---
title: "Dehydrohalogenation of 1° alkyl halide"
subject: "Engineering Chemistry"
chapter: 10
key: "semester_2/engineering_chemistry:dehydrohal-primary"
source: https://www.bctengineer.com/q/semester_2/engineering_chemistry/dehydrohal-primary
site: BCT Engineering
---

# Dehydrohalogenation of 1° alkyl halide

Engineering Chemistry · Chapter 10: Reaction Mechanism in Organic reactions

- Asked in: 2073 Bhadra, 2.5 marks
- Video: [Elimination reactions |Dehydrohalogenation |Class12 |Chemistry| Haloalkanes & Haloarenes |Lecture-16](https://www.youtube.com/watch?v=pKWpfHaNnXw) — Benzene Classes
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_2/engineering_chemistry/dehydrohal-primary
- Whole chapter: https://www.bctengineer.com/md/semester_2/engineering_chemistry/chapter-10

## Question

Explain the reaction mechanism of dehydrohalogenation of 1° alkyl halide.

## Short answer

Primary alkyl halide undergoes E2 elimination with alcoholic KOH: a strong base abstracts a β-hydrogen while the halogen leaves, forming an alkene in one concerted step; rate = k[RX][base].

## Answer

A primary alkyl halide undergoes dehydrohalogenation with alcoholic KOH by the **E2 mechanism** (one step), because a primary carbocation is too unstable for E1.

**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.

## Check yourself

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

1. Why do primary halides undergo E2 rather than E1?
   A. Primary halides have no beta-hydrogens
   B. E1 needs a strong base which primary halides lack
   C. Primary halides are too hindered for any elimination
   D. A primary carbocation is too unstable to form
   Answer: D. E1 requires a carbocation; primary ones are unstable, so the concerted E2 route is used. (Key point: Why E2 for primary halides)

2. How does E2 take place?
   A. In one step: base removes beta-H as C-Br breaks and the pi bond forms
   B. In two steps via a carbocation
   C. In two steps via a carbanion
   D. The halide first adds then loses H2
   Answer: A. E2 is concerted with a single transition state and no intermediate. (Key point: E2 steps)

3. What is the rate law for E2 elimination?
   A. Rate = k[RX], first order
   B. Rate = k[Base], first order
   C. Rate = k[RX][Base], second order
   D. Rate = k[RX]^2
   Answer: C. Base and substrate both take part in the single step, so E2 is bimolecular and second order. (Key point: E2 rate law)

## Close variants set in other exams

- [Dehydrohalogenation; mechanism and kinetics of E1 [partly illegible]](https://www.bctengineer.com/question.md?key=semester_2%2Fengineering_chemistry%3Ae1-kinetics) — Engineering Chemistry, chapter 10, asked 2×
- [Elimination reaction; dehydrohalogenation of tertiary halide](https://www.bctengineer.com/question.md?key=semester_2%2Fengineering_chemistry%3Ae1-tertiary-dehydrohal) — Engineering Chemistry, chapter 10, asked 3×

## Related in Engineering Chemistry

- [Mechanism for primary alkyl halide in aqueous medium; solvent](https://www.bctengineer.com/question.md?key=semester_2%2Fengineering_chemistry%3Asn2-primary-solvent) — Engineering Chemistry, chapter 10
- [SN reactions; hydrolysis of 3° alkyl halide](https://www.bctengineer.com/question.md?key=semester_2%2Fengineering_chemistry%3Asn-tertiary-hydrolysis) — Engineering Chemistry, chapter 10, asked 2×
- [Tertiary alkyl halide with aq. NaOH / alc. KOH](https://www.bctengineer.com/question.md?key=semester_2%2Fengineering_chemistry%3Atert-halide-naoh-koh) — Engineering Chemistry, chapter 10
- [3° alkyl halide with alcoholic KOH; solvent](https://www.bctengineer.com/question.md?key=semester_2%2Fengineering_chemistry%3Atert-alc-koh-solvent) — Engineering Chemistry, chapter 10
- [Hydrolysis of 3° alkyl halide by aq. NaOH; solvent](https://www.bctengineer.com/question.md?key=semester_2%2Fengineering_chemistry%3Atert-hydrolysis-solvent) — Engineering Chemistry, chapter 10
- [Primary alkyl halide with alcoholic and aqueous alkali](https://www.bctengineer.com/question.md?key=semester_2%2Fengineering_chemistry%3Aprimary-halide-alc-aq-alkali) — Engineering Chemistry, chapter 10