Alcohols

1. Overview

Alcohols contain the hydroxyl group, -OH, bonded to a saturated carbon atom. They sit at the centre of AS organic chemistry: alkenes, halogenoalkanes, carbonyl compounds, carboxylic acids and esters can be turned into alcohols, and alcohols into each of these. This section covers those reactions, classifying alcohols, the tri-iodomethane test, and why alcohols are weaker acids than water.

Key Definitions

• Alcohol: an organic compound in which each hydroxyl group, -OH, is bonded to a saturated carbon atom; an alcohol with one -OH group, no ring and no C=C bond has the general formula CₙH₂ₙ₊₁OH.
• Primary alcohol: an alcohol in which the carbon atom bonded to the -OH group is bonded to one other carbon atom (or to none, in methanol); for example, ethanol, CH₃CH₂OH.
• Secondary alcohol: an alcohol in which the carbon atom bonded to the -OH group is bonded to two other carbon atoms; for example, propan-2-ol, CH₃CH(OH)CH₃.
• Tertiary alcohol: an alcohol in which the carbon atom bonded to the -OH group is bonded to three other carbon atoms; for example, 2-methylpropan-2-ol, (CH₃)₃COH.
• Heating under reflux: heating a reaction mixture in a flask fitted with a vertical condenser, so that vapours condense and run back into the flask; no volatile reactant or product escapes, so the mixture can be heated for a long time.
• Distillation of a product as it forms: heating a reaction mixture with a condenser leading away from the flask, so that a volatile product boils off as soon as it forms and is collected away from the reagents.
• Dehydration: a reaction that removes a molecule of water from a compound; an alcohol loses its -OH group and a hydrogen atom from an adjacent carbon atom, forming an alkene.
• Esterification: the reversible reaction of an alcohol with a carboxylic acid, catalysed by concentrated sulfuric acid, that forms an ester and water; it is a condensation reaction.
• Tri-iodomethane test: warming a compound with aqueous iodine and aqueous sodium hydroxide; a yellow precipitate of tri-iodomethane, CHI₃, shows a CH₃CH(OH)- group in an alcohol (or a CH₃CO- group in an aldehyde or ketone).
• Alkoxide ion: the anion RO⁻ formed when an alcohol loses the hydrogen ion from its -OH group, for example the ethoxide ion, CH₃CH₂O⁻, from ethanol and sodium.

Content

Classifying alcohols

Count the other carbon atoms bonded to the carbon that carries the -OH group.

Class	Carbons on the C-OH carbon	Example
primary	one (none in methanol)	propan-1-ol, CH₃CH₂CH₂OH
secondary	two	propan-2-ol, CH₃CH(OH)CH₃
tertiary	three	2-methylpropan-2-ol, (CH₃)₃COH

When a molecule has more than one -OH group, classify each group on its own:

• Propane-1,2-diol, HOCH₂CH(OH)CH₃: a primary -OH on C1 and a secondary -OH on C2.
• Propane-1,2,3-triol, HOCH₂CH(OH)CH₂OH: two primary -OH groups and one secondary.
• 2-Methylbutane-2,3-diol, (CH₃)₂C(OH)CH(OH)CH₃: a tertiary -OH on C2 and a secondary -OH on C3.

How alcohols are made

Equations in this topic use [O] for an oxygen atom from an oxidising agent and [H] for a hydrogen atom from a reducing agent (topic 13.2). State symbols are shown only where the state matters.

Start from	Reagents and conditions	Alcohol formed
alkene	steam, H₂O(g), with an H₃PO₄ catalyst, heat and pressure	ethene gives ethanol
alkene	cold, dilute, acidified KMnO₄	a diol
halogenoalkane	NaOH(aq), heat under reflux	same class as the halogenoalkane
aldehyde	NaBH₄ or LiAlH₄	primary
ketone	NaBH₄ or LiAlH₄	secondary
carboxylic acid	LiAlH₄ in dry ether (NaBH₄ is too weak)	primary
ester	dilute acid or dilute alkali, heat under reflux	the alcohol part of the ester

CH₂=CH₂(g) + H₂O(g) ⇌ CH₃CH₂OH(g)

CH₃CH=CH₂ + [O] + H₂O → CH₃CH(OH)CH₂OH

CH₃CH₂CH₂Br + NaOH → CH₃CH₂CH₂OH + NaBr

CH₃CH₂CHO + 2[H] → CH₃CH₂CH₂OH

CH₃COCH₃ + 2[H] → CH₃CH(OH)CH₃

CH₃CH₂COOH + 4[H] → CH₃CH₂CH₂OH + H₂O

CH₃COOCH₂CH₃ + NaOH → CH₃COONa + CH₃CH₂OH

Steam adds to propene as Markovnikov's rule predicts, giving mostly propan-2-ol (topic 14.2). Cold KMnO₄ loses its purple colour; hot, concentrated KMnO₄ would break the C=C bond instead. LiAlH₄ reacts violently with water, so it is used in dry ether. Alkaline hydrolysis of an ester goes to completion and gives the salt of the acid; acid hydrolysis is reversible (topic 18.2).

Combustion

Alcohols burn in plenty of oxygen to give carbon dioxide and water. The reaction is strongly exothermic, so ethanol is used as a fuel.

CH₃CH₂OH(l) + 3O₂(g) → 2CO₂(g) + 3H₂O(l)

For CₙH₂ₙ₊₁OH, complete combustion uses 3n/2 mol of O₂ per mole of alcohol: when balancing, count the oxygen atom already in the alcohol. In a limited supply of oxygen, carbon monoxide or soot forms too:

CH₃CH₂OH(l) + 2O₂(g) → 2CO(g) + 3H₂O(l)

Substitution: alcohol to halogenoalkane

The -OH group is replaced by a halogen atom.

Reagent	Conditions
HX(g), e.g. HBr	heat
KCl with concentrated H₂SO₄ or H₃PO₄	heat
PCl₃	heat
PCl₅	room temperature
SOCl₂	room temperature

CH₃CH(OH)CH₃ + HBr → CH₃CHBrCH₃ + H₂O

CH₃CH₂OH + KCl + H₂SO₄ → CH₃CH₂Cl + KHSO₄ + H₂O

3CH₃CH₂OH + PCl₃ → 3CH₃CH₂Cl + H₃PO₃

CH₃CH₂OH + PCl₅ → CH₃CH₂Cl + POCl₃ + HCl

CH₃CH₂OH + SOCl₂ → CH₃CH₂Cl + SO₂ + HCl

• KCl and the concentrated acid make HCl in the flask.
• PCl₅ gives steamy white fumes of HCl at room temperature, a test for -OH (carboxylic acids and water also give it).
• With SOCl₂ both other products are gases, so the halogenoalkane is easy to separate.
• Every -OH group reacts, so a diol needs two PCl₅ (or two SOCl₂) per molecule.

Reaction with sodium

Sodium replaces the hydrogen of the -OH group, giving hydrogen gas and a sodium alkoxide:

2CH₃CH₂OH(l) + 2Na(s) → 2CH₃CH₂O⁻Na⁺ + H₂(g)

You see bubbles of a colourless gas that gives a squeaky pop with a lighted splint. The sodium sinks (it is denser than ethanol), fizzes steadily and slowly disappears. Sodium ethoxide stays dissolved in the excess ethanol.

Two -OH groups release one H₂ molecule, so 1 mol of a diol gives 1 mol of H₂.

Why alcohols are weaker acids than water

An acid is a proton donor (topic 7.2). Water and alcohols both lose H⁺ from an O-H bond, but only to a tiny extent. For ethanol:

CH₃CH₂OH ⇌ CH₃CH₂O⁻ + H⁺

Ethanol is a weaker acid than water: sodium reacts vigorously with water (it floats, melts and fizzes rapidly) but only steadily with ethanol. Neither reacts with carbonates, unlike a carboxylic acid (topic 18.1).

The reason is the alkyl group. Alkyl groups donate electrons (a positive inductive effect, topic 14.2). The ethyl group pushes electron density towards the oxygen atom, so:

1. The O-H bond is less polarised, so H⁺ is released less readily.
2. The negative charge on the oxygen of CH₃CH₂O⁻ is more concentrated than in OH⁻. The ethoxide ion is less stable and attracts H⁺ more strongly, so the equilibrium for ethanol lies further to the left than for water.

So CH₃CH₂O⁻ is a stronger base than OH⁻. In water, ethoxide ions take H⁺ from water molecules and the solution becomes alkaline:

CH₃CH₂O⁻ + H₂O → CH₃CH₂OH + OH⁻

Oxidation with acidified dichromate(VI) or manganate(VII)

The oxidising agent is K₂Cr₂O₇ or KMnO₄ acidified with dilute H₂SO₄. It removes two hydrogen atoms, the H of the -OH group and an H from the carbon bonded to it, and a C=O bond forms.

Alcohol	Distil the product off	Heat under reflux
primary	aldehyde	carboxylic acid
secondary	ketone	ketone
tertiary	no reaction	no reaction

CH₃CH₂CH₂OH + [O] → CH₃CH₂CHO + H₂O

CH₃CH₂CH₂OH + 2[O] → CH₃CH₂COOH + H₂O

CH₃CH(OH)CH₃ + [O] → CH₃COCH₃ + H₂O

• Colour change. Dichromate(VI), Cr₂O₇²⁻, is orange and is reduced to green Cr³⁺. Manganate(VII), MnO₄⁻, is purple and is reduced to almost colourless Mn²⁺.
• Tertiary alcohols have no hydrogen on the carbon bonded to -OH, so no C=O can form without breaking a C-C bond; the orange (or purple) colour stays. Ketones resist further oxidation for the same kind of reason.
• Distil for the aldehyde. An aldehyde boils lower than its alcohol, because its molecules cannot hydrogen-bond to each other. Add the oxidising agent a little at a time to the warm alcohol (kept in excess): the aldehyde boils off as it forms, before it can be oxidised further.
• Reflux for the acid. With an excess of oxidising agent and a vertical condenser (cooling water in at the bottom, top open), any aldehyde that evaporates condenses and runs back into the flask until it is oxidised to the acid.

Telling primary, secondary and tertiary alcohols apart

Warm a few drops of the alcohol with acidified K₂Cr₂O₇:

Alcohol	Acidified K₂Cr₂O₇	Distillate with Tollens' reagent
primary	orange to green	silver mirror (aldehyde)
secondary	orange to green	no change (ketone)
tertiary	stays orange	no product to test

Dichromate alone picks out tertiary alcohols. To tell primary from secondary, oxidise a larger sample with distillation and test the distillate: Tollens' reagent or Fehling's reagent reacts with an aldehyde but not with a ketone (topic 17.1).

Worked example 1: oxidising a primary alcohol to an acid

Question: 5.28 g of 3-methylbutan-1-ol, (CH₃)₂CHCH₂CH₂OH, is heated under reflux with an excess of acidified potassium dichromate(VI). After purification, 4.71 g of 3-methylbutanoic acid, (CH₃)₂CHCH₂COOH, is collected. (a) Write an equation using [O]. (b) Calculate the percentage yield. (c) Calculate the minimum mass of K₂Cr₂O₇ needed to oxidise all the alcohol to the acid. Use $A_{\text{r}}$: H 1.0, C 12.0, O 16.0, K 39.1, Cr 52.0.

(a) Equation. A primary alcohol needs 2[O] to become a carboxylic acid:

(CH₃)₂CHCH₂CH₂OH + 2[O] → (CH₃)₂CHCH₂COOH + H₂O

(b) Percentage yield. The alcohol is C₅H₁₂O, $M_{\text{r}}$ = 5(12.0) + 12(1.0) + 16.0 = 88.0. $$n = \frac{5.28}{88.0} = 0.0600\ \text{mol}$$ One mole of alcohol gives one mole of acid. The acid is C₅H₁₀O₂, $M_{\text{r}}$ = 5(12.0) + 10(1.0) + 2(16.0) = 102.0, so the theoretical mass is 0.0600 × 102.0 = 6.12 g. $$\text{yield} = \frac{4.71}{6.12} \times 100 = 77.0\%$$

(c) Minimum K₂Cr₂O₇. Count electrons with oxidation numbers (H = +1, O = −2).

• Alcohol: the five carbon atoms add up to −10 in C₅H₁₂O and to −6 in C₅H₁₀O₂, so each alcohol molecule loses 4 electrons.
• Dichromate: each Cr goes from +6 in Cr₂O₇²⁻ to +3 in Cr³⁺, so each Cr₂O₇²⁻ ion gains 2 × 3 = 6 electrons.

Electrons balance when 3 mol of alcohol (losing 12 mol of electrons) react with 2 mol of Cr₂O₇²⁻ (gaining 12 mol). $$n(\text{Cr}_2\text{O}_7^{2-}) = 0.0600 \times \tfrac{2}{3} = 0.0400\ \text{mol}$$ $M_{\text{r}}$(K₂Cr₂O₇) = 2(39.1) + 2(52.0) + 7(16.0) = 294.2, so the mass is 0.0400 × 294.2 = 11.8 g.

Answer: (b) 77.0%; (c) 11.8 g (3 s.f.).

Dehydration to an alkene

Dehydration is an elimination reaction: the -OH group and an H from the next carbon leave, and a C=C bond forms.

CH₃CH₂OH → CH₂=CH₂ + H₂O

Conditions: pass the alcohol vapour over a heated Al₂O₃ catalyst, or heat the alcohol with concentrated H₂SO₄ or concentrated H₃PO₄. The concentrated acid speeds up the reaction and also takes up the water, so it acts as a catalyst and as a dehydrating agent. If a question asks for its role, write "dehydrating agent".

More than one alkene forms when the H can come from different neighbouring carbons, or when the alkene has cis and trans isomers. Butan-2-ol, CH₃CH(OH)CH₂CH₃, can lose a hydrogen from C1 or from C3:

• from C1: but-1-ene, CH₂=CHCH₂CH₃
• from C3: but-2-ene, CH₃CH=CHCH₃, which exists as cis and trans isomers (topic 13.4)

That makes three alkenes in all. Methanol has only one carbon atom, so it cannot form an alkene.

Esterification

Heat the alcohol and the carboxylic acid with a few drops of concentrated H₂SO₄ as a catalyst. The reaction is reversible, and it is a condensation: two molecules join and a small molecule, water, is lost.

CH₃COOH + CH₃CH₂OH ⇌ CH₃COOCH₂CH₃ + H₂O

To name an ester, put the alcohol part first (ethyl) and the acid part second (ethanoate). Ethanol with propanoic acid gives ethyl propanoate, CH₃CH₂COOCH₂CH₃. Esters have fruity smells.

The tri-iodomethane test

Method: add aqueous iodine to the compound, then aqueous sodium hydroxide drop by drop until the brown colour just fades, and warm gently. A positive result is a pale yellow precipitate of tri-iodomethane, CHI₃.

What it shows: the alcohol contains the CH₃CH(OH)- group, written CH₃CH(OH)R, where R is H or a carbon-based group. Alkaline iodine first oxidises CH₃CH(OH)R to CH₃COR. The CH₃ group then ends up as CHI₃, and the rest of the molecule leaves as the carboxylate ion RCO₂⁻, which has one carbon fewer than the alcohol. Overall:

CH₃CH(OH)R + 4I₂ + 6OH⁻ → CHI₃ + RCO₂⁻ + 5I⁻ + 5H₂O

Alcohol	CH₃CH(OH)- group?	Organic ion formed
ethanol	yes (R = H)	methanoate, HCO₂⁻
propan-2-ol	yes (R = CH₃)	ethanoate, CH₃CO₂⁻
hexan-3-ol	no	none
2-methylpropan-2-ol	no: no H on the C-OH carbon	none

Ethanol is the only primary alcohol that gives the test; propan-1-ol, for example, does not. A tertiary -OH group never does, because the first step needs a hydrogen on the C-OH carbon. Aldehydes and ketones with a CH₃CO- group also give the test (topic 17.1).

Worked example 2: a diol with two kinds of -OH group

Question: Hexane-1,5-diol is HOCH₂CH₂CH₂CH₂CH(OH)CH₃. (a) Classify each -OH group. (b) Give the organic product of heating it under reflux with an excess of acidified K₂Cr₂O₇, and the colour change. (c) Does it give a yellow precipitate with alkaline I₂(aq)? If so, give the organic ion formed. (d) 1.77 g of the diol reacts completely with an excess of sodium. Calculate the volume of hydrogen formed at room conditions.

(a) Classify. C1 is bonded to one other carbon (C2), so that -OH is primary. C5 is bonded to two other carbons (C4 and C6), so that -OH is secondary.

(b) Oxidise under reflux. The primary end goes all the way to -COOH and the secondary end becomes a ketone:

HOCH₂CH₂CH₂CH₂CH(OH)CH₃ + 3[O] → CH₃COCH₂CH₂CH₂COOH + 2H₂O

The product is 5-oxohexanoic acid. The solution turns from orange to green.

(c) Tri-iodomethane test. Yes: C5 and C6 form a CH₃CH(OH)- group, with R = -CH₂CH₂CH₂CH₂OH. A yellow precipitate of CHI₃ forms. The primary -OH at the other end is not changed in this test, so RCO₂⁻ is HOCH₂CH₂CH₂CH₂CO₂⁻, the 5-hydroxypentanoate ion.

(d) Hydrogen volume. Both -OH groups react, so 1 mol of diol gives 1 mol of H₂. The diol is C₆H₁₄O₂, $M_{\text{r}}$ = 6(12.0) + 14(1.0) + 2(16.0) = 118.0. $$n = \frac{1.77}{118.0} = 0.0150\ \text{mol}$$ $$V = 0.0150 \times 24.0 = 0.360\ \text{dm}^3$$

Answer: (a) primary on C1, secondary on C5; (b) CH₃COCH₂CH₂CH₂COOH, orange to green; (c) yes, HOCH₂CH₂CH₂CH₂CO₂⁻ (with CHI₃); (d) 0.360 dm³ (3 s.f.).

Key Equations

R and R′ are alkyl groups; n is the number of carbon atoms.

Complete combustion (3n/2 mol of O₂ per mole of alcohol):

CₙH₂ₙ₊₁OH + (3n/2)O₂ → nCO₂ + (n + 1)H₂O

Oxidation (distil for the aldehyde, reflux for the acid):

RCH₂OH + [O] → RCHO + H₂O

RCH₂OH + 2[O] → RCOOH + H₂O

RCH(OH)R′ + [O] → RCOR′ + H₂O

Sodium (1 mol of H₂, 24.0 dm³ at room conditions, per 2 mol of -OH):

2ROH + 2Na → 2RO⁻Na⁺ + H₂

Substitution with PCl₅:

ROH + PCl₅ → RCl + POCl₃ + HCl

Dehydration (n of at least 2):

CₙH₂ₙ₊₁OH → CₙH₂ₙ + H₂O

Esterification (concentrated H₂SO₄ catalyst):

RCOOH + R′OH ⇌ RCOOR′ + H₂O

Tri-iodomethane test (R = H or a carbon-based group):

CH₃CH(OH)R + 4I₂ + 6OH⁻ → CHI₃ + RCO₂⁻ + 5I⁻ + 5H₂O

Common Mistakes to Avoid

• Mistake: heating under reflux to make an aldehyde, or writing "reflux" for every reaction. Fix: distil the aldehyde off as it forms; reflux only to make the acid. Reflux is for heating volatile substances for a long time, so a reaction done cold, such as with PCl₅, never needs it.
• Mistake: saying a tertiary alcohol is oxidised, reporting a precipitate, or giving the colour change backwards. Fix: a primary or secondary alcohol turns the orange solution green. 2-Methylpropan-2-ol leaves it orange.
• Mistake: naming concentrated H₂SO₄ only as a catalyst, or as an oxidising agent, when an alcohol is dehydrated. Fix: its role here is dehydrating agent: it removes water from ethanol to give ethene. In esterification its role is catalyst.
• Mistake: writing tri-iodomethane as CH₃I or CIH₃, or thinking every secondary alcohol gives the test. Fix: the precipitate is CHI₃, and only alcohols with CH₃CH(OH)- react. Hexan-3-ol, CH₃CH₂CH(OH)CH₂CH₂CH₃, gives no precipitate.
• Mistake: treating a diol as if it had one -OH group: giving it one class, or using one PCl₅ or one Na per molecule. Fix: classify each -OH separately, and allow one PCl₅ (or one Na) per -OH. Propane-1,2-diol has a primary and a secondary -OH; HOCH₂CH₂OH + 2PCl₅ → ClCH₂CH₂Cl + 2POCl₃ + 2HCl.
• Mistake: explaining acidity with "the electronegative oxygen" alone, or calling ethanol the stronger acid. Fix: the ethyl group donates electrons to oxygen, so the O-H bond is less polarised and CH₃CH₂O⁻ is less stable than OH⁻; ethanol is the weaker acid.
• Mistake: choosing an oxidising agent to turn a carbonyl compound into an alcohol, or NaBH₄ to reduce a carboxylic acid. Fix: reduce aldehydes and ketones with NaBH₄ or LiAlH₄, but carboxylic acids with LiAlH₄ only: propanoic acid gives propan-1-ol.

Exam Tips

• State reagents and conditions in full: "K₂Cr₂O₇ and dilute H₂SO₄, distil" or "heat under reflux"; "NaOH(aq), heat"; "LiAlH₄ in dry ether". Give the concentration or temperature when it decides the product, as in cold, dilute KMnO₄.
• Describe observations with the start and end colour and the state: "orange solution turns green", "pale yellow precipitate", "effervescence; the gas gives a squeaky pop", "steamy white fumes".
• Construct equations with [O] and [H] by counting: 1[O] for an aldehyde or ketone, 2[O] for a carboxylic acid, 2[H] to reduce an aldehyde or ketone, 4[H] to reduce an acid. Then balance with H₂O.
• Deduce a structure by writing what each result proves (green with dichromate: primary or secondary; yellow precipitate with alkaline iodine: CH₃CH(OH)-), then draw a structure that fits every result and the molecular formula.
• Draw a displayed formula with every bond shown, including O-H: write -O-H, not -OH.