[harkkam]

Guys suppose you have an primary amide which is attacked by a -OH/H₂O

   O
   !!
R-C-NH2       +    -OH/H20 ------->

1. First of all I believe that the -OH is more basic than the Leaving group NH2 since the NH2 does not have a negative charge, this drives the reaction toward the weaker base.

2. Why does the nitrogen of amide which becomes the amine. then remove a proton from the OH. Why doesn't the nitrogen attack the water solvent for that proton.

   O                                                 O
   !!                                                 !!
R-C-OH     +  :NH2          ------- >   R-C-O^-    + NH3

The nitrogen attacks the Oxygen of the Intermediate carboxylic Acid instead of the water solvent because the Carboxylic acid is more acidic due to resonance stabalization.

Are these correct assumptions to explain why we get a carboxlate salt. thank you

[StarvinMarvin]

When discussing such reactions you'd have to ask a following question:

which is the most stable leaving group? OH- or NH2?

Of course, OH-. Therefore, if the reaction would go as yuo drew it, it should immediately get back to the starting material. For the reaction to work, NH2 must leave the structure as a stable leaving group (either NH4+ or gaseous NH3). For this to happen, when OH- would attack the carboxylic carbon atom, NH2 would need to take its proton in order to leave as a thermodynamically favourable leaving group. What is more, proton removal is even more favoured as the carboxylate is then stabilised by charge delocalization. That is why the proton cannot come from the solvent. By the way, NH2 would constitute a powerful base which would then more likely attack the carboxylic acid instead of water (which is only relatively acidic).

[harkkam]

Right I meant NH3 as the leaving group, NH2 has two unbonded orbitals which is very basic and stronger than -OH.

The NH2 would have to be protonated first to +NH3 on the carboxy acid which would leave after -OH attacked.

So basically :NH3 would be a better leaving group than -OH since it has no charge

[StarvinMarvin]

NH3 is a better leaving group because once it cleaves, it leaves the system (because it is gaseous). Moreover, once the proton exchange takes place, OH- no longer wants to become a leaving group as it transformed into carboxylate stabilized by resonance.

[azmanam]

careful here.  you're confusing acid and basic mechanisms.

under basic conditions, OH- would be a nucleophile and NH2- would have to be the leaving group (no way to get C-NH3+) before nitrogen leaves.  The competition between leaving groups is OH- and NH2-.

under acidic conditions, H2O is the nucleophile... then proton transfer is possible and NH3 is the leaving group.  The competition between leaving groups is H2O and NH3.

[StarvinMarvin]

No, NH3+ cannot even be formed. NH3 (neutral) is formed by immediate deprotonation of carboxylic acid. Right?

[nj_bartel]

I think his problem lies in that you said ammonia was the leaving group, when it's actually NH₂^-.

NH3 is a better leaving group because once it cleaves, it leaves the system (because it is gaseous).

I think you sort of argued the point he was making in your last post though.

[StarvinMarvin]

I gave the thing some thought and found a solution to the question. As it turns out, the mechanism INVOLVES participation of water. Then again, nothing of the sort of NH2- is cleaved because it would be simply unfavourable. Initially, I thought the NH2 deprotonates OH group that attacked the amide in the first place, but instead it deprotonates water to become able to cleave as a neutral entity.

I attached a mechanism of this reaction taken from Eur. J. Org. Chem.

exact reference is: D. Zahn; Eur. J. Org. Chem.; 2004; 4020-4023

[nj_bartel]

Hm.  How would they provide evidence for that mechanism?  I don't see a way to do it isotopically. 

I always had the mechanism explained to me as this:

NH₂^- is indeed an awful, awful LG, but none the less, it still does happen to a very slight extent.  The tiny amount of NH₂^- that acts a LG immediately deprotonates the carboxylic acid formed forming ammonia, which, being gaseous, leaves the reaction.  The product ammonia leaving drives the reaction forward until there's essentially no amide left.

[StarvinMarvin]

According to the paper, the mechanism was given in accordance with Car-Parinello Molecular Dynamics simulation. They also referenced some works involving other ab initio computations which prove their findings.

Getting back to the NH2- business:

don't you think that once NH2- is cleaved (however unlikely that is) would it be prone (as a powerful base) to attack the carboxylic carbon atom from which the electron density is being withdrawn by two oxygen atoms instead of deprotonating it?

[azmanam]

my point was you need to be careful which intermediates/LG you discuss because they change depending on whether the reaction is under acidic or basic conditions. 

for clarification C-NH3+ refers to the tetrahedral amide intermediate with sufficient proton transfers to give nitrogen - still attached to the tetrahedral amide intermediate - with 3 protons (thus nitrogen has a positive charge).  sorry if I was confusing.

I've drawn the reaction under basic, neutral, and acidic conditions.  Under basic conditions, there's no way to get NH3 as a leaving group.  You'd have to either have way too many charges on the intermediate, or even if everything else was protonated, you'd have a positively charged intermediate in a basic mechanism.  This won't happen.  The base will remove that proton on the NH3 grop way faster than NH3 will leave as a leaving group.

The question was concerning hydrolysis under basic conditions.  I recognize the reference you've provided... but I'm very uncomfortable proposing RNH3+ as an intermediate in a mechanism under basic conditions...  The only reason it works here is the protonation states of the rest of the molecule result in the molecule having an overall net negative or net neutral charge.  For undergrads just starting out, this may be a bit confusing.  I would caution that some might try to have a molecule with a net positive charge... This is the species I contend will not form under basic conditions.  Even still, I contend that deprotonation of the RNH3+ is MUCH more favorable than the ROH in that intermediate (what I guess they're calling B2).  The pKa of RNH3+ is MUCH lower than ROH and the RNH3+ should be deprotonated MUCH faster, imho.

Most undergrad text would probably draw the mechanism as I've drawn it below.  op should check with the instructor for what the correct mechanism will be in an exam situation.

You might also be interested in this discussion on whether reaction mechanisms can even be proven: http://www.chemistry-blog.com/2009/05/13/survivor-mechanisms/

[nj_bartel]

The knowledge level is starting to jump way above my own, so I'm going to reply to selected parts.

don't you think that once NH2- is cleaved (however unlikely that is) would it be prone (as a powerful base) to attack the carboxylic carbon atom from which the electron density is being withdrawn by two oxygen atoms instead of deprotonating it?

Sure, there's a probability of that happening, but it seems to me like it would certainly be more likely for NH2- to deprotonate a carboxylic acid rather than act as a nucleophile on it.  At any rate, if it were to act as a nucleophile, we'd just be back at stage 1 and there would be another opportunity for it to act as a LG and be given another chance to irreversibly deprotonate the acid.


Azmanam -

Those are the only mechanisms for that I've come across, as an undergrad.  The idea of the net charge on a molecule being the factor in deciding a reasonable mechanism is an interesting thing I hadn't really thought of before.  As far as the blog post - I find it's difficult to have a philosophical debate if multiple areas of philosophy are going to be open for debate concurrently.  If you're going to argue over the plausibility of ever knowing a reaction mechanism, it isn't really useful, in my opinion, to leave open a linguistic debate on the meaning of know - define it and be done with it.  As for my side in the debate, I think that it is logically possible to Know (100% sure) a reaction mechanism - but we'd need to be able to physically observe the molecule interacting and slow the rate of interaction to be intelligible.  Then, I'm sure someone would argue like Descartes and say the eyes can deceive, but you have to trust some things.

[azmanam]

don't you think that once NH2- is cleaved (however unlikely that is) would it be prone (as a powerful base) to attack the carboxylic carbon atom from which the electron density is being withdrawn by two oxygen atoms instead of deprotonating it?

Sure, there's a probability of that happening, but it seems to me like it would certainly be more likely for NH2- to deprotonate a carboxylic acid rather than act as a nucleophile on it.  At any rate, if it were to act as a nucleophile, we'd just be back at stage 1 and there would be another opportunity for it to act as a LG and be given another chance to irreversibly deprotonate the acid.

Yes, this is an important observation, that I meant to touch on. 

I've come up with what I call The Code for carbonyl addition mechanisms.  There are only 3 elementary steps in a carbonyl addition mechanism.  a) proton transfer (these are always reversible) b) nucleophilic addition to a carbonyl (electrons go up onto oxygen) c) electrons collapse down from oxygen (and kick off a good leaving group).  All carbonyl addition mechanisms are composed of some combination of those 3 elementary steps - and only those elementary steps.  Go back and look at the pdf insert and the 3 mechanisms I've drawn, they're all combinations of those elementary steps.

The point I'm trying to make is that once you expel the leaving group (elementary step c) there are still 2 options available.  In the case of NH2- as a leaving group, it certainly can add back into the carbonyl (step b) or it can deprotonate the carboxylic acid (step a).  Step a moves the mechanism forward, and step b identifies that step in the mechanism as reversible.  For these carbonyl addition mechanisms, it's really important to identify which steps are reversible.  I'm actually pretty disappointed with myself that I didn't throw in the reversible arrows in my mechanisms.  All steps in all three mechanisms are reversible.

I find it's difficult to have a philosophical debate if multiple areas of philosophy are going to be open for debate concurrently.  If you're going to argue over the plausibility of ever knowing a reaction mechanism, it isn't really useful, in my opinion, to leave open a linguistic debate on the meaning of know - define it and be done with it.

Interesting point.  I think that was the point of one of the reviewers.  We can't even open the bigger debate until we close the smaller debate of what it means to know something.  The main authors don't try to define it and the reviewer has a problem with that.