---
title: "3° alkyl halide with alcoholic KOH; solvent"
subject: "Engineering Chemistry"
chapter: 10
key: "semester_2/engineering_chemistry:tert-alc-koh-solvent"
source: https://www.bctengineer.com/q/semester_2/engineering_chemistry/tert-alc-koh-solvent
site: BCT Engineering
---

# 3° alkyl halide with alcoholic KOH; solvent

Engineering Chemistry · Chapter 10: Reaction Mechanism in Organic reactions

- Asked in: 2072 Magh, 3+3 marks
- Video: [Why alkyl halide with aqueous KOH forms alcohol but with alcoholic KOH forms alkene?](https://www.youtube.com/watch?v=mbZlNvxRoPg) — Chemistry Concept
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_2/engineering_chemistry/tert-alc-koh-solvent
- Whole chapter: https://www.bctengineer.com/md/semester_2/engineering_chemistry/chapter-10

## Question

Explain the reaction mechanism for the reaction between 3° alkyl halide and alcoholic potassium hydroxide. What solvent favors the reaction mechanism?

## Short answer

Tertiary alkyl halide with alcoholic KOH proceeds by E1: slow ionization to carbocation, then fast loss of a β-proton to give alkene. Polar protic solvents like ethanol stabilize the carbocation and favor E1.

## Answer

#### Mechanism: E1 elimination
A tertiary halide, e.g. tert-butyl bromide, with alcoholic KOH gives an alkene.

**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.
(With a high concentration of strong base, E2 can also compete, but the usual textbook mechanism for tertiary halides is E1.)

#### Solvent
A **polar protic solvent** such as ethanol or water favours E1. It stabilises the carbocation and the leaving halide by solvation, so the ionisation step becomes faster. Alcohol (less water) favours elimination over substitution.

## 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 removal of H by the base, then fast loss of Br-
   D. Slow ionisation to a carbocation, then fast removal of a beta-hydrogen by the base
   Answer: D. 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 does a tertiary halide give with alcoholic KOH?
   A. An alkene, by elimination
   B. A tertiary alcohol
   C. An alkane
   D. A carboxylic acid
   Answer: A. For (CH3)3CBr the product is 2-methylpropene. (Key point: Product)

3. Which solvent favours E1?
   A. A polar aprotic solvent such as DMSO
   B. A non-polar solvent such as hexane
   C. A polar protic solvent such as ethanol or water
   D. Any solvent equally
   Answer: C. Polar protic solvents stabilise the carbocation and leaving halide by solvation. (Key point: Favourable solvent)

4. How does the polar protic solvent speed up E1?
   A. It makes the base stronger
   B. It stabilises the carbocation and Br- so the ionisation step is faster
   C. It removes the need for ionisation
   D. It prevents the formation of a carbocation
   Answer: B. Solvation lowers the energy of the slow ionisation step. (Key point: Why solvent helps)

## 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
- [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
- [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
- [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×
- [Dehydrohalogenation of 1° alkyl halide](https://www.bctengineer.com/question.md?key=semester_2%2Fengineering_chemistry%3Adehydrohal-primary) — Engineering Chemistry, chapter 10, asked 4×