Reduction of Camphor to Borneol and Isoborneol

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Abraham Baldwin Agricultural College *

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2041

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Chemistry

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Apr 3, 2024

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docx

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Hannah Bryson Reduction of Camphor to Borneol and Isoborneol Purpose In this experiment, camphor underwent reduced to form two isomers, borneol and isoborneol, using the reducing agent sodium borohydride. The reduction reaction reduces camphor into borneol, while the oxidation of borneol will revert back to camphor. Observations After adding ice cold water to the flask, a solid of crystals formed at the bottom of the flask. After filtration, we could see white crystals on the filter paper. When the CH4CL2 was added, the mixture became clear. After boiling off the solvent, a white crystal product was left. Data and Calculation Reaction mechanism: 4C 10 H 16 O + NaBH 4 + 4H 2 O ——> 4C 10 H 18 O + NaB(OH) Calculations: 0.28 g of Camphor Camphor = 152.23 g/mol 0.28 g/152.23 g/mol 0.00183932 mol of reactant (limiting reagent) 0.25g of sodium borohydride sodium borohydride = 37.83 g/mol 0.25 g/37.83 g/mol 0.00660851 mol of reactant 250mg of camphor camphor = 152.23 g/mol 0.00164225 moles of reactant which is also a limiting reagent
Hannah Bryson 250mg of NaBH4 NaBH4 = 37.83g/mol 0.0066085 moles of reactant Borneol and isoborneol predicted mole yield = 0.00164225 Borneol and isoborneol molecular weight = 154.25g/mol The theoretical yield of product: (0.00183932 * 154.25 g/mol) =0.284 g of borneol and isoborneol Actual yield of the product: = 0.35 g Percent yield of the product: ¿ Actual Theoretical × 100% ¿ 0.35 0.284 × 100% = 123% Melting point of the product: 10 /sec: 220-245 1 /sec: 218-227 Conclusion: Using the data collected, we can determine that the borneol/isoborneol product we obtained still contained impurities. First and foremost, our percentage yield was abnormally high; this indicated that there was an error within the procedure. It is likely that when evaporating the solvent, not all solvent was evaporated, thus skewing both the percent yield and the melting point. The melting point of borneol is 208 while the melting point of isoborneol is 212-214 . We did two tests of melting points on our product: one slow and one fast. The melting range of our product was determined to be 218-227 . In comparison to pure borneol and isoborneol, our melting point was higher. This indicates that there were still impurities left within our product. I do believe, overall, we were successful in obtaining the product; however, we messed up by not allowing all of the solvent to evaporate.
Hannah Bryson Post Lab Questions: Based on the correct answers to the previous question, did you use a deficit, just enough or an excess of NaBH4 compared to the substrate? Justify your answer. If it was an excess or a deficit, indicate by how much. Show your calculations, both on the numbers and the units. An excess of NaBH4 was used in comparison to the substrate. In the previous calculations, it was determined that 0.00183932 mol of reactant (limiting reagent) and 0.00660851 mol of NaBH4 were used. With this experiment being a 1:1 mole ratio, only 0.00183932 moles of NaBH4 were needed. This means that there was a excess of 0.00476919 moles of NaBH4. What was the double role of using cold water in the reaction? Use the correct answer to the previous question for inspiration. Using cold water is to slow down the reaction rate in order for the reaction to occur. If room temperature water were used in place of the cold water, the reaction would occur at a much more rapid place, thus causing the reaction to prematurely react with NaBH4. A basic solution resulted from the use of water in the reaction. If ice was not available and you used room temperature water, what undesirable reaction could take place? Justify your answer. Using room temperature water would cause the reaction to occur prematurely with NaBH4, potentially causing unwanted products to form. Cold water allows for the reaction to slow down and produce the desired products, borneol and isoborneol.
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