# How to derive these general formulae for number of stereoisomers of a compound with a possible plane of symmetry?

These formulae are used if the molecule has a possible plane of symmetry. One such example would be:

Here the carbons marked with an asterisk are stereogenic centres (the asterisk is not used to mark isotopes). We can clearly see that if carbon number 2 (in the entire longest chain) and the carbon number 4 have opposite stereogenic configuration, then the molecule will be achiral because it will have a plane of symmetry. For example, (2R,3S,4S)-pentanetriol will be a meso compound with a plane of symmetry. This compound clearly satisfies the criteria we have set for the type of molecules being discussed.

Let us take another example.

Again, we see that if the molecule has the same geometrical configuration at the first and the third double bond, then the molecule has a plane of symmetry. For example, (2,7)-diphenylocta-(2Z,4E,6Z)-triene clearly has a plane of symmetry.

Our goal is to find the total number of stereoisomers such compounds can have in total. We can assume that a molecule does not have both, a double bond and a chiral centre.

In our class notes, we have written these formulae:

For geometrical isomers (i.e. in case of polyenes),

1. If 'n' is even (here n is the number of double bonds):$$\text{Number of stereoisomers} = 2^{n-1}+2^{n/2-1}$$
2. If 'n' is odd, then: $$\text{Number of stereoisomers} = 2^{n-1}+2^{(n-1)/2}$$

And for optical isomers (molecules with chiral centres):

1. If 'n' is even (here n is the number of chiral centres): $$\text{Number of enantiomers} = 2^{n-1}$$ $$\text{Number of meso compounds} = 2^{n/2-1}$$ $$\text{Total number of optical isomers} = 2^{n-1}+2^{n/2-1}$$
2. If 'n' is odd: $$\text{Number of enantiomers} = 2^{n-1}-2^{(n-1)/2}$$ $$\text{Number of meso compounds} = 2^{(n-1)/2}$$ $$\text{Total number of optical isomers} = 2^{n-1}$$

How to derive these formulae?

• Aug 5 '16 at 18:26
• How many strings of length $n$ using only the characters "A" and "B" are palindromes?
– f''
Aug 6 '16 at 0:08
• @f'' I tried using the same method! But I'm getting a different formula for optical isomers when 'n' is odd (I get one that is identical to n=odd for geometrical isomers). Aug 6 '16 at 4:29
• Oh, I see. This case is trickier because of the central stereocenter. If the left and the right sides have identical configurations, then it is no longer a stereocenter because it has two identical substituents, so you actually only have half as many isomers of this kind as you would expect. This is why the total number of isomers is lower than you calculated by the amount $2^{(n-1)/2}$.
– f''
Aug 6 '16 at 5:40
• Will a mathematical approach be more appropriate? In that case, this question is worth sharing with the Math StackExchange community... Dec 26 '17 at 15:09

I managed to crack the formula for optical isomers with odd chiral centers, so I'll share my attempt here. Hopefully others may innovate on it and post solutions for other formulae.

## Pseudo-chiral carbon atoms - an introduction

The Gold Book defines pseudo-chiral/pseudo-asymmetric carbon atom as:

a tetrahedrally coordinated carbon atom bonded to four different entities, two and only two of which have the same constitution but opposite chirality sense.

This implies that, in your case:

If chiral carbons 2 and 4 both have configuration R (or both S), then the central carbon 3 will be achiral/symmetric, because now "two and only two of its groups which have the same constitution" will have the same chirality sense instead. (Your approach by "plane of symmetry" is wrong. Find more details on this question)

Hence, there can be two stereoisomers (r and s) possible on the 3rd carbon due to its pesudochirality. But, there will be only one if both substituents on left and right have the same optical configurations.

## Building up an intuition by manual counting

For optical isomers with odd number of chiral centers and similar ends, you can guess that, if there are $$n$$ chiral centers, then the middle ($$\frac{n+1}2$$-th) carbon atom will be pseudo-chiral. To build up an intuition, we'll manually count optical isomers for $$n=3$$ and $$n=5$$:

### Case $$n=3$$

Take the example of pentane-2,3,4-triol itself. We find four (=$$2^{n-1}$$) isomers:

$$\begin{array}{|c|c|c|}\hline \text{C2}&\text{C3}&\text{C4}\\\hline R&-&R\\\hline S&-&S\\\hline R&S&R\\\hline R&S&S\\\hline \end{array}$$

As expected from the relevant formula, we find that the first two ($$=2^\frac{n-1}2$$) are meso compounds, and the remaining two ($$=2^{n-1}-2^\frac{n-1}2$$) are enantiomers.

### Case $$n=5$$

Take the example of heptane-2,3,4,5,6-pentol:

We expect $$16~(=2^{n-1})$$ isomers, with the C4 carbon being pseudo-chiral. To avoid a really large table, we observe that the number of meso isomers is easily countable (<< number of enantiomers). Here is a table of those four (=$$2^\frac{n-1}2$$) meso isomers:

$$\begin{array}{|c|c|c|c|c|c|}\hline \text{C2}&\text{C3}&\text{C4}&\text{C5}&\text{C6}\\\hline R&R&-&R&R\\\hline R&S&-&S&R\\\hline S&R&-&R&S\\\hline S&S&-&S&S\\\hline \end{array}$$

Note that the total optical isomers are given by $$2^{n-1}$$ isomers (more on that below). Hence, the number of enantiomers is easily $$12(=2^{n-1}-2^\frac{n-1}2)$$.

## A formula for the number of meso isomers

As you must have observed from the table, the sequence of optical configurations, when read from the fourth carbon atom, is exactly the same towards both left and right. In other words, if we fix an arbitrary permutation for the optical configurations of the carbon atoms on the left (say RSS), then we will get only one unique permutation of the optical configurations on the right (SSR).

We know that each carbon on the left has two choices (R or S), and there are $$\frac{n-1}{2}$$ carbon atoms on the left. Hence, the total number of permutations will be $$2\times2\times2\cdots\frac{n-1}{2}\text{ times}=2^\frac{n-1}{2}$$.

Since, our description ("the sequence of optical configurations, when read from the fourth carbon atom, is exactly the same on both left and right") describes meso isomers, we have hence counted the number of meso isomers, which is $$2^\frac{n-1}{2}$$.

## A formula for the number of total isomers

We note that there are $$n$$ chiral carbons (including that pseudo chiral carbon). Again, each chiral carbon has $$2$$ choices. Hence, the maximum possible number of optical isomers is $$2\times2\times2\cdots n\text{ times}=2^n$$. This is the maximum possible, not the actual total number of isomers, which is much lower.

The reduction in number of isomers is because the string of optical configurations reads exactly the same from either terminal carbon. Example: RSsRS is the same as SRsSR. This happens because the compound has "similar ends"

Hence, each permutation has been over counted exactly twice. Thus, the actual total number of isomers is half of the maximum possible, and is $$=\frac{2^n}2=2^{n-1}$$.

## Conclusion

Hence, we have derived that, if 'n' (number of chiral centers) is odd for a compound with similar ends, then:

• $$\text{Number of meso isomers} = 2^{(n-1)/2}$$
• $$\text{Total number of optical isomers} = 2^{n-1}$$
• $$\text{Number of enantiomers} = 2^{n-1}-2^{(n-1)/2}$$
• Regarding your statement under the introduction part: "If chiral carbons 2 and 3 both have configuration R (or both S), then the central carbon 3 will be achiral/symmetric,..." : I think you concluded that such a molecule would be achiral due to the absence of a pseudo-asymmetric carbon. Your linked question deals with the opposite case where the molecule has a pseudo-asymmetric carbon but is achiral. Similarly, even if the molecule lacks a pseudo-asymmetric centre, it can be chiral. It doesn't matter for our counting purpose but I wanted to make a note of it. Jul 14 '20 at 8:10
• "If chiral carbons 2 and 3 ...then the central carbon 3 will be achiral/symmetric" - here, did you mean "if chiral carbons 2 and 4"? Similarly, isn't it the last two entries in the table for n=3 which are meso compounds? Also, in that same case, aren't the total isomers RSR, RRR, SSS, SRS(all four of these meso), and then RSS and RRS(both enantiomers)? Apr 25 '21 at 2:41
• @HarryHolmes Every pair of meso compounds can be considered to be one and same. RSR and RRR are effectively the same compound due to them being meso Apr 25 '21 at 7:44
• @SafdarFaisal: I saw the edit, thanks. About the meso compounds; when you pass plane polarized light through RSR and RRR separately, yes, you wouldn't be able to differentiate them, but how does that change the fact that the relative arrangement of atoms on the central carbon are different? Wouldn't they be technically different compounds, though you couldn't practically tell them apart? Apr 25 '21 at 9:14
• @SafdarFaisal: also, what does it mean when you left the configurations for $C_3$ blank? Apr 25 '21 at 9:16

We can calculate no. of optical isomers for straight chain molecules with even and odd no. of Chiral centres very easily by observing symmetries and using simple mathemetical logics. I will briefly elaborate on even no. of centres. Odd no. cases can also be calculated by similar idea.

### Molecule with even no. of Chiral Carbon Centres where molecule can be divided into two equal halves

Calculation of total no. of Optical isomers include calculation of Meso forms and Enantiomeric pairs. In that case, (and also for Odd no. of chiral carbon centres) calculation of meso forms are very easy.

For calculating no. of Meso isomers, consider a general straight chain molecule with even no. of chiral centres (say, $n = 2k$) as below,

The groups ($A_i$, $\forall i = 1,2,...,2k+1$) are arranged so as to represent a meso compound in general.
Now observe that between $k$th and $k+1$th Carbon centre, by an imaginary plane the molecule can be divided into two equal halves. That's why, as a whole the above molecule is a Meso compound. Now, to become a meso compound, an isomer of this compound also has to be symmetric w.r.t this imaginary plane.
So,we can only change configuration the upper $k$ or lower $k$ Carbon atoms to form a new Meso compound, because the configuration of the symmetric Carbon atom will automatically change to maintain symmetricity.
Now, we can change $0$ atom (i.e. our starting compound) $1$ Carbon atom or $2$ atoms or $3$ atoms and so on upto $k$, and thus we are actually calculating each isomers twice.Because , inversion of $k-1$ atoms is nothing but the mirror image of inversion of $1$ atom ($k$ th atom) which is already considered before.
Thus,total no of Meso compounds = $$\frac{\binom{k}{0} + \binom{k}{1} + .... + \binom{k}{k}}{2} = 2^{k-1} = 2^{\frac{n}{2} -1}$$ which gives us the result.

Now for Enantiomeric pairs,fix the first Chiral Carbon centre ($\ce{C_1}$), because if we don't fix it, we will end up counting mirror images also, which we don't want for counting pairs. Now,fix its symmetrically opposite carbon atom ($\ce{C_{2k}}$) as the relative configuration opposite to $\ce{C_1}$ i.e. $A_3$ on the left and $A_2$ on the right.Thus, remaining $n-2$ centres can be oriented in $2^{n-2}$ ways and all will be different optically active isomers and there are no mirror images also. Thus, Enantiomeric pairs will be = $2^{n-2}$.
So, total no. of Optical isomers = $2 \times 2^{n-2} + 2^{\frac{n}{2} -1}$ = $2^{n-1} + 2^{\frac{n}{2} -1}$

For, odd no. of carbon atoms also similar idea can be applied. If you take $n = 2k+1$, no. of meso isomers will be, $$\binom{k}{0} + \binom{k}{1} + .... + \binom{k}{k} = 2^k = 2^{\frac{n-1}{2}}$$ The logic behind it and the remaining part is left for the reader to think about...

• while calculating enantiomers, why did you fix relative configuration of C1 opposite to C2k? Jul 16 '20 at 4:06
• @shreya The fixing of relative configuration of C1 has to be done because we actually want to count distinct Enantiomeric pairs to be precise, and after calculating distinct pairs we can multiply it by 2 to get total enantiomers. Now this fixing concept is basically similar to the permutations of people around a round table if you think carefully. If you can understand permutations around a round table, this problem is actually very similar. And also the molecule overall should be dividable into two equal halves and hence the fixing should be done for C2k as well. Jul 17 '20 at 5:11