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Organic Chemistry I · Substitution and Elimination

SN1, SN2, E1, E2 chart and cheat sheet

Four mechanisms, one decision. This page puts everything on one sheet you can print: a side-by-side chart, a step-by-step flow chart, and quick tables for reagents and solvents. The big idea: look at the substrate first, then the reagent, then the solvent. That order settles almost every college exam question.

Already know the basics? Jump to the 6 practice questions or the worked examples.

The SN1 / SN2 / E1 / E2 comparison chart

SN1 vs SN2 vs E1 vs E2 at a glance
FeatureSN2SN1E2E1
Rate lawrate = k[RX][Nu]rate = k[RX]rate = k[RX][base]rate = k[RX]
Steps1, concerted2+: ionize, then nucleophile adds (then lose H⁺ if the nucleophile was neutral)1, concerted2: ionize, then a weak base removes a β-H
IntermediateNone (one transition state)CarbocationNone (one transition state)Carbocation
Best substrateMethyl > 1° > 2°. Never 3°.3°, allylic, benzylic. Some 2°. Not methyl or simple 1°.3° > 2° > 1°. Needs a β-hydrogen.3°, allylic, benzylic. Some 2°.
ReagentStrong nucleophile, ideally a weak baseWeak, neutral nucleophile (often the solvent)Strong base (bulky bases give E2 even with 1°)Weak base (often the solvent), usually with heat
SolventPolar aprotic (DMSO, DMF, acetone, CH₃CN)Polar protic (H₂O, ROH)Not the deciding factor; often the base's alcohol (EtO⁻ in EtOH)Polar protic (H₂O, ROH)
StereochemistryInversion at the reacting carbonRacemization, usually with a slight excess of inversionAnti-periplanar H and leaving group; stereospecificNo geometric requirement; more stable alkene (usually E) favored
RegiochemistryNucleophile goes where the leaving group wasNucleophile goes on the (possibly rearranged) cation carbonZaitsev (more substituted) alkene; Hofmann (less substituted) with a bulky baseZaitsev (more substituted) alkene
RearrangementsNoYes, hydride or alkyl shifts are possibleNoYes, hydride or alkyl shifts are possible

Quick memory hook: the 1 and 2 count the species in the slow step (molecularity), not the number of steps. SN1 and E1 share a carbocation, so they usually happen together. SN2 and E2 both need a strong reagent, so they compete with each other.

Decision flow chart: which mechanism?

Run these checks in order. Each step either gives an answer or sends you to the next one. The "major" pathway is a trend; real reactions often give some of the minor product too.

  1. Is the leaving group on an sp³ carbon? If it's on an alkene or aromatic carbon (vinyl or aryl halide), none of these four mechanisms works. Otherwise, go on.
  2. Classify the reagent with the reagent table below: strong nucleophile/weak base, strong nucleophile/strong base, strong bulky base, or weak nucleophile/weak base.
  3. Is the reagent a strong, bulky base (t-BuOK, DBU, LDA)? Then it's E2 on any substrate with a β-hydrogen, and the less substituted (Hofmann) alkene is often major.
  4. Classify the substrate carbon and follow its branch:
    • Methyl: SN2 only. There's no β-hydrogen for elimination and no stable cation for SN1.
    • Primary (1°): strong nucleophile → SN2. Strong unhindered base (HO⁻, RO⁻) → still mostly SN2, with some E2. Weak nucleophile → little or no reaction, unless the carbon is allylic or benzylic.
    • Secondary (2°): strong nucleophile/weak base in a polar aprotic solvent → SN2. Strong base (HO⁻, RO⁻) → E2 major. Weak nucleophile in a polar protic solvent → SN1 + E1, slow, mixtures (most likely when the carbon is allylic or benzylic).
    • Tertiary (3°): never SN2. Strong base → E2. Weak nucleophile/weak base (water, an alcohol) → SN1 + E1 together. Heat shifts the mix toward E1.
  5. Check the details before drawing the product: inversion for SN2, racemization and possible rearrangement for SN1/E1, anti-periplanar geometry for E2, and Zaitsev vs Hofmann for the alkene.

Reagent and solvent cheat tables

Reagent cheat table
Reagent typeExamplesWhat it usually does
Strong nucleophile, weak baseI⁻, Br⁻, RS⁻ (and HS⁻), N₃⁻, CN⁻SN2 at methyl, 1°, and 2° carbons. Rarely eliminates.
Strong nucleophile, strong baseHO⁻, CH₃O⁻, CH₃CH₂O⁻ (unhindered RO⁻)SN2 at methyl and 1°; E2 at 2° and 3°.
Strong, bulky base (poor nucleophile)t-BuOK (KOC(CH₃)₃), DBU, LDAE2, often giving the Hofmann (less substituted) alkene.
Weak nucleophile, weak baseH₂O, ROH (CH₃OH, CH₃CH₂OH)SN1 + E1 at 3° (and allylic, benzylic, some 2°). Slow or no reaction at 1°.
Solvent cheat table
Solvent typeExamplesFavorsWhy
Polar protic (has O–H or N–H)H₂O, CH₃OH, CH₃CH₂OH, CH₃COOHSN1, E1Hydrogen bonds stabilize the carbocation and the leaving group. They also cage anionic nucleophiles, which slows SN2.
Polar aprotic (no O–H or N–H)DMSO, DMF, acetone, CH₃CN (acetonitrile), HMPASN2Cations are solvated but anions are left "naked" and reactive, so SN2 speeds up.

Two more things to keep in mind. Heat favors elimination over substitution. And a good leaving group (I⁻, Br⁻, Cl⁻, tosylate, or water after protonating an alcohol) is needed for all four. HO⁻ itself is a poor leaving group, so a plain alcohol has to be activated first.

Worked examples

Example 1: secondary halide with ethoxide

2-Bromopropane is heated with sodium ethoxide (NaOCH₂CH₃) in ethanol. Which mechanism gives the major product?

  1. Substrate: the bromine is on a secondary sp³ carbon.
  2. Reagent: ethoxide is a strong, unhindered base and a strong nucleophile.
  3. Secondary branch of the flow chart: a strong base means E2 is major. Some SN2 ether (ethyl isopropyl ether) forms as a minor product.

Answer: E2, giving propene as the major product.

Example 2: tertiary halide in warm ethanol

2-Bromo-2-methylpropane (tert-butyl bromide) is warmed in ethanol with no added base. What happens?

  1. Substrate: tertiary, so SN2 is out.
  2. Reagent: ethanol is a weak nucleophile and a weak base. No strong base is present, so E2 is out too.
  3. Tertiary branch: weak nucleophile/weak base → the carbocation forms, and SN1 and E1 compete.

Answer: SN1 + E1, giving tert-butyl ethyl ether (SN1) and 2-methylpropene (E1). Warming shifts the mix toward the alkene.

Example 3: SN2 stereochemistry with cyanide

(R)-2-Bromobutane reacts with NaCN in DMSO. Give the mechanism and the product's configuration.

  1. Substrate: secondary. Reagent: CN⁻ is a strong nucleophile and a weak base. Solvent: DMSO is polar aprotic.
  2. That's the secondary SN2 case, so the cyanide attacks from the back and the carbon inverts.
  3. Check the label. In the product, –CN (carbon carrying N, N, N) takes bromine's place as priority 1, and ethyl, methyl, and H keep their ranks. The groups are inverted with the same priority order, so R becomes S.

Answer: SN2, giving (S)-2-methylbutanenitrile.

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Practice questions

Pick an answer, then check it. A wrong answer tells you what went wrong, so you can try again. Every question also has a full solution.

Question 11-Bromopropane is heated with potassium tert-butoxide (t-BuOK). What is the major pathway?

Show solution

Strong, bulky base → E2, even on a primary substrate. 1-Bromopropane gives propene.

Question 22-Bromo-2-methylbutane is heated in methanol with no other reagent. What do you expect?

Show solution

The tertiary C–Br bond ionizes in polar protic methanol. The carbocation is trapped by methanol (SN1 → 2-methoxy-2-methylbutane) or loses a β-hydrogen (E1 → mainly 2-methyl-2-butene, the Zaitsev alkene). Heat pushes the ratio toward the alkene.

Question 3Which two mechanisms have the rate law rate = k[substrate]?

Show solution

SN1 and E1 share a rate-limiting ionization that involves only the substrate, so both are first order. SN2 and E2 are single concerted steps that include the nucleophile or base, so both are second order.

Question 42-Bromobutane reacts with sodium methanethiolate (NaSCH₃) in DMF. What is the major pathway?

Show solution

RS⁻ ions are strong nucleophiles but weak bases, and DMF is polar aprotic. At a secondary carbon that means SN2, with inversion, giving 2-(methylthio)butane.

Question 5Bromocyclohexane is heated with sodium ethoxide in ethanol. What is the major product?

Show solution

A secondary halide with a strong base (ethoxide) undergoes mainly E2. The H removed must be anti-periplanar to the Br, which in a cyclohexane chair means both are axial. The product is cyclohexene.

Question 6You want NaCN to react with 1-bromobutane by SN2 as fast as possible. Which solvent should you pick?

Show solution

Polar aprotic solvents (DMSO, DMF, acetone, acetonitrile) speed up SN2 because they solvate Na⁺ but not CN⁻. Protic solvents slow SN2 by hydrogen-bonding to the nucleophile.

Answer key

  1. B, E2: a bulky strong base eliminates even at a 1° carbon.
  2. C, SN1 + E1: tertiary carbon, weak nucleophile, protic solvent.
  3. D, SN1 and E1: the slow step is ionization of the substrate alone.
  4. A, SN2: 2° carbon, strong nucleophile/weak base, polar aprotic solvent.
  5. B, cyclohexene by E2: 2° substrate plus a strong base.
  6. C, DMSO: polar aprotic solvents speed up SN2.

Frequently asked questions

What is the fastest way to decide between SN1, SN2, E1, and E2?

Check the substrate first (methyl, primary, secondary, tertiary), then the reagent (strong or weak, bulky or not), then the solvent (protic or aprotic). Strong bulky bases give E2, methyl and primary give SN2 unless the base is bulky, tertiary never gives SN2, and weak nucleophiles in protic solvents point to SN1 and E1.

Are SN1, SN2, E1, and E2 on the MCAT?

Yes. Nucleophilic substitution, and how substrate structure, nucleophile strength, and solvent affect it, falls under organic chemistry topics in the AAMC MCAT content outline. Elimination is less central, but knowing how a strong base changes the outcome helps with passage questions.

Why do SN1 and E1 usually happen together?

They share the same slow step: the leaving group leaves to form a carbocation. That cation can then be trapped by the nucleophile (SN1) or lose a neighboring proton (E1), so mixtures are normal. Higher temperature favors elimination.

Can I print this cheat sheet?

Yes. When you print the page, the practice questions and buttons are hidden, so you get the chart, flow chart, and tables on paper.

Sources

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