How can I determine the number of possible pairs of diastereomers here?

My first guess was two: RR with RS and SS with RS. But what about the potential chirality center middle carbon? There are two identical branches attached to it; however, those two branches can change R/S. So, is it a chirality center?


3 Answers 3


I basically agree with Ron's answer, but had to draw all of the possible structures to confirm it. The complication is that carbon-2 is non-stereogenic but it may be chirotopic depending on the configuration of the neighboring atoms. The IUPAC Gold Book calls this a pseudo-asymmetric carbon atom. Carbon-2 is not a stereocenter, because it does not have four different "things" attached to it. However, its configuration is important relative to carbons-1 and -3. Below are all of the possible combinations.

enter image description here

Out of the eight possibilities, we see there are only four unique compounds (names generated in ChemDraw):

A: (1​R,2​s,3​S)-1,3-dichloro-1,2,3-triphenylpropane

B: (1​S,3​S)-1,3-dichloro-1,2,3-triphenylpropane

C: (1​R,2​r,3​S)-1,3-dichloro-1,2,3-triphenylpropane

F: (1​R,3​R)-1,3-dichloro-1,2,3-triphenylpropane

The small r and s are essentially used to indicate the stereochemistry at carbon-2 relative to carbons-1 and -3. Note that in compounds B and F, the stereochemistry at carbon-2 is inconsequential as in both cases the phenyl group is syn to one chloride and anti to another chloride. I haven't been able to find an adequate explanation of assigning the r/s designations, but I believe that the R configured substituent receives higher priority than the S configured substituent, and Cahn–Ingold–Prelog system is followed from there.

  • $\begingroup$ Hmm. The solutions said the answer was two. Like my guess... I conclude that the solutions is mistaken. $\endgroup$
    – yolo123
    Nov 3, 2014 at 19:03
  • $\begingroup$ Carbon 2 is achirotopic, not chirotopic and it is called a stereocenter as it can have r/s configuration. It is not strictly a chiral center because two attached groups differ only by their configurations. $\endgroup$
    – EJC
    Oct 15, 2016 at 19:11
  • $\begingroup$ As far as I can tell (by reading Anslyn/Dougherty section 6.3.3) C-2 is chirotopic only in those isomers which are chiral, i.e. B and F. In the meso compounds A and C, C-2 lies within the plane of symmetry and is therefore achirotopic. So, both of you are correct. :) $\endgroup$ Mar 27, 2017 at 16:13
  • 2
    $\begingroup$ As for the r/s notation, future visitors to this question may be interested in: What does lowercase r-s notation mean? and IUPAC name for 1,2,3-trichlorocyclopropane? where the notation is explained in detail with reference to the Blue Book. $\endgroup$ Mar 27, 2017 at 16:17
  • $\begingroup$ @orthocresol I've changed it to reflect that C2 may or may not be in a chiral environment. $\endgroup$
    – jerepierre
    Mar 27, 2017 at 16:53

Let's number the 3 aliphatic carbons to make our discussion clearer.

enter image description here

Using the definition of a chiral carbon as a carbon that has 4 different substituents, carbons 1, 2 and 3 are all capable of being chiral. Carbons 1 and 3 are relatively straightforward chiral carbons, it's easy to see the 4 different substituents and assign "R" or "S" configurations. However carbon 2 is problematic.

If we assign R- and S- configurations to carbons 1 and 3 respectively, then there are 4 different groups attached to carbon 2, but the difference in 2 of the groups is just "R" or "S". In fact, in this situation there is actually a plane of symmetry bisecting the molecule and the molecule is achiral - it is a meso-isomer. The plane of symmetry contains carbon 2 and the hydrogen and phenyl group attached to carbon 2, and, as expected, the R- and S-substituents are enantiomeric. In such cases it is correct to say that carbon 2 is pseudo-asymmetric (thanks to @GaurangTandon for pointing this out!). The Cahn–Ingold–Prelog priority rules can still be used to describe carbon 2, the R substituent is given priority over the S substituent and lowercase "r" and "s" are used to describe carbon 2, the pseudo-asymmetric carbon.

In the case where we assign R- and R- (or S- and S-) configurations to carbons 1 and 3 respectively, then there is no longer a plane of symmetry bisecting the molecule. Carbon 2 is now prochiral (this means changing one substituent is all that is required to make carbon 2 chiral), but the molecule as a whole is now chiral.

Going in the order carbon 1-2-3 here are the various possibilities:

  1. 1R,3R is chiral and it has an enantiomer 1S,3S
  2. 1R,2r,3S is an achiral (there is a plane of symmetry bisecting the molecule), meso-compound and a diastereomer of 1R,3R;
  3. 1R,2s,3S is a second achiral, meso-compound.

So there are 3 diastereomers; one is chiral and exists as a d,l pair and the two other diastereomers are achiral, meso-compounds.

  • $\begingroup$ Ron, just to be sure, the question was asking for pairs. So, I'm listing them: 1. 1R,3R and 1R,2r,3S 2. 1R,3R and 1R,2s,3S 3. 1S,3S and 1R,2r,3S 4. 1S,3S and 1R,2s,3S 5. 1R,2s,3S and 1R,2r,3S Could you please back me up? I found 5 total. $\endgroup$
    – yolo123
    Nov 3, 2014 at 20:28
  • $\begingroup$ This is actually quite weird. This was a multiple choice question and there not odd numbers :O ! I should use another exercise package. Too many mistakes. $\endgroup$
    – yolo123
    Nov 3, 2014 at 20:33
  • $\begingroup$ @yolo123 I don't think that is right. Diastereomers don't come in "pairs". As I said in my answer, there are 3 diastereomers. Maybe the question was asking how many pairs of enantiomers? $\endgroup$
    – ron
    Nov 3, 2014 at 20:39
  • $\begingroup$ Oh boy! Why have I not thought of that...? Word for word: "How many pairs of diastereomers are possible in the following molecule? PhCH(Cl)CH(Ph)CH(Cl)Ph?" $\endgroup$
    – yolo123
    Nov 3, 2014 at 20:40
  • $\begingroup$ Definitely, this question is very ambiguous... Honestly, thanks a lot jerepierre and ron. Very informative. $\endgroup$
    – yolo123
    Nov 3, 2014 at 20:43

I don't think it is hindered enough to prevent rotation around the central carbon, so I think it only has two stereogenic centres, leading to four possible stereocentre combinations as you have identified: RR, SS, SR, RS

This results in more than two pairs of diastereomers though:

1) RR with SR
2) RR with RS
3) SS with SR
4) SS with RS

  • 2
    $\begingroup$ This analysis is incomplete because there are two possible compounds with configuration (1R,3S), depending on the relative (syn/anti) configuration at C-2. $\endgroup$
    – jerepierre
    Nov 3, 2014 at 17:04
  • $\begingroup$ @jerepierre Indeed, thanks for posting a better one :) $\endgroup$
    – Sam
    Nov 5, 2014 at 14:46

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