Justin
Registered: 09/26/09
Posts: 30
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Reply with quote | #1 |
Hi, I don't understand #15 molecule L.
It seems like you would make this molecule by rotating the rightmost carbon that has a double bond but not attached to the oxygen (=C--), similar to the way molecule M is formed. However, molecule L is not a conformational isomer, but a diastereomer instead. How come?
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gloomfilter
Registered: 09/27/09
Posts: 25
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Reply with quote | #2 |
Yeah, I thought so, too. But, the reason why is because this is the case of a stereoisomer without a typical stereocenter. The double bond prevents rotation and then each of the C's is attached to three different things. In comparison to the original molecule, two of the substituents are switched in their wedges(not all), therefore it is a stereoisomer -> diastereomer.
http://www.chem.ucla.edu/harding/IGOC/S/stereocenter.html
If you look at the figure in the middle, that is basically what is happening. |
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fredaf
Registered: 09/25/09
Posts: 12
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Reply with quote | #3 |
I don't see how we were suppossed to know whether the CH3s were changed to point outward or into the page. For example, for I thought molecule was a diastereomer and molecule L was not an isomer. |
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Justin
Registered: 09/26/09
Posts: 30
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Reply with quote | #4 |
I'm not sure what you mean by your question fredaf. Could you please rephrase it?
Molecule L is definitely an isomer since it contains the same atoms as (+)-abscisic acid.
Also, the CH3 are shown to be coming towards you and away w/ the broken and solid dashes. The other CH3s are planar w/ the page due to sp2 hybridization which favors the trigonal planar arrangement.
Please let me know if that answer makes sense or not, since I'm not quite sure what you're asking.
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Arman
Registered: 10/08/09
Posts: 2
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Reply with quote | #5 |
Molecule L is not a conformational isomer because you cannot rotate around the double bond without breaking it. If you build a model, it's clear that rotation around the double bond is not possible.
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jcsong
Registered: 09/30/09
Posts: 17
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Justin
Registered: 09/26/09
Posts: 30
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Reply with quote | #7 |
No, because the atoms are still attached in the same order, hence it cannot be constitutional.
A constitutional isomer requires the atoms to be bonded in a different order. Try writing it out, and if it's the same then it's not a constitutional isomer. (e.g. CH3CH2CH2CH3)
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Chemkilla09
Registered: 10/25/09
Posts: 4
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Reply with quote | #8 |
Molecule L has only one stereo center so therefore doesn't it only have an enantiomer because it is chiral? |
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Justin
Registered: 09/26/09
Posts: 30
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edchang90
Registered: 10/05/09
Posts: 22
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Reply with quote | #10 |
If we look at the double bonds on the right and all those carbons are stereocenters, does that mean there are 6 stereocenters for molecule L? Does that also mean that molecule K and molecule L have the same number of stereocenters? |
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monhernandez
Registered: 09/26/09
Posts: 10
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Reply with quote | #11 |
Also in determining the number of stereocenters a molecule has, we should always include the sp2 atoms that are stereocenters as well? I thought we only really needed to consider the sp3 atoms that were stereocenters unless told otherwise. |
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