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 = ale.\[KOH/\Delta \]
You need to login to perform this action.
You will be redirected in
3 sec
