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- 2071 Bhadra · 1+4 marks
- 2068 Bhadra · 1+4 marks
Elimination reactions; E1 vs E2 mechanism
Short answer
Elimination reactions remove H and X from an alkyl halide to form an alkene. E1 is two-step: carbocation formation then base attack; rate = k[RX]; weak base suffices; rearrangement possible. E2 is one-step concerted; rate = k[RX][base]; needs strong base; anti-periplanar geometry; no rearrangement. Examples: (CH₃)₃CBr (E1) vs CH₃CH₂Br (E2).
What are elimination reactions? Write the differences between E₁ and E₂ reaction mechanism taking suitable example.
Elimination reactions
An elimination reaction is a reaction in which two atoms or groups are removed from adjacent carbon atoms of a molecule, forming a double (or triple) bond, without being replaced by other groups. Dehydrohalogenation of an alkyl halide is the elimination of H and X (as HX) using a strong base such as alcoholic KOH to form an alkene.
Difference between E1 and E2
| Basis | E1 | E2 |
|---|---|---|
| Steps | Two steps, carbocation intermediate | One step, concerted |
| Rate law | Rate = k[RX], first order | Rate = k[RX][Base], second order |
| Molecularity | Unimolecular | Bimolecular |
| Substrate | 3° > 2° (stable carbocation) | 3° > 2° > 1° |
| Base | Weak base is enough | Strong, concentrated base needed |
| Solvent | Polar protic | Less polar / alcohol |
| Rearrangement | Possible | Not possible |
| Stereochemistry | No special geometry | Anti-periplanar H and X |
| Example | (CH₃)₃CBr in ethanol | CH₃CH₂Br + alc. KOH |
Examples
- E1: (CH₃)₃C–Br → (CH₃)₃C⁺ + Br⁻ (slow); then the base removes a β-hydrogen: (CH₃)₃C⁺ + OH⁻ → (CH₃)₂C=CH₂ + H₂O (fast).
- E2: HO⁻ + H–CH₂–CH₂–Br → [HO···H···CH₂···CH₂···Br]⁻‡ → H₂O + CH₂=CH₂ + Br⁻ (one step).
✓ Answer checked by Claude Opus 5.5 (high effort)
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