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I think I read somewhere that if an organic compound has n number of chiral centers, there must exist 2^n number of isomers. Is that true? And why?

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    $\begingroup$ In short, because math. Each center can be either R or S. What are your options with one center? two centers? three? Then again, there may be a symmetry which produces meso forms and makes the overall number less than $2^n$. $\endgroup$ – Ivan Neretin Mar 15 '17 at 6:33
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That 2^n number of isomers are exactly stereoisomers ( Not "total no. of isomers"). There is no direct formula for calculating the number of structural isomers.Total number of isomers = (number of structural isomers + number of stereoisomers).

Though there are direct formulas for obtaining number of stereoisomers, but there is no direct formula for finding out the total number of structural isomers.

As I said above, we do have formulas for finding out the number of stereoisomers. Here is the exact formulas which you can use to determine number of stereoisomers:

a) When the molecule is unsymmetrical and contains 'n ' chiral carbon atoms:

Total no. of stereoisomers = 2^n

b) When the molecule is unsymmetrical and has even number of stereogenic centres or chiral carbon atoms:

Total no. of stereoisomers = Number of optical isomers + Number of meso forms = 2(n-1) + 2(n/2-1)

c) When the molecule is symmetrical and has odd no. of stereogenic centres

Total no. of stereoisomers = [2(n-1)-2(n/2-1/2)] + 2(n/2-1/2)]

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    $\begingroup$ The excessive use of bold makes this hard to read. $\endgroup$ – MaxW Mar 15 '17 at 14:48

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