# NMR SPECTROSCOPY Essay Example

• Category:
Chemistry
• Document type:
Math Problem
• Level:
Masters
• Page:
2
• Words:
1246

Problem 1

n (+1/2) =e−ΔE/kT

## =1.978×10-25 joules

Population difference at 300 MHz is then given as =e−ΔE/kT

= e (-1.978×10-25)/(1.3806503 × 10-23 × 273)

= e (– 5.27 x 10-5)

= 0.999947301

For 500MHz Spectrometer

ΔE = hv = 6.626× 10-34 x 500×106

Population difference = e− ΔE/kT

= e (– 3.313 x 10-25/1.3806503 × 10-23 × 273)

= e -8.79 × 10-5

= 0.999912103

A 500MHz spectrometer gives better sensitivity than at 300MHz spectrometer. The sensitivity of NMR spectroscopy increases with increase in frequency.

Problem2.

Higher field instrument operates at a better magnetic field hence high frequency thus resulting in a strong radio frequency signal and as a result a greater population difference will be established between the nuclear spin states and thus a better signal to noise ratio than in the case of a low field instruments.

Problem 3

 γ(107T-1.s-1) Natural abundance (%) Larmor frequency(MHz)
 ϒ(107T-1.s-1) Natural abundance (%) Larmor frequency(MHz)

(c)

From the given results, it actually makes sense to call spectrometers by a single frequency. A 300MHz spectrometer corresponds with the applied magnetic field of 7,05T for 1H NMR and similarly a 50MHz spectrometer corresponds with the applied magnetic field of 11.75T for 1H NMR.

Problem 4

From 1H NMR of the given anisaldehyde, δ obtained from SDBS database is equal to 10.42.

Using a specific frequency of 400MHz and a reference frequency of 400Hz, we can obtain the resonance frequency from the equation

 ν–νref SpecFreq (MHz)

10.42= (v-400) ÷ (400)

V=4568 Hz

Problem 5

Ethylchloride C2H5Cl, (CDCl3, 300MHz)

Ethylbromide C2H5Br, (CDCl3, 90MHz)

From the given spectra, it is quite clear that δfor CH2 and CH3 groups of both compounds is the same and can be obtained directly from the two spectra.

CH2 group’s δ = 3.43 and

CH3 group’s δ = 1.7

The two carbons in each of the two compounds are in different structural environments and thus produce different signals in the NMR structure. The carbon attached to chlorine in the ethylchloride is ‘deshielded’ because of the electromagnetic nature of chlorine and this shifts its signal towards the left of the spectrum. A similar effect is observed in 1H NMR of ethylbromide. The two protons of the CH2 group neighboring the bromine are shifted to the left of the spectrum, while the hydrogens of the CH3 group that is far away from the bromine produce a signal towards the right of the spectrum.

From the formula

Vref =300 MHz for ethylchloride and δ = 3.43 for CH2 group, substituting in the above equation we can obtain v as follows

3.43= ((v-300) x 106) ÷ (300).

Evaluating this, we obtain v = 300.001029MHz

The same is repeated for CH3 group where δ = 1.7 and Vref =300MHz and thus V= 300.00051MHz

Vref =90 MHz for ethylbromide and δ = 3.5 for CH2 group, substituting in the above equation we can obtain v as follows

3.5= ((v-90) x 106) ÷ (90).

Evaluating this, we obtain v = 90.000315MHz

The same is repeated for CH3 group where δ = 1.7 and Vref =90MHz and thus V= 90.000153MHz

Problem 6

From the typical values obtained online, the coupling constant J

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

J = 1032.880 – 1025.526 = 7.5 Hz

Center peak = 1029.203 Hz

δ =1029.203/300 = 3.43

1H NMR (CDCl3, 300 MHz): δ = 3.43 (q, J = 7.35 Hz, 2H)

When the coupling constants of two (or more) adjacent hydrogen nuclei are different,

Signals are not simple triplets or quartets.

Problem seven

Hydroxyl methine resonate at 2.95 ppm the proton resonance is simple doublet

Problem 8

Alkanes (sp3 carbon) — 125 Hz

Alkenes (sp2 carbon) — 150 Hz

Alkynes (sp carbon) — 250 Hz

The resonance frequency is a function of the locally generated magnetic field which is very dependent on the molecular structure of the compound being analyzed. Resonance increases down from alkanes to alkynes, that is, alkene protons resonate at higher frequencies than alkane protons.

Problem 9

The aldehyde proton is attached directly to the carbonyl carbon hence is closer to the center of the deshielding zone.In addition, because of the electromagnetive effect of the carbonyl oxygen, the electron density at the carbonyl carbon is decreased and this results in further deshielding to resonance of δ of 9.00- 10.00 ppm. The downfield shift is due to additional effect of another nearby electron withdrawing oxygen atom next to it.

The α position is one bond away from a functional group; e.g., the carbonyl carbon. The α protons are far from the deshielding center as compared to the aldehydic proton. Carbonyl group has an effect on the chemical shift of the α-protons but this effect is less than the effect on the aldehyde proton. The signal of that of α-protons appears at approximately 2.4ppm.

Problem 10

Schoolery rule

If Y – CH2 – X then δ = 0.23 + ∆x + ∆y

The compound selected is CH2BrCl

δtheor = 0.23 + 2.33 + 2.53 = 5.09

δexp = 5.16

Topey rule δ = 5.28 + δgem + δcis + δtrans

The compound is Ph-CHa-CHb-COOH

δa = 5.28 + 1.38 + 0 + 0.98 = 7.64

δb= 5.28 + 0.80 + 0.36 = 6.44

Problem 11

Neighbouring hydrogen nuclei affects hydrogen’s under consideration and not nieghbouring carbon nuclei because an hydrogen nuclei has a high affinity for electrons as a result of high magnetic field as compared to carbon. Hydrogen nuclei has a high gyromagnetic ratio of 26.75 x107 T-1s-1 and a over 99% abundance whereas that of 13C is 6.73 x107 T-1s-1 and 1% abundance. 12C has no influence on the NMR spectrum because it has no orientation in any applied field as its nuclear spin number is zero.

Problem 12

N =5, n+1 = 5 + 1 = 6

It has six peaks in the ratio 1:5:10:10:5:1 it is sextet

Splitting tree is in here;

Problem 13

2H is also called 2Deuterium is a heavy hydrogen. At natural abundance it has very low sensitivity and therefore its magnetic field effect is very minimal to be felt by the neighbouring atoms and thus cannot be observed.

Problem 14

Higher resolution spectrometer (higher field strength instrument) gives a strong signal with better resolution and this provides useful information for analyzing fine structure in a spectrum.

Problem 15

The result is a quartet in the ratio 1:3:3:1, the resulting spectrum is as shown below.

Example of actual molecule that displays this pattern is propionamide, CH3CH2NH2C=O

Problem 16

, the AMX spin system splitting tree changes as shown below.AM=JAXWhen J

 0 0

he Splitting TreeT

JAX

JAM JAM

Problem 17

Spin system

(i) 2nI+1

n= 3 therefore, 2 × 2 × 1/2 = 2

AX2 Systems

(ii) n = 1 therefore, 2 × 1 × ½ = , AX System

(iii) n =3 therefore, 2 × 3 × ½ = 3

AX3 Systems

Problem 18

Peaks appear at 5.3 ppm for ethene protons

Br H

C=C

Stereochemistry

Br H

H …….. Et

Olifinic protons at δ 5.88 (1H, d, J = 11.6 Hz) and 6.05 (1H, dt, J = 11.6 and 7.5 Hz) are isomers

Reference

Gunther Herald. (1995).NMR spectroscopy, Basic principles, concepts, and applications in

Chemistry. Wiley

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