Reduction of Camphor Lab Report

Reduction of Camphor Lab Report 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 Boiling Melting Point (g/mol) Point (C°) (C°) Camphor 152.23 209 175 0.99 Sodium Borohydride 37.83 500 400 1.07 Borneol 154.25 213 208 1.01 Diethyl Ether 74.12 34.6 -116.3 0.713 Magnesium Sulfate 120.37 N/A 1124 2.66 Isoborneol 154.253 212.0 212.0 0.992 Methanol 34.042 64.7 -97.8 0.79 Water 18.015 100.0 0.0 1.00 Mechanism: Molar Mass Density (g/ml) Figure 1. The mechanism of the reduction of camphor. There are two possible products, each one resulting from the hydronium ion attacking from either the top face or bottom face of the site of reaction (Reference 1). Experimental: 0.100 grams of camphor were weighed out and placed in a 10 mL Erlenmeyer flask. Then 1.0 mL of methanol was added to the flask.This mixture was stirred using a glass stirring rod until the camphor was dissolved. 0.10 grams of NaBH4 were weighed out and added carefully in four different portions to the flask while it was being swirled. Once all of the borohydride was added, the solution was then boiled for 2 minutes using a hot water bath. The mixture was cooled to room temperature, and 3.5 mL of ice water were added. Via suction filtration, a solid formed and was collected after drying on the filter. This solid was then transferred to a 10 mL Erlenmeyer flask. 4.5 mL of ether were added to the flask in order to dissolve the solid, as well as some anhydrous MgSO4 to dry the solution. The MgSO4 was then filtered away from the solution into a 25 mL Erlenmeyer flask that had been weighed prior to the transfer. The original 10 mL Erlenmeyer flask was then rinsed using 1.0 mL of ether, which was then filtered into the 25 mL Erlenmeyer flask. The ether was evaporating using a warm water bath. The flask was then weighed, and the yield of the product was calculated. The melting point of the substance was identified, and an IR spectrum of the product was obtained. Results: Before the reaction took place, the amount of camphor delivered to the Erlenmeyer flask was weighed to be 0.104 grams. Once the reaction had finished, and the ether had been evaporated from the solution, the amount of camphor was weighed to be 0.0862 grams (Reference 2). The percent yield of camphor was determined to be 82.88% using the following equation: Percent Yield = (Actual Yield/ Theoretical Yield) * 100% (0.0862 / 0.104) * 100% = 82.88% 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 4.0 and the other at 3.6. These two peaks represent the two different possible products, borneol (4.0) and isoborneol (3.6). 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 = (1.00 / 6.07) = 0.1647 Isoborneol mole ratio = (5.07 / 6.07) = 0.8353 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 sp3 hybridized carbons within the molecule. Figure 3. The NMR of the product obtained by analyzing the product of the reaction performed in this experiment. There are two major peaks present, one at 3.6 with an integration of 5.07 and the other at 4.0 with an integration of 1.00. Discussion Two possible alcohol products, isoborneol and borneol, could be obtained from the reduction of camphor. To determine the ratios of the products obtained, the product was analyzed by IR spectroscopy, shown in Figure 2. The IR spectrum shows a broad O-H stretch at 3400 cm-1 and a C-H sp3 stretch at 3000-2900 cm-1. Isoborneol and borneol both produce these stretches, indicating that pure products were formed. In the analysis done using H-NMR spectroscopy, shown in Figure 3, the product exhibited two major peaks, one at 4.0 ppm and the other at 3.6 ppm. Isoborneol and Borneol both have an H-group beside the -OH group, but the deshielding of the -H is different because of the different stereochemistry in each product. Because the -H in borneol is more deshielded than that of isoborneol, the peak at 4.0 ppm represents borneol and the peak at 3.6 ppm represents isoborneol. The -OH group in borneol pulls more of the hydrogen’s electron density, as the hydrogen is closer to the -OH group in borneol compared to isoborneol. Isoborneol is more stable than borneol because it has less steric hindrance, and therefore it will be the major product (Reference 3). This corresponds with the calculation using the integration values from H-NMR spectroscopy, shown in the results section. The molar ratio of the product was 83.53% isoborneol and 16.47% borneol. The observed melting point was 198204 ℃ . The melting point of isoborneol is 212℃ , and the melting point of borneol is 208℃ , shown in Table 1. This lower and broader melting point range could be caused by impurities, such as ether or water. Another factor could be that the product consisted of a mixture of both isoborneol and borneol, and each has a different melting point. The percent yield was calculated as 82.88%, indicating the possibility of a small percentage of impurities in the product. Conclusion: In order to reduce camphor to isoborneol, sodium borohydride is added. Sodium borohydride turns the double bonded Oxygen on camphor into an -OH group. The hydride that is on the sodium borohydride is a good nucleophile, and therefore, is being oxidized in this reaction. This allows camphor to be reduced. On the NMR, there are two points that are either borneol or isoborneol. The peak at 4.0 is borneol because it is more deshielding than isoborneol. At the end of this experiment, a percent yield of 82.88% 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.