How to Choose Between SN1, SN2, E1, and E2
Deciding which mechanism dominates is one of the most common exam challenges in sophomore organic chemistry. The answer depends on four interacting factors: the substrate (what carbon the leaving group is on), the nucleophile or base (its strength and size), the solvent (polar protic vs. polar aprotic vs. nonpolar), and the temperature. No single factor determines the outcome in isolation — you have to evaluate all four together.
The Four Pathways: A Quick Reference
| Mechanism | Steps | Intermediate | Regiochemistry | Stereochemistry | Rearrangements? | Best Substrate |
|---|---|---|---|---|---|---|
| SN2 | 1 (concerted) | None | N/A | Inversion (Walden) | No | Methyl, 1°, (2° slow) |
| SN1 | 2 | Carbocation | N/A | Racemization | Yes | 3°, (2°, allylic/benzylic) |
| E2 | 1 (concerted) | None | Hofmann (bulky base); Zaitsev (small base) | Anti-periplanar required | No | Any with strong bulky base |
| E1 | 2 | Carbocation | Zaitsev | Not stereospecific | Yes | 3°, (2°) + heat + polar protic |
The Four Decision Factors Explained
1. Substrate (Most Important Factor)
The carbon bearing the leaving group determines what is structurally possible.
- Methyl (CH₃–X): Only SN2 is possible. No carbocation stability rules out SN1/E1, and no β-hydrogens rule out E2.
- Primary (1°): SN2 strongly preferred with a strong nucleophile. E2 with a bulky base. SN1/E1 essentially impossible — primary carbocations are too unstable.
- Secondary (2°): The ambiguous case. Can go SN2, SN1, E2, or E1 depending on the other three factors. This is where the decision tree matters most.
- Tertiary (3°): SN2 is impossible (too hindered). Goes E2 with any strong base, or SN1/E1 with weak nucleophile in polar protic solvent.
- Allylic / Benzylic: Both SN1 and SN2 are accessible due to resonance stabilization. Conditions determine which dominates.
2. Nucleophile / Base Strength and Bulk
- Strong, non-bulky nucleophile (NaOH, NaCN, NaI): Favors SN2 at accessible carbons. E2 competes at high temperature.
- Strong, bulky base (KOtBu, LDA): Cannot reach a hindered carbon for SN2. Strongly favors E2 at all substrate types. This is the clearest single indicator of elimination.
- Weak nucleophile / base (H₂O, alcohols): Cannot drive SN2 or E2. Favors SN1 (or E1 at high temperature) at substrates that can form stable carbocations.
3. Solvent
- Polar aprotic (DMSO, DMF, acetone, acetonitrile): Dissolves salts but does not solvate the nucleophile. Leaves the nucleophile bare and highly reactive — strongly favors SN2. Does not stabilize carbocations well, so disfavors SN1/E1.
- Polar protic (water, ethanol, methanol, acetic acid): Solvates and shields the nucleophile (slows SN2) while stabilizing carbocations and the leaving group through hydrogen bonding — favors SN1/E1 at substrates capable of forming stable cations.
- Nonpolar / weakly polar (ether, THF, CH₂Cl₂): Neither stabilizes carbocations nor dramatically enhances nucleophilicity. SN2 can proceed if the nucleophile is strong. SN1/E1 are disfavored.
4. Temperature
Increasing temperature always shifts the competition toward elimination. Substitution reactions have a negative ΔS (two particles become one), while elimination has a positive ΔS (one particle becomes two). At high temperature, the TΔS term dominates and elimination products are entropically favored. As a rule of thumb: heat favors elimination; room temperature favors substitution, all else being equal.
Common Exam Scenarios
- KOtBu + 2° substrate → E2. Bulky base is the override — it always pushes toward elimination regardless of temperature.
- NaI / acetone + 1° substrate → SN2. Polar aprotic + strong nucleophile + accessible carbon = textbook SN2.
- H₂O / EtOH + 3° substrate + heat → E1. Weak nucleophile + stable carbocation + high temperature = E1.
- NaOH / DMSO + 3° substrate → E2. Even a non-bulky strong base forces E2 at tertiary because SN2 is impossible.
- NaCN / DMSO + 2° substrate, 25°C → SN2 (slower than at 1°, but SN2 wins in aprotic solvent at low temperature).
Frequently Asked Questions
How do I know if a reaction is SN1 or SN2?
SN2 is favored by primary substrates, strong non-bulky nucleophiles, polar aprotic solvents, and room temperature. SN1 is favored by tertiary (or secondary) substrates, weak nucleophiles, and polar protic solvents that stabilize the carbocation intermediate. Secondary substrates can go either way depending on the other conditions.
When does elimination occur instead of substitution?
Elimination is favored when a strong, bulky base is used (KOtBu, LDA) — this promotes E2 regardless of substrate. Elevated temperature also shifts the balance toward elimination. E1 occurs alongside SN1 at tertiary or secondary substrates with weak bases in polar protic solvents at high temperature.
What is the best solvent for SN2?
Polar aprotic solvents are best for SN2: DMSO, DMF, acetone, and acetonitrile. These solvents dissolve ionic reagents but do not solvate the nucleophile, leaving it reactive. Polar protic solvents like water and ethanol slow SN2 by hydrogen-bonding to the nucleophile.
Can tertiary substrates undergo SN2?
No. Tertiary substrates are too sterically hindered for backside attack. SN2 is essentially impossible at tertiary carbons. Tertiary substrates react by SN1 with weak nucleophiles in polar protic solvents, or E2 with any strong base.
What is the difference between E1 and E2?
E2 is a concerted one-step mechanism: the base abstracts a β-hydrogen at the same time the leaving group departs. It requires a strong base and anti-periplanar geometry of H and the leaving group. E1 is a two-step mechanism: the leaving group ionizes first to form a carbocation, then a base removes a β-hydrogen. E1 requires a stable carbocation and polar protic solvent.
What does Zaitsev's rule predict?
Zaitsev's rule predicts that the more substituted alkene is the major elimination product when using a small, strong base. The more substituted alkene is more thermodynamically stable due to hyperconjugation from adjacent C–H σ-bonds. With a very bulky base (like KOtBu), the less hindered β-hydrogen is abstracted instead, giving the less substituted (Hofmann) product.
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