CUET UG Chemistry Organic Chemistry Reaction Mechanisms SN1 vs SN2 E1 vs E2 Markovnikovs Rule

Must-Know (15–20 detailed bullets)

• SN2 reaction is bimolecular nucleophilic substitution; rate = k[substrate][nucleophile]; example: CH₃Br + OH⁻ → CH₃OH + Br⁻ (NCERT Class 12, Haloalkanes and Haloarenes).
• SN1 reaction is unimolecular nucleophilic substitution; rate = k[substrate]; proceeds via carbocation intermediate; example: (CH₃)₃CBr → (CH₃)₃C⁺ + Br⁻ (slow step).
• SN2 mechanism involves complete inversion of configuration (Walden inversion); demonstrated in hydrolysis of optically active 2-bromobutane.
• SN1 leads to racemization due to planar carbocation attacked from both sides; e.g., hydrolysis of 2-bromo-2-methylpropane gives racemic mixture.
• Tertiary alkyl halides favor SN1 due to stable carbocations; primary favor SN2 due to minimal steric hindrance.
• Strong nucleophiles (e.g., OH⁻, CN⁻) favor SN2; weak nucleophiles (e.g., H₂O, CH₃OH) favor SN1.
• Polar protic solvents (e.g., water, alcohol) stabilize ions and favor SN1; polar aprotic solvents (e.g., DMSO, acetone) favor SN2 by enhancing nucleophilicity.
• E2 reaction is bimolecular elimination; rate = k[substrate][base]; requires anti-periplanar geometry; example: CH₃CH₂CH₂Br + OH⁻ → CH₃CH=CH₂ + H₂O + Br⁻.
• E1 reaction is unimolecular elimination; rate = k[substrate]; proceeds via carbocation, then deprotonation; example: (CH₃)₃CBr → (CH₃)₂C=CH₂ after loss of H⁺.
• E2 is favored by strong bases (e.g., OH⁻, RO⁻); E1 occurs with weak bases and good ionizing solvents.
• Zaitsev’s rule: major product in elimination is more substituted alkene; e.g., dehydrohalogenation of 2-bromobutane gives 2-butene (80%) over 1-butene (20%).
• Markovnikov’s Rule: in addition of HX to unsymmetrical alkene, H adds to carbon with more H atoms; e.g., propene + HBr → 2-bromopropane.
• Mechanism of Markovnikov addition involves formation of more stable carbocation; secondary > primary (e.g., protonation of propene gives 2° carbocation).
• Peroxide effect (Kharasch effect) reverses Markovnikov addition for HBr only; gives anti-Markovnikov product; e.g., propene + HBr (peroxide) → 1-bromopropane.
• SN2 transition state has pentacoordinate carbon with partial bond formation/breaking; energy diagram shows single peak.
• SN1 energy profile has two peaks: first for carbocation formation (rate-determining), second for nucleophile attack.
• E2 transition state involves simultaneous C-H and C-X bond breaking; requires anti-coplanar conformation for orbital overlap.
• Tertiary substrates undergo E2 even with weak base if heated; heat favors elimination over substitution.
• For primary alkyl halides with strong bulky base (e.g., tert-butoxide), Hofmann product (less substituted alkene) dominates due to steric control.
• Carbocation stability order: 3° > 2° > 1° > methyl; hyperconjugation and inductive effects explain this (verify from NCERT).

Difficulty Level

Intermediate — requires understanding of reaction kinetics, stereochemistry, and substrate effects; questions often mix concepts like solvent or base strength.

Common CUET Traps (3 bullets)

• Trap: Assuming all HX additions follow Markovnikov rule. Avoid: Remember HBr with peroxide gives anti-Markovnikov; HCl and HI do not show this effect.
• Trap: Thinking SN1 always gives complete racemization. Avoid: Partial racemization occurs due to ion pairing; full racemization is idealized.
• Trap: Confusing E2 stereochemistry requirements. Avoid: E2 requires anti-periplanar arrangement, especially in cyclohexane derivatives (leaving group and H must be trans-diaxial).

Practice MCQs (5 questions)

Q1. Which of the following reactions follows second-order kinetics?
A) Hydrolysis of tert-butyl chloride in water
B) Reaction of methyl bromide with aqueous KOH
C) Dehydration of ethanol to ethene
D) Bromination of benzene

Answer: B
Explanation: Methyl bromide undergoes SN2 reaction with OH⁻, rate depends on both [CH₃Br] and [OH⁻].
Why others fail: A is SN1 (first order); C is acid-catalyzed elimination (first order); D is electrophilic substitution (not nucleophilic).

Q2. What is the major product when propene reacts with HBr in the presence of benzoyl peroxide?
A) 1-Bromopropane
B) 2-Bromopropane
C) 1,2-Dibromopropane
D) Propane

Answer: A
Explanation: Peroxide effect leads to anti-Markovnikov addition of HBr.
Why others fail: B is Markovnikov product formed without peroxide; C is addition of Br₂, not HBr.

Q3. Which condition favors SN1 over SN2?
A) Primary alkyl halide
B) Strong nucleophile
C) Polar protic solvent
D) Low temperature

Answer: C
Explanation: Polar protic solvents stabilize carbocation intermediate in SN1.
Why others fail: A favors SN2; B favors SN2; low temperature favors substitution but doesn’t distinguish mechanism.

Q4. In the dehydrohalogenation of 2-bromo-2-methylbutane, the major product is:
A) 2-Methyl-1-butene
B) 2-Methyl-2-butene
C) 3-Methyl-1-butene
D) Pent-1-ene

Answer: B
Explanation: Zaitsev’s rule: more substituted alkene (trisubstituted) is major; 2-methyl-2-butene has three alkyl groups on C=C.
Why others fail: A and C are less stable (disubstituted); C is incorrectly numbered.

Q5. Which of the following statements is correct regarding the hydrolysis of (R)-2-bromooctane?
A) SN2 gives (S)-2-octanol
B) SN1 gives exclusively (R)-2-octanol
C) SN2 gives racemic mixture
D) SN1 gives (S)-2-octanol only

Answer: A
Explanation: SN2 inverts configuration; (R) reactant gives (S) product.
Why others fail: SN1 gives racemization, not pure enantiomer; C is wrong because SN2 does not racemize.

Last‑Minute Revision (15–20 one‑liners)

• ⚠️ SN2: rate = k[substrate][nucleophile]; inversion occurs.
• ⚠️ SN1: rate = k[substrate]; racemization due to planar carbocation.
• ⚠️ Tertiary halide + weak base → SN1/E1; primary + strong base → SN2/E2.
• ⚠️ Polar protic solvents (H₂O, ROH) favor SN1/E1; polar aprotic (DMF, DMSO) favor SN2.
• ⚠️ E2 requires anti-periplanar geometry — critical in cyclohexane rings.
• ⚠️ Markovnikov addition: H adds to C with more H; forms more stable carbocation.
• ⚠️ Peroxide effect applies to HBr only — not HCl or HI.
• ⚠️ Anti-Markovnikov product: 1-bromopropane from propene + HBr (peroxide).
• ⚠️ Zaitsev rule: more substituted alkene = major elimination product.
• ⚠️ Hofmann product: less substituted alkene favored by bulky bases.
• ⚠️ Carbocation stability: 3° > 2° > 1° due to hyperconjugation (9, 6, 3 α-H respectively).
• ⚠️ SN2 transition state: pentacoordinate carbon, backside attack.
• ⚠️ E2 transition state: concerted C-H and C-X bond breaking.
• ⚠️ Methyl and primary carbocations rarely form; rearrangements occur to stabilize.
• ⚠️ Hydride shift: e.g., 1° → 2° carbocation in neopentyl systems.
• ⚠️ Methanol (CH₃OH) is polar protic — favors SN1, not SN2.
• ⚠️ CN⁻ is strong nucleophile — favors SN2 unless substrate is tertiary.
• ⚠️ Bulky base like (CH₃)₃CO⁻ favors E2 over SN2 even in primary halides.
• ⚠️ Heat favors elimination (E1/E2) over substitution (SN1/SN2).
• ⚠️ Mnemonic: “SN1 = Stable Carbocation, SN2 = Steric Hindrance Matters.”