Absolute Configuration

[Rawrzor]

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When assigning priorities, am I supposed to base it off of atomic number or weight? BR and EK says atomic weight, but while browsing through the forums, I read atomic number. So far, I've done problems based on the weight.

There's a problem in BR Orgo, pg 222. #12, that I'm confused about. In the first chiral carbon with the methyl, I thought that the C attached to the aldehyde would be 1, then the methyl would be 2, and then the C to the left of that would be 3. That would be S...then invert it so it would be R...but it's apparently S. I got the second chiral configuration right, but if what I did above is wrong, then I'm wrongly doing this wrong too and getting the right answer somehow.

Attached a crude image for those without the book.

Thanks!

 

[kleinishere]

You're method of approach is totally correct but the error is in the prioritization.

#1 will definitely be the C to the right. Thanks to the O in the aldehyde group, this C is the highest priority.

#2, however, will be the C on the left of your diagram. This C is bonded to 2 other carbons and 2 H's making it a secondary carbon.

#3 will be the methyl, a primary carbon. It's only connected to the chiral carbon, and then 3 H's.


Now, do your clockwise circle suggesting "R", invert it as the lowest priority group (the unseen H) is pointing toward the viewer, and, voila, you'll arrive at "S."

 

[Rabolisk]

You use atomic number. If you have isotopes, then you do the tie-breaker by considering its weight.