In presence of organic peroxide, the halogen adds with the carbon in double bond having greater number of hydrogen atom.(antimarkovnikov addition) It is called "Kharash Effect or, Peroxide effect".
\(\mathrm{CH}_2=\mathrm{CH}-\left(\mathrm{CH}_2\right)_2 \mathrm{COOH} \xrightarrow{\mathrm{HBr}} \mathrm{BrCH}_2 \mathrm{CH}_2\left(\mathrm{CH}_2\right)_5 \mathrm{COOH}\)
hence, Option D is a correct answer.
$CH_3 - CH = CH - CH_3 \xrightarrow{{{O_3}}} A $$\xrightarrow[{Zn}]{{{H_2}O}} B.$
