0.0002 times as intense as 1Hspectra.

• For weak signals, signal to noise is increased by averaging out the signal from multiplescans. The noise averages to zero and the signal increases.

• Signal to noise increases as (# of scans)0.5. Given then relative sensitivity to 1H NMRyou can see that many scans could be needed.

• Special instrumental techniques have been devised for obtaining such weak spectra.

• Coupling (splitting) between carbons generally is not observed. The reason is the lownatural abundance of carbon (1.1%), making it unlikely that two 1 3C nuclei will residenext to each other in a given molecule.

• Attached protons, however, with a spin of 1/2, will couple with the 1 3C nucleus togenerate spin-coupling.

• The signal from the carbon will be split into (n + 1) peaks, where n is the number ofattached protons.

1 3C Chemical shfts

The range of chemical shifts is large compared to that in proton NMR; trends in carbon chemicalshifts parallel those for proton chemical shifts.

Chemical shifts in CMR are more sensitive to small changes in chemical environment.

Characteristics of chemical shifts for carbons:• simple methyl carbons in alkanes, or attached to an alkene, of aryl group δ 15-30• simple methylene, methine and quarternary carbons, carbons in alkanes attached to an

alkene of aryl group δ 20-60• characteristic chemical shifts for carbons attached to electronegative atoms• oxygen (in ethers or the single bonded oxygen in an ester) δ 40-80,• nitrogen (in amines or the single bonded nitrogen in an amide) δ 25-27• chlorine δ 20-60• bromine δ 10-50• iodine δ –15 -25• alkyne δ 65-85,• nitrile δ 110-130• alkene carbons to δ 100-150• aromatic carbons δ 110-170• amide (C=O) δ 160-180• ester (C=O) δ 160-190• ketone and aldehyde (C=O) 190-220

1 3C coupling and decoupling

Splitting of 1 3C resonances by protons (1 3C – 1H splitting) is so large relative to differences inchemical shifts that is can be difficult to distinguish between different peaks and peaks that are splitby coupling.• Spectra can be obtained in proton noise-decoupling mode wherein which the sample is

irradiated at a second frequency which promotes all of the protons in the molecule to high spinstates, disallowing the spin-coupling process. All of the split 1 3C peaks are thereby reduced tosharp singlets.

• While this operating mode provides less information than the non-decoupled mode, it iscommonly used because it results in a significant signal enhancement due to a phenomenaknown as the "Nuclear Overhauser Effect" (NOE).

• NOE enhancement is not uniform, hence the integration of a 1 3C NMR spectrum is typicallymeaningless.

Spectra in which proton coupling has been eliminated are called proton-decoupled carbon NMRspectra.

• The DEPT (distortionless enhancement by polarization transfer) technique yields separatespectra for methyl, methylene and methine carbons, and each line in these spectra correspondsto a line in the complete CMR spectrum; lines in the complete CMR spectrum that do notappear in the DEPT spectra are assumed to arise from quaternary carbons.

• CMR spectra are generally not integrated because the instrumental technique used for takingspectra, gives relative peak integrals that are governed by factors other than the number ofcarbons.

introduction to nmr spectroscopyww2.chemistry.gatech.edu/class/2311/marder/nmr.pdf · introduction...

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