Molar mass of $CH _4=16\,g / mol$
Weight of 2 moles of $CH _4=16 \times 2$
$=32\,g$
${C_2}{H_2}\xrightarrow{{{O_3}}}X\xrightarrow{{Zn/C{H_3}COOH}}Y$
$C{H_2} = CH{(C{H_2})_8}COOH + HBr\xrightarrow{{peroxide}}....$
ટ્રાન્સ $\left( Ph - CH = CH - CH _3\right) \rightarrow$ સીસ $\left( Ph - CH = CH - CH _3\right)$
$\begin{array}{*{20}{c}}
{\begin{array}{*{20}{c}}
{\,\,\,\,\,\,\,C{H_3}} \\
|
\end{array}} \\
{C{H_3} - C - C{H_3}} \\
| \\
H
\end{array}{\mkern 1mu} $ $\mathop {\xrightarrow{{C{H_3}OBr}}}\limits_{C{H_3}OH} $
ઉપરની પ્રક્રિયાને ધ્યાનમાં લો. અને મધ્યવર્તી ' $X$ ' ને ઓળખો.