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Hydrolysis of 3° alkyl halide by aq. NaOH; solvent

Short answer

Hydrolysis of tert-alkyl halide proceeds by SN1: slow ionization to planar carbocation then fast OH− attack gives alcohol; rate = k[RX]. Polar protic solvents (water, alchols) favor SN1 by stabilizing carbocation and leaving group through solvation.

Explain the reaction mechanism for the hydrolysis of 3° alkyl halide by aqueous NaOH. What types of solvent favors the reaction mechanism?

Hydrolysis of a 3° alkyl halide by aqueous NaOH

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

  • 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.

Solvent

Solvent: polar protic solvents (water, ethanol, methanol, formic acid). They stabilise the carbocation and the halide ion by solvation (ion–dipole interaction and hydrogen bonding with the leaving group), which lowers the energy of the transition state of the slow step.

✓ Answer checked by Claude Opus 5.5 (high effort)

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