Question
Geometrical isomerism is shown by
✓
Answer
a
(a) $\mathop {\mathop {\mathop {\mathop {HC}\limits_{|\,\,\,\,\,} }\limits_{C{H_3}} = \mathop {\mathop {CH}\limits_{|\,\,\,\,\,} }\limits_{C{H_3}} }\limits_{{\rm{(Cis)}}} }\limits_{{\rm{2 - butene}}} $ and $\mathop {\mathop {\mathop {\mathop {HC\,}\limits_{\,|\,\,\,\,\,\,\,} }\limits_{C{H_3}} = \mathop {\mathop {CH}\limits^{|\,\,\,\,\,} }\limits^{\,C{H_3}} }\limits_{{\rm{(Trans)}}} }\limits_{{\rm{2 - butene}}} $
(a) $\mathop {\mathop {\mathop {\mathop {HC}\limits_{|\,\,\,\,\,} }\limits_{C{H_3}} = \mathop {\mathop {CH}\limits_{|\,\,\,\,\,} }\limits_{C{H_3}} }\limits_{{\rm{(Cis)}}} }\limits_{{\rm{2 - butene}}} $ and $\mathop {\mathop {\mathop {\mathop {HC\,}\limits_{\,|\,\,\,\,\,\,\,} }\limits_{C{H_3}} = \mathop {\mathop {CH}\limits^{|\,\,\,\,\,} }\limits^{\,C{H_3}} }\limits_{{\rm{(Trans)}}} }\limits_{{\rm{2 - butene}}} $
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