How many strictly optically active stereoisomers exist for a perfectly symmetrical molecule having e — Isomerism and Stereochemistry Chemistry Question

💡 Solution & Explanation

The molecule is symmetrical with an odd number of chiral centers ($n=3$). The standard formula for determining the number of optically active isomers in such an odd system is $2^{n-1} - 2^{(n-1)/2}$. Substituting $n=3$, we get $2^{(3-1)} - 2^{(3-1)/2} = 2^2 - 2^1 = 4 - 2 = 2$ active enantiomers.