Does doubling the nucleophile double the rate?
"Does doubling the nucleophile double the rate?" is a free, interactive Class 12 Chemistry simulation for NEET and JEE, built on the lesson "SN1 and reactivity order" in Haloalkanes and Haloarenes. Each bromide can react by SN2, where OH⁻ pushes in from the back in one step, or by SN1, where the C–Br bond first breaks on its own. Slide [OH⁻] and watch which rate follows it. The rate constants are illustrative; the orders are NCERT's. SN2: rate = k[RX][Nu⁻], one step, CH₃X > 1° > 2° > 3° because bulky groups block the back side. SN1: rate = k[RX], via a carbocation, 3° > 2° > 1° > CH₃X because 3° carbocations are most stable. JEE Main asked 62 questions from this chapter across the 42 shifts Lumi has analysed (2017–2026).
From the lesson SN1 and reactivity order in Haloalkanes and Haloarenes. Change the values and watch what happens.
In JEE Main
42 shifts analysed
62 questions from Haloalkanes and Haloarenes across the 42 JEE Main shifts Lumi has analysed (2017–2026).
The idea behind it
NCERT §6.7.1
- SN1 (substitution nucleophilic unimolecular) reactions are usually run in polar protic solvents such as water, alcohol or acetic acid.
- (CH₃)₃CBr + OH⁻ → (CH₃)₃COH + Br⁻ follows first order kinetics: the rate depends on [tert-butyl bromide] only, not on [OH⁻].
- Step I: the polarised C–Br bond breaks slowly and reversibly into a carbocation and Br⁻; the energy comes from solvation of the halide ion by the protons of the protic solvent. Step II: the nucleophile attacks the carbocation quickly. The slow step sets the rate.
- The more stable the carbocation, the faster it forms and the faster the reaction. Tertiary carbocations are the most stable, so 3° halides react fastest: SN1 order 3° > 2° > 1° > CH₃X, the reverse of SN2.
- Allylic and benzylic halides are very reactive in SN1 because their carbocations are stabilised by resonance.
- For a given alkyl group, both mechanisms follow R–I > R–Br > R–Cl >> R–F; the large iodide ion leaves most readily.
- Bromobutanes: SN1 order n-C₄H₉Br < (CH₃)₂CHCH₂Br < CH₃CH₂CH(Br)CH₃ < (CH₃)₃CBr, and SN2 the exact reverse. Of the two primary bromides, (CH₃)₂CHCH₂Br is faster in SN1 because (CH₃)₂CH– has a larger electron-donating inductive effect.
- Benzylic bromides: SN1 order C₆H₅C(CH₃)(C₆H₅)Br > C₆H₅CH(C₆H₅)Br > C₆H₅CH(CH₃)Br > C₆H₅CH₂Br, and SN2 the reverse. Two phenyl groups stabilise a cation by resonance, but a phenyl group is bulkier than methyl, which slows SN2.
Questions
3 answers
- What does the "Does doubling the nucleophile double the rate?" simulation show?
- Each bromide can react by SN2, where OH⁻ pushes in from the back in one step, or by SN1, where the C–Br bond first breaks on its own. Slide [OH⁻] and watch which rate follows it. The rate constants are illustrative; the orders are NCERT's.
- Does JEE Main ask questions from Haloalkanes and Haloarenes?
- Yes. JEE Main asked 62 questions from this chapter across the 42 shifts Lumi has analysed (2017–2026).
- Is this simulation free?
- Yes. It is free and runs in the browser on a phone or a laptop. It belongs to the lesson "SN1 and reactivity order" in Haloalkanes and Haloarenes (Class 12 Chemistry).
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