[kriggy]

Hi there the question is to write the reaction mechanism for this reaction

I was thinking that after the nucleophilic addition the intermediate is geminal amino alcohol and then the OH gets protonated by acid and leaves forming carbocation that is compensated by electron pair of the amine which then loses a proton so E1 mechanism take place.
After quick google search I found out that the OH is protonated by the hydrogen atom from NH group. It seems weird that the acid catalyst doesnt participate in the last step. Could anyone comment on that or explain why the acid catalyst doesnt participate in the protonation of aminoalcohol?
THanks

[orgopete]

I agree with intermolecular mechanism. An intramolecular mechanism can be written with fewer steps, but I have been skeptical of these four-membered ring proton transfers. I have searched for bona fide proton transfer reactions in four-membered rings, but the smallest ring I can find is five. I think electron-electron repulsion should make these intramolecular proton transfers unlikely. Can anyone give an example proving a proton transfer in a four-membered ring?

[Dan]

I have been skeptical of these four-membered ring proton transfers. I have searched for bona fide proton transfer reactions in four-membered rings, but the smallest ring I can find is five.

There could be an intermolecular proton transfer from either of the two HNC(=S)NH, which would be either 5- or 7-membered ring transition states. That's just an observation, I don't know what the "correct" answer is.

[Vidya]

Hi there the question is to write the reaction mechanism for this reaction

I was thinking that after the nucleophilic addition the intermediate is geminal amino alcohol and then the OH gets protonated by acid and leaves forming carbocation that is compensated by electron pair of the amine which then loses a proton so E1 mechanism take place.
After quick google search I found out that the OH is protonated by the hydrogen atom from NH group. It seems weird that the acid catalyst doesnt participate in the last step. Could anyone comment on that or explain why the acid catalyst doesnt participate in the protonation of aminoalcohol?
THanks

The catalyst acid is added only in small amount which is used up only in first step and now H on the N with the positive charge is again very acidic and is strongly picked up by lone pairs of the O of the OH group

[orgopete]

I have been skeptical of these four-membered ring proton transfers. I have searched for bona fide proton transfer reactions in four-membered rings, but the smallest ring I can find is five.

There could be an intermolecular proton transfer from either of the two HNC(=S)NH, which would be either 5- or 7-membered ring transition states. That's just an observation, I don't know what the "correct" answer is.

This shows how writing some steps in a mechanism can be difficult. Since the adduct has an O(-) and NH2(+), one must surmise whether a proton transfer can occur to give an NH(=S)N(-) intramolecularly at low pH before an intermolecular reaction takes place? Another option is to draw a concerted transfer with a water molecule so it's oxygen can simultaneously protonate the oxygen and deprotonate the nitrogen. The cost in this case would be entropic. Not only does the semithiocarbizide attack the carbonyl group, but another water must align itself in a six-member end ring to make the transfer.

Besides trying to balance concerted v noncencerted, 4 v 6, or 4 v 5(7), the optimal pH of the reaction for a rate limiting step, one must protonate a less basic ROH (in order to generate H2O as a leaving group) in the presence of a more basic nitrogen. This can be a general problem for all nitrogen nucleophiles that form water under acidic conditions (enamine, oxime, hydrazine, etc.). These mechanisms are the most difficult to confidently write. To that end, I agree with the question being asked by the original poster.

[orgopete]

The catalyst acid is added only in small amount which is used up only in first step and now H on the N with the positive charge is again very acidic and is strongly picked up by lone pairs of the O of the OH group

I agree with the pKa differential just noted. I'm referring to a different aspect, the O-H-N angle. I suggest we might gain some insight from hydrogen bonds or other proton transfer reactions. In water, the O-H-O bond angle is ~180^o. In an acetate elimination or beta-keto acid decarboxylation, the bond angle is ~120^o. In a Cope elimination, the C-H-O bond angle is ~108^o.

The difficulty I see is that if the electron pairs take up a lot of space, then as that bond angle becomes more and more acute, the electrons would overlap. I perceive this as a much higher energy state, hence my interest in finding an example proving it occurs this way. With protic solvents, one can simply write intermolecular proton transfers (presumably with ~180^o angles). The downside is that one must write an additional steps in the mechanism.

[Vidya]

but there are many reactions in which H+ is transferred intramolecularly and energetically they all are favored