--- title: "SN reaction; inversion of configuration; SN1" subject: "Engineering Chemistry" chapter: 10 key: "semester_2/engineering_chemistry:sn-inversion" source: https://www.bctengineer.com/q/semester_2/engineering_chemistry/sn-inversion site: BCT Engineering --- # SN reaction; inversion of configuration; SN1 Engineering Chemistry · Chapter 10: Reaction Mechanism in Organic reactions - Asked in: 2074 Bhadra, 1+3+1 marks - Video: [Nucleophilic Substitution Reactions - SN1 and SN2 Mechanism, Organic Chemistry](https://www.youtube.com/watch?v=yrvV85H737o) — The Organic Chemistry Tutor - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_2/engineering_chemistry/sn-inversion - Whole chapter: https://www.bctengineer.com/md/semester_2/engineering_chemistry/chapter-10 ## Question What is meant by SN reaction? Explain the mechanism for the nucleophilic reaction that proceeds with an inversion of configuration. Why inversion product predominates more than retention product in SN₁ reaction. ## Short answer SN reactions replace a leaving group with a nucleophile. SN2 proceeds with inversion: backside attack of nucleophile, one-step concerted, Walden inversion (e.g., CH₃Br + OH⁻). In SN1, the planar carbocation can be attacked from either face, but the leaving group ion pair partially shields the front, making rear-side attack slightly easier, so inversion predominates. ## Answer #### SN reaction 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. a halide) of a substrate: R–X + Nu⁻ → R–Nu + X⁻. The two mechanisms are SN1 (unimolecular) and SN2 (bimolecular). #### Mechanism with inversion of configuration (SN2) **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). For a chiral halide, e.g. (R)-2-bromobutane + OH⁻ → (S)-butan-2-ol, the product has the opposite configuration. 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). #### Why inversion predominates in SN1 In SN1 the carbocation formed is **planar** (sp²), so the nucleophile can attack from either face: attack from the face opposite to the leaving group gives **inversion**, and attack from the same face gives **retention**. This gives a mixture (racemisation). Inversion slightly predominates because the departing leaving group stays near the front face as an ion pair and partly blocks it, so attack from the rear is easier. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What is an SN reaction? A. A nucleophile replaces a leaving group on a substrate B. Two groups are lost from adjacent carbons C. A reagent adds across a C=C bond D. An electrophile replaces a hydrogen on a ring Answer: A. R-X + Nu- gives R-Nu + X-. (Key point: SN reaction) 2. Which substitution mechanism proceeds with inversion of configuration? A. SN1, via a planar carbocation B. E1 C. SN2, via backside attack D. E2 Answer: C. SN2 is a concerted backside attack, giving Walden inversion. (Key point: Which mechanism inverts) 3. Why does the nucleophile attack from the back in SN2? A. Frontal attack is blocked by the carbocation's positive charge B. It minimises repulsion with the leaving group and overlaps best with the C-X sigma* orbital C. The leaving group is always larger than the nucleophile D. Backside attack lowers the number of steps to zero Answer: B. Backside is the position of least repulsion and best orbital overlap. (Key point: Why backside attack) 4. Why does inversion slightly predominate over retention in SN1? A. The nucleophile is attracted to the back by the positive charge on Br B. Rear attack is the only pathway allowed for a carbocation C. The carbocation is pyramidal with the rear face exposed D. The departing leaving group stays near the front face as an ion pair and partly blocks it Answer: D. The ion pair partly shields the front face, so attack from the rear is a little easier. 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