\(2-\)બ્યુટીન \(\xrightarrow{\text{B}{{\text{r}}_{\text{2}}}}\,\,C{{H}_{3}}\,-\,\,\underset{*}{\mathop{C}}\,HCl\,\,-\,\,\underset{*}{\mathop{C}}\,HCl\,\,-\,\,C{{H}_{3}}\) કુલ શકય આવશ્યક સમઘટ \(\,\,=\,\,\text{3}\)
તેથી \(x\) એ \(2\) થશે. એટલે કે \(1\) - બ્યુટીન અને \(2\) - બ્યુટીન અને \(2^{n-1} + 2^{n/2-1}\) દ્વારા \(2\) બ્યુટેન થશે.
$(i)\,\mathop {\begin{array}{*{20}{c}}
{\,\,\,\,\,C{H_3}} \\
| \\
{C{H_3} - C - C{H_3}} \\
| \\
{\,\,\,\,\,C{H_3}}
\end{array}}\limits_{(Neo - pen\tan e)\,(i)} $
$(ii\mathop {)\,\begin{array}{*{20}{c}}
{C{H_3} - CH - C{H_2} - C{H_3}} \\
{|\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,} \\
{\,\,\,\,\,C{H_{3\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}}}
\end{array}}\limits_{(Iso - pen\tan e)\,\,(ii)} $
$(iii)\,\mathop {C{H_3} - C{H_2} - C{H_2} - C{H_2} - C{H_3}}\limits_{(n - pen\tan e)} $