A) \[\underset{\underset{\,\,\,\,\,\,\,\,\,\,\,\,N{{H}_{2}}\,\,\,\,N{{H}_{2}}}{\mathop{\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,|\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,|}}\,}{\mathop{C{{H}_{3}}-CH-C{{H}_{2}}}}\,\]
B) \[\underset{\underset{\,\,\,\,\,\,\,\,\,\,OH\,\,\,\,\,OH}{\mathop{\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,|\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,|}}\,}{\mathop{C{{H}_{3}}-CH-C{{H}_{2}}}}\,\]
C) \[\underset{\underset{OH}{\mathop{|}}\,}{\mathop{C{{H}_{3}}-C=C{{H}_{2}}}}\,\]
D) \[C{{H}_{3}}C\equiv CH\]
Correct Answer: D
Solution :
\[C{{H}_{3}}C{{H}_{2}}C{{H}_{2}}OH\underset{160-{{180}^{o}}C}{\mathop{\xrightarrow{\text{conc}\text{.}\,\,{{H}_{2}}S{{O}_{4}}}}}\,C{{H}_{3}}CH=C{{H}_{2}}\] \[\xrightarrow{B{{r}_{2}}}C{{H}_{3}}-\underset{Br}{\mathop{\underset{|}{\mathop{CH}}\,}}\,-\underset{Br}{\mathop{\underset{|}{\mathop{C{{H}_{2}}}}\,}}\,\xrightarrow{\text{Alc}\,.\,KOH}\underset{\text{Propyne}}{\mathop{C{{H}_{3}}-C\equiv CH}}\,\]
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