SN1 vs SN2 vs E1 vs E2: how to tell them apart
Work in this order. Substrate class rules out most options — methyl and primary go SN2, tertiary go SN1/E1 or E2, secondary is genuinely ambiguous. Then the reagent: a strong nucleophile that is a weak base gives substitution, a strong bulky base gives elimination, and a weak neutral nucleophile (the solvent) gives SN1/E1. Then solvent and temperature break the remaining ties — polar protic and heat both favour the unimolecular, elimination-heavy side.
Why all four compete for the same starting material
Every one of these reactions begins with the same setup: a carbon carrying a leaving group, and a species with a lone pair. That species has two things it can attack. If it attacks the carbon bearing the leaving group, you get substitution. If it pulls a proton off the neighbouring carbon instead, you get elimination.
That single fork — attack carbon, or attack hydrogen — is the whole story. A lone pair that is a good nucleophile goes for carbon. A lone pair that is a good base goes for the proton. Many reagents are both, which is why real problems give mixtures and why exam questions ask for the major product.
The second fork is timing. Does the leaving group leave first, generating a carbocation (unimolecular, the "1" reactions), or does everything happen in one concerted step (bimolecular, the "2" reactions)? The substrate decides that, because it decides whether a carbocation is survivable.
Step 1 — classify the substrate. This eliminates most of the options.
Count the carbons attached to the carbon bearing the leaving group.
- Methyl and primary. A primary carbocation is far too unstable to form, so SN1 and E1 are off the table. The carbon is also wide open to backside attack. Expect SN2 — unless the base is strong and bulky, in which case E2.
- Tertiary. Backside attack is blocked by three alkyl groups, so SN2 is off the table. A tertiary carbocation is stable enough to form. Expect SN1, E1 or E2, and the reagent decides which.
- Secondary. Everything is possible. This is where you must actually use the reagent, and where exam questions live.
Two substrates that break the pattern. Allylic and benzylic halides form resonance-stabilised carbocations, so they do SN1 even when they look primary. Vinyl and aryl halides do neither — the C–X bond has partial double-bond character and the backside is shielded by the π system.
Step 2 — read the reagent
Sort the reagent on two independent axes: how good a nucleophile it is, and how good a base it is. They are not the same property.
| Reagent type | Examples | Pushes toward |
|---|---|---|
| Strong nucleophile, weak base | I⁻, Br⁻, RS⁻, N₃⁻, CN⁻ | SN2 |
| Strong nucleophile and strong base | HO⁻, RO⁻ (methoxide, ethoxide) | SN2 on primary, E2 on secondary and tertiary |
| Strong bulky base | t-BuO⁻, LDA, DBU | E2, and often the Hofmann (less substituted) product |
| Weak, neutral nucleophile | H₂O, ROH, RCOOH | SN1 / E1 — the solvent is doing the work |
The useful shortcut: anything negatively charged is strong and drives a bimolecular pathway; anything neutral is weak and lets the leaving group leave on its own first.
Step 3 — let solvent and temperature break the tie
- Polar protic (water, methanol, ethanol, acetic acid) hydrogen-bonds to anions, caging the nucleophile and stabilising the carbocation. It favours SN1 and E1.
- Polar aprotic (DMSO, DMF, acetone, acetonitrile) solvates the cation but leaves the anion naked and furious. It favours SN2.
- Heat favours elimination in every case. Elimination makes more particles from fewer, so the entropy term grows with temperature. If a question says "heat" or "Δ", it is nudging you toward E1 or E2.
The one-page decision table
| Substrate | Weak nucleophile / base (H₂O, ROH) | Strong nucleophile, weak base (I⁻, CN⁻) | Strong base (RO⁻, HO⁻) | Bulky strong base (t-BuO⁻) |
|---|---|---|---|---|
| Methyl / 1° | No reaction | SN2 | SN2 | E2 |
| 2° | SN1 / E1 | SN2 | E2 | E2 |
| 3° | SN1 / E1 | SN1 (slow) | E2 | E2 |
Worked examples
1-bromobutane + NaOH
Primary substrate, so no carbocation. Hydroxide is a strong nucleophile and a strong base, but with a primary carbon there is no steric barrier to backside attack and no β-branching to encourage elimination. SN2, giving butan-1-ol, with inversion at a carbon that happens not to be a stereocentre here.
2-bromo-2-methylbutane + sodium ethoxide in ethanol
Tertiary substrate rules out SN2. Ethoxide is a strong base, which rules out the leisurely SN1/E1 path. So E2. Now apply Zaitsev. Taking a proton from one of the methyl groups puts the double bond between C-1 and C-2, giving 2-methylbut-1-ene — disubstituted. Taking it from the CH₂ of the ethyl puts the double bond between C-2 and C-3, giving 2-methylbut-2-ene — trisubstituted. The more substituted alkene is more stable, so 2-methylbut-2-ene is the major product.
2-bromobutane + water, warmed
Secondary substrate, weak neutral nucleophile, polar protic solvent, heat. Nothing is strong enough to force a concerted mechanism, so the bromide leaves first. SN1 and E1 compete: butan-2-ol from water capturing the cation, plus but-2-ene from a solvent molecule taking a β-proton. Because the intermediate is a planar carbocation, the alcohol is racemic.
Mistakes that cost marks
- Calling a secondary substrate without reading the reagent. Secondary is a genuine fork. The reagent is the deciding evidence, not a tiebreak.
- Forgetting that SN2 inverts. If the carbon is a stereocentre, an SN2 gives the opposite configuration — not a racemate.
- Forgetting that SN1 racemises. The carbocation is planar, so the nucleophile arrives from both faces. Expect a racemic mixture, usually with slight excess of inversion from ion pairing.
- Applying Zaitsev to a bulky base. t-Butoxide cannot reach the internal proton it would need for the Zaitsev product, so it takes the accessible terminal proton and gives the Hofmann alkene.
- Ignoring the anti-periplanar requirement in E2. The β-hydrogen and the leaving group must be 180° apart. On a cyclohexane ring, that means both must be axial — which sometimes means the apparent Zaitsev product is geometrically impossible.
How Organic Chemistry AI helps here
The procedure above is not hard; applying it at speed, to an unfamiliar substrate, is. When you type or photograph a substitution or elimination problem, the app names the reaction type it has settled on and then walks the steps that justify it — which carbon it classified, what it decided the reagent was, and which selectivity rule picked the major product. That is the part worth reading, because it is the part you have to reproduce on the exam.
Adding the Substitution or Elimination topic hint when you type a problem tells the solver which family you believe you are in, which is useful when a question is ambiguous.
A typed problem with the Elimination topic hint. The hint rides along with the question and steers the solver without changing the text you see.
Frequently asked
Is a secondary substrate always ambiguous?
In principle yes, which is why exam questions use secondary substrates so often. In practice the reagent resolves it: a strong nucleophile that is a weak base (iodide, azide, cyanide) gives SN2, a strong base gives E2, and a neutral solvent nucleophile gives an SN1/E1 mixture.
How do I know whether E1 or E2 is happening?
Look at the base. E2 needs a strong base present in the flask; E1 happens when there is no strong base and the carbocation forms first, so a weak solvent molecule removes the proton afterwards. E1 also always comes with SN1 product alongside it, because the same carbocation can be captured instead of deprotonated.
Does a better leaving group change which mechanism runs?
It changes the rate far more than the pathway. Iodide leaves faster than bromide, which leaves faster than chloride, and fluoride barely leaves at all. A better leaving group speeds up every one of the four mechanisms, so it rarely flips the answer — but it can make an otherwise sluggish SN1 competitive.
Check your mechanism call in seconds
Type the substrate and reagent, or photograph the question, and read the reasoning step by step.
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