Oxidation

Another very important kind of reaction in which alcohol molecules can be involved is oxidation.

Review

It has been some time since you studied oxidation in CH 105. Let's briefly review oxidation: oxidation involve the loss of electrons. In some cases this might means the outright loss of electrons, as is the case when metals are oxidized to form cations. In other cases, the loss of electrons is only a partial loss of electrons. This can happen through the formation of covalent bonds to more electronegative elements such as oxygen.

In the case of organic compounds, we will be dealing with oxidation in terms of covalent bonds rather than outright charges. In the case of organic compounds, oxidation usually manifests itself in one of two ways (or both):

	the addition of more oxygen to a molecule, or
	the loss of hydrogen atoms with their electrons from a molecule.

The addition of oxygen usually involves the addition of one atom of oxygen at a time. The loss of hydrogen atoms usually involves the loss of two hydrogen atoms at one time. We will deal mostly with the loss of hydrogen during this lesson and the gain of oxygen in the next lesson.

Combustion

The extreme case of oxidation is combustion. Under such conditions alcohols, like many other organic compounds, are converted to CO₂ and H₂O. When combustion is complete, enough oxygen (O₂) reacts with the alcohol to convert all of the carbon to carbon dioxide and all of the hydrogen to water. Note that these equations are not balanced. Balancing them would point out that, mole for mole, different amounts of oxygen are necessary for the combustion of different alcohols.

Equations for the combustion of methanol and ethanol. [63rxn10.JPG]

Oxidation of Secondary Alcohols

Less severe oxidation is also possible. Let me start by talking about the oxidation of secondary alcohols. I want to start with secondary alcohols because that serves as a good point of comparison for primary and tertiary alcohols.

You should recall that secondary alcohols are alcohols such as the one shown here (and in Example 12 in your workbook). These are molecules in which the carbon to which the hydroxyl group is attached is bonded to two other carbon atoms. In this example those other carbon atoms are shown as unspecified alkyl groups, R and R'. The reason for using R and R' instead of just using R and R is to show that the alkyl groups do not necessarily have to be the same alkyl group on each side of the carbon.

H
|
R-C-R'
|
OH

The oxidation of a secondary alcohol involves the removal of two atoms of hydrogen from the compound. The hydrogen atoms must be on adjacent carbon and oxygen atoms. When these two atoms of hydrogen are removed, they are replaced by a double bond between carbon and the oxygen atoms from which they were removed. Compounds that are formed in this way, or compounds which have this type of a group (a carbonyl group which is bonded to two other alkyl groups) are called ketones. We will study ketones later in this lesson. For now, let's concentrate on what happens to the alcohol.

Equation showing oxidation of secondary alcohol. [63rxn11.JPG]

This reaction is an oxidation reaction. To see why, focus on what happens to the electrons around the carbon atom attached to the oxygen atom. (This is also shown in Example 12-c in your workbook.)

Equations with Lewis diagrams showing oxidation of secondary alcohol. [63rxn12.JPG]

The carbon atom starts with one of its valence electrons bonded to the oxygen. When the oxidizing agent removes the hydrogen atoms and their electrons, the electrons of the carbon atom rearrange to form a double bond to the oxygen. Because oxygen is more electronegative than carbon, any electrons bonded to oxygen can be considered "lost" as far as the carbon atom is concerned. Thus, in the process of changing from a single bond between carbon and oxygen to a double bond, the carbon atom has "lost" another electron to oxygen and has therefore been oxidized. Also, don't forget that an electron left with each of the departing hydrogen atoms. Again we have the loss of electrons.

This equation can be "completed" by showing the two hydrogen atoms that were removed from the original molecule along with their electrons. They are written in a way that does not tell you where they ended up, because where they ended up and what they are bonded to depends on what the oxidizing agent is. The H· emphasizes that the electrons came with the hydrogen. This becomes very important in biological systems in metabolic processes such as the electron transport system.

Equation with Lewis diagrams showing oxidation of secondary alcohol including removed Hs. [63rxn13.JPG]

Sometimes (such as in Example 12 in your workbook) those hydrogens might be written as [H]. The intent, again, is to indicate that those two hydrogen atoms are bonded to something, but we don't know what. It is best not to write H₂ (although you may sometimes see it that way) because that would imply that it comes off as hydrogen gas, which it doesn't. That is the reason for writing 2 [H] (as in Example 12) or 2 H· as in the example above.

A specific example of this reaction is shown below (and in Example 12-b in your workbook). The names are included to give you a preview of ketone nomenclature. Also note that the oxidizing agent and the conditions for the reaction can be specified above and below the arrow for the reaction.

2-butanol

K₂Cr₂O₇
¾¾¾¾¾¾®
H₂SO₄, heat

H H H
| | |
H-C-C-C-C-H
| || | |
H O H H

2-butanone

+ 2[H]

Oxidation of Primary Alcohols

Primary alcohols can be oxidized by the same process as secondary alcohols.

The process is the same, but the results are slightly different. Two hydrogen atoms are removed from the molecule. One is from the oxygen and the other is from the adjacent carbon atom, either one will do. Because the carbon atom involved in the reaction starts with two hydrogen atoms, it still has one hydrogen atom attached to it after the reaction is over. That makes the product an aldehyde instead of a ketone. We will also study aldehydes later in this lesson.

Equation showing the oxidation of a primary alcohol. [63rxn14.JPG]

A specific example of this is shown below (and in Example 13 in your workbook). The names shown give you a preview of aldehyde nomeclature.

1-pentanol

oxidation
¾¾¾¾¾¾®

H H H H O
| | | | ||
H-C-C-C-C-C + 2[H]
| | | | |
H H H H H

pentanal

Oxidation of Tertiary Alcohols

This kind of oxidation reaction falls flat on its face when a tertiary alcohol is used. To review the process involved with both primary and secondary alcohols -- the carbon and oxygen involved in the reaction each lose a hydrogen atom.

Now look at the structural formula of this tertiary alcohol. Look at the carbon atom bonded to the oxygen. It does not have a hydrogen atom bonded to it. Therefore it cannot lose a hydrogen atom. Therefore it cannot be oxidized in this way. (A similar diagram is shown in your workbook in Example 14.)

Equation for the non-reaction of oxidizing a tertiary alcohol. [63rxn15.JPG]

Top of Page

E-mail instructor: Sue Eggling