question_answer

Geometrical isomerism is found in

A) pemene-1      

B)        propene

C)        2-butene       

D)        butene-1

Correct Answer: C

Solution :

For a compound to exhibit geometrical isomerism it is necessary that it has (i) Atleast one carbon-carbon double bond. (ii) The two group attached to same carbon atom are different. From the choices find the one which satisfies both the conditions. (a)\[\underset{pentene-1}{\mathop{C{{H}_{3}}-C{{H}_{2}}-C{{H}_{2}}-C{{H}_{2}}=C{{H}_{2}}}}\,\] (b)\[\underset{propene}{\mathop{C{{H}_{3}}-CH=C{{H}_{2}}}}\,\] (d)\[\underset{1-butene}{\mathop{C{{H}_{3}}-C{{H}_{2}}-CH=C{{H}_{2}}}}\,\] In choices (a), (b) and (d), two groups attached to same carbon atom are different. So they will not show geometrical isomerism. \[\because \,\,2-\]butene satisfies both the conditions. \[\therefore \]It exists in two structural forms and shows geometrical isomerism.