Question
Explain the SN1 and SN2 mechanisms of nucleophilic substitution reactions with one example each. Compare them on the basis of stereochemistry, rate law, and substrate preference.
Solution — Step by Step
SN2 (Substitution Nucleophilic Bimolecular) is a one-step, concerted mechanism. The nucleophile attacks from the back side of the carbon bearing the leaving group, at the same time as the leaving group departs. No intermediate forms.
Example: CH₃Br + OH⁻ → CH₃OH + Br⁻
The transition state looks like:
HO⁻ ---→ C ←--- Br
(with 3 H atoms going from umbrella shape to inverted umbrella)Rate law: Rate = [substrate][nucleophile] — depends on both. “Bimolecular” in the name refers to the rate law, not the number of molecules in the mechanism.
Stereochemistry: Walden inversion — if the carbon is a chiral centre, the product has the opposite configuration (R → S or S → R). Like an umbrella flipping inside-out.
Substrate preference: Primary > Secondary >> Tertiary. Bulky groups around the carbon (tertiary) block back-side attack by steric hindrance.
SN1 (Substitution Nucleophilic Unimolecular) is a two-step mechanism with a carbocation intermediate.
Step 1 (slow, rate-determining): Leaving group departs → forms carbocation Step 2 (fast): Nucleophile attacks the carbocation from either face
Example: (CH₃)₃CBr + H₂O → (CH₃)₃COH + HBr (solvolysis in polar protic solvent)
Rate law: Rate = [substrate] — depends only on substrate concentration. The rate-determining step involves only the substrate.
Stereochemistry: Racemisation (partial). The carbocation is sp² hybridised and planar. The nucleophile can attack from either face with roughly equal probability, giving a mixture of R and S products (racemate). In practice, slightly more inversion than retention due to ion pair effects.
Substrate preference: Tertiary > Secondary > Primary (opposite of SN2). Tertiary carbocations are most stable due to hyperconjugation and inductive effect.
| Feature | SN1 | SN2 |
|---|---|---|
| Steps | 2 (via carbocation) | 1 (concerted) |
| Rate law | First order: [R-X] | Second order: [R-X][Nu] |
| Intermediate | Carbocation | None (transition state only) |
| Stereochemistry | Racemisation | Inversion (Walden) |
| Substrate preference | Tertiary | Primary |
| Solvent preference | Polar protic (stabilises carbocation) | Polar aprotic (free nucleophile) |
| Nucleophile strength | Less critical | Strong nucleophile needed |
Three factors determine SN1 vs SN2:
- Substrate: 1° → SN2; 3° → SN1; 2° → depends on other factors
- Nucleophile: Strong, unhindered nucleophile (CN⁻, I⁻, RS⁻) favours SN2; weak nucleophiles or solvents favour SN1
- Solvent: Polar protic (water, alcohols) favours SN1 (stabilises ionic intermediates); polar aprotic (DMF, DMSO, acetone) favours SN2 (doesn’t solvate/cage the nucleophile)
Why This Works
The fundamental difference: SN2 requires the nucleophile to physically approach the carbon from behind while the leaving group is still attached — steric bulk makes this increasingly impossible as we go from 1° to 3°.
SN1 avoids this problem by first removing the leaving group to give a planar carbocation, which any nucleophile can then attack. Tertiary carbocations are stable enough to form; primary ones are not (too unstable), so SN1 is not feasible for primary substrates.
Alternative Method
A memory trick for stereochemistry: “SN2 = 2 faces blocked on one side = inversion (Walden). SN1 = carbocation is flat = nucleophile has 1 face on each side = racemisation.”
Common Mistake
Students often write that SN1 always gives a 50:50 racemic mixture. In practice, the ratio is rarely exactly 50:50 — the leaving group hasn’t fully diffused away when the nucleophile attacks, so the backside is slightly more accessible, giving more inversion than retention. The board answer says “racemisation” or “partial racemisation.” Don’t write “always 50:50” — say “predominantly racemisation with some inversion.”
In JEE Main, a common question shows a chiral substrate and asks the stereochemical outcome. If it’s primary + strong nucleophile + aprotic solvent = SN2 = inversion. If it’s tertiary + weak nucleophile + protic solvent = SN1 = racemisation. Memorise both rules cold.