A) \[{{C}_{2}}{{H}_{4}},alc\,KOH/\Delta \]
B) \[{{C}_{2}}{{H}_{5}}Cl,aq\,KOH/\Delta \]
C) \[C{{H}_{3}}OH,aq\,KOH/\Delta \]
D) \[{{C}_{2}}{{H}_{2}},PB{{r}_{3}}\]
Correct Answer: A
Solution :
\[\underset{(A)}{\mathop{{{H}_{2}}C=C{{H}_{2}}}}\,\xrightarrow{HBr}C{{H}_{3}}C{{H}_{2}}Br\] \[\xrightarrow[\Delta ]{alc.KOH}\underset{(A)}{\mathop{{{H}_{2}}C=C{{H}_{2}}}}\,\] Hence, \[A={{C}_{2}}{{H}_{4}}\] \[B=alc\,KOH/\Delta \]
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