Organic Chemistry — Complete Exam Answers

Q(c): Five Examples of Electrophilic Addition Reactions in Alkenes

1. Addition of HBr (Hydrohalogenation) — Markovnikov's Rule

CH₂=CH₂ + HBr → CH₃–CH₂Br

• Step 1: The π bond attacks H⁺ from HBr → more stable secondary or tertiary carbocation forms (Markovnikov's rule).
• Step 2: Br⁻ attacks the carbocation → product.

CH₃–CH=CH₂ + HBr → CH₃–CHBr–CH₃ (2-bromopropane, major product)

2. Addition of H₂SO₄ (Acid-catalyzed Hydration)

CH₂=CH₂ + H₂SO₄ → CH₃–CH₂–OSO₃H → (H₂O) → CH₃CH₂OH

• Step 1: Protonation of the double bond → carbocation (Markovnikov selectivity).
• Step 2: HSO₄⁻ attacks carbocation → alkyl hydrogen sulfate.
• Step 3: Hydrolysis with water → alcohol + H₂SO₄ regenerated.

3. Halogenation (Addition of Br₂/Cl₂)

CH₂=CH₂ + Br₂ → BrCH₂–CH₂Br (1,2-dibromoethane)

• Step 1: The π cloud polarizes Br–Br → electrophilic Br⁺ adds to form a bromonium ion (cyclic, 3-membered ring with Br bridging).
• Step 2: Br⁻ attacks the bromonium ion from the back (anti addition) → anti vicinal dibromide.

Result: trans (anti) product; bromine water is decolorized — a classic test for unsaturation.

4. Halohydrin Formation (Addition of Br₂/H₂O)

CH₂=CH₂ + Br₂/H₂O → BrCH₂–CH₂OH (bromohydrin)

• Step 1: Bromonium ion forms (same as above).
• Step 2: In aqueous medium, H₂O (better nucleophile than Br⁻ here) attacks the more substituted carbon of the bromonium ion → anti addition.
• Step 3: Proton loss → β-bromohydrin.

5. Oxymercuration–Demercuration (Addition of H and OH, Markovnikov)

R–CH=CH₂ →(Hg(OAc)₂/H₂O, then NaBH₄)→ R–CH(OH)–CH₃

• Step 1: Hg(OAc)₂ acts as electrophile; forms mercurinium ion (like bromonium).
• Step 2: Water attacks the more substituted carbon (Markovnikov) → organomercury alcohol.
• Step 3: NaBH₄ reduces C–Hg bond → C–H. No rearrangement occurs.

Q(c) OR: E1cb Reaction Mechanism

• The leaving group is poor (e.g., F, OH)
• The β-hydrogen is acidic (adjacent to EWG like –NO₂, –CN, –C=O)
• Strong base present

Mechanism Steps:

Example: Elimination from 2-fluoroethanol

     OH                    OH
     |                     |
HO⁻ + H–CH₂–CH₂–F  →  ⁻CH₂–CH₂–F  →  CH₂=CH₂  +  F⁻
         (β-H abstracted)   (carbanion)

Better Classic Example: Aldol-type substrate

Base + H–CH(COCH₃)–CH₂F  →  [⁻C(COCH₃)–CH₂F]  →  CH₂=CHCOCH₃ + F⁻
                               (stabilized carbanion)

Q(d): Why Peroxide Effect (Kharasch Effect) is Observed with HBr but NOT with HF, HCl, or HI

Radical Chain Mechanism (for HBr):

ROOR → 2 RO• → RO• + H–Br → ROH + Br•

• Step 1: Br• adds to the less substituted carbon of the alkene → more stable secondary radical (anti-Markovnikov selectivity)
• Step 2: The carbon radical abstracts H from HBr → product + new Br•

Why only HBr?

Summary of Specific Reasons:

• HF: H–F bond (136 kcal/mol) is too strong. Step 2 (H abstraction from HF) is highly endothermic — the chain cannot propagate.
• HCl: H–Cl bond is still strong enough that Step 2 is endothermic or thermoneutral. Cl• is also too reactive and less selective — doesn't add preferentially.
• HBr: Both steps are exothermic. Br• is reactive enough to add to alkenes but not so reactive that it lacks selectivity. The chain propagates efficiently.
• HI: I• is too unreactive (stable, large, polarizable radical) — it does not add readily to the π bond (Step 1 is endothermic). The chain cannot initiate.

Conclusion: HBr occupies the "Goldilocks" position — its BDE and the reactivity of Br• make both propagation steps thermodynamically favorable, which is not true for HF, HCl, or HI.