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3° alkyl halide with alcoholic KOH; solvent

Short answer

Tertiary alkyl halide with alcoholic KOH proceeds by E1: slow ionization to carbocation, then fast loss of a β-proton to give alkene. Polar protic solvents like ethanol stabilize the carbocation and favor E1.

Explain the reaction mechanism for the reaction between 3° alkyl halide and alcoholic potassium hydroxide. What solvent favors the reaction mechanism?

Mechanism: E1 elimination

A tertiary halide, e.g. tert-butyl bromide, with alcoholic KOH gives an alkene.

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. (With a high concentration of strong base, E2 can also compete, but the usual textbook mechanism for tertiary halides is E1.)

Solvent

A polar protic solvent such as ethanol or water favours E1. It stabilises the carbocation and the leaving halide by solvation, so the ionisation step becomes faster. Alcohol (less water) favours elimination over substitution.

✓ Answer checked by Claude Opus 5.5 (high effort)

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