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E² and SN² mechanisms; factor governing SN²

Short answer

E2: concerted β-H removal and leaving group departure, e.g. CH3CH2Br + KOH(alc) → ethene, rate = k[RX][base]. SN2: backside attack, one-step displacement with inversion, rate = k[RX][Nu−], e.g. CH3Br + OH− → CH3OH. A key factor governing SN2 is substrate steric hinderance: CH3 > 1° > 2° > 3°.

Explain the reaction mechanism involved in E² and SN² reactions. Mention the one factor governing SN² reaction.

E2 mechanism

Mechanism: E2 (one step, concerted).

 B:⁻  H—CH₂—CH₂—Br   →  [B···H···CH₂···CH₂···Br]⁻‡  →  B–H + CH₂=CH₂ + Br⁻

The base removes the β-hydrogen while the C–Br bond breaks and the π bond forms at the same time (the H and Br are anti-periplanar). Example: CH₃CH₂Br + KOH(alc) → CH₂=CH₂ + KBr + H₂O

  • Rate = k[RX][Base]: second order (bimolecular).
  • Needs a strong base and a high concentration of it; there is no intermediate.

SN2 mechanism

Mechanism: SN2 (one step, concerted).

HO⁻ + H₃C—Br  →  [HO···CH₃···Br]⁻‡  →  HO—CH₃ + Br⁻
                 (transition state)
 HO⁻  →  C ← Br      backside attack, 180° from Br
         /|\
        H H H        inverted like an umbrella in the wind

Overall: CH₃Br + NaOH(aq) → CH₃OH + NaBr

  • Rate = k[CH₃Br][OH⁻]: second order (bimolecular).
  • The nucleophile attacks from the side opposite to the leaving group; the three H atoms flip over (Walden inversion).
  • Reactivity order: CH₃X > 1° > 2° > 3° (steric hindrance).

Factor governing SN2

The structure of the substrate (steric hindrance) is the main factor: the reactivity falls as CH₃X > 1° > 2° > 3°, because bulky alkyl groups block the backside attack. (Other factors: nucleophile strength, leaving group, solvent.)

✓ Answer checked by Claude Opus 5.5 (high effort)

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