Introduction to oxidation of alcohols
Oxidation of alcohols is a kind of organic reaction. Different types of alcohols oxidized to form aldehydes, ketones or acids. Thus this reaction is used to distinguish different types of alcohols, such as- primary, secondary or tertiary. Primary alcohols (R-CH2-OH) are oxidized to aldehydes (R-CHO) or to carboxylic acids (R-COOH), secondary alcohols (R1R2CH-OH) are oxidized to form ketones (R1R2C=O). But tertiary alcohols do not show any oxidation reaction at all.
Oxidation of different types of alcohols
Different types of alcohols are oxidized usually with a solution of sodium or potassium dichromate (VI) and acidified with dilute sulphuric acid. After oxidation reaction the orange solution turns into green solution as chromium (III) ions formed from dichromate ions (VI). The half equation for this reaction is:
Cr2O72- + 14H+ + 6e– → 2Cr3+ + 7H2O
Oxidation of primary alcohols
Depending on reaction conditions primary alcohols can be oxidized to either aldehydes or carboxylic acids. To form carboxylic acids, primary alcohols are first oxidized to aldehydes, then it is further oxidized to the carboxylic acid.
It is possible to form aldehydes only from the primary alcohols by using excess amount of alcohols and distil off the aldehydes as soon they are formed. Due to the presence of less oxidizing agents aldehydes can not oxidize further to form carboxylic acid. Removing aldehydes from the reaction mixture also reduce the possibility of further oxidation process. For example the oxidation of ethanol to form ethanal is as follows:
3CH3CH2OH + Cr2O72- + BH+ → 2CH3CHO + 2Cr3+ + 7H2O
By simplifying the reaction we may write:
CH3CH2OH + [O] → CH3CHO + H2O
Carboxylic acid formation
If oxidation of primary alochols are completed in presence of excess amount of oxidizing agent and intermediate product aldehydes, carboxylic acids are formed. The carboxylic acids can be distilled off after heating primary alcohols with oxidizing agent under reflux. For example the oxidation of ethanol to form ethanoic acid is as follows:
3CH3CH2OH + 2Cr2O72- + 16H+→ 2CH3COOH + 4Cr3+ + 11H2O
By simplifying the reaction we can write:
CH3CH2OH + 2[O] → CH3COOH + H2O
Oxidation of secondary alcohols
It doesn’t matter what reaction condition is applied, but secondary alcohols will always oxidized to form only one product ketones. By heating under reflux secondary alcohols are oxidized in presence of a solution of sodium or potassium dichromate (VI) and dilute sulphuric acid, to form ketones. For example a secondary alcohol, propan-2-ol is oxidized to form propanone.
(CH3)2CHOH + [O] → (CH3)2CO + H2O
Oxidation of tertiary alcohols
Tertiary alcohols cannot be oxidized at all in presence of even a strong oxidizing agent to form any aldehydes, ketones or acids. There are no reaction involved in presence of sodium or potassium dichromate (VI) with any tertiary alcohols. The reason behind this phenomena is absence of hydrogen at the carbon center which holds the alcohol group in tertiary alcohols. The oxidation reaction of alcohols takes place by removing the hydrogen present at the carbon center and from alcohol group (-OH) to form aldehydes, ketones or acids.