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Unimolecular elimination mechanism; difference from E2

Short answer

E1: unimolecular elimination. The C-X bond breaks slowly to give a carbocation; then a base rapidly removes a β-hydrogen, forming an alkene. Rate = k[RX]. E2: bimolecular, concerted; base attacks β-H as X leaves, transition state; rate = k[RX][base]; strong base required; no carbocation.

Write the mechanism of unimolecular elimination reaction. How does it differ from bimolecular elimination reaction?

E1 (unimolecular elimination) 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.

Difference from E2 (bimolecular elimination)

BasisE1E2
StepsTwo steps, carbocation intermediateOne step, concerted
Rate lawRate = k[RX], first orderRate = k[RX][Base], second order
MolecularityUnimolecularBimolecular
Substrate3° > 2° (stable carbocation)3° > 2° > 1°
BaseWeak base is enoughStrong, concentrated base needed
SolventPolar proticLess polar / alcohol
RearrangementPossibleNot possible
StereochemistryNo special geometryAnti-periplanar H and X
Example(CH₃)₃CBr in ethanolCH₃CH₂Br + alc. KOH

✓ Answer checked by Claude Opus 5.5 (high effort)

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