---
title: "Unimolecular elimination mechanism; difference from E2"
subject: "Engineering Chemistry"
chapter: 10
key: "semester_2/engineering_chemistry:e1-mechanism-differ-e2"
source: https://www.bctengineer.com/q/semester_2/engineering_chemistry/e1-mechanism-differ-e2
site: BCT Engineering
---

# Unimolecular elimination mechanism; difference from E2

Engineering Chemistry · Chapter 10: Reaction Mechanism in Organic reactions

- Asked in: 2069 Poush, 3+2 marks; 2067 Mangsir, 3+2 marks (asked 2 times)
- Video: [E2 vs. E1 - Mechanism](https://www.youtube.com/watch?v=jdajgFJHYsU) — Master Organic Chemistry
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_2/engineering_chemistry/e1-mechanism-differ-e2
- Whole chapter: https://www.bctengineer.com/md/semester_2/engineering_chemistry/chapter-10

## Question

Write the mechanism of unimolecular elimination reaction. How does it differ from bimolecular elimination reaction?

## Short answer

E1: unimolecular elimination. The C-X bond breaks slowly to give a carbocation; then a base rapidly removes a β-hydrogen, forming an alkene. Rate = k[RX]. E2: bimolecular, concerted; base attacks β-H as X leaves, transition state; rate = k[RX][base]; strong base required; no carbocation.

## Answer

#### E1 (unimolecular elimination) mechanism
**Mechanism: E1 (two steps).**
```text
Step 1 (slow): ionisation to carbocation
 (CH₃)₃C—Br  ---slow--->  (CH₃)₃C⁺ + Br⁻

Step 2 (fast): base removes a β-hydrogen
 OH⁻ + H—CH₂—C⁺(CH₃)₂ ---fast---> H₂O + CH₂=C(CH₃)₂
```
Overall: (CH₃)₃C–Br + KOH(alc) → (CH₃)₂C=CH₂ + KBr + H₂O (2-methylpropene)
- Rate = k[RX]: first order (unimolecular), as only the halide takes part in the slow step.
- Carbocation intermediate, so rearrangement is possible; competes with SN1.

#### Difference from E2 (bimolecular elimination)
| 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 |

## Check yourself

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

1. How does E1 dehydrohalogenation of (CH3)3C-Br proceed?
   A. One concerted step in which base removal of H and loss of Br happen together
   B. Fast attack of OH- on carbon, then slow loss of Br-
   C. Slow ionisation to a carbocation, then fast removal of a beta-hydrogen by the base
   D. Slow removal of H by the base, then fast loss of Br-
   Answer: C. E1 is two steps: the halide ionises slowly to a carbocation, then the base quickly removes a beta-H to form the alkene. (Key point: E1 mechanism steps)

2. What is the rate law for E1 elimination?
   A. Rate = k[RX], independent of base concentration
   B. Rate = k[RX][Base]
   C. Rate = k[Base]
   D. Rate = k[RX]^2
   Answer: A. Only the halide takes part in the slow ionisation step, so E1 is first order in RX and independent of the base. (Key point: E1 rate law)

3. How does E1 differ from E2 in number of steps?
   A. E1 has one step; E2 has two steps
   B. E1 has two steps with a carbocation; E2 has one concerted step
   C. Both have two steps
   D. Both are single concerted steps
   Answer: B. E2 is concerted with no intermediate, while E1 has a carbocation intermediate. (Key point: Steps difference)

4. Which substrate order applies to E1?
   A. 1 degree > 2 degree > 3 degree
   B. 1 degree is fastest
   C. All substrates react equally
   D. 3 degree > 2 degree (stable carbocation needed)
   Answer: D. E1 rate rises with carbocation stability, so tertiary halides react fastest. (Key point: Substrate preference)

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