A compound can be divided into two equal halves and contains even $n$ asymmetric carbon atoms. The number of stereo isomers is:

2^{n}

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2 (n + 1)

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2^{(n / 2 - 1)}

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

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Solution

The correct option is C $2^{n - 1} + 2^{(n / 2 - 1)}$
A compound can be divided into two equal halves and contains even n asymmetric carbon atoms. The number of stereo isomers is $2^{n - 1} + 2^{(n / 2 - 1)}$ .
$2^{n - 1}$ represents the number of d, l isomers and $2^{(n / 2 - 1)}$ represents the number of meso forms.

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