Sciencemadness Discussion Board

help with mechanisms

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cheeseandbaloney - 12-8-2010 at 15:40

I wasn't sure where to put this, so I thought I'd ask here first. For any of the organic chemists out there I would greatly appreciate help with some mechanisms. One example is after addition of dichlorocarbene across the diene in cyclopentene the usual product is obtained. But the subsequent reaction that causes the ring to expand into the product 1,6-dichlorocyclohexene is where I am stuck. I feel a little sloppy on carbon-carbon bond breaking and ring expansion mechanisms. I feel that I don't fully grasp the rules on electron transfers during C-C breakage. Also, if anyone can point me in the right direction for pages on mechanisms like these that could help give me a better and deeper understanding of these 'rules' would be appreciated.

p.s. - This is all for pure curiosity and the drive to learn organic chemistry. I am not in school so I do not have the privilege of asking a professor, so I turn to you. I've looked in my books, but the one I have with the question was bought used online and doesn't have the solutions manual. I've looked online but the only link available to download the solutions manual .pdf is broken.

DDTea - 12-8-2010 at 21:46

This will be hard to show without pictures, and this nomenclature might be arbitrary, but it should explain what's happening. After the initial reaction between dichlorocarbene and cyclopentene, a bicyclic compound is formed--a fused cyclopentane/cyclopropane. Let's call the two shared carbons 1 and 6 and the third member of the cyclopropane ring carbon 2 (I'm doing it this way because these will be their numbers in the resulting cyclohexene ring).

Cyclopropanes are very high energy species; they have a lot of ring strain, which is going to drive this mechanism. So, with that in mind, the electrons between carbons 1 and 6 attack carbon 2, displacing a chloride. The chloride takes its electrons and shifts to carbon 6--similar to a hydride shift--giving you 1,6-dichloro-1-cyclohexene.

This is a neat mechanism because of the principles involved. You can do some thermodynamic calculations if you want; they'll show you that the product is *much* more favored than the reactants here.

(edited for clarity).


[Edited on 8-13-10 by DDTea]

cheeseandbaloney - 12-8-2010 at 22:23

oh wow, I was just overlooking the chloride shifting over a carbon! derp. I knew the ring strain was what drove the reaction but didn't know exactly how to approach the first step. Interestingly enough, after building this reaction using a molecular model kit it made a lot more sense in 3d, especially for the double bond reformation. Thanks for the help!

edit: hope this picture works

[Edited on 8/13/2010 by cheeseandbaloney]

reaction.png - 1kB

Arrhenius - 13-8-2010 at 00:45

If you enjoy pondering or working through organic mechanisms I would suggest buying "The Art of Writing Reasonable Organic Reaction Mechanisms" by Robert Grossman. This is essentially the standard for mechanistic formality.

The ring expansion you've described is probably not as simple as DDTea described. If you were to draw a mechanism showing the C-Cl sigma bond breaking, populating the sigma* antibonding orbital of the cyclopropyl C-C bond and opening the ring this would not be correct - strictly speaking. In heterolytic mechanisms (moving electron pairs only - not radicals) bond cleavage - think SN2 type 'backside attack' mechanism - results from adding electron density to the antibonding orbital at the atom undergoing the bond forming/breaking event. Frontier orbitals would not predict good overlap for this process to take place. Also, keep in mind that the ring expansion you described might take place under various conditions, and each might utilize a unique mechanism.

Cyclopropanes undergo a rich diversity of reactions, many of which can be best rationalized by homolytic mechanisms. Solvolysis or thermolysis of 6,6-dichlorobicyclo[3.1.0]hexane is probably a radical mechanism, as evidenced by deuterium labeling experiments that show incorporation of deuterium at carbons 1 and 5 of the starting bicycle. The base induced ring expansion is probably best explained by an ionic mechanism (see below), but do not think of this as a 'chloride migration', because the chloride is probably not delivered intramolecularly.


untitled.gif - 19kB

cheeseandbaloney - 13-8-2010 at 01:16

Quote: Originally posted by Arrhenius  
If you enjoy pondering or working through organic mechanisms I would suggest buying "The Art of Writing Reasonable Organic Reaction Mechanisms" by Robert Grossman. This is essentially the standard for mechanistic formality.


It's funny you mention that book, because I recently purchased it out of this curiosity. I recently have started teaching myself MO theory and have slowly but surely begun to understand it after drawing and building myself some molecular orbital diagrams. Like visualizing when electron orbitals are 'out of phase' and are antibonding...etc. And I do understand that the conditions present can change the mechanism, (for example, in the case of acid and base hydrolysis, the hydroxide ion is not considered a catalyst, but a reagent therefore it 'promotes' the reaction and is not reversible. Whereas acid hydrolysis is reversible). Things like this.

Admittedly I'm still a little rusty, so Grossman's book gets a little too advanced for me towards the middle, but we all gotta start somewhere, right?

edit: oh! and thanks for the responses so far! Arrhenius, your pictures really puts a lot of the pieces in my head together visualizing the p orbitals like that. Reading can only help me grasp it to a certain degree before I have one of those "a ha!" moments haha

[Edited on 8/13/2010 by cheeseandbaloney]

DJF90 - 13-8-2010 at 03:17

There are several valuable resources out there to help teach organic chemistry. Starting with the most easily accessible, "The virtual organic chemistry textbook" courtesy of MSU is very helpful and is indeed what I started with before I got to university. It often covers mechanisms well, and shows a diverse (but by no means comprehensive) range of chemistry.

http://www2.chemistry.msu.edu/faculty/reusch/VirtTxtJml/intr...

Secondly is a book written by Peter Sykes: "A guidebook to mechanisms in organic chemistry". You can find this online as a free download if you wish. It covers key mechanistic considerations ideal for starting out.

Thirdly is a book you're bound to have seen mentioned before: "Organic Chemistry" by Clayden, Greeves, Warren and Wothers. Again, available for free download if you so wish. Core organic textbook at many universities, covers a broad range of material in significant detail, although it doesn't quite cut it for specialist topics.

Finally, "Designing Organic Syntheses; The disconnection approach", by Stuart Warren. This complements the above book and teaches ways to design synthetic routes. Very much recommended!

As a personal note, bear in mind that not every reaction has its mechanism elucidated. There are many reactions where we're just not sure how the transformation occurs. Similarly, there are reactions where several mechanisms have been postulated, and each holds its own merits. Try not to get hung up on not knowing the mechanism in these cases, as I have done myself before. You just have to accept thats the way it is.

Welcome to sciencemadness, and good luck with your endeavours.

DDTea - 13-8-2010 at 04:27

Arrhenius: nice call about the orbital geometry, I didn't think about that.


I'm trying to wrap my head around how this would affect the mechanism, but I can't quite see it: it's worth bearing in mind that in cyclopropane-type systems, the electron density is not concentrated between the atoms as one would expect from a normal sigma bond. There is a lot more pi character at play, so the electron density is concentrated outside of the ring, with the bonding orbitals arranged to form a hexagon rather than a triangle (i.e., imagine the nodes tangent to each vertex of the triangle). If you wanted to explain the type of orbitals in terms of valence-bond theory, you'd have to call them sp4 orbitals, which makes people uncomfortable :) This is confirmed by X-Ray Diffraction.

Here's another proposed mechanism: suppose the bond between the shared carbons breaks, moving to the bridging carbon and displacing a chloride. The result is a six-membered ring carbocation intermediate with charge delocalization. The expelled chloride ion can now bind to either of the two originally shared carbons. This would lead to a racemic product.


DJF90--interesting that you mention "Designing Organic Syntheses; the disconnection approach." I worked with a professor briefly a few years back who studied at Oxford. The two books he told me to look into as soon as possible were that one as well as "The Art of Writing Reasonable Organic Reaction Mechanisms." I never did buy them, though, as my interests shifted away from synthesis.

My turn to recommend a book: Anslyn & Dougherty's "Modern Physical Organic Chemistry." It's probably the best textbook I've ever come across. It's pricey for the amateur, but if you're serious about Organic Chemistry, it's well worth the money. There's more information in that book than you may ever need in your life.

Oh, and Arrhenius--what software did you use to draw your diagram?

[Edited on 8-13-10 by DDTea]

Ozone - 13-8-2010 at 06:47

I recommend:

http://www.amazon.com/Electron-Flow-Organic-Chemistry-Scudde...

Followed by:

http://www.amazon.com/Arrow-Pushing-Organic-Chemistry-Unders...

Cheers,

O3

Arrhenius - 13-8-2010 at 10:39

cheeseandbaloney: Note that the orbitals I've drawn are SP3 hybrid orbitals, which are not to be confused with P orbitals. There's a simple pneumonic LCAOMO - Linear Combinations of Atomic Orbitals give Molecular Orbitals. This idea gives rise to hybridized SP orbitals that are drawn with a large 'lobe' and small 'lobe' of the dumbbell. While the smaller lobe is not technically the sigma* orbital, it is simple to visualize the sigma C-Cl (grey big lobe) interacting out of phase with what is geometrically analogous to the sigma*C-C (white small lobe). Intuitively, an SP orbital is an S and a P orbital added together (not entirely this simple). In phase interaction of two bonding MO's (molecular orbitals) gives rise to a bond - as I've drawn in for two carbons of the cyclopropane above.

DDTea: Program is Chemdraw. I suggest you download ACD Labs' Chemsketch - free for 'student' version.

For those interested in molecular orbital theory and frontier orbitals I would suggest "Frontier Orbitals and Organic Chemical Reactions" by Ian Fleming

While I agree with DJF90, mechanisms are only our best guess given the information at hand, I would argue that a very strong understanding of mechanism is absolutely crucial if you wish to be an advanced organic chemist. It becomes much easier to predict reactivity or to design transformations - which may or may not be known - and to optimize reaction conditions when you have some idea of how the reaction is taking place. For instance, say you're doing a Mitsunobu Reaction and you observe a large amount of a product containing an alkene instead of the inverted alcohol. Realizing that this arises from elimination of the intermediate oxophosphonium ion might lead you to the idea of adding the azodicarboxylate very slowly so that the concentration of this reactive intermediate is very low compared to the nucleophile.


[Edited on 13-8-2010 by Arrhenius]

cheeseandbaloney - 13-8-2010 at 20:28

Thanks Arrhenius, and everybody! I feel I basically mastered everything in orgo 1 in my opinion, with resonance being one of the most important IMO. I've been independently studying organic chemistry for the past 3 years or so simply out of a need to learn it for some reason. I don't plan on going to a university for chemistry, just keeps me sane when I'm bored. Never did well in a school environment anyways.

But! I've gotten to the point where I want to be able to predict reaction mechanisms to reactions I've never seen before. It wasn't till recently that I realized I will hafta master MO theory before I can be more comfortable with doing this (it sure helped me visualize why a diels-alder reaction happens!). Might there be any simple rules to know when a C-C sigma bond will be broken? or hell, I guess if anybody wants to present me with some problems they feel will help further my knowledge, by all means!

DJF90

I actually own "Organic Chemistry" by Clayden, Greeves, Warren and Wothers, but the book I primarily taught myself with was "Organic Chemistry" by Paula Yurkanis Bruice before the other book was recommended to me. I feel books are a great investment, so keep up the suggestions!

edit:
03

I am considering buying one of the two books you suggested since I do not have either. Which of the two would you feel would situate me better for the time being?


[Edited on 8/14/2010 by cheeseandbaloney]

Arrhenius - 14-8-2010 at 09:17

There is no simple answer as to when a sigma bond will be broken. Certainly we can think that when a bond breaking event leads to a more stable product, this 'downhill' process is likely to take place. Conversely, we can put energy into a system (heat usually) to drive the reaction 'uphill' on a reaction coordinate. Typically it is extremely difficult to break a sigma bond that is not weakened or strained in some way that will drive its breaking. In your example above, relief of the ring strain associated with a cyclopropane is the driving force. In an SN2 reaction the weak bond to the leaving group is broken to form a stronger bond (but not always stronger). For instance, the Finkelstein reaction forms a weaker bond, but the driving force in this reaction is entropy not enthalpy! This is the case in several other common reactions. The Fischer esterification typically is very poorly favored enthalpically, but clearly can be driven to completion by entropically removing the water by azeotropic distillation or by using a large amount of sulfuric acid to desiccate the reaction.

Other reactions include pericyclic reactions, sigmatropic reactions, metathesis, transition metal catalyzed reactions, photochemical reactions. The list is long. You will probably not become so talented at mechanisms that you will be able to avoid having to learn the enormous number of reactions that are possible. It's easiest to solve mechanistic problems when they look familiar to a reaction you know - at least to have a better idea of when bond formation/cleavage is likely to be allowed.

Here as some relatively simple mechanism problems:


untitled2.bmp - 667kB

Sandmeyer - 14-8-2010 at 10:19

Interesting posts Arrhenius! :)

Quote:
While I agree with DJF90, mechanisms are only our best guess given the information at hand


Ideally, organic chemistry should be absolutely analytical, but it would mean that we are able to grasp the infinity. It is empirical, I guess for the same reason that we can not calculate pi. Experiment and extrapolation is the only means the organic chemists (humans) currrently have - in contrast to "God" (and possibly R. B. Woodward :D ).

kmno4 - 14-8-2010 at 12:22

At first - I am the last one who will propose any mechanisms just out of my head.
"While I agree with DJF90, mechanisms are only our best guess given the information at hand, I would argue that a very strong understanding of mechanism is absolutely crucial if you wish to be an advanced organic chemist."
I cannot agree with it. Guessing means and gives nothing. Reactions go (or not) without our guessing.
Only studying many experimental results can give you right to propose any mechanisms. Of course, some reactions are easy to explain, some are not.
"Very strong understanding of mechanism" is not the same as "very strong ability for writing nicely looking combination of arrows and lines". I have always felt respect for chemical reactions. The more I learn, the deeper respect I feel :D

Back to the point.
Uncatalysed, thermal ring expansion of mentioned gem-dichlorocyclopropane derivative goes most probably via concerted, disrotatory, electrocyclic ring opening at the
C-C bond and a concomitant ionization of a carbon-halogen bond
.
Five-membered ring is itself strained and such isomerisation goes in temperature below 200 C (with 100% yield). In contrast, 6- and 7-membered rings are stable upon reflux above 200 C.

These wisdoms come from review article (one of many about cyclopropanes):
http://dx.doi.org/10.1021/cr0100087
and from literature cited there.

cheeseandbaloney - 14-8-2010 at 13:06

Quote: Originally posted by kmno4  

"Very strong understanding of mechanism" is not the same as "very strong ability for writing nicely looking combination of arrows and lines". I have always felt respect for chemical reactions. The more I learn, the deeper respect I feel :D



I couldn't agree more with this statement, although I'm not sure how much weight my opinion holds since I am not a formally trained organic chemist. Although mastering certain mechanisms and why they happen genuinely give me a deeper understanding of manipulating the world on a quantum scale. Pretty arrows and lines only tell part of the story, but if I know why I'm drawing these pretty arrows to point where all those crazy dancing electrons are going, it helps me grasp how molecules bond and break a lot better.

Arrhenius

Thanks for the problems, still goofing around with the last one, but now I have something to do when I get off work. It's kinda funny, figuring out mechanisms is my version of sudoku in a way...

DJF90 - 14-8-2010 at 16:24

Arrhenius: The last reaction you depict, is the base in excess? And 1 eq. MsCl? The only thing I can come up with at this hour is enolate formation, micheal addition to the ethyl acrylate, followed by "displacement" of the tosylate. As a weak base is used, the thermodynamic endo product is obtained. Displacement of the tosylate would of course not be Sn2, but via the carboxonium ion, present because of the anomeric effect.

kmno4: 5 membered rings are actually not all that strained at all. Whoever taught you that should be shot. If a 5 or 6 membered ring can form, it most likely will.

[Edited on 15-8-2010 by DJF90]

Arrhenius - 14-8-2010 at 17:00

Haha. I had a feeling folks wouldn't see eye to eye on the importance of mechanism. I absolutely agree that mechanisms can be aggravating at times, because they're abstract and not capable of being proven. Nonetheless, I honestly feel it's important to understand them if you wish to build complex molecules, be it with existing methodology or not.

KMnO4
Quote:

Guessing means and gives nothing. Reactions go (or not) without our guessing.

Perhaps I chose my words poorly. Many reactions are very well established by experimental methods. I don't know how familiar you are with labelling studies, kinetic isotope effects, ultrafast spectroscopy, etc. but people work extremely hard to come close to proving mechanisms, and my hat's off to them (not my cup of tea!). I'll try to dig up some truly astonishing mechanisms - ones that are necessary to rationalize the result of a reaction. As for rationalizing how to make a reaction go, I think the Swern Oxidation is an excellent example. I'm sorry, but one wouldn't come up with the correct series of reagent additions to make that reaction go. I'm near certain that the mechanism preceded that reaction.

DJF90 IIRC there was 3eq of base and ~1eq MsCl. :D Enjoy! Let me know if you get stuck, or if anyone wants the answers. I think there's one excellent answer, and one somewhat acceptable answer for the last one (DJF90's). And yes, I second that - 5 & 6 member rings are not strained.

Ring strain:
3: -27kcal/mol
4: -26kcal/mol
5: 6kcal/mol
6: 0kcal/mol

[Edited on 15-8-2010 by Arrhenius]

Arrhenius - 14-8-2010 at 17:18

Here KMnO4, this one's for you:

untitled.gif - 4kB

Not a really astonishing one, but I think you all can solve this one.

DDTea - 14-8-2010 at 18:51

Quote: Originally posted by Arrhenius  

Many reactions are very well established by experimental methods. I don't know how familiar you are with labelling studies, kinetic isotope effects, ultrafast spectroscopy, etc. but people work extremely hard to come close to proving mechanisms, and my hat's off to them (not my cup of tea!).



...and for those interested in experimental methods of "proving" mechanisms (or at least lending support to one route over another), Anslyn & Dougherty's "Modern Physical Chemistry" excels here. There are several chapters devoted to kinetic isotope effects and linear free energy relationships.

If you want to see a fanatical love of data, look up some of the papers by C. Gardner Swain.

[Edited on 8-15-10 by DDTea]

[Edited on 8-15-10 by DDTea]

<sub>Edit by Nicodem: Fixed the missing "]" which caused a serious formating error.</sub>

[Edited on 11/4/2012 by Nicodem]

DJF90 - 15-8-2010 at 09:10

Quote:
I think there's one excellent answer, and one somewhat acceptable answer for the last one (DJF90's).


Fire away with the excellent answer...

Arrhenius - 15-8-2010 at 14:14

Here are my answers. In #4, the oxidopyrillium species is a well known in dipolar cycloaddition chemistry, hence I've drawn a concerted mechanism. The HOMO-LUMO pairing of the dipole and dipolarophile are such that it gives the 'conjugate addition' product.

answer 1.gif - 13kB

answer 2.gif - 16kB

Nicodem - 15-8-2010 at 14:44

I apologize to drop in this interesting discussion with a banal and off topic request, but Arrhenius, could you provide the reference for that presumable Claisen rearrangement of the allyl hydrazone you gave as an example a couple of posts above. It looks such a useful and ingenious reaction that I would like to read more about it.

To keep on the topic, I would like to stress that mechanism in chemistry is the interpretation of experience itself (like all theory is). It explains the route from A to B and why it does not give C instead. Therefore, without understanding mechanisms it is near to impossible to design a synthesis of more complex structures, let alone optimizing it. For example, if you try to rely on thermodynamics in organic synthesis you get nowhere, but if you combine everything, the knowledge of thermodynamics, activation energies (kinetics), the understanding of transition states and mechanisms in general, you are half way to predict the outcome of any reaction. And correctly predicting the outcome of half of experiments is already in the domain of the best synthetic chemists.
What is however important to always have in mind is that there is a difference in "the most likely mechanism" and "the current understanding of the mechanism" of a reaction. The first is just some fancy arrow pushing and the other is a theory based on experimental work. "The most likely mechanism" is a pedagogic approach on how to build a working hypothesis, and I think this is the kind that a few posters above had in mind when comparing it to guessing. I'm afraid that in schools the difference is rarely emphasized, just like the difference in hypothesis and theory in general is rarely thoroughly explained to the students.

Cheeseandbaloney, for beginners I would highly recommend:
Grossman R.B.: The Art of Writing Reasonable Organic Reaction Mechanisms (2ed., Springer, 2007)

It is an excellent book to have in paper (though if you are a poor amateur you can easily download an illegal PDF copy).

Arrhenius - 15-8-2010 at 17:51

Nicodem: It would technically fall under an aza Cope, not Claisen ;) Regardless: Mundal et al "Triflimide-catalysed sigmatropic rearrangement of N-allylhydrazones as an example of a traceless bond construction" Nature Chemistry 2, 294–297 (2010)

Here's the supp. info.

Unfortunately I don't access to this article. I believe there are several related papers.
And thank you for being the first to back my opinion that mechanisms are useful. You make a good point though.

Nicodem - 15-8-2010 at 23:54

Thanks. I don't have access to that paper either, but hopefully someone will post it in References.
Quote: Originally posted by Arrhenius  
It would technically fall under an aza Cope, not Claisen

This is a good example in the discrepancy between named reactions and mechanisms. Since the mechanism of Cope and Claisen rearrangements are identical (except for the O-heteroatom in Claisen), and there are many other such cases, it is sometimes hard to decide how to call something if the only difference is just a formal nature of reagents, conditions, different heteroatoms or other structural peculiarities. There are many reactions with identical mechanisms and different names (some reactions with mechanistically identical pathways have up to 3 or 4 names!). And there are many reactions with identical outcome, but very different mechanisms (and thus different reaction conditions), so you have chemists who mistakenly tend to use the same reaction name on mechanistically different transformations. Then there are also named reactions that actually can proceed trough more than one mechanism, depending on the reaction partners or conditions. It is confusing territory, mostly because reactions were named before enough experience accumulated to properly interpret it.
For example, I still do not get it why in this particular case calling it an aza-Cope would be more appropriate than aza-Claisen, but I guess the authors had their own opinion on this. Perhaps because Claisen rearrangements are an undergroup of Cope rearrangements? But then they use a catalyst that can only catalyse Claisen but not Cope rearrangements (protonation of the heteroatom and the heteroatom is a Claisen's peculiarity).

I noticed during my work that people usually do not comprehend the importance of understanding mechanisms (and all the theory in general) until they start to get more involved in practical lab work, particularly designing syntheses. It then becomes obvious that those who have limited knowledge in mechanisms become very handicapped in their practical experimental work. They tend to rely on asking others for advices on how to perform reactions, with which reagents, what conditions, what protection groups, etc. Usually they become an annoyance to the coworkers, especially when these realize the only reason they ask advice is because they are to lazy to learn mechanisms. Such people also tend to be completely unable to design rational synthetic routes and this is a terrible handicap if your job is to develop target oriented syntheses. Another consequence is in that they tend to become more sensitive to "synthetic frustrations" and as a consequence try to limit their research to a single chemical transformation (though this is only allowed in the academy - in the industry they simply fire you if try that). I worked with few such people and always felt kind of sorry for them, but once they get too old to learn it is too late to repair the situation (I see it on myself - I'm sorry I did not learn more theory when I was younger and the learning process was easier). So my advice to students would be to never give up on learning theory. Not doing so will make you a very unhappy and unsuccessful chemist, possibly unemployed or employed in some office of some governmental agency. Theory is after all practice condensed in knowledge, so you can not do without it if you have to do practical work.

Sandmeyer - 16-8-2010 at 09:52

Quote: Originally posted by Nicodem  
For example, I still do not get it why in this particular case calling it an aza-Cope would be more appropriate than aza-Claisen, but I guess the authors had their own opinion on this. Perhaps because Claisen rearrangements are an undergroup of Cope rearrangements? But then they use a catalyst that can only catalyse Claisen but not Cope rearrangements (protonation of the heteroatom and the heteroatom is a Claisen's peculiarity).


If we compare the products, namely 1,5-dienes, resulting from both Cope and Claisen rearrangements we see that in the 1,5-diene-product from Claisen, one double bond is always a C=O, that is not the case with Cope rearrangements. So, I would say that Arrhenius is right, it should be called aza-Cope, if the above example would be a aza-Claisen, as you claim, it would contain one carbonyl group as one "ene" of the 1,5-diene. As far as the catalyst goes, we can also see it as if they just shown that Cope rearrangement can also be catalysed by the same catalyst as Claisen. Many different reactions can be catalysed by the same catalyst, for example we don't change the name of Diels-Alder reaction into Friedel-Crafts just because a Lewis acid can catalyse it.

un0me2 - 16-8-2010 at 13:53

Quote: Originally posted by Arrhenius  
....

DDTea: Program is Chemdraw. I suggest you download ACD Labs' Chemsketch - free for 'student' version.

.......


There is a better "FREE" program available, No-Fee, Symyx Draw for Students/Academics & Home Use (http://www.symyx.com/micro/getdraw/). You have to register to download it, but it kicks shit out of Chemsketch (which always gives funky as hell structures when you press F9). It also has an export as picture function.

Sandmeyer - 16-8-2010 at 16:24

It's best to explain in pictures, following picture explains why I would call it aza-Cope and not aza-Claisen (sorry for the bad graphics, it is time to sleep):




scimad.gif - 11kB

I mentioned the presence of a carbonyl in product 1,5-diene as what distinguishes the aza-Claisen from aza-Cope, but hydrolysis of imine to a carbonyl is a separate step and not part of the pericyclic reaction. It follows that Claisen proceeds via O-allyl-enol species while aza-Claisen via analogous N-allyl-enamine species and so, logically, the reaction Arrhenius mentiones can not be considered as aza-Claisen since it does not proceed through a nitrogen analogue of allyl vinyl ether (Claisen), his example fits aza-Cope rearrangement more... All in my opinion of course... :)


[Edited on 17-8-2010 by Sandmeyer]

franklyn - 18-8-2010 at 12:54

A Guidebook to Mechanism in Organic Chemistry 6 ed
recommended by DJF90 - http://www.sciencemadness.org/talk/viewthread.php?tid=14305#...
http://rapidshare.com/files/126713899/A_Guidebook_to_Mechani...

Arrow-Pushing in Organic Chemistry
recommended by Ozone - http://www.sciencemadness.org/talk/viewthread.php?tid=14305#...
http://ifile.it/jugotx6/9780470171103.rar

Molecular Electronic Structure Theory Part 1 - Helgaker T, Jorgensen P, Olsen J.
This is entirely steeped in high level mathematical descriptions more suited to physics ( I don't understand it either )
http://rapidshare.com/files/128521981/Molecular_Electronic-S...
or instead
http://rapidshare.com/files/128523134/Molecular_Electronic-S...

.

Nicodem - 18-8-2010 at 13:44

No, no, Sandmeyer. That is not what I had in mind. I was comparing the pericyclic step of the transformation directly to the Claisen rearrangement in a strictly isosteric fashion (N-Boc instead of O and C=N instead of C=C). See the first equation below.

The paper is now kindly made available in the Wanted references thread by Solo. The authors are of the opinion that the protonation occurs before Boc removal on the basis that equivalent non-Boc hydrazones give poorer conversion (see the relevant part of the article on the right). Seems plausible, however this also means comparing the sigmatropic rearrangement to Claisen's can not be correct, because this type of rearrangement can only be catalysed by the protonation of the other nitrogen (the one initially Boc protected). The catalytic effect of the acid by protonation of the "imine's" nitrogen (which is the only protonable one prior to Boc removal), indicates the electron flow is most likely the opposite to the one in the Claisen-type of rearrangements (see the third equation bellow). So it can not be a Claisen, but it does not have that much to do with a Cope either, because Cope does not involve protonation or heteroatoms in that position. Anyway, the authors avoid messing up with such details and do not compare the pericyclic component of this one-pot tandem transformation neither with the Cope nor the Claisen rearrangement, or any other named reaction, but consider it as a [3,3]-sigmatropic rearrangement of its own type. So, perhaps it was Arrhenius' idea to call it a Cope-like rearrangement? Or perhaps they call it that way in one of the two preceding papers that I did not yet check. Or maybe it was Stevens who 30 years ago reported the sigmatropic rearrangement of allylhydrazones who compared it to Cope. Better not call it names anyway. :P

allylhydrazone.gif - 23kB

cheeseandbaloney - 18-8-2010 at 21:54

great discussion! love this shit.

un0me2

Is there a '3d' mode for Symyx Draw? I do like it better than chemsketch, after the adjustment period.

Arrhenius

I forgot to say, checkmate on reaction #4. I'm still steadily learning new important reactions though for expanding the 'moves' I could make. Any more you think I should tackle, be my guest. One of the most important 'moves' right now IMO is the reformation of the C=O bond. I've noticed it can be the driving force for a lot of reactions. I remember reading this is because it's very thermodynamically stable. It seems once you find 'patterns' like this and build on them one can start to become more comfortable in synthetic organic chem.

Ozone

I ended up buying the 2nd book you recommended. Still have a couple to go through at the moment.

Arrhenius - 19-8-2010 at 00:43

My advice is try to move beyond Functional Group Interconversions (FGIs). It takes a sharp memory to write mechanisms for functional group dancing. However, it takes considerable skill to rationalize rearrangements, especially sigmatropy, etc. That being said, there are certain FGIs with stereochemical outcome that absolutely demand your understanding the mechanism. Also, given the abundance of transition metal chemistry in organic synthesis, I would suggest you take a look into these mechanisms. I would also recommend you consider purchasing one of KC Nicolaou's books "Classics in Total Synthesis" - I have the first edition and it's pretty good. The strategy employed in these highlighted total syntheses is typically quite fantastic. Anyone up to try working through a few more problems?

I've tried to hit on a few practical reasons that make mechanisms and associated transition states important 1.) Rationalizing FGI outcome 2.) Multi-step mechanisms 3.) organometallic mechanisms 4.) highly strategic bond formation (see anything by E.J. Corey & co. among others).

set 2.gif - 18kB

Sandmeyer - 19-8-2010 at 16:12

Quote: Originally posted by Nicodem  
No, no, Sandmeyer. That is not what I had in mind. I was comparing the pericyclic step of the transformation directly to the Claisen rearrangement in a strictly isosteric fashion (N-Boc instead of O and C=N instead of C=C). See the first equation below.


I see, however, it is an incorrect way of reasoning; 1.) since the first equation is messed up, logically LUMO should reside on the C=N carbon, 2.) since Arrhanius explicitly provided benzaldehyde as example of carbonyl component and not 2-phenylacetaldehyde (enol ether of). As we know, benzaldehyde has no alpha hydrogens, consequently it can not be used as a direct carbonyl precursor for Claisen rearrangement of any kind, only for Cope -- as I've depicted above.


Quote: Originally posted by Nicodem  
The catalytic effect of the acid by protonation of the "imine's" nitrogen (which is the only protonable one prior to Boc removal), indicates the electron flow is most likely the opposite to the one in the Claisen-type of rearrangements (see the third equation bellow). So it can not be a Claisen, but it does not have that much to do with a Cope either, because Cope does not involve protonation or heteroatoms in that position.



Well, that's not the case, aza-Cope is catalyzed by protonation of imine nitrogen, logically, since the LUMO energy of iminium species is further lowered compared to that of the corresponding imine. We all agree that Cope and Claisen rearrangements are [3,3]-sigmatropic rearrangements, however, the reaction Arrhenius refers to can not rationally be considered a Claisen rearrangement since benzaldehyde can not enolize/enaminaze, I don't think this is such a far out idea, I think it is quite basic and obvious but for some reason you are avoiding it. :) You did compare the direction of the electron flow in the Arrhenius example with that of Claisen rearrangement, but more importantly you didn't mention that the electron flow in the Arrhenius example is fully consistent with that of aza-Cope rearrangement, i.e C=N carbon - LUMO, alkene - HOMO.

Nicodem - 19-8-2010 at 23:34

Quote: Originally posted by Sandmeyer  

I see, however, it is an incorrect way of reasoning; 1.) since the first equation is messed up, logically LUMO should reside on the C=N carbon, 2.) since Arrhanius explicitly provided benzaldehyde as example of carbonyl component and not 2-phenylacetaldehyde (enol ether of). As we know, benzaldehyde has no alpha hydrogens, consequently it can not be used as a direct carbonyl precursor for Claisen rearrangement of any kind, only for Cope -- as I've depicted above.

Ah yes. The flow of electrons is likely wrong in the first equation. I always have troubles with deciding which Pi-system interacts with its LUMO and which one with its HOMO. That orbital stuff was never appropriately thought to me (unfortunately we had very old professors). I never trust much my intuition on this and whenever I run on similar problem I go asking a friend to calculate energies with Gaussian (even though I have serious troubles trusting computational chemistry, but obviously it is more accurate than my intuition). I guess this is one of those things I was trying to explain in one previous post, that one should learn these things while young. It is too late trying to do so later when you work in the lab and have children at home.
Your second point is however something I can't really integrate in the discussion. The O-allylation of aldehydes does not constitute as part of the Claisen rearrangement, so neither can the formation of allylhydrazones be considered as part of the aza-Cope rearrangement. If you compare two reactions, the comparison simply must be isosteric - you can't base the comparison on the synthesis of the starting compound (named reactions are defined by the mechanism, though it sometimes turns out that more than one mechanism operates).

Quote:
You did compare the direction of the electron flow in the Arrhenius example with that of Claisen rearrangement, but more importantly you didn't mention that the electron flow in the Arrhenius example is fully consistent with that of aza-Cope rearrangement, i.e C=N carbon - LUMO, alkene - HOMO.

I only compared it to Claisen because that is what it appeared like before I had the article in hands. I soon gave up on that comparison, but truly can not call that an aza-Cope rearrangement either, though I agree that that the electron flow is consistent. How about not calling it anything but a [3,3]-sigmatropic rearrangements? The names of named reactions are often used confusingly in the literature. There is an inflation of names and an inflation of their use. You think the use of the name aza-Cope is consistent? Check this:

3-Aza-Cope Rearrangement of Quaternary N-Allyl Enammonium Salts. Stereospecific 1,3 Allyl Migration from Nitrogen to Carbon on a Tricyclic Template

Is that an aza-Cope or an aza-Claisen? To tell you the truth, now that you got me so confused, I can't be sure any more (even though a few days ago I would say Claisen before even blinking with an eye). So, what you call aza-Cope would be 2-aza-Cope and what I would call aza-Claisen would be 3-aza-Cope. But that mean it could be the opposite as well! Some might call your aza-Cope an 2-aza-Claisen. Yet, how about HOMO & LUMO roles, and the flow of electrons? I think that until IUPAC defines the proper use of heteroatomic prefixes in named reactions, one should avoid their use and stick to naming mechanisms utmost (but only when enough evidence is obtained).

[Edited on 20/8/2010 by Nicodem]

Nicodem - 21-8-2010 at 14:12

Apparently, and to my surprise, it has become an incurable habit to consider any hetero-Cope or "other than 3-oxa"-Claisen rearrangements as synonymous. At least that is what the author of the review on aza-Claisen in Top. Curr. Chem. 244 (2005) 149–213 says it's been going on in the literature.
Quote:
Introduction

3,3-Sigmatropic rearrangements are defined as uncatalyzed processes to migrate
a sigma bond of two connected allyl systems from position 1 to position 3.That
means both allyl systems suffer from an allyl inversion. Though described for
the first time in 1940, the Cope rearrangement can be considered as the basic
type of such a process, since C–C bonds only are reorganized during the course
of the reaction [1].More than two decades earlier, in 1912, L. Claisen first described
the rearrangement of aromatic allyl vinyl ethers to generate o-allyl
phenols [2]. This so-called Claisen rearrangement is characterized by the replacement
of the C3 carbon of the rearrangement system against a heteroatom
X. The basic Claisen rearrangement bears X=O; consequently, such a process can
be termed as a 3-oxa Cope rearrangement. Analyzing the literature, rearrangement
systems displaying other heteroatoms X in position 3 can be found as hetero
Claisen and 3-hetero Cope rearrangements.Focussing on systems with X=N,
names such as aza- and amino-Claisen as well as 3-aza-Cope rearrangement
occur in the literature.Furthermore, the term aza/amino Claisen rearrangement
is widely used for nitrogen introduction processes rearranging 1-aza-3-oxy-
Cope systems (imidates) to generate carbamates. Finally, the Fischer indole
synthesis represents a special type of aza-Claisen rearrangement incorporating
two N atoms in the 3 and 4 positions of the rearrangement system.

Intending to set a firm basis concerning the notion of the sigmatropic
rearrangements, the following review will use the term Claisen rearrangement
for 3,3-sigmatropic core systems incorporating a heteroatom X in position 3, i.e.,
an aza-Claisen-type rearrangement is characterized by X=nitrogen (Fig. 1).

So, regardless of the electron flow, structural peculiarities, etc., we can call the example at hands either as 2,3-diaza-Cope or 2,3-diaza-Claisen rearrangement. It makes no difference even if we see a difference. I just hate (stupidly) named reactions!


CopeClaisenNomenclature.gif - 46kB

Sandmeyer - 21-8-2010 at 15:26

Quote: Originally posted by Nicodem  

Ah yes. The flow of electrons is likely wrong in the first equation. I always have troubles with deciding which Pi-system interacts with its LUMO and which one with its HOMO. That orbital stuff was never appropriately thought to me (unfortunately we had very old professors). I never trust much my intuition on this and whenever I run on similar problem I go asking a friend to calculate energies with Gaussian (even though I have serious troubles trusting computational chemistry, but obviously it is more accurate than my intuition). I guess this is one of those things I was trying to explain in one previous post, that one should learn these things while young. It is too late trying to do so later when you work in the lab and have children at home.


No need for Gaussian, to figure out the direction of electron flow in this problem it is enough to know that enamines are nucleophilic on beta carbon while N-Boc hydrazones, in analogy to imines and contrast to enamines, are electrophilic. ;)

EDIT: The review Nicodem refers to, and that settled the matter of naming the reaction, seems like an interesting read, can someone please post it, thanks! :D


[Edited on 22-8-2010 by Sandmeyer]

Nicodem - 22-8-2010 at 02:16

Quote: Originally posted by Sandmeyer  
No need for Gaussian, to figure out the direction of electron flow in this problem it is enough to know that enamines are nucleophilic on beta carbon while N-Boc hydrazones, in analogy to imines and contrast to enamines, are electrophilic. ;)

I'm not so confident when it comes with less familiar systems like this hydrazone one. Also, it is not only the double bond systems that need comparison, but since it is about [3,3]-sigmatropic rearrangements the whole allylic and "heteroallylic" (enaminic, enolic, etc.) moiety needs to be considered to find what is the energy difference between the HOMO & LUMO. Also, the electron flow will determine what formal charge needs to be considered in the calculation / intuitive evaluation. It is not always trivial to comprehend which moiety reacts with its HOMO and which one with the LUMO (as the combination that has the lowest energy difference might not be obvious at first glance). Furthermore, things change with acid catalysis. So you end up with having to consider a bunch of orbital combinations and find out which HOMO-LUMO interaction is most favourable just to realize if the reaction is viable at all, possibly at any "normal" temperatures, and if protonation or Lewis acid coordination catalyses it. I wish someone with more knowledge of pericyclic reactions could describe a few personal tricks for intuitive evaluation of the reactions.

For example, what would you expect the order is in HOMO and LUMO energy of systems like RCH=N-NBoc-, RCH=CH-NBoc-, RCH2=CH-NMe-, RCH=CH-N<sup>+</sup>Me<sub>2</sub>-, RCH=N<sup>+</sup>H-NBoc-, RCH=N-O-, RCH=N<sup>+</sup>H-O-, etc.? I'm afraid I don't know. If the question was which of these fragments when combined with an allyl group would most likely rearrange at the lowest temperature, I could not answer easily without resorting to Gaussian or Schrödinger-Jaguar. And even then, interpreting the results leaves much too much uncertainties.


PS: Find the review attached. (BTW, nearly the whole seriez was made available in PDF if you know where to search)

Attachment: Recent Advances in Charge-Accelerated Aza-Claisen Rearrangements.pdf (1MB)
This file has been downloaded 972 times


Sandmeyer - 22-8-2010 at 11:00

Quote: Originally posted by Nicodem  

I could not answer easily without resorting to Gaussian or Schrödinger-Jaguar. And even then, interpreting the results leaves much too much uncertainties.


Resorting to... Schrödinger-Jaguar huh? Is that a beer?

cheeseandbaloney - 2-10-2010 at 21:43

Hey, another quick mechanistic question! Was flipping through some of the chapters in one of my orgo books ferking around with some of the problems. Came across what seemed like an easy one, but was an answer I didn't expect. What would you fellow chemists consider the major product of this reaction?

randomrxn.png - 2kB

Arrhenius - 12-10-2010 at 23:23

The major products would be cyclohexanone and methyl iodide. Mechanistically, the vinyl ether is protonated on *carbon* with formation of an oxocarbenium ion, then iodide attacks at the methyl carbon of the methyl vinyl ether. Hydrogen iodide mediate cleavage of aryl methyl ethers should proceed through the same mechanism. I do believe it can cleave other ethers, but this is definitely a brutally harsh reagent. Trimethylsilyl iodide (TMSI) is often employed to cleave ethers, especially methyl ethers, as a gentler alternative.

Cheers & keep reading!



cyclohexanone.gif - 3kB

[Edited on 13-10-2010 by Arrhenius]

cheeseandbaloney - 14-10-2010 at 00:43

yo Arrhenius! Why does the iodide specifically attack the methyl group and not the carbonyl carbon?

edit: wait, does it haff to do with the stability of products, or enthalpy?

[Edited on 10/14/2010 by cheeseandbaloney]

Ebao-lu - 15-10-2010 at 00:09

just a short copypaste from https://www.sciencemadness.org/whisper/viewthread.php?tid=14...
Quote:

O-desmethyl tramadol is prepared by treating tramadol as a free base under O-demethylating reaction conditions, e.g., reacting it with a strong base such as NaOH or KOH, thiophenol and diethylene glycol (DEG) with heating to reflux (Wildes, et al., J. Org. Chem., 1971, 36, 721)

What is the mechanism of this kind demethylation? I'm not able to view the original paper unfortunately
My propasal is that it is simply a necleophilic catalysis by thiolate, which is a stronger nucleophile then hydroxide. And at the same time, it is a better leaving group then alkoxide. So it is Ar-OMe + -SR =intermediate=> Ar-SR + OMe-
Ar-SR + 2OH- =intemrediate=> Ar-O- + SR- + H2O
But that intermediate is very unstable, thats why high temperature is required.
Is it correct ?


[Edited on 15-10-2010 by Ebao-lu]

Nicodem - 15-10-2010 at 09:18

Quote: Originally posted by cheeseandbaloney  
yo Arrhenius! Why does the iodide specifically attack the methyl group and not the carbonyl carbon?

edit: wait, does it haff to do with the stability of products, or enthalpy?

Addition of HI (or HCl, HBr...) on the carbonyl group is reversible with the equilibrium very highly in favour of the carbonyl rather than gem-halohydrins. The more likely further reaction of cyclohexanone under very acidic conditions is the acid catalysed aldol condensation, but this should be slower than the vinyl ether cleavage, so it should not represent a problem under properly chosen conditions.
If aqueous HI is used for the demethylation then the competing reaction is the addition of H2O on the intermediate O-methylcyclohexanonium carbocation. This is a normal hydrolysis of enol ethers which however leads to the same product (cyclohexanone) and methanol. If HI is concentrated enough, methanol ends up forming methyl iodide via SN2). Thus you have an interesting situation where the end products are the same regardless of the mechanistic pathway. This is actually a common situation where two or more mechanistic routes lead to identical products from the same substrate and the (major) pathway can only be demonstrated by observing (spectroscopically, chromatographically, etc.) or catching the intermediate (in this case methanol in the second pathway, but absent in the first).

Quote: Originally posted by Ebao-lu  
What is the mechanism of this kind demethylation?

Like most demethylations (if not all) it is just a normal SN2 substitution. Similarly as the demethylation mentioned the two posts above yours.

Ebao-lu - 26-10-2010 at 01:51

Btw, Nicodem and Arrhenius, the rearrangement you are talking about could be also semi-homolytic, like in Fisher's indole synthesis and Hoffman-Loeffler reaction. That means, any arrows should be avoided in the mechanism, because arrow is an electron pair, not a single electron. Usually such rearrangement mechanisms, if i remember correctly, are drawn like dotted line if i'm not mistaken, and BOC-protected aminogroup is just an analogue of protonated nitrogen in ability to form cathionradicals like RNH2(+*) or
BOCNH* <-> t-BuO-C(O-)=NH(+*) <-> t-BuO-C(O*)=NH


[Edited on 26-10-2010 by Ebao-lu]

[Edited on 26-10-2010 by Ebao-lu]

Nicodem - 26-10-2010 at 07:16

Quote: Originally posted by Ebao-lu  
Btw, Nicodem and Arrhenius, the rearrangement you are talking about could be also semi-homolytic, like in Fisher's indole synthesis and Hoffman-Loeffler reaction. That means, any arrows should be avoided in the mechanism, because arrow is an electron pair, not a single electron.

The crucial step in the Fisher's indole synthesis is a pericyclic reaction, namely a [3,3]-sigmatropic rearrangement, thus there is a circular flow of electrons (as pairs). Normal arrows are therefore used to depict what goes on. I never heard about articles claiming it involves a homolytic cleavage of bonds.
The Hofmann-Löffler reaction does involve a homolytic bond cleavage, but I see no similarities with the above [3,3]-sigmatropic rearrangements.

Ebao-lu - 26-10-2010 at 07:35

i'm sorry. the rearrangement you are talking about is not that kind, as there is no N-N bond cleavege at all! did not read it carefully first
Quote:
I never heard about articles claiming it involves a homolytic cleavage of bonds

here is one http://www.springerlink.com/content/x0h4047554788212/fulltex...
I have also seen most articles that claimed an arrow-drawn mechanism of fisher indole synthesis, that is maybe due to simplicity
Indeed, i think there is no _ideal_ synchronous sigmatropic rearrangement. Some are bit heterolytic(even having extremely short living ion intermediates and such), some bit homolytic.. depending on what is more beneficial.
Otherwise how a charge-attraction product formation in some diels alder reactions could be explained? Sometimes it is not thermodinamically favored over others, so that is definitely due to kinetics, or in other words activation energy difference. The latter may result from higher energy of transition state in case of unfavored product and lower in case of proper (charge to charge) orientation. If we draw them both then it is clear that it is mostly because of charge interaction, and sometimes there is no substantial repulsion in case of non proper product, there is just an attraction in proper. But those charges are derived from pi-system, that means it is not an ideal 4+2 reaction!


[Edited on 26-10-2010 by Ebao-lu]

[Edited on 26-10-2010 by Ebao-lu]

Nicodem - 27-10-2010 at 09:33

Quote: Originally posted by Ebao-lu  

Quote:
I never heard about articles claiming it involves a homolytic cleavage of bonds

here is one http://www.springerlink.com/content/x0h4047554788212/fulltex...
I have also seen most articles that claimed an arrow-drawn mechanism of fisher indole synthesis, that is maybe due to simplicity
Indeed, i think there is no _ideal_ synchronous sigmatropic rearrangement. Some are bit heterolytic(even having extremely short living ion intermediates and such), some bit homolytic.. depending on what is more beneficial.

That is an interesting paper, but it does not demonstrate the mechanism of the Fischer indolization involves a homolytic bond cleavage. The N-N bond is weak and easily homolyticaly cleaved and thus it is not particularly surprising they got an ESR evidence for the N-N cleavage under the reaction conditions. This however does not demonstrate this is part of the reaction pathway. Aditional evidence must be presented, for example, an experiment where spin traps or radical inhibitors inhibit the conversion rate to the indole. Or that the Fischer indolization can be driven photochemically (though this would only demonstrate the feasibility of the homolytic pathway, but not also that this is the mechanism in the acid catalysed reaction).
Though interested, I currently don't have the time to follow the references cited in that paper, but you are welcome to research this topic and report back with a short review.
Quote:
Otherwise how a charge-attraction product formation in some diels alder reactions could be explained? Sometimes it is not thermodinamically favored over others, so that is definitely due to kinetics, or in other words activation energy difference. The latter may result from higher energy of transition state in case of unfavored product and lower in case of proper (charge to charge) orientation. If we draw them both then it is clear that it is mostly because of charge interaction, and sometimes there is no substantial repulsion in case of non proper product, there is just an attraction in proper. But those charges are derived from pi-system, that means it is not an ideal 4+2 reaction!

The Diels-Alder reaction mechanism, as well as its regio- and stereoselectivity, is fully and thoroughly explained on the basis of orbital theory. No need to call for any other more exotic and off the trail explanation. Mechanistic hypotheses first need to fulfil the Occam's razor criteria. If they don't, the chances they are wrong or incomplete increase.

Ebao-lu - 28-10-2010 at 00:27

Maybe you are right as there is no sufficient evidence that N-N bond homolytic cleavage is part of a reaction pathway, but i don't mean that it is not 3+3 reaction, of course it is. I mean, that in this reaction's transition state there is practically no polarization of N-N bond, so it is cleaved homolytically with immediate formation of new C-C bond
I think computatuinal techniques can give answer wether there is polarization of N-N bond in fischer's indole synth transition state or not. Other methods like radical traps etc may not work, because recombination of radicals could be a best trap, especially if they are close or even synchronously formed and recombinated like in a 3+3 reaction. Spin complex formation like in those article could be just to make them bit more stable, live for a while or just be formed without synchronous recombination. Another possible method could be measuring the rate of indole formation and compare it with amplitude of those signals, in different conditions like temperature, catalyst etc. If they are in high correlation, that should mean it is a part of reaction pathway.

As for articles, i have not read them yet. Btw those guys are having another paper that i am trying to comprehend now where they claim partially intermolecular mechanism of Fischer http://www.springerlink.com/content/g52q4n254hg33r76/ . The nature of N-N cleavage seems to be debated before according to this article, but i think not enough. There should be, or maybe there are, articles that gave a definite answer on that question. Did not read cited refferences yet, besides some need access, which i don't possess.

My point still is, that synchronous rearangements([3+3] [4+2] etc) can be two types - homolytic and heterolytic (or better to say, polar/ non polar) ones. I dont say that N-N bond in fisher's indole synthesis is cleaved _before_ formation of C-C, both processes are synchronous. But the nature of this cleavage is homolytical, so there is no polarization in transition state. Drawing arrows in case of any rearrangements was always associated with polarization of atoms in transition state according to my understanding, but is there any polarization in transition state indeed? That would depend on substrate. Some substrates would develop substantial polarization in transition state(like in heteroDA reaction), some will not(like in non polar, symmetrical substrates). But nethertheless arrows are always used to depict mechanism of both polar and non polar transition states, and i think that is not correct(for non polar ones), so that is mostly for simplicity. I may be wrong, but i dont understans what else can arrows mean but the direction of electron pairs flow in transition state, and that means polarization

Maybe, theories that are accepted today can give their own explanation of why protonation or a lewis acid can promote fischer indole formation, but mine would always be the analogy with N-N homolytical bond cleavage, that is eased after protonation because of increased stability of protonated nitrogen radicals. And in principle i suppose it is quite the same as in accepted theories - the difference is that i don't need computer to calculate all those electronic orbitals in order to predict what would happen, i just use an approach of easing up N-N bond homolytic cleavage, since nitrogen radicals become more stable if protonated, and maybe also some impact of electrostatic repulsion between positive charged nitrogen atoms.
I wonder, what the computational model would say concerning this difference between protonated and not protonated hydrazone reactivity towards a sigmatropic rearrangement, and difference in activation energies between sigmatropic route and homolytic dissociation of diprotonated hydrazone. It should be quite a simple model for computation. It is quite possible, that activation energy for a sigmatropic rearrangement in this case may be even higher then homolytic dissociation, because in case of sigmatropic rearrangement's transition state all the atoms are changing their location and that should results in increasing activation energy, while in homolytic cleavage much less number of atoms are relocating in transition state. This may also elucidate the mechanism of Fisher indole synth

Maybe all i wrote above in concern of HOMO-LUMO and other related today's accepted theories sounds dumb, but my understanding helps me to understand mechanisms and i'm satisfied. Most physical or quantum chemists i knew were not very good in organic chemistry, their perfect knowledge of HOMO-LUMO and other related things(or just book illustration of that) did not allow them to predict outcome of reaction. So apparently, some simplification of classic theories is allowed.
If there is example where my simplified understanding would not allow to understand mechanism or products formed, i will definitely change my mind and avoid using it in future.

Quote:

The Diels-Alder reaction mechanism, as well as its regio- and stereoselectivity, is fully and thoroughly explained on the basis of orbital theory. No need to call for any other more exotic and off the trail explanation.

I am sure, this thorough explanation was given mostly for the simplest cases of D-A. More complicated D-A reaction outcomes are usually explained in a way similar to this, including charges: http://138.253.125.24/~ng/suzanna/Cycloaddition2.html .And that is the thing i was talking about in the beginning, about not ideal sigmatropic reactions. Perhaps that is just illustration for students for simplicity, and there are some more thorough explanations, i never seen them probably because the same reason - being satisfied with existing ones, and not good understanding of psi-functions and all that kind. I may be wrong btw..
Quote:
Mechanistic hypotheses first need to fulfil the Occam's razor criteria. If they don't, the chances they are wrong or incomplete increase.

Thanks, i prefer using Einstein's razor! (or Einstein's interpretation of Occam's razor, not sure) that says "Make everything as simple as possible, but not simpler."
I think i used it :cool: ->:). There was no sophistication in my D-A explanation, it was just simplification. Most terms i used are frequently used in organic chemistry like "activation energy" "charge attraction" (better interaction), "transition state" etc. I did not say that DA mechanism does not involve orbital interactions(as all reactions, not only DA involve it), i just neglected that as not critical in this particular case for understanding the regioselectivity and changed it to charge explanation, which does not violate or conflict with any existing rules or accepted mechanism, being just a simplification of that also in order not to draw orbitals on this board. I was sure you will also not recall about orbitals. But if you did - feel free to explain that my explanation does not fall under Occam's razor(if you want) - not to debate with you, just for you to convince me that my explanation is really a sophistication, or over simplification of existing facts and thus may become wrong(i agree that it is incomplete btw, but it was not a new mechanism proposal, just a simplifacation of existing one).. And i'll change my mind immediately if you do that, Nicodem
:)


[Edited on 29-10-2010 by Ebao-lu]

Copenhagen - 17-11-2010 at 15:46

From personal experience as a beginning organic student I found Arrow Pushing by Levy to be the most helpful to me. It was the easiest to read. It helped my mechanistic intuition immensely. I was able to cover up the products on a lot of the key reactions and correctly predict them by understanding the fundamentals that Levy teaches.

I also own the Art of Writing Reasonable Reaction Mechanisms and I agree that it is an awesome book on teaching understanding the mechanism. It includes everything that Arrow Pushing does not. For example it discusses the differences between various electrophiles and nucleophiles and considering the solvent conditions. "snowballs chance in hell" comes to mind. You won't make the mistake of drawing a hydroxide leaving group in an acidic medium for example.

Solvent conditions are very important and are not lucidly or adequately explained in any of the course textbooks that I have read. You memorize that SN2 goes with aprotic and SN1 goes with protic but you don't really learn why...

Arrow Pushing makes it easy to understand that a strong base nucleophile (electron dense) will be shielded by a protic solvent because the positively charged protons of the solvent will be attracted to negative-electron dense nucleophile and will be less able to react. An aprotic solvent won't produce such a problem for the incoming nucleophiles. Little things like this are so important in learning mechanisms.


The other books I have found of greater benefit after completing both semesters of organic as they tend to be go in to a lot of detail. Electron Flow in Organic Chemistry is also a good one but I find Arrow Pushing more to the point.

Nicodem - 29-11-2010 at 13:07

Quote: Originally posted by Ebao-lu  
My point still is, that synchronous rearangements([3+3] [4+2] etc) can be two types - homolytic and heterolytic (or better to say, polar/ non polar) ones. I dont say that N-N bond in fisher's indole synthesis is cleaved _before_ formation of C-C, both processes are synchronous. But the nature of this cleavage is homolytical, so there is no polarization in transition state.

I don't have time to go replying to your post point by point, but will only point out that the most important aspect of pericyclic mechanisms is in that the bond formation and breaking occurs in a concerted manner. This allows a much lower activation energy to reach the transition state than any other mechanisms, such as polar or radical ones, could ever allow.
Let me explain you with an of topic example. If you go to a luna park and observe how a Ferris wheel works, you can note that even though it is able to lift tons of people, it only works on a relatively weak electromotor (requires little activation energy to turn around). That is because the force to lift (break bonds) is compensated with the force given from the fall on the other side of the wheel (bond formation). This is because both sides of the wheel (the one going up and the one going down) are connected trough the rotating axis (concerted mechanism). Now if you would have to lift all those people up to the same height with a lift, give them time to see the view from their transition state perspective, and then take them down with the same lift, the process would not be concerted any more and it would require much more energy (full lift and wasted fall). This would be so even if comparing with a unbalanced, unevenly occupied Ferris wheel (not fully synchronous).
I'm specifically comparing pericyclic reactions with a wheel because geometry plays a crucial role - the transition state must be a cyclic structure so that the bonds can easily form and break in a concerted manner.
This is a very much oversimplified pedagogic model, but the goal is to show that when you propose a mechanistic pathway of a reaction, you must search for the one with the lowest activation energy, because that is how molecules "decide" if they will react or not. If they have enough kinetic energy to reach any possible transition state they will roll over the to the other side of the thermodynamic hill and give the respective product.

Sandmeyer - 23-12-2010 at 16:45

Ok, since Arrhenius have not posted any problems for a while I'll post a problem. Provide mechanism for transformation 1 to 2 in DMF. I'll give a solution proposal on request.

EDIT: It is hard to see from the picture but it should read: "1.) Br2 2.) Base"


MechanSciMad2.bmp - 18kB


Gool luck! :D

[Edited on 24-12-2010 by Sandmeyer]

cheeseandbaloney - 23-12-2010 at 21:57

yo sandmeyer! does it involve any tautomerization out of curiosity?

Sandmeyer - 24-12-2010 at 05:18

yO! Nope. And no radicalz either... Ho-Ho-Ho...

cheeseandbaloney - 25-12-2010 at 23:13

hmmmm so first hope your holidays went fantastically grand, second! Sandmeyer, goofed around with it for a bit tonight hope i'm on the right track:

mechanism.png - 18kB

Sandmeyer - 27-12-2010 at 01:05

Yeah, you've got it cheezy -- it's an Sn2' all-right... :D

Ebao-lu - 1-2-2011 at 06:15

Sorry, Nicodem, i dont know how all those thoughts were coming in my mind, but now i understand you are right. Arrow-drawn mechanism does not mean polarization of any atoms in transition state, which i first believed to be, what ever direction this flow goes. As well, polarization in cases of some hetero-DA reactions transition state does not mean there is no electron pairs flow or it is somewhat impaired. The problem of my ignorance should have come from observation, that most hetero-DA's reaction mechanisms i seen were drawn as electron pair flow in only one direction, the same that is taking place when reaction is not concerted. Then i decided that another direction of flow is less favorable, that electron pairs flow should include a polarization in transition state, and that it may be not ideal and displaced by polar interations. Now i understand that direction of electron pairs flow does not mean anything. Thanks. Pretty much loved your example about luna park.




[Edited on 1-2-2011 by Ebao-lu]

Ebao-lu - 1-2-2011 at 06:19

Could someone explain a mechanism of this sodium sulfite reduction?
The book where it was found is here


sodiumsulfite.JPG - 44kB

[Edited on 1-2-2011 by Ebao-lu]

[Edited on 1-2-2011 by Ebao-lu]

GreenD - 12-4-2011 at 07:20

Hi all,

at the risk of starting a new thread and looking like an idiot, I've decided to post my question here instead.

I am looking at a paper where the amidification of a carboxylic acid is done with POCl3;

R-COOH + POCl3 -> [intermediate] --- (R,R-NH) ---> R-C(O)N-R,R

At first glance I would assume one to use SOCl2, or oxalyl chloride to get to the acid chloride, and go by amidification from there. But instead POCl3 is used.

Under what circumstances would POCl3 be beneficial?

All I could find was that POCl3 is used extensively in biochemistry for some kind of coupling (sorry it escapes me) or for the dehydratioon of amides, giving nitriles...

Having seen the paper used POCl3 I had assumed that possibly the starting material was not an acid, but an amide. The amide made into a nitrile, and undergoes hydrolysis to make an acid, however this is not the case. The starting material IS and acid, and there is ONLY the addition of POCl3 and the corresponding amine to give an amide.

So - whats the deal with POCl3?

Nicodem - 12-4-2011 at 09:29

Quote: Originally posted by GreenD  
The starting material IS and acid, and there is ONLY the addition of POCl3 and the corresponding amine to give an amide.

Perhaps it was some cheap amine that can be used in large excess. Otherwise it would be a bit of irrational to use POCl3 when it leaves a bunch of amine consuming side products after the COOH to COCl transformation.
Quote:
So - whats the deal with POCl3?

It leaves non-volatile side products, while oxalyl chloride and SOCl2 are more suitable, since all you need to do to get a crude acyl chloride is to rotavap the reaction mixture (or distil via a distillation column in case of lower boiling acyl chlorides).

GreenD - 13-4-2011 at 15:10

Does SOCl2 or oxalyl chloride react with secondary cyclic amines? Indoles?

Reactive towards anything other than alcohols and acids? Chemistry book says "Make sure you know the mechanism - thionyl chloride reacts in many ways."

It acts as a lewis acid - and I believe an indolic amine is stable enough to go without side reaction.

Nicodem - 28-5-2011 at 00:28

Quote: Originally posted by GreenD  
Does SOCl2 or oxalyl chloride react with secondary cyclic amines? Indoles?

Reactive towards anything other than alcohols and acids? Chemistry book says "Make sure you know the mechanism - thionyl chloride reacts in many ways."

It acts as a lewis acid - and I believe an indolic amine is stable enough to go without side reaction.

Indole gets quantitatively C-acylated by oxalyl chloride already at 0 C, requiring no catalyst. I don't know about its reaction with SOCl2, but I would suspect it to react with it as well. However, indole is not a "secondary cyclic amine". Secondary amines, cyclic or not, vigorously react with oxalyl chloride or SOCl2 in a different way. They get N-acylated or N-sulfinated respectively, to give oxalamides or sulfinamides.
In general, oxalyl chloride and SOCl2 can react with any nucleophilic enough functional group. Under acidic catalysis they even react with weak pi-nucleophiles like benzene and alkenes.

Arrhenius - 5-7-2011 at 13:42

Ebao-lu:

This is an interesting reduction of tribromoimidazole - primarily because most of us don't often think about simple reagents like sodium sulfite. That being said sodium sulfite is a fairly versatile reducing agent, and is dirt cheap. I don't know the mechanism for sure, but I feel bad that no one's ventured a guess, so I will. This leaves unanswered as to why this reaction can achieve selective (or atleast somewhat selective) di-debromination. Given the yield, it's quite possible that this outcome is controlled by optimizing reaction time rather than pure thermodynamics or so.

What I've drawn is a combination of some literature precedent (see below). Brominated hydroxyphenols are easily debrominated with sodium sulfite, whereas their monomethyl ether counterparts are not reduced under the same conditions. This is likely due to the fact that the hydroxyphenol undergoes facile tautomerization to a keto or quinone form. For this reason, invoking addition of sulfite via a more reactive tautomer of imidazole seems necessary. Given that a solution of sodium sulfite is only mildly basic, a proton catalyzed tautomerization seems OK.

It seems that either sulfite or water can act as the end reductant, yielding either the bromosulfate ion or hypobromite, respectively.

sulfite reduction.gif - 11kB

Refs:
1.) Arch. Biochem. and Biophys., (1974), 161(2), 632-637.
2.) J. Am. Chem. Soc., (1952), 74(6), 1601–1602


Ebao-lu - 26-8-2011 at 04:04

Yes, Arrhenius i think you are close to truth. That should be some kind of ionic substitution "at positive halogen". It is known to take place in some chlorinations with CCl4(those that are not radical) and a-halosulfones reduction, SO2Cl2 chlorinations i suppose have similar mechanism
In this article it is even called "SNCl+" substitution, i don't know if it is widely accepted name, so here we should have SNBr+
http://www.sciencedirect.com/science/article/pii/S0040402001...

Thanks for information about brominated hydroxyphenols reduction with Na2SO3, that makes sence. Here we should have something similar.

[Edited on 26-8-2011 by Ebao-lu]

[Edited on 26-8-2011 by Ebao-lu]

How's that work?!!?

Arrhenius - 4-12-2011 at 23:17

Hey all. I'm a bit saddened by the Organic stickies. It seems people are more interested in benzaldehyde and Duff reactions than mechanisms!! :P Shocking! So I thought maybe we could work through some new problems. I'm a firm believer that you should know *how* your synthesis is working, not just how to do it.

Here are a few rules for writing any mechanism:
- Don't break the rules.
- Don't skip steps.
- Identify a nucleophile and an electrophile and:
1.) make a bond
2.) break a bond
3.) draw the result
- Special case (e.g. pericyclic rxns) don't have a nucleophile, so this won't work
- Try to balance reaction equations. Organic chemists lose sight of this, but this basic concept is still important!

Fluorescein:
First synthesized in 1871. Basically, just mix resorcinol, phthalic anhydride, some sulfuric acid, and heat it a bit - someone should write up a prep. The product is intensely fluorescent due to the electronic structure and extended conjugation in 2.
1.) Propose a mechanism for this reaction.
2.) For those unfamiliar, the 'delta' means heating. Why do you need to heat this reaction? What steps require this energy?
3.) Wikipedia calls this a "Friedel-Crafts" reaction. There's no AlCl3, FeCl3 or an acid chloride, so why is this still a Friedel-Crafts reaction?
fluorescein.bmp - 144kB

In the Grignard reaction of ethylmagnesium bromide with benzaldehyde, the anticipated secondary alcohol 3 is isolated as the major product, along with minor impurity 4. Indeed, this minor impurity is found in most reactions with organometal nucleophiles.
1.) Draw a mechanism for the formation of 4.
2.) What is the name of this reaction?
3.) What is the oxidation state of the aldehyde carbon in the starting benzaldehyde? and in the products? What is the oxidant or reducing agent that promotes this redox, and what is its oxidation state throughout the reaction?
benzyl alcohol.bmp - 70kB

Magpie and others have worked on the synthesis of furfural, 6, from pentoses. The Org Synth prep prepares furfural from corncobs.
1.)What is the mechanism (following the rules!) for this reaction? (if you want to cheat, it's hidden in Magpie's thread).
Furfural can be used to prepare butenolide, 7.
2.) What is the mechanism for the reaction of 6 to 7?
3.) What is the name of this reaction?
butenolide.bmp - 81kB

[Edited on 5-12-2011 by Arrhenius]

UnintentionalChaos - 30-1-2012 at 10:15

I have a question related to a recent synthesis of mine. I was trying to work through the reaction mechanism and came to a sticking point. Please see below. While I made phenylacetylene, from styrene, the common undergrad experiment utilizing trans-stilbene is a simpler system and the same mechanism should apply. First, the alkene is brominated, giving the trans/anti addition product. This is heated with an alkali hydroxide in a high boiling solvent (usually triethylene glycol) to induce a double dehydrohalogenation. I see this described usually as two E2 eliminations. If we do the first E2, we find that the product is always E, which lacks an antiperiplanar H and Br for the second elimination.

I've heard that synperiplanar can E2 as well, but the transition state is significantly higher energy. Does the reaction proceed this way, does the high temperature induce a double-bond isomerization, or does it go by a different mechanism alltogether? E1 would require spontaneous formation of a vinylic carbocation, which I find exceedingly unlikely.

alkyne.gif - 5kB

[Edited on 1-30-12 by UnintentionalChaos]

[Edited on 1-30-12 by UnintentionalChaos]

DJF90 - 30-1-2012 at 16:57

Double dehalogenation should only be possible when the product would be a terminal alkyne, as with styrene dibromide. The product you have drawn (1,2-diphenylacetylene) would likely have to be prepared via a sonogashira coupling. Do you have a reference stating that the reaction you have drawn is possible?

UnintentionalChaos - 30-1-2012 at 17:05

Quote: Originally posted by DJF90  
Double dehalogenation should only be possible when the product would be a terminal alkyne, as with styrene dibromide. The product you have drawn (1,2-diphenylacetylene) would likely have to be prepared via a sonogashira coupling. Do you have a reference stating that the reaction you have drawn is possible?


http://www.orgsyn.org/orgsyn/orgsyn/prepContent.asp?prep=cv3...

Quite possible, in fact I did it as an undergraduate organic chem lab.

This question is actually about phenylacetylene, I figured that this system posed a similar problem. With styrene dibromide to phenylacetylene, the two conformations for the first dehydrohalogenation (loss of the benzylic bromine) give a cis and a trans beta-bromostyrene. Yield of phenylacetylene exceeds 50%

[Edited on 1-31-12 by UnintentionalChaos]

UnintentionalChaos - 2-2-2012 at 22:43

I can't edit anymore. I found my answer in the lit. Looks like an E1cb.

http://144.206.159.178/FT/986/86265/1458858.pdf

GreenD - 5-3-2012 at 17:26

I need this article, or could someone explain the mechanism within? Or both?

http://pubs.acs.org/doi/abs/10.1021/ja01469a048

zoombafu - 5-3-2012 at 17:44

I just got this book to go along with my organic chemistry textbook

<a href="http://www.amazon.com/gp/product/1592577539/ref=as_li_ss_tl?ie=UTF8&tag=thehomche0e-20&linkCode=as2&camp=1789&creative=390957&am p;creativeASIN=1592577539">The Complete Idiot's Guide to Organic Chemistry</a><img src="http://www.assoc-amazon.com/e/ir?t=thehomche0e-20&l=as2&o=1&a=1592577539" width="1" height="1" border="0" alt="" style="border:none !important; margin:0px !important;" />

It does a really good job of explaining mechanisms.

Arrhenius - 28-3-2012 at 23:03

GreenD:

The authors consider this a Beckmann rearrangement. Given that in this paper a variety of metals including nickel, copper and cobalt are demonstrated to affect the isomerization of oximes to amides, the mechanism is likely Lewis acid activation - as opposed to redox. Bare in mind that Brønsted acids catalyze this transformation as well. The mechanism in this case is as follows:
oxime isomerization.bmp - 420kB

Oximes, in particular those derived from aldehydes (aldoximes), are configurationally stable. That is to say, the cis-/trans- stereochemistry about the C=N bond does not readily interconvert at room temperature. This is important when we consider the mechanism, because only when the aldoxime hydrogen and hydroxyl group are anti-periplanar can the E2 type elimination take place. I've drawn the orbitals involved to make this clear. The electron pair in the C-H bond (large grey lobe) must populate the anti-bonding (small white lobe) of the breaking N-O bond, and symmetry requires they be anti-periplanar. Nickel promotes this reaction by coordinating the oxygen, effectively weakening the N-O bond by pulling electron density away from the oxygen center. You could also think of this as a resonance structure (shown) where there is a positive charge on oxygen and a bond to nickel. From the resultant nitrile, the metal can coordinate nitrogen, making the carbon more electrophilic, and allow water to add. A proton transfer (written ~H+) gets us to the amide.

[Edited on 29-3-2012 by Arrhenius]

GreenD - 2-4-2012 at 18:49

Arrhenius, bravo! That was great. The explanation of the anti-periplanar geometry, as well as the complexing. I was also stumped how the oxygen was moving from the aldoxime to the amide, but it is actually from water. The stability of the aldoxime also cleared - thanks.

Here is numero 2. I've never encountered an alpha-alkelation before, and I have no idea where the hell to even start on this one.


Arrhenius - 4-4-2012 at 06:54

GreenD
It seems like you have some training in organic chemistry, but here are a few rules - mentioned previously - that I find extremely helpful for working through 'unknown' mechanisms. Additionally, I've drawn reversible arrows where I believe reactions to be reversible. This is often important to consider why reactions proceed to particular products. A majority of reactions involve a nucleophile and electrophile, and hence three rules for drawing a mechanism to determine reaction producst/intermediates are:
1.) identify a nucleophile and an electrophile
2.) make a bond, break a bond (i.e. use arrows in a valence allowed manner)
3.) draw the result

Here's how to apply these to your second question. I've also drawn partial charges, where applicable, to indicate nucleophilic/electrophilic sites. The key is not to get ahead of yourself, as this often leads to confusion if you try to skip a step (e.g. writing a proton transfer prematurely).

Step 1: The key realization to make here is that primary and secondary amines react with ketones and aldehydes in a somewhat enthalpically favorable reaction, resulting in either an imine/enamine (tautomers) which are moderately stable, or an iminium which is strongly electrophilic. Also, if an acid or base catalyst is present - in this case HCl - it will probably be utilized in the very first step of the mechanism; in this case we protonate a ketone/formaldehyde.

Now, the next bit depends on how you run the reaction. I probably drew it differently than your stepwise reactions suggest. Step 1, strictly speaking, probably forms the dimethyliminium ion of formaldehyde - this is much more electrophilic than formaldehyde itself. However, if the diethylketone were present I believe it would proceed through an enamine catalyzed Mannich reaction as I've shown. Alternatively, the base in step 2 deprotonates the diethylketone, and the resulting enolate adds to dimethylformaldiminium chloride. Enolates and enamines are 'isolelectronic', which is to say in both cases you can consider a resonance structure with a negative charge on the carbon alpha to the carbonyl to suggest it is nucleophilic. While it's of little consequence here, the enamine will be of the E (E=trans Z=cis) stereochemistry for thermodynamic reasons.

Step 3: Here, you must realize that amines react exhaustively with alkyl halides - when present in sufficient amounds - to form tetraalkylammonium halides.

Step 4: The formal positive charge on nitrogen weakens the strength of the C-N bond to the neighboring CH2. Another way to say this is that the NMe3+ becomes a better leaving group. Sodium hydroxide (pKa ~16) can deprotonate (only partially!) the ketone (pKa 20) to give the thermodynamic enolate, which then gives rise to a type of Hoffmann elimination. This is somewhat reversible because the product possesses a Michael acceptor, but is enrtropically driven according to Le Chatelier's principle because trimethylamine is removed from the reaction as a gas.

mannich2.bmp - 1.1MB

GreenD - 4-4-2012 at 08:00

Incredible!
Thank you so much. I will have to practice this a few times before I can "see" it if you know what I mean.

Quite an interesting reaction all together.

ZHANGNIUBI - 11-4-2012 at 07:56

speak of the mechanisms, I have many interesting problems, I would say that are hard but useful when people want to learn total synthesis.

these problems, may have been known by some of us, called Fukuyama group meeting problems.



未命名.jpg - 53kB

ZHANGNIUBI - 11-4-2012 at 08:01

I'd say that whenever you know how Fukuyama's group meeting problem's mechanism, you ll be really fascinated.

2bfrank - 31-5-2012 at 02:21

Im an IT dumbass, hence attached rather than post it directly. (been to busy to sort out how to do that, help would be good) but this mechanism was associated with a synthetic and medicinal unit I am currently undertaking. Its predominantly using an iminium/enamine type catalyst with a Michaels reaction and an Aldol reaction, I just thought it pretty cool. To me its poetic. Im still a little unclear, and taking my time with the stereo aspects to it, but thought it a Pretty cool, albeit a simple mech.

Cheers

Attachment: ol100857s.pdf (439kB)
This file has been downloaded 879 times


2bfrank - 6-6-2012 at 05:00

Quote: Originally posted by 2bfrank  
Im an IT dumbass, hence attached rather than post it directly. (been to busy to sort out how to do that, help would be good) but this mechanism was associated with a synthetic and medicinal unit I am currently undertaking. Its predominantly using an iminium/enamine type catalyst with a Michaels reaction and an Aldol reaction, I just thought it pretty cool. To me its poetic. Im still a little unclear, and taking my time with the stereo aspects to it, but thought it a Pretty cool, albeit a simple mech.

Cheers


Ive been trying to be clear how an enamine can form with just the basic catalyst. I have attached a schematic, I really would appreciate some feed back. E1cb of an OH usually requires conjugation, If I remember correctly, but the intermediate is purely electrostatic attraction between the iminium and the OH anion, and I do not know if its feasible. My apologize of attaching rather than somehow posting the mech.

Attachment: enamine via catalyctic base with saturated aldehyde.doc (260kB)
This file has been downloaded 626 times


PeetPb - 12-8-2012 at 12:44

hi, I've been searching all over the internet for the mechanism of oxidation of alkanes (aryl derivatives, like toluene) to carboxylic acid by KMnO4. I found that it should be a radical mechanism. The MnO4(-) displaces one hydrogen and makes a free radical, but beyond that I couldn't find anything else. If you could help me I'd really appreciate it. thanx a lot

Nicodem - 14-8-2012 at 23:42

Quote: Originally posted by PeetPb  
hi, I've been searching all over the internet for the mechanism of oxidation of alkanes (aryl derivatives, like toluene) to carboxylic acid by KMnO4. I found that it should be a radical mechanism. The MnO4(-) displaces one hydrogen and makes a free radical, but beyond that I couldn't find anything else. If you could help me I'd really appreciate it. thanx a lot

At least two mechanisms have been demonstrated that depend on the reaction media. The final product can also be determined by the reaction conditions and media. See these studies for details:

DOI: 10.1126/science.7569922
DOI: 10.1021/ja00705a642
DOI: 10.1021/jo971168e
DOI: 10.1021/ja00904a040

For a general review of KMnO4 oxidations and their mechanisms, see: DOI: 10.1016/j.tet.2008.10.038

flux157 - 18-2-2013 at 13:04

I have stumbled on a reaction on which a 1,2-alkyl halide is converted into a ketone by the reaction with aqueous KOH. This is described in Yuki et al, Japanese Patent 69:10,776; Chemical Abstracts 71, 30220e (1969) and also in this links: http://www.erowid.org/archive/rhodium/chemistry/tcboe/chapte... ; http://www.erowid.org/archive/rhodium/chemistry/tcboe/pictur...

Quote:

23.2 g (0.1 mole) of 1-(3,4-methylenedioxyphenyl)-1,2-dichloropropane is refluxed for 10hrs with 75g of 15% KOH (potassium hydroxide) solution. The mixture is cooled extracted with benzene. The benzene is removed in vacuum and the residue distilled to yield approximately 15.2 g of piperonylacetone (bp 149-151°C/10 mmHg). 85% yield from the dichloro compound.


If this is true, what is the mechanism behind it? Shouldn't the reaction of a 1,2-alkyl halide with aqueous KOH give a glycol instead of a ketone? If anyone can clear this out it would be great.

DraconicAcid - 18-2-2013 at 13:20

Quote: Originally posted by flux157  
I have stumbled on a reaction on which a 1,2-alkyl halide is converted into a ketone by the reaction with aqueous KOH. This is described in Yuki et al, Japanese Patent 69:10,776; Chemical Abstracts 71, 30220e (1969) and also in this links: http://www.erowid.org/archive/rhodium/chemistry/tcboe/chapte... ; http://www.erowid.org/archive/rhodium/chemistry/tcboe/pictur...

Quote:

23.2 g (0.1 mole) of 1-(3,4-methylenedioxyphenyl)-1,2-dichloropropane is refluxed for 10hrs with 75g of 15% KOH (potassium hydroxide) solution. The mixture is cooled extracted with benzene. The benzene is removed in vacuum and the residue distilled to yield approximately 15.2 g of piperonylacetone (bp 149-151°C/10 mmHg). 85% yield from the dichloro compound.


If this is true, what is the mechanism behind it? Shouldn't the reaction of a 1,2-alkyl halide with aqueous KOH give a glycol instead of a ketone? If anyone can clear this out it would be great.


It would give a glycol if it were SN2, but with the big aromatic group on the 1 carbon, it will probably eliminate. Or maybe both chlorines get substituted to form the glycol, but then elimination takes place to give 1-(3,4-methylenedioxyphenyl)-1-propen-2-ol, which would then tautomerize to the ketone.

cortex - 4-4-2013 at 14:02

I don’t understand where the Proton in the Benkeneser Reaction(or the Birch-Reduction) with Ephedrine comes from. I found that mechanism below, is it right? Does the proton come from the water added to stopp these reactions ?
In the normal Birch-reduction ist the alkanole acting as the donator but in the papers i found THF oder Et2O is used as a solvent and no alkanole or anything else is used as a co-solvent.
Can somebody explain this to me ?
The mechanism i found:




best regards



[Edited on 4-4-2013 by cortex]

simba - 4-4-2013 at 14:41

Quote: Originally posted by DraconicAcid  
Quote: Originally posted by flux157  
I have stumbled on a reaction on which a 1,2-alkyl halide is converted into a ketone by the reaction with aqueous KOH. This is described in Yuki et al, Japanese Patent 69:10,776; Chemical Abstracts 71, 30220e (1969) and also in this links: http://www.erowid.org/archive/rhodium/chemistry/tcboe/chapte... ; http://www.erowid.org/archive/rhodium/chemistry/tcboe/pictur...

Quote:

23.2 g (0.1 mole) of 1-(3,4-methylenedioxyphenyl)-1,2-dichloropropane is refluxed for 10hrs with 75g of 15% KOH (potassium hydroxide) solution. The mixture is cooled extracted with benzene. The benzene is removed in vacuum and the residue distilled to yield approximately 15.2 g of piperonylacetone (bp 149-151°C/10 mmHg). 85% yield from the dichloro compound.


If this is true, what is the mechanism behind it? Shouldn't the reaction of a 1,2-alkyl halide with aqueous KOH give a glycol instead of a ketone? If anyone can clear this out it would be great.


It would give a glycol if it were SN2, but with the big aromatic group on the 1 carbon, it will probably eliminate. Or maybe both chlorines get substituted to form the glycol, but then elimination takes place to give 1-(3,4-methylenedioxyphenyl)-1-propen-2-ol, which would then tautomerize to the ketone.


How is the glycol supposedly getting eliminated? That doesn't make sense to me. What I think it would happen is an elimination + substitution reaction. The first halogen gets eliminated, and the second gets substituted, giving the hydroxyalkene which is unstable and will tautomerize to the ketone, as you said.

HeYBrO - 30-8-2014 at 03:23

Can some one help me understand or elaborate what is going on in this mechanism? What happens after the radical is formed?
Original paper

Screen Shot 2014-08-30 at 9.21.19 pm.png - 109kB

[Edited on 30-8-2014 by HeYBrO]

CuReUS - 6-10-2014 at 08:48

Quote: Originally posted by cheeseandbaloney  
I recently have started teaching myself MO theory and have slowly but surely begun to understand it after drawing and building myself some molecular orbital diagrams. Like visualizing when electron orbitals are 'out of phase' and areantibonding...etc.
could you tell the name of the book from which you are learning.
Quote:
but the one I have with the question was bought used online and doesn't have the solutions manual. I've looked online but the only link available to download the solutions manual .pdf is broken.[/quote ] which book is it,i will try to find the solution manual:D




[Edited on 6-10-2014 by CuReUS]

JAVA - 9-11-2014 at 09:59

This is the conversion of pseudoephedrine to methyl amphetamine. The intermediate that is formed is a unstable aziridine. Hence the formation of a partial hydride that should attack on the (partial positive) benzylic carbon.

I think it's very difficult to measure this in the lab and verify the mechanism. The intermediates are so unstable that they react in nanoseconds, I suppose.

ScreenHunter_322 Nov. 09 18.44.jpg - 45kB

chemrox - 11-12-2014 at 22:15

As far as books are concerned two were mentioned and a third would be introductory: "Pushing electrons," Daniel Weaks, sort of a first primer on the subject. In addition I found a couple of text book chapters (attached).


Attachment: electronpushing.pdf (316kB)
This file has been downloaded 354 times

Attachment: Pushing Electrons.pdf (9.3MB)
This file has been downloaded 539 times


SecretSquirrel - 24-9-2015 at 08:45

I need help with the mechanism below. It is a Michael type addition of enamine to acetylenedicarboxylate.



proposed_mechanism.jpg - 22kB

My questions are:

1.) Is the first step of the mechanism correct?
2.) If it is, what goes on electron-wise in the brackets, so that we get the end product?


Thank you very much!



[Edited on 24-9-2015 by SecretSquirrel]

proposed_mechanism.jpg - 22kB

byko3y - 25-9-2015 at 05:28

acetylenedicarboxylate attacks enamine (the reaction is also closely related to stork enamine alkylation), then nitrogen becomes positively charged, while another carbon on alkyne group becomes negatively charged, thus the total charge is zero. Although I'm not sure if the last image is a correct configuration of the molecule.

[Edited on 25-9-2015 by byko3y]

SecretSquirrel - 25-9-2015 at 05:54

Are you sure about that? I think it is the other was around, because acetylenedicarboxylate is electron-poor and thus cannot atack the enamine.

I think the last structure is correct... well at least acording to NMR and x-ray analyses.

Nicodem - 25-9-2015 at 15:23

Quote: Originally posted by SecretSquirrel  

1.) Is the first step of the mechanism correct?
2.) If it is, what goes on electron-wise in the brackets, so that we get the end product?

1.) It can be considered correct though it is formally wrong for two reasons. As you have it now, the equation does not hold as you made up an extra hydrogen on the first intermediate. When the enaminone adds on the acetylenedicarboxylate triple bond, it is a nucleophilic conjugate addition, so you need to depict the electron flow all the way to the opposite carbonyl's oxygen (you end up with a protonated amine and a deprotonated enolate, but you can transfer the proton without actually depicting it as a separate mechanistic step - such proton transfers are considered trivial even when they are not).

2.) The intermediate in the brackets firstly equilibrates through proton transfer (from NH2+ to O-) to give a neutral imine form. This then equilibrates via tautomerization to give the more stable enaminone form. This can occur also by direct proton transfer from RCOCH to O-. It is arguable what path is more likely, because this is a complex matter of proton transfer kinetics and is better simply ignored by not depicting proton transfer steps.

SecretSquirrel - 26-9-2015 at 02:48

Thank you!

You mean something like this? Hope I got it right this time.






possible mechanism.jpg - 30kB

[Edited on 26-9-2015 by SecretSquirrel]

Nicodem - 27-9-2015 at 03:47

Yes, almost perfect. Just correct the geometry: the allene group should be linear.

You can further simplify by skipping the intermediate in the brackets. Like I said, proton transfers are considered trivial. Depending on the context, they can be omitted. Sometimes even tautomerization steps can be omitted and the more stable form depicted directly, just like you did in the step toward the imine (enol-ester tautomerization). Of course, things could become too boring as you could get away by depicting only the two reactants and the enaminone product when omitting all the proton transfers and both tautomerizations.

fluorescence - 1-12-2015 at 10:10

Hey Guys,

So for my advanced inorganic practical course I will have to prepare Dicyclohexanephosphine. And my instructor gave me the hint to prepare for a Grignard, the theoretical part will also be about Grignard reactions and that I will have to add the Grignard Compound to PBr3.

Makes sense. I'll probably prepare some Cyclohexanemagnesiumbromide and then make the Dicyclohexanephosphinechloride and reduce that with LiAlH4 to the Phosphine.

The question is how the mechanism works here. I'm pretty sure that I can remember from my lecture on Organometallics that Magnesium and Lithium organometallic compounds tend to form sixmembered ring structures for their reactions.

So I just assumed the following mechanism. The Phosphorus has a positive partial charge due to the more electronegative Bromine atoms and the C-Mg-Br Carbon is slightly negative so I think it could actually work like this.

I didn't care about lone pairs here so you could see everything.

Any suggestions ?



Mechanism.jpg - 26kB

CaptainPike - 17-5-2016 at 17:35

I wish someone would help me to understand this often talked about, base catalyzed reaction between benzaldehyde and nitromethane. I thought I understood that the addition of hydroxide was required to get anything started.

I wanted to perform this preparation and I used methanol as a medium (and possible phase boundary catalyst?) I made a solution of 40 mL MeOH, 20 mL benzaldehyde, and 14 mL nitromethane, and put in the freezer. This sat, sealed in a 100 mL bottle with air in the headspace for several days in my kitchen freezer before I was able to try the procedure.

When I added the AQUEOUS NaOH, drop wise… nothing happened. Everything was really cold so I just continued slowly until the whole, close to 9 g of hydroxide was in. This was my idea of a 200 mmol scale reaction.

This wasn't my first time doing this and I expected to see "a bulky white precipitate", start to appear immediately. Nothing ever precipitated. I can smell the benzaldehyde – it is clearly not gone anywhere.

The chilled down unholy trinity was pretty clear, not cloudy and definitely single phased and liquid when it went in to the reaction vessel, immersed in a salt/ice bath to be stirred. Could the methanol have somehow decomposed the nitromethane while waiting those few days in the freezer?

I did have someone else prepare the aqueous sodium hydroxide – it acted just like I was dripping distilled water into the benzaldehyde, nitromethane solution… Maybe there was a misunderstanding or come to think of it, discipline's GETTING PRETTY LAX AROUND HERE! I'll be seeing her later on tonight…

Crowfjord - 17-5-2016 at 17:58

Little to nothing should have reacted until base was added. Hydroxide deprotonates the nitroalkane to the nitronate, which adds to the aldehyde and forms a carbon-carbon bond. The negative charge is now on the oxygen.

I'm pretty sure that the nitro-alcoholate stays in solution. Slowly stir the solution into a mixture of concentrated HCl and ice. This should result in formation of the yellow nitrostyrene, which can be filtered off.

CaptainPike - 18-5-2016 at 18:26

will try this, thank you

then later (the next day):
there definitely precipitated a reddish, syrupy substance after addition to the icy HCl. Decanting the supernatant, this substance was insoluble in mildly hot pet ether. Some yellowish impurity (?) did migrate, coloring this solvent – I left the rest cool slowly… it doesn't look like anything will be crystallizing, it still looks tarry on the bottom of the small beaker.

There is that orange peel smell again, however, which became apparent immediately on addition to the acid.

[Edited on 19-5-2016 by CaptainPike]

sulfuric acid is the king - 26-2-2017 at 08:22

Can this work?

H.jpg - 43kB

myristicinaldehyde - 26-2-2017 at 08:32

The 2nd step would tautomerize to the imine- the acidic solution would then hydrolyze it to 2-butanone.

sulfuric acid is the king - 26-2-2017 at 11:51

Thanks!
But is it somehow theoreticaly possible to obtain butylamine from nitrobutene by Bechamp reduction?So,after first step?Some mix of butenamine and butylamine?

[Edited on 26-2-2017 by sulfuric acid is the king]

sulfuric acid is the king - 2-3-2017 at 14:33

BTW how can "Lithium aluminium hydride" make third,single bond molecule,but Fe/HCl can't?

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