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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) 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 Meso 2

And I got $6$ enantiomer pairs ($12$ optical isomers): enter image description here enter image description here enter image description here enter image description here enter image description here enter image description here

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?

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    $\begingroup$ No simple general formula exists, unless you introduce more restrictions than given by the literal interpretation of the sentence "for a symmetric molecule with an even number of chiral centres". For one, you can have optical isomers with no chiral centers. $\endgroup$ Commented Mar 23, 2022 at 9:02
  • $\begingroup$ @NicolauSakerNeto Sorry for not mentioning that, this formula is for acyclic molecules with chiral centres only, not considering $\pi$ bonds or rings (basically the simple case) $\endgroup$
    – Righter
    Commented Mar 23, 2022 at 9:06
  • $\begingroup$ I feel that the formulae should satisfy the condition: $($No. of meso isomers$)\ \times\ 2\ +\ ($No. of optical isomers$)\ =\ 2^n$. $$$$ This would give us: No. of optical isomers $= 2^n - 2^{\frac n 2}$, which matches with the example I took. Is this correct? $\endgroup$
    – Righter
    Commented Mar 23, 2022 at 14:12

3 Answers 3

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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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  • $\begingroup$ I detect a down vote, now if I could detect suggestions from the voter for improvement that would be awesome! $\endgroup$ Commented Nov 14, 2022 at 13:09
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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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  • $\begingroup$ So just reduce the number of enantiomers you counted by 4 i.e. there are 8 enantiomers and the formula is correct. Also when there are odd number of chiral centres (similar to above problem), the total number of stereoisomers = 2^(n-1). $\endgroup$ Commented Aug 12, 2022 at 16:11
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    $\begingroup$ Why don't you add your comment to own answer? $\endgroup$ Commented Aug 12, 2022 at 16:51
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Try to use these formulas for finding number of stereoisomers for a given compound:

  1. Geometrical Isomers:

If n is odd: $2^{n-1}+2^{\frac {n+1}{2}-1}$

If n is even: $2^{n-1}+2^{\frac {n}{2}-1}$

  1. Optical Isomers:

(2.1) If there is no meso compound:

Optically Active Isomers(O.A): $2^{n}$

Racemic Mixtures obtained: $\frac {O.A}{2}$

(2.2) If there is a meso compound:

(2.2.1) (For even values of n)

Optically Active Isomers: $2^{n-1}$

Meso Isomers: $2^{\frac {n}{2}-1}$

(2.2.2) (For odd values of n)

Optically Active Isomers: $2^{n-1}$

Meso Isomers: $2^{\frac {n-1}{2}}$

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