An alkene "A" on reaction with $O_3$ and $Zn-H_2O$ gives propanone and ethanal in equimolar ratio. The additio

Question

An alkene "A" on reaction with $O_3$ and $Zn-H_2O$ gives propanone and ethanal in equimolar ratio. The addition of HCl to alkene "A" gives "B" as the major product. The structure of product "B" is:

Accepted Answer

1. **Identify Alkene A (Reverse Ozonolysis):** Ozonolysis cleaves the carbon-carbon double bond to form carbonyl compounds. To identify the starting alkene, remove the oxygen atoms from the products (Propanone and Ethanal) and join the carbonyl carbons with a double bond.
   - Products: Propanone [$(CH_3)_2C=O$] and Ethanal [$O=CHCH_3$].
   - Alkene A: **2-Methylbut-2-ene** [$(CH_3)_2C=CHCH_3$] .
2. **Reaction with HCl (Electrophilic Addition):** The addition of HCl to the unsymmetrical alkene A follows **Markovnikov's Rule**.
   - **Step 1 (Protonation):** The proton ($H^+$) adds to the carbon with more hydrogen atoms ($C-3$) to form the more stable carbocation. A **tertiary ($3^\circ$) carbocation** is formed at $C-2$ rather than a secondary one at $C-3$ .
     $$ (CH_3)_2C=CHCH_3 + H^+ \rightarrow (CH_3)_2C^+ -CH_2CH_3 $$
   - **Step 2 (Nucleophilic Attack):** The chloride ion ($Cl^-$) attacks the stable tertiary carbocation.
     $$ (CH_3)_2C^+ -CH_2CH_3 + Cl^- \rightarrow (CH_3)_2C(Cl)CH_2CH_3 $$
3. **Conclusion:** The major product B is **2-Chloro-2-methylbutane**.

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