Question

How many geometrical isomers are possible in the following two alkenes ? (i) $\quad$$CH_3 - CH = CH - CH = CH - CH_3$ (ii) $\quad$$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

When the ends of alkene containing n double bonds are different, the number of geometrical isomers is $2^n$. Thus for $CH_3 - CH = CH - CH = CH - Cl$. 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 $CH_3 - CH = CH - CH = CH - CH_3$ Number of geometrical isomers $=2^{2-1}+2^{\frac{2}{2}-1}=2^{1}+2^{0}=2+1=3.$