Substitution and elimination · 9 min read

SN1, SN2, E1 or E2? How to decide which mechanism wins

An alkyl halide plus a nucleophile or base can go four ways. Which path wins depends on four things: the substrate, the reagent, the solvent and the temperature. Learn those four and you can predict most exam questions.

The four mechanisms at a glance

	SN2	SN1	E2	E1
Steps	One, concerted	Two (cation forms first)	One, concerted	Two (cation forms first)
Rate law	k[RX][Nu]	k[RX]	k[RX][Base]	k[RX]
Substrate	Me > 1° > 2°; not 3°	3° > 2°; not 1°	3° > 2° > 1°	3° > 2°
Reagent	Good nucleophile	Weak nucleophile	Strong base	Weak base
Solvent	Polar aprotic	Polar protic	Any, often alcohol	Polar protic
Stereochemistry	Inversion	Racemisation (often partial)	Anti-periplanar H and LG	No strict requirement
Rearrangement	No	Possible	No	Possible

See each mechanism drawn out: SN2, SN1, E2.

Step 1: classify the substrate

Methyl and primary: SN2 is easy because the carbon is open to backside attack. SN1 and E1 are very unlikely because the cation would be unstable. With a bulky strong base (such as tert-butoxide) E2 takes over.
Tertiary: no backside attack is possible, so SN2 is out. With a strong base you get E2. With a weak base and a polar protic solvent you get a mix of SN1 and E1.
Secondary: the borderline case. All four are possible, so the reagent and solvent decide.
Allylic and benzylic: these stabilise both cations and the SN2 transition state, so they are reactive by every route.

Step 2: nucleophile or base?

Nucleophilicity (how fast it attacks carbon) and basicity (how strongly it takes a proton) are related but not identical.

Good nucleophile, weak base: I⁻, Br⁻, N₃⁻, CN⁻, RS⁻, PhS⁻. These push toward SN2 on primary and secondary carbons.
Strong base and good nucleophile: HO⁻, MeO⁻, EtO⁻. Substitution on primary, elimination on tertiary, a mixture on secondary (heat favours elimination).
Strong, bulky base: t-BuOK, LDA, DBU. These are too hindered to attack carbon and give elimination.
Weak nucleophile, weak base: H₂O, ROH. These work only through carbocations (SN1/E1), so they need a substrate that can ionise.

Step 3: consider the solvent

Polar aprotic solvents (DMSO, DMF, acetonitrile, acetone) leave the nucleophile "naked" and fast, which boosts SN2. Polar protic solvents (water, alcohols) stabilise ions through hydrogen bonding, helping the leaving group depart and the carbocation form, which boosts SN1 and E1. The same protic solvent also cages anionic nucleophiles and slows SN2.

Step 4: temperature

Elimination produces more molecules from fewer, so it gains entropy. Higher temperature therefore favours elimination over substitution. If a question mentions heat, lean toward the alkene.

Leaving group matters in all four

A good leaving group is the conjugate base of a strong acid and is stable on its own: I⁻ > Br⁻ > Cl⁻ ≫ F⁻, and tosylate or mesylate are excellent. HO⁻ is a poor leaving group, which is why alcohols need acid or conversion to a sulfonate first. See the Appel reaction and the Mitsunobu reaction for two ways to activate an alcohol.

Regiochemistry of elimination

With a small base, the more substituted alkene (Zaitsev product) usually dominates because it is more stable. With a bulky base, the less hindered hydrogen is removed and the less substituted alkene (Hofmann product) dominates. A good example of a Hofmann-type outcome is the Hofmann elimination.

Worked examples

1-bromobutane + NaCN in DMSO. Primary, a good nucleophile that is a weak base, polar aprotic solvent: SN2.
2-bromo-2-methylpropane + NaOEt in EtOH, heat. Tertiary, strong base: E2.
2-bromo-2-methylpropane in warm water. Tertiary, weak nucleophile and base, polar protic: SN1 with some E1.
2-bromopropane + NaOEt in EtOH, heat. Secondary, strong base, heat: mostly E2, with some SN2.
Primary bromide + t-BuOK. Bulky base: E2 despite being primary.

Test yourself on these in Mechanism Lab's Quiz mode, where you pick the mechanism for a named reaction.

Practise in the interactive app Browse all 345 reactions