Alcohols sit at the crossroads of organic chemistry — they can be converted to alkenes, haloalkanes, aldehydes, ketones, and carboxylic acids, each by changing one reagent or one condition, making them the most synthetically versatile functional group in Module 7.
Use the PDF for classwork, homework or revision. It includes key ideas, activities, questions, an extend task and success-criteria proof.
A forensic breathalyser works by oxidising ethanol in breath to ethanoic acid using potassium dichromate — the orange dichromate turns green as it is reduced. A winemaker monitors the souring of wine into vinegar (ethanol → ethanoic acid) when wine is exposed to air. A polymer chemist dehydrates ethanol to ethene, which is then polymerised to polyethylene.
Before you read on: Write down what you think is the structural difference between the product of oxidation (ethanoic acid, CH₃COOH) and the product of dehydration (ethene, CH₂=CH₂). What bonds have been added to, or removed from, the ethanol molecule in each case?
Dehydration is the exact reverse of hydration — water is eliminated across the C–OH bond and an adjacent C–H bond, regenerating the C=C double bond — and the conditions that drive this reversal are the opposite of those that drive hydration.
Dehydration is an elimination reaction in which an alcohol loses a water molecule (H from one carbon, OH from the adjacent carbon) to form an alkene. It is the reverse of alkene hydration (L06, L10).
R–CH₂–CH₂OH → R–CH=CH₂ + H₂O
Ethanol → ethene:
CH₃CH₂OH → CH₂=CH₂ + H₂O
Conditions: conc. H₂SO₄, ~170°C
Propan-1-ol → propene:
CH₃CH₂CH₂OH → CH₃CH=CH₂ + H₂O
Conditions: conc. H₂SO₄ or conc. H₃PO₄, heat
When a secondary or tertiary alcohol is dehydrated, the H can be removed from either adjacent carbon, giving two possible alkene products. The major product is the more substituted alkene (more alkyl groups on the C=C carbons). For HSC, NESA accepts any alkene product, but the more substituted alkene is preferred.
Butan-2-ol dehydration:
CH₃CHOHCH₂CH₃ → CH₃CH=CHCH₃ + H₂O (but-2-ene — major, more substituted)
CH₃CHOHCH₂CH₃ → CH₂=CHCH₂CH₃ + H₂O (but-1-ene — minor, less substituted)
Dehydration of butan-2-ol — H removed from C3 and OH removed from C2 combine to form H₂O; C=C forms between C2 and C3 giving but-2-ene (major, more substituted product)
The reverse of the haloalkane substitution from L10 — instead of converting a haloalkane to an alcohol with NaOH, you can convert an alcohol to a haloalkane with HX — and the conditions and halide reactivity order are what HSC specifically tests.
When an alcohol is treated with a hydrogen halide (HX), the –OH group is replaced by the halogen, producing a haloalkane and water:
R–OH + HX → R–X + H₂O
Comparison — reverse reactions:
These are reverse reactions — HX pushes toward haloalkane; NaOH(aq) pushes toward alcohol.
Oxidation of alcohols is the single most important reaction for understanding how the organic functional group classes connect — primary alcohols become aldehydes and then carboxylic acids, secondary alcohols become ketones, and tertiary alcohols refuse to react — and the reason is structural, not arbitrary.
Oxidation of an alcohol involves removing two hydrogen atoms — one H from the C–OH carbon and one H from the –OH oxygen — to form the C=O (carbonyl) bond. For this to be possible, the C–OH carbon must have at least one H atom available to be removed.
Colour change = oxidation occurred (1° or 2° alcohol)
No change (stays orange) = no oxidation (3° or no alcohol)
Under acidic conditions. Colour change = oxidation occurred
(Note: neutral/basic → brown MnO₂ precipitate instead)
The product from oxidising a primary alcohol depends entirely on the equipment used:
Aldehyde has lower BP than alcohol — collected as it forms before excess oxidant can oxidise it further. Removes product from reaction mixture immediately.
All components stay in flask; excess oxidant has maximum contact time to fully oxidise aldehyde intermediate through to carboxylic acid.
| Alcohol Class | Oxidant | Equipment | Product | Colour Change (K₂Cr₂O₇) |
|---|---|---|---|---|
| Primary (1°) | K₂Cr₂O₇/H⁺ | Distillation | Aldehyde | Orange → Green |
| Primary (1°) | K₂Cr₂O₇/H⁺ (excess) | Reflux | Carboxylic acid | Orange → Green |
| Secondary (2°) | K₂Cr₂O₇/H⁺ | Reflux | Ketone | Orange → Green |
| Tertiary (3°) | K₂Cr₂O₇/H⁺ | Any | No reaction | Stays orange |
Putting dehydration, substitution, and oxidation together with the production reactions from L10 reveals that alcohols are the central hub of the organic reaction map — every arrow into and out of alcohol connects to a different functional group class.
Alcohol reaction hub — dashed arrows show production routes (L10); solid arrows show this lesson's reactions. Alcohol is the central node connecting every major functional group class.
"Dehydration adds water to the alcohol." Dehydration REMOVES water — the prefix "de-" means removal. The –OH and a neighbouring H leave together as H₂O. If water appears as a reactant in your equation, you have written a hydration reaction (the reverse process).
"Tertiary alcohols give ketones when oxidised." Tertiary alcohols do NOT react with K₂Cr₂O₇/H⁺ or KMnO₄/H⁺ — the solution stays orange (or purple). The absence of colour change IS the result for a tertiary alcohol. Only secondary alcohols give ketones on oxidation.
"KMnO₄ turns orange when it oxidises an alcohol." Orange is the colour of dichromate (Cr₂O₇²⁻). Permanganate (MnO₄⁻) is purple and becomes colourless (Mn²⁺) in acidic conditions, or forms a brown precipitate (MnO₂) in neutral/alkaline conditions. KMnO₄ never turns orange.
"You can get either aldehyde or carboxylic acid from a primary alcohol just by changing the oxidant." The key variable is the equipment — distillation vs. reflux — not just the oxidant choice. Reflux with any strong oxidant gives acid; distillation removes the aldehyde before it can be further oxidised regardless of which oxidant is used.
DEHYDRATION:
R-CH₂-CH₂OH → R-CH=CH₂ + H₂O
Conditions: conc. H₂SO₄ or H₃PO₄ · heat · distillation
Note: concentrated acid (NOT dilute)
SUBSTITUTION WITH HX:
R-OH + HX → R-X + H₂O
Conditions: HCl/HBr/HI · heat under reflux
Reactivity: HI > HBr > HCl
Always write H₂O as second product!
OXIDATION:
1° + K₂Cr₂O₇/H⁺ + distillation → ALDEHYDE (orange→green)
1° + K₂Cr₂O₇/H⁺ (excess) + reflux → CARBOXYLIC ACID (orange→green)
2° + K₂Cr₂O₇/H⁺ + reflux → KETONE (orange→green)
3° + K₂Cr₂O₇/H⁺ → NO REACTION (stays orange)
KMnO₄/H⁺: purple → colourless (NOT orange!)
For each transformation below, write: the reagent, the conditions (temperature/equipment), and the type of reaction.
An unknown alcohol X (molecular formula C₃H₈O) is tested with the following reagents:
Identify alcohol X, write its structural formula, and explain each piece of evidence. Then write the equation for the oxidation of X to its maximum oxidation product.
Question 1. A student adds excess acidified K₂Cr₂O₇ to pentan-1-ol under reflux. What is the organic product and what colour change is observed?
Question 2. Which conditions would convert ethanol to ethanal (acetaldehyde) rather than to ethanoic acid?
Question 3. A student treats 2-methylpropan-2-ol with acidified K₂Cr₂O₇. What does the student observe, and why?
Question 4. Which conditions would correctly convert butan-1-ol to butanal (and NOT butanoic acid)?
Question 5. A compound with molecular formula C₄H₈O does not give a positive Tollens' test and does not react with acidified K₂Cr₂O₇ (solution stays orange). What is the most likely identity of this compound?
Look back at what you wrote in the Think First section. What has changed? What did you get right? What surprised you?
Question 6 4 marks
Write balanced equations for three reactions of ethanol: (a) dehydration to ethene; (b) reaction with HBr to form a haloalkane; (c) oxidation to ethanoic acid. For each equation, state the reagents and conditions required.
Question 7 5 marks
The same primary alcohol — pentan-1-ol — can be converted to either pentanal or pentanoic acid using K₂Cr₂O₇/H₂SO₄. (a) Write the equation for each transformation. (b) State the specific conditions (including equipment) that produce each product. (c) Explain why using reflux rather than distillation changes the product from aldehyde to carboxylic acid.
Question 8 6 marks
A student performs a two-step synthesis: Step 1 — ethene is reacted with steam to produce ethanol. Step 2 — ethanol is converted to ethanoic acid. For each step, write the balanced equation, state the reagents and conditions, and explain the role of a specific condition in determining the product (including why reflux in step 2 gives acid rather than aldehyde).
Return to your prediction about the structural difference between ethanoic acid (oxidation) and ethene (dehydration), both starting from ethanol.
Oxidation (ethanol → ethanoic acid): Two H atoms are removed and replaced with a C=O and an OH group. The carbon count stays at 2, the chain stays intact, but the functional group changes from –OH to –COOH. The molecule gains an oxygen (from the oxidising agent). Net: H₂ removed, [O] added.
Dehydration (ethanol → ethene): The –OH group and an H from the adjacent carbon leave together as H₂O. The carbon count stays at 2, the molecule gets smaller (loses H₂O), and a C=C double bond forms where the C–OH and C–H bonds were. Net: H₂O removed, C=C formed.
The breathalyser, the winemaker, and the polymer chemist are each exploiting a different reactivity of the same –OH group: oxidation to –COOH, further oxidation to –COOH, and elimination of –OH as water respectively. One functional group, three industries, three completely different products.
Answer questions on Reactions of Alcohols — Dehydration, Substitution & Oxidation before your opponents cross the line. Fast answers = faster car. Pool: lessons 1–12.