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) \[{{C}_{2}}{{H}_{2}},C{{H}_{3}}CHO\]
Correct Answer: C
Solution :
\[{{C}_{2}}{{H}_{4}}+{{H}_{2}}S{{O}_{4}}\xrightarrow{80{{\,}^{o}}C}X\xrightarrow{{{H}_{2}}O\Delta }Y\] 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.
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