A) Potassium fumarate
B) Calcium carbide
C) Ethylene bromide
D) All of these
Correct Answer: D
Solution :
\[\begin{array}{*{35}{l}} CH-COOK \\ |\,|\,| \\ CH-COOK \\ \end{array}\,+2\,{{H}_{2}}O\xrightarrow{\text{Electrolyisis}}\] \[\underset{{}}{\mathop{\underbrace{\begin{array}{*{35}{l}} CH \\ |\,|\,| \\ CH \\ \end{array}}_{\text{Anode}}}}\,+2\,C{{O}_{2}}+\underbrace{2\,KOH+{{H}_{2}}}_{\text{Cathode}}\] \[Ca{{C}_{2}}+2\,{{H}_{2}}O\to Ca{{\left( OH \right)}_{2}}+CH\equiv CH\] \[\,\underset{Br\,\,\,\,\,}{\mathop{\underset{|\,\,\,\,\,\,\,\,\,}{\mathop{C{{H}_{2}}}}\,}}\,-\underset{\underset{Br\,\,\,\,}{\mathop{|\,\,\,\,\,\,\,\,\,}}\,}{\mathop{C{{H}_{2}}}}\,+2\,KOH\to CH\equiv CH+2\,KBr+2\,{{H}_{2}}O\]
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