Flashcards · 9701 · AS Topic 14

Hydrocarbons (AS Level) flashcards

11 cards on AS Topic 14, subtopics 14.1 to 14.2. Pick a subtopic or study them all.

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Initiation step for the chlorination of methane: Cl₂ → 2Cl• in UV light: homolytic fission of the Cl–Cl bond.
Propagation steps for the chlorination of methane: Cl• + CH₄ → •CH₃ + HCl •CH₃ + Cl₂ → CH₃Cl + Cl• Each step uses a radical and makes one, so the chain continues.
Termination steps for the chlorination of methane: Any two radicals combine: Cl• + Cl• → Cl₂ •CH₃ + Cl• → CH₃Cl •CH₃ + •CH₃ → C₂H₆
Why does free-radical substitution give a mixture of products?: It is poorly controlled: Cl• can attack the product too, giving further substitution (CH₂Cl₂, CHCl₃, CCl₄).
Why are alkanes generally unreactive?: The C–C and C–H bonds are strong and almost non-polar, so they do not attract nucleophiles or electrophiles.

Mechanism of electrophilic addition of Br₂ to an alkene (outline): The C=C (high electron density) induces a dipole in Br₂. The π electrons attack the δ+ bromine atom, Br–Br breaks heterolytically, a carbocation forms, then Br⁻ attacks the carbocation.
Test for a C=C bond: Shake with bromine water: orange-brown → colourless.
Conditions: alkene → alkane, and alkene → alcohol: Alkane: H₂, nickel catalyst, about 150 °C. Alcohol: steam, H₃PO₄ catalyst, about 300 °C and 60 atm.
Alkene with cold dilute acidified KMnO₄: Gives a diol; purple → colourless.
Alkene with hot concentrated acidified KMnO₄: what the fragments show: The C=C is broken. =CH₂ → CO₂ + H₂O; =CHR → carboxylic acid; =CR₂ → ketone. The fragments locate the double bond.
Markovnikov's rule and the reason behind it: When HBr adds to an unsymmetrical alkene, H adds to the carbon that already has more hydrogens, because this forms the more stable carbocation (tertiary > secondary > primary; alkyl groups are electron-donating).