# Oxidation state and chiral configurations

I'm doing this same problem found on the following link for a biochemistry homework.

http://www.chegg.com/homework-help/...ogs-used-enzymology-molecular-biolog-q1481931

My question is regarding both finding the oxidation states as well as finding the absolute configurations.

I am told in my lecture notes that to find the oxidation state you take "the number of bonds the atom forms with more electronegative atoms less the number of the bonds that it forms with less electronegative atoms. The charge should be added to this sum."

However, I am not getting the same answers as given here and so I'm not sure if the person who answered this is wrong or if I'm wrong. But, for example, the first Phosphorous I would have calculated the oxidation state to be +5 since it makes two bonds with Sulfur (which is more electronegative) and then 3 bonds with the oxygens. Is this not correct?

And for the number 2 labeled Nitrogen, I would have calculated the oxidation state to be -3 since it is bonded to three carbons which are less electronegative.

When figuring out the chiral center configurations I'm not getting the same answers as given either, and I'm assuming this has to do with my understanding of priorities.

For example, the bottom right carbon of attached to the OH group, I would have given R configuration because the carbon to the right of this chiral center is attached to an O and an N whereas the other carbon is attached to an O and a C. So the OH group would get priority 1, the carbon to the right priority 2, and the carbon to the left priority 3. This would give a counter clockwise motion, but since the hydrogen is coming out of the page I would flip it to R configuration instead of S.

Am I wrong?

AGNuke
Gold Member
OS of Phosphorous = +5
OS of Nitrogen = -2. It is only bonded to 2 Carbon Atoms, as far as I can see the molecule.

As for Chiral centre, lookout for sp3 hybrid C-atom (C with 4 valencies) and make sure that all of the 4 valencies are unique. Also, to determine the configuration, try making their Fisher Projection. It is tough, but try it once.

chemisttree
Homework Helper
Gold Member
And for the number 2 labeled Nitrogen, I would have calculated the oxidation state to be -3 since it is bonded to three carbons which are less electronegative.
That's what I get too. Actually you should say it is bonded to carbon three times rather than to three carbons.

For example, the bottom right carbon of attached to the OH group, I would have given R configuration because the carbon to the right of this chiral center is attached to an O and an N whereas the other carbon is attached to an O and a C. So the OH group would get priority 1, the carbon to the right priority 2, and the carbon to the left priority 3. This would give a counter clockwise motion, but since the hydrogen is coming out of the page I would flip it to R configuration instead of S.

C > N

Chemisttree, are you saying that carbon attached to an O and C should have higher priority because C is higher priority than Nitrogen? I thought highest priority is given to atoms with higher atomic weight, and Nitrogen has a greater atomic weight than carbon.

chemisttree