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Topic: Optical isomers of cyclic compounds  (Read 6765 times)

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Offline Rutherford

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Optical isomers of cyclic compounds
« on: November 01, 2012, 02:02:16 PM »
On the picture, there are 3 different isomers of 1,2-cyclobutane-dicarboxylic acid. I need to assign them their configurations, using the R/S and the cis/trans notations.

First to R/S. I assume that the missing atoms on few bonds are hydrogen atoms. For the top C chiral atom in the first isomer I already have the H atom put under the plane, so I assign to COOH priority 1 to CH2 2 and to CH 3, so I get an anti-clockwise direction which should be S, but it is R. Why?

Offline discodermolide

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Re: Optical isomers of cyclic compounds
« Reply #1 on: November 01, 2012, 02:05:22 PM »
I would think that they"missing atoms" are methyl groups!
With this info. Try the re-assignment.
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Offline Rutherford

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Re: Optical isomers of cyclic compounds
« Reply #2 on: November 01, 2012, 02:43:11 PM »
They can't be methyl groups if it is 1,2-cyclobutane-dicarboxylic acid. If so, do you mean that it is really S?

Offline discodermolide

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Re: Optical isomers of cyclic compounds
« Reply #3 on: November 01, 2012, 09:59:58 PM »
Here is what I get for the assignment
Note if you draw the bonds as you have done "missing" atoms imply methyl groups. If you mean hydrogen atoms then you have to put them in.
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Offline Rutherford

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Re: Optical isomers of cyclic compounds
« Reply #4 on: November 02, 2012, 10:01:50 AM »
I didn't draw it, it was in the solution this way. In your pictures, the top C chiral atom in the first isomer has a COOH group above the plane, so the hydrogen should be below the plane. Now, I see that I didn't assign the priorities properly, The CH part of the ring has bigger priority than the CH2 part, so it is R. I will try to solve the rest now, if I get stuck I will ask here. Thanks for the help till now.

Offline Rutherford

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Re: Optical isomers of cyclic compounds
« Reply #5 on: November 02, 2012, 01:01:46 PM »
Okay, I understood that all configurations you posted are correct. Now to the cis/trans. How is it determined here?

Offline discodermolide

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Re: Optical isomers of cyclic compounds
« Reply #6 on: November 02, 2012, 01:07:12 PM »
Make models and you will see. If both substituents are above the ring plane then they are cis. One above and one below they are trans.

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Offline Rutherford

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Re: Optical isomers of cyclic compounds
« Reply #7 on: November 02, 2012, 01:58:15 PM »
Then I would have just 1 cis and just 1 trans?

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Re: Optical isomers of cyclic compounds
« Reply #8 on: November 02, 2012, 02:02:20 PM »
Yes, in the molecules you drew just one of each.
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Offline Rutherford

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Re: Optical isomers of cyclic compounds
« Reply #9 on: November 02, 2012, 02:08:59 PM »
Then the R/S notation is more precise and I will stick to it. Thanks again and again  :).

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