A) \[C{{H}_{3}}-C{{H}_{2}}-C{{H}_{2}}-C{{H}_{3}}\]
B) \[C{{H}_{3}}C{{H}_{2}}\overset{\begin{smallmatrix} C{{H}_{3}} \\ | \end{smallmatrix}}{\mathop{CH}}\,C{{H}_{3}}\]
C)
D) \[C{{H}_{3}}C{{H}_{2}}-C\equiv CH\]
Correct Answer: C
Solution :
\[C{{H}_{3}}-\underset{\begin{smallmatrix} | \\ C{{H}_{2}}C{{H}_{3}} \end{smallmatrix}}{\overset{*}{\mathop{CH-}}}\,\]contains asymmetric carbon thus optically active.
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