Source: Chem 237 Marathon Worksheet Key (University of Washington)
Difficulty: Intermediate | Prerequisites: Chapters 6–8 (stereochemistry, alkyl halides, nucleophiles and bases)
Tags: SN1, SN2, E1, E2, radical halogenation, substitution, elimination, nucleophilic substitution, leaving group, carbocation, alkyl halide, stereochemistry, Walden inversion, Zaitsev rule
This chapter is where organic chemistry stops being a catalogue of reactions and starts being a puzzle. You have an alkyl halide (or similar substrate), a reagent, and you need to predict which of five possible pathways the reaction takes: SN1, SN2, E1, E2, or radical halogenation. The answer depends on a handful of variables you can read straight off the structure: the quality of the leaving group, the degree of substitution at the electrophilic carbon, and whether the reagent is a strong or weak nucleophile and base. Getting this decision logic down is the single most exam-relevant skill in CHEM 237.
When an alkyl halide meets a reagent, five pathways compete: SN1, SN2, E1, E2 and radical halogenation. You pick the winner by checking three things in order: is the leaving group any good, is the reagent a strong or weak nucleophile/base, and how crowded is the carbon bearing the leaving group. Master that checklist and most Chapter 9 problems collapse to about 30 seconds of reasoning.
SN1 (Unimolecular Nucleophilic Substitution)
A two-step substitution in which the leaving group departs first to form a carbocation, then the nucleophile attacks. Rate depends only on substrate concentration.
In simple terms: the leaving group leaves on its own, creating a flat, positively charged carbon that the nucleophile then hits from either face.
SN2 (Bimolecular Nucleophilic Substitution)
A one-step substitution in which the nucleophile attacks the electrophilic carbon at the same time as the leaving group departs. Rate depends on both substrate and nucleophile concentration.
Think of it as a backside attack: the nucleophile pushes in from the opposite side of the leaving group, like an umbrella flipping inside out.
E1 (Unimolecular Elimination)
A two-step elimination sharing the same carbocation intermediate as SN1. After the leaving group departs, a base removes a proton from an adjacent carbon to form an alkene.
In simple terms: the leaving group leaves first (just like SN1), but instead of a nucleophile attacking, a base plucks off a neighbouring hydrogen to make a double bond.
E2 (Bimolecular Elimination)
A one-step elimination in which a strong base removes a proton anti-periplanar to the leaving group, forming an alkene in a single concerted step.
Think of it as: the base grabs a hydrogen and the leaving group exits at the same time, but only if the H and the leaving group are on opposite sides of the bond (anti-coplanar geometry).
Leaving group
The atom or group that departs with the bonding electrons. Good leaving groups are weak bases (halides such as Br⁻, Cl⁻, I⁻ and tosylate). Bad leaving groups (e.g. OH⁻, NH₂⁻) shut down all four ionic mechanisms.
In simple terms: if the group cannot leave on its own as a stable anion, none of SN1/SN2/E1/E2 will happen.
Carbocation
A positively charged, sp²-hybridised carbon intermediate. Stability order: tertiary > secondary > primary > methyl. Relevant to SN1 and E1, which both go through this intermediate.
In simple terms: a carbon that has temporarily lost its leaving group and is desperate for electrons. The more alkyl groups surrounding it, the more stable it is.
Carbocation rearrangement (hydride / methyl shift)
A 1,2-shift of a hydrogen atom (hydride shift) or methyl group to an adjacent carbocation, converting it to a more stable carbocation. Occurs only in mechanisms that form carbocations (SN1, E1).
In simple terms: if a neighbouring carbon can donate a hydrogen or methyl to turn a secondary carbocation into a tertiary one, it will.
Anti-periplanar (anti-coplanar) geometry
The spatial requirement for E2: the C–H bond and the C–leaving group bond must be 180° apart (anti to each other) in a Newman projection. This controls which hydrogen is removed and therefore which alkene forms.
Think of it as: the H and the leaving group must point in exactly opposite directions for E2 to work.
Zaitsev's rule (Saytzeff's rule)
The more substituted (more stable) alkene is the major product of elimination, unless steric effects override.
In simple terms: the double bond forms preferentially toward the more crowded side of the molecule.
Radical halogenation
A chain reaction initiated by light (hv) or heat, in which a halogen radical abstracts a hydrogen from an alkane to form an alkyl halide. Selectivity depends on the halogen: bromine is highly selective for the most stable radical; chlorine is largely unselective; fluorine reacts based on statistical probability; iodine is generally unreactive.
Every Chapter 9 problem follows the same logic. Work through these checks in order and you will land on the correct mechanism nearly every time.
Step 1: Is there a good leaving group?
Good leaving groups: Br⁻, Cl⁻, I⁻, tosylate (OTs), mesylate (OMs), water (after protonation)
Bad leaving groups: OH⁻, OR⁻, NH₂⁻, F⁻ (borderline)
If the leaving group is bad, no reaction occurs. Full stop. This eliminates SN1, SN2, E1 and E2 entirely.
Step 2: Is the reagent a strong base, a strong nucleophile, both, or neither?
Strong base AND strong nucleophile (e.g. NaOH, NaNH₂, NaOCH₃): both E2 and SN2 are on the table. The substrate decides which wins.
Strong base but poor nucleophile (e.g. KOtBu, bulky alkoxides): E2 only. The steric bulk prevents backside attack.
Weak base but strong nucleophile (e.g. NaCN, NaI, NH₃, CH₃CO₂⁻): SN2 is possible. E2 is off the table.
Weak base AND weak nucleophile (e.g. H₂O, CH₃OH): only SN1 and E1 are possible, and only if the substrate can form a stable carbocation.
Step 3: What is the degree of substitution at the electrophilic carbon?
Primary (1°): SN2 is strongly favoured. SN1/E1 do not occur (primary carbocations are too unstable). E2 occurs if a strong base is present.
Secondary (2°): SN2 is feasible (slower than 1°, but still works). With a strong base, E2 competes. With a weak base/weak nucleophile, SN1/E1 can occur but are slower than for 3° substrates.
Tertiary (3°): SN2 is completely blocked by steric hindrance. Strong base gives E2. Weak base/weak nucleophile gives SN1 + E1 (these always compete with each other).
Step 4: Is it a radical reaction?
If the reagent is X₂ (Cl₂, Br₂, F₂, I₂) with light (hv) or heat, and the substrate is an alkane (no leaving group), the mechanism is radical halogenation. No leaving group is needed for this pathway.
Primary alkyl halides (1°)
SN2 is the dominant pathway. The electrophilic carbon is unhindered, so backside attack by a nucleophile proceeds easily.
E2 can compete if a strong, bulky base is present (e.g. KOtBu), but with a typical strong base/nucleophile like NaOH, SN2 wins.
SN1 and E1 do not occur. A primary carbocation is far too unstable to form.
With a weak base and weak nucleophile (e.g. H₂O), no reaction occurs, because SN2 needs a strong nucleophile and SN1 cannot form the carbocation.
Secondary alkyl halides (2°)
The borderline case. Both SN2 and E2 are possible with a strong nucleophile/base.
SN2 is feasible but slower than at a primary carbon. Reaction will invert stereochemistry (Walden inversion).
E2 becomes increasingly competitive, especially with bulky bases.
With weak base/weak nucleophile, SN1 and E1 become options, though they are slower than for tertiary substrates.
Tertiary alkyl halides (3°)
SN2 is completely impossible. Three bulky groups block the backside of the electrophilic carbon.
E2 is the only option with a strong base.
SN1 and E1 are the pathways with a weak base/weak nucleophile. These two always run in parallel because they share the same carbocation intermediate. The carbocation is tertiary and therefore stable enough to form.
No carbocation rearrangement is needed when the carbocation is already tertiary.
The reagent's character as a nucleophile versus a base is one of the most important variables in mechanism selection.
Strong base + strong nucleophile
Examples: NaOH (hydroxide), NaOCH₃ (methoxide), NaNH₂ (amide).
These reagents open the door to both SN2 and E2.
At a primary carbon, SN2 dominates.
At a secondary carbon, SN2 and E2 compete roughly evenly.
At a tertiary carbon, only E2 occurs (SN2 is sterically blocked).
SN1 and E1 are not required when a strong base/nucleophile is present.
Strong base + poor nucleophile (bulky)
Examples: KOtBu (potassium tert-butoxide).
The steric bulk around the oxygen makes it a terrible nucleophile but it remains a powerful base.
E2 is the only viable pathway.
The most stable (most substituted, Zaitsev) alkene is typically the major product.
Note for later courses: tert-butoxide can sometimes favour the less substituted (Hofmann) alkene for steric reasons, but this is not tested in Chem 237.
Weak base + strong nucleophile
Examples: NaCN (cyanide), NaI (iodide), NH₃ (ammonia), CH₃CO₂⁻ (acetate).
E2 is off the table because the reagent is too weak a base to abstract a proton in a concerted step.
SN2 is the dominant pathway at primary and secondary carbons.
At a tertiary carbon, SN2 is blocked, and the reagent is not basic enough for E2, so SN1/E1 become the options.
Weak base + weak nucleophile (solvent-type)
Examples: H₂O, CH₃OH (methanol), CH₃CH₂OH (ethanol).
Neither SN2 nor E2 can occur.
SN1 and E1 are the only options, but they require a substrate that can form a reasonably stable carbocation (secondary or tertiary).
At a primary carbon with a weak base/weak nucleophile, no reaction occurs at all.
Radical halogenation uses X₂ + light (hv) or heat. It replaces a C–H bond with a C–X bond via a radical chain mechanism (initiation, propagation, termination). The key exam question is: which hydrogen gets replaced?
The answer depends entirely on which halogen you are using.
Bromine (Br₂ / hv): highly selective
Bromine is selective for the most stable radical intermediate.
Stability order: 3° radical > 2° > 1° > methyl.
The major product is the most substituted (most crowded) alkyl bromide.
Selectivity overwhelms statistics. Even if there are many more primary hydrogens, the tertiary position wins.
Chlorine (Cl₂ / hv): largely unselective
Chlorine's selectivity for more stable radicals is modest, and it is roughly cancelled out by the statistical advantage of more abundant hydrogen types.
All possible monochlorinated products form in significant amounts.
Chlorination is generally not useful when you need a single product.
Fluorine (F₂ / hv): reacts by statistics
Fluorine is extremely reactive and essentially non-selective.
The major product is determined by whichever type of hydrogen is most plentiful (statistical control).
If there are nine equivalent methyl hydrogens and only one tertiary hydrogen, the primary alkyl fluoride is the major product.
Iodine (I₂ / hv): no reaction
Iodine is generally unreactive for radical halogenation.
The C–I bond is too weak relative to the H–I bond formed. The reaction is thermodynamically unfavourable.
SN2: inversion of configuration (Walden inversion)
The nucleophile attacks from the back side of the leaving group. The configuration at the electrophilic carbon is always inverted, like an umbrella flipping in the wind. If the starting material is R, the product is S (and vice versa).
SN1: racemisation (stereorandom)
The carbocation intermediate is sp²-hybridised and planar. The nucleophile can attack from either face with roughly equal probability. The product is a roughly 50:50 mixture of R and S configurations at the carbon that bore the leaving group. Other chiral centres in the molecule remain unchanged.
E2: anti-periplanar geometry required
The hydrogen being removed and the leaving group must be anti-coplanar (180° dihedral angle). This geometric constraint controls which hydrogen is abstracted and therefore which alkene forms. In a cyclohexane ring, both the H and the leaving group must be in axial positions on opposite faces.
If the leaving group is on a wedge bond, the hydrogen removed must be on a dash bond (and vice versa).
E1: no geometric requirement
Because the carbocation intermediate is planar, any adjacent hydrogen can be removed by the base. There is no anti-periplanar constraint. The product alkene follows Zaitsev's rule: the more substituted alkene is favoured.
Radical halogenation: stereorandom at the radical centre
The radical intermediate is also planar (sp²). If a chiral centre is created, a racemic mixture results.
"OH is a good leaving group." It is not. Hydroxide (OH⁻) is a strong base, which makes it a terrible leaving group. Alcohols do not undergo SN1, SN2, E1, or E2 unless the OH is first protonated to water (H₂O), which is a good leaving group. Multiple worksheet problems (2, 9, 15) test exactly this.
"A strong nucleophile means SN2 always wins." Not at a tertiary carbon. Steric hindrance completely blocks backside attack regardless of nucleophile strength. At a tertiary substrate with a strong nucleophile that is also a strong base, E2 takes over.
"SN1 can happen at a primary carbon if the nucleophile is weak." It cannot. A primary carbocation is far too unstable to form under normal conditions. Weak base + weak nucleophile + primary substrate = no reaction.
"E2 can use any hydrogen adjacent to the leaving group." Only in a superficial sense. E2 requires anti-periplanar geometry between the H and the leaving group. In rigid systems (cyclohexanes, bicyclics), this constraint limits which hydrogens are available, and it controls the stereochemistry of the resulting alkene.
"Carbocation rearrangements happen in all mechanisms." They happen only in SN1 and E1, because those are the only mechanisms that form a carbocation intermediate. SN2 and E2 are concerted (one step), so there is no carbocation to rearrange.
⚠️ The "no reaction" cases are tested heavily. If the leaving group is bad (OH, NH₂), write "no reaction" and explain why. Do not try to force a mechanism.
⚠️ Always state which mechanisms you considered and why you ruled them out, not just which one wins. Full marks require the reasoning.
⚠️ Stereochemistry is frequently tested alongside mechanism identification. If the problem says "include stereochemistry where relevant," you must address inversion (SN2), racemisation (SN1), and anti-periplanar geometry (E2).
⚠️ Carbocation rearrangements (hydride and methyl shifts) are a common trap. Whenever SN1 or E1 is the mechanism, ask yourself: can the carbocation rearrange to become more stable? If a secondary carbocation sits next to a tertiary centre, a hydride shift will occur.
⚠️ Know the radical halogenation selectivity hierarchy cold: Br₂ is selective for the most stable radical, Cl₂ is unselective, F₂ goes by statistics, I₂ does not react. This is a guaranteed exam question.
1. True or False: An alcohol (R–OH) treated with NaOH will undergo SN2.
False. OH⁻ is a bad leaving group, so no substitution or elimination occurs.
2. Fill in the blank: SN2 reactions always proceed with ______ of stereochemistry at the electrophilic carbon.
Inversion (Walden inversion).
3. True or False: A tertiary alkyl halide treated with a strong nucleophile will undergo SN2.
False. The tertiary carbon is too sterically hindered for backside attack.
4. Fill in the blank: In radical halogenation, ______ is the most selective halogen for the most stable radical, while ______ is generally unreactive.
Bromine; iodine.
5. True or False: E2 requires the H and leaving group to be syn-periplanar.
False. E2 requires anti-periplanar (180°) geometry, not syn.
This material connects directly to Chapter 6 (stereochemistry), because SN2 inversion and SN1 racemisation both require a solid grasp of R/S assignments and wedge-dash notation. If the stereochemistry chapter still feels shaky, revisit it before sitting the Chapter 9 exam.
The leaving-group concept builds on acid-base chemistry from general chemistry. Good leaving groups are the conjugate bases of strong acids (HBr, HCl, HI), while bad leaving groups are conjugate bases of weak acids (H₂O, NH₃). Thinking in terms of pKa is a reliable shortcut.
Elimination reactions (E1, E2) produce alkenes, which are the subject of the next set of chapters (additions to alkenes). Understanding how alkenes form here sets up the reverse question: what happens when you add reagents across a double bond?
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