R S Configuration (AAMC 7 Spoiler)

[MCATMountain]

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Hey guys, could someone please determine and explain the R/S configuration of Carbon #7? This is a question from AAMC 7. Thanks.

 

[PKA<0]

S the OH is priority 1. The carbonyl makes carbon 8 priority number 2. Carbon 6 is priority 3. H is in back.

 

[Saxappeal1]

The -OH is your first priority. Now you have to figure out what your second priority is. When you look at the molecule, both th left carbon and the right carbon have a methyl group first, so no help there. Then, there is a carbon attached to an oxygen with the both the left carbon chain the the right one, but the left has a carbonyl, which counts as COO not C-O, so the left carbon has more priority, so you go OH- counterclockwise, ergo S.

 

[maxn]

I believe it is S. Looking at the 4 groups bonded to carbon 7 there is an oxygen, a hydrogen, and two carbons(6 and 8). You need to order the groups from 1 to 3 based on priority(which is determined by atomic number). So oxygen would be first, carbon would be second and hydrogen would be last. To determine priority of the carbons, look at what is connected to those carbons. Carbon 6 is connected to 2 carbons and 2 hydrogens. Carbon 8 is connected to 4 carbons, so carbon 8 is higher priority than 6. Once you have determined each priority, draw a circle from 1 to 2 to 3. If your circle is clockwise, it is R. If it is counter-clockwise it is S.

 

[MCATMountain]

Thanks for the replies.

 

[Marine to MD]

I believe it is S. Looking at the 4 groups bonded to carbon 7 there is an oxygen, a hydrogen, and two carbons(6 and 8). You need to order the groups from 1 to 3 based on priority(which is determined by atomic number). So oxygen would be first, carbon would be second and hydrogen would be last. To determine priority of the carbons, look at what is connected to those carbons. Carbon 6 is connected to 2 carbons and 2 hydrogens. Carbon 8 is connected to 4 carbons, so carbon 8 is higher priority than 6. Once you have determined each priority, draw a circle from 1 to 2 to 3. If your circle is clockwise, it is R. If it is counter-clockwise it is S.

This. The carbonyl is irrelevant.

 

[HinduHammer]

I believe it is S. Looking at the 4 groups bonded to carbon 7 there is an oxygen, a hydrogen, and two carbons(6 and 8). You need to order the groups from 1 to 3 based on priority(which is determined by atomic number). So oxygen would be first, carbon would be second and hydrogen would be last. To determine priority of the carbons, look at what is connected to those carbons. Carbon 6 is connected to 2 carbons and 2 hydrogens. Carbon 8 is connected to 4 carbons, so carbon 8 is higher priority than 6. Once you have determined each priority, draw a circle from 1 to 2 to 3. If your circle is clockwise, it is R. If it is counter-clockwise it is S.

This. The carbonyl is irrelevant.

QFT - carbonyl irrelevant in this scenario.