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DAT Organic Chemistry · Review sheet · Free sample

SN1, SN2, E1, E2

Chapter: Substitution and Elimination (SN1, SN2, E1, E2)

Updated 2026-10-19

Mechanism comparison

FeatureSN2SN1E2E1
Rate lawk[RX][Nu]k[RX]k[RX][Base]k[RX]
StepsOne, concertedTwo, carbocationOne, concertedTwo, carbocation
Best substrateMethyl > 1° > 2° (3° no)3° > 2° (benzylic, allylic)3° > 2° > 1°3° > 2°
Nucleophile / baseStrong nucleophileWeak (often the solvent)Strong baseWeak base, heat
SolventPolar aprotic (DMSO, DMF, acetone, CH₃CN)Polar protic (H₂O, ROH)VariesPolar protic
StereochemistryInversion (backside attack)Racemization (mostly)Anti-periplanar H and LGMixture
RearrangementNoYes (hydride/methyl shifts)NoYes

Decision guide

  • Methyl or 1° + strong nucleophile → SN2. 1° + bulky strong base (t-BuOK) → E2.
  • 2° + strong base (HO⁻, RO⁻) → mostly E2. 2° + good nucleophile that is a weak base (I⁻, Br⁻, CN⁻, N₃⁻, RS⁻) in aprotic solvent → SN2.
  • 2° or 3° + weak nucleophile in protic solvent → SN1/E1 (heat favors E1).
  • 3° + strong base → E2. 3° never does SN2.
  • Zaitsev: E1/E2 give the more substituted alkene. Bulky bases (t-BuOK, LDA) give the less substituted (Hofmann) alkene.

Leaving groups and nucleophiles

  • Good leaving groups are weak bases: I⁻ > Br⁻ > Cl⁻ ≫ F⁻; tosylate (OTs), mesylate (OMs), triflate are excellent. OH⁻ is poor unless protonated (–OH₂⁺) or converted to OTs.
  • In polar protic solvents nucleophilicity rises DOWN a group (I⁻ > Br⁻ > Cl⁻ > F⁻) because small ions are tightly solvated.
  • In polar aprotic solvents nucleophilicity follows basicity (F⁻ > Cl⁻ > Br⁻ > I⁻).
  • Neopentyl halides are 1° but react very slowly by SN2 (β-branching blocks the back side).
  • Carbocation stability: 3° > 2° > 1° > methyl; allylic and benzylic cations are resonance-stabilized.

⚠️ Common traps

  • Predicting SN2 at a tertiary carbon.
  • Forgetting carbocation rearrangements in SN1/E1.
  • Calling every strong base a good SN2 nucleophile (bulky bases eliminate).
  • Ignoring the anti-periplanar requirement in cyclohexane E2 (H and leaving group must both be axial).

Mini-quiz

Pick an answer to see right away whether it is correct and why.

  1. (R)-2-Bromobutane reacts with NaCN in DMSO. What is the stereochemical outcome?
  2. tert-Butyl bromide is warmed in aqueous ethanol with no added base. Which pathways operate?
  3. 2-Bromo-2-methylbutane is treated with potassium tert-butoxide. What is the major product?
  4. Which is the best leaving group?
  5. Which rate law fits an SN2 reaction?
0 of 5 answered

Answer key: 1 B · 2 C · 3 B · 4 A · 5 C