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Topic: R/S configuration  (Read 1416 times)

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Offline yayalo17

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R/S configuration
« on: February 19, 2020, 04:51:06 PM »
In the first exercise the priorities are: 1 with phenol, 2 with benzene, 3 OH, 4 CH3, the one with lower priority is already on the dotted line, anti-clockwise, configuration S On the second, since we have a methyl and an isopropyl, all 2 are directly linked to C, therefore we consider the atomic number of the other elements the priority is: 1 isopropyl, 2 NH2.3 Methyl and 4 H, the lowest one is already on the dashed, configuration R




In these cases?



Thanks

Offline kriggy

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Re: R/S configuration
« Reply #1 on: February 20, 2020, 01:29:14 AM »
Your priorities are wrong:
The first example (from highest to lowest): OH > phenol > benzene > CH3,  its S so you got your answer correct

The second example: NH2 > iPr > Me > H, again S

The way it works is that you look at the one atom that is bonded to the chiral center to see if they can be distinguished: in the first example its 3 carbons and one oxygen. Oxygen has higher priority therefore its no.1, then you continue through the chain from the carbons in the direction of the highest priority. Its clear that the methyl group has only hydrogens while the other two continue with carbons. THerefore, the methyl group has lower priority and we can designate in no. 4. To distinguis between the remaining two, we just do the same and we end up with those two atom chains:

"C-C-C-O" for phenol and "C-C-C-C" for benzene. THe  oxygen has higher priority than carbon, therefore phenol is no. 2 and benzene no.3

THe other example is bit simpler: NH2 is no.1 while H is no. 4. Then you differentiate between the remaining two and you end up with chains "C-H" for methyl and "C-C" isopropyl, making the isopropyl no.2 and methyl no.3

hope it helps. You can try the remaining two by yourself

Offline yayalo17

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Re: R/S configuration
« Reply #2 on: February 20, 2020, 09:39:26 AM »
Your priorities are wrong:
The first example (from highest to lowest): OH > phenol > benzene > CH3,  its S so you got your answer correct

The second example: NH2 > iPr > Me > H, again S

The way it works is that you look at the one atom that is bonded to the chiral center to see if they can be distinguished: in the first example its 3 carbons and one oxygen. Oxygen has higher priority therefore its no.1, then you continue through the chain from the carbons in the direction of the highest priority. Its clear that the methyl group has only hydrogens while the other two continue with carbons. THerefore, the methyl group has lower priority and we can designate in no. 4. To distinguis between the remaining two, we just do the same and we end up with those two atom chains:

"C-C-C-O" for phenol and "C-C-C-C" for benzene. THe  oxygen has higher priority than carbon, therefore phenol is no. 2 and benzene no.3

THe other example is bit simpler: NH2 is no.1 while H is no. 4. Then you differentiate between the remaining two and you end up with chains "C-H" for methyl and "C-C" isopropyl, making the isopropyl no.2 and methyl no.3

hope it helps. You can try the remaining two by yourself
are these correct?
Thank you

Offline yayalo17

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R/S configuration
« Reply #3 on: February 21, 2020, 05:30:17 AM »

Are they correct?

Offline chenbeier

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Re: R/S configuration
« Reply #4 on: February 21, 2020, 05:41:43 AM »
The first one is correct. The second one I don't see a methyl group. So it is not chiral. If there is one the result is wrong. Priority OH, OCH3,CH3,H gives S.

Offline yayalo17

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Re: R/S configuration
« Reply #5 on: February 21, 2020, 10:07:38 AM »
The first one is correct. The second one I don't see a methyl group. So it is not chiral. If there is one the result is wrong. Priority OH, OCH3,CH3,H gives S.
the line above oxygen is a ch3, right?
on the dotted wedge there is not even a ch3?
even if it is not written.
Then how does OH> OCH3 prioritize?
they are both directly connected to oxygen so the next chain must be considered, therefore C has priority over H

Offline hollytara

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Re: R/S configuration
« Reply #6 on: February 21, 2020, 10:41:19 AM »
Both correct as written out.


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