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An alkene "A" on reaction with O
3

and Zn–H
2

O gives propanone and ethanal in equimolar ratio. Addition of HCl to alkene "A" gives "B" as the major product. The structure of product "B" is:

  1. A
    Cl-CH
    2

    -CH
    2

    -CH(CH
    3

    )
    2
  2. B
    n-pentyl chloride derivative
  3. C
    2-chloro-2-methylbutane
  4. D
    2-chloro-3-methylbutane

Solution & Step-by-step Explanation

Ozonolysis gives Propanone (CH
3

COCH
3

) and Ethanal (CH
3

CHO). Combining the pieces at the double bond: the alkene "A" is 2-methylbut-2-ene (CH
3

C(CH
3

)=CHCH
3

).

Addition of HCl to 2-methylbut-2-ene follows Markovnikov's rule. The Cl

attacks the more substituted carbon (C2) to form the most stable carbocation.
Product "B": 2-chloro-2-methylbutane.

Practice this question

Try it yourself before checking the explanation above.

An alkene "A" on reaction with O
3

and Zn–H
2

O gives propanone and ethanal in equimolar ratio. Addition of HCl to alkene "A" gives "B" as the major product. The structure of product "B" is:
A
Cl-CH
2

-CH
2

-CH(CH
3

)
2
B
n-pentyl chloride derivative
C
2-chloro-2-methylbutane
D
2-chloro-3-methylbutane

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