SN1/SN2/E1/E2
Decision Framework
Master the logic behind choosing the correct reaction pathway based on substrate, nucleophile/base, and solvent.
Organic Chemistry 1
Four Competing Pathways
When a substrate bearing a leaving group meets a nucleophile or base, four possible reactions compete. Two are substitution (the nucleophile replaces the leaving group) and two are elimination (a base removes a proton to form a double bond).
SN2
One-step substitution: backside attack, inversion of configuration
SN1
Two-step substitution via carbocation: racemization
E2
One-step elimination: concerted, anti-periplanar geometry
E1
Two-step elimination via carbocation: Zaitsev product
Three Variables Control Everything
You do not need to memorize every combination. Instead, evaluate three factors in order of importance:
1
Substrate structure: Is the carbon bearing the leaving group methyl, primary (\(1^\circ\)), secondary (\(2^\circ\)), or tertiary (\(3^\circ\))?
2
Nucleophile/base strength: Is it a strong nucleophile, a strong bulky base, or a weak nucleophile/base?
3
Solvent: Is the solvent polar aprotic (favors SN2) or polar protic (favors SN1/E1)?
Substrate: The First Filter
Nucleophile vs Base
Nucleophilicity describes the ability to attack carbon (substitution). Basicity describes the ability to abstract a proton (elimination). Some species are both; the distinction controls which pathway wins.
Strong Nuc / Weak Base
Favors SN2. Examples: \(\mathrm{I}^-\), \(\mathrm{RS}^-\), \(\mathrm{CN}^-\), \(\mathrm{N}_3^-\)
Strong Base / Strong Nuc
Favors E2 on bulky substrates, SN2 on methyl/\(1^\circ\). Examples: \(\mathrm{OH}^-\), \(\mathrm{RO}^-\)
Bulky Strong Base
Favors E2 almost exclusively. Examples: \(t\text{-BuO}^-\), LDA, DBU
Solvent: Aprotic vs Protic
| Polar Aprotic | Polar Protic | |
|---|---|---|
| Examples | \(\text{DMSO}\), \(\text{DMF}\), \(\text{acetone}\), \(\mathrm{CH_3CN}\) | \(\mathrm{H_2O}\), \(\mathrm{MeOH}\), \(\mathrm{EtOH}\), \(\mathrm{AcOH}\) |
| H-bond donors? | No | Yes |
| Effect on Nuc | Nucleophile stays free and reactive | Nucleophile is solvated, less reactive |
| Favors | SN2 | SN1 / E1 |
The Master Decision Flowchart
SN2 vs SN1: Key Differences
| SN2 | SN1 | |
|---|---|---|
| Mechanism | One step (concerted) | Two steps (carbocation) |
| Rate law | \(\text{rate} = k[\text{Sub}][\text{Nuc}]\) | \(\text{rate} = k[\text{Sub}]\) |
| Substrate | Methyl > \(1^\circ\) > \(2^\circ\) | \(3^\circ\) > \(2^\circ\) (never methyl/\(1^\circ\)) |
| Stereochemistry | Inversion | Racemization |
| Solvent | Polar aprotic | Polar protic |
| Nuc/base | Strong nucleophile | Weak nucleophile |
E2 vs E1: Key Differences
| E2 | E1 | |
|---|---|---|
| Mechanism | One step (concerted) | Two steps (carbocation) |
| Rate law | \(\text{rate} = k[\text{Sub}][\text{Base}]\) | \(\text{rate} = k[\text{Sub}]\) |
| Substrate | \(1^\circ\) through \(3^\circ\) | \(2^\circ\) and \(3^\circ\) only |
| Geometry | Anti-periplanar required | No geometry requirement |
| Base | Strong base required | Weak base / heat |
| Product | Zaitsev (or Hofmann if bulky base) | Zaitsev (most substituted) |
How to Decide Step by Step
1
Identify the substrate. Count the carbon groups on the carbon bonded to the leaving group: methyl, \(1^\circ\), \(2^\circ\), or \(3^\circ\).
2
If methyl or \(1^\circ\): SN2 is the default. A strong bulky base switches to E2.
3
If \(3^\circ\): SN2 is impossible. A strong base gives E2; a weak nucleophile in polar protic solvent gives SN1/E1.
4
If \(2^\circ\): Check nucleophile + solvent. Strong nuc + polar aprotic → SN2. Strong base → E2. Weak nuc + polar protic → SN1/E1.
5
Heat check: Higher temperature always favors elimination over substitution (entropy-driven).
Example 1: Simple SN2
Example 2: E2 with Bulky Base
Example 3: SN1 with Weak Nuc
The Tricky \(2^\circ\) Substrate
Secondary substrates are the battleground where all four pathways compete. The nucleophile/base and solvent become the deciding factors.
SN2
Strong nuc (\(\mathrm{CN}^-\), \(\mathrm{I}^-\)) + polar aprotic solvent (DMSO, DMF)
E2
Strong base (\(\mathrm{OH}^-\), \(\mathrm{RO}^-\), \(t\text{-BuO}^-\)) in any solvent, especially with heat
SN1/E1
Weak nuc/base (\(\mathrm{H_2O}\), \(\mathrm{ROH}\)) + polar protic solvent
Common Mistakes to Avoid
✕ SN2 on \(3^\circ\) Carbon
SN2 requires backside attack. A tertiary carbon is completely blocked by three bulky groups: SN2 never occurs on \(3^\circ\).
✕ SN1 on Methyl/\(1^\circ\)
SN1 requires a stable carbocation. Methyl and primary carbocations are far too unstable to form: SN1 does not occur.
✕ Ignoring the Solvent
Even with a strong nucleophile, a polar protic solvent can solvate it into weakness: the solvent can flip the mechanism.
✕ Forgetting Heat
Higher temperature always shifts the balance toward elimination (entropy favors two products from one reactant).
Master Summary Table
| SN2 | SN1 | E2 | E1 | |
|---|---|---|---|---|
| Substrate | Me, \(1^\circ\), \(2^\circ\) | \(3^\circ\), \(2^\circ\) | \(1^\circ\), \(2^\circ\), \(3^\circ\) | \(3^\circ\), \(2^\circ\) |
| Nuc/Base | Strong nuc | Weak nuc | Strong base | Weak base |
| Solvent | Polar aprotic | Polar protic | Any | Polar protic |
| Rate law | Bimolecular | Unimolecular | Bimolecular | Unimolecular |
| Stereo | Inversion | Racemization | Anti-periplanar | None specific |
Question 1 of 5
\(\mathrm{CH_3Br}\) is treated with \(\mathrm{NaI}\) in acetone. What is the major reaction pathway?
Methyl substrate is completely unhindered (SN2 loves it). \(\mathrm{I}^-\) is a strong nucleophile but a weak base. Acetone is polar aprotic. All three variables point squarely to SN2.
Question 2 of 5
\(\mathrm{(CH_3)_3CBr}\) is dissolved in ethanol (\(\mathrm{EtOH}\)) with no added base. What pathway dominates?
The substrate is tertiary, so SN2 is impossible. There is no strong base, so E2 is ruled out. \(\mathrm{EtOH}\) is a polar protic solvent and a weak nucleophile: the substrate ionizes to form a carbocation, and SN1 (with some E1) dominates.
Question 3 of 5
A secondary alkyl bromide is treated with \(\mathrm{NaOCH_3}\) in DMSO. Which pathway is most likely?
\(\mathrm{CH_3O}^-\) is both a strong nucleophile and a strong base. The key tiebreaker is the solvent: DMSO is polar aprotic, which keeps the nucleophile unsolvated and highly reactive. Polar aprotic + strong nuc on a \(2^\circ\) substrate tips the balance to SN2. (In a protic solvent, E2 would compete more strongly.)
Question 4 of 5
Which factor most strongly determines whether SN2 can occur?
SN2 requires a backside attack by the nucleophile on the electrophilic carbon. If that carbon is crowded with bulky groups (\(3^\circ\)), the nucleophile physically cannot reach it. Steric hindrance of the substrate is the single most important factor for SN2 viability.
Question 5 of 5
A \(2^\circ\) alkyl chloride is treated with \(t\text{-BuOK}\) at high temperature. What is the most likely product?
\(t\text{-BuO}^-\) is a strong, bulky base: too sterically hindered to act as a nucleophile, so substitution is disfavored. High temperature further favors elimination. The result is E2, producing the alkene (Zaitsev product).