Option (a) is correct (i) Mechanism : Follows $\mathrm{S}_{\mathrm{N}}2$ Mechanism Step I : Step II :Step III : $\mathrm{CH}_3 \mathrm{CH}_2-\mathrm{O}^{\overset{+}{\overset{\vert}{\mathrm{H}}}}-\mathrm{CH}_2 \mathrm{CH}_3 \longrightarrow$ $\mathrm{CH}_3 \mathrm{CH}_2-\mathrm{O}-\mathrm{CH}_2 \mathrm{CH}_3 + \mathrm{H}^{+}$ (ii) Para-nitro phenol has higher boiling point than ortho-nitrophenol due to intermolecular hydrogen bonding present in para-nitrophenol, which require more energy to break these bonds during boiling. In o-nitrophenol intramolecular hydrogen bonding is present to greater extent. Thus o-nitrophenol is stream volatile due to low boiling point. [OR] b(i) Friedel craft Alkylation reaction will take place. (ii) A conjugated diketone is produced Benzoquinone, (iii) Dehydration takes place and alkene is formed. $\mathrm{CH}_3 \mathrm{OH} \xrightarrow[573\,\mathrm{K}]{\mathrm{Cu}} \mathrm{HCHO}$ $R-\underset{\mathrm{OH}}{\underset{\vert}{\mathrm{C}}}H-R' \xrightarrow[573\,\mathrm{K}]{\mathrm{Cu}} R-\underset{\mathrm{O}}{\underset{\parallel}{\mathrm{C}}}-R$