Lab Report 3 Final Copy - Grade: A PDF

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Lab Report 3 Final Copy...

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Isoborneol Oxidation and Camphor Reduction

Lead Author: Hannah Strickland Editor: Reviewer:

Chemistry 238 Section G6

Experiment 3

Introduction: Oxidation and reduction reactions are a part of everyday life, in areas such as fires, rusting metal, and even a banana rotting.1 Not only are they important in everyday life, they are also important in the aspects of organic chemistry. In this experiment, oxidation and reduction were observed. Oxidation is the increase of carbon-oxygen bonds, or a decrease of carbon-hydrogen bonds. Reduction is the decrease of carbonoxygen bonds, or an increase of carbon-hydrogen bonds.2 Observation of oxidation was achieved by oxidizing isoborneol to camphor. This was done by using the oxidizing agent hypochlorous acid to turn the alcohol group into a ketone. This reaction is shown in figure 1. Reduction was achieved by reducing camphor to isoborneol and borneol. Camphor was reduced by using the reducing agent sodium borohydride. The ketone was reduced back to an alcohol. This reaction is shown in figure 2. If isoborneol is oxidized to camphor, and then camphor is reduced, it will form two products (isoborneol and borneol) due to the fact that there are two possibilities for a nucleophilic attack.

Figure 1: Figure one shows the mechanism for the oxidation of isoborneol to form camphor.

Figure 2: Figure two shows the mechanism for the reduction of camphor. A top face nucleophilic attack results in the product borneol and a bottom face nucleophilic stack results in the product isoborneol.

Table 1: Table of Reagents Compound

Boiling Point (°C)

Melting Point (°C)

Molecular Weight (g/mol)

Density (g/cm3)

Acetic Acid

117.9

16.7

60.052

1.05

Bleach

101.0

18.0

74.439

1.21

Boric Acid

300.0

170.9

61.831

1.50

Borneol

213.0

208.0

154.253

1.01

Camphor

209.0

175.0

152.253

0.99

34.6

-116.3

74.120

0.71

212.0

212.0

154.253

0.992

—

1124.0

120.361

2.66

64.7

-97.8

34.042

0.79

Sodium Bicarbonate

851.0

50.0

84.007

2.20

Sodium Bisulfite

851.0

150.0

104.061

1.48

Sodium Borohydride

500.0

400.0

37.833

1.07

Sodium Hydroxide

1,388.0

318.0

39.997

2.13

Sodium Sulfate

1429.0

884.0

142.040

2.66

100.0

0.0

18.015

1.00

Diethyl Ether Isoborneol Magnesium Sulfate Methanol

Water

Experimental: The first part of the experiment was the oxidation of isoborneol. First, 1.5 mL of acetic acid, 4.0 mL of bleach, and 0.509 g of isoborneol were obtained. The acetic acid and isoborneol were added to a 25 mL erlenmeyer flask and the isoborneol dissolved. Then, the bleach was added and the mixture was swirled. A precipitate immediately formed and the solution then turned a yellow-white color. The mixture was allowed to completely react for 10 minutes. Then, the solution was tested with a starch iodine test strip, which was negative. This indicated that all of the excess bleach had reacted. Then, 15 mL of water was added to the mixture and the mixture was transferred to a separatory funnel apparatus. Next, 15 mL of diethyl ether was added to the separatory funnel, along with 1 mL of sodium bisulfate. The separatory funnel was shaken three times, and off-gased after each time. Then, the aqueous layer was removed from the funnel and 10 mL of sodium bisulfate was added to the organic layer. Then, 10 mL of sodium bicarbonate was added to the funnel, the funnel was shaken, and the aqueous layer was removed. The aqueous layer was then tested with litmus paper. The test strip turned blue, indicating that the layer was basic. Next, anhydrous sodium sulfate was added to the organic layer to remove any water present. The organic layer was then decanted from the drying agent and into a beaker. Lastly, the beaker was placed in a warm bath at 37.5°C to allow the ether to evaporate. Then the beaker was weighed, a percent yield was calculated, the melting point was determined, and an IR spectrum was done on the product, camphor. This IR spectrum is shown in figure 3.

The second part of this experiment is the reduction of camphor. First, 0.123 g of camphor was obtained and placed in a 10 mL erlenmeyer flask, along with 0.5 mL of methanol. Next, 0.114 g of sodium borohydride was added in four parts to the mixture. The flask was then placed in a hot bath for 2 minutes. Then, 3.5 mL of ice water was added to the mixture. The mixture was then poured into a suction filtration apparatus to allow for drying. The solid from the suction filtration was transferred to a 10 mL preweighed flask and 4.5 mL of ether and some anhydrous magnesium sulfate were added. This mixture was then placed back into the suction filter apparatus and filtered again. This process was allowed to go on for five minutes. Then, the liquid portion from the suction filter apparatus was placed in a warm bath for 10 minutes to allow the ether to evaporate. Lastly, a percent yield was calculated, a melting point was determined, NMR was done, and an IR spectrum was done as well. Results: During this experiment the oxidation of isoborneol to camphor, and the oxidation of camphor to isoborneol and borneol were observed. The product of the oxidation of isoborneol formed camphor. An IR spectrum was done on the product of this reaction, this graph is shown in figure 3. The product of the reduction of camphor formed two products, isoborneol and borneol. These products were analyzed by using IR spectroscopy, shown in figure 4, and H-NMR, shown in figure 5. Next, the molar ratio calculations are shown. They are calculated by using the integration of the isoborneol peak and the borneol peak from the H-NMR graph, shown in figure 5. Finally, the percent yield calculations are shown for camphor and isoborneol/ borneol.

Figure 3: Figure three shows the IR spectrum for camphor. The carbon-hydrogen bond (3000-2800 cm-1) and the carbon-oxygen double bond (~1736 cm-1) are labeled, as well as an impurity (3500-3300 cm-1).

Figure 4: Figure four shows the IR spectrum for the products of the reduction of camphor, which are isoborneol and borneol. The carbon-hydrogen bond (3000-2800 cm-1) and the oxygen-hydrogen (35000-3200 cm-1) bond are labeled.

Figure 5: Figure five shows the H-NMR spectrum for the two products from the oxidation of camphor: isoborneol (3.75 ppm) and borneol (4.00 ppm). The integration of the isoborneol peak is 7.6351 and the integration for the borneol peak is 1.000. These two products are labeled with their differentiating peaks, along with what hydrogen causes these peaks labeled in red. NMR: Molar Ratios from Integration Calculations 1.00 (borneol integration) + 7.6351 (isoborneol integration) = 8.6531 Isoborneol: 7.6351 ÷ 8.6351 x 100% = 88.42% Borneol: 1.000 ÷ 8.6531 x 100% = 11.58% Camphor Percent Yield Calculations: 23.121 g (flask with product) — 22.479 g (flask) = 0.642 g Actual yield: 0.642 g camphor ÷ 152.23 g/mol camphor = 0.00423 mol Theoretical yield: 0.509 g isoborneol ÷ 154.253 g/mol isoborneol = 0.00330 mol Percent yield: 0.00423 ÷ 0.00330 x 100% = 128.18%

Isoborneol and Borneol Percent Yield Calculations: 30.924 g (flask with product) — 30.840 g (flask) = 0.084 g Actual yield: 0.084 g ÷ 154.253 g/mol isoborneol/borneol = 0.000545 mol Theoretical yield: 0.123 g camphor ÷ 152.23 g/mol camphor = 0.000808 mol Percent yield: 0.000545 ÷ 0.000808 x 100% = 67.45%

Discussion: During this experiment, isoborneol was oxidized by hypochlorous acid to form camphor. Then, camphor was reduced by sodium borohydride to form two products which were isoborneol and borneol. In the reaction of oxidizing isoborneol (shown in figure 1), the alcohol is oxidized to a ketone. This is a type of elimination.3 In the reaction of the reduction of camphor (figure 2) the ketone is reduced to an alcohol by sodium borohydride. This reaction will form two different products (isoborneol and borneol) depending on where the reducing agent attacks camphor. If the reducing agent approaches from the top (also known as an exo attack), then borneol is formed. If the reducing agent approaches from the bottom (also known as an endo attack), then isoborneol is formed. Both products are stereoisomers of each other.4 The product of oxidizing isoborneol was camphor. At the end of the first part of this reaction, the product of this oxidization was analyzed. This was done by an IR spectroscopy and determining melting point. The IR spectrum, shown in figure 3, shows a C-H sp3 stretch at 3000-2800 cm-1 and a C=O stretch at ~1736 cm-1, which are both present in camphor. The IR spectrum also shows an impurity stretch at 3500-3300 cm-1. Because the stretch is similar to an O-H stretch, this impurity most likely came from water or ether still present in the final product. The most likely factor was that the drying agent did not remove all of the water from the solution, or the ether did not completely evaporate in the warm bath. The melting point was also taken on the product. The melting point of the product was determined to be 174-179°C. The melting point of camphor, shown in table one, is 175°C. Due to the lower and broadened melting point of the product, other substances, such as water or ether, were most likely present with the camphor.5 The percent yield calculated, shown in the results, also confirmed that impurities were present. The percent yield calculated was 128.18%, which is impossible if the product was just camphor. Most likely, there was water and ether present in the product causing such a high percent yield over 100%. The product of reducing camphor was isoborneol and borneol. These products were analyzed in several ways. The first way was done by an IR spectroscopy, shown in figure 4. The IR spectrum shows a C-H sp3 stretch at 3000-2800 cm-1 and an O-H stretch at 35000-3200 cm-1. Both of these bonds are present in isoborneol and borneol, indicating that they are not impurity stretches. Another analysis of the products was done by H-NMR spectroscopy, shown in figure 5. Both isoborneol and borneol have an H group beside the -OH group. Due to the different stereochemistry in each product, the deshielding of each -H is different. The -H in borneol is more deshielded, placing it at 4.00 ppm. The -H in isoborneol is more shielded, placing it at 3.75 ppm. This difference is due to the location of the hydrogens. Because the hydrogen is closer to the -OH group in borneol, due to stereochemistry, it is going to be more deshielded. The -OH group in borneol essentially pulls more of the hydrogen’s electron density, because it is closer to it than the hydrogen in isoborneol. Because isoborneol has less steric hindrance it is going to be more stable than borneol, which has more steric hindrance.6 Another factor is that the -OH group is in the equatorial position in isoborneol, which is

the more stable form. Because isoborneol is more stable, it is going to be the major product. This can be calculated by using the integration of the according peaks on the H-NMR graph. This calculation is shown in the results section. The molar ratio of the product was 88.42% isoborneol and 11.58% borneol. Another analysis of the product was done by melting point. The melting point observed was 202-205°C. The melting point of isoborneol is 212°C, and the melting point of borneol is 208°C, both shown in table 1. The lower and broader melting point of the product obtained could be explained by the fact that the product was a mixture of isoborneol and borneol in the product, which both have different melting points. Another factor could also be impurities present in the product such as water or ether. Finally, a percent yield was calculated, which is shown in the results section. The percent yield calculated was 67.45%, which is a reasonable percent yield. There is a possibility that this percent yield could contain impurities along with the product. Conclusion: In this experiment, oxidation and reduction were observed by oxidizing isoborneol and reducing camphor. These were done through the process of mixing the chemicals with oxidizing and reducing agents. The products of the oxidation and reduction experiments were analyzed by IR spectroscopy, melting point, and H-NMR spectroscopy. Using the H-NMR integrations, the molar ratios of the two products from the reduction of camphor were calculated. The ratio was 88.42% isoborneol and 11.58% borneol. This ratio is explained by the stability of isoborneol over borneol. This experiment could be improved in several ways. The biggest complication faced was the occurrence of impurities in the products, which messed up data to a degree. If impurities, such as water and ether, were removed more efficiently from the final product then the results could have improved. References: 1Olson, M. V. oxidation-reduction reaction https://www.britannica.com/science/oxidationreduction-reaction (accessed Feb 9, 2017). 2Definitions of oxidation and reduction (redox) http://www.chemguide.co.uk/inorganic/ redox/definitions.html (accessed Feb 9, 2017). 3Oxidation of Isoborneol to Camphor http://www.brynmawr.edu/chemistry/Chem/ mnerzsto/Labs/Isoborneol-to-camphor-August-5-2015.pdf (accessed Feb 11, 2017). 4Preparation and Stereochemistry of Bicyclic Alcohols http://cms.cerritos.edu/uploads/ lwaldman/212Lab/212Experiments/212labexp07_stereochem_camphor_new.pdf (accessed Feb 11, 2017). 5Why do impure solids melt at lower temperatures: melting points explained http:// kirsoplabs.co.uk/lab-aids/impure-solids-melt-lower-temperatures/ (accessed Feb 11, 2017). 6What is shielding and deshielding in NMR? Can you give me an example? | Socratic https://socratic.org/questions/what-is-shielding-and-deshielding-in-nmr-can-yougive-me-an-example (accessed Feb 11, 2017)....