A) 4-chloro, 1 hydroxy butane
B) \[{{3}^{o}}\]-butyl alcohol
C) Secondary butyl amine
D) n-butyl alcohol
Correct Answer: C
Solution :
\[\underset{{}}{\mathop{C{{H}_{3}}-\underset{\,N{{H}_{2}}}{\mathop{\overset{H\,\,}{\mathop{\underset{|\,\,\,\,\,\,\,}{\overset{|\,\,\,\,\,}{\mathop{C\,\,\,-}}}\,}}\,}}\,C{{H}_{2}}-C{{H}_{3}}}}\,\] Secondary butyl amine is optically active.
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