Organic Chemistry Study Tips

By James Ashenhurst

A simple trick: The (R)-(S) Toggle

Last updated: October 14th, 2020 |

1-RS

The difficulty of assigning (R)/(S) to a stereocenter can be highly dependent on the way the molecule is drawn. For example, assigning the stereochemistry of A should be a cakewalk if you are the least bit familiar with the rules, while B  and C are a little more difficult. The thing is, you’re not always going to be given a structure with the 4th ranked substituent in the back. One thing I see students do a lot in this situation is try to redraw the structure to make it easier to assign (R)/(S) While this can be effective, it can easily lead to mistakes. Furthermore, when you’re dealing with cyclic systems like morphine [C], trying to draw the mirror image in a reasonable time frame might induce brain trauma.   [By the way, for B, you remember why the CH2CH2Br group is ranked 3rd, right? [Note 1]]

Instead of redrawing the whole molecule, just leave the hydrogen (or other 4th ranked substituent) in front and figure out whether the substituents ranked 1,2, and 3 go clockwise or counterclockwise. Then toggle (i.e. reverse direction CW to CCW or vice versa) to account for the fact that the 4th ranked substituent is in the front, and assign (R) or (S) based on that.  You end up with the same answer without having to redraw anything.

We also make these types of adjustments when we’re looking in mirrors. If I’m looking in a mirror at you holding out your right hand, I account for the mirror by “flipping” your image to know that you are really holding out your left. It’s the same thing with molecules.

2-RS
This trick seems obvious, but I’ve met enough people who don’t do it to make it worth mentioning. Determining (R)/(S) eventually becomes second nature, as does the ability to flip, rotate and invert chemical structures in your head. Much like London cab drivers, one of the interesting side effects of becoming an organic chemist is that it hyper-develops your brain’s spatial abilities.

[note 1 – Priority is determined at the point of first difference. Hence, of the two carbons attached to the stereocenters, CH2 and C=O, the C=O is given priority [O > H], regardlesss of whether a higher ranked element is further down the second chain. ]

Comments

Comment section

25 thoughts on “A simple trick: The (R)-(S) Toggle

  1. Is there a trick for rotations when the hydrogen (or other fourth ranked substituent ) is in the plane of the page?

  2. Hello James,

    Thanks for the wonderfully organized website. I had a question about the swap trick –

    We were told to visualize molecules in class and the trick is never taught even though the instructor is highly rated and knows it. Could this be because it does not work in all cases?
    H
    |
    C-Cl
    Wedge OCH3 Hash CH3
    The above is just supposed to be a tetrahedral structure with H pointing up and Cl pointing right (both are in the plane of the page). On the bottom left is an OCH3 with a wedge (coming out of page) and a CH3 on the bottom right with a hash (going into page). The priority is Cl>OCH3>CH3. This gives S.
    According to the swap rule, if the hydrogen is not pointing down we switch it to R, but that would be wrong because if you use a molecular model kit and orient the structure, you have S.
    Am I doing something wrong or is the Rule not a hard set rule and just correct >90% of the time?

    1. The example I gave has H in the front. I didn’t cover an example where it is in the plane of the page.

      This is a VERY old post. An even better trick I found was to do this: if H is in the plane of the page, do a single swap of H with whatever group is a dash. Then determine R/S. Whatever value you get will be the opposite of the true R/S. In your example, if you swap H with (dashed) CH3, and then determine R/S, it “looks like” R. But since we’ve done a single swap it’s the opposite of that – S – in accordance with what you say.

      Relevant post:
      https://www.masterorganicchemistry.com/2011/01/24/the-single-swap-rule/

      1. Great. Thank you!

        I should have read the “single swap rule” more carefully. You’ve confirmed for me that the “R/S toggle” method above doesn’t work in all cases. The single swap rule makes sense and is more universal. Thanks again!

  3. Does the R/S Chart you made up there work all the time? I get the swap rule and whole idea behind it, but this chart makes life so much easier….

  4. Is there a trick for when there is no hydrogen atom at all? For example what would be a quick way or assigning the configuration for this molecule: CH3CH2CBrClCH(CH3)2

    1. Yes, you always assign priority based on the first point of difference. Start by looking at the four atoms bonded to that carbon: C, C, Br, Cl. Br is priority #1, and Cl is priority #2. Between the two carbons, you have to travel farther down the chain to break the tie. So looking at the carbon on the left we see it is bonded to C, H, H. The carbon on the right is bonded to C, C, H. Since C > H the right hand carbon “wins” and is assigned priority #3 and the left hand carbon is priority #3.

      https://www.masterorganicchemistry.com/2017/01/17/determining-rs-2-the-method-of-dots/

  5. What is the priority between 2 groups which differ only by the stereochemistry at a single position? E.g. The designation of C1 (or C4) in Quinic acid. (The differences in the stereochemistry at C3 and C5 make C1 and C4 chiral! Hence when trying to assign the CIP designation for C1 it is necessary to prioritize C3 over C5 or vice versa.
    Thanks
    Jim

  6. Hi (I can’t find your name…sorry),

    There are so much of materials about wedge-dash notations, R-S system etc.
    But not even once it mentions (from the material that I have come across), how to determine which atom comes out of plane (forward) and which stays in the plane.

    Obviously most of the ref material seems to indicate that lowest priority atom/groups goes to the back. Why in the world there’s no reasoned check list on determining the plane of wedges/dashes in the simple theory books? At college y1 level.

    Please advice. Thanks.

    1. I say this a lot, but molecules are three-dimensional objects just like anything else.
      Say you have a car that you want to sell and are taking some pictures of it to show people online. There’s no “rule” saying whether you have to take the photo from standing in front of the car, on from the passenger side, or the driver’s side, or from the back.
      It’s the same car! You’re just taking pictures from different perspectives.
      It’s the same thing with molecules. There’s no one way you’re supposed to depict three-dimensional molecules on a page. Depending on your purposes, there can be multiple ways to do it, all correct.

  7. If 4th ranked substituent is not shown in the molecule how to decide whether it’s at front or at back to assign r and s configuration

    1. In a tetrahedral carbon drawn in a line diagram, two substituents are drawn in the plane of the page, one is drawn as a wedge, and one is drawn as a dash. Therefore if you only see three groups drawn to a carbon, one of which is a wedge, you assume that the fourth is a dash, and vice versa. That’s often the case with implicit (“hidden”) hydrogens.

  8. How do I assign R and S if my number #4 is not on dash or wedge, but in the plane? in other words, on a regualor bond line.

    Thanks.

  9. How do u assign R and S configuration to a stereoisomer with a stereo genic centre where the H is neither on a wedge or a dash, but just in the plane of the book(a straight line)

  10. I’m sorry for my ignorance, but in molecule B, shouldn’t the C double bonded to the O be priority # 1? A double bond is treated as bonded twice to that atom, therefore 2 ‘O’s should take preference over OH.

    Thanks

    1. Ranking is always determined at the first point of difference. The four atoms directly bonded to that carbon are O, C, C, and H. The oxygen takes first priority, H takes fourth. Between the two C atoms, we break the tie by moving further down the chain. The first C is bonded to C, H, H. The second C is bonded to O, O, C. [that’s double-counting the carbonyl, just like you said!]. Since O > C, the C(O)CH3 “wins” priority #2.

      See: https://www.masterorganicchemistry.com/2017/01/17/determining-rs-2-the-method-of-dots/

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