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Alcohols are organic compounds containing the hydroxyl functional group (-OH) bonded to a saturated carbon atom. They have the general formula CₙH₂ₙ₊₁OH and are among the most versatile molecules in organic chemistry — they can be oxidised, dehydrated, esterified, and used as fuels and solvents.
Like halogenoalkanes, alcohols are classified according to the number of carbon atoms bonded to the carbon carrying the -OH group:
| Classification | Carbon Bearing -OH | Example |
|---|---|---|
| Primary (1°) | Bonded to one other carbon (or none) | Ethanol, CH₃CH₂OH |
| Secondary (2°) | Bonded to two other carbons | Propan-2-ol, (CH₃)₂CHOH |
| Tertiary (3°) | Bonded to three other carbons | 2-Methylpropan-2-ol, (CH₃)₃COH |
This classification is critical because it determines the oxidation products.
The -OH group can form hydrogen bonds with other alcohol molecules and with water. Hydrogen bonding occurs because oxygen is highly electronegative, creating a significant δ+ on the hydrogen. This Oδ⁻-Hδ+ can interact with a lone pair on the oxygen of a neighbouring molecule.
Consequences of hydrogen bonding:
Within the alcohol homologous series, boiling points increase with chain length due to increasing van der Waals forces (more electrons and greater surface area), in addition to the hydrogen bonding that all alcohols exhibit.
| Alcohol | Formula | Boiling Point (°C) | Solubility in Water |
|---|---|---|---|
| Methanol | CH₃OH | 65 | Fully miscible |
| Ethanol | C₂H₅OH | 78 | Fully miscible |
| Propan-1-ol | C₃H₇OH | 97 | Fully miscible |
| Butan-1-ol | C₄H₉OH | 117 | Slightly soluble |
| Hexan-1-ol | C₆H₁₃OH | 157 | Very slightly soluble |
The trend in solubility is clear: as the hydrocarbon chain grows, the non-polar portion dominates over the polar -OH group, and the alcohol becomes less able to interact with water molecules.
Alcohols burn in air in an exothermic combustion reaction:
C₂H₅OH + 3O₂ → 2CO₂ + 3H₂O
Ethanol is used as a fuel (biofuel) and as a fuel additive. It burns more cleanly than petrol, producing less CO and particulates. Methanol is used as a racing fuel due to its high octane rating.
The oxidation of alcohols is one of the most important topics in A-Level organic chemistry. The oxidising agent is acidified potassium dichromate(VI), K₂Cr₂O₇ / H₂SO₄. In equations, this is written as [O] for simplicity.
The visible colour change is from orange (Cr₂O₇²⁻, dichromate) to green (Cr³⁺, chromium(III) ions).
graph TD
A["Unknown Alcohol"] --> B{"What class?"}
B -->|"Primary"| C["Can be oxidised in two stages"]
B -->|"Secondary"| D["Can be oxidised in one stage"]
B -->|"Tertiary"| E["Cannot be oxidised — dichromate stays orange"]
C -->|"Distil with limited [O]"| F["Aldehyde (R-CHO)"]
C -->|"Reflux with excess [O]"| G["Carboxylic Acid (R-COOH)"]
F -->|"Further [O] if not removed"| G
D -->|"Distil or reflux"| H["Ketone (R-CO-R')"]
H -->|"No further oxidation possible"| I["Ketone is the final product"]
Primary alcohols can be oxidised in two stages:
Stage 1: Primary alcohol → Aldehyde
CH₃CH₂OH + [O] → CH₃CHO + H₂O
To obtain the aldehyde, use distillation — heat gently with a limited amount of oxidising agent and collect the aldehyde as it distils off (aldehydes have lower boiling points than the parent alcohols because they cannot hydrogen bond with each other as effectively).
Stage 2: Aldehyde → Carboxylic acid
CH₃CHO + [O] → CH₃COOH
To obtain the carboxylic acid, use reflux — heat with excess oxidising agent under reflux so the aldehyde cannot escape and is fully oxidised.
Why does the apparatus matter? Distillation removes the aldehyde from the hot reaction mixture as soon as it forms, preventing further oxidation. Reflux condenses the vapour and returns it to the flask, keeping it in contact with the oxidising agent. The apparatus choice determines the product — this is a classic exam question.
Secondary alcohols are oxidised to ketones:
(CH₃)₂CHOH + [O] → (CH₃)₂CO + H₂O
Ketones cannot be further oxidised under these conditions (because there is no hydrogen on the carbonyl carbon to remove). Whether you distil or reflux, the product is always the ketone.
Tertiary alcohols cannot be oxidised by acidified dichromate. There is no hydrogen atom on the carbon bearing the -OH group, so the oxidation reaction cannot take place. The dichromate solution remains orange — there is no colour change.
Why can't tertiary alcohols be oxidised? Oxidation of an alcohol involves removing a hydrogen from the C-OH carbon and removing the O-H hydrogen. For this to happen, there must be a C-H bond on the same carbon as the -OH group. In tertiary alcohols, this carbon is bonded to three other carbon groups and no hydrogen — so the reaction cannot proceed.
| Alcohol Type | Product (distillation) | Product (reflux with excess) | Dichromate Colour Change |
|---|---|---|---|
| Primary | Aldehyde | Carboxylic acid | Orange → Green |
| Secondary | Ketone | Ketone (no further oxidation) | Orange → Green |
| Tertiary | No reaction | No reaction | Stays orange |
This difference in oxidation behaviour provides a chemical test to distinguish between primary, secondary, and tertiary alcohols.
Alcohols can be dehydrated to form alkenes by removing water (H₂O) from adjacent carbon atoms:
CH₃CH₂OH → CH₂=CH₂ + H₂O
Alternatively, pass alcohol vapour over a heated aluminium oxide (Al₂O₃) catalyst.
Dehydration is the reverse of the hydration of alkenes. It is an important route from alcohols back to alkenes.
When butan-2-ol is dehydrated, the water can be lost in two ways, producing a mixture of alkenes:
The major product is typically the more substituted alkene (but-2-ene in this case), following Zaitsev's rule. However, you should be aware that a mixture of products is formed.
Alcohols react with carboxylic acids in the presence of a concentrated sulfuric acid catalyst to form esters:
CH₃COOH + C₂H₅OH ⇌ CH₃COOC₂H₅ + H₂O
Esters are characterised by their pleasant, fruity smells and are used as flavourings, perfumes, solvents, and in the manufacture of polyesters.
The ester linkage is written as -COO-. In the name, the alcohol part comes first (as an alkyl group) followed by the acid part with the suffix changed to "-oate". For example, ethanol + ethanoic acid → ethyl ethanoate.
Methanol + propanoic acid → methyl propanoate
The "methyl" comes from methanol (1 carbon), and "propanoate" comes from propanoic acid (3 carbons, with -oic acid changed to -oate).
A practical method to identify the class of an unknown alcohol:
Add acidified potassium dichromate and warm.
Test the organic product.
Tollens' reagent contains the [Ag(NH₃)₂]⁺ complex ion. Aldehydes reduce Ag⁺ to Ag (metallic silver, which forms a mirror on the test tube wall). Ketones cannot do this because they lack the H on the carbonyl carbon needed for further oxidation.
Fehling's solution contains Cu²⁺ ions complexed with tartrate. Aldehydes reduce Cu²⁺ (blue) to Cu₂O (red/brick-red precipitate). Again, ketones do not react.
C₆H₁₂O₆ → 2C₂H₅OH + 2CO₂
CH₂=CH₂ + H₂O → CH₃CH₂OH
| Feature | Fermentation | Hydration of Ethene |
|---|---|---|
| Feedstock | Renewable (sugar) | Non-renewable (crude oil) |
| Process type | Batch | Continuous |
| Rate | Slow | Fast |
| Purity | Low (needs distillation) | High |
| Temperature | 25–35 °C | 300 °C |
| Carbon neutral? | Yes (CO₂ absorbed by plants) | No |
Edexcel 9CH0 specification Topic 7 — Organic Chemistry I, sub-topic 7.2 covers alcohols: classification as primary, secondary or tertiary; oxidation by acidified potassium dichromate(VI) — primary alcohols oxidise to aldehydes (under controlled conditions, distilling out the aldehyde) or to carboxylic acids (under reflux), secondary alcohols oxidise to ketones, tertiary alcohols are not oxidised by K2Cr2O7/H+ (and the negative result distinguishes them from primary/secondary alcohols); dehydration to alkenes with concentrated H2SO4 or H3PO4 catalyst at high temperature; combustion; and the synoptic role of alcohols as solvents and as fuel additives — refer to the official specification document for exact wording. Sub-topic 7.2 is examined directly on Paper 2 (oxidation products, observations) and Paper 3 (synoptic with esterification CP15, distillation CP6, NMR/IR identification). The data booklet provides characteristic IR values: O-H stretch (broad) 3200–3550 cm−1; C-O stretch 1000–1300 cm−1.
Question (8 marks):
(a) Predict the products and observations for the reaction of acidified potassium dichromate(VI) with each of the following alcohols, heated under reflux: (i) ethanol; (ii) propan-2-ol; (iii) 2-methylpropan-2-ol. (6)
(b) State the colour change observed for each positive reaction, and explain why tertiary alcohols are resistant to oxidation. (2)
Solution with mark scheme:
(a)(i) Ethanol — primary alcohol. Under reflux with excess oxidising agent, ethanol oxidises in two steps: first to ethanal (aldehyde), then further to ethanoic acid (carboxylic acid).
CH3−CH2−OH[O]CH3−CHO[O]CH3−COOH
M1 — primary alcohol oxidation to carboxylic acid identified.
A1 — final product: ethanoic acid (CH3−COOH).
(a)(ii) Propan-2-ol — secondary alcohol. Oxidation gives a ketone; ketones are not further oxidised by K2Cr2O7/H+.
CH3−CH(OH)−CH3[O]CH3−CO−CH3
M1 — secondary alcohol oxidation to ketone identified.
A1 — product: propanone (CH3−CO−CH3, also known as acetone).
(a)(iii) 2-methylpropan-2-ol — tertiary alcohol. No reaction.
M1 — no reaction occurs.
A1 — explanation: the C bearing -OH has no H atom (it is bonded to three methyl groups and one OH); oxidation requires removal of an H from the C-OH carbon, which is impossible for tertiary alcohols.
(b) Step 1 — colour change.
A1 — for both (i) and (ii), the orange (Cr2O72−) solution turns green (Cr3+). For (iii), the solution remains orange (no reaction).
Step 2 — explanation of tertiary resistance.
A1 — the oxidation mechanism removes the H atom on the C-OH carbon as the alcohol becomes the carbonyl; tertiary alcohols have no such H, so the mechanism fails.
Total: 8 marks (M3 A5).
Question (6 marks): A student carries out CP6 (preparation of an organic liquid by distillation) by oxidising butan-1-ol with hot acidified K2Cr2O7. Two procedures are tested:
(a) State the product of each procedure. (2)
(b) Write the equations for each procedure (use [O] to represent the oxidising agent). (2)
(c) Explain how the apparatus differs in each procedure and why this controls the product. (2)
Mark scheme decomposition by AO:
(a)
(b)
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