West Virginia University Green Oxidation of Borneol to Camphor with Oxone Lab

Green Oxidation of Borneol to Camphorwith Oxone
Rocky Lifton
Taylor Campbell
Garrett Craft
An oxidation reaction is essentially the process in which an atom loses
electrons. Of course, to compensate, another atom within the molecule must
undergo a reduction in which the atom gains electrons. This entire process is
termed as a “redox-reaction.” In organic chemistry however, it is common that these
conditions aren’t the set criteria, and rather, the removal of a single oxygen atom
from a molecule (or the addition of two hydrogen) will constitute a “redox-reaction.”
Figure 1.1 and 1.2 respectively illustrate typical oxidation, and reduction reactions.
The dangers of a typical oxidation were avoided with the usage of Oxone, as
opposed to a bleach oxidation. Some oxidations can result in serious risk of
carcinogenic effects. Oxone avoids those risks as well as being far more
environmentally friendly (thus duly labeled as Green.) In place of bleach, Oxone was
used to convert borneol to camphor; as opposed to bleach the reaction yields higher
amount of product.
Figure 1.1: Illustrates a basic Oxidation reaction utilizing
PCC to oxidize a primary alcohol to an aldehyde.
2) H 2O
Figure 1.2: Illustrates a basic Reduction reaction utlizing
DIBAL-H to reduce an ester to an aldehyde.
H 2O
H Cl
H 2O
H 2O
Figure 1.3: Illustrates the mechanism in which camphor is synthesized.
H 2O
H 3O
Figure 1.4: Illustrates possible side reaction during side reaction during synthesis of camphor.
Experimental Section:
Combine 0.2g borneol,
1mL EtoAc, Magne8c S8r
Bar, 50mg NaCl, 1mL DI
Add brine and shake
Combine the two organic
Take TLC of ini8al
Isolate the organic layer
Add MgSO4 decant, and
Use I2 Chamber to spot
Remove magne8c s8r
bar and add 1mL EtoAc
Camphor is now
acquired, weigh,
calculate percent yield,
take IR, take 13CNMR
S8r for 30 minutes
u8lizing magne8c s8r bar
Add 3mL DI water,
NaHSO3 test aqueous
layer with starch I2 paper
Table of Chemicals:
Utilized to acquire the
Chemical formula of
synthesized product
C10H18O. Has a molar mass
of 154.25g/mol. Melting
point of 208°C. Boiling
point of 213°C.
Utilized to represent a
Chemical formula of
basic oxidation. Product.
C10H16O. Molar mass of
152.24g/mol. Melting
point of 175-177°C.
Boiling point of 209°C.
Ethyl AcetateO
Utilized to begin reaction
Chemical formula of
with Borneol. Utilized to
C4H8O2. Molar mass of
extract organic layer in a
88.11g/mol. Melting point
9:1 ratio with Hexane.
of -83.6°C. Boiling point of
Utilized to extract organic
Chemical formula of C6H14.
layer in a 9:1 ratio with
Molar mass of
Ethyl Acetate.
86.18g/mol. Melting point
of -96 to -94°C. Boiling
point of 68.5-69.1°C.
Magnesium SulfateO
Utilized to dry organic
Chemical formula of
MgSO4. Molar mass of
120.37g/mol. Melting
point of 1,124°C.
Used in place of bleach as
Chemical formula of
the oxidizing agent.
KHSO5. Molar mass of
Sodium BisulfateO
Utilized to reduce trace
Chemical formula of
NaHSO4. Molar mass of
120.06g/mol. Melting
point of 58.5°C. Boiling
point of 315°C
Sodium Chloride
Utilized to isolate proper
Chemical formula of NaCl.
Molar mass of
58.44g/mol. Melting point
of 801°C. Boiling point of
No crystals of camphor were obtained during experimentation. The results
were collected from another lab group and utilized in this report for the sake of
consistency. Ideally the camphor crystals would result as a clear to whitish
crystalline structure. The camphor was recorded to have a melting point of 167°C.
The percent yield was found to be 54%. This was found by first converting both the
gram amounts of borneol, and camphor into moles.

Borneol = 0.2g / 154.24g = 0.0013mol. (theoretical yield)

Camphor = 0.1g / 152.24g = 0.0007mol (actual yield)
The actual yield (0.0007mols of camphor) was then divided by the theoretical yield
(0.0013mols of borneol) to find the percent yield.

0.0007mol / 0.0013mol = 0.5384
This yield was then multiplied by 100%.

0.5384 x 100% = 53.84%
Borneol was recorded to have an Rf value of 0.7. Camphor was recorded to have an
Rf value of 0.73. The Rf values were recorded by dividing the distance the pigment
moved from the solvent front over the dimensions of the TLC paper. In this case, the
TLC paper measured 6cm. The distance the pigment moved from the solvent front
for borneol was recorded to be 4.2cm. The distance the pigment moved from the
solvent front for camphor was recorded to be 4.4cm.

4.2cm / 6cm = 0.7 (borneol’s Rf value)

4.4cm / 6cm = 0.73 (camphor’s Rf value)
Figures 2.1 and 2.2 respectively illustrate the IR and NMR spectra for the
Figure 2.1: Illustrates the IR Spectra for the experiment product of camphor.
Figure 2.2: Illustrates the +HNMR Spectra for the experiment product of camphor.
The recorded melting point of the product was found to be 167°C. This is very
nearly consistent with the literature value of 175-177°C. The lowering in the range
of the melting point can be attributed to an impurity in the product as proved by
mixed melting point theory. Remaining water due to a lack of proper drying could
cause this impurity, or it could be left-over reactant (borneol) that remained in the
final product. The percent yield was found to be 54%. This could be due to the
previously discussed possibility of an impurity. An impurity could effect the entirety
of the reaction, resulting in a lessening of product. The IR spectra indicate no
alcoholic peaks, which is a sign that the drying agents worked perfectly. The
expected peak at around 1710 (recorded at 1740.79) indicates that camphor is
certainly our product. The +HNMR spectra coincide with the chemical structure of
camphor. The TLC helped to identify leftover borneol in the experiment. This was
done by taking an initial TLC when the reactants were present, and comparing it to
the product’s TLC. If there were remnants of borneol, it would be shown on the
second TLC reading.
An oxidation was completed in the experiment. Overall, the goal of oxidizing
borneol to camphor utilizing green oxidation techniques was successful. In future
experimentation the skill set of a reliable oxidation utilizing these green techniques
(Oxone in place of bleach) will be applicable. The experimentation was successful.
1. The PubChem Project. (n.d.). Retrieved February 24, 2016, from
2. Weldegrima, S. (2015). Experimental Organic Chemistry: Laboratory Manual
For CHM 2210L and CHM 2211L. Tampa, FL: University of South Florida.
3. March Jerry; (1985). Advanced Organic Chemistry reactions, mechanisms
and structure (3rd ed.). New York: John Wiley & Sons, inc.
4. Organic Redox Systems: Synthesis, Properties, and Applications, Tohru
Nishinaga 2016
9:04 PM Sun Sep 19
VPN 33%O
< AA usflearn.instructure.com + Assignment 4 Rubrics and... Mail - Nidal... Assignment 4 T USF Librarie... Ex Libris Dis... 10... Detailed Orgo II lab milestones for experimetnal rubrics.pdf Download i Info X Close o Page < 6 of 29 0 ZOOM + side reaction in this experiment. 1pt Experimental Include the Section summary of the step by step procedures in a form of flowchart for this experiment 2pts Include all other reagents/chemicals/solvents that aren't included in the lab manual or include any procedural change that is made during the actual lab. 2pts Table of Chemicals Physical properties of chemicals used in this experiment. 2pts Chemical properties of the chemicals used in this experiment. 1pt List all hazardous, toxicity, flammability and other safety related properties of the chemicals used in this experiment 1pt Results List physical properties (m.p., appearance and color of crystals of camphor) 1pt Show the calculation for the percentage yield of the reaction. 2pts List the Retention factor (Rt) of borneol and camphor. 1pt Include the IR and NMR spectra of the product 2pts Discussion Discuss m.p. Interpret the IR spectrum. Discuss the carbonyl peak of the product in the IR spectrum. Compare the IR spectrum of camphor with that of borneol and discuss the potential difference. 2pts and percentage yield of reaction by comparing them with the literature of theoretical values. 2pts Interpret the 1H NMR spectrum. Assign all the H peaks on the 1H NMR. 2pts Explain how following the reaction's course by TLC is helpful. 1pt Discuss how green (safe) the reaction was. 1pt Conclusion What has the information from the data revealed? Discuss how the reaction Did the lab accomplish what it set out to do? What did you learn from this experiment? 1 pt 6 9:03 PM Sun Sep 19 VPN 33% 0 Page < f 5 of 29 o ZOOM X Х Experiment 3 Criteria Milestone 1 Milestone 2 Milestone 3 Milestone 4 Milestone 5 Title Page Your name, Name of partners in Lab Group (list all), TA's Name, Title of experiment 1pt Introduction / Background Describe briefly green chemistry and the major advantages of it. 1pt Describe oxidation reactions in organic chemistry. Specify the theory, conditions, examples of oxidation reactions using different oxidizing agents. 2pts Describe the importance of using safe, benign and environmentally friendly oxidizing agents in oxidation reactions. 1pt Explain why and how the conditions (oxidizing agent and the solvents) are preferred to the traditional oxidation reaction conditions. Interpret the importance of this experiment with green chemistry perspectives. 2pts Introduction / Mechanism Describe in Label all the reactants, the detail the ions/reactive intermediates plausible and products involved in the mechanism of mechanism. the reaction in 1pt this experiment with arrow pushing including all the intermediates involved in the reaction. 1pt Introduction / Side Reaction(s) Show the reaction scheme for the potential Label all the reactants and products in the side reaction. 1pt 5 side reactior It 9:04 PM Sun Sep 19 VPN 33%O < AA usflearn.instructure.com + Assignment 4 Rubrics and... Mail - Nidal... Assignment 4 T USF Librarie... Ex Libris Dis... 10... Detailed Orgo II lab milestones for experimetnal rubrics.pdf Download i Info X Close o Page < 7 of 29 0 ZOOM + along with the techniques in this experiment can be applied to other situations. 1pt References 2 References 1pt Proper Citations 1pt Overall Format/ Organization / Language Are all tables, figures, schemes labeled appropriately? 1pt Are all sections clearly labeled? Is proper scientific language used? 1pt Page layout of the report: Is spacing double? Is the font Times New Roman, 12pt? Is margin 1 inch? 1pt 55 50 45 - 40 - 35 - 30 % Transmittance 25 - 20 15 10 - 5 - 4000 3800 3600 3400 3200 3000 2800 2600 2400 2000 1800 1600 1400 1200 1000 800 600 2200 Wavenumbers (cm-1) 9/14/2021 3:05:00 PM ΜΟ1(s) 1.80 1.85 1.73 MO4(m) M03(m) 2.84 2.34 3.38 3.09 3.03 2.98 2.91 2.56 10.17 1.37 ΠΟ.39 C0.51- 11.92 1.00 0.19 חזון חד חן יוון 14 13 12 11 10 9 00 7 9 5 4 3 2 1 Ο Chemical Shift (ppm) ΜΟ1(s) MO6(m) ΜΟ7(m) 1.80 1.85 1.73 MO4(m) MO5(m) MO3(m) 2.84 2.34 3.38 3.20 3.14 3.09 3.03 2.98 2.63 2.56 0.17 1.37 ΤΙΤ 14 13 12 11 10 9 8 7 6 5 4 3 2 1 O -1 Chemical Shift (ppm MO1(s) 1.80 1.85 1.73 M04(m) MO3(m) 2.84 2.34 2.98 2.91 2.56 3.38 71.35 1.32 1.31 1.23 _0.17 1.37 10.39 00.51 C1.92 01.00 [0.19 14 13 12 11 10 9 8 7 6 5 4 3 2 1 0 Chemical Shift (ppm) MO1(s) M06(m) M07(m) 1.85 - 1.80 MO4(m) M05(m) MO3(m) 2.84 2.34 3.38 3.20 3.14 3.09 3.03 2.56 2.98 2 2.63 1.35 1.32 1.31 1.23 0.17 1.37 0.39 10.51 (1.92 C1.00 0.19 14 13 12 11 10 9 8 7 6 5 4 3 1 0 -1 Chemical Shift (ppm C:\Eft\Data\Org II, Sec07,2021\Exp03\Sec07, hood6.h1 (Hz) (ppm) 1.23 1 73.8 2 1.31 78.4 ZI"T" Height 0.0570 0.0617 0.0242 0.0597 3 1.32 79.5 4 1.35 81.0 5 1.73 0.8399 6 1.80 103.8 108.2 111.1 1.0000 7 1.85 0.8865 8 2.34 140.7 0.2643 0.0457 9 2.40 144.0 . 10 0.1388 2.55 2.56 11 0.1389 12 2.63 0.0461 153.0 153.7 157.9 165.1 168.1 170.6 1 13 2.75 14 2.80 15 2.84 0.0660 0.0407 0.2357 0.0958 0.0959 16 174.5 2.91 2.93 17 18 2.98 3.03 3.09 0.1257 0.0976 19 175.6 178.9 182.1 185.6 188.2 192.1 200.7 202.9 3.14 22 3.20 اما ما ما ما ما ما را به 3.35 0.1003 0.0730 0.0447 0.0208 0.0325 0.0325 0.0261 0.0178 3.38 25 3.43 3.49 206.0 209.5 210.9 3.51 No. Value 1 0.194 Non-Negative Value 0.194 1.000 2 1.000 3 1.919 Absolute Value 1.444e+6 7.437e+6 1.428e+7 3.790e +6 2.907e+6 1.015e +7 1.919 (ppm) [1.16 .. 1.43] (1.66 .. 1.77] [1.77 .. 1.90] [2.28 .. 2.40] [2.47 .. 2.65] [2.66 .. 3.25] [3.29.. 3.57] 0.510 TIIT 5 0.510 0.391 1.365 0.391 6 1.365 7 0.169 1.254e+6 0.169

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