1 NMR SPECTROSCOPY CHEM 212. Introduction to Spectroscopy 2 Spectroscopy is the study of the...

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NMR SPECTROSCOPY

CHEM 212

Introduction to Spectroscopy

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

1. Electromagnetic 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


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5. 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):

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

7. 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 = ___

l

c E = hn =

___

l

hc


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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:

Energ

y

The Spectroscopic Process5

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

hn hn hn

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

Energ

y

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. E

xcita

tion

Types of Spectroscopy6

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UVX-rays IRg-rays RadioMicrowave

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

Vis

ible

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

Basis of NMR Spectroscopy7

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


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

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

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

Spin Quantum Numbers of Common Nuclei

Element 1H 2H 12C 13C 14N 16O 17O 19F 31P 35Cl

Nuclear 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


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

mH H

mI = +½ I = -½

Creating Non-Degenerate Nuclei

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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

Nuclear Magnetic Resonance

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When a nuclei of spin

+½ encounters a

photon where

n = E/h, the two “couple”

The nuclei “flips” its spin state

from +½ to –½ and is

now opposed to B0

The nuclei “relaxes”

with the re-emission of a photon and returns to the + ½ spin state

Nuclear Magnetic Resonance

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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 Shift13

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

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The 1H NMR Spectrum15

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

Si

CH3

H3C CH3CH3


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


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 Signals18

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-xyleneCH3

CH3

H

H

CH3

CH3

H

Z

CH3

CH3

Z

H

Same Compound


Examples

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

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Remember:

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


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• 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


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

a. Electronegativity b. Hybridizationc. Proton acidity/exchange


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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


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

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


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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

CH3Cl CH2Cl2 CHCl3

d of H 3.05 5.30 7.27

-CH2Br -CH2CH2Br -CH2CH2CH2Br

d of H 3.30 1.69 1.25


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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)


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What we observe is slightly different:Type of H Carbon

hybridizationName of H Chemical Shift,

d

R-CH3, R2CH2, R3CH sp3 alkyl 0.8-1.7

C=C-CH3 sp3 allyl 1.6-2.6

CC-H sp acetylenic 2.0-3.0

C=C-H sp2 vinylic 4.6-5.7

Ar-H sp2 aromatic 6.5-8.5

O=C-H sp2 aldehydic 9.5-10.1

Chemists refer to this observation as magnetic anisotropy


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Magnetic Anisotropy – Aromatic Protonsa. In a magnetic field, the six 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.


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Similarly this effect operates in alkenes:


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


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Intensity of Signals—Integration

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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


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Spin-Spin Splitting41

Consider the spectrum of ethyl alcohol: Why does each resonance “split” into

smaller peaks?

HO

CH2

CH3


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


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

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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 :


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 the resonance for Ha slightly downfield (the magnetic field of a proton is tiny compared to B0)

C C

HbHa

R

HbHb

Ha

All 3 Hb protons + ½

resonance for Ha in absence of 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

C C

HbHa

R

HbHb

Ha

2 Hb protons + ½

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

2 Hb protons - ½


or or


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

C C

HbHa

R

HbHb

Ha

All 3 Hb protons - ½



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



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


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 1

1 doublet 2 1 1

2 triplet 3 1 2 1

3 quartet 4 1 3 3 1

4 quintet 5 1 4 6 4 1

5 sextet 6 1 5 10 10 5 1

6 septet 7 1 6 15 20 15 6 1

1H NMR—Spin-Spin Splitting54

Common patterns:X CH3

X

X

X

X

tert-butyl - singletmethyl - singlet

iso-propyl – septet - doublet

ethyl – quartet - triplet

n-propyl – triplet - quintet - triplet


Another Example:

BrC

CBr

Br

HaHb

Hb

Br

Br

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 signals

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

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

If 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:

H H

H H

Aliphatic compounds that have free rotation about each bond

G

Ha

Hb

Hc

H

H

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


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

-CH2- -CH3


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:

C

CX

H

H

H

The 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:

2Jgem = 0 – 1 Hz

3Jtrans = 11- 18 Hz

3Jcis = 6 - 15 Hz

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

3JAB3JAB

3JAC

C

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

3J = 6-8

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

3Jtrans = 4-83Jcis = 6-12

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


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 Determination68

13C NMR72

• 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 NMR73

13C NMR74

• 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 NMR75

• 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 NMR76

13C NMR77

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

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X C Z

Y

methine

X C Y

H

methylene

X C H

H

methyl

H HH

Shoolery Values for MethyleneX or Y Substituent

ConstantX 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

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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

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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

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Shoolery Tables

Work for aromatics as well (.pdf posted)

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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!

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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)

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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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