Flashcards · 9620 · AS Topic 20

Alkanes (AS Level) flashcards

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Alkane: A saturated hydrocarbon (contains only C and H, with only single C–C bonds). General formula CₙH₂ₙ₊₂ for chains.
Petroleum (crude oil) and a fraction: Petroleum: a mixture consisting mainly of alkane hydrocarbons. Fraction: part of the distillate with boiling points in a narrow range.
How does fractional distillation of crude oil work?: Heated crude oil vapour enters a column that is hot at the bottom and cool at the top. Each compound condenses where the temperature falls below its boiling point: short chains high up, long chains low down.
Why do longer-chain alkanes have higher boiling points?: More electrons and a bigger contact area, so stronger van der Waals forces between molecules, needing more energy to separate them.

Cracking: Breaking C–C bonds in alkanes to give smaller molecules.
Thermal vs catalytic cracking: conditions and main products: Thermal: high pressure, high temperature; gives a high percentage of alkenes. Catalytic: slight pressure, high temperature, zeolite catalyst; gives motor fuels and aromatic hydrocarbons.
Economic reasons for cracking: Distillation gives more of the heavy fractions than are needed and too little of lighter ones such as motor fuels. Cracking turns low-value surplus into higher-value products, including alkenes for polymers.
Octadecane, C₁₈H₃₈, cracks into hexadecane and one other molecule. Name it.: C₁₈H₃₈ − C₁₆H₃₄ = C₂H₄: ethene.

Products of complete and incomplete combustion of alkanes: Complete (plenty of oxygen): CO₂ and H₂O. Incomplete (limited oxygen): CO and/or C (soot), and H₂O.
Pollutants from the internal combustion engine: NOₓ (N₂ and O₂ from the air combine at high temperature), CO (toxic) and carbon particles from incomplete combustion, and unburned hydrocarbons.
What does a catalytic converter do? Give one equation.: Removes CO, NOₓ and unburned hydrocarbons, forming CO₂, N₂ and H₂O. CO and hydrocarbons are oxidised; NO is reduced. 2CO + 2NO → 2CO₂ + N₂
How is SO₂ removed from power station flue gases?: A base (CaO or CaCO₃) neutralises the acidic gas: CaO + SO₂ → CaSO₃ CaCO₃ + SO₂ → CaSO₃ + CO₂
Balanced equation for complete combustion of methanol: CH₃OH + 1½O₂ → CO₂ + 2H₂O

Conditions and mechanism type for the chlorination of methane: UV light; free-radical substitution.
Initiation step in the chlorination of methane: Cl₂ → 2Cl• (UV light splits the Cl–Cl bond by homolytic fission).
The two propagation steps in the chlorination of methane: CH₄ + Cl• → HCl + •CH₃ •CH₃ + Cl₂ → CH₃Cl + Cl•
Termination steps in the chlorination of methane: Two radicals combine: 2Cl• → Cl₂; Cl• + •CH₃ → CH₃Cl; 2•CH₃ → C₂H₆.
Why does chlorinating methane give a mixture of products?: Cl• can remove H from any chloromethane that still has C–H bonds, giving CH₂Cl₂, CHCl₃ and CCl₄. Excess methane favours CH₃Cl.