Alcohols — Classification and Properties

Spec Mapping — OCR H432 Module 4.2.1 — Alcohols, covering classification of alcohols as primary, secondary or tertiary; the influence of hydrogen bonding on physical properties (volatility, boiling point, water solubility, viscosity); the basis for the differential reactivity of the three classes towards oxidation; and the preparatory framework for the alcohol chemistry developed across Lessons 2–3 (refer to the official OCR H432 specification document for exact wording).

Alcohols are one of the most strategically important homologous series in organic chemistry. They sit at the centre of the synthetic web that joins alkenes, haloalkanes, carbonyls and carboxylic acids — you can oxidise an alcohol up to an aldehyde, ketone or carboxylic acid; dehydrate it down to an alkene; substitute the hydroxyl for a halide; or condense it with a carboxylic acid to make an ester. Before you can use any of those reactions confidently you must (i) recognise the hydroxyl group as the functional group, (ii) classify the alcohol as primary, secondary or tertiary by counting alkyl groups on the C-OH carbon, and (iii) understand how hydrogen bonding drives the boiling-point, solubility and viscosity trends that distinguish alcohols from alkanes of similar mass. This lesson develops all three skills, with worked examples and the synoptic links you will need when you meet oxidation (Lesson 2), dehydration and substitution (Lesson 3), nucleophilic substitution (Lessons 4–6), and the analytical fingerprint of alcohols in IR and mass spectroscopy (Lessons 9–10).

Key Definition: an alcohol is an organic compound containing the hydroxyl functional group (-OH) bonded directly to a saturated (sp³) carbon atom. The general formula for a saturated aliphatic alcohol is $C_nH_{2n+1}OH$ , equivalent to $C_nH_{2n+2}O$ .

The hydroxyl functional group

The defining feature of an alcohol is the -OH group attached to an sp³ carbon. The oxygen carries two lone pairs and is more electronegative than either carbon or hydrogen, so the C-O and O-H bonds are both polar:

$\overset{\delta+}{\text{C}} - \overset{\delta-}{\text{O}} - \overset{\delta+}{\text{H}}$

The combined dipoles make the oxygen a strong hydrogen-bond acceptor (lone pairs available) and the hydroxyl hydrogen a strong hydrogen-bond donor (highly deshielded). Every property we discuss below — boiling point, water solubility, viscosity, reactivity towards oxidation — flows directly from this polarised C-O-H motif.

Note that phenol (C₆H₅OH) and the enols (HO-C=C) are NOT classified as alcohols in OCR A-Level terms because the -OH is bonded to an sp² (aromatic or alkenyl) carbon. Their acidity and reactivity are markedly different and are treated separately in Module 6 (aromatic chemistry).

Classifying alcohols: primary, secondary, tertiary

Alcohols are classified by counting the number of alkyl (or aryl) groups bonded to the carbon carrying the -OH:

Class	Alkyl groups on C-OH carbon	General formula	Carbon environment
Primary (1°)	0 or 1	R-CH₂-OH	-CH₂- with one H less than methanol
Secondary (2°)	2	R₂CH-OH	-CHR-
Tertiary (3°)	3	R₃C-OH	-CR₃, no H on C-OH

Primary (1°) H | R-C-OH | H e.g. ethanol

Secondary (2°) R' | R-C-OH | H e.g. propan-2-ol

Tertiary (3°) R' | R-C-OH | R'' e.g. 2-methylpropan-2-ol

flowchart TD
  A[Identify the C-OH carbon] --> B{How many other C atoms bonded to it?}
  B -->|0 or 1| C[Primary - 1 degree<br/>R-CH2-OH<br/>methanol, ethanol, butan-1-ol]
  B -->|2| D[Secondary - 2 degrees<br/>R2CH-OH<br/>propan-2-ol, cyclohexanol]
  B -->|3| E[Tertiary - 3 degrees<br/>R3C-OH<br/>2-methylpropan-2-ol]
  C --> F[Oxidises to aldehyde then carboxylic acid]
  D --> G[Oxidises to ketone]
  E --> H[Resists oxidation - colour stays orange]

Worked example 1 — classifying isomers of butanol

The molecular formula C₄H₁₀O has four structural isomers that are alcohols (plus three ethers). Classify each:

Name	Structure	C count on C-OH	Class
Butan-1-ol	CH₃CH₂CH₂CH₂OH	1	Primary
Butan-2-ol	CH₃CH₂CH(OH)CH₃	2	Secondary
2-Methylpropan-1-ol	(CH₃)₂CHCH₂OH	1	Primary
2-Methylpropan-2-ol	(CH₃)₃COH	3	Tertiary

Notice that 2-methylpropan-1-ol is still primary even though it has a branched chain — the branching is one carbon away from the -OH and so plays no role in classification. The branch matters for naming (IUPAC requires the longest chain to include the -OH), not for class.

Worked example 2 — methanol

Methanol, CH₃OH, has zero alkyl groups on the C-OH carbon. By the strict rule it should be a class of its own, but by convention OCR mark schemes accept methanol as primary. Some textbooks (and the IUPAC Gold Book) refer to it as the zero-degree alcohol; you should be able to write "primary (methanol is the limiting case)" if pressed.

Why classification matters

Classification controls how the alcohol reacts with the dichromate(VI) oxidising agent introduced in Lesson 2:

Primary alcohols oxidise stepwise — first to aldehyde (if you distil) and then to carboxylic acid (if you reflux).
Secondary alcohols oxidise to ketones only; over-oxidation requires breaking a C-C bond, which acidified dichromate cannot do.
Tertiary alcohols have no C-H bond on the C-OH carbon and so cannot lose the H that an oxidation requires — the solution stays orange.

Classification therefore underpins one of the most frequently examined practical experiments at OCR A-Level: distinguish three unknown alcohols using K₂Cr₂O₇/H₂SO₄ and Tollens' or Fehling's test for the resulting carbonyl.

Physical properties — the role of hydrogen bonding

Three intermolecular forces act between alcohol molecules in the liquid phase:

London dispersion forces — present in all molecules; strength scales with the surface area and number of electrons. Long alkyl chains have stronger London forces than short chains.
Permanent dipole-dipole forces — present because C-O and O-H bonds are polar.
Hydrogen bonds — the dominant interaction. The O-H hydrogen of one molecule donates a hydrogen bond to a lone pair on the oxygen of a neighbouring molecule, building a transient hydrogen-bonded network in the liquid.

Key Definition: a hydrogen bond is an electrostatic attraction between a hydrogen atom covalently bonded to a small, highly electronegative atom (N, O or F) and a lone pair on a second N, O or F atom on a neighbouring molecule. Hydrogen bonds are 5-25 kJ mol⁻¹ — about 5-10× weaker than a covalent bond but 5-10× stronger than typical London forces.

Boiling points

Compare molecules of similar relative molecular mass:

Molecule	Mᵣ	b.p. (°C)	Strongest IMF
Ethane, C₂H₆	30	-89	London
Methanal, HCHO	30	-19	Permanent dipole
Methanol, CH₃OH	32	+65	Hydrogen bonding
Propane, C₃H₈	44	-42	London
Ethanal, CH₃CHO	44	+20	Permanent dipole
Ethanol, C₂H₅OH	46	+78	Hydrogen bonding
Butane	58	-1	London
Propan-1-ol	60	+97	Hydrogen bonding

Ethanol boils 120 °C higher than propane despite having essentially the same mass. The difference is the energy required to break the hydrogen-bond network when the molecules escape into the gas phase — alkane → alcohol replaces a weak London-only network with a strong hydrogen-bonded one.

Trends along the homologous series

Alcohol	Mᵣ	b.p. (°C)
Methanol	32	65
Ethanol	46	78
Propan-1-ol	60	97
Butan-1-ol	74	118
Pentan-1-ol	88	138
Hexan-1-ol	102	158

Each extra -CH₂- adds about 20 °C — fairly typical for adding 14 g mol⁻¹ of carbon-and-hydrogen surface area to which London forces can attach. The hydrogen-bond contribution stays approximately constant (one -OH per molecule throughout).

Branching lowers the boiling point

For a fixed Mᵣ, branched alcohols boil at lower temperatures than their straight-chain isomers because compact, near-spherical molecules pack less efficiently and have lower surface area for London contact:

Isomer of C₄H₁₀O	b.p. (°C)
Butan-1-ol (straight, 1°)	118
Butan-2-ol (2°)	99
2-Methylpropan-1-ol (1°, branched)	108
2-Methylpropan-2-ol (3°)	82

Tertiary alcohols sit at the bottom of the ladder: not only are they more compact, but the -OH is sterically shielded by three alkyl groups, reducing the effectiveness of the hydrogen-bond donor.

Water solubility

Short-chain alcohols are miscible with water in all proportions because the -OH can hydrogen-bond directly to water. As the alkyl chain lengthens, the hydrophobic "tail" dominates and solubility collapses:

Alcohol	Solubility in water at 20 °C (g per 100 g H₂O)
Methanol	Miscible
Ethanol	Miscible
Propan-1-ol	Miscible
Butan-1-ol	7.4
Pentan-1-ol	2.2
Hexan-1-ol	0.6
Octan-1-ol	0.05

Diols and triols (ethane-1,2-diol, glycerol) extend the solubility further — every extra -OH compensates for a few -CH₂- groups in the chain.

Viscosity and density

Viscosity increases with chain length and with the number of -OH groups. Glycerol (propane-1,2,3-triol), with three -OH groups, is famously viscous because each molecule can participate in many hydrogen bonds simultaneously.
Density of liquid alcohols is below that of water (ethanol 0.789 g cm⁻³, propan-1-ol 0.803 g cm⁻³). This is why a layer of butan-1-ol floats above the aqueous phase in a separating-funnel extraction.
Volatility decreases as chain length grows — the inverse of boiling point.

Worked example 3 — ordering boiling points

Place these molecules in order of increasing boiling point and justify: propane, propanal, propan-2-ol, propan-1-ol, propane-1,2,3-triol (glycerol).

Rank	Molecule	b.p. (°C)	Reasoning
1	Propane	-42	London only
2	Propanal	+48	London + permanent dipole, no H-bonding (no O-H)
3	Propan-2-ol	+82	H-bonded, but secondary -OH sterically shielded
4	Propan-1-ol	+97	H-bonded, primary -OH most accessible
5	Glycerol	+290	Three -OH groups, extensive 3D H-bond network

The 50 °C jump from propanal to propan-2-ol shows the strength of one hydrogen bond per molecule; the 190 °C jump from propan-1-ol to glycerol shows the cumulative effect of three.

Worked example 4 — identifying class from a structural formula

Classify (a) 2-methylbutan-2-ol, (b) 3-methylbutan-1-ol, (c) cyclohexanol, (d) 2-phenylethanol (C₆H₅CH₂CH₂OH).

(a) The C-OH carbon bears two methyls and an ethyl — three alkyls — tertiary. (b) The C-OH carbon bears one -CH₂CH(CH₃)₂ substituent — one alkyl — primary. (c) The C-OH carbon is part of the ring and bears two ring carbons — two alkyls — secondary. (d) The C-OH carbon bears one -CH₂C₆H₅ substituent — one substituent (the aromatic ring counts as one alkyl, attached via the sp³ carbon) — primary.

Worked example 5 — predicting water solubility

Which of pentan-1-ol, propane-1,2-diol, and hexan-3-ol is most soluble in water? Justify.

Propane-1,2-diol is the most soluble — miscible with water — because it carries two -OH groups on a three-carbon chain, so the hydrophilic-to-hydrophobic ratio is highest. Pentan-1-ol (one -OH on a five-carbon chain) and hexan-3-ol (one -OH on a six-carbon chain) are increasingly hydrophobic; pentan-1-ol is sparingly soluble (~2 g per 100 g) and hexan-3-ol is less soluble still (~0.5 g per 100 g).

Synoptic Links

Synoptic Links — Connects to:

ocr-alevel-chemistry-basic-organic / nomenclature-and-isomerism (IUPAC naming of branched alcohols; structural isomerism in C₄H₁₀O).

ocr-alevel-chemistry-basic-organic / functional-groups-and-homologous-series (alcohols as one of the named homologous series).

ocr-alevel-chemistry-acids-redox-bonding / electronegativity-and-polarity (C-O and O-H bond polarisation, the origin of the hydrogen bond).

ocr-alevel-chemistry-carbonyls-polymers-spectroscopy / carbonyls-aldehydes-ketones (the oxidation products of primary and secondary alcohols).

PAG anchor: PAG 5 — Synthesis of an organic liquid is foreshadowed here: many PAG 5 protocols use an alcohol as starting material (e.g. preparation of 1-bromobutane from butan-1-ol, lesson 4) or as the target (e.g. propan-2-ol from propanone by reduction, Year 13). PAG 7 — Qualitative analysis of organic functional groups uses the alcohol classification you have just learned to drive the dichromate/Tollens'/Fehling's flowchart (Lesson 2).

Specimen question modelled on the OCR H432 paper format

Question (9 marks): Compound X is a saturated alcohol with relative molecular mass 88. Its boiling point is 99 °C and it dissolves to the extent of 12.5 g per 100 g of water at 20 °C. When heated under reflux with acidified potassium dichromate(VI), the orange solution turns green and a sweet-smelling neutral product Y is isolated; no carboxylic acid is formed even on prolonged reflux. (a) Deduce the molecular formula of X. (b) Identify a possible structure for X, explaining how the data support your choice. (c) Predict the structure and class of product Y and comment on what the absence of acid formation tells you about the structure of X.

AO breakdown

Mark	AO	Awarded for
1	AO2	Deduce molecular formula C₅H₁₂O from Mᵣ = 88
2	AO1	State X must be a saturated alcohol with the general formula C_nH_(2n+2)O
3	AO2	Recognise that the colour change (orange → green) shows oxidation; tertiary alcohols are excluded
4	AO3	Note that no acid forms on prolonged reflux — so X is not primary
5	AO3	Conclude X is secondary
6	AO2	Propose a plausible secondary C₅H₁₂O structure (e.g. pentan-2-ol or pentan-3-ol)
7	AO2	Use solubility (12.5 g per 100 g) and b.p. (99 °C) consistency check
8	AO2	Identify Y as the corresponding ketone
9	AO3	Justify that secondary alcohols oxidise to ketones only — no further oxidation because the carbonyl C carries no C-H bond

AO split: AO1 = 1, AO2 = 5, AO3 = 3.

Mid-band response

(a) Mᵣ = 88; (12 × 5) + (1 × 12) + 16 = 88. So X is C₅H₁₂O.

(b) The orange-to-green colour change shows X is oxidised by acidified dichromate(VI), so X is not tertiary. The fact that no carboxylic acid forms even on extended reflux means X is also not primary (primary alcohols give acids on reflux). X must therefore be a secondary alcohol. A plausible structure is pentan-2-ol, CH₃CH(OH)CH₂CH₂CH₃, which has the right Mᵣ and is a secondary alcohol. The boiling point of 99 °C is consistent with a five-carbon alcohol (b.p. of pentan-2-ol is 119 °C, of pentan-1-ol 138 °C — the value here suggests perhaps a branched isomer).

(c) Y is the oxidation product of a secondary alcohol — a ketone, pentan-2-one (CH₃COCH₂CH₂CH₃). Ketones cannot be oxidised further by dichromate because the carbonyl carbon already carries no C-H bond, so the oxidising agent has nothing left to remove. This explains why no acid forms even after long reflux.

Examiner commentary: M1-M2 awarded for molecular formula and homologous-series link. M3-M5 awarded for the deduction chain (not tertiary → not primary → secondary). M6 awarded for a plausible structure (pentan-2-ol acceptable, pentan-3-ol equally good). M7 missing — the answer notes b.p. inconsistency but does not commit. M8-M9 awarded for the ketone identification with mechanism-level reasoning. Around 8/9 — misses the boiling-point cross-check.

Top-band response

(a) Mᵣ = 88 = (12 × n) + (2n + 2) + 16 with n = 5; molecular formula C₅H₁₂O, consistent with the general alcohol formula C_nH_(2n+2)O.

(b) Three deductions from the data narrow the structure:

(i) The colour change orange → green corresponds to reduction of Cr₂O₇²⁻ (orange, Cr(+6)) to Cr³⁺ (green, Cr(+3)), so X is oxidised. Tertiary alcohols are excluded — they have no α-hydrogen on the C-OH carbon and do not reduce dichromate.

(ii) No carboxylic acid forms on extended reflux. Primary alcohols give an aldehyde (which is normally distilled out to preserve it) and on reflux are further oxidised to carboxylic acid. The absence of acid even after extended reflux excludes a primary alcohol.

(iii) Combining (i) and (ii): X must be secondary.

The candidate structures of C₅H₁₂O are pentan-2-ol, pentan-3-ol, and 3-methylbutan-2-ol. All three give a ketone on oxidation, all three are secondary, and all three are consistent with the data; without further information (e.g. an NMR or IR fragment), the data are insufficient to distinguish them. The boiling point of 99 °C is most consistent with 3-methylbutan-2-ol (b.p. 112 °C, branched secondary) or pentan-3-ol (b.p. 116 °C); pentan-2-ol is slightly higher at 119 °C. The 12.5 g per 100 g solubility is consistent with the five-carbon secondary alcohol family.

(c) Y is the corresponding ketone — pentan-2-one, pentan-3-one or 3-methylbutan-2-one. Ketones are unreactive towards further dichromate oxidation because the carbonyl carbon carries no C-H bond; oxidation would require C-C bond cleavage, which dichromate cannot achieve. This is the mechanistic basis for the secondary-to-ketone stopping point.

Examiner commentary: Full 9/9. The discriminators: (i) explicit Cr(+6) → Cr(+3) tracking, (ii) systematic exclusion of tertiary and primary classes from the data, (iii) recognition that the data cannot uniquely fix the structure and an honest enumeration of candidates, (iv) explicit mechanism-level justification for the secondary-to-ketone stopping point. The phrase "C-C bond cleavage" demonstrates A-Level depth on the limit of dichromate oxidation.

Common errors and mark-loss patterns

Pedagogical observations — not fabricated statistics:

Many candidates classify methanol as "not an alcohol" or as zero-degree rather than accepting the conventional "primary" assignment. OCR mark schemes accept primary; do not lose a mark for over-precision.
A frequent pitfall is to count carbons in the whole molecule rather than only the carbons directly bonded to the C-OH carbon. Branches further away play no role in classification.
Many candidates say "alcohols are soluble in water because they are polar" — the mark scheme demands the specific reason: hydrogen bonding between the alcohol -OH and water.
A common error is to write that "the O-H bond is broken when an alcohol boils." Boiling breaks intermolecular hydrogen bonds, not the intramolecular covalent O-H bond.
Candidates often forget to compare with the alkane control when asked why alcohols have higher boiling points. The mark scheme wants both forces present (alcohol has hydrogen bonds AND London) versus alkane (London only).
A persistent error is the branched-isomer boiling-point trend. Branched alcohols boil lower than straight-chain isomers of equal Mᵣ because they pack less efficiently and have lower London-force surface area — not, as some candidates write, because branching weakens the hydrogen bond.
Candidates sometimes mistake viscosity for boiling point — they are correlated but distinct. Viscosity reflects flow resistance in the liquid; boiling point reflects the temperature at which the vapour pressure equals atmospheric.

Going further

The partition coefficient ( $\log P$ or octanol-water $\log K_{ow}$ ) is a quantitative measure of an alcohol's hydrophilicity: the ratio of its concentration in octan-1-ol (a hydrophobic reference) to its concentration in water at equilibrium. Methanol has $\log P = -0.77$ (more soluble in water than in octanol); decan-1-ol has $\log P = +4.6$ (effectively insoluble in water). Drug-discovery teams optimise $\log P$ to be in the "Lipinski window" (around 1-3) so that a candidate drug can both dissolve in the bloodstream and cross lipid membranes. Hydrogen-bonded clusters in liquid water and alcohols are studied by neutron diffraction and vibrational spectroscopy (the broad O-H stretch in the IR, lesson 9). The role of hydrogen bonding in protein folding and DNA base pairing (Watson and Crick, 1953) is the synoptic anchor to A-Level Biology and is worth raising in Oxbridge interviews. Recommended reading: Atkins' Physical Chemistry, chapter on intermolecular forces; Clayden, Greeves and Warren Organic Chemistry, alcohols chapter; the IUPAC nomenclature rules at iupac.org. Oxbridge interview prompt: "Why does the boiling point of HF (19 °C) drop below that of H₂O (100 °C), even though F is more electronegative than O?"

A-Level misconceptions

The errors that distinguish A from A*:

Methanol is primary by convention, not "not classifiable". Mark schemes accept either "primary" or "no alkyl substituent on the C-OH carbon".
Phenol is not an alcohol in OCR terms — the -OH is on an sp² aromatic carbon, the chemistry is acidic, and the IR fingerprint is different.
Hydrogen bonding is the dominant intermolecular force, not the only one. London forces and permanent dipoles also operate; the trends along the homologous series are largely London-driven.
Tertiary alcohols cannot be oxidised by acidified dichromate, but they can be oxidised by stronger reagents (chromic acid, KMnO₄ under forcing conditions, ozone). A-Level "cannot be oxidised" is shorthand for "not by acidified dichromate at A-Level conditions".
Branched alcohols have lower boiling points than straight-chain isomers, because of London-force packing, not because the hydrogen bond is weaker.
The hydrogen-bonded liquid network is dynamic, not static. Hydrogen bonds form, break and reform on a picosecond timescale; the time-averaged number is what we mean by "the network".
Solubility in water and miscibility are not the same thing. Two liquids are miscible (mix in all proportions); a solid dissolves to a maximum extent in a liquid.
The hydroxyl proton in -O-H is not the same as the methyl protons in -CH₃ in NMR. The O-H is broad, exchangeable, and shifts with concentration; the C-H signals are sharp and concentration-independent. (Lesson 10 picks this up in mass spec; NMR is treated in the carbonyls course.)

Summary

An alcohol is an sp³ carbon bearing a hydroxyl group, classified as primary, secondary or tertiary by counting the alkyl substituents on the C-OH carbon. Hydrogen bonding between the -OH group and a neighbouring oxygen lone pair drives the elevated boiling points (~150 °C above the corresponding alkane), the water miscibility of short-chain alcohols, the high viscosity of polyols like glycerol, and the analytical IR fingerprint that Lesson 9 will exploit. Classification controls oxidation chemistry (Lesson 2): primary alcohols oxidise stepwise to aldehyde and carboxylic acid, secondary to ketone only, and tertiary not at all under standard conditions. The next lesson develops the dichromate-oxidation framework — reagent, conditions, colour change, and the distillation-versus-reflux choice that selects aldehyde or carboxylic acid as the major product.

Reference: OCR A-Level Chemistry A (H432) Module 4.2.1 (a)-(c) (refer to the official OCR H432 specification document for exact wording).

Lesson 1: Alcohols — Classification and Properties