---
title: "Nucleophile vs electrophile; dehydrohalogenation mechanism"
subject: "Engineering Chemistry"
chapter: 10
key: "semester_2/engineering_chemistry:nucleophile-dehydrohal-tbutyl"
source: https://www.bctengineer.com/q/semester_2/engineering_chemistry/nucleophile-dehydrohal-tbutyl
site: BCT Engineering
---

# Nucleophile vs electrophile; dehydrohalogenation mechanism

Engineering Chemistry · Chapter 10: Reaction Mechanism in Organic reactions

- Asked in: 2075 Baisakh, 2+3 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/nucleophile-dehydrohal-tbutyl
- Whole chapter: https://www.bctengineer.com/md/semester_2/engineering_chemistry/chapter-10

## Question

Differentiate between nucleophile and electrophile. Explain the mechanism of dehydrohalogenation of 2-Bromo-2-methylpropane.

## Short answer

Nucleophiles are electron-rich species (Lewis bases) that donate electrons to electrophiles, which are electron-deficient (Lewis acids). Dehydrohalogenation of 2-bromo-2-methylpropane (tertiary halide) follows E1: slow C–Br cleavage giving (CH₃)₃C⁺, then OH⁻ removes a β-H to yield 2-methylpropene.

## Answer

#### Nucleophile vs electrophile
| Basis | Nucleophile | Electrophile |
|---|---|---|
| Nature | Electron-rich species with a lone pair; donates electrons | Electron-deficient species; accepts an electron pair |
| Meaning | "Nucleus-loving", attacks positive centres | "Electron-loving", attacks negative centres |
| Charge | Negative or neutral | Positive or neutral |
| Acts as | Lewis base | Lewis acid |
| Examples | OH⁻, CN⁻, Cl⁻, NH₃, H₂O | H⁺, NO₂⁺, AlCl₃, carbocations, BF₃ |

#### Dehydrohalogenation of 2-bromo-2-methylpropane
The substrate is a tertiary halide, (CH₃)₃CBr. With alcoholic KOH it reacts by E1.

**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.

## Check yourself

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

1. Which describes a nucleophile?
   A. An electron-deficient species that accepts a pair, a Lewis acid
   B. An electron-rich species that donates an electron pair, a Lewis base
   C. A species that always carries a positive charge
   D. A species that attacks negative centres
   Answer: B. Nucleophiles (OH-, CN-, NH3) have lone pairs and donate them to positive centres. (Key point: Nucleophile vs electrophile)

2. Which of these is an electrophile?
   A. OH-
   B. NH3
   C. NO2+
   D. CN-
   Answer: C. NO2+, H+, BF3, AlCl3 and carbocations are electron-deficient electrophiles. (Key point: Identify the electrophile)

3. How does (CH3)3CBr react with alcoholic KOH?
   A. By E1: slow ionisation to a carbocation, then removal of beta-H
   B. By E2 in one concerted step
   C. By SN2 backside attack
   D. By SN1 to give tert-butyl alcohol
   Answer: A. It is a tertiary halide, so E1 with alcoholic KOH gives 2-methylpropene. (Key point: Mechanism for 2-bromo-2-methylpropane)

4. What is formed from (CH3)3C-Br and alcoholic KOH?
   A. (CH3)3C-OH
   B. CH3CH2CH=CH2
   C. (CH3)3C-CH3
   D. (CH3)2C=CH2 (2-methylpropene) with KBr and H2O
   Answer: D. Loss of HBr gives 2-methylpropene. (Key point: Product of the reaction)

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