1/ \(2C{H_4}\xrightarrow{{{t^o}}}{C_2}{H_2} + 3{H_2}\)
2/ $2C_2H_2\xrightarrow{{{t^o,xt,p}}}CH \equiv C - CH = C{H_2}$
3/ $CH \equiv C - CH = C{H_2}+H_2\xrightarrow{{{Pd/PbCO_3,t^o}}}C{H_2} = CH - CH = C{H_2}$
4/ $nC{H_2} = CH - CH = C{H_2}\xrightarrow{{{t^o,xt,p}}}{( - C{H_2} - CH = CH - C{H_2} - )_n}$