chemical shift and coupling constants. unpaired nuclear spins are of importance in nmr. the actual...

CHEMICAL SHIFT AND COUPLING CONSTANTS

Unpaired nuclear spins are of importance in NMR.

The actual spectral data acquired by NMR is the free-induction decay, or FID.

FT

3.603.603.653.653.703.703.753.753.803.803.853.853.903.90

Time, sec Frequency, Hz

FID

An applied magnetic field B0, the strength of which is measured in

tesla (T), and the frequency n of radiation used for resonance,

measured in hertz (Hz), or megahertz (MHz)—(1 MHz = 106 Hz).

NMR Spectrometer

Schematic diagram of a nuclear magnetic resonance spectrometer.

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• An NMR spectrum is a plot of the intensity of a peak against its chemical shift, measured in parts per million (ppm).

1H NMR—The Spectrum

Protons in different environments absorb at slightly different frequencies, so they are distinguishable by NMR.

High frequency:High frequency: The shift of an NMR signal to the left on the chart paper.

Low frequency:Low frequency: The shift of an NMR signal to the right on the chart paper.

1H-NMR spectrum of methyl acetate.• NMR absorptions

generally appear as sharp peaks.

• Increasing chemical shift is plotted from left to right.

• Most protons absorb between 0-10 ppm.

• The terms “upfield” and “downfield” describe the relative location of peaks. Upfield means to the right. Downfield means to the left.

• NMR absorptions are measured relative to the position of a reference peak at 0 ppm on the d scale due to tetramethylsilane (TMS). TMS is a volatile inert compound that gives a single peak upfield from typical NMR absorptions.

1D 1H NMR Spectra of Skeletal Muscle Tissue

Chemical Shift

When an atom is placed in a magnetic field, its electrons circulate about the direction of the applied magnetic field.

This circulation causes a small magnetic field at the nucleus which opposes the externally applied field.

The magnetic field at the nucleus (the effective field) is therefore generally less than the applied field by a fraction , B = Bo (1- )

The electron density around each nucleus in a molecule varies according to the types of nuclei and bonds in the molecule. The opposing field and therefore the effective field at each nucleus will vary. This is called the chemical shift phenomenon.

= (/2)Blocal = (/2)(1-)

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• In the vicinity of the nucleus, the magnetic field generated by the circulating electron decreases the external magnetic field that the proton “feels”.

• Since the electron experiences a lower magnetic field strength, it needs a lower frequency to achieve resonance. Lower frequency is to the right in an NMR spectrum, toward a lower chemical shift, so shielding shifts the absorption upfield.

1H NMR—Position of Signals

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Nuclear Magnetic Resonance Spectroscopy


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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.

•Protons near electronegative atoms are deshielded, so they absorb downfield.


Chemical Shift is field dependent

The chemical shift of a nucleus is the difference between the resonance frequency of the nucleus and a standard, relative to the standard. This quantity is reported in ppm and given by the symbol delta, .

Choice of Solvents: D2O, H2O, CDCl3, DMSO

Reference Compounds: Aqueous solution: DSS (4,4-dimethyl-4-silapentane-1-sulfonic acid) (TSP) Trimethylsilyl propionate

(CH3)4Si, usually referred to as TMS used in CDCl3, DMSO

The magnitude of the screening depends on the atom. For example, carbon-13 chemical shifts are much greater than hydrogen-1 chemical shifts.

Tetramethylsilane (TMS)

CH3

Si CH3

CH3

CH3

13

• Protons in different environments absorb at slightly different frequencies, so they are distinguishable by NMR.

• The frequency at which a particular proton absorbs is determined by its electronic environment.

• The size of the magnetic field generated by the electrons around a proton determines where it absorbs.

• Modern NMR spectrometers use a constant magnetic field strength B0, and then a narrow range of frequencies is applied to achieve the resonance of all protons.

• Only nuclei that contain odd mass numbers (such as 1H, 13C, 19F and 31P) or odd atomic numbers (such as 2H and 14N) give rise to NMR signals.

The difference in resonance frequencies among the various hydrogen nuclei within a molecule due to shielding/deshielding is generally very small.

The difference in resonance frequencies for hydrogens in CH3Cl compared to CH3F under an applied field of 7.05T is only 360 Hz, which is 1.2 parts per million (ppm) compared with the irradiating frequency.

360 Hz300 x 106 Hz

1.2 = 1.2 ppm106=

ChemicalChemicalShiftsShifts

11H-NMRH-NMR

RCH2 OR

(CH3 )4Si

ArCH3

RCH3

RC CH

RCCH3

ROHRCH2 OH

ArCH2 R

O

O

RCH2 RR3 CH

R2 NH

RCCH2R

R2 C=CRCHR2

R2 C=CHR

RCH

O

RCOH

O

RCH2 ClRCH2 BrRCH2 I

RCH2 F

ArHO

O

R2 C=CH2

RCOCH3

RCOCH2R

ArOH

9.5-10.1

3.7-3.9

3.4-3.6

Type of Hydrogen

0 (by definition)

Type of Hydrogen

Chemical Shift ()

1.6-2.62.0-3.0

0.8-1.01.2-1.41.4-1.7

2.1-2.3

0.5-6.0

2.2-2.6

3.4-4.0

Chemical Shift ()

3.3-4.0

2.2-2.52.3-2.8

0.5-5.0

4.6-5.05.0-5.7

10-13

4.1-4.73.1-3.3

3.6-3.84.4-4.5

6.5-8.5

4.5-4.7

Chemical shift depends on the (1) electronegativity of nearby atoms, (2) hybridization of adjacent atoms, and (3) diamagnetic effects from adjacent pi bonds.Electronegativity

CH3OH

CH3F

CH3Cl

CH3BrCH3I

(CH3)4C(CH3)4Si

CH3-XElectroneg-ativity of X

Chemical Shift ()

4.03.5

3.1

2.82.5

2.1

1.8

4.263.47

3.05

2.68

2.16

0.86

0.00

Chemical ShiftHybridization of adjacent atoms.

RCH3, R2CH2, R3CH

R2C=CHR, R2C=CH2

RCHO

R2C=C(R)CHR2

RC CH

Allylic

Type of Hydrogen(R = alkyl)

Name ofHydrogen

Chemical Shift ()

Alkyl

Acetylenic

Vinylic

Aldehydic

0.8 - 1.7

1.6 - 2.6

4.6 - 5.7

9.5-10.1

2.0 - 3.0

Chemical ShiftDiamagnetic effects of pi bonds

A carbon-carbon triple bond shields an acetylenic hydrogen and shifts its signal to lower frequency (to the right) to a smaller value.

A carbon-carbon double bond deshields vinylic hydrogens and shifts their signal to higher frequency (to the left) to a larger value.

RCH3

R2C=CH2

RC CH

Type of H Name

Alkyl

VinylicAcetylenic

0.8- 1.0

4.6 - 5.72.0 - 3.0

Chemical Shift ()

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• The chemical shift of a C—H bond increases with increasing alkyl substitution.

1H NMR—Chemical Shift Values

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• Protons in a given environment absorb in a predictable region in an NMR spectrum.


Chemical Shift Magnetic induction in the bonds of a carbon-carbon

triple bond shields an acetylenic hydrogen and shifts its signal lower frequency.

Chemical Shift Magnetic induction in the bond of a carbon-carbon

double bond deshields vinylic hydrogens and shifts their signal higher frequency.

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• In some cases, such as the benzene molecule, the circulation of the electrons in the aromatic orbitals creates a magnetic field at the hydrogen nuclei which enhances the Bo field. This phenomenon is called deshielding.

• 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.

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• Four different features of a 1H NMR spectrum provide information about a compound’s structure:a. Number of signalsb. Position of signalsc. Intensity of signals.d. Spin-spin splitting of signals.

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• 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.

1H NMR—Number of Signals

• To determine equivalent protons in cycloalkanes and alkenes, always draw all bonds to hydrogen.

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• 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.


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• Proton equivalency in cycloalkanes can be determined similarly.


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Other Factors Affecting Chemical Shift1.Solvent2.Presence of electronegative atoms.3.pH.4.Temperature5.Hydrogen bond6.Conformational Changes7.Presence of ligand

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• The chemical shift of a C—H can be calculated with a high degree of precision if a chemical shift additivity

table is used.• The additivity tables starts with a base chemical shift value

depending on the structural type of hydrogen under consideration:

Calculating 1H NMR—Chemical Shift Values

CH3 CH2

CH

Methylene Methine

0.87 ppm 1.20 ppm 1.20 ppmBase Chemical Shift

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• Consider the spectrum below:

1H NMR—Spin-Spin Splitting

Ethyl Bromide

Spin-Spin Coupling

Spin-Spin Coupling (Splitting)

Observation: A nucleus with a magnetic moment may interact with other nuclear spins resulting in mutual splitting of the NMR signal from each nucleus into multiplets.

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The Origin of 1H NMR—Spin-Spin Splitting

The frequency difference, measured in Hz, between two peaks of the doublet is called the coupling constant, J.

J

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• Spin-spin splitting occurs only between nonequivalent protons on the same carbon or adjacent carbons.


Let us consider how the doublet due to the CH2 group on BrCH2CHBr2 occurs:

• When placed in an applied field, (B0), the adjacent proton (CHBr2) can be aligned with () or against () B0. The likelihood of either case is about 50% (i.e., 1,000,006 vs 1,000,000).

• Thus, the absorbing CH2 protons feel two slightly different magnetic fields—one slightly larger than B0, and one slightly smaller than B0.

• Since the absorbing protons feel two different magnetic fields, they absorb at two different frequencies in the NMR spectrum, thus splitting a single absorption into a doublet, where the two peaks of the doublet have equal intensity.

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Let us now consider how a triplet arises:

• When placed in an applied magnetic field (B0), the adjacent protons Ha and Hb can each be aligned with () or against () B0.

• Thus, the absorbing proton feels three slightly different magnetic fields—one slightly larger than B0(ab). one slightly smaller than B0(ab) and one the same strength as B0 (ab).

When determining the spin-spin coupling, look at the number of protons on the adjacent carbon. For the methyl group, look at the methylene group. There are

2 protons, so using the N+1 rule tells us that the peak should be a triplet in a 1:2:1 ratio.

CH2 b

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For the methylene group, look at the methyl group. There are 3 protons, so using the N+1 rule tells us that the peak should be a quartet in a 1:3:3:1

ratio.

CH3 a

4. The Coupling Constant, J - the separation between peaks in a multiplet measured in units of Hz.

3.253.253.303.303.353.353.403.403.453.453.503.50

1.501.501.551.551.601.601.651.651.701.701.751.75

CH3CH2Br CH2 b

CH3 a

Jba

Jab

Jba

Jba

Jab

a b

Jba = Jab = 7.3 Hz

•The energy of the interactions between two spins A and B can be found by the relationship:

E = JAB * IA * IBE = JAB * IA * IB

• IA and IB are the nuclear spin vectors, and are proportional to A and B, the magnetic moments of the two nuclei. JAB is the scalar coupling constant. So we see a very important feature of couplings. It does not matter if we have a 60, a 400, or an 800 MHz magnet, the coupling constants are always the same!!!

Lets do a more detailed analysis in term of the energies. Lets think a two energy level system, and the transitions for nuclei A. When we have no coupling (J = 0), the energy involved in either transition (A1 or A2) is equal (no spin-spin interaction).

A XA X

A1

A2

A1

A2

Bo E

J = 0J > 0

E4

E3

E2

E1

•When J > 0, the energy levels of the spin system will be either stabilized or destabilized. Depending on the relative orientations of the nuclear moments, the energies for the A1 and A2 transition will change giving two different frequencies (two peaks for A).

A1

A2

A1 = A2

A2 A1

A2 A1

A1 A2

A1 A2

J = 0J > 0 J < 0

•As mentioned before, the choice of positive or negative J is a definition. However, we see that we won’t be able to tell if we have a positive or negative J, because the lines in the spectrum corresponding to the different transitions basically

change places. Unless we are interested in studying the energies, this is not important for structure elucidation…

Now look at some simple examples.

Examine the size of the peaks in the splitting.

Hb is augmenting external field causing a

larger energy gap.

Hb decrementing external field causing a

smaller energy gap.

Ha is being excited. Hb is causing spin-

spin splitting by slightly increasing or

decreasing the magnetic field

experienced by Ha.

Spin-Spin Splitting

Figure 13.15b, p.512

Two neighboring atoms assist external field.

More energy needed to excite. Peak is “downfield”.

One neighbor assists, one hinders. No effect.

Both neighbors oppose. Less energy

needed to excite, “upfield”.

Again Ha is flipping, resonating. The two Hb are causing spin-

spin splitting by slightly changing the

magnetic field experienced by Ha.

Recall that for the two Hb atoms the two states

(helping and hindering the external field) are almost equally likely.

This give us the 1 : 2 : 1 ratio.

Figure 13.15c, p.512

All Hb augment

Two augment, one decrement.

One augment, two decrement.

All decrement.

Ha being excited.

Three equivalent Hb causing spin spin splitting.

Three neighboring Hb’s causing splitting when Ha is

excited.

Spin-Spin Splitting in 1H NMR Spectra Peaks are often split into multiple peaks due to magnetic

interactions between nonequivalent protons on adjacent carbons, The process is called spin-spin splitting

The splitting is into one more peak than the number of H’s on the adjacent carbon(s), This is the “n+1 rule”

The relative intensities are in proportion of a binomial distribution given by Pascal’s Triangle

The set of peaks is a multiplet (2 = doublet, 3 = triplet, 4 = quartet, 5=pentet, 6=hextet, 7=heptet…..)

1 1 1 1 2 1 1 3 3 1 1 4 6 4 1 1 5 10 10 5 11 6 15 20 15 6 1

singlet

doublet

tripletquartet

pentet

hextet

heptet

p. 491

Attempt to anticipate the splitting patterns in each molecule.

Magnitude of Coupling Constant, J

The magnitude of the coupling constant, J, can vary from 0 to about 20 Hz.

This represents an energy gap (E = h) due to the interaction of the nuclei within the molecule. It does not depend on the strength of the external

field.

J is related to the dihedral angle between bonds. J largest for 0 (eclipsed) or 180 (anti), smallest for 90, intermediate for gauche.

Table 13.4, p.511

anti gauche

vinyl systems

Figure 13.17, p.513

Naturally if there are two non-equivalent nuclei they split each other.

Figure 13.19, p.513

Three nonequivalent nuclei. Ha and Hb split each other. Also Hb and Hc split each other.

Technique: use a tree diagram and consider splittings sequentially.

Figure 13.20, p.514

More complicated system

Figure 13.21, p.514

Not equivalent (R1 is not same as R2) because there is no rotation

about the C=C bond.

Return to Vinyl Systems

JAB = 11-18 Hz, BIG

JAC = 0 – 5 Hz, SMALL

JBC = 5 – 10 Hz, MIDDLE

Each of these patterns is different from the

others.

Analysis

Now examine the left most signal….

Fast Exchange

OH

H H

H

HH

ethanol

Expect coupling between these

hydrogens. Three bond separation.

There is no coupling observed

especially in acid or base.

Reason: exchange of weakly acidic

hydrogen with solvent.

The spectrometer sees an “averaged

hydrogen”. No coupling and broad

peak.

• J is independent of magnetic field strength• Mutually coupled hydrogen nuclei have the same J value.

The magnitude of J gives information about structure.

C C

H HC C

H

H

C CH H

C CH

H

J = 7 Hz J = 15 Hz

J = 10 Hz J = 2 Hz

Assignment

35203540 Hz 27602780 Hz 24802500 Hz 3.95 ppm 22002220 Hz 21202140 Hz

35203540 Hz 27602780 Hz 24802500 Hz 3.95 ppm 22002220 Hz 21202140 Hz

10Hz

Karplus equation for determining dihedral angleCoupling consts.

J, Hz

H1’-H2’ 5.9

H2’-H3’ 5.5

H3’-H4’ 3.0

H4’-H5’ 4.1

H4’-H5” 3.5

H5’-H5” 12.3

H2’-C2’-OH 6.6

H3’-C3’-OH 4.7

H5’-C5’-OH 7.2

H5”-C5’-OH 4.4

decoupled

13C NMR

13C has spin states similar to H. Natural occurrence is 1.1% making 13C-13C spin spin

splitting very rare.H atoms can spin-spin split a 13C peak. (13CH4 would

yield a quintet). This would yield complicated spectra.

H splitting eliminated by irradiating with an additional frequency chosen to rapidly flip (decouple) the H’s averaging their magnetic field to zero.

A decoupled spectrum consists of a single peak for each kind of carbon present.

The magnitude of the peak is not important.

chemical shift and coupling constants. unpaired nuclear spins are of importance in nmr. the actual...

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