OxfordAQA A-Level Chemistry: Alcohols (9620) – Revision Notes

Start with the alcohols study guide if any idea here is new: it explains each step with worked examples. This page is the short version of section 3.3.5 Alcohols from the OxfordAQA International AS and A-level Chemistry (9620) specification, Version 5.3, for International AS and A-level exams May/June 2020 onwards, covering 3.3.5.1 Oxidation of alcohols, 3.3.5.2 Elimination and Required practical 4. All of it is International AS. The Unit 2 exam (Organic 1 and Physical 1) tests it, and the practical and synoptic Unit 5 exam may test any content.

Test yourself afterwards with the alcohols practice questions. All 9620 topics: course page for 9620 chemistry. Record progress on the printable checklist as you go, and sit a free diagnostic (about 10 minutes) to rank your weakest areas.

Key definitions

Term	Meaning
Primary alcohol	The C–OH carbon is bonded to one other carbon (methanol: none); it carries two H atoms
Secondary alcohol	The C–OH carbon is bonded to two other carbons; it carries one H atom
Tertiary alcohol	The C–OH carbon is bonded to three other carbons; it carries no H atom
Aldehyde	–CHO group (C=O on a terminal carbon); suffix -al
Ketone	C=O flanked by two carbon atoms, –CO–; suffix -one
[O]	Oxygen from the oxidising agent; acceptable in oxidation equations
Reflux	Continuous boiling with a vertical condenser, so vapour returns to the flask
Elimination (dehydration)	Acid-catalysed loss of a water molecule, turning an alcohol into an alkene

3.3.5.1 Oxidation of alcohols

Oxidising agent: potassium dichromate(VI) made acidic with dilute sulfuric acid. Positive result: orange (Cr₂O₇²⁻) → green (Cr³⁺).

Class	Example	Product	Equation with [O]
Primary, distil as it forms	hexan-1-ol	hexanal	CH₃(CH₂)₄CH₂OH + [O] → CH₃(CH₂)₄CHO + H₂O
Primary, reflux with excess oxidant	hexan-1-ol	hexanoic acid	CH₃(CH₂)₄CH₂OH + 2[O] → CH₃(CH₂)₄COOH + H₂O
Aldehyde, further oxidation	hexanal	hexanoic acid	CH₃(CH₂)₄CHO + [O] → CH₃(CH₂)₄COOH
Secondary	heptan-2-ol	heptan-2-one	CH₃CH(OH)(CH₂)₄CH₃ + [O] → CH₃CO(CH₂)₄CH₃ + H₂O
Tertiary	3-ethylpentan-3-ol	none (stays orange)	—

Method in steps: predicting an oxidation product

1. Circle the C–OH carbon. Count the carbons bonded to it to find the class.
2. Primary: keep the chain; –CH₂OH → –CHO (-al) with distillation, or –COOH (-oic acid) with reflux.
3. Secondary: –CH(OH)– → –CO– (-one); keep the locant of the old –OH carbon.
4. Tertiary: nothing forms, and the orange colour remains.
5. Write [O] equations: one [O] per oxidation step. Alcohol → aldehyde or ketone also gives H₂O; aldehyde → acid gives no water.
6. Count C, H and O on each side before moving on.

Distil or reflux: the must-know distinction

	Distil immediately	Reflux, then distil
Product from a primary alcohol	Aldehyde	Carboxylic acid
Amounts	Alcohol in excess (oxidant limited)	Oxidising agent in excess
Condenser	Sloping, leading to a collecting vessel	Vertical, open at the top
Reason	Without O–H groups, aldehyde molecules cannot hydrogen bond to each other; so it boils below the alcohol and escapes before a second oxidation	Condensed vapour drips back, so any aldehyde keeps meeting excess dichromate and ends up as the acid

Aldehyde or ketone? The two tests

Reagent	Conditions	Aldehyde	Ketone
Fehling’s solution (blue, Cu²⁺)	Warm in a water bath	Red precipitate (Cu₂O)	Stays blue
Tollens’ reagent ([Ag(NH₃)₂]⁺)	Warm in a water bath	Silver mirror	No change

The aldehyde is oxidised (RCHO + [O] → RCOOH); the metal ion is reduced (Cu²⁺ → Cu⁺, Ag⁺ → Ag). Water bath, not a flame: organic liquids are flammable. More on the silver complex in the 9620 transition metal chemistry guide.

Why tertiary alcohols resist oxidation

Oxidation removes the H from the –OH and an H from the C–OH carbon to make C=O. A tertiary C–OH carbon has no H atom, so it cannot form C=O without breaking a C–C bond. That does not happen with acidified dichromate(VI).

3.3.5.2 Elimination

Reagent and conditions: heat with a concentrated acid catalyst, such as concentrated H₂SO₄ or concentrated H₃PO₄.

General equation: alcohol → alkene + H₂O.

Mechanism in three arrows

1  lone pair on O  → H⁺            gives R–OH₂⁺ (protonated alcohol)
2  C–O bond electrons → O          H₂O leaves; carbocation forms
3  adjacent C–H bond electrons → C–C   C=C forms; H⁺ released (catalyst back)

Method in steps: how many alkenes?

1. Mark the C–OH carbon (it becomes the carbocation).
2. List each neighbouring carbon that carries at least one H.
3. Equivalent neighbours give the same alkene; different neighbours give different alkenes.
4. For each alkene, check each double-bond carbon: two different groups on both carbons means E and Z isomers both form.

Worked reminders.

• 3-ethylpentan-3-ol: the three neighbours are all CH₂ groups of ethyl chains, so they are equivalent. One alkene: 3-ethylpent-2-ene, CH₃CH=C(CH₂CH₃)₂. C3 carries two ethyl groups, so no E/Z isomers.
• Hexan-2-ol: neighbours are C1 and C3. Loss of H from C1 gives hex-1-ene. Loss from C3 gives hex-2-ene, which has E and Z forms. Total: three alkenes.

Why it matters

Alkenes made by elimination can be polymerised to addition polymers without monomers derived from crude oil, provided the alcohol does not come from crude oil itself.

Required practical 4: distil a product from a reaction

Set-up and reasons, ready to quote:

• Anti-bumping granules: smooth boiling.
• Thermometer bulb level with the side arm: measures the temperature of the vapour leaving.
• Condenser water in at the bottom, out at the top: the jacket stays full of cold water.
• Electric heater or water bath: no naked flame near flammable vapour.
• Open (or vented) collecting vessel: never heat a closed system.
• Collect only the fraction distilling near the product’s boiling point.

Percentage yield: convert both the starting mass and the product mass to moles, use the mole ratio from the equation, then (actual ÷ theoretical) × 100. More mole work: the 9620 moles and equations guide.

Quick self-test

1. Classify 4-methylpentan-2-ol.
2. Name the oxidising agent the specification gives for alcohols, and state the colour change when it works.
3. Write an equation, using [O], for the oxidation of octan-1-ol to octanoic acid.
4. Explain why 3-ethylpentan-3-ol leaves acidified dichromate(VI) orange.
5. Hexanal is wanted from hexan-1-ol. Give the key feature of the method and its reason.
6. What is seen when an aldehyde is warmed with Tollens’ reagent?
7. What is seen when cyclohexanone is warmed with Fehling’s solution?
8. Give a catalyst and the condition for eliminating water from an alcohol.
9. Name every alkene formed when 4-methylpentan-2-ol is heated with concentrated phosphoric acid.
10. Calculate the maximum mass of hexanoic acid (Mr 116.0) from 5.10 g of hexan-1-ol (Mr 102.0).
11. 4.08 g of hexan-1-ol (Mr 102.0) gives 2.40 g of hexanal (Mr 100.0). Calculate the percentage yield.
12. In step 1 of the elimination mechanism, where does the curly arrow start and finish?

Answers

1. Secondary: the C–OH carbon (C2) is bonded to C1 and C3.
2. Acidified potassium dichromate(VI); orange to green.
3. CH₃(CH₂)₆CH₂OH + 2[O] → CH₃(CH₂)₆COOH + H₂O
4. The C–OH carbon has no hydrogen atom, so a C=O cannot form; it is not easily oxidised.
5. Use the alcohol in excess and distil off the hexanal as it forms, because it boils lower than the alcohol (no hydrogen bonding between its molecules) and escapes before it is oxidised to the acid.
6. A silver mirror forms.
7. No change: the solution stays blue.
8. Concentrated sulfuric acid or concentrated phosphoric acid, with heat.
9. 4-methylpent-1-ene, (E)-4-methylpent-2-ene and (Z)-4-methylpent-2-ene.
10. n = 5.10 ÷ 102.0 = 0.0500 mol; mass = 0.0500 × 116.0 = 5.80 g.
11. n(alcohol) = 0.0400 mol; n(hexanal) = 0.0240 mol; yield = 0.0240 ÷ 0.0400 × 100 = 60.0 %.
12. It starts at an oxygen lone pair (on –OH) and ends at H⁺.

Where marks are usually lost

• Classifying by counting every branch in the molecule instead of only the carbons bonded to the C–OH carbon.
• Writing “the dichromate turns green” for a tertiary alcohol.
• Giving reflux as the method for an aldehyde, or “distil” without “as it forms”.
• Explaining the aldehyde’s lower boiling point by “weaker covalent bonds” rather than no hydrogen bonding between its molecules.
• Writing –COH for an aldehyde group, or leaving H₂O out of alcohol → aldehyde equations.
• Stating “no reaction” for a ketone instead of “stays blue” or “no silver mirror”.
• Calling the Fehling’s precipitate brown, orange or black.
• Starting a curly arrow at H⁺ or at a positive charge.
• Not showing H⁺ released at the end, so the acid is not shown to be a catalyst.
• Missing E/Z isomers, or a second positional isomer, among elimination products.

Official syllabus

OxfordAQA, International AS and A-level Chemistry (9620) specification, Version 5.3, for International AS and A-level exams May/June 2020 onwards: section 3.3.5 Alcohols (International AS), 3.3.5.1 Oxidation of alcohols and 3.3.5.2 Elimination, with Required practical 4.