Question

What are X and Y in the reaction $ {{C}_{2}}{{H}_{4}}+{{H}_{2}}S{{O}_{4}}\xrightarrow{80{{\,}^{o}}C}X({{H}_{2}}O./\Delta a)Y $

• A. $ {{C}_{2}}{{H}_{6}},{{C}_{2}}{{H}_{5}}OH $
• B. $ {{C}_{2}}{{H}_{2}},{{C}_{2}}{{H}_{5}}SH $
• C. $ {{C}_{2}}{{H}_{5}}OS{{O}_{3}}H,{{C}_{2}}{{H}_{5}}OH $
• D. $ {{C}_{2}}{{H}_{2}},C{{H}_{3}}CHO $

Answer 1

Answer: (C)

$ {{C}_{2}}{{H}_{4}}+{{H}_{2}}S{{O}_{4}}\xrightarrow{80{{\,}^{o}}C}X\xrightarrow{{{H}_{2}}O\Delta }Y $ Ethylene reacts with $ {{\text{H}}_{\text{2}}}\text{S}{{\text{O}}_{\text{4}}} $ to form ethyl hydrogen sulphate as $ \text{ }\underset{C{{H}_{2}}}{\overset{C{{H}_{2}}}{\mathop{\text{ }\!\!|\!\!\text{ }\!\!|\!\!\text{ }}}}\,+HHS{{O}_{4}}\xrightarrow{{}}\underset{(X)}{\mathop{\underset{C{{H}_{2}}HS{{O}_{4}}}{\overset{C{{H}_{3}}}{\mathop{|}}}\,}}\, $ So, X is ethyl hydrogen sulphate and X (Ethyl hydrogen sulphate) on hydrolysis gives ethyl alcohol and $ {{\text{H}}_{\text{2}}}\text{S}{{\text{O}}_{\text{4}}} $ is regenerated as - $ \underset{C{{H}_{2}}HS{{O}_{4}}+HOH}{\overset{C{{H}_{3}}}{\mathop{|}}}\,\xrightarrow{\Delta }\underset{\begin{smallmatrix} \text{Ethyl}\,\text{alcohol} \\ \,\,\,\,\,\,\,\,\,\,\,\,\,\text{(}\gamma \text{)} \end{smallmatrix}}{\mathop{\underset{C{{H}_{2}}OH}{\overset{C{{H}_{3}}}{\mathop{|}}}\,}}\,+{{H}_{2}}S{{O}_{4}} $ So, the Y is ethyl alcohol.