Which of the following compounds is capable of showing geometrical isomerism?

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

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

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

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None of the above

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Solution

The correct option is C $C H_{3} - C H = C = C = C H - C H_{3}$
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.

a, b are even cumulenes so they will not show geometrical isomerism. In even cumulenes, groups on the terminal carbons exist in a perpendicular plane. now for compounds to exhibit geometrical isomerism, groups on the terminal carbons must be in the same plane.

Odd numbered cumulenes show geometrical isomerism. When there is three consecutive double bonds, terminals lie in the same plane giving geometrical isomerism.
hence c will show geometrical isomerism.

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Similar questions

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}

M e C H = C H - C H = C = C H - C H = C H_{2}

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

Write IUPAC names of the following compounds:
$a . C H_{3} C H = C (C H_{3})_{2}$

$b . C H_{2} = C H - C \equiv C - C H_{3}$

$c .$

$d .$

$e .$

$f .$

$g . C H_{3} - C H = C H - C H_{2} - C H = C H - C H | C_{2} H_{5} - C H_{2} - C H = C H_{2}$

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

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