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Author Topic: racemic mixture vs optical activity  (Read 15947 times)

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113zami

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racemic mixture vs optical activity
« on: July 24, 2008, 09:18:30 AM »

http://i256.photobucket.com/albums/hh198/trna/kap0002.jpg

http://i256.photobucket.com/albums/hh198/trna/kap0001.jpg


the answer to #20 is E and the answer to #26 is D

recemic mixture = optically inactive, easy enough...but 

for #26 why isn't the answer A?

according to the correct answer from #20 meso structures are optically inactive, and in #26 I and III have the same exact structure and arrangment as the ones in #20 so I don't understand why can't they form a recemic mixture too ie be optically inactive just like the molecules in #20??

I understand the dafinition of recemic mixture: contains equal quantities of two enantiomers and I understand that I and III are meso compounds not enantiomers but the end result is still the same, since a recemic mixture = not optically active... so why can't A be the answer for #26??
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Hemicar

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Re: racemic mixture vs optical activity
« Reply #1 on: July 24, 2008, 09:37:17 AM »

Well A can not be the answer for 26 because 1 and 3 are meso compounds, the meso compounds exists in one form (try to rotate, twist or find mirror image and you will have the same form) and for racemic mix. we have to have 2 forms (2 different enantiomers)
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azmanam

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Re: racemic mixture vs optical activity
« Reply #2 on: July 24, 2008, 10:36:08 AM »

ARG! I had a whole post written out, and then I lost it all...

I don't think #20 is a very good question, personally.  Those two compounds look like enantiomers, but are really the same compound.  One is just rotated 180 degrees from the other.

In fact, that's my little mnemonic for remembering meso compounds: enantiomers that are really the same molecule.

Quote
in #26 I and III have the same exact structure and arrangment as the ones in #20 so I don't understand why can't they form a recemic mixture too ie be optically inactive just like the molecules in #20?

Because, like the molecules in #20, I and III are really the same compound.  I and III are the same meso compound.  II and IV are enantiomers, which mixed together would give a racemic mixture.  I and III, as a meso compound, WILL be optically inactive, but because they are not enantiomers (they're identical) they cannot give a racemic mixture.

I think you may be getting confused as to WHY both racemic mixtures and meso compounds are optically inactive.  It is true that they are, but for different reasons.

In a racemic mixture, there are two compounds (enantiomers) that are present in a 1:1 ratio.  It's not always easy, but it is possible to separate those two compounds into two batches of enantiomerically pure compounds.  If you take a sample of one of the enantiomerically pure compounds and measure its optical rotation, the plane polarized light would interact with one molecule and rotate.  It would interact with the next molecule and rotate a bit more, but in the same direction thereby reinforcing the rotation from the previous molecule.  The plane polarized light would interact with the millions of molecules in the enantiomerically pure sample and the light would be rotated a bit more each time - in the same direction - to give a non-zero value for the optical rotation at the end of the experiment.

You can do the same with a sample of the other enantiomer.  The plane polarized light would interact with one molecule of the other enantiomer and rotate.  But because this sample is the opposite enantiomer, the plane polarized light would rotate in the OPPOSITE direction.  It would interact with the next molecule and rotate again.  Same situation for the millions of molecules in this sample of the OTHER enantiomer.  At the end, if the concentration of the two samples was the same, you would have another non-zero value for the optical rotation.  This value would have the same magnitude as the first sample, but the opposite sign (i.e. if the first enantiomerically pure sample gave an optical rotation of -75 degrees, the second sample (if it is the same concentration) will give an optical rotation value of +75 degrees.  Same magnitude, opposite sign).

Now, if you mix the two enantiomers together again in a 1:1 ratio, you will regenerate the racemic mixture.  Now there is an equal number of molecules of one enantiomer as there are of the other enantiomer.  When you introduce plane polarized light, the plane polarized light will interact with one molecule of one enantiomer and rotate.  Then it has an equal probability of interacting with a molecule of the OTHER enantiomer.  If it interacts with a molecule of the OTHER enantiomer, the plane polarized light will rotate back the OTHER way, canceling out the first rotation.  After the plane polarized light has interacted with all the millions of molecules of one enantiomer and all the millions of molecules of the other enantiomer, the net effect on the optical rotation will be nil.  That is, for all the rotations in one direction, the plane polarized light will have as many rotations in the other direction and the end value for the optical rotation will be zero.  Thus racemic mixtures are optically inactive.

NOW, for meso compounds, the story is the same but different.  Take a look at question 20 again.  The two compounds look like enantiomers.  If they really were enantiomers and you took the optical rotation of a mixture of the compounds, the net effect would be nil just like the racemic mixture above.  BUT, if they really were enantiomers, you might be able to separate them and find the non-zero values for each enantiomer separately.  The difference here is that while the two compounds LOOK like enantiomers, they are really the SAME compound.  Because they are the same compound they cannot be separated. 

When you attempt to take an optical rotation of a meso compound, the plane polarized light interacts with one molecule that might happen to be rotated to look like the compound on the left in question 20.  This might cause the plane polarized light to rotate.  But then the plane polarized light will interact with a molecule that might be rotated to look like the compound on the right.  This might rotate the plane polarized light back in the OTHER direction.  In bulk, the net effect of a meso compound is an optical rotation value of zero.  When the plane polarized light gets to the end of the experiment, it will have rotated a net value of zero degrees. 

Thus, meso compounds are optically inactive and can never be made to be optically active by separating compounds like a racemic mixture can.  A racemic mixture, being a mixture of two different enantiomers, can be separated to give 2 batches of enantiomerically pure compound.  The separate enantiomerically pure compounds will give a non-zero value for optical rotation.  Mixed together, the racemic mixture of 2 separate enantiomers will have the effect of canceling out the rotation of the separate enantiomers and a racemic mixture will give an optical rotation value of zero.  A meso compound cannot be separated and will always give an optical rotation value of zero.  Both meso compounds and racemic mixtures are optically inactive.

(whew)

http://www.cem.msu.edu/~reusch/VirtTxtJml/sterism2.htm#isom12b
http://www.cem.msu.edu/~reusch/VirtTxtJml/sterism3.htm#isom17
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113zami

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Re: racemic mixture vs optical activity
« Reply #3 on: July 25, 2008, 06:20:43 AM »

WOW AWESOME, THANK YOU SO SO MUCH azmanam for taking the time to explain it so well :-*  :-*
I get it!
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imanooblar

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Re: racemic mixture vs optical activity
« Reply #4 on: April 06, 2010, 06:18:23 PM »

hey azmanam,

sorry to bring back an old topic, but was looking up meso compounds/optical rotation and was led to this thread. I in no way intend to criticize your comment, but just wanted to clarify some things, since I'm having a lot of confusion with this and just wanted to get it sorted out =p

In your post, you stated that meso compounds are "enantiomers that are really the same molecule," however in the link that you provided, it states that enantiomers are actually "achiral diastereomers." If they were the same molecule, then could any molecule floating around be a meso compound because depending on the way it's oriented, it could be construed as meso.

Also, in your post you mentioned that the reason why meso compounds are optically inactive is because one compound rotates plane polarized light one way, while the other compound rotates plane polarized light the other way. In my book it states that meso compounds are optically inactive because it is the chiral centers in a single compound that offset each other creating an optically inactive molecule. So say for example if I had an RS and and SR (which is a meso compound) http://www.cem.msu.edu/~reusch/VirtTxtJml/Images/tartaric.gif, would it be the R and S in the compound that offsets each other, or from what I understand the way you explained it, the compound on the left (in the meso compound picture on the right hand side) offsetting the compound on the right.

Hopefully this post makes sense. Anyways, hopefully you see this. By the way, your answer for the enantiomers and optical activity couldn't have been said any better, just got confusing when it got to meso compounds.
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azmanam

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Re: racemic mixture vs optical activity
« Reply #5 on: April 07, 2010, 12:15:09 AM »

My explanation of meso compounds included first approximations and mnemonics.  I'm glad it's been able to help you at least in part.  My mnemonic for determining if the relationship between two molecules is meso or not (and that's important.  It's a comparison mnemonic when looking at two molecules to determine if the pair are enantiomers, diastereomers, or meso) is: the two molecules look like they should be enantiomers, but when you rotate one of them you discover they're really the same molecule. 

This is not in contradiction to the website.  The phrase used by the site ('achiral diastereomers') is not incorrect, just slightly out of context.  When drawing all the possible diastereomers of a molecule with more than one stereocenter, you'd expect all of the stereoisomers to be chiral.  Each new compound would be a diastereomer to each other.  But in the case of meso compounds, one of the diasereomers from the list will in actuality be achiral.  It is one achiral compound which is a valid diastereomer to the other, chiral, stereoisomers.  It is an achiral diastereomer.  Hope that helps.

As to what parts are physically rotating the light and canceling what out, that's where the first approximations come in.  I don't actually know - on the molecular level - what part of the mixture is interacting with the light and to what extent.  If it makes more sense to you that one stereocenter cancels out another stereocenter, then I'm ok with that mental image.  The point that you should focus on is that meso compounds give an optical rotation of zero.
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imanooblar

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Re: racemic mixture vs optical activity
« Reply #6 on: April 07, 2010, 08:23:05 AM »

thanks for a quick reply azmanam.

After looking at the link you provided further, I can see how it is an achiral diastereomer.

So could they be both a form of enantiomer or achiral diastereomer?

From wikipedia, it states that enantiomers are non-superimposable mirror images of each other, which makes meso compounds seem to fit in that category. If so, do meso compounds have the same chemical properties?

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azmanam

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Re: racemic mixture vs optical activity
« Reply #7 on: April 07, 2010, 08:41:12 AM »

meso compounds are not enantiomers.  There is only 1 meso compound (for a given set of stereoisomers).  (RS)-tartrate is the same as (SR)-tartrate.  They're the same compounds.  They're not enantiomers.  They ARE superimposable mirror images (and thus not enantiomers).  They only look like enantiomers.  They look like enantiomers, but on further analysis are NOT - they're actually two representations of the same compound. (RS)-tartrate is achiral and a diastereomer of (RR)- and (SS)-tartrate.  (RS)-tartrate and (RR)-tartrate are diastereomers.  (RR) and (SS) tartrate are enantiomers (RS) and (SR)-tartrate are neither diastereomers nor enantiomers.  (RS)- and (SR)-tartrate are the same compound, just drawn differently.

The wiki definition is correct, but meso compounds do not fit that definition.  Meso compounds ARE superimposable on their mirror image.  I apologize if my mnemonic confused more than helped.  Here's the facts:

*meso compounds are achiral
*meso compounds are superimposable on their mirror images
*meso compounds have an optical rotation of 0
*meso compounds do not have an enantiomer - the 'enantiomer' is really a different representation of the same compound
*when drawing all possible stereoisomers for a compound, a meso compound (if formed) will be diastereomeric to the other stereoisomers

hope this helps.  Keep asking if you're still confused.  We want to make sure you understand by the time you leave :)
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imanooblar

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Re: racemic mixture vs optical activity
« Reply #8 on: April 07, 2010, 09:38:10 AM »

haha alright thanks azmanam. I think I am understanding it a little better. Now I guess I have two more questions.

If meso compounds are the same compounds, only rotated, then why even have the name meso compounds. I guess my question is why would you have a name for two compounds, when they are the exact same compound.

Also, from my understanding, diastereomers are compounds with opposite configurations and are not mirror images of each other. So from what we discussed earlier, meso compounds are neither enantiomers or diastereomers? They're in their own separate category? And since they're essentially the same molecule, I'm guessing they have the same chemical properties, like enantiomers.

Just confusing because achiral means non super imposable on their mirror images and meso comounds are, which is where achiral diastereomers confused me.
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azmanam

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Re: racemic mixture vs optical activity
« Reply #9 on: April 07, 2010, 03:01:27 PM »

Let's back up a bit.  Take diethyl tartrate - the prototypical compound when discussing the concept of meso.

There are only 3 stereoisomers of diethyl tartrate.  1 & 2 are enantiomers.  1 & 3 are diastereomers.  2 & 3 are diastereomers. 

3 is a meso compound.

There are 2 ways to draw compound 3.  (well, ok, there are way more than 2 ways to draw compound 3, but only 2 ways to keep all the C-C bonds in the same orientation)  The two ways to draw compound 3 are drawn below.  Flipping the compound (like a pancake) vertically results in drawing the compound the same way, but reversing the stereocenters.  3 and (flipped3) look like they should be enantiomers.  But look at the IUPAC names and numberings.  this might make it more clear.  They have the same IUPAC name.  In one case, carbon 1 is at top.  In the other case, carbon 1 is at bottom.  They're still the same compound 3.  They're two ways of drawing the same one compound.  We've been talking about meso as if it were a compound that refers to the relationship between two compounds.  That's not really accurate.  A compound is meso.  This all started by looking at the original problem the op was working on.  I said I didn't like the problem because two of the compounds were the same meso compound.  Meso is not technically a relationship word.  It's a descriptor of one compound.  It's just typically introduced in the context of drawing all possible stereoisomers.  When we draw all possible stereoisomers for tartrate, we mistakenly draw 3 and (flipped3) and think they're two different compounds.  In actuality, it is one compound that is meso.

Does this help any? :)
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orgopete

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Re: racemic mixture vs optical activity
« Reply #10 on: April 07, 2010, 07:44:25 PM »

I believe an element of this discussion that is missing is the role of chirality centers and chirality.

In a compound with a single chirality center, you can have two forms of a chiral compound or if both isomers are present together, a racemic mixture.

In a compound with two chirality centers, you have 22=4 compounds possible. These are the diastereomers. Diastereomers that are the mirror images of each other are enantiomers. (Those that are not mirror images are diastereomers.) However, symmetry can reduce a compound with two chirality centers to a single compound. In that case, the mirror image gives a structure that can be superimposed on itself to give the same compound. This is the meso isomer(s).
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imanooblar

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Re: racemic mixture vs optical activity
« Reply #11 on: April 08, 2010, 03:09:11 PM »

@ azmanam
thanks!! i think i finally understand it now, the diagrams really helped.

@orgopete
everything made sense up until you mentioned that enantiomers are diastereomers that are the mirror images of each other. is that correct? so enantiomers are under the category of diastereomers? i always thought they were different.
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orgopete

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Re: racemic mixture vs optical activity
« Reply #12 on: April 09, 2010, 02:28:57 AM »

First a correction of my earlier post.
Quote
In a compound with two chirality centers, you have 22=4 compounds possible. These are the diastereomers. Diastereomers that are the mirror images of each other are enantiomers.
That should have said they are "stereoisomers", see http://en.wikipedia.org/wiki/Diastereomers. Compounds that are mirror images are enantiomers, those that are not are diastereomers.

Hence, it should have said that stereoisomers that are mirror images of each other are enantiomers and those that are not are diastereomers.

For example, 2,3-dichloropentane can have 4 isomers (2n, 22 isomers), RR, SS, RS, and SR. The enantiomers-diastereomer relationships azmanan described could be applied to these isomers.

If this same analysis were applied to 2,3-dichlorobutane, you would again get 4 isomers, RR, SS, RS, and SR. However, because RS and SR 2,3-dichlorobutanes are symmetric, they can be superimposed and are thus the same compound. It is referred to as a meso compound.
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