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3° alkyl halide with aqueous caustic potash

Short answer

Tertiary alkyl halide reacts with aqueous KOH via SN1: rate-determining formation of a planar tert‑butyl carbocation, then rapid attack by hydroxide to give tertiary alcohol with racemization; water (polar protic) stabilizes the ions.

Describe the mechanism involved in the reaction between a tertiary alkylhalide and aqueous caustic potash.

A tertiary alkyl halide reacts with aqueous KOH by the SN1 mechanism to give a tertiary alcohol.

Mechanism: SN1 (two steps).

Step 1 (slow, rate determining): ionisation
   CH₃                   CH₃
CH₃—C—Br   ---slow--->  CH₃—C⁺  +  Br⁻
   CH₃                   CH₃       (planar carbocation)

Step 2 (fast): nucleophile attacks the carbocation
   CH₃                      CH₃
CH₃—C⁺  + OH⁻ ---fast--> CH₃—C—OH
   CH₃                      CH₃

Overall: (CH₃)₃C–Br + KOH(aq) → (CH₃)₃C–OH + KBr

  • Rate = k[(CH₃)₃C–Br]: first order; independent of [OH⁻].
  • The tertiary carbocation is stabilised by hyperconjugation and the +I effect of three methyl groups, so tertiary halides react by SN1.
  • Water (polar protic solvent) stabilises the ions and favours the reaction.

✓ Answer checked by Claude Opus 5.5 (high effort)

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