CH_3-CH=CH_2 + HBr xrightarrow{ } ........., | vedclass

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(B) The reaction of propene $(CH_3-CH=CH_2)$ with hydrogen bromide $(HBr)$ follows Markovnikov's rule.According to Markovnikov's rule,the negative par

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$CH_3-CH(Br)-CH_3$

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(B) The reaction of propene $(CH_3-CH=CH_2)$ with hydrogen bromide $(HBr)$ follows Markovnikov's rule.According to Markovnikov's rule,the negative part of the addendum $(Br^-)$ attaches to the carbon atom of the double bond that has the lesser number of hydrogen atoms.Therefore,the reaction is: $CH_3-CH=CH_2 + HBr \rightarrow CH_3-CH(Br)-CH_3$.The product formed is $2-$bromopropane.

$CH_3-CH=CH_2 + HBr \xrightarrow{\ \ } .........$,the product formed is

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