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NEET CHEMISTRYHard

An alkene "A" on reaction with O3O_3 and Zn−H2OZn-H_2O gives propanone and ethanal in equimolar ratio. Addition of HClHCl to alkene "A" gives "B" as the major product. The structure of product "B" is:

A

Cl−CH2−CH2−CH(CH3)2Cl-CH_2-CH_2-CH(CH_3)_2

B

H3C−CH2−CH(CH2Cl)−CH3H_3C-CH_2-CH(CH_2Cl)-CH_3

C

H3C−CH2−C(Cl)(CH3)2H_3C-CH_2-C(Cl)(CH_3)_2

D

H3C−CH(Cl)−CH(CH3)2H_3C-CH(Cl)-CH(CH_3)_2

Step-by-Step Solution

Ozonolysis of alkene A gives propanone (CH3COCH3CH_3COCH_3) and ethanal (CH3CHOCH_3CHO). The alkene is 2-methylpent-2-ene. Addition of HClHCl follows Markovnikov's rule to give 2-chloro-2-methylpentane.

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