OCR A Level Chemistry A: Alcohols (H432) – Revision Notes

These revision notes condense section 4.2.1, Alcohols, of the OCR A Level Chemistry A (H432) specification, Version 3.1 (May 2026), for first assessment in 2017. All of 4.2.1(a)(i) to (e) is here, from Module 4, Core organic chemistry, with the PAG7 link. H432 is the full A level. The study guide has the full explanations and worked examples.

Afterwards, try the practice questions. The OCR chemistry hub lists every H432 topic, the printable checklist tracks your progress, and the free diagnostics show where your gaps are.

Key terms

Term	Meaning
Alcohol	Compound containing the –OH (hydroxyl) functional group on a saturated carbon; homologous series CₙH₂ₙ₊₁OH
Hydrogen bond	Attraction between a δ+ H on O–H and a lone pair on O of another molecule
Primary alcohol	C–OH carbon bonded to one other carbon (methanol: none)
Secondary alcohol	C–OH carbon bonded to two other carbons
Tertiary alcohol	C–OH carbon bonded to three other carbons
[O]	Symbol for oxygen from the oxidising agent, used to balance oxidation equations
Elimination	Removal of a small molecule (here H₂O) from a larger one, forming a C=C bond
Substitution	One atom or group (here –OH) replaced by another (here a halogen)

4.2.1(a)(i): Physical properties in one table

Property	Explanation
Polarity	O is more electronegative than C and H, so O–H and C–O bonds are polar
Higher boiling point than an alkane of similar Mᵣ	Similar London forces, plus hydrogen bonds between –OH groups; more energy to separate molecules
Low volatility	Same reason: hydrogen bonds hold molecules in the liquid
Short-chain alcohols soluble in water	–OH forms hydrogen bonds with water molecules
Solubility falls with chain length	Larger non-polar chain cannot hydrogen-bond and disrupts hydrogen bonding in water
Boiling point rises with chain length	More electrons, stronger London forces

Link to 2.2.2(l) for hydrogen bonding and 4.1.2(c) for London forces in alkanes.

Method: comparing boiling points

1. Pick compounds of similar Mᵣ, so London forces are about equal.
2. Name the extra force in the alcohol: hydrogen bonding.
3. Say hydrogen bonds are stronger than London forces.
4. Conclude: more energy needed to overcome the intermolecular forces, so a higher boiling point.

4.2.1(a)(ii): Classification shortcut

Count the hydrogens on the C–OH carbon as a check:

Class	H atoms on C–OH carbon	Example
Primary	2 (3 in methanol)	heptan-1-ol, CH₃(CH₂)₆OH
Secondary	1	2-methylpentan-3-ol
Tertiary	0	2-methylheptan-2-ol

4.2.1(b): Combustion

General equation: CₙH₂ₙ₊₁OH + (3n/2) O₂ → n CO₂ + (n + 1) H₂O

Alcohol	Balanced equation
methanol	2CH₃OH + 3O₂ → 2CO₂ + 4H₂O
propan-1-ol	2C₃H₇OH + 9O₂ → 6CO₂ + 8H₂O
butan-2-ol	C₄H₉OH + 6O₂ → 4CO₂ + 5H₂O

A half such as 4.5O₂ still balances the equation; double everything when whole numbers are asked for. Incomplete combustion gives CO or C.

4.2.1(c): Oxidation with Cr₂O₇²⁻/H⁺

Reagent: potassium dichromate with sulfuric acid (K₂Cr₂O₇/H₂SO₄). Colour change when oxidation happens: orange to green.

Alcohol	Conditions	Product	[O] equation (general)
Primary	Gentle heat; distil the aldehyde out as it is made	Aldehyde	RCH₂OH + [O] → RCHO + H₂O
Primary	Excess oxidising agent, heat under reflux	Carboxylic acid	RCH₂OH + 2[O] → RCOOH + H₂O
Secondary	Heat under reflux	Ketone	R₂CHOH + [O] → R₂CO + H₂O
Tertiary	Heat	None; the colour remains orange	No equation

Must-know distinctions

• Distil vs reflux. Distillation removes the aldehyde from the oxidising agent; reflux keeps it in the flask until it becomes the acid.
• Aldehyde vs alcohol boiling point. Aldehydes have no O–H, so no hydrogen bonds between their molecules. They boil lower and distil out first.
• Ketone vs acid. A secondary alcohol stops at the ketone. Only primary alcohols reach a carboxylic acid.
• Why tertiary resists. No H atom on the C–OH carbon to remove.

PAG7 test

Warm the sample with a few drops of acidified potassium dichromate in a water bath. Green means a primary or secondary alcohol or an aldehyde (6.3.1(c)(vi)); orange remaining fits a tertiary alcohol.

Small worked reminder

Butan-1-ol, CH₃CH₂CH₂CH₂OH:

• distilled as formed: CH₃CH₂CH₂CH₂OH + [O] → CH₃CH₂CH₂CHO + H₂O (butanal)
• reflux, excess oxidant: CH₃CH₂CH₂CH₂OH + 2[O] → CH₃CH₂CH₂COOH + H₂O (butanoic acid)

Check: count H atoms. Alcohol 10 H; aldehyde 8 H + water 2 H. Balanced.

4.2.1(d): Dehydration to alkenes

• Reagent: concentrated H₃PO₄ or concentrated H₂SO₄ as acid catalyst; heat.
• General: CₙH₂ₙ₊₁OH → CₙH₂ₙ + H₂O
• Mechanism not required.

Method: predict the alkene(s)

1. Find the C–OH carbon.
2. Note every adjacent carbon bearing one or more H atoms.
3. Make a C=C to each different neighbour in turn; each gives a product.
4. Check each product for E/Z isomerism (4.1.3).

Example: 2-methylheptan-2-ol, (CH₃)₂C(OH)CH₂CH₂CH₂CH₂CH₃, gives 2-methylhept-1-ene and 2-methylhept-2-ene. The second has two CH₃ groups on one C=C carbon, so it has no E/Z isomers.

4.2.1(e): Substitution to haloalkanes

• Reagents: halide ions in acid, for example NaBr with H₂SO₄, heated.
• In-situ step: NaBr + H₂SO₄ → NaHSO₄ + HBr
• Substitution: ROH + HBr → RBr + H₂O
• Mechanism not required.

Example: heptan-1-ol gives 1-bromoheptane, CH₃(CH₂)₆Br.

Reaction map

From an alcohol	Reagent	Product type
Any alcohol	O₂, burn	CO₂ + H₂O
Primary	Cr₂O₇²⁻/H⁺, distil	Aldehyde
Primary	excess Cr₂O₇²⁻/H⁺, reflux	Carboxylic acid
Secondary	Cr₂O₇²⁻/H⁺, reflux	Ketone
Any with H on a neighbouring carbon	H₃PO₄ or H₂SO₄, heat	Alkene
Any	NaBr/H₂SO₄, heat	Bromoalkane

Quick self-test

1. Classify each alcohol: (a) 2-methylhexan-3-ol, (b) 3-ethylhexan-3-ol, (c) 2-ethylbutan-1-ol.
2. Write a balanced equation, with whole numbers, for the complete combustion of heptan-1-ol.
3. Calculate the amount of O₂, in mol, for complete combustion of 0.200 mol of propan-1-ol.
4. Explain why ethanol mixes with water but hexane does not.
5. Name the product when 2-methylpropan-1-ol is warmed with acidified dichromate and the product is distilled off as it forms.
6. Write the [O] equation for oxidising 2-methylpropan-1-ol fully to the carboxylic acid.
7. 2-methylbutan-2-ol is warmed with acidified potassium dichromate. What is seen, and why?
8. Name every alkene that forms when pentan-3-ol is heated with concentrated H₃PO₄.
9. Give the reagents and an equation for making 2-bromobutane from butan-2-ol.
10. 3.52 g of pentan-2-ol (Mᵣ 88.0) is oxidised and 2.75 g of pentan-2-one (Mᵣ 86.0) is collected. Calculate the percentage yield.

Answers

1. (a) Secondary: the C–OH carbon (C3) is bonded to C2 and C4. (b) Tertiary: C3 is bonded to C2, C4 and the ethyl carbon. (c) Primary: the OH is on a –CH₂OH at the end of the chain.
2. 2C₇H₁₅OH + 21O₂ → 14CO₂ + 16H₂O
3. propan-1-ol needs 4.5 mol O₂ per mole, so 0.200 × 4.5 = 0.900 mol.
4. Ethanol’s –OH forms hydrogen bonds with water molecules. Hexane is non-polar and cannot hydrogen-bond, so it cannot replace the hydrogen bonds between water molecules.
5. 2-methylpropanal, (CH₃)₂CHCHO.
6. (CH₃)₂CHCH₂OH + 2[O] → (CH₃)₂CHCOOH + H₂O
7. The mixture stays orange. It is a tertiary alcohol with no H on the C–OH carbon, so it is not oxidised.
8. Only pent-2-ene, as (E)- and (Z)-pent-2-ene. Both neighbours of C3 are equivalent, so removing H from C2 or C4 gives the same structure.
9. NaBr and H₂SO₄, heat. CH₃CH(OH)CH₂CH₃ + HBr → CH₃CHBrCH₂CH₃ + H₂O
10. n(pentan-2-ol) = 3.52 ÷ 88.0 = 0.0400 mol; theoretical mass = 0.0400 × 86.0 = 3.44 g; yield = 2.75 ÷ 3.44 × 100 = 79.9%.

Where marks are usually lost

• Comparing boiling points without matching Mᵣ, or naming “stronger London forces” as the reason an alcohol boils higher.
• Writing that covalent O–H bonds break when an alcohol boils.
• Drawing a hydrogen bond to the H of a C–H bond; only the H on O–H takes part.
• Classifying by the number of H atoms on the whole molecule, or by the position number in the name.
• Leaving out water as a product in [O] equations, or putting 2[O] in the ketone equation.
• Writing “reflux” for making the aldehyde.
• Saying tertiary alcohols “do not react” without the observation: the solution stays orange.
• Naming only one alkene when the C–OH carbon has two different neighbours carrying H.
• Writing “Br₂” or “bromine water” as the reagent for making a bromoalkane from an alcohol.

Official syllabus

OCR A Level Chemistry A (H432) specification, Version 3.1 (May 2026), for first assessment in 2017, section 4.2.1 Alcohols, learning outcomes (a) to (e). Published by OCR.