9701 · 15.1
Halogenoalkanes
Halogenoalkanes have a polar carbon-halogen bond, making the carbon atom an easy target for attack. Depending on the conditions and the molecule's shape, this leads to either substitution (swapping the halogen) or elimination (forming a double bond).
Need to know
What you need to know
- **With aqueous alkali (e.g., NaOH(aq))**: Forms an alcohol. Reagent: NaOH(aq). Conditions: Warm. Nucleophile: $OH^−$. Example: $CH_3CH_2Br + OH^− \rightarrow CH_3CH_2OH + Br^−$.
- **With ethanolic cyanide (e.g., KCN)**: Forms a nitrile, extending the carbon chain by one carbon. Reagent: KCN in ethanol. Conditions: Reflux. Nucleophile: $CN^−$. Example: $CH_3CH_2Br + CN^− \rightarrow CH_3CH_2CN + Br^−$.
- **With ethanolic ammonia (NH₃)**: Forms a primary amine. Reagent: Excess concentrated NH₃ in ethanol. Conditions: Heat in a sealed tube. Nucleophile: $NH_3$. Example: $CH_3CH_2Br + NH_3 \rightarrow CH_3CH_2NH_2 + HBr$. (The HBr then reacts with more ammonia: $HBr + NH_3 \rightarrow NH_4Br$).
Explanation
The Halogenoalkane Reaction Maze
- Halogenoalkanes: C–X polar bond — C δ+, X δ−. | Sim hint: View tetrahedral shape around the C–X bond.
- SN2: one step, backside attack, inverted configuration. | Sim hint: Primary halogenoalkanes favour SN2 (good leaving group I⁻ > Br⁻ > Cl⁻).
- SN1: two steps via carbocation — racemisation possible. | Sim hint: Tertiary halogenoalkanes favour SN1 (stable carbocation).
- Elimination competes with substitution — especially with hot ethanolic OH⁻. | Sim hint: Compare substitution vs elimination conditions in exam answers.