The objective of this lab was straightforward. We were given an unknown compound and we were to perform an IR spectroscopy and as well as NMR spectroscopy. With the IR spectroscopy, I was able to name the functional groups I have on my compound and further confirmed my assumptions by looking at the NMR spectroscopy after. The unknown number I was given was number 203. The molecular weight of the compound was 121. From the molecular weight, I calculated the molecular formula using the rule of 13.
The hydrogen deficiency index was also calculated so help construct the structure of my compound. My initial step in figuring out my molecular formula was to use the rule of 13. The molecular weight of my unknown compound was 121. Since the molecular weight is an odd number, Nitrogen is present in the compound. By dividing the molecular weight by 13, we get 9 and remainder of 4. The maximum Carbons the compound had are 9 and 13 Hydrogens. Since Nitrogen was present, I had to add it to my formula and subtract a Carbon and two Hydrogens from the base formula.
The Essay on Molecular Weight of Volatile Liquid Using Dumas Method
We have to stir the water in beaker to get the perfect temperature. * The rubber band which is used to tight the flask may release more vapour because the rubber band is not properly tight. The Dumas method is the method used to calculate the molecular weight of the substance. * This method has a condition that the substance used must be a liquid. The liquid that has boiling point above room ...
Hence the final molecular formula would be C8H11N with a HDI of 4. Looking at the IR spectroscopy, there is a sharp peak at around 3000 cm-1, which indicates C-H bonds. At around 1600 cm-1, we see a peak which tells us there are C=C that might corresponds to an aromatic benzene ring, which I proved using the NMR. The Nitrogen is also present on the IR spectroscopy leaving its mark with a peak at around 3200 – 3500 cm-1. Lastly, I finalized my structure using the NMR spectroscopy.
With the information I have collected and calculated, the last thing I had to do was to figure out the arrangement of my structure. I knew that the molecular formula of my unknown compound is C8H11N, with 4 HDI. First, at around 1. 2ppm, there was a high peak, which indicated my methyl group, and at 2. 5ppm, my CH2 group was present. Moving down the spectrum at around 3. 5ppm, my NH2 is present. At 7ppm, my aromatic benzene ring is present as well. The only thing I had to figure out is the positions of my NH2 and my CH2CH3: Ortho, Para, or Meta.
Ortho is the most logical position for my unknown compound, which turned out to be 2-ethylaniline. At 7ppm, there were a total of 9 peaks, but the Ortho position has a total of 10 peaks. Since peaks on the NMR spectroscopy sometimes do overlap, I assumed that one of the peaks overlapped with one, which is safe to assume instead of the Meta position, which has 8 peaks. Using everything all of the data from this experiment, I have come to a conclusion that my unknown compound (#203) is actually 2-ethylaniline with a molecular weight of 121 and molecular formula of C8H11N.
