Alcohol Oxidation: Types and Equations Revision Notes | A Level AQA Chemistry

1. Why Alcohol Oxidation Matters

Alcohol oxidation sits at the heart of organic synthesis — it is the route from simple alcohols to aldehydes, ketones, and carboxylic acids, all of which appear repeatedly across A-Level chemistry and in industrial processes.

The key insight you need to carry through this lesson is this: the type of alcohol determines what product forms, and whether oxidation is even possible. Get the classification right, and the rest follows logically.

2. Classifying Alcohols: Primary, Secondary, and Tertiary

The classification of an alcohol depends entirely on the carbon atom that carries the $\text{-OH}$ group — specifically, how many other carbon atoms are directly bonded to it.

A secondary alcohol has the $\text{-OH}$ carbon bonded to two other carbons. Propan-2-ol ($\text{CH}_3\text{CH(OH)CH}_3$) is the classic example. A tertiary alcohol has the $\text{-OH}$ carbon bonded to three other carbons — 2-methylpropan-2-ol is the standard A-Level example.

3. The Oxidising Agent: Acidified Potassium Dichromate(VI)

The standard oxidising agent used in A-Level alcohol oxidation is acidified potassium dichromate(VI), written as $\text{K}_2\text{Cr}_2\text{O}_7$ in dilute sulphuric acid. In equations, the oxidant is represented as $[\text{O}]$ — this notation is explicitly acceptable in AQA mark schemes and keeps equations manageable.

The colour change is the observable evidence of oxidation: the solution turns from orange (dichromate(VI) ion, $\text{Cr}_2\text{O}_7^{2-}$) to green (chromium(III) ion, $\text{Cr}^{3+}$). If no colour change occurs, the alcohol has not been oxidised — a result you would observe with a tertiary alcohol.

4. Oxidation of Primary Alcohols: Two Stages

Primary alcohols are the most interesting case because oxidation can stop at two different points, depending on the experimental conditions. The first product is an aldehyde; if oxidation continues, the aldehyde is further oxidised to a carboxylic acid.

Think of it as a two-rung ladder: primary alcohol at the bottom, aldehyde in the middle, carboxylic acid at the top. The conditions you choose determine which rung you reach.

The condition that controls which product forms is critical for exam questions. To obtain the aldehyde, warm the primary alcohol with acidified potassium dichromate(VI) and distil the aldehyde from the mixture as it forms so that it is removed before further oxidation. To obtain the carboxylic acid, heat under reflux with excess oxidising agent, keeping the reaction mixture hot for long enough for complete oxidation.

Worked example: Writing the Equation for Oxidation of Ethanol to Ethanal

Write an equation for the oxidation of ethanol to ethanal using [O] notation. Include state symbols.

1. $\text{CH}_3\text{CH}_2\text{OH}$ — ethanol, a primary alcohol — Primary alcohols are oxidised to aldehydes in the first stage. The product will be ethanal.
2. $\text{CH}_3\text{CHO}$ — ethanal — The -CH2OH group loses two H atoms to become -CHO. The carbon count stays the same.
3. $\text{CH}_3\text{CH}_2\text{OH} + [\text{O}] \rightarrow \text{CH}_3\text{CHO} + \text{H}_2\text{O}$ — One [O] is consumed; the two H atoms removed combine with it to form water. Check: C: 2=2, H: 6=6, O: 2=2. Balanced.
4. $\text{CH}_3\text{CH}_2\text{OH} + [\text{O}] \rightarrow \text{CH}_3\text{CHO} + \text{H}_2\text{O}$ — Using [O] notation, the symbolic oxygen balances the oxidation equation. In practice exam answers often omit state symbols when [O] notation is used, because [O] is not a real isolated reagent with a state symbol.

Answer: $\text{CH}_3\text{CH}_2\text{OH} + [\text{O}] \rightarrow \text{CH}_3\text{CHO} + \text{H}_2\text{O}$

Worked example: Writing the Equation for Oxidation of Ethanol to Ethanoic Acid

Write an equation for the full oxidation of ethanol to ethanoic acid using [O] notation. Include state symbols.

1. Ethanol ($\text{CH}_3\text{CH}_2\text{OH}$) $\rightarrow$ ethanoic acid ($\text{CH}_3\text{COOH}$) — Full oxidation of a primary alcohol goes all the way to the carboxylic acid. Two [O] atoms are needed in total.
2. Ethanol has 1 O; ethanoic acid has 2 O. Net gain = 1 O. But we also produce water, so we need 2 [O] overall. — Track oxygen carefully: $\text{CH}_3\text{CH}_2\text{OH}$ has 1 O atom; $\text{CH}_3\text{COOH}$ has 2 O atoms. The extra O comes from the oxidising agent.
3. $\text{CH}_3\text{CH}_2\text{OH} + 2[\text{O}] \rightarrow \text{CH}_3\text{COOH} + \text{H}_2\text{O}$ — Check atoms — C: 2=2, H: 6=6 (4 in acid + 2 in water), O: 1+2=3 on left; 2+1=3 on right. Balanced.
4. $\text{CH}_3\text{CH}_2\text{OH} + 2[\text{O}] \rightarrow \text{CH}_3\text{COOH} + \text{H}_2\text{O}$ — With [O] notation, this is the balanced oxidation equation used at A-level. The symbolic [O] terms represent oxygen supplied by the oxidising agent.

Answer: $\text{CH}_3\text{CH}_2\text{OH} + 2[\text{O}] \rightarrow \text{CH}_3\text{COOH} + \text{H}_2\text{O}$

5. Oxidation of Secondary Alcohols: Ketones

Secondary alcohols follow a simpler story: they are oxidised to ketones and the reaction stops there. A ketone has the carbonyl group ($\text{C=O}$) flanked by two carbon groups on either side, which means there is no hydrogen on the carbonyl carbon available for further oxidation.

The standard example is propan-2-ol oxidised to propanone. The colour change of acidified potassium dichromate(VI) from orange to green confirms oxidation has occurred, but no further reaction takes place even with excess oxidising agent under reflux.

Worked example: Writing the Equation for Oxidation of Propan-2-ol to Propanone

Write an equation for the oxidation of propan-2-ol to propanone using [O] notation. Include state symbols.

1. $\text{CH}_3\text{CH(OH)CH}_3$ — propan-2-ol, a secondary alcohol. Product: propanone ($\text{CH}_3\text{COCH}_3$). — Secondary alcohols are oxidised to ketones. The -CHOH- group becomes -CO-.
2. $\text{CH}_3\text{CH(OH)CH}_3 + [\text{O}] \rightarrow \text{CH}_3\text{COCH}_3 + \text{H}_2\text{O}$ — One [O] is consumed; the H from O-H and the H from C-H combine with it to give water. Check: C: 3=3, H: 8=8 (6 in propanone + 2 in water), O: 2=2. Balanced.
3. $\text{CH}_3\text{CH(OH)CH}_3 + [\text{O}] \rightarrow \text{CH}_3\text{COCH}_3 + \text{H}_2\text{O}$ — This balanced [O] equation is the form usually accepted in A-level organic chemistry answers.

Answer: $\text{CH}_3\text{CH(OH)CH}_3 + [\text{O}] \rightarrow \text{CH}_3\text{COCH}_3 + \text{H}_2\text{O}$

Notice in the structure of propanone that the carbonyl carbon has no C-H bond — it is bonded to two methyl groups and the oxygen. This is precisely why ketones resist further oxidation: there is no hydrogen on the carbonyl carbon for the oxidising agent to remove.

6. Tertiary Alcohols: Resistance to Oxidation

Tertiary alcohols do not react with acidified potassium dichromate(VI) under normal laboratory conditions. The carbon bearing the $\text{-OH}$ group has no C-H bond available — it is bonded to three other carbon groups — so the oxidising agent has nothing to act on.

The observable result is that the orange colour of the acidified dichromate(VI) solution does not change when a tertiary alcohol is added. This colour persistence is itself a useful diagnostic test to distinguish tertiary alcohols from primary and secondary ones.

7. Comparing All Three Types: A Unified View

Now that each type has been covered individually, it is worth drawing the comparison explicitly — because AQA 'compare' questions will ask you to address similarities and differences in a single, structured response.

Primary alcohols undergo two stages of oxidation: first to an aldehyde, then to a carboxylic acid. Secondary alcohols undergo one stage: oxidation to a ketone. Tertiary alcohols undergo no oxidation under standard conditions. The trend is directly linked to the number of C-H bonds on the carbon bearing the $\text{-OH}$ group: primary has two, secondary has one, tertiary has none.

8. Exam Technique and Mark-Winning Strategies

Alcohol oxidation questions appear in multiple formats on AQA papers: equation writing, 'describe and explain' questions about observations, and extended 'compare' or 'evaluate' questions. Each format rewards specific habits.

For equation questions, always check three things before moving on: the equation is balanced (count every atom), state symbols are included, and the correct number of $[\text{O}]$ atoms is used (one for aldehyde or ketone formation; two for full oxidation to a carboxylic acid). A missing state symbol or an unbalanced equation can cost you an accuracy mark even if your chemistry is correct.

For 'explain' questions about why tertiary alcohols are not oxidised, the mark-winning phrase is: the carbon bearing the hydroxyl group has no C-H bond, so the oxidising agent cannot remove hydrogen. Saying 'it is more stable' or 'it is less reactive' without this mechanistic reason will not score the explanation mark.

The thread running through this entire lesson is structural logic: the number of C-H bonds on the carbon bearing $\text{-OH}$ dictates how many times the oxidising agent can act. Primary alcohols offer two opportunities, secondary offer one, and tertiary offer none. Keep that structural picture in mind and every equation, observation, and exam answer follows directly from it.