E1 vs E2 vs substitution: the full decision map

Once elimination joins substitution in the mechanism menu, four pathways compete on the same substrate: SN1, SN2, E1, E2. Predicting which one runs is not memorization — it is a small set of decisions, each of which shifts the balance in a predictable way. This guide extends the SN1 vs SN2 decision tree to include both elimination mechanisms, so you can look at an alkyl halide plus a set of conditions and know the product before the arrow-pushing starts.

The four mechanisms in one paragraph each

SN2 — concerted backside displacement. Second-order kinetics, strong nucleophile pushes into the electrophilic carbon while the leaving group departs, inversion of stereochemistry. Wins on methyl and primary substrates.

SN1 — stepwise via a carbocation. First-order in substrate, weak nucleophile (often the solvent), racemization at the reacting carbon. Wins on tertiary substrates in polar protic solvents.

E2 — concerted anti-periplanar elimination. Second-order kinetics, strong base removes a β-hydrogen while the leaving group departs from the α-carbon in the same step. Requires the β-H and the leaving group to sit anti to each other. Gives an alkene.

E1 — stepwise elimination via the same carbocation as SN1. First-order in substrate, base is weak or absent. The cation loses a β-proton to whatever mild base is around (usually the solvent), giving an alkene.

Note the pairing: SN2 and E2 both need a strong reagent (nucleophile or base, often the same species playing either role) and both go in one step. SN1 and E1 share a carbocation intermediate and are decided by what happens after ionization — nucleophilic capture (SN1) or β-proton loss (E1).

The base–nucleophile axis

The single most useful shortcut is to sort your reagent into one of four boxes:

• Strong base, strong nucleophile — HO⁻, RO⁻ (alkoxides). Drives both SN2 and E2. Substrate class then picks which one wins.
• Strong base, poor nucleophile — bulky bases: LDA, DBU, t-BuO⁻, DBN, Hünig’s base. E2 only. The bulk kills SN2’s backside attack while leaving β-hydrogen deprotonation untouched.
• Weak base, strong nucleophile — I⁻, RS⁻, N₃⁻, CN⁻, Br⁻. SN2 only. The base is too weak to abstract β-hydrogens on the timescale that competes with substitution.
• Weak base, weak nucleophile — water, alcohols, carboxylic acids. Solvolysis regime. Substrate class decides: primary and methyl → SN2 (slow) or no reaction; tertiary → SN1/E1 mixture.

The idea that “strong base = E2, strong nucleophile = SN2” is over-simplified because most species are both. Sorting by “nucleophile-and-base,” “base-not-nucleophile,” “nucleophile-not-base,” or “neither” gives you a cleaner grid.

The substrate axis

Overlay that grid with substrate class:

• Methyl and primary — SN2 dominates with any decent nucleophile. E2 only appears if you deliberately pick a hindered base. Never SN1 or E1 (primary carbocations are too unstable).
• Secondary — the referee’s nightmare. All four mechanisms compete. Base–nucleophile axis decides.
• Tertiary — SN2 is dead (steric blockade). With a strong base, E2 dominates. With a weak base in a polar protic solvent, SN1 and E1 both run and you get a mixture of alcohol/ether and alkene.
• Neopentyl and other β-branched primaries — SN2 is painfully slow (steric hindrance at the β-position); E2 works with any decent base; SN1/E1 are still ruled out.

Heat and Zaitsev

Temperature is the last knob. Elimination has a positive activation entropy (splitting one molecule into two) more than substitution does, so higher temperature favours elimination. This is the reason a lot of textbook procedures warn: if you want SN2, run the reaction cold. If you want E2, reflux.

Zaitsev’s rule (also spelled Saytzeff): when E2 has a choice between two β-hydrogens, the more substituted alkene predominates because it is thermodynamically more stable (hyperconjugation). Exception: bulky bases like t-BuO⁻ flip the preference to the less substituted (Hofmann) product, because the base cannot reach the more crowded β-H.

For E1, Zaitsev’s rule holds strongly — the carbocation intermediate lives long enough to lose whichever proton gives the most stable alkene.

The full decision tree

Given an alkyl halide substrate and a reagent, run this in order:

1. What class is the substrate? Methyl or primary → almost certainly SN2 (unless the reagent is a bulky base, in which case E2). Tertiary → not SN2. Secondary → keep going.
2. Is the reagent a strong base?
   • Yes → E2 is in play. If it is also a strong nucleophile → E2 vs SN2 competition; substrate and temperature decide. If it is a bulky base → E2 wins outright.
   • No → E2 is out. SN2 vs SN1/E1 depending on substrate.
3. Solvent? Polar aprotic (DMSO, DMF, acetonitrile) → speeds up SN2 and E2 equally; anionic reagents effectively stronger. Polar protic (water, alcohols) → SN1/E1 regime for tertiary substrates.
4. Temperature? Cold and short → substitution favoured. Hot and refluxing → elimination favoured.
5. Any special substrate features? β-Branching → SN2 dies. Very good leaving group (TfO⁻, TsO⁻) → SN1/E1 faster in polar protic. Allylic/benzylic → SN1 accelerated because the cation is stabilized.

A canonical worked example

If the base had been the bulky t-butoxide instead of ethoxide, the same substrate would still eliminate but the regiochemistry would flip to 2-methyl-1-butene (Hofmann), because the t-butoxide is too fat to squeeze onto the more substituted β-hydrogen.

Anti-periplanar geometry (the reason E2 sometimes surprises you)

E2 requires the β-H and the leaving group to be anti-periplanar — 180° dihedral — because the σ-bond breaking to release the leaving group has to overlap with the C–H σ-bond being deprotonated. In an acyclic substrate this is almost always accessible via bond rotation. In a rigid ring (menthyl and neomenthyl chlorides are the classic McMurry Ch. 11 example), only one geometry is available, and the alkene it makes may not be the Zaitsev product.

If you see a cyclohexyl substrate in an E2 problem, draw the chair. The β-H that can go anti-periplanar to the axial leaving group is the one that reacts. If the leaving group is equatorial, no β-H is anti-periplanar to it in that chair; the molecule has to flip to the other chair (paying a conformational cost) before E2 can happen.

Where E1 shows up

E1 is less common in practice because most useful eliminations are set up with a strong base (E2 conditions). But E1 wins when:

• The substrate is tertiary, allylic, or benzylic (stable cation).
• The base is very weak (solvent-strength base).
• The temperature is elevated.
• The solvent is polar protic (stabilizes the cation).

Solvolysis of tertiary alkyl halides in ethanol at 80 °C is the textbook case: the products are a mixture of SN1 ether and E1 alkene, in ratios that depend on the specific substrate. Because both mechanisms share a rate-determining ionization step, they always run together in that ratio.

Practice problems

Where to go next

• To count degrees of unsaturation in whatever alkene you produce, use the degrees of unsaturation calculator — it flags a π-bond addition from the parent alkane.
• The functional groups cheatsheet is a quick check when a problem hands you a substrate you cannot immediately classify.
• Once you have alkenes in hand, the next major mechanism family is electrophilic addition to π-bonds — the mirror image of what you just did.

The core insight: every reagent gets sorted on two axes (base strength, nucleophile strength) and every substrate on one (methyl/primary/secondary/tertiary). Six inputs — substrate class, base strength, nucleophile strength, solvent, leaving group, temperature — deterministically pick one of four mechanisms.