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