---
title: "Hydrolysis of 3° alkyl halide by aq. NaOH; solvent"
subject: "Engineering Chemistry"
chapter: 10
key: "semester_2/engineering_chemistry:tert-hydrolysis-solvent"
source: https://www.bctengineer.com/q/semester_2/engineering_chemistry/tert-hydrolysis-solvent
site: BCT Engineering
---

# Hydrolysis of 3° alkyl halide by aq. NaOH; solvent

Engineering Chemistry · Chapter 10: Reaction Mechanism in Organic reactions

- Asked in: 2069 Poush, 4+1 marks
- Video: [Hydrolysis of Halogenoalkanes](https://www.youtube.com/watch?v=ujJutNEs2Ww) — MaChemGuy
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_2/engineering_chemistry/tert-hydrolysis-solvent
- Whole chapter: https://www.bctengineer.com/md/semester_2/engineering_chemistry/chapter-10

## Question

Explain the reaction mechanism for the hydrolysis of 3° alkyl halide by aqueous NaOH. What types of solvent favors the reaction mechanism?

## Short answer

Hydrolysis of tert-alkyl halide proceeds by SN1: slow ionization to planar carbocation then fast OH− attack gives alcohol; rate = k[RX]. Polar protic solvents (water, alchols) favor SN1 by stabilizing carbocation and leaving group through solvation.

## Answer

#### Hydrolysis of a 3° alkyl halide by aqueous NaOH
**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
- 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.

#### Solvent
**Solvent:** polar protic solvents (water, ethanol, methanol, formic acid). They stabilise the carbocation and the halide ion by solvation (ion–dipole interaction and hydrogen bonding with the leaving group), which lowers the energy of the transition state of the slow step.

## Check yourself

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

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

2. 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)

3. Which solvent type favours SN1, and why?
   A. Polar protic, because it stabilises the carbocation and halide ion by solvation
   B. Polar aprotic, because it leaves the nucleophile 'naked'
   C. Non-polar, because it avoids solvating the ions
   D. Polar aprotic, because it hydrogen-bonds to the carbocation
   Answer: A. Polar protic solvents (water, alcohols) stabilise ions through ion-dipole interaction and hydrogen bonding, speeding ionisation. (Key point: Solvent for SN1)

4. What stabilises the tertiary carbocation?
   A. Resonance with the Br atom
   B. Hydrogen bonding with OH-
   C. Backside attack by water
   D. Hyperconjugation and +I effect of three methyl groups
   Answer: D. Alkyl groups donate electron density to the cationic carbon. (Key point: Why tertiary uses SN1)

## 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
- [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×
- [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
- [SN² mechanism for hydrolysis; solvent; no carbocation](https://www.bctengineer.com/question.md?key=semester_2%2Fengineering_chemistry%3Asn2-hydrolysis-solvent-carbocation) — 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