Enantiomeric excess \((e. e.) = \frac{6g -4g}{(6+ 4)} \times 100 = 20\, \%\)
$\begin{array}{*{20}{c}}
{R = - CH - Cl}\\
{|\,\,}\\
{\,\,\,\,C{H_3}}
\end{array}\,\,\,\,,\,\,\,\,\,\,\,\,\,\,\begin{array}{*{20}{c}}
{S = - CH - Cl}\\
{|\,\,}\\
{Br}
\end{array}$
$CH_3CH=CHCH_2CHBrCH_3$