Part 2: Combustion, dehydration and substitution

Select conditions to change an alcohol into an alkene or haloalkane, and identify every possible alkene product.

Balance oxygen already in the alcohol

Complete combustion of an alcohol gives CO₂ and water. Count carbon and hydrogen first; then remember the oxygen already present in the fuel when finding the O₂ coefficient.

For propan-1-ol, C₃H₈O, products 3CO₂ + 4H₂O contain ten O atoms; the alcohol supplies one, so 4½O₂ supplies the other nine. Doubling gives an all-integer equation.

Remove water across neighbouring carbons

Heat an alcohol with a concentrated acid catalyst such as H₃PO₄ or H₂SO₄ to eliminate water and form an alkene. Remove OH from its carbon and H from an adjacent carbon, placing C=C between them. This is dehydration, not removal of H₂. The detailed mechanism is not required by this OCR AS outcome.

Propan-1-ol gives propene. Butan-2-ol can eliminate towards either adjacent carbon, giving but-1-ene and but-2-ene; but-2-ene has E and Z forms. Check both positional and stereoisomer possibilities. The acid promotes elimination and is regenerated.

Replace OH using a halide and acid

Alcohols can react with a halide ion in acidic conditions to form a haloalkane. A common bromide preparation heats an alcohol with NaBr and sulfuric acid to generate HBr in situ, then isolates the organic product. Give both halide and acid: NaBr alone is not the complete reagent set.

The net organic equation is R–OH + HBr → R–Br + H₂O. This changes the functional group without extending the carbon chain. A mechanism is not required for this alcohol substitution outcome; the mechanism required for haloalkane hydrolysis is taught separately.

Choose reflux or distillation for the purpose

Reflux uses a vertical condenser to return vapour to the flask, allowing heating without losing volatile reactants. Distillation carries vapour through a sloping condenser to a separate receiver, collecting the required boiling fraction. Cooling water enters the condenser at the bottom and leaves at the top.

Use anti-bumping granules added before heating and appropriate electric heating for flammable liquids; do not seal heated apparatus. Chromium(VI) oxidants are hazardous and require the prescribed controls and waste collection. The Organic Synthesis lessons explain the complete purification sequence.

Worked elimination: enumerate neighbours, then remove duplicates

For pentan-3-ol, the OH-bearing carbon is carbon 3. Either neighbouring CH₂ can supply the H eliminated with OH, but the two directions are equivalent by symmetry. Both yield pent-2-ene after correct renumbering. That alkene has an E/Z pair, so there is one alkene connectivity and two geometrical forms.

For 2-methylpropan-2-ol, the three neighbouring methyl groups are equivalent. Removing H from any one gives CH₂=C(CH₃)₂. This alkene has no E/Z pair because one end is CH₂ and the other bears identical methyl groups. A tertiary alcohol therefore does not automatically give several different alkene connectivities.

Work within the AS direct-elimination model: remove OH from one carbon and H from an adjacent carbon without rearranging the skeleton. Detailed rearrangement chemistry is not needed for this outcome. A question specifying “structural isomers” counts connectivities; one specifying all alkene isomers may also need eligible E/Z forms.

The same alcohol can follow different routes

With acidified dichromate, a suitable alcohol is oxidised: the C–O bonding changes towards a carbonyl. With a concentrated acid catalyst and heat, dehydration removes water and forms C=C. With halide plus acid, substitution replaces OH by a halogen. The word “acid” alone therefore does not identify which transformation is intended.

For ethanol, controlled oxidation produces ethanal, dehydration produces ethene and reaction with HBr produces bromoethane. Write the intended product first, then select the complete reagent set and conditions. Adding all reagents you remember can introduce conflicting chemistry rather than strengthen the answer.

The alcohol-to-haloalkane route needs acidic conditions, whereas hydrolysis of the haloalkane back to alcohol uses aqueous alkali. Combining acid with that hydroxide preparation simply consumes the nucleophile through neutralisation.

Worked combustion: include the fuel’s own oxygen

A 0.460 g sample of ethanol, M = 46.0 g mol⁻¹, burns completely. n(ethanol) = 0.0100 mol. From CH₃CH₂OH + 3O₂ → 2CO₂ + 3H₂O, it consumes 0.0300 mol O₂ and produces 0.0200 mol CO₂.

At RTP with molar gas volume 24.0 dm³ mol⁻¹, the oxygen volume is 0.720 dm³ and the carbon dioxide volume is 0.480 dm³. The equation’s three moles of water per mole of ethanol do not justify treating the water as a gas at RTP. State phases and conditions before applying volume ratios.

For a calorimetry calculation, the amount burned is found from the burner’s loss of mass, not its starting mass or the mass of water heated. The Energetics lesson applies that amount to q = mcΔT and explains why a simple flame calorimeter usually captures less energy than the fuel releases.

Quick checks

Original Finesse questions. Reveal the indicative worked solutions after attempting each question; these are not official OCR A mark allocations.

Sources