A) \[{{C}_{2}}{{H}_{5}}Cl,\text{ }KOH~(alc.),\text{ }\Delta \]
B) \[2{{C}_{2}}{{H}_{5}}OH,\text{ }conc.\text{ }{{H}_{2}}S{{O}_{4}},140{{\,}^{o}}C\]
C) \[{{C}_{2}}{{H}_{5}}Cl,\text{ }Mg\] (dry ether)
D) \[{{C}_{2}}{{H}_{2}}\text{ }dil.\text{ }{{H}_{2}}S{{O}_{4}},\text{ }HgS{{O}_{4}}\]
Correct Answer: B
Solution :
Ethyl chloride reacts with sodium ethoxide to form diethyl ether as \[{{C}_{2}}{{H}_{5}}\,O{{C}_{2}}{{H}_{5}}\xrightarrow{{}}\] \[\underset{diethyl\,ether}{\mathop{{{C}_{2}}{{H}_{5}}-O-{{C}_{2}}{{H}_{5}}}}\,+HCl\] Diethyl ether is also obtained by reaction of ethyl alcohol with cone \[{{\text{H}}_{\text{2}}}\text{S}{{\text{O}}_{\text{4}}}\]at \[\text{140}{{\,}^{o}}\text{C}\text{.}\] \[C{{H}_{3}}C{{H}_{2}}O\,C{{H}_{2}}C{{H}_{3}}\xrightarrow[140{{\,}^{o}}C]{{{H}_{2}}S{{O}_{4}}}\] \[\underset{\operatorname{diethyl}\,\,ether}{\mathop{{{C}_{2}}{{H}_{5}}-O-{{C}_{2}}{{H}_{5}}}}\,+{{H}_{2}}O\]
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