1. Classification and Oxidation of Alcohols
Alcohols contain the hydroxyl functional group (−OH). Primary (1°), secondary (2°), and tertiary (3°) alcohols are classified by the number of carbon atoms attached to the carbon bearing the −OH group.
| Type | Carbons on the C-OH carbon | Examples |
|---|---|---|
| Primary (1°) | One (or none, for methanol) | Ethanol, CH₃CH₂OH; propan-1-ol, CH₃CH₂CH₂OH |
| Secondary (2°) | Two | Propan-2-ol, CH₃CH(OH)CH₃; butan-2-ol, CH₃CH(OH)CH₂CH₃ |
| Tertiary (3°) | Three | 2-methylpropan-2-ol, (CH₃)₃COH |
The standard laboratory oxidising agent is acidified potassium dichromate(VI) (K₂Cr₂O₇/H₂SO₄). When oxidation occurs, orange dichromate ions (Cr₂O₇²⁻) are reduced to green chromium(III) ions (Cr³⁺):
\[ \text{Cr}_2\text{O}_7^{2-} + 14\text{H}^+ + 6e^- \rightarrow 2\text{Cr}^{3+} + 7\text{H}_2\text{O} \quad (\text{Orange} \rightarrow \text{Green}) \]
Primary Alcohols (1°)
- Partial Oxidation to Aldehyde: Use excess alcohol and distillation apparatus with gentle heating. The aldehyde has a lower boiling point than the alcohol (no hydrogen bonding between aldehyde molecules) and distils off immediately before further oxidation: \[ \text{CH}_3\text{CH}_2\text{OH} + [\text{O}] \xrightarrow{\text{distil}} \text{CH}_3\text{CHO} + \text{H}_2\text{O} \]
- Full Oxidation to Carboxylic Acid: Heating under reflux with excess acidified dichromate. Vapours condense and drip back into the flask to ensure complete oxidation: \[ \text{CH}_3\text{CH}_2\text{OH} + 2[\text{O}] \xrightarrow{\text{reflux}} \text{CH}_3\text{COOH} + \text{H}_2\text{O} \]
Secondary & Tertiary Alcohols
- Secondary Alcohols (2°): Oxidised to ketones under reflux: \[ \text{CH}_3\text{CH(OH)CH}_3 + [\text{O}] \rightarrow \text{CH}_3\text{COCH}_3 + \text{H}_2\text{O} \] Ketones resist further oxidation.
- Tertiary Alcohols (3°): Resist oxidation completely. There is no hydrogen atom attached to the carbon atom bearing the −OH group. The solution remains orange.
Distil or Reflux?
For a primary alcohol, the set-up decides the product:
- Distil: use excess alcohol. The aldehyde is distilled off as it forms, so it is not oxidised further.
- Reflux: use excess oxidising agent. Vapour condenses and drips back into the flask, so the aldehyde is oxidised to the carboxylic acid.
Required practical 4: distil a product from a reaction
Practise the distillation set-up on the RP4 page. Ethanol is oxidised to ethanal, and the ethanal is collected by distillation.
2. Elimination (Dehydration) to Form Alkenes
Heating an alcohol with an acid catalyst (concentrated sulfuric acid or concentrated phosphoric acid) at 170°C eliminates water to produce an alkene:
\[ \text{C}_2\text{H}_5\text{OH} \xrightarrow{\text{conc } \text{H}_2\text{SO}_4, 170^\circ\text{C}} \text{CH}_2=\text{CH}_2 + \text{H}_2\text{O} \]
Example: propan-2-ol gives propene when heated with concentrated phosphoric acid or concentrated sulfuric acid:
\[ \text{CH}_3\text{CH(OH)CH}_3 \xrightarrow{\text{conc H}_3\text{PO}_4 \text{ or conc H}_2\text{SO}_4\text{, heat}} \text{CH}_3\text{CH}{=}\text{CH}_2 + \text{H}_2\text{O} \]
Mechanism: Acid-Catalysed Elimination
- A lone pair on the O of the -OH group takes an H⁺ from the acid.
- The C-O bond breaks and water leaves. This gives a carbocation.
- An H⁺ is lost from the next carbon. The electrons from the C-H bond form the C=C double bond.
The H⁺ is given back at the end, so the acid is a catalyst.
Polymers Without Crude Oil
Alcohols can be made by fermentation of sugars from plants. Dehydrating these alcohols gives alkenes. The alkenes can be used to make addition polymers, without using monomers from crude oil.
Beyond the OxfordAQA specification: details of fermentation are background only (see section 4).
3. Distinguishing Aldehydes from Ketones
| Test Reagent | Active Chemical Species | Observation with Aldehydes | Observation with Ketones |
|---|---|---|---|
| Tollens' Reagent | Ammoniacal silver nitrate [Ag(NH₃)₂]⁺ | Silver mirror forms on flask walls (\(\text{Ag}^+ + e^- \rightarrow \text{Ag}(s)\)) | No reaction (solution remains colourless) |
| Fehling's Solution | Aqueous copper(II) tartrate complex | Blue solution forms brick-red precipitate of \(\text{Cu}_2\text{O}(s)\) | No reaction (solution remains deep blue) |
Warm each test in a water bath. Aldehydes are oxidised by these mild oxidising agents. Ketones are not, so there is no change. For all the organic tests in one place, see Organic Analysis and RP5.
4. Industrial Production of Ethanol
Beyond the OxfordAQA specification
9620 does not require fermentation or the hydration of ethene. This comparison is background only.
| Comparison Factor | Fermentation of Glucose | Direct Hydration of Ethene |
|---|---|---|
| Feedstock | Carbohydrates (sugar cane, starch) - Renewable | Crude oil fraction cracking - Non-renewable |
| Reaction Equation | C₆H₁₂O₆ → 2C₂H₅OH + 2CO₂ | CH₂=CH₂ + H₂O ⇌ C₂H₅OH |
| Conditions | Yeast enzymes, anaerobic, 35°C, atmospheric pressure | Conc H₃PO₄ catalyst, 300°C, 60 atm steam |
| Process Type | Slow batch process | Fast continuous process |
| Product Purity | Impure aqueous solution (max ~15%), requires fractional distillation | Pure ethanol (>95% purity) |
Go deeper
These UK AQA pages cover the same ideas in more detail. They may include content that is not in 9620.
5. Practice Questions
Show Answer & Mark Scheme
Reagent: Potassium dichromate(VI) acidified with dilute sulfuric acid. (1 mark)
Apparatus: Distillation apparatus (heating with a Liebig condenser set downwards for distillation). (1 mark)
Reason: Ethanal has a lower boiling point (21°C) than ethanol because ethanal molecules cannot form hydrogen bonds. Ethanal distils off immediately as it forms, preventing it from being further oxidised to ethanoic acid. (1 mark)
Show Answer & Mark Scheme
Mark Scheme:
- Curly arrow from a lone pair on the O of -OH to H⁺. (1 mark)
- Curly arrow from the C-O bond to the O of the protonated -OH group, so water leaves. (1 mark)
- Carbocation intermediate, CH₃C⁺HCH₃. (1 mark)
- Curly arrow from a C-H bond on the next carbon to the C-C bond, with loss of H⁺, giving CH₃CH=CH₂. (1 mark)
Show Answer & Mark Scheme
Mark Scheme:
- CH₃CH₂CH₂OH + 2[O] → CH₃CH₂COOH + H₂O. (1 mark for the correct product, 1 mark for the balanced equation)
- Orange to green. (1 mark)
Show Answer & Mark Scheme
Mark Scheme:
- Warm each with Fehling's solution: propanal turns the blue solution to a brick-red precipitate. (1 mark)
- Propanone: no change (stays blue). (1 mark)
- 2-methylpropan-2-ol is a tertiary alcohol, with no H atom on the carbon bonded to -OH, so it is not easily oxidised. (1 mark)