SN1 vs SN2: How to Tell Which Mechanism You're Looking At
These four letters cause more confusion than almost anything else in organic chemistry — not because the mechanisms are hard individually, but because students never learn a reliable way to decide which one they're looking at under exam pressure. Here's a clean way to tell them apart, every time.
The basics first
SN1 = Substitution, Nucleophilic, Unimolecular (rate depends on only 1 species)
SN2 = Substitution, Nucleophilic, Bimolecular (rate depends on 2 species)
That's literally what the numbers mean — not "1 step" and "2 steps," which is the most common misconception. It's about how many molecules are involved in the rate-determining step.
SN2: the one-step story
SN2 happens in a single, smooth motion — the nucleophile attacks from the opposite side of the leaving group at the exact same time the leaving group departs. No intermediate forms.
Rate = k[substrate][nucleophile]
alt="SN2 mechanism diagram showing backside attack and inversion" style="max-width:100%;height:auto;display:block;margin:16px auto;">Notice the three plain bonds flip completely from pointing up to pointing down — that's the Walden inversion. Both species show up in the rate equation because they're both involved in the one and only step.
SN1: the two-step story
SN1 happens in two stages:
- The leaving group leaves first, forming a carbocation intermediate
- The nucleophile then attacks that carbocation
Because step 1 (bond breaking) is slow and step 2 (bond forming) is fast, only the substrate concentration affects the rate.
Rate = k[substrate]
alt="SN1 mechanism diagram showing carbocation intermediate and racemization" style="max-width:100%;height:auto;display:block;margin:16px auto;">The flat carbocation in the middle is the whole story — once the leaving group is gone, the nucleophile has no preferred side to attack from, so you get a roughly equal mix of both configurations instead of one clean inverted product.
The 5 questions that tell you which one you're looking at
1. What type of carbon is the leaving group attached to?
| Carbon type | Mechanism |
|---|---|
| Primary (1°) | SN2 |
| Secondary (2°) | Could be either — check other clues |
| Tertiary (3°) | SN1 |
Why: SN2 needs the nucleophile to physically approach from the back side. A tertiary carbon is too crowded (steric hindrance). Tertiary carbons form very stable carbocations, which is exactly what SN1 needs.
2. Is the nucleophile strong or weak?
Strong nucleophile (OH⁻, CN⁻, RO⁻) → pushes toward SN2
Weak nucleophile (H₂O, ROH) → pushes toward SN1
3. Is the solvent polar protic or polar aprotic?
Polar protic (water, alcohols) → favors SN1
Polar aprotic (acetone, DMSO) → favors SN2
4. What's the leaving group?
Good leaving groups (I⁻, Br⁻, Cl⁻, tosylate) work for both mechanisms — this clue alone won't decide it.
5. Does the product show racemization or inversion?
SN2 → inversion (the Walden inversion)
SN1 → racemization (roughly 50:50 mix)
A worked example
Question: (CH₃)₃C–Br reacts with H₂O in water. What mechanism?
Tertiary carbon + weak nucleophile + protic solvent — all three point to SN1.
Question: CH₃CH₂Br reacts with NaOH in acetone. What mechanism?
Primary carbon + strong nucleophile + aprotic solvent — clean SN2.
The one sentence to remember
SN1 loves stability (tertiary carbon, weak nucleophile, protic solvent). SN2 loves speed and access (primary carbon, strong nucleophile, aprotic solvent, one clean shove from the back).
If most of your clues point one direction, trust that direction — real exam questions almost never give you a genuinely 50/50 ambiguous case.
Struggling with a specific mechanism question from your textbook? Drop it in the comments or contact me — happy to walk through it.
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