OCR A Level Chemistry A: Alcohols (H432)

This study guide teaches section 4.2.1, Alcohols, of the OCR A Level Chemistry A (H432) specification, Version 3.1 (May 2026), for first assessment in 2017. It sits in Module 4, Core organic chemistry, and covers every lettered outcome from 4.2.1(a)(i) to 4.2.1(e), with the additional guidance and the PAG7 link. H432 is the full A level; the specification assesses Module 4 in components 02 (Synthesis and analytical techniques) and 03 (Unified chemistry).

Other H432 topics are on the OCR chemistry hub; record your progress with the printable checklist and test yourself with the free diagnostics. After this page, use the revision notes and the practice questions.

Outcomes in 4.2.1

Outcome	You must be able to	Additional guidance
4.2.1(a)(i)	Explain polarity, water solubility and low volatility of alcohols using hydrogen bonding	See also 2.2.2(l) and 4.1.2(c)
4.2.1(a)(ii)	Classify alcohols as primary, secondary or tertiary	None
4.2.1(b)	Write equations for the combustion of alcohols	None
4.2.1(c)(i)	Oxidise primary alcohols to aldehydes or carboxylic acids, and control which forms	Use [O] in equations; PAG7 (see also 6.3.1 c)
4.2.1(c)(ii)	Oxidise secondary alcohols to ketones	Use [O] in equations
4.2.1(c)(iii)	Explain why tertiary alcohols resist oxidation	None
4.2.1(d)	Eliminate H₂O with an acid catalyst and heat to form alkenes	Mechanism not required
4.2.1(e)	Substitute with halide ions in acid to form haloalkanes	Mechanism not required

Hydrogen bonding itself is taught in the bonding and structure guide. Naming rules and skeletal formulae are in the basic concepts of organic chemistry guide.

Three isomers of C₆H₁₃OH run through this page:

• A: 4-methylpentan-1-ol, (CH₃)₂CHCH₂CH₂CH₂OH
• B: hexan-3-ol, CH₃CH₂CH(OH)CH₂CH₂CH₃
• C: 2,3-dimethylbutan-2-ol, (CH₃)₂C(OH)CH(CH₃)₂

4.2.1(a)(i): Polarity, solubility and volatility

Oxygen is more electronegative than both carbon and hydrogen. So the O–H bond is polar (O δ−, H δ+), and the C–O bond is polar too. The δ+ hydrogen of one –OH group attracts a lone pair on a neighbouring molecule’s oxygen. That attraction is a hydrogen bond.

Alcohol molecules therefore have two kinds of intermolecular force:

• hydrogen bonds between –OH groups
• London forces (induced dipole–dipole interactions) along the hydrocarbon chain, as in alkanes

Why small alcohols dissolve in water

An alcohol’s –OH group can hydrogen-bond to water molecules, so the alcohol can slot into water’s hydrogen-bonded network. Methanol, ethanol and propan-1-ol are fully miscible with water.

As the chain gets longer, solubility falls. The hydrocarbon part cannot hydrogen-bond, and a larger non-polar section breaks up more of the hydrogen bonding between water molecules than the single –OH group can replace. Isomer A, with six carbons, is only slightly soluble.

Why alcohols are less volatile than alkanes

Compare molecules of similar Mᵣ, because they have a similar number of electrons and so similar London forces. The difference in boiling point then comes from the hydrogen bonds.

Compound	Mᵣ	Boiling point
pentane, C₅H₁₂	72.0	36 °C
butan-1-ol, C₄H₉OH	74.0	118 °C

Butan-1-ol boils over 80 °C higher. Boiling it means breaking hydrogen bonds as well as London forces, and hydrogen bonds are the stronger of the two. More energy is needed, so the alcohol is less volatile.

Within the alcohol series, boiling point still rises with chain length, because London forces grow with the number of electrons. Branching lowers it, as it does for alkanes in 4.1.2(c): branched molecules have less surface contact.

4.2.1(a)(ii): Primary, secondary and tertiary

Find the carbon carrying the –OH group and count the carbon atoms attached to it.

Class	Carbon neighbours of the C–OH carbon	Group in the molecule
Primary	one (none for methanol)	–CH₂OH
Secondary	two	>CH(OH)
Tertiary	three	>C(OH)–, carrying no H

Worked example 1: classify A, B and C

• A, (CH₃)₂CHCH₂CH₂CH₂OH: the OH carbon is a CH₂ joined to one other carbon. Primary.
• B, CH₃CH₂CH(OH)CH₂CH₂CH₃: the OH carbon is C3, joined to C2 and C4. Secondary.
• C, (CH₃)₂C(OH)CH(CH₃)₂: the OH carbon is C2, joined to two methyl carbons and C3. Tertiary.

A ring alcohol such as cyclobutanol is secondary, because its C–OH carbon is bonded to two ring carbons. A useful check: primary has two H atoms on the C–OH carbon, secondary has one and tertiary has none. That H count matters again for oxidation.

4.2.1(b): Combustion

Alcohols burn in plenty of oxygen to give carbon dioxide and water. For any alcohol in the series CₙH₂ₙ₊₁OH:

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

Remember the oxygen atom already in the alcohol when you balance. All three isomers have the same molecular formula, so they share one equation:

C₆H₁₃OH + 9O₂ → 6CO₂ + 7H₂O

With a limited oxygen supply, combustion is incomplete and carbon monoxide or soot can form instead of CO₂.

Worked example 2: products of burning B

2.55 g of hexan-3-ol burns completely. Find the mass of CO₂ and of water formed. (Mᵣ: C₆H₁₃OH 102.0, CO₂ 44.0, H₂O 18.0)

1. n(C₆H₁₃OH) = 2.55 ÷ 102.0 = 0.0250 mol
2. From the equation, 1 mol alcohol gives 6 mol CO₂ and 7 mol H₂O.
3. n(CO₂) = 6 × 0.0250 = 0.150 mol, so mass = 0.150 × 44.0 = 6.60 g
4. n(H₂O) = 7 × 0.0250 = 0.175 mol, so mass = 0.175 × 18.0 = 3.15 g

4.2.1(c): Oxidation with acidified dichromate

The usual oxidising agent is acidified dichromate, Cr₂O₇²⁻/H⁺, made from potassium dichromate and sulfuric acid (K₂Cr₂O₇/H₂SO₄). As the alcohol is oxidised, the orange Cr₂O₇²⁻ is reduced to green Cr³⁺. In equations, the specification asks you to write [O] for the oxidising agent and to balance with it.

4.2.1(c)(i): primary alcohols

A primary alcohol is oxidised first to an aldehyde and then to a carboxylic acid. For isomer A:

(CH₃)₂CHCH₂CH₂CH₂OH + [O] → (CH₃)₂CHCH₂CH₂CHO + H₂O (4-methylpentanal)

(CH₃)₂CHCH₂CH₂CH₂OH + 2[O] → (CH₃)₂CHCH₂CH₂COOH + H₂O (4-methylpentanoic acid)

You control which product forms through the conditions:

Target	Conditions	Why it works
Aldehyde	Gentle heating, alcohol in excess (limited oxidising agent); distil the product off as it forms	Aldehyde molecules cannot hydrogen-bond to each other, so the aldehyde boils lower than the alcohol and escapes before it can be oxidised further
Carboxylic acid	Excess acidified dichromate; heat under reflux	The vertical condenser returns any aldehyde vapour to the flask, where it remains with the oxidising agent and is converted to the acid

The boiling points for the two-carbon series show the idea: ethanal 20 °C, ethanol 78 °C, ethanoic acid 118 °C.

4.2.1(c)(ii): secondary alcohols

A secondary alcohol is oxidised to a ketone, normally by heating under reflux with acidified dichromate. A ketone is not oxidised further by this reagent. For isomer B:

CH₃CH₂CH(OH)CH₂CH₂CH₃ + [O] → CH₃CH₂COCH₂CH₂CH₃ + H₂O (hexan-3-one)

4.2.1(c)(iii): tertiary alcohols

In oxidation, the H on the –OH group and an H on the C–OH carbon are removed (they leave in the H₂O). A tertiary alcohol has no H on that carbon, so it resists oxidation. Warmed with acidified dichromate, isomer C leaves the mixture orange.

The test-tube test (PAG7)

The specification links this outcome to PAG7 and to 6.3.1(c)(vi), where acidified dichromate is used to identify primary and secondary alcohols and aldehydes. Add a few drops of acidified potassium dichromate to a little of the unknown and warm it in a water bath.

Observation	Conclusion
Orange to green	Primary or secondary alcohol, or an aldehyde
Stays orange	Tertiary alcohol (or a compound such as a ketone)

Worked example 3: yield of a ketone

6.63 g of hexan-3-ol is heated under reflux with excess acidified dichromate. After purification, 4.42 g of hexan-3-one is collected. Calculate the percentage yield. (Mᵣ: C₆H₁₄O 102.0, C₆H₁₂O 100.0)

1. n(hexan-3-ol) = 6.63 ÷ 102.0 = 0.0650 mol
2. The ratio is 1 : 1, so the maximum n(hexan-3-one) = 0.0650 mol
3. Theoretical mass = 0.0650 × 100.0 = 6.50 g
4. Percentage yield = 4.42 ÷ 6.50 × 100 = 68.0%

4.2.1(d): Elimination of water to form alkenes

Heating an alcohol with an acid catalyst, such as concentrated H₃PO₄ or concentrated H₂SO₄, removes a molecule of water. The OH comes off one carbon and an H comes off a neighbouring carbon, leaving a C=C double bond. This is an elimination (dehydration) reaction. The mechanism is not required.

Worked example 4: alkenes from C and B

Isomer C, (CH₃)₂C(OH)CH(CH₃)₂. The C–OH carbon has two kinds of neighbour: the CH₃ groups and the CH carbon.

• Lose H from a CH₃ group: CH₂=C(CH₃)CH(CH₃)₂, 2,3-dimethylbut-1-ene
• Lose H from the CH carbon: (CH₃)₂C=C(CH₃)₂, 2,3-dimethylbut-2-ene

(CH₃)₂C(OH)CH(CH₃)₂ → C₆H₁₂ + H₂O

Neither alkene shows E/Z isomerism, because in each a carbon of the C=C carries two identical groups.

Isomer B, hexan-3-ol. Removing H from C2 gives hex-2-ene; removing H from C4 gives hex-3-ene. Both can exist as E and Z isomers (see 4.1.3 for E/Z), so up to four alkenes can form.

4.2.1(e): Substitution to form haloalkanes

An alcohol reacts with halide ions in the presence of acid to give a haloalkane. The –OH group is replaced by a halogen atom. The mechanism is not required.

For a bromoalkane, heat the alcohol with sodium bromide and sulfuric acid. The acid and the bromide produce hydrogen bromide in the flask:

NaBr + H₂SO₄ → NaHSO₄ + HBr

The HBr then substitutes the alcohol. For isomer A:

(CH₃)₂CHCH₂CH₂CH₂OH + HBr → (CH₃)₂CHCH₂CH₂CH₂Br + H₂O (1-bromo-4-methylpentane)

The haloalkane keeps the carbon skeleton and the position of the functional group, so isomer C gives 2-bromo-2,3-dimethylbutane. The reactions of haloalkanes follow in 4.2.2.

Common errors

• Comparing an alcohol’s boiling point with an alkane of very different Mᵣ, so the London forces are not matched.
• Saying “the O–H bond breaks” on boiling. Boiling breaks hydrogen bonds between molecules, not covalent bonds.
• Counting hydrogens instead of carbons when classifying. Classification counts carbons bonded to the C–OH carbon.
• Forgetting the oxygen in the alcohol when balancing a combustion equation.
• Writing “reflux” for the aldehyde. Reflux sends the aldehyde back for further oxidation; distil it off instead.
• Writing 2[O] for a secondary alcohol, or claiming a ketone goes on to an acid.
• Forgetting a second alkene when the C–OH carbon has two different neighbours.

Next steps

Next: the revision notes, then the practice questions. For mole and yield methods, see the amount of substance guide.

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.