Answer:
\[\underset{Ethanoic\,acid}{\mathop{C{{H}_{3}}COOH}}\,\xrightarrow{aq.NaOH}\underset{Sod.acetate}{\mathop{C{{H}_{3}}COONa}}\,\xrightarrow[\Delta
,630K]{Soda\,\text{lime}}\]
\[\underset{Methane}{\mathop{C{{H}_{4}}}}\,\xrightarrow[hv]{C{{l}_{2}}}C{{H}_{3}}Cl\xrightarrow[(Wurtz\,reaction)]{Na,dry\,ether}\underset{Ethane}{\mathop{C{{H}_{3}}C{{H}_{3}}}}\,\xrightarrow[hv]{C{{l}_{2}}}\]\[\underset{Chloroethane}{\mathop{C{{H}_{3}}C{{H}_{2}}Cl}}\,\xrightarrow[\Delta
]{alc.KOH}\underset{Ethene}{\mathop{{{H}_{2}}C=C{{H}_{2}}}}\,\xrightarrow{B{{r}_{2}},CC{{l}_{4}}}\]
\[\underset{1,2-Dibromoethane}{\mathop{BrC{{H}_{2}}-C{{H}_{2}}Br}}\,\xrightarrow[\Delta
]{alc.KOH}\underset{Bromoethene}{\mathop{{{H}_{2}}C=CHBr}}\,\xrightarrow[in.liq.N{{H}_{3}}]{NaN{{H}_{2}}}\]
\[\underset{Ethyne}{\mathop{HC\equiv
CH}}\,\xrightarrow[\text{(Polymerisation)},\,\,873\,K]{\text{Red}\,\text{hot}\,\text{iron}\,\text{tube}}\underset{\text{Benzene}}{\mathop{{{C}_{6}}{{H}_{6}}}}\,\]
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