An alkene A on reaction with O_{3} and Zn-H_{ | vedclass

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(C) $1$. Ozonolysis of alkene $A$ yields propanone $(CH_{3}COCH_{3})$ and ethanal $(CH_{3}CHO)$.$2$. By reversing the ozonolysis reaction,we join the

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$CH_{3}-CH_{2}-C(Cl)(CH_{3})_{2}$

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(C) $1$. Ozonolysis of alkene $A$ yields propanone $(CH_{3}COCH_{3})$ and ethanal $(CH_{3}CHO)$.$2$. By reversing the ozonolysis reaction,we join the carbonyl carbons with a double bond: $(CH_{3})_{2}C=CH-CH_{3}$. This is $2-methylbut-2-ene$.$3$. Addition of $HCl$ to $2-methylbut-2-ene$ follows Markovnikov's rule,which proceeds via the formation of the most stable carbocation.$4$. Protonation of the double bond at the $C-3$ position leads to a tertiary carbocation: $(CH_{3})_{2}C^{+}-CH_{2}-CH_{3}$.$5$. Subsequent attack by the chloride ion $(Cl^{-})$ on the tertiary carbocation yields $2-chloro-2-methylbutane$,which is $CH_{3}-CH_{2}-C(Cl)(CH_{3})_{2}$.

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:

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