Directions : In the following question more than one of the answers given may be correct. Select the correct answer and mark it according to the code:
Geometrical isomerism is exhibited by (1) 2-chloro but-2-ene (2) but-2-ene (3) 3-methyl pent-2-ene (4) 2-methyl but-2-eneA) 1, 2 and 3 are correct.
B) 1 and 2 are correct.
C) 2 and 4 are correct.
D) 1 and 3 are correct.
Correct Answer: A
Solution :
The main conditions for exhibiting geometrical isomerism are (i) presence of double bond. (ii) presence of different groups on double bonded C-atom. (iii) presence of atleast one similar group on adjacent double bonded carbon atoms.\[\underset{2-chloro\text{ }but-2-ene}{\mathop{{{H}_{3}}C-\underset{\begin{smallmatrix} | \\ H \end{smallmatrix}}{\mathop{C}}\,=\overset{\begin{smallmatrix} Cl \\ | \end{smallmatrix}}{\mathop{C}}\,-C{{H}_{3}},}}\,\] \[\underset{butene-2}{\mathop{{{H}_{3}}C-\overset{\begin{smallmatrix} H \\ | \end{smallmatrix}}{\mathop{C}}\,=\overset{\begin{smallmatrix} H \\ | \end{smallmatrix}}{\mathop{C}}\,-C{{H}_{3}}}}\,\] \[\underset{3-methyl\text{ }pent-2-ene}{\mathop{{{H}_{3}}C-C{{H}_{2}}--\overset{\begin{smallmatrix} C{{H}_{3}} \\ | \end{smallmatrix}}{\mathop{CH}}\,=CH-C{{H}_{3}}}}\,\] \[\underset{2-methyl\text{ }but-2-ene}{\mathop{{{H}_{3}}C-CH=\overset{\begin{smallmatrix} C{{H}_{3}} \\ | \end{smallmatrix}}{\mathop{C}}\,=CH-C{{H}_{3}}}}\,\] (2-methyl but-2-ene does not show geometrical isomerism because same\[(-C{{H}_{3}})\] group are present on double bonded carbon). \[\therefore \] Only 2-chloro but-2-ene, butene-2 and 3 methyl pent-2-ene will exhibit geometrical isomerism.
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