16. Alcohols
🎯 Learning outcomes
After completing this lesson, you should be able to:
- recall the reactions (reagents and conditions) by which alcohols can be produced, including addition of steam to alkenes, reaction of alkenes with cold dilute acidified KMnO₄, substitution of a halogenoalkane, reduction of carbonyl compounds and carboxylic acids, and hydrolysis of an ester
- describe the reactions of alcohols with oxygen, halogenating agents, sodium, oxidising agents, heated catalysts (dehydration) and carboxylic acids (esterification)
- classify alcohols as primary, secondary and tertiary, and state characteristic distinguishing reactions
- deduce the presence of a CH₃CH(OH)– group from a positive iodoform test
- explain the acidity of alcohols compared with water
1🏷️Classifying alcohols
Alcohols contain the –OH functional group. They’re classified by how many carbon groups are attached to the carbon bonded to the –OH group.
Primary (1°)
–OH carbon attached to one other carbon (or none)
Secondary (2°)
–OH carbon attached to two other carbons
Tertiary (3°)
–OH carbon attached to three other carbons
Alcohols can also have more than one –OH group: a diol has two (e.g. ethane-1,2-diol, used in antifreeze), and a triol has three (e.g. propane-1,2,3-triol, also called glycerol).
Primary, secondary and tertiary alcohols undergo oxidation very differently — this is one of the most important distinguishing tests in this topic (see Section 4).
2🧪Preparing alcohols
Alcohols can be made from a variety of starting materials, using six key reactions:
| Method | Reagents / conditions |
|---|---|
| Steam addition to an alkene | H₂O(g), H₃PO₄ catalyst |
| Alkene + cold, dilute, acidified KMnO₄ | forms a diol (two –OH groups added across the C=C) |
| Substitution of a halogenoalkane | NaOH(aq), heat (hydroxide ion as nucleophile) |
| Reduction of an aldehyde or ketone | NaBH₄, or LiAlH₄ in dry ether |
| Reduction of a carboxylic acid | LiAlH₄ in dry ether (more forcing conditions than for aldehydes/ketones) |
| Hydrolysis of an ester | dilute acid or dilute alkali, heat |
Reducing carbonyl compounds
Aldehydes and ketones can be reduced back to alcohols by three common methods:
- H₂ gas with a nickel catalyst — heterogeneous catalytic hydrogenation
- NaBH₄ (sodium tetrahydridoborate(III)) in alkaline methanol — a mild, convenient reducing agent
- LiAlH₄ (lithium tetrahydridoaluminate(III)) in dry ether — a stronger reducing agent, needed to reduce carboxylic acids
LiAlH₄ is a hazardous reagent: it reacts violently with any water, releasing heat and hydrogen gas that usually ignites. The dry ether solvent is also highly volatile and flammable — this combination means LiAlH₄ is handled with care and avoided where a milder reagent like NaBH₄ will do the job.
NaBH₄ and LiAlH₄ act as nucleophiles, attacking the δ+ carbon of a C=O bond. Because they need a polarised carbon to attack, they do not reduce C=C double bonds — unlike catalytic hydrogenation with H₂/Ni, which reduces C=C readily.
Hydrolysis of an ester
An ester can be split back into an alcohol and a carboxylic acid (or its salt) by two routes:
- Acid hydrolysis — dilute acid (e.g. dilute H₂SO₄) and heat under reflux; reversible, gives the carboxylic acid directly
- Base (alkaline) hydrolysis — dilute NaOH(aq) and heat under reflux; irreversible, gives the carboxylate salt (must be acidified afterwards to obtain the free acid)
3🔥Combustion, substitution & reaction with sodium
Combustion
Like other organic compounds, alcohols burn in a plentiful supply of oxygen to form carbon dioxide and water:
Substitution to halogenoalkanes
The –OH group of an alcohol can be swapped for a halogen atom using several different reagents:
- HX (a hydrogen halide), or KBr with H₂SO₄ or H₃PO₄ (generating HBr in situ)
- PCl₃ with heat
- PCl₅ (reacts readily even in the cold, giving a visible mist of steamy HCl fumes)
- SOCl₂ (sulfur dichloride oxide / thionyl chloride) — a particularly clean method, since both by-products (SO₂ and HCl) are gases that escape, leaving a pure halogenoalkane
Reaction with sodium
Alcohols react with sodium metal in a similar way to water, but less vigorously, releasing hydrogen gas and forming a sodium alkoxide:
4🌈Oxidation of alcohols
Alcohols are oxidised using acidified potassium dichromate(VI), K₂Cr₂O₇, or acidified potassium manganate(VII), KMnO₄, as the oxidising agent. Crucially, primary, secondary and tertiary alcohols behave differently:
| Alcohol type | Product (distil) | Product (reflux) |
|---|---|---|
| Primary | aldehyde | carboxylic acid |
| Secondary | ketone | ketone (no further oxidation) |
| Tertiary | no reaction — cannot be oxidised | |
Mild oxidation with acidified K₂Cr₂O₇ gives a distinctive colour change from orange to green when a primary or secondary alcohol is oxidised (as Cr₂O₇²⁻, orange, is reduced to Cr³⁺, green). A tertiary alcohol produces no colour change, since it cannot be oxidised.
Remember why tertiary alcohols resist oxidation: oxidation of an alcohol requires removing a hydrogen atom directly from the carbon bonded to –OH. A tertiary alcohol’s central carbon has no such hydrogen (it’s bonded to three other carbon groups instead), so there’s nothing for the oxidising agent to remove.
5💧Dehydration & esterification
Dehydration to an alkene
Heating an alcohol with a suitable catalyst removes a molecule of water, forming an alkene:
This is an elimination reaction — the reverse, conceptually, of the steam-addition reaction used to prepare alcohols from alkenes in Section 2.
Esterification
An alcohol reacts with a carboxylic acid, using a concentrated acid catalyst (typically conc. H₂SO₄ or conc. H₃PO₄), to form an ester and water:
Esterification is a reversible reaction — the position of equilibrium can be pushed towards the ester by removing water as it forms, or towards hydrolysis by adding excess water (see Section 2 for the reverse reaction).
6🟡The iodoform test
The iodoform (tri-iodomethane) test detects a CH₃CH(OH)– group in an alcohol, CH₃CH(OH)–R. Warming the alcohol with alkaline aqueous iodine, I₂(aq), produces a yellow precipitate of tri-iodomethane, CHI₃, along with the carboxylate ion RCO₂⁻.
Ethanol, CH₃CH₂OH, can be written as CH₃CH(OH)H — it has the required group (R = H) and gives a positive iodoform test. Propan-1-ol, CH₃CH₂CH₂OH, does not have this group (the –OH carbon only carries one CH₃-containing chain, not a methyl group directly on the –OH carbon itself) and gives a negative result.
The iodoform test isn’t limited to alcohols — methyl ketones (CH₃CO–R) give the same positive result, since I₂/NaOH first oxidises a CH₃CH(OH)– alcohol group to the corresponding CH₃CO– ketone group before the substitution and cleavage steps occur.
7⚖️Acidity of alcohols vs. water
Once dissolved in water, ethanol molecules dissociate only to a very small degree:
Compare this with the dissociation of water itself:
Even though water’s own dissociation is tiny, the concentration of H⁺(aq) from ethanol is even lower — so ethanol is a weaker acid than water.
The electron-donating alkyl (ethyl) group attached to the negatively charged oxygen in the ethoxide ion pushes electron density towards that oxygen (a positive inductive effect). This concentrates extra negative charge on the oxygen, making the ethoxide ion a stronger base — it accepts an H⁺ ion more readily than hydroxide, OH⁻, does.
The hydroxide ion, by contrast, has only a hydrogen atom attached to its negatively charged oxygen — no electron-donating alkyl group to intensify the charge. This makes OH⁻ a comparatively weaker base than the ethoxide ion, meaning water releases H⁺ ions slightly more readily than ethanol does.
A more stable, more effective base holds on to H⁺ more strongly. Since the ethoxide ion is a better base than hydroxide (due to the inductive effect), its conjugate acid — ethanol — must be the weaker acid of the two.
📌 Exam tips & quick summary
✅ Do remember
- 1°/2°/3° alcohols classified by how many carbons attach to the C–OH carbon
- NaBH₄/LiAlH₄ reduce C=O but not C=C (nucleophilic mechanism); LiAlH₄ is needed for the harder reduction of –COOH
- oxidation: primary → aldehyde (distil) → carboxylic acid (reflux); secondary → ketone only; tertiary → no reaction
- K₂Cr₂O₇ colour change orange → green confirms oxidation has occurred
- iodoform test: CH₃CH(OH)– group + I₂/NaOH → yellow CHI₃ precipitate
- ethanol is a weaker acid than water — the ethyl group’s inductive effect stabilises negative charge on the ethoxide ion, making it a stronger base than OH⁻
🚫 Do not confuse
- distillation (stops at aldehyde) vs. reflux (goes all the way to carboxylic acid) — the apparatus setup is part of the answer, not just the reagent
- tertiary alcohols resist oxidation because they lack an H on the –OH carbon — not because the reagent is “too weak”
- a stronger base (ethoxide) implies a weaker conjugate acid (ethanol) — don’t mix up the direction of this relationship
🔁 Quick recap
- Classification: 1° (RCH₂OH), 2° (R₂CHOH), 3° (R₃COH); also diols/triols
- Key reactions: combustion, halogenation, Na, oxidation, dehydration, esterification
- Oxidation ladder: 1° → aldehyde → acid; 2° → ketone; 3° → no reaction
- Iodoform: CH₃CH(OH)– + I₂/NaOH → yellow CHI₃
- Acidity: ethanol < water, due to the ethyl group’s positive inductive effect