Distillation Using The Hickman Still PDF

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Name: Hina Warsi

NSHE ID: 1008421439

Date: June 3, 2011

Title of Experiment: Distillation Using the Hickman Still Diagram of Apparatus:

Result: My hypothesis regarding the composition of the first test tube —Fraction 1 will contain solely hexane liquid. The second test tube will contain a mixture of both Hexane and Toluene liquids; lastly, the third test tube —Fraction 2 will contain solely Toluene liquid. Conclusion: At 72 ºC (Time: 14 min) we collected our fraction 1 which contains Hexane liquid. At temperature 84 ºC (Time: 20 min) we collected the intermediate fraction which was a mixture of both Hexane and Toluene liquid. Lastly all the remaining liquid was collected at 110 ºC (Time: 28 min) which contains Toluene liquid. Toluene can be separated from the Hexane and Toluene mixture using Hickman’s distillation because Hexane has a lower boiling point causing it to condense and separate itself faster. Post Lab Questions: 1. Diagram (Tº/ Time (min)) Type

Fraction 1 (70-72 ºC)

Intermediate (82-100ºC)

Minutes 0 2 4 6 8 10 12 14 16 18 20 22

Temperature (ºC) 24º C 25º C 26º C 27º C 28º C 60º C 68º C 72º C 76º C 80º C 84º C 92º C

24 26 28

Fraction 2 (110ºC)

104º C 108º C 110º C

Graph of Time vs. Temperature 110 º C 100 º C 90 º C 80 º C 70 º C Temperature (º C) 60 º C 50 º C 40 º C 30 º C 20 º C 10 º C 0ºC 4

8

12 14 16 Time (Minutes)

20

24

28

2. What would be your expectations regarding the composition of Fractions 1 & 2? (Hint: Are these fractions pure or not?) Both fraction A and fraction B are pure since the distillation of both occurred at different temperatures. 3. What are the boiling points of hexane and toluene? Hexane’s boiling point is 69 º C & the boiling point of Toluene is 110.6 º C. 4. Which component will be distilled from the mixture first? Hexane.

Name: Hina Warsi

NSHE ID: 1008421439

Date: June 24 , 2011

Title of Experiment: Diels Alder Reaction –Experiment #10 Results: 1. Observations: a. The vial and its contents were liquid after the addition of the 3-Sulfolene, Maleic Anhydride and all of the Xylene. Once everything had been heated and cooled, it was still in liquid form. After the solution had the gone through the micro column it was diluted and appeared to be a light yellowish pigment. Soon after adding a couple drops of hexane solution to the Craig Tube containing the yellowish solution, a precipitate of small crystals formed and then after placing it in the hot water bath, the crystals in the solution were no longer visible, yet there was a dingy precipitate at the bottom of the tube. However after placing the tube in a cold water bath, the crystals had once again become visible and then it was centrifuged to separate the crystals from the excess solution.

Amount Used Melting Point Actual Mass Yield Theoretical Mass Yield

Maleic Anhydride

3-Sulfolene

0.096 g 53 º C —60 º C -------

0.259 g 64 º C—66 º C -------

Product: Cis-4-Cyclohexene-1,2Dicarboxylic Anhydride ---103 º C 0.014 g 0.1489 g

Work Shown of Theoretical Yield & Percent Yield: (0.096g of Maleic Anhydride) * (1 mole/ 98 grams) * (1 mole of Cis-4-Cyclohexene-1, 2-Dicarboxylic Anhydride/ 1 mole of Maleic Anhydride) * (152 g / 1 mole) = 0.1489 g Percent Yield: 0.014 g / 0.1489 g * 100% = 9.4 % 2. Which reagent is limiting and which one is the excess? Limiting Reagent: Maleic Anhydride Excess Reagent: 3-Sulfolene

Conclusion: I would say that our results met our expectations only to the extent that we didn’t have to repeat the experiment to accumulate our product. Our actual mass yield was quite small in comparison with our theoretical mass yield (Simply put, what we recovered was much smaller than what we should have ideally recovered). These results could be due to experimental errors that took place during the experiment. Since this experiment takes a lot of time to complete, we might have turned up the heat at a faster rate on the hot plate than it needed to be, therefore losing or burning our product. Also while transferring the solution from the vial to the micro filtration we actually lost some to our pipette because of the lack of Toluene solution that was mixed with it, which made our solution jam inside the pipette and crystallize. Additionally we might have not added enough hexane to our filtered solution in the Craig Tube which could have resulted in less crystals. Another thing that might have contributed to experimental error could have been the lack of time that the solution in the Craig Tube was in the hot water bath. And when we centrifuged our Craig Tube, some crystals were left at the bottom of the centrifuge tube and that resulted in the loss of final product. Post Lab Questions: 1. Draw the mechanism for the conversion of 3-Sulfolene to 1,3-butadiene

2. What reagents (diene and dienophile) would you need to prepare 1-methyl-4propylcylohexane?

Name: Hina Warsi

NSHE ID: 1008421439

Date: June 22 , 2011

Title of Experiment: Elimination: E1 Reaction—Experiment # 8 Results: See Results Page Conclusion: I would say yes, our results to the experiment met our expectations. When comparing the E1 reaction to the E2 reaction, there are many differences. In the E1 reaction, a uni-molecular process creates a carbo-cation intermediate instead of the process occurring simultaneously as in E2. In E1, it is only necessary for the base to affect the rate of the reaction of the substrate instead of looking at both the base and the concentration. In the chromatogram, the peaks are labeled by the boiling points from lowest to highest since GLC is based on the temperature at which it spends its time in the vapor phase. Therefore, the peaks following the boiling temperature is 1-butene, trans-2-butene and cis-2-butene. In our results, it showed that the trans2-butene is the most substituted, and thus the major product. In addition to this, since there is a cis and trans product during the reaction, it shows stereo selectively. It is because of this that the trans is the more stable one in the E1 elimination reaction. Post Lab Question: 1. Discuss the outcome of the E1 reaction: Which alkene was formed as the major product? Why? Which was formed as the minor product? Why? a. In this experiment the major product that is formed is the trans-2-butene. Trans-2butene is the major product because of its structure and because it is energetically favored [since it is a Zaitsev’s product and exhibits steric hindrance]. The minor product that is formed is 1-butene. It is the minor product because it is unstable when compared to the other two, and it is less substituted.

Name: Hina Warsi

NSHE ID: 1008421439

Date: June 22 , 2011

Title of Experiment: Elimination: E2 Reaction—Experiment # 9 Results: See Results Page Conclusion: I would say yes, our results to the experiment met our expectations. The E2 elimination reaction occurs simultaneously, and this is because the rate of the reaction depends on the concentration of the substrate and base (the catalyst). The results showed that the trans-2-butene was a major product and formed the most isomers. This makes sense because the trans-2-butene is more substituted. Therefore, the trans-product is more stable than the cis-product. As the result, the cis and trans reaction had a more anti-periplanar geometry compared to the other one where the leaving group was different as it was less substituted, making it easier to form a certain isomer than the other. Post Lab Question: 2. Discuss the outcome of the E2 reaction: Which was formed as the major product? Why? Which was formed as the minor product? Why? What percent of 1-Butene product was produced? Would you expect this percent to be high or low? a. In our experiment, the major product that formed was trans-2-butene. It was the major product, because the product formation of it was 56%. In addition to that, trans-2-butene is the major product because its structure is more substituted and energetically favored since it forms a Zaitsev product and has steric hindrance. 1butene was the minor product that was forming, since it has a lower percent composition.

Name: Hina Warsi

NSHE ID: 1008421439

Date: June 15, 2011

Title of Experiment: Extraction—Experiment #5 Results: Compound

Molecular Weight

Grams Used

Mass Recovered

Percent Yield

Experimental Melting Point

Unfiltered Benzoic Acid Filtered Benzoic Acid 9- Fluorenone Crystallized 9- Fluorenone

122.12 g

--------

--------

--------

120 º C—127 º C

Literature Melting Point 122.4 º C

122.12 g

0.104 g

0.085 g

43.3%

114 º C—117 º C

122.4 º C

180.21 g --------

0.104 g 0.045 g

0.045 g 0.019 g

81.7% 42.2%

67 º C—69 º C 69 º C—72 º C

83.5 º C 84 º C

Work Shown: Benzoic Acid Benzoic Acid w/ Watch Glass – Watch Glass (Actual Yield / Theoretical Yield) * 100 %

57.821 g – 57.776 g = 0.045 g (0.045 g / 0.104 g) * 100 % = 43.3%

9-Fluorenone 9- Fluorenone w/ Vial – Vial (Actual Yield / Theoretical Yield) * 100 %

23.006 g – 22.921 g = 0.085 g (0.085 g / 0.104 g) * 100 % = 81.7 %

Crystallized 9- Fluorenone Crystallized 9- Fluorenone w/ Watch Glass – Watch Glass 53.012 g – 52.993 g = 0.019 g (0.045 g / 0.019 g) * 100 % = 42.2 % (Actual Yield / Theoretical Yield) * 100 % Reactions:

Conclusion: After obtaining the results, it is clear that extraction is used to purify compounds— our experimental data is relatively close to the theoretical data. Our obtained melting points are relatively close to the literature melting point [in the given range], which means that we were able to purify our compounds accordingly. Simply put, I believe that the extractions met our expectations, although the percent yield is satisfactory, errors could have resulted throughout the experiment. Errors such as; when separating the aqueous layer and organic layer, some of the organic layer may have been transferred into the flask containing only the aqueous layer, and the transfer of our 9-Flourene liquid to the cotton pipette, some of the solution was lost since the pipette was pouring out the liquid too fast, etc. However, aside from the minor errors that took place throughout the experiment we were still able to collect a sufficient amount of compound back. Post Lab Questions: 1. You have been given a mixture (1:1) of the two compounds shown below, diethyl ether, aqueous NaHCO3, aqueous NaOH, and aqueous HCl. Using a flowchart, outline your procedure and explain how to separate, isolate and purify these two compounds.

2. How could the purity of each compound be assessed? a. The purity of different compounds can be assessed by comparing the melting points of each.

Name: Hina Warsi

NSHE ID: 1008421439

Date: June 8, 2011

Title of Experiment: Gas Chromatography (Gas-Liquid Chromatography) Analysis of a Mixture –Experiment 2 Results: For results, refer to chromatographs. Conclusion: The results met the expectation that Hexane would have a retention rate shorter than that of Toluene. These expectations arose from the hypothesis stated in the pre lab. Since Hexane’s boiling point was lower than Toluene it was able to transform into a vapor much faster and therefore was the first to show up on the chromatograph. However since Toluene has a much higher boiling point than Hexane it took longer for it transform into a vapor and therefore held a longer retention time. Post Lab Questions: 1. Using your chromatographs for prepared 1:1 mixture, Fraction #1 and Fraction #2, calculate the percent composition and retention time for each of the above mixtures. Refer to chromatographs 2. Using the results for each fraction, give a comment on the effectiveness of your distillation Our distillation was effective due to the fact that Fraction 1 was completely Hexane and Fraction 2 was solely Toluene. This is evident because after having our distillation complete last week we had our three different liquids in three separate test tubes, and after returning to lab on Monday, it was apparent that Hexane had completely evaporated while there was a great amount of Toluene still present. When referring to the chromatographs that were printed, you can see that Hexane had a retention rate of 3.5 minutes while Toluene was 5.6 minutes. Also, the chromatographs showed that Fraction 1 and Fraction 2 only one peak existed for each compound, further proving that in our test tube containing Hexane; only Hexane had to have been present for all of it to have evaporated.

Name: Hina Warsi

NSHE ID: 1008421439

Date: June 24, 2011

Title of Experiment: Competitive Aromatic Nitration—Experiment #11 Post Lab Questions: 1. Explain why the –NH-CO-CH3 group is an ortho/para director while the –COOCH3 group is a meta director using carbo cations intermediates. a. The –NH-CO-CH3 group is an ortho/para director because when the electrophile attacks at the ortho/para position, an intermediate is formed which has a resonance structure where the positive charge is stabilized by donation of electron pairs from nitrogen. Also, in this resonance form, all the carbons in the ring are an octect. This resonance form is very stable, thus favored. When the electrophile attacks the meta position, this resonance structure is not formed. Furthermore, ortho/para attacks results in more resonance structures than the meta attack. b. The –COOCH3 is a meta directing group because when the electrophile attacks the ortho/para position, an intermediate is formed with a resonance structure that places a positive charge next to the δ+ carbon and a δ− oxygen. The two similar charges repel each other, raising the energy of the intermediate, thus making the intermediate less stable. Because it is unstable, the intermediate is less likely to form. When an electrophile attacks the meta position, this unstable intermediate is not formed. Thus, –COOCH3 directs the attacking eletrophile towards the meta position

2. Which product do you expect to form in the nitration reaction: methyl metanitrobenzoate or para-nitroacetanilide. Why? a. The product I would expect to form is para-nitroacetanilide because the –NH-COCH3 substituent is an activating group while the –COOCH3 group on the methyl nitrobenzoate is a deactivating group. Whether a substituent is an activating group or a deactivating group depends on two things: Inductive effects and resonance effects. Resonance effect is withdrawal or donation of electrons through p-orbital overlap. Inductive effect is withdrawal or donation of electrons due to electronegativity. Since N is more electronegative than C on the ring, -NH-COCH3 has a slight electron withdrawing inductive effect, but since it has a longer of electrons it has a strong resonance donating effect. Overall, NH-CO-CH3 donates electrons to the ring making it more stable, lowering electronegativity, thus making the benzene ring more reactive. –COOCH3 has a strong resonance electron withdrawing effect due to its partial δ+ charge. It also has a strong electron withdrawing inductive effect. Overall, –COOCH3 takes electrons away from the benzene ring, making it less reactive.

3. Name the possible side products of this reaction and explain how you would remove them from your desired products. a. Possible side products include: i. Ortho-nitroacetanilide, a small amount of meta-nitroacetanilide, metanitrobenzoate, a very small amount of ortho-nitrobenzoate and paranitrobenzoate. These side products can be removed by recystallization; recrystallization that involves very slow cooling that would allow for larger more pure crystals to form.

Name: Hina Warsi

NSHE ID: 1008421439

Date: June 20, 2011

Title of Experiment: Nucleophilic Substitution: SN1 and SN2 Reactions—Experiment # 7 Results: See Results Page Conclusion: I would say that our lab results did meet our expectations. I say this because within our experiment the ones containing Bromine compounds showed to be a better leaving group in the SN1 reaction, since it was the weaker base, also because it reacted faster with 1% AgNO3 unlike the compounds that contained Chlorine. It was also showed that the more substituents a carbon had that was attached to the Halogen, the faster the SN1 reaction took place. Yet, in the SN2 reactions, our results led us to believe that the organic halides with the least amount of steric hindrance or with the smaller nucleophiles had quicker reaction rates. Post Lab Questions: 1. For the SN1 reaction, rank all the organic chlorides in order of increasing reactivity. How does the structure of the organic compound affect the rate of the SN1 reaction? Least Reactive Chlorides  Most Reactive Chlorides: SN1 Reaction Chlorobenzene, 1-Chlorobutane, 2-Chlorobutane, Benzyl Chloride, t-Butyl Chloride The structures that have the most stable carbocations will have the fastest reactions. 2. For the SN2 reaction, rank all the organic bromides in order of increasing reactivity. How does the structure of the organic chloride affect the rate of the S N2 reaction? Least Reactive Bromides  Most Reactive Bromides: SN2 Reaction Bromo-Benzene, t-Butyl Bromide, 2-Bromo-Butane, 1-Bromo-Butane, Iso-Butyl Bromide, Ally Bromide ≈ Benzyl Bromide The structures that maintain less hindrance around the site of attack will undergo SN2 reaction the fastest. 3. Discuss the effect of the leaving group in the S N1 reaction by comparing your results for t-butyl bromide and t-butyl chloride. Both, T-Butyl Bromide and t-Butyl Chloride have stable 3° carbocations and that is why the reactions occurred within 1 to 2 seconds of the reaction, so it was too fast for comparison. Although it was too fast for comparison, a rule of thumb is

that a weak base is a better leaving group. So in this case, Br – happen to be the better leaving group than Cl-

4. Why are the Allyl bromide and benzyl bromide very reactive in both the SN1 and the SN2 reaction? The resonance structures of both Allyl Bromide and Benzyl Bromide make both of the organic halides more stable in SN1 and SN2 reactions. 5. Why are the bromobenzene and chlorbenzene so unreactive? Bromobenzene and Chlorobenzene are nonreactive because if they were reactive an incoming nucleophile would have trouble reacting. The nucleophile would need to attack the area of the double bond and this wouldn’t be favored in the reaction due to the high amount of energy needed to break the bond. 6. Complete the following reaction showing the SN2 mechanism

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Procedures for Safety Equipment Fire Extinguisher Small Fire in a CONTAINER  place wire gauze screen with ceramic fiber or watch glass over the container Method doesn’t work and no one able to help? Use the Fire Extinguisher! PASS 1. Keep your back to the exit 10 to 20 ft away. 2. Pull the pin –discharges the extinguisher 3. Aim low –point nozzle at the base of the fire 4. Squeeze the lever above the handle 5. Sweep from side to side *MAKE SURE THE SAFETY OFFICER INSPECTS THE FIRE SITE afterwards Safety Shower Clothing on fire or chemical spills  Use safety shower Concentrated Acid on clothing  TAKE OFF CLOTHING before shower 1. Pull triangle handle 2. Shower for at least 15 minutes 3. Medical attention need for serious injury Mercury Spill Kit IF you break a mercury thermometer 1. Wear the gloves from the kit 2. Sprinkle sulfur (or Zinc) powder (that’s in the kit) on the spilled mercury 3. Collect with dustpan and brush and place in DESIGNATED plastic bag 4. Thermometer is disposed in...