---
title: "Elimination reaction; dehydrohalogenation; Saytzeff's rule"
subject: "Engineering Chemistry"
chapter: 10
key: "semester_2/engineering_chemistry:elimination-saytzeff"
source: https://www.bctengineer.com/q/semester_2/engineering_chemistry/elimination-saytzeff
site: BCT Engineering
---

# Elimination reaction; dehydrohalogenation; Saytzeff's rule

Engineering Chemistry · Chapter 10: Reaction Mechanism in Organic reactions

- Asked in: 2074 Bhadra, 1+2+2 marks; 2072 Asoj, 1+2+2 marks (asked 2 times)
- Video: [Saytzeff Rule (Dehydrohalogenation rex) Haloalkane & Haloarenes| Class12th| #neet #organicchemistry](https://www.youtube.com/watch?v=I7gbz-AKeFw) — Alchemy chemistry by Bhawna
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_2/engineering_chemistry/elimination-saytzeff
- Whole chapter: https://www.bctengineer.com/md/semester_2/engineering_chemistry/chapter-10

## Question

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

## Answer

#### 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).**
```text
 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:
```text
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.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. What happens in an elimination reaction?
   A. Two atoms or groups are removed from adjacent carbons, forming a double or triple bond
   B. A group is replaced by another group on the same carbon
   C. Two molecules add together to form a larger saturated molecule
   D. A hydrogen is exchanged between two carbons without forming a pi bond
   Answer: A. Elimination removes two groups from neighbouring carbons without replacing them, so a C=C (or triple) bond forms. (Key point: Elimination reaction)

2. How does a primary alkyl halide undergo dehydrohalogenation?
   A. By E1: slow loss of Br- to give a primary carbocation
   B. By E2: base removes beta-H as C-Br breaks, in one concerted step
   C. By SN2: OH- attacks carbon and replaces Br
   D. By E1cB: base first forms a carbanion
   Answer: B. A primary carbocation is too unstable, so elimination is concerted E2 with anti-periplanar H and Br. (Key point: E2 mechanism of primary halide)

3. According to Saytzeff's rule, which alkene is the major product?
   A. The least substituted alkene
   B. The alkene formed by removing H from the carbon with most hydrogens
   C. The more highly substituted, more stable alkene
   D. An equal mixture of all possible alkenes
   Answer: C. Saytzeff: the major product is the more substituted alkene, stabilised by hyperconjugation. (Key point: Saytzeff's rule)

4. 2-Bromobutane with alcoholic KOH gives mainly which product?
   A. But-2-ene (about 80%)
   B. But-1-ene (about 80%)
   C. Butan-2-ol
   D. Butane
   Answer: A. But-2-ene is more substituted and more stable, so it is the major product; but-1-ene is minor. (Key point: Saytzeff example)

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