Total number of stereoisomers for:


I tried the following combinations:

So, total number of stereoisomers I got is 12, but my teacher says it's 16. I don't know how to figure out what other four isomers are.

  • 3
    $\begingroup$ Just list all four-letter combinations of R and S, and that would be it. $\endgroup$ Dec 24 '18 at 14:00
  • 4
    $\begingroup$ You have just missed a few, for example you also need RSSS as well as SSSR and so forth. $\endgroup$
    – porphyrin
    Dec 24 '18 at 14:03

If you have studied combinatorics, and if you know the fundamental principal of counting, then this is basically making a four letter word with two letters- R and S. Each place has two choices, and you have four spaces to fill, so you get 2×2×2×2= 16 optical isomers for this molecule. Since the terminal alkyl groups are different, so there is no chance of a meso-compound. And as Ivan rightly stated in the comment, to list the isomers, make all possible combinations of R and S(you have missed a few as pointed out by porphyrin in the comments)


In such a case (1), once you have made sure that not 2 configurations are the same (because of symmetry) for n chiral centers, then the number of stereoisomers is 2^n, because there are 2 possibilities for each center, then 2 possibilities for the next one and so on. Overall, you get 2 possibilities for 1 chiral center, 2 * 2 possibilities for 2 chiral centers, and then 2^3, 2^4 etc.

You gave 14 configurations, so you obviously missed 2 configurations to get to 16 (i.e. the closest power of 2).

(1) There are cases where the overall shape of the molecule adds asymmetry but it is not the case here.


If u apply simple counting for every chiral atoms present in the molecule there are 2 possible stereoisomers so by fundamental principle of counting you can directly say that total number of stereoisomers are 2^n where n represents the total number of chiral atoms it's just a convenient formula based on the method described but others in their answers above


There would be 16 stereoisomers as suggested by other's answers.But there would be less no
of enantiomer as this molecule posses plane of symmetry in certain configurations

  • $\begingroup$ Aren't you talking about 2,3,5,6-tetrachloroheptane instead? $\endgroup$
    – mykhal
    Jan 24 '19 at 17:35

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