Question

Match the Column I and Column II (Based on the possible number of geometrical isomers).

• A. a-q; b-p; c-r; d-s
• B. a-p; b-r; c-r; d-s
• C. a-s; b-p; c-r; d-q
• D. a-q; b-r; c-p; d-s

Answer 1

The correct option is A a-q; b-p; c-r; d-s

a)There are two double bonds in this molecule across which geometrical isomerism is possible. So, total number of geometrical isomers= $2^2=4$
b)
It has 2 geometrical isomers.
c)It has three double bonds and has same groups at the end
$\therefore$Number of geometrical isomers is given by $2^{n-1}+2^{p-1}$
n=3 p=$\frac{n+1}{2}=\frac{4}{2}=2$
$2^{3-1}+2^{2-1}=6$
d)It has three double bonds and two ends are different $C_6{H}_{5}andC{H}_3$
$\therefore$Number of geometrical isomers=$2^n=2^3=8$