Nmr spectroscopy
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NMR Spectroscopy. CHEM 212. Introduction to Spectroscopy. Spectroscopy is the study of the interaction of matter with the electromagnetic spectrum Electromagnetic radiation displays the properties of both particles and waves

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NMR Spectroscopy

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Nmr spectroscopy

NMR Spectroscopy

CHEM 212

Introduction to spectroscopy

Introduction to Spectroscopy

Spectroscopy is the study of the interaction of matter with the electromagnetic spectrum

  • Electromagnetic radiation displays the properties of both particles and waves

  • 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

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

    E = hn

    The constant of proportionality is Plank’s constant, h

Introduction to spectroscopy1

Introduction to Spectroscopy

  • 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):

  • Amplitude describes the wave height, or strength of the oscillation

  • 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





n =

 E = hn =



Introduction to spectroscopy2

Introduction to Spectroscopy

  • Remember atoms and molecules are quantum mechanical particles

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

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

  • 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:


The spectroscopic process

The Spectroscopic Process

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

excited state

5. Detection: Photons that are reemitted and detected by the spectrometer correspond to quantum mechanical energy levels of the molecule


4. Relaxation

3. Excitation


rest state

rest state

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





Types of spectroscopy

Types of Spectroscopy








Frequency, n (Hz)






Wavelength, l

~0.01 nm

10 nm

1000 nm

0.01 cm

100 m

Energy (kcal/mol)

> 300






Basis of nmr spectroscopy

Basis of NMR Spectroscopy

Nuclear Spin States

  • 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 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

Basis of nmr spectroscopy1

Basis of NMR Spectroscopy

  • 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

Basis of nmr spectroscopy2

Basis of 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—degeneratein energy




I = +½

I = -½


Creating non degenerate nuclei

Creating Non-Degenerate Nuclei

Nuclear magnetic resonance

Nuclear Magnetic Resonance

Nuclear magnetic resonance1

Nuclear Magnetic Resonance

  • 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

Origin of the Chemical Shift

The proton 1 h nmr spectrum

The Proton (1H) NMR Spectrum

The 1 h 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

The 1 h nmr spectrum1

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 1 h nmr spectrum2

The 1H NMR Spectrum

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

    • Number of signals

    • Position of signals

    • Intensity of signals.

    • Spin-spin splitting of signals

The number 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

The number of signals1

The Number of Signals

  • 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

Same Compound

The number of signals2

The Number of Signals

  • Examples

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

The number of signals3

The Number of Signals

  • 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.

The number of signals4

The Number of Signals

  • Proton equivalency in cycloalkanes can be determined similarly:

The number of signals5

The Number of Signals

  • Enantiotopic Protons – when substitution of two H atoms by Z forms enantiomers:

    • The two H atoms are equivalent and give the same NMR signal

    • These two atoms are called enantiotopic

The number of signals6

The Number of Signals

  • Diastereotopic Protons - when substitution of two H atoms by Z forms diastereomers

    • The two H atoms are not equivalent and give two NMR signals

    • These two atoms are called diastereotopic

Chemical shift position of signals

Chemical Shift – Position of Signals

  • Remember:

Chemical shift position of signals1

Chemical Shift – Position of Signals

  • 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

 Downfield, deshielded

Upfield, shielded 

Chemical shift position of signals2

Chemical Shift – Position of Signals

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

    • Electronegativity

    • Hybridization

    • Proton acidity/exchange

Chemical shift position of signals3

Chemical Shift – Position of Signals

  • Electronegative groups comprise most organic functionalities:

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


    -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

Chemical shift position of signals4

Chemical Shift – Position of Signals

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

Chemical shift position of signals5

Chemical Shift – Position of Signals

  • The magnitude of the deshielding effect is cumulative:

    As more chlorines are added d becomes larger

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

Chemical shift position of signals6

Chemical Shift – Position of Signals


  • 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)

Chemical shift position of signals7

Chemical Shift – Position of Signals

  • What we observe is slightly different:

Chemists refer to this observation as magnetic anisotropy

Chemical shift position of signals8

Chemical Shift – Position of Signals

  • Magnetic Anisotropy – Aromatic Protons

    • In a magnetic field, the six  electrons in benzene circulate around the ring creating a ring current.

    • The magnetic field induced by these moving electrons reinforces the applied magnetic field in the vicinity of the protons.

    • The protons thus feel a stronger magnetic field and a higher frequency is needed for resonance. Thus they are deshielded and absorb downfield.

Chemical shift position of signals9

Chemical Shift – Position of Signals

  • Similarly this effect operates in alkenes:

Chemical shift position of signals10

Chemical Shift – Position of Signals

  • In alkynes there are two perpendicular sets of p-electrons—the molecule orients with the field lengthwise—opposing B0shielding the terminal H atom

Chemical shift position of signals11

Chemical Shift – Position of Signals

Chemical shift position of signals12

Chemical Shift – Position of Signals

Intensity of signals integration

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

Intensity of signals integration1

Intensity of Signals—Integration

Intensity of signals integration2

Intensity of Signals—Integration

Spin spin splitting

Spin-Spin Splitting

  • Consider the spectrum of ethyl alcohol:

  • Why does each resonance “split” into smaller peaks?

Spin spin splitting1

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 splitting2

Spin-Spin Splitting

  • 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

Nmr spectroscopy

  • 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 splitting3

Spin-Spin Splitting

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

Spin spin splitting4

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

Spin spin splitting5

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)

All 3 Hb

protons + ½

resonance for Ha in absence of spin-spin splitting

Spin spin splitting6

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



2 Hb

protons + ½

resonance for Ha in absence of spin-spin splitting

Spin spin splitting7

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



2 Hb

protons - ½

resonance for Ha in absence of spin-spin splitting

Spin spin splitting8

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

All 3 Hb

protons - ½

resonance for Ha in absence of spin-spin splitting

Spin spin splitting9

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

resonance for Ha in absence of spin-spin splitting

Spin spin splitting10

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

resonance for Ha in absence of spin-spin splitting

Spin spin splitting11

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 then + 1 rule

1 h nmr spin spin splitting

1H NMR—Spin-Spin Splitting

  • Common patterns:

methyl - singlet

tert-butyl - singlet

ethyl – quartet - triplet

n-propyl – triplet - quintet - triplet

iso-propyl – septet - doublet

1 h nmr spin spin splitting1

1H NMR—Spin-Spin Splitting

1 h nmr spin spin splitting2

1H NMR—Spin-Spin Splitting

  • Another Example:

1 h nmr spin spin splitting3

1H NMR—Spin-Spin Splitting

  • Another Example:

1 h nmr spin spin splitting4

1H NMR—Spin-Spin Splitting

Three general rules describe the splitting patterns commonly seen in the 1H NMR spectra of organic compounds:

  • Equivalent protons do not split each other’s signals

  • A set of n nonequivalent protons splits the signal of a nearby proton into n + 1 peaks

  • Splitting is observed for nonequivalent protons on the same carbon or adjacent carbons

    If Ha and Hb are not equivalent, splitting is observed when:

1 h nmr spin spin splitting5

1H NMR—Spin-Spin Splitting

  • Magnetic influence falls off dramatically with distance

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

Aliphatic compounds that have free rotation about each bond

Aromatic compounds where each proton is held in position relative to one another

1 h nmr spin spin splitting6

1H NMR—Spin-Spin Splitting

  • 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

This J

Is equal to this J



1 h nmr spin spin splitting7

1H NMR—Spin-Spin Splitting

  • 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:

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

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

The cis-relationship, is over an intermediate distance

1 h nmr spin spin splitting8

1H NMR—Spin-Spin Splitting

  • For this substituted ethylene we see the following spectrum:

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

2Jgem = 0 – 1 Hz


3Jtrans = 11- 18 Hz



3Jcis = 6 - 15 Hz

1 h nmr spin spin splitting9

1H NMR—Spin-Spin Splitting

  • 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

1 h nmr spin spin splitting10

1H NMR—Spin-Spin Splitting

  • Some common J-values

3J = 6-8

3Jtrans= 11-18

3J = 8-11

3Jtrans = 4-8

3Jcis = 6-12

3Jcis= 6-15

3Ja,a = 8-14

3Ja,e = 0-7

3Je,e = 0-5

3J = 5-7

3Jortho = 7-10 Hz

4Jmeta = 1-3 Hz

5Jpara = 0-1 Hz

3Jallyl= 4-10

3Jtrans = 4-8

3Jcis = 6-12

1 h nmr spin spin splitting11

1H NMR—Spin-Spin Splitting

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

1 h nmr spin spin splitting12

1H NMR—Spin-Spin Splitting

  • A combined example:

1 h nmr spin spin splitting13

1H NMR—Spin-Spin Splitting

  • 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

Structure Determination

Structure determination1

Structure Determination

Structure determination2

Structure Determination

Structure determination3

Structure Determination

13 c 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

13 c nmr1


13 c nmr2


  • 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.

13 c nmr3


  • 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.

13 c nmr4


13 c nmr5


Shoolery tables

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

Shoolery values for methylene

Shoolery Values for Methylene

Shoolery values for methine

Shoolery Values for Methine

Shoolery tables1

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 tables2

Shoolery Tables

  • Work for aromatics as well (.pdf posted)

Running an nmr experiment

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!

Running an nmr experiment1

Running an NMR Experiment

  • 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)

Running an nmr experiment2

Running an NMR Experiment

  • 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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