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E2

Organic ChemistryElimination reactionsReaction mechanismsAlkyl halides

Summary

E2 is bimolecular elimination: in one concerted step, a strong base removes a β-hydrogen as the leaving group departs, forming an alkene. The rate depends on both substrate and base, and the H and leaving group must be anti-periplanar.

Detail

Meaning: E2 is a one-step, concerted elimination. A strong base pulls off a β-hydrogen while the leaving group leaves the α carbon and a C=C π bond forms. No carbocation intermediate forms. The H and leaving group must be anti-periplanar; on a cyclohexane ring, both must be axial (trans-diaxial). The more substituted alkene usually predominates (Zaitsev), but a bulky base such as tert-butoxide often favors the less substituted (Hofmann) alkene.

Equation: rate = k[R–X][B], where k is the rate constant (m³ mol⁻¹ s⁻¹, often given as M⁻¹s⁻¹), [R–X] is the substrate concentration and [B] is the base concentration (mol/m³, often M). The reaction is second order overall.

On the MCAT: Passages ask you to choose among SN1, SN2, E1 and E2 from the substrate, base or nucleophile, solvent and temperature. A strong base and heat favor E2, and a tertiary halide with a strong base typically gives E2 because SN2 is blocked by crowding. Doubling the base concentration doubling the rate points to E2 or SN2. Stereochemistry questions ask which hydrogen can be removed in a Newman projection or chair, and therefore which alkene isomer forms.

Common confusion: Students assume E2 follows the SN2 substrate trend. It does not: crowding blocks SN2, while E2 is generally faster for more substituted halides (3° > 2° > 1°). Remember that E2 needs the anti-periplanar H, not just any β-hydrogen.

Sources

  • Organic Chemistry (Clayden)
  • Organic Chemistry (Wade)
  • Organic Chemistry (Klein)

Written for the MCAT and checked before it was published. For study reference only. Spotted an error? Let us know.

E2: MCAT Glossary