What this chapter covers16

Alcohols sit in the middle of AS organic chemistry. Alkenes and halogenoalkanes turn into them, and they turn into almost everything else: aldehydes, ketones, carboxylic acids, esters, alkenes and halogenoalkanes again. One small group, –OH, does all of this. It is polar, it has a hydrogen that can be removed as a proton, and it sits on a carbon whose own hydrogens decide how far oxidation can go. Learn what that carbon carries and most of this topic follows from it.

The verbs tell you how much to write. 16.1.1 says recall: six reagent-and-condition sets, and a mark usually needs the condition as well as the reagent — steam rather than water, cold and dilute manganate(VII) rather than manganate(VII). 16.1.2 says describe: equations, reagents, conditions and what you would see. 16.1.3(b) says state, so give the reagent and both colours, with no explanation needed. 16.1.4 says deduce: from a result, work back to the structure. 16.1.5 is the only statement that says explain, and the explanation it wants is the inductive effect of the alkyl group.

The page is in three parts. The first classifies alcohols and gives the six ways of making them. The second works through their reactions, with oxidation in the most detail because it appears in more questions than anything else in the topic. The third covers the tests that tell alcohols apart, and brings every reaction together in one map.

The hydroxyl group and the alcohols16.1.3(a)

An alcohol is an organic compound in which an –OH group is bonded to an sp³ carbon atom — a carbon carrying only single bonds. The –OH group is the hydroxy group (the older word, still common in textbooks, is hydroxyl). It is the functional group that decides almost everything on this page: where the molecule reacts, what it turns into, and how it behaves towards water.

The alcohols with one –OH group and no ring or double bond form a homologous series with general formula C_nH_2n+1OH, which can also be written C_nH_2n+2O. Each member differs from the next by CH₂. Names come from the parent alkane: drop the final –e, add –ol, and put a locant in front of –ol to show which carbon carries the –OH group.

The last four rows are the four alcohols with formula C₄H₁₀O. They are structural isomers: two are positional isomers of each other on a straight chain, two sit on a branched chain. The column on the right matters more than the name, because the type of alcohol decides what oxidation does to it. The next section explains the classification.

Why the –OH group matters physically

Oxygen is much more electronegative than carbon or hydrogen, so both the C–O and the O–H bonds are polar. The O–H hydrogen carries a significant δ+ charge and the oxygen carries two lone pairs, so alcohol molecules hydrogen-bond to each other and to water. That is why ethanol (M_r 46) boils at 78 °C while propane (M_r 44) boils at −42 °C, and why the small alcohols mix with water in all proportions. As the hydrocarbon chain lengthens, the non-polar part takes over: butan-1-ol is only partly soluble and hexan-1-ol barely dissolves.

Primary, secondary and tertiary alcohols16.1.3(a)

Classify an alcohol by looking at the carbon atom that carries the –OH group, and counting how many other carbon atoms are bonded to it. Hydrogen atoms on that carbon are what count in practice, because oxidation removes one of them — so a quick check is to count those too.

Methanol, CH₃OH, fits none of the three definitions because its carbon has no other carbon attached, but it has three hydrogens on the –OH carbon and behaves like a primary alcohol in every reaction on this page. Most textbooks and mark schemes simply group it with the primary alcohols.

Use the model below to test the rule on any alcohol. It reads the molecule, finds each –OH carbon, counts its carbon neighbours and hydrogens, and states the class together with what that class predicts for oxidation.

Alcohols with more than one –OH group16.1.3(a)

The syllabus asks for the classification to include molecules with more than one alcohol group. The rule does not change: classify each –OH group separately, by the carbon it sits on. A molecule can therefore be primary at one end and secondary in the middle.

Name such compounds by keeping the final –e of the alkane and adding –diol or –triol, with a locant for every –OH: ethane-1,2-diol, propane-1,2,3-triol.

The same logic applies to every reaction in part two. A diol reacts with sodium at both –OH groups, forms a diester with two molecules of a carboxylic acid, and — the practical consequence — its many hydrogen bonds make it viscous and high-boiling: ethane-1,2-diol boils at 197 °C, more than 100 °C above propan-1-ol, which has almost the same M_r.

Six ways to make an alcohol16.1.1

Outcome 16.1.1 lists six routes. Three start from families met earlier (alkenes, halogenoalkanes) and three start from carbonyl compounds and their relatives, which are studied properly in topics 17 and 18. The syllabus wants the reagents and conditions for each; the equations follow easily once those are known.

Pick a route and a starting compound below. The model writes the equation, names the alcohol and classifies it. It is a useful way to check that route (d) can never give a tertiary alcohol — the carbonyl carbon only gains one hydrogen — and that route (e) always gives a primary one.

Hydration of alkenes with steam16.1.1(a)

Ethanol is made industrially by adding water across the double bond of ethene. The water is supplied as steam and the catalyst is phosphoric(V) acid, H₃PO₄, adsorbed on a solid silica support so that it stays in the reactor.

CH₂=CH₂(g) + H₂O(g) ⇌ CH₃CH₂OH(g)    ΔH = −45 kJ mol⁻¹

The conversion per pass is only about 5%. The ethanol is condensed out of the mixture and the unreacted ethene is recycled, so the overall conversion reaches about 95%.

The mechanism

The animation shows the reaction as an electrophilic addition, just like the addition of HBr met in topic 14. The π electrons of the C=C bond attack a δ+ hydrogen of the acid, giving a carbocation and the dihydrogenphosphate ion, H₂PO₄⁻. A lone pair on a water molecule attacks the carbocation. The oxygen now carries three bonds and a positive charge, and it loses H⁺ back to H₂PO₄⁻, which regenerates the acid. The catalyst is used and returned in each cycle.

Unsymmetrical alkenes

With propene, the H⁺ can add to either end of the double bond. Adding it to the CH₂ end gives a secondary carbocation, stabilised by the electron-donating (positive inductive) effect of two alkyl groups; adding it to the CH end gives a primary carbocation. The secondary carbocation forms far faster, so the major product is propan-2-ol and the minor product propan-1-ol. The reasoning is identical to the one used for HBr in 14.2.3.

CH₃CH=CH₂ + H₂O → CH₃CH(OH)CH₃  (major)

Diols from cold dilute manganate(VII)16.1.1(b)

Shaken with a cold, dilute, acidified solution of potassium manganate(VII), an alkene is oxidised to a diol: one –OH group is added to each carbon of the former double bond. The purple colour of the manganate(VII) ion fades to colourless as it is reduced.

CH₂=CH₂ + [O] + H₂O → HOCH₂CH₂OH

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

[O] stands for oxygen supplied by the oxidising agent. This shorthand is what CIE expects in organic oxidation equations: the full ionic equation for manganate(VII) is not required. Check an [O] equation by counting atoms on each side, treating [O] as one O atom.

The two products above illustrate the previous section: ethane-1,2-diol has two primary –OH groups; propane-1,2-diol has one primary and one secondary.

Hydrolysis of halogenoalkanes16.1.1(c)

Heating a halogenoalkane under reflux with aqueous sodium hydroxide replaces the halogen atom with an –OH group. The hydroxide ion is the nucleophile, attacking the δ+ carbon of the C–X bond; the halide ion leaves.

CH₃CH₂Br + NaOH(aq) → CH₃CH₂OH + NaBr

(CH₃)₃CCl + OH⁻ → (CH₃)₃COH + Cl⁻

The –OH goes onto the same carbon the halogen left, so the class of alcohol matches the class of halogenoalkane: primary gives primary, tertiary gives tertiary. Primary halogenoalkanes react by S_N2, tertiary by S_N1 (15.1.5–15.1.6), and the rate follows the C–X bond strength, fastest for iodides.

Reduction of aldehydes and ketones16.1.1(d)

Aldehydes and ketones are reduced to alcohols by sodium tetrahydridoborate, NaBH₄ (sodium borohydride), or by lithium tetrahydridoaluminate, LiAlH₄ (lithium aluminium hydride). In each case the reducing agent delivers a hydride ion, H⁻, to the δ+ carbonyl carbon, and a proton then goes onto the oxygen. In equations, [H] stands for hydrogen supplied by the reducing agent.

CH₃CHO + 2[H] → CH₃CH₂OH    ethanal → ethanol (primary)

CH₃COCH₃ + 2[H] → CH₃CH(OH)CH₃    propanone → propan-2-ol (secondary)

This is oxidation in reverse. Oxidation removed two hydrogen atoms (one from the O, one from the C) to make the C=O; reduction puts two back. So an aldehyde gives a primary alcohol and a ketone gives a secondary alcohol. No carbonyl compound gives a tertiary alcohol this way, because the carbonyl carbon ends up with the hydrogen it gained.

Reduction of carboxylic acids16.1.1(e)

A carboxylic acid can be reduced all the way to a primary alcohol, but only by LiAlH₄ in dry ether; NaBH₄ is not strong enough. Four hydrogen atoms are needed: the C=O becomes CH₂ and one oxygen leaves as water.

CH₃COOH + 4[H] → CH₃CH₂OH + H₂O

CH₃CH₂COOH + 4[H] → CH₃CH₂CH₂OH + H₂O

The carbon chain is unchanged, and the carbon that was the –COOH carbon is now the –CH₂OH carbon, so the product is always a primary alcohol with the same number of carbon atoms as the acid. Balance [H] equations by counting: the acid RCOOH and the alcohol RCH₂OH differ by two H in the product, plus one H₂O formed, so 4[H].

Hydrolysis of esters16.1.1(f)

An ester is made from an alcohol and a carboxylic acid (section 21). Hydrolysis runs that reaction backwards: heating the ester under reflux with water, catalysed by acid or driven by alkali, splits it into its two parts and releases the alcohol.

With dilute acid

Heat under reflux with dilute hydrochloric or sulfuric acid. The H⁺ is a catalyst and the reaction is reversible, so an equilibrium mixture forms. A large excess of water pushes it towards hydrolysis.

CH₃COOCH₂CH₃ + H₂O ⇌ CH₃COOH + CH₃CH₂OH

With dilute alkali

Heat under reflux with dilute sodium hydroxide. The acid formed is immediately converted to its carboxylate salt, which cannot react with the alcohol, so the reaction goes to completion. This makes alkaline hydrolysis the better way to get all of the alcohol out.

CH₃COOCH₂CH₃ + NaOH → CH₃COONa + CH₃CH₂OH

Combustion16.1.2(a)

Alcohols burn in a plentiful supply of oxygen to give carbon dioxide and water. The flame of ethanol is clean and almost invisible in daylight, which is one reason it is used as a fuel on its own and blended into petrol.

CH₃OH + 1½O₂ → CO₂ + 2H₂O

CH₃CH₂OH + 3O₂ → 2CO₂ + 3H₂O

CH₃CH₂CH₂OH + 4½O₂ → 3CO₂ + 4H₂O

You should not need to memorise these: you should be able to balance any of them. The quick way is to balance carbon, then hydrogen, then count the oxygen needed on the right and remember that the alcohol already supplies one oxygen atom. For C_nH_2n+1OH the result is always

C_nH_2n+1OH + ³ⁿ⁄₂ O₂ → nCO₂ + (n+1)H₂O

In a limited supply of oxygen, incomplete combustion gives carbon monoxide or carbon (soot) instead of some of the CO₂. Because alcohols already contain oxygen, they need less air per carbon than the alkane of the same chain length, and they burn with less soot.

Pick an alcohol below. The model builds and balances the complete-combustion equation from the formula, shows the oxygen bookkeeping step by step, and plots enthalpy of combustion against the number of carbon atoms. Each CH₂ adds a near-constant 650 kJ mol⁻¹, the same increment seen in the alkanes.

Reaction with sodium16.1.2(c)

Sodium reacts with alcohols in the same way as with water, but more gently. The O–H bond breaks, hydrogen gas is released, and a sodium alkoxide is formed: an ionic compound of Na⁺ and the alkoxide ion RO⁻.

2CH₃CH₂OH + 2Na → 2CH₃CH₂O⁻Na⁺ + H₂

compare: 2H₂O + 2Na → 2NaOH + H₂

The product from ethanol is sodium ethoxide. If the excess ethanol is evaporated, it is left as a white solid. Dissolved in water, the ethoxide ion takes a proton from water and the solution is strongly alkaline.

The reaction is used as a test for an –OH group (it also works with water and carboxylic acids, so it is not specific to alcohols) and, in the laboratory, as a safe way to destroy small scraps of sodium. The reaction is slower than with water because ethanol is a slightly weaker acid than water; the next section explains why.

Acidity of alcohols compared with water16.1.5

Alcohols, like water, can lose a proton from the O–H group. Both are extremely weak acids: neither turns blue litmus red, and a solution of ethanol in water is neutral.

H₂O ⇌ H⁺ + OH⁻

CH₃CH₂OH ⇌ H⁺ + CH₃CH₂O⁻

The two equilibria look alike, and in each the negative charge of the anion sits on an oxygen atom. The difference is what is attached to that oxygen: hydrogen in the hydroxide ion, an ethyl group in the ethoxide ion. Measured on the same scale, ethanol has pK_a about 16 and water 15.7. The higher the pK_a, the weaker the acid, so ethanol is a slightly weaker acid than water.

The explanation the syllabus expects

Alkyl groups have a positive inductive effect: they push electron density along the σ bonds towards the atom they are attached to. In the ethoxide ion the ethyl group pushes electron density onto the oxygen, which already carries the negative charge. That increases the charge density on the oxygen, making the ion attract H⁺ more strongly. So the ethoxide ion is a stronger base than the hydroxide ion, it recaptures protons more readily, and the ethanol equilibrium lies further to the left.

Substitution to form halogenoalkanes16.1.2(b)

The –OH group of an alcohol can be replaced by a halogen atom. The syllabus names five reagent sets. In each one the –OH is first converted into something that leaves more easily than OH⁻, which is a poor leaving group on its own.

Choose a reagent and an alcohol below. The model writes the balanced equation, names the halogenoalkane, and checks that the atoms balance. It also flags the one combination to avoid: concentrated sulfuric acid with iodide.

The test for an –OH group

Add phosphorus(V) chloride to the dry compound. If steamy white fumes of hydrogen chloride are given off — they turn moist blue litmus red and form white smoke of NH₄Cl with ammonia — an –OH group is present. The compound must be dry, because PCl₅ reacts with water in the same way. Carboxylic acids also give the fumes, so this test shows an –OH group, not specifically an alcohol.

Oxidation: the reagent and the colour change16.1.2(d)

Primary and secondary alcohols are oxidised by warming them with an acidified solution of an oxidising agent. The syllabus names two:

Dichromate is the reagent of choice for the tests in part three, because its orange-to-green change is clear and it is milder, so the reaction can be controlled. Manganate(VII) is stronger and tends to take primary alcohols straight through to the carboxylic acid.

What is removed

Oxidation of an alcohol removes two hydrogen atoms: the one on the oxygen, and one from the carbon carrying the –OH. The C–O bond becomes a C=O bond. Writing the oxygen from the oxidising agent as [O], the two hydrogens leave as water:

R₂CH–OH + [O] → R₂C=O + H₂O

Everything else on this page follows from that one step. A primary alcohol has two hydrogens on the –OH carbon, so after losing one it still has one: it forms an aldehyde, which can be oxidised again. A secondary alcohol has one, and it forms a ketone with none left. A tertiary alcohol has none to lose, so it cannot be oxidised this way at all.

Primary alcohols to aldehydes: distillation16.1.2(d)(i)

To stop the oxidation of a primary alcohol at the aldehyde, the aldehyde has to be taken out of the flask as soon as it forms, before it meets any more oxidising agent. Two things make this possible:

• the alcohol is in excess, so there is not enough oxidising agent to go further;
• the aldehyde has a much lower boiling point than the alcohol, because aldehyde molecules cannot hydrogen-bond to each other. Heating the mixture gently with a distillation apparatus fitted drives the aldehyde off as it forms, and it is condensed and collected.

CH₃CH₂OH + [O] → CH₃CHO + H₂O    ethanol → ethanal

CH₃CH₂CH₂OH + [O] → CH₃CH₂CHO + H₂O    propan-1-ol → propanal

The aldehyde is always the most volatile of the three, which is exactly why distillation works.

Primary alcohols to carboxylic acids: reflux16.1.2(d)(ii)

To take a primary alcohol all the way to the carboxylic acid, the opposite arrangement is used: an excess of the oxidising agent, and heating under reflux. The condenser stands vertically, so any aldehyde that boils off condenses and runs back into the flask, where it is oxidised further. After refluxing, the acid is separated by distillation.

CH₃CH₂OH + 2[O] → CH₃COOH + H₂O

This is the sum of two steps, and seeing it that way explains the 2[O]:

CH₃CH₂OH + [O] → CH₃CHO + H₂O

CH₃CHO + [O] → CH₃COOH

Secondary and tertiary alcohols16.1.2(d)

A secondary alcohol is oxidised to a ketone. Ketones are not oxidised further by acidified dichromate, so the conditions do not matter: distilling or refluxing gives the same product, and heating under reflux is normally used to make sure all of the alcohol reacts.

CH₃CH(OH)CH₃ + [O] → CH₃COCH₃ + H₂O    propan-2-ol → propanone

CH₃CH₂CH(OH)CH₃ + [O] → CH₃CH₂COCH₃ + H₂O    butan-2-ol → butanone (butan-2-one)

A tertiary alcohol has no hydrogen on the carbon carrying the –OH. To make a C=O bond, a C–C bond would have to break, and dichromate cannot do that under these conditions. So 2-methylpropan-2-ol, warmed with acidified potassium dichromate(VI), shows no reaction: the solution stays orange.

The model below makes the prediction for any alcohol and either set of conditions. Change the conditions for a primary alcohol and watch the product change; do the same for a secondary alcohol and it does not. It then writes the [O] equation and checks it balances.

Dehydration to alkenes16.1.2(e)

An alcohol can lose a molecule of water — the –OH group and a hydrogen atom from a neighbouring carbon — to give an alkene. This is an elimination reaction. The syllabus names two sets of conditions:

CH₃CH₂OH → CH₂=CH₂ + H₂O

The animation shows the acid-catalysed route as the reverse of hydration: the acid protonates the –OH, water leaves (a much better leaving group than OH⁻), and a hydrogen on the next carbon is lost as H⁺ to form the C=C bond. The mechanism is not required by 16.1.2(e), but it explains why the acid is a catalyst and why the hydrogen must come from a neighbouring carbon.

When more than one alkene can form

If the –OH carbon has hydrogen-bearing carbons on both sides, and those sides are different, more than one alkene is possible. Butan-2-ol is the standard example. Removing a hydrogen from C1 gives but-1-ene; removing one from C3 gives but-2-ene, which exists as cis and trans (geometric) isomers because each carbon of its double bond carries two different groups. Butan-2-ol therefore gives three alkenes.

CH₃CH₂CH(OH)CH₃ → CH₂=CHCH₂CH₃ + H₂O    but-1-ene

CH₃CH₂CH(OH)CH₃ → CH₃CH=CHCH₃ + H₂O    cis- and trans-but-2-ene

The model below generates every alkene an alcohol can give. It finds each carbon next to the –OH carbon that carries a hydrogen, forms the double bond there, removes duplicates, and tests each product for cis–trans isomerism.

Esterification16.1.2(f)

An alcohol reacts with a carboxylic acid, when heated with a few drops of concentrated sulfuric acid as a catalyst, to form an ester and water. The reaction is reversible and reaches equilibrium.

CH₃COOH + CH₃CH₂OH ⇌ CH₃COOCH₂CH₃ + H₂O

ethanoic acid + ethanol ⇌ ethyl ethanoate + water

The mixture is heated under reflux and the ester is then distilled off. Esters have a characteristic fruity smell; small ones are volatile and used as solvents, flavourings and in perfumes. Ethyl ethanoate itself is a common solvent (nail-varnish remover, glues).

The two jobs of the concentrated sulfuric acid

• It is a catalyst: it provides H⁺ and speeds up an otherwise very slow reaction.
• Being concentrated, it absorbs some of the water formed, which moves the equilibrium to the right and improves the yield.

Naming an ester and drawing it

The name has two words. The first comes from the alcohol (ethanol → ethyl); the second from the acid (ethanoic acid → ethanoate). The formula is usually written acid part first — CH₃COOCH₂CH₃ — which is the reverse of the name, and is where most errors start.

Build any ester from an alcohol and an acid below. The model joins them, removes water, names the product, gives its molecular formula and checks that ester + water has exactly the atoms of acid + alcohol.

Telling primary, secondary and tertiary apart16.1.3(b)

The characteristic distinguishing reaction named in the syllabus is mild oxidation with acidified potassium dichromate(VI). Add a few drops of the alcohol to the acidified dichromate solution and warm the tube in a hot water bath for a few minutes.

This one test separates tertiary alcohols from the other two, but a primary and a secondary alcohol give the same colour change. To tell those two apart, you have to identify the product: collect the distillate from a mild oxidation (excess alcohol, distil) and test it for an aldehyde.

Both of those tests work because aldehydes are reducing agents and ketones are not — the same fact that made a primary alcohol oxidisable in two steps and a secondary alcohol in one. The tests themselves belong to topic 17, where the chemistry of the reagents is examined.

Run the full scheme below on any alcohol. The model shows what the dichromate tube does, then — if the tube turns green — what the distillate gives with Tollens' reagent, and finally the conclusion, with each step's reasoning.

The tri-iodomethane test16.1.4

Warm an alcohol with iodine solution and aqueous sodium hydroxide — together, "alkaline I₂(aq)". If a pale yellow precipitate forms, with a characteristic antiseptic smell, the alcohol contains the group CH₃CH(OH)–. The precipitate is tri-iodomethane, CHI₃ (older name iodoform).

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

R can be H or an alkyl group. The reaction removes the CH₃ carbon as CHI₃; the rest of the molecule is left as a carboxylate ion, RCO₂⁻, with one carbon fewer than the alcohol.

What happens, in two stages

1. The alkaline iodine is an oxidising agent. It oxidises the CH₃CH(OH)– group to CH₃CO–, a methyl carbonyl group.
2. The three hydrogens on that CH₃ are replaced by iodine, giving CI₃CO–. The CI₃ group then breaks away from the rest of the molecule as CHI₃, and the remaining carbonyl carbon becomes the carboxylate ion RCO₂⁻.

Three conclusions follow. Ethanol is the only primary alcohol that gives the test. Every positive secondary alcohol has the –OH on carbon 2 of the chain, next to a terminal CH₃. No tertiary alcohol gives it, because stage 1 needs the H on the –OH carbon.

The model below looks for the CH₃CH(OH)– group in any alcohol, predicts the result, and — when it is positive — builds the carboxylate ion and writes the balanced equation, checking atoms and charge.

The reactions of alcohols together16.1.1–16.1.2

Almost every question on this unit is a route question in disguise: what forms from what, and with which reagent. The map below puts the reactions of this page in one place, centred on ethanol.

The route finder below works on the same map. Choose a starting compound and a target, and it finds the shortest sequence of steps, with the reagents and conditions for each. Try propan-1-ol to propan-2-ol: there is no one-step route, and the answer goes through the alkene.

Ten things that lose marks in this topic16

Every one of these appears somewhere above. Together they make a checklist to run through before answering a topic 16 question.

Self-test16.1

Thirty questions across every outcome in 16.1. Each answer comes with its reasoning, so a wrong choice teaches as much as a right one.

Definitions to learn16.1

Data used on this page16.1

These are reference values used by the models and tables on this page. They are typical literature values, not the official data booklet, and sources differ in the last figure. Where an examination question gives data, use the data it gives.