A) \[C{{H}_{2}}=CH-CH=C{{H}_{2}}\]
B) \[HC\equiv C-C{{H}_{3}}\]
C) \[C{{H}_{2}}=CH-C{{H}_{3}}\]
D) \[C{{H}_{3}}-C{{H}_{2}}-C{{H}_{3}}\]
Correct Answer: B
Solution :
\[CH\equiv CH\xrightarrow{NaN{{H}_{2}}}\underset{A}{\mathop{CH\equiv {{C}^{-}}N{{a}^{+}}}}\,\]\[\xrightarrow[-NaBr]{C{{H}_{3}}Br}CH\equiv \underset{B}{\mathop{C}}\,-C{{H}_{3}}\]
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