$Ca{C_2} + 2{H_2}O \to Ca{(OH)_2} + {C_2}{H_2}$
${C_2}{H_2} + {H_2} \to {C_2}{H_4}$
$n({C_2}{H_4}) \to {( - C{H_2} - C{H_2} - )_n}$
$64.1\, kg$ $Ca{C_2}$માંથી મેળવેલ પોલિઇથિલિનનો જથ્થો ......$kg$ છે.
${C_2}{H_2} + {H_2} \to \mathop {{C_2}{H_4}}\limits_{{\rm{28}}\,g} $
$64\,g$ of $Ca{C_2}$ gives $28\,g$ of ethylene
$\therefore $ $64\,kg$ of $Ca{C_2}$ will give $28\,kg$ of polyethylene

$(i)$ $C{H_3} - C \equiv C - C{H_3}$
$(ii)$ $C{H_3} - C{H_2} - C{H_2} - C{H_3}$
$(iii)$ $C{H_3} - C{H_2} - C \equiv CH$
$(iv)$ $C{H_3} - CH = C{H_2}$
$C{H_2} = CH - C{H_3} + HBr \to C{H_3}CHBr - C{H_3}$
$\begin{array}{*{20}{c}}
{C{H_3}\,\,\,\,\,\,\,} \\
{|\,\,\,\,\,\,\,\,\,\,\,\,\,} \\
{{H_3}C - C - CH = C{H_2}} \\
{|\,\,\,\,\,\,\,\,\,\,\,\,\,} \\
{C{H_3}\,\,\,\,\,\,\,\,\,}
\end{array}$ $\xrightarrow{{{H_2}O/{H^ \oplus }}}{\mkern 1mu} \mathop A\limits_{Major\,product} \, + \,\mathop B\limits_{Minor\,product} $

