A) But-1-ene
B) But-2-ene
C) 2, 3-dichlorobutane
D) Ethene
Correct Answer: B
Solution :
Key Idea: For a compound to exhibit geometrical isomerism it must have following two things (i) It should have at least one carbon-carbon double bond. (ii) The two groups attached to same carbon atom should be different. Now draw all the choices to find answer \[\underset{But-1-ene}{\mathop{{{H}_{2}}C=C{{H}_{2}}}}\,C{{H}_{2}}C{{H}_{3}}\] (No geometrical isomerism because two groups attached to carbon atom are same) \[\underset{But-2-ene}{\mathop{{{H}_{3}}CCH=CHC{{H}_{3}}}}\,\] It will show geometrical isomerism as it full fills both conditions. \[\underset{2,\text{ }3-dichlorobutane}{\mathop{{{H}_{3}}C\underset{\begin{smallmatrix} | \\ Cl \end{smallmatrix}}{\mathop{CH}}\,=\underset{\begin{smallmatrix} | \\ Cl \end{smallmatrix}}{\mathop{CH}}\,C{{H}_{3}}}}\,\] (No geometrical isomerism as it does not complete any condition.) Ethene \[{{H}_{2}}C=C{{H}_{2}}\] No geometrical isomerism as group attached to carbon atoms are same.
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