Part B. Reduction of Camphor to Isobomeol Reductions. The camphor obtained in Pa
ID: 1084910 • Letter: P
Question
Part B. Reduction of Camphor to Isobomeol Reductions. The camphor obtained in Part A should not contain borneol. If it does, show your infrared spectrum to your instructor and ask for advice. If the amount of camphor obtained in Part A, or after the sublimation if you did this, is less than 0.25 g. obtain some camphor from the supply shelf to supplement your yield. If the amount is more than 0.25 g, scale up the reagents appropriately from the following amounts. Add 1.5 mL of meth anol to the cam- phor contained in a 50-mL flask. Stir with a glass stirring rod until the camphor has dis- solved. In portions, cautiously and intermittently add 0.25 g of sodium borohydride to the solution with a spatula. When all of the borohydride is added, boil the contents of the flask on a warm hot plate (low setting) for 2 minutes Isolation and Analysis of Product. Allow the reaction mixture to cool for several minutes and carefully add 10 mL of ice water. Collect the white solid by filtering it on a Hirsch funnel and, by using suction, allow the solid to dry for a few minutes. Transfer the solid to a dry Erlenmeyer flask. Add about 10 mL of methylene chloride to dissolve the product. Once the product has dissolved (add more solvent, if necessary), add about 0.5 g of granular anhy drous sodium sufate to dry the solution. When dry, the solution should not be cloudy. If the solution is still cloudy, add some more granular anhydrous sodium sulfate. Transfer the solu- tion from the drying agent into a preweighed dry flask. Evaporate the solvent in a hood, as described previously.Explanation / Answer
The reaction is between camphor and sodium borohydride and the product formed is isoborneol.
Mass of camphor used = 0.25 g
Molar mass of camphor = 152.23 g/mol
Moles of camphor = mass/molar mass = 0.25/152.23 = 0.0016 mol
Moles of isoborneol formed = moles of camphor used = 0.0016 mol
Molar mass of isoborneol = 154.25 g/mol
Mass of isoborneol produced = moles*molar mass = 154.25*0.0016 = 0.253 g
Theoretical yield of isoborneol = 0.253 g