Difficulty: Intermediate to Advanced | Prerequisites: CHM 255 Chapters 7–8, understanding of carbocation stability, nucleophile/base strength, stereochemistry (R/S), and Newman projections.
This is the section where you need to move from knowing what the products are to understanding how the reactions happen, step by step. Exam 3 will ask you to draw curved-arrow mechanisms for SN1, SN2, E1, E2, and radical reactions. Knowing the mechanism tells you the stereochemistry, the rate law, and why certain conditions favour certain pathways. This is arguably the most testable material on the exam.
SN1 and E1 are two-step mechanisms that share a common first step (formation of a carbocation). SN2 and E2 are one-step concerted mechanisms. Radical mechanisms follow an initiation-propagation-termination chain. Each mechanism has distinctive stereochemical and kinetic signatures that exams test directly.
Rate-determining step (RDS)
The slowest step in a multi-step mechanism. It controls the overall rate of the reaction. In SN1 and E1, the RDS is the departure of the leaving group. Think of it as the bottleneck.
Carbocation
A positively charged carbon with an empty p-orbital. Formed in SN1 and E1 when the leaving group departs. Stability: 3° > 2° > 1° > methyl. A planar, sp²-hybridised intermediate.
Backside attack
In SN2, the nucleophile approaches the electrophilic carbon from the side directly opposite the leaving group (180°). This geometry causes inversion of configuration.
Concerted mechanism
A reaction where bond-breaking and bond-forming happen simultaneously in a single step, with no intermediate. SN2 and E2 are concerted.
Anti-periplanar
The geometry required for E2: the beta-H and the leaving group are 180° apart when viewed in a Newman projection. Both lie in the same plane but on opposite sides.
Homolytic cleavage
Bond breaking where each atom retains one electron, producing two radicals. This is how radical reactions are initiated (e.g. Br–Br → 2 Br·).
Example: A tertiary alkyl bromide reacts with H₂O to form an alcohol.
Step 1 (RDS): Leaving group departure
The C–Br bond breaks heterolytically. The bromine leaves with both bonding electrons, forming Br⁻ and a tertiary carbocation. This is the slow, rate-determining step. Reactivity: 3° > 2° >> 1° (methyl does not undergo SN1).
Step 2: Nucleophilic attack
The weak nucleophile (H₂O) attacks the planar carbocation from either face. Because the carbocation is flat (sp²), the nucleophile has equal access to both sides, which is why SN1 produces a racemic mixture at a chiral centre.
Step 3: Deprotonation (acid-base step)
The oxonium ion (protonated alcohol) loses a proton to a water molecule acting as a base, yielding the neutral alcohol product.
Key features: two or three steps, carbocation intermediate, racemisation, favoured by polar protic solvents (which stabilise the carbocation and the departing anion), unimolecular rate law (rate = k[substrate]).
Example: A primary alkyl chloride reacts with CN⁻ in DMF to form a nitrile.
Single concerted step:
The strong nucleophile (CN⁻) attacks the electrophilic carbon from the backside (180° from the leaving group). As the new C–CN bond forms, the C–Cl bond breaks simultaneously. The chloride departs as Cl⁻ (the leaving group).
The transition state has five groups around the central carbon: the incoming nucleophile, the departing leaving group, and the three remaining substituents, all partially bonded. This is a single energy maximum, not an intermediate.
Stereochemistry: inversion of configuration (Walden inversion). If the substrate is R, the product is S, and vice versa.
Key features: one step, no intermediate, backside attack, inversion, favoured by polar aprotic solvents (DMF, DMSO, acetone), bimolecular rate law (rate = k[substrate][nucleophile]). Best with methyl and 1° substrates; 2° is borderline; 3° does not undergo SN2 due to steric hindrance.
Example: A tertiary alkyl bromide reacts with H₂O (weak base) to form an alkene.
Step 1 (RDS): Leaving group departure
Identical to SN1. The C–Br bond breaks heterolytically, forming a carbocation. This is the slow step.
Step 2: Deprotonation of a beta-hydrogen
A base (H₂O in this case) removes a proton from a carbon adjacent to the carbocation (the beta carbon). The electrons from the C–H bond form a pi bond with the carbocation carbon, generating the alkene.
The product follows Zaitsev's rule: the more substituted alkene is the major product.
Key features: two steps, shares the carbocation intermediate with SN1 (which is why E1 and SN1 compete), weak bases, polar protic solvents, Zaitsev selectivity. E1 always accompanies SN1 with 2° and 3° substrates.
Example: A primary alkyl chloride reacts with a strong base (e.g. alkoxide, ⁻OH) to form an alkene.
Single concerted step:
The strong base removes a beta-hydrogen at the same time as the leaving group departs. The electrons from the C–H bond form the new C=C pi bond as the C–Cl bond breaks. All four bond changes happen simultaneously.
Geometry requirement: the beta-H and the leaving group must be anti-periplanar (180° dihedral angle). This can be confirmed by drawing a Newman projection and verifying the H and leaving group are anti to each other.
Product selectivity: Zaitsev's rule applies. The more substituted alkene is the major product (unless a bulky base like tert-butoxide is used, which can favour the less substituted, Hofmann product).
Key features: one step, no intermediate, anti-periplanar geometry required, strong bulky bases, bimolecular rate law (rate = k[substrate][base]). Works with 1°, 2°, and 3° substrates, but competes with SN2 at 1° and 2°.
Initiation: Br₂ undergoes homolytic cleavage with heat (Δ) or light (hν), producing two bromine radicals (Br·).
Propagation (two steps, repeating cycle):
A Br· radical abstracts a hydrogen from the alkane, forming HBr and a carbon radical. The radical forms at the most substituted position (3° > 2° > 1°) because more substituted radicals are more stable.
The carbon radical reacts with another Br₂ molecule, forming the alkyl bromide product and regenerating a Br· radical to continue the chain.
Termination: Two radicals combine (Br· + Br·, R· + Br·, or R· + R·), ending the chain.
Initiation: Same as above, Br₂ → 2 Br· with heat.
Propagation:
A Br· radical abstracts the allylic hydrogen (the H on the carbon next to the double bond), forming an allylic radical. This radical is resonance-stabilised: the unpaired electron is delocalised across the allylic system.
The allylic radical reacts with Br₂ to form the allylic bromide and regenerate Br·.
The resonance stabilisation of the allylic radical is the reason this position is preferentially attacked. You should be able to draw the two resonance structures showing the radical on each end of the allylic system.
Mechanism | Steps | Substrate | Nucleophile/Base | Solvent | Stereochemistry |
|---|---|---|---|---|---|
SN1 | 2–3 | 3° > 2° | Weak nucleophile | Polar protic | Racemisation |
SN2 | 1 | Methyl > 1° > 2° | Strong nucleophile | Polar aprotic | Inversion |
E1 | 2 | 3° > 2° | Weak base | Polar protic | Zaitsev product |
E2 | 1 | 3° > 2° > 1° | Strong base | Polar aprotic or protic | Zaitsev product (anti-periplanar H) |
Students often draw a carbocation intermediate for SN2 or E2. These are concerted, one-step mechanisms. There is no intermediate, only a transition state.
Students sometimes draw E2 with the H and leaving group on the same side (syn). E2 requires anti-periplanar geometry. Check with a Newman projection.
Students confuse the stereochemical outcomes. SN1 = racemisation (flat carbocation, attack from both sides). SN2 = inversion (backside attack only). E1 and E2 do not create new stereocentres, but E2 can control alkene geometry (E/Z) through the anti-periplanar requirement.
Students forget that SN1 and E1 share the same first step. If conditions favour one, the other will also occur as a competing pathway.
⚠️ You will be asked to draw full curved-arrow mechanisms for at least one (likely two) of SN1, SN2, E1, E2. Practice drawing arrows from the electron source to the electron sink.
⚠️ The summary table above is the single most useful thing to memorise for the exam. Substrate + nucleophile/base strength + solvent = the mechanism.
⚠️ Radical mechanisms: know the three phases (initiation, propagation, termination) and be ready to draw each step with single-headed (fishhook) arrows.
⚠️ Stereochemistry questions will test SN1 (racemic) vs SN2 (inversion). If you see a wedge-dash substrate, you must draw the correct stereochemical outcome.
True or false: SN1 and E1 have the same rate-determining step. (True. Both start with leaving group departure to form a carbocation.)
Fill in the blank: SN2 produces ______ of stereochemistry, while SN1 produces a ______ mixture. (Inversion; racemic.)
True or false: E2 can proceed with the H and leaving group in a gauche conformation. (False. Anti-periplanar is required.)
Fill in the blank: In radical mechanisms, single-headed (fishhook) arrows show the movement of ______ electron(s). (One.)
True or false: A methyl substrate can undergo SN1. (False. Methyl carbocations are far too unstable.)
Q: A chiral secondary alkyl bromide is treated with H₂O. Draw the mechanism and predict the stereochemical outcome.
A: SN1 mechanism. Step 1: the Br⁻ departs (RDS), forming a planar carbocation. Step 2: H₂O attacks the carbocation from either face. Step 3: deprotonation gives the alcohol. The product is a racemic mixture (both R and S enantiomers formed in roughly equal amounts). An E1 alkene product will also be present.
Q: 1-Chlorobutane is treated with NaCN in DMSO. What is the mechanism? Draw the stereochemical outcome if the starting material is (R)-1-chlorobutane.
A: SN2. CN⁻ is a strong nucleophile, DMSO is polar aprotic, and the substrate is primary. One concerted step with backside attack. The product is (S)-pentanenitrile (inversion of configuration).
Q: Draw the full radical mechanism for the bromination of propane at the secondary position.
A: Initiation: Br₂ → 2 Br· (heat or light). Propagation step 1: Br· abstracts the 2° H from propane, forming a 2° carbon radical and HBr. Propagation step 2: the 2° radical reacts with Br₂, forming 2-bromopropane and regenerating Br·. Termination: any two radicals combine. Use single-headed (fishhook) arrows throughout.
Q: Why does E2 require anti-periplanar geometry?
A: The concerted mechanism requires orbital overlap between the C–H sigma bond (breaking), the C–C bond (becoming the pi bond), and the C–LG sigma bond (breaking). This overlap is maximised when all four atoms (H, C, C, LG) are in the same plane and the H and LG are anti (180°). Draw a Newman projection to confirm.
These mechanisms underpin nearly every reaction for the rest of the course. SN2 appears in Williamson ether synthesis, acetylide alkylation, and Gabriel synthesis. E2 returns in dehydrohalogenation and alcohol dehydration. Radical mechanisms reappear in polymerisation and some oxidation reactions. Understanding mechanisms also connects to kinetics (rate laws from the RDS) and thermodynamics (Hammond postulate for selectivity).
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