OChem 232 Experiment #28

Rick Politician July 30, 2005 Jesse Hofseth, Luke, Idid Experiment #28 An Oxidation-Reduction Scheme: Borneol, Camphor, Isoborneol Purpose: This lab aims to demonstrate to us that we can react an oxidizing agent with a secondary alcohol to produce a ketone, and that a ketone can be reacted with sodium borohydride to produce the isomer of the original secondary alcohol. This lab also allows us to practice our extraction, crystallization, evaporation, and melting points techniques. Data and Results: Part A - borneol ? camphor 0.340g of initial racemic borneol 0.340g * 1mol/154.25g = 0.002204mol of borneol 0.170g of final camphor 0.340g * 1mol/152.23 = 0.001117mol of camphor 0.001117mol of camphor / 0.002204mol of borneol = 50.7% yield Melting Point from Jesse??? Part B - camphor ? isoborneol Initial camphor 0.106g Initial sodium borohydride 0.107 Erlenmeyer Flask 40.449 Product and Flask 40.520 Weight of isoborneol product = 0.071g 0.106g * 1mol/152.23 = 0.000696mol of camphor 0.071g * 154.25mol/g = 0.00046.mol of isoborneol 0.00046mol of camphor / 0.000696mol of camphor = 66.09% yield Discussion: Part A of the experiment went fairly well with a yield of 50.7%. There are many places we could have lost some of our product including: not mixing the initial solution and methylene chloride completely, not extracting all of the methylene chloride from the mixture, or some of our camphor getting stuck on the anhydrous sodium sulfate. For Part B, our group tried twice, but did not manage to successfully crystallize the isoborneol. For an unknown reason, we could not dissolve all of the sodium borohydride in the 0.5Ml of methanol. Instead, it took almost 20mL of methanol to dissolve the sodium borohydride, even as the solution approached boiling. I wonder if we were using a similarly named, but wrong, solvent such as methylene chloride. When I discussed the crystallization with other groups, they had no problem dissolving the solutes. The source of this problem still remains a mystery. Conclusions: The melting point I measured of 119.5 to 121.9 degrees C is very close to the given melting of 2,4,6-Tribromoaniline of 122 degrees C. The second closest given had a melting point of 87 degrees Celsius. Thus, one can deduce that our resulting substituted benzene ring is likely the 2,4,6-Tribromoaniline. Since the measured melting point range has a breadth of only 2.4 degrees and the range is only 0.1 degrees Celsius from the given value, it would appear that the experimentally manufactured 2,4,6-Tribomoaniline in fairly pure. Questions: 1. 2. Our class did not experimentally determine which substituent was least activating. According to Table 12.2 in our textbook, it appears the Anisole is the least activating. Since the alkoxy group makes Anisole a strongly activating substituent group, one might predict that the bromination would take place at the ortho or para sites. 3. Spectroscopy - ??? Chromatography - The large bromine molecules could significantly increase the molecular weight of the products. More massive compounds tend to move slower down the column chromatography alumina. One could run samples of each possible compound and the product through a column using identical volumes of a standard solvent. The time it takes each compound to travel down the column could be timed. The measured time of the compounds that most closely resembles that of the measured time of the product could be understood to be the product. This method of identification does not take into consideration the differing attraction to the alumina each unique compound may have. Add Br2 to product dissolved in water - If there available sites on the ortho and para spots, the some of the Br ions will induce a substitution reaction at the 2 and 6 positions More ??