A) \[C{{H}_{2}}Br-CBr=C{{H}_{2}}\xrightarrow{Zn/C{{h}_{3}}OH}\]
B) \[HC\equiv C-C{{H}_{2}}-COOH\underset{{{40}^{o}}C}{\mathop{\xrightarrow{Aq.{{K}_{2}}C{{O}_{3}}}}}\,\]
C) \[C{{H}_{2}}Br-C\equiv C-C{{H}_{2}}Br\underset{\text{Heat}}{\mathop{\xrightarrow{Zn}}}\,\]
D) \[2C{{H}_{2}}=CH-C{{H}_{2}}I\underset{{}}{\mathop{\xrightarrow{{}}}}\,\]
Correct Answer: C
Solution :
\[C{{H}_{2}}Br-C\equiv C-C{{H}_{2}}Br\underset{\Delta }{\mathop{\xrightarrow{Zn}}}\,\]\[C{{H}_{2}}=C=C=C{{H}_{2}}\]
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