Difficulty: Intermediate–Advanced | Prerequisites: SN1, SN2 (Chapter 3), E1 and E2 elimination notes
This is the capstone topic for Chapters 3 and 4. You now know four reaction pathways (SN1, SN2, E1, E2), and the exam will hand you a substrate, a reagent, and conditions, then ask you to predict which mechanism dominates and what the product is. The decision depends on three variables: the substrate (methyl, primary, secondary, or tertiary), the reagent (strong nucleophile, strong base, weak nucleophile/base), and the solvent/conditions. If you can systematically work through these three factors, you can answer nearly every competition question. This set of notes brings together the decision logic.
Strong nucleophile + primary or methyl substrate = SN2. Strong base + secondary or tertiary substrate = E2. Weak nucleophile/base + tertiary substrate = SN1/E1 mixture. The reagent's character (nucleophilic vs basic) and the substrate's steric environment are the two biggest factors.
Nucleophile
A species that donates an electron pair to an electrophilic carbon. Good nucleophiles are typically small, negatively charged, and not sterically hindered (e.g. CN⁻, SH⁻, I⁻). In simple terms, a nucleophile attacks carbon.
Base
A species that abstracts a proton. Strong bases are often bulky or hindered (e.g. tert-butoxide, DBU, LDA), or are strong but non-bulky (NaOH, NaOMe, NaOEt). In simple terms, a base grabs a hydrogen.
Nucleophile vs base distinction
Some reagents can act as both. The key question is whether the species tends to attack carbon (nucleophilic, favours substitution) or abstract a proton (basic, favours elimination). Small, polarisable anions (CN⁻, RS⁻, I⁻) tend to be better nucleophiles. Large, hindered anions (tert-butoxide) tend to be better bases.
Strong base
A base strong enough to force E2 (e.g. NaOMe, NaOEt, KOtBu, NaOH, NaNH₂). These are the conjugate bases of weak acids.
Weak base / weak nucleophile
Species like H₂O, ROH, or other neutral molecules. These cannot force E2 or SN2. With tertiary substrates they promote SN1/E1 by acting as bystanders while the leaving group departs on its own.
Substrate classification
Methyl (CH₃–X), primary (RCH₂–X), secondary (R₂CH–X), tertiary (R₃C–X). This classification is the first branch point in every mechanism-prediction question.
Work through these questions in order:
What is the substrate? Methyl, primary, secondary, or tertiary?
What is the reagent? Strong nucleophile, strong base, both, or weak/neutral?
What conditions? Polar protic solvent, heat, concentration?
Strong nucleophile (e.g. NaCN, NaSH, NaI): SN2 dominates. The carbon is unhindered, so backside attack is easy.
Strong base that is also a nucleophile (e.g. NaOMe): SN2 (major), with some E2 (minor).
Weak nucleophile/base (e.g. H₂O): SN2 is possible but slow. SN1 and E1 are very unlikely because primary carbocations are too unstable.
Strong, non-bulky nucleophile/base (e.g. NaOMe): a mixture of SN2 and E2. The exact ratio depends on temperature and steric details.
Strong, bulky base (e.g. KOtBu): E2 dominates. The bulky base cannot perform backside attack easily, so it abstracts a proton instead.
Weak nucleophile/base (e.g. H₂O): SN1 and E1 compete. These are typically slow with secondary substrates unless the solvent is very polar protic.
Strong base (any): E2 dominates. SN2 is impossible at a tertiary carbon due to steric hindrance.
Weak nucleophile/base (e.g. H₂O): SN1 and E1 compete. The leaving group departs to form a relatively stable tertiary carbocation, and then either a nucleophile attacks (SN1) or a proton is lost (E1). Heat favours E1.
To shift from SN1/E1 to E2: use a stronger base. A strong base forces the concerted E2 pathway instead of waiting for the leaving group to depart on its own.
To shift from E2 to E1: use a weaker base. Without a strong base to force concerted elimination, the reaction defaults to the stepwise E1 path (if the substrate can form a stable carbocation).
To favour elimination over substitution: increase temperature, use a bulky base, or use a tertiary substrate.
To favour substitution over elimination: use a good nucleophile that is not a strong base (e.g. CN⁻, I⁻), keep the temperature low, use a primary substrate.
Small, polarisable, negatively charged species that are not sterically hindered tend to be better nucleophiles (CN⁻, SH⁻, I⁻). They favour substitution.
Large, hindered, strongly basic species tend to be better bases (tert-butoxide, diisopropylamine). They favour elimination.
Some species are both good nucleophiles and strong bases (e.g. NaOMe, NaOEt). With these, the substrate decides: primary substrates give SN2; tertiary substrates give E2; secondary substrates give a mixture.
Students often try to apply SN2 to tertiary substrates. SN2 never occurs at a tertiary carbon; the steric hindrance is too great.
A common error is treating every negatively charged species as a strong base. CN⁻ is negatively charged but is a much better nucleophile than it is a base. Character matters more than charge.
Students sometimes forget that SN1 and E1 compete with each other (they share the same first step). A question asking for "the product" of a tertiary substrate with a weak nucleophile may have both substitution and elimination products.
Changing "use a stronger base" and "use a weaker base" answers get swapped frequently. Remember: strong base forces E2; weak base defaults to E1 (if the substrate supports it).
Pharmaceutical synthesis routinely requires choosing between substitution and elimination. Medicinal chemists select reagents and conditions to steer reactions towards the desired product. For instance, when building a drug molecule that requires a new C–C bond at a primary carbon, they choose a strong nucleophile (like cyanide) under SN2 conditions, deliberately avoiding elimination.
⚠️ "By what mechanism does this reaction proceed?" is one of the highest-frequency question types. Have the decision framework memorised: substrate first, then reagent.
⚠️ "How would you change the conditions to favour E2/E1/SN2?" is another classic. Know the lever: base strength.
⚠️ Questions that give you a substrate + reagent and ask for the product(s) require you to predict the mechanism first, then draw the product. Do not skip the mechanism step.
⚠️ Nucleophile vs base identification (e.g. "Is CN⁻ more likely a nucleophile or a base?") appears as a standalone short-answer question. Small and polarisable = nucleophile. Large and hindered = base.
Fill in the blank: A tertiary alkyl halide treated with a strong base undergoes ______ elimination. (E2.)
True or false: SN2 can occur at a tertiary carbon. (False.)
Fill in the blank: To change a reaction from E2 to E1, you would use a ______ base. (Weaker.)
True or false: CN⁻ is more likely to act as a base than a nucleophile. (False. CN⁻ is a good nucleophile.)
Fill in the blank: A primary alkyl halide with NaCN undergoes ______ substitution. (SN2.)
Q: A secondary alkyl bromide is treated with NaSH. By what mechanism is this reaction likely to occur?
A: SN2. NaSH provides HS⁻, which is a strong nucleophile (small, polarisable, charged). Secondary substrates can undergo SN2 with a good nucleophile.
Q: A primary alkyl chloride is treated with NaOMe. What is the most likely mechanism?
A: SN2 (major), with E2 as a minor pathway. NaOMe is both a good nucleophile and a strong base, but the primary substrate favours SN2.
Q: A tertiary alkyl bromide is treated with H₂O. By what mechanism does the reaction proceed?
A: E1 or SN1. Water is a weak nucleophile and weak base. The tertiary substrate ionises to form a carbocation (the rate-determining step), and then either substitution (SN1) or elimination (E1) follows.
Q: How would you change the conditions of a tertiary alkyl bromide + H₂O reaction to favour E2?
A: Use a stronger base instead of water (e.g. NaOEt or KOtBu). A strong base forces the concerted E2 mechanism.
Q: A secondary alkyl iodide is treated with a bulky amine (diisopropylamine). What mechanism dominates?
A: E2. The reagent is a strong, bulky base. It cannot easily perform backside attack on a secondary carbon, so it abstracts a beta-proton instead.
Q: What are the most likely product(s) when a secondary alkyl tosylate is treated with H₂O?
A: A mixture of the substitution product (an alcohol, from SN1, with possible racemisation) and the elimination product (an alkene, from E1). With a secondary substrate and a weak nucleophile, SN1 and E1 compete.
Q: A secondary alkyl iodide is treated with NaOMe. The desired product is the E2 elimination alkene. How could you increase the proportion of elimination?
A: Switch to a bulkier base such as KOtBu. Bulky bases favour elimination over substitution because they cannot easily reach the electrophilic carbon for backside attack.
This framework brings together everything from Chapters 3 and 4 into a single decision tree. It is the foundation for retrosynthetic analysis later in the course, where you will need to choose reagents and conditions to build target molecules. The nucleophile-vs-base distinction also reappears in carbonyl chemistry (Chapter 7+), where nucleophilic addition competes with enolisation.
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