Is this formula for number of optical isomers correct?

For a symmetric molecule with an even number of chiral centres (for acyclic molecules with chiral centres only, not considering $\pi$ bonds or rings), the formulae (also mentioned in [this question](https://chemistry.stackexchange.com/q/57017/108935)) are:



> 
> No. of meso isomers $=2^{\frac n 2 -1}$
> 
> 
> No. of optical isomers $=2^{n-1}$
> 
> 
> 


However, when I went to test it out on $\ce{CH3-CHD-CHD-CHD-CHD-CH3}$ ($n=4$), my results were not matching with the formula.


I got $2$ meso isomers:
[![Meso 1](https://i.stack.imgur.com/1uzOQ.png)](https://i.stack.imgur.com/1uzOQ.png)
[![Meso 2](https://i.stack.imgur.com/vdGhX.png)](https://i.stack.imgur.com/vdGhX.png)


And I got $6$ enantiomer pairs ($12$ optical isomers):
[![enter image description here](https://i.stack.imgur.com/TNvzo.png)](https://i.stack.imgur.com/TNvzo.png)
[![enter image description here](https://i.stack.imgur.com/ywsA2.png)](https://i.stack.imgur.com/ywsA2.png)
[![enter image description here](https://i.stack.imgur.com/Bp7WV.png)](https://i.stack.imgur.com/Bp7WV.png)
[![enter image description here](https://i.stack.imgur.com/8DLd1.png)](https://i.stack.imgur.com/8DLd1.png)
[![enter image description here](https://i.stack.imgur.com/Zkj6f.png)](https://i.stack.imgur.com/Zkj6f.png)
[![enter image description here](https://i.stack.imgur.com/9BEMG.png)](https://i.stack.imgur.com/9BEMG.png)


The formula for meso isomers seems to match; $2^{{4 \over 2} -1} = 2^1 = 2$


But the formula for optical isomers doesn't match; $2^{4-1} = 2^3 = 8 \neq 12$


Where did I go wrong? Or is the formula wrong? If so, what is the correct formula?




You disproved the formula for the number of optical isomers. You are better off to count off $2^n$ different combinations of $R/S$ configurations, then note that the mesos are counted twice (because the intended mirror-image pairs are just one isomer for meso compounds).


So the true number of optical isomers is $2^n-2^{(n/2)-1}$, out of which $2^{(n/2)-1}$ are the meso compounds and the rest consist of enatiomeric pairs.


Thus for $n=4$ you have $2^4-2^1=14$ total optical isomers, which matches your two meso compounds and 12 others that are six pairs of enantiomers.


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I think that the enantiomeric pairs 1 and 4; 2 and 3 are identical. Just rotate the molecule by 180 degrees about an axis perpendicular and passing through the centre of CH3 - CH3 line. This counting problem arises when symmetrical molecule can be numbered from two ends. Hope this helps..


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