E2 Elimination
E2 Elimination: The base removes a β-hydrogen anti-periplanar to the leaving group in the very same step that the C–leaving-group bond breaks, forming the…
What happens
The base removes a β-hydrogen anti-periplanar to the leaving group in the very same step that the C–leaving-group bond breaks, forming the new π bond and the alkene in one concerted step, no carbocation ever forms.
Reaction summary
CH₃CH(Br)CH₂CH₃ + strong base → CH₃CH=CHCH₃ (major) + HBr
Reagents and conditions
- Starting material: 2-Bromobutane
- Reagents: Strong base (e.g. KOtBu)
- Products: 2-Butene (major, Zaitsev product)
Mechanism
- Anti-periplanar elimination: The base removes a β-hydrogen oriented 180° from the leaving group; as that C–H bond breaks, the C–Br bond breaks simultaneously and a new π bond forms.
- Alkene product: The more substituted (more stable) alkene is usually the major product (Zaitsev's rule), unless a bulky base favors the less hindered Hofmann product instead.
Selectivity
- Regioselectivity: Zaitsev's rule: the more substituted alkene is the major product with small bases, a sterically bulky base (e.g. potassium tert-butoxide) favors the less hindered Hofmann product instead.
- Stereoselectivity: Requires anti-periplanar geometry between the leaving group and the β-hydrogen being removed.
Common mistakes
- Forgetting E2 needs a strong base and happens in one concerted step, unlike E1, there's no carbocation intermediate and thus no rearrangement.
- Picking a β-hydrogen that isn't anti-periplanar to the leaving group, the geometry requirement isn't optional.
Memory trick
"E2 = 2 things break together, 1 step, anti-periplanar": base removes H and the leaving group leaves, simultaneously, from opposite faces.