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Elimination reaction; dehydrohalogenation; Saytzeff's rule

Short answer

Elimination removes two groups from adjacent carbons to form a double bond. Dehydrohalogenation of a primary alkyl halide follows E2: base removes β-hydrogen and halide leaves simultaneously. Saytzeff’s rule states the more substituted alkene is the major product.

What do you mean by elimination reaction? Write the mechanism for dehydrohalogenation in primary alkylhalide. Show your acquaintance to Saytzeff's rule.

Elimination reaction

An elimination reaction is a reaction in which two atoms or groups are removed from adjacent carbon atoms of a molecule, forming a double (or triple) bond, without being replaced by other groups. Dehydrohalogenation of an alkyl halide is the elimination of H and X (as HX) using a strong base such as alcoholic KOH to form an alkene.

Dehydrohalogenation of a primary alkyl halide

Mechanism: E2 (one step, concerted).

 B:⁻  H—CH₂—CH₂—Br   →  [B···H···CH₂···CH₂···Br]⁻‡  →  B–H + CH₂=CH₂ + Br⁻

The base removes the β-hydrogen while the C–Br bond breaks and the π bond forms at the same time (the H and Br are anti-periplanar). Example: CH₃CH₂Br + KOH(alc) → CH₂=CH₂ + KBr + H₂O

  • Rate = k[RX][Base]: second order (bimolecular).
  • Needs a strong base and a high concentration of it; there is no intermediate.

Saytzeff's rule

Saytzeff (Zaitsev) rule: in dehydrohalogenation, when more than one alkene can form, the major product is the more highly substituted (more stable) alkene, i.e. the hydrogen is removed from the β-carbon that has the fewest hydrogens. Example: 2-bromobutane + alc. KOH:

CH₃CH₂CH(Br)CH₃ --alc. KOH--> CH₃CH=CHCH₃ (but-2-ene, ~80%, major)
                              + CH₃CH₂CH=CH₂ (but-1-ene, ~20%, minor)

More alkyl groups on C=C give greater stability by hyperconjugation.

✓ Answer checked by Claude Opus 5.5 (high effort)

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