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