Question

No. of geometrical isomers for the molecules if R is same?

• A. 2
• B. 3
• C. 6
• D. 5

Answer 1

The correct option is B 3

The given compound is a symmetrical compound. It has same 'R' terminal groups.
The formula used to determinne the number of geometrical isomers in compound having same terminal groups is,
(i) When even number of double bond present: $2^{(n-1)}+2^{\left[\right(n/2)-1]}$
where, n is the number of double bond present in the molecule.
(ii) When odd number of double bond present: $2^{(n-1)}+2^{\left[\right(n-1\left)/2\right]}$
where, n is the number of double bond present in the molecule.

In the given molecule we have two double bond so,
No. of geometrical isomers = $2^{(n-1)}+2^{\left[\right(n/2)-1]}$
= $2^{(2-1)}+2^{\left[\right(2/2)-1]}$
=$2^1+2^0$
=2+1
=3
Hence, symmetrical compound with 2 double bonds has 3 geometrical isomers.