1
$\begingroup$

Terpin (4-(2-hydroxypropan-2-yl)-1-methylcyclohexan-1-ol) has two $\mathrm{sp^3}$ carbons with different substituents on a cycloalkane, so it has two cis-/trans-isomers, and from what I was taught, is a stereoisomer.

Terpin

The way I was taught to find stereoisomers was to count the number of double bonds that can be E/Z and the stereocentres, and do $2^n.$ There are no stereocentres, and no double bonds, so the number of stereoisomers would only be 1 (only one molecule). This disagrees with the cis-/trans-isomers number which would be 2.

I can't find anything online about these kinds of cases. Is this general rule of counting stereocentres and double bonds with E/Z not apply in every case? And is this a case that happens a lot that when I'm solving problems, I should always keep this in mind?

$\endgroup$
  • $\begingroup$ I don't get your point - there are 2 geometric isomers, neither of them is optically active, that's it... $\endgroup$ – Mithoron Oct 19 at 14:16
3
$\begingroup$

The simple rules don't work with complicated examples. There are non-chiral molecules with "stereocenters", and chiral molecules lacking "stereocenters". The textbook example for your question is 1,4 dichloro cyclohexane (or 1-chloro 4-bromo cyclohexane to get even closer to your example):

https://chem.libretexts.org/@api/deki/files/8171/14diclc6.gif?revision=1

William Reusch explains this and other cyclohexane cases here: https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Chirality/Stereoisomers/Stereoisomerism_in_Disubstituted_Cyclohexanes

The way I was taught to find stereoisomers was to count the number of double bonds that can be E/Z and the stereocentres, and do $2^n$.

You established that this rule does not work here. If you added an oxo group to one of the ring carbons (turning it into a ketone and making the two branches of the cyclohexane different from one another), you would get the expected pattern: two stereocenters, four stereoisomers.

I found a more exact formulation of the rule on another one of William Reusch's pages:

[William Reusch] As a general rule, a structure having $n$ chiral centers will have $2^n$ possible combinations of these centers. Depending on the overall symmetry of the molecular structure, some of these combinations may be identical, but in the absence of such identity, we would expect to find $2^n$ stereoisomers.

Source: https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Chirality/Stereoisomers/Compounds_with_Several_Stereogenic_Centers

So what about 1-chloro 4-bromo cyclohexane or terpin? If you use the following definition of stereocenter, both have two stereocenters:

[Gamini Gunawardena] If the interchange of two ligands on an atom in a molecule results in a stereoisomer of the molecule, the atom is called a stereocenter or stereogenic center.

Source: https://chem.libretexts.org/Ancillary_Materials/Reference/Organic_Chemistry_Glossary/Stereocenter

You have 4 combinations, but there are pairs of identical molecules when you consider the two chair conformations inter-converting.

Other examples where the simple $2^n$ rule does not work are the following:

  1. Meso compounds: Stereochemistry of meso compounds

  2. Axial chirality: M/P nomenclature for axial-chirality

  3. Planar chirality: Why is trans-cyclooctene chiral?

  4. Molecular knots: Molecular knots absolute configuration

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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