CH_2 = CH - (CH_2)_5COOH xrightarrow[HBr]{tex | vedclass

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Question

(B) In the presence of peroxide,the addition of $HBr$ to an unsymmetrical alkene follows the anti-Markovnikov rule (Kharasch effect). The bromine atom

Vedclass · Accepted Answer

$BrCH_2 - (CH_2)_6COOH$

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(B) In the presence of peroxide,the addition of $HBr$ to an unsymmetrical alkene follows the anti-Markovnikov rule (Kharasch effect). The bromine atom attaches to the terminal carbon atom of the double bond. $CH_2 = CH - (CH_2)_5COOH + HBr \xrightarrow{	ext{Peroxide}} BrCH_2 - CH_2 - (CH_2)_5COOH$,which is $BrCH_2 - (CH_2)_6COOH$.

$CH_2 = CH - (CH_2)_5COOH \xrightarrow[HBr]{\text{Peroxide}} Z$,where $Z$ is:

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