Oxidation of the Alkenes

This lesson covers: 

1. Oxidation of alkenes with cold, dilute acidified potassium manganate(VII)
2. Oxidation of alkenes with hot, concentrated acidified potassium manganate(VII)
3. How oxidation products indicate alkene structure

Cold, dilute oxidation forms diols

When alkenes react with cold, dilute acidified potassium manganate(VII) (KMnO_4(aq)), they undergo a chemical change known as oxidation.

• The pale purple KMnO₄ solution changes to colourless as the permanganate ions (MnO₄^-) are reduced to manganese(II) ions (Mn²⁺).
• In this reaction, each carbon atom in the C=C double bond gains an -OH group. This forms a compound called a diol, which has two alcohol groups.

For instance, ethene reacts with KMnO_4(aq) to produce ethane-1,2-diol:

H₂C=CH₂ + [O] + H₂O ➔ HO-CH₂CH₂-OH

This reaction is significant because it can be used to test for the presence of a C=C double bond in a compound, which is a characteristic feature of alkenes.

Hot, concentrated oxidation breaks C=C bonds

Using hot, concentrated acidified KMnO_4(aq) results in a more intense oxidation of alkenes, leading to the complete breaking of the C=C double bond:

1. The diol intermediate loses its -OH groups and gets further oxidised into ketones, aldehydes or carboxylic acids.
2. The nature of the final oxidation products depends on the types of groups attached to the carbons of the C=C bond.

Oxidation products with hot concentrated KMnO₄

The oxidation products formed when an alkene reacts with hot, concentrated acidified KMnO₄ depend on the groups bonded to each carbon atom in the C=C double bond:

1. Two hydrogens (=CH₂) - If a carbon atom in the double bond is bonded to two hydrogen atoms, it will oxidise to carbon dioxide (CO₂).
2. One hydrogen + one alkyl group (=CHR) - If a carbon atom has one hydrogen atom and one alkyl group (R) bonded to it, it will oxidise to form a carboxylic acid (RCOOH).
3. Two different alkyl groups (=CR¹R²) - If a carbon atom has two different alkyl groups (R¹ and R²) bonded to it, it will oxidise to form a ketone (R¹R²C=O).

When an unsymmetrical alkene containing two different groups on either side of the C=C double bond is oxidised, two different oxidation products will be formed, one from each side of the double bond.

For example, oxidising (CH₃)₂C=CH₂ would give a ketone, (CH₃)₂C=O, from one side of the double bond and carbon dioxide (CO₂) from the other side.

Using oxidation products to deduce alkene structure

The harsh oxidation conditions break and alter the original structure of the alkene.

However, by examining the oxidation products, we can infer structural information about the original alkene:

• Carbon dioxide - This indicates that the double bond was between two terminal carbons in the chain.
• Ketone - Suggests that each of the double bonded carbons was originally attached to two alkyl groups.
• Carboxylic acid - Implies that one of the double bonded carbons was initially bonded to an alkyl group, and the other to a hydrogen atom.

So the specific oxidation products provide vital clues about the identity of unknown alkenes.

Worked example 1 - Determining alkene structure from oxidation products

Given the oxidation products as carbon dioxide and propanoic acid, determine the possible structure of the original alkene.

Step 1: Interpretation of oxidation products

1. Carbon dioxide - Indicates a carbon atom in the double bond was bonded to two hydrogen atoms.
2. Propanoic acid (CH₃CH₂COOH) - Suggests the other carbon in the double bond was bonded to an ethyl group (CH₃CH₂) and a hydrogen atom.

Step 2: Constructing the alkene structure

• The alkene must have four carbon atoms.
• The carbon that produced carbon dioxide was a terminal carbon (at the end of the chain) with two hydrogens.
• The carbon that produced propanoic acid was a middle carbon, attached to an ethyl group and a hydrogen atom.

A possible structure of the original alkene could be but-1-ene:

Worked example 2 - Determining alkene structure from oxidation products

Given the oxidation products as propanone and butanone, determine the possible structure of the original alkene.

Step 1: Interpretation of oxidation products

1. Propanone (CH₃COCH₃) - Indicates one of the carbons in the double bond was bonded to two methyl groups (CH₃).
2. Butanone (CH₃CH₂COCH₃) - Suggests the other carbon in the double bond was bonded to a methyl group (CH₃) and an ethyl group (CH₃CH₂).

Step 2: Constructing the alkene structure

• The alkene must have seven carbon atoms.
• The carbon that produced propanone was a middle carbon, attached to two methyl groups
• The carbon that produced butanone was a middle carbon, attached to a methyl group and an ethyl group.

A possible structure of the original alkene could be 2,3-dimethylpent-2-ene: