Organic Chemistry Lab 5 - Lab Report

Lab 5: Acid-Base Extraction Introduction: The given unknowns from experiment 4 were found to be aspirin and phenacetin. In order to separate these two compounds from a mixture, acid-base extraction must be performed. In order to separate aspirin and phenacetin, aspirin must become an aqueous solution, while phenacetin remains in an organic layer. An acid such as aspirin can be separated from a neutral compound such as phenacetin, with the use of a base, creating an ionic, aqueous solution. After separating the solution by density, separatory funnel can be used to remove the layers into different containers. To obtain the dry mass, acid can be added to the aqueous solution, which causes it to precipitate, and the precipitate can be collected through vacuum filtration. For the organic layer, a drying agent such as sodium sulfate, can be added to remove water. In order to test the effectiveness of separation, a TLC plate can be spotted to see if there is any impurity in the product. Results: Unknown #13 Aspirin Crude Weight: 1.35g % Yield: 90% Rf of Spot: X / Y = 0.62 Phenacetin Crude Weight: 1.26g % Yield: 84% Rf of Spot: X / Y = 0.55 This TLC plate spotted with Aspirin, Phenacetin, aqueous layer, and organic layer (left to right). Compounds Melting range (°C) Aqueous layer 133.3-134.7 Aqueous layer + Aspirin (1:1 mixture) 133.8-134.7 Organic layer 133.1-134.4 Organic layer + Phenacetin (1:1 mixture) 132.9-133.9 Observations: Based on the separation scheme, the two compounds in the unknown mixture #13 were aspirin and phenacetin. This proved to be accurate while separating the two substances, as there was one organic layer, and one aqueous layer that was formed with the use of a strong base. This meant that one compound was surely aspirin, as it is acidic. In addition, the separation of the aqueous layer and the organic layer was visible, with the more dense organic layer on the bottom. After separation was successful, a TLC plate run in order to determine the purity of the crystal. The TLC had matching distances traveled by the spot for the unknown compounds, and the known compounds. This indicated that the crystal was pure. Discussion:. After separation of the two compounds, measurements were taken in order to see how effective the separation process was. Overall, the percent yield of aspirin was 90%, and the yield of phenacetin was 84%. In addition to this, a TLC plate was run, in order to see if there were any impurities within the solution. The TLC plate indicated that the two unknown compounds matched with the known compounds of phenacetin and aspirin. This indicated that the spots were not impure, as there was only one compound detected for each spot. Lastly, a melting point observation was taken for the two separated compounds, as well as a one to one mixture with the separated compounds and known compounds. The melting point ranges of each matched with the literature values. In addition, when a 1:1 mixture was formed, a melting point depression was not observed. This indicates that the mixture of the separated substance and the known substance were the same. Conclusion: All in all, the experiment was successful. The objective was met in separating the two substances effectively. The percent yields are a good indicator that the separation was effective. A possible error that may have caused a slight decrease in the percent yield may have been when the two layers were being separated in the separatory funnel. It is very possible that a small portion of the organic layer remained in the solution containing the aqueous layer. The TLC, however, indicated that the separation was successful, as only one compound was found in each spot. In addition, each spot matched with the spot of the known compounds. As such, it can be concluded that aspirin and phenacetin were the two compounds in the mixture. Post lab questions: Lab manualA) Using too much sample would create spots that are too big, which can cause adjacent spots to overlap. B) If not enough sample is used, the spot may not be visible. C) If the solvent is too-polar, this can cause the sample to move too far down the TLC plate for effective separation. This can cause the spot to move up to the solvent front, preventing an Rf calculation. D) If a crystalline material that is insoluble in eluent is attempted to be spotted, it won’t properly spot on the TLC plate because it will not dissolve in the eluent. In order for a successful spot, the material must be soluble in an eluent. Pavia1) Since solute A has a distribution coefficient of 1.0 between water and diethyl ether, a 100 mL solution of 5g of A in water in two 25mL portions of ether would result in a smaller amount of A remaining in the water than if the solution were extracted with a 50mL portion of ether. 0.1g more of A will remain in a solution extracted with one portion of ether. 25mL portion 1: K= C2/C1 = ((5-x)g/25mL diethyl ether) / (xg/100mL water) 1=((5-x)*100)/25x 125x=500; x=4 25mL portion 2: (4-x)g/25mL diethyl ether) / (4g/100mL) 1=((4-x)*100) / (4/100mL water) 125x=400; x=3.2g of compound A will remain in the water layer. 50mL portion: (5-xg/50mL diethyl ether) / (xg/100mL water) 1=((5-x)*100)/50x 150x=500; x=3.3g of A will remain in the water layer. References: 1) Schaller, Chris. “Acid-Base Extraction.” Chemistry LibreTexts, Libretexts, 5 June 2019, chem.libretexts.org/Ancillary_Materials/Demos%2C_Techniques %2C_and_Experiments/General_Lab_Techniques/Acid-Base_Extraction. 2) Clark, Jim. “CIE A Level Chemistry Support: Learning Outcome 11.2(f).” Chemguide, 2013, www.chemguideforcie.co.uk/section112/learningf.html.