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Introduction to spectroscopic structure identification using H-NMR and IR
Spectroscopy
By
Dhara Parikh
03/18/2022
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Experimental section
In this experiment, the unknown organic compound was given along with their molecular weight,
and the objective was to identify the correct compound using 1HNMR and IR spectroscopy. First,
a small, diluted sample was taken in a disposable test tube using a glass Pasteur pipette to make
the sample for NMR and IR spectroscopy. Next, 1.0 ml of deuterated chloroform was added to the
same test tube and mixed. The prepared solution was then added to the clean NMR tube using the
glass Pasteur pipette such that the tube was approximately 4 cm filled. NMR tube was closed and
set aside. NMR spectra were developed using Eft 60 MHz pulsed Fourier transform NMR
spectrometer. The top of the lid was opened, and the eject button on the left side was pressed,
resulting in the sample chamber popping up. Then the NMR tube was clean with the Kim wipe so
that no fingerprints were present on the tube. Now, holding the NMR tube with the Kim wipe, the
tube was placed in the sample chamber. On releasing the eject button, the tube sinks into the probe,
and then the spinner is started by flipping the lever down. Once the spinning stopped, the P_NMR
program was opened on the computer. Next, the file with the course name was opened, and all the
details regarding the name initials, date, and compound description were entered. After that,
another program, NUTS, was opened, which showed the NMR spectra developed for the sample
and that spectra were printed out for further analysis. NMR spectra on Varian 400 MHz MR
spectrometer were developed using a similar procedure with the help of a lab assistant.
After obtaining the 1 H-NMR spectra, the IR spectra were obtained using a Thermo-Nicolet iS-5
Infrared spectrophotometer. First, Omnic software was opened on the computer for IR spectra, and
the background spectrum was collected. Next, the ATR stage was cleaned using Kim wipes, and
a sample drop was placed at the stage center. Once the sample was placed, the press was pulled
down, and the IR spectra were collected on the computer using the collect button. Once the IR
spectra were obtained, it was printed out for further analysis. IR and 1H NMR spectra for unknown
compounds B and R are shown below.
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a.
IR Data:
Key peaks
Structural unit
3100cm-1
O-H stretch
1709cm-1
C=O stretch
2700cm-1
Sp3 C-H stretch
b.
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H NMR Spectral Data:
Chemical
Integration
Multiplicity
Assignment
shift
(Number
ẟ(ppm)
of H’s)
10.4
1
Singlet
-OH
2.3
3
Singlet
-CH3
Figure1: IR (a) and NMR (b) spectra of unknown compound B
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a.
IR Data:
Key peaks
3318cm
-1
2944cm-1 &
Structural unit
O-H stretch
Sp3 C-H stretch
2832cm-1
1021cm-1
Sharp peak
indicating
presence of
oxygen
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b.
H NMR Spectral Data:
Chemical
shift
ẟ (ppm)
Integration
(Number
of H’s)
Multiplicity
Assignment
3.6
3
Singlet
-CH3
2.5
1
Singlet
-OH
Figure 2: IR (a) and 1H NMR (b) spectra for unknown compound R
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Discussion:
Compound B: Acetic acid
IR and NMR spectra of figure 1 are consistent with acetic acid, and its structure is shown below:
There are two prominent peaks in the IR spectra of acetic acid; the first peak is wide and broad at
3100cm-1, which is consistent with the O-H functional group. At the same time, another peak is at
1709cm-1, which corresponds to a carbonyl group. Based on both the peaks, we can say that the
compound has a carboxylic acid as its functional group. Moreover, a few small peaks around
2700cm-1 suggest the C-H stretch of sp3 hybridized carbon.
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H NMR spectra of figure 1, two peaks indicate that the given compound has the two types of
hydrogen. The integration ratio of the two peaks is 3:1. Both peaks are singlet, so there is no
vincinal coupling of the hydrogen. Peak with the integration of 1 hydrogen has a chemical shift of
10.4 ppm. The carboxylic acid group has a chemical shift between 10-12, consistent with the peak
with one hydrogen. So, we can say that hydrogen corresponds to the carboxylic acid group.
Another peak with the integration of 3 Hydrogen is at 2.3 ppm, which corresponds to the methylene
hydrogen, so we can say that three hydrogens are of the methyl (-CH3) group. Moreover, the given
molecular weight of compound B is 60g/mol, which is also consistent with the acetic acid.
Therefore, based on all the information, we can conclude that compound B is acetic acid which
contains a carboxylic acid group, three hydrogens of the methyl group, and one free hydrogen of
carboxylic acid group.
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Compound R: Methanol
IR and NMR spectra in figure 2 are consistent with the methanol, and its structure is shown below:
In the IR spectra of figure 2, a strong, broad peak at 3318cm-1 suggests O-H stretch. Another sharp
peak at 1021 suggests the presence of oxygen. Moreover, there is no peak at 1600-1700cm-1,
suggesting that there is no carbonyl group. Also, there are short sharp peaks at 2944 and 2832,
which suggest the C-H stretch due to sp3 hybridized carbon. Therefore, we can say from all the
above peaks that the compound has an alcohol functional group.
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H NMR spectra of figure 2 have two speaks that suggest two types of hydrogen present in the
given compound. The integration ratio of both peaks is 3:1. The peaks are singlet, so there is no
vincinal coupling of the hydrogen. The peak with the integration of 3 hydrogens has the chemical
shift at 3.6ppm. The methylene hydrogen of the alcohol group falls between 3.4-4.0 ppm, which
is consistent with the peak with the integration of 3 hydrogens. Another peak with the integration
of 1 hydrogen has the chemical shift of 2.5 ppm. The chemical shift for alcohol ranges from 05ppm. As 2.5ppm falls in the category, the peak with one hydrogen corresponds to the hydroxy
hydrogen. Given that the molecular weight of the compound R is 32g/mol, which is consistent
with methanol. Hence from all the information, we can say that compound R has an Alcohol
functional group, three hydrogens, the methyl group, and one hydroxyl hydrogen.
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