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SN reactions; hydrolysis of 3° alkyl halide

Short answer

SN reactions involve a nucleophile replacing a leaving group. Hydrolysis of a 3° alkyl halide by aqueous NaOH follows SN1: slow formation of a planar carbocation, then fast nucleophilic attack by OH⁻ to give tert-butyl alcohol. Rate = k[RX], independent of [OH⁻].

What do you mean by SN reactions? Explain reaction mechanism for the hydrolysis of 3° alkyl halide by aqueous sodium hydroxide.

SN reactions

A nucleophilic substitution (SN) reaction is a reaction in which a nucleophile (an electron-rich species with a lone pair, e.g. OH⁻, CN⁻, NH₃) replaces the leaving group (e.g. halide) of a substrate. R–X + Nu⁻ → R–Nu + X⁻. The two mechanisms are SN1 (unimolecular) and SN2 (bimolecular).

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.

✓ Answer checked by Claude Opus 5.5 (high effort)

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