Flashcards · 9620 · AS Topic 21

Halogenoalkanes (AS Level) flashcards

10 cards on AS Topic 21, subtopics 3.3.3.1 to 3.3.3.2. Pick a subtopic or study them all.

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Nucleophile: An electron pair donor that attacks an electron-deficient (δ+) atom.
Why is the carbon of a C–X bond δ+?: The halogen is more electronegative than carbon, so the bonding pair is pulled towards X.
Halogenoalkane with OH⁻, CN⁻ and NH₃: conditions and products: OH⁻: warm aqueous NaOH/KOH → alcohol. CN⁻: KCN in ethanol and water, heat under reflux → nitrile (one carbon longer). NH₃: excess, in ethanol, heated in a sealed vessel → primary amine.
Order of rate of substitution of halogenoalkanes, and why: R–F < R–Cl < R–Br < R–I. The C–I bond has the lowest bond enthalpy, so it needs least energy to break. Polarity cannot explain this: C–F is most polar but slowest.
Nucleophilic substitution with OH⁻: the curly arrows: Arrow from a lone pair on the O of OH⁻ to the δ+ carbon; arrow from the C–X bond to X. Products: alcohol and X⁻.
Why is excess ammonia used to make a primary amine from a halogenoalkane?: The amine product is also a nucleophile. Excess NH₃ makes it more likely that the halogenoalkane reacts with ammonia rather than with the amine.

Conditions for elimination from a halogenoalkane: KOH (or NaOH) dissolved in ethanol, heated. Product: an alkene, plus H₂O and X⁻.
OH⁻ as a nucleophile vs OH⁻ as a base: Nucleophile: bonds to carbon (substitution, gives an alcohol; favoured by warm aqueous conditions). Base: accepts H⁺ from a carbon next to the C–X carbon (elimination, gives an alkene; favoured by hot ethanolic conditions).
Elimination mechanism: the three curly arrows: Lone pair on the O of OH⁻ to an H on the carbon next to C–X; C–H bond to between the two carbons (forming C=C); C–X bond to X.
Products of 2-bromopropane with KOH: Substitution: propan-2-ol. Elimination: propene. Both reactions happen at the same time (concurrently).