NMR Essay Example

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    Chemistry
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14NMR Spectroscopy

NMR SPECTROSCOPY

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Problem 1

Determine the ratio of the population of nuclear spins in the spin down state (- NMR) to the population of nuclear spins in the spin up state (NMR 1) . The Boltzmann equation is used.

NMR 2

For a 1H nuclei NMR 3= 300MHz

The NMR 4hNMR 5

hNMR 6= (6.626 x NMR 7) x (3oo x NMR 8Hz) = 1.978×10-25 joules

Boltzmann equation for 1H nuclei at 7.05 T and 300K

NMR 9

= NMR 10 = 0.99995

n (+NMR 11) = 0.999995 n (-NMR 12)

for the 1H nuclei NMR 13= 500MHz

hNMR 14= (6.626 x NMR 15) x (5oo x NMR 16Hz) = 3.313×10-25 joules

Boltzmann equation for 1H nuclei at 7.05 T and 300K

NMR 17

= NMR 18 = 0.99991

n (+NMR 19) = 0.99995 n (-NMR 20)

n (+NMR 21) = 0.99991 n (-NMR 22)

NMR spectroscopy with the higher frequency is more sensitive hence the recommended.

Problem 2

When running An NMR spectrum, a better S/N ratio is obtained on a higher field instrument. This is because the sensitivity of NMR is determined by the strength of the magnetic field. The strength of the magnetic field brings the difference in the determination of the energy difference and the population difference. The formula for the energy difference is

NMR 23hNMR 24

Where h- planck’s constant

NMR 25— the frequency of the field

From the calculation for the energy difference, the higher the frequency of the field, the higher the energy difference hence a better S/N ratio is obtained.

Problem 3

  1. Larmar frequency

Larmar frequency (ω) = gyromagnetic ratio (γ) x applied field strength (Bo)

= 11.74ToB

NMR 26 (NMR 27

Natural abundance (%)

Larmor frequency

-31.8154

295.6132

127.2616

  1. Resonance frequency (υ)

υ = NMR 28

where γ- gyromagnetic ratio and Bo- applied field strength (T)

Bo= 7.05 T

NMR 29 (NMR 30

Resonance frequency (NMR 31)

  1. Given the answers above, it is clear that spectrometers are named by a single frequency. This is the resonance frequency for the 1H nucleus at the given applied field strength (Bo).

Problem 4

Chemical shift (δ) = NMR 32

Where ν- frequency of anisaldehyde

νref- frequency of tetramethylsilane which is 300MHz

δ= 10.42 from the SDBS database

δ= 10.42 = NMR 33

10.42 = NMR 34

NMR 35= 4568 MHz

Problem 5

From the spectra, the δ for both compounds for CH2 and CH3 are the same.

CH2
δ =3.43

CH3
δ= 1.7

The different signals in NMR structure are brought about by the different environments that the two carbons are in. the chlorine in ethylchloride deshields the carbon here hence the shift moves downfield. In the ethylbromide, the CH2 group has its protons that are neighboring the bromine shifted left but the H’s of the CH3 group produce an upshield effect.

(δ) = NMR 36

For the ethylchloride

300MHz- NMR 37

3.43 – δ for the CH2

3.43= ((v-300) x 106) ÷ (300)

NMR 38= 300.00102MHz

1.7 – δ for the CH3

1.7= ((v-300) x 106) ÷ (300)

NMR 39= 300.0005MHz

For the ethylbromide

90MHz- NMR 40

3.5 – δ for the CH2

3.5= ((v-90) x 106) ÷ (90)

NMR 41= 90.000315MHz

1.7 – δ for the CH3

1.7= ((v-90) x 106) ÷ (90)

NMR 42= 90.000153MHz

Problem 6

From the values obtained online;

J(Hz) = (δ [peak 1]) – (δ [peak 2]) x (operating frequency(MHz))

Peak 1= 1.26ppm

Peak 2= 1.14ppm

Frequency= 60MHz

J= (1.26-1.14) x 60MHz

= 75.6Hz – 68.4Hz = 7.2Hz

The center line of the two peaks is the position of the chemical shift.

Central peak = [(75.6Hz + 68.4Hz) / 2] = 72Hz

This means, chemical shift δ = 72Hz / 60 = 1.2ppm

1H NMR (chloroethane, 60MHz): δ = 1.2ppm (q, J= 7.2Hz, 2H)

The signals of two or more adjacent Hydrogen nuclei, with different coupling constants, are not simple triplets or quartets. The resonance frequencies also depend on the molecular mass of the compounds.

Problem 7

The chemical shifts of alcohols (OH) are determined by the hydrogen bonding. Downfield shifts are experienced in hydrogen bonded groups. In secondary alcohols, the number of H-bonds are limited as compared to the primary alcohols thus there is a larger downfield experienced.

An example is the 1-phenyl-4,4-dimethyl-1-pentyn-3-ol shown below:

NMR 43

Problem 8

The resonance frequency of alkenes is larger than that of the alkanes. This is because of the shielding that is brought about by the diamagnetic fields and the paramagnetic fields induced by the applied field perpendicular to the double bond. This happens due to the presence of a double bond thus this effect is not seen in alkanes. In alkynes, the induced diamagnetic field, parallel to the triple bond, is larger thus the shielding is also larger hence the observed large resonance frequencies and corresponding chemical shifts.

Problem 9

  1. Aldehydic proton

The C=O bond has a strong deshielding effect thus the aldehyde proton is most affected by this causing it to have a downfield shift. The spectra from an aldehyde group is found to be in doublets; this is brought about by spin coupling to ring protons.

  1. Ester protons

α- protons of an ester will experience a slight upfield shift. This is due to their placement a bond away from the carbonyl carbon which is the major de-shielding area. The chemical shift of the ester is also affected by the stereo-isomer of the ester. It is therefore advisable to get the different shifts for the different stereo isomers of the ester.

Problem 10

Shoolery’s additivity rule

δ= 0.23 + ΔX + ΔY

example CNCH2Br

δ= 0.23 + ΔX + ΔY

δ= 0.23 + ΔCN + ΔBr

δ= 0.23 + 1.70 + 2.33

δ= 4.26ppm

Problem 11

Only neighboring hydrogen nuclei affect the hydrogen nuclei affects their neighboring groups and the carbon nuclei does not have the same effect. This is because, only the 1H has nuclear spin numbers (I = ½). The carbon in the compound is 12C and has no nuclear spin numbers I = 0. Therefore, only the hydrogen atom has the magnetic properties to have any effect on its neighboring groups.

Problem 12

Splitting tree

The CH2 adjacent to the Cl group will be split in to 2 peaks since it is adjacent to 1 Hydrogen atom

The methyl group also splits into two peaks since it is adjacent to 1 Hydrogen atom

The same case applies to the OCH3

The CH however splits into 7 peaks since it is adjacent to 6 hydrogen atoms.

NMR 57NMR 56NMR 55NMR 54NMR 53NMR 52NMR 51NMR 50NMR 49NMR 48NMR 47NMR 46NMR 45NMR 44

Problem 13

Deuterium (2H) with spin = 1 does not show in the proton NMR. This is because it has a different magnetic moment from that of Hydrogen and therefore it is not detected. This factor explains why deuterium is used as a solvent since water cannot be used because it will interfere with the signal of the molecule that is being tested.

Problem 14

Running spectra on higher field strength instruments is advisable as higher field strength instruments have more sensitivity thus the results are more accurate, fine structures are obtained and they are obtained within a shorter time period.

Problem 15

∑JAXmX

NMR 58

NMR 59

NMR 60

NMR 61JAX

NMR 62

NMR 63

NMR 64

NMR 65JAX

NMR 66

NMR 67

NMR 68

NMR 69JAX

NMR 70

NMR 71

NMR 72

NMR 73JAX

NMR 74

NMR 75

NMR 76

NMR 77JAX

NMR 78

NMR 79

NMR 80

+NMR 81JAX

NMR 82

NMR 83

NMR 84

NMR 85JAX

NMR 86

NMR 87

NMR 88

NMR 89JAX

A ratio of 1:3:3:1 results as shown below

NMR 90

Problem 16

JAX=-JAM

NMR 91

0

0

NMR 92

NMR 93NMR 94

NMR 95NMR 96NMR 97NMR 98NMR 99NMR 100JAX

NMR 101NMR 102NMR 103

Problem 17

Spin system

(i) in cyclopropene, the 2 vinyl and the 2 allycil protons are magnetically equivalent to each other. Therefore, the spectram will have two identical triplets- A2X2 system. This is a first order spectrum.

(ii) 1,1- difluoroallene- the compound has two pairs of equivalent protons that are not magnetic equivalent. The magnetic non- equivalent are given the AA’ designation. This produces an A2X2 spin system. This is a first order spectrum.

(iii) the compound has 3 pairs of unsymmetrical and unequivalent protons hat are magnetic equivalent. This produces an AX3 System. It is a second order spectrum.

Problem 18

Stereochemistry- Cis

NMR 104O

a e NMR 105H

NMR 106

The double bond will have a cis stereo isomer at a mutual coupling of 11.6 Hz.

Coupling pattern

3Jab= 3.8Hz

3Jbc= 8.8Hz

3Jbd= 9.2Hz

3Jde= 4.3Hz

NMR 107NMR 108NMR 109

NMR 110NMR 111

NMR 112NMR 113NMR 114NMR 115NMR 116NMR 117NMR 118NMR 119NMR 120

NMR 121NMR 122NMR 123

NMR 124NMR 125NMR 126NMR 127NMR 128NMR 129NMR 130

NMR 131NMR 132NMR 133NMR 134NMR 135NMR 136NMR 137

NMR 138NMR 139NMR 140NMR 141NMR 142NMR 143NMR 144NMR 145NMR 146NMR 147NMR 148NMR 149NMR 150NMR 151NMR 152NMR 153NMR 154NMR 155NMR 156NMR 157NMR 158NMR 159NMR 160NMR 161NMR 162NMR 163NMR 164NMR 165NMR 166NMR 167

References

Akitt, J., (1973). NMR and Chemistry; An introduction to Nuclear Magnetic resonance spectroscopy. Chapman and Hall, London, 78-156.

Becker, E. D., (1980). High Resolution NMR. Theory and Chemical applications. Academic press, 2, 45-78.

Black, K.D. & Gunstone, F.D. (1990). The synthesis and spectroscopic properties of some polyol esters and ethers. McGraw Hill, New York, 56, 169-173.

Edwards, C. J, (2000). Principles of NMR. Process NMR associates LLC, Pit Rd, Danbury, 87, 21-27

Freeman, R., (1987). A Handbook of Nuclear Magnetic Resonance. Wiley, New York, 89-167.

Guther, H., (1980). NMR Spectroscopy- An introduction. John Wiley, New York, 133-187.

Hershenson, H. M. (1965). Nuclear Magnetic Resonance and Electron Spin Resonance Spectra. Academic Press, New York, 145-177.

Memory, J.D., (1982). NMR of Aromatic Compounds. John Wiley, New York, 89-107.

Pople, J. A, Schneider, G. & Berstein, J., (1979). High Resolution Nuclear Magnetic Resonance. McGraw Hill, New York, 458-500.

Roberts, D. J., (1979). Nuclear Magnetic Resonance, Applications to organic chemistry. McGraw Hill, New York, 157-200.

Sohar, P., (1984). Nuclear
Magnetic Resonance Spectroscopy. CRC Press, Boca Raton, 67-88

Zurcher, R.F, (1966). The Cause and Calculation of Proton Chemical Shifts in Non-conjugated Organic Compounds. Wiley, New York, 47-78.