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Ir Spectrum Lab Report

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For both of the unknown compounds the techniques of IR Spectroscopy and both Carbon and Hydrogen NMR were used to identify the compounds. In IR Spectroscopy infrared waves are shot through a compound and a spectrum of peaks are created on a computer. Depending on where the peaks are located and what shapes they are one can identify the functional groups of the compound. In Carbon NMR (nuclear magnetic resonance) Spectroscopy the number of unique carbons the compound has are found. Each peak on the spectrum is a unique carbon. Depending on the location on the spectrum different functional groups that that carbon is a part of can be identified. It can also bring some enlightenment on the symmetry of the molecule. If the Carbon NMR says that …show more content…

The unknown was Number 13. The chemical formula is C4H8O. Starting analyzing from the IR Spectrum did not reveal much information. The IR Spectrum is shown in Figure 7 and summarized in Table 4. From the IR Spectrum it was known that there were alkanes and a C-O stretch in the molecule.

Next, again was Carbon NMR, shown in Figure 8. There were two unique carbons. This gave insight on the symmetry of the molecule because from the chemical formula it is known that there are four carbons in the molecule. Since only two are unique that meant that the molecule has symmetry. Carbon 1 was located at 68 ppm. That meant that it could be an alcohol or an ether; however, no alcohol showed up in the IR Spectrum, so it must be an ether. Carbon 2 was an alkane. This is summarized in Table …show more content…

The structure of the molecule worked perfectly with what was known from the IR, CNMR, and HNMR spectra. It has a C-O stretch and alkanes as suggested from the IR Spectrum. It has an ether as shown by Carbon NMR and has a hydrogen adjacent to a single bonded oxygen as revealed by the Hydrogen NMR. The molecule also had symmetry which was a necessity. Lastly, it has the correct chemical formula C4H8O. It was known which carbons and hydrogens went where for a couple of different reasons. First, Carbon 1 was an ether which meant it had to be part of the ether and adjacent to the oxygen. That left Carbon 2 to take the alkane spot. Hydrogen A was adjacent to a single bonded oxygen which meant it needed to be with Carbon 1. Again, this left Hydrogen B to fill the alkyl spot because it was bonded as a CH2. All of this is shown in Figure 11. Unlike, Unknown 1 the splitting patterns in Unknown 13 did not offer much insight. This was because both of the hydrogens were triplets. However, information from the splitting pattern was not needed to figure out

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