A) acetylene
B) methane
C) ethane
D) ethylene
Correct Answer: D
Solution :
Dehydrohalogenation of ethyl iodide takes place by alcoholic KOH and gives ethylene as: \[H-\underset{H}{\overset{H}{\mathop{\underset{|}{\overset{|}{\mathop{C}}}\,}}}\,-\underset{I}{\overset{H}{\mathop{\underset{|}{\overset{|}{\mathop{C}}}\,}}}\,-H+\underset{Alc.}{\mathop{KOH}}\,\] \[\underset{\text{Ethylene}}{\mathop{C{{H}_{2}}=C{{H}_{2}}}}\,+KI+{{H}_{2}}O\]
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