A) 2-butene
B) 2-butyne
C) 2-butanol
D) Butanone
E) Butanol
Correct Answer: A
Solution :
\[\underset{\text{cis 2 butene}}{\mathop{\begin{matrix} C{{H}_{3}} \\ \ \ H \\ \end{matrix}>C=C<\begin{matrix} C{{H}_{3}} \\ H\ \ \ \\ \end{matrix}}}\,\] \[\underset{\text{Trans 2 butene}}{\mathop{\begin{matrix} C{{H}_{3}} \\ \ \ H \\ \end{matrix}>C=C<\begin{matrix} H\ \ \, \\ C{{H}_{3}} \\ \end{matrix}}}\,\] Cis-trans isomerism shown by compound which have double or triple bond by which they restrict their rotation, since 2 butyne have no hydrogen on triple bonded carbon. \[C{{H}_{3}}-C\equiv \underset{\text{2 butyne}}{\mathop{C-C{{H}_{3}}}}\,\] [It does not show cis-trans]
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