Determine the total number of geometrical isomers possible for the...

For compound (I): $\mathrm{C_6H_5-CH=CH-CH=CH-CH=CH-CH_3}$

The number of double bonds capable of showing geometrical isomerism is $n = 3$.

Since the terminal groups are different ($\mathrm{C_6H_5}$ and $\mathrm{CH_3}$), the molecule is unsymmetrical.

The number of geometrical isomers is given by $2^n$.

Substituting $n = 3$, the number of geometrical isomers is $2^3 = 8$.

For compound (II): $\mathrm{C_6H_5-CH=CH-CH=CH-C_6H_5}$

The number of double bonds capable of showing geometrical isomerism is $n = 2$.

Since the terminal groups are identical ($\mathrm{C_6H_5}$), the molecule is symmetrical.

For a symmetrical polyene with an even number of double bonds, the number of geometrical isomers is given by $2^{n-1} + 2^{(n/2)-1}$.

Substituting $n = 2$, the number of geometrical isomers is $2^{2-1} + 2^{(2/2)-1} = 2^1 + 2^0 = 2 + 1 = 3$.

The total number of geometrical isomers for (I) and (II) are 8 and 3 respectively.

Answer: 8 and 3