By Fatskills Exam Guides Team — the exam nerds behind 28,500+ quizzes and 2.1M practice questions across 500+ global exams.
IIT JEE (Main + Advanced)
"Mastering SN1, SN2, E1, and E2 mechanisms unlocks 8–10 marks in IIT JEE—enough to push you from a 90th to a 99th percentile rank. These reactions explain everything from drug synthesis to plastic manufacturing, but in the exam, they’re disguised as ‘predict the product’ or ‘choose the correct solvent’ questions. Today, you’ll learn the exact steps to solve them—no guesswork, just logic."
Before diving in, ensure you understand:1. Nucleophiles vs. Bases – Nucleophiles attack carbon; bases attack protons.2. Leaving Groups (LG) – Weak bases (e.g., I⁻, Br⁻, OTs⁻) are good LGs; strong bases (e.g., OH⁻, NH₂⁻) are poor LGs.3. Carbocation Stability – 3° > 2° > 1° (hyperconjugation & inductive effects).
Question: Predict the major product when 1-bromopropane reacts with NaCN in DMSO. Solution:1. Substrate: 1° (R-CH₂-Br) → SN2 or E22. Nu⁻/Base: CN⁻ (strong Nu⁻, weak base) → SN23. Solvent: DMSO (polar aprotic) → Favors SN24. Mechanism: SN2 → Backside attack → Inversion5. Product: CH₃-CH₂-CH₂-CN (propanenitrile) What we did and why: 1° substrate + strong Nu⁻ + polar aprotic solvent → SN2 dominates.
Question: Predict the major product when 2-bromo-2-methylpropane reacts with water. Solution:1. Substrate: 3° (R₃C-Br) → SN1 or E12. Nu⁻/Base: H₂O (weak Nu⁻, weak base) → SN1/E13. Solvent: H₂O (polar protic) → Favors SN1/E14. Mechanism: SN1 → Carbocation → H₂O attacks → Alcohol - OR E1 → Carbocation → H₂O deprotonates → Alkene5. Major Product: Alkene (E1) (Zaitsev’s rule → more substituted alkene) - Minor Product: Alcohol (SN1) What we did and why: 3° substrate + weak Nu⁻/base + polar protic solvent → E1 dominates (Zaitsev’s rule).
Question: Which of the following will give only substitution (no elimination) when treated with NaOCH₃ in CH₃OH? (A) CH₃-CH₂-Br (B) (CH₃)₂CH-Br (C) (CH₃)₃C-Br (D) CH₃-Br Solution:1. Substrate: - (A) 1° → SN2/E2 possible - (B) 2° → SN2/E2 possible - (C) 3° → E2 (strong base) - (D) Methyl → Only SN22. Nu⁻/Base: OCH₃⁻ (strong base) → E2 possible for 1°/2°/3°3. Solvent: CH₃OH (polar protic) → Slows SN2, but OCH₃⁻ is strong enough4. Key Insight: Methyl halides (CH₃-X) cannot undergo E2 (no β-hydrogens).5. Answer: (D) CH₃-Br (only SN2, no elimination) What we did and why: Methyl substrate + strong base → No E2 possible (only SN2).
"Listen up—this is your 60-second cheat sheet for SN1/SN2/E1/E2:1. Substrate first: - Methyl/1° → SN2 or E2 (if strong base) - 2° → All possible (check Nu⁻/base/solvent) - 3° → SN1 or E1 (unless strong base → E2)2. Nu⁻/Base next: - Strong Nu⁻ (CN⁻, RS⁻) → SN2 - Strong base (t-BuO⁻, LDA) → E2 - Weak Nu⁻/base (H₂O, ROH) → SN1/E13. Solvent matters: - Polar protic (H₂O, ROH) → SN1/E1 - Polar aprotic (DMSO, DMF) → SN2/E24. Mechanism shortcuts: - SN2 → Inversion, no rearrangements - SN1 → Racemization, rearrangements possible - E2 → Anti-periplanar, Zaitsev’s rule - E1 → Zaitsev’s rule, rearrangements possible5. Exam traps: - Alcoholic KOH = E2 - Neopentyl = No SN2 - Methyl halides = No E2
Now go crush those 10 marks!
Join 4M+ learners. Unlock unlimited quizzes, wrong-answer tracking, flashcards + reminders, study guides, and 1-on-1 challenges.