AlcoholsEdexcel A-Level Chemistry: Revision notes
Section 1
Classifying alcohols
Alcohols contain the hydroxyl group, –OH. They are classified by the number of carbon atoms bonded to the carbon that carries the OH:
- Primary: the C–OH carbon is bonded to one other carbon (e.g. propan-1-ol, CH₃CH₂CH₂OH)
- Secondary: bonded to two other carbons (e.g. propan-2-ol, CH₃CH(OH)CH₃)
- Tertiary: bonded to three other carbons (e.g. 2-methylpropan-2-ol, (CH₃)₃COH)
The class decides how the alcohol is oxidised, so it is the key to most questions on this topic.
Methanol and ethanol are both primary. Count the carbons attached to the C–OH carbon, not the length of the chain.
Section 2
Combustion and halogenation
Alcohols burn completely in excess oxygen to give carbon dioxide and water: C₂H₅OH + 3O₂ → 2CO₂ + 3H₂O.
The OH group can be replaced by a halogen to make a halogenoalkane:
- Chloroalkane: PCl₅ at room temperature. CH₃CH₂OH + PCl₅ → CH₃CH₂Cl + POCl₃ + HCl (misty fumes of HCl)
- Bromoalkane: KBr with 50% concentrated sulfuric acid, which makes HBr in situ: KBr + H₂SO₄ → KHSO₄ + HBr, then ROH + HBr → RBr + H₂O
- Iodoalkane: red phosphorus and iodine, which form PI₃ in situ: 3ROH + PI₃ → 3RI + H₃PO₃
The sulfuric acid is diluted to about 50% so it does not oxidise HBr to bromine.
Learn the three reagent sets as a trio: PCl₅ (Cl), KBr with 50% H₂SO₄ (Br), red P with I₂ (I).
Section 3
Oxidation with acidified potassium dichromate(VI)
The oxidising agent is potassium dichromate(VI) in dilute sulfuric acid, with the orange Cr₂O₇²⁻ ion reduced to green Cr³⁺. [O] represents the oxygen supplied by the oxidant.
- Primary alcohol → aldehyde → carboxylic acid: CH₃CH₂OH + [O] → CH₃CHO + H₂O, then CH₃CHO + [O] → CH₃COOH
- Secondary alcohol → ketone: CH₃CH(OH)CH₃ + [O] → CH₃COCH₃ + H₂O
- Tertiary alcohol: not oxidised, so the solution stays orange (no hydrogen on the C–OH carbon)
To stop at the aldehyde, distil it off as it forms. To reach the carboxylic acid, heat under reflux with excess oxidant.
The aldehyde is detected by warming with Benedict's or Fehling's solution: blue solution to a brick-red precipitate of copper(I) oxide. Ketones give no change.
An orange-to-green colour change shows only that the alcohol is primary or secondary. It does not show which; use Benedict's or Fehling's on the product to decide.
Section 4
Dehydration to alkenes
Heating an alcohol with concentrated phosphoric acid removes water in an elimination reaction and forms an alkene: CH₃CH₂OH → CH₂=CH₂ + H₂O.
Phosphoric acid is preferred to concentrated sulfuric acid because it is less oxidising and gives fewer side products. You do not need to know the mechanism.
Section 5
Preparing and purifying a liquid organic compound
Organic preparations use a standard sequence of techniques:
- Heating under reflux: a condenser fitted vertically returns vapour to the flask, so a reaction can be heated for a long time without losing volatile reactants or products
- Distillation: separates liquids by boiling point, e.g. removing a product with a lower boiling point as it forms
- Extraction with a solvent in a separating funnel: shake, allow the layers to separate and run off the lower layer; the less dense layer is on top. Release the pressure by opening the tap
- Drying with an anhydrous salt such as anhydrous sodium sulfate or calcium chloride, then decant or filter
- Boiling temperature determination: collect the fraction that distils at the known boiling point of the product; a narrow range indicates purity
Section 6
Core practicals 5 and 6
Core Practical 5, oxidation of ethanol: ethanol is added to warm acidified potassium dichromate(VI). Distillation collects ethanal (boiling point about 21 °C) before it is oxidised further; reflux with excess oxidant gives ethanoic acid. Test the products with Benedict's or Fehling's solution (aldehyde positive) and observe orange to green.
Core Practical 6, chlorination of 2-methylpropan-2-ol: shake the tertiary alcohol with concentrated hydrochloric acid at room temperature: (CH₃)₃COH + HCl → (CH₃)₃CCl + H₂O. Separate the upper organic layer, wash with sodium hydrogencarbonate solution to remove acid, dry with anhydrous sodium sulfate and distil, collecting the fraction at about 51 °C.
Practical questions ask why each step is done. Pair every step with its purpose, e.g. sodium hydrogencarbonate removes excess acid and the drying agent removes water.
That's the notes covered.
Carry on to the next subtopic.
Exam questions on Alcohols
- A technician is given four structural isomers of molecular formula C₄H₁₀O: butan-1-ol, butan-2-ol, 2-methylpropan-1-ol and 2-methylpropan-2-ol. She will warm each with acidified potassium dichromate(VI) solution and compare the results.Explain why 2-methylpropan-2-ol does not change the colour of acidified potassium dichromate(VI) solution when warmed.2 marks
- A chemist converts propan-1-ol into three different halogenoalkanes: 1-chloropropane, 1-bromopropane and 1-iodopropane. A different reagent or reagent mixture is needed for each halogen.Write two equations to show how potassium bromide and 50% concentrated sulfuric acid convert propan-1-ol into 1-bromopropane. The first equation shows the formation of the reactive halogenating species.2 marks
- A student carries out Core Practical 5 on the oxidation of ethanol using acidified potassium dichromate(VI). In one experiment she wants to collect ethanal; in a second experiment she wants to make ethanoic acid.Describe how she should carry out the experiment to collect ethanal, and explain why this method prevents further oxidation.3 marks
Written by the Exaim team, led by Shaun Daswani (Head of Upper Secondary, Improve ME Institute; MSc Financial Mathematics, Imperial College London; BSc, UCL) and Jason Daswani (operational lead, Improve ME Institute; LSE).