# R, S Configuration and Branching - Which Path?

Which path do I take when we have a branched chiral center? None of the textbooks I've consulted adequately covers this or even provides an example problem in which there was branching. Consider the above:

1) Chiral carbon with the following attachments H, an carboxylic acid group, a methyl group, and then the ring.

2) H gets the lowest priority.

3) The methyl group gets the second lowest priority since it has just one carbon and then hydrogens; the other paths have either more carbons or oxygens!

4) Between the carboxylic acid group and the path along the ring, we have an oxygen and a hydrogen going for the carboxylic acid group. But which path do I take? After comparing the carbon in the carboxylic acid group to the carbon on the ring, do I compare the double-bonded oxygen to the next carbon on the ring or the single-bonded oxygen? How does one choose in cases of branching such as these?

No, you have a carbon figuratively attached to 3 oxygen atoms vs. a carbon figuratively attached to 3 carbon atoms; the former has higher priority.

• Even if it would be only one Oxygen compared to a branched Carbon, the former will still have higher priority. – Martin - マーチン Jun 3 '14 at 4:42

As @ron mentioned $\ce{-CO_{2}H}$ has higher priority so the numeration given by @ADG in the fischer projection is incorrect because he assumes that the ring has higher priority than $\ce{-CO_{2}H}$. You can also think in the ring priority as $(C,C,C)$ and $\ce{-CO_{2}H}$ priority as $(C,O,O)$. As you may notice$(C,O,O)$ has higher priority than $(C,C,C)$ (the total atomic number of the substituent group in the first case is higher).
See the following figure: (Source)

In this case I don't think you need to continue the path, but if that's the case you can see how it works in the figure.

• The $\ce{CO2H}$ moiety should be thought of as $\ce{(C,O,O,O)}$ because the carbon is bonded via three bonds to oxygen. – Martin - マーチン Jun 4 '14 at 2:48

convert it to fischer projection: you can see it is (S)
Edit: Added Explanation Going according to atomic numbers.

Atomic no.1

1.$\ce{H}$- lowest priority.

Atomic no. 6

2.$\ce{CH3}$ - next priority.Has $\ce{H}$
3.$\ce{C6H4..}$ - next priority. Has $\ce{C}$
4.$\ce{COOH}$ - next priority.Has $\ce{O}$