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(D) When a molecule has $n$ asymmetric carbon atoms and can be divided into two identical halves,it possesses a plane of symmetry.For such molecules,t

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$2^{n-1} + 2^{(n/2-1)}$

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(D) When a molecule has $n$ asymmetric carbon atoms and can be divided into two identical halves,it possesses a plane of symmetry.For such molecules,the number of optically active isomers (enantiomeric pairs) is $2^{(n/2-1)}$.The number of meso forms is $2^{(n/2-1)}$.The total number of stereoisomers is the sum of optically active forms and meso forms,which is $2^{(n-1)} + 2^{(n/2-1)}$.

$A$ molecule is divided into two identical halves and contains $n$ asymmetric carbon atoms. What is the total number of stereoisomers?

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