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Reduction of Camphor Lab Report
Organic Chemistry II Lab (CH 238)
University of Alabama at Birmingham
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Reduction of Camphor Lab Report
Matthew Estacio - Writer Maddie Looney -Reviewer Nora Cipkowski - Editor
Introduction: The purpose of the lab involved applying several different concepts. Melting point and IR data were used to analyze product purity and composition. NMR integration values were used to determine relative product percentages. Then, the relative percentages of different reaction products were explained using structural data. The reduction reaction mechanism was drawn and analyzed. The experiment reduced camphor into isoborneol by adding sodium borohydride, forming boric acid and sodium chloride as byproducts. Camphor contains an electrophilic carbonyl group that can be reduced by sodium borohydride. The hydride within sodium borohydride acts as a reducing agent because it is a very strong nucleophile. Two different alcohol products can be obtained because the carbonyl is planar, consisting of two faces that can be attacked.
Table 1: Table of Reagents Compound Molar Mass (g/mol)
Boiling Point (C°)
Melting Point (C°)
Density (g/ml)
Camphor 152 209 175 0. Sodium Borohydride 37 500 400 1. Borneol 154 213 208 1. Diethyl Ether 74 34 -116 0. Magnesium Sulfate 120 N/A 1124 2. Isoborneol 154 212 212 0. Methanol 34 64 -97 0. Water 18 100 0 1.
Mechanism:
Before the reaction took place, the amount of camphor delivered to the Erlenmeyer flask was weighed to be 0 grams. Once the reaction had finished, and the ether had been evaporated from the solution, the amount of camphor was weighed to be 0 grams (Reference 2). The percent yield of camphor was determined to be 82% using the following equation: Percent Yield = (Actual Yield/ Theoretical Yield) * 100% (0 / 0) * 100% = 82%
The melting point of the product was determined to be between 198-204°C
The IR spectrum (Figure 2) and NMR (Figure 3) of the product were obtained and analyzed. The IR spectrum showed a broad peak at 3400 and strong, sharp peaks at 2900-3000. The NMR of the product exhibited two major peaks, one at 4 and the other at 3. These two peaks represent the two different possible products, borneol (4) and isoborneol (3). The integrations of the two peaks were used to determine the mole fraction of each product. The following equation was used to calculate these values: Mole Ratio = (single peak integration / sum of equivalent peak integrations) Borneol mole ratio = (1 / 6) = 0. Isoborneol mole ratio = (5 / 6) = 0.
Figure 2. The IR spectrum obtained from analyzing the product of the reaction performed in this experiment. There is a broad peak present at about 3400, which is characteristic of an alcohol group. The strong peak at around 2900 most likely represents sp 3 hybridized carbons within the molecule.
At the end of this experiment, a percent yield of 82% was achieved. Despite this being a very good percent yield, this shows that there are still impurities. To improve this experiment, the evaporation process could be slowed down, so that all of the solvent is properly evaporated leaving less impurities. This would have made the results even more accurate than they were in this particular trial.
References: 1.) Ege, S. (1998). Chem2o06 - 1997/98 - experiment 7. Retrieved February 05, 2021, from chemistry.mcmaster/~chem2o6/labmanual/expt7/2o6exp7 2.) Spider, C. (2017). (±)-camphor. Retrieved February 05, 2021, from chemspider/Chemical-Structure.2441.html 3.) National Center for Biotechnology Information. "PubChem Compound Summary for CID 6321405, Isoborneol" PubChem , pubchem.ncbi.nlm.nih/compound/Isoborneol. Accessed 5 February, 2021.
Reduction of Camphor Lab Report
Course: Organic Chemistry II Lab (CH 238)
University: University of Alabama at Birmingham
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