CHM 205 Post Lab 2023 with verified questions and answers

Exp 0: Define melting point range Span of temperature from the point at which the crystals first begin to liquefy to the point at which the entire sample is liquid. Exp 0: What is the proper heating rate used to determine a compound's melting point range? What would happen if you exceeded this heating range? Proper heating rate: 2 degrees C per minute Exceeding this heating rate would introduce errors, such as premature liquefaction or putrefaction of the compound. liquefaction: process that generates liquid from solid putrefaction: process of decay of organic matter Exp 0: Describe what happened in each case. You are trying to determine the melting point of a compound and (a) the compound turns from white to brown before melting or (b) the compound slowly disappears from the capillary tube before melting. (a) The compound was denatured forming a different, undesired compound. This suggests that an error occurred and the crystal must be purified again or reported with the temperature followed by a "d". (b) An impurity is present. If this sublimation event occurs, the crystal must be repurified and the melting point determined again, but this time the capillary tube should be sealed before measurement. denature: take away or alter the natural qualities of Exp 0: What are the melting points of...? (a) ferrocene and acetylferrocene (b) adipic acid and citric acid (a) ferrocene: 173-176 degrees C acetylferrocene: 81-83 degrees C (b) adipic acid: 151-154 degrees C citric acid: 153-154 degrees C Exp 0: Adipic acid and citric acid have very similar melting points. Given two unknown samples, one being adipic acid and the other citric acid, how could you determine which was which? - calculate mixed melting point - compare boiling points - titrate each and compare the results to known pKa values for acids Exp 1: Why is it important to use only the minimal amount of hot solvent needed to achieve a recrystallization? What would be the effect of using too much solvent while dissolving a solid for recrystallization? Too much solvent will make it difficult for the crystals to recrystallize as there will be too much liquid for the crystals to dry quick enough if at all. The crystals will remain partially dissolved in the solvent and will not precipitate. Exp 1: What do you think would happen to your recrystallization solution if you did not use a pre-heated filter during the "hot" gravity filtration step? The crystals would have crystallized on the filter paper itself instead of going through the filter paper into the beaker. Exp 1: Benzyl alcohol has a boiling point of 205 degrees C. It has solubility characteristics that would seem to make it a good choice for use as a recrystallization solvent for fluorenol (mp 153-155 degrees C), but in fact benzyl alcohol is a poor choice of solvent for this recrystallization. Why? The boiling point of the solvent, benzyl alcohol, is above the melting point of the fluorenol so as the solution cools, the liquid fluorenol will separate. Exp 1: Explain how each of the following contaminants is removed during the recrystallization process: (a) sand (inorganic contaminant) (b) colored organic contaminants (c) a very soluble contaminant (a) Sand can be removed through basic filtration as it will not be able to go through the filter paper. (b) Colored organic contaminants will be removed by adding charcoal and subsequent hot gravity filtration. (c) A very soluble contaminant can be removed by hot gravity filtration. Exp 1: Why is acetone a popular solvent for cleaning glassware? 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 major concern when using acetone? It dissolves almost any organic compound and is a very polar molecule which will ease removing things from glassware. It would not make a good solvent as it will be soluble with the compound. Using acetone is a major concern because it is extremely flammable. Exp 2: Suppose that just before adding a drying agent to an organic solvent 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? No, if there are still water droplets in your organic layer, they must be removed with a pipet before the addition of the drying agent because the drying agent can only remove minute amounts of water. Exp 2: Devise a general extraction scheme for separating: (a) an organic base mixed with an organic neutral compound (b) an organic acid and a phenol (a) Add aqueous HCl to the solution, which converts the base to an ammonium chloride salt which is rather soluble. Thus, the neutral compound remains in the organic layer and the base is now in the aqueous layer. (b) First, add MtBE and then a weak base, such as sodium bicarbonate. This will isolate the stronger organic acid, which is now bound to the sodium ion in the aqueous layer. The layers should be separated, then a strong acid, such as HCl, should be added to the aqueous layer in order to return the acid to its original organic state. Now that both the phenol and the organic acid are dissolved in the organic solvents, the solvents can be evaporated and the dry compounds isolated. Exp 2: When two immiscible solvents are mixed, two layers form. Liquid-liquid extractions commonly use organic 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 two layers was the aqueous layers? The first and simplest method is to add water to the solution and observe which layer it dissolves in. Whichever layer this is will be the aqueous layer. Another simple method is to add a simple salt, such as NaCl, to the solution. Whichever layer the salt dissolves into is the aqueous layer. You could add an organic compound to the solution and whichever layer it dissolves in will be the organic layer. Exp 2: 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 an acid (RCOOH) or a base (RNH2)? Why? Potassium carbonate is basic so it would perform much better as a drying agent for a basic compound. If the potassium carbonate were added to the acid, an acid-base reaction would occur, producing water instead of dehydrating the sample. Exp 3a: ortho-Hydroxyacetophenone has a melting point of 4-6 degrees C. para-Hydroxyacetophenone has a melting point of 109-111 degrees C. Explain the basis for the sizable difference in melting points of these two compounds. Difference in melting points is due to differences in their chemical structure. ortho-Hydroxyacetophenone has a hydroxyl group proximal to a functional group containing a double bonded oxygen. This placement of the functional group results in INTRAmolecular hydrogen bonding and thus a lower affinity for other identical molecules. para-Hydroxyacetophenone has a hydroxyl group located opposite of the functional group containing the double bonded oxygen. This placement results in INTERmolecular hydrogen bonding and thus a large affinity for other identical molecules around it. This results in the formation of a lattice causing higher melting points. Exp 3a: 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 two unknown halogenated alkenes, but you see only one spot with an Rf of 0.91. The solvent used was ethyl acetate. (b) You run TLC on a mixture of a thiol and an amine, and again you get only one spot with an Rf of 0.15. The solvent was a mixture of petroleum ether and dichloromethane. (c) When you put your TLC plate in the developing chamber, the solvent in the chamber covered the spots on the TLC plate baseline. (a) The solvent used is too polar causing the solute to be highly attached to the mobile phase and only slightly absorbed to the stationary phase. This results in minimal separation of the compounds and both compounds will travel far. To fix this, use a less polar solvent. (b) Our solvent is not polar enough causing the solute to hardly move up the chromatography paper. To fix this, use a solvent that is more polar. (c) If the solvent covers the spots on the TLC, then your experiment is ruined because the spotted solutes will disperse into the solvent. If this happens, the solute will likely not climb the chromatography paper at all and the experiment will have to be repeated so as to only add enough solvent to cover the bottom of the chromatography paper but not the spotted solutes. Exp 3a: You run a TLC plate spotted with three compounds: naphthalene, o-toluic acid, and fluorenol. Predict the relative Rf values. The factor which influences the Rf value in the case of TLC is the polarity. The decreasing polarity follows as: o-toluic acid, fluorenol, and naphthalene. The compound with the larger Rf is less polar because it interacts strongly with the polar adsorbent on the TLC plate. Thus, the decreasing Rf values follows as: naphthalene fluorenol, o-toluic acid. Exp 3a: For each of the following pairs, predict which compound will have the larger Rf value if both are run on a SiO2 TLC plate in 10% acetone/hexane: 4-decanone or 4-decanol; xylene or benzoic acid; cycloheptane or cycloheptanone. 4-decanone is a ketone which will travel farther and have a higher Rf value. Subsequently, 4-decanol is an alcohol and is thus more polar, causing it to adsorb to the silica gel and move up less. Xylene is less polar than benzoic acid and thus will travel farther and have a higher Rf value than benzoic acid. Cycloheptane is less polar and thus will travel farther and have a higher Rf value than cycloheptanone. Exp 3a: Arrange the following solvents in order of increasing polarity: ethyl acetate, dichloromethane, iso-propanol, ethanol, toluene, and heptane. heptane toluene dichloromethane ethyl acetate n-propanol ethanol Exp 3a: Name three important things to keep in mind while spotting TLC plates. Make sure the spots are higher than the developing solvent. Keep the diameter of the drops small when spotting the solute onto the silica gel. Larger diameter drops will produce blurry, indistinct lines. Spot multiple times in order to achieve a high concentration of sample in as small a spot as possible. Exp 3b: Why does ferrocene elute from the column first? Why was the solvent changed in the middle of the column procedure? It is less polar than the acetylferrocene and thus more attracted to the relatively nonpolar mobile phase rather than the polar stationary phase. The mobile phase was switched to MtBE in the middle of the column because acetylferrocene needed a more polar mobile phase in order to be eluted from the column. Exp 3b: Ferrocene eluted from the column first, but it ended up as the higher of the two spots during TLC analysis. Explain. TLC and column chromatography are almost exact opposites in that the flow of solvent is down in column chromatography and up in TLC. Thus, due to the direction of solvent flow in the two methods, column chromatography will have its least polar compounds toward the bottom and its most polar compounds toward the top while TLC will have its least polar compounds toward the top and its most polar compounds toward the bottom. This is why ferrocene was at the bottom of the column chromatography and at the top of the TLC. Exp 3b: Why is it important to limit the sand bath temperature to ~60 degrees C when evaporating solvent from your collected fractions? If the temperature of the sand bath is too high, the solute may melt or vaporize. For example, the melting point of acetylferrocene is around 80 degrees C, so if the sand bath were set much higher than the acetylferrocene compound would melt and negatively affect the yield we have to calculate. Exp 3b: In running column chromatography 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 a column on a mixture of two unknown halogenated alkenes, but they both elute from the column at the same time. The solvent used was methylene chloride (dichloromethane). (b) You run an alumina column on a mixture of a thiol and an amine, but you never get any compound off the column. The solvent was a mixture of petroleum ether and dichloromethane. (a) When this occurs, you can conclude that the two halogenated alkenes have too similar of retardation factors for a solvent to separate. To fix this problem, the student should use two different solvents, starting with a less polar solvent and switching to a more polar solvent. The use of two different solvents would result in a separation of the two mixed compounds. (b) If neither compound elutes off the column, it can be concluded that the solvent is not polar enough to strip the compounds from their affinity to the stationary phase. To fix this, a more polar solvent should be used. However, it cannot be too polar otherwise the compounds will elute off at the same time. Exp 3b: You run a column to separate a mixture of three compounds: naphthalene, o-toluic acid, and fluorenol. The column is run with a solvent system based on hexanes, but which becomes more polar with the addition of more and more dichloromethane over time. Predict the elution order. Acids are generally more polar than alcohols, which are generally more polar than alkenes. The elution order will go from less to more polar compounds: naphthalene fluorenol o-toluic acid. Exp 3b: Suppose you are given 150 mg of an unknown mixture that contains three compounds, and you are told to separate the compounds using column chromatography, using alumina as a stationary phase. How would you decide what solvent or solvent system to use? The first step would be to run multiple TLC plates spotted with the mixture and various mobile phases to determine which solvent results in the best separation of the mixture. Once one or two good solvents have been found, column chromatography can be run. The first solvent should be the least polar of the chosen solvents with the more polar of the chosen solvents being introduced midway into the elution process. Exp 3b: The compounds used in experiment 3b were both highly colored, so it was easy to see their progress as they moved down the column during elution. Most organic compounds, however, are colorless in solution. How might you monitor the process of a column chromatography given a colorless sample? If the bands were colorless, a very slow drip rate would have to be set up and TLC plates would have to be run as the elution is occurring in order to determine which compounds are being eluted. Exp 4: Define the following terms: (a) monomer (b) repeating unit (c) condensation polymerization (d) cross-linked polymer (a) monomer: a molecule that can be bonded to other identical molecules to form a polymer (b) repeating unit: part of a polymer whose repetition would produce the complete polymer chain, except for the end-groups, by linking the repeating units together successively along the chain (c) condensation polymerization: two monomers react to form a larger unit while a small molecule like water or HCl is eliminated (d) cross-linked polymer: a chemical bond formed between adjacent chains of a complex molecule Exp 4: Most organic polymers are named based on the monomers used in their production. However, poly(vinyl alcohol), PVA, is not made from vinyl alcohol, but instead from vinyl acetate that is hydrolyzed after formation of the polymer. Why can't we use vinyl alcohol in this polymerization? PVA is made from vinyl acetate rather than vinyl alcohol because vinyl alcohol is an unstable compound and is rapidly converted into a more stable form, acetaldehyde. Exp 5a: Note that each vertical line in the graph grid equals 5% and the scale runs both ways for benzene and toluene: 100% benzene - 0% benzene 0% toluene - 100% toluene (a) What is the % composition of the vapor in equilibrium with a boiling liquid that has a composition of 30% benzene and 70% toluene? (b) If a sample of vapor is found to have a composition of equal parts benzene and toluene, what is the composition of the boiling liquid that produced this vapor? (a) 35% toluene and 65% benzene - find point on liquid line that makes 30% benzene and 70% toluene and go across to find equal point on vapor line (b) 82% toluene and 18% benzene - find point on vapor line that makes 50% benzene and 50% toluene and go across find point on liquid line Exp 5a: (a) Explain meaning of horizontal line AB. (b) What physical transformation does the line DE represent? (a) It represents an isothermal line that means temperature is constant for the time being. (b) It represents vapor to liquid transformation without change in composition. Exp 5a: Under what conditions can a good separation be achieved by simple distillation? Can only be achieved if there is a large difference ( 100 degrees C) in boiling points of the liquids to be separated Exp 5a: Explain why the boiling point of your cyclohexane:toluene mixture rose slowly throughout your simple distillation procedure. As the distillation proceeds, the vapor becomes richer in the more volatile cyclohexane and the liquid remaining in the flask becomes enriched in toluene. As the composition of the liquid mixture changes to a higher mole percentage of toluene, the total vapor pressure decreases, which results in a higher temperature being required for the mixture to begin boiling. Therefore, as the percent of toluene in the mixture increases, the boiling point will also increase. Exp 5b: Why do we use steam distillation to isolate eugenol rather than purify it by simple distillation? Eugenol has a boiling point of ~250 degrees C. This temperature would have to be obtained in simple distillation, risking the decomposition of the sample which would produce poor and impure yield. Instead, steam distillation lowers the boiling point of the compound to ~100 degrees C since the initial mixture is heterogeneous. The vapor pressures are independent of mole fractions, leading to a quicker boiling point. Exp 6: How could you isolate the (R)-(+)-a-phenylethylamine from the mother liquor that remained after you crystallized and filtered off the S-(-)-a-phenylethylamine? The solution could be heated to near boil and then cooled in an ice bath. This would cause the amine salts to form needle-like crystals, which could be vacuum-filtered out of the solution. Finally, the addition of NaOH would convert the compound back to its enantiomer form. Exp 6: The formation of a white solid is often observed after a-phenylethylamine comes into contact with carbon dioxide in the atmosphere. What is that white compound? L-phenylalanine Exp 6: If the boiling point of the pure S-(-)-a-phenylethylamine is 186-188 degrees C, what is the boiling point of its enantiomer? What optical rotation would be observed if 3.72 g of the (S)-amine were mixed with 3.72 g of the (R)-amine? Enantiomers have the same physical properties, thus the boiling point of its enantiomer would also be 186-188 degrees C. Equal quantities of (S) and (R) form a racemic mixture, so the optical rotation of the mixture would be 0. Exp 6: Propose a method for how you would resolve (+-)-2-chloropropanoic acid using (-)-amphetamine as the resolving agent. The mixture would first be boiled with the resolving agent. This solution would then be cooled in an ice bath so that crystals form. The crystals would then be reheated and vacuum filtered from the solution. Finally, the diastereomeric crystals would be reconverted back to enantiomers by adding HCl. The final enantiomer would be heated, separated into the organic layer, and dried.