# Systematic Approach to Finding Number of Isomers [duplicate]

I have this compound

I want to know how many isomers (including stereoisomers) may have this constitution. Is there a systematic approach towards solving such types of problems related to the total number of isomers?

Note: The above compound has 64 stereoisomers (from textbook).

## marked as duplicate by Mithoron, airhuff, aventurin, Todd Minehardt, pentavalentcarbonMay 26 '18 at 0:01

• Yes there is a systematic approach: there are 6 things which may be one way or another, hence $2^6$ isomers. – Ivan Neretin May 23 '18 at 15:01
• @IvanNeretin Do you mean the chiral centers and the double bonds? – Abhijith S Raj May 23 '18 at 15:17
• I mean the double bonds and the atoms in the ring with attached side chains (the latter are also chiral centers, but that's irrelevant). – Ivan Neretin May 23 '18 at 15:19
• @IvanNeretin Ok Thankyou. But what about total number of isomers including the structural isomers also? – Abhijith S Raj May 23 '18 at 15:21
• That number is a great deal greater and can't be estimated easily. – Ivan Neretin May 23 '18 at 15:28

For instance, the given framework has 4 stereo centers, hence theoretically, you have $2^4 = 16$ stereoisomers for that compound alone. If your book says it has 64 stereoisomers, it probably has counted geometric isomers of conjugated double bonds (E/Z isomers) as well. All these 64 isomers are illustrated as compound 1 in the scheme below where each red asterisk represent a chiral center. Note that, since the ring has three different substitution groups, none would lead to symmetry in the molecule. Thus, all theoretical amount is equal to the actual number of isomers. Now, question is whether you need to count the stereoisomers for this framework only or not. If it is only for this framework, then the number is 64. But, if you want to change the framework by moving the substitution around, that would be a different story.
• If, for example, you want to move substituents around the ring then this is the same as the number of ways of choosing 5 objects out of 8 which is $8!/(5!3!) = 56$, then you have the stereo-centres to consider also, $2^4$ for each of these 56 and then multiply by the number of E,Z configurations. Quite a few thousand. – porphyrin May 24 '18 at 7:36