NMR Spectroscopy

Topics: Oxygen, Chemistry, Atom Pages: 17 (4135 words) Published: February 20, 2014
NMR Project

Sample 35 and 40

Chemical Engineering

Chemical Analysis

Aim
The two samples are 35 and 40. Analytical techniques are vital in everyday science and so NMR procedure is an absolute essential to deduce compounds and chemical structures; this assignment requires the identification of two samples using an NMR software program. Background

NMR spectroscopy is a well established method for structure determination of various organic molecules ranging from small molecules all the way up to large bio-molecules such as proteins, DNA and polysaccharides. The chemical environment surrounding each atom as well their proximity to other atoms in a molecule can be studied with various experiments, enabling high-resolution 3D-structure determinations as well as verification of the chemical structure. NMR spectroscopy is well suited for wood characterization, both in solution and in the solid-state.

Sample Number # 40
Sample 40 has been provided with sample information stating; an Infrared reading of 3350 / 1040 cm-1, a boiling point ranging between +107⁰C and +109⁰C,

Figure 1: Sample spectrum
Parameters:
Frequency = 100MHz Accumulation = 20
Because there is a need to report the location of an NMR signal/peak in a spectrum relative to a reference signal from a standard compound added to the sample, we use tetramethylsilane, (CH3)4Si, usually referred to as TMS. TMS is unreactive and easily removed from the sample after the measurement. By zooming in on peak E, we see that this signal has a chemical shift at 0ppm; therefore this is just the TMS signal, which doesn’t need to be analysed.

Spectrum was not clear; in order to view the spectrum in more detail changed the parameters.

Figure 2: sample NMR spectrum at higher frequency
Parameters:
Frequency= 400MHz Accumulations = 80

From the information given about this sample, it has an i.r band at 3350 cm-1 and 1040cm-1. Comparing with literature value, i.r band of 3200 – 3400 cm-1 (broad) suggest that it has an O – H bond. But this is however just an early assumption. Keeping this in mind, more detail analysis of each peak will be made to obtain a correct compound. Therefore the sample 33 should contain the following below bonds.

E.g.
Integration:

Table 2:
Peak Peak Height cm
Integration ratio
No of protons
A 2
6
1H
B 4
1
2H
C 2
2
1H
D 12
1
6H

Sample 33 has an Integration Ratio of 6:1:2:1 giving a total of 10 protons. The area under the peak is proportional to the number of protons that the peak represents. The integral measures the area of the peak and so the integral gives the relative ratio of the number of H for specific peak. In general, the more protons the more intense the peak / signal so I will now integrate my sample and simply measure the heights of the integral. I have a 6:1:2:1 ratio as seen below, although the integrals look too large to be 12:2:4:2 so this is most likely 6:1:2:1 as it is just the smallest simplest ratio it could be. Peak D

The integral ratio of peak D is 6:1, therfore an enviroment around 1 of the Carbon, within the compound is attached to 6 protons One way that this could exist is that of symmetrical branches or 2 branches. This may be a starting chain for my compound however the chain cannot be purely based on this as it is a ratio and so any scale factor could change the number of protons in that enviroment.

E.g.

Chemical Shift
The chemical shift (ppm) of each signal is tabulated in Table 1:

Peaks
Splitting Pattern
Chemical Shift (ppm)
A
Doublet
0.89
B
Nonet
1.7
C
Doublet...
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