The product formed in the first step of the reaction of $\begin{array}{*{20}{c}} {Br\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,} \\ {|\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,} \\ {C{H_3} - C{H_2} - CH - C{H_2} - CH - C{H_3}} \\ {\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,|} \\ {\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,Br} \end{array}$ with excess $Mg / Et _{2} O \left( Et = C _{2} H _{5}\right)$ is
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