Question

How many geometrical isomers are possible in the following two alkenes?
$\left(i\right){CH}_3-CH=CH-CH=CH-{CH}_3$
$\left(ii\right){CH}_3-CH=CH-CH=CH-Cl$

• A. $4$ and $4$
• B. $4$ and $3$
• C. $3$ and $3$
• D. $3$ and $4$

Answer 1

The correct option is D $3$ and $4$

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 $n$
Number of geometrical isomers
$=2^{2-1}+2^{\frac{2}{2}-1}=3$

When the ends of alkene containing n double bonds are different the number of geometrical isomers is $2^n$ thus for the given

Number of geometrical isomers $=2^2=4$

Hence, the correct option is $D$