---
title: "Mechanism and kinetics of SN2; stereochemical inversion [partly illegible]"
subject: "Engineering Chemistry"
chapter: 10
key: "semester_2/engineering_chemistry:sn2-kinetics-inversion"
source: https://www.bctengineer.com/q/semester_2/engineering_chemistry/sn2-kinetics-inversion
site: BCT Engineering
---

# Mechanism and kinetics of SN2; stereochemical inversion [partly illegible]

Engineering Chemistry · Chapter 10: Reaction Mechanism in Organic reactions

- Asked in: 2071 Magh
- Video: [Everything you Need to Know About SN2 Reaction in 10 Minutes: Kinetics, Mechanism, Inversion, TS...](https://www.youtube.com/watch?v=jEYygIjm91Y) — Conquer Chemistry
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_2/engineering_chemistry/sn2-kinetics-inversion
- Whole chapter: https://www.bctengineer.com/md/semester_2/engineering_chemistry/chapter-10

## Question

Describe mechanism and kinetics of nucleophilic bimolecular substitution reaction. Why does SN2 reaction take place with stereochemical inversion?

## Short answer

SN2 is a concerted single-step displacement: nucleophile attacks backside, bonds to leaving group break simultaneously; rate = k[RX][Nu⁻], second order. Stereochemical inversion occurs because the nucleophile approaches from the side opposite the leaving group, causing substituents to flip (Walden inversion).

## Answer

#### Mechanism
**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

#### Kinetics
The reaction takes place in one step, and both reactants take part in the rate-determining step, so:
Rate = k [R–X] [Nu⁻]
It is second order overall (first order in each reactant). Doubling either concentration doubles the rate.

#### Stereochemical inversion
In SN2 the nucleophile attacks the carbon from the **backside**, 180° from the leaving group. This is the position of least repulsion between the lone pair of the incoming nucleophile and the electron-rich leaving group, and it gives the best overlap with the C–X σ* orbital. As the new bond forms, the other three groups flatten and then flip to the opposite side, so the configuration is inverted (Walden inversion).

## Check yourself

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

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

2. What happens to the SN2 rate if the concentration of either reactant is doubled?
   A. The rate doubles
   B. The rate stays the same
   C. The rate quadruples for either one
   D. The rate halves
   Answer: A. The rate is first order in each reactant. (Key point: Effect of doubling concentrations)

3. Which describes the SN2 mechanism?
   A. Two steps with a carbocation
   B. Two steps with a carbanion
   C. One concerted step with a transition state and no intermediate
   D. A radical chain reaction
   Answer: C. Bond making and breaking happen together. (Key point: SN2 mechanism)

4. What causes stereochemical inversion in SN2?
   A. Frontal attack followed by rotation
   B. Formation of a planar carbocation
   C. Rearrangement of the alkyl group
   D. Backside attack, after which the other three groups flip over
   Answer: D. Walden inversion arises from rear attack. (Key point: Cause of inversion)

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