Mini Chemistry
Part of Mini EducationSwitch subject

Alcohols: Oxidation and Dehydration

Learn and apply Alcohols: Oxidation and Dehydration in the published Chemistry course sequence.

  • GCE A-Level H2 Chemistry 9476-2027
On this page

Alcohols: Oxidation and Dehydration: Orientation

Alcohol questions are usually “classify → choose conditions → predict product”: oxidation depends on 1°/2°/3° and distil vs reflux, while dehydration is an elimination to an alkene. This lesson also finishes with phenol as a deliberate comparison. Phenol is not an alcohol: its -OH group is attached directly to a benzene ring, so its acidity and ring reactions must be learned separately.

This topic gets cleaner if you cross-check Organic Mechanisms: Curly Arrows, Electrophiles, Nucleophiles while navigating from the Organic Chemistry hub.

Definitions (Must Know)

A. Alcohol classification

  • Primary (1°) alcohol: RCH₂OH
  • Secondary (2°) alcohol: R₂CHOH
  • Tertiary (3°) alcohol: R₃COH

B. Oxidation (organic shorthand)

In organic chemistry, oxidation is often shown as adding [O] (increase in C–O bonds / decrease in C–H bonds).

C. Dehydration

Dehydration of an alcohol removes H₂O to form an alkene (an elimination reaction).

D. Phenol

In phenol, C₆H₅OH, the -OH group is bonded directly to a benzene-ring carbon. Do not classify phenol as a primary, secondary or tertiary alcohol; those labels apply when -OH is attached to an aliphatic carbon.

Detailed Explanations

A. Why oxidation products depend on alcohol class

Primary and secondary alcohols have a hydrogen on the carbon bearing -OH that can be removed during oxidation. Because tertiary alcohols have no C–H bond on that carbon, therefore they resist oxidation under typical K₂Cr₂O₇/H⁺ conditions.

B. Oxidation products by alcohol type

Primary (example: ethanol): CH₃CH₂OH + [O] → CH₃CHO + H₂O CH₃CHO + [O] → CH₃COOH

Secondary (example: propan-2-ol): CH₃CH(OH)CH₃ + [O] → CH₃COCH₃ + H₂O

Tertiary (example: 2-methylpropan-2-ol):

  • no oxidation with K₂Cr₂O₇/H⁺ under typical syllabus conditions

C. Workflow: distillation vs reflux (how to control oxidation)

  1. Identify alcohol class (1°, 2°, 3°).
  2. If it is primary, decide whether the question wants aldehyde (controlled) or carboxylic acid (complete).
  3. Choose apparatus wording:
    • distil to remove aldehyde as it forms (stops further oxidation)
    • reflux with excess oxidant to push to carboxylic acid

Mini example: Ethanol + K₂Cr₂O₇/H⁺ under reflux gives ethanoic acid; distillation gives ethanal.

  • To stop at an aldehyde: distil the aldehyde off as it forms (prevents further oxidation).
  • To form a carboxylic acid: heat under reflux with excess oxidising agent (keeps the product in contact with oxidant).

D. Dehydration to form alkenes

Example (ethanol → ethene): CH₃CH₂OH → CH₂ = CH₂ + H₂O

Conditions to quote (any one route):

  • concentrated H₃PO₄, heat

E. Other reactions of alcohols

  • Combustion: complete combustion forms CO₂ and H₂O.
  • With sodium: 2ROH + 2Na → 2RONa + H₂; the alcohol acts as a very weak acid.
  • With hydrogen halides: ROH + HX → RX + H₂O, using the stated acid and suitable conditions.
  • With phosphorus(V) chloride: ROH + PCl₅ → RCl + POCl₃ + HCl; steamy fumes of hydrogen chloride are observed.
  • Tri-iodomethane reaction: ethanol and alcohols containing CH₃CH(OH)R give a pale-yellow precipitate of CHI₃ with alkaline iodine on warming.

F. Phenol is not simply “an aromatic alcohol”

Phenol has its -OH group attached directly to a benzene ring. The phenoxide ion is stabilised by delocalisation, so phenol is more acidic than water and ordinary alcohols, though less acidic than carboxylic acids.

  • It reacts with sodium and with aqueous NaOH to form phenoxide, but not with aqueous carbonate.
  • Bromine water reacts readily without a halogen carrier, giving 2,4,6-tribromophenol as a white precipitate and decolourising the bromine water.
  • With dilute nitric acid it forms a mixture of 2-nitrophenol and 4-nitrophenol; more vigorous nitration can give 2,4,6-trinitrophenol.
  • Phenoxide can react with an acyl chloride to form a phenyl ester.

Keep phenol and alcohol conditions separate: their different acidity and activated aromatic ring lead to different reactions.

Worked Examples

Modelled example 1

Controlled Oxidation of Ethanol

Core

Problem

Ethanol is warmed with acidified potassium dichromate(VI), and the product is distilled as it forms. State the organic product and one observation.
Study the worked solution

  1. Classify the alcohol

    Method

    Identify ethanol as a primary alcohol.

    Reason

    Primary alcohols first oxidise to aldehydes.

    Working

    CH₃CH₂OH → [O] CH₃CHO

  2. Use the apparatus cue

    Method

    Distil ethanal from the reaction mixture as it forms.

    Reason

    Removing the aldehyde limits its further oxidation to ethanoic acid.

    Working

    Organic product: ethanal, CH₃CHO.

  3. State the oxidant observation

    Method

    Record an orange-to-green colour change.

    Reason

    Acidified dichromate(VI) ions are reduced as ethanol is oxidised.

    Working

    Orange → green.

Guided practice 2

Distinguish Phenol from Ethanol

About 7 min

Problem

Separate samples of phenol and ethanol are tested with aqueous sodium hydroxide and then with bromine water. Predict and explain the different results.

Compare acidity and ring reactivity

Hints

Hint 1: compare conjugate bases
The phenoxide ion is stabilised by delocalisation; ethoxide is not.
Hint 2: compare bromination
The phenolic –OH group activates the benzene ring towards substitution.
View solution step by step

  1. Compare reaction with sodium hydroxide

    Method

    State that phenol forms sodium phenoxide but ethanol does not react appreciably with aqueous sodium hydroxide.

    Reason

    Phenol is more acidic because negative charge in phenoxide is delocalised; ethanol is too weak an acid for this reaction to proceed appreciably.

    Working

    C₆H₅OH + OH⁻ → C₆H₅O⁻ + H₂O

  2. Compare reaction with bromine water

    Method

    State that phenol decolourises bromine water and forms a white precipitate of 2,4,6-tribromophenol, while ethanol gives no such reaction.

    Reason

    The –OH group donates electron density into the benzene ring, allowing rapid substitution without a halogen carrier.

    Working

    Phenol: orange-brown bromine water → colourless solution + white precipitate.

Common misconception 3

Oxidation of Propan-2-ol

Find and correct the mistake

Learner claim

A learner says heating any alcohol under reflux with acidified potassium dichromate(VI) produces a carboxylic acid. Correct the claim for propan-2-ol and name its product.

Use the alcohol class

Alcohol class
Product class

View solution step by step

  1. Classify before using the conditions

    Method

    Classify propan-2-ol as a secondary alcohol.

    Reason

    The carbon bearing OH is attached to two carbon groups.

    Working

    CH₃CH(OH)CH₃: secondary alcohol.

  2. Predict the oxidation product

    Method

    Form the ketone propanone.

    Reason

    Secondary-alcohol oxidation converts the C–OH centre into C=O without producing a carboxylic acid under these conditions.

    Working

    CH₃CH(OH)CH₃ → [O] CH₃COCH₃; product: propanone.

Examiner practice 4

Use Three Reactions of Propan-2-ol

6 marks

Problem

For propan-2-ol, state the organic or gaseous product and a relevant observation when it reacts separately with (a) sodium, (b) phosphorus(V) chloride and (c) alkaline iodine on warming. Explain why the final test is positive. [6 marks]

Try this before viewing the solution

View solution step by step

  1. React with sodium

    2 marks

    Method

    Form sodium propan-2-oxide and hydrogen gas; describe effervescence.

    Reason

    Sodium replaces the hydrogen of the hydroxyl group.

    Working

    2(CH₃)₂CHOH + 2Na → 2(CH₃)₂CHONa + H₂

  2. React with phosphorus(V) chloride

    2 marks

    Method

    Form 2-chloropropane and observe steamy hydrogen chloride fumes.

    Reason

    PCl₅ replaces –OH by Cl and also forms POCl₃ and HCl.

    Working

    (CH₃)₂CHOH + PCl₅ → (CH₃)₂CHCl + POCl₃ + HCl

  3. Apply the tri-iodomethane test

    2 marks

    Method

    State that a pale-yellow precipitate of CHI₃ forms.

    Reason

    Propan-2-ol contains the required CH₃CH(OH)R unit and is oxidised under the test conditions to a methyl ketone.

    Working

    Positive test: pale-yellow tri-iodomethane precipitate.

Challenge 5

Dehydrate Ethanol to Ethene

Minimal support

Synthesis transfer

Starting from the target conversion ethanol → ethene, state one complete reagent-and-condition route and identify the small molecule eliminated.

Choose one coherent route

Valid route
Small molecule removed

Hints

Hint 1: reaction name
Dehydration means removal of the elements of water.
Hint 2: route choice
Use concentrated H₃PO₄ with heat.
View solution step by step

  1. Select a complete route

    Method

    Heat ethanol with concentrated phosphoric(V) acid.

    Reason

    The acid and heat promote elimination from the alcohol.

    Working

    Reagent and condition: concentrated H₃PO₄, heat.

  2. Track the eliminated molecule

    Method

    Remove water and form the C=C bond of ethene.

    Reason

    Loss of H and OH from adjacent positions gives the alkene.

    Working

    CH₃CH₂OH → CH₂ = CH₂ + H₂O.

Mind Stretchers

Mind stretcher 1Extension

Explain why tertiary alcohols resist oxidation under these conditions.

Show Hint

Classify the carbon bearing OH before choosing an oxidation product.

Show Answer

Mark scheme:

  • Oxidation would require breaking a C–C bond or removing a hydrogen from the carbon bearing the -OH.
  • Tertiary alcohols have no C–H bond on that carbon, so oxidation does not proceed under typical conditions.

Mind stretcher 2: Distinguishing controlled oxidation conditionsExtension

Question. Explain why propan-1-ol may give propanal when product is distilled, but propanoic acid under reflux with excess oxidant.

Show Hint

Ask whether the first oxidation product remains in contact with the oxidising mixture.

Show Answer

Distillation removes volatile propanal as it forms and limits further oxidation. Under reflux, propanal remains with excess acidified oxidant and is further oxidised to propanoic acid.