3
$\begingroup$

I am having trouble understanding why the structure

$$\ce{C(CHClMe)4}$$

has only five stereo isomers. According to my textbook, the stereo isomers correspond to the following series of absolute configurations:

  • (R,R,R,R)
  • (R,R,R,S)
  • (R,R,S,S)
  • (R,S,S,S)
  • (S,S,S,S)

But my question is, does not the order in which these appear change the total? Isn't (R,S,R,S) different from (R,R,S,S)?

$\endgroup$
4
  • $\begingroup$ Just to make sure, is this the structure of your compound? $\endgroup$
    – andselisk
    Jun 3, 2020 at 12:37
  • $\begingroup$ Did you try drawing it out or playing with a model? $\endgroup$
    – Zhe
    Jun 3, 2020 at 12:42
  • $\begingroup$ @andselisk Yes it is $\endgroup$
    – John Tony
    Jun 3, 2020 at 12:48
  • $\begingroup$ 4 1.RRRR 2.RRRS 3.RRSS 4.RSSR SSSR will be same as RRRS on rotating and 3 and 4 will be different . $\endgroup$ May 17, 2021 at 15:19

1 Answer 1

7
$\begingroup$

To answer your question, if you haven't figured it out by now, RRSS is the same as RSRS and it is the only one of the five that is optically inactive (Compare with nonactin). Structure A numbers the positions of chiral chlorine-bearing carbons. Structure B is RSRS and C is RRSS. Clearly, they are superimposable and, therefore, identical. A similar exercise with RRRS and RSSS will give a similar result.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.