Question

Find out the total number of optically active isomers of pentane-1,2,3,4,5-pentaol?

• A. 3
• B. 2
• C. 8
• D. 6

Answer 1

The correct option is B 2

Number of asymmetric C atoms = 2
When the molecule is symmetrical and has even number of chiral carbons.
$\text{Total number of Optical isomers}=2^{n-1}+2^{p-1}$
where n is the number of asymmetrical C atoms and $p=\frac{n}{2}$
$p=\frac{2}{2}=1$
$\text{Total number of Optical isomers}=2^{2-1}+2^{1-1}$
$\text{Total number of Optical isomers}=2^1+2^0=3$