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Tertiary alkyl halide with aq. NaOH / alc. KOH

Short answer

With aqueous NaOH, tertiary alkyl halide undergoes SN1: carbocation formation then OH⁻ attack, giving tert-butyl alcohol. With alcoholic KOH, E1 elimination occurs: the same carbocation loses a β-proton to OH⁻, forming 2-methylpropene. Aqueous conditions favor substitution; alcoholic conditions favor elimination.

Write the mechanism of the reaction of tertiary alkyl halide with a) Aqueous NaOH b) Alcoholic KOH

a) With aqueous NaOH: substitution (SN1)

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 + NaOH(aq) → (CH₃)₃C–OH + NaBr

Product: tert-butyl alcohol (substitution). Aqueous NaOH, with water as solvent, favours substitution.

b) With alcoholic KOH: elimination (E1)

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)

Product: alkene. Alcoholic KOH favours elimination because the alkoxide/hydroxide acts as a base rather than a nucleophile, and in the absence of much water the proton loss from the carbocation dominates.

✓ Answer checked by Claude Opus 5.5 (high effort)

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