---
title: "Bromomethane with aqueous NaOH"
subject: "Engineering Chemistry"
chapter: 10
key: "semester_2/engineering_chemistry:bromomethane-naoh"
source: https://www.bctengineer.com/q/semester_2/engineering_chemistry/bromomethane-naoh
site: BCT Engineering
---

# Bromomethane with aqueous NaOH

Engineering Chemistry · Chapter 10: Reaction Mechanism in Organic reactions

- Asked in: 2073 Bhadra, 2.5 marks
- Video: [Simply Mechanisms 4a: SN2. Reaction of bromoethane with aqueous KOH](https://www.youtube.com/watch?v=z_WCjX9guao) — FranklyChemistry
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_2/engineering_chemistry/bromomethane-naoh
- Whole chapter: https://www.bctengineer.com/md/semester_2/engineering_chemistry/chapter-10

## Question

How does the reaction of bromomethane occur with aqueous caustic soda?

## Short answer

Bromomethane reacts via SN2: hydroxide ion attacks the backside of carbon in a concerted step, displacing bromide and forming methanol with Walden inversion; rate = k[CH3Br][OH⁻].

## Answer

Bromomethane (CH₃Br) is a methyl halide. Its carbon is even less hindered than that of a primary halide, so it reacts with aqueous NaOH by the **SN2 mechanism** to give methanol.

**Mechanism: SN2 (one step, concerted).**
```text
HO⁻ + H₃C—Br  →  [HO···CH₃···Br]⁻‡  →  HO—CH₃ + Br⁻
                 (transition state)
```
```text
        H                 H                  H
         \                |                 /
 HO⁻ +  H—C—Br  →  [HO····C····Br]⁻  →  HO—C—H  +  Br⁻
         /               / \                \
        H               H   H                H
 backside attack      transition state    inverted product
```
Overall: CH₃Br + NaOH(aq) → CH₃OH + NaBr
- Rate = k[CH₃Br][OH⁻]: second order (bimolecular).
- The nucleophile attacks from the side opposite to the leaving group; the three H atoms flip over like an umbrella in the wind (Walden inversion).
- Reactivity order: CH₃X > 1° > 2° > 3° (steric hindrance).

## Check yourself

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

1. Which mechanism does bromomethane follow with aqueous NaOH?
   A. SN1, via a methyl carbocation
   B. E1 elimination
   C. SN2, because the carbon is unhindered
   D. E2 elimination
   Answer: C. A methyl halide is unhindered and would give an unstable carbocation, so it reacts by SN2. (Key point: Mechanism for CH3Br)

2. How does the nucleophile attack in an SN2 reaction?
   A. From the same side as the leaving group, giving retention
   B. From the side opposite to the leaving group (180 degrees), giving inversion
   C. From either face of a planar carbocation
   D. Sideways at 90 degrees to the C-X bond
   Answer: B. Backside attack is least repulsive and overlaps best with the C-X sigma* orbital, so the configuration is inverted (Walden inversion). (Key point: Backside attack in SN2)

3. What is the rate law for SN2 hydrolysis of CH3Br by OH-?
   A. Rate = k[CH3Br], first order
   B. Rate = k[OH-], first order
   C. Rate = k[CH3Br]^2[OH-]
   D. Rate = k[CH3Br][OH-], second order
   Answer: D. Both substrate and nucleophile are in the single rate-determining step, so the reaction is second order overall. (Key point: SN2 rate law)

## Related in Engineering Chemistry

- [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
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