---
title: "Dehydrohalogenation of t-butyl bromide; E1 vs E2"
subject: "Engineering Chemistry"
chapter: 10
key: "semester_2/engineering_chemistry:tbutyl-dehydrohal-e1-e2"
source: https://www.bctengineer.com/q/semester_2/engineering_chemistry/tbutyl-dehydrohal-e1-e2
site: BCT Engineering
---

# Dehydrohalogenation of t-butyl bromide; E1 vs E2

Engineering Chemistry · Chapter 10: Reaction Mechanism in Organic reactions

- Asked in: 2070 Bhadra, 3+2 marks
- Video: [E1 and E2 Elimintaion Reactions | Mechanism](https://www.youtube.com/watch?v=L9VtDeIkPCg) — Najam Academy
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_2/engineering_chemistry/tbutyl-dehydrohal-e1-e2
- Whole chapter: https://www.bctengineer.com/md/semester_2/engineering_chemistry/chapter-10

## Question

Explain the reaction mechanism for the dehydrohalogenation of tertiary butyl bromide. Differentiate between E¹ and E² mechanism.

## Short answer

Tertiary butyl bromide undergoes E1 via slow carbocation formation then fast base removal, yielding 2-methylpropene; Rate = k[RX]. Unlike E1, E2 is a concerted one-step mechanism with rate = k[RX][base], requiring strong base and anti-coplanar H and X.

## Answer

#### Dehydrohalogenation of tert-butyl bromide
tert-Butyl bromide, (CH₃)₃C–Br, is a tertiary halide. When heated with alcoholic KOH (or in a polar protic solvent such as ethanol), it undergoes E1 elimination through a stable 3° carbocation to give 2-methylpropene.

**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.
- The carbocation is an intermediate, so rearrangement is possible, and E1 competes with SN1.

#### 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 |

## 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. Slow ionisation to a carbocation, then fast removal of a beta-hydrogen by the base
   C. Fast attack of OH- on carbon, then slow loss of Br-
   D. Slow removal of H by the base, then fast loss of Br-
   Answer: B. 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 product of t-butyl bromide with alcoholic KOH?
   A. tert-Butyl alcohol
   B. Propene
   C. But-1-ene
   D. 2-Methylpropene
   Answer: D. Loss of HBr gives (CH3)2C=CH2. (Key point: Product)

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

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)

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