CHM205/Organic Chemistry Lab 1 (UM)
CHM205/Organic Chemistry Lab 1 (UM) CHM205
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Table of Contents n a a o Organic Chemistry I CHM Z05 0 Pipers and Melting Points 5 I Recrystallization ll 2 Extraction 19 3a Thin Layer Chromatography 31 3b Column Chromatography 39 4 Making Polymers 45 5a Simple and Fractional Distillation 55 Sb Steam Distillation 65 6 Resolution of Enantiomers 73 Organic Chemistry II CHM 206 7 SN1 and SN Reactions 87 8 The DielsAlder Reaction 93 9 The El Reaction Synth of Cyclohexene 99 10 Nitration of Methyl Benzoate A Macroscale Synthesis 109 Ma The Fischer Esteri cation Synth of an Ester 117 HI ID of an Unknown Ester Transesteri cation and hp 123 I2 OFF an Insect Repellent 129 I3 The Grignard Reaction Synth of Benzoic Acid 139 14 Aldol Condensation 147 I5 Oxidative Cleavage Synth of Adipic Acid quotS155 IR Infrared Spectroscopy An Exercise 16739 NMRa Nuclear Magnetic Resonance Spectroscopy Exercise 179 NMRb Nuclear Magnetic Resonance Spectroscopy ASecondVisit Y S 191 Experiment 1 Recrystallization IntroductionTheory For Part A we used 050 grams of unknown 9 We added a total of 105 mL of hot water After the crystals were formed and washed they weighed 0142 grams We divided 0142 grams by 050 grams which yielded 0284g which meant the percent recovery was 284 We found the melting point range of the pure crystals when tested quickly to be around 120 C A second slow test ranged between 125 130 C This led us to believe that our unknown was either Benzamide or Benzoic acid When we found the mixed melting point of our unknown with Bezoic acid we found that the melting point was 105 110 C which was substantially lower than the literature range of 121 122 C However when we performed the mixed melting point with Benzamide we got a range of 126 130 C which was marginally lower than its literature value of 127 130 C Therefore we believe that our unknown was Benzamide For Part B we used 0075 grams of our unknown We recovered 0069 grams of pure crystals We divided 0069 g by 0075 g which yielded a percent recovery of 92 The melting point of the crystals alone showed a range of 120 128 C Procedure See Carbon Copy for Procedural steps Calculations PartA Mass of unknown 050 g Mass of filter paper 0896g Mass of filter paper with crystals 1038g Mass of crystals 0142 g Recovery macro 014205 x 100 284 PartB Mass of unknown 0075 g Mass of Craig Tube 5872 g Mass of Craig Tube with crystals 5941 g Mass of crystals 0069 g Recovery macro 00690075 x 100 92 Discussions amp Error Analysis For Part A the weight of our unknown was largely wrong some of the compound was lost during the transfer from the weigh paper to the ask The crystals may not have been completely dry when we weighed them Both of these errors have definitely affected the percent recovery For Part B again the original weight could have been wrongly measured and the crystals may not have been dried Some of the substance could have also been lost when we were transferring the crystals to the weigh paper The difference in the error margin definitely points out the transfer error we are aware to have happened in Part A The procedure of Part A also opens up many more instances of error to be made by the experimenter in comparison to Part B This explains the wide difference between our error margin between 284 and 92 Furthermore in terms of the melting point analysis we used the fast and slow testing method for each mixed melting point in order to give two trials for each test and make the temperature results as accurate as possible In this way we were able to be more sure with our determination of which substance unknown 9 happened to be Conclusion It is hard to say if the experiment was successful without knowing the true identity of our unknown However even with the loss of substance in Part A the results of the mixed melting point were very clear It was easy to follow and understand the purpose of every step of the recrystallization The skills we leamed in this experiment will be helpful in any later experiments in which we will need to perform recrystallization This experiment has successfully shown two different procedures to recrystallization and has taught us the importance of procedural precision with our mistakes in Part A Luckily in respect to the experimental goals of this experiment this did not jeopardize the classi cation of our substance however we are now aware that these mistakes cannot be repeated Post Lab Questions Exp 0 l De ne melting point range based on the two Values that are used to describe a compound s melting point The melting point range is the range of temperatures at which a solid will become a liquid The first temperature indicates where the intermolecular forces begin to break and the last temperature indicates the point at which all of the forces have broken 2 What is the proper heating rate used to determine a compound s melting point range What do you think would happen if you exceeded this heating rate The proper rate of heating is l2 Cminute If you exceed this rate it would be dif cult to accurately determine the melting point range because melting would happen so quickly 3 Suppose you are trying to determine the melting point of a compound and the following happens Describe what happened in each case How would you either report or correct this a The compound tums from white to brown before melting If the compound tums from white to brown before it melts this probably means it is decomposing into a different product first which is indicated by the color change b The compound slowly disappears from the capillary tube before melting The compound often shrinks or appears to shrink before melting It should melt normally after this 4 Find literature melting points for the following pairs of compounds a Ferrocene about 170174 C and Acetylferrocene 8183 C b Adipic acid 151153 C and Citric acid 153155 C 5 Adipic acid and citric acid have very similar melting points Given two unknown samples one being citric acid and the other adipic acid how could you determine which was which without using spectroscopy Explain in detail You could determine the melting temperatures of the two samples individual using the MelTemp The sample with the lower melting point range would be adipic acid Or you could take a sample of what you know to be adipic acid and mix it with one of the unknowns Then determine the melting point of the mixture If the melting point range is not lowered or broadened that means the unknown used in this mixture was adipic acid The same could be done for citric acid 6 Using your data for Pasteur pipets from Part A of Experiment 0 and literature density if needed explain how you would a Add about 08 mL of chloroform to a vial The easiest way to do this would be to add 08 mL to a graduated cylinder Then simply transfer the liquid to the vial using a pipet plastic or glass with a bulb If you were simply estimating the amount you could use a glass pipet with 1 mL and 05 mL marks on it Then simply fill the pipet with liquid to a point that appears to be 08 mL b Add about 12 grams of methanol to a vial Weigh an empty clean and dry graduated cylinder Add 50 drops of methanol to the cylinder and weigh it again Subtract the first weight from the second weight to find the mass of ethanol Divide this mass by 50 to find the weight of 1 drop of methanol Use this to determine how many drops are needed to get 12 grams of methanol c Determine the volume and mass of 38 drops of methylene chloride CHgCl2 Weigh a clean and dry 10 ml graduated cylinder Add 38 drops of methylene chloride Weigh the cylinder again Subtract the first weight from the second weight to get the mass of the 38 drops Find the volume by reading the graduated cylinder PostLab Questions Exp 1 L The solubility of an unknown compound X in ethanol is found to be 021 g per 10mL at 0 C and 14 g per 10mL at 78 C What is the minimum amount of ethanol needed to recrytallize a 20 g sample of compound X After recrystallization of the 20 g sample how much of the compound X will remain dissolved in the cold solvent ie How much of compound X is not recovered after the recrystallization x mLx 1429 mL 1429 mL x 030009 g The minimum amount of ethanol needed is 1429 mL and 030009 grams would remain soluble 2 Why is it important to use only the minimal amount of hot solvent needed to achieve a recytallization What would be the effect of using too much solvent while dissolving a solid for recrystallization The addition of too much solvent would hinder later recrystallization and cause the loss of product The solute would still be soluble at lower temperatures which are used when cooling the solution for crystals so more solute would remain in solution with the solvent 3 What do you think would happen to your recrytallization solution if you did not use a preheated lter during the hot gravity ltration step If we had not used a preheated lter during the hot gravity ltration step I think that crystals would have formed when the liquid came in contact with the room temperature lter paper and cooled The crystals would be too large for ltration so there would be a loss of product The slower crystals form the more pure they will be The rapid formation of crystals from contact with the lter would mean they were less pure 4 Benzyl alcohol has a bp of 205 C It has a solubility characteristics that would seem to make it a good choice for use as a recrytallization solvent for urorenol mp 152 155 C but in fact benzyl alcohol is a poor choice of solvent for this recrystallization Why do you think this is Hint you do not need any other info to answer the question Benzyl alcohol is not a good choice for a solvent in this case because its boiling point is higher than uorenol s melting point This means that by the time benzyl alcohol reached its boiling point the uorenol would have already melted 5 Explain how each of the following contaminants is removed during the recrystallization process a Sand inorganic contaminant Sand would be removed through hot gravity ltration because it is an insoluble impurity It would not dissolve in the solution and therefore would not be able to pass through the lter paper b Colored organic contaminants Colored organic contaminants are removed by adding charcoal and then performing hot gravity ltration The colored organic contaminants are strongly absorbed on the surface of the carbon in charcoal The charcoal would not pass through the lter during ltration so neither would the colored organic contaminants c A very soluble contaminant A very soluble contaminant would be removed by heating the solvent to near boiling which would allow the solute and contaminant to dissolve Then the solution would be ltered and slowly cooled at which point the solubility of the compound would decrease the solution would become saturated with the compound and it would begin to crystallize If the crystallization happens slowly contaminants will be excluded 6 Your TA may have told you to use acetone to clean your glassware after you nished your experiment Acetone is a very popular solvent for cleaning glassware Given the properties of acetone listed in MtC why do you think this is DO you think acetone would make a good recrystallization solvent Why or why not From a safety standpoint what do you suppose would be a mai or concem when using acetone Acetone is able to dissolve most organic compounds and is a polar solvent It will be able to clean the glassware by dissolving the organic and polar substances used in the lab It is not a good recrystallization solvent because of its low boiling point If it was heated it would quickly vaporize and leave behind the impure solute in the ask Acetone could be a safety hazard because it is very ammable Even its vapor could ignite if it came in contact with heat Experiment 1 Recrystallization IntroductionTheory Initially the experimenter tested his skills of measuring mass and volume through the process of determining the average number of drops in 1 mL of H20 which is approximately drops However this result is susceptible to human error by the fact that measuring a volume such as this with the human eye is difficult In Part B of the experiment the experimenter had to calibrate a Pasteur pipet create a filtering pipet and create a filtertip pipet This procedure introduced the experimenter to the process of preparing both a ltering pipet and a filtertip pipet and rinsing each of these pipets with methanol and then hexane For Part C Procedure See Carbon Copy for Procedural steps Data amp Observations For Part A we used 0508 grams of unknown 9 We added a total of 105 mL of hot water After the crystals were formed and washed they weighed grams We divided 0435 grams by 0508 grams which yielded 08648g which meant the percent recovery was 8648 We found the melting point range of the pure crystals to be 139 142 C This led us to believe that our unknown was either 2Chlorobenzoic acid or Salicylamide When we performed the mixed melting point with salicylamide we got a range of 138142 C which was lower than its literature value of 140144 C When we found the mixed melting point of our unknown with 2Chlorobenzoic acid we found that the melting point was 139141 C which was nearly the same as the literaturerangeofl38 140 C Therefore we believe that our unknown was 2Chlorobenzoic acid For Part B we used 0073 grams of our unknown We recovered 0036 grams of pure crystals We divided 0036 g by 0073 g which yielded a percent recovery of 493 Discussions amp Error Analysis For Part A the weight of our unknown could have been wrong some of the compound could have been lost in the transfer from the weigh paper to the ask The crystals may not have been completely dry when we weighed them Both of these errors would have affected the percent recovery It is also possible that our melting point range was off we could have heated the compounds too quickly For part B again the original weight could have been wrong and the crystals may not have been dried Some of the substance could have also been lost when we were transferring the crystals to the weigh paper Conclusion It is hard to say if the experiment was successful without knowing the true identity of our unknown However there were not any major mistakes and the results of the mixed melting point seem to be pretty clear It was easy to follow and understand the purpose of every step of the recrystallization The skills we leamed in this experiment will be helpful in any later experiments where we will need to perform recrystallization We are now more comfortable with the process Post Lab Questions Exp 0 l De ne melting point range based on the two Values that are used to describe a compound s melting point The melting point range is the range of temperatures at which a solid will become a liquid The first temperature indicates where the intermolecular forces begin to break and the last temperature indicates the point at which all of the forces have broken 2 What is the proper heating rate used to determine a compound s melting point range What do you think would happen if you exceeded this heating rate The proper rate of heating is l2 Cminute If you exceed this rate it would be dif cult to accurately determine the melting point range because melting would happen so quickly 3 Suppose you are trying to determine the melting point of a compound and the following happens a the compound tums from white to brown before melting b the compound slowly disappears from the capillary tube before melting Describe what happened in each case How would you either report or correct this a If the compound tums from white to brown before it melts this probably means it is decomposing into a different product first which is indicated by the color change b The compound often shrinks or appears to shrink before melting It should melt normally after this 4 Find literature melting points for the following pairs of compounds a ferrocene about 170174 C and acetylferrocene 8183 C b adipic acid 151153 C and citric acid 153155 C 5 Adipic acid and citric acid have Very similar melting points Given two unknown samples one being citric acid and the other adipic acid how could you determine which was which without using spectroscopy Explain in detail You could determine the melting temperatures of the two samples individual using the MelTemp The sample with the lower melting point range would be adipic acid Or you could take a sample of what you know to be adipic acid and mix it with one of the unknowns Then determine the melting point of the mixture If the melting point range is not lowered or broadened that means the unknown used in this mixture was adipic acid The same could be done for citric acid 6 Using your data for Pasteur pipets from Part A of Experiment 0 and literature density if needed explain how you would a add about 08 mL of chloroform to a Vial The easiest way to do this would be to add 08 mL to a graduated cylinder Then simply transfer the liquid to the Vial using a pipet plastic or glass with a bulb If you were simply estimating the amount you could use a glass pipet with 1 mL and 05 mL marks on it Then simply fill the pipet with liquid to a point that appears to be 08 mL b add about 12 grams of methanol to a Vial Weigh an empty clean and dry graduated cylinder Add 50 drops of methanol to the cylinder and weigh it again Subtract the first weight from the second weight to find the mass of ethanol Divide this mass by 50 to find the weight of 1 drop of methanol Use this to determine how many drops are needed to get 12 grams of methanol c determine the Volume and mass of 38 drops of methylene chloride CH2Cl2 Weigh a clean and dry 10 ml graduated cylinder Add 38 drops of methylene chloride Weigh the cylinder again Subtract the first weight from the second weight to get the mass of the 38 drops Find the Volume by reading the graduated cylinder PostLab Questions Exp 1 L x mLx 1429 mL 1429 mL x 030009 g The minimum amount of ethanol needed is 1429 mL and 030009 grams would remain soluble A The addition of too much solvent would hinder later recrystallization and cause the loss of product The solute would still be soluble at lower temperatures which are used when cooling the solution for crystals so more solute would remain in solution with the solvent 3 If we had not used a preheated lter during the hot gravity ltration step I think that crystals would have formed when the liquid came in contact with the room temperature lter paper and cooled The crystals would be too large for ltration so there would be a loss of product The slower crystals form the more pure they will be The rapid formation of crystals from contact with the lter would mean they were less pure i Benzyl alcohol is not a good choice for a solvent in this case because its boiling point is higher than uorenol s melting point This means that by the time benzyl alcohol reached its boiling point the uorenol would have already melted 1 a Sand would be removed through hot gravity ltration because it is an insoluble impurity It would not dissolve in the solution and therefore would not be able to pass through the lter paper b Colored organic contaminants are removed by adding charcoal and then performing hot gravity filtration The colored organic contaminants are strongly absorbed on the surface of the carbon in charcoal The charcoal would not pass through the lter during filtration so neither would the colored organic contaminants c A very soluble contaminant would be removed by heating the solvent to near boiling which would allow the solute and contaminant to dissolve Then the solution would be filtered and slowly cooled at which point the solubility of the compound would decrease the solution would become saturated with the compound and it would begin to crystallize If the crystallization happens slowly contaminants will be excluded 6 Acetone is able to dissolve most organic compounds and is a polar solvent It will be able to clean the glassware by dissolving the organic and polar substances used in the lab It is not a good recrystallization solvent because of its low boiling point If it was heated it would quickly vaporize and leave behind the impure solute in the ask Acetone could be a safety hazard because it is very ammable Even its vapor could ignite if it came in contact with heat Experiment 2 Extraction IntroductionTheory To isolate and leam proper techniques of liquidliquid extraction Additionally further advancing our grasp on lab safety ad chemical interactions Procedure See Carbon Copy for Procedural steps Calculations Unknown 16 Weight of Acid 0037 g Fast Melting Point 119129 C Unknown Benzoic acid 113121 C Unknown transcinnamic acid 108 C Therefore the acid is identified as Benzoic acid Weight of Phenol 0052g Fast Melting Point 7172 C Unknown I3naphthol 6976 C Unknown 4tbutylphenol 8082 C Therefore the phenol is identified as 3naphthol Weight of Neutral Compound 0037 g Fast Melting Point 7173 C Unknown 9 uorene 6465 C Therefore the neutral compound is identified as 9 fluorene Acid KD CORGCAQU CHTBE CH20 W1 Tare mass with chips 18968 g W Mass after drying 19068 g W2W119068 18968 0100 g KD CHTBE CH20 0102 1913 785 Recovery Acid 0037 0150 X 100 2466 Recovery Phenol 0052 0150 X 100 3466 Recovery Neutral Compound 037 0150 X 100 24666 Discussion and Error Analysis There were many places for error in this eXperiment We could have weighed incorrectly or lost some of the sample when transferring it We may not have completely removed the layers that we were supposed to Our products may have been wet which would result in incorrect percent recoveries Our percent recoveries were quite low which may indicate some kind of error Something must have been done incorrectly because we recovered very little solid substances of phenol and neutral compounds were very minimal which made the testing of the substances limited with miXed boiling points Conclusion This was a complicated and lengthy eXperiment that left a lot of room for error It was important to follow the directions careful and be vigilant throughout the process Our percent recoveries for the acid and phenol were substantial with which we were pleased with however our recovery of the neutral compound did not make any sense and we are unsure as to where the error in recovery occurred Overall however testing the melting points for each substance was made difficult due to the differences with the published values which proved the presence of substantial impurities that presented certain doubts for identifying the substances Regardless this eXperiment was a good way to practice the skills and procedures that will be performed in this laboratory PostLab Questions Exp 2 l Truamatic acid has a distribution coefficient of 58 between MtBE and water If 75 mg of traumatic acid is added to a centrifuge tube containing 3 mL of water and 2 mL of MtBE how much traumatic acid would be in each layer after through mixing KD Z CMtBeCH20 30mL x00l54 g Acid in water layer 0075x00596 g Acid in MtBE layer 2 Given a solution of 10 mg of traumatic acid dissolved in 100 mL of water and using the partition coefficient given for traumatic acid in Qul show that extracting the 100 mL of aqueous solution with two 10 mL portions of MtBE would recover more of the traumatic acid from the aqueous solution than one extraction using 20 mL of MtBE Procedure 1 two I 0 mL MtBE extractions KD Z CMtBECH20 I 00mL I 00mL x000632 g in aqueous layer x000400 g in aqueous layer 0010x000368 g extracted 000632x000232 g extracted Extracted 6 mg of acid percent recovery 60 Procedure 2 one 20 mL MtBE extraction 1 00mL x000462 g in aqueous layer 0010x00053 8g extracted Percent recovery54 Procedure I recovers more acid 3 Give an equation for each of the organic salts below showing how you would recover the parent compounds 4 Suppose that just before adding a drying agent to an organic solent that was used to extract an aqueous solution you notice that there are still tiny water droplets in the organic layer Can you proceed with adding drying agent What should you do You cannot proceed with the drying agent because there is more water present than the drying agent is capable of handling It can only remove miniscule amounts of water The water droplets would be too much You should use a pipet to remove the droplets Some of the organic layer may be taken out with the droplets If this happens allow the layers to separate in the pipet and then remove the dense aqueous layer from the pipet leaving only the organic layer This can then be retumed to the original container 5 Devise a general extraction scheme for separating the following pairs a an organic base mixed with an organic neutral compound BN First you should dissolve the base and neutral compound in a solvent like MtBE Then a strong acid is added to the solution and the solution is shaken This should result in a separation of an aqueous and organic layer The aqueous layer can be removed and stored in a separate container This extraction step should be repeated twice more Next a strong base like NaOH should be added to cause a precipitate to form Then the mixture can be vacuum filtrated and dried which results in a sample of organic base The organic solution should be put into a tared conical vial with a boiling chip and then placed in a hot sand bath with air ow until the liquid organic solvent boils off This will leave only the neutral compound which should be weighed to find the mass and percent recovery Finding the melting point and performing mixed melting points will determine the identity of the compound b an organic acid HA and a phenol ArOH Dissolve the mixture in an organic solvent like MtBE Then add a weak base which will deprotonate the acid and make it soluble in the aqueous layer The solution should be shaken and the layers should be allowed to separate Remove the aqueous layer and repeat the extraction twice more A strong acid like HCl should be added to the aqueous solution which will result in a precipitate A vacuum filtration should be performed and the precipitate can be dried leaving a sample of the organic acid A strong base should be added to the organic layer allowing it to be solubilized in aqueous solution repeat this twice more HCl should be added to this which results in a precipitate Filter and dry this precipitate leaving a sample of phenol The percent recovery can be found by weighing the sample and melting points can be performed to determine the identity of the compounds 6 When 2 immiscible solvents are mixed 2 layers form Liquidliquid extractions commonly use organic solvents that are less dense than aqueous solvents that are less dense than aqueous solutions and therefore form the top layer An important exception to this rule is chlorinated solvents which are often more dense than aqueous solutions and form the bottom layer Suppose you were unsure which layer was which during your extraction What simple method do you think you could use to determine which of the 2 layers was the aqueous You could extract some of one layer using a pipet and add it to a container of water If it dissolves in water then it is the aqueous layer if it does not then it is the organic layer 7 Potassium carbonate is an excellent drying agent but it should not be used with some classes of organic compounds Would it be a better choice to use in drying an ether solution containing and acid RCOOH or a base RNHg Why Since potassium carbonate is an alkaline base it may react with the acid RCOOH instead of drying it It is a better choice for a solution containing a base RNH2 because it will not react with it MCAT Practice Test Questions 8 a Soil contains organic matter called humus Humus is classi ed into humic and nonhumic materials Humic materials are operationally divided into 3 main fractions humic acid fulvic acid and humin Fraction B is thus more likely Fraction B is most likely humic acid because it is only soluble at alkaline pHs So it would be soluble in NaOH and insoluble in HCl like Fraction B 9 b A mixture of sand benzoic acid and naphthalene in ether is best separated by Filtration would remove the sand which is not soluble The benzoic acid should be removed next using basic extraction because naphthalene is neutral It should be removed last using evaporation 10 a Suppose an extraction with methylene chloride dl4gmL is performed with the desired compound initially in brime d 1 gmL In a separatory funnel which layer will be the organic layer Bottom layer Since brine is a salt water mixture that means that it is the aqueous layer which makes methylene chloride the organic layer Since methylene chloride has a higher density it will be the bottom layer Experiment 3a Thin Layer Chromatography IntroductionTheory To leam chromatographic separation of compounds based on polarity Additionally to improve out TLV techniques spotting developing visualizing Rf calculation and how to use TLC 0 identify the components of an OTC pain killer Thin layer chromatography is a technique used to separate mixtures and gives a quick answer to how many compounds are in a mixture Thin layer chromatography can be performed on a sheet of glass metal or plastic coated with a thin layer of solid absorbent Procedure See Carbon Copy for Procedural steps Data amp Observations Part A TLC Part B TLC For our first TLC Plate we had one spot of orthohydroxyacetophenone left one spot of parahydroxyacetophenone right and one spot that was a mixture of the two centre After performing thin layer chromatography and putting the plate under the UV light we found that the ortho compound traveled the farthest the least soluteadsorbent interactions and the para compound traveled the least the most interactions The spot that was a mixture of the two separated so that there was one dot that was the same distance as the ortho and one that was the same distance as para We found that the Rf for ortho was 050 and the Rf for para was 036 In Part B we had one TLC plate with a spot of from left to right a spot of acetaminophen caffeine our unknown acetylsalicylic acid and ibuprofen After performing thin layer chromatography and putting our plates under the UV light and then within iodine we found that our unknown sample matched both acetaminophen and caffeine both of which came out highly concentrated in the iodine staining At this point the compounds present narrow the unknown to either Goody s or Exedrin brands Comparing the relative amounts between the acetaminophen and caffeine the caffeine showed lighter iodine staining whereas the acetaminophen showed very clear dark spots The unknown was thus identified as being Goody s due to its lower amounts of caffeine which matched out TLC plate results Calculations Ortho traveled 44 cm solvent traveled 88 cm Rf 4488 050 Para traveled 32 cm Rf 3288 036 Discussion and Error Analysis This was a relatively simple experiment The procedure was very straight forward and did not require too much attention to detail or precision That being said mistakes are still possible by using two much solvent and emerging the baseline thus dissolving the solutes not correctly labeling the solutes being spotted and incorrectly measuring the travel distances for the Rf values and not making our spots concentrated enough just to name a few From our clear results it does not seem that any of these mistakes are prevelent in our experiment The only slight discrepancy during the observation of our Part B results was that the pure acetaminophen compound spot shifted into the column of the caffeine compounds initially confusing our reading of the TLC plate as shown with an arrow in the second Part B image Once that was clari ed the results were very clear and successful in isolating the unknown compound Conclusion This experiment made the difference between RfVa1ueS visibly clear using the UV light and iodine It was easy to see that more polar solutes would have more interactions with solvent and would therefore not travel very far meaning that they would have lower Rf values Conversely less polar solutes would have less binding interactions with the solvent and would be able to travel farther resulting in higher Rf values It was also relatively easy to identify our unknown It would have been more difficult if our unknown had been a mixture of multiple unknowns but it still would have been possible to identify it because the compounds would have separated on the TLC plate and the individual spots could have been compared to the known compounds PostLab Questions Exp 3 1 oHydroxyacetophenone has a melting point of 46 C pHydroxyacetophenone has a melting point of 109111 C Explain the basis for the sizable difference in melting points of these two compounds The difference between the melting points of oHydroxyacetophenone and p Hydroxyacetophenone is explained by their structure and intermolecular forces P Hydroxyacetophenone has a higher dipole moment which means that it is more polar Higher polarity means there are stronger intermolecular forces which elevate the melting point 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 2 unknown halogenated alkenes but you see only one spot with a Rf of 015 The solvent used was ethyl acetate Since the maximum Rf value is 10 an Rf value of 091 is considered very high and this means that the compound traveled much farther than the solvent This indicates that the solvent was too polar A way to correct this problem would be to use a less polar solvent so that the soluteadsorbent interactions increase and the solutes move more slowly b You run TLC on a mixture of a thiol and an amine and again you get only one spot with a Rf of 015 The solvent was a mixture of petroleum ether and dichloromethane An Rf value of 015 means that the compound traveled slowly because of the soluteadsorbent interactions To fix this you could use a more polar solvent c When you put your TLC plate in the developing chamber the solvent in the chamber covered the spots on the TLC plate baseline If the solvent covers the spots this means that the solute is going to dissolve in the solvent To fix this you can use less solvent or raise the baseline on the TLC plate 3 You run a TLC plate spotted with 3 compounds naphthalene otoluric acid and uorenol Predict the relative Rf values Napthalene would have the highest Rf value because it would have the least amount of soluteadsorbent interactions because it is an unsaturated hydrocarbon Fluorenol would have the next highest value because it would have a lower amount of soluteadsorbent interactions Otoluic acid would have the lowest Rfvalue because it would have the most soluteadsorbent interactions because it is a carboxylic acid 4 For each of the following pairs predit which compound will have a larger Rf value if both are run on a SiOg TLC plate in 10 acetonehexane 4decanone or 4decanol xylene or benzoic acid cycloheptane or cycloheptanone 4decanone would have a higher Rf value because it is a ketone compared to 4decanol which is an alcohol which means that 4decanone will have less adsorbent interactions Xylene would have the larger Rf value because it is a carboxylic acid which means that it would have more soluteadsorbent interactions that carboxylic acid Cycloheptane would have a large Rf value then cycloheptanone because it is a hydrocarbon and would have a small amount of interactions 5 Arrange the following solvents in order of increasing polarity ethyle acetate dichloromethane npropanol ethanol toluene heptane Increasing polarity heptane toluene dichloromethane ethyl acetate npropanol ethanol 6 Name 3 important things to keep in mind while spotting TLC plate Three things to remember do not let the baseline and dots dip into the solvent make sure to have small spots so that they are well defined and will lead to better separation and make sure that the spots are highly concentrated spot them a few times MCAT Questions See binder for Questions 7 a Spot with the smallest Rf 8 d Decrease 9 C H20 BF3 CH3CH22OH Experiment 3b Column Chromatography IntroductionTheory The goal of this experiment is to chromatographically separate compounds based on polarity Colum chromatography is used for purifying liquids and solids It works by loading an impure sample into a column of adsorbent then an organic solvent ows down through the column and components of the sample separate form each other by partitioning between the stationary packing material and mobile eluent Different compounds with different polarities partition to different extents This means they move through the column at different rates Then the eluent is collected in fractions Procedure See Carbon Copy for Procedural steps Data and Observations We used 101 mg of ferroceneacetyl ferrocene and 102 mg of alumina into a conical vial We also included 3 drops of MtBE After stirring until it was a dry powder we transferred it to the micro column and added another layer of sand on top of it We then added petroleum ether until the yellow band eluted and put the yelloworange solution that eluted into a pretared Erlenmeyer ask with a boiling chip After that we added MtBE until the red band eluted We put the red solution into another pretared Erlenmeyer ask with a boiling chip For the TLC we used a solvent of 15 MtBE and 85 petroleum ether We set up the TLC plate with samples of all 5 collected fractions The yelloworange and red spots both matches the spots before them on the TLC plate showing that the compounds were pure and allowing us to combine fractions 1amp2 and 3amp4 when isolating each compounds We evaporated the rest of the solution in the ask in a sand bath We then found the melting point for each The yelloworange solution had a melting point range of 169173 C and the red solution had a melting point range of 7983 C The melting points thus reinforced the identification of the yelloworange fraction as being ferrocene and the red fraction as being acetylferrocene Calculations Amnt alumina 0102 g Amount Ferrocene Acetylferrocene 0101 g Weigt Flast 1 with boiling chips 37 849 g Weight Flask 1 with yelloworange fraction 37894 g Weigt Flast 2 with boiling chips 20767 g Weight Flask 2 red fraction 21237 g Rf of yellow orange solution 25cm34cm0735cm Rf of red solution 09cm34cm0265cm Weight Ferrocene 0045 g Weight Acetylferroce 0047 g Flask 1 Recovery 37849g37894g099g X 100 99 recovery Flask 2 Recovery 20767g2l237g097g X 100 97 recovery Melting point Ferrocene 169172 C Melting Point Acetylferrocene 7983 C Discussion and Error Analysis There were several places where error could have occurred We could have improperly weighed our components or lost some of them during transfer from the weigh paper We also may not have stirred our mixture of ferroceneacetyl ferrocene alumina and MtBE until it was a dry powder We may also have missed the exact time that the bands started and stopped eluting It is also possible that our solutions may not have evaporated fully in the sand bath For our red solution there seemed to be some stuck on the side of the ask that did not properly evaporate We may also have missed the exact melting point when using the Meltemp apparatus Conclusion It is easy to see the connection between TLC and column chromatography Both procedures depend on the solutes interactions with the solvent and adsorbent For TLC more interactions higher polarity means a smaller distance traveled on the TLC plate For column chromatography more interactions means the compound elutes more slowly Both are effective ways to see the separation of the compounds With the TLC plate it may be less obvious if a UV light is necessary compared to the colored bands in column chromatography Both of the procedures are relatively straightforward PostLab Questions Exp 3b 1 Why does ferrocene elute form the column first Why was the solvent hanged in the middle of the column procedure Ferrocene is not very polar so it dissolved in the lower polarity solvent and moved down the column first The solvent was changed to a higher polarity solvent so that acetylferrocene which is more polar would dissolve more quickly and make its way down the column 2 Ferrocene eluted fomr the column first but it ended up as the higher of the two spots during TLC analysis Explain Ferrocene eluted from the column first because it was a lower polarity solute so it had less interaction with the solvents and absorbents Because of its lower polarity it also had less interaction with the solvent and adsorbent on the TLC plate which allowed it to move the farthest The two processes are very similar but in opposite directions 3 Why is it important to limit the sand bath temperature to around 60 C when evaporating solvent from your collected fractions It is important to limit the temperature to 60 C because if it is too high the compound will melt It is essential that the solute is isolated from the solvent through evaporation 4 In running column chromatography under the following conditions you et results that are less than ideal Consider the situation and suggest a correction a You run a column on a mixture of 2 unknown halogenated alkenes but they both elute from the column at the same time The solvent used was methylene chloride dichloromethane Methylene chloride was too polar so both compounds were carried down the column To x this use a less polar solvent like petroleum ether so that one of the compounds will elute faster b You run a alumina column on a mixture of a thiol and an amine but you never get any compound off the column ie Neither compounds elute The solvent was a mixture of petroleum ether and dichloromethane The petroleum ether and dichloromethane mixture is a low polarity solvent which resulted in neither compound making it down the column The solution would be to use a more polar solvent 5 You run a column to separate a mixture of 3 compounds naphthalene oturic acid and uorenol The column is run with a solvent system based on hexane but which becomes more polar with the addition of more and more dichloromethane over time Predict the elution order Naphthalene would elute first because it is the lowest polarity and would dissolve in the low polarity solvent Fluorenol would elute next because it is more polar than naphthalene but not otoluic acid If dichloromethane was added uorenol would elute from the more polar solvent Otoluic acid would elute last because it is the most polar More dichloromethane would need to be added to the solvent 6 Suppose you are given 150 mg of an unknown mixture that contains 3 compounds and you are told to us CC using alumina as a stationary phase How would you decide what solvent or solvent system to use You would have to test different solvents starting with a nonpolar solvent If one of the compounds is nonpolar this would cause it to elute Then you could slowly increase the polarity of the solvent seeing if different bands begin to elute from the column If you increase the polarity too quickly multiple compounds may move down the column at the same time 7 The compounds used in Exp 3b were both highly colored so it was easy to see their progress as they moved down the column during elution More organic compounds are colorless in solution How might you monitor the progress of a CC given a colorless sample You could monitor the progress based on volume and change test tubes every time a certain volume eluted You could also monitor the polarity of what is being eluted by performing TLC at certain increments MCAT Review Notes Questions 7 You could monitor the progress based on volume and change test tubes every time a certain volume eluted You could also monitor the polarity of what is being eluted by performing TLC at certain increments 8 d Naphthalene Experiment 4 Making Polymers IntroductionTheory In this experiment many polymerization reactions occurred including the condensation reaction to form Nylon106 radial polymerization reactions and the formation of a crosslinked polymer A polymer is a larger molecule made of monomers and are present in our everyday lives In the condensation reaction 2 monomers came together and water is eliminated in the process Nylon106 is formed by condensation of sebacoyl chloride the tencarbondiacid chloride and hexamethylene diamine the six carbondiacid chloride and finally slime is formed from the crosslinked polymerization of polyvinyle alcohol PVA and borax sodium borate decahydrate Procedure See Carbon Copy for Procedural steps Data and Observations The measurements for all three of the experimental parts were achieved without much experimental difficulty Since amount of recovery is not being analyzed at the end of each experiment the accuracy was paid attention to but not a central concem of the experiment We started with the polystyrene experiment first due to the hour it takes for the heating process The making of Nylon106 was relatively straight forward the only part that needs attention is maintaining the two separate layers of solution before stringing out the nylon while wrapping it around a test tube and being careful not to have contact with the skin Making slime was very easy and just required the mixing of the two solutions until a consistent semisolid consistency was achieved Qualitative Data Table of Solubility for Polystyrene Nylon106 amp Slime in Acetone Toluene and Ethanol Acetone Tuolene Ethanol Polystyrene Yes Yes Yes Nylon106 No No No Slime No No Yes Discussion and Error Analysis There were a few errors that could have occurred in this experiment We could have tested different amounts of each product in the three solutions or even pieces that were too big for the concentration of that particular solution to allow for it to dissolve What seemed conceming was the fact that there was not a general consensus on the results as a whole between our groups within the class Every group seemed to get slightly different observations of solubility and even sometimes appearances of bubbles during the process This we did not observe however overall we saw that Polystyrene was soluble in all three solutions and slime dissolved in ethanol Conclusion The creation of Polystyrene Nylon106 and slime were interesting in their simplicity and their complex intemal mechanisms The three different processes of polymerization was valuable in understanding the differences between the condensation reaction the radial polymerization reaction and the formation of a crosslinked polymer The presence of variant observations however was slightly worrying as the error analysis did not focus much on the creation of the substances as they were very straight forward but rather just the testing which too was simple We thus had difficulty identifying which area of the experiment would lead to the different observations of solubility especially when we felt very strongly that our experiment was conducted with accuracy and precision PostLab Questions Exp 3b 1 De ne the following terms a Monomer A molecule that may react chemically to another molecule of the same type to form a larger molecule such as dimer trimer tetramer polymer etc b Repeating unit The group of atoms that is derived from a monomer and repeats throughout a polymer Also known as monomeric unit and its repetition produces a complete polymer chain c Condensation polymerization A process by which two molecules join together resulting loss of small molecules which is often water The type of end product resulting from a condensation polymerization is dependent on the number of functional end groups of the monomer which can react The monomers for condensation polymerization have two main characteristics Firstly instead of double bonds these monomers have functional groups like alcohol amine or carboxylic acid groups Secondly each monomer has at least two reactive sites which usually means two functional groups d Crosslinked polymer A chemical bond atom or group of atoms that connects two adjacent chains of atoms in a large molecule such as a polymer or protein 2 You synthesized Nylon106 using interfacial polymerization Draw a representation of what your experiment looked like Clearly label the contents and identify each layer 3 Most organic polymers are named based on the monomers used in their production However polyVinyl 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 The monomer vinyl alcohol will not polymerize via addition polymerization because of the unfavorable formation of an unstable radical It is normally formed from the polymerization and hydrolysis of polyvinyl acetate Vinyl Alcohol Ethenol Therefore Structure of PVA 4 Vinylidene chloride CHgCCLg is copolymerized with Vinyl chloride to make Saran Wrap Write a structure that includes at least 2 units for the copolymer that is formed Polyvinylidene chloride PVDC 5 lsobutylene CHgCCH g is used to prepare cold ow rubber Write a structure for the addition polymer formed from this alkene Polyisobutylene 6 Maleric anhydride reacts with ethylene glycol HOCHgCHgOH to produce an alkyl resin Write the structure of the condensation polymer produced 391 cationic vinyl H jH H H3 pol3939merization 3 I I I C hr 4 C C 4 PL8 8 I I H H3 H CH3 i5 b t391e poljrisobut39le ne 1 quot j p 1 ill 9 3 III p 941 P5 39 PM CI H CI H H Cl CC CC Z n quot39 039 H C39 CI CI CI Cl Vinylidene Chloride Vinyl Chloride Saran PVA R CH CH2OH CH CH2OH CH CH2OH CH Experiment 5b Steam Distillation IntroductionTheory The purpose of this experiment is to nd the total pressure of the immiscible system via its component vapor pressures Steam Distillation technique provides a powerful way to isolate compounds that are not miscible with water andor may have boiling points too high to allow for use of normal distillation The compound eugenol is being isolated here Eugenol is a compound of the essential oil from cloves It has a boiling point of 254 a melting point of 12 C 10 C and is not particularly hazardous Cloves have a minute amount of caryophyllene Procedure See Carbon Copy for Procedural steps Data and Observations We started with 1036 grams of ground cloves and put them in a 25 mL RB ask Then we added 15 mL of D1 H20 and a magnetic stir bar We heated the system for 10 minutes increased the temperature and used a long neck glass pipet to collect 5 mL of distillate Then we tumed off the heat To perform the isolation we collected 5 mL of CH2Cl2 in an Erlenmeyer ask and added 2 mL of this to a centrifuge tube We capped the tube and shook it for 30 seconds Then we transferred the oil layer lower to a clean and dry 5 mL conical vial 1 We repeated this extraction twice adding 1 mL of CH2Cl2 each time Then we combined all of the extraction in conical vial 1 and removed the H20 droplet We added 34 microspatulatipfuls of Na2SO4 into the conical vial Next we transferred the oil layer into a clean and dry 5 mL conical vial 2 that originally weighed 23736 g We evaporated the CH2Cl2 under the hood using a sand bath and a gentle stream of air After evaporation the conical vial weighed 23768 g At the end of the experiment and IR spectrum was obtained Calculations Weight of cloves 1036 g Tare weight of conical vial 23736 g Weight after evaporation of conical vial with eugenol 23768 Weight of recovery substance 0032g Recovery 00321036 X 100 308 Discussion and Error Analysis There were several possible small errors We could have had the temperature of the system too high which could have caused bumping and frothing We could have made a mistake when collecting 5 mL of distillate because we had to mark the 5 mL mark on a centrifuge tube by adding 5 mL of water to another centrifuge tube and comparing the mark This was not very precise It is also possible that we did not transfer the entire oil layer into the conical vial during the extraction We may have left some water in the conical vial It is possible that the dry oil layer was not transferred entirely into conical vial 2 The dry oil layer may not have completely evaporated Conclusion The experiment seemed to be quite successful We were able to isolate eugenol without decomposing it There were no major errors The only problem we encountered was the analysis with the IR spectrum where we were explained by our TA that there were inconsistencies which we could not explain as we could not observe where any error could have been made We thus used another groups IR analysis Their IR Spectrum looks Very similar to the graph for the IR Spectrum of crude eugenol in the lab manual For example the first 3 Values starting from the left labeled on the chart in the book for crude eugenol are 3511 1637 and 1605 and for pure eugenol are 3525 2939 and 2844 The corresponding points on our graph had the Values 347649 293573 and 163851 This shows that they were relatively successful in isolating the crude eugenol PostLab Questions Exp 5b 1 We used steam distillation because eugenol is not miscible in water Also steam distillation is performed below 100 C which allows the eugenol to be isolated without decomposing which would happen at higher temperatures 2 a 10931 P H20 0069 atm P 0z39l x0056 mol H20 0069 atm0931 atm x000415 mol oil 000425 mol oil x 169 g oill mol 0z39l0701 g b 5 of 2g01g oil x 1 mol 0z39l169 g oz39l5917 x 104 mol oil 5917 x 104 mol oz39lx0069 atm0931 atm x000798 mol H20 x 18 g H201 mol H200144g H20 3 1 atm0692 atm P H20 0308 atm P 0z39l x g oill g H200308 atml 14 gmol0ctane0692 atml8gm0lH20 x 282g oil lglg282gx l00262 water 282gl g282gx l00738 octane 4 l atm0965 atm P H20 0035 atm P am39lz39ne x g anilinel g H200035 atm93gmolam39lz39ne0965 atml8gm0lH20 x 0187g aniline 5 The ortho isomer is steam Volatile because it has a dipole moment so it is polar The para isomer is symmetrical so it does not have a dipole moment and is not Volatile Experiment 5b Steam Distillation IntroductionTheory The purpose of this experiment is to nd the total pressure of the immiscible system via its component vapor pressures Steam Distillation technique provides a powerful way to isolate compounds that are not miscible with water andor may have boiling points too high to allow for use of normal distillation The compound eugenol is being isolated here Eugenol is a compound of the essential oil from cloves It has a boiling point of 254 a melting point of 12 C 10 C and is not particularly hazardous Cloves have a minute amount of caryophyllene Procedure See Carbon Copy for Procedural steps Data and Observations We started with 1036 grams of ground cloves and put them in a 25 mL RB ask Then we added 15 mL of D1 H20 and a magnetic stir bar We heated the system for 10 minutes increased the temperature and used a long neck glass pipet to collect 5 mL of distillate Then we tumed off the heat To perform the isolation we collected 5 mL of CH2Cl2 in an Erlenmeyer ask and added 2 mL of this to a centrifuge tube We capped the tube and shook it for 30 seconds Then we transferred the oil layer lower to a clean and dry 5 mL conical vial 1 We repeated this extraction twice adding 1 mL of CH2Cl2 each time Then we combined all of the extraction in conical vial 1 and removed the H20 droplet We added 34 microspatulatipfuls of Na2SO4 into the conical vial Next we transferred the oil layer into a clean and dry 5 mL conical vial 2 that originally weighed 23736 g We evaporated the CH2Cl2 under the hood using a sand bath and a gentle stream of air After evaporation the conical vial weighed 23768 g At the end of the experiment and IR spectrum was obtained Calculations Weight of cloves 1036 g Tare weight of conical vial 23736 g Weight after evaporation of conical vial with eugenol 23768 Weight of recovery substance 0032g Recovery 00321036 X 100 308 Discussion and Error Analysis There were several possible small errors We could have had the temperature of the system too high which could have caused bumping and frothing We could have made a mistake when collecting 5 mL of distillate because we had to mark the 5 mL mark on a centrifuge tube by adding 5 mL of water to another centrifuge tube and comparing the mark This was not very precise It is also possible that we did not transfer the entire oil layer into the conical vial during the extraction We may have left some water in the conical vial It is possible that the dry oil layer was not transferred entirely into conical vial 2 The dry oil layer may not have completely evaporated Conclusion The experiment seemed to be quite successful We were able to isolate eugenol without decomposing it There were no major errors The only problem we encountered was the analysis with the IR spectrum where we were explained by our TA that there were inconsistencies which we could not explain as we could not observe where any error could have been made We thus used another groups IR analysis Their IR Spectrum looks Very similar to the graph for the IR Spectrum of crude eugenol in the lab manual For example the first 3 Values starting from the left labeled on the chart in the book for crude eugenol are 3511 1637 and 1605 and for pure eugenol are 3525 2939 and 2844 The corresponding points on our graph had the Values 347649 293573 and 163851 This shows that they were relatively successful in isolating the crude eugenol PostLab Questions Exp 5b 1 Why do we use steam distillation to isolate eugenol rather than purify it by simple distillation We used steam distillation because eugenol is not miscible in water Also steam distillation is performed below 100 C which allows the eugenol to be isolated without decomposing which would happen at higher temperatures 2 Given that for 2 immiscible liquids like those used in this experiement Calculatie the following for an oil mwt 169 gmol that distills during steam distillation at a boiling temperature of 98 C at 1 atm pressure Hint the vapor pressure of water at that temperature is needed Google it a The mass of the oil that codistills with each one gram of water at 98 C 10931 P H20 0069 atm P 0z39l x0056 mol H20 0069 atm0931 atm x000415 mol oil 000425 mol oil x 169 g oill mol 0z39l0701 g b Suppose you had 2 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 5 of 2g01g oil x 1 mol 0z39l169 g oil5917 x 104 mol oil 5917 x 104 mol oz39lx0069 atm0931 atm x000798 mol H20 x 18 g H201 mol H200144g H20 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 codistills with each gram of water and the percent composition of the vapor that is produced sduring the steam distillation Need vapor pressure of water at the steam distillation temp Note if you multiply the second equation from question 3 above by the mw of both the oil and water you get mass oilmass water P 0i1MW0i1P WaterMWWater l atm0692 atm P H20 0308 atm P 0z39l x g oill g H200308 atml 14 gmol0ctane0692 atml8gm0lH20 x 282g oil lglg282gx l00262 water 282gl g282gx l00738 octane 4 One synthetic preparation of aniline involves reducing nitrobenzene with iron and HC1 and then steam distilling the resultant aniline at 99 C How much aniline codistills with each gram of water using the steam distillation l atm0965 atm P H20 0035 atm P am39lz39ne x g anilinel g H200035 atm93gm0lam39lz39ne0965 atml8gm0lH20 x 0187g aniline Experiment 6 Resolution of Enantiomer IntroductionTheory In this experiment a racemix mixture of 1phenylethylamine is being separated into its 2 enantiomers 1phenylethylamine has a melting point of 325 C and a boiling point of 185 C and corrosive properties This experiment teaches to separate enantimoers and leam to use polarimeter Procedure See Carbon Copy for Procedural steps Data and Observations The rst step in this experiment was making the crystals We added 64 mL of phenylethylamine and 50 mL of methanol in a 125 mL Erlenmeyer ask In a 250 mL Erlenmeyer ask we added 78 g of tartaric acid and 50 mL of methanol We added the amine solution rst solution to the tartaric acid solution second solution into a 250 mL ask We warmed the solution on a hot plate until it began to boil We cooled it at room temperature and then put it into an ice bath to crystallize it Then we reheated the mixture to boiling to dissolve the needlelike crystals Then we stoppered the ask and stored it in our locker for two weeks After we opened the ask we filtered the crystals and washed them with 56 mL of cold methanol We saved a small amount and dried it to perform a melting point analysis In a 125 mL Erlenmeyer ask we added the rest of the crystals and 30 mL of 10 aqueous NaOH We warmed the ask for 5 minutes and then cooled it We transferred the solution to a separatory funnel Next we withdrew the lower aqueous layer and discarded it We poured the top amine layer into an Erlenmeyer ask containing K2CO3 A clean and dry 10 mL graduated cylinder was weighed We transferred the amine solution into the cylinder We combined our amine solution with another groups so that we had 4 mL of amine Then we reweighed the cylinder Methanol was added to the cylinder up to the 7 mL mark and mixed it with a glass pipet Lastly we measured the observed rotation of the amine Calculations Dry graduated cylinder 27464 g Cylinder with 4 mL of amine and 298 mL of methanol 41462 g Concentration of solution 41462 27464 298 13998 298 4697 gmL Melting Point 193195 C Observed rotation 13 74 A589 nm Temp 20 C Speci c Rotation aD20 1374 4697 gmL 29 Optical Rotation a pure Samine enantiomer 403 ee optical purity 29403 X 100 0072 100 0072 9993 Total enantiomer 0072 12 X 9993 50037 Total enantiomer 12 X 9993 49965 Discussion and Error Analysis One possible error may have occurred when performing the melting point It was a long process so it was possible that the machine was set too high and the temperature was higher than the thermometer indicated It is also possible that the crystals were not completely pure especially when we combined with our partners The main error may have occurred when transferring the lower aqueous layer It is possible that not the entire layer was removed The K2CO3 may still not have removed all of the water that remained Conclusion The entirety of this experiment took a relatively long time but it was not very difficult Making the crystals was straightforward The most difficult part was removing the aqueous layer because it was important to be precise After performing our calculations we determined that the amine contains 50037 of the enantiomer and 49965 of the i racemate PostLab Questions Exp 6 then a What is the optical purity of the amine 403 x 100 8759 b Calculate S and R enantimoers of the anime 2x 1241 9380 total enantiomers 2x 1241 621 total enantiomers 2 How could you isolate the Ra phenylethylamine from the mother liquor that remained after you crystallized and filtered off the Sozphenylethylamine You could evaporate the liquid from the solution This would leave behind the crystal of the diastereomeric 1phenylethylammonium hydrogen tartrate Then you could react that salt with a base to obtain the R orphenylethylamine 3 The formation of a white solid is often observed after the a phenylethy1amine comes into contact with carbon dioxide in the atmosphere What is that White compound Give an equation that shows its formation The white solid is a carbonate salt NH2 OH H lt39o C02 9 j 4 If the boiling point of the pure Sa phenylethy1amine is 186 188 C what is the boiling point of its enantiomer What optical rotation would be observed if 372 g of the Samine were mixed with 372 of the Ramine The boiling point of the enantiomer would be 186188 C because enantiomers from the same substance have the same physical properties The optical rotation would be 0 because a mixture of equal amounts of two enantiomers leads to a racemic mixture Racemic mixtures cannot rotate plane polarized light 5 a Propose a method for how you would resolve 2chloropropanoic acid RCOOH using amphetamine RHNg as the resolving agent You could mix the j 2chloropropanoic acid with the amphetamine and heat the solution This would create diastereomers that could be separated by allowing the solution to crystallize One of the diastereomers would form crystals and the other would be left in the solution After ltering the crystals they could react with an acid like HCL This would convert the diastereomer back into the enantiomer b See Question in book The same procedure could be used for these enantiomers 6 Calculate the speci c rotation of a substance that is dissolved in a solvent 03 5 gmL and that has an observed rotation of 23 as determined with a 05 dm cell 05dm035 gmL 13 l43 7 Calculate the observed rotation of a substance that is dissolved in solution at 25 gmL and is 75 optically pure Assume a 10 dm cell is used The specific rotation of the optically pure substance is 400 40 x100 75 x 30 mL x 75 8 Final Exam Details Monday 16th from 8930pm in LCl 10 MCAT Question 9 Which of the follow do not show optical purity d A solution containing 2 M R2butanol and 2 M S2butanol Organic Chemistry Lab Notes 0 Pipets amp Melting Points Part A Proper Technique using Pasteur Pipets and Plastic Pipets Pasteur pipets o deliver 1 mL of liquid in 28 drops 0 14 mL delivered in 7 drops 0 keep Pasteur pipet vertical or lose drops and contaminate bulb with sample horizontally Part B Preparation of Filter Pipets filter pipet o filtration of sample isolate solid or liquid or remove unwanted solid from liquid 0 loose cotton into constriction of pipet with metal wire just before the narrow end filter tip pipet 0 remove solids from very small amounts of liquid by leaving solid behind as opposed to trapping it o pushed to the very tip of the pipet Part C Determining of Melting point of unknown compound melting point intensive property identify compound 0 distinct mp narrow range 2 3 C purity of sample impurities lower melting point and broaden range 0 how does melting point of a liquid relate to the freezing point Rate of heating 9 1 2 Cmin during the final 10 C prior to melting 0 1st Fast test run 0 2nd more precise o narrowing down the unknown to a few options Part D Mixed Melting Points if 2 compound are mixed together 9 a change in melting point temperature and range form solid mixture will not be observed if the unknown and known are identical compounds Chapter 1 Filtration Gravity filtration slow 0 Removing drying agent from 50 mL solution 0 Fluted paper Vacuum filtration faster than gravity 9 Buschner funnel with filter paper 0 Microfiltration lt20 30O mg 9 Hirsch funnel better in minimizeing product losses 0 For smaller amounts lt50mg 9 Craig Tube used Removing drying agent from microscale solutions lt10mL 9 Pasteur pipets with cotton plugs help minimize loss of material Chapter 2 Identifying Compounds Phase Diagrams O Melting points for solids o Mp temperature at which solid and liquid phases of pure substance coexist at a pressure of 1 atm 0 Presence of impurities lowers mp good indicator of purity o The purer the material the higher its mp and the narrow the mp range o When mixed with other compounds decreases the mp I Called Melting Point Depression o Biggest errors in melting point heating too fast 9 slow accurate o Traces of solvent or other volatile materials can be present in solid 9 leads to sweating before mp observed Exp 0 1 Questions 1 Always add acid to water True What is the difference between a filtering pipet and a filter tip pipet Filtering pipet removes solid from liquid but filter tip pipet removes liquid from solid What is the melting point What is a fast melt and how is this used in relation to the slow melt A melting point is the range at which a substance is converted from solid to liquid it is also the freezing point Fast melt is much faster but not as accurate than slow melt and gives a single number for the melting point which can be used as a reference to do the slow melt more accurately Why is it important to cool you sample slowly when recystallization It will allow for larger more pure crystals for form Fill in chart procedural chart 100mg A contaminated with 20 mg B 9 add solvent Add hot water until almost all is dissolved 9 heat Remove impurities in hot solution with charcoal and bring to low boil 9 cool Allow to cool and crystals to form may be induced 9 filter B and some A in mother liquor OR pure A crystals Unknown X gave a slow melt of 120123 When mixed with B the mp remained the same when mixed with D the mp was significantly lower a What is X mixed with B called Positive melt b What is X mixed with D called Negative melt c What is the identity of your compound B mp 117 120 C What does the dielectric constant tell you about a solvent Would you choose a solvent with a large or small dielectric constant to dissolve a polar compound The dielectric constant is how well the solvent conducts electrical current Large E 9 like dissolves like Draw Buchner funnel what size sample would you use the funnel for Around 15 mg What type of tubing is used for the filtration Rubber tubing 1 Recrystalization Recrystalization process of purification involves dissolution of solid in hot solvent filtration of heated solution crystal formation isolation of crystalline compound must understand factors that dictate solubility and crystal formation of solid amp ability to manipulate hot solutions heater never over 4 9 could crack plate are not explosion proof solvent vapors can self ignite autoignition temperature Part A Macroscale Recrystallization heat DI water to boiling point weight impure unknown add hot water and heat until solid dissolves o no more than 25mL o small insoluble impurities can have negative in uence on recrystallization 9 hot gravity filtration o soluble colored impurities come from conjugated pi bonds strongly absorbed on surface of carbon 9 removed with charcoal hot gravity filation cool solution add charcoal powder boiling chips o never add charcoal to boiling solution boils over filter o pre warmed funnel 9 during hot gravity filtration is to avoid formation of crystals that can occur if the solution you are filtering cools down when it comes into contact with room temp filter funnel o done quickly and all together after cooling crystals form o dependent on I concentration of solute I temperature of solvent I what solvents are used easily soluble solute compound less likely to form crystals o induced crystal formation add seed crystals scratching cool with ice bath reheat filtrate to evaporate and concentrate filtrate o the slower the crystals grow more pure they are collect crystals 9 suction filtration Buchner funnel o wash crystals with cold water remove excess water press down 0 put on dry filter paper 9 in desiccator with Drierite CaSO4 overnight weight crystals 9 MP o 2 mixed MPs o total weight amp Recovery Part B Microscale Crystallization using a Craig Tube Craig Tube used in microscale recrystallization solids 10100 mg Weigh unknown into test tube 05 15mL dropwise near boiling water until solid dissolved can spin rod to mix Cool add charcoal pellet and boiling chips reheat to boil Hot gravity filtration pre warmed filter pipet o Filter boiling chips insoluble impurities and charcoal o Filtrate should be clear amp colorless if not perform second filtration Why residual carbon in hot filtrate discouraged O Cool Craig tube to room temp if no crystals then induce with seed crystals scratchingice bath If still none 9 reduce volume of filtrate evaporation around 20 or until crystals appear in heated solution Not vacuum filtration 9 Centrifugation 0 Craig tube inverted rapid rotation directs solvents past narrow interface leaves crystals behind o Tare weight of craig tube used 9 recovery when dry crystal 9 Identify compound Chapter 1 Solvents solvents play crucial role o dilute reagents control the reaction temperature to recystallize compounds and in chromatographic application difference between melting and dissolving o melting phase transition a solid undergoes when heated I regular crystal structure of solid is lost when solid melts but intermolecular distance between molecules doesn39t really change o dissolution when solid is dissolved in a solvent solvent molecules surround each individual molecule or ion I same happens when liquid is dissolved in a solvent ability of solvent to dissolve solute depends on intermolecular forces between solvent and solute molecules Chapter 3 Purification Techniques to purify compound use differing solubilies of compounds in different solvents and that impurities will be present in smaller amount than desired product o dissolve material in hot solvent and cool solution slowly o solubility of dissolved material decreases with decreasing temperature solution will become oversaturated and solid will separate form solution o when crystals are formed he same individual molecules will fit in crystal lattice structure 9 crystallization equilibrium process that produces very pure material solubility of compound depends on intermolecular interactions between the compound and the solvent and on the polarity of both ideally the solvent should be poor solvent at room temperature and very good solvent at high temperature Essential steps o Dissolve impure solid at high temp o Filter solution for insoluble fraction o Cool solution slowly to control crystallization process o Filter formed crystals o Dry crystals Microscale 10 100mg 9 Craig tube o Crystals put in tube and into centrifuge I Force the solvent mother liquor out of craig tube into centrifuge tube and crystals remain in craig tube 2 Extraction liquid liquid extraction separation of compounds using 2 immiscible solvents 0 depends on how differently the compounds of sample mixture partition themselves between the 2 immiscible liquids 1 Compound mixture dissolved in suitable solvent solvent immiscible with first solvent added diethyl ether and water 2 Contents mixed immiscibles allowed to separate 0 less dense upper 0 more dense lower 0 components of initial mixture distributed amongst 2 immiscible solvents 9 determined by partition coefficient Kg 9 relative solubility I compound that is more soluble in the less dense solvent will be in upper layer I compound more soluble in more dense solvent will be in lower layer 2 immiscible layers separated transferred and component in that solvent isolated by solvent evaporation and or crystallization 0 usually used to separate organic product from reaction mixture after aqueous work up K1 C2C1 Organic solvents used in liqu liqu extraction 0 Petroleum ether ethyl acetate methylene chloride toluene 0 Diethyl ether effective extraction solvent with low boiling point 37degcel BUT highly volatile and extreme ammable 0 Replace with methyl tert butyl ether MtBE Extraction under Basic and Acidic Conditions 0 Depends on relative solubility that each compound has in 2 immiscible solvents 0 Change in pH of solvent addition of acid or base changes solubility of organic compound in solvent 9 NaOH or NaHCO3 o The increase in water solubility I due to acid base reaction that converts less water solution carboxylic acid into more soluble sodium carboxylate I water solubility of other organic compounds having acidic hydrogens like phenols can be increased by adding base 0 aqueous solution of NaHCO3 converts carboxylic acids to their sodium carboxylates but its not strong enough base to form sodium salts of phenolic compounds I therefore 2 weak organic acids can be partitioned or separated pKa values differ by 5 units and aqueous base used in deprotination is no basic enough to remove the hydrogen from the weaker organic acid I if aqueous NaOH used instead in carboxylic acid and phenol mixture deprotonates both and are found in aqueous base layer separation would not be feasible using NaOH as base I the less water soluble organic amine in the presence of acid is converted to the more water soluble organo ammonium ion 9 dilute hydrochloric acid used for extraction of basic organic substances or remove basic impurities Liqu liqu extraction 9 separate weak organic acid from weak organic base 0 Can also use aqueous acid or base with ether as immiscible organic solvent Organic compound that is not weak acid or weak base alkanes alkenes ketones etc can be separated from weak organic acids or bases Part A Determination of the Partition Coefficient for Benzoic Acid Benzoic acid in centrifuge tube Add MtBE dissolve compound DI water vortex mixture Uncap and 2 layers separate 0 Top layer MtBE 0 Bottom layer water 0 Quick and simple technique to identify each layer Add DI water and see where it goes Remove bottom layer with Pasteur pipet Organic layer in conical vial add anhydrous sodium sulfate drying agent 0 Removes traces of moisture in organic solvents Recap vial leave in sodium sulfate for 5 mins 0 Transfer dried organic phase MtBE treated with drying agent into tared dry concial vial with boiling chips To remove compounds absorbed to solid sodium sulfate rince MtBE 0 Transfer this MtBE to conical vial evaporate organic solvent in fume hood 9 warm sand bath air stream 0 Reheat vial to remove last solvent traces 9 Determine amount benzoic acid from organic phase calculate distribution coefficient Part B Separation of a Sample Mixture Containing 3 Components OR a base carboxylic acid neutral compound a carboxylic acid phenol neutral compound Remember 0 1 Shake contents of biphasic mixture 0 2 When neutralizing with base NaOH add dropwise always stirring and checking pH with litmus paper do not add too much base 0 3 Isolation of basic compound milk like or oil like during addition of NaOH 9 oily base recovery induced crystallization 0 4 When aqueous solutions have been neutralized can pour down sink MtBE goes in non halogenated waste container Since we have a phenol 9 used NaHCO3 instead of HCl dissolve unknown mixture in MtBE in centrifuge tube NaHCO3 mix when layers separated bottom layer transfer to centrifuge tube repeat with 2 portions of 5 NaHCO3 combine with first portion back extract aqueous portion used to extract the organic solution 0 add MtBE to centrifuge tube with combined aqueous portions cap shake 0 draw off MtBE layer when separated add to original MtBE solution combined NaHCO3 extraction treat with 10 NaOH until basic 9 precipitate should from amine from unknown mixture 0 isolate precipitate with vacuum filtration with Hirsch funnel was cold DI water 0 weight Recovery MP Carboxylic acid component isolated in similar way 0 Extract organic solution 3 times with 5 NaOH combining into centrifuge tube 0 Back extract with MtBE acidify aqueous extract with 10 NaHCO3 solution 9 precipitate do same as above Isolating Neutral component in original organic layer 0 Wash organic layer with DI water dry with sodium sulfate 0 Remove organic layer from drying agent evaporate MtBE solvent with gentle heating o When neutral compound isolated dry 9 mass MP identify Why must compounds be dried Oily Base Recovery Recovering Organic Base Unknown by adding NaOH getting non crystallized precipitate oily or milky After making aqueous mixture basic pH paper extract 3 times with MtBE o Remove each ether layer carefully 0 Back extract combined ether layers with DI water dry combined ether layers with sodium sulfate 5 mins Transfer dried organic solution via dry Pasteur pipet to tared and dry concial vial with boiling chips To remove absorbed compounds in solid sodium sulfate rinse with MtBE Evaporate organic solvent in hood w sand bath reheat vial to remove last traces of solvent Dry vial until constant weight of solid obtained Syringe Pipet calibrated to deliver volume Chapter 3 Extraction when compounds show more affinity for one environment than another 0 partition between stationary and mobile phases Liquid liquid extraction o Based on varying solubilities of different solutes in immiscible solvents I LOW DENSITY 9 top I Organic solvent almost always lighter than aqueous EXP chlorinated solvents I Separatory funnel used o Ratio of concentration of solute in each layer is defined by the aprtition coefficient K the distribution constant 0 K C2 C1 Eg Organic compound A has a partition coeff between water and ethyle acetate equatl to 8 and 10g of A is dissolved in 100 mL of water If 20 mL of ethyle acetate are added you can calculate how much A remains in the water layer x g K810g xg20mLxg100mL x38g This means the original 10 g A 62 g are in the ethyl acetate layer and 38 g are in the aqueous layer The second extraction of the aqueous layer with another 20 mL of ethyl acetate leads to K838g yg20mLyg100mL y15g The second extraction results in 23 g of A in the organic layer Combining both ethyl acetate layers yields 85 g A and a higher recovery can be obtained by continuing the extractions or by using ethyl acetate in each extraction Chapter 3 Drying the Organic Fractions chemical drying agents 9 anhydrous inorganic salts which can accpeet water into their crystal structure to form hydrates MgSO4 Anhydrous H2O 9 MgSO4 x H2O ranked according to their rate of water absorption capacity effectiveness and inertness o polar solvents ether and ethyle acetate retain more water need high capacity drying agents 0 non polar solvents petroleum ether and chloroform are easier to dry Exp 2 Questions 1 Separation in liquid liquid phase 2 solvents relies on the fact that the 2 are immiscible True 2 You should discard the unused solvent fractions after each extraction False 3 You should pray down you bench with the green solution before leaving True 4 Acids and bases in their neutral form are not soluble in organic solvents False 5 Nonpolar solvents need drying agents with a high drying capacity False 6 What is the name of the organic solvent used in this lab Will it be on the top or bottom of the aqueous layer MtBE Top layer 7 How many pka units do two weak acids have to be apart in order to separate them What solvent would you use to do this and why 5 pka units a compound like diethyl ether would be good because it has a low boiling point but it extremely ammable and volatile and is usually replaced with MtBE for that reason but is also organic 8 What is the equation for the partition coefficient and what odes each variable tell you Kd C2 C1 9 C2 is the concentration of the solute in solvent 2 C1 is the concentration of the solute in solvent 1 Kd related how miscible a solute is in one solvent versus another 9 What do you back extract think about what the partition coefficient tells you when is this step performed Back extraction is done to remove as much solute from the solvent as possible which can39t be done in one shot Further if a compound like an acid dissociates its base must be reprotonated 10 Following table is about extraction Compound Solvent 1 Solvent 2 Solvent 3 Solvent 4 HA HA A HA HA HB HB HB HB HB HC HC HC C HC HB D D HD HD a What it is a bad idea to use solvent 2 before solvent 1 Solvent 1 would be better before 2 because it would only isolate 1 compound D whereas solvent 2 would dissolve 2 b Which compound is most likely a base HC c Which solvent is the organic solvent S4 d Which compound is your neutral HB e What is the order of solvents you39d use to extract each compound 4312 11 An extraction with 40 mL of organic solvent with 150 mL of water resulted in recovering 8 g from the organic layer How much sample did you originally add to the extraction with a partition coefficicent of 4 4 840 x150 15020 x x 75 g in aqueous layer total 85 75 g 155 g 12 A second extraction is done on the aqueous layer form part a with 20 mL of organic solvent How much sample would you recover form the organic layer 4 x20 75150 mL x 4g Z5 425g 13 What is the total percent yield of the extraction 4812g155gx100 774 3 Chromatography a Thin Layer Chromatography Chromatography separate mixtures of compounds for either analytical identification and quantification or for preparative purification mixture of compounds dissolved in liquid and solution mobile phase is transported along immobile non soluble adsorbent stationary phase solute molecules interact with stationary phase in repetitive adsorption desorption steps 0 different degrees of interactions with adsorbent different mobilities separation 0 compound with stronger interactions with stationary phase moves SLOWER 0 compound with weak interactions FASTER 0 Uses thin layer adsorbent stationary phase silica gel bound to solid support glass plastic aluminum sheet 0 Silica gel amorphous form of silicon oxide SiO2 that has hydrated surfaces I Porous material large surface area 9 allows extensive molecular interactions 0 solvent will rise on plate due to capillary forces When silica gel particles are packed to thin layer microscopic gaps between particles network of capillaries 0 Think layer adsorbent stationary phase 0 Ascending solvent mobile phase 0 Dynamic process in which molecules are constantly adsorbed to the surface and dissolve back in the mobile solvent Structural differences of compounds 9 manifest differences in reversible binding between solute and binding sites on adsorbent 0 Each will travel a specific distance 0 Use UV light and 12 to analyze location of colored spots Different compounds vary in structures 9 functional groups 0 Impacts polarity 9 impacts capacity for interaction intermolecular forces with the adsorbent stationary phase 0 Gives different ascending rates distances relative to solvent Distance depends on o 1 Polarity of compound solute o 2 Polarity of solvent mobile phase 0 3 Polarity of adsorbent stationary phase there is reversible binding between the moving compound solute and the polar sites on the surface of a polar adsorbent 0 binding due to intermolecular forces 0 same true for interaction of solvent and adsorbent more polar solute 9 greater the interactions less distance solute travels less polar solute 9 weaker the interactions greater distance solute travel Functional Groups from most polar greatest affinity to polar adsorbent Carboxylic acids RCO2H Alcohols Amines Thiols ROH RNH2 RSH Aldehydes ketones esters RCO H RCOR RCOOR Halogenated hydrocarbons RX RC1 RBr Unsaturated hydrocarbons RCHCHR Saturated hydrocarbons RH polarity of a solvent in uences rate distance that solute ascends plate o increase in polarity of solvent 9 increases interaction of solvent with the polar sites of adsorbent 9 reduces solute adsorbent interactions 9 solute spends less time interacting with adsorbent and travel further Elutropic Series decreasing polarity of common elution solvents most polar on top water methanol MeOH ethanol EtOH npropanol PrOH acetone MeCOMe ethyl acetate MeCOOCH2Me diethyl ether MeCH2OCH2Me dichloromethane Ch2 C12 toluene C6H5 CH3 cyclohexane C6H12 petroleum ether hexanes RH common adsorbents silica gel amp alumina 0 activity of adsorbent in uences the rate or distance a solute will migrate 0 most active adsorbents bind the solute molecules more tightly o greatly affected by their water content most active least water Retardation factor Rfvalue of solute ratio of distance travelled by compound to distance travelled by solvent 0 Rf 0010 Part A TLC Analysis of ortho Hydroxyacetophenone and para Hydroxyacetophenone NB Iodine vapors are hazardous kept in hood NB smaller more concentrated spots are better short contact time better separation Spotting o hydroxyacetophenone p hydroxyacetophenone both Use 3070 ethyl acetatepetroleum ether into development jar allow solvent to soak the wick o Wick saturated air in jar with solvent and enhances chromatography by approximating equilibrium conditions When 5mm below top mark solvent line evaporate solvent Visualizing Chromatogram observe under shortwave ultraviolet UV light 254 nm 0 circle darker spots observe with iodine cross dark spots appear where adheres 0 after 1 min measure distance from original sample to centre of spots 0 calculate Rf values both compounds have same functional groups but very different RfVal11 S Why Part B TLC Analysis of Analgesic Components and an Unknown Mixture determine components present in an unknown mixture 0 mixtures Aspirin caffeine acetaminophen ibuprofen 0 components acetylsalicylic acid acetaminophen caffeine ibuprofen crush unknown sample in test tube ethyl acetate dissolved spot TLC plate w unknown and 2 standards caffeine acetylsalicyclic acid 0 develop with ethyl acetate as solvent 0 since caffeine is a minor component in some mixtures Anacin Goodys Exedrin spot more heavily to see it Waste 0 TLC capillaries MP tube jars 0 Plates regular trash 0 Filter paper dry in hood then trash Chapter 3 Chromatography TLC used for several uses Establish 2 compounds are identical Determine the number of component in a mixture Determine the appropriate solvent for a column chromatography separation Check effectiveness of a separation achieved by column chromatography by recrystallization or by extraction 0 Monitor progression of reaction 3 major steps 0 injection I spotting is used sample applied to plate before any solvent allowed to ascend the adsorbent layer 0 separation I developing or running plate as solvent ascends the plate sample is partitioned between moving liquid phase and stationary solid phase 0 detection I different methods may be used to visualized the separated spots OOOO Spotting o Spots must be small and concentrated spot multiple times 0 Prepare sample by dissolving it in a chosen volatile solvent usually acetone or methylene chloride Developing 0 Choosing developing solvent depends on materials to be separated 0 Analyste moving up TLC plate is either in mobile phase solvent or stationary and the competition between the 2 phases determines how fast the different spots will move 0 More polar solvents will result in higher RfValL1 S and faster movement up the plate 0 Commonly used solvents I Methylene chloride dichloromethane and toluene intermediate polarity 9 good for many functional groups I Hexanes and petroleum ether 9 good for hydrocarbons I Hexanes or petroleum ether with varying proportions of toluene or ether 9 gives solvent mixtures of moderate polarity useful for common functional groups I More polar materials require ethyl acetate acetone or methanol 0 Use filter paper in far wait for filter paper to be saturated 9 create atmosphere saturated with the solvent to enhance the speed and quality of development of the plate I Never open or lead to uneven development Visualization 0 UV lamp 9 compound will show as darker o Reagent is iodine 9 forms either brown or yellow complexes I Most effective if analyte contain aromatic groups I Mark before spots disappear iodine sublimes in air 0 If iodine doesn39t work I Potassium permanganate solution then heated I Anisaldehyde in dilute sulfuric acid solution effective for carbonyl functionalities I Ceric ammonium molybdate in dilute sulfuric acid effective for hydroxyl compounds I Ninhydrin solution used for amines and amino acids Analysis and applications of TLC 0 Rf distance traveled by substance distance traveled by solvent front I If 2 compounds have the same traveled distance same 0 Can monitor progress of reaction I Is the starting material still present I Any new compounds present I Is there only one product or several I How long is the optimum reaction time meaning product is formed in good yield but no side reactions are interfering yet Exp 3a Questions 1 All chromatographic methods work the same True 2 TLC can be used to monitor the progress of a reaction True 3 Do not remove iodine from hood ever True 4 You should put a watch glass on top of the beaker to keep the air saturated with solvent True 5 The larger the spot the better the resolution on the TLC plate False 6 Define the following a 2 silica gel silicon oxide stationary phase on TLC plate b Spotting Using a capillary tube to put a spot of the solution containing the unknown onto the TLC plate detection of the distance traveled by the solute can be done with UV and Iodide c UV and Iodine Short wave UV light and iodine used to visualize the distance traveled by the solute d Rgvalue The retardation factor the ratio of the distance traveled by the solute to the distance traveled by the solvent 7 What would cause your samples to move the same distance with the solvent front What would case your samples not to move at all If the stationary phase was not selective enough it would not slow one solute down versus another and the solvent would remain a mixture without having dissolved solutes separately It would not move at all if the capillary action was hindered 8 What are the 3 major steps in TLC Injection 9 putting the sample on the TLC plate and letting the sample run up Detection 9 figuring out the relative distances traveled 9 Amanda Bynes have accidentally mixed up Vit C Ibuprofen and Benzoylmethylecgonine She decided to try and determine the identities with TLC SiO plate with hexane solvent However she made several mistakes Explain why the problem arose and how you would fix it a Each of the samples streaked across the plate The stationary phase was no polar enough to slow any solute from reaching the top b All of her samples ended up at the same location at the top The stationary phase wasn39t differentially selective enough she should choose a different stationary phase c She didn39t see any spots when she went to visualize the plate She put her original spots below the level of the liquid in the jar and the samples washed off Do it again with the spots at a higher initial level d The solvent is moving too slow up the TLC plate The air is not saturated she needs to seal the jar The capillary action isn39t working e The solvent front stopped half way up the plate The solvent is too attracted to the stationary phase one or the other has to be changed 3 Chromatography b Column Chromatography Column Chromatography preparative method for separating and isolating compounds from mixtures Used for obtaining compounds from natural sources or for purifying products form reaction mixtures Essentially like upsidedown TLC plate 0 Instead of thin adsorbent attached to support column is filed with a larger amount of adsorbent and the mixture is loaded on top of it o TLC capillary forces Column eluent percolates through by gravity I As eluent moving down column carries soluble compounds with it I Compounds with strong interactions SLOW I Compounds with weak interactions FAST Under right conditions compounds will separate is distinctive bands coming out individually Solvent is evaportated from each Procedure of Column Chromatography Dry column packing layers 0 sand 0 dry absorbent 0 dry component mixture 0 sand 2 3mm Dry column loading 0 Weigh Ferroceneacetyleferrocene mixure put in conical vial 0 Add alumina and few drops of MtBE stir until solvent evaporates and solids behave like dry powder 0 Surround layer with sand in micro column Eluting the column 0 Petroleum ether mobile phase 1 add on top of sand 9 perculates 0 Use air pressure to facilitate the solvent flow through column 0 Do not let mobile phase drop below upper layer of sand 0 Yellow orange band travels down 9 collect solvent 1 to elute yellow orange band 9 collect solvent 2 colorless Change solvent to MtBE 9 collect solvent 3 red orange band 9 collect solvent 4 colorless 9 collect solvent 5 colorless fold back 12 inch latex tubing when connecting to apparatus 9 provides at end and resistance to gripping pressure TLC Analysis of fractions and product isolation 0 Use solvent 15 MtBE 85 petroleum ether 0 Transfer to tared ask w boiling chips and evaporate contents in sand bath at 60 C Weight compound MP recovery Chapter 3 Chromatography based on solid liquid phase partitioning stationary phase may be almost any material that does not dissolve in associated liquid phase 0 solids most commonly silica gel SiO2x H20 or alumina Al2O3xH2O steps for separation O O O O O 0 Exp 3b Questions selecting column size depends on size of sample More sample difficult to separate 9 larger column Diameter of column related to amount of sample Length of column related to difficulty of separation Generally total weight of absorbent used should be 20 times weight of crude sample 9 100200 mg only 2 3g absorbent needed for 20g sample a 250g absorbent selecting stationary phase and filling column activated alumina as powders heated with no water can be neutral acidic or basic silica gel has 1020 water dry packing method for very small columns Pasteur pipet organic solvent selected as eluent is added to the column to saturate the absorbend wet packing method slurry made with absorbent mixed with organic solvent then poured into column and tapped to remove bubbled once filled should never be allowed to run dry 9 homogeneity of column is compromised as cracks can impact separation selecting eluents selected by doing test runs on TLC plates stationary phase must match absorbent used in column as closely as possible silica gel TLC used if silica gel will be column packing the more polar the analyte the more polar the eluent should be to effectively separate components of mixture 9 Rf 02 eluents can be changed throughout boiling point should be high enough so that while running column rapid evaporation after the fractions are collected bp between 50 120 C solvents in increasing polarity non polar 9 hydocarbons hexanes cyclohexanes petroleum ether to toluene and carbon tetrachloride slightly polar diethyl ether dichloromethane chloroform polar acetone and ethyl acetate very polar ethanol and methanol loading column running column organic compound will adsorb onto or adhere to the fine particles of absorbent if solution is colored colored band should be visible at top of column eluent continuously added rate controlled by stopcock as different fractions are eluted off column polarity of eluent can be increased analyzing eluted fractions TLC plates run for each fraction based on this can be combined and then isolated and purified isolating different components 1 Eluent is continuously added to the column True 2 Activated alumina contains 1020 water Flase 3 A Pasteur popette column can be used to separate a mixture with similar R values False 4 Let the column run dry when you want to elute the next compound False 5 The more polar the analyte the less polar the eluent should be Flase 6 Define the purpose of the following a A1zQ3LHzQ Alumina it is the stationary phase b Cotton and sand Used to c Gradient eluent Its an eluent that interacts more with one of the components of the mobile phase than another 7 Explain what conditions these should be based on the compound information a A 3 compound mixture with similar RfVal1l S A larger column need to interact more with the stationary phase b Large sample with very different RfVal1l S Small stationary phase the goal is always to have a s little as needed 8 Explain how TLC and Column Chromatography are similar in relation to the sample interactions Both will slow down the more polar compound more than the less polar compound so the least polar solvent will elute faster and more further up a TLC plate but a more polar solvent will elute last and amore a shorter distance on a TLC plate 9 Column chromatography is upside down TLC Explain what this means Instead of capillary action the mobile phase is placed on top of the stationary phase and moves down with gravity and air pressure 10 Amanda meets up with Lindsay to ask her how to pack the perfect dry column Explain the steps Lindsay would tell Amanda add about 23 mm of sand add activated alumina until the column is about half filled add the mixture to be separated put another 23 mm of sand on top 11 She doesn39t pay attention to what she collects in her test tube Use the TLC plate to explain why she got the samples she did based on each lane Vit C Rf 040 Ibuprofen Rf 075 Benzoylmethylecgonine Rf 080 a Plate 1 she didn39t change test tubes the entire time of collection Thus all the samples ended up in the same tube b Plate 2 only one sample got in the test tube that was tested on plate 2 It was probably vit c so she collected eluent correctly c Plate 3 2 pills got mixed into this tube she didn39t switch her test tubes correctly It contains ibuprofen and benzoyleth d Plate 4 either she didn39t collect any compounds in the test tube or the spots got submerged in the solution and were dissolved off the plate before getting to move through the stationary phase e How would you separate the top two spots more try a different solvent less polar one f What order would the samples elute from a column benzoylmeth first then ibuprofen then vitC 4 Making Polymers Polymer large molecule chemically linked of many smaller monomers Formed through synthetic process polymerization o Polymers cellulose is a polymer of glucose o Synthetic polymers polystyrene polyethylene and PVC o Natural polymers cotton silk wool 3 different polymerizations o condensation polymerization o chain addition polymerization o cross linked polymerization Condensation Polymerization 2 monomers react to from a larger unit eliminating water or HCl Eg polyamides polyesters Nylon 106 polyamide sebacoyl chloride 10 carbon diacidchloride hexamethylene diamine 6 carbon diamine Amide linkage forms between an acid halide and an amine with the linkage of HCl Instead of dicarboxylic acid use diacid chloride because acid halide is more reactive NaOH small amount added to neutralize HCl released each time a new amide bond formed Preparation polystyrene Styrene strong odor 9 HOOD Styrene in test tube benzoyl peroxide Dissolve solid heat in boiling water bath for 1 hour 9 mixture solidified or become viscous Break test tube with screw clamp remove polystyrene Test solubility o Acetone o Toluene o Ethanol Chain Addition Polymerization Polystyrene polymerization of styrene Uses initiator that forms a reactive species benzyl peroxide o Generates radicals when heated to 80 90 C o Radicals then react with styrene monomers to begin the chain addition process o Active species continues to add monomer units to end until all are used up Preparation Nylon106 20 NaOH solution 5 hexamethylene diamine solution in beaker 5 sebacoyl chloride pored down inside wall of hexamethylene diamine to oat sebacoyl chloride solution on top of solution already there nylon forms between interface between 2 layers wooden stick slowly lifts film from interface do not pull too fast or rope will break wind around test tube rotate around into a yard do not touch nylon 9 abrasive to skin cut rope and put in water thickness of rope collapses try between 2 pieces of filter paper probably monomers still left in beaker 0 mix contents vigorously with glass stirring rod 0 pour mixture in ice water filter with gravity filtration and wash nylon on filter paper with water 0 test solubility I acetone I toluene I ethanol Differences between Chain addition amp Condensation polymerization Chain addition 0 Vinyl type monomers simplest ethylene everything else just has substituent additions like styrene vinyle chloride acrylonitrile 0 Have simple hydrocarbon backbone 0 Functional group is pendant to polymer chain 0 Empirical formula is identical to monomer Condensation 0 Functional groups are part of polymer backbone and formed as the monomer units link together 0 Small molecule is produced as two monomer units link together empirical formulas of monomers nad polymers are different Cross linked Polymer Slime reaction of polyvinyl alcohol PVA with borax sodium borate decahydrate under aqueous conditions Gives elastic gel 0 Borate ion BOH394 coordinated to the alcohol groups in PVA to create a 3D network linking PVA chains together 0 Bonds in cross links between borate ion and alcohol functional groups of PVA are weak break easily under weight of gel 0 Leaving slime on at surface slowly atten out as the molecular chains slide over each other rearrange themselves and reconnect Preparing Slime Polyvinyl alcohol borax solution 9 stir mixture Solid and liquid chemical wastes in waste container in hood NONE out of sink Saftety Distillation a Simple and Fractional Distillation repeated simple distillation and combination and recombination of different distillate and condensation fractions 9 eventually could separate mixture of 2 volatile liquids with close boiling points 0 more volatile component would be more pure 0 tedious require large volume of initial liquid mixture fractional distillation 0 fractional column has extensive surface area that allows exchange of heat between ascending vapor and descending liquid condensation o as condensate of 2 component system system with 2 volatile liquids accepts heat from vapor condensate is partially vaporize and vapor enriched in more volatile component I therefore vapor loses heat to condensate 9 part of vapor condense and condensate enriched in less volatile component 0 vapor that condenses at top of column pure very enriched in more volatile comp I condensate that has returned to distillation ask pure or enriched with less volatile component separation 9 depends 0 on differences in boiling points of 2 liquids 0 rate of distillation 0 heat source 0 insulation of the column 0 how efficient the column is achieve good separation 9 even heating required along with low rate of distillation 0 thermal equilibrium and high re ex ratio 0 type of column column packing process of packing all determine efficiency or ability that column has for separating the components of liquid mixture Gas chromatography 9 to determine how well distillation configuration can separate 2 volatile liquids o For each fraction 9 composition by volume of each of the liquids being distilled Fumes o Liquids dripping out should be directly collected into graduated cylinder 0 If falls through air 2 problems vapors are released amp more volatile liquid evaporates more and introduces error Heating closed vessel 0 Collection port open to the atmosphere 0 PVnRT can have explosive effect due to pressure Part A Simple Distillation Slightly varying bps no need for fractional column 0 Liquid brought to boil 9 moves up still heaad and into condenser cooled back to liquid Cyclohexanetoluene mixture 121 by vol and 2 boiling chips to round bottom ask Flask on heating mantle on lab jack plugged into rheostat Connect distillation still head to ask May use glycerol very sparingly in order to prevent leakage during distillation 2 hoses connected to water cooled condenser lowerentry upperout attached to distillation head and fit bent adapter to end of condenser non mercury thermometer below bottom of side arm of distillation head 9 in ow of distillation vapor during distillation Tube 1 2 o raise rheostat controlling heat source to 30 extra 10 to start bring solution boil 0 adjust to maintain rate of distillation to 1 drop per 23seconds 0 record temperature every 05 mL o at 5 mL fraction collection transfer to another tube Tube 34 0 Raise heat gradually 35 40 o Cap all vial immediately prevent evaporation As distillation progresses temperature will need to increase because liquid mixture in distillation ask is enriched with the less volatile component Total of 20 mL stop Part B Fractional Distillation Boiling points less than 25 C apart Vapor moves up with cooling trays 9 condenses back to liquid 9 only vapor with lowest bp moves into condenser converted back to liquid o Cooling trays Better separation because provide theoretical plates so re uxing liquid can condense o More volatile will push up o Low bps will be at the bottom column pre packed with copper mesh 0 packing material 9 allows rising vapors to contact descending condensate large surface area offers little resistance to vapor wrap distillation volum with plastic tubing to insulate column o insulation 9 reduces heat loss reduces efficiency in separation of high boiling point liquids cyclohexanetoluene mixture in round bottom ask boiling chips o ask in heating mantle and plugged into rheostat o wrap top of ask and column with aluminum foil to obtain efficient heat transfer Tube 1 raise heat to 35 55 for first few mins Tube 2 raise heat to 40 Tube 34 raise heat gradually to 50 Maintain distillation to 1 drop every 23 seconds o Record temp every 05 mL Part C Gas Chromatography amp Correlation Factor Measure standard mixture of cyclohexanetoluene 9 34 uL microliters Rinse syringe a couple of times before using other solutions Chart recorder must be started just before the injection is made amp place cap back on the standard mixture 0 Not placing cap back affect data o Not rinsing GC syringe affect sample being analyzed o Effect of 1 minute delay between injection time and time of start of recorder have on the observed GC retention times or the appearance of chromatogram Calculate peak areas by triangulation o Correction factors for each component 9 peak areas from chromatogram from standard mixture and relative volume of standard mixture I Interested in relative response of 2 components number of ways to calculate them I Eg Injecting equal amountso feach liquid each component can be normalized corrected to one of the compounds in your 2 component mixture I 2 peaks shown in chromatogram 9 Ac amp AT I Half of 1 uL is volume of cyclohexane injected Ac area I Half of 1 uL is volume of cyclohexane injected AT area 0 Since cyclohexane and toluene have different molar heat capacities I Will have different responses with observed peak areas I Cyclohexane has higher molar heat capacity I Toluene will be standard 9 correction factor both relative to toluene AT AT 1 Ac AT 0 Volume percent of the 2 components in each distillation fraction I Vc Corrected Areac Corrected Ac Corrected AT x 100 I VT Corrected AreaT Corrected Ac Corrected AT x 100 Part D Gas Chromatography Analysis of Distillate For fractions of simple and fractional distillation Convert peak areas calculated by triangulation from chromatogram for each fraction 9 to relative volume using correction factors Chapter 2 Boiling point for Volatile Liquids heating a liquid results in an increase of vapor pressure of the liquid to the point where it equals the applied pressure usually atmospheric o where liquid observed to boil o normal boiling point is measured at 760 mmHg 760 torr or 1 atm boiling point can be measured during a distillation about 5 mL of liquid is sufficient for microscale distillation Chapter 3 Distillation Distillation process of vaporizing a liquid condensing the vapor and collecting the condensate in another container 0 Technique is very useful for separating a liquid mixture when the components have different boiling points or when one component will not distil o Principle method of purifying liquid in lab On molecular level o Boiling point corresponds to temperature at which vapor pressure of compounds equals the external pressure exerted on liquid o Therefore liquid boils when molecules are able to escape the surface of the liquid 0 The liquid is heated the more energy added to liquid phase 9 molecules of lowest boiling component will have enough energy to escape into gas phase o Equilibrium will establish itself while other molecules in gas phase will recondesne into liquid phase as their lose their energy Simple Distillation 0 Used to purify a compound that is almost pure less than 10 impurity or if impurities are non volatile 0 Eg Distilling sea water Fractional Distillation o Used to separate 2 liquid with rather similar boiling points Exp 5a Questions 1 A simple distillation is used to purify a compound that is almost pure already True 2 Temperature always remains the same throughout the distillation process False 3 Raoults law states that the sum of the vapor pressures the miscible components in a liquid mixture is equal the total vapor pressure of the liquid True 4 In fractional distillation the solvent with the lower vapor pressure will be removed first True 5 In gas chromatography the mobile phase is a noble gas True 6 Explain the difference between fractional and simple distillation and when you would use one over the other Simple is good for separating compounds that are almost pure and could be used to separate 2 volatile liquids but would have to be done repeatedly Fractional distillation requires less material than simple distillation and can be used more easily to separate two volatile liquid with similar boiling points 7 Would you expect ferrocene or acetylferrocene to come off the GC column first and why 8 Explain why you need a correction factor when using the GC along with how do you calculate it Because of Raoult s law triangulate the areas of the peaks and divide themby the area of the toluene peak this gives the correction factor which is multiplied with the areas of the peaks 5 Distillation b Steam Distillation Raoults law 9 if 2 liquids are miscible o the vapor pressure above the liquid ill be the sum of the 2 individual vapor pressures 0 partial vapor pressure for each component PA and PB depends on the vapor pressure of the pure component P A and P B and the mole fraction of that component NA and NB 39 Protal PA PB P A NA P B NB When 2 immiscible liquids are mixed in a heterogenous mixture 0 Each liquid exerts its own vapor pressure to give the total pressure of the system regardless of the amount of each component I PTotal POA PCB 0 Mole fraction doesn39t facor in 9 since 2 components are NOT miscible When PTotoa1 1 atmophere 9 system boils 0 Since total pressure comprises of each component system will boil at lower temperature than either components individually 0 Composition of vapor given off when boiling constant since total pressure is independent of amounts of 2 liquids present Even though 2 immiscible liquids may be mixed 9 behave if it they are distilled separately 0 They do give rise to homogenous vapor that co distills since all gases mix homogenously in any proportion 0 Once distillate condenses oil and water no longer miscible and 2 liquids separate isolation quick Steam distillation involves temperatures below 100 9 isolates compounds that are 0 1 Not miscible with water oils and tars 0 2 Have boiling points too high to allow for normal distillation where compounds may decompose at high temps Cloves Eugenia Caryophyllata 0 Rich in compound eugenol 4 allyl 2 methoxyphenol 0 Is a terpene alos a phenol or aromatic hydroyl compound 0 Bp of eugenol is 250 C would decompose it directly distilled Procedure Steam Distillation 0 Apparatus microscale round bottom ask on aluminum block and water condenser 0 Grind cloves with mortar and pestle record exact weight 0 mix cloves with DI water and in roundbottom ask with magnetic stir bar 0 make sure all securely clamped and tight will have to manipulate hickman distillation head Distillation hot plate on low and stir mixture and start low ow of water aluminum block must be at least 130 C to start distilling o if heat too strongly mixture will bump and froth and push clove bits into the Hickman head once liquids collects in Hickman head must be collected with pipet and transferred to centrifuge tube until you have 5 mL 0 normal for distillate to be cloudy Isolation of the Eugenol should have 2 layers in centrifuge tube more oil volume o oil on top or bottom Collect CH2Cl2 dichloromethane in Erlenmeyer ask for extraction keep stoppered as its volatile Used calibrated Pasteur pipet to add dichloromethane to centrifuge tube o Extracts neutral eugenol into organic layer 0 On top or bottom Cap tube shake and vent allow layers to separate Transfer lower dichloromethane layer to conical vial repeat twice and extract all water Dry dichloromethane with anhydrous soldium sulfate for about 510 mins then transfer to clean vial Rinse drying agent with small fresh dichloromethane Evaporate dichloromethane in hood using gentle heat and soft stream of air do not evaporate past point at which all solvent has evaporated o Remember product is a volatile oil can be evaporated eventually o Get to where volume stops changing rapidly Weigh the conical vial 9 recovery IR Spectrum of oil and compare to pure eugenol from library spectrum Chapter 3 Steam Distillation Steam Distillation o When an organic compound is co distilled with water o Advantage is that the desired material distills at a temperature below 100 C even thought the compound can have a high bp The laws of physics change for immiscible compounds each compound exerts its own vapor pressure independing of the other components and independent of molar fraction 1 step external source of steam used to heat water to boil and bubbled through compound to be distilled Steam mixture condensed in water condenser 0 organic and aqueous parts separate o distillate collected 2nd step steam generated in ask as compound to be distilled o steam will codistill with this compound Exp 5b Questions 1 Steam distillation is when an organic compound is codistilled with water True 2 Immiscible compounds vapor pressures are dependent on other components in the mixture False 3 When you mix dichloromethane and water the bottom layer will be the dichloromethane True 4 Steam distillation allows you to distill compounds at around 100 C even if its boiling point is higher True 5 What kind of condenser jacket will you be using in this lab Should you seal the top or leave it open A microscale water condenser leave it open never heat a closed container 6 What is the name of the compound you39ll be extracting Where does this compound collect and how should you exact it Eugenol in the port to the right of the Hickman head collect it with a pipet 7 What two items do you use to heat the round bottom ask Aluminum plate heating plate 8 What happens when you heat the solution too rapidly Define the term and explain what happens It bumps up on the sides of the container and can get stuck in the Hickman head 6 Resolution of Enantiomers rxns of racemic mixtures or achiral compounds almost always lead to a racemic product or achiral product 0 conversely 9 biological system can synthesize a single enantiomer from achiral or racemic starting materials because the enzymes that catalyze these rxns are chiral Enantiomers 0 Non superimposable mirror image isomers have same physical and chemical characteristics EXCEPT I Rotate plane polarized light in equal but opposite directions Rotate Left levorotatory Right dextrorotatory I React at different rates with other chiral compounds 0 Enantiomers cannot be separated frome ach other by physical separation 9 distillation crystallization extraction or chromatography 0 Racemic mixtures can be resolved separated by converting enantiomers chemically to diastereomer have different physical and chemical properties I Separating diastereomers physically then chemically reconverting them back into enantiomer Racemix 1 phenylethylamine alpha mehtylebzylamine or 1 phenylethanamine 0 Resolved with resolving agent stereoisomer of tartaric acid 9 I L tartaric acid amp 2R3R tartaric acid 0 yields a pair of diastereomeric salts 1 phenylethylammonium hydrogen tartrates 0 after mixing the solution becomes opaque and amine salts precipitate I the S 2R3R and R 2R3R salts are diastereomers I have different solubilities in methanol and can be separated with fractional crystallization I obtaining proper crystals 9 dense prisms 0 when solution is heated needles redissolve but prisms do not I to obtain prismatic crystals methanol solution is heated to near boil and rapidly cooled in ice bath to yield both types of crystals I then rewarmed to near boiling to dissolve the needles and set to cool slowly to form the prisms 0 the less soluble amine diastereometic salt is separated by vacuum filtration from the other salt which is dissolved in methanol I after separation and isolation of less soluble salt acidic treatment of amine with sodium hydroxide converts it to disodium tartrate and the nearly enantiomerically pure free S amine I disodium tartrate is water soluble but the amine is insoluble and forms a layer on top of the alkaline solution I amine is separated from the aqueous solution using a separatory funnel and dried with anhydrous potassium carbonate 0 optical rotation of crude amine will be measured using polarimeter Polarimetry Used to measure the optical rotation of product Light of single wavelength usually D line of sodium with wavelength5893 nm Then passes through Nicol prism CaCO3 9 filters out all light waves except those vibrating in single plane plane polarized light 0 The light is rotated 9 optical rotation 0 Specific rotation also depends on temperature standard temp 20 degrees and wavelength of light o Separate enantiomers rotate the plane of polarized light in equal amounts but in opposite directions and that an equimolar mixture of 2 enantiomers racemix mixture or racemate show no rotation zero optical rotation o no correlation exists between the absolute configuration R or S and the direction and that compounds rotate plane polarized light Degrees of the observed rotation alpha depends on o Structure of the compound o Concentration of the solution of the compound c 0 Length of the polarimeter tube 1 in which solution is placed specific rotation observed rotation dm length x concentration of sample gmL o Theoretical yield of the resolved pure S enantiomer is half of the mass of the stating racemic mixture Eg If 21g of amine are dissolved in 70 mL of methanol and placed in a 10 dm polarimeter tube with a total capcity of 70 mL If the observed polarimeter reading is 875 and the temperature is 20 degrees celcius then sr 875 100dm21g70mL 292 degrees The optical purity of the sample is the percent excess of one enantiomer compared ot the amount present in the racemix mixture A pure enantiomer is said to be 100 optically pure while a racemix mixture is 0 pure Optical purity measured specific rotation rotation of pure enantiomer x 100 Eg Samine enantiomer sr 403 degrees Then optical purity of amine 292403 x 100 725 This means that the amine contains 725 enantiomer and 275 of the racemate OR total enantiomer 725 12 x 275 863 total enantiomer 12 x 275 138 Safety Phenylethylamine is toxis and ammable o Avoid contact with skin and ames Methanol is toxic if ingested and ammable o Handle with care and avoid ames Na0H is caustic o Avoid contact with skin Procedure for Resolution of Enantiomers o Dissolve racemic mixture in methanol and mix in Erlenmeyer ask o Dissolve tartaric acid in methanol in ask may be cloudy o Add amine solution to tartaric acid solution in large ask o Warm solution until boil then cool in ice bath to crystallize the tartrate salts I 2 kinds of crystals are observed I prisms mp 190200 I needles mp 168170 o reheat mixture almost to boil to dissolve needle crystals I residual prismatic crystals act as seed crystals for amine diastereomeric salt I stopper the ask and store 0 vacuum filter prismatic crystals and wash with cold methanol I pour filtrate into organic waste jug I save small amount of crystal and let them dry for melting point analysis 0 rest of the crystal in ask and add aqueous sodium hydroxide and warm for 5 mins then cool and transfer in separatory funnel I with draw lower aqueous layer discard I pour top amine layer into centrifuge tube with anhydrous potassium carbonate drying agent I if amine is still cloudy after drying it can be clarified further by centrifugation 0 weight empty graduated cylinder transfer amine to it and reweigh to determine the yeld of amine in g I gather 4 mL worth with combining with others I fill to 7mL with methanol and mix well I transfer solution to polarimeter cell with no air bubbles trapped in cell I measure optical rotation of amine solution 9 quite small about 10 to 11 for about 3 g of amine diluted to 7mL with methanol in a 10cm 1dm tube 0 pure S amine has specific rotation of 403 Alternate NMR technique for determining optical purity 0 NMR technique that uses an optically pure chiral resolving agent to produce diastereomers of the R and S configuration of the amine 0 Sample prep for NMR is very involved and impractical Exp 6 Questions 1 What is a racemic mixture Is it optically active It is an equal mixture of 2 enantiomers it is not optically active 2 What 2 things does the specific rotation depend on the length of the polarimeter tube the concentration of sample in gmL 3 What is optical purity The percent excess of one enantiomer compared to the amount present in the racemix mixture
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