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Elimination reaction; dehydrohalogenation of tertiary halide

Short answer

An elimination reaction removes two atoms or groups from adjacent carbons, forming a double bond. Dehydrohalogenation of a tertiary alkyl halide proceeds by E1: slow ionization to a carbocation, then fast base removal of a β-hydrogen yielding an alkene. Rate = k[RX] (first-order).

What is elimination reaction? Explain the reaction mechanism for the dehydrohalogenation of tertiary alkyl halide.

Elimination reaction

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.

Dehydrohalogenation of a tertiary alkyl halide

Tertiary halides eliminate by the E1 mechanism, in a polar protic solvent such as ethanol.

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.

✓ Answer checked by Claude Opus 5.5 (high effort)

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