$H_2C = CH - CH_2 - I \xrightarrow[CC{{l}_{4}}]{HI(excess)}$
${C_2}{H_2}\xrightarrow{{{O_3}}}X\xrightarrow{{Zn/C{H_3}COOH}}Y$
ટ્રાન્સ $\left( Ph - CH = CH - CH _3\right) \rightarrow$ સીસ $\left( Ph - CH = CH - CH _3\right)$
${C_2}{H_2}\xrightarrow[{HgS{O_4}/{H_2}S{O_4},60{\,^o}C}]{{{H_2}O}}X \rightleftharpoons C{H_3}CHO$