Which of the following compounds can exhibit geometrical isomerism

C H_{3} C H = C H C H_{3}

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C_{6} H_{5} C H = C H_{2}

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H O O C - C H = C H - C O O H

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Both (a) and (c)

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Solution

The correct option is D Both (a) and (c)
Conditions for geometrical isomers:
1) Free rotation must be restricted.
2) Both the terminals must be attached with different groups.
3) Terminals must not be in perpendicular planes.
In option b, one of the terminal carbon is attached with same group thus it will not show geometrical isomerism.
In both (a) and (c), terminal carbons are attached with different group thus they will show geometrical isomerism. Hence, the correct answer is option (d).

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Similar questions

C H_{3} - C H = C H - C H = C H - C_{2} H_{5}

B r - C H = C H - C H_{2} - C H = C H - C H_{2} - C H = C = C = C H - C H = C (C H_{3})_{2}

C H_{3} - C H = C H - C H = C H - C_{2} H_{5}

C H_{3} - C H = C = C H - C H = C = C = C H - C H = C = C H - C H_{3}

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