Mastering concerted bimolecular elimination — the anti-periplanar requirement, regioselectivity, and how to distinguish E2 from SN2.
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The E2 (elimination bimolecular) reaction is a one-step, concerted elimination where all three events happen simultaneously:
Because both the substrate and the base are involved in the rate-determining step, E2 follows second-order kinetics.
Unlike SN1 or E1, there is no carbocation intermediate. The base removes a β-hydrogen while:
All at once — this is why E2 is called a concerted mechanism.
For E2 to proceed, the β-hydrogen and leaving group must be anti-periplanar — exactly 180° apart in the same plane. This geometric requirement arises from orbital alignment: the σ C–H bond must overlap with the σ* C–X antibonding orbital to allow simultaneous bond breaking and π bond formation.
Free rotation around C–C bonds allows molecules to adopt the required anti conformation easily. Draw a Newman projection to confirm the geometry before predicting the product.
The leaving group and the β-hydrogen must both be axial. If the leaving group is equatorial, a chair flip must occur before elimination can take place. This is one of the most commonly tested E2 concepts — always draw the chair conformation and identify axial substituents before predicting the E2 product.
When multiple β-hydrogens exist, which alkene forms as the major product?
The more substituted alkene is favored. More substituted alkenes have more stabilizing hyperconjugation interactions and are thermodynamically more stable. When a small, unhindered base is used, it preferentially removes the β-H that leads to the most substituted alkene.
Forms when a bulky base is used (e.g., tert-butoxide, LDA, triethylamine). Steric hindrance prevents the bulky base from approaching the more substituted β-hydrogen, so it instead removes the less hindered, less substituted β-hydrogen — giving the less substituted alkene (Hofmann product).
Rate = k[substrate][base]
Unlike SN1/E1, no carbocation stability is required. The rate depends directly on base strength — not on how well the substrate stabilizes a positive charge.
E2 shows a single energy barrier with one transition state — a concerted, four-center geometry where the base, β-hydrogen, α-carbon, and leaving group are all partially bonded simultaneously. There is no intermediate.
β-Branching also increases E2 relative to SN2, even with small nucleophiles, due to steric hindrance around the carbon center.
A strong base is required: OH⁻, OR⁻, tert-butoxide, LDA, triethylamine. Weak bases favor E1 instead (or SN1/SN2 depending on the substrate).
Polar aprotic solvents increase base strength (nucleophile/base left "naked") and favor E2. Polar protic solvents can weaken the base through hydrogen bonding, pushing the reaction toward E1 or SN1.
If you can answer these four questions for any substrate, you can solve almost any E2 exam problem.
Practice drawing chair conformations. Practice predicting Zaitsev vs. Hofmann. Practice distinguishing E2 from SN2. The roadmap PDF (available from the overview guide) walks through the full decision tree for all four reactions.
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