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How do we assign priority groups of tartaric acid? Counting outwards from any of the chiral carbons we get the same atom.

enter image description here

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For simplicity, here is tartaric acid without any stereochemistry:

$$\ce{HOOC-CHOH-CHOH-COOH}$$

And here is a depiction that includes the stereochemical information:

 O=C-OH
   |
 H-C-OH
   |
HO-C-H
   |
 O=C-OH

(using the Fischer convention)

The chiral carbons are the central ones; we need to assign priorities to each of the atoms bound to them. For both, the discussion is similar, because they are chemically practically identical.

  1. $\ce{O}$ from $\ce{OH}$.
  2. $\ce{C}$ (carboxylic acid group)
    1. $\ce{O}$ from the double bond
    2. $\ce{O}$ from the $\ce{OH}$ group
    3. $\ce{(O)}$ (a ghost oxygen) from the double bond.
  3. $\ce{C}$ ($\ce{CHOH}$ group)
    1. $\ce{O}$ from the $\ce{OH}$ group
    2. $\ce{C}$ from the distal carboxy group
    3. $\ce{H}$
  4. $\ce{H}$

These lists are written out and the first point of difference is noted; here it is the higher priority of 2.2 (an oxygen atom) with respect to 3.2 (a carbon atom) — and this does not even require extensive knowledge on how to open double bonds.

Thus, the priorities of the groups on the two carbon atoms are:

 O=C-OH      b         .           O=C-OH
   |         |         :             |
 H-C-OH    d-C-a    ···c···        H-R-OH
   |         |         |     –>      |
HO-C-H    ···c···    a-C-d        HO-R-H
   |         :         |             |
 O=C-OH      ·         b           O=C-OH

And we realise that I have drawn (2R,3R)-tartaric acid.

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  1. First you have to rank the molecules around the two chiral centres of tartaric acid according to Cahn Ingold Prelog principle. (I have added the principle in below after this 3 points)

  2. After ranking the molecules according to Cahn Ingold Prelog principle, then orient the molecule such that the lowest priority group is directed away from you. The three remaining groups then project toward you.

If you intend to find the configuration of the molecule (i.e. R or S), then let me add this point:

  1. If the three groups projecting towards you are ordered from highest priority to lowest priority (i.e from rank 1 to rank 3) in clockwise manner, then the configuration is "R". If the three groups projecting towards you are ordered from highest priority to lowest priority (i.e from rank 1 to rank 3) in counterclockwise manner, then the configuration is "S". I mentioned "from rank 1 to rank 3" - because the 4th one is oriented away from us and is not considered while checking the order.

Cahn Ingold Prelog principle:

Rule to prioritize the groups attached to a chiral carbon:

  1. Prioritize the four atoms, or groups of atoms, attached to the chiral center based on the atomic number of the atom that is bonded directly to the chiral center. The higher the atomic number, the higher the priority.

    • Like say you have carbon and lets the 4 atoms attached to it are: $\ce{H},\ce{OH},\ce{Cl},\ce{Br}$.
    • So here the carbon is chiral, as all the 4 atoms attached to it are different and now for ranking, we rank it on the basis of atomic number. Here as Br has highest atomic number, It would be ranked 1, then Cl would be ranked 2, then OH will be ranked 3 (oxygen will be considered as it is directly attached to the chiral carbon) and then hydrogen will be ranked last.
  2. Atoms participating in double/triple bonds are considered to be bonded to an equivalent number of similar "phantom" atoms by single bonds. This is done just for the purpose of the rule. See this pic for better understanding: rule 2 image

  3. If two or more of the atoms that are bonded directly to the chiral center are the same, then prioritize these groups based on the next set of atoms (i.e. atoms adjacent to the directly bonded atoms)

    • Lets take an example here. Say to the carbon I have $\ce{CH3},\ce{COOH},\ce{Cl},\ce{H}$ attached. Here first we will compare the atomic number of the atoms which are directly attached to the chiral carbon i.e. C, Cl and H. From these as Cl has highest atomic number, hence it is ranked 1. Then, for the carbon, as for both the groups i.e $\ce{CH3}$'s and $\ce{COOH}$'s, it is the carbon atom which is directly connected to the chiral carbon. Hence, we need to look for the next set of atoms. $\ce{CH3}$ next set of atoms are H, H & H; while $\ce{COOH}$'s next set of atoms are O, H and O-C (see pic provided in rule 2). So from here easily we can say $\ce{COOH}$ is ranked 2, then $\ce{CH3}$ (i.e. rank 3) and then H.
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  • $\begingroup$ Hi there, thank you for your answer. I was indeed asking about how to rank the molecules around the two chiral centre's of the tartaric acid. I am fine with the finding the configuration of a molecule once the groups have been ranked. Could you help me with ranking the groups please? $\endgroup$ – Jonathan Smith Mar 20 '17 at 11:20
  • $\begingroup$ I have added the rules. $\endgroup$ – Yb609 Mar 20 '17 at 11:51

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