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Chapter 14:Nuclear Magnetic Resonance

Spectroscopy

I. Introduction of NMR Spectroscopy (14.1)II. 1H NMR: Number of Signals (14.2)III. 1H NMR: Position of Signals (14.3)IV. Chemical Shift of Protons on sp2 and sp

Hybridized Carbons (14.4)V. 1H NMR: Intensity of Signals (14.5)VI. 1H NMR: Spin-Spin Splitting (14.6)VII. More Complex Examples of Splitting

(14.7)VIII.Spin-Spin Splitting in Alkenes (14.8)IX. Other Facts About 1H NMR Spectroscopy

(14.9)X. Using 1NMR to Identify an Unknown (14.10)XI. 13C NMR Spectroscopy (14.11)

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I. Introduction to NMR Spectroscopy

1H, 13C, 31P,19F, 15N

Some nuclei have nuclear spin.

A. Basis of NMR Spectroscopy

A spinning proton produces a magnetic field.

Absence of external magnetic field: spins

oriented randomly

Presence of external magnetic field Bo : more

spins align with Bo

Bo

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RF Radiation

hν∆E = hν

Bo

• Two nuclear spin states separated by ∆E

• Lower energy state – nuclear spin magnetic moment aligned with Bo

• Higher energy state – nuclear spin magnetic moment aligned against Bo

• Application of RF radiation pulse causes spin to flip from lower to higher energy state.

• Units of Bo are telsa (T); units of ν are hertz (Hz = s-1).

• ν is directly proportional to Bo:

ν = γBo

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Fig. 14.1

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How Fourier transform NMR works:

x

y

z

x

y

z

Apply RF pulse

along +x

After RF pulse, net magnetization relaxes

into +z direction.

Spins precess about +z. Net magnetization (blue

arrow) lies along +z.

Signal is detected by looking along +y with an RF receiver:

Fourier transform

time domain frequency domain

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B. 1H NMR Spectroscopy

1H NMR spectrum of CH3OC(CH3)3

δ scale 10 8 6 4 2 0Chemical Shift (ppm)

CH3O –

(CH3)3C –

TMS

upfield

downfield

Inte

nsity

• TMS is tetramethylsilane, (CH3)4Si, a chemical shift reference.

• t-Butyl methyl ether has sharp absorptions at 1.2 ppm and 3.2 ppm.

• (MHz)er spectromet NMR of TMS from downfield (Hz) (ppm)

ννδ =

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II. Number of Signals in 1H NMR Spectra

• Protons in different chemical environments give different NMR signals.

• Equivalents protons have identical NMR signals.

Examples:

Figure 14.2: Number of NMR Signals of Representative Organic Compounds

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Substitution test for chemical shift nonequivalence: substitute each H with another atom (e.g. Cl).

Example:

CH3 methylcyclobutane

CH2ClCH3

Cl

(chloromethyl)cyclobutane 1-chloro-1-methylcyclobutane

CH3 CH3

Cl Clcis and trans-1-chloro-2-methylcyclobutanes

CH3 CH3Cl Cl

cis and trans-1-chloro-3-methylcyclobutanes

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A. Determining Equivalent Protons in Alkenes and Cycloalkanes

CH

H

HbHa

Cl

N

C N C N

1,1-dicyanoethene 1-chloro-2-cyanoetheneH’s are equivalent Ha and Hb nonequivalent

1 NMR signal 2 NMR signals

H

H

H

HH

H C

Ha

Hb

HcHc

HbN

cyclopropane cyanocyclopropaneAll H’s equivalent 3 types of H’s

1 NMR signal 3 NMR signals

How many NMR signals does methoxyethene give?

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B. Enantiotopic and Diastereotopic Protons

Use substitution test to determine whether CH2 protons are equivalent.

1. Enantiotopic Protons

F

H3CH

H

F

H3CCl

H

Substitute eachH with Cl

F

H3CH

Cl+

fluoroethane S-1-chloro-1- R-1-chloro-1-fluoroethane fluoroethane

enantiomers

CH2 protons of fluoroethane are enantiotopic.Enantiotopic protons are equivalent and give the same NMR signal.

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2. Diastereotopic Protons

F

H

H

H

HH Substituteeach CH2

H with Cl

F

H

H

H

ClH F

H

H

Cl

HH+

S-3-fluoro- 3S,4R-3-chloro- 3R,4R-3-chloro-cyclobutene 4-fluorocyclo- 4-fluorocyclo-

cyclobutene cyclobutene

diastereomers

CH2 protons of S-3-fluorocyclobutene are diastereotopic. Diastereotopic protons are always nonequivalent and give different NMR signals.

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III. Position of Signals in 1NMR Spectrum

A. Shielding and Deshielding Effects

Orbital motion of the electron creates a magnetic field that opposes B0. Electron shieldsthe nucleus from B0.

Figure 14.3: How Chemical Shift is Affected by Electron DensityAround a Nucleus

• Shielding shifts absorptions upfield.

• Deshielding shifts absorptions downfield.

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B. Characteristic Chemical Shift Values

Type of Proton δ(ppm)sp3 C-HRCH3R2CH2R3CH

0.9 – 2~0.9~1.3~1.7

Z = C,O, N 1.5 – 2.5

-C≡C-H ~2.5Z = N, O, X 2.5 – 4

4.5 – 6

6.5 – 8

9 – 1010 – 12

RO-H or R2N-H 1 – 5

C C H

Z

CH

Z

C CH

H

RC

H

O

RC

OH

O

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IV. Chemical Shift of Protons on sp2 and sp Hybridized Carbons

Circulating π electrons affect the chemical shifts of H’s bound to multiply bonded carbons.

A. Protons on Benzene Rings

HH

B0

Circulating π electrons

induced magnetic field

• Circulating π electrons create a ring current.• Induced magnetic field reinforces B0 in

vicinity of aromatic protons.• Aromatic protons are deshielded and absorb

downfield at 7.3 ppm.

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B. Protons on Carbon-Carbon Double Bonds

C CH

H

B0

Induced magnetic fieldreinforces B0 in vicinity of vinyl protons.

• Vinyl protons are deshielded and absorb downfield at 4.5 – 6 ppm.

C. Protons on Carbon-Carbon Triple Bonds

C

C

H

RB0

• Induced magnetic fieldopposes B0 in vicinity of alkynyl proton.

• Alkynyl protons are shielded and absorb upfield at ~2.5 ppm.

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V. Intensity of 1H NMR Signals

The area under an NMR signal is proportional to the number of protons that give rise to that signal.

10 8 6 4 2 0Chemical Shift (ppm)

TMS

1H NMR spectrum of C5H12O

60

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Determining structure from integrals:

1. Add integrals: 60 + 20 = 802. Divide by number of H’s in formula: 80/12 = 6.7

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3. Divide each integral by 6.7:

20/6.7 = 3.0 (3 protons) 60/6.7 = 9.0 (9 protons)

4. Assign substructures:

3.2 ppm signal arises from 3 equivalent protons (1 CH3 group). This group must be attached to O, thereby giving –OCH3.

1.2 ppm signal arises from 9 equivalent protons, consistent with 3 CH3 groups.

Of the formula C5H12O, only C is left.

5. Draw the structure:

H3C O C

CH3

CH3

CH3

t-butyl methyl ether

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VI. 1H NMR: Spin-Spin Splitting• Spin-spin splitting occurs between

nonequivalent protons on the same carbon or adjacent carbons.

• Protons that split each other’s NMR signals are coupled. The splitting of the signal (in Hz) is called the coupling constant.

• Coupling with n adjacent protons splits a signal into n+1 peaks.

Chemical Shift (ppm)

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A. Why Spin-Spin Splitting Occurs

Number of Adjacent SplittingAdjacent Proton SpinsProtons

0 none singlet

1 doublet

2 triplet

3 quartet

Conclusion: n adjacent protons split NMR signal into n+1 peaks.

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n n+1 Peak Pattern Name

0 1 1 singlet

1 2 1 1 doublet

2 3 1 2 1 triplet

3 4 1 3 3 1 quartet

4 5 1 4 6 4 1 quintet

5 6 1 5 10 10 5 1 sextet

6 7 1 6 15 20 15 6 1 septet

B. More on Spin-Spin Splitting and the n + 1 Rule

JJThe degree to which an NMR signal is split by spin-spin coupling is measured by the coupling constant, J.

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C. More Features of Spin-spin Splitting

1. Equivalent protons do not spin couple.

Br-CH2-CH2-Br One singlet only

2. Splitting is observed for nonequivalent protonson the same carbon or adjacent carbons.

CH3Ha

Hb

HbHa

ClC N C N

Ha splits Hb into a doublet, and vice versa.

3. Splitting is usually not observed if the protons are separated by more than three σ bonds.

H2CC

CHCH3

O

HbHa

3 proton triplet

2 proton quartetHa and Hbare not coupled.

3 proton singlet

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VII. More Complex Spin-Spin Splitting Patterns

More complex splitting occurs when the absorbing proton is coupled to nonequivalent protons on two (or more) adjacent carbons.

Figure 14.6: 1H NMR Spectrum of 2-Bromopropane

When adjacent nonequivalent protons form an equivalent set, the n+1 rule still holds.

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Figure 14.7: 1H NMR Spectrum of 1-Bromopropane

When n protons on one adjacent carbon and m protons on another adjacent carbon are not equivalent, the signal is split into (n+1)(m+1) peaks.

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Figure 14.8: Splitting Diagram for Hb Protons of 1-Bromopropane

• Hb signal is split into 12 peaks, a triplet of quartets.

• If Jab >> Jbc, all 12 peaks are observed.

• If Jab ~ Jbc, peaks overlap and fewer peaks are observed.

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VIII. Spin-Spin Splitting in Alkenes

Characteristic coupling constants for disubstituted alkenes:

geminal H’s cis H’s trans H’sJgeminal < Jcis < Jtrans0-3 Hz 5-10 Hz 11-18 Hz

C CHb

Ha

C CHa Hb

C CHa

Hb

Figure 14.9: 1H NMR Spectra for Alkenyl Protons of (E)- and (Z)-3-Chloropropenoic Acids

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Monosubstituted alkenes give more complex splitting patterns.

Figure 14.10: The 1H NMR Spectrum of Vinyl Acetate

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C CHc

Hb O

Hd

CO

CH3

Jbc = 1.2 Hz (geminal)Jcd = 6.5 Hz (cis)Jbd = 14 Hz (trans)

Figure 14.11: Splitting Diagram for Alkenyl Protons of Vinyl Acetate

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IX. More on 1NMR SpectroscopyA. OH Protons

Figure 14.12: 1H NMR Spectrum of Ethanol

• OH proton of an alcohol usually does not split NMR signal of adjacent protons.

• Signal due to an OH proton not split by adjacent protons.

• OH protons exchange rapidly among ROH molecules.

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B. Cyclohexane Conformations

H

H

H

H

Rapid conformational interconversion averages axial and equatorial proton chemical shifts.

C. Protons on Benzene Rings

Figure 14.13: 1H NMR Spectra of Aromatic Protons

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X. Using 1H NMR to Identify an Unknown

Problem 14.24: Propose a structure for a compound of formula C7H14O2 with an IR absorption at 1740 cm-1 and the NMR data in the table.

Absorption δ (ppm) Integral

singlet 1.2 26triplet 1.3 10

quartet 4.1 6

Step 1: Identify functional group(s) and number of different types of protons.

1740 cm-1 IR absorption: C=ONMR data: 3 sets of nonequivalent protons

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Step 2: Determine number of protons that give rise to the three NMR absorptions.

Integral total = 26 + 10 + 6 = 42 units

Divide by number of protons:42/14 = 3.0 integral units per H

Divide integral for each absorption by this number and round to nearest digit:

26/3.0 = 8.7 9 H’s at 1.2 ppm10/3.0 = 3.3 3 H’s at 1.3 ppm6/3.0 = 2.0 2 H’s at 4.1 ppm

Step 3: Use splitting patterns to determinewhich C’s are bonded to each other.

9 proton singlet at 1.2 ppm arises from 3 CH3 groups bonded to C.3 proton triplet at 1.3 ppm must be CH3bonded to CH2.2 proton quartet at 4.1 ppm must be CH2bonded to CH3.

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Step 4: Draw out the pieces and put them together to get the structure.

-O- CH3CH2-

H3C C

CH3

CH3

C O

H3C C

CH3

CH3

CO

OCH2CH3

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XI. 13C NMR Spectroscopy

• Physical basis same as 1H NMR spectroscopy.

• 13C natural abundance is 1.1%.

• 1H - 13C spin splitting suppressed.

• All 13C peaks are singlets.

• Broader chemical shift range than 1H NMR.

• Peak intensity not proportional to number ofabsorbing carbons.

A. Number of Signals in 13C NMR Spectra

Number of signals equals number of different types of carbon in a molecule.

O

5 signals 7 signals 3 signals

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B. Position of Signals in 13C NMR Spectra

Type of Carbon δ (ppm)

sp3 C-H 5 – 45

―C≡C― 65 – 100

Z = N, O, X 30 - 80

100 – 140

120 – 150

160 – 210

CH

Z

C C

C

C O

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ppm

Figure 14.14: Representative 13C NMR Spectra