Question
Given a haloalkane, a reagent, and a solvent — how do we systematically determine whether the reaction follows SN1, SN2, E1, or E2?
Solution — Step by Step
This is always the first filter:
- Methyl or 1-degree: SN2 or E2 only. Carbocations at methyl/primary carbon are too unstable for SN1/E1.
- 3-degree: SN1 or E1 (and E2 if a strong base is present). SN2 is impossible — too much steric hindrance.
- 2-degree: All four mechanisms are possible. We need more information.
The same species can act as a nucleophile (attacks carbon) or a base (attacks hydrogen). We classify by behaviour:
- Strong nucleophile, weak base (, , ): favours SN2 (good at attacking carbon, poor at abstracting protons)
- Strong base, bulky (t-BuO, LDA): favours E2 (too bulky for SN2, excellent at removing protons)
- Strong base + good nucleophile (, ): SN2 and E2 compete (temperature decides — heat favours E2)
- Weak nucleophile, weak base (water, ROH): favours SN1/E1 (waits for carbocation to form)
- Polar protic (water, ethanol, acetic acid): stabilises carbocations and leaving groups — favours SN1/E1
- Polar aprotic (DMSO, DMF, acetone): does not solvate nucleophile — favours SN2/E2
Higher temperature favours elimination (E1 or E2) over substitution (SN1 or SN2). If the question mentions “heating” or “reflux,” tilt toward elimination.
graph TD
A[Identify substrate degree] --> B{Methyl or 1-degree?}
B -->|Yes| C{Strong nucleophile?}
C -->|Yes| D[SN2]
C -->|Strong bulky base| E[E2]
B -->|No| F{3-degree?}
F -->|Yes| G{Strong base present?}
G -->|Yes| H[E2]
G -->|No / Weak| I{Heat?}
I -->|Yes| J[E1]
I -->|No| K[SN1]
F -->|2-degree| L{Reagent + Solvent analysis}
L --> M[All 4 possible - use Steps 2-4]Why This Works
The four mechanisms form two pairs:
- SN2 and E2 — both bimolecular, both need a strong reagent, both work with unhindered substrates
- SN1 and E1 — both unimolecular, both go through carbocations, both need ionisation-friendly conditions
Within each pair, the competition is between substitution and elimination:
- SN2 vs E2: nucleophilicity wins (SN2) or basicity wins (E2). Bulky bases tilt to E2. Heat tilts to E2.
- SN1 vs E1: both happen simultaneously from the same carbocation. Higher temperature increases the E1 fraction.
Alternative Method
The “four-box” method is a popular exam shortcut:
| Strong/concentrated reagent | Weak/dilute reagent | |
|---|---|---|
| Methyl/1-degree | SN2 (or E2 if bulky base + heat) | No reaction (carbocation unstable) |
| 3-degree | E2 | SN1 + E1 (SN1 at low temp, E1 at high temp) |
For 2-degree substrates, use the full analysis from Steps 2-4.
JEE Advanced loves giving 2-degree substrates because all four mechanisms compete. The answer always depends on the specific nucleophile/base, solvent, and temperature. Read every detail in the question — nothing is mentioned without reason.
Common Mistake
Students often treat the four mechanisms as independent choices. In reality, SN1 and E1 always compete (same carbocation intermediate), and SN2 and E2 always compete (same reagent acting as nucleophile or base). The question is really two decisions: (1) Does the reaction go through a carbocation or not? (2) Does the reagent attack carbon or hydrogen?