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Topic: Chirality Question  (Read 12226 times)

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Offline coolguy99

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Chirality Question
« on: September 24, 2009, 12:29:54 AM »
There was an extra-credit problem on my exam, and I just wanted to see if I got it right, haha

It was..

2,3,4-tribromopentane

how many stereoisomers are possible?
how many meso compounds are there?

I said.. 4 stereoisomers possible, because of the two chiral centers.. and 1 meso compound.

Yes/no?.. I was on the fence if chiral centers were the determinant for max possible isomers, or stereocenters..

Offline Markov

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Re: Chirality Question
« Reply #1 on: September 24, 2009, 03:28:15 AM »
I am afraid it is slightly more complicated. Although I don't have the time and possibly not even the skills to sort it out right away for you. I'd have to dig and draw for a little bit.

But as a starting point, ask yourself if there is really only two stereocenters possible?

Offline HW

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Re: Chirality Question
« Reply #2 on: September 24, 2009, 05:11:22 AM »
I think there are 3 stereoisomers at which one of them is meso compound.

Offline victor

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Re: Chirality Question
« Reply #3 on: September 24, 2009, 05:53:00 AM »
Based on its stereocenters which are located at C2 and C4, we can obtain that there are 4 stereoisomers possible. But if we draw it (I draw it using Fischer projection to make it easy), we can see that there're two meso compounds. So the list would be:
1. 2S,4S
2. 2R,4R
3. 2S,4R (meso)
4. 2R,4S (meso)

Well that's what I think.

Victor

Offline Dan

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Re: Chirality Question
« Reply #4 on: September 24, 2009, 09:17:48 AM »
Based on its stereocenters which are located at C2 and C4,

It's not that simple, you have a pseudo-asymmetric centre at C3...
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Offline Markov

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Re: Chirality Question
« Reply #5 on: September 24, 2009, 11:47:27 AM »
Based on its stereocenters which are located at C2 and C4,

It's not that simple, you have a pseudo-asymmetric centre at C3...

That was my point. I think you will find that there are 4 isomers.

Two are "the two enantiomeric" forms mentioned earlier, where the C2 and C4 are "2S,4S" and "2R, 4R", the C3 will not behave as a stereocenter. Think of it as C3 bears two identical groups.

In the "meso" forms (2R, 4S) and (2S, 4R), however, the C3 is really a stereocenter.   
That gives two new isomers, which I think you should refer to as diasteromeric.

It is a really intriguing question, and I am far from 100% sure to be right here. I am staying tuned...

Offline Dan

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Re: Chirality Question
« Reply #6 on: September 24, 2009, 12:10:46 PM »
That was my point. I think you will find that there are 4 isomers.

Two are "the two enantiomeric" forms mentioned earlier, where the C2 and C4 are "2S,4S" and "2R, 4R", the C3 will not behave as a stereocenter. Think of it as C3 bears two identical groups.

In the "meso" forms (2R, 4S) and (2S, 4R), however, the C3 is really a stereocenter.   
That gives two new isomers, which I think you should refer to as diasteromeric.

It is a really intriguing question, and I am far from 100% sure to be right here. I am staying tuned...

Yup, spot on. Four isomers, specifically:

1. 2S,4S enantiomer of 2
2. 2R,4R enantiomer of 1
3. 2S,3s,4R (=2R,3s,4S) meso
4. 2S,3r,4R (=2R,3r,4S) meso

In 1 and 2 C3 is not a stereogenic centre, in 3 and 4 it is pseudo-asymmetric (r/s)
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Offline orgopete

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Re: Chirality Question
« Reply #7 on: September 25, 2009, 11:26:37 AM »
Three centers, eight possible compounds. However, there are four pairs of meso compounds. I have drawn all possibilities. If you flip over any structure, you will get the identical meso structure shown in the rectangle.
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Offline Dan

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Re: Chirality Question
« Reply #8 on: September 25, 2009, 12:12:02 PM »
However, there are four pairs of meso compounds. I have drawn all possibilities.

The last two pairs you drew are not meso. In the last four structures you drew, you have drawn two different compounds both of which are chiral (by definition, a chiral molecule cannot be meso).

You have a pair of enantiomers along with two (meso) diastereoisomers.
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Offline orgopete

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Re: Chirality Question
« Reply #9 on: September 25, 2009, 02:26:23 PM »
When I take the upper structure of the last two pairs, I get the lower structure. Since I got the lower structure, I thought it was a meso compound. Did I err in that flip?
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Offline Dan

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Re: Chirality Question
« Reply #10 on: September 26, 2009, 01:24:28 PM »
Your structures are not wrong, but your use of the word meso is. A molecule is meso if its absolute configuration is invariant on reflection (achiral) despite having at least two stereogenic centres - this means that if you have a chiral molecule it cannot fit the definition of meso. For your last two pairs you have boxed structures that are related by rotations, not reflection - the pair is just the same molecule from different perspectives with no reflection applied. Note that if you take the 3rd pair and draw the mirror image, you get the 4th pair - which is non-superimposable on the 3rd. The 1st and 2nd boxes are achiral and both are meso, the 3rd and 4th boxes are chiral and  an enantiomeric pair. Does this make sense?
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Offline orgopete

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Re: Chirality Question
« Reply #11 on: September 29, 2009, 02:39:45 PM »
Agreed.

I was trying to answer the question as a student might answer the question. Since there are three chirality centers, therefore there out to be eight possible compounds. This is similar to a compound with two chirality centers and if symmetrical, then a meso compound exists for two of the apparent isomers. They turn out to be the same compound.

I had looked at it from the way that symmetry reduces the number of compounds by formation of meso-structures. In that, two of the compounds also degenerate into a single compound, as you have noted or by a flip as I have noted. However, as you point out, these compounds are not the enantiomers of each other, which are listed side by side. The enantiomers are chiral and therefore not meso.
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