Nmr2 pl

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Nuclear Magnetic Resonance Spectroscopy

Presented By

Uttam Prasad Panigrahy

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•Nuclear magnetic resonance spectroscopy is a powerful analytical technique used to characterize organic molecules by identifying carbon-hydrogen frameworks within molecules.

•Two common types of NMR spectroscopy are used to characterize organic structure: 1H NMR is used to determine the type and number of H atoms in a molecule; 13C NMR is used to determine the type of carbon atoms in the molecule.

•The source of energy in NMR is radio waves which have long wavelengths, and thus low energy and frequency.

•When low-energy radio waves interact with a molecule, they can change the nuclear spins of some elements, including 1H and 13C.

Introduction to NMR Spectroscopy

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• When a charged particle such as a proton spins on its axis, it creates a magnetic field. Thus, the nucleus can be considered to be a tiny bar magnet.

• Normally, these tiny bar magnets are randomly oriented in space. However, in the presence of a magnetic field B0, they are oriented with or against this applied field. More nuclei are oriented with the applied field because this arrangement is lower in energy.

• The energy difference between these two states is very small (<0.1 cal).


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• In a magnetic field, there are now two energy states for a proton: a lower energy state with the nucleus aligned in the same direction as B0, and a higher energy state in which the nucleus aligned against B0.

•When an external energy source (hn) that matches the energy difference (DE) between these two states is applied, energy is absorbed, causing the nucleus to “spin flip” from one orientation to another.

•The energy difference between these two nuclear spin states corresponds to the low frequency RF region of the electromagnetic spectrum.


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•Thus, two variables characterize NMR: 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).


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•The frequency needed for resonance and the applied magnetic field strength are proportionally related:

•NMR spectrometers are referred to as 300 MHz instruments, 500 MHz instruments, and so forth, depending on the frequency of the RF radiation used for resonance.

•These spectrometers use very powerful magnets to create a small but measurable energy difference between two possible spin states.


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• 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.


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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

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•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.


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• The chemical shift of the x axis gives the position of an NMR signal, measured in ppm, according to the following equation:


• By reporting the NMR absorption as a fraction of the NMR operating frequency, we get units, ppm, that are independent of the spectrometer.

• 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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• 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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•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.


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

1H NMR—Chemical Shift Values

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


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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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• The presence of nearby atoms or groups will effect the base chemical shift by a specific amount:

• The carbon atom bonded to the hydrogen(s) under consideration are described as alpha () carbons.

• Atoms or groups bonded to the same carbon as the hydrogen(s) under consideration are described as alpha () substituents.

• Atoms or groups on carbons one bond removed from the a carbon are called beta () carbons.

• Atoms or groups bonded to the carbon are described as alpha () substituents.

Calculating 1H NMR—Chemical Shift Values

(Hydrogen under consideration)C C H

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Nuclear Magnetic Resonance SpectroscopyCalculating 1H NMR—Chemical Shift Values

Added Chemical Shifts Substituent Type of Hydrogen -Shift -Shift

C C CH3 0.78 --- CH2 0.75 -0.10 CH --- ---

RC C C

Y

[Y = C or O] CH3 1.08 --- Aryl- CH3 1.40 0.35 CH2 1.45 0.53 CH 1.33 --- Cl- CH3 2.43 0.63

CH2 2.30 0.53 CH 2.55 0.03 Br- CH3 1.80 0.83 CH2 2.18 0.60 CH 2.68 0.25 I- CH3 1.28 1.23 CH2 1.95 0.58 CH 2.75 0.00 OH- CH3 2.50 0.33 CH2 2.30 0.13 CH 2.20 --- RO- (R is saturated) CH3 2.43 0.33 CH2 2.35 0.15 CH 2.00 ---

R–CO

O

or ArO CH3 2.88 0.38 CH2 2.98 0.43 CH 3.43 --- (ester only)

R–C

O

CH3 1.23 0.18 where R is alkyl, aryl, OH, CH2 1.05 0.31 OR', H, CO, or N CH 1.05 ---

(Hydrogen under consideration)C C HH

H

H

HCl

Base Chemical Shift = 0.87 ppm

no substituents = 0.00

one - Cl (CH3) = 0.63

TOTAL = 1.50 ppm

(Hydrogen under consideration)C C HH

H

H

HCl


one - Cl (CH2) = 2.30

no substituents = 0.00

TOTAL = 3.50 ppm

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• 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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• In a magnetic field, the loosely held electrons of the double bond create a magnetic field that reinforces the applied field in the vicinity of the protons.

• The protons now feel a stronger magnetic field, and require a higher frequency for resonance. Thus the protons are deshielded and the absorption is downfield.


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• In a magnetic field, the electrons of a carbon-carbon triple bond are induced to circulate, but in this case the induced magnetic field opposes the applied magnetic field (B0).

• Thus, the proton feels a weaker magnetic field, so a lower frequency is needed for resonance. The nucleus is shielded and the absorption is upfield.


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1H NMR—Chemical Shift Values)

1H NMR of Methyl Acetate

C

O

R O

H3C C O

Base Chemical Shift = 0.87 ppm one = 2.88 ppm TOTAL = 3.75 ppmO

CH3

C

O

R

Base Chemical Shift = 0.87 ppm one = 1.23 ppm TOTAL = 2.10 ppm

2,3-Dimethyl-2-Butene

(Hydrogen under consideration)


one (CH3) = 0.78 ppm

TOTAL = 1.65 ppm

H2C CH

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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.

1H NMR—Intensity of Signals

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1H NMR—Intensity of Signals

Methyl -Dimethylpropionate

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• Consider the spectrum below:

1H NMR—Spin-Spin Splitting

Ethyl Bromide

1H - 1H CouplingYou’ll notice in the spectra that we’ve seen that the signals don’t appear as single lines, sometimes they appear as multiple lines. This is due to 1H - 1H coupling (also called spin-spin splitting or J-coupling). Here’s how it works: Imagine we have a molecule which contains a proton (let’s call it HA) attached to a carbon, and that this carbon is attached to another carbon which also contains a proton (let’s call it HB). It turns out that HA feels the presence of HB. Recall that these protons are tiny little magnets, that can be oriented either with or against the magnetic field of the NMR machine. When the field created by HB

reinforces the magnetic field of the NMR machine (B0 ) HA feels a slightly stronger field, but when the field created by HB opposes B0, HA feels a slightly weaker field. So, we see two signals for HA depending on the alignment of HB. The same is true for HB, it can feel either a slightly stronger or weaker field due to HA’s presence. So, rather than see a single line for each of these protons, we see two lines for each.

C C

HBHA

HA HBHA is split into two lines because it feels the magnetic field of HB.

HB is split into two lines because it feels the magnetic field of HA.

For this line, HB is lined up with the magnetic field

(adds to the overall magnetic field, so the line

comes at higher frequency)

For this line, HB is lined up against the magnetic field(subtracts from the overall magnetic field, so the line

comes at lower frequency)

More 1H - 1H Coupling

What happens when there is more than one proton splitting a neighboring proton? We get more lines. Consider the molecule below where we have two protons on one carbon and one proton on another.

C C

HBHA

HA'HA + HA' HBHA and HA' appear at the same chemical shift because they are

in identical environments They are also split into two lines (called a doublet) because they

feel the magnetic field of HB.

HB is split into three lines because it feels the magnetic

field of HA and HA'

Note that the signal produced by HA + HA' is twice the size

of that produced by HB

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

Rules for Spin-Spin Splitting• Equivalent protons do not split each other

• Protons that are farther than two carbon atoms apart do not split each other

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1H NMR—Spin-Spin Splitting

Splitting is not generally observed between protons separated by more than three bonds.

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

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

The Origin of 1H NMR—Spin-Spin Splitting

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

J

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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).

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• Because the absorbing proton feels three different magnetic fields, it absorbs at three different frequencies in the NMR spectrum, thus splitting a single absorption into a triplet.

• Because there are two different ways to align one proton with B0, and one proton against B0—that is, ab and ab—the middle peak of the triplet is twice as intense as the two outer peaks, making the ratio of the areas under the three peaks 1:2:1.

• Two adjacent protons split an NMR signal into a triplet.

• When two protons split each other, they are said to be coupled.

• The spacing between peaks in a split NMR signal, measured by the J value, is equal for coupled protons.

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Nuclear Magnetic Resonance Spectroscopy1H NMR—Spin-Spin SplittingWhenever two (or three) different sets of adjacent protons are equivalent to each other, use the n + 1 rule to determine the splitting pattern.

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Nuclear Magnetic Resonance Spectroscopy1H NMR—Spin-Spin SplittingWhenever two (or three) different sets of adjacent protons are equivalent to each other, use the n + 1 rule to determine the splitting pattern.

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Nuclear Magnetic Resonance Spectroscopy1H NMR—Spin-Spin Splitting

Whenever two (or three) different sets of adjacent protons are not equivalent to each other, use the n + 1 rule to determine the splitting pattern only if the coupling constants (J) are identical:

a a

b

c

Free rotation around C-C bonds averages coupling constant to J = 7Hz

Jab = Jbc

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Nuclear Magnetic Resonance Spectroscopy1H NMR—Spin-Spin Splitting

Whenever two (or three) different sets of adjacent protons are not equivalent to each other, use the n + 1 rule to determine the splitting pattern only if the coupling constants (J) are identical:

a

b

c

c

Jab = Jbc

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Nuclear Magnetic Resonance Spectroscopy1H NMR—Structure Determination

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1H NMR—Structure Determination

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Nuclear Magnetic Resonance Spectroscopy1H NMR—Structure Determination

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1H NMR—Structure Determination

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The Nature of NMR Absorptions

• 1H NMR of methyl acetate has 2 equivalent kinds of H’s so shows 2 peaks

• Electrons in neighboring bonds shield or expose nuclei from magnetic field

H’s on C next to electron withdrawing C=O

H’s on C next to electron withdrawing O

Intensity of 1H NMR peak is proportional to # of equivalent H’s

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13C NMR of methyl acetate has 3 kinds of C’s so shows 3 peaks Electrons in neighboring bonds shield nuclei from magnetic field

C next door to electron withdrawing C=OC next to electron

withdrawing OC of C=O

Intensity of 13C NMR peak is not related to # of equivalent C’s

1H 13C

Example:• How many signals would you expect each to have

in its 1H and 13C spectra?

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CH3

CH3C C

CH3

CH3

O

CH3

CH3

Solution:• How many signals would you expect each to have

in its 1H and 13C spectra?

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CH3

CH3C C

CH3

CH3

O

CH3

CH3

1H 13C

CH3

CH3C C

CH3

CH3 CH3

CH3C C

CH3

CH3

1 2

3 5

O

CH3

CH3

H H

HH

HH

HH

C

CC

C

CC

O

CH3

CH3

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Shift = relative energy of resonance13.3 Chemical Shifts

Downfield = deshielded (more exposed to magnet)

Upfield = shielded

(more protected from magnet)

tetramethylsilane [TMS]Reference point Si

CH3

CH3H3C

CH3

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13.4 Signal Averaging & FT-NMRCarbon-13: (only carbon isotope with a nuclear spin)Natural abundance =1.1%

of C’s so sample is very dilute in this isotope

Single run

Average of 200 runs

Sample measured using repeated accumulation of data and averaging of signals, incorporating pulse and the operation of Fourier transform (FT-NMR)

All signals are obtained simultaneously using a broad pulse of energy and resonance recorded

Frequent repeated pulses give many sets of data that are averaged to eliminate noise

Fourier-transform of averaged pulsed data gives spectrum

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13.5 13C NMR Spectroscopy• Each signal shows different types of environments of carbon• 13C resonances are 0 to 220 ppm downfield from TMS • C’s shift downfield (deshield) if next to electron-withdrawing

– Like O, N, X (halogens)

sp3 C signal ~ 0 to 90

sp2 C ~ 110 to

220

C(=O) at low field, 160 to 220

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13C NMR Example: 2-butanone

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13C NMR Example: p-bromoacetopheone

Learning Check:

70

Assign resonances in the given 13C spectrum of methyl propanoate

CH3 O C CH2 CH3

O1 2 3 4

Solution:

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Assign resonances in the given 13C spectrum of methyl propanoate

CH3 O C CH2 CH3

O1 2 3 4

13.14

13C NMR Spectroscopy

1H and 13C NMR compared:

both give us information about the number ofchemically nonequivalent nuclei(nonequivalent hydrogens or nonequivalentcarbons)

both give us information about theenvironment of the nuclei (hybridization state,attached atoms, etc.)

it is convenient to use FT-NMR techniques for1H; it is standard practice for 13C NMR


13C requires FT-NMR because the signal for acarbon atom is 10-4 times weaker than thesignal for a hydrogen atom

a signal for a 13C nucleus is only about 1% asintense as that for 1H because of themagnetic properties of the nuclei, and

at the "natural abundance" level only 1.1% ofall the C atoms in a sample are 13C (most are12C)


13C signals are spread over a much widerrange than 1H signals making it easier toidentify and count individual nuclei

Figure 13.20 (a) shows the 1H NMR spectrumof 1-chloropentane; Figure 13.20 (b) showsthe 13C spectrum. It is much easier to identifythe compound as 1-chloropentane by its 13Cspectrum than by its 1H spectrum.

Figure 13.20(a) (page 511)

ClCH2

1H

CH3ClCH2 2CH2CH2CH2CH3

2 3

10.0 9.0 8.0 7.0 6.0 5.0 4.0 3.0 2.0 1.0 0

Chemical shift (d, ppm)

Figure 13.20(b) (page 511)

ClCH2

13C

2CH2CH2CH2CH3

a separate, distinctpeak appears foreach of the 5carbons CDCl33

200 180 160 140 120 100 80 60


40 20 0

13.15

13C Chemical Shifts

are measured in ppm (d)

from the carbons of TMS

13C Chemical shifts are most affected by:

• hybridization state of carbon

• electronegativity of groups attached to carbon

Examples (chemical shifts in ppm from TMS)

23

138

sp3 hybridized carbon is more shielded than sp2


OH

61

O

202

sp3 hybridized carbon is more shielded than sp2


OH

23 61

an electronegative atom deshields the carbonto which it is attached


O

138 202

an electronegative atom deshields the carbonto which it is attached

Table 13.3 (p 513)

Type of carbon Chemical shift (d),ppm

RCH3 0-353

R2

R3

R4

2CH2

3CH

4C

15-40

25-50

30-40

Table 13.3 (p 513)


RCH3 0-35


RC CR 65-903

R22CH2

R3

15-40

25-50

R22C CR2 100-1502

3CH110-175

R4 4C 30-40

Table 13.3 (p 513)


RCH2

RCH2

RCH2

RCH2

RCH2

2Br

2Cl

2NH2

2OH

2OR

20-40

25-50

35-50

50-65

50-65

Table 13.3 (p 513)


RCH2 20-40


O2Br

RCH2

RCH2

2ClRCOR

160-18525-5035-50 O2NH2

RCH2

RCH2

2OH

2OR

50-65

50-65

RCR 190-220

13.16

13C NMR and Peak Intensities

Pulse-FT NMR distorts intensities of signals.Therefore, peak heights and areas can bedeceptive.

Figure 13.21 (page 514)

CH 3 7 carbons give 7signals, butintensities are notequal

OH

200 180 160 140 120 100 80 60 40 20 0


13.17

13C—H Coupling

Peaks in a 13C NMR spectrum are typicallysinglets

13 C—13C splitting is not seen because theprobability of two 13C nuclei being in the samemolecule is very small.

13C—1H splitting is not seen because spectrumis measured under conditions that suppressthis splitting (broadband decoupling).

13.18

Using DEPT to Count the Hydrogens

Attached to 13C

Distortionless Enhancementof Polarization Transfer

Measuring a 13C NMR spectrum involves

1. Equilibration of the nuclei between the lowerand higher spin states under the influence ofa magnetic field

2. Application of a radiofrequency pulse to givean excess of nuclei in the higher spin state

3. Acquisition of free-induction decay dataduring the time interval in which the equilibriumdistribution of nuclear spins is restored

4. Mathematical manipulation (Fourier transform)of the data to plot a spectrum


Steps 2 and 3 can be repeated hundreds of timesto enhance the signal-noise ratio

2. Application of a radiofrequency pulse to givean excess of nuclei in the higher spin state

3. Acquisition of free-induction decay dataduring the time interval in which the equilibriumdistribution of nuclear spins is restored


In DEPT, a second transmitter irradiates 1Hduring the sequence, which affects the appearanceof the 13C spectrum.

some 13C signals stay the same

some 13C signals disappear

some 13C signals are inverted

Figure 13.23 (a) (page 516)O

CCH2

CHCH

2CH2CH2CH3

CH2 2

CHO CH2

CH2

2

CH2 3

C

C

200 180 160 140 120 100 80 60


40 20 0

Figure 13.23 (a) (page 516)O

CCH2

CHCH

2CH2CH2CH3

CH3 3

CH

CH2CH22 CH22

2

200 180 160 140 120 100 80 60


40 20 0

Figure 13.23 (b) (page 516)O

CCH2

CHCH

2CH2CH2CH3

CH3 3

CH

CH and CH3

unaffected

C and C=O nulledCH2CH22 CH

2

CH2 inverted100 80 60

2

40 20 0


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13.6 DEPT 13C NMR• DEPT (distortionless enhancement by polarization transfer)

Normal spectrum shows all C’s then: Obtain spectrum of all C’s except quaternary

(broad band decoupled) Change pulses to obtain separate information for

CH2, CH Subtraction reveals each type

100

Normal spectrum shows all C’s:(Difficult to Assign so many C’s)

DEPT-90: shows only CH’sQuaternary C’s don’t show

(Can now narrow our assignments)

DEPT 13C NMRCH3

CC

CC

CCH3

CH3

H

H H

H H

H OH

1

23

45

67

8

5 26

101


DEPT-135: Positive =shows CH’s and CH3’s

DEPT 13C NMR

Negative =shows CH2’s(Can narrow assignments even further)

CH3

CC

CC

CCH3

CH3

H

H H

H H

H OH

1

23

45

67

8

5 26

1

7,8

102


DEPT-135: Positive =shows CH’s and CH3’s

DEPT 13C NMR

Negative =shows CH2’s(Can narrow assignments even further)

CH3

CC

CC

CCH3

CH3

H

H H

H H

H OH

1

23

45

67

8

5 2 1

7,8

6

24

103

13.7 Uses of13C NMR: Example

• Evidence for product of elimination of 1-chloro-methyl cyclohexane

CH3CH3

ClCH2

or

KOH

ethanol

Expect 7 different C’s;5 sp3 resonances 20-502 sp2 resonances 100-150

Expect 5 different C’s;3 sp3 resonances 20-502 sp2 resonances 100-150

CH3

104

13.8 1H NMR & Proton Equivalence

• Proton NMR is much more sensitive than 13C and the active nucleus (1H) is nearly 100 % of the natural abundance

• Shows how many kinds of nonequivalent hydrogens are in a compound

• Theoretical equivalence can be predicted by seeing if replacing each H with “X” gives the same or different outcome

• Equivalent H’s have the same signal while nonequivalent are different– There are degrees of nonequivalence

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Nonequivalent H’s• If replacement of each H with “X” gives a different constitutional isomer

then the H’s are in constitutionally heterotopic environments and will have different chemical shifts

• – they are nonequivalent under all circumstances

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Equivalent H’s• Two H’s that are in identical environments (homotopic) have the same

NMR signal• Test by replacing each with X if they give the identical result, they are

equivalent (homotopic)

107

Enantiotopic Distinctions• If H’s are in environments that are mirror images of each other, they are

enantiotopic

• Replacement of each H with X produces a set of enantiomers

• The H’s have the same NMR signal (in the absence of chiral materials)

108

Diastereotopic Distinctions• In a chiral molecule, paired hydrogens can have different

environments and different shifts• Replacement of a pro-R hydrogen with X gives a different diastereomer

than replacement of the pro-S hydrogen • Diastereotopic hydrogens are distinct chemically and spectrocopically

*

Learning Check:

109

• Identify sets of H’s as

Unrelated (U), homotopic (H), enantiotopic (E), or diasterotopic (D)

Solution:• Identify sets of H’s as

Unrelated (U), homotopic (H), enantiotopic (E), or diasterotopic (D)

110

DD

D

E

D H

111

13.9 Chemical Shifts in 1H NMR• Proton signals range from 0 to 10• Electronegative atoms cause downfield shift H’s on

sp3 C

Higher field

H’s on sp2 C

Lower field

Shifts in 1H NMR

112

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