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CHEM 3AL: Organic Chemistry Lab

by: Carolyn Smullin

CHEM 3AL: Organic Chemistry Lab CHEM 3AL

Marketplace > University of California Berkeley > CHEM 3AL > CHEM 3AL Organic Chemistry Lab
Carolyn Smullin
GPA 3.78

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These notes include answers to the post-lab questions assigned each week.
Organic Chemistry Lab
Organic Chemistry Lab
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This 11 page Bundle was uploaded by Carolyn Smullin on Thursday February 4, 2016. The Bundle belongs to CHEM 3AL at University of California Berkeley taught by in Spring 2014. Since its upload, it has received 432 views.

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Date Created: 02/04/16
Experiment #2: Investigating Solubility and Acid-Base Reactions Discussion 1. Analysis: a. Part 1: Solute miscibility with a nonpolar solvent: Like Dissolves Like We mixed hexane (nonpolar) with: - diethyl ether - miscible - ethyl acetate – miscible -acetone - miscible -ethanol - miscible -dichloromethane – miscible -water – immiscible b. Part 2: Solute miscibility with a polar solvent: We mixed water (polar) with : -Diethyl ether - immiscible -Ethyl acetate - immiscible -Dichloromethane -immisicble -Acetone - miscible -Ethanol - miscible -Toluene – immiscible c. Part 3: water solubility of alcohols Add water to the following alcohols Ethanol – after 25 drops ethanol still dissolved in water (miscible) 1-propanol – after 25 drops still dissolved in water (miscible) 1-butanol – after 10 drops solubility was reached (partition formed) (immiscible) 1-pentanol – after 5 drops solubility was reached (partition formed) (immiscible) This is because small alcohols are completely soluble in water. Whatever propor tions you mix them in, you will get a single solution. However, solubility falls as the length of the hydrocarbon chain in the alcohol increases. Once you get to four carbons and beyond, the fall in solubility is noticeable, and you may well end up with tw o layers in your test tube d. Part 4: Temperature Dependence of Solubility We formed a mixture of benzoic acid and water. The two were immiscible (acid crystals remained solid at bottom of tube.) However, when we heated the tube in the sand bath, after coming to a boil the solution was clear and the crystals had dissolved. But when we allowed the solution to cool to room temperature, a precipitate formed – this shows that solubility increased with temperature and reduced when the temperature dropped. e. Part 5: Solubility of an Acid We mixed benzoic acid with water, then added the following acids: NaOH (base) – acid dissolves NH4OH (base) – acid dissolves NaHCO3 (base) – acid floats, doesn’t dissolve, bubbles HCl (acid) – acid sinks to bottom, doesn’t dissolve Diethyl ether (base) – two liquids are immiscible and the acid floats Solubility of an acid in water increased with addition of alcohol. f. Part 6: Solubility of a Base We mixed diisobutyla mine (base) with water and added: NaOH (base) – immiscible, mixes with stirring then separates again HCl (acid) – produces heat and creates a solid layer where two meet Diethyl either (base) – immiscible, mixes with stirring then separates again Solubility of base in water increased with addition of an acid (clear, no partition) but with addition of a base, a partition formed (immiscible) g. Part 7: Solubility of an Acid -Base Mixture We mixed diisobutylamine (base) with benzoic acid crystals in water. The more base we added to the acidic solution, the less miscible the so lution became and the more defined the partition. With addition of drops of base we added, the more bubbles formed, meaning the two solutions are not mixing at all even when stirred vigorously. Solubility of acids and bases together decreases with addit ion of acid or base. 2. Chemical Reactions a. All of our reactions consisted of us manipulating the intermolecular forces present in the pure compounds. No bonds were broken. In the case of solute miscibility, the intermolecular forces in the solvent interacted with the intermolecular forces of the solutes to form a solution, miscible or not. In the ca se of the acid-base solubility, base and acid reacts. Base takes proton from acid, so benzoic acid is deprotonated to produce sodium benzoate, which is the one dis solving in water since it is an ionic compound. So, benzoic acid appears to be dissolving, but is actually being deprotonated. 3. Further Predictions NOTE: like dissolves like does not always work. If solute molecules disrupt the solvent in such a way as to produce a structure that was LOWER IN ENTROPY, then the mixture would be less likely to exist than the separate liquids. But if the two mix and the entropy of the mix is more positive than any negative change in enthalpy, then the mix can occur and the two liquids are miscible. In order for two substances to be soluble , the solute and solvent must form sufficient intermolecular bonds with each other that can overcome the bonds between each of the substances individually. a. 1,2-dichlorethane in hexane: hexane is nonpolar, and dichloro ethane is polar. WOULD be soluble, because they are both entirely covalent and thus the dispersion forces between them would be sufficient for the two to mix together. b. 1,2-dichloroethane in water: The chlorinated compounds are polar organic molecules, but because they cannot hydrogen bond with water are immiscible with water. WOULD NOT be soluble, because the dispersion forces that could occur between the two sub stances are much weaker than the hydrogen-bonds between the water molecules. c. Dibromomethane in water : WOULD NOT be soluble, same r eason as with the dichloroethane. d. Methanol in water: short chain alcohols (methanol and ethanol) are miscible with water as the OH can hydrogen bond with water molecules. In addi tion the small hydrocarbon part makes them very much less hydrophobic than a molecule with a longer hydrocarbon chain. As the chain length increases the molecule becomes more hydrophobic and the solubility in water decreases. WOULD be soluble, because the -OH group on such a small molecule would form hydrogen -bonds with the water, allowing plenty of bonding between the methanol and water molecules. e. 1,4-butanediol in water: WOULD be soluble, but not as soluble as the methanol, because again it's the -OH group forming the hydrogen-bonds with the water, but there's a higher organic vs alcohol - group ratio, so it’s harder for the hydrogen -bonds to keep the rest of the organic dissolved. Questions: 1. Polar Protic: ethanol Polar Aprotic: acetone 2. Two organic molecules in this experiment that in their pure liquid states interact mainly through ID-ID forces: - Hexane - Diethyl ether - Ethanol 3. When 157 mg of dipentylamine was added to 122 mg benzoic acid and 2 mL water, a homogenous solution formed. Draw structure of species present in water - H2O will H-bond to the NH on dipentylamine and to the OH on benzoic acid. 4. When 1.57 g of dipentylamine was added to 122 mg of benzoic acid and 2 mL of water, two distinct layers were observed. Explain. - the limiting reagent is benzoic acid. Benzoic acid will dissolve in water along with dipentylamine but the dipentylamine would be left over. So the two layers are water and dipentylamine. 5. Water-acetone: miscible, H-bonding. Water-ethanol: miscible, polar and polar, H-bonding. Water-diethyl ether: NOT MISCIBLE – diethyl ether on top because it is less dense than water. Water-dichloromethane: NOT MISCIBLE – dichloromethane on bottom because it is more dense. Water-ethyl acetate: NOT MISCIBLE – ethyle acetate on top. Hexane-acetone: MISCIBLE - hexane is just a large non polar hydrocarbon compound. Acetone has CH3 on the two sides and this will make it miscible in hexane. Acetone dissolves both polar compounds and nonpolar comounds well . Hexane-toluene: MISCIBLE, nonpolar and nonpolar. 6. “Benzoic acid dissolved in water upon the addition of a saturated solution of sodium bicarbonate”: INCORRECT - Base takes proton from acid, so benzoic acid is deprotonated to produce sodium benzoate, which is the one disso lving in water since it is an ionic compound. S o, benzoic acid appears to be dissolving, but is actually being deprotonated and its conjugate base sodium benzoate is dissolving. Experiment #3: Mixed Melting Points Discussion Part 1: In order to find another student with an unknown solid iden tical to my own, I would take an equal ratio of my solid and theirs, grind and mix, and then place in a melting point capillary. I would then take a melting point measurement, and if the melting point was depressed from what my solids melting point range was, then I knew the solids were not identical. I continue d this until I found a solid which, when mixed with my own, produced a melting point range nearly identical to my pure solid. Part 2: See Worksheet Part 3: See worksheet Questions: 1. Sand will not change the melting point range of a crystalline compound X because it is an insoluble impurity. 2. Baking soda will change the melting point range of the compound because it is a soluble impurity, thus it will depress and broaden the MP range. 3. The reason that the mixed sample had two melting ra nges is that the two compounds were insoluble in each other, and so each maintained its original melting point range in its pure state. The solution was not at the eutectic point. 4. You cannot rerun a MP range with the same sample that is already melted, bec ause you will not know when the MP range begins and ends. OR: resoldifying a compound will throw off MP data because the compound may have decomposed during the heating process. 5. A solid usually melts over a range of temperatures rather than at one specifi c temperature. In addition, the melting point range is affected by a number of factors in addition to that of purity. Particle size, amount of material used, density of packing in the capillary tube, thickness of the capillary tube, and the rate of heating in the melting point apparatus, are all factors that should be carefully considered to ensure an accurate melting point. Experiment #4 Discussion: 1. In our table of known data, we are given values for the dipole moments, solubility in water, density relative to water, and boiling point temperatures of numerous molecules. In the lab, we measured dipole moments of an unknown liquid via a charged rod, measure solubility in water by mixing the unkno wn in water and observing if a l ayer formed, measured density relative to water by determining which of the two was the top layer of the partition, and measured BP by immersing a ther mometer into a boiling solution of our unknown in a sand bath. From these observations, we went through our list of knowns and found the best match to our unknown. 2. ? Questions: 1. We can accurately determine the density of a liquid by these steps: - First weigh the beaker - Then add an exact amount of liquid - Weigh the beaker now with the liquid - Subtract the initial weight from the second weight to determine weight of liquid - Now divide this mass by the volume of the liquid you added - This is your density (mass/volum e) 2. The thermometer is not placed directly in liquid, rather in the reflux lin e (vapor to liquid equilibrium). Hold the thermometer above the liquid because you want to measure the temperature of the hot vapor, not the liquid (impurities will alter the BP w hen the probe is in the solution, but impurities in vapor do not affect BP.) 3. I think you would need the BP in order to accurately distinguish the four unknowns, because with just dipole, solubility, and density information we still do not know much. - Cis has a higher BP than Trans. You need polarity and solubility and density data to identify if the BP are too close. Run tests with the controls if you have them in lab and compare. 4. Since glucose is soluble in water, we would expect the addition of glucose to water to lower the melting point/freezing point of water, and increase the boiling point (due to a decrease in vapor pressure.) This is because glucose is a soluble, nonvolatile impurity. Lab 6: recrystallization of an unknown solid and decolor ization of brown sugar Discussion: 1. After testing the various solvents for solubility, our group selected the solvent that dissolved the solid upon heating, and recrystallized the solid when removed from heat and placed in an ice bath. We had unknown solid D, and chose water as our solvent. 2. After recrystallizing solid in our solvent of choice, water, we took a MP range of our recovered crystals. We compared the solubility of our unknown solid and MP range of the crystals with our known data, and determined the identity to be camphoric acid. 3. The solubility data was helpful in narrowing down the possibilities. Since there were two compounds with MP ranges similar to ours, we compared the solubility in w ater and ethyl to accurately determine which of the two our solid was. 4. Decolorization of brown sugar using activated charcoal: we weighed out brown sugar and made a brown sugar and water solution (color of amber). We then added 10 mg of charcoal powder to the solution, mixed it, and boiled it for 5 min. After this, we passed the warm solution through a filter, added water to the filtered solution, and filtered it a second time. The solution turned clear, like water. The charcoal attracted the large particle s through IDID and DID interactions to remove the colored impurities. Questions: 1. To determine if all of the brown sugar solution material (including the sugar) is adsorbed onto the activated carbon, you could take a MP of the filtered solution and see if i t is the same as pure water, or depressed due to presence of a soluble impurity (sugar.) 2. It is not a good idea to use a large excess of activated charcoal to remove colored impurities, because if there is more charcoal than impurities, some charcoal will r emain in the solution and solution will be black. 3. Student starts with 100 mg unknown solid, recrystallizes it and weighs the crystals. The crystals weigh 120 mg and the MP is very broad and much lower than expected: this is because the student probably did not wash the crystals with ice cold solvent and then let them dry before weighing and taking a MP. The solvent covers the crystals and acts like a soluble impurity, depressing the MP. 4. Student starts with 100 mg unknown solid, recrystallizes it, and weighs the crystals. The crystals weigh 27 mg. What happened: student may have used too much solvent and so not many crystals recovered ; student didn’t allow for a complete recrystallization to occur (letting it stay in ice bath long enough, letting the crystal s dry before weighing). They can fix this by obtaining a second crop of crystals: evaporate some of the solvent from the mother liquor and again cool the solution. Experiment #7: TLC Discussion: 1. Criteria used to narrow down list of four solvents to two to use in second part of experiment: - Solvents that led to large Rf values for all the compounds - Solvents in which the compounds were all far from eachother 2. We compared the unknown to the four known analgesic compounds on the TLC plates developed in our two chosen solvents. We looked for which analgesic our unknown was most similar to (Rf values, visualization, etc.) and found that to be aspirin. 3. I predicted that ibuprofen would have the highest Rf value, then aspirin, then acetaminophen, and caffeine would have the smallest. That lab results proved this prediction to be true. This is because our mobile phase was nonpolar, and ibuprofen is the least polar of the compounds, meaning that it will travel the furthest on the plate. As the compounds become more polar, their Rf value decreases. Questions: 1. It is important to have the spotting line on your TLC plate above the solvent level line in the developing chamber, because if the developing solvent is above the level of the origin point (i.e. where you spotted your material), some of the material that is absorbed onto the TLC plate will be washed off the plate before the experiment even begins. 2. Since we observed only o ne spot on the origin line in a solvent mixture of 90:10 hexane: ethyl acetate, we should next try a mixture of 90:10 ethyl acetate: hexane to make our mobile phase slightly more polar. Our compound is obviously polar (as it has strong interactions with stationary phase on origin line) and so if we increase polarity of the mobile phase we should see some movement of the compound away from the origin line. 3. Observing one spot does not necessarily mean you have only one product. You must undergo various visualization techniques to determine that conclusion. A product may show up under an iodine chamber, but not a UV lamp if it is not conjugated, and so on. It is likely that at least one product is present, but it could be the case that more than one product ha s a similar polarity to the product viewed, appearing as one spot. Thus this cannot prove that there is only one spot. 4. Only conjugate compounds (multiple bonds separated by a single bond) show up under UV light as conjugated double bonds absorb UV light. 5. In a developing solvent of 90:10 ethyl acetate: hexane, we would expect the most non polar compound to have traveled the furthest on our plate. The least polar spot would travel the furthest because it would not react strongly with the polar stationary phas e. Experiment #8 Discussion 1. Plate 4: - We observed our 3 pure H/S samples and adulterant sample - Found which were UV, I2, and both active - Noticed similarities between adulterant path and pure samples Plate 5: - Compared adulterant to a pure H/S that was most similar to the adulterant, and another groups pure H/S sample believed to be part of our adulterant - Spots aligned/matched up - Thus adulterant sample is composed of our pure H/S and the other groups pure H/S 2. By comparing Rf values and noting appearance/odor, we were able to determine what 2 compounds made up our adultered sample Questions: 1. After co-spotting two samples with the same Rf value and visualization data, we obtain a dot with a shadow beneath it. This tells us the two samples were not the same. Even though two samples may have the same Rf data and visualization techniques, they are not necessarily the same. Even though their previous plates gave the same Rf values, their assumption that the two samples were the same was incorrect. This could be cause d by experimental error in their TLC data. They also only used UV light in their visualization techniques, and should additionally use I2 for better accuracy. 2. One student used hexane to extract essential oil and another ethyl acetate. Will tlc plate be the same? - No, because hexane is a saturated hydrocarbon (nonpolar) while diethyl ether contains an ether group (polar). The ether group will extract the polar impurities, while hexane would not. 3. In a developing solvent 90% hexane: 10% ethyl acetate, the most nonpolar molecule will travel the furthest (as polar will interact strongly with polar silica gel.) In 100% hexane, same thing will occur but Rf values will be smaller (samples will not travel as far because solvent is more nonpolar . 4. 3,7-dimethyl-1,6-octadien-3-ol is racemic. We will not see 2 spots on a tlc plate because the two are enantiomers and have the same polarity. 5. Citral is composed of two isomeric alkenes, cis and trans. Because the two are diastereomers, we would expect to see 2 spots on a TLC pl ate. Diastereomers are oriented differently in space and have different polarities, so their IMF will be different. Thus, how they react with the polar stationary phase and nonpolar mobile phase will be different based on polarity. Experiment #9: What do you take for Pain? Discussion: 1. In this experiment, we reacted sodium naproxen (soluble in water) with 95% ethanol. The solubility of naproxen in 95% ethanol is 1g/25 mL. What we did is we protonated sodium naproxen to isolate naproxen. The sodium will dissolve in the ethanol to become an aqueous acidic layer while the naproxen precipitates out. 2. We recrystallized the solid using a co -solvent system of ethanol and water. We chose these two solvents because in order to carry out a co -solvent mechanism, you must have one solvent in which the compound is very soluble in and the other in which it has limited solubility. The two solvents must be miscible. Thus, water and ethanol are a perfect pair. Naproxen is soluble in ethanol, but insoluble in water. We could not use diethyl ether or chloroform because they are not miscible in water. 3. N/A Questions: 1. The Opadry coating is removed using ethanol as opposed to water because sodium naproxen (sodium salt) is very soluble in water and would disintegra te. By using ethanol, the tablet stays intact and we assure a higher percent yield in the end. 2. The results from the TLC plate did not prove the product was pure. Although there was only one spot under UV light and I2 chamber, impure substances could have similar interactions with the developing solvent and visualization techniques as the compound we spotted, and so are not visible. TLC would not be able to distinguish between chemicals with similar interactions and properties. 3. In order to isolate aspirin, acetaminophen, and caffeine from Excedrin, one could follow this procedure: a) First, dissolve the pill in dichloromethane. Because acetaminophen is insoluble in dichloromethane and both aspiri n and caffeine are soluble, acetaminophen will precipitate out and the other two will form a liquid and can be removed using a Pasteur pipette. b) Because caffeine has similar reactivity to diisobutylamine, it will precipitate when strong acid is added, such as HCl or HBr. Aspirin, however, will not precipitate, so the solid caffeine can be removed. c) At this point, aspirin is in a solution of dichloromethane and acid. One can obtain solid aspirin through heating and then cooling to perform recrystallization. 4. To purify the caffeine by recrystallization, I would use hot water. Caffeine dissolves in hot water, but not in room temperature water, which makes recrystallization very easy. Ethanol could also be used, but caffeine is not as soluble in ethanol and may pro duce less pure results. 5. Sulfuric acid was used to neutralize the sodium salt of naproxen rather than acetic acid because it is much stronger. The hydrogen ion in sulfuric acid will dissociate more readily and therefore will more effectively protonate (neut ralize) the naproxen. Experiment #10 Discussion: 1. If a precipitate formed, it was indicative that the reaction took place. In part 1 with NaI/Acetone, the reactions that occurred were SN2 (NaI/Acetone good nucleophile for SN2). So if a precipitate did not form, it meant that no SN2 reaction occurred, and the electrophile was no suitable for SN2. In part 2 with AgNO3/Ethanol, the reactions that occurred were SN1 (AgNO3/Ethanol is a bad nucleophile, only does SN1). So if a precipitate did not form, it meant no SN1 reaction occurre d, and that the electrophile was not suitable. 2. Some molecules did not react win part one because they were not suitable electrophiles for SN2 reactions (due to 3-beta branching, 3-alpha branching, and ring strain.) All of the molecules reacted in part two. However, we predicted that those electrophiles that were primary carbocations would not react, and yet they did. 3. Br is the better leaving group between Br and Cl, so the activation energy for the rate determining step will be less in both SN1 and SN2 reac tions - ie reactions with Br will be faster than with Cl as the leaving group. Questions: 1. It is important to use dry tubes for the reaction involving NaI/Acetone because the reaction is an SN2 mechanism. SN2 prefers to react in a polar aprotic solvent. Ace tone is polar aprotic, but water is polar protic, so the presence of water would slow down the SN2 reaction (if not hinder it by H -bonding with Acetone.) 2. To obtain a good yield of product from a reaction of 1 -bromopropane with NaI to give 1 - iodopropane, I would use a polar aprotic solvent. A polar aprotic solvent would favor an SN2 reaction and favor the production of 1 -iodopropane. 3. 2-bromopropane reacts with NaI/acetone nearly 10^4 times faster than bromocyclopropane due to steric hinderance and ring stra in in the transition state of bromocyclopropane. The secondary carbon on bromocyclopropane is more sterically hindered than that on 1 -bromopropane, and so nucleophile has a harder time attacking the carbon. 4. See worksheet. 5. See worksheet.


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