A) \[{{(C{{H}_{3}})}_{4}}C\]
B) \[{{({{C}_{2}}{{H}_{5}})}_{2}}CH\cdot C{{H}_{3}}\]
C) \[{{C}_{2}}{{H}_{5}}CH(C{{H}_{3}})\cdot {{C}_{3}}{{H}_{7}}\]
D) \[C{{H}_{3}}CH\overline{\,\,\,\,\,\,\,}\overline{\,\,\,\,\,\,\,}CH\cdot C{{H}_{3}}\]
Correct Answer: C
Solution :
\[{{C}_{2}}{{H}_{5}}\underset{{{C}_{3}}{{H}_{7}}}{\overset{H}{\mathop{-\underset{|}{\overset{|}{\mathop{{{C}^{*}}}}}\,}}}\,-\,\,\,\,C{{H}_{3}}\] 3-methyl hexane has a chiral carbon atom, hence, is optically active. Though it shows very little rotation in plane of plane polarized light.
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