All four mechanisms compete for one substrate and reagent — run the levers below in order and let each narrow the field.
The substrate ladder: methyl → 3°, SN2 fades out, cation pathways switch on.
1. Start With the Substrate: Methyl, 1°, 2°, or 3°
Classify the carbon bearing the leaving group first — its substitution class rules out half the mechanisms before the reagent matters.
- Methyl / 1°: no cation forms, so default SN2 — E2 only under a strong bulky base.
- Tertiary (3°): backside attack blocked, so no SN2; weak reagents give SN1/E1, a strong base gives E2.
- Secondary (2°): all four possible — levers 2–4 decide.
Two extremes: methyl is open to SN2; 3° shuts it down but forms a cation readily.
2. Then Read the Reagent: Strong or Weak, Nucleophile or Base, Bulky or Small
Score the reagent on two axes — nucleophile strength (attacks carbon) versus base strength (grabs a β-proton) — plus whether it is bulky or small.
- Strong Nu, weak base (I–, N3–, CN–) → SN2.
- Strong bulky base (t-BuO–, LDA, DBU) → E2 (Hofmann).
- Strong small base + good Nu (HO–, EtO–) → SN2 or E2.
- Weak Nu/base (H2O, ROH) → SN1/E1 only.
1° + strong Nu, weak base → clean SN2.
3. Decide Substitution vs Elimination: Good Nucleophile or Good Base?
A strong nucleophile substitutes; a strong bulky base eliminates — and on a 3° center it can only give the alkene by E2.
3° + strong base → E2; SN2 is impossible, the base outruns SN1.
4. Turn Up the Heat: Temperature Tips Toward Elimination
Elimination gains entropy (one molecule → two), so heat always shifts the competition toward elimination.
3° + weak Nu, warm water → SN1 alcohol (shown) + E1 alkene; more heat = more alkene.
5. Check the Solvent: Aprotic Pushes SN2, Protic Pushes SN1/E1
Polar aprotic solvents (DMSO, DMF) leave the nucleophile "naked" for SN2; polar protic ones (water, alcohols) stabilize ions for SN1/E1.
Same 2° halide: protic ethanol nudges ionization; a naked Nu in aprotic solvent forces SN2.
6. Worked Decisions: Putting the Four Levers Together
Each case starts at the substrate and stops as soon as one mechanism is forced.
- 1° + NaN3 in DMSO: strong Nu, aprotic → SN2.
- 3° + KOtBu, warm: SN2 blocked, bulky base → E2 (Hofmann).
- 3° + H2O, heat: weak Nu, protic, warm → SN1 + E1, alkene-rich.
- 2° + NaOEt, heat: strong small base + heat → mostly E2.
Bulky base: tert-butoxide can't reach the carbon, so it goes E2.
7. Summary
Substrate (methyl/1° → SN2, 3° → SN1/E1/E2, 2° defers) · reagent (strong Nu → SN2, bulky base → E2, weak → SN1/E1) · heat → elimination · solvent (aprotic → SN2, protic → SN1/E1) · traps: SN2 inverts, cations rearrange, E2 needs anti-periplanar geometry, SN1/E1 travel together.
Quiz yourself
Tap a question to reveal the answer — free, no login.
The substrate. Its substitution class rules whole mechanisms in or out before the reagent matters: methyl/1° kills SN1/E1, while 3° kills SN2. Only after that do you read the reagent, heat, and solvent.
SN2. A 2° center allows all four, but azide is a strong nucleophile and weak base, and the aprotic solvent leaves it naked and reactive — substitution beats elimination.
E2 to give isobutylene. The 3° center blocks SN2, and tert-butoxide is a strong, bulky base too big to substitute — so it strips a β-hydrogen. Heat reinforces elimination.
They share the same rate-determining step: loss of the leaving group to form a carbocation. Once the cation exists, a nucleophile can capture it (SN1) or a base can remove a β-proton (E1). You get a mixture, and adding heat shifts it toward the E1 alkene.
Draw this on the whiteboard
Open the OChem Board whiteboard — benzene rings, curved arrows, wedge/dash bonds and a clickable periodic table built in. No account needed.