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Topic: Optical Rotation Question  (Read 8232 times)

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Offline Cheemistree8889

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Optical Rotation Question
« on: April 14, 2007, 08:30:46 PM »
If I have an tetra-substitued adamantane derivate does it exhibit optical rotation?  I know that the stereogenic point in the molecule isn't an atom, but I'm unsure if the "enatiomer" is actually and enatiomer seeing the substituent is on a different carbon.  Oh yeah, the adamantane derivative has a carboxylic acid substituent on the top and left carbon, a Br on the right most carbon and a methyl group on the bottom most carbon.  The enatiomer I drew has the Br and the carboxylic acid switching from left to right.  I hope I made it clear.
« Last Edit: April 14, 2007, 10:31:45 PM by Cheemistree8889 »

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Re: Optical Rotation Question
« Reply #1 on: April 15, 2007, 12:56:36 PM »
Yes, those are the correct enantiomers, and they would exhibit optical rotation.

However, you should reevaluate your statement about the number of asymmetric carbon atoms in the molecule!

Offline Cheemistree8889

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Re: Optical Rotation Question
« Reply #2 on: April 15, 2007, 01:10:48 PM »
Am I correct to say that the substituted carbons are asymmetric?  I have trouble stating this because in the case of the brominated carbon and the methylated carbon the R and S configurations cannot be determined (the carboxylic acid groups are both 2 carbons away).

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Re: Optical Rotation Question
« Reply #3 on: April 15, 2007, 01:47:36 PM »
You can still assign priority to the groups.  Just count out the priority of the next center, etc.

Offline Cheemistree8889

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Re: Optical Rotation Question
« Reply #4 on: April 15, 2007, 09:40:59 PM »
Ok I think I got it. Thanks alot

Offline Custos

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Re: Optical Rotation Question
« Reply #5 on: April 16, 2007, 01:57:46 AM »
Are you sure they are enantiomers? Two of the substituents are the same (COOH) - The mirror image seems superimposable to me....

Offline Cheemistree8889

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Re: Optical Rotation Question
« Reply #6 on: April 16, 2007, 08:12:52 AM »
If you align the COOH's there is no superimposable structure..if you have a picture you could send it maybe I'm wrong.

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Re: Optical Rotation Question
« Reply #7 on: April 16, 2007, 12:09:03 PM »
Are you sure they are enantiomers? Two of the substituents are the same (COOH) - The mirror image seems superimposable to me....

D'OH!  I messed up.  I thought one of the acids was an alcohol.  They are achiral.

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Re: Optical Rotation Question
« Reply #8 on: April 16, 2007, 02:15:44 PM »
so how about the top carbon that's next to a COOH? doesn't it have four different substituents: COOH, C-C-Br, C-C-COOH, and C-C-CH3

would the top carbon on the left side be S? and the left carbon next to the COOH should be R

then the two other bridge carbons would also be chiral

does that make sense?

Math and alcohol don't mix, so... please, don't drink and derive!

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Re: Optical Rotation Question
« Reply #9 on: April 16, 2007, 03:58:07 PM »
It does, but there is a plane of symmetry through the Me and Br which makes it meso.

Offline Custos

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Re: Optical Rotation Question
« Reply #10 on: April 16, 2007, 07:51:35 PM »
If you align the COOH's there is no superimposable structure..if you have a picture you could send it maybe I'm wrong.
It's a bit hard to do in a picture - the best way is for you to make the two models (as an aside, every organic chemist should have a model kit for making structures) and see that they are superimposable. The other way of thinking about it is that the bridgehead positions in adamantane are spatially related as a tetrahedron (all the carbons are symmetrically distributed so you can ignore those). So a tetra(bridgehead)-substituted adamantane can be pictured just like tetrahedral methane. If you have 4 different groups it can be chiral. If two are the same it is achiral.

Offline Cheemistree8889

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Re: Optical Rotation Question
« Reply #11 on: April 16, 2007, 09:41:01 PM »
So that isn't an enatiomer because it is achiral?  I'm very confused now...

Offline Custos

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Re: Optical Rotation Question
« Reply #12 on: April 17, 2007, 02:07:08 AM »
That's right. The compound on the left and it's mirror image are one and the same, so the molecule is achiral. For them to be enatiomers the original compound and its mirror image have to be non-superimposable - that is, you can't get from one to the other just by rotating them around in space.

Try this. Get some plasticine and matches. Colour your matches with felt pens. Make a model where you have four different matches stuck in the plasticine pointing to the vertices of a tetrahedron. Also make the mirror image. No matter how you rotate the models around you can't get them to line up exactly. But if two of the matches are the same colour you can.

The first example is two enantiomers. The second example is when the compound is achiral, like your question, because it has a plane of symmetry.

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