Source: Organic Chemistry, The Ohio State University (Baldwin)
Tags: SN1, SN2, E1, E2, reaction pathway, decision flowchart, nucleophile vs base, substitution vs elimination, strong nucleophile, weak nucleophile, leaving group, steric hindrance
Difficulty: Intermediate to Advanced | Prerequisites: SN1, SN2, E1, and E2 mechanisms (Parts 1 and 2 of these notes, plus Chapter 6)
Big picture: This is the capstone skill for Chapters 6 and 7. The exam will hand you a substrate, a reagent, a solvent, and sometimes a temperature, and ask you to predict the mechanism and the product(s). You need a systematic method for sorting through four possible pathways. The decision tree below is based on Dr. Baldwin's flowchart and covers the logic step by step. If you can work this flowchart reliably, you can answer the majority of mechanism-prediction problems on the exam.
Start by classifying the nucleophile/base as strong or weak. Strong reagents mean SN2 or E2. Weak reagents mean SN1 or E1. Then look at the substrate (1°, 2°, or 3°), whether the reagent is more nucleophilic or more basic, whether it is bulky, and whether heat is applied. Each combination leads to one dominant pathway.
Strong nucleophile/base
A species that actively attacks the substrate or abstracts a proton. If the reagent is strong, the mechanism is bimolecular (SN2 or E2). Examples: HO⁻, RO⁻, CN⁻, N₃⁻, SR⁻, H₂N⁻, R₃N, LDA, t-BuO⁻. In simple terms, these are reactive enough to force something to happen in a single step with the substrate.
Weak nucleophile/base
A species that is not reactive enough to force a bimolecular mechanism. The substrate must ionise first (forming a carbocation), so the mechanism is unimolecular (SN1 or E1). Examples: H₂O, CH₃OH, CH₃SH. In simple terms, these sit and wait for the substrate to fall apart on its own.
Nucleophilicity vs. basicity
Nucleophilicity measures how well a species donates electrons to an electrophilic carbon (attacks carbon). Basicity measures how well it abstracts a proton. Some species are good nucleophiles but weak bases (e.g. RS⁻, CN⁻), and some are strong bases but poor nucleophiles (e.g. t-BuO⁻, LDA). This distinction decides whether SN2 or E2 wins.
Leaving group type (1°, 2°, 3°)
Refers to the substitution of the carbon bonded to the leaving group. A primary (1°) carbon has one alkyl group attached, secondary (2°) has two, and tertiary (3°) has three. This classification is central to the decision flowchart because it determines steric accessibility.
Polar protic solvent
A solvent with O–H or N–H bonds that can hydrogen-bond (e.g. water, methanol, ethanol). Stabilises ions through solvation. Favours SN1 and E1.
Polar aprotic solvent
A solvent that is polar but lacks O–H or N–H bonds (e.g. DMSO, DMF, acetone, THF). Does not stabilise anions well, leaving the nucleophile "naked" and more reactive. Favours SN2 and E2.
This is the systematic approach from the Baldwin guide:
Identify whether the nucleophile/base is strong or weak.
Strong → the reaction is either SN2 or E2
Weak → the reaction is either SN1 or E1
Confirm you have a good leaving group and determine the substrate type (1°, 2°, 3°).
No good leaving group → no reaction
The substrate type narrows down which pathway is viable
For strong reagents: decide if the species is more nucleophilic or more basic.
Good nucleophile (not bulky, charged or neutral) → SN2
Strong base (especially if bulky) → E2
For weak reagents: decide if the reaction is heated.
No heat → SN1 dominates
Heat (Δ) → E1 dominates
Cross-reference the substrate type with the reagent type to find the major pathway (see the flowchart below).
Confirm that other variables match (solvent, stereochemistry, products).
Branch 1: Weak nucleophile/base (H₂O, CH₃OH, CH₃SH)
Substrate | Condition | Major Pathway |
|---|---|---|
1° | Any | No reaction (carbocation too unstable to form) |
2° or 3° | Heat (Δ) | E1 |
2° or 3° | No heat | SN1 |
Branch 2: Strong nucleophile/base
Substrate | Reagent Character | Major Pathway |
|---|---|---|
1° | Nucleophilic (CN⁻, N₃⁻, SR⁻, R₃P, R₃N, R₂S) | SN2 |
1° | Bulky base (t-BuO⁻, LDA) | E2 |
2° | Nucleophilic | SN2 (but slower than at 1°; E2 competes) |
2° | Basic | E2 |
3° | Nucleophilic | No reaction by SN2 (too hindered); E2 instead |
3° | Basic | E2 |
Weak nucleophiles/bases (wait for ionisation):
H₂O, CH₃OH (MeOH), CH₃SH
Strong nucleophiles (good at attacking carbon, drive SN2):
CN⁻, N₃⁻, SR⁻ (thiolates), R₃P, R₃N, R₂S
Strong nucleophiles that are also strong bases (can drive SN2 or E2):
HO⁻, RO⁻ (alkoxides like MeO⁻, EtO⁻), H₂N⁻, R₃N
Bulky bases (too large for SN2, drive E2 with Hofmann selectivity):
t-BuO⁻ (potassium tert-butoxide), LDA (lithium diisopropylamide)
Note: nBuLi (n-butyllithium) almost always acts as a base, not a nucleophile.
SN2 cannot occur on a 3° substrate because of steric hindrance
E2 is faster than SN2 on 2° substrates due to steric hindrance
SN2 occurs much faster on 1° substrates, unless a bulky base is used (then E2)
SN2 proceeds with inversion of stereochemistry (Walden inversion)
E2 requires an antiperiplanar hydrogen
Bulky bases form the least substituted alkene (Hofmann product)
Small bases follow Zaitsev's rule (most substituted alkene)
Both SN2 and E2 prefer polar aprotic solvents
Temperature decides which dominates: heat → E1, no heat → SN1
Both E1 and SN1 follow Zaitsev's rule for the elimination product
SN1 and E1 form the most stable carbocation possible (watch for rearrangements)
Rearrangements: no more than 2 shifts per molecule
Carbocations cannot form at 1° carbons (no SN1 or E1 at primary substrates)
Both SN1 and E1 prefer polar protic solvents
SN1 gives racemic products (no stereochemical control)
Use reversible (equilibrium) arrows when drawing SN1/E1 mechanisms
Example A: 1° alkyl bromide + NaOH in THF
NaOH is a strong nucleophile and strong base
1° substrate → SN2 is strongly favoured
THF is a polar aprotic solvent (consistent with SN2)
Product: substitution with inversion, Br replaced by OH
Answer: SN2
Example B: 2° alkyl bromide + MeOH (no strong base added)
MeOH is a weak nucleophile
2° substrate → carbocation can form
No heat mentioned → SN1 dominates
Product: methyl ether (OMe replaces Br), racemic at the reaction centre
Answer: SN1
Example C: 2° alkyl chloride + CH₃SH
CH₃SH is a weak nucleophile
2° substrate → SN1 conditions
Product: thioether, watch for rearrangement if a more stable carbocation is accessible
Answer: SN1
Example D: 1° substrate + OTs (tosylate LG) + MeOH
MeOH is weak, 1° substrate → normally no reaction (can't form 1° carbocation)
Check: is there anything special about the substrate (allylic, benzylic, resonance)?
If ordinary 1° → no reaction
Example E: 3° alkyl bromide + MeOH + heat
MeOH is weak, 3° substrate, heat applied
Both SN1 and E1 occur; heat favours E1
Major pathway: E1, with SN1 as minor
Product: most substituted alkene (Zaitsev), check for rearrangement before drawing
"A 2° substrate always goes SN1." Only with a weak nucleophile. With a strong nucleophile, 2° substrates can go SN2 (if the reagent is nucleophilic) or E2 (if the reagent is basic). The reagent matters as much as the substrate.
"SN2 and E2 prefer polar protic solvents." The opposite. SN2 and E2 prefer polar aprotic solvents. Polar protic solvents stabilise the nucleophile and slow down bimolecular reactions. SN1 and E1 prefer polar protic solvents.
"If it's a strong base, it's always E2." On a 1° substrate with a non-bulky strong nucleophile/base like NaOH, SN2 is the dominant pathway. E2 takes over when the base is bulky or the substrate is 2°–3°.
"You can just memorise the products without understanding the mechanism." The exam will ask you to draw the mechanism, predict stereochemistry, and explain why one pathway wins. The flowchart is a tool for reasoning, not a substitute for understanding the steps.
⚠️ The decision flowchart is the single most important tool for Chapters 6–7 exam problems. Practise it until it is automatic.
⚠️ Always start by classifying the nucleophile/base. This is the fork in the road that everything else follows from.
⚠️ For 2° substrates with strong reagents, the answer depends on whether the reagent is more nucleophilic or more basic. This is the trickiest case and the most commonly tested.
⚠️ Solvent choice is testable: polar protic → SN1/E1; polar aprotic → SN2/E2.
⚠️ If the problem says "heat" or shows Δ, that is a signal for E1 (with weak nucleophile) or E2 (with strong base).
⚠️ For SN1/E1 problems, always check for carbocation rearrangements before drawing products.
⚠️ For SN2 products, draw inversion of configuration. For SN1 products, draw a racemic mixture.
⚠️ For E2 products with a small base, draw the Zaitsev product. For E2 with a bulky base, draw the Hofmann product.
Fill in the blank: Strong nucleophile + 1° substrate → _______.
True or False: SN1 and E1 prefer polar aprotic solvents.
Fill in the blank: Weak nucleophile + 3° substrate + heat → _______.
True or False: SN2 can occur at a tertiary carbon.
Fill in the blank: A bulky base with a 2° substrate gives the _______ (Zaitsev/Hofmann) product via E2.
Q: A 3° alkyl bromide is treated with NaCN in DMSO. What happens?
A: NaCN is a strong nucleophile. However, the substrate is 3°, so SN2 is blocked by steric hindrance. CN⁻ is not a particularly strong base, so E2 is also sluggish. The dominant pathway is E2, though the reaction may be slow. If any SN1 occurs (unlikely in DMSO, which is polar aprotic, not protic), it would be minor.
Q: A 2° alkyl bromide is treated with NaOCH₃ in methanol. Predict the product(s).
A: NaOCH₃ is a strong base and strong nucleophile. The substrate is 2°, where E2 and SN2 compete. Because the base is not bulky, both pathways are possible, but E2 is generally faster at 2° due to steric hindrance. The major product is the Zaitsev alkene (E2). Some SN2 substitution product (with inversion) may form as a minor product.
Q: A 1° alkyl chloride is treated with t-BuOK in THF. What is the major product?
A: t-BuO⁻ is a bulky, strong base. Even though the substrate is 1°, the bulk of the base prevents SN2. E2 dominates, giving the Hofmann (least substituted) alkene as the major product.
Q: 2-Bromopentane is dissolved in water and heated. Predict the major product and mechanism.
A: Water is a weak nucleophile and polar protic solvent. 2° substrate with heat → E1 is the major pathway. The major product is the more substituted alkene (Zaitsev's rule): 2-pentene. Some SN1 product (2-pentanol, racemic) will also form as a minor product.
Q: Explain why SN2 cannot occur at a tertiary carbon.
A: SN2 requires the nucleophile to attack the electrophilic carbon from the backside (180° from the leaving group). At a tertiary carbon, three bulky alkyl groups block this approach. The steric hindrance is too great for the nucleophile to reach the carbon in the transition state, so the reaction does not proceed.
This decision framework is the foundation for every reaction chapter that follows. Addition reactions to alkenes (Chapters 8–9) produce the same kinds of substrates that undergo substitution and elimination, so the chemistry cycles. In synthesis problems later in the course, you will need to run this logic in reverse: "What starting material and conditions give me this product?" The mechanism-prediction skill built here becomes a retrosynthetic tool.
SN1 vs SN2, E1 vs E2, substitution vs elimination, reaction pathway decision, nucleophile strength, base strength, nucleophilicity vs basicity, substrate classification, primary secondary tertiary, steric hindrance, polar protic solvent, polar aprotic solvent, inversion of configuration, racemic mixture, Zaitsev's rule, Hofmann product, bulky base, decision flowchart, reaction prediction, CHEM 2510, organic chemistry chapters 6 and 7, Baldwin, Ohio State