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Lab reports

by: Geraldine Orlando Castellanos

Lab reports CHM 205

Geraldine Orlando Castellanos

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Lab reports from experiments 3-6
Organic Chemistry 1
Tegan Eve
Chemistry, orgo
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This 17 page Bundle was uploaded by Geraldine Orlando Castellanos on Thursday February 18, 2016. The Bundle belongs to CHM 205 at Florida State University taught by Tegan Eve in Fall 2015. Since its upload, it has received 116 views. For similar materials see Organic Chemistry 1 in Chemistry at Florida State University.

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Date Created: 02/18/16
Geraldine Orlando ID#: C11462857 Ms. Piumi Y. Layanage & Dr. Tegan Eve CHM205-P2 October 20 , 2015 Experiment 3a: Thin Layer Chromatography Results Discussion Conclusion Post Lab questions 1) o-Hydroxyacetophenone has a melting point of 4-6 ⁰C. p- Hydroxyacetophenone has a melting point of 109-111⁰C! Explain the basis for the sizable difference in melting points of these two compounds. 2) In running TLC under the following conditions, you get results that are less than ideal. Consider the situation in each case and suggest a correction: a. You run TLC on a mixture of a two unknown halogenated alkenes, but you see only one spot with an R of f.91. The solvent used was ethyl acetate b. You run a TLC on a mixture of a thiol and an amine, and again you get only one spot with an R off0.15. The solvent used was a mixture of petroleum ether and dichloromethane. c. When you put your TLC plate in the developing chamber, the solvent in the chamber covered the spots on the TLC plate baseline. 3) You run a TLC plate spotted with three compounds: naphthalene, o-toluic acid, and fluorenol. Predict the relative R falues. 4) For each of the following pairs, predict which compound will have the larger Rfvalue if both are run on a SiO TL2 plate in 10% acetone/hexane: 4-decanone or 4-decanol; xylene or benzoic acid; cycloheptane or cycloheptanone. 5) Arrange the following solvents in order of increasing polarity: ethyl acetate, dichloromethane, iso-propanol, ethanol, toluene, and heptane. 6) Name three important things to keep in mind while spotting TLC plates. MCAT Questions 7) Four compound I, II, III, and IV, are separated by TLC. Compound III is the most polar, II the least polar, and I and IV have intermediate polarity. The solvent system is 85:15 ethanol: methylene chloride. Which spot belongs to compound III? Geraldine Orlando ID#: C11462857 Ms. Piumi Y. Layanage & Dr. Tegan Eve CHM205-P2 th October 20 , 2015 8) What would be the effect on R valuesfif the TLC described below were run with hexane rather that ether as the eluant? 9) Which of the following correctly orders the solubility, from the greatest to the least, of NaCl in the solvents BF , H3O, a2d CH (CH )O3? 2 Data and Calculations Part A:  o­hydroxyacetophenone: = 0.31 p­hydroxyacetophenone: = 0.43 Part B: Unknown #3 – rose 0.8, 2.3, & 3.7 cm Acetaminophen – rose 2.0 cm Aspirin – rose 3.7 cm Caffeine – rose 0.8 cm Results and Discussion The retardation factor (R ) is defined as the ratio of the distance traveled by the  compound to that of the solvent front in a given time. In part A, the R fwas calculated to be 0.31 for o­hydroxyacetophenone and 0.43 for p­hydroxyacetophenone.  For part B, our unknown must have three compounds in it because there were three  different levels of spotting. The levels matched those of aspirin (3.7 cm), caffeine (0.8  cm) and acetaminophen (2.0 cm). Conclusion P­hydroxyacetophenone (p­hydrox) had a higher R fthan o­hydroxyacetophenone (o­ hydrox). This indicates that p­hydrox is less polar than o­hydrox and interacts less with the adsorbent allowing it to rise higher up the TLC plate. For part B, because our unknown matched with acetaminophen, aspirin, and caffeine.  As such, we can conclude that unknown #3 is either Excedrin, or Goody’s, because  both drugs contain all three components.  During this experiment, various errors could have occurred such as: not allowing the  TLC plate to sit in the solvent long enough, placing too large of a drop of the unknown  compound in the “lanes”, or forgetting to circle the spots when they glow under the UV  light. Not allowing the TLC plate to sit long enough in the solvent might not allow the  compounds to climb to their highest potential which could alter our observations and  results. Making the drops of the compounds too large in the lanes makes it difficult to  measure the highest point to which it climbs because the area of the spot becomes too large and it may also merge into the spot of another compound. It is necessary to  circle how high the compounds rise because after a few minutes, the plate will dry and  then you will be unable to see the compounds under the UV light anymore. Without  knowing how high the compounds rise, you will not be able to calculate the R for you  may forget which other compounds the spots aligned with. Geraldine Orlando ID#: C11462857 Ms. Piumi Y. Layanage & Dr. Tegan Eve CHM205-P2 October 20 , 2015 Post Lab Questions 1. P­hydrox has a higher melting point than o­hydrox because it can form  hydrogen bonds with other molecules. These bonds require more energy to  break because they are so strong.  On the other hand, o­hydrox exhibits steric strain from the interactions between its  hydroxyl and acetyl groups. This makes the bonds weaker and less difficult to break.  Because p­Hydrox is more polar, it creates a larger dipole­dipole moment and requires more heat to melt. 2. TLC under various conditions a, b, and c: a. The two unknowns must share similar properties, which cause their R fto be either  identical, or too similar to distinguish. A simple solution would be to use a less polar  solvent that results in more intricate, precise interactions between the compounds and  the solvents, and allows them to appear as two distinct spots.  b. Thiols and amines also share similar properties, which causes them to react with the solvent and adsorbent in the same way. Using a more polar solvent will allow the  solute to react more precisely with the solvent and adsorbent to create distinctly  separate sports. c. Allowing the solvent to cover the spots on the baseline would wash them away.  3. The solvent must be below the baseline so the capillary action of the solvent with  the plate can move the solute up.  The R fvalue is inversely proportional to the polarity of the solute. As such, the more  polar the solute is, the lower the R fvalue because it won’t climb as high on the plate.  So according to their relative polarities, and thus, their R values, the three compounds  can be arranged as: o­toluic acid < fluorenol < naphthalene. 4. Comparing R fvalues:  a. 4­decanone because it is more polar b. Xylene because it has a lower melting point c. Cycloheptane because it has more substituent groups, thus making it more polar 5. Heptane < toluene < dichloromethane < ethyl acetate < n­propanol < ethanol 6. Three important things to keep in mind while spotting the TLC plates: a. Make the spots as small as possible so they are well defined and do not merge  when climbing up the plate. b. Do not touch the TLC plate with your fingers because the oils can interfere with the  path of the solution. Geraldine Orlando ID#: C11462857 Ms. Piumi Y. Layanage & Dr. Tegan Eve CHM205-P2 October 20 , 2015 c. Draw the pencil mark lightly so that the adsorbent layer does not get torn which  would affect the capillary action of the plate. This marking also ensures that you keep  the solvent below the baseline when placing the TLC plate in the beaker MCAT Questions 7.A 8.D 9.A Experiment 5a: Simple and Fractional Distillation Data and Calculations Part A: Simple Distillation Average of Volume 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 (mL) temperatu re (°C) SF1 °C) 85 90 92 95 98 99 100 101 101 102 86.2 10 10 10 10 10 10 105.8 SF2 °C) 106 107 107 107 4 4 5 6 6 6 10 10 10 11 11 11 110.7 SF3 °C) 112 112 113 114 8 9 9 0 0 0 11 11 11 11 11 11 118.7 SF4 °C) 120 121 121 121 5 6 7 8 9 9 105.35 Part B: Fractional Distillation Average of Volume 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 (mL) temperatu re (°C) FF1 °C) 64 68 70 73 74 75 75 77 78 79 73.3 FF2 °C) 80 80 80 81 81 81 81 81 81 81 80.7 FF3 °C) 81 81 81 81 82 83 84 85 86 87 83.1 FF4 °C) 88 89 89 90 92 95 96 97 98 100 93.4 82.625 Part C: Gas Chromatography Helium Air Flow: 50 m/min Injector Temperature: ~140°C Column Temperature: ~120°C ~140°C Detector Temperature: Detector Current: 100 mA Part D: Gas Chromatography Analysis of Distillates Gas Chromatography 1 (Standard)  Area cyclohexane = (10.7) (0.4) = 4.28cm 2 2  Area toluene = (9.4) (0.3) = 2.82 cm  Correction factor cyclohexane= 4.28/2.82 = 1.52  Correction factor toluene = 2.82/2.82 = 1  Corrected area cyclohexane = 4.28/1.52 = 2.82 cm 2  Corrected area toluene = 2.82 cm 2  Cyclohexane volume % = 2.82/5.64 * 100 = 50%  Toluene volume % = 2.82/5.64 *100 = 50% Gas Chromatography 2 (SF 1) 2  Area cyclohexane = (11.6) (0.3) = 3.48cm  Area toluene = (3.5) (0.3) = 1.05 cm 2 2  Corrected area cyclohexane = 3.48/1.52 = 2.29cm  Cyclohexane volume % = 2.29/3.34 * 100 = 68.6% Gas Chromatography 3 (SF 2)  Area cyclohexane = (13.3) (0.4) = 5.32cm 2 2  Area toluene = (5.6) (0.4) = 2.24cm  Corrected area cyclohexane = 5.28/1.52 = 3.47cm 2  Cyclohexane volume % = 3.47/ 5.71* 100 = 60.7% Gas Chromatography 4 (SF 3)  Area cyclohexane = (11.3) (0.3) = 3.39cm 2  Area toluene = (7.5) (0.4) = 3cm 2 2  Corrected area cyclohexane = 3.39/1.52= 2.23cm  Cyclohexane volume % = 2.23/5.23 * 100 = 42.6% Gas Chromatography 5 (SF 4) 2  Area cyclohexane = (7.5) (0.3) = 2.25cm  Area toluene = (11.8) (0.4) = 4.72cm 2 2  Corrected area cyclohexane = 2.25/1.52= 1.48cm  Cyclohexane volume % = 1.48/6.2 * 100 = 23.9% Gas Chromatography 6 (FF 1)  Area cyclohexane = (14) (0.4) = 5.6cm 2 2  Area toluene = (3) (0.3) = 0.9cm  Corrected area cyclohexane = 5.6/1.52= 3.68cm 2  Cyclohexane volume % = 3.68/4.58 * 100 = 80.2% Gas Chromatography 7 (FF 2) 2  Area cyclohexane = (13.5) (0.2) = 2.7cm  Area toluene = (4.2) (0.4) = 1.68cm 2 2  Corrected area cyclohexane =2.7 /1.52= 1.77cm  Cyclohexane volume % = 1.77/3.46 * 100 = 51.21% Gas Chromatography 8 (FF 3)  Area cyclohexane = (11.3) (0.4) = 4.52cm 2 2  Area toluene = (6.4) (0.3) = 1.92cm  Corrected area cyclohexane = 4.52/1.52 = 2.97cm 2  Cyclohexane volume % = 2.97/4.89 * 100 = 60.7% Gas Chromatography 9 (FF 4) 2  Area cyclohexane = (6.4) (0.2) = 1.28cm  Area toluene = (12.4) (0.4) = 4.96cm 2  Correction factor cyclohexane= 1.28/1.52= 0.84 2  Corrected area cyclohexane= 1.28/1.52= 0.84cm  Cyclohexane volume % = 0.84/5.8 * 100 =14.5 % Discussion From the experiments performed in this lab session, we saw that in simple distillation, the compounds of the mixture were separated by evaporation and condensation of the liquids. The temperature range for this process was wider than for the fractional distillation. In the case of fractional distillation the process took longer to start, but once it started to condense it went faster. The temperature range was smaller, and the average from the overall process was closer to the theorical boiling point value for cyclohexane, which is approximately 80°C. Also according to the results obtained from the calculation for volume percentage, we may say that there was some source of error at the moment of the distillation or when the gas chromatography took place, because the percentage has a sudden increase instead of a continuous decrease in cyclohexane percentage, like shown in the simple distillation graph of average temperature vs volume percent. Conclusion We can conclude that fractional distillation is more efficient than simple distillation to separate the compounds in our mixture, because the volume percent of cyclohexane was more in the first fraction compared to the simple distillation fraction. Also by using Gas Chromatography we were able to measure more accurately the actual amount of each compound in all of the fractions, and that way we were able to make a graphical representation of the percent decrease in cyclohexane, which was expected because it was the first compound to evaporate from the mixture, because it has a lower boiling point. Post Lab Questions 1. Refer to MtC Figure 3-17 to answer the following questions. 100% benzene  0% benzene 0% toluene  100% toluene a. What is the % composition of the vapor in equilibrium with a boiling liquid that has a composition of 30% benzene and 70% toluene? The composition percent of the vapor will be 6% benzene and 94% toluene. b. If a sample of vapor is found to have a composition of equal parts benzene and toluene, what is the composition of the boiling liquid that produced this vapor? The composition of the boiling liquid is 17% benzene and 83% toluene. 2. a. Explain the meaning of the horizontal line AB in Figure 3-17. The line in the graph represents the process of going from liquid to vapor. A refers to the 50/50 mixture of benzene/toluene, which is then heated and B represents the vapor phase values. b. What physical transformation does the line DE represent in Figure 3-17? After the condensation process, point D gives liquid E, and this liquid is richer in benzene than C. 3. Again referring to Figure 3-17, what is the approximate boiling point of a liquid that is 20% benzene and 80% toluene? The approximate boiling point is 100.62°C 4. Under what conditions can a good separation be achieved by simple distillation? The conditions to get a good separation using simple distillation can only happen if there is a large difference in the boiling points of the liquids that make up the mixture. 5. The table below has the vapor pressures of cyclohexane and toluene at various temperatures. Cyclohexane Toluene Temp mmHg Temp mmHg (°C) (°C) 30 122 30 37 40 185 40 59 50 272 50 92 60 389 60 139 70 544 70 204 80 743 80 291 90 996 90 407 100 1311 100 556 110 747 a. Calculate the mole fraction of each component in solution if you mix 8.4 g of cyclohexane and 9.2 g of toluene. Cyclohexane= 8.4 g =0.0998mol 84.15948 g/mol 9.2g Toluene= =0.0998mol 92.13842g/mol 0.0998mol NC= 0.1996 =0.5 N = 0.0998mol=0.5 T 0.1996 b. Assuming that this is an ideal mixture that follows Raoult’s las, calculate the partial vapor pressure of cyclohexane in the mixture at 50°C ∂ pressure=0.5∗272 mmHg=136mmHg c. Estimate to the nearest degree the boiling point temperature of this solution. 760torr 1atm mmHg→atm=743mmHg∗ ∗ =0.978atm (760mmHg )( 760torr) The closes boiling point temperature of the solution is 80°C. 6. Explain why the boiling point of your cyclohexane:toluene mixture rose slowly throughout your simple distillation procedure. As distillation takes place, the vapor becomes richer in the more volatile compound, which is cyclohexane, the liquid remaining in the flask is enriched in toluene. When the composition percent of toluene increases, the boiling point increases too. MCAT Review Question 7. Fractional distillation would most likely be used to separate which of the following compounds? c. Aniline (boiling point of 184°C) and benzyl alcohol (boiling point of 22°C) EXPERIMENT 5B: STEAM DISTILLATION Results Cloves initial mass: 1.03g Mass of Eugenol recovered: 0.374g % Recovery: 36.31% Results In the experiment performed during this lab session, we were able to separate the natural oil Eugenol from the mixture using steam distillation. The condensate liquid from the steam distillation was collected and then separated by using different methods of isolation, to end only with the natural oil. The percent recovery was low, but we need to take into account that the initial weight had also other parts from the clove, not only eugenol. From out IR analysis we got a graph that did not look very similar to the pure eugenol IR analysis. Our graph was more similar to the student example shown in the lab manual, which has a big difference in the first part of the graph. However, the final part of the graph has numbers that are close to the ones found in the pure eugenol analysis. This difference between the IR of pure eugenol and our result may be due to the presence of other compounds in our final distillate. Conclusion Steam distillation is a good process to separate compounds with high melting points. In the case of eugenol, this was the best method to separate it, because it has a boiling point of 250°C. For that reason, simple and fractional distillation cannot be used in this case. Also to get a more accurate percent recovery of eugenol, we should know the different compounds that are found in cloves, and their proportion to get a better initial mass for pure eugenol. Post-lab Questions 1. Why do we use steam distillation to isolate eugenol rather than purify it by simple distillation? We used steam distillation to isolate eugenol because its boiling point is 250°C, if we used a simple distillation process it would most likely decompose.   2. Given that for two immiscible liquids like those used in this experiment (an oil and water). Calculate the following for and oil that distills during steam distillation at a boiling temperature of 98°C at 1 atm of pressure: a. The mass of the oil that co-distills with each one gram of water at 98°C VP H2O 701mmHg VP Oil9mmHg VF oil.078 VF H2O= 0.92 1g water = 0.056 moles water Total moles vapor = 0.061 Moles of oil in vapor = 0.061 x 0.078 = 0.0047 moles oil Mass of oil that co-distills = 0.0047mole x 169 g/mole = 0.8g b. Suppose you had two grams of a spice that contained 5% by mass of this oil. How much water would be required to recover all of the oil by steam distillation? Amount of oil = 0.00059 moles Mole fraction of oil in vapor is 0.078 Total moles vapor = 0.007586 Moles of water= 0.007 moles water Grams of water = 0.007 moles x 18 g/mole = 0.13g water 3. Pure octane has a boiling point of 125.7°C, but can be steam distilled with water at a temperature of 90°C. Calculate the mass of octane that co-distills with each gram of water and the percent composition of the vapor that is produced during the steam distillation. VP = 517 mmHg H2O VP octane 243 mmHg VF octane.320 VF = 0.68 H2O 1g water =0.056 moles water Total moles vapor = 0.082 Moles of octane in vapor = 0.0262 moles octane Mass of octane that co-distills = 0.0262mole x 114.23 g/mole = 3.0g VF octane243/760 * 100 = 32% VF H2O= 517/760 * 100 = 68% 4. One synthetic preparation of aniline involves reducing nitrobenzene with iron and HCl, and then steam distilling the resultant aniline at 99°C. How much aniline co-distills with each gram of water during the steam distillation? VP H2O= 728 mmHg VP aniline32 mmHg VF aniline.042 VF H2O= 0.96 1g water = 0.056 moles water Total moles vapor = 0.058 Moles of aniline in vapor = 0.058 x 0.042 = 0.0024 moles aniline Mass of aniline that co-distills = 0.0024mole x 93.13 g/mole = 0.23g 5. Nitration of phenol can be accomplished by using nitric acid at room temperature, BUT the result is a mixture of p-nitrophenol and o- nitrophenol. Luckily, the para and ortho isomers can be separated by steam distillation because the o-nitrophenol is steam volatile but the p- nitrophenol is not. Explain why this is. Because o-nitrophenol is more polar, therefore react more with H 2 molecules, and can this facilitates the steam distillation process.   Experiment 4: Making Polymers Results & Discussion Polymer Acetone Toluene Ethanol Polystyrene Soluble Soluble Insoluble Nylon 10,6 Insoluble Insoluble Insoluble Nylon 10,6 Insoluble Insoluble Insoluble excess Slime Insoluble Insoluble Insoluble This experiment had 3 parts. First we mixed styrene and benzoyl peroxide in order to create polystyrene. The substance was placed on a test tube and on a boiling water bath for almost an hour. Then we took the hard content out by breaking the test tube. After we got it, we analyze it properties, and we got a hard compound similar to plastic that breaks like hard candy. We tested the solubility of this compound in acetone, toluene and ethanol. We obtained the results stated above. For part 2 of the experiment, we created a Nylon 10,6 was created by mixing NaOH, hexamethylene diamine, and sebacoyl chloride. The nylon appeared between the layers, and we extract part of it by placing it around of a test tube. Then we mixed the remaining content and got a big clump of nylon. We analyze the solubility of both nylons and we got that both of them are insoluble in the three solvents (Acetone, toluene, and ethanol). Last part consisted on creating a slime compound, which was made by combining poly (vinyl alcohol) and borax. The mixture was stirred with a stirring rod until it had a gooey appearance. Theit was taken out of the beaker to examine the properties and solubility. We got that the slime obtained was elastic, green and had bubbles inside. Also we found that it is insoluble in the three solvents, and the only difference they showed was a decrease of the color intensity. Conclusion From this lab we were able to perform three different types of polymerization, which are chain addition, condensation and cross-linked. The observations made in this experiment showed us that the only soluble compound formed during the lab session was polystyrene, but it was insoluble on ethanol. The other two compounds created showed insolubility in all of the cases. Post Lab Questions 1. Define the following terms – you may need to consult your lecture text or other suitable source a. Monomer: Molecule that can be bound to another molecule to form a polymer b. Repeating unit: part of a polymer that when repeated will create a complete polymer chain by linking the repeated units together. c. Condensation polymerization: Type of polymerization where molecules are bound together and lose a small molecule as by- product. d. Cross-linked polymer: this polymerization process promotes a difference in the resulting polymers’ properties. 2. You synthesized Nylon- 10, 6, using interfacial polymerization. Draw a representation of what your experiment looked like. Clearly label the contents and identity of each layer. 3. Most organic polymers are named based on the monomers used in their production. However poly (vinyl alcohol) (PVA) is not made from vinyl alcohol, but instead from vinyl acetate that is hydrolyzed after formation of the polymer. Why can’t we use vinyl alcohol in this polymerization? Draw the structure of vinyl alcohol to help you answer this question. Vinyl alcohol cannot be used in this polymerization process because it will form an unstable radical under the reaction. For this reason we use vinyl acetate, to create an stable polymer 4. Vinylidene chloride, CH =2CCl is2copolymerized with vinyl chloride to make Saran Wrap. Write a structure that includes at least two units for the copolymer that is formed. 5. Isobutylene, CH =C (2H ) , is3 2ed to prepare cold-flow rubber. Write a structure for the addition polymer formed from this alkene 6. Maleic anhydride reacts with ethylene glycol, HOCH CH OH, to produce 2 2 an alkyl resin. Write the structure of the condensation polymer produced.


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