# 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) 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:

And I got $$6$$ enantiomer pairs ($$12$$ optical isomers):

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?

• 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. Commented Mar 23, 2022 at 9:02
• @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) Commented Mar 23, 2022 at 9:06
• 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? Commented Mar 23, 2022 at 14:12

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.

• I detect a down vote, now if I could detect suggestions from the voter for improvement that would be awesome! Commented Nov 14, 2022 at 13:09

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..

• 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). Commented Aug 12, 2022 at 16:11

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