How many geometrical isomers are possible in the following two alkenes ? (i)…
How many geometrical isomers are possible in the following two alkenes ?
(i) $\mathrm{CH}_3-\mathrm{CH}=\mathrm{CH}-\mathrm{CH}=\mathrm{CH}-\mathrm{CH}_3$
(ii) $\mathrm{CH}_3-\mathrm{CH}=\mathrm{CH}-\mathrm{CH}=\mathrm{CH}-\mathrm{Cl}$
4 and 4
4 and 3
3 and 3
3 and 4.
Solution
When the ends of alkene containing $n$ double bonds are different, the number of geometrical isomers is $2^n$. Thus for
$\mathrm{CH}_3-\mathrm{CH}=\mathrm{CH}-\mathrm{CH}=\mathrm{CH}-\mathrm{Cl}$.
Number of geometrical isomers $=2^2=4$ When the ends of alkene containing $n$ double bonds are same, then the number of geometrical isomers $=2^{n-1}+2^{p-1}$
where $p=\frac{n}{2}$ for even $n$ and $\frac{n+1}{2}$ for odd $n$, thus for
$\mathrm{CH}_3-\mathrm{CH}=\mathrm{CH}-\mathrm{CH}=\mathrm{CH}-\mathrm{CH}_3$
Number of geometrical isomers
$=2^{2-1}+2^{\frac{2}{2}-1}=2^1+2^0=2+1=3$.