A) \[{{C}_{2}}{{H}_{5}}Cl,\,KOH\,(alc.),\,\,\Delta \]
B) \[2{{C}_{2}}{{H}_{5}}OH,\,\,conc.\,{{H}_{2}}S{{O}_{4}},\,\,140{}^\circ C\]
C) \[{{C}_{2}}{{H}_{5}}Cl,\,Mg\,(dry\,ether)\]
D) \[{{C}_{2}}{{H}_{5}}\,dil.\,{{H}_{2}}S{{O}_{4}},\,HgS{{O}_{4}}\]
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 o ethyl alcohol with cone \[{{H}_{2}}S{{O}_{4}}\] at\[140{}^\circ C\]. \[C{{H}_{3}}C{{H}_{2}}O\,\,C{{H}_{2}}C{{H}_{3}}\xrightarrow[140{}^\circ C]{{{H}_{2}}S{{O}_{4}}}\] \[\underset{diethyl\,\,ether}{\mathop{{{C}_{2}}{{H}_{5}}-O-{{C}_{2}}{{H}_{5}}+{{H}_{2}}O}}\,\]
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