---
title: "Tertiary alkyl halide with aq. NaOH / alc. KOH"
subject: "Engineering Chemistry"
chapter: 10
key: "semester_2/engineering_chemistry:tert-halide-naoh-koh"
source: https://www.bctengineer.com/q/semester_2/engineering_chemistry/tert-halide-naoh-koh
site: BCT Engineering
---

# Tertiary alkyl halide with aq. NaOH / alc. KOH

Engineering Chemistry · Chapter 10: Reaction Mechanism in Organic reactions

- Asked in: 2073 Bhadra, 2.5+2.5 marks
- Video: [Reaction of Alkyl Halide with alc. KOH - Halogen Derivatives of Alkane- Chemistry Class 12](https://www.youtube.com/watch?v=kFNqt7jW2E4) — Ekeeda
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_2/engineering_chemistry/tert-halide-naoh-koh
- Whole chapter: https://www.bctengineer.com/md/semester_2/engineering_chemistry/chapter-10

## Question

Write the mechanism of the reaction of tertiary alkyl halide with
a) Aqueous NaOH
b) Alcoholic KOH

## Short answer

With aqueous NaOH, tertiary alkyl halide undergoes SN1: carbocation formation then OH⁻ attack, giving tert-butyl alcohol. With alcoholic KOH, E1 elimination occurs: the same carbocation loses a β-proton to OH⁻, forming 2-methylpropene. Aqueous conditions favor substitution; alcoholic conditions favor elimination.

## Answer

#### a) With aqueous NaOH: substitution (SN1)
**Mechanism: SN1 (two steps).**
```text
Step 1 (slow, rate determining): ionisation
   CH₃                   CH₃
CH₃—C—Br   ---slow--->  CH₃—C⁺  +  Br⁻
   CH₃                   CH₃       (planar carbocation)

Step 2 (fast): nucleophile attacks the carbocation
   CH₃                      CH₃
CH₃—C⁺  + OH⁻ ---fast--> CH₃—C—OH
   CH₃                      CH₃
```
Overall: (CH₃)₃C–Br + NaOH(aq) → (CH₃)₃C–OH + NaBr

Product: tert-butyl alcohol (substitution). Aqueous NaOH, with water as solvent, favours substitution.

#### b) With alcoholic KOH: elimination (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)

Product: alkene. Alcoholic KOH favours elimination because the alkoxide/hydroxide acts as a base rather than a nucleophile, and in the absence of much water the proton loss from the carbocation dominates.

## Check yourself

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

1. What happens when a tertiary alkyl halide reacts with aqueous NaOH?
   A. E2 elimination to give an alkene
   B. SN1 substitution to give a tertiary alcohol
   C. SN2 substitution with inversion
   D. No reaction
   Answer: B. Water favours substitution, giving (CH3)3C-OH via SN1. (Key point: Aqueous NaOH reaction)

2. What is the main product of a tertiary halide with alcoholic KOH?
   A. A tertiary alcohol
   B. An ether only
   C. An alkene, by E1 elimination
   D. An alkane
   Answer: C. Alcoholic KOH acts as a base and favours elimination to the alkene. (Key point: Alcoholic KOH reaction)

3. What is the slow step in both the SN1 and E1 pathways?
   A. Ionisation of the halide to a carbocation
   B. Attack of OH- on the carbocation
   C. Removal of the beta-hydrogen by the base
   D. Formation of the transition state with the base
   Answer: A. Both begin with slow ionisation; the second step is fast. (Key point: Rate-determining step)

## Related in Engineering Chemistry

- [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
- [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
- [3° alkyl halide with alcoholic KOH; solvent](https://www.bctengineer.com/question.md?key=semester_2%2Fengineering_chemistry%3Atert-alc-koh-solvent) — Engineering Chemistry, chapter 10
- [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
- [Elimination reaction; dehydrohalogenation of tertiary halide](https://www.bctengineer.com/question.md?key=semester_2%2Fengineering_chemistry%3Ae1-tertiary-dehydrohal) — Engineering Chemistry, chapter 10, asked 3×
- [3° alkyl halide with aqueous caustic potash](https://www.bctengineer.com/question.md?key=semester_2%2Fengineering_chemistry%3Atert-aq-koh) — Engineering Chemistry, chapter 10