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Claisen
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[*] posted on 25-3-2011 at 00:14
Some Questions


1. The product P is ?



2. The total no. of optical isomers of the following compound are -




Attempt -

1. I cannot recall any specific reaction I studied in the past.
2. I always have a problem in finding out the no. of optical isomers in CYCLIC compounds. I got the answer as 10.

[Edited on 25-3-2011 by Claisen]

[Edited on 25-3-2011 by Claisen]
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Claisen
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[*] posted on 25-3-2011 at 01:54


I am attaching the image in which I marked the chiral atoms.
Chiral atoms are shown with big dots. They are 5 in total so 10 optical isomers. Is that right?

154j0j5.jpg - 8kB
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[*] posted on 25-3-2011 at 03:30


Reaction 1 is a Diels Alder with the first compound acting as the diene and the second as the dienophile.

http://en.wikipedia.org/wiki/Diels_Alder
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[*] posted on 25-3-2011 at 08:12


Quote: Originally posted by Claisen  
They are 5 in total so 10 optical isomers. Is that right?


If there are 5 stereocenters the number of enantiomers will be 2 to the power of 5, i.e. 32.
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Nicodem
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[*] posted on 25-3-2011 at 12:39


1. The product P is most like 2-naphthol and the diene at the left plays no role. The Diels-Alder reaction requires conditions harsher than the conditions required for the aromatization by tautomerization (especially if there are traces of any bases or acids), so there would be little to no Diels-Alder product there. As an interesting side note: where there is a requirement for a double tautomerization to attain a state of aromaticity, the compounds are actually quite stable (e.g., cyclohex-2-ene-1,4-dione does not tautomerize that easily to hydroquinone unless acid or base catalysis is viable).

2. There are 10 stereocenters so there are potentially 2^10 stereoisomers (1024) if all are undefined, but it appears to me that at least the double bonds configuration is defined (or at least not explicitly undefined), which makes the answer potentially 2^7 (128). The number of optical isomers is much lower, but I can't really bother to look up how you calculate the permutations.

PS: Please keep all the homework questions in one thread.
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Claisen
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[*] posted on 25-3-2011 at 14:23


I never studied Dies-Alder reaction. I know it now.

@Nicodem:
How did you get 10 stereocenters? Can you explain?

Quote:
PS: Please keep all the homework questions in one thread.

ok from now on I will post all my questions in this thread.

[Edited on 25-3-2011 by Claisen]
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[*] posted on 25-3-2011 at 19:10


I counting 11 including the Halogens so im interested in Claisens question as well Nicodem.




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Nicodem
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[*] posted on 26-3-2011 at 03:17


A stereogenic center is each atom or bond that can have two or more configurations with the same connectivity. This includes double bonds (E/Z diastereomers), non-rotable single bonds which can have two configurations (atropisomers), atoms substituted with four or more different substituents and/or lone electron pairs (classic chirality), but there are also other types.
The compound in question has three double bonds and seven carbon atoms substituted with four different substituents, thus having 10 stereogenic centers altogether. Each of these centers can have two configurations of the substituents, but retaining the same connectivity. The number of stereoisomers is thus easy to calculate. But that is not what the question asks and I must admit that I don't know the answer. I have no clue on how to calculate the number of optical isomers (enantiomers) without actually helping myself with drawing structures.

PS: These are not really homework questions aren't they? Because if they are, the first is a trick question and the second is too hard to answer unless someone knows an easy formula to calculate the number of enanatiomers from the number of stereogenic centers (there are too many permutations to solve this problem by drawing structures!). Though it could be that the teacher is actually ignorant of the topics he is trying to teach - it would not be the first time.




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[*] posted on 30-3-2011 at 10:33


Quote: Originally posted by Nicodem  
The number of stereoisomers is thus easy to calculate. But that is not what the question asks and I must admit that I don't know the answer. I have no clue on how to calculate the number of optical isomers (enantiomers) without actually helping myself with drawing structures.

On second thought, in structures that don't have any plane of symmetry the number of stereoisomers actually equal the number of optical isomers (there can be no meso compounds unless there are planes of symmetry present). I can find no such plane of symmetry on a first glance, so I think the number of optical isomers is actually also 2^10 (or 2^7 if double bonds configurations are taken as defined).
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Claisen
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[*] posted on 30-3-2011 at 12:33


@ Nicodem
You are wrong.
The answer provided is 8=2^3 which means 3 asymmetric carbon atoms. I have searched so many books but I never found how to count them in cyclic rings.

Quote:
These are not really homework questions aren't they?


Quite right. These are those questions from my test which I was not able to answer.
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[*] posted on 30-3-2011 at 13:56


Quote: Originally posted by Nicodem  
1. The product P is most like 2-naphthol and the diene at the left plays no role. The Diels-Alder reaction requires conditions harsher than the conditions required for the aromatization by tautomerization (especially if there are traces of any bases or acids), so there would be little to no Diels-Alder product there. As an interesting side note: where there is a requirement for a double tautomerization to attain a state of aromaticity, the compounds are actually quite stable (e.g., cyclohex-2-ene-1,4-dione does not tautomerize that easily to hydroquinone unless acid or base catalysis is viable).



I think you are wrong here.
Examiners love pattern matching questions where functionality is hidden within structure and it tests an important scientific skill.
Can the candidate strip away the irrelevant bits and see the key reactions that build molecular structure?
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[*] posted on 31-3-2011 at 08:59


Quote: Originally posted by ScienceSquirrel  
I think you are wrong here.
Examiners love pattern matching questions where functionality is hidden within structure and it tests an important scientific skill.
Can the candidate strip away the irrelevant bits and see the key reactions that build molecular structure?

Are you saying that examiners are completely ignorant of the most basic chemical mechanisms? Perhaps, but someone who knows what a Diels-Alder reaction is, should know even better what tautomerization is. If the examiner teaches so blindly, he teaches nothing. Same goes for examiners who ask about products of reactions never described in the literature. Paper chemistry only looks good on paper, but when it comes to reality... It would be quite pathetic if an examiner would flank a student for giving the correct answer to that question!
Quote: Originally posted by Claisen  
@ Nicodem
You are wrong.
The answer provided is 8=2^3 which means 3 asymmetric carbon atoms. I have searched so many books but I never found how to count them in cyclic rings.

Hm..., let me guess. You got that answer from the same person who composed the question no. 1? That would explain a lot of things. If you live in one of those countries where you have to pay for your schooling, make a reclamation and change school.

As for counting stereogenic centres, that's very simple and I explained how to do that somewhere upthread. It makes no difference if the compound is cyclic or acyclic.

Edit: I would not want to do injustice to your teacher, so I leave open the possibility that you rewrote his questions wrongly. For example, the first question would make much more sense if the reactant would be cumarin.

[Edited on 31/3/2011 by Nicodem]
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Claisen
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[*] posted on 31-3-2011 at 11:04


Quote: Originally posted by Nicodem  

The compound in question has three double bonds and seven carbon atoms substituted with four different substituents, thus having 10 stereogenic centers altogether.


I see that you counted the double bonds too. But they don't form any optical isomers. They form geometrical isomers. I still don't get how you count 7 carbon atoms with four different substituents. In cyclic compounds its not only about substitutents but something else too (I think so).

How many stereogenic carbons do you count in allegra-
http://upload.wikimedia.org/wikipedia/commons/e/ec/Fexofenad... ?

I am attaching my question paper. You may look (pg. 3) for my first question and its options.

All questions given are standard questions made by a panel of experts in organic chemistry.

Attachment: IITFST2QP1.pdf (309kB)
This file has been downloaded 479 times

I will get the solutions tomorrow so I will post them if you want to take a look at it.

[Edited on 31-3-2011 by Claisen]

[Edited on 31-3-2011 by Claisen]
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ScienceSquirrel
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[*] posted on 31-3-2011 at 12:12


Have a look at the paper Nicodem, I am right as far as the examiners are concerned. :D
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[*] posted on 31-3-2011 at 12:17


Quote: Originally posted by Claisen  
I see that you counted the double bonds too. But they don't form any optical isomers. They form geometrical isomers.

Double bonds are stereogenic and they give rise to stereoisomers (2 per each bond), but they do not give rise to optical isomers by themselves. But you can not neglect that each double bond diastereomer is also an stereoisomer and thus each undefined double bonds doubles the number of optical isomers of a given compound.
Take a simple compound like 4-phenylbut-3-en-2-ol as an example. The position 2 is an atom substituted with all different substituents and is thus a classical stereogenic center giving rise to two optical isomers (enantiomers). However, the double bond can also give rise to two stereoisomers. Taken together you have 4 discrete optical isomers. These can be in two different relationships: either being in diastereomeric or enantiomeric relation. If you draw them on paper you will see that each compound has its corresponding enantiomer (which means they are all optical isomers).
Quote:
I still don't get how you count 7 carbon atoms with four different substituents.

I advise you to go from the first to the last carbon and check all four substituents on each. Mark each carbon that has all four of them different. It is rather simple, it just calls for discipline to go trough all of them carbons and not forget any.
Quote:
In cyclic compounds its not only about substitutents but something else too (I think so).

Cyclic compounds can have conformational isomers arising from the energy difference due to sterical interactions among the substituents. Thus a conformation where a certain substituent is preferentially in either equatorial or axial configuration can be substantially preferred. However, this is generally a dynamic isomerism (meaning that the interconversion barrier is commonly too low for the isomers to be separable entities at room temperature). A similar thing is observed also in acyclic compounds where certain rotamers are lower in energy than others, again due to steric interactions. But rotamers are also not optical isomers, even though they can have enantimeric relationships, as they are most commonly not isolable (there are notable exceptions - see atropisomerism).

Quote:
How many stereogenic carbons do you count in allegra-
http://upload.wikimedia.org/wikipedia/commons/e/ec/Fexofenad... ?

Only one.
Quote:
I am attaching my question paper. You may look (pg. 3) for my first question and its options.

All questions given are standard questions made by a panel of experts in organic chemistry.

Quite sad. A panel of experts in organic chemistry that is unaware of keto-enol tautomerism and aromatic stabilization. The half life of most enols at room temperature was found to be 30 min in the total absence of acids, bases or protic solvents, so the reverse reaction must be of similar rate when the enol form is trapped in an aromatic energy hole. The Diels-Alder cyclization between enones and unactivated dienes generally requires heating to 100-150 °C. You tell me if there is any chance of getting any product (B).
But what is perhaps worse is that they composed a question about a reaction that is not even documented in the literature. I wander then, how do they know the correct answer? Guesswork? Product (B), which would be the product of a potential Diels-Alder cyclization does not even show up in SciFinder!
What a lazy bunch of experts!
Quote: Originally posted by ScienceSquirrel  
Have a look at the paper Nicodem, I am right as far as the examiners are concerned. :D

I wish I could share your sense of humour, but when it comes to the degradation of science I can't get myself to laugh.

[Edited on 31/3/2011 by Nicodem]




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[*] posted on 31-3-2011 at 12:46


Lighten up, it is an examination paper.
You pass exams by giving the examiners what they want.
There are lots of flaws in examination papers but if you want that magic bit of paper that lets you get to the next stage you just jump through the hoops!
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Claisen
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[*] posted on 31-3-2011 at 13:01


@Nicodem - You are right about the product. They must have overlooked something.

I marked up the chiral atoms that I counted. Please tell me which ones are missing/incorrectly marked.



154j0j5.jpg - 7kB
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[*] posted on 1-4-2011 at 02:03


Quote: Originally posted by Claisen  
I marked up the chiral atoms that I counted. Please tell me which ones are missing/incorrectly marked.

You missed the two marked in red. Next time try going trough each carbon and mark them as either stereogenic or non-stereogenic, so that you know you did not accidentally forget any.

stereo.gif - 6kB




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Claisen
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[*] posted on 1-4-2011 at 06:11


Thats where my problem lies.
Each of those marked in red have two -CH2 groups on either sides.
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Claisen
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[*] posted on 1-4-2011 at 10:37


As Nicodem requested, I am posting my questions in this thread (not a homework/test question, but a doubt)

I am interested in finding the hybridisation of Cu in the complex [Cu(NH3)4]2+.
Here Cu is in +2 O.S.
so its outer most configuration is 3d9 4s0 4p0
Now since there is one unpaired electron in d-orbital and lack of availability of space/another electron for its pairing, Cu should undergo sp3 hybridisation to form a tetrahedral complex. But I read that it is dsp2 (square planar). Where does that one electron go to empty the d orbital? Is VBT unable to explain this?

Thanks.
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[*] posted on 1-4-2011 at 11:48


Quote: Originally posted by Claisen  
Is VBT unable to explain this?
More or less, yes. Valence bond theory is an incomplete approximation of the full quantum mechanics working behind the scenes.
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[*] posted on 1-4-2011 at 12:00


Then how do we explain the dsp2 hybridisation, through CFT?
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[*] posted on 1-4-2011 at 16:56


CFT? What is CFT?

The square planar geometry of the copper complex arises from complex quantum mechanics, but a simple explanation is this:

(and I doubt you need this unless you're in Inorganic chemistry right now):

The tetrahedral geometry of d-metal complexes has a tripley degenerate state for it's HOMO (highest occupied molecular orbital).

When you have 9 d electrons, you end up filling these 3 states with 5 electrons

(due to 4 being filled in the HOMO - 1 orbital - the eg)

In quantum mechanics, degenerate orbitals that are not completely filled (i.e. 5 electrons in a tripley degenerate state that can hold 6) the complex will go through a distortion to lower the energy of that unpaired electron.

Thus, due to the energy levels of the square planar orbital, you find that now there is a single orbital - the HOMO that is not degenerate with any others and holds that one single unpaired electron.

I'll try to link you to the distortion diagram...

I can find papers on it but no diagrams. Sorry!


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[*] posted on 2-4-2011 at 00:16


Thanks GreenD...that is enough for me.
CFT=Crystal Field Theory.
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[*] posted on 4-4-2011 at 12:35


Quote: Originally posted by Claisen  
Thats where my problem lies.
Each of those marked in red have two -CH2 groups on either sides.

You got the definition of the stereogenic center wrong. If all four substituents are different then you have a stereogenic center (as simple as that). It does not matter if two, three or all substituents start with the methylene group or the same type of atom or group. It is about configurations that can have mirror images that are not superimposable, thus for the chirality it is irrelevant if there are two CH2 groups attached.
See for example the structure of menthol and the configurations at its three stereogenic carbons (http://en.wikipedia.org/wiki/Menthol).
As an interesting curiosity: You can even have optical isomers of ethanol, for example, if it is alfa-deuterated (yes, even isotopes induce chirality and such enantiomers often even have measurable optical activity!).




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