What is the product formed when CH_2 = CH(CH_ | vedclass

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(A) The reaction of $CH_2 = CH(CH_2)_2COOH$ with $HBr$ follows Markovnikov's rule. According to Markovnikov's rule,the negative part of the addendum $

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$CH_3-CH(Br)-(CH_2)_2COOH$

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(A) The reaction of $CH_2 = CH(CH_2)_2COOH$ with $HBr$ 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 fewer hydrogen atoms. In the given molecule $CH_2 = CH(CH_2)_2COOH$,the double bond is between $C_1$ and $C_2$. $C_1$ has $2$ hydrogens and $C_2$ has $1$ hydrogen. Therefore,$Br^-$ will attach to $C_2$ and $H^+$ will attach to $C_1$. The resulting product is $CH_3-CH(Br)-(CH_2)_2COOH$.

What is the product formed when $CH_2 = CH(CH_2)_2COOH$ reacts with $HBr$?

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