Q. The number of $meso$ forms in the following compound is $HOOC.CH\left( CH_{3} \right).CH(OH).$ $CH(Cl).CH(OH)CH\left( CH_{3} \right).$ $COOH$
A) 3
B) 4
C) 8
D) 16
Explanation
(b)
Number of meso structures in compound having odd number of chiral carbon atoms and symmetrical molecule =$2^{\left( \frac{n}{2} - \frac{1}{2} \right)}$
Given, $n = 5$
∴ Number of $meso$ forms =$2^{\frac{5}{2} - \frac{1}{2}} = 2^{2} = 4$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.