---
title: "3° alkyl halide with aqueous caustic potash"
subject: "Engineering Chemistry"
chapter: 10
key: "semester_2/engineering_chemistry:tert-aq-koh"
source: https://www.bctengineer.com/q/semester_2/engineering_chemistry/tert-aq-koh
site: BCT Engineering
---

# 3° alkyl halide with aqueous caustic potash

Engineering Chemistry · Chapter 10: Reaction Mechanism in Organic reactions

- Asked in: 2072 Magh, 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-aq-koh
- Whole chapter: https://www.bctengineer.com/md/semester_2/engineering_chemistry/chapter-10

## Question

Describe the mechanism involved in the reaction between a tertiary alkylhalide and aqueous caustic potash.

## Short answer

Tertiary alkyl halide reacts with aqueous KOH via SN1: rate-determining formation of a planar tert‑butyl carbocation, then rapid attack by hydroxide to give tertiary alcohol with racemization; water (polar protic) stabilizes the ions.

## Answer

A tertiary alkyl halide reacts with aqueous KOH by the **SN1 mechanism** to give a tertiary alcohol.

**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 + KOH(aq) → (CH₃)₃C–OH + KBr
- Rate = k[(CH₃)₃C–Br]: first order; independent of [OH⁻].
- The tertiary carbocation is stabilised by hyperconjugation and the +I effect of three methyl groups, so tertiary halides react by SN1.
- Water (polar protic solvent) stabilises the ions and favours the reaction.

## Check yourself

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

1. Which mechanism applies to a tertiary halide and aqueous KOH?
   A. SN2
   B. E2
   C. E1cB
   D. SN1
   Answer: D. Tertiary halides in water ionise to carbocations, giving substitution. (Key point: Mechanism with aqueous KOH)

2. How does hydrolysis of (CH3)3C-Br by aqueous NaOH occur?
   A. Slow ionisation to a planar carbocation, then fast attack by OH-
   B. Concerted backside attack by OH- with a single transition state
   C. Fast ionisation, then slow attack by OH-
   D. Base removes a beta-H first, then OH- adds to the double bond
   Answer: A. A tertiary halide follows SN1: slow rate-determining ionisation, then fast nucleophilic attack on the planar carbocation. (Key point: SN1 mechanism steps)

3. What is the product from (CH3)3C-Br and aqueous KOH?
   A. (CH3)2C=CH2
   B. CH3CH2CH2CH2OH
   C. (CH3)3C-OH (tertiary alcohol)
   D. (CH3)3C-O-C(CH3)3
   Answer: C. OH- attacks the carbocation to give tert-butyl alcohol. (Key point: Product)

4. What is the rate law for SN1 hydrolysis of (CH3)3CBr?
   A. Rate = k[RX][OH-]
   B. Rate = k[RX], independent of [OH-]
   C. Rate = k[OH-]
   D. Rate = k[RX]^2
   Answer: B. The slow step involves only the halide, so doubling [OH-] does not change the rate. (Key point: SN1 rate law)

## Related in Engineering Chemistry

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