A) \[{{\left( C{{H}_{3}} \right)}_{2}}C=CHC{{H}_{3}}\]
B) \[C{{H}_{3}}C\equiv CH\]
C) \[{{\left( C{{H}_{3}} \right)}_{2}}CHC{{H}_{2}}C{{H}_{3}}\]
D) \[C{{H}_{3}}-C{{H}_{2}}-\underset{\begin{smallmatrix} | \\ C{{H}_{3}} \end{smallmatrix}}{\mathop{C}}\,=C{{H}_{2}}\]
Correct Answer: A
Solution :
\[{{\left( C{{H}_{3}} \right)}_{3}}C-C{{H}_{2}}OH\xrightarrow[170{}^\circ C]{conc.\,{{H}_{2}}S{{O}_{4}}}{{\left( C{{H}_{3}} \right)}_{2}}C\underset{(X)}{\mathop{=}}\,CHC{{H}_{3}}\]
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