When accurately determining the number of strictly optically active isomers for a perfectly symmetri — Isomerism and Stereochemistry Chemistry Question

💡 Solution & Explanation

For symmetrical molecules with an even number of chiral centers ($n$), the total number of stereoisomers is $2^{n-1} + 2^{(n/2 - 1)}$. The number of optically active forms specifically is $2^{n-1}$, while the number of meso forms specifically is $2^{(n/2 - 1)}$. Therefore, $2^{n-1}$ correctly calculates the active isomers alone.