# Organic Chemistry/Introduction to reactions/Redox reactions

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### Oxidation and reduction

Two important types of reactions in organic chemistry are oxidation and reduction.
In oxidation reactions, the oxidized species loses electron density.
In reduction reactions, the reduced species gains electron density.
Of course, these two actions happen in unison as one species is reduced and the other is oxidized. The term redox was coined from the fragments red (reduction) and ox (oxidation). A standard mnemonic for the terms is “OILRIG”: oxidation is loss, reduction is gain.

### Oxidation

Oxidation was first observed when oxygen drew electrons off of metals, which were then referred to as "oxidized". (Oxygen is more elecronegative than most other elements.) The term was then applied later to the part of any reaction where electrons are drawn off. Other elements that commonly oxidize in organic reactions include halogens like chlorine and bromine.

### Reduction

Reduction of a chemical species results in the gain of electrons for that species. This does not necessarily include any change in formal charge; any time an atom increases its electron density even a little bit it is said to be reduced. For example, if an oxygen is removed from a carbon and replaced by a hydrogen (assume the oxygen is also bonded to another atom), the formal charge of the carbon does not change. However, the carbon "sees" a greater share of the electrons from the single bond to hydrogen than it did for the single bond to oxygen. That is because hydrogen is less electronegative than oxygen and gives up its electrons a bit more easily than oxygen does. So a carbon bonded to hydrogen can take up more of its electron density than the same carbon bonded to oxygen.

### Oxidation numbers

It's possible to assign an “oxidation number” to each atom in a molecule. There are a two different approaches to this. For organic molecules it is generally possible to find all the oxidation numbers using a set of simplified rules. There is no single best set of rules, but as an example, given in order of decreasing priority:
* The sum of oxidation numbers in any species is equal to its overall charge.
* Oxidation number of hydrogen is +1. Exceptions: it is -1 when bound to a metal and 0 when bound only to itself.
* Oxidation number of fluorine is -1, unless bound to itself.
* Oxidation number of oxygen is -2, unless bound to itself or a more electronegative element, such as fluorine.
From this it is possible to find, for example, that the oxidation state of carbon in methanal:
We find that:
* Oxygen has oxidation number -2.
* Hydrogen has oxidation number +1
* The overall charge, and so the total oxidation number, is zero.
Denoting the oxidation number of carbon as c, we have
1 × (−2) + 2 × 1 + c = 0 = c
Carbon has oxidation number zero. Verify for yourself that in methane it has −4, and in ethane −3.
An alternative and more general approach is to take the structure of the molecule, and break the bonds such that:
* In a heteronuclear bond (between different atoms) the more electronegative atom gains all the electrons.
* In a homonuclear bond (both atoms are the same) the electrons are divided evenly between the atoms.
As usual, a single bond holds two electrons and a double bond four. (There are further complications which we will not go into here.) The charge of each atom after this process is its oxidation number. Given that oxygen is more electronegative than carbon, which is more electronegative than hydrogen, verify that this gives the same results as before for methanal, methane, and ethane.
Both methods give the result that in a neutral species containing only a single element (such as H$_2$ or graphene) all atoms have oxidation state 0.
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