One of the most important chemical properties of alcohols is their behaviour during oxidation. By reacting alcohols with a suitable oxidising agent, we can synthesise valuable organic products such as aldehydes, ketones, and carboxylic acids. In this lesson, we explore these reactions and the crucial roles played by the laboratory setup: distillation versus reflux.
🔑 Key Principle
The outcome of oxidising a primary alcohol depends entirely on the reaction conditions. Simple distillation isolates the intermediate aldehyde, while heating under reflux with an excess of the oxidising agent ensures complete oxidation to a carboxylic acid.
The Oxidising Agent and Colour Change
The standard oxidising mixture used in A-Level organic chemistry is acidified potassium dichromate(VI), written as \( \text{K}_2\text{Cr}_2\text{O}_7 / \text{H}_2\text{SO}_4 \). This contains the orange dichromate(VI) ion, \( \text{Cr}_2\text{O}_7^{2-} \), in dilute sulfuric acid.
When an alcohol is oxidised, the dichromate(VI) ion acts as the oxidising agent and is reduced itself to the chromium(III) ion, \( \text{Cr}^{3+} \). This redox reaction results in a distinct, easily observed colour change:
Colour change: Orange (\( \text{Cr}_2\text{O}_7^{2-} \)) → Green (\( \text{Cr}^{3+} \))
Distillation and Reflux
To control the extent of oxidation and avoid losing volatile compounds, we use two different apparatus setups:
A technique used to separate a volatile product from a reaction mixture. The mixture is heated, and the components vaporise. The component with the lowest boiling point (typically the aldehyde) vaporises first, travels down a condenser, and is collected as a liquid, stopping further chemical exposure.
A technique involving the continuous boiling and condensation of a reaction mixture. A condenser is fitted vertically above the reaction flask. Volatile vapours rise, condense, and drip back down into the flask to react further. This allows heating for extended periods without losing reactants or products.
In the context of organic pathways, oxidation is characterised by the gain of oxygen atoms or the loss of hydrogen atoms from an organic molecule.
In organic equations, we represent the oxygen supplied by the oxidising agent using the shorthand notation [O]. When writing balanced equations, treat [O] as a reactant. Always remember to check whether water, \( \text{H}_2\text{O} \), is produced as a co-product.
Oxidation Pathways
The product formed during oxidation depends entirely on the classification of the starting alcohol:
1. Primary Alcohols (1°)
Primary alcohols can undergo two stages of oxidation:
- Stage 1 (Partial Oxidation): Yields an aldehyde.
Conditions: Acidified potassium dichromate(VI), gentle heating, and distillation (to distil off the aldehyde immediately, preventing further oxidation).
\( \text{R-CH}_2\text{OH} + [\text{O}] \rightarrow \text{R-CHO} + \text{H}_2\text{O} \) - Stage 2 (Complete Oxidation): Yields a carboxylic acid.
Conditions: Excess acidified potassium dichromate(VI), heating under reflux (to keep the volatile aldehyde in the flask so it is fully oxidised).
\( \text{R-CH}_2\text{OH} + 2[\text{O}] \rightarrow \text{R-COOH} + \text{H}_2\text{O} \)
2. Secondary Alcohols (2°)
Secondary alcohols are oxidised to ketones under reflux. Ketones cannot be oxidised further under mild conditions, because doing so would require breaking a strong carbon-carbon bond.
Conditions: Acidified potassium dichromate(VI), heating under reflux.
3. Tertiary Alcohols (3°)
Tertiary alcohols are resistant to oxidation by acidified potassium dichromate(VI). The reaction mixture remains orange, and no reaction takes place.
🔑 Key Principle: Why Tertiary Alcohols Resist Oxidation
For oxidation to occur, the carbon atom holding the \( -\text{OH} \) group must be bonded to at least one hydrogen atom (which is lost along with the hydroxyl hydrogen). In a tertiary alcohol, the \( \text{C-OH} \) carbon is bonded to three alkyl groups and has zero hydrogen atoms attached. It is impossible to form a carbon-oxygen double bond without breaking a stable carbon-carbon bond.
- The oxidation of ethanol to form ethanal.
- The oxidation of ethanol under reflux to form ethanoic acid.
- The oxidation of propan-2-ol to form propanone.
Solution:
- Ethanol to Ethanal (Distillation):
This is a partial oxidation reaction where 2 hydrogen atoms are removed to form a water molecule.
\[ \text{CH}_3\text{CH}_2\text{OH} + [\text{O}] \rightarrow \text{CH}_3\text{CHO} + \text{H}_2\text{O} \] - Ethanol to Ethanoic Acid (Reflux):
This is complete oxidation. Two units of oxidising agent are used: one to form the aldehyde intermediate, and a second to introduce oxygen to form the carboxylic acid.
\[ \text{CH}_3\text{CH}_2\text{OH} + 2[\text{O}] \rightarrow \text{CH}_3\text{COOH} + \text{H}_2\text{O} \] - Propan-2-ol to Propanone:
Oxidation of a secondary alcohol. Two hydrogen atoms are lost to form the ketone double bond.
\[ \text{CH}_3\text{CH(OH)CH}_3 + [\text{O}] \rightarrow \text{CH}_3\text{COCH}_3 + \text{H}_2\text{O} \]
When drawing reflux apparatus in exams, ensure the condenser is open at the top. Sealing the top of a reflux condenser creates a closed system under heating, leading to a build-up of gaseous pressure and risk of explosion. Make sure water enters the condenser at the bottom and exits at the top to ensure efficient cooling.
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