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Organic Chemistry: Alcohols

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  Edexcel IAL Chemistry

Organic Chemistry: Alcohols

The big idea: An alcohol is just a hydrocarbon chain with an –OH group stuck on it, and almost everything interesting it does — burning, swapping the –OH for a halogen, losing water to form an alkene, or getting oxidised — comes down to that one –OH group being reactive while the rest of the chain just comes along for the ride.

Quick Summary

Here's the whole chapter in one scan before you dive into the detail:

What an alcohol is

CnH2n+1OH — a hydrocarbon chain plus a hydroxyl (–OH) group.

Primary, secondary, tertiary

Classified by how many carbons are attached to the C bonded to –OH.

Combustion

Alcohol + O₂ → CO₂ + H₂O (complete combustion).

Halogenation

–OH is swapped for –Cl, –Br or –I using different reagents for each.

Dehydration

Heat with concentrated phosphoric acid → alkene + water.

Oxidation

1° → aldehyde → carboxylic acid. 2° → ketone. 3° → no reaction.

1. Alcohols — Introduction

1 What actually makes something an "alcohol"?

Every alcohol contains a hydroxyl functional group, –OH, bonded to a carbon chain. That's the whole definition. The general formula is CnH2n+1OH, which is exactly the formula for an alkane (CnH2n+2) but with one hydrogen swapped out for an –OH group. That's a useful way to think about it: alcohol = alkane skeleton + one H replaced by OH.

Naming follows the pattern alkane name + "ol" — so the 2-carbon alkane ethane becomes ethanol, propane becomes propanol, and so on. If a molecule has two –OH groups, it's called a diol (and with three, a triol).

AlcoholStructural formulaCarbons
MethanolCH₃OH1
EthanolCH₃CH₂OH2
PropanolCH₃CH₂CH₂OH3
ButanolCH₃CH₂CH₂CH₂OH4
Think of it like this

If alkanes are the "boring but stable" backbone of organic chemistry, the –OH group is like a little handle bolted onto that backbone. Nearly every reaction in this chapter is really a reaction happening at that handle — the rest of the carbon chain is basically just watching.

2 Primary, secondary, tertiary — and why it matters

This classification is one of the most important ideas in the whole chapter, because it controls which reactions can happen later (especially oxidation, and the HCl chlorination shortcut). The rule is simple: look at the carbon atom that the –OH is directly bonded to, and count how many other carbons are attached to it.

  • Primary (1°) alcohol — the OH-bearing carbon is attached to one other carbon (or none, as in methanol). Example: propan-1-ol, CH₃CH₂CH₂OH.
  • Secondary (2°) alcohol — the OH-bearing carbon is attached to two other carbons. Example: propan-2-ol, CH₃CH(OH)CH₃.
  • Tertiary (3°) alcohol — the OH-bearing carbon is attached to three other carbons. Example: 2-methylpropan-2-ol, (CH₃)₃COH.
H H H H OH H | | | | | | H--C--C--C--OH H--C--C---C--H | | | | | | H H H H H H PRIMARY: propan-1-ol SECONDARY: propan-2-ol (OH-carbon touches (OH-carbon touches 1 other carbon) 2 other carbons) OH | H3C -- C -- CH3 | CH3 TERTIARY: 2-methylpropan-2-ol (OH-carbon touches 3 other carbons)
How to classify fast
Circle the carbon bonded to –OH → count its carbon neighbours → 1 = primary, 2 = secondary, 3 = tertiary
Practice Question 1
Butan-2-ol has the structure CH₃CH(OH)CH₂CH₃. Is it primary, secondary or tertiary? Explain your reasoning.
Practice Question 2
Draw and name the diol that would form from butane if two hydrogens on adjacent carbons (carbons 1 and 2) were replaced with –OH groups. What would it be called?

2. Reactions of Alcohols

1 Combustion

Alcohols burn in air (complete combustion) just like alkanes do — the fuel is fully oxidised to carbon dioxide and water, releasing energy as heat. This is why ethanol is used as a biofuel.

General equation
alcohol + oxygen → carbon dioxide + water

For ethanol specifically:

CH₃CH₂OH + 3O₂ → 2CO₂ + 3H₂O
Common mistake

Students often forget to balance the extra oxygen already present inside the alcohol molecule (the O in –OH). Always balance carbon and hydrogen first, then work out oxygen last — the O in the fuel counts toward your total oxygen balance.

2 Converting alcohols into halogenoalkanes

This whole family of reactions does one thing: it replaces the –OH group with a halogen atom (this is called halogenation). But — and this is the tricky bit examiners love — the exact method and reagent is different for each halogen. You need three separate recipes memorised.

2a Chlorination — using PCl₅

Phosphorus(V) chloride (PCl₅) is added directly to the alcohol at room temperature. The reaction is vigorous, so no heating is needed. It produces two inorganic by-products: phosphoryl chloride (POCl₃) and hydrogen chloride gas (HCl) — and it's the appearance of steamy white fumes of HCl that makes this reaction so useful as a test for the –OH group in an unknown compound.

Chlorination equation
CH₃CH₂CH₂OH + PCl₅ → CH₃CH₂CH₂Cl + POCl₃ + HCl

There's also a shortcut that only works for tertiary alcohols: simply shaking the tertiary alcohol with concentrated hydrochloric acid at room temperature is enough — no PCl₅ needed.

(CH₃)₃COH + HCl → (CH₃)₃CCl + H₂O
Watch out

The HCl-shaking shortcut only works for tertiary alcohols. Primary and secondary alcohols are far less reactive toward HCl alone and won't undergo this reaction — for those, you still need PCl₅.

2b Bromination — using KBr + H₂SO₄

Unlike chlorination, bromination doesn't use a ready-made halogenating reagent — it's generated in situ (in the reaction mixture itself). A warmed mixture of potassium bromide and 50% concentrated sulfuric acid is used with the alcohol. Because this happens in two conceptual stages, examiners expect two separate equations.

Stage 1 — the inorganic reactants react together to generate hydrogen bromide:

2KBr + H₂SO₄ → K₂SO₄ + 2HBr

Stage 2 — the HBr generated then reacts with the alcohol, for example with butan-1-ol:

CH₃CH₂CH₂CH₂OH + HBr → CH₃CH₂CH₂CH₂Br + H₂O
Concentration matters!

You must use 50% sulfuric acid, not fully concentrated. If the acid is too concentrated, it will oxidise the bromide ions all the way to bromine (Br₂) instead of just protonating them to HBr — giving you the wrong products entirely.

2c Iodination — using red phosphorus + iodine

This one uses a mixture of red phosphorus and iodine heated under reflux with the alcohol. Again this is really two linked equations, because the phosphorus and iodine react together first to form phosphorus(III) iodide, which is the actual halogenating agent.

Step 1
2P + 3I₂ → 2PI₃
Step 2 (e.g. with ethanol)
3C₂H₅OH + PI₃ → 3C₂H₅I + H₃PO₃

The by-product here is phosphorous acid (H₃PO₃) — note this is a different acid from the phosphoryl chloride seen in chlorination, so don't mix them up in your equations.

The three halogenation methods, side by side
  • Chlorine: PCl₅, room temperature (or HCl shaking — tertiary only)
  • Bromine: KBr + 50% H₂SO₄, warmed (generates HBr in situ)
  • Iodine: red P + I₂, heat under reflux (generates PI₃ in situ)

3 Dehydration to alkenes

Instead of swapping the –OH for something else, dehydration removes it completely — along with a hydrogen from the next-door carbon — to form a carbon-carbon double bond. This is an elimination reaction (mirroring the elimination reactions you'll see with halogenoalkanes), and it's done by heating the alcohol with concentrated phosphoric acid, which acts as a catalyst.

Dehydration equation (ethanol → ethene)
CH₃CH₂OH → CH₂=CH₂ + H₂O
Why doesn't phosphoric acid appear in the equation?

It's a catalyst — it speeds up the reaction but isn't consumed by it, so it doesn't appear as a reactant or product. The note also mentions a practical reason: the water produced by the reaction dilutes the concentrated phosphoric acid, which is a nice detail examiners sometimes ask about.

Practice Question 3
Write the equation for the chlorination of propan-1-ol using PCl₅, and explain what observation would confirm that the –OH group is present.
Practice Question 4
2-methylpropan-2-ol is shaken with concentrated hydrochloric acid at room temperature and a reaction occurs. Would you expect the same result if you tried this with butan-1-ol instead? Explain.

3. Oxidation of Alcohols

1 The oxidation "ladder"

This is the single most examined idea in this chapter, so get it rock solid. What an alcohol can be oxidised into depends entirely on whether it's primary, secondary, or tertiary — because oxidation happens at the carbon bonded to the –OH, and that carbon needs a spare hydrogen to lose for oxidation to be possible.

  • Primary alcohols → oxidised to aldehydes → can be oxidised further to carboxylic acids. (Two separate oxidation steps — a "ladder" you can stop halfway up.)
  • Secondary alcohols → oxidised to ketones only. There's nowhere further to go — ketones cannot be oxidised any further under these conditions.
  • Tertiary alcoholsdo not oxidise at all under these conditions.
Why can't tertiary alcohols be oxidised?

Oxidation here really means "remove a hydrogen from the carbon holding the –OH, and form a C=O double bond." A tertiary alcohol's OH-carbon is already fully bonded to three other carbons plus the OH — it has no hydrogen left on that carbon to remove. No spare hydrogen, no oxidation. This single fact explains the entire pattern.

PRIMARY ALCOHOL --[O]--> ALDEHYDE --[O]--> CARBOXYLIC ACID (e.g. ethanol) (ethanal) (ethanoic acid) SECONDARY ALCOHOL --[O]--> KETONE (stops here — no further oxidation) (e.g. propan-2-ol) (propanone) TERTIARY ALCOHOL --[O]--> NO REACTION

2 The oxidising agent: acidified K₂Cr₂O₇

The standard oxidising agent used is acidified potassium dichromate(VI), K₂Cr₂O₇ — an orange solution. "Acidified" means it's dissolved in dilute acid (typically dilute sulfuric acid), because the reduction of dichromate ions requires H⁺ ions supplied by that acidic medium.

Colour change to remember
Orange Cr₂O₇²⁻ (dichromate) → Green Cr³⁺ (chromium III) as the alcohol is oxidised

This colour change (orange → green) is itself a useful visual clue in practicals: if you see the orange solution staying orange, the alcohol you added probably wasn't oxidised (hint: tertiary alcohol).

Starting alcoholProduct
Ethanol (primary)Ethanal (aldehyde) → Ethanoic acid (carboxylic acid)
Propan-2-ol (secondary)Propanone (ketone) — stops here
2-methylpropan-2-ol (tertiary)No reaction

3 Controlling how far the oxidation goes

Since aldehydes are only a "stopping point on the way" to carboxylic acids, chemists have to choose their apparatus carefully depending on what product they actually want.

TechniqueWhat it doesUsed for
Heating under refluxVapours condense and drip back into the flask, so nothing escapes — the reaction can go all the wayFull oxidation: primary → carboxylic acid, secondary → ketone
Distillation with additionOnly the oxidising agent is heated; alcohol is added slowly, and product distils off immediately as it formsPartial oxidation: primary alcohol → aldehyde only (stopping before the acid stage)

The physical reason distillation works to "trap" the aldehyde is simple: aldehydes have a lower boiling point than the alcohol they came from, so as soon as one forms, it boils off and is collected before it has a chance to be oxidised further by the excess oxidising agent still in the flask.

REFLUX SET-UP DISTILLATION SET-UP (full oxidation) (stop at aldehyde) open top alcohol added | dropwise --> [condenser] | | [oxidising mixture, [flask - heated] heated] alcohol + oxidiser | stays & refluxes [aldehyde distils off immediately] | [collected in receiver flask]
Memory hook

Reflux = keep it in the pot, let it cook fully. Distillation = grab the product the moment it's made, before it can react further.

4 Testing for aldehydes: Fehling's solution

Fehling's solution is an alkaline solution containing Cu²⁺ ions, which act as a mild oxidising agent — but only strong enough to oxidise aldehydes, not ketones. This selectivity is exactly why it's such a useful diagnostic test to distinguish the two.

  • When warmed with an aldehyde: the aldehyde is oxidised to a carboxylic acid (which becomes a carboxylate salt in the alkaline conditions), and the Cu²⁺ ions are reduced to Cu⁺ ions.
  • The result: the clear blue solution turns opaque due to a red precipitate of copper(I) oxide, Cu₂O forming.
  • When warmed with a ketone: nothing happens — ketones can't be oxidised, so the solution stays clear blue. This is a negative test.
Fehling's — remember the colours
Clear blue (Cu²⁺) → Opaque red precipitate (Cu₂O) = POSITIVE for aldehyde

5 Testing for aldehydes: Tollens' reagent

Tollens' reagent — also called ammoniacal silver nitrate solution — is an aqueous alkaline solution of silver nitrate dissolved in excess ammonia. It works on the same principle as Fehling's, but with silver ions instead of copper ions.

  • When warmed with an aldehyde: the aldehyde is oxidised to a carboxylic acid (forming a salt in the alkaline conditions), and Ag⁺ ions are reduced to solid Ag atoms.
  • These silver atoms deposit as a thin, shiny layer on the inside of the test tube — the famous "silver mirror" test.
  • With a ketone: no reaction, no mirror forms — negative test, same logic as Fehling's.
Tollens' — remember the result
Ag⁺ (colourless solution) → Ag atoms (silver mirror on tube) = POSITIVE for aldehyde
A trap examiners love

Both Fehling's and Tollens' reagents are themselves the oxidising agents (Cu²⁺ and Ag⁺ respectively), and in doing their job they get reduced. Don't accidentally write that the aldehyde "reduces" the copper/silver — that's correct! — but also don't forget to say the aldehyde itself is oxidised to a carboxylic acid. Both halves of the redox reaction need to be in your answer for full marks.

Practice Question 5
A student has two unlabelled bottles: one contains propan-2-ol (a secondary alcohol) and one contains propanal (an aldehyde). Describe a chemical test, including the observation, that would let the student tell them apart.
Practice Question 6
Explain why, when preparing propanone from propan-2-ol using acidified potassium dichromate(VI), it is not necessary to distil the ketone off as soon as it forms — unlike when preparing an aldehyde from a primary alcohol.

What to Memorise

General formula
CnH2n+1OH — an alkane with one H replaced by OH.
Classification rule
Count carbons attached to the OH-carbon: 1 = primary, 2 = secondary, 3 = tertiary.
Chlorination
PCl₅ at room temp → steamy HCl fumes (test for –OH). Tertiary only: shake with HCl.
Bromination
KBr + 50% H₂SO₄ (warmed) generates HBr in situ, which then reacts with the alcohol.
Iodination
Red phosphorus + I₂ under reflux generates PI₃ in situ, which reacts with the alcohol.
Dehydration
Heat with concentrated phosphoric acid (catalyst) → alkene + water.
Oxidation ladder
1° → aldehyde → carboxylic acid. 2° → ketone (stop). 3° → nothing.
Oxidising agent
Acidified K₂Cr₂O₇: orange (Cr₂O₇²⁻) → green (Cr³⁺) when oxidation occurs.
Reflux vs distillation
Reflux = full oxidation. Distillation with addition = stop at the aldehyde.
Fehling's test
Blue Cu²⁺ → red Cu₂O precipitate = positive for aldehyde. No change = ketone.
Tollens' test
Ag⁺ → silver mirror = positive for aldehyde. No mirror = ketone.
Combustion
Alcohol + O₂ → CO₂ + H₂O (complete combustion, heat given out).

Concepts Checklist

Tick these off honestly — if you can't tick it with confidence, go back and re-read that bit.

Exam Tips & Common Traps

Trap: Mixing up the three halogenation reagents

Examiners frequently ask you to identify the reagent/conditions for a named halogenation. Don't just remember "you swap OH for a halogen" — you need the specific reagent for each: PCl₅ for chlorine, KBr + 50% H₂SO₄ for bromine, red P + I₂ for iodine. Mixing these up (e.g. writing PCl₅ when asked for bromination) is one of the most common lost marks in this topic.

Trap: Forgetting tertiary alcohols don't oxidise

A very common exam question gives you an unknown alcohol, treats it with acidified dichromate, and asks you to explain "no colour change was observed." If you don't immediately think "tertiary alcohol, no spare hydrogen on the OH-carbon," you'll miss easy marks. Always link the observation back to the structural reason.

Trap: Confusing reflux and distillation purposes

Students often think reflux and distillation are interchangeable "heating methods." They're not — the choice of apparatus is the entire mechanism for controlling how far oxidation goes. Reflux keeps everything together so oxidation can complete (aldehyde → acid); distillation removes the aldehyde the instant it forms so it can't be oxidised further. If a question describes preparing an aldehyde, the apparatus must be distillation, not reflux.

What examiners actually look for
  • Fully balanced equations, including state symbols where asked — don't skip inorganic by-products like POCl₃ or H₃PO₃.
  • Correct use of "oxidised" vs "reduced" — remember the alcohol/aldehyde is oxidised, while Cr₂O₇²⁻, Cu²⁺, or Ag⁺ are reduced.
  • Precise observations, not vague ones: "orange to green" not just "colour changes"; "red precipitate," not "it turns red."
  • Linking structure to reactivity: always be ready to explain why (e.g. why tertiary alcohols don't oxidise, why 50% not concentrated H₂SO₄ is used for bromination).
Quick self-test before the exam

Cover the whole guide and try to answer: (1) What's the general formula of an alcohol? (2) How do you tell 1°, 2°, 3° apart? (3) Name the reagent for each halogenation. (4) What forms when you dehydrate an alcohol, and with what reagent? (5) What does each type of alcohol oxidise to, and why? (6) What's the colour change for acidified dichromate, Fehling's, and Tollens'? If you can answer all six without peeking, you're in great shape.

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