Master the three core transformations of alcohols: eliminate water to make alkenes, swap –OH for a halogen, or strip electrons to build aldehydes, ketones and carboxylic acids — and learn exactly which conditions control which product.
Four printable worksheets that build from the foundations up to exam-style questions — start at whatever level suits you.
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
Worked Equations:
CH₃CH₂OH + HBr → CH₃CH₂Br + H₂O (bromoethane)
CH₃CH₂CH₂OH + HI → CH₃CH₂CH₂I + H₂O (1-iodopropane)
Comparison — reverse reactions:
R–OH + HX → R–X + H₂O (alcohol → haloalkane — this lesson)
R–X + NaOH(aq) → R–OH + NaX (haloalkane → alcohol — L10)
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.
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 |
|---|---|---|---|---|
| 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.
Problem: For each reaction below, write the balanced equation, name the product, and state all conditions: (a) dehydration of propan-1-ol; (b) reaction of butan-1-ol with HBr; (c) oxidation of propan-2-ol to its maximum oxidation product.
Problem: Three unlabelled alcohols — P, Q, R — all have molecular formula C₄H₁₀O. When treated with acidified K₂Cr₂O₇: P stays orange. Q turns green; distillation of the product gives a compound that produces a silver mirror with Tollens' reagent. R turns green; the product does NOT give a silver mirror and does NOT turn litmus red. (a) Classify P, Q, R as 1°/2°/3°. (b) Identify each compound. (c) Write the oxidation equation for Q under reflux conditions.
Complete the Learn phase to unlock Practice.
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.
Q1. A student adds excess acidified K₂Cr₂O₇ to pentan-1-ol under reflux. What is the organic product and what colour change is observed?
Q2. Which conditions would convert ethanol to ethanal (acetaldehyde) rather than to ethanoic acid?
Q3. A student treats 2-methylpropan-2-ol with acidified K₂Cr₂O₇. What does the student observe, and why?
Q4. Which conditions would correctly convert butan-1-ol to butanal (and NOT butanoic acid)?
Q5. 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?
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.
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.
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).
Q1 — B: Pentan-1-ol is a primary alcohol. Excess K₂Cr₂O₇/H⁺ under REFLUX gives carboxylic acid (pentanoic acid). Orange → green. A would result from DISTILLATION. C is wrong — pentan-1-one would require a secondary alcohol. D has wrong colour change — purple→colourless is KMnO₄.
Q2 — B: To produce the aldehyde from a primary alcohol, the aldehyde must be removed before excess oxidant can oxidise it further — achieved by DISTILLATION. A (excess oxidant, reflux) fully oxidises to ethanoic acid. D describes dehydration to ethene — not oxidation.
Q3 — C: 2-methylpropan-2-ol is a tertiary alcohol — the C–OH carbon is bonded to three other carbons with no H atoms. K₂Cr₂O₇/H⁺ stays orange = no oxidation. This orange/no-change result is the diagnostic test for tertiary alcohols.
Q4 — C: To stop at the aldehyde stage, the aldehyde must be removed from the reaction mixture as soon as it forms — before the oxidant can oxidise it further to butanoic acid. Distillation achieves this because butanal has a lower boiling point than butan-1-ol.
Q5 — D: No reaction with K₂Cr₂O₇/H⁺ (stays orange) → not a primary or secondary alcohol and not an aldehyde. No positive Tollens' → not an aldehyde. C₄H₈O with a C=O that is not an aldehyde = ketone. Butan-2-one (CH₃COCH₂CH₃) is the C₄ ketone.
Q6 (4 marks): (a) CH₃CH₂OH → CH₂=CH₂ + H₂O. Conditions: conc. H₂SO₄, heat (~170°C), distillation. (b) CH₃CH₂OH + HBr → CH₃CH₂Br + H₂O. Conditions: HBr, heat under reflux. (c) CH₃CH₂OH + 2[O] → CH₃COOH + H₂O. Conditions: excess K₂Cr₂O₇/H₂SO₄, reflux.
Q7 (5 marks): (a) Pentanal: CH₃CH₂CH₂CH₂CH₂OH + [O] → CH₃CH₂CH₂CH₂CHO + H₂O. Pentanoic acid: + 2[O] → CH₃CH₂CH₂CH₂COOH + H₂O. (b) Pentanal: K₂Cr₂O₇/H₂SO₄, gentle heating, DISTILLATION. Pentanoic acid: excess K₂Cr₂O₇/H₂SO₄, REFLUX. (c) Pentanal has a lower boiling point than pentan-1-ol. Under distillation, pentanal boils off and is removed from the reaction mixture as soon as it forms — before excess oxidant can oxidise it further. Under reflux, all volatile components are condensed and returned to the flask, giving the oxidant maximum contact time to oxidise the aldehyde all the way to the carboxylic acid.
Q8 (6 marks): Step 1: CH₂=CH₂ + H₂O ⇌ CH₃CH₂OH. Conditions: H₂O (steam), H₃PO₄ catalyst, ~300°C, ~65 atm. High pressure used because the left side has 2 moles of gas — by Le Chatelier, high pressure shifts equilibrium right toward ethanol. Step 2: CH₃CH₂OH + 2[O] → CH₃COOH + H₂O. Conditions: excess K₂Cr₂O₇/H₂SO₄, reflux. Role of reflux: keeps all volatile components (including the aldehyde intermediate, ethanal) in the reaction flask, giving the excess oxidant maximum contact time to oxidise ethanal through to ethanoic acid. If distillation were used instead, ethanal would be removed before further oxidation.
Return to your prediction about the structural difference between ethanoic acid (oxidation) and ethene (dehydration), both starting from ethanol. Oxidation: two H atoms removed, C=O and OH group added — functional group changes from –OH to –COOH. Dehydration: –OH and an H from adjacent carbon leave as H₂O, C=C double bond forms. How well did your structural prediction compare?
Coming up in Lesson 13: The oxidation products of primary alcohols (aldehydes) and secondary alcohols (ketones) become the focus — understanding their distinct structural properties, reactivity with Tollens' and Fehling's reagents, and how to distinguish between them experimentally.
What reagent and equipment convert a primary alcohol to an aldehyde (stopping before carboxylic acid)?
Why do tertiary alcohols not react with K₂Cr₂O₇/H⁺?
Give the HX reactivity order and explain why HI is the fastest.
What acid concentration is required for dehydration and why does concentration matter?
A student observes that KMnO₄ turns orange when oxidising an alcohol. What is wrong with this observation?
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