Kaplan - Assigning Priority in Fischer Projections

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ManlyMan246

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I thought that I learned Fischer projections pretty well during Orgo. You look at each substituent, and prioritize them based on the atomic number of the nearest atom, and work your way outwards in case of ties.

For the following compound:
Code:
        HC=O
          |
    H3C---C---Br
          |
          OH
Kaplan has assigned the aldehyde the 2nd priority, although I am pretty sure it should have 3rd priority after Bromine and the hydroxyl group. Is this just a mistake on Kaplan's part or am I missing something?

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I thought that I learned Fischer projections pretty well during Orgo. You look at each substituent, and prioritize them based on the atomic number of the nearest atom, and work your way outwards in case of ties.

For the following compound:
Code:
        HC=O
          |
    H3C---C---Br
          |
          OH
Kaplan has assigned the aldehyde the 2nd priority, although I am pretty sure it should have 3rd priority after Bromine and the hydroxyl group. Is this just a mistake on Kaplan's part or am I missing something?

Your question is independent of the type of drawing...in any type of drawing the priority rules stay the same when numbering substituents. I am pretty sure Kaplan is wrong and you are correct. Oxygen is a higher priority than Carbon. I've done a ton of problems and never seen a time where the priority rule doesn't apply.
 
EK did the same thing with the same example. I'm gonna trust what I've learned and assume they're both wrong.
 
hilarious that Ek are such copy cats!
I'm using TPR and often Ek uses the same examples.........gotto be more than coincidence.......
 
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exactly.....you'd think if your gonna copy something at least take a second to check it out........
 
I thought that I learned Fischer projections pretty well during Orgo. You look at each substituent, and prioritize them based on the atomic number of the nearest atom, and work your way outwards in case of ties.

For the following compound:
Code:
              HC=O
              |
    H3C---C---Br
            |
            OH
Kaplan has assigned the aldehyde the 2nd priority, although I am pretty sure it should have 3rd priority after Bromine and the hydroxyl group. Is this just a mistake on Kaplan's part or am I missing something?

I had to do a quick double take at this...and you are correct Kaplan actually does have it incorrect. I went back to my o.chem text book and re-read the material.

For those of you that are trying to get it straight here is the info :)

Look up "Cahn-Ingold-Prelog convention"

1) atoms with higher atomic numbers receive higher priorities

I>Br>Cl>S>F>O>N>C-13>C-12>Li>H-3 (tritium)>H-2 (deuterium)>H (plane old hydrogen)

****Don't let IUPAC naming fool you!!! -OH always has priority in the naming of simple structures...not in assigning -R and -S conventions!!!!*****

---in cases of ties use the next atoms along the chain of each group as a tiebreaker [ i.e. -CH(CH3)2 vs -CH2CH3 you would go with the isopropyl...the 1st one]
---treat double and triple bonds as if each were a bond to a separate atom

H
l (the C has these bonds C-H, C-R, C-O, C-O)
R--C=O (the O has these bonds C-O, C-O)

2) Use a 3-D model and put the 4th priority group in the back then see if your 1-2-3 is clockwise (R...the regular way it should work) or counterclockwise (S...the sinister way it goes)...if your 4th group is pointing forward then do the same thing except switch your answer (i.e. if the 4th group is a wedge then if you find and R it is a S or if you find a S it is a R)

Sooooooooooo...for this example that the original poster gave us

1 --> -Br
2 --> -OH
3 --> -C (Aldehyde)
4 --> -C (Methly)

Therefore 1-2-3 group goes in the clockwise direction giving you a R sign. However, the CH3 is pointing forward in this Fischer Projection so we must assign it the opposite letter R --> S. So it is an S.


HC=O
|
H3C---C---Br
|
OH
 
ehhh what the heck the original compound he posted got all jacked when i replied .... just look at his the 1st one ...
 
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