[azmanam]

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.

[azmanam]

Also, if water is catalytic

water is catalytic... geez.  I meant if acid is catalytic... (smacks self on head)

[Heory]

 

[movies]

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)

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.

[Heory]

 

[philmont702a]

last post.

[Heory]

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?

[Heory]

 
So many posts in the thread! Nice to see many people took part in the discussion, hehe.

[movies]

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.

[Heory]

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?

[movies]

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:


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. 

[Heory]

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?

[Heory]

 
It seems that we will lengthen the thread to 100 posts

[movies]

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; 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?

[Heory]

  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?