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Chemistry Forums for Students => Organic Chemistry Forum => Organic Chemistry Forum for Graduate Students and Professionals => Topic started by: Heory on November 05, 2009, 08:54:50 PM
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I feel that this board is so desolate. I'd like to share an interesting reaction copied from my textbook as another problem for this week. Does anybody want to have a try?
Provide a complete arrow pushing mechanism for the following reaction. Predict product.
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Is PH=3 triphenylphosphine?
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Is PH=3 triphenylphosphine?
NO. H3O+
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Oh so the pH is supposed to be a pH of 3 with the HBr or before hand?
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H3O+ and PH=3
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I don't get what you are saying...
Please be more descriptive...
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Sorry I'm not good at English so I cann't decribe it very well by words.
Well, "1)H2SO4, H2O,PH=3; 2)48%HBr, H2O", understand?
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It is not clear whether PH=3 means pH=3 or is some kind of pseudostructural symbol describing presence of reagent.
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I didn't think that my question would be so obscure.
PH is hydrogen ion concentration
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this??
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NO
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This ????
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The amino attact the epoxide ?then the HBr generate carboncation?and the C=C make an addition?
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I need some hint!!!!
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Now This structure in the first step??
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No. Please deliberate on the problem before posting you answer. Don't just guess.
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Well,but i have no patient……guss it is easier than to think it in my hard mind
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So I see what looks like a donor/acceptor epoxide that looks like a latent dipole and an alkene waiting to be a dipolarophile. I suspect you'll get a 3+2 dipolar cycloaddition. Followed by lactam formation to give a stable trans 6/5 fused ring system? I chose the regioselectivity of the double bond because the more substituted side is "more carbocation like" and I predicted that side to interact with the "more carbanion/enolate like" side of the epoxide. I chose the facial selectivity of the double bond to put the methyl group away from the aryl group on the one side and the alkylamine group away from the ester on the other.
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Very novel idea, azmanam! Is this kind of 1,3-dipolar cycloaddition involving an epoxide supported by any document?
I quite agree with the resonance structure(very good idea! ), though actually the reaction didn't take place in that way. My frist hint will be that we shall consider why the ester group was designed next to the epoxy group.The ester group is not just an EWG.
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sure, you can use epoxides.
http://dx.doi.org/10.1002/anie.197705721
You can also add a carbene to an aldehyde to get the same 1,3-dipole
http://dx.doi.org/10.1021/ja00382a039
But that's really not the pathway? huh. I was convinced
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what a pity i have no access to the data base. The reaction didn't react in that way, mabe because of the different reaction conditions.
Dood night! See your new idea tomorrow.
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ok, try number two is initial formation of the amide followed by acid promoted epoxide opening to give a 7-membered ring.
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Until Heory wakes back up, perhaps we can discuss this one amongst ourselves? I don't really like my second guess, because it just doesn't 'look right.' ya know? I don't see any driving reason why someone would want to make a 7-membered lactamether with a benzyllic bromide. I suspect amide formation will happen first under strongly acidic conditions, but from there, I don't know what direction to turn. Is that the end of the aqueous acid step? Does the epoxide and alkene really survive the acidic conditions? What else might we do after amide formation?
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Well, I'm thinking your second proposal is wrong based on his hint alone. Still looking a little deeper to see if I can point you somewhere else. Idk, this doesn't seem too reasonable, but what about intramolecular lactone formation (4-membered ring) followed by amide formation to get the B-lactam? It would at least be something someone might aim to make.
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I was just drawing out a b-lactam forming process, too! mine goes through a bit different pathway than what you're describing, I think. That's only step 1, though. I don't know what to do w/ step 2. Just addition of HX to an alkene?
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You did go through a different mechanism, but yeah, same end product. I wasn't really sure where to take it from there. Just brominating the alkene definitely seems too simple, but carbocation formation doesn't really leave attack open to anything except that bromide.
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Some thing may be wrong i'm confusing about this question
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nj_bartel, good idea. Beta-lactam is indeed the necessary part of antibiotics thus bound to be aimed to make by someone. But the structure you desire for wasn't formed.
azmanam, there's another path for step 1. BTW, I like your arrow pushing in the formation of the beta-lactam. Every time I draw a nucleophlic replacement of a ester by a nucleophlic reagent, I have to divide it into 2 steps. You used the most concise pattern.
Ulfsaar, good idea. But the first step is not pinacol rearrangement and I don't understand the following rearrangement. Would you please be more descriptive?
Since we were all off the track, I'll give the second hint:
Remember the reaction was performed in aqueous solution. Bonus: name the reaction in the first step.
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HINT 3:
Which named reaction can be used to achieve the transformation RCHO--->RCH2CHO?
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Dazon reaction? a good idea!!
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Right!
Go on, Ulfsaar. I believe that you can solve this problem.
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ammmm
i just think a [3,3] rearrangement ……
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ammmm
i just think a [3,3] rearrangement ……
imine is not stable under acidic condition.
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Had to break out my new "Strategic Applications"[1] book for this one. Darzens followed by Eshweiler-Clark Cyclization[2].
[1] http://www.chemistry-blog.com/2009/10/12/book-review-strategic-applications-of-named-reactions-in-organic-synthesis/
[2] http://dx.doi.org/10.1021/jo01018a011
http://dx.doi.org/10.1021/jo01348a006
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Closer now. You meant that imine was destroyed after its formation. But when pH=3, imine cann't be formed, or its formation was overwhelmed by another named reaction. You had a right idea, but wrong way. Look for another way for the following process. Bonus: name it.(not Eshweiler-Clark)
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Come on! Just have a try. You needn't post your answer only when you are quite certain that you are right.
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Something like this?
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Something like this?
No. actually imine was not the intermidate of the reaction. After Darzen reaction, there's another named reaction, which coundn't be seem from your picture.
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about azmanam’s latest answer:
The right idea was the nuleophilic attack on the aldehyde's carbonyl carbon, but the product was far from the right one. An important functional group was ignored in the reaction immediately after Darzen reaction.
we still have a long way to go to finish step 1 (pH=3), let alone come to the end of the whole reaction, and stereochemistry should be considerd. We have just completed about 1/5 of the whole work.
BTW, 48%HBr was just used to increase acidity of the reaction system. Br- didn't react.
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so something happens with the hemiaminal? some intervention of the p-methoxyaryl group? Surely CO2 doesn't come back and react?
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after Darzen reaction, only 3 functional groups existed and you always chose the wrong 2 groups of them to react at first. Choose another 2 groups to react first(a named reaction).
Sorry for my poor Enlish. If I'm not descriptive enough, please let me know.
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Ok. Here we go:
At the point where we have the amine and the aldehyde (those should be the two dominant structures after the Darzens rxn)
Possible Nucleophiles (in what I feel are order of strength):
N atom of amine
O atom of water
benzyllic carbon of aldehyde due to enol character
the carbon atoms ortho and para to the OMe group on the benzene ring
distal carbon of alkene in amine (the monosubstitued side)
O atom of aldehyde
Possible Electrophiles (also in order of strength, imho):
H of H3O+
carbonyl carbon of aldehyde
internal carbon atom of alkene of amine (the disubstituted side)
the carbon bearing the OMe(H+) group for a possible ipso substitution
One of those from each list have to react, right? Is it the alkene and the aldehyde?
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Yes! Name it and go on.
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Prins followed by hydride shift followed by 5-exo cyclization
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yes, prins.
the hydride didn't shift in presence of WATER (large amount). BTW, why didn't you choose the methyl group to shift? In a former problem, when I treated the pinacol reaction with a hydide shift, you said only the alkyl group could meet the orbital overlap requirement.
after prins reaction, N was the right nuleophile but the epoxy group was a wrong electrophile.
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For the prinse reaction, was there any stereochemistry ?
Use Newman projection.
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methyl group wont shift because you'd sacrifice a tertiary carbocation to form a secondary carbocation. hydride shifts so that you trade a tertiary carbocation for a tertiary carbocation. in the previous post, the shift was restricted by cyclic constraint (the carbocation and shifting groups were within a 6-membered ring), and you need overlap of the sigma bond with the empty p-orbital to get shift to occur (same with last week's C-Si bond problem). In this case, free rotation about the C-C single bond allows for both the methyl group or the hydride to have good overlap for shifting.
The reason I didn't have N attack straight away was the disfavored formation of a 4-membered ring, and I liked that I could make a 3- or a 5-membered ring after shifting the carbocation. There are typically 3(/4) things that happen after a prins. Attack by water (with elimination to allylic alcohol sometimes), attack by a nucleophile (like N), or coordination of another aldehyde to form a dioxane.
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A OH was introduced and the other OH was kicked away by the amine group.
Consider a larger ring that would be formed.
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If the hydride shifted, a carbocation would be formed next to the OH group, then what would occur must be pinacol reaction, not the epoxide formation. Am I right?
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piperidine formation?
What text did you see this in? Did they give a reference?
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Could you please provide the citation for this reaction? The reactivity is surprising under the conditions, assuming that the piperidine is formed.
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The product for step 1 is right. There's another named reaction for step 2. Because we haven't finish the whole reaction yet, I think it's not proper to tell where I copied it from.
we shall consider the stereochemistry of step 1 now, or have a discussion on what you think are unreasonalbe.
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What are you defining as step one above? Is the piperidine the end of step one, or the prins the end of step one?
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philmont702a, are we in the same region
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step 1 : 1) pH=3
step 2: 2)48%HBr
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Yes, I understand the steps. Are you defining the piperidine as the completion of step one, or the prins product as the end of step one?
No, we are not in the same region unless you're in the US.
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the red structure of azmanam's picture is the right product of step 1 (pH=3).
We have the same local time, so I thought we were.
In my text book, only 2 products of each step were given, full mechanism including stereochemistry was not given. I think the stereoselectivity does exist. I will post my own idea later, we can have a disscussion.
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Step 2 includes two major reactions, which have little relationship with each other, and one of which is a named reaction. Have to go to bed, see you tomorrow.
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From the piperidine structure in red, the HBr is presumed to react with the alcohol and likely to form an electrophilic carbocation or eliminate to form the tetra-substituted alkene. From there, there are a number of options. What do you think?
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re philmont702a:
path 1: need very high temperature to surmount the high activation barrier, thus not favoured. And the resulted amine will be destroyed by hydolysis, won't it?
path2: the most reasonable of the three. But the much stained 5-memebered ring was formed through an very sterically hindered way, thus overwhelmed by another path.
path3: 7-membered ring's is not very likely to be formed.
Hint: F-C rxn is right. Under the very acidic condition, oxygen atom of ArOMe was protonated, therefore...Go on
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Go on, azmanam. Certainly you can finish the problem.
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Actually neither product of philmont702a' path 2 is the framework of opiod-type molecules. Check it carefully.
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Actually, it looks a lot like the molecular skeleton of part of morphine. It's not morphine and it's not likely going to be taken on to morphine, but it is the same carbon skeleton.
To do a FC, you need a carbocation, I don't like this pathway much, but here's my guess.
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azmanam, your product is so strained that I think its formation is impossible. Sorry I made a mistake, the first product of philmont702a's path 2 is is the framework of opiod-type molecules, and maybe that way proposed by him was right, which was different from my own idea. I don't have the standard anwswer for the problem, we can have a disscussion. But both the products are certain. My only reference is my text book Medicinal Chemistry, so I had to take a photo. I copied it from the synthesis of pentazocine, which is of the opiod-type.Maybe we are too boring with this problem now, so I post it.
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The following is my own idea, I'm not quite sure whether it's right. For F-C rxn, 6-membered ring is most easily to be formed.
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??? ??? ??? ??? ???
That is so strange^ I can't understand why does not the amino group make addition to the carbonly group instead of the Prins reaction?In a special words that is 囧
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Your product in red, and my molecule in blue (showing philmont's pdt's similarity to morphine) are the same except for the loss of the methyl group from the aryl ring.
OMe (or OH in your case) is a srong o,p director. 6 membered ring from straightaway FC rxn at the meta position is unlikely. I like the p-directed FC followed by alkyl shift better.
I'm at home w/o chemdraw, but we can talk about stereochem when I get to work. I don't think you've drawn the newman projection of the aldehyde piece correctly. You have the aryl group directly attached to the aldehyde carbon, but there is a methylene in between. The methylene makes the aryl chain much smaller than it looks and I suspect will not impart much selectivity on the prins step. I suspect the stereodivergent step will be at the FC stage.
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Yes, in my newman projection, the right group is Bn not Ar, but still Bn is the largest group that binds the aldehyde's carbonyl carbon.
Of course I agree that OH is a srong o,p director, but if the cabocation attacked p-position, strain of TS of the 5-membered ring would be quite large, or even the TS couldn't be organised, I think. OH is a srong o,p director, doesn't mean that activity of m-position is decreased by OH. Actually OH activated all position of the benzene ring, but o,p is more reactive than m.
Selectivity on the Prins step (maybe not very favorable) decides that on the F-C step.
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ortho/para directing: The OH is definitely an ortho/para director. Thus, it will react preferentially at those positions FIRST, then if put under enough duress may react further.
5-membered ring: The rates for 5 membered rings are faster than 6, which are faster than 7, etc. This goes for a multitude of reactions from RCM to Dieckmann condensations.
Stereochem: Though I can see from your newman projection why you might think that stereochemistry might be imparted on this reaction, that same newman projection can be rotated and the strain would not be significantly different. If any selectivity is observed, it is minimal, and definitely diastereoselective, not enantioselective. The stereochemistry you are predicting for the product is more probable to come from the cyclization step of the friedel-crafts reaction. However, there appears to be no significant director for the stereochemistry of that step.
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;)bad network.
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Here are the relevant references. One is in German, and we don't have access to it anyway (though I think it is the key reference we need!) and the other doesn't discuss the mechanism in any detail, only to say the product is one with the new alkyl group meta to the OH.
http://dx.doi.org/10.1002/ange.19470590703
http://dx.doi.org/10.1002/jhet.5570060108
I disagree that the 5-membered spirocycle is disfavored. I made a model of it and it fits together quite nicely. Chem3D shows the piperidine in a nice chair, the 5-membered ring in a nice envelope and - in the lowest energy conformer - the correct C-C bond (the one from aryl to tertiary carbon) to be situated nicely to migrate.
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;)
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I'm going to stir up the mud again. Going back to the aldehyde and amine (the product of the Darzen's reaction), you can get to the product through imine formation. If I would have drawn the immediate product of the Eshweiler-Clark cyclization correctly, we would have[1]. I really believe the lone pair on nitrogen is more nucleophilic than the pi bond of the alkene in this first step. I believe imine formation happens followed by 6-endo-trig cyclization (favored under Baldwin's rules. This transition state is much more ordered. That gets us to the same carbocation which can be trapped by water to get us to the same piperidine/alcohol for the HBr step.
And, no, I really truly don't believe the meta position will attack anything. The ortho and para positions are much, much more nucleophilic. All it takes is a chair flip and rotation of the methylene linker to put the aryl group in close proximity to the carbocation without much steric interference. Build a model, it fits quite nicely. At that point, analogous to Bayer-Villiger type mechanisms, the more electron rich carbon atom (the tertiary one) will migrate. It also happens to be nicely oriented to provide maximum overlap of the C-C sigma bond and the pi* orbital for migration.
[1] http://www.chemicalforums.com/index.php?topic=37220.msg142811#msg142811
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??? ??? ??? ??? ???
That is so strange^ I can't understand why does not the amino group make addition to the carbonly group instead of the Prins reaction?In a special words that is 囧
I agree. Why not like this? Iminium ion cyclizations are certainly well precedented!
EDIT: Looks like azmanam beat me to it!
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I agree. Why not like this? Iminium ion cyclizations are certainly well precedented!
Whilst iminium ion cyclisations are precedented, is there not a danger that the pH is so low that there is not really any significant RNH2 around, rather RNH3+, and so not nucleophilic enough?
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I've been mulling that as well. The pH for the first step is set very specifically to pH=3. With a relatively high pH for acidic conditions, the equilibrium between protonated ammonium and free amine is faster than, say, if the reaction was run at pH=-7. Also, once imine forms, even if the equilibrium lies to the left, the irreversible intramolecular cyclization will LeChatelier the reaction toward the piperidine. Also, if water is catalytic, then only a catalytic amount of nitrogen will be protonated. That would be in competition with protonated aldehyde. I don't know if acid is catalytic or not, but something else to consider.
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Also, if water is catalytic
water is catalytic... geez. I meant if acid is catalytic... (smacks self on head)
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:)
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I really disagree that the iminium would not form under these conditions. The combination of amine + aldehyde + acid is common to a number of transformations that go through iminium intermediates such as the Mannich reaction, Pictet–Spengler cyclization, Leuckart–Wallach reaction, and the Eschweiler–Clarke methylation.
Clearly from the numerous reactions with iminium ions, the iminium is accessible under acidic aqueous conditions, even if it is not necessarily the dominant species in the pot. It is really all about relative rates (Curtin–Hammett control of the reaction). As azmanam said, if we were at pH = –7, then the amine would probably be completely protonated, but at pH 3 the equilibrium should still be rapid and free amine should be present.
I would call what I drew an aza-Prins cyclization, not an Escweiler–Clarke cyclization (I have never heard that before, only the E–C methylation). Certainly, Matt Shair called a similar transformation in the synthesis of cortistatin an aza-Prins cyclization. (see here (http://pubs.acs.org/doi/abs/10.1021/ja8071918))
You wouldn't expect any enantioselectivity in the cyclization since you are using achiral reagents (any stereochemistry from the epoxide has already be destroyed). However, the cyclic transition state in the aza-Prins cyclization would likely give you a good chance for decent selectivity since one transition state would have two groups in pseudo-equatorial positions while the other would place them pseudo-axial.
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:)
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last post.
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I browsed several related books but only found arrow pushing of step 1 (pH=3) in a praxis book. It's Prins, not aza-prins and no stereoselectitity was proposed in that book. I agree with your point 3. I expected the selectivity because I really didn't want the synthesis so terriable, and I believed at least there's a slight selectivity .
Actually I really don't agree with point 1 and 2. Since every body of us is convinced by his own idea, and this question has been "beaten to death" by philmont702a, I'am not going to make further persuasion. OK, let's start a new problem.
BTW philmont702a, did you see the message I sent to you?
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:D
So many posts in the thread! Nice to see many people took part in the discussion, hehe.
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Heory, are you familiar with the Curtin–Hammett principle? You should really read up on it if you are not. What philmont posted in point 2 is completely consistent with this principle.
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No, I didn't know the name of the principle, but maybe I knew the principle itself. Does it means that the E-C or Aza-prins is a driving force (because of intramolecular process?) to draw the equilibration to the reaction you desired for?
What we have different ideas on was not the principle, but which one of the prins and the aza-prins is a better driving force. I choose the former with the help of imine's hydolysis and a more electrophilic carbon, and you choose the latter. At least you would agree that prins is another path to the product, wouldn't you?
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No, it is a general principle of chemical kinetics. The idea is based on the fact that the rate of any reaction is a function of not only the concentration of the reactive intermediate, but also the rate constant for the subsequent transformation. Suppose we have a reaction of a compound A to product C with the rate law rate1 = k1 * [A] and then we have another pathway that leads to a different product (call it D) by the similar rate law rate2 = k2 * [ B]. Next, suppose that A and B are in rapid equilibrium (much faster than either of the forward reactions to products) with one another and that equillibrium rather heavily favors B. Since the rates of the two reactions depend on both the concentrations and the magnitude of k1 and k2, there is a situation where even though B is disfavored in the equilibrium, the pathway through intermediate A in fact dominates and C is the major product, not D. This would occur when k1 >> k2.
Here is a handy reaction coordinate diagram I lifted from Wikipedia:
(https://www.chemicalforums.com/proxy.php?request=http%3A%2F%2Fupload.wikimedia.org%2Fwikipedia%2Fcommons%2F9%2F98%2FCurtin-Hammett_principle.png&hash=d1f876baeec0875dfdf4f283b759c86ec6f7e229)
The way this is usally summed up for the organic chemist is that the major product of the reaction can come from the minor component of an equilibrium mixture provided that the minor component is much more reactive than the major component. The product distribution is controlled by the ∆∆G value in the picture (the difference in the two transition state energies), not by the underlying equilibrium.
This is especially important in the cases of imines and iminium ions. I know that you are thinking that the protonated aldehyde is more reactive anyway, but that is not the whole story because you have to consider the other components in the reaction, especially the fact that it is an intermolecular reaction, and that there are other competitive pathways such as addition of the N to the aldehyde. Additionally, in this case you also need to consider that the imine can tautomerize to a conjugated enamine that would be quite stable.
A particular example with iminiums that I alluded to before is the Mannich reaction. If the Mannich reaction were controlled by the equillibrium instead of the relative rates of reaction of the two intermediates, then we would never see the Mannich reaction, just aldol reaction. A famous example of this from total synthesis would be Robinson's route to tropinone. Perhaps a more pertinent example would be Heathcock's syntheses of the daphniphyllum alkaloids through an olefin/iminium ion cyclization cascade.
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Thanks for your fully interpretation!
Here A is the protonated imine and B is the protonated aldehyde, C is the aza-prins product and D is the prins product, am I right?
The product distribution is controlled by the ∆∆G value in the picture (the difference in the two transition state energies).
∆∆G=∆G2-∆G1-∆G, if ∆∆G>0, product C is favoured and if ∆∆G<0, product D is favoured. I think ∆G's magnititude is very large(∆G=RTlnK, K=[B.]/[A.], [B.]>>[A.]), and ∆G1's magnititude is not smaller than ∆G2 (prins vs. aza prins)thus ∆∆G<0, D (prins product) is favored.
imine can tautomerize to a conjugated enamine that would be quite stable, but which is neither the reactant of aza-prins nor easily tautomerize back to imine due to its stability you claimed.
I haven't done any experiments on Mannich reaction, is it performed at pH<=3 and in aqueous solution?
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:D
It seems that we will lengthen the thread to 100 posts
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Not quite. ;)
In this depiction, A would be the imine and B would be the aldehyde. Remember that the protonated forms are activated and therefore higher in energy.
The van't Hoff equation you used applies to the equilibrium only (∆G in the figure). It is slightly more complicated than what you calculated because you didn't account for everything in the rate law. Also, the magnitude of ∆G cannot be very large or else the two species would not be in rapid equilibrium with one another. Anyway, it doesn't really tell us anything since we already said the equilibrium is shifted heavily in one direction so there must already be the gap in ∆G as shown in the figure.
Note that the transition state energied, ∆G1(double-dagger) and ∆G2(double-dagger) cannot be determined with the van't Hoff equation! What you wrote is not consistent with the diagram posted above (you reversed the magnitudes of the TS energies), so it is not the same discussion. You are also ignoring the fact that one reaction is bimolecular while the other is unimolecular. In general, intramolecular reactions are much faster than polymolecular reactions.
Please refresh yourself on the definition of tautomers (http://"http://en.wikipedia.org/wiki/Tautomer"); two tautomers must be in rapid equilibrium. Enamine/imine tautomerism would set up another Curtin–Hammett situation where a less prevalent, but more reactive, intermediate undergoes further reaction.
Since you don't know the Mannich reaction, let's use a simpler example. Consider the acid-catalyzed aldol reaction. This transformation involves the intermediacy of an enol, which is a rather unstable tautomer of a ketone. The equilibrium between ketone and enol is shifted far toward the ketone side, but the reaction still occurs through the reactive enol. Do you see how that applies here?
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;D Thank you Mr. Movies! Maybe I was not quite understood when using English. I've read your words and I'll consider it carefully.
"Also, the magnitude of ∆G cannot be very large or else the two species would not be in rapid equilibrium with one another." Could you please explain why, for I cannot see there is any relationship between ∆G and reaction rate. BTW, will you help on my Problems on Mechanism?
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No problem - I just want to help you understand the concepts!
For the magnitude of ∆G, if it is too large a difference then interconversion between the two would require a lot of energy, so one of the species would not be readily accessible. You would have to really exaggerate the drawing above for this to be the case, but the energy difference has to be within an reasonable range.
For the mechanism problems, do you mean the other thread you have running on this subforum? I glanced at those, but I will take a closer look if I have a few minutes this afternoon.
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I made a video to help understand the definition of tautomerization.
http://www.youtube.com/watch?v=rH43Izm3nfQ