Propose a mechanism for the hydrolysis of N,N

Question

Propose a mechanism for the hydrolysis of N,N-dimethylacetamide

(b) under acidic conditions.

Accepted Answer

Hello everyone. So in this video, we want to propose a mechanism for this acid cally hydrolysis of this compound here. So I went ahead to actually already draw the mechanism and instead of actually drawing it out with you guys, I already drawn it so that we can go ahead and explain each step in more detail. So I'm gonna go ahead and paste this mechanism that I already drew in advance. All right. So the keyword in this problem here is we're being told that this is a acid callis hydrolysis. So keyword here is acid. It means we're under acidic conditions and any acidic condition sort of mechanism, we always perate first. And then for a base cally hydrolysis or base basic conditions, we always have to go ahead and de perate first. So in our case, again, we are under acidic conditions. So that's why we have Perin it in our first step. So after pronation, what we get is this intermediate here. And we see here, we have two different colored arrows. We have the green colored arrows and then we have the red. So the red arrows here are all going to be what we have going on in the actual mechanism and the blue is just going to or the green rather is going to be any of our arrows that are just for our resonance structures. So we see here that we also have these resonant arrows indicating the resonance as well as our curvy brackets here off to the right and the left. So again, these three intermediates here are just going to be resonant structures, But the only one that will proceed in our mechanism will be the one in the middle here. So once we get to this step here, we have this thermal charge of plus one on the Carbonis oxygen. So then that goes ahead and we have our pipe bo breaking just so we can form a long pair on the oxygen. So no longer has a positive one charge on our oxygen. So instead we get a positive charge on the central carbon here. This makes this central carbon with this plus one from a charge electro. So we do have an nucleic attack from our h2o molecule. And we also see here for the resonance side that the long pairs on the nitrogen on our substituent here is gonna go ahead and form a pi bond to relieve the carbon of its positive one formal charge. So we get this intermediate here where we do have a newly formed pi bond. And the nitrogen atom here on our substituent does gain a plus one formal charge. But again, the only intermediate that follows through with the mechanism will be the one with a positively charged carbon. All right. So we see here that we have a tetrahedral intermediate. Now we have our O H, we have our C H three, we have the N E T two and then we have our H2O. So what happens now is that we see a deep pronation step of from H2O of our proton on our H2O that's added to the central carbon here. So we get this intermediate here. And what happens is that the bone pairs on the nitrogen that is actually basic is gonna go ahead and pro itself with a floating H plus ion here. And then we see that we have a very good leaving group. We have this H N E T two. So we're gonna go ahead and actually break away this bonds just so this can leave as a leaving group. Once this group leaves, we see that we have another intermediate here. And we do see we have these resonance arrows, but in our case, these resonance do contribute to our final product. So that's why we have a in these red arrows and not in the other color. So once we get this intermediate here, we again see a positive charge central carbon here. So what's gonna happen is that the lawn pairs on the alcohol group off to the right is gonna go ahead and come down to form a pi bond, which is what we see here. So we know that the carbon neo groups like to reform. So then one of the lone pairs on the auction is gonna go ahead and come down to form a carbo neal group. Then this pi bond that was newly formed in the previous step is gonna go ahead and form a long pair on the positively charged oxygen atom. So once we do that, we see that we have an alcohol group off to the right and a somewhat almost complete Carboni group. So once we get to this step here, we're gonna take the leaving group that had just left the H N E T two, the long pairs on the nitrogen will go ahead and pro itself removing this proton from our Carboni group to give us an actual Carboni group and alcohol group off to the terminal right end. So then this, that's how we got this mechanism here. Again, keyword in this problem is going to be acid cally hydrolysis that deals with acidic conditions. And that's why the first step is going to be a pronation step. And all of our intermediates cannot have any negative charges, only positive charges. So all of this here is going to be my final answer for this problem. Let's go ahead and put a box around it as well. So this highlighted box here is going to summarize that this is going to be the final answer for this mechanism problem.

Propose a mechanism for the hydrolysis of N,N-dimethylacetamide
(b) under acidic conditions.

Key Concepts

Hydrolysis Mechanism

Acidic Conditions

Nucleophilicity

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