Characterization of Unknowns by Means of Infrared, Proton,

Carbon, and UV/Vis Spectroscopy

Introduction:

Being able to identify the signature features that a compound possesses by different

means of spectroscopy is very important in determining if the desired compound has been

isolated. This is detrimental to not only the industry, but the research level as well. Every

molecular compound possesses features that can be characterized by different means of

spectroscopy due to their structure. UV/Vis spectroscopy is a tool that is capable of determining

the UV radiation frequencies within a molecule which shows details about the compound’s

structure. This is accomplished by matching UV frequencies with the conjugated double bonds

that reside in the compound. The UV/Vis matches the wavelength frequencies with the

absorption of this radiation. UV/Vis spectroscopy was used to only analyze unknown #3 due to

its dark red color. The solvent used to dissolve this unknown was methanol. To accomplish

characterizing this unknown by means of UV/Vis spectroscopy involved getting a cuvette of the

solvent to be used as a blank in the machine. Though this, a baseline reading was obtained to

compensate for the noise the blank caused. Nuclear magnetic resonance spectroscopy is very

versatile in determining what an unknown compound looks like. This technique operates by

displaying the elements located around hydrogen atoms. Proton NMR can be interpreted by the

determining the integral of the peaks, splitting of peaks, and the locations in the compound that

are shifted. NMR alone can be used to determine what an unknown compound may look like.

Lastly, Infrared spectroscopy was used to identify the specific functional groups within the

unknown compound. These functional groups vibrate at varying degrees and at varying

frequencies. This type of spectroscopy is useful, but not as detrimental as proton NMR. These

four different types of spectroscopy were used in determining the structures of unknowns 1-3.

Procedure:

Infrared spectroscopy was used to determine all unknown compounds. A background was

run before the unknowns were even tested, and this was done to ensure a correct reading. Once a

background was ran, the surface of the machine was wiped with methanol to ensure no substance

was on the crystal. After machine was cleaned, a small amount of sample was placed on the

crystal to being the infrared spectroscopy. Once infrared was successfully ran properly, surface

was again cleaned, and this process was repeated for the rest of the unknown compounds.

UV/Vis spectroscopy was only used for unknown 3 in this experiment. The solvent used for the

blank and the unknown was methanol. A quartz cuvette of the methanol blank, and the unknown

compound mixed with the methanol were both placed in the spectroscopy machine. The blank of

methanol was used to get a baseline reading. Before beginning the NMR spectroscopy for the

unknowns, a solubility test was needed in order to choose the right solvent for the corresponding

unknown. This was done by mixing 10 mg of each unknown with 5 mL of the corresponding

solvent for the test. The solvents tested include: DMSO, water, chloroform, and methanol. From

the solubility test, methanol was used for unknowns 1 and 3. Methanol did not dissolve unknown

2, so DMSO was used instead of methanol. Once the right solvents were chosen, around 5mL of

the solvent was mixed well with around 10 mg of the unknown compound. This mixture was

placed inside an NMR tube and then analyzed inside the NMR machine for 5 minutes. Carbon

NMR spectroscopy was done the exact same way as the proton NMR, but took a different

amount of time to complete analysis. Carbon NMR spectroscopy took around 19 hours to

complete, due to the amount of scans entered into the computer prior to beginning the NMR

sequence.

Results:

Before any spectroscopy could take place, a solubility test needed to be initiated in order

to dissolve to correct substance. Below is the table showing the solubility the unknowns had in

different solvents.

Unknown

Solvent 1 2 3

Water N N N

Methanol Y N Y

DMSO Y Y Y

Below is the table showing the degree of unsaturation for the unknown compounds:

Unknown 1 2 3

Molecular Formula C8H17N3 C10H16N2O8Na4 C20H12O5

Degree of

Unsaturation

2 4 15

Unknown #1:

Below are the IR, proton NMR, and carbon NMR for the first unknown compound.

(Figures 1, 2, 3)

Figure 1: Unknown#1 IR C

Structure Broad Curve C-H N-H N=C

Frequency 3000’s 3000 3100 1550

Figure 2: Unknown#1 Proton NMR C

Peak 1.2 1.5 2.2 5.0

Structure CH3 CH2 CH2 NH2

Figure 3: Unknown#1 Carbon NMR

1.2 1.5 1.5

5.0

2.2

PPM 140 60 50 41

Structure N=C=N CH2 N-CH3 CH2

Unknown#2:

Below are the IR, proton NMR, and carbon NMR for the second unknown compound.

(Figures 4, 5, 6)

Figure 4: Unknown#2 IR C

Structure COOH C=O C-H

Frequency 3300 1700 3000

Figure 5: Unknown#2 Proton NMR C

PPM 2.1 3.4 4.2

Structure OH CH2 CH2

Multiplicity Singlet Singlet Singlet

*Not triplets

2.1 3.4

4.2

Figure 6: Unknown#2 Carbon NMR

PPM 49 50 160

Structure CH2 CH2 COOH

Unknown#3:

Below are the IR, proton NMR, UV/Vis, and carbon NMR for the last unknown

compound. (Figures 7, 8, 9, 10)

49

50

160

Figure 7: Unknown#3 IR

Structure C=O C-H C-O-C Aromatic

Frequency 1700 3000 Broad 1750 1500

Figure 8: Unknown#3 UV/Vis C

Peak 270

Structure C=C conjugation

Figure 9: Unknown#3 Proton NMR C

PPM 6.5 4.9 7.7 8

Structure Aromatic H OH Aromatic H Aromatic H

Figure 10: Unknown#3 Carbon NMR

Discussion:

The structure of the first unknown was EDAC. The IR spectra showed a C=O bond at

1700, a C-H stretch at 3000, N-H stretch at 3100, and an N=C stretch at 1550. The proton NMR

shows a methyl peak at 1.2 and two methine peaks at 1.5 and 2.5 respectively. The peak at 1.2

points towards a methyl group due to chemical shifting. There are multiple peaks at 1.5 that

points to multiple methine groups in a chain. Lastly, the peaks at 2.5 point to methyl groups

directly attached to a nitrogen, again due to chemical shifting.

The structure of the second unknown was EDTA. The IR spectra for this unknown shows

a COOH stretch at 3300, a C=O stretch at 1700, and a N-H stretch at 3300. The proton NMR

shows a OH shift at 2.1, and methine shifts at 3.4 and 4.2 respectively. The shift at 2.1 is shown

in the alcohol groups at the end of each carboxylic acid in EDTA. The shift at 3.4 is shown as the

methine carbons connected to the nitrogen and carboxylic acid in EDTA. Lastly, the shift at 4.2

is shown in the structure as the two connected carbons between the nitrogens in EDTA. The

carbon NMR shows important shifts at 49, 50, and 160 ppm. The carbons at 49 ppm are the

carbons in the middle of EDTA between the two adjacent nitrogens. The carbons at 50 ppm are

the carbons connected between the nitrogen and the carbon of each of the four carboxylic acids

in EDTA. The carbons at 160 ppm are the carbons involved in the double bond of the carboxylic

acid.

The structure of the last unknown was fluorescein. The IR spectra shows frequencies at

1500, 1700, 3000, and 1750 which correspond to an aromatic stretch, C=O stretch, C-H stretch,

and ester stretch respectively. The IR spectra displayed more than one aromatic ring in the

molecule. The UV/Vis spectra showed the specified conjugation inside of the fluorescein

molecule. The proton NMR shows a peak at 4.9 and is shown in the alcohol group in the ring on

the far right of fluorescein. The peak at 6.5 is shown in the hydrogens attached to the ring on the

left of fluorescein, as well as the hydrogen on the ring on the right side of fluorescein. The peak

at 7.7 shows the hydrogens on the bottom ring of fluorescein due to the nearby acid. The last

peak at 8.0 is shown again in the bottom ring at the hydrogen connected to the acid. Since the

proton NMR displayed a significant amount of the molecule, all that was left was the carbon

NMR. The carbon NMR showed peaks mainly in the mid 100’s ppm, which correspond to

carbons inside a ring with a degree of conjugation. It also displayed carbons in the upper 100’s

ppm, which pointed towards a carbonyl, carboxylic acid, and a carbon attached to an alcohol.

Due to the saturation of the molecule, the carbon NMR displayed similar peaks for many

carbons.