--- title: "SN¹ retention/inversion vs SN² inversion; mechanism" subject: "Engineering Chemistry" chapter: 10 key: "semester_2/engineering_chemistry:sn1-retention-inversion" source: https://www.bctengineer.com/q/semester_2/engineering_chemistry/sn1-retention-inversion site: BCT Engineering --- # SN¹ retention/inversion vs SN² inversion; mechanism Engineering Chemistry · Chapter 10: Reaction Mechanism in Organic reactions - Asked in: 2071 Bhadra, 2+3 marks - Video: [Stereochemistry of SN¹ & SN² Products • Organic Chemistry Mechanism ](https://www.youtube.com/watch?v=pGMaPlnY3sk) — Dr Marwan Ahmed - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_2/engineering_chemistry/sn1-retention-inversion - Whole chapter: https://www.bctengineer.com/md/semester_2/engineering_chemistry/chapter-10 ## Question Explain why SN¹ reaction gives both retention and inversion isomers but SN² gives only inversion isomer. Write the mechanism of given chemical reaction. (CH₃)₃C−Br + NaOH → (CH₃)₃COH + NaBr ## Short answer SN1 gives both retention and inversion because the planar carbocation can be attacked from either face; SN2 undergoes exclusive inversion due to backside attack. (CH3)3C–Br + NaOH follows SN1: slow ionization to (CH3)3C+ then fast OH– attack forming (CH3)3COH. ## Answer #### SN1 gives both retention and inversion, SN2 only inversion - **SN1:** the first step gives a planar carbocation, which is attacked by the nucleophile from either face with almost equal probability: from the rear gives inversion and from the front gives retention. The product is a mixture, nearly racemic, with slightly more inversion because the leaving group shields the front. - **SN2:** a single step in which the nucleophile attacks only from the back of the carbon (opposite to the leaving group). The three other groups flip like an umbrella, so only the inverted product forms (Walden inversion). No planar intermediate exists. #### Mechanism of (CH₃)₃C–Br + NaOH → (CH₃)₃C–OH + NaBr **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. Why does SN1 at a chiral carbon give both inversion and retention? A. The carbocation is planar so the nucleophile can attack either face B. The nucleophile attacks only from the back in a concerted step C. The leaving group never leaves the carbon D. The carbocation is tetrahedral and holds its configuration Answer: A. The sp2 carbocation has two equivalent faces; rear attack gives inversion and front attack gives retention. (Key point: Planar carbocation in SN1) 2. How does the nucleophile attack in an SN2 reaction? A. From the same side as the leaving group, giving retention B. From either face of a planar carbocation C. From the side opposite to the leaving group (180 degrees), giving inversion D. Sideways at 90 degrees to the C-X bond Answer: C. 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. Why does SN2 give only the inverted product? A. The nucleophile attacks equally from both faces B. No planar intermediate exists; attack happens only from the rear C. A carbocation forms and then collapses D. The leaving group blocks the rear face Answer: B. A concerted backside attack allows no frontal pathway. (Key point: Why SN2 gives only inversion) 4. 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. Base removes a beta-H first, then OH- adds to the double bond D. Slow ionisation to a planar carbocation, then fast attack by OH- Answer: D. A tertiary halide follows SN1: slow rate-determining ionisation, then fast nucleophilic attack on the planar carbocation. (Key point: SN1 mechanism steps) ## Related in Engineering Chemistry - [SN reaction; inversion of configuration; SN1](https://www.bctengineer.com/question.md?key=semester_2%2Fengineering_chemistry%3Asn-inversion) — Engineering Chemistry, chapter 10 - [Bromomethane to methanol; SN² inversion](https://www.bctengineer.com/question.md?key=semester_2%2Fengineering_chemistry%3Asn2-bromomethane-inversion) — Engineering Chemistry, chapter 10 - [SN² mechanism and difference from SN¹](https://www.bctengineer.com/question.md?key=semester_2%2Fengineering_chemistry%3Asn2-vs-sn1) — Engineering Chemistry, chapter 10 - [Mechanism and kinetics of SN2; stereochemical inversion [partly illegible]](https://www.bctengineer.com/question.md?key=semester_2%2Fengineering_chemistry%3Asn2-kinetics-inversion) — 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 - [Tertiary alkyl halide with aq. 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