The quick table
| Substrate | Strong nucleophile, weak base I⁻, Br⁻, CN⁻, N₃⁻, RS⁻ | Strong base OH⁻, RO⁻ | Bulky strong base t-BuO⁻, LDA | Weak nucleophile and base H₂O, ROH, heat |
|---|---|---|---|---|
| Methyl, CH₃X | SN2 | SN2 | SN2 | Very slow |
| Primary, RCH₂X | SN2 | SN2 (E2 if heated) | E2 | Very slow |
| Secondary, R₂CHX | SN2 (polar aprotic solvent) | Mostly E2 | E2 | SN1 and E1, slowly |
| Tertiary, R₃CX | SN1 / E1 | E2 | E2 | SN1, with E1 when heated |
1. Look at the carbon with the leaving group
SN2 needs the nucleophile to reach the back of that carbon, so it's fastest for methyl and primary, slower for secondary, and doesn't happen at tertiary carbons. SN1 and E1 go through a carbocation, so they need a tertiary (or stabilised) carbon. Primary carbocations are too unstable to form.
2. Is the reagent a nucleophile or a base?
Strong bases (hydroxide, alkoxides) pull off a β-hydrogen, so they push towards E2, especially with secondary and tertiary substrates. Good nucleophiles that are weak bases (iodide, cyanide, azide, thiolates) favour SN2. Bulky bases such as potassium tert-butoxide can't reach the carbon, so they give E2 even with primary substrates.
3. Check the solvent
Polar protic solvents (water, alcohols) stabilise carbocations and the leaving group, which helps SN1 and E1. Polar aprotic solvents (acetone, DMSO, DMF) leave the nucleophile "naked" and reactive, which helps SN2.
4. Heat favours elimination
Elimination makes more molecules than substitution, so its entropy change is more positive. Raising the temperature tips mixtures towards E1 or E2.
Worked examples
Bromoethane + NaOH(aq), warm. Primary, good nucleophile → SN2, giving ethanol.
2-Bromo-2-methylpropane in ethanol, warm. Tertiary, weak nucleophile, protic solvent → SN1 (2-ethoxy-2-methylpropane), with some E1.
2-Bromo-2-methylpropane + NaOEt, heat. Tertiary with a strong base → E2, giving 2-methylpropene.
2-Bromobutane + NaOEt in ethanol, heat. Secondary with a strong base → mainly E2. The more substituted alkene, but-2-ene (mostly E), is the major product: Zaitsev's rule.
2-Bromobutane + KOt-Bu. A bulky base removes the more accessible hydrogen, so the less substituted alkene, but-1-ene, is major: the Hofmann product.
Stereochemistry clues
- SN2 inverts the stereocentre (Walden inversion).
- SN1 at a stereocentre gives a racemic mixture, because the carbocation is flat.
- E2 needs the H and the leaving group anti-periplanar (180° apart).
- SN1 and E1 can include carbocation rearrangements (hydride or methyl shifts).
The mechanisms
See every step of SN1 and E1 in the reaction library.
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