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Chemistry 2131:
Organic Chemistry for the Life Sciences (3)

Oxidation of Alkenes

• next we are going to look at the oxidation of alkenes. Before starting on that, we will look at how to tell whether a reaction is an oxidation, reduction or neither.
• the simple redox you are familiar with probably dealt with simple organic molecules. We think of oxidation as a reaction where electrons are removed and the compound becomes more positively charged. Likewise, reduction involves the gain of electrons and a more negative charge.
• I always remember it by "reduction reduces charge"
• but how can we apply these concepts to organic chemistry? There are different approaches to this question. Your text book gives one possible tack to take. The method I will describe here is one I find easier to deal with.
• what we have to do is determine the oxidation number for the reactants and products. Then we simply compare them.
• there are a number of simple steps to determine the oxidation number.
  
  1. Assign an oxidation level to each carbon atom that undergoes a change between the reactant and product. How to do this? For every bond from carbon to a less electronegative element (including hydrogen), and for every negative charge on carbon, assign a value of -1. For every bond from carbon to another carbon, or for any unshared electron pair on carbon, assign a value of 0. For every bond from carbon to a more electronegative atom and for every positive charge, assign a value of +1. Add up these numbers for each carbon atom. This sum is the oxidation level for that carbon atom.
  2. Determine the oxidation number, N_OX for both the reactant and the product, by adding the oxidation levels for each carbon that changes.
  3. Calculate the difference between the oxidation number of the products and the reactant (N_OXproduct - N_OXreactant). If the difference is a positive number the transformation is an oxidation. If the difference is 0, the transformation is neither oxidation nor reduction. If the difference is negative, the transformation is a reduction.
• it is a good idea to go back and check your work by doing balanced half reactions as decribed in the text. There are a few rules to follow:
  
  1. write a half reaction showing the organic reactant and product.
  2. complete the balance by adding oxygen atoms in the form of H₂O and hydrogen in the form of H+ (for basic reactions use OH- and H₂O)
  3. use "dummy electrons" to balance the charges. If electrons appear on the left side of the equation it is reduction, if they appear on the right it is oxidation.

2. Oxidation of Alkenes:

• let's look at an oxidation of alkenes, that mediated by permanganate (MnO₄-). The product is called a glycol. This reaction takes place under alkaline conditions.
• to demonstrate the stereochemistry of the reaction, we'll use the example of the cycloalkene, cyclopentene.
• the first step in the reaction will seem familiar. The pi electrons from the double bond attack one of the double bonded oxygens on permanganate, the pi electrons of the C-O double bond go onto the Mn atom, and the pi electrons from the other C-O double bond attack the other carbon atom of the C-C double bond.
• thus, the reaction intermediate is a cyclic manganate ester.
• an important note on the stereochemistry of the reaction is that this intermediate is in cis, the cyclic ester is on one face of the molecule.
• further reaction of this intermediate with water (or hydroxide) give the cis diol and MnO₂. The mechanism for this step is a little fuzzy, but it can be simplified by saying that the oxygens that are part of the cyclic ring remove protons from water.
• this reaction then has syn steroechemistry.
• you can also use osmium tetroxide for this reaction.
• check to make sure that the reaction is in fact an oxidation reaction