Question

The number of geometrical isomers in case of a compound with the structure

Answer 1

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Number of geometrical isomers = 2n-1 + 2[n+1/2]-1 C6H6 - CH = CH - CH = CH - CH = CH - C6H5 (n =3, odd) Number of geometrical isomers = 22 + 22-1 =22 + 21 = 4 + 2 = 6

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