Chapter 3 Nuclear Magnetic Resonance Spectroscopy

• Many atomic nuclei have the property of nuclear spin. When placed between the poles of a magnet, the axis of rotation of the spinning nuclei precess around the axis of the field. This precession can be detected by irradiation with energy in the radiofrequency region of the spectrum. When the precession and irradiation frequencies are equal the system is said to be in resonance.

Nuclear Magnetic Resonance (NMR)- Absorption of radiowave EM energy by nuclei in a magnetic field.

• Background and Theory– Electron spin observed - 1922– By 1926, became apparent that nuclear

spin also existed.– 1939, Rabi observes absorption of radio

frequency (RF) energy by nuclei of H2

gas; gets Nobel prize in physics in 1944.

Nuclei are charged and if they have spin, they are magnetic

No Field

Applied Magnetic Field = B

Energy of transition = energy of radiowaves

Higher energy state: magnetic field opposes applied field

Lower energy state: magnetic field aligned with applied field

3.1 How NMR Works

Nuclear Spin

• Do all nuclei have spin movement?• Nuclear spin depends on its spin angular

momentum number I If the protons and neutrons comprising these nuclei

are not paired(one is odd number, the other even number), the overall spin of the charged nucleus generates a magnetic dipole along the spin axis. I=1/2, 3/2, 5/2 …

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

Nuclear Spin (cont.)

• If both the protons and neutrons are even numbers, I=0, there is no nuclear spin movement.

Example: 12C, 16O, 32S

• If both the protons and neutrons are odd number, I=1, 2, 3… , the nucleus will spin.

Example: 2H, 14N

NMR Background & Theory (cont.) • Observed in bulk samples- 1946

– Bloche (Stanford) for H in H2O

– Purcell (Harvard) for H in CnH2n+2 (paraffin)

– Bloche & Purcell got Nobel prize in 1952 in Physics.

• Basic NMR relation

Bo

omega = radio frequency in megaHerz (MHz)

gamma, magnetogyricconstant - depends onnucleus, e.g. 1H, 13C...

appliedmagnetic

field

NMR Theory (cont.) • In 1949 & 50, it was observed that ethanol gave 3

signals. Not due to experimental error. HO CH2 CH3

3 signals from H’s on 3 different groups or chemical environments.Since signal is proportional to # of H’s, the peaks can be assigned.

20,000,000 Hz20,000,050 Hz

20,000,085 Hz

• These variations in frequency () are due to different magnetic environments. Explanation: The electrons surrounding the H nucleus provide shielding of applied magnetic field so that:-

Bo(1-)

sigma - depends on density of electronsaround nucleus

NMR Theory (cont.)

Shift in , due to shielding by electrons, is called the

• 60 MHz NMR spectrum of CH3CH2OH

60,000,000 Hz

60,000,220 Hz

60,000,320 Hz

60,000,072 Hz

CH3 Si CH3

CH3

CH3

(parts per million)

• The variations in frequency () is depended on the externally applied magnetic field of strength Bo

CH3Br 60MHz(Bo=14100) =162 Hz 100MHz(Bo=23500) =270 Hz• To correct the effect of Bo, Chemical shift of

sample was compared with a standard TMS (tetramethyl silane)

• Chemical Shift

δ on the left side of TMS +

δ on the right side of TMS -

TMS

• Only one peak on NMR spectrum• High electronic density of H in TMS. Almost all t

he H peaks of organic compounds appear on the left of the TMS peak

• TMS is volatile compound(bp 26.5oC), very easy to remove. It dissolves in most organic solvents

• δ of TMS is set as 0

Calculation of chemical shift1,1,2-trichloropropane

3.3 Factors affecting chemical shift

• Depends on adjacent group

For protons on carbon attached to an electronegative atom or group X, the chemical shift increases with the electronegativity of X. This is due to the inductive effect on the shielding of the protons and is apparent in the methyl halides.

Depends on atom attached Fig 1 NMR Chemical Shifts and Splitting Patterns

Compound CH3X CH3F CH3OH CH3Cl CH3Br CH3I CH3C-3 CH4 (CH3)4Si

Element X F O Cl Br I C H Si

Electronegativity of X

4.0 3.5 3.1 2.8 2.7 2.5 2.1 1.8

Chemical shift, ppm

4.26 3.40 3.05 2.68 2.16 0.9 0.23 0.0

ppm

•Depends on adjacent group (cont.)

The inductive (deshielding) effect of a substituent on a proton decreases as the separation between the proton and substituent is increased.

CH3 - CH2 - CH2 - OH

0.92 1.57 3.58

• Depends on carbon group attached:

C C

C/H

HC/H

O

1.6 - 3 ppm

C C

C/H

HC/H

C

C

O

H

C/H

H

C/H

C

on a carbonyl (aldehyde):

to carbonyl (aldehydes & ketones):

9 - 10 ppm

to C=C (allylic):

1.6 - 3 ppm

1.6 - 3 ppm

to aromatic ring (benzylic):

• Depends in hybridization:

C C

C/H

H

C C H

means mustbe C or H, notO, N, or X

0.2 to 2 ppm

C-C:

C=C:

4.5 to 7 ppm

C=C:1.6 to 3 ppm

HAromatic: 6.8 to 8 ppm

C/H C

C/H

HC/H

=

Depends on anisotropy(各向异性）

• Protons on an aromatic ring appears at very low field (7.27), due to the aromatic ring current.

• In addition, substituents with significant(diamagnetic) anisotropy e.g. NO2 and C=O deshield the ortho protons even more

Chemical Shifts of aliphatic protons

H bond

• δ increases while H bond was formed among molecules. That is due to the decrease of electronic density around the nucleus

• Example: δ of OH in n-butene-2-ol

1% 1

100% 5

Peak areas

• The relative peak areas are given by the changes in height of the integration curve. These may be obtained by counting lines if the spectrum is recorded on ruled paper

• The signal strength, or peak area, as measured by electronic integration, is directly proportional to the number of identical protons in the sample which produce the signal.

3.4 Spin-spin coupling

3.4 Spin-spin coupling

• The signal of a proton may split into several peaks in high resolution spectrum, as shown before. This is caused by spin coupling. The distance among these peaks (Hz) is Coupling Constants.

• Spin coupling is due to the spin of H on the adjacent C

Actual H = H0 + H’

Actual H = H0 - H’

N+1 rule

• Proton will split into n+1 peaks by n protons on the adjacent carbon. The intensity of each peak can be expressed with (a + b)n

•

Coupling Constants

• The spacing of the lines (splitting) in the multiple peaks is called coupling constants J (Hz). They are independent on B and ν

• Spinning coupling can be attribute to ortho, meta and same carbon etc.

• Coupling constants can help us to conclude the structure of compound

Δ ν/J >10 weak coupling (primary spectra)Δ ν/J <10 strong coupling (advanced spectra)

Primary spectra

• The number of split peaks depend on H of adjacent carbon. n+1

• Area ratio of split peaks follow (x+1)n

• Advanced spectra are very complicated.

Interpretation of NMR

• Elemental analysis, UV, IR results before interpreting NMR

1 Check the signal of TMS (whether it’s sharp, symmetric and at original point)

2 Some peaks may from solvent (CDCl3 may have peak at δ7.25)

3 Determine the H number from area

Interpretation of NMR

4 Analyze signal peak (CH3O, CH3N, CH3-ph, CH3-C=O, CH3-C=C, CH3CR3 etc) first. Then, analyze the methyl (methylene) peak with spin coupling

5 Determine the signal of COOH, CHO, OH at low field region

6 The active H atom in OH, NH, COOH can be verified with addition of D2O

Interpretation of NMR

7 The signal of benzene ring in aromatic compounds is complicated

C4H10O, 6:1:2:1

C8H14O4 6:4:4

C4H7O2Cl

C3H6O3, δ7.8 peak can exchange with D2O

C11H13O2N, 5:2:6H