Module 1 - Solvent Extraction and Partition Coefficient Part A

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CHEM 2203 Experiment 4A: Calculation and Comparison of the Solubility of 9-Fluorenone in a Mixture of Methylene Chloride and Water through the Implementation of Solvent Extraction, Establishing Equilibrium Distribution, and Isolation to Determine the Partition Coefficient at the Microscale Level Objective: The goal of this experiment is to separate the created solution into an aqueous mixture and an organic mixture and record all findings, to eventually calculate the partition coefficient. The Techniques that are new will include separation of a solid precipitation from a liquid by using a glass filtered pipet and heater block, weighing and measuring substances in mL and mg, and usage of automatic delivery pipettes. Experimental Procedure: The procedure was followed as described in Mayo, pages 144-145, with the modifications listed in the Canvas document. Additional modifications to the procedure included the following: Use of 2mL Methylene Chloride. Additional Modification: approx. 0.094g 9-fluorenone was substituted for the intended .075g. Reaction Scheme: There was no reaction included in this experiment. Data and Results: 9-Fluorenone Methylene Chloride Water Mol. Formula C 13 H 8 O CH 2 Cl 2 H 2 O Amount (mg) 94 - - Amount (mL) - 1 1 mmoles .522 - - Molar Mass (g/mol) 180.2 84.93 18.02 Density (g/mL) - 1.33 1 Mp/Bp (ºC) 80-83 40 100 Observations:
1. The 9-fluorenone was initially a bright yellow grainy substance. The water and methylene used were both clear in color with the methylene being slightly more opaque. 2. After the combination of the water, 9-fluorenone, and the methylene in a 5mL conical vial the solid completely dissolved into the solvents and couldn’t be seen. 3. The 9-fluorenone turned the bottom organic layer of methylene a slightly yellowish color and did not alter the color of the water at all. 4. The tared test tube with a boiling stone inside of it weighed in at approximately 11.697g 5. The heat of the test tube turned the liquid methylene chloride + 9-fluorenone solution into a yellowish liquid substance due to the evaporation of a majority of the methylene. 6. After rolling off the test tube upon a lab table the liquid completely disappeared and left a yellowish, grainy substance that appeared similar to the initial specimen of 9-fluorenone. 7. The initial weight of the test tube after the first round of heating and cooling was 11.759g 8. The constant weight of the test tube after the second round of heating and cooling was 11.759g Calculations: 1. Constant weight of test tube with boiling stone and 9-fluorenone - initial weight of the test tube and boiling stone: 11.759g - 11.697g = 0.062g 2. Initial weight of 9-fluorenone - (Constant weight of test tube with boiling stone and 9-fluorenone - initial weight of the test tube and boiling stone): 0.094g - 0.062g = .032g 3. (Constant weight of test tube with boiling stone and 9-fluorenone - initial weight of the test tube and boiling stone) / (Initial weight of 9-fluorenone - (Constant weight of test tube with boiling stone and 9-fluorenone - initial weight of the test tube and boiling stone): 0.062g/mL / 0.032g/mL = 1.9375 Questions: 1.
a. Carbon to carbon bonds are nonpolar because their electronegativity difference is 0. Carbon to hydrogen bonds are nonpolar because the electronegativity difference between them is close to .4, which falls into the nonpolar bond range. b. These non-polar regions interact with solvent molecules with london-dispersion forces or dipole-induced dipole forces. London dispersion forces may be at work because the nonpolar molecules, carbon and hydrogen, are attracted to each other and temporarily induced dipole forces. This is because while 9-fluorenone is very nonpolar, the water is very polar, causing a temporary attraction between them. c. Carbon to oxygen bonds are polar because they have an electronegativity difference of approximately .9, which falls into the polar range. The oxygen will pull more of the electrons towards it, creating a negative charge for itself and a positive charge for the attached carbon. d. This polar region interacts with solvent molecules through hydrogen bonding. The oxygen that is double bonded to the carbon would exhibit hydrogen bonding because it is a one of the three highly electronegative elements (O, N, F), has active lone pairs of electrons, and is interacting with a polar solvent containing hydrogen. e. Based on the answers to parts (a)-(d), I believe that 9-fluorenone would be more soluble in methylene chloride (a more non-polar organic solvent) than water (a very polar solvent). This is because the non-polar organic solvent is able to be more attracted to the large amount of non-polar bonds in the 9-fluorenone while the water only has the one oxygen atom to work with.
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f. My experimental partition coefficient was consistent with my prediction in part (e). This is because both prove that the 9-fluorenone will dissolve better in the non-polar organic solvent. 2. a. If substituted for methylene chloride in this partition coefficient experiment, chloroform would form the bottom layer due to its density. b. Chloroform would form as the bottom layer because it has a higher density than water. Chloroforms density is 1.49g/mL while water’s density is only 1.0g/mL. c. Despite the similarities, methylene chloride is more commonly used than chloroform as an extraction solvent because it is less dangerous and potentially harmful to humans than chloroform. While both are somewhat dangerous, methylene chloride poses less of a risk to experimenter health. 3. a. If this experiment is performed with anthracene in place of 9-fluorenone, the partition coefficient would definitely be higher than the partition coefficient that was obtained for the 9-fluorenone. b. This is explained by the lewis structures of both solutes. While 9-fluorenone has one polar bond, anthracene has no polar bonds and is made up entirely of non-polar molecules. This would mean that while 9-fluorenone minimally hydrogen bonds with the water molecules, anthracene would avoid this entirely. Therefore, the only forces at work would be the dipole-induced dipole forces between the anthracene and the methylene chloride, eventually resulting in a 100% yield of the solute in the non-polar organic layer. 4. After the mixture is thoroughly mixed there should be approximately 46mg of the organic compound in the methylene chloride layer and 10mg of the organic compound in the water layer. This is because:
Works Cited 1. Mayo, D. W.; Pike, R. M.; Forbes, D. C. Microscale Organic Laboratory with Multistep and Multiscale Syntheses, 5th ed.; John Wiley & Sons, Inc., 2011; pp 141-145. 2. Modifications for Expts 4A: Partition Coefficient Part A, Canvas document.