
In the above reaction, the reactant undergoes $1,4$ Michael addition first then the formed intermediate undergoes reduction in the presence of $NaBH _{4}$ to form the desired product.
$CH _{3} CH = CHCH \left( CH _{3}\right)_{2} \stackrel{ HBr }{\longrightarrow}$



$I.$ $DDT$ $II.$ Gammerane
$III.$ Carbon tetrachloride $IV.$ Chlorobenzene