(D) The given compound is $CH_3-CH=CH-CH=CH-CH=CH-Cl$. There are $n = 3$ double bonds that can exhibit geometrical isomerism. The number of geometrica

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(D) The given compound is $CH_3-CH=CH-CH=CH-CH=CH-Cl$. There are $n = 3$ double bonds that can exhibit geometrical isomerism. The number of geometrical isomers for a polyene with $n$ double bonds (where the ends are different) is given by the formula $2^n$. Here,$n = 3$,so the number of geometrical isomers $= 2^3 = 8$.