A) Both the compounds form same product on treatment with alcoholic KOH.
B) Both the compounds form same product on treatment with aq. NaOH.
C) Both the compounds form different product on reduction.
D) Both the compounds are optically active.
Correct Answer: A
Solution :
| \[Cl-C{{H}_{2}}-C{{H}_{2}}-Cl\,(Ethylene\,\,chloride)\,\,\xrightarrow[\Delta ]{\text{alc}\text{. }KOH}\]\[C{{H}_{2}}=CH-Cl\] |
| \[C{{H}_{3}}-CHC{{l}_{2}}\,(Ethylene\,\,chloride)\,\,\xrightarrow[\Delta ]{\text{alc}\text{. }KOH}\,C{{H}_{2}}\] |
| \[=CH-Cl\] |
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