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 :
\[\underset{(A)}{\mathop{{{H}_{2}}C=C{{H}_{2}}}}\,\xrightarrow{HBr}C{{H}_{3}}C{{H}_{2}}Br\]\[\xrightarrow{Alc.\,\,KOH(B)}{{H}_{2}}C=C{{H}_{2}}\] Hence,\[A\]and\[B\]are respectively\[{{C}_{2}}{{H}_{4}}\]and \[alc.\text{ }KOH\]. \[Alc.\text{ }KOH\] causes the dehydrohalogenation (ie, removal of hydrogen and halogen) reactions.
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