Reduction of aldehydes and ketones.

The reduction of aldehydes, certainly an important topic in itself, is a good jumping off point for some practice assigning oxidation numbers. The two most common reagents for the reduction of aldehydes and ketones are sodium borohydride (NaBH₄) and lithium aluminum hydride (LiAlH₄), which are both hydride donors. In NaBH₄ or LiAlH₄, hydrogen is bound to an element less electronegative than itself (The electronegativity of lithium is 1.0, and boron is 2.0, while the electronegativity of hydrogen is 2.1). In terms of oxidation-reduction nomenclature, the bonding electrons in these reducing agents are under the control of hydrogen, it being more electronegative in this context. Hydrogen bonded to lithium or boron is in the position of having gained one electron (oxidation state -1).

Don't make this harder than it needs to be. All we are doing is looking at the bonds and deciding whose control the electrons are under. Imagine such a bond forming, which element gained and which element lost. If an element gains control over electrons, being more electronegative, its oxidation number moves in the negative direction. The element is reduced.

Let's see if we can do this for the aldehyde, the other reagent in the redox reaction. What does the terminal carbon of the aldehyde look like in the initial state? Within the carbonyl of an aldehyde, carbon is in the position of having one bond to hydrogen, a double bond with oxygen, and a single bond to another carbon. Carbon is more electronegative than hydrogen (carbon = 2.5, hydrogen = 2.1), less electronegative than oxygen (carbon = 2.5, oxygen = 3.5), and equal, of course, to itself in electronegativity. The electrons in the single bond with hydrogen, in the terms of oxidation-reduction, are under the control of carbon, so carbon has gained one electron having bonded with hydrogen. The bonding electrons with oxygen, though, are under the control of oxygen, so carbon has lost two electrons in having formed a double bond with oxygen. The carbon-carbon electrons are shared equally. So, in this position, our carbon has gained one electron and lost two, so the oxidation number of carbon in an aldehyde is +1.

Finally, let us look at the primary alcohol formed by reduction of the aldehyde, the carbon has two bonds to hydrogen and only one bond now to oxygen, so the oxidation state here is -1. Each bond to hydrogen represents carbon gaining control over one electron, and its bond with oxygen represents the loss of one electron. Therefore, in going from aldehyde to primary alcohol, in being reduced, the oxidation state of carbon has changed from +1 to -1. The hydrogen of the reducing agent, though, has been oxidized. It is now part of the alcohol, and its oxidation state has changed from -1 to +1. Carbon has been reduced, while hydrogen has been oxidized.