A) \[C{{H}_{3}}C\equiv CH\]
B) \[C{{H}_{3}}\text{ }\underset{\begin{smallmatrix} | \\ OH \end{smallmatrix}}{\mathop{C}}\,=C{{H}_{2}}\]
C) \[C{{H}_{2}}=C=C{{H}_{2}}\]
D) \[C{{H}_{3}}\text{ }\underset{\begin{smallmatrix} | \\ OH \end{smallmatrix}}{\mathop{CH}}\,-\underset{\begin{smallmatrix} | \\ OH \end{smallmatrix}}{\mathop{C{{H}_{2}}}}\,\]
Correct Answer: A
Solution :
\[C{{H}_{3}}C{{H}_{2}}C{{H}_{2}}OH\xrightarrow[\begin{smallmatrix} \,\,\,\,\,\,\,\,\,\,{{170}^{o}} \\ (dehydration) \end{smallmatrix}]{{{H}_{2}}S{{O}_{4}}}\underset{\begin{smallmatrix} propene \\ \,\,\,\,'A' \end{smallmatrix}}{\mathop{C{{H}_{3}}CH=C{{H}_{2}}}}\,\] \[\xrightarrow{B{{r}_{2}}}\underset{1,\text{ }2-dibromopropane}{\mathop{\overset{\begin{smallmatrix} Br \\ | \end{smallmatrix}}{\mathop{C{{H}_{3}}CH}}\,-\underset{\begin{smallmatrix} | \\ Br \end{smallmatrix}}{\mathop{C{{H}_{2}}}}\,}}\,\xrightarrow[(Dehydrohalogenation)]{Alc.\,KOH}\] \[\underset{\begin{smallmatrix} propyne \\ \,\,\,\,'C' \end{smallmatrix}}{\mathop{C{{H}_{3}}C\equiv CH}}\,\]
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