---
title: "Elimination reactions; E1 vs E2 mechanism"
subject: "Engineering Chemistry"
chapter: 10
key: "semester_2/engineering_chemistry:e1-vs-e2"
source: https://www.bctengineer.com/q/semester_2/engineering_chemistry/e1-vs-e2
site: BCT Engineering
---

# Elimination reactions; E1 vs E2 mechanism

Engineering Chemistry · Chapter 10: Reaction Mechanism in Organic reactions

- Asked in: 2071 Bhadra, 1+4 marks; 2068 Bhadra, 1+4 marks (asked 2 times)
- Video: [Intro to Elimination Reactions | E1 & E2 | 7.4 Organic Chemistry](https://www.youtube.com/watch?v=oQPM9pAo0V8) — Chad's Prep
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_2/engineering_chemistry/e1-vs-e2
- Whole chapter: https://www.bctengineer.com/md/semester_2/engineering_chemistry/chapter-10

## Question

What are elimination reactions? Write the differences between E₁ and E₂ reaction mechanism taking suitable example.

## Short answer

Elimination reactions remove H and X from an alkyl halide to form an alkene. E1 is two-step: carbocation formation then base attack; rate = k[RX]; weak base suffices; rearrangement possible. E2 is one-step concerted; rate = k[RX][base]; needs strong base; anti-periplanar geometry; no rearrangement. Examples: (CH₃)₃CBr (E1) vs CH₃CH₂Br (E2).

## Answer

#### Elimination reactions
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.

#### Difference between E1 and E2
| Basis | E1 | E2 |
|---|---|---|
| Steps | Two steps, carbocation intermediate | One step, concerted |
| Rate law | Rate = k[RX], first order | Rate = k[RX][Base], second order |
| Molecularity | Unimolecular | Bimolecular |
| Substrate | 3° > 2° (stable carbocation) | 3° > 2° > 1° |
| Base | Weak base is enough | Strong, concentrated base needed |
| Solvent | Polar protic | Less polar / alcohol |
| Rearrangement | Possible | Not possible |
| Stereochemistry | No special geometry | Anti-periplanar H and X |
| Example | (CH₃)₃CBr in ethanol | CH₃CH₂Br + alc. KOH |

#### Examples
- **E1:** (CH₃)₃C–Br → (CH₃)₃C⁺ + Br⁻ (slow); then the base removes a β-hydrogen: (CH₃)₃C⁺ + OH⁻ → (CH₃)₂C=CH₂ + H₂O (fast).
- **E2:** HO⁻ + H–CH₂–CH₂–Br → [HO···H···CH₂···CH₂···Br]⁻‡ → H₂O + CH₂=CH₂ + Br⁻ (one step).

## 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. A group is replaced by another group on the same carbon
   B. Two atoms or groups are removed from adjacent carbons, forming a double or triple bond
   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: B. Elimination removes two groups from neighbouring carbons without replacing them, so a C=C (or triple) bond forms. (Key point: Elimination reaction)

2. How do the rate laws of E1 and E2 differ?
   A. E1: k[RX] first order; E2: k[RX][Base] second order
   B. E1: k[RX][Base]; E2: k[RX]
   C. Both are first order in RX only
   D. E1 is second order and E2 is first order
   Answer: A. E1 is unimolecular (halide only); E2 is bimolecular (halide and base). (Key point: Rate law difference)

3. Which pairing is correct?
   A. E1: strong concentrated base needed; E2: weak base is enough
   B. Both need a strong concentrated base
   C. Neither depends on the base
   D. E1: weak base is enough; E2: strong, concentrated base needed
   Answer: D. E1 base acts after the slow ionisation, so weak is fine; E2 base takes part in the rate-determining step. (Key point: Substrate and base requirements)

4. Which pairing of E1 and E2 features is correct?
   A. E1: no rearrangement; E2: carbocation intermediate
   B. E1: needs anti-periplanar H and X; E2: rearrangement possible
   C. E1: rearrangement possible; E2: needs anti-periplanar H and X
   D. Both always give rearranged products
   Answer: C. E1 goes through a carbocation so rearrangement can occur; E2 is concerted and needs H and X anti-periplanar. (Key point: Rearrangement and geometry)

## Related in Engineering Chemistry

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- [Unimolecular elimination mechanism; difference from E2](https://www.bctengineer.com/question.md?key=semester_2%2Fengineering_chemistry%3Ae1-mechanism-differ-e2) — Engineering Chemistry, chapter 10, asked 2×
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- [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×

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