Halogenoalkanes: properties, reactions, mechanisms
Halogenoalkanes: Properties, Reactions & Mechanisms
Objective: Understand the key properties of halogenoalkanes, the major substitution and elimination reactions they undergo, and the underlying mechanisms. Use analogies to make the concepts memorable.
1. Properties of Halogenoalkanes
- General formula: R–X where X = F, Cl, Br, I.
- Polarity: The C–X bond is polar (δ⁺ on C, δ⁻ on X) → makes them good electrophiles.
- Reactivity trend: I > Br > Cl > F (due to bond strength & polarizability).
- Solubility: Often soluble in organic solvents, sparingly soluble in water.
- Analogy: Think of the halogen as a “hand‑off” point where a new partner (nucleophile or base) can grab the carbon.
2. Major Reactions
2.1 Substitution Reactions
Two main types: SN1 (unimolecular) and SN2 (bimolecular).
- SN2 – “Back‑side attack” in one step. Key features:
- Occurs with primary or secondary halogenoalkanes.
- Base or nucleophile attacks from the side opposite the leaving group.
- Inversion of configuration (Walden inversion).
- Rate ∝ [Nu]·[Halogenoalkane].
- SN1 – “Carbocation intermediate” in two steps. Key features:
- Favoured by tertiary halogenoalkanes.
- First, C–X bond breaks → formation of a stable carbocation.
- Second, nucleophile attacks the carbocation (often leading to racemization).
- Rate ∝ [Halogenoalkane] (independent of nucleophile concentration).
2.2 Elimination Reactions
Two main types: E2 (bimolecular) and E1 (unimolecular).
- E2 – “Concerted” removal of H and X in one step.
- Requires a strong base.
- Occurs with primary or secondary halogenoalkanes.
- Produces an alkene with anti‑syn (E2) geometry.
- E1 – “Carbocation intermediate” similar to SN1.
- Favoured by tertiary halogenoalkanes.
- First, C–X bond breaks → carbocation.
- Second, base removes a β‑hydrogen → alkene.
3. Mechanistic Details
3.1 SN2 Mechanism (Step‑by‑Step)
- Base/Nucleophile approaches the electrophilic carbon from the side opposite the leaving group.
- Simultaneous bond formation (Nu–C) and bond breaking (C–X).
- Transition state: pentavalent carbon, partial bonds.
- Product: Inverted configuration.
3.2 SN1 Mechanism (Step‑by‑Step)
- Leaving group departs, forming a stable carbocation.
- Carbocation is attacked by the nucleophile from either side.
- Result: mixture of stereoisomers (racemization).
3.3 E2 Mechanism (Step‑by‑Step)
- Base abstracts a β‑hydrogen while the leaving group departs.
- Simultaneous formation of C=C double bond.
- Transition state: anti‑syn geometry (E2).
3.4 E1 Mechanism (Step‑by‑Step)
- Leaving group departs → carbocation.
- Base removes β‑hydrogen → alkene.
- Possible rearrangements of the carbocation before elimination.
4. Reaction Summary Table
| Reaction Type | Key Conditions | Typical Substrate | Outcome |
|---|---|---|---|
| SN2 | Strong nucleophile, polar aprotic solvent | Primary / Secondary | Inverted product |
| SN1 | Weak nucleophile, polar protic solvent | Tertiary | Racemized product |
| E2 | Strong base, polar aprotic solvent | Primary / Secondary | Alkene (E2) |
| E1 | Weak base, polar protic solvent | Tertiary | Alkene (E1) |
5. Exam Tips (💡)
Remember:
- Identify the substrate type (primary, secondary, tertiary) to decide between SN2/E2 or SN1/E1.
- Check the solvent: polar aprotic → SN2/E2; polar protic → SN1/E1.
- Look for the leaving group quality: I > Br > Cl > F.
- For mechanisms, sketch the transition state: show bond breaking/ forming arrows.
- Use the mnemonic “S N 1 & 2, E 1 & 2” to recall the reaction types.
6. Quick Quiz (🧪)
- Which reaction is favoured for a tertiary halogenoalkane in a polar protic solvent?
- What is the stereochemical outcome of an SN2 reaction?
- Why does a strong base favour an E2 mechanism over an SN2?
Revision
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