3
$\begingroup$

If E/Z is a type of stereoisomerism, then you should be able to use the formula $2^n,$ where $n$ is the number of stereogenic centres. How is it that there are 16 possible E/Z isomers according to the answer in my textbook, but there are no stereogenic centres in all-trans-retinol?

Structure of all-trans-retinol

Improve this question
$\endgroup$
2
  • 5
    $\begingroup$ E/Z is not about centres at all. $\endgroup$
    – Ivan Neretin
    Commented Jan 1, 2023 at 8:56
  • $\begingroup$ See E/Z notation $\endgroup$
    – Poutnik
    Commented Jan 1, 2023 at 9:49

2 Answers 2

Reset to default
5
$\begingroup$

A molecule may include

  • any natural number including zero of double bonds (and for each of them, individually, one can check if their configuration can be described as E, Z, or if this label is not applicable here [e.g., ethene])
  • any natural number including zero of stereogenic centres (either R, or S)
  • any natural number including zero of helical axes (either P, or M)

These three properties are independent of each other.

Improve this answer
$\endgroup$
0
1
$\begingroup$

The molecule contains four conjugated double bonds $\ce{C=C}$, which are all drawn in the E (trans) disposition. And there is no chiral carbon atom, so no R or S atom. But one, two, three or all of the double bonds $\ce{C=C}$ may be in the Z (cis) disposition. This produces $2^4 = 16$ different molecules which are all isomers. They are EEEE, ZEEE, EZEE, EEZE, EEEZ, then EEZZ, EZEZ, EZZE, ZEEZ, ZEZE, ZZEE, then EZZZ, ZZZE, ZZEZ, ZEZZ, and ZZZZ.

Improve this answer
$\endgroup$

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.