The product of the following reaction: CH_2=C | vedclass

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Question

(B) The reaction involves the electrophilic addition of $HBr$ to an alkene. According to Markovnikov's rule,the proton $(H^+)$ adds to the carbon atom

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$CH_3-CH(Br)-CCl_3$

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(B) The reaction involves the electrophilic addition of $HBr$ to an alkene. According to Markovnikov's rule,the proton $(H^+)$ adds to the carbon atom with more hydrogen atoms,and the bromide ion $(Br^-)$ adds to the more substituted carbon atom to form a more stable carbocation intermediate. In $CH_2=CH-CCl_3$,the $CCl_3$ group is strongly electron-withdrawing due to the inductive effect. However,the addition follows the standard Markovnikov rule where $H^+$ attaches to the terminal $CH_2$ group to form the stable carbocation $CH_3-CH^+-CCl_3$. Then,the $Br^-$ ion attacks this carbocation to yield the final product: $CH_3-CH(Br)-CCl_3$.

The product of the following reaction: $CH_2=CH-CCl_3 + HBr$ is:

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