---
title: "SN reactions; hydrolysis of 3° alkyl halide"
subject: "Engineering Chemistry"
chapter: 10
key: "semester_2/engineering_chemistry:sn-tertiary-hydrolysis"
source: https://www.bctengineer.com/q/semester_2/engineering_chemistry/sn-tertiary-hydrolysis
site: BCT Engineering
---

# SN reactions; hydrolysis of 3° alkyl halide

Engineering Chemistry · Chapter 10: Reaction Mechanism in Organic reactions

- Asked in: 2072 Asoj, 1+4 marks; 2070 Bhadra, 2+3 marks (asked 2 times)
- Video: [Nucleophilic Substitution Reactions | SN1 Reaction and SN2 Reaction](https://www.youtube.com/watch?v=VpFGcpYgwy8) — 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/sn-tertiary-hydrolysis
- Whole chapter: https://www.bctengineer.com/md/semester_2/engineering_chemistry/chapter-10

## Question

What do you mean by SN reactions? Explain reaction mechanism for the hydrolysis of 3° alkyl halide by aqueous sodium hydroxide.

## Short answer

SN reactions involve a nucleophile replacing a leaving group. Hydrolysis of a 3° alkyl halide by aqueous NaOH follows SN1: slow formation of a planar carbocation, then fast nucleophilic attack by OH⁻ to give tert-butyl alcohol. Rate = k[RX], independent of [OH⁻].

## Answer

#### SN reactions
A **nucleophilic substitution (SN) reaction** is a reaction in which a nucleophile (an electron-rich species with a lone pair, e.g. OH⁻, CN⁻, NH₃) replaces the leaving group (e.g. halide) of a substrate. R–X + Nu⁻ → R–Nu + X⁻. The two mechanisms are SN1 (unimolecular) and SN2 (bimolecular).

#### 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.

## Check yourself

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

1. What defines an SN reaction?
   A. Two groups are removed from adjacent carbons
   B. A nucleophile replaces the leaving group of a substrate
   C. An electrophile adds across a double bond
   D. A radical abstracts hydrogen from the substrate
   Answer: B. In R-X + Nu- -> R-Nu + X-, the nucleophile with a lone pair replaces the leaving group. (Key point: Nucleophilic substitution)

2. 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)

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

4. Why do tertiary halides react by SN1?
   A. Three methyl groups give no steric hindrance to backside attack
   B. The C-Br bond in tertiary halides is non-polar
   C. A tertiary carbocation is the least stable and so forms fastest
   D. The tertiary carbocation is stabilised by hyperconjugation and the +I effect of three methyl groups
   Answer: D. Alkyl groups stabilise the carbocation through hyperconjugation and +I effect; they also hinder backside attack. (Key point: Why 3 degree halides use 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
- [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
- [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
- [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
- [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