Both convert R–LG into R–Nu; four clues on the substrate and conditions decide the pathway and the stereochemistry.
Open carbon + strong Nu → SN2; stable cation + weak Nu → SN1.
1. SN2 Happens in One Concerted, Backside Attack
Bimolecular (rate = k[substrate][nucleophile]): the nucleophile attacks 180° opposite the leaving group in a single step, inverting the carbon.
2. A Crowded Carbon Shuts Down the SN2
Each alkyl group added to the reacting carbon blocks the backside approach, so the rate collapses from methyl down to tertiary.
SN2 rate: methyl > 1° > 2° » 3° — the exact opposite of SN1.
3. SN1 Ionizes First, Through a Flat Carbocation
Unimolecular (rate = k[substrate]): the leaving group departs first to a flat sp² carbocation, favored by the 3°/resonance substrates SN2 can't touch.
4. SN1 Racemizes While SN2 Inverts
SN2's one-sided attack inverts a single enantiomer; SN1's flat cation is hit on both faces, scrambling it to a racemic mixture.
5. Nucleophile, Solvent, and Leaving Group Break the Tie
For borderline 2° carbons: strong anionic Nu + polar aprotic solvent push SN2; weak neutral Nu + polar protic solvent push SN1; both need a good leaving group (I⁻, OTs).
6. Summary
SN2: one step · backside · inversion · methyl/1° · strong Nu · aprotic. SN1: two steps · flat carbocation · racemic · 3° · weak Nu · protic. Read the substrate first.
A textbook SN2: primary ethyl bromide + a strong nucleophile (cyanide) in acetone → propanenitrile, with inversion.
Worked example
- The substrate is 1° — a 1° carbocation is far too unstable for SN1.
- Cyanide is a strong, unhindered nucleophile → clean backside SN2.
Answer. Butanenitrile, by SN2 (rate depends on both the substrate and CN−).
Quiz yourself
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SN1. Its rate law is rate = k[substrate], with no nucleophile term, because the slow step is the substrate ionizing to a carbocation before the nucleophile ever gets involved. In SN2 the nucleophile is in the rate-determining transition state, so doubling it would double the rate.
It went by SN1. The flat, sp² carbocation intermediate can be attacked from either face at nearly equal rates, scrambling the stereocenter into a ≈50:50 mixture. A pure SN2 would have inverted the center to give a single enantiomer instead.
Same feature, opposite consequences: three methyl groups crowd the reacting carbon. That steric bulk blocks the backside approach an SN2 needs (no room for the nucleophile), but it also stabilizes the resulting tertiary carbocation, which is exactly what the SN1 rate-determining step requires.
A polar aprotic solvent such as DMSO, DMF, or acetone. It dissolves the ionic reagents but can't hydrogen-bond to the anionic nucleophile, leaving it "naked" and highly reactive for backside attack. A polar protic solvent would instead solvate the nucleophile (slowing SN2) and stabilize a carbocation (favoring SN1).
Draw this on the whiteboard
Open the OChem Board whiteboard — benzene rings, wedge/dash bonds, and a clickable periodic table built in. No account needed.