How many geometrical isomers are possible in the following two alkenes ? (i) CH ₃- CH = CH - CH = CH - CH ₃ (ii) CH ₃- CH = CH - CH = CH - Cl

Step-by-step solution

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= n 2 for even n and n+1 2 for odd n , thus for CH ₃- CH = CH - CH = CH - CH ₃ Number of geometrical isomers =2²⁻¹+2^ 2 2 -1 =2^1+2^0=2+1=3 .