Organic Chemistry I · Decision guide
Substitution vs elimination: how to choose SN1, SN2, E1, or E2
Substitution replaces a leaving group with a nucleophile. Elimination removes a leaving group and a neighboring hydrogen to make a double bond. Choose between them by checking the substrate first, then the reagent and conditions. Often, the honest prediction is a major pathway plus competing products.
1. Check the substrate and leaving group
Find the carbon bonded to the leaving group: the alpha carbon. Count its carbon neighbors to classify it as methyl, primary, secondary, or tertiary. The neighboring carbons are beta carbons; ordinary E1 or E2 alkene formation needs a beta hydrogen.
This guide concerns typical substitution and elimination at saturated, sp³ carbons. A halogen directly attached to an alkene or aromatic ring needs different rules. Allylic and benzylic positions are adjacent to those systems; resonance can help them react.
Look for a usable leaving group, such as bromide, iodide, or a sulfonate. An alcohol’s OH group is a poor leaving group under basic conditions. Protonation can make it leave as water; alternatively, convert it to a better leaving group before applying this guide.
| Substrate | First possibilities |
|---|---|
| Methyl | SN2 with a suitable nucleophile. No beta carbon, so no ordinary E2. |
| Primary | Usually SN2; a strong bulky base can favor E2 when a beta H is available. |
| Secondary | SN2 and E2 compete. SN1/E1 can become viable under ionizing conditions, especially with resonance stabilization. |
| Tertiary | SN2 is blocked by crowding. Strong base suggests E2; ionizing solvent with a weak nucleophile/base suggests SN1/E1. |
These are trends. A neopentyl substrate is primary but unusually slow in SN2 because nearby branching blocks attack. Primary allylic or benzylic substrates can behave differently from ordinary primary substrates because their carbocations can be resonance-stabilized.
2. Separate nucleophilicity from basicity
A nucleophile donates electrons to an electrophilic atom; a base accepts a proton. One reagent can do both. Iodide and thiolates are useful nucleophiles that are relatively weak bases, so they often favor SN2 at accessible carbons. Hydroxide and small alkoxides are strong bases and good nucleophiles: expect more competition, especially at secondary centers.
A bulky strong base, such as tert-butoxide, has difficulty reaching a crowded carbon but can remove an accessible beta H. That often favors E2, even for primary substrates. Reagent concentration also matters: the usual SN2 rate depends on substrate and nucleophile concentrations, while the E2 rate depends on substrate and base concentrations.
3. Use solvent and heat as supporting evidence
Polar aprotic solvents, such as DMSO or acetone, often help anionic nucleophiles perform SN2 because they do not strongly hydrogen-bond to those anions. Polar protic solvents, such as water or ethanol, solvate anions and can support ionization to a sufficiently stable carbocation. That makes SN1/E1 plausible, not automatic.
Strong base can still give E2 in a protic solvent. Heating often increases elimination relative to substitution, but “heat means E2” is unreliable. Base strength, substrate, and available geometry still decide which pathways are possible.
4. Draw the mechanism and check the product
SN2: draw nucleophile-to-carbon and carbon–leaving-group-to-leaving-group arrows in one step. Backside attack inverts the reacting center; an R/S label need not switch if substituent priorities change. No carbocation forms, so carbocation rearrangements do not occur.
E2: draw three simultaneous arrows: base lone pair to beta H, C–H bond to the C–C bond, and C–leaving-group bond to the leaving group. The reacting C–H and C–leaving-group bonds normally must be anti-periplanar.
SN1/E1: first move the leaving-bond electrons to the leaving group, forming a carbocation. Nucleophile capture gives substitution; removal of a beta H gives elimination. A hydride or alkyl shift may rearrange the carbocation first. SN1 at a stereocenter often gives both configurations, but ion pairing can prevent an exact 50:50 mixture.
For elimination, draw all accessible alkenes. A small base often favors the more substituted Zaitsev product; bulky bases often favor the less substituted Hofmann product. Geometry and conjugation can override that shortcut. Cyclohexane E2 normally needs trans-diaxial H and leaving group.
Common questions
Does a negative charge mean strong base?
No. Iodide is negatively charged but a weak base. Consider the conjugate acid and the reagent’s structure, not charge alone.
Can more than one mechanism happen?
Yes. SN1/E1 commonly share an intermediate, and SN2/E2 often compete. Without enough conditions, report plausible pathways instead of inventing a single certain product.
What should I check last?
Count atoms, formal charges, and carbon valences. Confirm that every curved arrow starts at electrons and that the proposed stereochemistry fits the mechanism.