` Oxidation of Borneol to Camphor using Oxone® and catalytic NaCl: A Green Experiment for the Undergraduate Organic Chemistry Laboratory
GREEN OXIDATION OF BORNEOL TO CAMPHOR
WITH OXONE®
Introduction
The objective of this experiment is to oxidize the natural terpene, (1S)-borneol, to (1S)-camphor using green oxidation conditions. Oxone®, a product of DuPont, is a “triple salt” (2KHSO5.KHSO4.K2SO4) containing potassium peroxymonosulfate, KHSO5, as the active oxidizing agent. This compound combined with a catalytic amount of sodium chloride serves to oxidize the secondary alcohol of borneol to the ketone of camphor. Borneol and camphor are members of a large class of natural products called terpenes. Read about these unique and important compounds in the accompanying essay. The isolated camphor product will be dried, using a
homemade desiccator, sublimed, and characterized by melting point, IR, and 1H NMR spectroscopy. Oxone® is a new and powerful oxidizing agent finding ever-expanding use as a safe alternative to traditional oxidizing agents such as chromium reagents and organic peracids (eg. MCPBA). Under the conditions used in this experiment, it is known to oxidize chloride ion to molecular chlorine.1 The species that oxidizes borneol’s alcohol group is most likely hypochlorous acid (HOCl) present
in the “chlorine water” that exists in the reaction mixture. The exact mechanism of the reaction is presently unknown but may involve the steps illustrated below. Oxidation chemistry has historically been environmentally harmful. The commonly used and reliable “Jones oxidant” (a chromium-based reagent) is carcinogenic and has a corrosive action on the hands and mucous membranes.2 Additionally, chromate salt by-products from a chromic acid oxidation are harmful to the environment and thus pose a disposal problem. The salts produced in the Oxone®/NaCl experiment, on the other hand, are environmentally benign (mainly potassium sulfate), and the health and safety hazards are very low. Consider each of the 12 Principles of Green Chemistry learned and how it might apply to this experiment.
Camphor has the distinction of being one of the first natural terpene products isolated from nature with a rich and colorful history. It can be found in the camphor tree (cinnamomum camphora), an evergreen tree found in Asia, and from many other plant sources. Camphor can be synthetically produced from the oil of turpentine. Humans have found a wide range of uses for camphor including medicinal applications (antimicrobial, anesthetic, cough suppressant), plasticizer, embalming fluid, pyrotechnics, moth repellent, and preservative in pharmaceuticals and cosmetics. See if you can find camphor on the label of any of your products at home.
Watch the associated video:
Oxidation of Borneol
As you watch the video create bullet point procedural list (3 pts)
Complete the table below. Show all calculations (5 pts)
| mmoles (1S)-Borneol | mmoles Sodium Chloride | mmoles Oxone® | Theoretical yield of product | Crude % yield |
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Why is this considered a “green” oxidation? You may want to compare this oxidation method with others in your textbook (2 pts
Write the balanced equation for the reaction performed. (2 pts)
Star all stereogenic centers (chiral carbons) in the borneol starting material and camphor product. (2 pts)
Why does the carbonyl stretching frequency in the infrared spectrum of camphor occur at 1740 cm-1 whereas that of acetophenone (C6H5COCH3) is found at 1680 cm-1 and that for hexanone is found at 1720 cm-1 (3 pts)
Name three ways you could distinguish your product from the starting material. Name the technique and the observations that your would use to make the distinction. (3 pts)
Part 2 Reduction:
This experiment involves the use of a reducing agent (sodium borohydride) for converting a ketone (camphor) to a secondary alcohol (isoborneol) as illustrated in the second step of the two-step oxidation/reduction sequence shown below. The spectra of borneol, camphor, and isoborneol will be compared to detect structural differences and to determine the extent to which the final step produces a pure alcohol isomeric with the starting material.
In principle, the reduction of camphor can give two diastereomeric alcohols, corresponding to reaction of borohydride at the two faces of the C=O bond. Reaction at the top (“exo”) face regenerates the original starting material, borneol. Reaction at the bottom (“endo”) face yields the other diastereomer, isoborneol. You will use NMR to determine the preferred product
Watch the second video
As you watch the video create bullet point procedural list (3 pts)
Complete the table below. Show ALL calculations (3 pts)
| mmoles (1S)-Camphor | mmoles NaBH4 | Theoretical yield of product | Mass of product | Crude % yield |
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Below are NMR spectra of the reduction product obtained using LiAlH4 as a reducing agent. Use this data to calculate the relative proportions of the two isomers. (2 pts)
This is the same data for the reduction using NaBH4 as the reducing agent. The picture is not as pretty but the integrations are shown. Calculate the relative portions of the two isomers. (2 pts)
Could optical rotation be used to determine the relative amounts of borneol and isoborneol in the product mixture? Explain why or why not (3 pts)
Use the data above is exo or endo attack preferred? Why do you think that this preference exists? (3 pts)
