A) Nil
B) Four
C) Two
D) Three
E) One
Correct Answer: E
Solution :
\[\underset{\text{2, 3-dimethyl butane}}{\mathop{C{{H}_{3}}-\underset{C{{H}_{3}}\ }{\mathop{\underset{|}{\mathop{C}}\,H-}}\,\underset{C{{H}_{3}}\ }{\mathop{\underset{|}{\mathop{C}}\,H-}}\,C{{H}_{3}}+C{{l}_{2}}}}\,\to \]\[\underset{\text{2, 3-dimethyl chloro butane}}{\mathop{C{{H}_{3}}-\underset{C{{H}_{3}}\ }{\mathop{\underset{|}{\mathop{C}}\,H-}}\,\underset{C{{H}_{3}}\ }{\mathop{\underset{|}{\overset{\bullet }{\mathop{C}}}\,H-}}\,C{{H}_{2}}Cl}}\,\] Due to the presence of chiral carbon it shows the optical activity and its mirror image are non-superimposable so it shows one enantiomer pair. \[\] \[\begin{matrix} \underset{(\text{I})}{\mathop{C{{H}_{3}}-\underset{C{{H}_{3}}\ }{\mathop{\underset{|}{\mathop{C}}\,H-}}\,\,\underset{C{{H}_{3}}\ }{\mathop{\underset{|}{\mathop{C}}\,H-}}\,C{{H}_{2}}Cl}}\, & \underset{(\text{II})}{\mathop{ClC{{H}_{2}}-\underset{C{{H}_{3}}\ }{\mathop{\underset{|}{\mathop{C}}\,H-}}\,\underset{C{{H}_{3}}\ }{\mathop{\underset{|}{\mathop{C}}\,H-}}\,C{{H}_{3}}}}\, \\ \end{matrix}\]
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