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An alkene "A" on reaction with O3O_3 and Zn−H2OZn-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:

A

(Option 1 Structure Missing)

B

(Option 2 Structure Missing)

C

4

D

(Option 4 Structure Missing)

Step-by-Step Solution

  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 [(CH3)2C=O(CH_3)_2C=O] and Ethanal [O=CHCH3O=CHCH_3].
  • Alkene A: 2-Methylbut-2-ene [(CH3)2C=CHCH3(CH_3)_2C=CHCH_3] .
  1. Reaction with HCl (Electrophilic Addition): The addition of HCl to the unsymmetrical alkene A follows Markovnikov's Rule.
  • Step 1 (Protonation): The proton (H+H^+) adds to the carbon with more hydrogen atoms (C−3C-3) to form the more stable carbocation. A tertiary (3∘3^\circ) carbocation is formed at C−2C-2 rather than a secondary one at C−3C-3 . (CH3)2C=CHCH3+H+→(CH3)2C+−CH2CH3(CH_3)_2C=CHCH_3 + H^+ \rightarrow (CH_3)_2C^+ -CH_2CH_3
  • Step 2 (Nucleophilic Attack): The chloride ion (Cl−Cl^-) attacks the stable tertiary carbocation. (CH3)2C+−CH2CH3+Cl−→(CH3)2C(Cl)CH2CH3(CH_3)_2C^+ -CH_2CH_3 + Cl^- \rightarrow (CH_3)_2C(Cl)CH_2CH_3
  1. Conclusion: The major product B is 2-Chloro-2-methylbutane.
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