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Mechanism of E1 and E2 reactions

Short answer

E1 proceeds via carbocation intermediate, e.g. (CH3)3CBr → (CH3)2C=CH2, rate = k[RX]. E2 is concerted with simultaneous β-H removal and C–Br clevage, e.g. CH3CH2Br → CH2=CH2, rate = k[RX][base].

Describe the reaction mechanism of E₁ and E₂ with the suitable example.

E1 mechanism

Mechanism: E1 (two steps).

Step 1 (slow): ionisation to carbocation
 (CH₃)₃C—Br  ---slow--->  (CH₃)₃C⁺ + Br⁻

Step 2 (fast): base removes a β-hydrogen
 OH⁻ + H—CH₂—C⁺(CH₃)₂ ---fast---> H₂O + CH₂=C(CH₃)₂

Overall: (CH₃)₃C–Br + KOH(alc) → (CH₃)₂C=CH₂ + KBr + H₂O (2-methylpropene)

  • Rate = k[RX]: first order (unimolecular), as only the halide takes part in the slow step.
  • Carbocation intermediate, so rearrangement is possible; competes with SN1.

E2 mechanism

Mechanism: E2 (one step, concerted).

 B:⁻  H—CH₂—CH₂—Br   →  [B···H···CH₂···CH₂···Br]⁻‡  →  B–H + CH₂=CH₂ + Br⁻

The base removes the β-hydrogen while the C–Br bond breaks and the π bond forms at the same time (the H and Br are anti-periplanar). Example: CH₃CH₂Br + KOH(alc) → CH₂=CH₂ + KBr + H₂O

  • Rate = k[RX][Base]: second order (bimolecular).
  • Needs a strong base and a high concentration of it; there is no intermediate.

✓ Answer checked by Claude Opus 5.5 (high effort)

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