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SN¹ retention/inversion vs SN² inversion; mechanism

Short answer

SN1 gives both retention and inversion because the planar carbocation can be attacked from either face; SN2 undergoes exclusive inversion due to backside attack. (CH3)3C–Br + NaOH follows SN1: slow ionization to (CH3)3C+ then fast OH– attack forming (CH3)3COH.

Explain why SN¹ reaction gives both retention and inversion isomers but SN² gives only inversion isomer. Write the mechanism of given chemical reaction. (CH₃)₃C−Br + NaOH → (CH₃)₃COH + NaBr

SN1 gives both retention and inversion, SN2 only inversion

  • SN1: the first step gives a planar carbocation, which is attacked by the nucleophile from either face with almost equal probability: from the rear gives inversion and from the front gives retention. The product is a mixture, nearly racemic, with slightly more inversion because the leaving group shields the front.
  • SN2: a single step in which the nucleophile attacks only from the back of the carbon (opposite to the leaving group). The three other groups flip like an umbrella, so only the inverted product forms (Walden inversion). No planar intermediate exists.

Mechanism of (CH₃)₃C–Br + NaOH → (CH₃)₃C–OH + NaBr

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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