DAT Review · Organic Chemistry
Substitution and Elimination: SN1, SN2, E1, E2
On this page 5 sections
In 30 seconds
Scope: Predicting substitution vs. elimination, mechanism identification, stereochemical outcomes, and factors that shift SN1↔SN2 and substitution↔elimination. This is one of the highest-yield topics; expect 4–6 questions. You MUST know the decision grid cold: substrate type, nucleophile/base strength, solvent, and temperature are your four levers.
The college version
Core Review
SN2: Bimolecular Nucleophilic Substitution
One concerted step. The nucleophile attacks from the BACKSIDE (opposite the leaving group) as the leaving group departs. This results in complete inversion of configuration (Walden inversion).
Rate = k[Nu][RX] (second-order, bimolecular).
Substrate reactivity: Methyl > 1° > 2° >> 3° (NO reaction at 3° — steric hindrance blocks backside attack). Allylic and benzylic substrates are also good.
Nucleophile: Strong nucleophile required. Good nucleophiles: I⁻, Br⁻, Cl⁻, CN⁻, N₃⁻, RS⁻, HO⁻, CH₃O⁻, RNH₂.
Leaving group: Good leaving groups are weak bases (conjugate bases of strong acids): I⁻ > Br⁻ > Cl⁻ > F⁻ (terrible). Tosylate (OTs), mesylate (OMs), and triflate (OTf) are excellent.
Solvent: Polar aprotic solvents (DMSO, DMF, acetone, acetonitrile, HMPA) enhance nucleophilicity by NOT hydrogen-bonding to the nucleophile. Polar protic solvents slow SN2 by solvating (caging) the nucleophile.
SN1: Unimolecular Nucleophilic Substitution
Two steps: (1) Leaving group departs → planar carbocation intermediate (rate-determining). (2) Nucleophile attacks the carbocation from either face.
Rate = k[RX] (first-order, unimolecular).
Substrate reactivity: 3° > 2° >> 1° > methyl. Carbocation stability order: 3° (most stable due to hyperconjugation and inductive effects) > 2° (resonance-stabilized, e.g., allylic/benzylic) > 1° > methyl.
Nucleophile: Weak nucleophiles are sufficient — often the solvent itself acts as nucleophile (solvolysis). H₂O, ROH, and RCOOH are common.
Solvent: Polar protic solvents (water, alcohols, carboxylic acids) stabilize the carbocation intermediate and the leaving group through hydrogen bonding.
Stereochemistry: The planar carbocation can be attacked from either face → racemization (mixture of R and S products, though not always exactly 50:50 due to ion-pair effects).
Carbocation rearrangements: 1,2-hydride shifts and 1,2-alkyl shifts occur when they produce a MORE stable carbocation. ALWAYS check for this! A 2° carbocation adjacent to a 3° carbon will rearrange to a 3° carbocation.
E2: Bimolecular Elimination
One concerted step. A strong base abstracts a β-proton while the leaving group departs and the π bond forms. All bonds break/form simultaneously.
Rate = k[RX][Base] (second-order).
Geometry requirement: The β-hydrogen and the leaving group must be anti-periplanar (180° dihedral angle) for optimal orbital overlap. On cyclohexane rings, this means both must be axial.
Regiochemistry — Zaitsev's rule: With unhindered bases (NaOH, NaOEt), the more substituted (more stable) alkene is the major product.
Hofmann product: With bulky bases (t-BuOK, LDA), the LESS substituted alkene predominates because the bulky base cannot access the more hindered β-hydrogen.
Substrate reactivity: 3° > 2° > 1° (more substituted = more alkene stability in the transition state).
Stereochemistry: Anti elimination (H and LG on opposite sides of the forming π bond).
E1: Unimolecular Elimination
Two steps: (1) Leaving group departs → carbocation (same as SN1). (2) Base abstracts a β-proton to form the alkene.
Rate = k[RX] (first-order).
Competes directly with SN1. Heat favors elimination over substitution (entropy: elimination produces more molecules).
Zaitsev product predominates. Carbocation rearrangements are possible, just like SN1.
The Master Decision Guide
| Substrate | Strong Nu, Weak Base | Strong, Bulky Base | Weak Nu, Protic Solvent | Strong Nu, Strong Base, Heat |
|---|---|---|---|---|
| Methyl | SN2 | — | — | — |
| 1° | SN2 | E2 (with t-BuOK) | No reaction | E2 (dominates with heat) |
| 2° | SN2 (with good Nu) | E2 (Hofmann) | SN1/E1 (slow, mix) | E2 (Zaitsev) |
| 3° | — | E2 (Hofmann) | SN1/E1 (mix) | E2 (Zaitsev) |
Key qualitative rules:
- Strong Nu + weak base → SN2
- Strong, bulky base → E2
- Weak Nu + polar protic solvent + 3°/2° → SN1/E1 mix
- Heat tilts the balance toward elimination
- Negatively charged nucleophiles/bases favor SN2/E2; neutral species (H₂O, ROH) favor SN1/E1
Common Traps
- Forgetting t-BuOK gives the Hofmann product. Students default to Zaitsev for all E2 reactions.
- Applying SN2 to 3° substrates. Steric hindrance blocks backside attack completely.
- Ignoring carbocation rearrangements. The DAT loves showing a 2° substrate that looks like it gives one product but rearranges to a 3° carbocation and gives a different product.
- Confusing nucleophilicity and basicity. NaOEt is both a strong nucleophile AND strong base — it can do both SN2 and E2 depending on substrate and temperature.
Eli explains
The same idea, in plain words
Explain it like I’m 10
Imagine trying to replace a teammate in a game. SN2: you tap them from behind and immediately take their spot (inversion!). SN1: they walk off the field, leaving a hole for a moment (carbocation), and then you jump in from either side (racemization). E2: both happen at once — you pull them off while someone else pulls a flag from their pocket. E1: they leave, creating a hole, and a flag falls off by itself. Heat makes the "flag-pulling" (elimination) more likely.
Key takeaways
- SN2 = INVERSION of configuration. If the starting material is R and the nucleophile has the same CIP priority as the leaving group, the product is S.
- SN1 = RACEMIZATION (planar carbocation). Both enantiomers form.
- E2 requires ANTI-PERIPLANAR geometry. Cyclohexane substrates must have the leaving group axial.
- Zaitsev (more substituted alkene) with small bases; Hofmann (less substituted) with bulky bases like t-BuOK.
- Carbocation rearrangements only in SN1/E1 (not SN2/E2). Always scan for 1,2-hydride or 1,2-alkyl shifts that produce a more stable carbocation.
- (R)-2-bromobutane + NaSH in DMSO → ? Answer: (S)-2-butanethiol. Strong nucleophile (HS⁻), polar aprotic solvent, 2° substrate = SN2 with inversion. R → S.
- 2-bromo-2-methylbutane + H₂O, heat → ? Answer: Mixture of 2-methyl-2-butene (major, Zaitsev, E1) and 2-methyl-2-butanol (SN1), plus possibly 2-methyl-1-butene (minor, Hofmann from E1). 3° substrate, weak nucleophile, protic solvent, heat = SN1/E1 mix.
- 2-bromobutane + t-BuOK → ? Answer: 1-butene (major, Hofmann) and trans-2-butene + cis-2-butene (minor, Zaitsev). Strong bulky base + 2° substrate = E2. The bulky base preferentially abstracts the less hindered β-hydrogen (on C-1), giving the less substituted alkene as the major product.
Study tools & related lessonsYou’ll learn to · Related
You’ll learn to
- Predict major product(s) from SN1, SN2, E1, and E2 pathways
- Determine mechanism from given conditions (substrate, reagent, solvent, temperature)
- Explain stereochemical outcomes: inversion (SN2), racemization (SN1), anti-periplanar (E2)
- Apply Zaitsev's and Hofmann's rules for elimination regiochemistry
- Recognize when carbocation rearrangements occur
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